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7501 | https://home.strw.leidenuniv.nl/~keller/Teaching/OAI_2015/QAI-waves1.pdf | Wave Optics: Diffraction & Interference • Huygens sources & wave propagation – Math: Green’s function • Fraunhofer vs. Fresnel interference • Applications of diffraction & interference: – Diffraction limit of telescopes & microscopes – Fresnel diffraction behind halfscreen – Fresnel zone plates – Interference behind realistic double slit – Interference behind realistic N slit Huygens sources help to visualize field propagation Rays & Phase fronts Hecht, Fig. 4.26 Huygens sources: quantitative formulation . . Q: What is ?
Plane waves propagate undisturbed 1 2 ≈ ! 2 = " #! $ Q: Prove this equals ES Diffraction: integration over Huygens sources Hecht, Fig. 10.1 , Q: When is this description valid?
• Spectral coherence (monochromatic) • Spatial coherence (one coherent source) Q: How about backward light?
• Factor (1+cosθ)/2 Diffraction behind an aperture & ≈/( Diffraction angle Fresnel Fraunhofer NOTE: long-wavelength water wave is not representative for optics Diffraction pattern changes with distance from slit Hecht Fig.10.5 In near field: Shape changes with distance from slit In far field: - fixed shape - only scaling ∝L 10.2 Fraunhofer diffraction 10.3 Fresnel diffraction Q: What is far enough?
Fraunhofer regime when L >> D2/λ ), ′ , ≈ + − ′ $ + ⋯ 2 ≈ + $ −2 ) + ′$ 2 ), ′ , ), … ≈ /) $ , … 0$"/./) . / $ … / Fresnel term only for ≪ $/ Fourier transform in Fraunhofer regime: ≫ $/ Many examples of Fraunhofer & Fresnel diffraction • Fresnel diffraction behind slit & half screen • Fresnel zone plate • Fraunhofer diffraction behind slit • Fraunhofer diffraction behind circular aperture • Rayleigh criterium for resolution • Diffraction limit of telescope & microscope • Fraunhofer interference behind Young’s double slit • Interference behind N identical slits 1.22 1.22 sin D D λ λ θ θ = ⇒ ≈ Interference fringes in Fresnel regime Pattern behind screw Pattern behind half opaque screen Fresnel diffraction in 2D and 3D F D ≡ "/ /$ F 2D: Fresnel integral "/ /$ $ / GH 3D: Trivial integral Fresnel diffraction in 2D: Fresnel number & Cornu spiral Hecht Fig.10.54 & 10.55 E ≡ "/ I /$ I F FJ Q: What is w1 and w2 ?
A: "/ I $ K∆ "/ The semi-infinite opaque screen Hecht Fig.10.59 Fresnel diffraction behind circular aperture Ecentre= "/ /$ $ / G #!
Hecht Fig.10.41 & 10.42 Fresnel lens: focusing with black-and-white pattern !!
Fresnel number NF=b2/Lλ counts number of “Fresnel zones” How to increase the intensity in P by blocking Fresnel zones in the beam !
Fraunhofer diffraction behind slit of width w ) ∝ ∆ N/$ 0N/$ 0 OPQ R/ N/$ 0N/$ ), ′ , & As graphical construction yields Δ −sin & & ∝ 0 OPQ R/ N/$ 0N/$ −T OPQ U U with V $ K sin & T Diffraction minima and maxima behind slit Opening angle of first minimum: first min.
sin( ) b λ θ = Q: What is spacing maxima & minima A: Only minima equidistant Fraunhofer diffraction behind rectangular hole 0 1 1 0 0 2 2 sin sin sin ; sin y x P x y x x x y y y E E k b k b β β β β β θ β θ = = = Diffraction behind square hole What is orientation of hole?
Diffraction behind rectangular hole Field profile factorizes !!
Fraunhofer diffraction behind round hole q R q s P ds Field per unit Area 2 1 0 2 ( sin ) sin A P E R J kR E kr kR π θ θ = 0 Area 1 2 1 1 exp( sin ) ( sin ) exp( sin ) 1 sin A P E E isk dA r J kR ivkR v dv kR θ π θ θ θ − ∝ ∝ − ≡ ∫∫ ∫ Bessel function Properties of Bessel function J1(x) ↓ = = = 1 0 1 ( ) 1 lim 2 ( ) 0 als 3.832,7.016,10.173 x J x x J x x 2 1 0 2 ( sin ) sin P J kR I I kR θ θ = 1.22 1.22 0 als sin 3.832 sin 2 I kR R D λ λ θ θ = = ⇒ = = Central disk = Airy disk Rings = Airy rings Compare with diffraction from slit of rectangle: 0 als sin I b λ θ = = Why is diffraction angle of round hole larger?
Rayleigh criterion for resolution Q: When can you resolve two maxima? Lord Rayleigh: - If intensity between peaks drops 19% - If one diffraction max. coincides with other diffraction minimum For round holes: For rectangular holes: 1.22 D λ θ ∆ > D λ θ ∆ > Resolution of telescope Well distinguishable Barely distinguishable What is angular resolution of telescope with D = 4 cm (for λ=500 nm)?
Answer: θres ≈15 µrad (diameter moon ≈0.5º ≈8 mrad) Note: Fluctuations in earth atmosphere (“seeing”) typically limit resolution for D > 20 cm Resolution of microscope Similar criterium as for telescope: Note: also valid for large angles & immersion objectives min 1.22 1.22 2 sin 2 . .
s n i N A λ λ = = 1.22 1.22 1.22 / 2 . .
res res res x f D D f N A λ λ λ θ θ ≈ ⇒ ≈ ≈ ≈ Ernst Abbe (1840-1905) Abbe diffraction limit for microscopes ( ) λ θ = × 1.22 2. .sin in d n Young’s experiment with a double slit Interference of two diffracted waves Q: Why did it take until 1803?
• Spectral coherence • Spatial coherence Young’s double slit experiment Hecht, Fig. 9.8 Path length differences are crucial 1. Young’s experiment with limited spectral coherence Constructive interference for all colors only in the center (θ=0) Signaal Different colors have different fringe periods Total intensity A=0.6 cm, V=0.59 A=0.8 cm, V=0.36 A=1.0 cm, V=0.15 2. Young’s experiment with limited spatial coherence distance between slits increases a b b b b θ ( ) ( ) α β β α − = − ∝ = ∝ ∫ ∑ ∫ 0 Openingen / 2 1 0 0 / 2 0 exp( ) exp( ) exp( ) sin sin sin P y b N y y n b E E ik y dy E ik y dy ik an N E b Interference Diffraction Fraunhofer diffraction behind N slits slit y0=0 slit y1=a slit y2=2a slit y3=3a Again: factorization of diffraction from single slit & mutual interference Fraunhofer diffraction behind N small slits (b →0) ( ) ( ) ( ) α θ α α ∝ = 1 2 sin ; sin y N E k a Q: What happens at α= mπ?
A: Diffraction maximum (orde m with a.sinθ=mλ) Q: How does peak intensity scale with N?
( ) ( ) α π α α → = sin lim sin m N N A: Intensiteit ∝N2 (coherent sources) Better angular resolution with multiple sources/detectors Westerbork telescope array: λ ≈21 cm (ν ≈1.4 GHz) Angular resolution of array P&P Fig. 16-14 .
( ) ( ) ( ) tot dif N slit I I I θ θ θ − = × |
7502 | https://mathoverflow.net/questions/482374/sets-of-integers-with-same-sum-and-same-sum-of-reciprocals | Skip to main content
Sets of integers with same sum and same sum of reciprocals
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Let A, B be two distinct sets of natural numbers. Is it possible to have ∑a∈Aa=∑b∈Bb and ∑a∈A1/a=∑b∈B1/b at the same time?
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edited Nov 20, 2024 at 1:51
Wlod AA
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For example 2,15,20 and 3,4,30. Another example is 2,8,18,24 and 3,4,9,36.
– Thomas Bloom
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Nov 13, 2024 at 4:48
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The smallest example in terms of the sum seems to be 4,10,12 and 5,6,15.
– Michael Lugo
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Nov 14, 2024 at 14:53
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Yes. Take a large N. Let E:={n≤N:p∣n⟹p≤N100}, where p stands for a prime.
Note that
∣∣∣[N]∖E∣∣∣≤∑N100≤p<NNp≤CNlogN,
implying
|E|≥(1−o(1))N.(1)
Now,
loglcm(E)=log∏p≤N100p⌊logpN⌋≤∑p≤N100logN≤N10.
In other words,
lcm(E)≤eN/10≤2N/5.(2)
Therefore, for any A⊆E, we have ∑n∈A1n=plcm(E) for some p≤Nlcm(E)≤2N/4.
Finishing up, each A⊆E gives rise to a pair (∑n∈An,lcm(E)∑n∈A1n), which lives in [N2]×[2N/4], the size of which is at most 2N/3. Since there are 2|E|≥2N/2 many subsets of E, the pigeonhole principle guarantees two (distinct) with the same pair.
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edited Nov 14, 2024 at 1:02
answered Nov 13, 2024 at 4:40
mathworker21mathworker21
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What led you to think of this construction?
– semisimpleton
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@semisimpleton It's not a construction, or did you mean the construction of E? I just thought there were a lot of sets so two of them must have the same pair (∑aa,∑a1/a). But there were too many possibilities for ∑a1/a, so I needed to trim a bit. One way to do that is throw out large primes.
– mathworker21
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Nov 22, 2024 at 1:00
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Yes, that's what I meant to ask. Thanks!
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For any non-zero (a,b,c) with b2≠ac, take (x/c,x/b,x/a) where (1/a+1/b+1/c)x=a+b+c. If the resulting numbers aren't all integers, then multiply all six by a common factor to clear fractions.
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answered Nov 13, 2024 at 16:26
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Simple and superb!
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A simplified version of mathworker21's argument.
Denote N=6T−1, then the set A of divisors of N has |A|=τ(N)=T2 elements. Consider a subset B of A, there are 2T2 choices of B. Denote w(B):=(sum of elements B, sum of reciprocals of elements of B).
The sum of elements of B does not exceed σ(N)=(1+2+…+2T−1)(1+3+…+3T−1)<2T⋅3T=6T. The sum of their reciprocal multiplied by N is an integer and does not exceed the same value 6T. Thus, the number of distinct pairs w(B) does not exceed 6T⋅6T=36T that is much less than 2T2, therefore, there exist many subsets with the same w(B).
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answered Nov 14, 2024 at 8:34
Fedor PetrovFedor Petrov
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Nice, and this is smarter than what I did. You get eT versus eT2 and I had to work to get ecT versus eCT.
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Here is a 4-parameter family of solutions with |A|=|B|=3.
Consider the optic equation 1a+1b=1c; it is known that the positive integer solutions to this equation are parametrised by
abc=km(m+n),=kn(m+n),=kmn,
with k,m,n∈Z+. Hence the set {a,b,−c} has sum k(m2+mn+n2) and sum of reciprocals 0.
Now for any m,n,m′,n′∈Z+, let
m2+mn+n2m′2+m′n′+n′2=k′k.
Define a,b,c as above, and a′,b′,c′ analogously. Then
a+b−c=k(m2+mn+n2)=k′(m′2+m′n′+n′2)=a′+b′−c′,
and
1a+1b−1c=0=1a′+1b′−1c′,
so the sets A={a,b,c′} and B={a′,b′,c} satisfy the desired conditions.
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answered Nov 13, 2024 at 16:55
chronondecaychronondecay
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If one has a solution in rational numbers one can scale up to get integers.
Rather than starting with equal sums of reciprocals. one could start with two equations such as 1+5+10=3+13 and 2+4+15=21. Then for any x,y the sets A={x,5x,10x,21y} and B={3x,13x,2y,4y,15y} have equal sum. Setting the sums of the reciprocals equal fixes the ratio xy=r for some rational number r=PQ. Set x=P,y=Q for an integer solution. If the elements have a relatively small least common multiple (so if they come from the divisors of an integer with small prime divisors). The final scaled up integer solution will have correspondingly smaller members. parametric solutions are possible, which might not differ from what has already been suggested.
If there are two solutions (A,B) and (A′,B′) then (A∪A′,B∪B′) is a solutions So one might want primitive solutions which do not split this way. Of course if A∩B′≠∅ and/or A′∩B≠∅
then the common elements can be removed and that might give a primitive solution.
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answered Nov 25, 2024 at 18:49
Aaron MeyerowitzAaron Meyerowitz
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7503 | https://pubmed.ncbi.nlm.nih.gov/8905326/ | Comparison of the pharmacological profiles of the hypnotic drugs, zaleplon and zolpidem - PubMed
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. 1996 Oct 10;313(1-2):35-42.
doi: 10.1016/0014-2999(96)00510-9.
Comparison of the pharmacological profiles of the hypnotic drugs, zaleplon and zolpidem
D J Sanger1,E Morel,G Perrault
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1 Synthélabo Recherche, CNS Research Department, Bagneux, France.
PMID: 8905326
DOI: 10.1016/0014-2999(96)00510-9
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Comparative Study
Comparison of the pharmacological profiles of the hypnotic drugs, zaleplon and zolpidem
D J Sanger et al. Eur J Pharmacol.1996.
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. 1996 Oct 10;313(1-2):35-42.
doi: 10.1016/0014-2999(96)00510-9.
Authors
D J Sanger1,E Morel,G Perrault
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1 Synthélabo Recherche, CNS Research Department, Bagneux, France.
PMID: 8905326
DOI: 10.1016/0014-2999(96)00510-9
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The BZ1 (omega 1)-selective compound, zolpidem, is a clinically effective hypnotic drug with a pharmacological profile which differs from those of benzodiazepine anxiolytics and hypnotics. Zaleplon (CL 284,846) has recently been described as a hypnotic agent which also has BZ1 (omega 1) receptor selectivity. The pharmacological effects of zolpidem and zaleplon were therefore compared in mice and rats. Both drugs blocked tonic convulsions induced in mice by pentylenetetrazole and electroconvulsive shock and clonic convulsions induced by isoniazid. Zaleplon was more potent than zolpidem but the maximal effect of zolpidem for increasing the latency to isoniazid-induced convulsions was greater than that of zaleplon. Little tolerance developed to the anticonvulsant effect of zaleplon against isoniazid-induced seizures following twice daily administration of 10 or 30 mg/kg for 10 days. Both compounds reduced locomotor activity and produced motor deficits in the rotarod and loaded grid tests in mice. However, while zaleplon produced all three effects at similar doses, zolpidem showed the greatest potency for reducing locomotion. Zaleplon and zolpidem also decreased locomotion and produced a rotarod deficit in rats. Again, the difference between the doses giving rise to these two effects was greater for zolpidem than for zaleplon. In a drug discrimination procedure using rats trained to discriminate a dose (5 mg/kg) of chlordiazepoxide, zaleplon produced partial substitution for chlordiazepoxide at doses which greatly reduced response rates. These results show that zaleplon and zolpidem have similar pharmacological profiles, presumably related to their BZ1 (omega 1) receptor selectivity. However, the difference between doses producing motor deficits (rotarod, loaded grid) and those giving rise to other effects (anticonvulsant, decreased locomotion) was greater for zolpidem than for zaleplon. This difference may be related to a greater in vivo intrinsic activity of zolpidem as indicated by the different efficacies of the two drugs to antagonise isoniazid-induced convulsions.
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Clinical Care of Balamuthia Infection
Key points
Treatment options
Clinicians
For 24/7 diagnostic assistance, specimen collection guidance, shipping instructions, and treatment recommendations, please contact CDC Emergency Operations Center at 770-488-7100.
Balamuthia can cause rare but serious illnesses. They include a nearly always fatal brain infection called granulomatous amebic encephalitis, or GAE. Even when patients with GAE are treated with medications, most do not survive. Much more information is needed in treating patients with GAE due to Balamuthia.
Treatment recommendations
Effective treatment for infections caused by Balamuthia mandrillaris has not been established. The recommendations are based on a small number of Balamuthia survivor case reports.
The duration of treatment for people with a Balamuthia infection also has not been established. In case reports of survivors, duration of treatment has ranged from several weeks to several months, or even years. The decision to stop treatment should be made on a case-by-case basis and include consideration of the patient's clinical status and review of laboratory and radiographic findings.
Recommended Medications to Treat GAE Caused by Balamuthia
Recommended Treatment for GAE Caused by Balamuthia infection
| Drug | Dose | Notes |
---
| Pentamidine (IV) | 4 mg/kg given once per day | Although pentamidine has been used successfully in combination with the drugs listed below, pentamidine is very toxic and doesn’t cross the normal, intact blood-brain barrier well. Its use must be a clinical decision. |
| Sulfadiazine (oral) | 1.5 g every 6 hours in adults; 200 mg/kg/day in 4–6 doses in pediatric patients (maximum 6 g/day) | |
| Flucytosine (oral) | 37.5 mg/kg every 6 hours (maximum 150 mg/kg/day) | |
| A mold-active azole (e.g., voriconazole, posaconazole, or isavuconazole) | Dosing will vary based on drug and patient. Consult a clinical pharmacist with dosing questions. | Fluconazole and itraconazole are NOT recommended due to poor in vitro efficacy. |
| Azithromycin (oral or IV) | 20 mg/kg/day in 1 dose (max 500 mg/day) in pediatric patients; 500 mg/day in 1 dose for adults | |
| Clarithromycin (oral, alternative to azithromycin) | 14 mg/kg/day in 2 doses (max 2 g/day) | |
| Miltefosine (oral)a | Up to 45 kg body weight: 100 mg daily (i.e., one 50 mg cap po with breakfast and dinner) For pediatric cases, 2.5 mg/kg/day up to 100 mg daily 45 kg body weight and higher: 150 mg daily (i.e., one 50 mg cap po with breakfast, lunch, and dinner) | Miltefosine is now commercially available. Visit impavido.com for more information. |
| Nitroxoline | Contact CDC for dosing | Nitroxoline is an investigational drug that may be effective for Balamuthia infections. It is not FDA-approved in the United States, but available for treatment of free-living ameba infections through CDC’s expanded access Investigational New Drug program. Contact the CDC Emergency Operations Center at 770-488-7100 for more information. |
For pediatric cases, 2.5 mg/kg/day up to 100 mg daily
45 kg body weight and higher: 150 mg daily (i.e., one 50 mg cap po with breakfast, lunch, and dinner)
aThe standard miltefosine dosing recommended for the treatment of leishmaniasis is presented in the table. The oral preparation is the only formulation available. A higher dose would lead to increased nausea, vomiting, or diarrhea. Miltefosine is mildly nephrotoxic but is not cleared by the kidneys, so dosing does not need to be adjusted for patients with impaired renal function. Because little data are available about the effective dose for amebic infection, the risk for nephrotoxicity should be balanced with the risk of death from GAE or disseminated disease.
Survivors
Although Balamuthia GAE is often fatal, there are several recorded cases of Balamuthia infection in the United States in which the patient survived after long-term treatment with multiple drugs. Survivors have also been reported in other countries. Some U.S. survivors had excisional biopsies of one of their multiple brain lesions. At least one survivor had an accompanying cutaneous lesion. Early diagnosis and treatment might increase the chances for survival.
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1.04: Fixed-Point Binary Representation of Numbers
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1.03: Sources of Error
1.05: Floating-Point Binary Representation of Numbers
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( \newcommand{\kernel}{\mathrm{null}\,})
Lesson 1: Introduction to Binary Representation of Numbers
Learning Objectives
After successful completion of this lesson, you should be able to:
1) enumerate different bases of number systems and be able to convert base-2 numbers to base-10 numbers.
Introduction
In everyday life, we use a number system that has a base of 1010 (decimal system). When a number is written in this system, we can quickly decipher it as we see such numbers throughout the day.
For example, look at the number 257.36257.36. Each digit in 257.36257.36 has a value of 00 through 99 and has a place value where the position of the number with respect to the decimal point determines its value. Each position in the decimal number corresponds to an integer exponent of 1010. In the number 257.36257.36, the 22 is in the place value of hundred, that is 102102, while the 33 is in the place of one-tenths, that is, 10−110−1. The number 257.36257.36 can be written as
257.36=2×102+5×101+7×100+3×10−1+6×10−2=200+50+7+0.3+0.06=257.36
257.36=2×102+5×101+7×100+3×10−1+6×10−2=200+50+7+0.3+0.06=257.36
In a binary system, we have a similar system as the base-1010, but the only difference is that the base is made of only two digits, 00 and 11. So it is a base-22 system. A number like (1011.0011)(1011.0011) in base-22 represents an equivalent decimal number as
(1011.0011)2=((1×23+0×22+1×21+1×20)+(0×2−1+0×2−2+1×2−3+1×2−4))10=((1×8+0×4+1×2+1×1)+(0×0.5+0×0.25+1×0.125+1×0.0625))10=(8+0+2+1+0+0+0.125+0.0625)10=(11.1875)10
(1011.0011)2=((1×23+0×22+1×21+1×20)+(0×2−1+0×2−2+1×2−3+1×2−4))10=((1×8+0×4+1×2+1×1)+(0×0.5+0×0.25+1×0.125+1×0.0625))10=(8+0+2+1+0+0+0.125+0.0625)10=(11.1875)10
Note the dot (.) that separates the integer and fractional part in the binary representation or any other representation is called the radix point.
Example 1.4.1.11.4.1.1
Find the equivalent of (1101)2(1101)2 in base-1010.
Solution
(1101)2=(1×23+1×22+0×21+1×20)10=(1×8+1×4+0×2+1×1)10=(8+4+0+1)10=(13)10
(1101)2=(1×23+1×22+0×21+1×20)10=(1×8+1×4+0×2+1×1)10=(8+4+0+1)10=(13)10
Example 1.4.1.21.4.1.2
Find the equivalent of (0.1101)2(0.1101)2 in base-1010.
Solution
(0.1101)2=(1×2−1+1×2−2+0×2−3+1×2−4)10=(0.5+0.25+0+0.0625)10=(0.8125)10
(0.1101)2=(1×2−1+1×2−2+0×2−3+1×2−4)10=(0.5+0.25+0+0.0625)10=(0.8125)10
Example 1.4.1.31.4.1.3
Find the equivalent of (1101.1101)2(1101.1101)2 in base-1010.
Solution
From Example 1.4.1.11.4.1.1,
(1101)2=(13)10
(1101)2=(13)10
From Example 1.4.1.21.4.1.2,
(0.1101)2=(0.8125)10
(0.1101)2=(0.8125)10
Hence
(1101.1101)2=(13.8125)10
(1101.1101)2=(13.8125)10
Audiovisual Lectures
Title: Binary Representation of Numbers
Summary: Learn about the binary representation of numbers - how to convert a base-2 number to a base-10 number.
Lesson 2: Fixed-Point Binary Representation of Numbers
Learning Objectives
After successful completion of this lesson, you should be able to:
1) convert a base-10 number to a base-2 number in a fixed format.
Introduction
To understand the binary system, we need to convert binary numbers to decimal numbers and vice-versa. We have already seen an example of how binary numbers can be converted to decimal numbers. Let us see how we can convert a decimal number to a binary number. For example, take the decimal number 11.187511.1875.
First, look at the integer part: 1111.
Divide 1111 by 22. This division gives a quotient of 55 and a remainder of 11. Since the remainder is 11, a0=1a0=1.
Divide the quotient 55 by 22. This division gives a quotient of 22 and a remainder of 11. Since the remainder is 11, a1=1a1=1.
Divide the quotient 22 by 22. This division gives a quotient of 11 and a remainder of 00. Since the remainder is 00, a2=0a2=0.
Divide the quotient 11 by 22. This division gives a quotient of 00 and a remainder of 11. Since the remainder is 11, a3=1a3=1.
Since the quotient now is 00, the process is stopped. The above steps are summarized in Table 1.4.2.11.4.2.1.
Table 1.4.2.11.4.2.1. Converting a base-1010 integer to binary representation.
| | Quotient | Remainder |
| 11/211/2 | 55 | 1=a01=a0 |
| 5/25/2 | 22 | 1=a11=a1 |
| 2/22/2 | 11 | 0=a20=a2 |
| 1/21/2 | 00 | 1=a31=a3 |
Hence
(11)10=(a3a2a1a0)2=(1011)2
(11)10=(a3a2a1a0)2=(1011)2
For any integer, the algorithm for finding the binary equivalent is given in the flow chart in Figure 1.4.2.11.4.2.1.
Now let us look at the decimal part, that is, 0.18750.1875.
Multiply 0.18750.1875 by 22. This multiplication gives 0.3750.375. The number before the decimal is 0,0, and the number after the decimal is 0.3750.375. Since the number before the decimal is 00, a−1=0a−1=0.
Multiply the number after the decimal, that is, 0.3750.375 by 22. This multiplication gives 0.750.75. The number before the decimal is 0,0, and the number after the decimal is 0.750.75. Since the number before the decimal is 00, a−2=0a−2=0.
Multiply the number after the decimal, that is, 0.750.75 by 22. This multiplication gives 1.51.5. The number before the decimal is 1,1, and the number after the decimal is 0.50.5. Since the number before the decimal is 11, a−3=1a−3=1.
Multiply the number after the decimal, that is, 0.50.5 by 22. This multiplication gives 1.01.0. The number before the decimal is 1,1, and the number after the decimal is 00. Since the number before the decimal is 11, a−4=1.a−4=1.
Since the number after the decimal is 00, the conversion is complete. The above steps are summarized in Table 1.4.2.21.4.2.2.
Table 1.4.2.21.4.2.2. Converting a base-1010 fraction to binary representation.
| | Number | Number after decimal | Number before decimal |
| 0.1875×20.1875×2 | 0.3750.375 | 0.3750.375 | 0=a−10=a−1 |
| 0.375×20.375×2 | 0.750.75 | 0.750.75 | 0=a−20=a−2 |
| 0.75×20.75×2 | 1.51.5 | 0.50.5 | 1=a−31=a−3 |
| 0.5×20.5×2 | 1.01.0 | 0.00.0 | 1=a−41=a−4 |
Hence
(0.1875)10=(a−1a−2a−3a−4)2=(0.0011)2
(0.1875)10=(a−1a−2a−3a−4)2=(0.0011)2
Having calculated
(11)10=(1011)2
(11)10=(1011)2
and
(0.1875)10=(0.0011)2,
(0.1875)10=(0.0011)2,
we have
(11.1875)10=(1011.0011)2.
(11.1875)10=(1011.0011)2.
In the above example, when we were converting the fractional part of the number, we were left with 00 after the decimal number and used that as a place to stop. In many cases, we are never left with a 00 after the decimal number. For example, finding the binary equivalent of 0.30.3 is summarized in Table 1.4.2.31.4.2.3.
Table 1.4.2.31.4.2.3. Converting a base-1010 fraction to approximate binary representation.
| | Number | Number after decimal | Number before decimal |
| 0.3×20.3×2 | 0.60.6 | 0.60.6 | 0=a−10=a−1 |
| 0.6×20.6×2 | 1.21.2 | 0.20.2 | 1=a−21=a−2 |
| 0.2×20.2×2 | 0.40.4 | 0.40.4 | 0=a−30=a−3 |
| 0.4×20.4×2 | 0.80.8 | 0.80.8 | 0=a−40=a−4 |
| 0.8×20.8×2 | 1.61.6 | 0.60.6 | 1=a−51=a−5 |
As you can see, the process will never end. In this case, the number can only be approximated in binary format, that is,
(0.3)10≈(a−1a−2a−3a−4a−5)2=(0.01001)2
(0.3)10≈(a−1a−2a−3a−4a−5)2=(0.01001)2
Example 1.4.2.11.4.2.1
Convert (13.875)10(13.875)10 to base 22.
Solution
For (13)10(13)10, conversion to binary format is shown in Table 1.4.2.41.4.2.4.
Table 1.4.2.41.4.2.4. Conversion of base-1010 integer to binary format.
| | Quotient | Remainder |
| 13/213/2 | 66 | 1=a01=a0 |
| 6/26/2 | 33 | 0=a10=a1 |
| 3/23/2 | 11 | 1=a21=a2 |
| 1/21/2 | 00 | 1=a31=a3 |
So
(13)10=(1101)2.
(13)10=(1101)2.
Conversion of (0.875)10(0.875)10 to binary format is shown in Table 1.4.2.51.4.2.5.
Table 1.4.2.51.4.2.5. Converting a base-1010 fraction to binary representation.
| | Number | Number after decimal | Number before decimal |
| 0.875×20.875×2 | 1.751.75 | 0.750.75 | 1=a−11=a−1 |
| 0.75×20.75×2 | 1.51.5 | 0.50.5 | 1=a−21=a−2 |
| 0.5×20.5×2 | 1.01.0 | 0.00.0 | 1=a−31=a−3 |
So
(0.875)10=(0.111)2
(0.875)10=(0.111)2
Hence
(13.875)10=(1101.111)2
(13.875)10=(1101.111)2
Audiovisual Lectures
Title: Base 10 to Base 2 Conversion
Summary: See a method to convert a base-10 to a base-2 number.
But what is the mathematics behinds this process of converting a decimal number to binary format?
Let zz be the decimal number written as
z=x.y
z=x.y
where
xx is the integer part, and yy is the fractional part.
We want to find the binary equivalent of xx. So, we can write
x=an2n+an−12n−1+...+a020
x=an2n+an−12n−1+...+a020
If we can now find a0,...,ana0,...,anin the above equation, then
(x)10=(anan−1...a0)2
We now want to find the binary equivalent of y. So, we can write
y=b−12−1+b−22−2+...+b−m2−m
If we can now find b−1,...,b−min the above equation, then
(y)10=(b−1b−2...b−m)2
Let us look at this using the same example as before.
Example 1.4.2.2
Convert (11.1875)10 to base 2.
Solution
To convert (11)10 to base 2, what is the highest power of 2 that is part of 11. That power is 3, as 23=8 to give
11=23+3
What is the highest power of 2 that is part of 3? That power is 1, as 21=2 to give
3=21+1
So
11=23+3=23+21+1
What is the highest power of 2 that is part of 1? That power is 0, as 20=1, to give
1=20
Hence
(11)10=23+21+1=23+21+20=1×23+0×22+1×21+1×20=(1011)2
To convert (0.1875)10 to base 2, we proceed as follows. What is the smallest negative power of 2 that is less than or equal to 0.1875? That power is −3 as 2−3=0.125.
So
0.1875=2−3+0.0625
What is the next smallest negative power of 2 that is less than or equal to 0.0625? That power is −4 as 2−4=0.0625.
So
0.1875=2−3+2−4
Hence
(0.1875)10=2−3+0.0625=2−3+2−4=0×2−1+0×2−2+1×2−3+1×2−4=(0.0011)2
Since
(11)10=(1011)2
and
(0.1875)10=(0.0011)2
we get
(11.1875)10=(1011.0011)2
One can show the above process algebraically for any general number, but it is out-of-scope for this course.
Lesson 3: Writing a Fixed-Point Binary Number in Given Word Length
Learning Objectives
After successful completion of this lesson, you should be able to:
1) write a base-2 number in a given word length.
Introduction
A fixed-point binary number may need to be stored in a given number of bits called the word length. Some of the bits in the word would be used for the integer part and the rest of them for the fractional part. In addition, a bit would also be used to signify the sign of the number. This is best illustrated through an example.
Example 1.4.3.1
Given that a fixed-point positive binary number is stored in eight bits word length, where the first six bits are used for the integer part and the next two for the fractional part, how would the (13.875)10 be stored? We are given that (13.875)10=(1101.111)2.
Solution
Since there are six bits for the integer part and two for the fractional part, we will first rewrite the number as
(13.875)10=(1101.111)2≈(001101.11)2
This occupies the right number of bits for the integer part (6 bits) and the fractional part (2 bits). We write 1101 as 001101 to occupy the 6 places without changing it value. We approximate the fractional part .111 to .11 as we have only 2 bits for it.
Multiple Choice Test
(1). (25)10=(?)2
(A) 100110
(B) 10011
(C) 11001
(D) 110010
(2). (1101)2=(?)10
(A) 3
(B) 13
(C) 15
(D) 26
(3). (25.375)10=(?.?)2
(A) 100110.011
(B) 11001.011
(C) 10011.0011
(D) 10011.110
(4). Representing √2 in a fixed-point binary register with 2 bits for the integer part and 3 bits for the fractional part gives a round-off error of most nearly
(A) −0.085709
(B) 0.0392
(C) 0.1642
(D) 0.2892
(5). An engineer working for the Department of Defense is writing a program that transfers non-negative real numbers to integer format. To avoid overflow problems, the maximum non-negative integer that can be represented in a 5-bit binary integer word is
(A) 16
(B) 31
(C) 63
(D) 64
(6). For a numerically controlled machine, integers need to be stored in a memory location. The minimum number of bits needed for an integer binary word to represent all integers between 0 and 1024 is
(A) 8
(B) 9
(C) 10
(D) 11
For complete solution, go to
Problem Set
(1). Convert the following.
a) (19)10=(?)2
b) (75)10=(?)2
Answer
: a) 100111 b) 1001011
(2). Convert the following.
a) (110111)2=(?)10
b) (11001)2=(?)10
Answer
: a) 55 b) 25
(3). Convert the following.
a) (0.375)10=(?)2
b) (0.075)10=(?)2
Answer
: a) 0.011 b) 0.000100110011...
(4). Convert the following.
a) (0.110001)2=(0.?)10
b) (0.0111)2=(0.?)10
Answer
: a) 0.765625 b) 0.4375
(5). Convert the following.
a) (19.375)10=(?.?)2
b) (75.075)10=(?.?)2
Answer
: a) 10011.011 b) 1001011.00010011.......
(6). Convert the following.
a) (110111.110001)2=(?.?)10
b) (11001.0111)2=(?.?)10
Answer
: a) 55.765625 b) 25.4375
1.03: Sources of Error
1.05: Floating-Point Binary Representation of Numbers |
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Working with Equations
9 Transposing Equations
Click play on the following audio player to listen along as you read this section.
Have you ever come across a situation during your math studies where you’re required to solve for a variable which doesn’t seem to be in the right place? Take a look at the following example to see what I mean.
[latex]\Large \text{A} = {\text{B}}^{2} \times 0.7854 \times \text{H}[/latex]
In a perfect world, you would like to solve for “A” and at the same time be given the values of both “B” and “H.”
But what if you were given “A” and you had to solve for “B”? How would you go about doing this?
The idea here would be to move the variables around and isolate “B.” What this means is that “B” is on one side of the equation by itself, and everything else is on the other side. Take a look at the equation again when this has been done.
[latex]\Large \text{B} = \sqrt{\dfrac{\text{A}}{.7854 \times \text{H}}}[/latex]
Changing the formula around is referred to as “transposing” an equation.
It’s not as simple as just moving stuff around though. There are rules to get to this point, and those rules and their application are what we are going to deal with in this part of the chapter.
The most important thing to remember when transposing equations is that whatever is done to one side of the equation must also be done to the other side of the equation.
If you look at it mathematically, this makes sense. We already determined that an equation is two mathematical expressions that are separated by an equal sign.
What this means is the addition, subtraction, division, and multiplication variables and constants on one side of the equation are equal to all the addition, subtraction, division, and multiplication variables on the other side of the equation.
So if you decide to add 5 to one side, you must add 5 to the other side. What this does is keep the equation equal.
Take a look at the following example.
[latex]\Large \begin{array}{c}10+7=9+8 \ \text{This works out to be:} \ 17=17\end{array}[/latex]
Here we have an equation that is true. Now add 5 to the left hand side of the equation, and you’ll see that, in order to keep the equation true, you’ll have to add 5 to the right hand side of the equation.
[latex]\Large \begin{array}{c} 10 + 7 + \mathbf{5} = 9 + 8 + \textbf{?} \ 22 = 9 + 8 + ? \ 22 = 9 + 8 + \mathbf{5} \ 22 = 22 \end{array}[/latex]
Keep in mind that this is an example where we added to one side. If we had subtracted, divided, or multiplied, things would be different. We would have to do the same thing to the other side.
Transposing Equations Using Addition and Subtraction
Start with a simple equation.
[latex]\Large \begin{array}{c}7+2=8+1 \ \text{This works out to be:}\ 9=9\end{array}[/latex]
Then add 4 to one side of the equation and solve.
[latex]\Large 7+2+4=8+1+ \text{ ?}[/latex]
As the left hand side of the equation has had 4 added to it, the right hand side of the equation also has to have 4 added to it. We get:
[latex]\Large \begin{array}{c}7+2+ \mathbf{4} = 8 + 1 + \textbf{?} \ 13 = 9 + \text{ ?} \ 13 = 9 + \mathbf{4} \ 13 =13 \end{array}[/latex]
As stated before, the rule is that whatever you do to one side you must do to the other. In this case, if we add 4 to one side then we have to add 4 to the other side to keep it equal.
Subtraction would work the same way. If we were to subtract 4 from one side we would have had to subtract 4 from the other side in order to keep it equal.
Let’s try out this concept with an equation that has an unknown variable in it. Take a look at the following equation and solve for “J.”
[latex]\Large 7+ \text{J}=5+8[/latex]
In order to solve for “J” we must isolate “J” on one side of the equation, and it actually doesn’t matter which side we isolate “J” on. If we follow our rule, in order to isolate “J” we would have to get rid of the 7 on the left side. The question becomes, how is that done? What we essentially have to do is move the 7 from the left side to the right side.
Once again, we always have to keep in mind that whatever we do to one side, we must do to the other.
Start with this. Would you agree that 7 − 7 = 0?
What if we subtracted 7 from the left hand side of the equation? What we would be left with is just “J” on the left side, which solves the problem of isolating “J.”
[latex]\Large \begin{array}{c} \mathbf{(7-7)} +\text{J} = 5+8 \ 0 + \text{J} = 5+8 \ \text{J}= 5+8 \end{array}[/latex]
Mathematically this is not correct yet as we only dealt with the left side of the equation. What we need to do now is to do the same thing to the right side and then solve the equation. So we end up subtracting 7 from the right side of the equation to make everything equal once again.
[latex]\Large \begin{array}{c}(7-7)+\text{J} = 5+8-7 \ 0 + \text{J} = 13-7 \ \text{J} = 13 - 7 \ \text{J}=6 \end{array}[/latex]
You can always check that your answer is correct by taking the answer and putting it back in the equation to replace the variable.
[latex]\Large \begin{array}{rl}\text{Replace J with 6} & \ \downarrow & \ 7 + \text{J} & = 5+8 \ 7+6 & = 5+8 \ 13 & = 13 \end{array}[/latex]
Example
Solve for G.
[latex]\Large 27+ \text{G} = 43+49[/latex]
Step 1: Isolate the variable you are trying to find. In this case “G.” In order to do this, we must remove the 27 from the left side of the equation. This is done by subtracting 27 from the left side and then also from the right side.
[latex]\Large \begin{array}{c}27+\text{G} = 43+49 \ \mathbf{(27-27)}+\text{G} = 43+49 - \mathbf{27}\end{array}[/latex]
Step 2: Work through the equation.
[latex]\Large \begin{array}{c}(27-27) +\text{G} = 43+49-27 \ 0 + \text{G} = 92-27 \ \text{G}=65\end{array}[/latex]
Step 3: Check your answer.
[latex]\Large \begin{array}{rl}\text{Replace G with 65} &\ \downarrow & \ 27 + \text{G} & = 43 + 49 \ 27 + 65 & = 43 + 49 \ 92 & =92\end{array}[/latex]
Example
Solve for H.
[latex]\Large \text{H}-16=13+19[/latex]
Step 1: Isolate the variable you are trying to find. In this case “H.” In order to do this we must remove the 16 from the left side of the equation. This is done by adding 16 to the left hand side of the equation and then also adding 16 to the right hand side of the equation.
[latex]\Large \begin{array}{c}\text{H}-16 = 13+19 \ \text{H} + \mathbf{(-16+16)}=13 + 19 \mathbf{ +16}\end{array}[/latex]
Step 2: Work through the equation.
[latex]\Large \begin{array}{c}\text{H}+(-16+16)=13+19+16 \ \text{H}+0=32+16 \ \text{H}=48 \end{array}[/latex]
Step 3: Check your answer.
[latex]\Large \begin{array}{l}\text{Replace H with 48} \ \downarrow \ \text{H}-16 = 13+19 \end{array}[/latex]
[latex]\Large \begin{array}{c} 48-16=13+19 \32=32 \end{array}[/latex]
Practice Questions
Now it’s time to do a couple of practice questions for yourself. Make sure to check the video answers to see how you did.
Question 1
Solve for S.
[latex]\Large 142+\text{S}=198+257[/latex]
Question 2
Solve for Y.
[latex]\Large \text{Y} - 22 = 51+53[/latex]
Transposing Equations Using Multiplication and Division
Transposing equations using multiplication and division uses the same basic principle as addition and subtraction in that whatever you do to one side you must do to the other.
Before we dive into this part of the chapter, we should refresh our memories a little and talk about reciprocals as it relates to math. Take a look at the following two numbers with one being a whole number and the other a fraction.
[latex]\Large 4 \text{ and } \dfrac{1}{4}[/latex]
Don’t forget that when we write the number 4, we could also write it as:
[latex]\Large\dfrac{4}{1}[/latex]
Do you remember what will happen if we multiplied those two together?
[latex]\Large 4 \times \dfrac{1}{4} = \dfrac{4}{4} = 1[/latex]
Reciprocals are numbers that when multiplied together equal 1. This is an important concept when transposing using multiplication and division. If we divide a number by its reciprocal and end up with one, we have essentially removed that number from the equation.
Here is an example. Solve for K.
[latex]\Large 8 \times \text{K} = 14 \times 17[/latex]
Similar to transposing equations with addition and subtraction, the first thing to do is to isolate the variable we are trying to find. In this case, “K.” This means that the 8 needs to be removed from the left hand side of the equation. Multiplying 8 by its reciprocal will give us a value of 1. Perfect!
[latex]\Large 8 \times \dfrac{1}{8} = \dfrac{8}{8} = 1[/latex]
Now adding this information to the equation, we would end up with this:
[latex]\Large \begin{array}{c} 8 \times \dfrac{1}{8} = \dfrac{8}{8} = 1 \ \mathbf{\dfrac{1}{8}} \times 8 \times \text{K}= 14 \times 17 \ \dfrac{8}{8} \times \text{K} = 14 \times 17\ 1\times \text{K}=14\times 17 \ \text{K}=14\times 17 \end{array}[/latex]
This would leave us with 1 × K on the left hand side of the equation which would end up being just “K” when multiplied together. This is exactly what we are looking for. We are not quite finished yet though. Now back to the golden rule.
Whatever you do to one side of the equation you must do to the other. Therefore, as we multiplied the left hand side of the equation by ⅛ we need to multiply the right hand side of the equation by ⅛.
[latex]\Large \dfrac{1}{8} \times 8 \times \text{K} = 14 \times 17 \times \dfrac{1}{8}[/latex]
So now if we followed this through we would end up with:
[latex]\Large \begin{array}\ \mathbf{\dfrac{1}{8}} \times 8 \times \text{K} = 14 \times 17 \times \mathbf{ \dfrac{1}{8}} \ \dfrac{8}{8} \times \text{K} = \dfrac{238}{8} \ \text{K} = 29.75 \end{array}[/latex]
We should do a check of the answer like we did previously to see if we are correct.
[latex]\Large \begin{array}{rl}\text{Replace K with 29.75} & \ \downarrow & \ 8 \times \text{K} & = 14 \times 17 \ 8 \times 29.75 & = 14 \times 17 \ 238 & = 238\end{array}[/latex]
Example
We’ll try another example, but this time we’ll deal with fractions that will have to be moved around. We’ll also start to do this in steps as we have done with previous questions.
Solve for L.
[latex]\Large \dfrac{4}{9} \times \text{L} = 12 \times 12[/latex]
Step 1: Isolate L. In this case, we have to move the 4/9 from one side to the other.In order to do this we have to multiply both sides by the reciprocal of 4/9. This will essentially remove 4/9 from the left hand side of the equation. The reciprocal is of 4/9 is 9/4.
[latex]\Large \dfrac{4}{9} \times \dfrac{9}{4} = \dfrac{36}{36} = 1[/latex]
Therefore we get:
[latex]\Large \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4}[/latex]
Step 2: Solve the equation
[latex]\Large \begin{array}{c} \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4} \ 1 \times \text{L} = 144 \times \dfrac{9}{4} \ \text{L} = \dfrac{1296}{4} \ \text{L}=324 \end{array}[/latex]
Step 3: Confirm your answer
[latex]\Large \begin{array}{rl}\text{Replace L with 324} & \ \downarrow & \ \dfrac{4}{9} \times \text{L} & = 12 \times 12 \ \dfrac{4}{9} \times 324 & = 12 \times 12 \ \dfrac{1296}{9} & = 144 \ 144 & = 144 \end{array}[/latex]
Example
This one is a bit more challenging. You’ll note in the question that there are no numbers, only letters. Take a look at the question, and see if you can come up with any ideas on how you would solve this equation.
Solve for “D” in the following equation.
[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}[/latex]
Step 1: Determine which variable you must isolate. In this case it’s given for us and it’s “D.”
[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\textbf{D}}[/latex]
The challenge here is that “D” is in the denominator (bottom) of the fraction. If we were to isolate it but it was still in the denominator of a fraction we would not have something that we could work with.
When “D” is isolated it must not only be by itself on one side of the equation but also appear as a whole number and not as the denominator of a fraction.
Step 2: To make this process easier break the equation down to look like the following.
[latex]\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}}[/latex]
Remember that:
[latex]\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{A}}{\text{B}}[/latex]
We haven’t really changed anything mathematically we have just made the equation into something that is easier to work with.
Step 3: Isolate “D.” In order to do this we must remove “C” from the right hand side of the equation. Do this by multiplying both the right and left hand side by 1/C.
[latex]\Large \mathbf{\dfrac{1}{\text{C}}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}} \times \mathbf{\dfrac{1}{\text{C}}}[/latex]
Now this might start looking a little bit complicated but if you follow it through you’ll start to see the answer form. Look at the right hand side of the equation. It now has both and C/1 and a 1/C. Multiplying those reciprocals together you get 1 and effectively take out the “C” on the right hand side.
[latex]\Large \dfrac{\text{C}}{1} \times \dfrac{1}{\text{C}}= \dfrac{\text{C}}{\text{C}}=1[/latex]
So we end up with:
[latex]\Large \dfrac{1}{\text{C}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= 1 \times \dfrac{1}{\text{D}}[/latex]
If we wanted to simplify this we could just do the following:
[latex]\Large \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}}[/latex]
Step 4: Get D into the numerator of the equation. What we need to do will take a little bit of math and some patience. We need to follow the same rules that we have been following.
The idea here is to multiply each side by D/1. This will make the right hand side of the equation equal to 1 but now the left hand side of the equation will have D in the numerator.
[latex]\Large \begin{array}{c}\dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}} \times \dfrac{\text{D}}{1} \ \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C}\times \text{B}} = 1 \end{array}[/latex]
What you’ll notice is that it actually creates more work but we have managed to get “D” into the numerator of the equation. The only problem is that we also have a bunch of other stuff on the same side as “D” and now our job becomes getting rid of all that stuff.
Now all we have to do is follow the rules and get the A, B and C over to the right hand side and isolate D. I’ll do this simply and in one quick calculation.
[latex]\Large \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} \times \dfrac{\text{C} \times \text{B}}{\text{A}} = \dfrac{\text{C} \times \text{B}}{\text{A}}[/latex]
What we end up with is:
[latex]\Large \text{D}= \dfrac{\text{C} \times \text{B}}{\text{A}}[/latex]
There is a lot of math involved there but if you follow the rules and go one step at a time you’ll eventually get there. My suggestion here is that you go over what we just went through a couple times before moving on. Understanding the math involved here is quite important when it comes to transposing formulas and equations.
At this point you might be wondering if there was a shortcut and as luck would have it there is for this procedure which I’ll go through now.. We’ll start with an example.
Take the following equation:
[latex]\Large \dfrac{3}{4}=\dfrac{3}{4}[/latex]
This equation is true. Now take each of the fractions and flip them around.
[latex]\Large \dfrac{4}{3} = \dfrac{4}{3}[/latex]
It seems that if you flip both fractions around then the equation is still true. Mathematically you are doing the same thing to one side as you are doing to the other. There is a lot of math involved here but the main point is that if you were to do all the math you would end up with the same answer.
How we can use this to help us simplify our question is to just do the following. Remember we started with:
[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}[/latex]
We needed to isolate D but the main problem (and the thing that causes us to do a lot of work) was that D was in the denominator and we needed it in the numerator.
Through our math wizardry we can get D into the numerator by just flipping around the left hand fraction. Whatever we do to that side we then must do to the right side. What we end up with is:
[latex]\Large \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}}[/latex]
All we need to do now is get C out of the left hand side by multiplying it by C/1 and then doing the same thing to the left hand side.
[latex]\Large \dfrac{\text{C}}{1} \times \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}} \times \dfrac{\text{D}}{1}[/latex]
We end up with…
[latex]\Large \dfrac{\text{C} \times \text{B}}{\text{A}} = \text{D}[/latex]
Now to take all that in in one shot might be a little much so you may want to go back and reread that whole explanation. But always keep in mind the math that goes along with that.
Example
In this example, we are going to use the formula for the area of a circle. Calculating the area of a circle is actually something we are going to do further into this volume, but for now we are just going to deal with the formula itself.
[latex]\Large \begin{array}{c} \text{Area of a circle} \ \text{A} = {\text{D}}^{2} \times 0.7854 \end{array}[/latex]
[latex]\Large \begin{array}{cl} \text{Where:} & \text{A = area of the circle} \ & \text{D = diameter of the circle} \end{array}[/latex]
Solving for the area of the circle would be pretty straight forward but as we are dealing with transposing equations in this section what we’ll do is solve for the diameter (D).
Step 1: Identify the variable you are trying to solve for.
[latex]\Large \text{A} = \mathbf{{\text{D}}^{2}} \times 0.7854[/latex]
From this you can see we have a couple problems we have to deal with. The first one is that we have to isolate D. The second one is that D has the exponent 2 attached to it so we have to somehow eliminate that.
Step 2: Isolate D.
For this we have to move the 0.7854 from the right hand side of the equation to the left hand side. We simply follow the rules we have used up to this point.
[latex]\Large \begin{array}{c} \mathbf{\dfrac{1}{0.7854}} \times \text{A} = {\text{D}}^{2} \times 0.7854 \times \mathbf{\dfrac{1}{0.7854}} \ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times \dfrac{0.7854}{0.7854} \ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times 1 \ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \end{array}[/latex]
That takes care of the first problem. Now we have to tackle the D2 issue.
Step 3: Remove the exponent from D.
Once again we go back to our original rule. Whatever you do to one side you must to the same thing to the other side.
We have to refer to the video that you watched earlier in this section regarding exponents and square roots. We’ll do a quick refresher before we solve for D.
Remember that:
[latex]\Large \begin{array}{c} {\text{D}}^{2} = \text{D} \times \text{D} \ \text{and... } \ \sqrt{\text{D} \times \text{D}} = \text{D} \end{array}[/latex]
So what we see is that if you square root a number that is squared (has an exponent of 2) you end up with the number itself. I’ll quickly show you with numbers.
[latex]\Large \begin{array}{c} {4}^{2} = 4 \times 4 \ {4}^{2} = 16 \ \sqrt{16} = \sqrt{4 \times 4} \ \sqrt{16} = 4 \end{array}[/latex]
Having gone through all that we can now solve the problem.
[latex]\Large \begin{array}{c} \sqrt{\dfrac{\text{A}}{0.7854}}= \sqrt{{\text{D}}^{2}} \ \sqrt{\dfrac{\text{A}}{0.7854}} = \text{D} \end{array}[/latex]
Practice Questions
Now it’s time to do a couple of practice questions for yourself. Make sure to check the videos answer to see how you did.
Question 1
Solve for B.
[latex]\Large \dfrac{\text{A} \times \text{B}}{\text{C}}=\dfrac{\text{D} \times \text{E}}{\text{F}}[/latex]
Question 2
Solve for C.
[latex]\Large \dfrac{\text{A} \times \text{B}}{\text{C}}=\dfrac{\text{D} \times \text{E}}{\text{F}}[/latex]
License
Math for Trades: Volume 2 Copyright © 2021 by Chad Flinn and Mark Overgaard is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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Research
Three simple rules to...
Three simple rules to ensure reasonably credible subgroup analyses
CCBYNC Open access
Research Methods & Reporting
Three simple rules to ensure reasonably credible subgroup analyses
BMJ
2015;
351
doi:
(Published 04 November 2015)
Cite this as: BMJ 2015;351:h5651
Article
Related content
Metrics
Responses
Peer review
James F Burke, assistant professor12,
Jeremy B Sussman, assistant professor23,
David M Kent, professor of medicine45,
Rodney A Hayward, professor of medicine23
Author affiliations
1Department of Neurology, University of Michigan School of Medicine, Ann Arbor, MI 48109-2800, USA
2VA Center for Clinical Management and Research, Ann Arbor
3Department of Internal Medicine, University of Michigan School of Medicine
4Department of Internal Medicine, Tufts University School of Medicine, Boston, MA, USA
5Predictive Analytics and Comparative Effectiveness Center, Institute for Clinical Research and Health Policy Studies, Tufts Medical Center, Tufts University School of Medicine
Correspondence to: J F Burke jamesbur@umich.edu
Accepted 2 October 2015
The limitations of subgroup analyses are well established—false positives due to multiple comparisons, false negatives due to inadequate power, and limited ability to inform individual treatment decisions because patients have multiple characteristics that vary simultaneously. In this article, we apply Bayes’s rule to determine the probability that a positive subgroup analysis is a true positive. From this framework, we derive simple rules to determine when subgroup analyses can be performed as hypothesis testing analyses and thus inform when subgroup analyses should influence how we practice medicine.
Summary points
Limitations of subgroup analyses are well established—false positives due to multiple comparisons, false negatives due to inadequate power, and limited ability to inform individual treatment decisions because patients have multiple characteristics that vary simultaneously. It remains uncertain when subgroup analyses should influence clinical practice
Categorical subgroup analyses should not be part of a typical clinical trial’s hypothesis testing analysis unless the prior probability for a subgroup effect being present is at least 20% and preferably higher than 50%
Rarely should more than one to two primary categorical subgroup analyses be performed
In trials with exceptional power to identify subgroup effects, hypothesis testing analyses of subgroups should be justified a priori
A table or figure reporting about a dozen subgroup analyses is a near ubiquitous feature of major clinical trial publications.1 2 The motivation behind these analyses is clear and compelling—to determine which patients most benefit from treatment, based on specific risk factors. However, the limitations of these analyses are well established—false positives due to multiple comparisons, false negatives due to inadequate power, and limited ability to inform individual treatment decisions because patients have multiple characteristics that vary simultaneously.3 When, if ever, should subgroup analyses, tested using subgroup treatment interactions, influence how we practice medicine?
Contrary to common belief, the well documented unreliability of subgroup analyses are not inherent; the same problems would arise for clinical trials themselves if we routinely performed underpowered trials examining haphazardly selected interventions. If properly selected (based on previous empirical evidence and current scientific theory), an adequately powered subgroup analysis can be a valid hypothesis testing endeavour. Trialists, reviewers, and editors should carefully consider such issues when making the essential scientific distinction between primary (that is, hypothesis testing) and secondary (that is, hypothesis generating) subgroup analyses.4 A positive, hypothesis testing analysis can directly influence patient care whereas a positive hypothesis generating analysis only calls for confirmatory research.
There are excellent general discussions of subgroup analyses and checklists to evaluate their credibility,3 5 6 7 but in this paper, we will quantitatively explore the key pragmatic question of when a subgroup analysis should be considered hypothesis testing versus hypothesis generating. We used simulation modelling to derive simple quantitative rules of thumb that can be applied by trialists, reviewers, and editors to ensure that subgroup analyses are properly contextualised and used by readers to quickly evaluate the credibility of a specific subgroup finding.
Predictive value of subgroup analyses as diagnostic tests: an analogy
Interpretation of a subgroup analysis is analogous to rigorously interpreting a diagnostic test. Before ordering a diagnostic test, a clinician considers the probability the person has the condition (the prior probability) and the accuracy of the test (often measured with sensitivity and specificity). With this information, the probability that a positive test is a true positive versus false positive can be estimated using Bayes’s rule: posterior odds=sensitivity÷(1−specificity)×prior odds.
Bayes’s rule can be seamlessly applied to the context of subgroup analysis, and informs why a shotgun approach to subgroup analysis fails. The sensitivity of a subgroup analysis is its statistical power: the probability of finding a true difference between groups if one exists. Most large clinical trials are powered to find a clinically meaningful difference between treatment and control groups around 80-90% of the time. Compared with the power for the trial’s main effect, most subgroup analyses have much less statistical power to identify subgroup effects. Power might often be closer to 20-30% for subgroup effect sizes similar in magnitude to the main treatment effect sizes (that is, a relative odds ratio for a subgroup treatment that is equal to the odds ratio for the overall treatment)8 9 Thus, the sample size needed to adequately contrast treatment effects measured in two different subgroups is much larger than the sample needed to distinguish an overall treatment effect from the null. Just as statistical power can be thought of as the sensitivity of a trial, the specificity of clinical trials is generally set to be 95%, based on the conventional significance threshold of P<0.05.
Finally, an estimate of the prior probability is needed to interpret a subgroup analysis. In both diagnostic testing and subgroup analyses, prior probability estimates are often unsettling given their inherent uncertainty and subjectivity, but failing to grapple with this tends to bias us towards falsely accepting new evidence as truth.10 Existing criteria to judge the credibility of subgroup analyses emphasise the importance of prior probability and specifically require that a hypothesis and its direction of effect are specified a priori and that the subgroup effect is supported by within-study empirical and biological evidence.6 In most cases, prior probability can be roughly estimated by thinking about the strength of previous theoretical or empirical evidence that the factor in question is likely to modify the relative treatment effect.11 Just as power for subgroups is usually much lower than for the main effect, so are the prior probabilities.
Expensive trials can only be justified if there is a reasonable probability of success based on prior data. In contrast, subgroup analyses with low priors are commonly conducted, perhaps because they are perceived as being essentially free, but as is shown below, conducting multiple subgroup analyses is statistically costly. We can also use empirical data as a rough starting point for thinking about prior probability. Of roughly 1200 subgroup analyses of recent clinical trials published in high impact journals, 83 (7%) were reportedly positive.1 Assuming a 5% false positive rate, only a fraction of these analyses were likely true positives. This observation is supported by the observation that less than 15% of these subgroup analyses met four of 10 criteria for credibility. So, a high-end starting point for the prior probability for the average published subgroup analysis is probably around 5%, which can be adjusted on a case by case basis, based on the prior empirical and theoretical evidence.
As per Bayes’s rule, low prior probabilities greatly increase the chance of a positive result being a false positive finding, and low power greatly increases this problem. Back to our analogy, the same phenomenon explains why ordering insensitive diagnostic tests with a low pretest probability leads to most positive test results being false positives (table 1⇓).12
Table 1
Comparison between diagnostic tests and subgroup analyses
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Application to subgroup analyses
Using this framework that has been applied in related contexts,13 we can calculate the probability of a positive subgroup analysis (treatment subgroup interaction) being a true positive versus false positive using Bayes’s rule.13 A positive analysis, in this context, refers to a significant difference in treatment effect between groups such that some groups can be demonstrated to have a greater or lesser relative treatment effect than other groups. Figure 1⇓ illustrates the association between prior probability and positive predictive value (the chance that a trial reporting a statistically significant result is not reporting a false positive) when subgroup statistical power is varied. In rare scenarios where there is excellent subgroup power, positive results can be highly reliable, but even in this ideal situation the positive predictive value drops precipitously when prior probability drops below 20% to 30% or when multiple subgroup analyses are performed without adjusting for multiple comparisons.
Fig 1 Association between prior probability and positive predictive values for subgroup analyses. The base case represents an uncommonly well powered, categorical subgroup analysis (that is, evenly divided subgroups, the effect the overall trial is powered for is entirely present in one subgroup, there is no effect is present in the other subgroup, and the trial has 90% power overall to find its primary effect), resulting in 37% power to find the subgroup effect. The probability of a true positive finding can be either reduced or improved by changing these assumptions. The lower line (multiple comparisons) illustrates how positive predictive values decline if 10 subgroup analyses are performed with 37% power to identify each subgroup effect and with no adjustment for multiple comparisons (as opposed to the one analysis illustrated in the base case). Conversely, the higher line (large subgroup effect) illustrates a scenario where the positive predictive values increase as power is improved—the effect size in the subgroup is twice as large as the base case (subgroup power=95%)
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Table 2⇓ quantifies the positive predictive value for statistically significant subgroup findings (that is, treatment subgroup interaction effects) over a wide range of prior probabilities and number of subgroup comparisons. Although we know of no established conventions for acceptable positive predictive values, for illustrative purposes we apply a minimum threshold of an 80% positive predictive value (that is, accepting a one in five chance (20%) of a positive finding being false). As this threshold is arbitrary and should differ depending on context,14 our results are presented so that readers can choose whatever threshold they wish. These analyses are based on P values of 0.05. When P values for subgroups are lower than 0.05, the positive predictive values will be higher. Like all rules of thumb, these three rules are proposed as only a starting point for thinking through individual cases.
Table 2
Positive predictive values (%) for significant subgroup findings according to prior probability and number of subgroup comparisons
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Rules of thumb for performing primary one-at-a-time subgroup analyses
Rule of thumb 1
Categorical subgroup analyses should not be part of a typical clinical trial’s primary (hypothesis testing) analysis unless the prior probability for a subgroup effect being present is at least 20% and preferably higher than 50%. Even under optimal circumstances, a subgroup analysis of a categorical variable will rarely have greater than 50% statistical power to detect a moderate subgroup effect, and more often is closer to 20%.9 In general, for a modestly powered subgroup effect (that is, 20% power) unless there is a strong prior probability (that is, 50%), the chance of a positive subgroup effect being a true positive finding will be less than 80% (table 2). Even when subgroup effect power is excellent (that is, 80%), to reach an 80% true positive threshold requires prior probability of 20%.
One seeming exception to this rule is that we recommend routinely conducting multivariable risk based analyses of trials with positive overall results and of negative trials in which the intervention has known harms (such as major surgery). We have so far only considered relative subgroup effects (for example, one group benefits from treatment while another group does not benefit). Risk based analyses examine how both the relative and absolute benefit of interventions varies between patients at high, medium, and low pretreatment risk. Even if a credible relative subgroup exists, it does not imply that treatment decisions should be different across subgroup levels, since treatment could still be worthwhile (or not worthwhile) in both groups. Further, risk based analyses can inform care even when the prior probability for a risk based subgroup effect is low.11 Even if no such effect exists, the absolute risk reduction with treatment will be higher in patients at high pretreatment risk and these analyses can quantify the magnitude of those differences.
Rule of thumb 2
Rarely should more than one to two primary categorical subgroup analyses be performed. When one study examines treatment differences across multiple risk factors (multiple comparisons), the likelihood that a study with a reported subgroup effect has at least one false positive result increases, thereby eroding the reliability of all positive findings. As shown in figure 1, the reliability of a positive finding falls considerably across the spectrum of prior probability when increasing the number of subgroup analyses. Even with an above average prior probability for a given subgroup comparison (that is, 20%) and excellent statistical power to find a subgroup effect (that is, 80%), the positive predictive value would fall from 80% if one comparison is made to 38% if 10 comparisons are made (table 2). Another, potentially less appreciated, consequence of performing multiple underpowered subgroup comparisons is that readers can be misled into believing that these negative results provide reliable evidence that the treatment is similarly effective in all patients, potentially masking important subgroup effects owing to inadequate power.9
Correction for multiple comparisons decreases the risk of false positives, but does not eliminate the broader problem. Firstly, correcting for multiple comparisons decreases statistical power, thus increasing the risk of false negative findings further. Secondly, as investigators add subgroup analyses with less and less evidence or theory to support them, the prior probability for the average subgroup analysis inevitably falls, further reducing the positive predictive value (fig 2⇓).
Fig 2 Effect of decreasing prior probability of each additional subgroup on positive predictive value (the likelihood that a positive finding is a “true positive”) adjusting for multiple comparisons. The two lines represent how the positive predictive value decreases under different assumptions about how prior probability decreases. In the blue line (slowly declining prior probability), prior probability starts at 0.4 and falls per extra subgroup either linearly (0.04 per subgroup). In the dotted line (rapidly declining prior probability), prior probability falls inversely to the number of subgroups
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Rule of thumb 3
In trials with exceptional power to identify subgroup effects, hypothesis testing subgroup analyses should be justified a priori. Various trial circumstances can lead to increased power to identify subgroup effects (table 3⇓). If, for example, the effect size difference between subgroups is larger than that found for the overall trial population, the probability of a true positive subgroup effect increases (fig 1). Although such effects are likely uncommon, they are not unheard of—and they are the effects that are of most clinical importance. For example, large subgroup effects were anticipated and then discovered for higher degrees of carotid stenosis in symptomatic endarterectomy trials,15 and in time to reperfusion for stroke16 and myocardial infarction.17 18 Similarly, in biologically directed treatments, such as for cancer treatments, it is possible that a treatment will be effective in one subgroup and ineffective in another.
Table 3
Changes in power to identify subgroup effects after changes in specification of subgroup effect
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In addition to anticipated subgroup effect size, equalising the proportion of individuals in each subgroup and testing a continuous variable (such as baseline blood pressure level) along its continuum will substantially improve statistical power.8 19 But even with excellent power, prior probability is an important determinant of whether a significant subgroup effect is a true positive. Just as an excellent diagnostic test (sensitivity 99%, specificity 95%) results in a 20% positive predictive value if the prior probability of disease is 1%, only 20% of statistically significant subgroup effects will be true positives if the subgroup effect only had a 1% chance of occurring to begin with, even with 99% power.
This demonstrates why looking at multiple variables might be fine for hypothesis generating analyses, but that primary (hypothesis testing) analyses must be based on previous empirical evidence or theory, generally be few in number, and be specified a priori. For unique cases with improved power to detect subgroup effects, after formally estimating subgroup effect power before the trial and considering the prior probability, table 2 can inform whether power is adequate to consider such an analysis as a hypothesis testing analysis versus a hypothesis generating analysis.
Conclusion
By not following a sound scientific process, conventional subgroup analyses increase the risk of both false positive and false negative findings. Careful consideration of the likelihood of a subgroup effect being present (prior probability) and the statistical power of the subgroup analysis (sensitivity) need to inform whether a subgroup analysis should be part of the primary (hypothesis testing) or exploratory (hypothesis generating) analyses. We recommend that hypothesis testing analyses include no more than one or two prespecified subgroup analyses founded on adequate prior probability and power so that positive findings are more reliable and can thus be used to target treatment.
This approach would substantially restrict the number of subgroup analyses performed. It can be argued that subgroup analyses that do not meet these criteria should never be performed because false positives will greatly outnumber true positives and could be integrated into clinical decisions in spite of the best intentions of researchers. However, there is also a reasonable argument supporting a limited role for exploratory hypothesis generating analyses of subgroups. For example, there was little reason to think that diabetics would fare better with coronary artery bypass than with percutaneous interventions before an exploratory subgroup analysis of the BARI trial.20 Although still somewhat controversial,21 the balance of evidence argues that this is a real subgroup effect that would not have been discovered without an exploratory analysis.22 23 At least, if such analyses are performed, they should be specifically designated as exploratory, broadly understood to be hypothesis generating and reported separately from hypothesis testing analyses. For clinicians, these exploratory analyses should be ignored until confirmed or refuted by subsequent studies.11
Subgroup analyses have historically misinformed as much as they have informed.3 The three simple rules outlined here can help guide more meaningful and accurate analyses and reporting of randomised controlled trials, better guide clinical implementation, and avoid repeating mistakes of the past.
Notes
Cite this as: BMJ 2015;351:h5651
Footnotes
We thank Tim Cole for a careful reading and useful comments on an earlier draft of this manuscript.
Contributors: The authors have experience designing and implementing research methods to individualise treatment decision making, including simulation analyses. RAH conceptualised the article and revised the manuscript. JBS performed the analyses and wrote the initial draft. JBS and DMK provided substantial intellectual feedback and revised the manuscript.
Competing interests: We have read and understood the BMJ Group policy on declaration of interests and declare no competing interests.
Provenance and peer review: Not commissioned; externally peer reviewed.
This is an Open Access article distributed in accordance with the Creative Commons Attribution Non Commercial (CC BY-NC 4.0) license, which permits others to distribute, remix, adapt, build upon this work non-commercially, and license their derivative works on different terms, provided the original work is properly cited and the use is non-commercial. See:
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7508 | https://mathoverflow.net/questions/107355/minimizing-product-subject-to-linear-constraints | Skip to main content
Minimizing product subject to linear constraints
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I am looking for a solver that allows me to solve an optimization problem of the form
minimizesubject tox1x2⋯xn(some linear constraints)
I've used Gurobi before. However, I couldn't find the way to include products in the objective function as well as in the constraints.
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edited Mar 10, 2021 at 14:34
Rodrigo de Azevedo
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asked Sep 17, 2012 at 2:39
GuillermoGuillermo
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What type of variables are xi?
– RobPratt
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This is a hard problem (maximizing the product is a bit better one, as sometimes one can take log of the objective function, and it becomes concave...). Your best shot might be to use the sum of squares approach for polynomial optimization, as implemented e.g. in YALMIP.
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answered Sep 17, 2012 at 5:34
Dima PasechnikDima Pasechnik
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Does YALMIP support multiobjective optimization (possibly only linear)?
– joro
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not that I know for sure, but I doubt this. Multiobjective optimization is very hard, in general, as you'd be computing a multidimensional object.
– Dima Pasechnik
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Recently discovered minizinc
MiniZinc is a medium-level constraint modelling language. It is high-level enough to express most constraint problems easily, but low-level enough that it can be mapped onto existing solvers easily and consistently. It is a subset of the higher-level language Zinc. We hope it will be adopted as a standard by the Constraint Programming community.
FlatZinc is a low-level solver input language that is the target language for MiniZinc. It is designed to be easy to translate into the form required by a solver.
There are several backends for the translated problem (MIP, SAT, etc).
Here is how something similar to your question will look like in minizinc:
```
var int: a;
var int: b;
constraint a + b <= 10;
constraint a>0;
constraint b>0;
solve maximize ab;
output [ show([a,b]) ];
[5, 5]
```
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answered Sep 17, 2012 at 6:20
jorojoro
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Hm, in your case you need "solve minimize ....;"
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7509 | https://apps.p-95.com/app_direct_i/drivesamplesize/_/ | ===============
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This web application performs sample size calculations for cohort and test negative design studies on brand-specific vaccine effectiveness (VE), and relative VE of one vaccine over another. The sample size calculations are adjusted for confounders, and subject dropout. For sample sizes in test negative design studies, this application also provides the size of the catchment that is required to obtain the required sample size.
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References
The calculations for minimum detectable VE are done using the epi.sscohortc function of the R package epiR which internally uses the approach described in Woodward M (2005). Epidemiology Study Design and Data Analysis. Chapman & Hall/CRC, New York, pp. 381 - 426.
The calculations for expected lower and upper limits of confidence intervals for VE are based on simulations. The confidence intervals are obtained using the approach described in Morris, J. A., & Gardner, M. J. (1988). Statistics in medicine: Calculating confidence intervals for relative risks (odds ratios) and standardised ratios and rates. British medical journal (Clinical research ed.), 296(6632), 1313. Please note that this method does not use an exact confidence interval.
Cohort study (incidence rate ratio)
We define absolute VE of a vaccine as 1 - Ratio of incidence rate of being a case in vaccinated subjects over unvaccinated subjects . Similarly, we define relative VE as 1 - Ratio of incidence rate of being a case in subjects vaccinated with one vaccine over another vaccine .
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The calculations for minimum detectable VE are done using the epi.sscohortt function of the R package epiR which internally uses the approach described in Lwanga S, Lemeshow S (1991). Sample Size Determination in Health Studies. World Health Organization, Geneva.
The calculations for expected lower and upper limits of confidence intervals for VE are based on simulations. The confidence intervals are obtained using the approach described in Rothman KJ (2012) Epidemiology: An Introduction. 2nd Ed., Oxford University Press, Oxford. Please note that this method does not use an exact confidence interval.
Test negative design studies (odds ratio)
We define absolute VE of a vaccine as 1 - odds of being vaccinated among cases divided by the odds of being vaccinated among control subjects . Similarly, we define relative VE as 1 - odds of being vaccinated with a particular vaccine versus the reference vaccine among cases divided by odds being vaccinated with a particular vaccine versus the reference vaccine among controls .
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The calculations for minimum detectable VE are done using the epi.sscc function of the R package epiR which internally uses the approach described in Dupont WD (1988) Power calculations for matched case-control studies. Biometrics 44: 1157 - 1168.
The calculations for expected lower and upper limits of confidence intervals for VE are based on simulations. The confidence intervals are obtained using the approach described in Morris, J. A., & Gardner, M. J. (1988). Statistics in medicine: Calculating confidence intervals for relative risks (odds ratios) and standardised ratios and rates. British medical journal (Clinical research ed.), 296(6632), 1313. Please note that this method does not use an exact confidence interval.
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7510 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4840009/ | Gravity-assisted drainage imaging in the assessment of pediatric hydronephrosis - PMC
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. 2016 Mar-Apr;10(3-4):96–100. doi: 10.5489/cuaj.3237
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Gravity-assisted drainage imaging in the assessment of pediatric hydronephrosis
Matthew R Acker
Matthew R Acker, MD, FRCSC
1 Department of Urology, University of Miami, Miami, FL, U.S.;
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1, Roderick Clark
Roderick Clark, MD
2 Division of Urology, University of Western Ontario, London, ON, Canada;
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2,✉, Peter Anderson
Peter Anderson, MD, FRCSC
3 Department of Urology, Dalhousie University, Halifax, NS, Canada
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3
Author information
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1 Department of Urology, University of Miami, Miami, FL, U.S.;
2 Division of Urology, University of Western Ontario, London, ON, Canada;
3 Department of Urology, Dalhousie University, Halifax, NS, Canada
✉
Correspondence: Dr. Roderick Clark, Division of Urology, University of Western Ontario, London, ON, Canada; Roderick.Clark@lhsc.on.ca
Copyright: © 2016 Canadian Urological Association or its licensors
PMC Copyright notice
PMCID: PMC4840009 PMID: 27217854
See commentary "The challenges of diagnosing obstructive hydronephrosis in children" on page 101.
See "The challenges of diagnosing obstructive hydronephrosis in children" on page 101.
Abstract
Introduction:
As early detection of hydronephrosis increases, we require better methods of distinguishing between pediatric patients who require pyeloplasty vs. those with transient obstruction. Gravity-assisted drainage (GAD) as part of a standardized diuretic renography protocol has been suggested as a simple and safe method to differentiate patients.
Methods:
Renal scans of 89 subjects with 121 hydronephrotic renal units between January 2004 and March 2007 were identified and analyzed.
Results:
Of all renal units, 65% showed obstruction. GAD maneuver resulted in significant residual tracer drainage in eight renal units, moderate drainage in 12 renal units, and some improvement in 40 units after the GAD maneuver. Of the eight renal units with significant residual tracer drainage, only two proceeded to pyeloplasty. After pyeloplasty, nine children had improved time to half maximum (T 1/2 Max) and 13 were unchanged.
Conclusions:
Our study was limited due to its retrospective design and descriptive analyses, but includes a sufficient number of subjects to conclude that GAD as part of a diuretic renography protocol is an effective and simple technique that can help prevent unnecessary surgical procedures in pediatric patients.
Introduction
With the adoption of antenatal sonography as a routine screening tool, early detection of hydronephrosis has increased.1,2 When pelvicaliceal system distention is detected, it often leads to further investigation, including diuretic renography. This imaging modality can help identify patients with ureteropelvic junction obstruction (UPJO) and compromised renal function.3 Early identification of significant UPJO allows for interventions such as pyeloplasty, which can improve kidney drainage and preserve renal function. Those patients with mechanical obstruction suggestive of UPJO on diuretic renography, significant flank pain, or recurrent urinary tract infections are often considered for pyeloplasty, particularly if there is evidence of decline in differential renal function (DRF) on serial diuretic renal scans.4 Selecting which patients will benefit from pyeloplasty remains a challenge. There is a lack of standardized protocols for both performing and interpreting diuretic renography. Additionally, many centres image patients in the supine position, which may underestimate drainage compared to the upright position, where the effects of gravity can help facilitate drainage (Fig. 1).5
Fig. 1.
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Balloon analogy for describing gravity-assisted drainage. UPJ: ureteropelvic junction.
Methods
The present study was approved by the Research Ethics Board at the IWK Health Centre. Ultrasonography was used to identify the presence of hydronephrosis, after which patients were referred for diuretic renal scans. Diuretic renography involves serial imaging of an injected radioactive tracer as it is filtered or secreted within the kidney through to its excretion in urine. Each child received simultaneous intravenous injection of 5 MBq/kg Tc 99m mercapto-acetyl-triglycine (MAG-3) and 1 mg/kg (to a maximum of 40 mg) of furosemide. After radiotracer injection, the patient is placed supine over a gamma camera that takes serial images of the radiotracer transit through the kidney, typically over a period of 20–30 minutes. Computer analysis of the uptake and excretion phases is performed to quantitate differential renal function (right vs. left), time to half maximum tracer activity (T ½ max), as well as the degree and significance of any obstruction. T ½ max of less than 10 minutes indicates the absence of obstruction, while a T 1/2 max of greater than 20 minutes is compatible with obstruction. A T 1/2 max between 10 and 20 minutes is considered equivocal.6 A gravity-assisted drainage (GAD) image was then performed at the end of the renal scan. The GAD image is a single, static image obtained after positioning the child in the upright position for five minutes to promote additional drainage of tracer from the collecting system. Percent drainage was calculated after the GAD maneuver. Residual tracer drainage >50% was considered significant, while 30–49% was considered moderate, 10–29% showing some improvement, and <10% was classified as no change. Followup was limited to the first postoperative renal scan, typically performed 3–6 months post-pyeloplasty.
Dismembered pyeloplasty was offered to patients who had an obstructed renal unit and a low GAD score on serial imaging, an obstructed renal unit with deterioration in renal function on serial imaging, or were experiencing symptoms or complications (e.g., flank pain or pyelonephritis) related to their UPJO. For patients who underwent pyeloplasty, percent change in dynamic renal function was assessed through comparison of preoperative and most recent postoperative diuretic renography, including T ½ max calculations and quantification of GAD.
Our study included children who were referred for GAD renography between January 2004 and March 2007. Scans were excluded from analysis if they showed duplicated collecting systems, ectopic ureter or ureterocele (n= 25), megaureter or ureterovesical junction obstruction (n=21), vesicoureteral reflux (n= 19), incomplete studies (n=7), presence of horseshoe kidney (n=4), non-functioning renal unit (n=6), or posterior urethral valves (n=3). Descriptive statistics were used in this observational study.
Results
Renal scans of 174 consecutive children with sonographically detected hydronephrosis who were referred for GAD renography between January 2004 and March 2007 were reviewed; 85 scans were excluded from analysis, as per exclusion criteria. Data from 18 of the remaining renal scans were missing complete T 1/2 max data and were excluded from analyses. After exclusion, 89 subjects with 121 hydronephrotic renal units remained for analysis (Fig. 2). Among the scans remaining, 76 were left-sided, 21 were right-sided, and 12 showed bilateral hydronephrosis (for 24 renal units), corresponding to a 2.7:1 ratio of left- to right-sided obstruction. Twenty-two pyeloplasty procedures were performed in our analyses (Table 1).
Fig. 2.
Open in a new tab
Summary of patient management. GAD: gravity-assisted diuretic renography.
Table 1.
Descriptive statistics by renal unit
| | | BaselineT 1/2 normal (<10 min) | Baseline T 1/2 equivocal (10–20 min) | BaselineT 1/2 abnormal (>20 min) | Total |
:---: :---: :---: |
| Sex | Mean time | 8.47 min | 16.39 min | Not reached | 14.10 |
| Female | 4 | 1 | 21 | 26 |
| Male | 13 | 24 | 58 | 95 |
| Pyelopasty | Yes | 0 | 4 | 22 | 26 |
| No | 17 | 21 | 57 | 95 |
| Laterality | Left | 8 | 15 | 53 | 76 |
| Right | 1 | 4 | 16 | 21 |
| Bilateral | 8 | 6 | 10 | 24 |
Open in a new tab
Of all renal units, 79 (65%) showed renographic evidence of obstruction (T 1/2>20 minutes). The average T ½ max in normal unobstructed renal units (T 1/2<10 minutes) was 8.56 minutes. For equivocal renal units (T 1/2 max 10–20 minutes), the average time to drainage was 15.60 minutes. For the 79 obstructed renal units (T 1/2 max >20 minutes) the GAD maneuver resulted in significant (>50%) residual tracer drainage in eight renal units (10%), moderate (30–49%) drainage in 12 renal units (15%), some improvement (10–29% change) in 40 units (51%), and no improvement (<10%) in 19 units (24%) (Fig. 2). Of the eight renal units with a significant residual tracer drainage, only two proceeded to pyeloplasty, as they had multiple episodes of hydronephrosis that necessitated eventual operative management.
Pre- and post-pyeloplasty data on GAD percent change for obstructed renal units (T 1/2 max >20) was analyzed for all 22 individuals (Fig. 2). Post-pyeloplasty T 1/2 max was normal in two individuals, equivocal in eight individuals and unchanged (T 1/2 max >20) in 12 individuals. Following the pyeloplasty procedure, the GAD maneuver resulted in significant (>50%) residual tracer drainage in no renal units (0%), moderate (30–49%) drainage in three renal units (14%), some improvement (10–29% change) in six units (27%), and no improvement in 13 units (59%) (Fig. 3).
Fig. 3.
Open in a new tab
Pre- vs post-pyeloplasty gravity-assisted drainage change in obstructed (T 1/2 max >20 min) renal units.
Discussion
GAD as part of a diuretic renography protocol is an effective, simple procedure that may provide additional diagnostic information in pediatric patients with hydronephrosis being considered for surgical management. In our retrospective study, eight renal units (10%) showed a significant (>50%) improvement in drainage with the GAD maneuver, while 12 (15%) showed moderate improvement (30–49%) in residual tracer activity following this simple preoperative maneuver. This suggests that 25% of these “obstructed” renal units were not truly obstructed, but simply required gravity to assist drainage of residual urine from an unobstructed, capacious collecting system. We demonstrated that this simple maneuver during diuretic renography obviated the need for pyeloplasty in six of eight “obstructed” renal units.
The GAD procedure has been previously reported4,7 and has been recommended by at least one clinical practice guideline,8 although it has not become standard of practice at all institutions. This procedure is easy to implement, carries no risk to the patient, and is demonstrated to have acceptable sensitivity and specificity to differentiate children who truly have obstructed renal units from those who do not.4 Our results are also consistent with other reports on the advantage of the GAD maneuver for improving the diagnostic accuracy of diuretic renography. Amarante et al (2003) performed a retrospective review of 24 patients that showed 43% of kidneys were classified as adequately drained on a post-void GAD image, as compared to 13% in the pre-void image.9
Other studies have also demonstrated the need for diagnostic accuracy using diuretic renography before undertaking pyeloplasty. Koff et al (2005) demonstrated that obstructive appearance on diuretic renography was ultimately determined to be non-obstructed in 40% of renal units, and that this finding was particularly evident in children under the age of two years.10 They highlight the need for caution in the use of diuretic renography to determine if surgical management is required.10 Yang et al (2010) showed that an early observation period is beneficial before undertaking pyeloplasty, especially given that many subjects with asymptomatic Grade 1 and 2 hydronephrosis have benign conditions that do not often improve with intervention. They demonstrated some benefit from pyeloplasty in individuals with more severe (≥ Grade 3 hydronephrosis) obstruction,11 although the literature has been mixed in this area.12,13 Our series showed overall little benefit of pyeloplasty, with return to normal function in only two renal units. Helmy et al (2012) showed improvement in hydronephrosis on all subjects who underwent pyeloplasty, although 65% continued to show persistent obstruction on diuretic renography.14 A systematic review in this area by Castagnetti et al (2008) concluded that symptomatic patients diagnosed postnatally, and those with moderate rather than severely compromised preoperative function seem to have the greatest chance of functional improvement after surgery.15
There are several limitations to our study. Our data were obtained retrospectively, which limited the type of information we could collect and the types of statistical analyses we could perform. Our followup period was also limited to the first postoperative renal scan in patients who underwent a pyeloplasty. Future studies should follow patients longitudinally to determine long-term renal function. It is possible that the small number of participants has biased our observations, although our results are similar to trends noted in previous studies.4,7 Since this is an observational study, we did not perform calculations to determine appropriate sample size, but felt that our analysis of these 174 scans before exclusion criteria represented an appropriate and practical sample for our analyses.
Conclusion
GAD imaging during diuretic renography is safe and effective, and provides valuable, additional diagnostic information in pediatric patients with hydronephrosis and possible UPJO. It has become the standard practice at our institution.
Acknowledgments
Dr. Clark was supported by the Dalhousie University Faculty of Medicine 2013 Department of Urology Ralph Pickard Bell Endowment.
Footnotes
Competing interests: The authors declare no competing financial or personal interests.
This paper has been peer-reviewed.
References
1.Tripp BM, Homsy YL. Neonatal hydronephrosis—the controversy and the management. Pediatr Nephrol Berl Ger. 1995;9:503–9. doi: 10.1007/BF00866741. [DOI] [PubMed] [Google Scholar]
2.Nguyen HT, Herndon CDA, Cooper C, et al. The Society for Fetal Urology consensus statement on the evaluation and management of antenatal hydronephrosis. J Pediatr Urol. 2010;6:212–31. doi: 10.1016/j.jpurol.2010.02.205. [DOI] [PubMed] [Google Scholar]
3.Belarmino JM, Kogan BA. Management of neonatal hydronephrosis. Early Hum Dev. 2006;82:9–14. doi: 10.1016/j.earlhumdev.2005.11.004. [DOI] [PubMed] [Google Scholar]
4.Wong DC, Rossleigh MA, Farnsworth RH. Diuretic renography with the addition of quantitative gravity-assisted drainage in infants and children. J Nucl Med Off Publ Soc Nucl Med. 2000;41:1030–6. [PubMed] [Google Scholar]
5.Houben CH, Wischermann A, Börner G, et al. Outcome analysis of pyeloplasty in infants. Pediatr Surg Int. 2000;16:189–93. doi: 10.1007/s003830050720. [DOI] [PubMed] [Google Scholar]
6.Shulkin BL, Mandell GA, Cooper JA, et al. Procedure guideline for diuretic renography in children 3.0. J Nucl Med Technol. 2008;36:162–8. doi: 10.2967/jnmt.108.056622. [DOI] [PubMed] [Google Scholar]
7.Rossleigh MA, Leighton DM, Farnsworth RH. Diuresis renography. The need for an additional view after gravity-assisted drainage. Clin Nucl Med. 1993;18:210–3. doi: 10.1097/00003072-199303000-00005. [DOI] [PubMed] [Google Scholar]
8.Gordon I, Piepsz A, Sixt R, Auspices of Pediatric Committee of European Association of Nuclear Medicine Guidelines for standard and diuretic renogram in children. Eur J Nucl Med Mol Imaging. 2011;38:1175–88. doi: 10.1007/s00259-011-1811-3. [DOI] [PubMed] [Google Scholar]
9.Amarante J, Anderson PJ, Gordon I. Impaired drainage on diuretic renography using half-time or pelvic excretion efficiency is not a sign of obstruction in children with a prenatal diagnosis of unilateral renal pelvic dilatation. J Urol. 2003;169:1828–31. doi: 10.1097/01.ju.0000062640.46274.21. [DOI] [PubMed] [Google Scholar]
10.Koff SA, Binkovitz L, Coley B, et al. Renal pelvis volume during diuresis in children with hydronephrosis: Implications for diagnosing obstruction with diuretic renography. J Urol. 2005;174:303–7. doi: 10.1097/01.ju.0000161217.47446.0b. [DOI] [PubMed] [Google Scholar]
11.Yang Y, Hou Y, Niu ZB, et al. Long-term followup and management of prenatally detected, isolated hydronephrosis. J Pediatr Surg. 2010;45:1701–6. doi: 10.1016/j.jpedsurg.2010.03.030. [DOI] [PubMed] [Google Scholar]
12.Gordon I. Diuretic renography in infants with prenatal unilateral hydronephrosis: An explanation for the controversy about poor drainage. BJU Int. 2001;87:551–5. doi: 10.1046/j.1464-410X.2001.00081.x. [DOI] [PubMed] [Google Scholar]
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15.Castagnetti M, Novara G, Beniamin F, et al. Scintigraphic renal function after unilateral pyeloplasty in children: A systematic review. BJU Int. 2008;102:862–8. doi: 10.1111/j.1464-410X.2008.07597.x. [DOI] [PubMed] [Google Scholar]
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Change of basis and inner product in non-orthogonal basis
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I have a vector, originally expressed in the standard coordinates system, and want to perform a change of basis and find coordinates in another basis, this basis being non-orthogonal.
Let B={e 1,e 2}B={e 1,e 2} be the standard basis for R 2 R 2.
Let B′={e′1,e′2}B′={e 1′,e 2′} be a non-orthogonal basis for R 2 R 2.
Let v v be some vector in R 2 R 2.
The standard inner product is
⟨a,b⟩=∑n i=0 a i b i.⟨a,b⟩=∑i=0 n a i b i.
I want to define an inner product in the non-orthogonal basis B′B′ so that ⟨e′1,e′2⟩B′=0⟨e 1′,e 2′⟩B′=0 since ∑n i=0 e′1 i e′2 i≠0∑i=0 n e 1 i′e 2 i′≠0.
Basically, I want to use this new inner product to get the component/coordinates of the vector v v on the basis B′B′.
linear-algebra
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edited Dec 1, 2020 at 10:07
Mateen Ulhaq
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joseph M'Bimbi-Benejoseph M'Bimbi-Bene
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1 By the way, ⟨,⟩⟨,⟩ may be achieved with \langle, \rangle. They play more nicely with the TeX layouts than < and > .Patrick Stevens –Patrick Stevens 2015-08-25 07:32:35 +00:00 Commented Aug 25, 2015 at 7:32
What you want seems confused to me. You do not need any inner product to perform a change of basis, and you certainly shouldn't be using two different inner products unless you have a very specific reason to do so.Marc van Leeuwen –Marc van Leeuwen 2015-08-25 07:45:19 +00:00 Commented Aug 25, 2015 at 7:45
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What you want to do is change the basis of the vectors you are working with, then take the inner product on that. since e′1,e′2 e 1′,e 2′ are a basis for R 2 R 2, every vector v v = v 1 e′1+v 2 e′2 v 1 e 1′+v 2 e 2′ for unique values v 1,v 2 v 1,v 2. Then you set ⟨v,w⟩=v 1 w 1+v 2 w 2⟨v,w⟩=v 1 w 1+v 2 w 2
To find an explicit method of converting a vector into this new coordinate system, look into change-of-basis matrices. if e′1=a 1 e 1+a 2 e 2,e′2=b 1 e 1+b 2 e 2 e 1′=a 1 e 1+a 2 e 2,e 2′=b 1 e 1+b 2 e 2, take the matrix with
(a 1 a 2 b 1 b 2)(a 1 b 1 a 2 b 2)
and take its inverse. Then if ⟨v,w⟩′⟨v,w⟩′ is your new inner product, you have ⟨v,w⟩′=⟨M−1 v,M−1 w⟩⟨v,w⟩′=⟨M−1 v,M−1 w⟩.
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edited Aug 25, 2015 at 8:02
Marc van Leeuwen
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Let u,v be any lin ind vectors in the plane. Define u.u=a>0,u.v=0=v.u,v.v=b>0. Extend this linearly for all pairs of vectors. It is easily verified that this is an inner product.
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answered Aug 25, 2015 at 7:44
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If you write down a 2×2 2×2 matrix P P whose columns are formed by the coordinates of e′1,e′2 e 1′,e 2′ in the standard basis, then multiplication (on the left) by P P does the following in terms of coordinates. If x,y∈R x,y∈R these can be interpreted as coordinates of a vector v v with respect to the new basis e′1,e′2 e 1′,e 2′, in other words one takes v=x e′1+y e′2 v=x e 1′+y e 2′; then (r s)=P(x y)(r s)=P(x y) gives the coordinates of v v in the standard basis e 1,e 2 e 1,e 2, that is, v=r e 1+s e 2 v=r e 1+s e 2. To see why this is true, note that by definition of P P it is true when (x,y)=(1,0)(x,y)=(1,0) and also when (x,y)=(0,1)(x,y)=(0,1) (one gets v=e′1 v=e 1′ respectively v=e′2 v=e 2′); the general case now follwos by taking linear combinations of these special cases.
But what you want is the opposite: you want to find the coordinates in the new basis of a vector given by its coordinates (r s)(r s) in the standard basis. Therefore you need to compute the inverse matrix P−1 P−1, and the desired coordinate change is given by (left) mluitplication by P−1 P−1.
Note that no inner product plays a role in any way when doing change of basis.
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answered Aug 25, 2015 at 8:00
Marc van LeeuwenMarc van Leeuwen
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obtaining the components on {e 1,e 2}{e 1,e 2} was just an illustration, i really wanted to define an inner product on this basis. And thank you, your first answer gave me the solution joseph M'Bimbi-Bene –joseph M'Bimbi-Bene 2015-08-25 11:32:11 +00:00 Commented Aug 25, 2015 at 11:32
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7512 | https://www.quora.com/Im-doing-Brilliant-orgs-complex-algebra-course-and-complex-conjugates-are-coming-up-a-lot-What-makes-them-so-important-Is-it-just-that-z-conjugate-z-z-%C2%B2-or-is-there-more-to-it | Something went wrong. Wait a moment and try again.
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I'm doing Brilliant.org's complex algebra course and complex conjugates are coming up a lot. What makes them so important? Is it just that zconjugate(z) ≡||z||² or is there more to it?
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There’s a lot more to it, although this is one of those situations where the full importance isn’t clear until you get into much more advanced mathematics. But I can give you a taste of one important part.
The first thing I’m going to throw around is the word “field”. A field is, essentially, a mathematical structure which supports the four basic arithmetic operations: addition, subtraction, multiplication, and division. The integers do not form a field (because you can’t divide an integer by an integer and get an integer, in general), but the rational numbers, real numbers, and complex numbers
There’s a lot more to it, although this is one of those situations where the full importance isn’t clear until you get into much more advanced mathematics. But I can give you a taste of one important part.
The first thing I’m going to throw around is the word “field”. A field is, essentially, a mathematical structure which supports the four basic arithmetic operations: addition, subtraction, multiplication, and division. The integers do not form a field (because you can’t divide an integer by an integer and get an integer, in general), but the rational numbers, real numbers, and complex numbers all do.
Now here’s the big claim: From the perspective of the field of real numbers, complex conjugation is a symmetry of the field of complex numbers.
Pick any non-real complex number z and its complex conjugate z∗. From these two numbers, construct a true equation using only addition, subtraction, multiplication, division, and any real numbers that you like.
As a simple example, let’s pick z=i. Here are some equations that it satisfies:
z2+1=0
zz∗=−1
2z+z∗=z
Keep going, pick any that you want. The equation just has to be true.
If you swap z and z∗ in any of these equations, you get a different equation which is still true.
In the first example, z∗2+1=0. In the second example, z∗z=−1. And so on. Keep going until you’re satisfied.
It doesn’t matter what complex number you pick, and it doesn’t matter what equation you pick. As long as you restrict yourself to equations involving only that number, its complex conjugation, field operations, and real numbers, this always works.
So there is a deep sense in which if you only know about the field of real numbers, and then try to understand the field of complex numbers using only the field of real numbers, you can’t tell the difference between z and z∗, for any z.
As a matter of interest, you may already have noticed something similar when working with quadratic equations. If you have a quadratic equation with rational coefficients, the irrational roots always seem to have the same structure: a±√b.
Essentially the same thing is going on here. From the perspective of the field of rational numbers, there is no difference between √b and −√b.
Of course, you can tell the roots apart if you allow inequalities, since −√2<√2. This is why we stuck to equalities only. The complex numbers don’t have an ordering like that, so it doesn’t apply anyway.
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One quick answer: Consider any polynomial of degree n and having real-number coefficients. Counting multiplicity (i.e., repeated zeros), the fundamental theorem of algebra states that there are exactly n zeros, which are, at most, complex numbers.
What is interesting about the case where the zeros includes nonreal (complex) numbers, is that the complex zeros come in conjugate pairs. For a simple example, consider P(x) = x^3 + x. Set this to zero for the zeros:
x^3 +x = 0 —-> x(x^2 + 1) = 0 —-> x = 0 or x^2 + 1 = 0. Focus on this quadratic to get x = i or x = -i. The zeros of P are 0, i, -i, and
One quick answer: Consider any polynomial of degree n and having real-number coefficients. Counting multiplicity (i.e., repeated zeros), the fundamental theorem of algebra states that there are exactly n zeros, which are, at most, complex numbers.
What is interesting about the case where the zeros includes nonreal (complex) numbers, is that the complex zeros come in conjugate pairs. For a simple example, consider P(x) = x^3 + x. Set this to zero for the zeros:
x^3 +x = 0 —-> x(x^2 + 1) = 0 —-> x = 0 or x^2 + 1 = 0. Focus on this quadratic to get x = i or x = -i. The zeros of P are 0, i, -i, and the sole nonreal (complex) ones are conjugates of each other.
Douglas Magowan
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Some properties of complex conjugates.
¯z+¯w=¯¯¯¯¯¯¯¯¯¯¯¯¯z+w¯z¯w=¯¯¯¯¯¯¯zwz+¯z=2 Re(z)z−¯z=2 Im(z)
As mentioned in the OP:
¯z∥=∥z∥z¯z=∥z∥2
if ϕ(z) is a holomorphic function then ϕ(¯z)=¯¯¯¯¯¯¯¯¯¯ϕ(z). That means e¯z=¯¯¯¯¯ezand log(¯z)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯log(z)
If cnzn+cn−1zn−1+⋯+c1z+c0=0 is a polynomial with real coefficients, complex roots will be found in conjugate pairs.
But, at the introductory level, the most important thing a
Some properties of complex conjugates.
¯z+¯w=¯¯¯¯¯¯¯¯¯¯¯¯¯z+w¯z¯w=¯¯¯¯¯¯¯zwz+¯z=2 Re(z)z−¯z=2 Im(z)
As mentioned in the OP:
¯z∥=∥z∥z¯z=∥z∥2
if ϕ(z) is a holomorphic function then ϕ(¯z)=¯¯¯¯¯¯¯¯¯¯ϕ(z). That means e¯z=¯¯¯¯¯ezand log(¯z)=¯¯¯¯¯¯¯¯¯¯¯¯¯¯log(z)
If cnzn+cn−1zn−1+⋯+c1z+c0=0 is a polynomial with real coefficients, complex roots will be found in conjugate pairs.
But, at the introductory level, the most important thing about complex conjugates is that they allow you to turn a complex factor into a real one.
e.g. 1−i2+i=(1−i)(2−i)(2+i)(2−i)=1–3i5=15−35i
David Roberts
degree in Economic Statistics, Metropolitan College, London (Graduated 1974)
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Author has 2.4K answers and 2M answer views
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6y
I’d add one point to the other worthy answers here - because they come in pairs.
(a + bi)(a - bi) is simply a² + b². You factorize a² + b² and you get a pair of complex conjugates. That is why complex numbers extend the field of the primes: numbers of the form a² + b² are never prime.
Let’s take 89. Which is 64 + 25. So it must factorize! In fact its two factors are (8 + 5i)(8 - 5i). Or 5 which has (2 + i)(2 - i).
Just solve a cubic equation, as Cardan did 500 years ago, with only one real root. You’ll find that the other two are complex conjugates.
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7513 | https://brainly.com/question/17757693 | [FREE] Formulas: Surface Area of a Cube SA = 6s^2 Calculate the surface area of a 2 cm cube. Do not round the - brainly.com
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Formulas:
Surface Area of a Cube S A=6 s 2
Calculate the surface area of a 2 cm cube. Do not round the answer. You do not need units in your answer.
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Surface area of a cube is 24.
What is the Surface area?
The total area covered by the surfaces of an object is described as the surface area. There are different 3D shapes that have different surface areas which can be easily calculated using formulas. Surface area is classified into two categories:
Lateral surface area or curved surface area
Total surface area
The total surface area consists of all the faces of a 3D shape, which includes flat surfaces and curved surfaces, while the lateral surface area is calculated to find the area that is enclosed by the curved surface of the shape. It does not include the area of bases. A sphere is a 3D figure which has only one round surface with no flat base.
For above information, s= 2 cm, so we put the value in SA= 6 s 2 then SA= 6 x 2 x2 = 24
Thus, the Surface area of a cube is 24.
Learn more about Surface area, here:
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The surface area of a 2 cm cube can be calculated using the formula S A=6 s 2. Substituting s=2 cm results in a surface area of 24. Therefore, the answer is 24.
Explanation
To calculate the surface area of a cube, we use the formula:
S A=6 s 2
where s represents the length of one side of the cube.
In this case, the side length of the cube is 2 cm. We will substitute 2 cm into the formula:
S A=6(2)2
First, we calculate (2)2:
(2)2=4
Then, we multiply this result by 6:
S A=6×4=24
Thus, the surface area of the 2 cm cube is 24.
It’s important to note that the total surface area accounts for all six identical square faces of the cube. Each face has an area of s 2, and since a cube has 6 faces, we multiply the area of one face by 6 to find the total surface area.
Examples & Evidence
For example, if the side length was increased to 3 cm, the surface area would be calculated as follows: S A=6(3)2=54. This shows how changing the side length affects the surface area.
The formula used for calculating the surface area of a cube, S A=6 s 2, is a well-established mathematical concept in geometry.
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The answer is 24
Explanation
You have to multiply 2 times 2 times 6 which equal 24.
Answered by ferryanaherrington27 •3 answers•126 people helped
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### Free Biology solutions and answers
Community Answer Use the formula S = 6s2 to find the surface area of the cube. The surface area of the cube is ____ square inches.
Community Answer 5 The equation SA = 6s2 can be used to calculate the surface area, SA, of a cube with side length s. What is the surface area, in square meters, of a cube with a side length of 0.5 meter?
Community Answer The formula for finding the surface area of a cube is SA = 6s2. If s = 4 cm, which THREE statements are correct?
Community Answer 1 A cube has a surface area of 24 units2. Find the length of the side. Remember, the formula for surface area of a cube is SA = 6s2 Question 5 options: 2 units 1 unit 4 units 6 units
Community Answer 2 Given that the surface of the area cube is given by the formula SA equals 6S2 what is the value of essay if S equals 1/2 cm
Community Answer The formula for the surface area of a cube is SA=6s2, where is a is the length of one side of the cube if s=1/4 of a unit, what is the surface area, in square units, of the cube?
Community Answer 1 When a food handler can effectively remove soil from equipment using normal methods, the equipment is considered what?
Community Answer 1 Gene got Medicare before he turned 65 and enrolled into a Medicare Advantage plan. He calls in February the month before his 65th birthday and is unhappy with his current plan. On the date of the call, what can Gene do about his coverage?
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7514 | https://www.aims-international.org/aims12/12A-CD/PDF/K311-final.pdf | Price Dependent Quadratic Demand Inventory Models with Variable Holding Cost and Inflation Rate ISBN: 978-81-924713-8-9 R. Venkateswarlu Gitam University (rangavajhala_v@yahoo.co.in) M. S. Reddy BVSR Engineering College (naveenasrinu@gmail.com) An attempt is made to develop an inventory model for perishable items when the demand rate is a quadratic function of price and the rate of deterioration is a linear function of time. It is also assumed that the holding cost is a linear function of time. Under instantaneous replenishment with zero lead-time, EOQ is determined for optimizing the total profit under inflation rate. The sensitive analysis is presented with numerical example at the end. 1. Introduction It is true that the unit price and other inventory related costs are dependent on time. However, most of the inventory models in the literature have considered unit price and inventory related costs to be independent of time and constant over the period under consideration. Buzacott modified the classical EOQ model incorporating constant inflation rate under different pricing policies. Misra , Gupta et al , Vrat and Padmanabhan are some of the authors who have studied inventory models with special reference to inflation rate. It is well known that the demand rate of any product is always in dynamic state. This variation is due to time or price or even with instantaneous level of inventory. An economic lot size model for price dependent demand under quantity and freight discounts was developed by Burwell . An inventory system of ameliorating items for price dependent demand rate was considered by Mondal et al . You developed an inventory model with price and time dependent demand. Ajanta Roy has developed an inventory model for deteriorating items with price dependent demand and time varying holding cost. Inventory modelers so far have considered two types of price dependent demand scenarios, linear and exponential. The linear price dependent demand implies a uniform change in the demand rate of the product per unit price whereas exponential price dependent demand implies a very high change in demand rate of the product per unit price. These two scenarios are quite unusual in realistic situations. Thus quadratic price dependent demand may be an alternative approach to the existing two scenarios. So, it is reasonable to assume that the demand rate, in certain commodities, due to seasonal variations may follow quadratic function of time [i.e., D(t) = a + bp + cp2; a ≥ 0, b ≠ 0, c ≠ 0 ]. The functional form given above explains the accelerated growth/decline in the demand patterns which may arise due to seasonal demand rate (Khanra and Chaudhuri ). We may explain different types of realistic demand patterns depending on the signs of a and b. Bhandari and sharma have studied a Single Period Inventory Problem with Quadratic Demand Distribution under the Influence of Marketing Policies. Khanra and Chaudhuri have discussed an order-level inventory problem with the demand rate represented by a continuous quadratic function of time. Sana and Chaudhuri have developed a stock-review inventory model for perishable items with uniform replenishment rate and stock-dependent demand. Kalam et al have studied the problem of production lot-size inventory model for Weibull deteriorating item with quadratic demand, quadratic production and shortages. An order level EOQ model for deteriorating items in a single warehouse system with price depended demand in non-linear (quadratic) form has been studied by Patra et al . Venkateswarlu and Mohan studied inventory model for time varying deterioration and price dependent quadratic demand with salvage value. Venkateswarlu and Reddy developed time dependent quadratic demand inventory model under inflation. Recently, Venkateswarlu and Reddy studied inventory models when the demand is time dependent quadratic demand and the delay in payments is acceptable. In this paper, we try to develop an integrated model which contains both the perishability and inflation phenomena with price dependent quadratic demand situation. The inventory deterioration is assumed to be constant. The solutions of the models are presented and also discussed the sensitivity of the models at the end. 2. Assumptions and Notations The mathematical model is developed on the following assumptions and notations: i) The Selling rate D(p) at time t is assumed to be , 0 , 0 , ) ( 2 b a cp bp a p D .
0 c Where, ‘a’ is the initial rate of demand ‘b’ is the rate with which the demand rate increases and ‘c’ is the rate with which the change in the rate demand rate itself increases. ii) Replenishment rate is infinite and lead time is zero. iii) p is the selling price per unit. 1402 Twelfth AIMS International Conference on Management iv) The rate of inflation is constant v) The unit cost and other inventory related cost are subjected to the same rate of inflation, say k. This implies that the ordering quantity can be determined by minimising the total system cost over the planning period. vi) A(t) is the ordering cost at time t. vii) ) 1 0 ( is the constant rate of deterioration. viii) C(t) denotes unit cost at time t. ix) I (t) is the inventory level at time t. x) Q(t) is the ordering quantity at time t=0 xi) ‘h’ is per unit holding cost excluding interest charges per unit per year. 3. Formulation and Solution of the Model The objective of the model is to determine the optimum profit for items having price dependent quadratic demand and the rate of deterioration follows a linear function of time with no shortages. The inventory level depletes as the time passes due to demand and deterioration during (0,t1) and due to demand only during the period (t1, T). If I(t) be the inventory level at time t, the differential equations which describes the inventory level at time t are given by 1 2 0 ), ( ) ( .
.
) ( t t cp bp a t I t dt t dI (1) T t t cp bp a dt t dI 1 2 ), ( ) ( (2) Together with I(t1)=0 and I(T)=0. The solution of equations (1) and (2) 1 3 3 1 2 1 1 2 0 6 6 2 ) ( ) ( ) ( ) ( t t t t t t t t t cp bp a t I T t t t t cp bp a t I 1 1 2 ) )( ( ) ( Let us consider the Ordering Quantity is Q. If T t then Q I ) 0 ( 6 ) ( 3 2 T T cp bp a Q Let ) (t C denotes the unit cost at timet . i.e., kt e C t C 0 ) ( where 0 C is the unit cost at time zero. Let ) (t A denotes the Ordering cost at timet . i.e., kt e A t A 0 ) ( where 0 A is the ordering cost at time zero. Total system cost during the planning period ‘’ is the sum of the Material cost, ordering cost and Carrying cost. Assume that = mT, Where ‘m’ is an integer for the number of replenishments to make during the period’’, and ‘T’ is time between replenishments. The Ordering cost during the period (0,) is ] ) 1 ( ........
..........
) 3 ( ) 2 ( ) ( ) 0 ( T m A T A T A T A A ] .........
) 1 ( 0 ) 3 ( 0 ) 2 ( 0 ) 1 ( 0 ) 0 ( 0 kT m kT kT kT kT e A e A e A e A e A ) .........
1 ( ) 1 ( 3 2 0 kT m kT kT kT e e e e A The Ordering Cost is 1 1 0 kT k e e A where mT . The Material cost during the period (0,) is Twelfth AIMS International Conference on Management 1403 ] ) 1 ( ........
..........
) 3 ( ) 2 ( ) ( ) 0 ( [ T m C T C T C T C C Q ] .........
[ ) 1 ( 0 ) 3 ( 0 ) 2 ( 0 ) 1 ( 0 ) 0 ( 0 kT m kT kT kT kT e C e C e C e C e C Q ) .........
1 ( ) 1 ( 3 2 0 kT m kT kT kT e e e e QC 1 1 0 kT k e e QC Similarly, The Carrying Cost/holding cost during the period (0,) is T dt t I h t C 0 ) ( ) ( But we have 1 1 ) ( 0 kT k e e C t C in the period (0,) The Carrying Cost/holding cost is T kT k dt t I h e e C 0 0 ) ( 1 1 1 1 0 0 ) ( ) ( 1 1 t T t kT k dt t I dt t I e e h C 2 ) )( ( 12 2 ) ( 1 1 2 1 2 4 1 2 1 2 0 t T cp bp a t t cp bp a e e h C kT k The total cost over the period (0,) is = Ordering cost + Material cost + Carrying cost 1 1 0 0 0 0 ) ( ) ( 1 1 1 1 1 1 t T t kT k kT k kT k dt t I dt t I e e h C e e QC e e A 1 1 ) ( ) ( 1 1 0 0 0 0 kT k t T t e e dt t I dt t I h C QC A 1 1 2 ) )( ( 12 2 ) ( 6 ) ( 2 1 2 4 1 2 1 2 0 3 2 0 0 kT k e e t T cp bp a t t cp bp a h C T T cp bp a C A If shortages are not allowed then the Sales revenue per cycle is given by T T dt cp bp a p dt p D p 0 2 0 ) ( ) ( ) ( 2 cp bp a pT The total profit ) , ( T p f per unit time = (1/T) (Sales revenue – Total cost) ) 1 ( 1 2 ) )( ( 12 2 ) ( 6 ) ( ) ( ) , ( 2 1 2 4 1 2 1 2 0 3 2 0 0 2 kT k e T e t T cp bp a t t cp bp a h C T T cp bp a C A cp bp a p T p f 1404 Twelfth AIMS International Conference on Management The total profit is maximum if 0 )) , ( ( , 0 )) , ( ( T T p f p T p f i.e., 0 ) 1 ( 1 2 ) ( 12 2 ) 2 ( ) 1 ( 1 6 ) 2 ( 3 2 )) , ( ( 2 1 4 1 2 1 0 3 0 2 Tk k Tk k e T e t T t t h C pc b e T e T T C pc b cp bp a p T p f and 0 ) 1 ( ) 1 ( 2 ) ( 12 2 ) 1 ( ) 1 ( 6 ) 1 ( ) 1 ( ) 1 ( 1 ) ( 2 1 ) 1 ( 1 2 ) ( 12 2 ) 1 ( 1 6 ) 1 ( 1 )) , ( ( 2 2 1 4 1 2 1 0 2 2 3 0 2 2 0 1 0 2 0 2 2 2 1 4 1 2 1 0 2 2 3 0 2 2 0 Tk k Tk Tk k Tk Tk k Tk Tk k Tk k Tk k Tk k e T e ke t T t t h C cp bp a e T e ke T T C cp bp a e T e ke A e T e t T h C T C cp bp a e T e t T t t h C cp bp a e T e T T C cp bp a e T e A T T p f The Optimal value of T is obtained solving equation f (p, T) by MATHCAD ) 1 ( 1 2 ) ( 12 2 ) 2 ( ) 1 ( 1 6 ) 2 ( 6 2 )) , ( ( 2 1 4 1 2 1 0 3 0 2 2 Tk k Tk k e T e t T t t h C c e T e T T C c cp b p T p f 2 2 1 4 1 2 1 0 2 2 1 0 2 3 2 1 4 1 2 1 0 2 2 2 3 0 2 3 3 0 2 2 2 0 2 2 2 0 3 0 2 2 ) 1 ( ) 1 ( 2 ) ( 12 2 ) 1 ( 1 ) ( ) 1 ( 1 2 ) ( 12 2 2 ) 1 ( ) 1 ( 6 ) 1 ( 1 6 2 ) 1 ( 1 2 1 ) 1 ( ) 1 ( ) 1 ( 1 )) , ( ( Tk k Tk Tk k Tk k Tk k Tk Tk k Tk k Tk k Tk Tk k e T e ke t T t t h C cp bp a e T e t T h C cp bp a e T e t T t t h C cp bp a e T e ke T T C cp bp a e T e T T C cp bp a e T e T C cp bp a e T e ke A e T e A T T p f 2 2 1 4 1 2 1 3 0 1 2 0 2 2 1 4 1 2 1 3 0 2 ) 1 ( ) 1 ( 2 ) ( 12 2 6 2 ) 1 ( ) 1 ( ) ( 2 1 2 ) 1 ( 1 2 ) ( 12 2 6 2 )) , ( ( Tk k Tk Tk k Tk k e T e ke t T t t h T T C cp b e T e t T h T C cp b e T e t T t t h T T C cp b T p T p f Twelfth AIMS International Conference on Management 1405 Also satisfying the following condition , 0 )) , ( ( 2 2 p T p f , 0 )) , ( ( 2 2 T T p f And 0 )) , ( ( )) , ( ( )) , ( ( 2 2 2 2 2 2 T p T p f T T p f p T p f It is found that the optimality conditions are satisfied for the following two cases viz., (i) b < 0 and c > 0 which gives retarded growth in demand model (ii) b < 0 and c < 0 gives accelerated decline in demand model 3.1 Numerical Example To demonstrate the effectiveness of the models developed, a numerical example is taken with the following values for the parameters. a =150, b = 5, c =0.01, A0 = 250, C0=4, θ =0.1, i = 0.05, k = 0.04 The MATHCAD output is presented in Table-1 and Table-2 which contains the optimum values of time (T), ordering quantity (Q) and total profit f(p,T) of the system for various values of inflation parameter (k) and deterioration parameter (θ). These tables provide certain important insights about the problem under study. Some observations are as follows: The behaviour of both the models developed here is almost similar in nature but the rate of change is slightly different. The optimal values of cycle time, ordering quantity and total cost increases with an increase in the inflation rate parameter ‘k’. For some particular values of θ, when the inflation rate k increases from 0.05 to 0.10, the cycle time and ordering quantity increases while the total profit f(p,T) also increases in both the models. For some particular values of k, when θ increases from 0.05 to 0.10, the cycle time and ordering quantity decreases whereas total profit f(p,T) increases in both the models. 3.2 Sensitive Analysis We now study sensitivity of the models developed to examine the implications of underestimating and overestimating the parameters individually and all together on optimal value of cycle time, ordering quantity and total system profit. The results are shown in Table-3 and Table-4. The following observations are made from these two tables: (i) The ordering quantity (Q), the unit price (p) and the total system profit f(p,T) increases (decreases) with the increase (decrease) in the value of the parameter ‘a’ where as the cycle time (T) is inversely related with the parameter ‘a’. (ii) Increase (decrease) in the values of the parameters ‘b’ and ‘c’ decrease (increase) the price per unit, ordering quantity and the total profit f(p,T) while the cycle time increases (decreases) with ‘b’ and decreases (increases) with ‘c’. However the rate of increase/decrease is marginal in case ‘p’ and ‘T’. The ordering quantity (Q), the unit price (p), and the total system profit f(p,T) increases (decreases) with the decrease (increase) in the value of the parameters ‘’ and ‘C0’. (iii) The optimum value of the total profit, ordering cost and cycle time is marginal but the unit price remain constant to the changes in the parameters A0 (iv) The total profit of the system is more sensitive than the cycle time and ordering quantity when the values of all parameters are under-estimated or over-estimated by 15%. Table 1 Retarded Growth Model (I.E., A>0, B<0 and C>0) S.No a b c k p T f(p, T) Q 1 150 -5 0.01 0.05 15.787 9.158 1008 2050 2 150 -5 0.01 0.06 15.785 9.295 1009 2117 3 150 -5 0.01 0.07 15.784 9.445 1009 2192 4 150 -5 0.01 0.08 15.782 9.612 1010 2279 5 150 -5 0.01 0.09 15.780 9.801 1011 2380 6 150 -5 0.01 0.10 15.778 10.017 1011 2499 1406 Twelfth AIMS International Conference on Management Table 2 Retarded Decline Model (I.E., A>0, B<0 and C<0) S.No a b c k p T f(p, T) Q 1 150 -5 -0.01 0.05 14.421 9.136 978.360 1980 2 150 -5 -0.01 0.06 14.419 9.271 978.956 2044 3 150 -5 -0.01 0.07 14.417 9.421 979.550 2129 4 150 -5 -0.01 0.08 14.416 9.587 980.141 2200 5 150 -5 -0.01 0.09 14.414 9.774 980.730 2297 6 150 -5 -0.01 0.1 14.412 9.988 981.316 2411 Table 3 Retarded Growth Model (A>0, B<0 And C>0) Parameters % change Change in p (%) Change in T (%) Change inf (pct.) (%) Change in Q (%) a -15% -15.6078 1.375846 -24.6693 -21.8049 -5% -5.23215 0.414938 -8.21538 -7.21951 5% 5.263825 -0.38218 8.234127 7.121951 15% 15.87382 -1.04826 24.70238 21.31707 b -15% 20.05448 0.054597 9.52381 9.414634 -5% 5.846583 0.010919 3.174603 3.121951 5% -5.20682 -0.01092 -3.16954 -3.12195 15% -14.0812 -0.03276 -9.53165 -9.41463 c -15% -0.76645 -0.02184 -0.19841 -0.29268 -5% -0.25971 -0.01092 -0.09921 -0.09756 5% 0.266042 0.010919 0.099206 0.097561 15% 0.798125 0.021839 0.198413 0.243902 C0 -15% -0.03801 1.397685 0.793651 3.073171 -5% -0.01267 0.414938 0.297619 0.878049 5% 0.012669 -0.38218 -0.19841 -0.82927 15% 0.04434 -1.05918 -0.69444 -2.29268 -15% -0.08868 6.693601 0.099206 4.682927 -5% -0.03167 2.096528 0 1.512195 5% 0.031672 -1.98733 0 -1.5122 15% 0.088681 -5.6235 -0.09921 -4.29268 A0 -15% 0 -1.23389 0.396825 -2.63415 -5% 0 -0.40402 0.099206 -0.87805 5% 0 0.404018 -0.09921 0.878049 15% 0 1.190216 -0.29762 2.585366 All -15% -0.12035 8.113125 -14.2977 -8.43902 -5% -0.03801 2.522385 -4.72877 -2.68293 5% 0.04434 -2.36951 4.761905 2.536585 15% -15.6078 1.375846 14.0873 7.414634 Table 4 Retarded Decline Model (A>0, B<0 and C<0) Parameters % change Change in p (%) Change in T (%) Change in f (pct.) (%) Change in (%) a -15% -4.79856 0.415937 -25.4286 -22.5253 -5% 4.777755 -0.3831 -8.47623 -7.42424 5% 14.27779 -1.06173 8.446789 7.373737 15% 15.85188 -0.04378 25.41396 22.0202 b -15% 4.798558 -0.01095 9.877755 9.494949 Twelfth AIMS International Conference on Management 1407 -5% -4.39637 0.010946 3.233983 361.3636 5% -12.1489 0.021891 -3.27742 -3.18182 15% 0.603287 0.010946 -9.83237 -9.54545 c -15% 0.194161 0 0.228137 0.252525 -5% -0.2011 -0.01095 0.076046 0.10101 5% -0.58942 -0.02189 -0.07605 -0.10101 15% -0.04161 1.368214 -0.22814 -0.25253 C0 -15% -0.01387 0.404991 0.75463 3.030303 -5% 0.013869 -0.3831 0.251543 0.909091 5% 0.041606 -1.03984 -0.25154 -0.80808 15% -0.09708 6.698774 -0.75463 -2.22222 -15% -0.03467 2.101576 0.071242 4.69697 -5% 0.027737 -1.98117 0.023713 1.565657 5% 0.090146 -5.63704 -0.02371 -1.46465 15% 0 -1.20403 -0.07114 -4.29293 A0 -15% 0 -0.39405 0.368167 -2.57576 -5% 0 0.394046 0.122756 -0.80808 5% 0 1.160245 -0.12265 0.858586 15% -0.12482 8.099825 -0.36806 2.575758 All -15% -0.04854 2.517513 -14.3071 -8.43434 -5% 0.041606 -2.36427 -4.73956 -2.67677 5% 0.145621 -6.65499 4.66495 2.575758 15% -4.79856 0.415937 14.06844 7.474747 4. References 1. Buzacott, J.A., 1975. Economic order quantities with inflation. Operat. Res.Q., Vol.26, pp553-558. 2. Misra, R.B., 1979. A note on optimal inventory management under inflation, Naval Res. Logist., Vol.26, pp.161-165. 3. Gupta, R., Vrat, P. and Swarup, K. (1985), “Inventory management development – A review with special reference to perishability, inflation and stock-dependent consumption”, Towards continuing Education, Vol. 6, Oct. – Nov., pp.232-237. 4. Padmanabhan, G. and Vrat, P. 1990. An EOQ model for items with stock dependent consumption rate and exponential decay. Engineering Costs and Production Economics, 18(3), pp.241-246. 5. Burwell T.H., Dave D.S., Fitzpatrick K.E., Roy M.R., (1997) Economic lot size model for price-dependent demand under quantity and freight discounts, International Journal of Production Economics, 48(2), pp 141- 155. 6. Mondal, B., Bhunia, A.K., Maiti, M., (2003) An inventory system of ameliorating items for price dependent demand rate, Computers and Industrial Engineering, 45(3), pp 443-456. 7. You, S.P. (2005) Inventory policy for products with price and time dependent demands, Journal of the operations research society, 56, pp 870- 873. 8. Ajanta Roy, (2008), An inventory model for deteriorating items with price dependent demand and time-varying holding cost, AMO-Advance modeling and optimization, Vol.10, No.1. 9. R.M.Bhandari and P.K.Sharma.(2000) A single period inventory problem with quadratic demand distribution under the influence of Market policies’, Eng. Science Vol. 12. No.2, pp. 117-127. 10. S.Kharna and K.S.Chaudhuri.(2003) ‘A note on order-level inventory model for a deteriorating item with time-dependent quadratic demand’ Computers and Operations research, Vol.30, pp.1901-1916. 11. Shibshankar Sana and K.S.Chaudhary.(2004) A Stock-Review EOQ Model with Stock-Dependent Demand, Quadratic Deterioration Rate’, Advanced Modelling And Optimization, vol.6, No.2, pp 25-32. 12. A. Kalam, D. Samal, S. K. Sahu & M. Mishra (2010) A Production Lot-size Inventory Model for Weibull Deteriorating Item with Quadratic Demand, Quadratic Production and Shortages, International Journal of Computer Science & Communication, Vol. 1, No. 1, Jan. 2010, pp. 259-262. 13. Patra, S.K. Lenka, T.K. and Ratha, P.C. (2010), An Order Level EOQ Model for Deteriorating Items in a Single Warehouse System with Price Depended Demand in Non-Linear (Quadratic) Form, International Journal of Computational and Applied Mathematics, Vol.5 No.3, pp. 277–288 14. R.Venkateswarlu and R. Mohan (2013) An Inventory Model for Time Varying Deterioration and Price Dependent Quadratic Demand with Salvage Value, Journal of Computational and Applied Mathematics, 07/2013; 1(1):21-27. 1408 Twelfth AIMS International Conference on Management 15. R.Venkateswarlu and M.S.Reddy (2014a) Time Dependent Quadratic Demand Inventory Models Under Inflation, Global Journal of Pure and Applied Mathematics, Volume 10, Number 1 (2014), pp. 77-85. 16. R.Venkateswarlu and M.S.Reddy (2014b) Time Dependent Quadratic Demand Inventory Models when Delay in Payments is Acceptable, International Journal of Modern Engineering Research (IJMER), Vol. 4, No.3, pp 60-71. |
7515 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6768166/ | An Equilibrium Model for the Calculation of Activity and Osmotic Coefficients in Aqueous Solutions - PMC
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J Res Natl Bur Stand (1977)
. 1984 May-Jun;89(3):251–263. doi: 10.6028/jres.089.014
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An Equilibrium Model for the Calculation of Activity and Osmotic Coefficients in Aqueous Solutions
Robert N Goldberg
Robert N Goldberg
1 National Bureau of Standards, Gaithersburg, MD 20899
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1 National Bureau of Standards, Gaithersburg, MD 20899
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PMCID: PMC6768166 PMID: 34566127
Abstract
A procedure is described for the calculation of activity and osmotic coefficients which is based upon a knowledge of the equilibria in solution and assumed single-ion activity coefficients. The procedure permits one to introduce chemical equilibria of various types (ion-pairing, complexation, hydration, and hydrolysis) into a model which can be used to calculate values of the excess Gibbs energy and the activity and osmotic coefficients. Both the Debye-Hückel theory and Pitzer’s expression are used to calculate the electrostatic contribution to the single-ion activity coefficients. Calculations have been performed on aqueous sulfuric acid, acetic acid, hydrofluoric acid, cadmium chloride, copper sulfate, and sodium carbonate. Properties which have been calculated are the excess Gibbs energy, the osmotic coefficient, the mean ionic activity coefficient, and Frank’s single-ion activity coefficient function. Agreement between calculated and measured properties has been obtained up to molalities of ≈ 1.0 mol kg−1
Keywords: acetic acid, activity coefficient, cadmium chloride, copper sulfate, equilibrium, excess Gibbs energy, hydrofluoric acid, models of solutions, osmotic coefficient, sodium carbonate, sulfuric acid
1. Introduction
Equilibrium models have been used [1–15]1 both for the prediction and for the correlation of activity and osmotic coefficient data in aqueous electrolyte solutions. These equilibrium models are particularly appropriate when one is dealing with solutions which exhibit association, complexation, hydration, or hydrolysis. When applied to such solutions, they are superior to the use of a model that assumes the electrolyte in solution to be a fully dissociated strong electrolyte. A variety of approaches has been used in these equilibrium models for treating these various types of equilibria in solution and for the calculation of the activity and osmotic coefficients. Several different types of functions for the calculation of the electrostatic contribution to the activity coefficient of the ions in solution have also been used.
The purposes of this paper are to 1) describe a procedure for (a) the calculation of activity and osmotic coefficients in aqueous solutions that uses a generalized approach for treating the equilibria in solution, and (b) the calculation of the Gibbs energy properties, 2) clarify the distinction between the stoichiometric and species quantities, frequently a source of confusion in the literature, and 3) explore the effects of parameter variations in the model on calculated values of the thermodynamic properties as applied to several representative types of electrolyte solutions. Using the equilibrium model, one can also calculate the values of the activity coefficients of individual ions. This permits one to then calculate values for Frank’s single-ion activity coefficient function.
2. The Model
A fundamental idea used here is the distinction between the stoichiometric components of which a system is composed and the particle constituents, or species, which are introduced to account for the properties of a solution. Stoichiometric quantities will be designated by “st” when not otherwise clearly indicated. Species or particle quantities will always be designated by a “ ^ ” placed over the quantity. For example, for the solution formed from 0.1 moles of pure H 2 SO 4(ℓ) and 1 kg/M 1 moles of H 2 O(ℓ) the stoichiometric molality of H 2 SO 4 is m(H 2 SO 4)=0.1 mol kg−1, m()=0.l mol kg−1, and m(H+) = 0.2 mol kg−1 if one views the electrolyte as being completely dissociated. However, if one considers the equilibrium: (aq)+H+(aq) = (aq), adopts a K of 992 for it , and uses a Debye-Hückel type expression (see eq (5)) for the activity coefficients of the ions with a B parameter equal to zero, one calculates (H+) = 0.134 mol kg−1, =0.0342 mol kg−1, =0.0658 mol kg−1, and (H 2 SO 4)=0.0 mol kg−1. Physical quantities other than amounts of substance can also be viewed as particle quantities. In particular it is the distinction between G ex and Ĝ ex which forms the basis of the model presented herein.
The general system to be considered is formed from 1 kg/M 1 moles of water and n ℓ(ℓ= 2 to N c) moles of other components. In terms of a particle model, the system can be viewed as being formed from moles of water and (i=2 to N s) moles of particles or species. The amount of water (n 1) will always be designated by a subscript “1.” The components and the species other than water will be designated by subscripts “ℓ” and “i,” respectively, with i and ℓ⩾2.
Each component ℓ in the solution is represented as where C and A are the reference cations and anions, respectively. The charges of the cations and anions are and , respectively; the ion-numbers are and and , respectively. It is important to note that the choice of reference species is arbitrary; e.g., for a solution of aqueous sulfuric acid, the reference species could be selected as either two H+ ions and one ion or as one H+ ion and one ion. In the former case, sulfuric acid would be represented as and in the latter case as . While the choice of reference species is arbitrary, it will not affect the amounts of the various species one calculates given the same set of equilibrium constants and allowed species. However, as will be seen later, the choice does affect the values of several of the stoichiometric thermodynamic properties.
The Gibbs energy of the solution in the stoichiometric representation is given by
(1)
where is the molality of water in pure water and and are, respectively, the chemical potentials of the reference cation and anion of the ℓ th component in the solution. In terms of an equilibrium or species model of the solution, the Gibbs energy is given by
(2)
If the equilibrium model is accurate, the Gibbs energy of the solution calculated using eqs (1) and (2) will be identical and the chemical potentials and the activities of the i th species will be the same in both the stoichiometric and species representations. The development to follow will start with a description of the solution in terms of the equilibria assumed to be present in solution, an assumed expression(s) for the activity coefficients of the solute species in solution, and the calculation of the activity of the water using the Gibbs-Duhem equation. It is important to note that there may be several equilibrium models of a solution which yield agreement with calculated properties. Thus agreement between calculated and measured properties does not, in the absence of direct molecular information, prove the correctness of the model used.
The equilibria in solution are described by a series of chemical equations:
(3)
where S jk is the j th species in the k th equilibrium, N k is the number of species in the k th equilibrium, N e is the total number of equilibria, and t jk is the number of moles of species, S jk participating in a given equilibrium; t jk is positive if S jk is a product and is negative if S jk is a reactant. The equilibrium constants are:
(4)
where â jk is the activity of species S jk. Since â i is equal to the product of the molality of the i th species () and its activity coefficient (), the complete formulation of the equations which describe the equilibria in solution requires that some assumptions(s) be made concerning the form of the ’s (i⩾2) in solution. In this paper two different expressions for the ’s will be used:
(5)
and
(6)
Equation (5) is the Debye-Hückel equation with an excluded volume or “ion-size” parameter B in the denominator. Equation (6) is the leading term of Pitzer’s expression for ; he has set b equal to 1.2. A m and A ϕ are Debye-Hückel constants, where A m = 3 A ϕ, = 1.17642 kg 1/2 mol−1/2 at 298.15 K. Equations (5) and (6) can be extended by the addition of the expression:
(7)
where λ ij and μ ijk are, respectively, the interaction parameters for pairs and triplets of particles. We shall later return to the subject of the extensions of eqs (5) and (6) and to several other aspects of the choice of an expression for . It should be noted that I is calculated as
(8)
We have formulated the equilibrium equations using an extent of reaction variable ξ, Thus, the amount of the i th species in solution is given by
(9)
where ξ k is the extent of reaction variable for the k th equilibrium and δ ij is the Kronecker delta; r is an integer which serves to identify the reference cation and anion of each of the components in the solution. The first summation term on the right side of eq (9) specifies the amount of species i which is formed in the absence of any equilibria in solution; the second summation term specifies the contributions, which may be positive or negative, to the amount of species i in solution from the equilibria in solution. If the water is a participant in the equilibria in solution, must also be calculated using the above equation. Hydration numbers can be introduced directly using
(10)
where (eq (9)) is the amount of water calculated using eq (9) and is the number of waters of hydration attached to the i th species. If the water participates in the equilibria in solution or if hydration numbers are introduced, kg−1 will not be equal to . The are given by
(11)
The activities of the species are given by
(12)
Thus in a model where hydration is introduced, both and â i will be affected by changes in both and (i⩾2).
To obtain a numerical solution of eq (4), it is necessary to make some initial guess for the activity of water if it is a participant in the equilibria in solution; we have generally used a value of unity. Thus, having formulated the simultaneous nonlinear eq (4), which necessarily include eqs (8) to (12) and eq (5) or (6), one is left with a numerical problem to obtain a self-consistent solution of these equations. It is assumed that, while such a solution may be difficult to obtain for large systems, a unique solution does exist and that one now has values for I, and for i⩾l and for and for i⩾2.
The activity of the water can now be calculated by application of the Gibbs-Duhem equation stated in terms of excess properties of the species
(13)
Use of eqs (11) and (13) leads to
(14)
It is also necessary to adopt some conventions concerning the limits of â 1 and . The conventions used herein are â 1→1 and, for i⩾2, →1 as ; also, (i⩾2) is defined to be equal to unity3. These conventions, together with the definitions of the activity â i, the activity coefficient , and the definition of the excess Gibbs energy of the i th species given in the following equations
(15)
(16)
and
(17)
lead to
(18)
The introduction of eqs (5) and (18) into eq (14) yields
(19)
Similarly, one obtains, using eq (6) instead of eq (5), the relationship
(20)
We now consider an ideal reacting solution in which (i⩾2) is equal to unity at all temperatures, pressures, and compositions and which also allows for the presence of both equilibria and hydration in solution. Application of the Gibbs-Duhem equation to such a solution leads to
(21)
and this in turn with eq (15) leads to
(22)
Since an excess property is defined as the difference between the real and the ideal, it follows from eqs (15) and (22) that
(23)
If the equilibrium model is an accurate representation of the solution, a 1 is equal to â 1. We thus have a procedure for the computation of the activity of the water which starts with an equilibrium model of the solution and an assumed expression for the activity coefficients in the solution. Since it is necessary to make an initial guess as to the value of the activity of the water, the calculation should be repeated using the value of â 1 from the previous iteration until convergence to within a given tolerance in its value is obtained. The (stoichiometric) osmotic coefficient is calculated as
(24)
where m ℓ is the stoichiometric molality of component ℓ and V ℓ is equal to (). Note that v ℓ is unity for a non-electrolytic component and that, as stated earlier, the value of the osmotic coefficient is dependent upon the choice of the reference species selected for each component in the solution.
The stoichiometric activity coefficients can be calculated using the principle that the chemical potential is independent of any representation of it. Equating â i, to a i, the stoichiometric single-ion activity coefficient is given by
(25)
where is the total stoichiometric molality of the i th species in solution. The mean ionic activity coefficient of component ℓ is given by
(26)
As was the case for the osmotic coefficient, the value of the mean ionic activity coefficient is dependent upon the reference species selected for a given component.
Other stoichiometric Gibbs energy properties can be calculated in addition to the activity of the water, the osmotic coefficient, and the mean ionic activity coefficient of the component. Additional properties of the water are calculated as follows:
(27)
(28)
(29)
and
(30)
Note that eqs (29) and (30) are definitions of the activity coefficient of the water (γ 1), and of the rational osmotic coefficient (ϕ x)respectively. These two quantities have not been frequently used in the literature. The properties of the solutes are:
(31)
(32)
(33)
(34)
and
(35)
where the mean ionic molality of component ℓ is defined as
(36)
Finally, the total properties of the solution are
(37)
and
(38)
Equation (27) is the stoichiometric analogue of eq (23). Equations (29) and (30) are, respectively, the definitions of the activity coefficient of the water and of the rational osmotic coefficient. The steps used in the overall computational procedure are summarized in figure 1.
Figure 1.
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Steps for the calculation of the amounts of species in solution and of the stoichiometric Gibbs energy properties of the solution.
It is interesting to consider the consequences of the ideal behavior of all of the solute particles in the species representation, i.e., = 1 for i⩾2. Application of eqs (21), (23), and (24) to such a solution leads to
(39)
Application of eqs (25) and (26) leads to
(40)
If the stoichiometric and species representations are identical, then ϕ and γ ℓ± will be equal to unity. This is the case when there are no chemical interactions (i.e., association, hydration, or hydrolysis) present in solution and the stoichiometric reference species chosen are the only ones present in solution.
An additional interesting feature of this model is that properties of individual ions and species are calculated (eq (25)). Unlike the stoichiometric properties given in eq (24) and eqs (26) to (35), there are presently no experimental data available with which one can compare these calculated values. Frank has defined a quantity δ± characteristic of single-ion properties for a binary electrolyte:
(41)
The definition is easily extended to multicomponent systems as was done for the case of the mean ionic activity coefficient (see eq (26)) and values of δ ℓ± can also be calculated if the equilibria in solution are known.
An alternative way of viewing this model uses the definition of the excess Gibbs energy
(42)
and its analogue for the species representation
(43)
Since the Gibbs energy is independent of representation,
(44)
Similar equations hold for , ℓ ⩾ 1. From the definition of ideality,
(45)
Equations analogous to the foregoing expression and to eq (39) exist for the stoichiometric components of a solution. Thus, introduction of eqs (22) and (45) and their stoichiometric analogues into eq (44) yields
(46)
Note that in the above equation.
The three terms in { } on the right side of eq (46) correspond, respectively, to Ĝ ex, Ĝ id, and G id. Inspection of the terms for Ĝ id shows that the differences between these quantities involve two factors: 1) a difference in Gibbs energies, i.e., terms multiplied by n i and , and 2) entropie terms, i.e., the (RT Σ n i ℓnm i) terms for the solute particles and the (RT Σ m i) terms for the solvent. If one views Ĝ ex as the electrostatic or ionic contribution to G ei, the stoichiometric excess Gibbs energy is seen to also consist of energetic and entropie contributions which are formally accounted for with this model. While eq (46) could be used to compute G ex directly, it is numerically preferable to use the computational scheme outlined earlier (see Fig. 1).
In this paper, two different expressions have been used for (see eqs (5) and (6)). While classical thermodynamics has little to say about the correctness of either of these or any other choices for , it does impose one important constraint on such a choice, namely that
(47)
or equivalently,
(48)
Thus, while it is tempting to try to assign a different value of B or b to each species in a solution in eqs (5) and (6), respectively, to do so would violate eqs (47) and (48). However, the extension of eqs (5) and (6) using eq (7) does not violate this thermodynamic constraint. Note that the use of eqs (5) or (6) for does not allow for the introduction of specific-ion effects attributable to long-range electrostatic interactions. These effects can be introduced by the use of eq (7). Specific-ion effects attributable to chemical equilibria are accounted forin the equilibrium part of the model. Expressions other than those in eqs (5) and (6) could be used to represent the electrostatic part of .
The long range electrostatic contributions to the Gibbs energy properties are introduced via eq (5) or (6) and the use of an equilibrium model. Other interactions accounted for in this model include: 1) chemical interactions, 2) hydration, and 3) volume exclusion effects. The attractive chemical interactions are accounted for by the use of the equilibrium constants for the processes which describe the equilibria in solution. These processes can involve ion-pairing, complexation, and hydrolysis. The effects of hydration are accounted for either by the introduction of equilibrium constants for specific reactions involving hydration or by the use of hydration numbers for each species in solution. The use of hydration numbers reduces the value of in eq (10) which in turn has consequent effects on the , and other properties. Volume exclusion effects are represented by the B or b parameter in eqs (5) or (6). Short range repulsive forces between particles can also be accounted for using the λ ij and/or μ ijk parameters. It is worth noting that there are similarities in the effects that changes in certain parameters have on thermodynamic properties. Specifically, an increased value off B (or b) is similar to the introduction of a positive λ ij or μ ijk and also to the introduction of hydration effects. Physically this should be the case since the excluded volume for a hydrated ion is larger than for one that is not hydrated. Also, a negative λ ij is similar to an association between particles i and j. The remainder of this paper will discuss the application of this model to several aqueous salt solutions containing representative types of chemical interactions.
3. Results and Discussion
We now compare the results of calculations using this model with experimental data and examine the results of perturbing the various input parameters in the model, i.e., the single-ion activity coefficient expressions, the assumed equilibrium constants, and the assumed state of hydration of the species in solution. To do this, calculations have been made on the following aqueous electrolyte solutions: sulfuric acid, acetic acid, hydrofluoric acid, cadmium chloride, copper sulfate, and sodium carbonate.
The results of calculations on aqueous sulfuric acid are shown in figures 2 through 4, where the chemical equilibrium considered is
(A)
The value of K A is 99 at 298.15 K . It is seen that the calculated values of γ±, ϕ, and G ex are relatively insensitive to moderate variations in K A up to molalities of 0.2 mol kg−1. However, as seen in figure 3, significant changes in the osmotic coefficient are produced by perturbing the value of the B parameter. Use of a value of B equal to 2.5 in a Debye-Hückel expression for (see eq (5)) produces good agreement with the experimental osmotic coefficients (the squares in fig. 2d) up to a molality of ≈O.2 mol kg−1. The effects of variations in hydration numbers is shown in figure 4. The minimum in the osmotic coefficient curve cannot be produced by variation in the B parameter, but, as seen in figure 4, a minimum is observed when hydration is introduced.
Figure 2.
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Clockwise from above, calculated activity coefficients, osmotic coefficients, and excess Gibbs energies of aqueous H 2 SO 4 at 298.15 K as a function of molality, K a is varied: 99 for the solid line, 89 for the dashed line, an 109 for the dotted line. Equation (6), Pitzer’s expression for , was used in calculating each of the curves. The ions were considered to be unhydrated.
Figure 4.
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Calculated osmotic coefficients of aqueous H 2 SO 4 at 298.15 K in which the extent of hydration of the ions is varied. All three curves were calculated using a value of 99 for K A and a Debye-Hückel expression for with B equal to 2.0. The dotted line was calculated assuming that all of the ions were not hydrated. The solid and dashed line were calculated, respectively, by assuming all three ions (H+, , and ) to be hydrated with three and four waters each. The squares are from the correlation of Pitzer, Roy, and Silvester .
Figure 3.
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Comparison of calculated osmotic coefficients for aqueous H 2 SO 4 at 298.15 K. The squares are the correlated values of Pitzer, Roy, and Silvester . The solid line was calculated using a value of K A =99 and Pitzer’s expression for . The dashed and dotted lines were calculated using the same value of Kj and values of B equal to zero and 2.5 respectively, in a Debye-Hückel expression for , (eq (5)). The ions were considered to be unhydrated.
The equilibrium considered in the description of aqueous acetic acid (HAc) is
(B)
For aqueous hydrofluoric acid, the equilibria are
(C)
and
(D)
For aqueous cadmium chloride the stepwise equilibria considered are:
(E)
(F)
(G)
and
(H)
The equilibria considered for aqueous copper sulfate are:
(I)
(J)
(K)
For aqueous sodium carbonate the equilibria considered are:
(L)
(M)
and
(N)
All of the above equilibrium constants refer to a temperature of 298.15 K with the exception of the value for process (B) which refers to 273.15 K. The Gibbs energies of formation given in the NBS Tables of Chemical Thermodynamic Properties were used to calculate the above values for equilibria (C), (D), (K), (L), and (M). The value for process (B) was calculated from the data for acetic acid tabulated by Larson and Hepler ; the values for processes (E), (F), (G), and (H) are those given by Reilly and Stokes ; and the values for processes (I), (J), and (K) are those given by Pitzer . The Debye-Hückel constants recommended by Clarke and Glew were used in all calculations.
The results of the calculations are shown in figures 5 through 9. For acetic acid, since the species ionic strength () is very low (it has a value of 0.044 mol kg−1 at I st = 1.0 mol kg−1), the choice of the expression for makes very little difference. Near agreement with the experimental osmotic coefficients is obtained to a molality of ≈0.2 mol kg−1. The difference between the measured osmotic coefficients and the calculated ones can be attributed to the formation of dimers and trimers of acetic acid and cannot be explained either by the introduction of hydration or by the use of different B parameters in eq (5) or by the choice of a different expression for .
Figure 5.
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The osmotic coefficients of aqueous acetic acid at 273.15 K. The squares are the experimental osmotic coefficients reported by Harris, Thompson, and Wood . The solid curve was calculated using a value of K B = 5.96× 10 4 which was obtained from the values of Δ G°, ΔH°, and for process (B) given by Larson and Hepler . The solid curve was also calculated using either Pitzer’s or a Debye-Hückel where B varied from zero to 2.0. In this figure and. in all subsequent ones, the ions were considered to be unhydrated.
Figure 9.
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The osmotic coefficients of aqueous Na 2 CO 3 at 289.15 K. The squares are from the correlation of Vanderzee . The solid line was calculated using Pitzer’s expression for and values of K l , K m, and K N from reference . The dashed line was calculated using these same values of K and a Debye-Hückel with B equal to 1.0.
Figure 6.
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The osmotic coefficients of aqueous HF at 298.15 K. V alues of K c = 1442 and K D = 2.63 were used to calculate the solid curve together with Pitzer’s expression for . The squares are from the correlation of Hamer and Wu .
Figure 7.
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The activity coefficients of aqueous CdCl 2, at 298.15 K, The successive formation constants given by Reilly and Stokes were used in doing the calcuiations. The solid line was calculated using Pitzer’s expression for and the dashed line was calculated using the Debye-Hückel with B set equal to 2.0. The squares are based on the measurements of Reilly and Stoke .
Figure 8.
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The osmotic coefficients of aqueous CuSO 4 at 298.15 K. The squares are the experimental data of Miller et al . The three curves were obtained using values of K E = 250 and K f =K G = 5. The solid line was calculated using Pitzer’s expression for , the dashed and dotted lines were obtained using the Debye-Hückel with B set equal to 2.0 and 5.0, respectively.
For aqueous HF, CdCl 2, CuSCX,, and Na 2 C0 3, agreement of calculated with measured properties can be obtained to molalities between 0.6 and 1.0 mol kg−1 by the variation of the B parameter in eq (5) and, for the case of CdCl 2, using only eq (6) for . Neither the use of eq (5) nor eq (6) is able to produce the minima in the calculated values of ϕ or γ ± which is observed for many electrolyte solutions. These minima can be produced, however, either by the introduction of hydration or by the use of eq (7) to extend the equations for . It should be noted that for Na 2 Co 3 solutions the osmotic coefficient does not approach the usual limit of unity as m st approaches zero mol kg−1; instead it approaches a value of 1.395 . This is a consequence of eq (39) and the presence of equilibria (L), (M), and (N),
Calculated values of δ± are shown in figure 10. The fact that values of δ± for acetic acid and for copper sulfate are essentially unity is attributable to a very near cancellation of terms in eqs (25) and (36). It should be noted that the value of δ± for Na 2 CO 3, unlike the other systems shown in figure 10, does not approach a value of unity as the molality approaches zero mol kg−1; the minimum value of δ± for Na 2 CO 3 occurs at a molality of 0.0040 mol kg−1. While there are presently no experimental values of δ± available with which to compare our calculated values, this property is potentially measureable .
Figure 10.
Open in a new tab
Calculated values of δ± for several electrolyte solutions: (a) corresponds to acetic acid and to CuSO 4, (b) corresponds to H 2 SO 4, (c) corresponds to HF; (d) corresponds to Na 2 CO 3; and (e) corresponds to CdCl 2. These curves were calculated using the values of the equilibrium constants cited above (a value of K A equal to 99 was used). Pitzer’s expression for (eq (6)) was used for acetic acid, hydrofluoric acid, and cadmium chloride. The Debye-Hückel expression for (eq (5)) was used for sulfuric acid, copper sulfate and sodium carbonate with respective values of B equal to 2.5, 5.9, and 1.0.
In summary, an equilibrium model for aqueous solutions has several important applications: 1) the Gibbs energy properties can be reliably estimated at low molalities if the appropriate equilibrium constants are known, 2) an equation of state can be generated which is appropriate for a particular type of solution, 3) amounts of species in a given solution can be calculated, 4) single-ion activities can be calculated, and 5) as was done here, effects of variations in the equilibria, state of hydration, and electrostatic contributions to the Gibbs energy properties can be investigated. A natural extension of this model is the calculation of enthalplies, heat capacities, and volumes of aqueous solutions.
Acknowledgments
The author thanks Drs. Graham Morrison, Ralph L. Nuttall, and, in particular, Robert H. Wood, for useful and stimulating discussions on the topics of this paper.
GLOSSARY
Roman
a
activity
b
constant equal to 1.2 in Pitzer’s equation
h
hydration number
m
molality/mol kg−1
n
amount or moles of substance
p
pressure
i jk
number of moles of species S jk participating in a given reaction
z
charge
A
an anion
A m ,A ϕ
Debye-Hückel constants; A m = 3 A ϕ = 1.17642 kg−l/2 mol−1/2 at 298.15 K
B
parameter in the Debye-Hückel equation
C
a cation
I
ionic strength
G
Gibbs energy
K
equilibrium constant
M
molar mass/kg mol−1
N s , N e, N c,N k
number of species in solution, equilibria, components, and, respectively, species in the k th equilibrium
R
gas constant
S jk
the j th species in the k th equilibrium
T
temperature
X
mole fraction
Greek
γ
activity coefficient
δ ij
Kronecker delta; δ u = 1 if i=j, δ ij = 0 if i≠j
δ±
Frank’s single-ion activity coefficient function
λ ij
pair-wise interaction parameter
μ ijk
triplet interaction parameter
v
ion number
ξ
extent of reaction variable
ϕ
osmotic coefficient
ϕ x
rational osmotic coefficient
Superscripts
ex
excess
id
ideal
st
stoichiometric
o
standard value quantity
the property of the pure substance
^
a species quantity
Subscripts
a
an anion
c
a cation except when used with capital Roman N as N c see Roman N e
i, j, k
a species or used as indices; also see N k under Roman
ℓ
a component
m
see A m under Roman
r
a reference cation or anion
s
see N s under Roman
I
water
±
mean ionic
Φ
see A ϕ under Roman
Biography
About the Author, Paper: Robert N. Goldberg is with the NBS Chemical Thermal Dynamics Division. The research reported on was carried out at NBS under the sponsorship of the NBS Office of Standard Reference Data and the U.S. Department of Energy.
Footnotes
1
Figures in brackets indicate literature references at the end of this paper.
2
Throughout this paper the activity will always have units of mol kg−1. The equilibrium constants, formed as products and quotients of activities, will have units of mol kg−l, or kg mol−1, or, for symmetrical reactions, will be dimensionless. For sake of brevity, the units for equilibrium constants will be omitted.
3
This is the convention most frequently used in the description of aqueous electrolyte solutions. It is based upon the molality scale. A different convention based upon the mole fraction scale is commonly used for the description of non-electrolyte solutions.
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7516 | https://pubmed.ncbi.nlm.nih.gov/39858896/ | Vibrio vulnificus-A Review with a Special Focus on Sepsis - PubMed
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. 2025 Jan 10;13(1):128.
doi: 10.3390/microorganisms13010128.
Vibrio vulnificus-A Review with a Special Focus on Sepsis
Marcello Candelli1,Marta Sacco Fernandez1,Cristina Triunfo1,Andrea Piccioni1,Veronica Ojetti2,Francesco Franceschi1,Giulia Pignataro1
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Affiliations
1 Emergency, Anesthesiological and Reanimation Sciences Department, Fondazione Policlinico Universitario A. Gemelli-IRCCS of Rome, 00168 Rome, Italy.
2 Department of Internal Medicine, UniCamillus International Medical University of Rome, 00131 Rome, Italy.
PMID: 39858896
PMCID: PMC11768060
DOI: 10.3390/microorganisms13010128
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Vibrio vulnificus-A Review with a Special Focus on Sepsis
Marcello Candelli et al. Microorganisms.2025.
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. 2025 Jan 10;13(1):128.
doi: 10.3390/microorganisms13010128.
Authors
Marcello Candelli1,Marta Sacco Fernandez1,Cristina Triunfo1,Andrea Piccioni1,Veronica Ojetti2,Francesco Franceschi1,Giulia Pignataro1
Affiliations
1 Emergency, Anesthesiological and Reanimation Sciences Department, Fondazione Policlinico Universitario A. Gemelli-IRCCS of Rome, 00168 Rome, Italy.
2 Department of Internal Medicine, UniCamillus International Medical University of Rome, 00131 Rome, Italy.
PMID: 39858896
PMCID: PMC11768060
DOI: 10.3390/microorganisms13010128
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Abstract
Vibrio vulnificus (V. vulnificus) is a Gram-negative, halophilic bacillus known for causing severe infections such as gastroenteritis, necrotizing fasciitis, and septic shock, with mortality rates exceeding 50% in high-risk individuals. Transmission occurs primarily through the consumption of contaminated seafood, exposure of open wounds to infected water, or, in rare cases, insect bites. The bacterium thrives in warm, brackish waters with high salinity levels, and its prevalence is rising due to the effects of climate change, including warming ocean temperatures and expanding coastal habitats. High-risk populations include individuals with underlying conditions such as chronic liver disease, diabetes, or immunosuppression, which heighten susceptibility to severe outcomes. The pathogenicity of V. vulnificus is mediated by an array of virulence factors, including hemolysins, proteases, and capsular polysaccharides, as well as mechanisms facilitating iron acquisition and immune system evasion. Clinical manifestations range from localized gastrointestinal symptoms to life-threatening systemic infections such as septicemia. Rare but severe complications, including pneumonia and meningitis, have also been reported. Treatment typically involves the use of doxycycline in combination with third-generation cephalosporins, although the emergence of multidrug-resistant strains is an escalating concern. Alternative therapeutic approaches under investigation include natural compounds such as resveratrol and the application of antimicrobial blue light. For necrotizing infections, prompt and aggressive surgical intervention remains essential to improving patient outcomes. As global temperatures continue to rise, understanding the epidemiology of V. vulnificus and developing innovative therapeutic strategies are critical to mitigating its growing public health impact.
Keywords: Vibrio vulnificus; antibiotic; microbiology; sepsis; virulence factor.
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The authors declare no conflict of interest.
Figures
Figure 1
V. vulnificus infection. Created in…
Figure 1
V. vulnificus infection. Created in BioRender ( , URL created on 28…
Figure 1
V. vulnificus infection. Created in BioRender ( URL created on 28 December 2024).
Figure 2
Climate change increased V. vulnificus …
Figure 2
Climate change increased V. vulnificus diffusio. Created in BioRender ( , URL…
Figure 2
Climate change increased V. vulnificus diffusio. Created in BioRender ( URL created on 28 December 2024).
Figure 3
Main virulence factors of V.…
Figure 3
Main virulence factors of V. vulnificus .
Figure 3
Main virulence factors of V. vulnificus.
Figure 4
Prisma flow diagram.
Figure 4
Prisma flow diagram.
Figure 4
Prisma flow diagram.
See this image and copyright information in PMC
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References
Leng F., Lin S., Wu W., Zhang J., Song J., Zhong M. Epidemiology, pathogenetic mechanism, clinical characteristics, and treatment of Vibrio vulnificus infection: A case report and literature review. Eur. J. Clin. Microbiol. Infect. Dis. 2019;38:1999–2004. doi: 10.1007/s10096-019-03629-5. - DOI - PubMed
Horseman M.A., Surani S. A comprehensive review of Vibrio vulnificus: An important cause of severe sepsis and skin and soft-tissue infection. Int. J. Infect. Dis. 2011;15:e157–e166. doi: 10.1016/j.ijid.2010.11.003. - DOI - PubMed
WHO . Risk Assessment of Vibrio vulnificus in Raw Oysters: Interpretative Summary and Technical Report. WHO; Geneva, Switzerland: 2008.
Liang J.H., Liang W.H., Deng Y.Q., Fu Z.G., Deng J.L., Chen Y.H. Vibrio vulnificus infection attributed. to bee sting: A case report. Emerg. Microbes Infect. 2021;10:1890–1895. doi: 10.1080/22221751.2021.1977589. - DOI - PMC - PubMed
CDC Vibrio Infection (Vibriosis). Vibrio and Oysters. 16 May 2024. [(accessed on 6 December 2024)]; Available online:
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7517 | https://hishamezzat.weebly.com/uploads/9/0/6/0/9060375/chapter_6cation.pdf | Group V Cation Page 132 Group IV Cations (Alkaline Earth Metals) Barium (II), Strontium (II), Calcium (II) This group of cations also called the alkaline earth group is composed of three elements: barium, strontium and calcium. These three elements occur within the same group of the periodic table and their cations are therefore closely related in their chemical reactions. This renders a clean separation from each other not feasible. However, for qualitative analysis, the carbonates, chromates, sulphates and oxalates are found useful. Thus, all three metal cations are precipitated as carbonates, then they are separated by precipitating barium as chromate, strontium as sulphate and calcium as oxalate with the help of certain Group V Cation Page 133 reagents or masking agents. Since the values of the solubility product constants of each of these cations with the three precipitating anions are too close to permit sharp separation and identification, it is necessary, for confirmation of these precipitates and their corresponding cations to employ the flame tests, as they are more specific especially by comparison with known solutions of these cations. Group reagent Ammonium carbonate solution in presence of ammonium hydroxide and ammonium chloride. The reagent has to be used in neutral or slightly alkaline media and is decomposed by acids (even by acetic acid) with the formation of carbon dioxide. The white precipitates formed with the group reagent are : barium carbonate BaCO3, strontium carbonate SrCO3 and calcium carbonate CaCO3. Group V Cation Page 134 Residue Centrifugate BaCO3 SrCO3 CaCO3 Group VI white white white - Wash with hot water - Dissolve in acetate buffer (acetic acid/ammonium acetate) (3) - Heat to boiling - Add a slight xss of K2CrO4 solution - Centrifuge Residue Centrifugate BaCrO4 Sr(Ac)2, Ca(Ac)2, xss K2CrO4 yellow - Wash with hot water - Alkalinize with conc.NH4OH - Dissolve in conc. HCl - Add (NH4)2CO3 solution - Divide into 2 portions or solid Na2CO3 (4) - Heat, Centrifuge + Conc. H2SO4 Evaporate almost to dryness and apply flame test Residue Centrifugate BaSO4 White ppt. SrCO3 CaCO3 xss CrO4 2- (acid insoluble) Apple-green white yellow color - Wash with hot water (reject) (5) - Dissolve in acetate Separation of Group IV cations Ba2+ , Sr2+, Ca2+ - Add solid NH4Cl and dissolve - Add Conc. NH4OH + (NH4)2CO3 solution (1) - Heat at 60-70oC in a w.b. (2) -Centrifuge Group V Cation Page 135 buffer or acetic acid - Add Na2S2O3 solid or triethanolamine (6) - Add (NH4)2SO4 solution - Heat on a b.w.b. - Centrifuge Residue Centrifugate SrSO4 Ca complex with Na2S2O3 white (acid insoluble) or triethanolamine or - Add ammonium - Add (NH4)2CO3 oxalate solution - Ignite the ppt. (7) and a little HAc with the filter paper - Warm on a w.b. SrS + C SrCO3 CaC2O4 White - Moisten with few -Dissolve in conc. - Dissolve in conc. drops of conc. HCl HCl HCl SrCl2 CaCl2 Apply the flame test - Apply flame test - Evaporate almost to dryness and apply flame test crimson color crimson color brick-red color Group V Cation Page 136 Notes : (1) Ammonium chloride is added to prevent the precipitation of Mg(OH)2 or Mg(CO3) by decreasing the concentration of (OH-) from ammonia solution and (CO3 2-) from ammonium carbonate solution through common ion effect. So, in ammonium carbonate solutions having the (CO3 2-) regulated by the presence of both NH4 + ions and free NH3 in moderately high concentrations, Mg2+ ion remains in solution. If NH4 + ion is present and NH3 is absent, the (CO3 2-) will be decreased to such a low value that CaCO3 will not be completely precipitated. (2) Commercial ammonium carbonate always contains ammonium hydrogen carbonate (ammonium bicarbonate) NH4HCO3 and ammonium carbamate NH2COONH4. These impurities form soluble salts with group IV cations, therefore heating at 60-70oC is necessary to convert these salts to the insoluble carbonates 2HCO3 - CO3 2- + CO2 + H2O NH2COO- + H2O NH4 + + CO3 2- (3) The alkaline earth carbonates are readily soluble in acidic solutions, so acetic acid could be used to dissolve the Group V Cation Page 137 carbonates before adding potassium chromate solution but actually acetate buffer (acetic acid/ammonium acetate) is used to control the acidity at pH 4-5 so that only BaCrO4 is precipitated, this is because of the alkaline earth chromates barium chromate is the least soluble and calcium chromate is the most soluble. The concentration of chromate ion can be easily regulated through controlling the [H+] so that in the buffer system BaCrO4 precipitates leaving Sr2+ and Ca2+ in solution. But if [H+] is greatly increased (pH < 4), as by adding a strong acid, barium chromate will dissolve. 2BaCrO4 + 2H+ 2Ba2+ + Cr2O7 2- + H2O (CrO4 2- and Cr2O7 2- are interchangeable depending upon the pH of the medium). If the solution is not acidic enough (pH > 5) strontium may be partially precipitated. (4) Alkalinization with NH4OH is necessary before the addition of (NH4)2CO3 to prevent its decomposition in acidic medium as the reagent is decomposed by acids with the formation of CO2. In this step Na2CO3 can also be used as the centrifugate containing group V cations has been already removed in a Group V Cation Page 138 previous step and thus there is no need to control (CO3 2-) by common ion effect. (5) The excess potassium chromate has to be removed otherwise it will react with triethanolamine giving a dark color which makes the observation of the SrSO4 and CaC2O4 precipitates difficult in the later tests. (6) Since group IV cations are difficult to separate, masking agents may be used. In order to separate Ca2+ and Sr2+ sodium thiosulphate (Na2S2O3) or triethanolamine N(C2H5OH)3 can be used as they form more stable complexes with Ca2+ than with Sr2+, under these conditions, SrSO4 precipitates but CaSO4 does not. (7) The SrSO4 precipitate and filter paper (cellulose) are charred together where SrSO4 is reduced to SrS that dissolves in conc. HCl. Reactions of Group IV cations - The atoms of barium , strontium and calcium have two valence electrons which are easily lost to form cations with an oxidation state of +2. - Their salts are normally white powders and form colorless solution unless the anion is colored. Group V Cation Page 139 - Solid calcium chloride is hygroscopic and is used as a drying agent. 1- Ammonia solution: No precipitates are formed because of their relatively high solubility. If the alkaline solutions are exposed to the atmosphere, some carbon dioxide is absorbed and a turbidity appears due to carbonate formation. So if a slight turbidity occurs on adding ammonia solution this may be due to small amounts of ammonium carbonate often present in an aged reagent. Generally, the hydroxides of the three elements of group IV may be considered as completely ionized in aqueous solutions. 2- Ammonium carbonate solution: White precipitates of carbonate are formed Ba2+ + CO3 2- BaCO3 Sr2+ + CO3 2- SrCO3 Ca2+ + CO3 2- CaCO3 The precipitates are soluble in acetic acid and in dilute mineral acids and slightly soluble in solutions of ammonium salts of strong acids. Group V Cation Page 140 3- Dilute sulphuric acid: White precipitates are formed. Ba2+ + SO4 2- BaSO4 Sr2+ + SO4 2- SrSO4 Ca2+ + SO4 2- CaSO4 - Of the sulphates of alkaline earth metals, barium sulphate is the most insoluble and calcium sulphate is the most soluble. The solubilities of BaSO4 and SrSO4 are so close that the separation of Ba2+ and Sr2+ ions by adding a soluble sulphate is impractical. But separation of SrSO4 from CaSO4 is possible especially on adding Na2S2O3 or triethanolamine to decrease the concentration of Ca2+ ion through complex formation. As the stability of the complexes increases with increased charge density, thus the complexes of Ca2+ ion are more stable than those of Sr2+. - The solubility of the alkaline earth metal sulphates is slightly increased by the addition of strong acids. The more soluble the metal sulphate, the greater is its increase in solubility on addition of a strong acid. Thus, while the solubility of BaSO4 is only slightly greater in acidic solution than in water, the solubility of CaSO4 is greatly enhanced. Group V Cation Page 141 - BaSO4 is almost insoluble in dilute acids and in ammonium sulphate solution but appreciably soluble in boiling concentrated sulphuric acid. BaSO4 + H2SO4 (conc) Ba2+ + 2HSO4 - - SrSO4 is insoluble in ammonium sulphate even on boiling. - CaSO4 dissolves in hot concentrated sulphuric acid . CaSO4 + H2SO4 2H+ + [Ca(SO4)2]2- The same complex is formed on heating with ammonium sulphate. 4- Ammonium oxalate solution: White precipitates are formed. Ba2+ + (COO)2 2- Ba(COO)2 Sr2+ + (COO)2 2- Sr(COO)2 Ca2+ + (COO)2 2 Ca(COO)2 Group V Cation Page 142 Caclium oxalate is the least soluble of the alkaline earth metal oxalates. Since oxalic acid is a moderately weak acid, the alkaline earth oxalates are dissolved by solutions of strong acids. - Barium oxalate is readily dissolved by acetic acid, strontium oxalate is sparingly soluble in it and calcium oxalate is insoluble in it. 5- Saturated calcium sulphate solution: - Barium gives an immediate white precipitate of barium sulphate. Ba2+ + SO4 2- BaSO4 - Strontium gives a white precipitate of strontium sulphate, formed slowly on cold but more rapidly on boiling (distinction from barium). - With calcium ions no precipitate is formed (difference from barium and strontium). 6- Potassium chromate solution: Of the alkaline earth chromates, barium chromate is the least soluble and calcium chromate is the most soluble. Group V Cation Page 143 The concentration of the chromate ion can be regulated through control of [H+] so that in an acetic acid - ammonium acetate buffered solution (pH 4-5), only BaCrO4 precipitates on adding an alkali chromate leaving Sr2+ and Ca2+ ions in solution. But if [H+] is greatly increased, as by using a strong acid, barium chromate will dissolve and the solution becomes reddish-orange owing to the formation of dichromate. 2BaCrO4 + 2H+ 2Ba2+ + Cr2O7 2- + H2O - The solubility products for SrCrO4 and CaCrO4 are much larger than for BaCrO4 and hence they require a larger [CrO4 2-] ion concentration to precipitate them. The addition of the buffer system lowers the chromate ion concentration sufficiently to prevent the precipitation of SrCrO4 and CaCrO4 but keeps it high enough to precipitate BaCrO4. - Barium chromate (yellow precipitate) is insoluble in dilute acetic acid (distinction from strontium and calcium) but readily soluble in mineral acids. Ba2+ + CrO4 2- BaCrO4 Group V Cation Page 144 - Strontium chromate (yellow precipitate) is soluble in acetic acid (distinction from barium) and in mineral acids. Sr2+ + CrO4 2- SrCrO4 - Calcium forms no precipitate with potassium chromate solution neither from dilute solutions nor from concentrated solutions in the presence of acetic acid. 7- Flame test: (Dry test) Flame tests are used for the detection and confirmation of cations of this group. The alkaline earth elements have loosely bound valence electrons that are capable of being excited even at the low temperature of a Bunsen flame, with emission of visible light. In the flame, volatile salts evaporate and dissociate into ions. The color of the emitted light is characteristic of the element involved. A platinum wire free from any adhering substances that will color the flame is used for the flame test. The platinum wire is moistened with concentrated hydrochloric acid before being Group V Cation Page 145 dipped into the substance. The best flames are obtained with the more volatile halides. Volatile barium salts, when heated in a non-luminous Bunsen flame impart an apple-green (yellowish-green) color to the flame, strontium salts a crimson and calcium salts a brick red color. |
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Pigeonhole Principle/Solutions
These are the solutions to the problems related to the Pigeonhole Principle.
Contents
[hide]
1 Introductory
1.1 I1
1.2 I2
2 Intermediate
2.1 M1
2.2 M2
2.3 M3
3 Olympiad
3.1 O1
3.2 O2
3.3 O3
3.4 O4
3.5 O5
3.6 O6
Introductory
I1
The Martian must pull 5 socks out of the drawer to guarantee he has a pair. In this case the pigeons are the socks he pulls out and the holes are the colors. Thus, if he pulls out 5 socks, the Pigeonhole Principle states that some two of them have the same color. Also, note that it is possible to pull out 4 socks without obtaining a pair.
I2
Consider the residues of the elements of , modulo . By the Pigeonhole Principle, there exist distinct such that , as desired.
Intermediate
M1
The maximum number of friends one person in the group can have is n-1, and the minimum is 0. If all of the members have at least one friend, then each individual can have somewhere between to friends; as there are individuals, by pigeonhole there must be at least two with the same number of friends. If one individual has no friends, then the remaining friends must have from to friends for the remaining friends not to also have no friends. By pigeonhole again, this leaves at least other person with friends.
M2
For the difference to be a multiple of 5, the two integers must have the same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the same remainder. Thus, the answer is 1 (E).
M3
Multiplying both sides by , we have Now, we wish to find a between 1 and such that is within of some integer. Let denote the fractional part of . Now, we sort the pigeons into the holes If any pigeon falls into the first or last hole, we are done. Therefore assume otherwise; then some two pigeons for . Assume, without loss of generality, that . Then we have that must fall into the first or last hole, contradiction.
Olympiad
O1
By the Triangle Inequality Theorem, a side of a triangle must be less than the sum of the other two sides. This is equivalent to say that every side must be greater than the subtraction of the other two sides.
Consider the following statements:
If we have 2 line segments of lengths (so that and ); then if we have at least a line segment of length left to check (so that ), we will get that and are sides of a triangle. This is true because . This means if we check, for example, and then any number less than or equal to will satisfy the condition of them being sides of a triangle.
If we have 3 line segments of lengths to check inside an interval with a form ![Image 48: $[k,2k$, so that we will find that they are sides of a triangle. This is true because , , and .
If we have 2 line segments of lengths so that , then are sides of a triangle if and . This is a generalization of 1.
Now let's consider the intervals ![Image 59: $[1,2$ , ![Image 60: $[2,4$ and .
If we have 3 lines segment of lengths ![Image 62: $a,b,c \in [1,2$, then they are sides of a triangle because of 2.
If we have 3 lines segment of lengths ![Image 63: $a,b,c \in [2,4$, then they are sides of a triangle because of 2.
To analyze the case of having 3 lines segment of lengths , we can have two subcases. In both of them we will assume that 2 line segments' lengths are in ![Image 65: $[1,2$ and 2 line segments' lengths are in ![Image 66: $[2,4$ (otherwise it wouldn't be necessary to check because we would have 3 lenghts in the same interval). Also, that the difference between the two lengths in ![Image 67: $[2,4$ is greater than or equal to 1. (So that we can't apply 1.)
A. If we can find so that , we are done because there is a ![Image 70: $c \in [2,4$ so that (Using 3. and the assumptions)
B. If we can't find so that , that means the lengths are . But if we look at them, they are 3 sides of a triangle.
Finally, as we wish to distribute 7 lengths in 3 intervals, the Pigeonhole Principle can be used to guarantee that at least 3 line segments' lengths belong to a same interval, and therefore to satisfy the condition.
O2
For the difference to be a multiple of 7, the integers must have equal modulo 7 residues. To avoid having 15 with the same residue, 14 numbers with different modulo 7 residues can be picked (). Thus, two numbers are left over and have to share a modulo 7 residue with the other numbers under the pigeonhole principle.
O3
Label the numbers in the set , consider the 100 subsets and for each of these subsets, compute its sum. If none of these sums are divisible by , then there are sums and residue classes mod (excluding ). Therefore two of these sums are the same mod , say (with ). Then , and the subset suffices.
O4
We can split up the circle into nine parts as shown below, where the smaller circle in the middle has a diameter of .
By the pigeonhole principle, at least one of these parts must contain or more points. Any two points within the smallest circle must have a distance of less than . For each of the outer regions, the maximum distance between points within the same region will be when the two points are on opposing corners, or when they are placed on the two outer corners.
First, we determine which of these lengths is longer. The segment shown in blue below has length . Using to approximate this value, we get . This value is slightly above , which means it is closer to than . Therefore, is longer than so we only need to verify that the length of is less than .
This distance is equal to the hypotenuse of a right triangle with legs of length and , as shown below.
After using the pythagorean theorem, we find the red line above has length . Approximating this value using , we get .
Therefore, no two points within the same region can have a distance of or more. As stated earlier, at least one of these regions must have or more points, so there must be a pair of points with a distance of less than .
O5
The pigeonhole principle is used in these solutions (PDF).
O6
In the worst case, consider that senator hates a set of 3 senators, while he himself is hated by a completely different set of 3 other senators. Thus, given one senator, there may be a maximum of 6 other senators whom he cannot work with. If we have a minimum of 7 committees, there should be at least one committee suitable for the senator after the assignment of the 6 conflicting senators.
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7519 | https://onlinelibrary.wiley.com/doi/10.1155/2018/1893562 | Clinical Oral Health Recommended Care and Oral Health Self‐Report, NHANES, 2013‐2014 - Wiener - 2018 - Advances in Public Health - Wiley Online Library
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Advances in Public Health
Volume 2018, Issue 1 1893562
Research Article
Open Access
Clinical Oral Health Recommended Care and Oral Health Self-Report, NHANES, 2013-2014
R. Constance Wiener,
Corresponding Author
R. Constance Wiener
rwiener2@hsc.wvu.edu
orcid.org/0000-0003-0371-6699
Department of Dental Practice and Rural Health, School of Dentistry, 104A Health Sciences Addition, P.O. Box 9415, West Virginia University, Morgantown, WV 26506-9448, USA wvu.edu
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Nilanjana Dwibedi,
Nilanjana Dwibedi
Department of Pharmaceutical Systems and Policy, West Virginia University School of Pharmacy, Robert C. Byrd Health Sciences Center [North], P.O. Box 9510, Morgantown, WV 26506-9510, USA wvu.edu
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Chan Shen,
Chan Shen
Departments of Health Services Research and Biostatistics, University of Texas MD Anderson Cancer Center, 1400 Pressler St., Houston, TX 77030, USA mdanderson.org
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Patricia A. Findley,
Patricia A. Findley
Rutgers University, School of Social Work, 536 George Street, New Brunswick, NJ 08901, USA rutgers.edu
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Usha Sambamoorthi,
Usha Sambamoorthi
orcid.org/0000-0001-8311-1360
Department of Pharmaceutical Systems and Policy, West Virginia University School of Pharmacy, Robert C. Byrd Health Sciences Center [North], P.O. Box 9510, Morgantown, WV 26506-9510, USA wvu.edu
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R. Constance Wiener,
Corresponding Author
R. Constance Wiener
rwiener2@hsc.wvu.edu
orcid.org/0000-0003-0371-6699
Department of Dental Practice and Rural Health, School of Dentistry, 104A Health Sciences Addition, P.O. Box 9415, West Virginia University, Morgantown, WV 26506-9448, USA wvu.edu
Search for more papers by this author
Nilanjana Dwibedi,
Nilanjana Dwibedi
Department of Pharmaceutical Systems and Policy, West Virginia University School of Pharmacy, Robert C. Byrd Health Sciences Center [North], P.O. Box 9510, Morgantown, WV 26506-9510, USA wvu.edu
Search for more papers by this author
Chan Shen,
Chan Shen
Departments of Health Services Research and Biostatistics, University of Texas MD Anderson Cancer Center, 1400 Pressler St., Houston, TX 77030, USA mdanderson.org
Search for more papers by this author
Patricia A. Findley,
Patricia A. Findley
Rutgers University, School of Social Work, 536 George Street, New Brunswick, NJ 08901, USA rutgers.edu
Search for more papers by this author
Usha Sambamoorthi,
Usha Sambamoorthi
orcid.org/0000-0001-8311-1360
Department of Pharmaceutical Systems and Policy, West Virginia University School of Pharmacy, Robert C. Byrd Health Sciences Center [North], P.O. Box 9510, Morgantown, WV 26506-9510, USA wvu.edu
Search for more papers by this author
First published: 26 June 2018
Citations: 4
Academic Editor: Ronald J. Prineas
This article is part of Special Issue:
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1. Introduction
2. Methods
3. Results
4. Discussion
5. Conclusion
Disclosure
Conflicts of Interest
Authors’ Contributions
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Abstract
Purpose. The purpose of this study was to determine the concordance of self-reported responses to oral health questions versus clinically evaluated recommended need for oral healthcare by calibrated dentists to determine usefulness of the questions for epidemiological studies. We additionally examined other factors associated with concordant self-reports versus clinical evaluations. Materials and Methods. We used a cross-sectional study design with 4,205 participants, ages 30 years and above, who had complete oral health self-perception data and dental referral data in the NHANES 2013-14. Calibrated dentists completed clinical oral healthcare assessments. The assessments were dichotomized to (1) recommendation for immediate care and (2) routine oral health care. Self-reported oral health needs were measured with 6 items (an overall oral health self-perception question, oral pain within the previous year, impact on job/school, suspected periodontal disease, tooth appearance, and tooth mobility). The key item of interest was the overall oral health self-perception question. Results. Concordance with clinically evaluated recommended need for oral healthcare varied from 52.0% (oral pain) to 65.4% (overall oral health self-perception). Many subgroup differences were observed. Conclusions. The overall self-perception of oral health and the clinical evaluation of oral healthcare need were substantially concordant; other self-reported measures were moderately concordant. This is useful information and points to the need for a minimum set of measures that can provide actionable information and capture the need for clinical dental care.
1. Introduction
The World Dental Federation (FDI) policy-makers adopted a new definition of oral health in 2016. In addition to addressing well-being and the absence of disease or infirmity, they defined oral health as being multifaceted, fundamental to health and quality of life, and subject to an individual’s circumstances . The FDI policy-makers described oral health as involving speaking, smiling, tasting, touching, chewing, swallowing, and emoting . The burden of poor oral health and its consequences have resulted in a call for oral health to be included in all health policies ; a call derived from the voices of the people for overall better care, better health, and lower cost . There are many known factors (social, psychosocial, economic, and cultural) that interact holistically with biological factors and have pivotal roles in overall health outcomes subject to an individual’s circumstances . Likewise, social, psychosocial, economic, and cultural factors also impact self-perception of health. However, in terms of clinical diagnoses and/or assessments, self-perception questions and clinical examinations may not have adequate agreement . In a clinical setting, the discordance between patient’s self-report of symptoms or lack thereof and a healthcare provider’s clinically derived diagnosis/assessment is often resolved. However, on a population level, using data to learn about ways to improve quality requires measures (1) that are of importance, (2) that are efficient and do not involve a lot of time, (3) that measure what is intended, and (4) that are helpful in informing policy . As such, to address a population’s oral health needs for policy determination, it is important to know the agreement between questions involving oral health self-perceptions/self-report of needs versus clinically evaluated oral healthcare need so that the fewest and the best questions can be used in population research.
A number of researchers have examined oral health self-reports and oral health outcomes. For example, researchers found agreement between the self-reported number of missing teeth and the clinically determined number of missing teeth in adults, ages 70 years and above .
However, researchers also determined that self-reports of periodontal disease had good specificity but low sensitivity with clinical determinations among Veterans . Among healthcare professionals, self-reports of periodontal surgery were associated with clinically determined periodontal disease measured in bone loss . And, in a study in which researchers completed a full mouth clinical assessment for periodontal disease, the self-report of periodontal disease was in agreement with the clinical results . In circumstances where only self-reports are available, valid correspondence with oral health needs is important to advance knowledge and to inform both treatment planning and policy development. Self-reported symptoms and health status matter. For example, since self-reported smokers were more than twice as likely to report poor oral health than nonsmokers and more likely to seek dental care symptomatically , report oral-facial pain , or report having higher dental needs , their dental treatment planning requires the consideration of their self-report.
However, there is a lack of consistency in epidemiological studies using self-reports with reference to oral health, due to the differences in which researchers ask oral health self-report questions, the end-points/outcomes for research that are considered, and the samples that are chosen. In summary, establishing which self-report questions have the best concordance with clinical evaluations has the potential to improve efficiency, improve reliability of epidemiological studies without the expense of clinical assessment, provide useful information for policy development, and ultimately improve oral healthcare without excessive measurement.
The purpose of this study was to determine the concordance of self-reported oral health questions versus the clinical evaluation of oral healthcare need by calibrated dentists to determine useful epidemiological questions. The determination of operant, valid questions about oral health is needed so that patient’s behaviors/symptoms/conditions can be determined efficiently and diplomatically. Our focus is to provide data-driven evidence on the oral health questions that were relatively more concordant with the clinical determinations for the need of immediate or routine dental care. Tension exists for both the provider and patient when required to collect extraneous data which wastes time, is not helpful, and does not improve health outcomes .
The present study received West Virginia University Institutional Review Board acknowledgement (protocol number 1606141771). The conceptual framework for this study was the Multidimensional Conceptual Model of Oral Health in which clinical oral health need is identified as oral tissue damage . In the model, tissue damage and oral disease (oral pain and discomfort, oral functional limits, and oral disadvantage) are factors for self-rated oral health.
2. Methods
2.1. Data Source
The data source for the present study was National Health and Nutrition Examination Surveys (NHANES) 2013-14 , which is available to researchers from the NHANES website. The Centers for Disease Control and Prevention researchers for the NHANES used stratified, multistage probability sampling designs for the surveys. The NHANES participants were civilians who were noninstitutionalized and who lived in the U.S., including Washington, DC. The researchers for the NHANES oversampled smaller subgroups to increase estimate accuracy.
Data for the full mouth periodontal examination were collected in a mobile examination center by calibrated licensed dentists who used #5 reflecting mirrors, Hu Friedy PCP-2 (Hu Friedy, Chicago, IL) periodontal probes with markings of 2-4mm; 6-mm, and 10-12 mm parallel to the tooth’s long axis for the periodontal examination, and #23 dental explorers for the dental examination . A reference examiner conducted 20-25 examination replications per year to verify calibration. The examiners reported if there was a need for a participant to seek dental care, or if the participant needed to continue routine care. Participants for the periodontal examination in the NHANES, 2013-14 were ages 30 years and above. Participants for the dental examination in the NHANES, 2013-2014 were ages 1 year and above.
The participants in the NHANES, 2013-2014, also responded to interview questions involving the status of their teeth and gingiva, demographic information, and questions regarding health and nutrition. Details of the NHANES study are available at the NHANES website, .
Eligibility for this study’s data set included complete data for the dentists’ oral health recommendations and responses from questions about oral health self-perception and oral pain in adults aged 30 years and above. The final sample size consisted of 4,205 adults.
2.2. Multidimensional Measures of Self-Reported Oral Health
We used six self-reported oral health measures: overall oral health self-perception; oral pain; impact on work/school; suspected periodontal disease; tooth appearance; and tooth mobility. The key oral health self-perception question was as follows: Overall, how would (you/survey participant [SP]) rate the health of (your/his/her) teeth and gums?” The possible responses were “Excellent, Very Good, Good, Fair, and Poor.” The responses to these questions were dichotomized to Excellent/Very Good/Good and Fair/Poor.
The question about oral pain was as follows: “How often during the last year (have you/ has SP) had painful aching anywhere in (your/his/her) mouth?” The impact on work/school question was as follows: “How often during the last year (have you/has SP) had difficulty doing (your/his/her) usual jobs or attending school because of problems with (your/his/her) teeth, mouth or dentures? The possible responses were “Very Often, Fairly Often, Occasionally, Hardly Ever, or Never.” The responses for these questions were dichotomized to (1) Very often/Fairly often; and (2) Occasionally and Hardly Ever/Never.
The periodontal question was as follows: “People with gum disease might have swollen gums, receding gums, sore or infected gums or loose teeth” followed by asking “(Do you/Does SP) think (you/s/he) might have gum disease?” The tooth appearance question was as follows: “During the past three months, (have you/has SP) noticed a tooth that doesn’t look right?” And the tooth mobility question was the mobile tooth question: the possible responses to these questions were yes or no.
The “How often during, suspected periodontal disease, appearance of a tooth or teeth not looking right during the previous three months, and a loose tooth/teeth not due to injury” were also used .
2.3. Concordance/Discordance between Self-Reports and Recommended Oral Health Care
We grouped adults into two groups: (1) the concordant group (self-reported responses which were in agreement with the clinical evaluation of oral healthcare need such that a self-report of concern/need and clinical evaluation of immediate need agreed or a self-report of no concerns/needs and clinical evaluation of routine care agreed); and (2) the discordant group (self-reported responses and clinical evaluation of oral healthcare need were not in agreement).
2.4. Outcomes
The primary outcome was the concordance of the overall oral health self-perception question with the clinical evaluation of oral healthcare need. We determined the percentage of agreement between the self-perception of fair or poor care and the clinical evaluation of oral healthcare need.
We were also interested in the specificity of the overall health self-perception question versus clinical evaluation of oral healthcare need. We determined the percentage of agreement between the self-perception of excellent/very good/good and the clinical evaluation of routine care.
2.5. Statistical Analyses
Due to the complex nature of NHANES, SAS® version 9.4 (SAS Institute, Inc., Cary, NC) was used with the supplied weights in the data set. The analyses also accounted for stratification, primary sampling unit values, and eligibility. We used chi-square tests to assess the statistical significance of unadjusted associations. We also performed logistic regressions on concordance between clinical evaluation of recommended care and self-reported oral health measures after controlling for sex, race/ethnicity, age, education, federal poverty level, insurance coverage, obesity, alcohol use, smoking status, physical activity, presence of chronic conditions (cancer, cardiovascular disease, and diabetes), general health status, and dental visits.
The level of statistical significance for alpha was set at 0.05. Strength of concordance was set at 0-20% as poor; 21-20% as slight; 41-60% as moderate, 61-80% as substantial; and 81-100% as almost perfect, based upon similar guidelines for the Kappa coefficient by Landis and Koch .
3. Results
In Table 1, we report the weighted percentages for the clinical evaluation of oral healthcare need versus the self-reported responses to questions about oral health status (overall oral health self-perception, oral pain, impact on work/school, suspected periodontal disease, tooth appearance which “does not looking right”, and tooth mobility). The percentages in the columns are for immediate or routine oral healthcare need for each self-reported response. Each response to the questions about oral health status was statistically significant, that is, more people who reported fair/poor oral health self-perception were more likely to have a clinical determination of needing immediate care; more people reporting pain were more likely to have a clinical determination of needing immediate care; more people who reported that there was an impact on work/school due to an oral condition were more likely to have a clinical determination of needing immediate care; more people reporting a suspected periodontal disease were more likely to have a clinical determination of needing immediate care; and more people who reported that a tooth’s appearance did not look right were more likely to have a clinical determination of needing immediate care.
Table 1. Oral health and recommended care versus variables of interest. Adults aged 30 years or older in National Health Examination and Nutrition Survey 2013-2014.
| | Immediate Care | Routine Care | Chi-sq | Prob | Sig |
--- --- --- |
| N | wt % | N | wt % |
| ALL | 2,411 | 50.2 | 1,794 | 49.8 | | | |
| |
| Overall oral health self-perception | | | | | | | ∗∗∗ |
| Fair/Poor | 1,089 | 79.3 | 253 | 39.6 | 368.373 | < .001 | |
| Ex/Vg/Good | 1,322 | 20.7 | 1,541 | 60.4 | | | |
| Oral Pain1 | | | | | | | ∗∗∗ |
| Yes | 221 | 66.6 | 83 | 49.1 | 27.625 | < .001 | |
| No | 2,190 | 33.4 | 1,711 | 50.9 | | | |
| Impact on work/school | | | | | | | ∗∗∗ |
| Yes | 295 | 66.6 | 106 | 48.7 | 39.768 | < .001 | |
| No | 2,116 | 33.4 | 1,688 | 51.3 | | | |
| Suspected periodontal disease2 | | | | | | | ∗∗∗ |
| Yes | 558 | 67.3 | 208 | 46.7 | 63.326 | < .001 | |
| No | 1,853 | 32.7 | 1,586 | 53.3 | | | |
| Tooth appearance does not look right | | | | | | | ∗∗∗ |
| Does not look right | 581 | 82.3 | 99 | 45.7 | 184.747 | < .001 | |
| Looks right | 1,830 | 17.7 | 1,695 | 54.3 | | | |
| Tooth mobility3 | | | | | | | ∗∗∗ |
| Mobile | 527 | 71.9 | 181 | 46.8 | 106.808 | < .001 | |
| No mobility | 1,884 | 28.1 | 1,613 | 53.2 | | | |
Note: based on 4,205 participants, who were 30 years and older and who had no missing data for the dentists’ oral health recommendations and responses from questions about oral health self-perception and oral pain. Ex/Vg/Good, Excellent/Very Good/Good. 1 Aching anywhere in the mouth during the last year. 2 If participant thought he or she “might have gum disease” (NHANES, 2017). 3 Participant was asked if “any teeth [were] becoming loose without an injury” (NHANES, 2017].
Table 2 has the concordance of the self-reported oral healthcare measures with the clinical evaluation of oral healthcare need in which the concordant group was in agreement with the self-report of a need with a clinical evaluation of oral healthcare need, or was in agreement with the self-report of no need with a clinical evaluation of routine oral healthcare; and the discordant group was in disagreement with the clinical evaluation of oral healthcare need. Clinical evaluation of oral healthcare need and the self-report for overall oral health self-perception had the highest concordance at 65.4%. The lowest concordance was with oral pain (aching anywhere in the mouth during the last year) at 52.0%.
Table 2. Concordance of self-reported oral health measures and oral health recommended care. National Health and Nutrition Examination Surveys 2013-2014.
| Total | N 4,205 | Wt % |
---
| Overall oral health self-perception | | |
| Concordant | 2,630 | 65.4 |
| Discordant | 1,575 | 34.6 |
| |
| Oral pain1 | |
| Concordant | 1,932 | 52.0 |
| Discordant | 2,273 | 48.0 |
| |
| Impact on work/school | |
| Concordant | 1,983 | 52.6 |
| Discordant | 2,222 | 47.4 |
| |
| Suspected periodontal disease2 | | |
| Concordant | 2,144 | 55.6 |
| Discordant | 2,061 | 44.4 |
| |
| Tooth appearance “does not look right” within the previous 3 months | |
| Concordant | 2,276 | 57.8 |
| Discordant | 1,929 | 42.2 |
| |
| Tooth mobility.3 | | |
| Concordant | 2,140 | 55.7 |
| Discordant | 2,065 | 44.3 |
Note: based on 4,205 participants, who were 30 years and older and who had no missing data for the dentists’ oral health recommendations and responses from questions about oral health self-perception and oral pain. 1 Aching anywhere in the mouth during the last year. 2 If participant thought he or she “might have gum disease” (NHANES, 2017). 3 Participant was asked if “any teeth [were] becoming loose without an injury” (NHANES, 2017].
The bivariate associations of concordant self-reported oral health with clinical evaluation of oral healthcare need are in Table 3. There were significant differences in concordance when considering sex, race/ethnicity, education, federal poverty level, insurance coverage, and diabetes for both overall oral health self-perception and oral pain. There were also significant differences in concordance when considering body mass index, smoking, cardiovascular disease, self-reported general health, and dental visit for the relationship with oral pain.
Table 3. Weighted % of concordance between clinical oral health recommended care and self-reported oral health measures adults aged 30 years or older in National Health and Examination Nutrition Survey, 2013-14.
| | Overall Oral health self-perception | Oral Pain1 | Impact on Job or school | Suspected Periodontal Disease | Tooth Appearance “does not look right” | Tooth mobility3 |
--- --- ---
| ALL | 65.4 | 52.0 | 52.6 | 55.6 | 57.8 | 55.7 |
| |
| Sex | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| Female | 69.2 | 57.1 | 57.8 | 59.8 | 62.4 | 60.4 |
| Male | 61.4 | 46.5 | 47.0 | 51.2 | 52.9 | 50.8 |
| Race/Ethnicity | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| Non-Hispanic White | 68.0 | 57.0 | 56.9 | 60.4 | 61.5 | 60.4 |
| Non-Hispanic Black | 57.5 | 40.0 | 42.1 | 45.0 | 49.5 | 45.8 |
| Hispanic | 61.9 | 39.3 | 42.3 | 43.4 | 50.8 | 43.9 |
| Other | 58.6 | 44.2 | 45.3 | 48.0 | 48.1 | 46.7 |
| Age groups | | ∗ | | | ∗ | |
| 30 - 44 years | 67.4 | 54.4 | 55.8 | 56.0 | 60.7 | 57.0 |
| 45 - 54 Years | 68.7 | 51.8 | 52.0 | 56.4 | 59.8 | 57.2 |
| 55 - 64 Years | 64.4 | 47.6 | 48.5 | 55.1 | 53.2 | 52.2 |
| 65, or older | 65.5 | 55.9 | 54.9 | 57.0 | 59.4 | 61.1 |
| Education | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| Less than high school | 67.2 | 41.2 | 42.2 | 46.2 | 52.1 | 44.0 |
| High school graduate | 61.1 | 41.6 | 43.3 | 50.6 | 51.2 | 50.7 |
| Some College | 64.7 | 50.3 | 51.4 | 54.3 | 55.8 | 53.6 |
| College | 68.2 | 65.1 | 64.3 | 64.3 | 66.7 | 66.2 |
| Federal Poverty Level | ∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| 0 - < 1.25 | 63.3 | 38.4 | 39.4 | 45.4 | 52.4 | 44.4 |
| 1.25 to < 2.00 | 62.6 | 44.1 | 44.5 | 50.7 | 53.6 | 48.7 |
| 2.00 - < 4.00 | 62.9 | 48.1 | 50.7 | 52.7 | 54.0 | 54.3 |
| 4.00 and above | 69.1 | 64.6 | 63.6 | 65.3 | 65.5 | 65.3 |
| Missing | 67.2 | 52.5 | 52.4 | 52.9 | 55.1 | 54.4 |
| Insurance coverage | ∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| Yes | 66.4 | 55.7 | 55.9 | 58.8 | 60.4 | 59.2 |
| No | 59.7 | 35.5 | 34.0 | 37.9 | 43.6 | 36.7 |
| Obesity | | ∗∗∗ | ∗∗∗ | | ∗ | ∗ |
| No | 66.7 | 55.4 | 55.7 | 57.0 | 60.0 | 58.3 |
| Yes | 63.5 | 47.2 | 48.2 | 53.5 | 54.8 | 52.0 |
| Alcohol use | | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| Non-Drinker | 64.2 | 51.9 | 54.6 | 55.9 | 57.2 | 54.5 |
| Moderate use | 68.3 | 58.7 | 59.3 | 61.6 | 63.2 | 62.1 |
| Heavy use | 62.1 | 42.8 | 41.8 | 46.4 | 50.2 | 48.5 |
| Missing | 63.7 | 46.0 | 46.1 | 51.3 | 54.4 | 49.9 |
| Smoking | | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| Current | 66.1 | 40.0 | 40.3 | 42.1 | 49.7 | 43.2 |
| Former | 63.1 | 51.9 | 51.3 | 57.1 | 57.3 | 56.0 |
| Never | 66.4 | 56.1 | 57.3 | 59.5 | 60.8 | 59.8 |
| Physical activity | | | | ∗ | ∗ | ∗ |
| Yes | 65.8 | 52.7 | 53.3 | 56.6 | 58.8 | 56.6 |
| No | 64.3 | 50.1 | 50.6 | 53.0 | 55.2 | 53.2 |
| Cancer | | | ∗ | | | ∗ |
| Yes | 68.3 | 56.3 | 58.1 | 58.8 | 59.6 | 60.3 |
| No | 65.0 | 51.3 | 51.7 | 55.1 | 57.6 | 55.0 |
| Cardiovascular disease | | ∗ | | | | |
| Yes | 65.4 | 52.5 | 53.8 | 59.2 | 59.6 | 53.3 |
| No | 65.5 | 52.0 | 52.5 | 55.2 | 57.7 | 56.0 |
| Diabetes | ∗ | ∗∗∗ | ∗∗∗ | | ∗ | ∗∗ |
| Yes | 60.4 | 43.4 | 42.7 | 52.2 | 51.0 | 48.6 |
| No | 66.4 | 33.6 | 54.4 | 56.3 | 59.1 | 57.1 |
| General health | | ∗∗∗ | ∗∗∗ | ∗ | ∗∗ | ∗∗∗ |
| Excellent/very good | 65.9 | 59.6 | 59.1 | 59.3 | 61.7 | 62.8 |
| Good | 63.1 | 48.6 | 50.0 | 54.1 | 54.6 | 52.5 |
| Fair/poor | 69.1 | 44.3 | 44.7 | 52.1 | 56.6 | 48.7 |
| Missing | 65.5 | 49.5 | 53.1 | 53.0 | 57.6 | 53.8 |
| Dental visit | | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ | ∗∗∗ |
| 1 year or less | 65.7 | 59.3 | 59.7 | 60.2 | 61.2 | 62.6 |
| More than 1 year | 64.9 | 39.1 | 39.9 | 47.6 | 51.8 | 43.6 |
Note: based on 4,205 participants, who were 30 years and older and who had no missing data for the dentists’ oral health recommendations and responses from questions about oral health self-perception and oral pain. Asterisks represent significant group differences in concordance versus discordance based on Rao-Scott Chi-square tests. 1 Aching anywhere in the mouth during the last year. 2 If participant thought he or she “might have gum disease” (NHANES, 2017). 3 Participant was asked if “any teeth [were] becoming loose without an injury” (NHANES, 2017). ∗∗∗ p < .001; ∗∗ .001 ≤ p < .01; ∗ .01 ≤ p < .05.
The adjusted odds ratios (AOR) and 95% confidence intervals (CI) from logistic regressions on concordance are in Table 4. Overall, females were more likely to have concordance than males. Non-Hispanic White individuals were more likely to have concordance than racial minorities. Participants with insurance, who were not obese, or who were never-smokers were more likely to be concordant. Reported fair/poor general health was associated with high concordance between clinical oral health recommended care and oral health self-perception.
Table 4. Adjusted odds ratios (AORs) and 95% confidence intervals (CIs) from logistic regressions on concordance between recommended care and self-reported oral health measures. Adults Aged 30 and older in National Health and Examination Nutrition Survey, 2013-14.
| | Overall Oral Health Self-Perception | Oral Pain1 | Impact on job or school | Suspected Periodontal Disease2 | Tooth appearance “does not look right” | Tooth mobility3 |
--- --- ---
| | AOR | AOR | AOR | AOR | AOR | AOR |
| | [95%CI] | [95%CI] | [95%CI] | [95%CI] | [95%CI] | [95%CI] |
| Sex | | | | | | |
| Female | 1.46∗∗∗ | 1.57∗∗∗ | 1.58∗∗∗ | 1.45∗∗ | 1.54∗∗∗ | 1.59∗∗∗ |
| | [1.28, 1.67] | [1.33, 1.84] | [1.28, 1.95] | [1.12, 1.86] | [1.29, 1.82] | [1.35, 1.87] |
| Male (ref) | | | | | | |
| Race/ethnicity | | | | | | |
| Non-Hispanic Black | 0.56∗∗∗ | 0.51∗∗∗ | 0.59∗∗∗ | 0.54∗∗∗ | 0.60∗∗∗ | 0.59∗∗∗ |
| | [0.43, 0.72] | [0.40, 0.66] | [0.46, 0.77] | [0.44, 0.66] | [0.49, 0.75] | [0.50, 0.70] |
| Hispanic | 0.66∗ | 0.54∗∗∗ | 0.61∗∗∗ | 0.52∗∗∗ | 0.63∗∗ | 0.57∗∗∗ |
| | [0.46, 0.94] | [0.40, 0.66] | [0.46, 0.77] | [0.38, 0.73] | [0.47, 0.85] | [0.45, 0.71] |
| Other | 0.52∗∗∗ | 0.47∗∗∗ | 0.52∗∗∗ | 0.50∗∗∗ | 0.47∗∗∗ | 0.47∗∗∗ |
| | [0.35, 0.78] | [0.34, 0.65] | [0.36, 0.75] | [0.35, 0.73] | [0.33, 0.67] | [0.35, 0.61] |
| Non-Hispanic White (ref) | | | | | | |
| Age in years | | | | | | |
| 30 - 44 years (Ref) | | | | | | |
| 45 - 54 years | 0.99 | 0.76∗ | 0.73 | 0.92 | 0.83 | 0.85 |
| | [0.80, 1.22] | [0.60, 0.97] | [0.53, 1.00] | [0.73, 1.16] | [0.66, 1.05] | [0.64, 1.14] |
| 55- 64 years | 0.80 | 0.56∗∗∗ | 0.57∗∗ | 0.78 | 0.58∗∗∗ | 0.62∗ |
| | [0.57, 1.14] | [0.40, 0.77] | [0.38, 0.84] | [0.56, 1.09] | [0.41, 0.82] | [0.40, 0.97] |
| 65, or older | 0.81 | 0.74 | 0.69 | 0.78 | 0.72 | 0.83 |
| | [0.54, 1.22] | [0.54, 1.02] | [0.45, 1.08] | [0.57, 1.05] | [0.49, 1.04] | [0.61, 1.12] |
| Insurance coverage | | | | | | |
| Yes (ref) | | | | | | |
| No | 0.71∗∗ | 0.51∗∗∗ | 0.55∗∗∗ | 0.54∗∗∗ | 0.58∗∗∗ | 0.55∗∗∗ |
| | [0.57, 0.89] | [0.40, 0.64] | [0.44, 0.70] | [0.44, 0.66] | [0.49, 0.69] | [0.46, 0.65] |
| Self-reported General Health | | | | | | |
| Fair/poor | 1.51∗∗∗ | 0.97 | 0.98 | 1.17 | 1.26 | 0.98 |
| | [1.22, 1.86] | [0.72, 1.31] | [0.76, 1.27] | [0.99, 1.39] | [0.91, 1.74] | [0.77, 1.26] |
| Excellent/very good/good (ref) | | | | | |
| Physical Activity | | | | | | |
| No | 0.96 | 1.02 | 0.98 | 0.84 | 0.86 | 0.95 |
| | [0.79, 1.17] | [0.86, 1.21] | [0.76, 1.26] | [0.69, 1.02] | [0.74, 1.01] | [0.83, 1.08] |
| Yes (ref) | | | | | | |
| Obese | | | | | | |
| Obese | 0.74∗∗ | 0.62∗∗∗ | 0.65∗∗∗ | 0.78 | 0.67∗∗∗ | 0.67∗∗ |
| | [0.61, 0.90] | [0.51, 0.76] | [0.52, 0.81] | [0.60, 1.01] | [0.55, 0.82] | [0.52, 0.87] |
| No (Ref) | | | | | | |
| Smoking status | | | | | | |
| Current smoker | 0.93 | 0.57∗∗∗ | 0.56∗∗∗ | 0.46∗∗∗ | 0.65∗∗ | 0.56 |
| | [0.67, 1.29] | [0.43, 0.77] | [0.39, 0.80] | [0.31, 0.67] | [0.47, 0.88] | [0.67, 1.18] |
| Former smoker | 0.88 | 0.85 | 0.85 | 0.86 | 0.90 | 0.89 |
| | [0.64, 1.21] | [0.64, 1.14] | [0.60, 1.20] | [0.67, 1.09] | [0.64, 1.27] | [0.67, 1.18] |
| Never smoker (ref) | | | | | | |
| Dental visit | | | | | | |
| More than 1 year | 0.97 | 0.48∗∗∗ | 0.50∗∗∗ | 0.70∗∗∗ | 0.73∗∗∗ | 0.51∗∗∗ |
| | [0.73, 1.29] | [0.40, 0.57] | [0.42, 0.59] | [0.60, 0.82] | [0.61, 0.87] | [0.41, 0.65] |
| 1 year or less (ref) | | | | | | |
Note: based on 4,205 participants, who were 30 years and older and who had no missing data for the dentists’ oral health recommendations and responses from questions about oral health self-perception and oral pain. Asterisks represent significant group differences in concordance compared to the reference group based on logistic regressions. 1 Aching anywhere in the mouth during the last year. 2 If participant thought he or she “might have gum disease” (NHANES, 2017). 3 Participant was asked if “any teeth [were] becoming loose without an injury” (NHANES, 2017).∗∗∗ p < .001; ∗∗ .001 ≤ p < .01; ∗ .01 ≤ p < .05.
4. Discussion
When using multidimensional measures of self-reported oral health, we found that the greatest concordance with clinical evaluation of oral healthcare need was with the question for overall oral health self-perception. Clinical evaluation of oral healthcare need and the self-report for overall oral health self-perception had a substantial concordance at 65.4%. The question may be a useful tool in oral health epidemiological studies, similar to the usefulness of the overall self-rated general health question in systemic epidemiology [17–19].
Another noteworthy finding is the moderate concordance of the appearance of teeth with clinically evaluated oral healthcare need. Although we do not know whether participants were self-conscious of the color, or shape rather considering than carious/periodontal condition of their tooth/teeth when they answered the question, the literature does include “pressures to conform” as a factor influencing body image and self-awareness . The media present images of the perfect smile and ultra-white teeth with which to compare one’s teeth. Reports in the media include the obsession of many people with ultra-white teeth , and those cultural influences may be affecting the participants’ responses to this particular question.
Although not a focus of this study, additional analysis indicated that the specificity of the overall oral health self-perception question was 60.4%; and, the specificities of the other measures were between 50.9% and 53.3%. These findings have implications for referral patterns. Future research is needed to explore the reasons behind the low specificity. Additionally, when these measures are used in epidemiological research, caution is necessary in interpreting results associated with these oral health questions.
The subgroup analyses also included variations in concordance between the clinical evaluation of oral healthcare need and self-reports. Some subgroups were consistently concordant (example: female, racial minorities) on all of the measures; other groups were not. These findings suggest that when researchers use the self-reported measures on some subpopulations (smokers, middle-aged adults), the self-reported measures may not be as reliable in indicating clinical need.
4.1. Similar Studies
There is a lack of recent, similar studies with which to compare this study due to the differences in which the questions for self-report are asked, the end-points/outcomes considered, and populations chosen for the research. For example, in a study of black women (median age 38 years), there were similar self-report questions; however, only periodontal disease status and intensity (and not all other clinical evaluations of oral healthcare needs) were considered . Similarly, in another study, there was moderate agreement with the women’s self-report of the removal of periodontally involved teeth and (clinically determined) severe periodontitis (Kappa=0.25; 95%CI, 0.17, 0.31); however, the study’s focus was periodontal disease and not overall oral health .
4.2. Study Strengths
This current study has several strengths. The researchers used a large, current, nationally representative study for the data source. Several self-report questions were included in the research. The dental examiners who conducted the research to establish the NHANES 2013-2014 data source were calibrated, licensed dentists who determined if a dental need existed or if routine care should be maintained. “Overall oral health need” was used in this study. This is consistent with the 2016 FDI World Dental Federation members’ emphasis upon the new definition for oral health; that is, oral health is multifaceted such that speech, sensing (smell, taste, and touch), and muscle action (chewing, swallowing, and emoting) can occur with confidence and without pain/discomfort/disease of the craniofacial complex . Included in the definition are the influences of physical and mental well-being (recognized as a continuum influenced by individual and cultural values/attitudes); biopsychosocial attributes of life leading to quality life; and change (circumstantial, perceptual, experiential, etc.) .
4.3. Study Limitations
There are challenges to the use of broad questions concerning oral health in research. Measures need to be valid and consistently used by researchers. In a study in New Zealand and Australia, Locker’s single question for global oral health rating was slightly altered and validated with caries, tooth loss, periodontal disease, and the short form of the Oral Health Impact Profile (OHIP-14) in adults, ages 35-44 years . Altered questions make comparisons difficult. Additionally, the FDI definition suggests that age, sex, and culture will influence oral health self-perception. Self-perception questions are less involved than clinical oral evaluations; however, they must be considered proxies that vary by population and questions posed. A consensus-based set of measures for oral healthcare is being developed with patient perception as a major feature; therefore, having the appropriate measures may improve research and quality of care .
In addition to the limitations imposed by definition variability, there are other limitations. One includes the nature of the observational study design’s purpose to establish association rather than causation. Studies in which self-report is used also have the potential for social desirability bias and therefore misclassifications. Although many covariates were used in this study, there is also the potential for having missed an important confounding factor.
4.4. Clinical Considerations
The ultimate goal of oral health research is to provide the information for oral healthcare practitioners to learn the evidence-based practices to provide the best preventive and restorative care for their patients, to improve oral healthcare quality, and eliminate redundancy and waste. To maximize these effects, research studies need good study designs with more uniform/standardized questions and terminologies which accurately reflect the patient presentation. Having useful questions to direct the conversation not only is more efficient, but also is more respectful and considerate of the patient’s time and circumstances.
5. Conclusion
The overall self-perception of oral health and the clinical evaluation of oral healthcare need were substantially concordant; other self-reported measures were moderately concordant. This is useful information and points to the need for a minimum set of measures that can provide actionable information and capture the need for clinical dental care.
Disclosure
The content is solely the responsibility of the authors and does not necessarily represent the official views of the National Institutes of Health.
Conflicts of Interest
The authors report no conflicts of interest.
Authors’ Contributions
All authors contributed to the conception and design of the research. R. Constance Wiener and Usha Sambamoorthi conducted the statistical analyses. R. Constance Wiener wrote the first draft. All authors contributed to the manuscript and approved the final version.
Acknowledgments
Research reported in this publication was supported by the National Institute of General Medical Sciences of the National Institutes of Health under Award no. U54GM104942, WVCTSI.
Open Research
Data Availability
The data used to support the findings of this study are available from the corresponding author upon request.
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Copyright © 2018 R. Constance Wiener et al.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Research funding
National Institute of General Medical Sciences. Grant Numbers: U54GM104942, WVCTSI
Publication History
Issue Online: 26 June 2018
Version of Record online: 26 June 2018
Manuscript accepted: 22 May 2018
Manuscript revised: 11 May 2018
Manuscript received: 15 February 2018
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Cauchy-Schwarz Inequality
The Cauchy–Schwarz inequality is a fundamental inequality in mathematics that relates the inner product of two vectors to their magnitudes. It states that for any two vectors x and y in an i…
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Cauchy–Schwarz inequality
In mathematics, the
Cauchy–Schwarz inequality
, also known as the
Cauchy–Bunyakovsky–Schwarz inequality
, is a usefulinequality encountered in many different settings, such as linear algebra, analysis, probability theory, vector algebra and other
areas. It is considered to be one of the most important inequalities in all of mathematics.
The inequality for sums was published by Augustin-Louis Cauchy (1821), while the corresponding inequality for integrals wasfirst proved by Viktor Bunyakovsky (1859). The modern proof of the integral inequality was given by Hermann AmandusSchwarz (1888).
Statement of the inequalityProofs
First proofSecond proofMore proofs
Special cases
Titu's lemmaR
(ordinary two-dimensional space)R
n
-dimensional Euclidean space)L
2
Applications
AnalysisGeometryProbability theory
GeneralizationsSeealsoNotesReferencesExternal links
The Cauchy–Schwarz inequality states that for all vectors and of an inner product space it is true thatwhere is the inner product. Examples of inner products include the real and complex dot product; see the examples in innerproduct. Equivalently, by taking the square root of both sides, and referring to the norms of the vectors, the inequality is writtenas
ContentsStatement of the inequality
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Moreover, the two sides are equal if and only if and are linearly dependent (meaning they are parallel: one of the vector'smagnitudes is zero, or one is a scalar multiple of the other).
If and , and the inner product is the standard complex inner product, then the inequality may berestated more explicitly as follows (where the bar notation is used for complex conjugation):orLet and be arbitrary vectors in a vector space over with an inner product, where is the field of real or complex numbers.We prove the inequalityand that equality holds if and only if either or is a multiple of the other (which includes the special case that either is the zerovector).If , it is clear that there is equality, and in this case and are also linearly dependent, regardless of , so the theorem istrue. Similarly if . One henceforth assumes that is nonzero.LetThen, by linearity of the inner product in its first argument, one hasTherefore, is a vector orthogonal to the vector (Indeed, is the projection of onto the plane orthogonal to .) We can thusapply the Pythagorean theorem towhich gives
First proof
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and, after multiplication by and taking square root, we get the Cauchy–Schwarz inequality. Moreover, if the relation inthe above expression is actually an equality, then and hence ; the definition of then establishes a relation oflinear dependence between and . On the other hand, if and are linearly dependent, then there exists such that (since ). ThenThis establishes the theorem.Let and be arbitrary vectors in an inner product space over .In the special case the theorem is trivially true. Now assume that . Let be given by , thenTherefore, , or .If the inequality holds as an equality, then , and so , thus and are linearly dependent. On theother hand, if and are linearly dependent, then , as shown in the first proof.There are many different proofs
of the Cauchy–Schwarz inequality other than the above two examples.
When consultingother sources, there are often two sources of confusion. First, some authors define
⟨⋅
,
⋅⟩
to be linear in the second argument ratherthan the first. Second, some proofs are only valid when the field is and not .
Titu's lemma (named after Titu Andreescu, also known as T2 lemma, Engel's form, or Sedrakyan's inequality) states that forpositive reals, one has
Second proofMore proofs
Special cases
Titu's lemma
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It is a direct consequence of the Cauchy–Schwarz inequality, obtained upon substituting and This form isespecially helpful when the inequality involves fractions where the numerator is a perfect square.In the usual 2-dimensional space with the dot product, let and . The Cauchy–Schwarz inequality isthatwhere is the angle between and The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, whichoccurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates and aswhere equality holds if and only if the vector is in the same or opposite direction as the vector or if one ofthem is the zero vector.In Euclidean space with the standard inner product, the Cauchy–Schwarz inequality isThe Cauchy–Schwarz inequality can be proved using only ideas from elementary algebra in this case. Consider the followingquadratic polynomial in Since it is nonnegative, it has at most one real root for , hence its discriminant is less than or equal to zero. That is,which yields the Cauchy–Schwarz inequality.For the inner product space of square-integrable complex-valued functions, one hasA generalization of this is the Hölder inequality.
R
2
(ordinary two-dimensional space)R
n
(
n
-dimensional Euclidean space)L
2
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7521 | https://webbook.nist.gov/cgi/cbook.cgi?ID=C7647145&Mask=40 | sodium chloride
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sodium chloride
Formula: ClNa
Molecular weight: 58.443
IUPAC Standard InChI:InChI=1S/ClH.Na/h1H;/q;+1/p-1 Copy
IUPAC Standard InChIKey:FAPWRFPIFSIZLT-UHFFFAOYSA-M Copy
CAS Registry Number: 7647-14-5
Chemical structure:
This structure is also available as a 2d Mol file or as a computed3d SD file
View 3d structure (requires JavaScript / HTML 5)
Other names: Salt
Permanent link for this species. Use this link for bookmarking this species for future reference.
Information on this page:
Ion clustering data
References
Notes
Other data available:
Gas phase thermochemistry data
Condensed phase thermochemistry data
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Gas phase ion energetics data
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THz IR spectrum
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Data at other public NIST sites:
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Ion clustering data
Go To:Top, References, Notes
Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
Data compiled by:Michael M. Meot-Ner (Mautner) and Sharon G. Lias
Note: Please consider using the reaction search for this species. This page allows searching of all reactions involving this species. Searches may be limited to ion clustering reactions. A general reaction search form is also available.
Clustering reactions
+ = (•)
By formula: Na++ClNa = (Na+•ClNa)
| Quantity | Value | Units | Method | Reference | Comment |
--- --- --- |
| Δ r H° | 177. | kJ/mol | MS | Chupka, 1959 | gas phase; Knudsen cell |
| Quantity | Value | Units | Method | Reference | Comment |
| Δ r S° | 73.6 | J/molK | MS | Chupka, 1959 | gas phase; Knudsen cell |
References
Go To:Top, Ion clustering data, Notes
Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
Chupka, 1959
Chupka, W.A., Dissociation Energies of Some Gaseous Halide Complex Ions and the Hydrated Ion K(H2O)+, J. Chem. Phys., 1959, 40, 2, 458, . [all data]
Notes
Go To:Top, Ion clustering data, References
Symbols used in this document:
Δ r H°Enthalpy of reaction at standard conditions
Δ r S°Entropy of reaction at standard conditions
Data from NIST Standard Reference Database 69: NIST Chemistry WebBook
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Magnetic Dipole
Table of Content
Magnetic Dipole Moment
Magnetic dipole in a Uniform Field Magnetic Field
Units of M
Work done in Rotating a Magnetic Dipole in a Magnetic Field
Potential Energy of a Magnetic Dipole in a Magnetic Field
Current Loop as a Magnetic Dipole
The Magnetic Dipole Moment of a Revolving Electron
Related Resources
The limit of a closed loop of electric current is defined as a magnetic dipole. The structures, which tend to align along the direction of magnetic field, are known as magnetic dipoles. A current loop creates as well as responds to magnetic fields. We first examine the torque felt by a dipole in a magnetic field. Consider the current loop in a magnetic field in the figure given below. The two forces which are equal and opposite do not act to bring about a movement in the dipole, but only rotate it.
A combination of two isolated, equal and opposite magnetic poles separated by a small distance constitutes a magnetic dipole. The distance between two magnetic poles is known as magnetic length and is denoted by 2l.A line passing through the north and south poles of a magnetic di-pole gives the axial line of the magnetic dipole.
A line passing through a point mid-way between the two poles of a magnetic dipole and perpendicular to its axial line gives the equatorial line of the magnetic dipole. The axial line and the equatorial line of the magnetic di-pole are shown in below figure.
Magnetic Dipole Moment
Magnetic dipole moment is the measure of the ability of a dipole to convert itself and come into alignment with a given external magnetic field. It may also be defined as the maximum expanse of torque caused by magnetic force on a dipole that ascends per unit value of surrounding magnetic field in vacuum. In case of a uniform magnetic field, the magnitude of the dipole moment is proportional to the maximum quantity of torque on the dipole. This generally occurs when the dipole is at right angles to the magnetic field.
Magnetic dipole in a Uniform Field Magnetic Field
Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle θ with the field as shown in figure. If ‘m’ is the strength of each pole, the two poles N and S experience two equal and opposite forces ‘mB’. These forces constitute a couple which tends to rotate the dipole. Suppose the couple exerts a torque of magnitude.
= (either force)(perpendicular distance between the two forces)
= mBAC = mB2l sin θ
Or= MB sin θ
Here, M = 2ml = magnetic moment of the dipole
If B = 1 and θ = 90º
= M
Magnetic moment M of the magnetic dipole is defined as the moment of the couple acting on it when it is placed at right angles to a uniform field of unit strength.
In such case,
Force on each pole = m1
Perpendicular distance between two forces = NS = 2l
So,= m2l
Thus, magnetic moment of a magnetic dipole is also defined as the product of its pole strength and the magnetic length.
Magnetic moment of a magnet is a vector quantity and is directed along SN inside the magnet.
Here,= unit vector along SN.
Sinceand, both, are vectors the torque acting on a magnet suspended in the magnetic field can be expressed as the cross-product ofandas follows.
The direction ofcan be obtained by applying right hand thumb rule.
Units of M
Units of M can be obtained from the relation
= MB sin θ
So, M =/B sin θ
Therefore, M can be measured in terms of joule per tesla (JT-1) or (NmT-1)
Since ,1 tesla = Wb/m 2
Thus, Unit of M = J/(Wb/m 2) = Jm 2/Wb
Since 1 J = 1 Nm
Therefore, Unit of M = 1 Nm T-1= 1 Nm/(Wb/m 2) = 1 Nm 3 Wb-1
Since unit of pole strength = Am
Thus, Unit of M = (Am) m = Am 2
Work done in Rotating a Magnetic Dipole in a Magnetic Field
When a magnetic dipole of magnetic moment M is oriented in a magnetic field of strength, B, making an angle ‘θ’. With its lines of force, it experiences a torquegiven by
Or
Letbe the small abgular displacement given to the dipole, work done dW is given by
Since the angle betweenandis zero.
So, dW = MB sin θ dθ
Work done in displacing the dipole from an angle θ 1 to θ 2 is
W = MB [(-cos θ 2) – (- cos θ 1)]
Or, W = MB (cos θ 1–cos θ 2)
Potential Energy of a Magnetic Dipole in a Magnetic Field
Potential energy of a magnetic dipole, in a magnetic field, is defined as the amount of work done in rotating the dipole from zero potential energy position to any desired position.
It is convenient to choose the zero potential energy position to be the one when the dipole is at right angles to the lines of force of magnetic field.
So,θ 1= 90º and θ 2= θ
Making these substitution in equation (3), we get,
Potential energy = W = MB (cos 90º–cos θ)
Or, W = – MB cosθ
In vector form,
Special cases:
(a) When θ = 0º, cos θ = 1
So, W = – MB
(b) When θ = 90º, cos θ = 0
So, W = – MB(0) = 0
(c) When θ = 180º,cos θ = -1
So, W = – MB (-1) = MB
θ = 0º position corresponds to the position of dipole aligning itself such that its magnetic moment vectoris parallel to field. In this case, potential energy is – MB, a minimum value. We know that every body in state of equilibrium must possess minimum potential energy. That is why the magnetic dipole exists in equilibrium whenis parallel to.
Refer this simulation to know more about alignment of magnetic domain
[Note: Please click on “red circle”to run the simulation]
This illustrates the fact that domains line up with a magnetic field and some metals, such as iron, partially retain the new domain alignment.
Current Loop as a Magnetic Dipole
Ampere found that the distribution of magnetic lines of force around a finite current carrying solenoid is similar to that produced by a bar magnet. This is evident from the fact that a compass needle when moved around these two bodies show similar deflections. After noting the close resemblance between these two, Ampere demonstrated that a simple current loop behaves like a bar magnet and put forward that all the magnetic phenomena is due to circulating electric current. This is Ampere’s hypothesis.
The magnetic induction at a point along the axis of a circular coil carrying current is,
B =µ o nI a 2/2(a 2+x 2)3/2
The direction of this magnetic field is along the axis and is given by right hand rule. For points which are far away from the centre of the coil,x>>a, a 2 is small and it is neglected. Hence for such points,
B =µ o nI a 2/2x 3
If we consider a circular loop, n = 1, its area A = π a 2
So, B =µ o IA/2πx 3
The magnetic induction at a point along the axial line of a short bar magnet is
B = (µ o/4π)(2M/x 3)
Or, B =(µ o/2π)(M/x 3)
Comparing equations (1) and (2), we find that
M = IA
Hence a current loop is equivalent to a magnetic dipole of moment M = IA
The magnetic moment of a current loop is defined as the product of the current and the loop area. Its direction is perpendicular to the plane of the loop.
Watch this Video for more reference
The Magnetic Dipole Moment of a Revolving Electron
According to Neil Bohr’s atom model, the negatively charged electron is revolving around a positively charged nucleus in a circular orbit of radius r. The revolving electron in a closed path constitutes an electric current. The motion of the electron in anticlockwise direction produces conventional current in clockwise direction.
Since the electron is negatively charged, conventional current flows in a direction opposite to its motion. The magnetic moments associated with the two types of motion of electron in an orbit are shown in figure. An electron revolving in an orbit of radius ‘r’ is equivalent to a magnetic shell of magnetic moment ‘M’ given by
M = iA
Here ‘i’ is the current to which the rotating electron is equivalent to and A is the area of the orbit. If ‘’ is the time period of rotation of electron.
i = charge/period = e/= e/(2π/ω)
or i = eω/2π
Here, ‘ω’ is the angular velocity of the electron.
So, M = (eω/2π)(πr 2) = eωr 2/2
According to Bohr’s theory, an electron can revolve in an orbit in which its angular momentum is an integral multiple of h/2π where ‘h’ is Planck’s constant.
mr 2 ω= n (h/2π)
Or,r 2 ω = n (h/2πm)
Substituting for r 2 ω in equation (4), we get,
M = n (eh/4πm)
Since an atom may have a large number of electrons, the resultant magnetic moment of the atom is the vector sum of the magnetic moments of various electrons.
The term eh/4πm is called Bohr’s magneton. It is the smallest value of magnetic moment which an electron possess. Any atom can possess magnetic moment which is an integral multiple of Bohr’s magneton. Thus, the magnetic moment is also quantised in the atomic and sub-atomic level.
In addition to the magnetic moment due to its orbital motion, the electron possesses magnetic moment due to its spin. Hence the resultant magnetic moment of an electron is the vector sum of its orbital magnetic moment and its spin magnetic moment.
An isolated magnetic pole is non-existent. Whenever we break a magnet we obtain two complete magnets.
A magnet when freely suspended, always points geographic north-south direction.
Magnetic length is nearly 7/8 times the geometric length.
A uniform magnetic field is represented by equidistant parallel lines of force.
Magnetic moment of a magnet is a vector quantity. Its direction is from S to N.
Magnitude of magnetic moment is equal to the product of pole strength and magnetic length.
It is not possible to separate out the two poles of a magnet.
Magnetic dipole is not a system composed of two poles because the existence of monopoles is not possible.
A loop of single turn is also a magnetic dipole. One face of the loop behaves as North Pole and the other face behaves as South Pole. The face of the coil, in which current is anticlockwise, behaves as north pole and the face in which current is clockwise, behaves as south pole.
Problem (JEE Main):
A uniformly charged disc of radius r and having charge q rotates with constant angular velocity ω. The magnetic dipole moment of this disc is 1/n (qωr 2). Find the value of n.
We know that, M/L = q/2m
So, M = qL/2m = q(Iω) / 2m
= [q (1/2 mr 2)ω] / 2m
= ¼ qωr 2
Therefore, from the above observation we conclude that, the value of n will be 4.
Question 1
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be,
(a)√3 W (b) W
(c)√3 (W/2) (d) 2W
Question 2
The magnetic induction, in air, at a distance d from an isolated point pole of strength m unit will be,
(a) m/d (b) m/d 2
(c) md (d) ma 2
Question 3
A magnet of moment M is suspended in a uniform magnetic field B. The maximum value of torque acting on the magnet is
(a) MB (b) ½ MB
(c) 2 MB (d) 0
Question 4
Magnetic field due to a short magnet of magnetic moment M at a distance d from its centre in the end on position is
(a) M/d 2(b) M/d 3
(c) 2 M/d 2(d) 2 M/d 3
Question 5
A magnetic dipole is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle 180º, then the work done is
(a) MB (b) 2 MB
(c) MB/2 (d) zero
Q.1Q.2Q.3Q.4Q.5
a b a d b
Related Resources
You might like tomagnetic moment.
For getting an idea of the type of questions asked, refer thePrevious Year Question Papers.
Click here to refer the mostUseful Books of Physics.
To get answer to any question related to magnetic dipoleclick here.
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7524 | https://www.youtube.com/watch?v=jPRUMMA_Ex8 | Use change of base formula log_2 x = ln x/ln 2 to show -1/(x(ln 2)(log_2 x)^2 = - ln2/(x(ln x))^2
Ms Shaws Math Class
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Posted: 6 Mar 2021
Transcript:
hi everyone we're going to show that negative 1 divided by x times ln of 2 times log base 2 of x squared equals negative ln of 2 divided by x ln x squared this goes with a problem that we did so this is just a simplification of an answer that we have so all we're going to have to do is change this part right here using the change of base so if you have log base a of x we can use a change of base and say this is ln of x divided by l n of a so that's what we're going to do we're going to write this as negative 1 divided by x times ln of 2 times uh and now let's change the base here this is going to be ln of x divided by ln of 2 squared now from there i can write this as negative 1 x divided by x times l n of 2 and this is going to be times ln of x squared and this is in the denominator so this is going to be ln of 2 squared and i just broke up this squared now all you have to do is bring this up in the numerator and so you're going to get negative ln of 2 squared divided by x times ln of 2 times ln of x squared all right now this one we have two of these i can write this as two of these instead of square so this cancels so what we're left with is negative ln of 2 divided by x times ln x squared and that's it thank you have a nice day bye-bye you |
7525 | https://www.benasciutticasa.it/post/62f6578bbe7ea0d96191661e/teli-da-bagno-vs-asciugamani-qual-%C3%A8-la-differenza-.html | Il tuo carrello
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Teli da bagno vs asciugamani: qual è la differenza?
11/07/2022
È abbastanza semplice.
La differenza è la loro dimensione, con i teli da bagno più grandi degli asciugamani da bagno .
Ti chiedi come questo influisca sulla tua routine quotidiana del bagno?
Asciugamani da bagno
La dimensione standard di un asciugamano da bagno dipende molto dal produttore ma diciamo che varia da 50 cm a 70 cm di larghezza e i 90 cm e i 115 cm di lunghezza.
Perché un asciugamano da bagno?
• È perfettamente leggero, quindi più facile da manovrare quando ci si asciuga dopo il bagno o la doccia o appena usciti dalla piscina.
• È meno costoso da lavare rispetto a un telo da bagno, utilizzando meno acqua e meno detersivo risulta anche più ecologico.
• Quando si tratta di avvolgere i capelli bagnati, questa misura si adatta perfettamente senza avere ingombro inutile.
• Risulta essere il preferite dalle famiglie perché si adatta sia ad adulti che a bambini, non essendo troppo grande e pesante.
• È molto più facile da riporre e piegare in particolare quando lo spazio è al limite. Si posiziona semplicemente anche su un gancio in modo ordinato, anche durante l'asciugatura.
Teli da bagno
La dimensione standard di un telo da bagno sono 70 cm fino a 100 cm di larghezza e 150 cm fino a 180 cm
Perché i teli da bagno?
• Con una generosa superficie di assorbimento, dopo una doccia o all’uscita dalla vasca lo sforzo per manovrarlo è notevolmente minore.
• Questa dimensione avvolge generosamente tutto il tuo corpo, lasciandoti ben coperto
• Offre un lusso adorabile, soprattutto se offri alloggio o hai spesso ospiti a casa tua.
• Le persone più alte o più grandi apprezzano quelle sensazioni avvolgenti che un telo da bagno può fornire.
• Questa dimensione funge anche da telo da piscina o da spiaggia, essendo abbastanza generoso per rilassarsi.
Ora che conosci i vantaggi sia degli asciugamani da bagno che dei teli da bagno, tu cosa scegli ?
Acquista qui la nostra vasta gamma di collezioni di asciugamani da bagno e qui se preferisci il telo bagno
È abbastanza semplice.
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Inserisci la partita IVA se non disponi di codice fiscale o se il codice fiscale azienda corrisponde alla partita IVA. |
7526 | https://ocw.mit.edu/courses/res-6-012-introduction-to-probability-spring-2018/dd00154c73cf3d5b7c57ef9b80b7f05e_N3I2ZLbh6zQ.pdf | MITOCW | MITRES6_012S18_L01-06_300k We will now continue and derive some additional properties of probability laws which are, again, consequences of the axioms that we have introduced.
The first property is the following.
If we have two sets and one set is smaller than the other-- so we have a picture as follows.
We have our sample space.
And we have a certain set, A.
And then we have a certain set, B, which is even bigger.
So the set B is the bigger blue set.
So if B is a set which is larger than A, then, naturally, the probability that the outcome falls inside B should be at least as big as the probability that the outcome falls inside A.
How do we prove this formally?
The set B can be expressed as a union of two pieces.
One piece is the set A itself.
The second piece is whatever elements of B there are, that do not belong in A. What are these elements?
They are elements that belong to B. And they do not belong to A, which means that they belong to the complement of A.
So we have expressed the set B as the union of two pieces.
Now this piece is A. This piece here is outside A. So these two pieces are disjoint.
And so we can apply the additivity axiom, and write that the probability of B is equal to the probability of A plus the probability of the other set.
And since probabilities are non-negative, this expression here is at least as large as the probability of A. And this concludes the proof of the property that we wanted to show.
Indeed, the probability of A is less than or equal to the probability of B.
The next property we will show is the following.
It allows us to write the probability of the union of two sets for the case now, where the two sets are not necessarily disjoint.
So the picture is as follows.
We have our two sets, A and B. These sets are not necessarily disjoint.
And we want to say something about the probability of the union of A and B.
Now the union of A and B consists of three pieces.
One piece is this one here.
And that piece consists of those elements of A that do not belong to B. So they belong to B complement.
This set has a certain probability, let's call it little a and indicate it on this diagram.
So a is the probability of this piece.
Another piece is this one here, which is the intersection of A and B. It has a certain probability that we denote by little b.
This is the probability of A intersection B.
And finally, there's another piece, which is out here.
And that piece has a certain probability c.
It is the probability of that set.
And what is that set?
That set is the following.
It's that part of B that consists of elements that do not belong in A. So it's B intersection with the complement of A.
Now let's express the two sides of this equality here in terms of little a, little b, and little c, and see whether we get the same thing.
So the probability of A union B. A union B consists of these three pieces that have probabilities little a, little b, and little c, respectively.
And by the additivity axiom, the probability of the union of A and B is the sum of the probabilities of these three pieces.
Let's look now at the right hand side of that equation and see whether we get the same thing.
The probability of A plus the probability of B, minus the probability of A intersection B is equal to the following.
A consists of two pieces that have probabilities little a and little b.
The set B consists of two pieces that have probabilities little b and little c.
And then we subtract the probability of the intersection, which is b.
And we notice that we can cancel here one b with another b.
And what we are left with is a plus b plus c.
So this checks.
And indeed we have this equality here.
We have verified that it is true.
One particular consequence of the equality that we derived is the following.
Since this term here is always non-negative, this means that the probability of A union B is always less than or equal to the probability of A plus the probability of B. This inequality here is quite useful whenever we want to argue that a certain probability is smaller than something.
And it has a name.
It's called the union bound.
We finally consider one last consequence of our axioms.
And namely, we are going to derive an expression, a way of calculating the probability of the union of three sets, not necessarily disjoint.
So we have our sample space.
And within the sample space there are three sets-- set A, set B, and set C.
We are going to use a set theoretic relation.
We are going to express the union of these three sets as the union of three disjoint pieces.
What are these disjoint pieces?
One piece is the set A itself.
The second piece is going to be that part of B which is outside A. So this is the intersection of B with the complement of A.
The third piece is going to be whatever is left in order to form the union of the three sets.
What is left is that part of C that does not belong to A and that does not belong to B. So that part is C intersection with A complement and B complement.
Now this set here, of course, is the same as that set because intersection of two sets is the same no matter in which order we take the two sets.
And similarly, the set that we have here is the same one that appears in that expression.
Now we notice that these three pieces, the red, the blue, and the green, are disjoint from each other.
So by the additivity axiom, the probability of this union here is going to be the sum of the probabilities of the three pieces.
And that's exactly the expression the we have up here. |
7527 | https://www.geeksforgeeks.org/dsa/express-number-sum-consecutive-numbers/ | Express a number as sum of consecutive numbers
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Given a number N, write a function to express N as sum of two or more consecutive positive numbers. If there is no solution, output -1. If there are multiple solution, then print one of them.
Examples:
```
Input : N = 10
Output : 4 + 3 + 2 + 1
Input : N = 8
Output : -1
Input : N = 24
Output : 9 + 8 + 7
```
```
Sum of first n natural numbers = n (n + 1)/2
Sum of first (n + k) numbers = (n + k) (n + k + 1)/2
If N is sum of k consecutive numbers, then
following must be true.
N = [(n+k)(n+k+1) - n(n+1)] / 2
OR
2 N = [(n+k)(n+k+1) - n(n+1)]
```
Below is the implementation based on above idea.
Try it on GfG Practice
C++
````
// C++ program to print a consecutive sequence
// to express N if possible.
include
using namespace std;
// Print consecutive numbers from
// last to first
void printConsecutive(int last, int first)
{
cout << first++;
for (int x = first; x<= last; x++)
cout << " + " << x;
}
void findConsecutive(int N)
{
for (int last=1; last<N; last++)
{
for (int first=0; first<last; first++)
{
if (2N == (last-first)(last+first+1))
{
cout << N << " = ";
printConsecutive(last, first+1);
return;
}
}
}
cout << "-1";
}
// Driver code
int main()
{
int n = 12;
findConsecutive(n);
return 0;
}
````
// C++ program to print a consecutive sequence
// C++ program to print a consecutive sequence
// to express N if possible.
// to express N if possible.
```
include #include
```
using namespace std; using namespace std
// Print consecutive numbers from
// Print consecutive numbers from
// last to first
// last to first
void printConsecutive(int last, int first) void printConsecutive int last int first
{
cout << first++; cout<< first ++
for (int x = first; x<= last; x++) for int x = first x<= last x ++
cout << " + " << x; cout<<" + "<< x
}
void findConsecutive(int N) void findConsecutive int N
{
for (int last=1; last<N; last++) for int last = 1 last< N last ++
{
for (int first=0; first<last; first++) for int first = 0 first< last first ++
{
if (2N == (last-first)(last+first+1)) if 2 N == last - first last + first + 1
{
cout << N << " = "; cout<< N<<" = "
printConsecutive(last, first+1); printConsecutive last first + 1
return; return
}
}
}
cout << "-1"; cout<<"-1"
}
// Driver code
// Driver code
int main() int main
{
int n = 12; int n = 12
findConsecutive(n); findConsecutive n
return 0; return 0
}
Java
````
// Java program to print a consecutive sequence
// to express N if possible.
import java.util.;
class GFG
{
// Print consecutive numbers from
// last to first
static void printConsecutive(int last, int first)
{
System.out.print(first++);
for (int x = first; x<= last; x++)
System.out.print(" + " + x);
}
static void findConsecutive(int N)
{
for (int last = 1; last < N; last++)
{
for (int first = 0; first < last; first++)
{
if (2N == (last-first)(last+first+1))
{
System.out.print(N+ " = ");
printConsecutive(last, first+1);
return;
}
}
}
System.out.print("-1");
}
// Driver code
public static void main(String[] args)
{
int n = 12;
findConsecutive(n);
}
}
// This code is contributed by umadevi9616
````
Python3
````
Python3 program to print a consecutive
sequence to express N if possible.
Print consecutive numbers
from last to first
def printConsecutive(last, first):
print (first, end = "")
first += 1
for x in range(first, last + 1):
print (" +", x, end = "")
def findConsecutive(N):
for last in range(1, N):
for first in range(0, last):
if 2 N == (last - first) (last + first + 1):
print (N, "= ", end = "")
printConsecutive(last, first + 1)
return
print ("-1")
Driver code
n = 12
findConsecutive(n)
This code is contributed by Shreyanshi Arun.
````
C#
````
// C# program to print a consecutive sequence
// to express N if possible.
using System;
class GfG
{
// Print consecutive numbers from
// last to first
static void printConsecutive(int last, int first)
{
Console.Write(first++);
for (int x = first; x <= last; x++)
Console.Write(" + "+x);
}
static void findConsecutive(int N)
{
for (int last = 1; last < N; last++)
{
for (int first = 0; first < last; first++)
{
if (2 N == (last - first)
(last + first + 1))
{
Console.Write(N + " = ");
printConsecutive(last, first + 1);
return;
}
}
}
Console.Write("-1");
}
// Driver code
public static void Main ()
{
int n = 12;
findConsecutive(n);
}
}
// This code is contributed by vt_m
````
PHP
````
php
// PHP program to print a consecutive
// sequence to express N if possible.
// Print consecutive numbers from
// last to first
function printConsecutive($last, $first)
{
echo $first++;
for ($x = $first; $x<= $last; $x++)
echo " + " , $x;
}
function findConsecutive($N)
{
for ($last = 1; $last < $N; $last++)
{
for ($first = 0; $first < $last; $first++)
{
if (2 $N == ($last - $first)
($last + $first + 1))
{
echo $N , " = ";
printConsecutive($last, $first + 1);
return;
}
}
}
echo "-1";
}
// Driver Code
$n = 12;
findConsecutive($n);
// This code is contributed by nitin mittal
?
````
JavaScript
````
// Javascript program to print a consecutive
// sequence to express N if possible.
// Print consecutive numbers from
// last to first
function printConsecutive(last, first)
{
document.write(first++);
for (let x = first; x<= last; x++)
document.write( " + " + x);
}
function findConsecutive(N)
{
for (let last = 1; last < N; last++)
{
for (let first = 0; first < last; first++)
{
if (2 N == (last - first)
(last + first + 1))
{
document.write(N + " = ");
printConsecutive(last, first + 1);
return;
}
}
}
document.write("-1");
}
// Driver Code
let n = 12;
findConsecutive(n);
// This code is contributed by _saurabh_jaiswal
````
Output:
12 = 3 + 4 + 5
Reference :
K
kartik
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7528 | http://www.mrsmcnickle.com/2fractions.html | Mrs. McNickle's Math Page
HomeAdditionSubtractionMultiplication and DivisionFractionsArea and PerimeterRoundingTime
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Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
Represent and interpret unit fractions in the form 1/n as the quantity formed by one part when a whole is partitioned into n equal parts.
Represent and interpret fractions, including fractions greater than one, in the form of m/n as the result of adding the unit fraction 1/n to itself m times.
Read and write fractions, including fractions greater than one, using standard form, numeral-word form and word form.
Plot, order and compare fractional numbers with the same numerator or the same denominator.
Identify equivalent fractions and explain why they are equivalent.
Partitioning is the concept of "equal sharing," or splitting a shape or group of objects into equally-sized groups.
These shapes do not represent halves, thirds, or fourths, because they are not partitioned equally. Students can think about these as being "unfair" divisions of an object.
THese objects are partitioned equally. It's important for students to understand that each section represents an equal fraction of the whole, as labeled above.
Fractions can be represented on a number line through shading, as shown. 4/4 (four-fourths) is the fraction represented on this number line.
Fractions can also be represented on a number line by plotting a single dot. 6/8 is plotted on this number line.
Students should be able to recognize fractions that are represented in different ways, like this fraction (3/4) represented on a number line and a circle diagram. |
7529 | https://books.google.com/books/about/Theory_of_Ordinary_Differential_Equation.html?id=bPJQAAAAMAAJ | Sign in
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| |
| Theory of Ordinary Differential Equations Earl A. Coddington, Norman Levinson McGraw-Hill, 1955 - Boundary value problems - 429 pages This book has developed from courses given by the authors and probably contains more material than will ordinarily be covered in a one-year course. It is hoped that the book will be a useful text in the application of differential equations as well as for the pure mathematician. Prerequisite for this book is a knowledge of matrices and the essentials of functions in a complex variable. The book thoroughly addresses linear equations, and touches on the use of the Riemann-Stieltjes integral, and the Lebesgue integral, and the theorems required from integration theory. The problems, in some cases, give additional material not considered in the text. More » |
From inside the book
Contents
| | |
--- |
| EXISTENCE AND UNIQUENESS OF SOLUTIONS | 1 |
| | |
| EXISTENCE AND UNIQUENESS OF SOLUTIONS continued | 42 |
| | |
| | |
--- |
| LINEAR DIFFERENTIAL EQUATIONS | 62 |
| | |
| Copyright | |
| | |
27 other sections not shown
Other editions - View all
‹›
Common terms and phrases
a₁ adjoint analytic assumed asymptotic boundary condition c₁ Chap characteristic roots Clearly coefficients column components considered constant matrix continuous function convergent critical points denoted diagonal diagonal matrix differential equation eigenfunctions eigenvalues entire function equicontinuous exists finite number fixed follows formal solution formula function f fundamental matrix given Green's formula Green's function hence HINT implies inequality integral interval L(C+ Lemma Let f limit-point linear system linearly independent linearly independent solutions nonsingular matrix P₁ Parseval equality periodic orbit periodic solution plane Poincaré-Bendixson theorem polynomial Prob problem Lx Proof of Theorem proves regular singular point result satisfies self-adjoint sequence Show solution of 1.1 solutions of Lx successive approximations Suppose t₁ Theorem 2.1 tion unique solution valid vanish vector λε μο φη
References to this book
| | | |
---
| | Differential Equations with Discontinuous Righthand Sides: Control Systems A.F. Filippov Limited preview - 1988 | |
| | | |
---
| | Dynamic Noncooperative Game Theory Tamer Basar,Geert Jan Olsder No preview available - 1999 | |
All Book Search results »
Bibliographic information
| | |
--- |
| Title | Theory of Ordinary Differential Equations International series in pure and applied mathematics TMH edition Tata McGraw-Hill edition Theory of Ordinary Differential Equations, Earl A. Coddington |
| Authors | Earl A. Coddington, Norman Levinson |
| Edition | 10, reprint |
| Publisher | McGraw-Hill, 1955 |
| Original from | the University of Michigan |
| Digitized | Nov 20, 2007 |
| ISBN | 0070992568, 9780070992566 |
| Length | 429 pages |
| | |
| Export Citation | BiBTeX EndNote RefMan | |
7530 | https://artofproblemsolving.com/wiki/index.php/Telescoping_series?srsltid=AfmBOorb7AOnUzab4r8UCsFkBzc4FjJNsGCSXnvfkLpYXqC7Niu-FCS4 | Art of Problem Solving
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Telescoping series
In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. This is often done by using a form of for some expression .
Contents
[hide]
1 Example 1
2 Solution 1
3 Example 2
4 Solution 2
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 See Also
Example 1
Derive the formula for the sum of the first counting numbers.
Solution 1
We wish to write for some expression . This expression is as .
We then telescope the expression:
.
(Notice how the sum telescopes— contains a positive and a negative of every value of from to , so those terms cancel. We are then left with , the only terms which did not cancel.)
Example 2
Find a general formula for , where .
Solution 2
We wish to write for some expression . This can be easily achieved with as by simple computation.
We then telescope the expression:
.
Problems
Introductory
When simplified the product becomes:
(Source)
The sum can be expressed as , where and are positive integers. What is ? (Source)
Which of the following is equivalent to (Hint: difference of squares!)
(Source)
Intermediate
Let denote the value of the sum can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine . (Source)
Olympiad
Find the value of , where is the Riemann zeta function
See Also
Algebra
Summation
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7531 | https://economics.princeton.edu/wp-content/uploads/2025/01/Extreme-Points-in-Multi-Dimensional-Screening-Niemeyer.pdf | EXTREME POINTS IN MULTI-DIMENSIONAL SCREENING PATRICK LAHR AND AXEL NIEMEYER Abstract. This paper characterizes extreme points of the set of incentive-compatible mechanisms for screening problems with linear utility.
Extreme points are exhaustive mechanisms, meaning their menus cannot be scaled and translated to make additional feasibility constraints binding. In problems with one-dimensional types, extreme points admit a tractable description with a tight upper bound on their menu size. In problems with multi-dimensional types, every exhaustive mechanism can be transformed into an extreme point by applying an arbitrarily small perturbation. For mechanisms with a finite menu, this perturbation displaces the menu items into general position. Generic exhaustive mechanisms are extreme points with an uncountable menu.
Similar results hold in applications to delegation, veto bargaining, and monopoly problems, where we consider mechanisms that are unique maximizers for specific classes of objective functionals. The proofs involve a novel connection between menus of extreme points and indecomposable convex bodies, first studied by Gale (1954).
JEL Codes: D82, D44, D86, C78, C65 Keywords: Multi-Dimensional Types, Extreme Points, Exposed Points, Indecomposable Convex Bodies, Multi-Good Monopoly Problem, Linear Delegation, Linear Veto Bargaining Date: December 3, 2024.
Acknowledgments: We thank Felix Bierbrauer, Peter Caradonna, Simone Cerreia-Vioglio, Gregorio Curello, Laura Doval, Mira Frick, Alkis Georgiadis-Harris, Daniel Gottlieb, Michael Greinecker, Nima Haghpanah, Deniz Kattwinkel, Andreas Kleiner, Elliot Lipnowski, Fabio Maccheroni, Alejandro Manelli, Jeffrey Mensch, Benny Moldovanu, Roger Myerson, Efe Ok, Tom Palfrey, Martin Pollrich, Luciano Pomatto, Justus Preußer, Kota Saito, Larry Samuelson, Fedor Sandomirskiy, Mario Schulz, Ina Taneva, Omer Tamuz, Tristan Tomala, Aleh Tsyvinsky, Rakesh Vohra, and Mark Whitmeyer, as well as seminar audiences at ASU, Columbia, the Paris Game Theory seminar, Paris 1, UCR, and USC for helpful discussions and comments.
Lahr: ENS Paris-Saclay, Department of Economics. Email: patrick.lahr@ens-paris-saclay.fr.
Niemeyer: Caltech, Division of the Humanities and Social Sciences. Email: niemeyer@caltech.edu.
1 arXiv:2412.00649v1 [econ.TH] 1 Dec 2024 1. Introduction Much of the mechanism design literature assumes that agents’ preferences can be described by a single dimension of private information. Under this assumption, the theory has delivered remarkably clean predictions for optimal mechanisms across various applications. However, in many environments, agents’ preferences are more realistically modeled assuming multi-ple dimensions of private information, for instance, in allocation problems with multiple heterogeneous goods or collective decision problems with several alternatives. Despite their importance, much less is known about multi-dimensional settings. Several results highlight an inherent complexity of optimal mechanisms in these settings, but explicit descriptions have not been obtained outside of a few special cases.1 In this paper, we study the structure of optimal mechanisms for a class of mechanism design problems featuring one- and multi-dimensional types. Specifically, we consider linear screening problems. A principal makes an allocation that affects their own and an agent’s utility. Both parties’ utilities are linear in allocations and depend on the agent’s type, where the allocation space and type space are convex sets in Euclidean space. Linear screening covers a range of problems with and without transfers, for example, monopoly and bilateral trade problems or delegation and veto bargaining problems.
Our main results characterize the extreme points of the set of incentive-compatible (IC) mechanisms for linear screening problems. Since the principal maximizes a linear functional— their expected utility—over the set of IC mechanisms, an optimal mechanism can always be found among the extreme points. While every optimal mechanism is a mixture over optimal extreme points, generic objective functionals are uniquely maximized at an extreme point.2 Moreover, essentially every extreme point is the unique maximizer of some objective functional.3 Thus, determining the structure of optimal mechanisms across instances of the principal’s problem is tantamount to determining the structure of the extreme points.
The extreme-point approach has seen successful applications in a number of other mecha-nism design settings, but with the sole exception of Manelli and Vincent (2007) (MV), it has not been applied to settings with multi-dimensional types.4 Although MV laid important groundwork for the monopoly problem, our characterizations reveal more explicit insights into the structure of extreme points and apply to a broader class of problems.5 Our main insight is that in every one-dimensional problem, the set of extreme points admits a tractable description, whereas in every multi-dimensional problem, the set of extreme points is virtually as rich as the set of all incentive-compatible mechanisms. An important observation is that every extreme point is exhaustive: the allocations made by the mechanism—its menu—cannot be scaled and translated to make additional feasibility 1See, for example, Rochet and Chon´ e (1998), Manelli and Vincent (2006, 2007), Hart and Reny (2015), Daskalakis et al. (2014), Daskalakis et al. (2017), or Hart and Nisan (2019).
2We show that the set of IC mechanisms is norm-compact and convex. The first claim then follows from Choquet’s theorem. The second claim follows from a theorem by Lau (1976) (where genericity is in a topological sense).
3This claim follows from a theorem by Straszewicz and Klee (Klee Jr, 1958). More precisely, the mechanisms that are uniquely optimal for some instance of the principal’s problem, i.e., exposed points, are dense in the set of extreme points.
4See, for example, Border (1991), Manelli and Vincent (2010), Kleiner et al. (2021), Nikzad (2022, 2024), or Yang and Zentefis (2024).
5We provide a detailed discussion of our relation to MV in Section 9.
2 constraints binding.6 In one-dimensional problems, extreme points admit a tight upper bound on their menu size on top of exhaustiveness. In contrast, in multi-dimensional problems, every exhaustive mechanism can be transformed into an extreme point by applying an arbitrarily small perturbation. For exhaustive mechanisms with a finite menu, this perturbation simply displaces the menu items into general position. In particular, generic exhaustive mechanisms are extreme points.
1.1. Discussion. A common explanation for the difficulty with multi-dimensional screening is that binding incentive constraints depend on the choice of mechanism, making it a priori unclear which constraints will be binding in an optimal mechanism; our results corroborate this explanation. The perturbation described in the previous paragraph modifies the binding incentive constraints of an exhaustive mechanism for an arbitrarily small set of types. Thus, since exhaustive mechanisms are defined only in terms of binding feasibility constraints, the qualitative properties that distinguish extreme points from other mechanisms are essentially only properties of binding feasibility constraints. In contrast, for all one-dimensional problems, properties of binding incentive constraints impose significant restrictions on the structure of the extreme points, e.g., by limiting their menu size to no more than a few allocations in typical applications.
A potential concern is that our results characterize the structure of optimal mechanisms across all instances of the principal’s problem, i.e., for arbitrary utility functions and beliefs about the agent’s type, while in some applications, the principal’s utility function is known.
For example, when a monopolist maximizes revenue, certain extreme points are suboptimal for every belief of the monopolist about the agent’s valuations. Our main insights remain the same in sample applications where the principal’s utility is fixed and state-independent, such as in the monopoly problem. In particular, with multi-dimensional types, we show that the extreme points that are (uniquely) optimal for some belief of the principal are again virtually as rich as the set of all IC mechanisms.
Our results offer some insights into the capabilities and limitations of the classical mechanism design paradigm. An important pillar for the success of the theory is that, in many applications, it makes predictions for optimal mechanisms that are independent of the specific details of the environment. We confirm that such predictions are obtainable for all one-dimensional linear screening problems, whereas they are largely unattainable for all multi-dimensional linear screening problems. When the structure of the optimal mechanism depends too finely on the model parameters, it is difficult to derive tangible practical guidance and testable implications from the theory since parameters such as type distributions may be unknown or unobservable in practice.
We emphasize that we do not provide a full solution to multi-dimensional linear screening in that we do not identify the optimal mechanism for each instance of the principal’s problem and show how this mechanism varies across instances. However, given the overwhelming complexity of the structure of extreme points, it seems implausible that such comparative statics exercises are feasible in full generality.
1.2. Technical Contributions. We obtain our results by establishing a connection between extreme points of the set of IC mechanisms and extremal elements of certain spaces of convex sets. Instead of studying the set of IC mechanisms or the agent’s associated indirect utility 6A feasibility constraint is an affine restriction on the set of feasible allocations, i.e., a halfspace.
3 functions,7 we study the space of all menus that the principal could offer the agent. By the well-known taxation principle, any IC mechanism is the agent’s choice function from some menu of allocations and vice versa. Since preferences are linear, offering the agent a menu is payoff-equivalent to offering the agent the menu’s convex hull. Thus, we can establish a bijection between payoff-equivalence classes of IC mechanisms and certain convex sets contained in the allocation space. We show that this bijection preserves convex combinations (in the sense of Minkowski) and therefore preserves extreme points. Analogous bijections hold onto the set of indirect utility functions.
The extremal elements of the space of compact convex sets in Euclidean space are relatively well understood in the mathematical literature and are referred to as indecomposable convex bodies, first studied by Gale (1954). Most of our results are derived from translating these mathematical insights into economic insights via the connection between IC mechanisms and menus in the form of convex sets. Two kinds of complications arise in this translation. First, feasibility requires that menus are contained in the space of allocations; these constraints are not generally considered in the literature on indecomposability. Second, certain menus are equivalent from the agent’s perspective when the type space is restricted, i.e., when the agent’s preferences are constrained to a subset of all linear preferences.
Indecomposable convex bodies in the plane are points, line segments, and triangles, but they are so plentiful and complex in higher dimensions that a complete description has not been obtained and is not to be expected.8 However, what is known in the mathematical literature is enough to obtain the relevant economic insights we present in this paper. The complexity of indecomposable convex bodies in two- versus higher dimensions mirrors the dichotomy between one- and multi-dimensional screening problems since, with linear utility and up to redundancies, an allocation space of a given dimension always corresponds to a type space of one dimension less. (Transfers would here be counted as an allocation dimension of its own.) 1.3. Structure of the Paper. Section 2 introduces relevant notation and mathematical definitions. Section 3 introduces the model. Section 4 gives a characterization of extreme points in terms of mechanisms that make an inclusion-wise maximal set of incentive and feasibility constraints binding. Section 5 clarifies the role of feasibility constraints by defining and characterizing exhaustive mechanisms. Section 6 presents our core results for one- versus multi-dimensional problems, along with several supporting results. Section 7 introduces the relevant mathematical tools and sketches the proof of our core results in the context of a delegation problem among lotteries over finitely many alternatives, with an emphasis on the special role of the three-alternative case.9 Section 8 discusses applications to monopolistic selling and veto bargaining, including essentially complete characterizations of undominated mechanisms in the sense of Manelli and Vincent (2007) for these settings. Section 9 provides 7For the indirect-utility approach, see e.g. Rochet (1987), Rochet and Chon´ e (1998), Manelli and Vincent (2006, 2007), and Daskalakis et al. (2017).
8Schneider (2014) writes (p. 166): “Most [(in the sense of topological genericity)] convex bodies in Rd, d ≥3, are smooth, strictly convex and indecomposable. It appears that no concrete example of such a body is explicitly known. This is not too surprising, since it is hard to imagine how such a body should be described.” We note that algebraic characterizations of indecomposable polytopes are known; see McMullen (1973), Meyer (1974), and Smilansky (1987). We provide a characterization along these lines in Appendix B.
9Problems with three alternatives have been considered as the simplest departure from the two-alternative case often studied in the literature on mechanism design without transfers; see B¨ orgers and Postl (2009).
4 an extensive discussion of the related literature, including multi-dimensional screening, extreme points in mechanism design, delegation and veto bargaining, and the mathematical foundations underlying this paper. Section 10 concludes.
Appendix A collects several auxiliary results, including the translation between the set of IC mechanisms and a certain space of convex sets. Appendix B deals with the geometry of the set of finite-menu mechanisms and provides an algebraic characterization of finite-menu extreme points (which generalizes the main result in Manelli and Vincent (2007)). Appendix C provides a complete characterization of extreme points for one-dimensional problems omitted from the main text for brevity. Appendix D contains the proofs for all results in the main text.
2. Notation and Mathematical Definitions Let X be a subset of a topological vector space E. ∆(X) denotes the set of Borel probability measures on X. int X denotes the interior of X, bndr X denotes the boundary of X, and cl X denotes the closure of X. conv X denotes the convex hull, cone X denotes the conical hull, and aff X denotes the affine hull.
Suppose X ⊆E is convex. ext X denotes the set of extreme points of X, i.e., those x ∈X for which x = λx′ + (1 −λ)x′′ and λ ∈(0, 1) implies x = x′ = x′′. exp X denotes the set of exposed points of X, i.e., those x ∈X for which there exists a continuous linear functional f : E →R such that f(x) > f(x′) for all x′ ∈X, x ̸= x′. Every exposed point is extreme, but the converse is not generally true. A face f of X is a convex subset of X such that for all x ∈f, x′, x′′ ∈X, and λ ∈(0, 1), x = λx′ + (1 −λ)x′′ implies x′, x′′ ∈f. The set X ⊆E is a polytope if it is the convex hull of finitely many (extreme) points.
We use the following standard terminology for convex sets in Euclidean space. A convex body K ⊂Rd is a non-empty compact convex set. A polyhedron P ⊆Rd is the finite intersection of closed halfspaces. A polyhedral cone is a cone that is also a polyhedron.
A polytope in Euclidean space is a bounded polyhedron. Every face f of a polyhedron P can be represented as f = arg maxa∈P a · θ for some θ ∈Rd. A face f is proper if f ̸= P. A vertex v of P is a face of dimension 0, i.e., an extreme point of P.10 A facet F of P is a face of P such that dim F = dim P −1. If P ⊆Rd is d-dimensional, then the facet-defining hyperplane of F is the unique supporting hyperplane H = {y ∈Rd | y · nH ≤cH} of P such that F ⊆H, where nH is the outer (unit) normal vector to P on F.
3. Model and Preliminaries 3.1. Allocations and Types. There is a principal and an agent. The principal chooses an allocation a ∈A ⊂Rd, where A is a d-dimensional polytope. The principal’s preferences over allocations depend on the agent’s private information, their type θ ∈Θ ⊂Rd \ {0}, where the set {λθ | θ ∈Θ, λ ∈R+} of all rays through the type space Θ is a d-dimensional polyhedral cone. We say that the type space is unrestricted if cone Θ = Rd. An agent of type θ ∈Θ derives utility a · θ from allocation a ∈A. Given the agent’s type θ ∈Θ, the principal derives utility a · v(θ) from allocation a ∈A, where v : Θ →Rd is a bounded objective function that captures the conflict of interest between both parties. There may be a veto allocation ¯ a ∈ext A that the agent can enforce unilaterally.
10The dimension of a convex set X ⊆Rd, denoted dim X, is the dimension of its affine hull.
5 Remark. The model subsumes several screening problems as special cases; see Sections 7 and 8 for examples. In particular, we subsume problems with transferable utility by interpreting one allocation dimension as a numeraire for which the principal and the agent have a known marginal utility. That is, Θ = ˜ Θ × {−1}, v(θ) = (. . . , 1) for all θ ∈Θ, and A = ˜ A × [0, κ], where κ ∈R is the total endowment of the numeraire.
Since utility is linear, we can identify types on the same ray from the origin because they have the same preferences over the allocations in A. We select normalized types in the unit sphere Sd−1 = {y ∈Rd : ||y|| = 1} as canonical representatives, i.e., Θ ⊆Sd−1.
In applications, we occasionally make other selections, e.g., when considering transferable utility. Thus, in our model, a d-dimensional allocation space A always corresponds to a (d −1)-dimensional type space Θ.11 3.2. Mechanisms. The principal designs a (direct and measurable) mechanism x : Θ →A to screen the agent.12 A mechanism asks the agent to report their type θ and then implements an allocation x(θ). By the revelation principle, it is without loss of generality for the principal to focus on mechanisms that are incentive-compatible (IC) and individually rational (IR): x(θ) · θ ≥x(θ′) · θ ∀θ, θ′ ∈Θ; (IC) x(θ) · θ ≥¯ a · θ ∀θ ∈Θ.
(IR) IC means that the agent has no incentive to misreport their type. IR means the agent has no incentive to veto the principal’s choice. To simplify the analysis, we assume that there exists a type ¯ θ ∈Θ for whom the veto allocation is one of their favorite allocations, i.e., ¯ a ∈arg maxa∈A a · ¯ θ. If no veto allocation exists, IR is satisfied by convention.
An optimal mechanism is any solution to the principal’s problem sup x:Θ→A Z Θ (x(θ) · v(θ)) dµ s.t.
(IC) and (IR), (OPT) where µ ∈∆(Θ) is the principal’s belief about the agent’s type. We assume that µ admits a bounded probability density, i.e., is absolutely continuous.
We say that a set of (IC) and (IR) mechanisms is a candidate set for optimality if it contains an optimal mechanism for every objective function v and belief µ of the principal.
3.3. Menus and Payoff-Equivalence. Instead of designing a mechanism, the principal can equivalently offer the agent a menu (or delegation set) M ⊆A, with ¯ a ∈M, from which the agent may choose their favorite allocation. That is, x(θ) ∈arg max a∈M a · θ 11Contrary to other notions of one-dimensionality in the mechanism design literature (see e.g. B¨ orgers, 2015, Chapter 5.6), a one-dimensional type space need here not imply a linear order on the underlying preferences. For example, Θ = S1 may be a circle.
12It is without loss of generality to consider deterministic mechanisms: every randomized allocation in ∆(A) can be replaced with its barycenter since both principal and agent have linear utility. In applications, we may think of the allocation space A as a set of lotteries over an underlying finite set of alternatives. In this case, a mechanism can be interpreted as a stochastic mechanism.
6 defines an IC and IR mechanism x : Θ →A (if maximizers exist). The value function U(θ) = θ · x(θ) is the agent’s indirect utility function associated with the mechanism x.
Mechanisms defined by the same menu are payoff-equivalent, i.e., the associated indirect utility functions are the same. For IC mechanisms, it can be shown that payoff-equivalence is equivalent to equality almost everywhere (Corollary A.5).13 Thus, payoff-equivalent mechanisms yield the principal the same expected utility since the belief µ is absolutely continuous.
We define the (essential) menu menu(x) = cl \ {x′(Θ) | x′ satisfies (IC) and (IR) and is payoff-equivalent to x} associated with an IC and IR mechanism as (the closure of) the set of allocations that are commonly made by all mechanisms in its payoff-equivalence class. For example, if the menu size | menu(x)| is finite, then the menu simply consists of the allocations that are made by the mechanism with strictly positive probability (cf. Daskalakis et al., 2017, Definition 7).
We henceforth identify payoff-equivalent mechanisms, i.e., x = x′ if x(θ) = x′(θ) for almost every θ ∈Θ, and write X for the set of payoff-equivalence classes of IC and IR mechanisms.14 In Appendix A.2, we show that X is L1-compact and convex. Therefore, a solution to (OPT) exists and can be found among the extreme points of X (Bauer’s maximum principle).
4. Binding Incentive and Feasibility Constraints In this section, we provide a characterization of the extreme points of the set of IC and IR mechanisms in terms of binding incentive and feasibility constraints. Optimal mechanisms solve a linear optimization problem, and therefore, identifying the binding constraints is crucial for finding a solution. This perspective will prove useful in the subsequent sections.
An (IC) constraint is represented by a pair of types (θ, θ′) ∈Θ × Θ, and we define IC(x) = ( (θ, θ′) ∈Θ × Θ arg max a∈menu(x) θ′ · a ⊆arg max a∈menu(x) θ · a ) (1) as the set of binding IC constraints of mechanism x ∈X. This definition considers a constraint as binding if type θ is indifferent to mimicing type θ′ regardless of how θ′ breaks ties.15 To define feasibility constraints, recall that the allocation space A is a polytope. Thus, there exists a finite set F of facet-defining hyperplanes H = {y ∈Rn | y · nH = cH} of A.
That is, A = T{H−: H ∈F}, where H−= {y ∈Rn | y · nH ≤cH} are the associated halfspaces containing A. Each halfspace corresponds to an affine restriction on the space of available allocations, and no restriction is redundant given the others; see Figure 1 for an illustration.
We define F(x) = {H ∈F | menu(x) ∩H ̸= ∅} (2) as the set of binding feasibility constraints of mechanism x.
13Almost everywhere equality is with respect to the spherical measure (since Θ ⊆Sd−1). For a Borel subset B ⊆Sd−1, the spherical measure is proportional to the Lebesgue measure of the set {λθ | θ ∈B, λ ∈[0, 1]}.
14See Appendix A.1 for a brief discussion of tie-breaking.
15Since ties are null events, IC(x) coincides for every type θ and almost every deviation θ′ with defining an IC constraint as binding if x(θ) · θ = x(θ′) · θ. The latter definition of binding constraints is not robust to tie-breaking.
7 H2 H1 H5 H4 H3 A menu(x) Figure 1. The menu of a mechanism x ∈X which is not an extreme point.
The allocation space A is a polytope defined by five facet-defining hyperplanes F = {H1, . . . , H5}. The four allocations marked with dots are the menu of the mechanism. The two rightmost allocations in the menu can be translated horizontally while maintaining the orientation of all dotted lines, keeping the set of binding constraints unchanged. This is clear for the feasibility constraints, and can be seen for the incentive constraints because each type—represented by a direction in R2—still chooses the same menu item(s).
Individual rationality constraints are irrelevant for the formulation of the following result; see Appendix A.3 for an explanation.
Theorem 4.1. A mechanism x ∈X with finite menu size is an extreme point of X if and only if there is no other mechanism x′ ∈X such that F(x) ⊆F(x′) and IC(x) ⊆IC(x′).
Proof. See Appendix D.1.
□ Remark. The inclusions F(x) ⊆F(x′) and IC(x) ⊆IC(x′) in Theorem 4.1 can equivalently be replaced by the equalities F(x) = F(x′) and IC(x) = IC(x′).
A mechanism with finite menu size is an extreme point if and only if it is the only mechanism that makes a given inclusion-wise maximal set of constraints binding; Figure 1 illustrates. Of the two types of constraints, binding feasibility constraints are easier to analyze and will be treated separately in the next section.
Let us briefly discuss the proof of Theorem 4.1. If x = λx′ + (1 −λ)x′′ is a finite menu mechanism in X, where λ ∈(0, 1) and x′, x′′ ∈X, then IC(x) = IC(x′) ∩IC(x′′).16 Thus, an important object for understanding extreme points is the set {x′ ∈X | IC(x) ⊆IC(x′)} of mechanisms that make an inclusion-wise larger set of IC constraints binding than a given finite-menu mechanism x ∈X. We show that this set is a polytope and a face of X; in particular, x ∈ext X if and only if x ∈ext{x′ ∈X | IC(x) ⊆IC(x′)}. Extreme points of a 16For almost all type pairs, this is immediate from the definition of the (IC) constraints. See Lemma D.1 for a complete argument.
8 polytope are uniquely determined by their incident facets, i.e., binding constraints. Thus, x ∈ext X if and only if x is uniquely determined by its binding feasibility constraints within the face, which completes the proof.
The result does not extend to mechanisms with infinite menu size because the relevant face is no longer a polytope.17 All our subsequent results will nevertheless accommodate mechanisms of infinite menu size.
Remark. The required steps for the proof outlined in the previous paragraph generalize the main results of Manelli and Vincent (2007, Theorems 17, 19, 20, and 24) about extreme points of the multi-good monopoly problem to arbitrary linear screening problems; see Appendix B.
5. Exhaustive Mechanisms In this section, we introduce and characterize exhaustive mechanisms and show that every extreme point is exhaustive. Exhaustiveness allows us to isolate the role of binding feasibility constraints in determining which mechanisms are extreme points. Our main results in the next section will clarify the role of binding incentive constraints.
Definition 5.1. Mechanisms x, x′ ∈X are positively homothetic if there exists λ ∈R++ and t ∈Rd such that x = λx′+t. Mechanisms x, x′ ∈X are homothetic if they are positively homothetic or one of them is constant. A mechanism x ∈X is exhaustive if there does not exist a mechanism x′ ∈X positively homothetic to x such that F(x) ⊆F(x′).
Two mechanisms are (positively) homothetic if one can be obtained from the other by scaling (with a strictly positive scalar) and translation. In geometric terms, a positive homothety leaves invariant the “shape” and “orientation” of menus. In economic terms, a positive homothety leaves invariant the agent’s ordinal preferences over menu items and, in particular, the binding incentive constraints. Positive homothethy defines an equivalence relation on X and every equivalence class of positively homothetic mechanisms contains an exhaustive mechanism, but this mechanism need not be unique; see Figure 2.
Theorem 5.2. Every extreme point x ∈ext X is exhaustive.
Proof. See Appendix D.2.
□ For mechanisms with finite menu size, Theorem 5.2 is a corollary of Theorem 4.1. If x is not exhaustive, then there exists a mechanism x′ ∈X positively homothetic to x such that F(x) ⊆F(x′). IC(x) = IC(x′) follows immediately from the definition of positive homothety. Therefore, x is not uniquely pinned down by its binding constraints. If x has a finite menu, then Theorem 4.1 completes the proof by contraposition. In general, the argument in Appendix D.2 shows that a mechanism that leaves slack in the feasibility constraints can be decomposed into mechanisms homothetic to itself.
We proceed by characterizing the set of exhaustive mechanisms more explicitly. This characterization is important since every property of exhaustive mechanisms is also a property 17For example, one can show the existence of strictly incentive-compatible extreme points x, x′ ∈ext X, i.e., IC(x) = IC(x′) = ∅, that make the same feasibility constraints binding, i.e., F(x) = F(x′). In the linear delegation problem discussed in Section 7, this amounts to showing that there exist smooth and indecomposable convex bodies, i.e., extended menus, that touch the same facets of the unit simplex, which follows by arguments similar to those in the proof of Theorem 6.6; see Schneider (2014, Theorems 2.7.1 and 3.2.18).
9 menu(x) menu(x′) Figure 2. Two menus of exhaustive mechanisms homothetic to each other.
of extreme points and hence of optimal mechanisms.18 Recall that nH is the normal vector of the facet-defining hyperplane H ∈F of the allocation space A.
Theorem 5.3. A mechanism x ∈X is exhaustive if and only if one of the following holds: (1) There exists a ∈ext A such that menu(x) = {a}.
(2) (a) span{nH}H∈F(x) = Rd and (b) T H∈F(x) H = ∅.
Proof. See Appendix D.2.
□ That is, a non-constant mechanism is exhaustive if and only if the facet-defining hyperplanes corresponding to the binding feasibility constraints satisfy two conditions: (a) their normal vectors span the ambient space and (b) they have an empty intersection. These conditions ensure that the mechanism can neither be translated or scaled relative to a point in a way that would make additional feasibility constraints binding. Figure 3 illustrates.
An equivalent formulation of condition (2) in Theorem 5.3 is that F(x) contains d + 1 hyperplanes of which (a) d intersect in a single point and (b) the last does not. In particular, if the facet-defining hyperplanes of the allocation space A are in general position, then a non-constant mechanism x ∈X is exhaustive if and only if |F(x)| ≥d + 1.19 If d = 2, then the facet-defining hyperplanes are always in general position; thus, a non-constant mechanism x ∈X is exhaustive if and only if |F(x)| ≥3.
We illustrate the characterization of exhaustiveness and its economic implications with two examples.
18While every optimal mechanism is a mixture over optimal extreme points, exhaustiveness is not necessarily preserved under convex combinations. Thus, technically not every optimal mechanism for a given instance (v, µ) of the principal’s problem need be exhaustive. However, topologically generic linear objective functionals are uniquely maximized at an extreme point (Lau, 1976). Thus, optimal mechanisms are generically exhaustive.
19The hyperplanes in F are in general position if every subset of more than d hyperplanes in F has an empty intersection.
10 menu(x) T H∈F(x′) H menu(x′) Figure 3. Illustrations of conditions (a) and (b) from Theorem 5.3. Left: condition (a) is violated by a menu touching two parallel facets of a rectangle.
The menu can be translated horizontally until it touches an inclusion-wise larger set of facets. Right: condition (b) is violated by a menu touching only two facets of a pentagon. The menu can be scaled relative to the intersection point of the two facet-defining lines until it touches an inclusion-wise larger set of facets.
Example 5.4. Let A = {a ∈Rd + | Pd i=1 ai ≤1} be the d-dimensional unit simplex embedded in Rd. The unit simplex is the allocation space when considering lotteries over finitely many alternatives or when dividing time or a budget across a finite set of options (see Sections 7 and 8 for applications). By Theorem 5.3, a non-constant mechanism x : Θ →A is exhaustive if and only if it makes all d + 1 feasibility constraints binding.
A facet of the unit simplex, i.e., feasibility constraint, is characterized by those lotteries in which some alternative is chosen with probability 0. Therefore, in economic terms, a non-constant exhaustive mechanism must allow the agent to avoid any particular alternative with probability 1.
Example 5.5. Let A = [0, 1]d be the unit cube in Rd. The unit cube is the allocation space in a problem with d goods, one of which could be money. For example, consider a bilateral trade problem where k goods are owned by the principal, d −k goods are owned by the agent, and the principal proposes a menu of possible trades to the agent. By Theorem 5.3, a non-constant mechanism x : Θ →A is exhaustive if and only if it makes d non-parallel feasibility constraints and at least one additional feasibility constraint binding.
A facet of the unit cube, i.e., feasibility constraint, is characterized by those allocations in which some good is either allocated to the principal with probability 1 or to the agent with probability 1. Therefore, in economic terms, an exhaustive mechanism must offer the agent a menu designating at least one good for which the menu contains an option where the agent receives the good with probability 1 and an option where the principal receives the good with probability 1. In addition, for every other good, there must be an option where at least one of the two parties receives the good with probability 1. (The latter condition is automatically satisfied if the menu must include the status quo in which every agent keeps their endowment.) 11 6. Extreme Points in One- versus Multi-Dimensional Type Spaces In this section, we show that the extreme points of the set of IC and IR mechanisms have a simple structure in every problem with one-dimensional types but are virtually as rich as the set of exhaustive mechanisms in every problem with multi-dimensional types. Recall that, in our model, a d-dimensional allocation space A always corresponds to a (d −1)-dimensional type space Θ. Also recall that F is the set of feasibility constraints defining the allocation space A.
Theorem 6.1. Suppose d = 2. Then, every extreme point x ∈ext X is exhaustive and satisfies | menu(x)| ≤|F|.
Proof. See Appendix D.3.
□ Remark. The bound is tight for the unrestricted type space and attained by allocating to each type one of their most preferred extreme points of the allocation space A.
Theorem 6.1 is the essential insight of a complete characterization of the extreme points for problems with one-dimensional types (Theorem C.1 in Appendix C): extreme points can be succinctly described as choice functions from a limited number of menu items, akin to the well-known posted-price result for the monopoly problem (Myerson, 1981; Riley and Zeckhauser, 1983). The complete characterization shows that a mechanism x ∈X is an extreme point if and only if menu(x) lacks a certain geometric structure, which we call a flexible chain.
In the multi-dimensional case, the structure of extreme points is fundamentally different and markedly more complex. To make this point, we equip the set of IC and IR mechanisms X with the L1-norm ||x|| = Z Θ ||x(θ)|| dθ.
(3) We say that a property holds for most elements of a subset of a topological space if it holds on a dense set that is also a countable intersection of relatively open sets (i.e., a dense Gδ); this is a standard notion of topological genericity.
Theorem 6.2. Suppose d ≥3. Then, every extreme point is exhaustive and most exhaustive mechanisms are extreme points.
Theorems 6.1 and 6.2 together show that properties of binding incentive constraints further discipline the set of exhaustive mechanisms if and only if the type space is one-dimensional.
Exhaustiveness is a property of binding feasibility constraints alone. Thus, our results corroborate the heuristic understanding in the mechanism design literature that the difficulty with multi-dimensional screening lies in identifying the incentive constraints that are binding in an optimal mechanism.
6.1. Additional Results. In the remainder of this section, we present additional results for the multi-dimensional case that further strengthen Theorem 6.2. We separately discuss extreme points of finite and infinite menu size as well as uniquely optimal mechanisms. All proofs are in Appendix D.3.
We first provide a genericity condition under which an exhaustive mechanism of finite menu size is an extreme point. For this, we say that a set of points M ⊆A is in general position if every hyperplane in Rd intersects M in at most d points.
12 Theorem 6.3. Suppose d ≥3. If x ∈X is exhaustive and menu(x) is finite and in general position, then x ∈ext X.
That is, every exhaustive mechanism with a finite menu can be transformed into an extreme point by perturbing its menu into general position. By carrying out such perturbations, we obtain the following genericity result: Theorem 6.4. Suppose d ≥3. For every k ∈N, the set of extreme points of menu size k is relatively open and dense in the set of exhaustive mechanisms of menu size k.20 Thus, extreme points remain prevalent among exhaustive mechanisms even when restricting attention to mechanisms that make only a limited number of allocations.
It is easy to show that mechanisms with a finite menu size are dense in the set of all mechanisms. Consequently, we have: Corollary 6.5. Suppose d ≥3. The set of extreme points of finite menu size is dense in the set of exhaustive mechanisms.
We next turn to mechanisms of infinite menu size.
Theorem 6.6. Suppose d ≥3. Most exhaustive mechanisms are extreme points of uncountable menu size.
Remark. The proof of Theorem 6.6 establishes the stronger claim that most exhaustive mechanisms are continuous functions (for which the menu is a connected subset of the allocation space). While examples of extreme points with uncountable menu size have been documented in the literature (Manelli and Vincent, 2007; Daskalakis et al., 2017), the existence and prevalence of continuous extreme points is novel.
Exhaustive mechanisms can also be approximated by mechanisms that are uniquely optimal for some objective and prior of the principal. That is, even the most parsimonious candidate sets are dense in the set of exhaustive mechanisms. The formal result is a consequence of a theorem due to Straszewicz and Klee (Klee Jr, 1958), which asserts that the exposed points of a norm-compact convex set are dense in its extreme points.
Corollary 6.7. Suppose d ≥3. For every exhaustive mechanism x ∈X and every ε > 0, there exists a mechanism x′ ∈ext X such that ||x −x′|| < ε and such that x′ is uniquely optimal for some objective function v : Θ →Rd and belief µ ∈∆(Θ).
In Section 8, we show that the gist of our results continues to hold if we only consider those extreme points that are unique maximizers for specific objectives of the principal such as revenue-maximization. That is, candidate sets remain complex even if the principal’s objective is a priori known and fixed and only their belief is considered a free parameter.
Remark. We have given an essentially, though not entirely, complete characterization of the extreme points of the set of IC and IR mechanisms. For example, menus that are not in general position and allow some affine dependencies among menu items can still be extreme points. In Appendix B (Theorem B.6), we provide a complete algebraic characterization of finite-menu extreme points. Using the connection to the relevant mathematical concepts to 20An alternative statement is that the set of extreme points of menu size k is relatively open and dense in the set of exhaustive mechanisms of menu size ≤k; see the proof.
13 be established in the next section, the reader can consult the references provided in Section 9 for additional conditions. A complete characterization of all extreme points is not to be expected (see Footnote 8 in the introduction).
7. Proof Ideas: The Case of Linear Delegation In this section, we explain the methodology behind our results.
Our approach is to translate between extreme points of the set of (IC) and (IR) mechanisms and extreme points of the set of all menus. Menus can be identified with convex bodies in allocation space, allowing us to draw upon a mathematical literature that has characterized extremal—there called indecomposable—elements of spaces of convex bodies. We illustrate this transfer of results from mathematics to economics through what we consider to be the simplest multi-dimensional screening problem; detailed proofs and generalizations are relegated to Appendices A and D.
7.1. Linear Delegation. We proceed in the context of the linear delegation problem and discuss the necessary adjustments for other problems at the end of this section: • A = {a ∈Rd + | Pd i=1 ai ≤1} is the unit simplex, i.e., the allocation space when considering lotteries over m = d + 1 alternatives or when dividing time or a budget across the alternatives (a lists the probabilities or shares of the first d alternatives); • Θ = Sd−1 is the unrestricted type space, i.e., the agent can have all possible von Neumann-Morgenstern preferences over A; • the principal’s objective function v : Θ →Rd is an arbitrary bounded function, i.e., the principal relies on the agent’s information in order to make an informed decision; • there is no veto alternative ¯ a for the agent.
The linear delegation problem features multi-dimensional types whenever there are m ≥4 alternatives and thus differs from classical formulations of delegation problems a la Holmstr¨ om (1977, 1984), which assume one-dimensional allocation and type spaces and single-peaked preferences; see Section 9 for further discussion.
Next to being a natural application of our model, there are two systematic reasons for considering the linear delegation problem: (1) In the linear delegation problem, incentive constraints are completely independent from feasibility constraints in the sense that every mechanism that makes an inclusion-wise maximal set of incentive constraints binding is an extreme point up to positive homothety (Lemma 7.1). This independence simplifies our arguments and renders the connection between extremal menus and indecomposable convex bodies most transparent.
(2) Every linear screening problem is linear delegation with a restricted type space (modulo IR constraints). This is because every linear screening problem can be represented with the unit simplex as its allocation space through an appropriate type space restriction.21 With such reformulations, however, cone Θ is no longer full-dimensional, and because of this additional complexity, we do not use reformulations to linear delegation in our general proofs.
21Consider a problem with allocation space A and type space Θ. Any allocation polytope A ⊂Rd is the image of a higher-dimensional simplex S ⊂Rn under a linear map f : Rn →Rd (Gr¨ unbaum et al., 1967, Chapter 5.1). An appropriate type space in Rn corresponding to the simplex is given by f T (Θ), where f T is the transpose of f.
14 7.2. From Mechanisms to Menus. So far, we have followed the literature in that we have stated our results in terms of direct mechanisms. However, IC mechanisms can equivalently be understood as the agent’s choice functions from different (closed) menus M ⊆A.
We call a closed set M ⊆A an extended menu if every allocation in A \ M is strictly preferred by at least one type θ ∈Θ to every allocation in M. In other words, if M is an extended menu, then there is no allocation that can be added to M without necessarily changing the agent’s choice function. Since the agent has linear utility and we are considering the unrestricted type space in this section, every menu M ⊆A is extended by passing to its convex hull conv(M), which is a convex body in allocation space.
It is straightforward to show that the map which assigns to every mechanism x ∈X the extended menu conv(menu(x)) ⊆A is a bijection between the space X of (payoff-equivalence classes of) IC mechanisms and the space M of convex bodies in allocation space.
In Appendix A (Theorem A.2), we show that the bijection between X and M commutes with convex combinations and, therefore, preserves the linear structure of the underlying spaces. For convex bodies M, M ′ ⊂Rd, this linear structure is given by Minkowski addition and positive scalar multiplication, defined as λM + ρM ′ = {λa + ρa′ | a ∈M, a′ ∈M ′}, (4) where λ, ρ ∈R+. In particular, extreme points of one space map to extreme points of the other space.
We also show in Appendix A that convergence of extended menus with respect to the Hausdorff distance implies convergence of the corresponding mechanisms in L1 (Lemma A.7).
Thus, any statement about compactness or denseness in the former space carries over to the latter.
7.3. Indecomposability and Exhaustiveness. We next explain how extreme points of the set of extended menus M can be understood in terms of the notion of indecomposability from the mathematical literature and the notion of exhaustiveness defined in Section 5.
A menu M ∈M is an extreme point of M if and only if it does not admit either of the following decompositions: (1) M = λM ′ + (1 −λ)M ′′ for λ ∈(0, 1) and M ′, M ′′ ∈M homothetic to M; (2) M = λM ′ + (1 −λ)M ′′ for λ ∈(0, 1) and M ′, M ′′ ∈M not homothetic to M.22 We call (1) a homothetic decomposition and (2) a non-homothetic decomposition.
Lemma D.2 in Appendix D shows that a mechanism x ∈X is exhaustive if and only if the associated extended menu M ∈M admits no homothetic decomposition. We can straightfor-wardly extend the definition of exhaustiveness to extended menus because exhaustiveness is solely a property of the feasibility constraints of the allocation space that are intersected by the menu of a mechanism.
Non-homothetic decompositions are closely related to the notion of decomposability from the mathematical literature. A convex body K ⊂Rd is decomposable if there exist convex bodies K′, K′′ ⊂Rd not homothetic to M such that M = K′ +K′′. By scaling the summands, decomposability is equivalent to the existence of convex bodies K′, K′′ ⊂Rd not homothetic to K such that K = λK′+(1−λ)K′′ with λ ∈(0, 1). A convex body that is not decomposable is indecomposable.
22The two cases are mutually exclusive: if one of M ′ or M ′′ is homothetic to M, then so is the other.
15 If an extended menu M ∈M is indecomposable, then M has no non-homothetic decompo-sition. The converse does not generally hold because the summands λM ′ and (1 −λ)M ′′ of a non-homothetic decomposition in our model are required to be subsets of A, i.e., feasible extended menus.23 However, when the allocation space is a simplex, indecomposability is necessary and sufficient for the absence of non-homothetic decompositions.
Lemma 7.1. In the linear delegation problem, an extended menu M ∈M is in ext M if and only if M is indecomposable and exhaustive.
Proof. See Appendix D.4.
□ Before proceeding with a characterization of the indecomposable convex bodies, we briefly discuss the economic meaning of indecomposability. Recall that an extreme point of finite menu size is determined by its binding incentive and feasibility constraints (Theorem 4.1).
Indecomposability of the associated extended menu ensures that (up to payoff-equivalence) there is no other, non-constant mechanism that makes an inclusion-wise larger set of incentive constraints binding; exhaustiveness ensures the same for the feasibility constraints. Thus, by Lemma 7.1 and in the linear delegation problem, the role of incentive and feasibility constraints in whether or not a mechanism is an extreme point can be completely separated.
Indeed, in other linear screening problems, extreme points need not make inclusion-wise maximal sets of incentive constraints binding. (Nevertheless, it is helpful to analyze feasibility constraints separately from the incentive constraints, as we have done in Section 5.) 7.4. Characterizing Extreme Points. Given Lemma 7.1, it remains to characterize indecomposable and exhaustive extended menus. Indecomposability has been characterized in the mathematical literature.
Theorem (Meyer, 1972; Silverman, 1973). A convex body M ⊂R2 is indecomposable if and only if it is a point, line segment, or triangle.
Figure 1 depicts the proof idea for convex polygons. The figure shows a quadrilateral and two deformations of the quadrilateral that translate the right-most, vertical facet-defining line either to the left or to the right. The resulting deformed quadrilaterals yield a non-homothetic decomposition of the original quadrilateral. Similar deformations can be found for any polygon, but triangles are the only polygons for which these deformations yield homotheties of the triangle. Thus, (degenerate) triangles are the only indecomposable convex polygons.
The extension to all plane convex bodies requires a more involved argument.
Theorem (Shephard, 1963). Let d ≥3. The set of indecomposable convex bodies in Rd is Hausdorff-dense in the set of all convex bodies in Rd.
Shephard identifies a large class of indecomposable polytopes, with the simplest being the simplicial polytopes, i.e., polytopes of which every proper face is a simplex. Roughly speaking, a simplicial polytope S is indecomposable because each two-dimensional face of S is a triangle and any decomposition of S into non-homothetic polytopes would also have to decompose every face of S individually, which is impossible because triangles are indecomposable. Simplicial polytopes are Hausdorff-dense in the space of all convex bodies.
23In the absence of feasibility constraints, every convex body trivially has homothetic decompositions, e.g.
through translations into opposite directions. This is why homothetic decompositions are ruled out in the definition of indecomposability.
16 v1 v2 v3 v4 w v1 v2 v3 v′ 4 w Figure 4. Illustration of how to perturb a polytope into a nearby simplicial polytope. Left: a pyramid with apex w and base v1-v2-v3-v4. Right: a simplicial polytope obtained from the pyramid by pulling the vertex v4 to a new vertex v′ 4 such that the five vertices are in general position. This procedure can be iteratively applied to the vertices of any polytope to obtain a nearby simplicial polytope. (Incidentally, a pyramid is already indecomposable.) First, every convex body is arbitrarily close to a polytope. (Take the convex hull of a finite set of points on the body’s boundary that is ε-dense in the boundary.) Second, every polytope can be transformed into a simplicial polytope by perturbing its vertices into general position; Figure 4 illustrates.
Exhaustiveness admits a simple economic characterization in the linear delegation problem, which follows immediately from Theorem 5.3 (recall Example 5.4). We state the characteri-zation in terms of mechanisms, but it can equivalently be stated in terms of the associated extended menus: • A constant mechanism x is exhaustive if and only if it dictates an alternative: there exists a ∈ext A such that menu(x) = {a}.
• A non-constant mechanism x ∈X is exhaustive if and only if it grants a strike: for every alternative k = 1, . . . , m there exists a lottery a ∈menu(x) in which alternative k is chosen with probability 0. That is, the agent is given the option to strike out any one of the alternatives. Geometrically speaking, this means that menu(x) touches all facets of the allocation simplex.
The following characterization result follows at once from the previous arguments and the bijection between the set of mechanisms X and the set of extended menus M.
Theorem 7.2. Consider the linear delegation problem: (1) With m = 3 alternatives, x ∈X is in ext X if and only if one of the following holds: (a) x dictates an alternative; (b) x grants a strike and has menu size at most three.
(2) With m ≥4 alternatives, ext X is dense in the set of mechanisms that grant a strike.
Thus, the theory predicts simple solutions for linear delegation problems with three alter-natives, but with four or more alternatives and up to approximation, the only distinguishing property of extreme points is that they dictate or grant a strike.
We remark that optimality in the linear delegation problem, even when there are only three alternatives, may require the use of stochastic mechanisms that offer the agent lotteries 17 over the alternatives.24 Lotteries can be interpreted as risky courses of action or as budget or time shares. Optimality may even require lotteries with full support, i.e., interior points of the simplex. Intuitively, lotteries give the principal more leeway in screening the agent and make it more difficult for the agent to align the allocation with their own preferences.
7.5. General Linear Screening Problems. We finally discuss the necessary adjustments to our approach when considering (IR) constraints, allocation spaces different from the simplex, and restricted type spaces.
In the context of the linear delegation problem, IR would mean that the menu of a mechanism must contain the veto alternative ¯ a ∈ext A. Any decomposition of a given convex body that contains ¯ a must also contain ¯ a. Thus, introducing IR constraints simply amounts to considering extreme points of the set of IC mechanisms that also satisfy IR. The same conclusion obtains in other linear screening problems; see Appendix A.3.
Suppose the allocation space A differs from the simplex. If an extended menu M ∈M is indecomposable, then it does not admit a non-homothetic decomposition. However, the converse is no longer true.
This is inconsequential for the denseness results for multi-dimensional problems since we only get additional, extremal but decomposable extended menus. For one-dimensional type spaces, these additional extreme points drive the bound on the menu size from three up to the number of feasibility constraints of the allocation space (Theorem 6.1). We provide a complete characterization of extremal extended menus for one-dimensional type spaces and arbitrary allocation spaces A (Theorem C.1 in Appendix C).
This characterization builds on a mathematical result due to Mielczarek (1998).
Suppose the type space is restricted, i.e. cone Θ ̸= Rd. Extending a menu now entails more than taking the convex hull because there are certain directions in the allocation space along which all types are worse off. Geometrically speaking, these directions form the polar cone of the type space. To prove our result, it is a technical convenience to extend menus beyond the boundaries of the allocation space and work with closed convex sets that share the polar cone as a common recession cone. Indecomposability for closed convex sets with a common recession cone is analogous to indecomposability for convex bodies and has been discussed in Smilansky (1987).
8. Specific Objectives: Multi-Good Monopoly and Linear Veto Bargaining Our previous analysis considered candidates for optimality that the principal must a priori consider when uncertain about both their objective function and the distribution of the agent’s types; we now fix the principal’s objective, e.g., revenue maximization, and characterize the mechanisms that remain relevant for optimality as the type distribution varies.
In applications to the multi-good monopoly problem and the linear veto bargaining problem, to be defined below, we show that the set of mechanisms that are uniquely optimal for some type distribution is dense in the set of undominated mechanisms. A mechanism is undominated if there is no other mechanism that yields the principal an unambiguously higher utility. We provide characterizations of undominated mechanisms, showing that they are almost as rich as the set of all (IC) and (IR) mechanisms. Thus, the gist of our main results holds when 24See Kov´ aˇ c and Mylovanov (2009) and Kleiner et al. (2021) for a discussion about the optimality of stochastic mechanisms in the classical one-dimensional delegation model.
18 restricting attention to extreme points that are unique maximizers for specific objectives of the principal. We discuss the two applications after introducing undominated mechanisms.
8.1. Undominated Mechanisms. For multi-dimensional problems, we have identified exhaustive mechanisms as a reference set in which the extreme points lie dense. However, with a fixed objective v, not every extreme point remains relevant for optimality. For example, an extreme point might minimize expected revenue for some type distribution µ.
The appropriate reference set now becomes the set of undominated mechanisms, originally defined for the multi-good monopoly problem by Manelli and Vincent (2007).
Definition 8.1. A mechanism x ∈X is dominated by another mechanism x′ ∈X if x′(θ) · v(θ) ≥x(θ) · v(θ) for almost all θ ∈Θ, with strict inequality on a set of types of positive measure. A mechanism x ∈X is undominated if it is not dominated by any other mechanism x′ ∈X.
Manelli and Vincent (2007) show for the monopoly problem that every undominated mechanism is optimal for some belief about the agent’s type. Their benchmark result can be extended from revenue maximization to arbitrary objectives: Theorem 8.2. For every undominated mechanism x ∈X, there exists a type distribution µ ∈∆(Θ) such that x is an optimal mechanism for a principal with belief µ.
Proof. See Appendix D.5.
□ Conversely, every mechanism that is optimal for some fully supported type distribution µ ∈∆(Θ) must clearly be undominated.
A priori, not every undominated mechanism is a necessary candidate for optimality.
(Undominated mechanisms need not be extreme or exposed points). However, in the following applications and as long as types are multi-dimensional, we show that every undominated mechanism is arbitrarily close to a mechanism that is uniquely optimal for some type distribution, i.e., arbitrarily close to a mechanism that is a necessary candidate for optimality.
8.2. Multi-Good Monopoly. The multi-good monopoly problem is the following linear screening problem: • A = [0, 1]m × [0, κ], where the first m allocation dimensions are the probabilities with which good i = 1, . . . , m is sold to the agent, and the last allocation dimension is the payment by the agent (and κ is some sufficiently large constant, which is without loss of generality whenever valuations are bounded); • Θ = [0, 1]m×{−1}, i.e., the consumer has valuations in [0, 1] for each good i = 1, . . . , m and money is the numeraire;25 • ¯ a = (0, . . . , 0, 0), i.e., the consumer can leave without paying anything; 25Due to linear utility, we implicitly assume that the goods are neither substitutes nor complements for the agent. This assumption is made in most papers on the multi-good monopoly problem. We could incorporate substitutes and complements by allowing the agent to have one valuation for each possible bundle B ⊆{1, . . . , m}. The allocation space is then the unit simplex over 2m deterministic allocations, i.e., all possible bundles, plus an extra dimension representing money as before. Free disposal, i.e., the agent being willing to pay weakly more for inclusion-wise larger bundles, and a fixed marginal utility for money can be modeled as a family of affine restrictions on the type space.
19 • v(θ) = ¯ v = {0, . . . , 0, 1} for all θ ∈Θ, i.e., the principal maximizes expected revenue (and goods can be produced at zero cost).26 In line with standard terminology in mechanism design with transfers, we abuse our language by referring to a ∈[0, 1]m as an allocation and t ∈[0, κ] as the transfer. Instead of probabilities, allocations can also be interpreted as quantities or as quality-differentiated goods with multiple attributes (for which the consumer has unit demand).
We next show that a large class of mechanisms in the monopoly problem is undominated.
A pricing function is a continuous convex function p : [0, 1]m →R+ such that p(0) = 0 that assigns a price to each possible allocation.27 The marginal price for good i = 1, . . . , d at allocation a ∈[0, 1]m with ai < 1 is the directional derivative ∇eip(a) of p at a in the coordinate direction ei (which exists by the convexity and continuity of p). The mechanism x ∈X obtained from a pricing function is the agent’s choice function from the menu M = {(a, p(a)) | a ∈[0, 1]m}.
Lemma 8.3. In the multi-good monopoly problem, every mechanism x ∈X can be obtained from a pricing function p with marginal prices in [0, 1]. If a mechanism x ∈X can be obtained from a pricing function p with marginal prices ∇eip uniformly bounded away from 0 and 1 for every good i = 1, . . . , d, then it is undominated.
Proof. See Appendix D.5.
□ In plain words, a mechanism that, on the margin, prevents low-valuation types from buying additional quantity while enabling high-valuation types to buy additional quantity is undominated. Such a mechanism features “no-distortion at the top” (the highest type receives the efficient allocation) and “exclusion at the bottom” (the lowest type receives nothing), which are well-known properties of optimal mechanisms in screening problems with transfers. In particular, such a mechanism features these two properties separately in each allocation dimension. Not all undominated mechanisms have marginal prices bounded away from zero and one, but the gap to the mechanisms that do admit this bound is negligible.28 Corollary 8.4. In the multi-good monopoly problem, the set of undominated mechanisms is dense in the set of all (IC) and (IR) mechanisms.
Proof. See Appendix D.5.
□ For a rough intuition for the richness of undominated mechanisms, consider the following trade-off. When the principal increases the price for some allocations, revenue increases from those types who continue to choose these allocations. However, some types that have previously chosen an allocation at the lower price may now opt for a cheaper allocation, decreasing revenue from the types that switch. This trade-off rules out a dominance relationship between many mechanisms.
26The literature makes the zero-cost assumption for simplicity. It can easily be relaxed to a constant marginal cost for each good. With decreasing marginal costs, extreme points also remain the relevant candidates for optimality (see the discussion in Manelli and Vincent (2007)). With increasing marginal costs, one has to follow the approach taken by Rochet and Chon´ e (1998).
27Convexity and continuity are without loss of generality because the agent has linear utility. p(0) = 0 reflects the (IR) constraint. See also Hart and Reny (2015, Appendix A.2).
28For an example, see the mechanism depicted in Figure 2 in Manelli and Vincent (2007).
In the bottom-right “market segment,” the marginal price for good one is 1.
20 Given the characterization of undominated mechanisms, the same arguments as in Section 7 can be applied to conclude that extreme points are dense in the set of undominated mechanisms and, therefore, in the set of all mechanisms by Corollary 8.4. In the following result, the first part is well-known (see, for example, Manelli and Vincent (2007, Lemma 4)).
Theorem 8.5. Consider the multi-good monopoly problem: (1) With m = 1 good, a mechanism x ∈X is in ext X and undominated if and only if x is a posted-price mechanism with price p ∈(0, 1), i.e., x(θ) = ( (1, p) if θ1 ≥p (0, 0) otherwise.
(2) With m ≥2 goods, the set of mechanisms x ∈X that are uniquely optimal for some belief µ ∈∆(Θ) is dense in X.
Proof. See Appendix D.5.
□ Remark. The proof shows that statement (2) remains true if the belief µ is required to have full support on Θ.
The second part says that any incentive-compatible and individually rational mechanism can be turned into a mechanism that is uniquely optimal for some belief of the seller by applying an arbitrarily small perturbation. The claim about uniquely optimal mechanisms is not an application of Straszewicz’ theorem upon showing denseness of the extreme points in the set of undominated mechanisms. While Straszewicz’ theorem guarantees that exposed points are arbitrarily close to extreme points, these points may be exposed by linear functionals unrelated to revenue maximization. Our proof modifies the theorem to obtain the desired result.
8.3. Linear Veto Bargaining. We now discuss the following linear veto bargaining problem: • A = {a ∈Rd + | Pd i=1 ai ≤1} is the unit simplex, i.e., the allocation space when considering lotteries over m = d + 1 alternatives or when dividing time or a budget across the alternatives (a lists the probabilities or shares of the first d alternatives); • Θ = Sd−1 is the unrestricted domain, i.e., the agent can have all possible von Neumann-Morgenstern preferences over A; • there is a veto alternative ¯ a ∈ext A for the agent (e.g., the status quo in a political context), and we set ¯ a = (0, . . . , 0) without loss of generality; • the principal’s preferences are given by a Bernoulli utility vector ¯ v independently of the agent’s information, i.e., v(θ) = ¯ v for all θ ∈Θ, and we assume for simplicity that (1) ¯ v ∈R++, i.e., the veto alternative is the principal’s least preferred alternative, and (2) arg maxi ¯ vi is a singleton, i.e., the principal has a unique favorite alternative.
The problem can be seen as a delegation problem with a state-independent objective and an IR constraint. Therefore, the extreme points of linear veto bargaining are exactly the extreme points of linear delegation that satisfy IR (Lemma A.9). Linear veto bargaining can also be seen as a no-transfers analogue of the monopoly problem since both problems feature state-independent objectives with an IR constraint.
Lemma 8.6. In the linear veto bargaining problem, a mechanism x ∈X is undominated if and only if menu(x) contains the veto alternative and the principal’s most preferred alternative.
21 Proof. See Appendix D.5.
□ The richness of undominated mechanisms in the veto bargaining problem comes from a trade-off similar to that in the monopoly problem. By adding an alternative to the menu of a mechanism, some types prefer the new alternative over their previous choice. Among those who switch, some types will do so in the principal’s favor, i.e., switch away from alternatives that the principal likes less than the new alternative. Other types will not switch in the principal’s favor, i.e., switch away from alternatives that the principal likes more than the new alternative. A similar trade-off arises when removing an alternative from the menu.
These trade-offs prevent a dominance relationship between mechanisms that allocate the principal’s most preferred alternative.
As before, given the characterization of undominated mechanism above, the same arguments as in Section 7 can be applied to conclude that the extreme points are dense in the set of undominated mechanisms whenever there are four or more alternatives. The claim about uniquely optimal mechanisms again requires additional arguments.
Theorem 8.7. Consider the linear veto bargaining problem: (1) With m = 3 alternatives, a mechanism x ∈X is undominated and in ext X if and only if menu(x) contains the veto alternative, the principal’s most preferred alternative, and at most one other lottery over the alternatives.
(2) With m ≥4 alternatives, the set of mechanisms x ∈X that are uniquely optimal for some belief µ ∈∆(Θ) is dense in the set of undominated mechanisms.
Proof. See Appendix D.5.
□ 9. Related Literature This paper relates to several areas of research, including multi-dimensional screening, extreme points in mechanism design, delegation and veto bargaining, and the mathematical literature on indecomposability. We will explain the relation to these four areas after first discussing Manelli and Vincent (2007), whose work most closely relates to ours.
9.1. Manelli and Vincent (2007) (MV). In the context of the multi-good monopoly problem, MV provide the first—and, prior to this paper, only—analysis of extreme points in multi-dimensional mechanism design, with two main contributions. First, they provide an algebraic characterization of finite-menu extreme points in terms of whether or not a certain linear system associated with a given mechanism has a unique solution. This characterization is based on auxiliary results about the facial structure of the set of incentive-compatible mechanisms. Second, they define the notion of undominated mechanisms and show that every undominated mechanism maximizes expected revenue for some distribution of types.
In comparison to MV, we consider arbitrary linear screening problems with or without transfers and subsume the multi-good monopoly problem as a special case. We contribute explicit, non-algebraic extreme-point characterizations (Section 6). These characterizations reveal the precise structure of the set of extreme points and, therefore, the structure of the possible solutions to linear screening problems. Along the way, we obtain generalizations of MVs results in our more general framework; see Appendix B and Theorem 8.2.
In comparison to MV, we also characterize undominated extreme points and uniquely optimal mechanisms. While MV show for the monopoly problem that all undominated 22 mechanisms are potentially optimal, it has not been known which undominated mechanisms are necessary candidates for optimality, i.e., which extreme points are undominated and uniquely optimal for some type distribution. A priori, one might conjecture that parsimonious candidate sets are significantly smaller than the set of undominated mechanisms. We show that this is not the case: the relevant exposed points are dense in the set of undominated mechanisms. Moreover, we provide new results about undominated mechanisms, showing that these mechanisms are themselves virtually as rich as the set of all IC and IR mechanisms.
Finally, we note that MV have shown for the monopoly problem, modulo minor details, that the extreme points of menu size k ∈N are relatively open and dense in the IC mechanisms of menu size k, provided k is smaller than the number of goods for sale plus one (their Remark 25). We show that this substantial qualifier on k is not necessary and that the result holds for arbitrary multi-dimensional screening problems with linear utility (Theorem 6.4).
9.2. Multi-dimensional Screening and Mechanism Design. The literature on multi-dimensional screening—and on the multi-good monopoly problem in particular—is much too large to be summarized here in detail. We focus on recent developments and point to a survey by Rochet and Stole (2003) for work up to the early 2000s.29 Recent work focuses mostly on the multi-good monopoly problem and can be classified into several approaches for gaining insights into multi-dimensional screening problems or for circumventing the severe difficulties associated with their classical formulations: • provide conditions for the optimality of common mechanisms such as separate sales or bundling (McAfee et al., 1989; Manelli and Vincent, 2006; Fang and Norman, 2006; Pavlov, 2011; Daskalakis et al., 2017; Menicucci et al., 2015; Bergemann et al., 2021; Haghpanah and Hartline, 2021; Ghili, 2023; Yang, 2023); • provide duality results that can be used to certify the optimality of a given mechanism (Daskalakis et al., 2017; Kleiner and Manelli, 2019; Cai et al., 2019; Kolesnikov et al., 2022; Kleiner, 2022); • identify specific structural properties of optimal mechanisms (e.g., subadditive pricing, monotonicity, no randomization) and show when such structure arises (McAfee et al., 1989; Manelli and Vincent, 2006; Hart and Reny, 2015; Babaioff et al., 2018; Ben-Moshe et al., 2022; Bikhchandani and Mishra, 2022); • quantify the worst-case performance (approximation ratio) of common mechanisms or classes of mechanisms (Hart and Nisan, 2017; Hart and Nisan, 2019; Li and Yao, 2013; Babaioff et al., 2017; Rubinstein and Weinberg, 2018; Hart and Reny, 2019; Babaioff et al., 2020; Ben-Moshe et al., 2022); • identify mechanisms with the optimal worst-case performance for a mechanism designer with Knightian uncertainty over the set of type distributions (Carroll, 2017; Deb and Roesler, 2023; Che and Zhong, 2023); • derive asymptotic optimality results for a large number of i.i.d. goods (Armstrong, 1999; Bakos and Brynjolfsson, 1999) or for the speed of convergence to first-best as the principal gains increasingly precise information about the agent’s type (Frick et al., 2024).
Our paper is orthogonal to these developments. We do not focus on specific properties and classes of mechanisms or attempt to escape intractabilities. Instead, we shed light on 29A sample of important early work includes Adams and Yellen (1976), Schmalensee (1984), McAfee et al.
(1989), Wilson (1993), Armstrong (1996), and Rochet and Chon´ e (1998).
23 where these intractabilities originate and identify the limits of the qualitative predictions that can be drawn within the standard Bayesian framework. Moreover, to the best of our knowledge, multi-dimensional screening without transfers has not been studied, with the exception of Kleiner (2022), whose duality approach to a multi-dimensional delegation problem is complementary to our extreme points approach.
Besides the implications for optimal mechanism design, we contribute to the literature on implementability with multi-dimensional type spaces (e.g., Rochet, 1987; Saks and Yu, 2005; Bikhchandani et al., 2006) by characterizing extreme points of the set of incentive-compatible mechanisms. By Choquet’s theorem, every non-extreme point can be represented as a mixture over extreme points.
In general, little is known about optimal multi-dimensional mechanism design with multiple agents (Palfrey, 1983; Jehiel et al., 1999; Chakraborty, 1999; Jehiel et al., 2007; Kolesnikov et al., 2022); see the conclusion for further discussion.
9.3. Extreme Points in Mechanism Design. A number of papers have approached mechanism design problems by studying the extreme points of the set of incentive-compatible mechanisms. However, aside from the previously discussed work by Manelli and Vincent (2007), this approach has only been applied to one-dimensional problems. For instance, Border (1991) uses extreme points—hierarchical allocations—in a characterization of the set of feasible interim allocation rules. Building on Border’s insights, Manelli and Vincent (2010) demonstrate the equivalence of Bayesian and dominant strategy incentive-compatibility in standard auction problems. A similar approach is discussed in Vohra (2011, Chapter 6).
Kleiner et al. (2021) present characterizations of the extreme points of certain majorization sets and show how these majorization sets naturally arise as feasible sets in many economic design problems.
In the context of mechanism design, their results immediately imply a characterization of the extreme points of the set of feasible and incentive-compatible interim allocation rules in one-dimensional symmetric allocation problems, providing a new perspective on Border’s theorem as well as BIC-DIC equivalence. Their approach is tailored to one-dimensional problems, elegantly handling both the IC constraints (monotonicity for one-dimensional types) and the Maskin-Riley-Matthews-Border feasibility constraints (majorization with respect to the efficient allocation rule).
In subsequent work, Kleiner et al. (2024) characterize certain extreme points of the set of measures defined on a compact convex subset of Rd that are dominated in the convex order by a given measure. Their result is a multi-dimensional analogue of results obtained in Kleiner et al. (2021) about the set of monotone functions that majorize a given monotone function (see also Arieli et al., 2023). These results apply to information design but have no obvious applications to mechanism design.
Nikzad (2022, 2024) builds on the majorization approach, allowing for additional constraints on the majorization sets. These constraints may, for example, correspond to fairness or efficiency constraints in a revenue-maximization problem. Yang and Zentefis (2024) provide a complementary analysis to Kleiner et al. (2021) based on characterizations of extreme points of sets of distributions characterized by first-order stochastic dominance conditions rather than second-order stochastic dominance conditions (majorization).
Extreme point approaches have also been used in mechanism design without transfers (e.g., Ben-Porath et al., 2014; Niemeyer and Preusser, 2024), and several other mechanism design papers use extreme points as a technical tool (e.g., Chen et al., 2019).
24 9.4. Delegation and Veto Bargaining. Much of the literature on optimal delegation has focused on one-dimensional allocation (action) and type spaces with single-peaked preferences; see Holmstr¨ om (1977, 1984), Melumad and Shibano (1991), Martimort and Semenov (2006), Alonso and Matouschek (2008), Amador and Bagwell (2013), Kolotilin and Zapechelnyuk (2019), and Kleiner et al. (2021).
The applications of our results to delegation differ from the classical literature in two ways.
First, allocations in our delegation problem are lotteries over finitely many alternatives.30 Second, both the principal and agent have arbitrary vNM preferences over these alternatives; that is, our problem features an unrestricted rather than single-peaked preference domain and therefore multi-dimensional types (and allocations). We can allow more general allocation spaces, provided the agent’s utility remains linear.
A small number of papers consider multi-dimensional type or allocation spaces. Koessler and Martimort (2012) study optimal delegation in a setting with a one-dimensional type space and two allocation dimensions across which the principal and the agent have separable quadratic preferences. Frankel (2016) links multiple independent, one-dimensional delegation problems.
Frankel shows that “halfspace delegation,” i.e., imposing a quota on the weighted average of actions across problems, is optimal for normally distributed states and approximately optimal for general distributions as the number of linked problems goes to infinity. See also Frankel (2014) for a robust mechanism design approach. Kleiner (2022) studies optimal delegation with both multi-dimensional type and allocation spaces. Kleiner’s duality-based approach is complementary to our extreme-point approach.
Veto bargaining is a classical problem in political science, originally studied in Romer and Rosenthal (1978). The case with incomplete information about the agent’s (vetoer’s) preferences has only recently been studied using a mechanism design approach by Kartik et al. (2021).31 Their model features one-dimensional private information. Amador and Bagwell (2022) and Saran (2022) study related one-dimensional delegation problems with IR constraints where the principal does not necessarily have state-independent preferences.32 Similarly, our model can nest linear delegation problems with IR constraints.
9.5. Mathematical Foundations. Gale (1954) introduced the notion of an indecomposable convex body and announced the first results about indecomposability. Gale’s results were later proven and published in Shephard (1963), Meyer (1972)/Silverman (1973), and Sallee (1972). These and other papers have provided many novel results that go beyond Gale’s original presentation. McMullen (1973), Meyer (1974), and Smilansky (1987) provide algebraic characterizations of indecomposable polytopes. Smilansky (1987) discusses indecomposable polyhedra. Related results characterize extremal convex bodies within a given compact convex set in the plane (Grza´ slewicz, 1984; Mielczarek, 1998); see Theorems 6.1 and C.1 for the application in our paper. Decomposability is related to deformations of polytopes, which we briefly use in Appendix B; Castillo and Liu (2022, Section 2) provide a concise treatment. Textbook references on indecomposability include Schneider (2014, Chapter 3.2), Pineda Villavicencio (2024, Chapter 6), and Gr¨ unbaum et al. (1967, Chapter 15).
30Delegation over a finite set of alternatives is also studied in the project selection literature; see Armstrong and Vickers (2010), Nocke and Whinston (2013), Che et al. (2013a), and Guo and Shmaya (2023).
31See also Ali et al. (2023).
32See also the ”balanced” delegation problem in Kolotilin and Zapechelnyuk (2019).
25 Characterizations of indecomposable convex bodies can alternatively be seen, via support function duality, as characterizations of the extremal rays of the cone of sublinear (i.e.
convex and homogeneous) functions. A subset of the results known in the literature on indecomposable convex bodies have been independently obtained in studies of the extremal rays of the cone of convex functions by Johansen (1974) (for two-dimensional domains) and Bronshtein (1978) (for d-dimensional domains).33 We finally mention a result due to Klee (1959, Proposition 2.1, Theorem 2.2), which shows that for most (in the sense of topological genericity) compact convex subsets of an infinite-dimensional Banach space, the extreme points of the set are dense in the set itself.
This follows since such sets have an empty interior, support points are dense in the boundary, hence in the set itself, and since most such sets are strictly convex, so that every support point is an extreme point. However, the set of IC mechanisms is a specific compact convex subset of an infinite-dimensional Banach space, which, in particular, is not strictly convex.
The content of our results is that whenever the type space is multi-dimensional, the extreme points are nevertheless dense in a certain part of the set.
10. Conclusion We have characterized extreme points of the set of incentive-compatible (IC) mechanisms for screening problems with linear utility. For every problem with one-dimensional types, extreme points admit a simple characterization with a tight upper bound on their menu size.
In contrast, for every problem with multi-dimensional types, we have identified a large set of IC mechanisms—exhaustive mechanisms—in which the extreme and exposed points lie dense.
Consequently, one-dimensional problems allow us to make predictions that are independent of the precise details of the environment, whereas such predictions are largely unattainable for multi-dimensional problems.
One might hope that restricting attention to specific instances of a given multi-dimensional screening problem allows more robust predictions regarding optimality. We have shown that such predictions remain elusive in applications to monopoly and veto bargaining problems, where the principal’s objective is fixed and state-independent and only the principal’s belief about the agent’s type is considered a free parameter.
While our focus has been on screening problems, where there is only a single (representative) agent, one should expect implications of our results for multi-agent settings. In multi-agent settings, Bayesian incentive compatibility of a given multi-agent mechanism is the same as separately requiring incentive compatibility with respect to each agent’s interim-expected mechanism (see, e.g., B¨ orgers, 2015, Chapter 6). These interim-expected mechanisms, one for each agent, must then be linked towards an ex-post feasible mechanism via an appropriate analogue of the Maskin-Riley-Matthews-Border conditions.34 Thus, if the extreme points in a multi-agent problem were simpler than the extreme points characterized here for the one-agent case, then this reduction in complexity would have to come from these additional conditions. This is not the case for problems with one-dimensional types (see, e.g., Kleiner et al., 2021) and is not to be expected for problems with multi-dimensional types.
33We thank Andreas Kleiner for pointing us to these references.
34Maskin and Riley (1984), Matthews (1984), Border (1991). Recent treatments include Che et al. (2013b), Gopalan et al. (2018), and Valenzuela-Stookey (2023); see these papers for further references and discussions of potential limitations of the reduced-form approach.
26 Our main methodological contribution is to link extreme points of the set of incentive-compatible mechanisms to indecomposable convex bodies studied in convex geometry. This methodology, where we study incentive-compatible mechanisms by analyzing the space of all menus from which the agent could choose, is potentially useful in other areas of economic theory. Examples that come to mind are menu choice a la Dekel et al. (2001) and the random expected utility (REU) model of Gul and Pesendorfer (2006).
27 Appendix A. Preliminaries & Auxiliary Results This appendix gathers general tools we use throughout the proofs of our results from the main text. Appendix A.1 shows that there are bijections between mechanisms, menus, and indirect utility functions that commute with convex combinations (in the sense of Minkowski).
The commutativity with convex combinations is essential for our subsequent analysis because we will study extremal menus and then translate back to extremal mechanisms, as explained in Section 7. Appendix A.2 introduces the relevant topological structure for the three sets of objects. Appendix A.3 discusses how individual rationality (IR) constraints are incorporated into our analysis.
A.1. Mechanisms, Menus, and Indirect Utility Functions. Recall that we have identi-fied payoff-equivalent mechanisms and that X is the set of payoff-equivalence classes of (IC) and (IR) mechanisms.
Let U = {U : θ 7→x(θ) · θ | x ∈X} denote the set of all indirect utility functions induced by the mechanisms in X. It is a direct consequence of (IC) that an indirect utility function is HD1 (homogeneous of degree 1) on cone Θ because types on the same ray from the origin have the same ordinal preferences.
Thus, we extend indirect utility functions U ∈U to Rd by setting U(λθ) = λU(θ) for all θ ∈Θ and λ ≥0 and U(z) = ∞for all z / ∈cone Θ.
A menu is simply a subset M ⊂A that the principal offers the agent and from which the agent chooses their favorite allocation. However, different menus can induce payoff-equivalent choice functions, i.e., payoff-equivalent IC mechanisms, for the agent. Thus, we define the notion of an extended menu, which is the inclusion-wise largest representative of a payoff-equivalence class of menus.
To define extended menus, let Θ◦= {y ∈Rd | ∀θ ∈Θ, y · θ ≤0} (5) denote the polar cone of Θ. The polar cone of type space is the set of all directions in allocation space A along which no type’s utility ever strictly improves. If cone Θ = Rd, then the only such direction is the trivial direction 0. By definition, we may add to every menu M ⊂A the polar cone Θ◦and instead offer the agent the Minkowski sum M + Θ◦without affecting the agent’s indirect utility. We may also take the closed convex hull of M, which does not affect indirect utility either since utility is linear. By requiring ¯ a ∈M, i.e., the veto allocation is in M, we ensure that the agent does not veto the menu.
Definition A.1. An extended menu is a closed convex set M ⊂Rd such that M = conv ext M + Θ◦, ext M ⊂A, and ¯ a ∈M. The set of all extended menus is denoted by M.
If the type space is unrestricted, i.e., cone Θ = Rd, then extended menus are convex bodies in A; otherwise, they are unbounded closed convex sets. That extended menus offer infeasible allocations when the type space is restricted has no physical meaning and is merely a convenient way to identify payoff-equivalent menus. We note that extended menus are uniquely pinned down by their extreme points.
Figure 5 illustrates the construction of extended menus. The depicted allocation and type spaces fit a one-good monopoly problem. The horizontal allocation dimension a1 is the probability of sale, and the vertical allocation dimension a2 is the payment. The type space 28 v1 v2 v3 v4 v5 Θ◦ cone Θ A M Figure 5. An example of an extended menu for a restricted type space. The type cone, cone Θ, is the 45◦cone, shaded in dark gray. The polar cone Θ◦ is the 135◦cone, also shaded in dark gray and with extremal rays that are orthogonal to the extremal rays of cone Θ. The allocation space A is the square.
An exemplary menu {v1, v2, v3, v4, v5} is depicted using dots. Its extension is the polyhedron M shaded in light gray and with boundary given by the dotted lines. M is obtained by taking the convex hull of {v1, v2, v3, v4, v5} and adding the polar cone Θ◦. Here, v2 and v3 “vanish” in the polar cone because v2 and v3 are dominated by the other three allocations for every type in Θ.
is Θ = [0, 1] × {−1}, where the first component is the agent’s valuation for the good. The extremal rays of the polar cone Θ◦are allocation directions in which (1) the agent gets the good with lower probability for the same payment, and (2) the agent gets the good with higher probability but for a marginal price that makes the type (1, −1), who is willing to pay most, just indifferent. The extended menu M corresponds to a mechanism where some types never transact (v1), some types buy a cheap lottery that sometimes allocates the good (v5), and all other types buy the good with probability 1 at a more expensive price (v4).
As in Section 7, we equip M with the operations of Minkowski addition and positive scalar multiplication.
Theorem A.2. The following functions are bijections that commute with convex combinations: • Φ1 : X →M where x 7→(conv menu(x) + Θ◦); • Φ2 : M →U where M →(θ 7→supy∈M y · θ); 29 • Φ3 : U →X where U 7→Q θ∈Θ(∂U(θ) ∩A).35 That is, Φ1 maps (payoff-equivalence classes of) IC and IR mechanisms to the extension of their menus; Φ2 maps extended menus to their support functions; Φ3 maps indirect utility functions to their subdifferential.
Proof. Define an auxiliary map Φ′ 1 : x 7→(cl conv x(Θ) + Θ◦), where x : Θ →A is an IC and IR mechanism, and claim that Φ′ 1(x) = M ∈M. We have ¯ a ∈M for otherwise there would exist a type that strictly prefers ¯ a to x(Θ) by the linearity of utility and the definitions of the closed convex hull and the polar, which contradicts that x satisfies (IR). We also have ext M = ext(cl conv x(Θ) + Θ◦) ⊆ext(cl conv x(Θ)) ⊂A, where the first inclusion follows since Θ◦is a cone and the second inclusion follows since x(Θ) ⊆A and A is compact and convex. Finally, M is closed and convex because it is a sum of a compact convex set and a closed convex set.
The map Φ2 : M →U is well-defined: for each M ∈M, its support function supa∈M a · θ is an indirect utility function in U because arg maxa∈M a · θ is non-empty for all θ ∈Θ and every selection from the argmax is an IC and IR mechanism.
We next show that Φ2 ◦Φ′ 1 is the map that assigns to each IC and IR mechanism x its indirect utility function U ∈U. Let U ∈U be the indirect utility function associated with x ∈X. As desired, we have U(θ) = ( supa∈x(Θ) a · θ = supy∈cl conv x(Θ)+Θ◦y · θ = supy∈M y · θ if θ ∈cone Θ ∞= supy∈M y · θ otherwise.
In the first case, the first equality is (IC) and the second equality follows from the definitions of the closed convex hull and the polar cone. The second case also follows by definition of the polar cone.
The map Φ2 : M →U is injective because support functions uniquely determine closed convex sets (Hiriart-Urruty and Lemar´ echal, 1996, Theorem V.2.2.2).
Thus, if x and x′ are payoff-equivalent IC and IR mechanisms, then Φ′ 1(x) = Φ′ 1(x′) because Φ2 is injective and (Φ2 ◦Φ′ 1)(x) = (Φ2 ◦Φ′ 1)(x′) (by the definition of payoff-equivalence). Thus, Φ′ 1 can be defined on the set of payoff-equivalence classes X in the obvious way.
Φ′ 1 : X →M and Φ2 : M →U are bijective because Φ2 ◦Φ′ 1 is bijective and Φ2 is injective.
We next show that Φ1 = Φ′ 1. Let x = (Φ′ 1)−1(M) ∈X be any representative mechanism from the payoff-equivalence class associated with M ∈M. Then, exp M ⊆menu(x) since every mechanism x′ that is payoff-equivalent to x must necessarily allocate to each type θ ∈Θ with a uniquely preferred option a ∈exp M that option. Moreover, menu(x) ⊆ cl ext M because every type can find a favorite allocation in ext M. By Theorem 2.3 in Klee (1959), ext M ⊆cl exp M ⊆menu(x) since menu(x) is compact. Thus, menu(x) = cl ext M.
Consequently, M = conv menu(x) + Θ◦since M is closed, as desired.
We next verify that the inverse of the composition Φ2 ◦Φ1 : X →U is given by Φ3 : U →X.
Take any (IC) and (IR) mechanism x with associated extended menu M ∈M and associated 35∂U(θ) denotes the subdifferential of U at θ ∈Θ. The proof and Corollary A.4 below confirm that the subdifferential of an indirect utility function U ∈U is well-defined because U is convex.
30 indirect utility function U ∈U. For all θ ∈Θ, we have x(θ) ∈arg max a∈x(Θ) a · θ ⊆arg max a∈M a · θ = ∂U(θ), where the first step is (IC), the second step is immediate from the definition of M, and the third step is a property of support functions (Rockafellar and Wets, 2009, Corollary 8.2.5).
It remains to show commutativity with convex combinations. That Φ2 : M →U commutes with convex combinations is a property of support functions (Hiriart-Urruty and Lemar´ echal, 1996, Theorem V.3.3.3).36 That Φ3 : U →X commutes with convex combinations follows from the linearity of the gradient map, which is almost everywhere well-defined. Thus, Φ1 : X →M must also commute with convex combinations.
□ Theorem A.2 is fundamental to our approach because the bijections between X, U, and M map extreme points to extreme points.37 We prove our main results by investigating the extreme points of M. Occasionally, however, we shall work with mechanisms or indirect utility functions, if this simplifies our arguments.
We say that (x, M, U) are associated if they are isomorphic in the sense of Theorem A.2.
Given Theorem A.2, the definitions of (positive) homothety and exhaustiveness straight-forwardly extend from X to M and U. For example, if x ∈X with associated M ∈M, then F(M) := F(x) = {H ∈F | H ∩ext M ̸= ∅}.
(6) We note a few corollaries of Theorem A.2.
Corollary A.3. Let x ∈X and M ∈M be associated. Then, menu(x) = cl ext M.
We have proven this claim as part of the proof of Theorem A.2. Note that ext M is closed if ext M is finite or d = 2.
The following characterization of indirect utility functions is analogous to the one by Rochet (1987, Proposition 2) for settings with transfers.
Corollary A.4. U ∈U if and only if the following conditions are satisfied: (1) U is sublinear (i.e., convex and HD1).
(2) U is continuous on its effective domain cone Θ = {z ∈Rn : U(z) < ∞}.
(3) For all θ ∈cone Θ, U(θ) ≥¯ a · θ.
(4) For all θ ∈cone Θ, ext ∂U(θ) ⊂A.
Proof. By the previous result, U ∈U is the support function of an extended menu M ∈M.
Conversely, every closed sublinear function Rd →R ∪{∞} is the support function of a closed convex set. (A sublinear function that is continuous on a closed effective domain is closed.) It remains to show that the remaining properties hold if and only if M ∈M. The effective domain of a sublinear function (in our case: cone Θ) and the recession cone of the associated closed convex set (in our case: Θ◦) are mutually polar cones (Hiriart-Urruty and Lemar´ echal, 1996, Proposition V.2.2.4). Continuity comes for free since cone Θ is polyhedral (Rockafellar, 1997, Theorem 10.2). It is easy to see that (3) holds if and only if ¯ a ∈M (Hiriart-Urruty 36Remark on the cited theorem: in general, the sum of two closed convex sets need not be closed, but it is always closed if the two sets have the same recession cone, which is here Θ◦.
37Extreme points are usually only defined for convex subsets of vector spaces, which M is not. However, we can embed M into a vector space by Theorem A.2, which justifies the use of the term “extreme point.” 31 v4 v1 v2 v3 Θ◦ cone Θ A M Figure 6. The extended menu M = conv{v2, v3} + Θ◦of a mechanism x that can be decomposed pointwise almost everywhere, i.e., up to payoff-equivalence, but not pointwise everywhere. Define x as follows: assign to each type in the interior of cone Θ their favorite allocation between v2 and v3 and two the types on the extremal rays of cone Θ the allocations v1 and v4. M can be decomposed by translating the vertex v3 horizontally; thus x can be decomposed up to payoff-equivalence, i.e., pointwise almost everywhere. However, x cannot be decomposed pointwise everywhere.
and Lemar´ echal, 1996, Proposition V.2.2.4). Finally, ext M ⊂A if and only if ext ∂U(θ) ⊂A for all θ ∈Θ follows from Corollary 8.2.5 in Rockafellar and Wets (2009).
□ We also note the following sanity check that almost everywhere equivalence indeed coincides with payoff-equivalence for (IC) and (IR) mechanisms. This justifies modeling the set X of payoff-equivalence classes of mechanisms in L1.
Corollary A.5. Let x and x′ be mechanisms that satisfy (IC) and (IR). Then, x and x′ are payoff-equivalent if and only if x = x′ almost everywhere.
Proof. If x and x′ are payoff-equivalent, then there exists an indirect utility function U ∈U such that x, x′ ∈∂U by Theorem A.2. Thus, x = x′ almost everywhere since the subdifferential of a convex function is almost everywhere a singleton. Conversely, suppose x = x′ almost everywhere. Let x ∈∂U and x′ ∈∂U ′. Then, ∇U = ∇U ′ almost everywhere. Thus, U = U ′ + c for c ∈R, and c = 0 because U and U ′ are sublinear. Thus, x and x′ are payoff-equivalent.
□ 32 Remark. We briefly comment on a subtle difference between extreme points of the set of payoff-equivalence classes of IC and IR mechanisms, i.e., ext X, versus extreme points of the set of IC and IR mechanisms themselves. For the former, a mechanism is an extreme point if it does not coincide with a convex combination of two other mechanisms up to payoff-equivalence, i.e., for almost every type. For the latter, a mechanism is an extreme point if it does not coincide with a convex combination of two other mechanisms for every type.
For an extremal equivalence class with associated indirect utility function U ∈ext U, every element of Q θ∈Θ ext(∂U(θ) ∩A) is an extreme point of the set of (IC) and (IR) mechanisms.
There can exist additional extreme points of the set of (IC) and (IR) mechanisms such that their payoff-equivalence classes are not extreme points of the set of payoff-equivalence classes X. These additional extreme points can only exist if the type space Θ is restricted and only if types on the boundary of cone Θ break ties to the boundary of A; see Figure 6 for an example. Since we assume that the prior distribution µ is absolutely continuous, these additional extreme points are irrelevant for optimality.
A.2. Topologies and Compactness. We now define topologies on the three sets, X, M, and U, discuss the relation between these topologies, and show that the three sets are compact under their respective topologies.
We equip the set X of payoff-equivalence classes of mechanisms with the L1-norm ||x|| = Z Θ ||x(θ)|| dθ.
(7) We equip the set U of indirect utility functions with the sup-norm ||U|| = sup θ∈cone Θ: ||θ||≤1 U(θ).
(8) We equip the set M of extended menus with the Hausdorff distance d(M, M ′) = inf {ε > 0 : M ⊆M ′ + εB and M ′ ⊆M + εB} , (9) where B = {z ∈Rd : ||z|| ≤1} is the unit ball in Rd.
Thus, (X, || · ||) and (U, || · ||) are normed spaces.38 We also have: Lemma A.6. (M, d) is a metric space and d(M, M ′) ≤d(conv ext M, conv ext M ′).
Proof. We have d(M, M ′) = inf {ε > 0 : M ⊆M ′ + εB and M ′ ⊆M + εB} = inf ε > 0 : conv ext M + Θ◦⊆conv ext M ′ + Θ◦+ εB and conv ext M ′ + Θ◦⊆conv ext M + Θ◦+ εB ≤inf ε > 0 : conv ext M ⊆conv ext M ′ + εB and conv ext M ′ ⊆conv ext M + εB = d(conv ext M, conv ext M ′), where the inequality is because Z1 ⊆Z2 implies Z1 + Z3 ⊆Z2 + Z3 for Z1, Z2, Z3 ⊂Rd.
It remains to show that (M, d) is a metric space. Since conv ext M, conv ext M ′ ⊆A, we have d(M, M ′) ≤d(conv ext M, conv ext M ′) < ∞. Thus, d is a metric on M since extended 38For (X, || · ||), recall that payoff-equivalent mechanisms are almost everywhere equal by Corollary A.5.
33 menus are closed and since the Hausdorff distance is an extended metric on the space of all closed subsets of Rn.
□ The topologies on U and M are equivalent and finer than the topology on X.
Lemma A.7. Consider sequences (xn)n∈N ⊂X, and (Mn)n∈N ⊂M, (Un)n∈N ⊂U such that (xn, Mn, Un) are associated for all n ∈N. Then, the following hold: (1) Mn →M if and only if Un →U.
(2) If Un →U, then xn →x.
Proof. Claim (1) is Theorem 6 in Salinetti and Wets (1979). For claim (2), let Dn ⊂cone Θ be the set of points where Un is differentiable and let D ⊂cone Θ be the set of points where U is differentiable. Let D∗= D ∩T n∈N Dn. Indirect utility functions are convex and therefore almost everywhere differentiable. Moreover, the countable union of nullsets is null; thus cone Θ \ D∗is null. Theorem VI.6.2.7 in Hiriart-Urruty and Lemar´ echal (1996) implies that ∇Un(θ) →∇U(θ) for all θ ∈cone Θ \ D∗. Moreover, by Theorem A.2, xn →x pointwise almost everywhere. The Dominated Convergence Theorem implies convergence in L1.
□ The following lemma is crucial to apply Bauer’s maximum theorem, Choquet’s theorem, and the Straszewicz-Klee theorem, and hence for the interpretation of our results about extreme points.
Lemma A.8. X, M, and U are compact and convex.
Proof. Convexity of X is immediate because (IC) and (IR) are linear constraints and because A is convex. By Theorem A.2, M and U are also convex.
For compactness, consider any sequence {Mn}n∈N ⊂M. By Blaschke’s selection theorem, {cl conv ext Mn}n∈N has a convergent subsequence {cl conv ext Mnk}k∈N with compact convex limit K ⊆A. Let M = K + Θ◦. It is readily verified that M ∈M. By Lemma A.6, the subsequence {Mnk}k∈N convergences to M ∈M. Thus, M is compact. By Lemma A.7, X and U are also compact.
□ A.3. Individual Rationality (IR). The following result is an analogue of the familiar observation in mechanism design with transfers that if IR holds for “the lowest type,” then IR holds for every type. In our setting, however, a “lowest type” need not exist and is instead a type ¯ θ ∈Θ who likes the veto allocation most, i.e., ¯ a ∈arg maxa∈A a · ¯ θ.
Lemma A.9. Suppose x = λx′ + (1 −λ)x′′ almost everywhere, where x, x′, x′′ are (IC) mechanisms and λ ∈(0, 1). If x satisfies (IR), then x′ and x′′ satisfy (IR).
Thus, the extreme points of the set of (IC) and (IR) mechanisms are simply the extreme points of the set of (IC) mechanisms that satisfy (IR).
Proof. Let Θ∗= {θ ∈Θ | ¯ a ∈arg max a∈A a · θ} be the set of types who like the veto allocation most. We have assumed in Section 3 that Θ∗ is non-empty.
Let f ∗= \ θ∈Θ∗ arg max a∈A a · θ.
34 Since A is a polytope, f ∗is a face of A. If x is an (IC) and (IR) mechanism, then x(Θ)∩f ∗̸= ∅.
Since f ∗is a face of A, if x = λx′ + (1 −λ)x′′ for (IC) mechanisms x′, x′′ and λ ∈(0, 1), then x′(Θ) ∩f ∗̸= ∅and x′′(Θ) ∩f ∗̸= ∅.
For the sake of contradiction, suppose x′ does not satisfy (IR). Then, ¯ a · θ > x′(θ) · θ for some θ ∈Θ. By (IC), x′(θ) · θ ≥a∗· θ for a∗∈x′(Θ) ∩f ∗. Thus, ¯ a · θ > a∗· θ, which contradicts the definition of f ∗. Thus, x′ satisfies (IR). Analogously, x′′ satisfies (IR).
□ Appendix B. Extreme Points of Finite-Menu Mechanisms and Deformations This appendix characterizes for any given IC mechanism with finite menu size the set of all IC mechanisms that make an inclusion-wise larger set of IC constraints binding. This set is important in our analysis: whenever an IC mechanism can be written as a convex combination of two other IC mechanisms, then these two mechanisms must make at least the same incentive constraints binding as the given mechanism.
We use this characterization to prove Theorem 4.1 and Theorem C.1. Moreover, we can use the characterization to generalize results by Manelli and Vincent (2007, Theorems 17, 19, 20, and 24) (MV) to arbitrary linear screening problems. In particular, we get an algebraic characterization of finite-menu extreme points (Theorem B.6). We discuss the exact relation to MV at the end of this section.
Throughout this section, we restrict attention to extended menus M ∈M of finite size, i.e., | ext M| < ∞. Let MFin ⊂M denote the set of all extended menus of finite size. These are polyhedra since they can be written as the convex hull of their extreme points plus the polar of the type space (which is a polyhedral cone). In light of Lemma A.9, we can ignore IR constraints. We make two closely connected definitions.
Definition B.1. The normal fan NM of an extended menu M ∈MFin is the collection {NCf} of the normal cones NCf = {θ ∈cone Θ | f ⊆arg max a∈M a · θ} to the faces f of M. The normal fan NM′ is coarser than the normal fan NM, denoted NM′ ≼NM, if each normal cone in NM′ is a union of some set of normal cones in NM.
Since the agent has linear utility, the set of each type’s most preferred alternatives is a face of M. The normal fan hence summarizes which types’ most preferred alternatives lie on which faces of the extended menu. The normal fan yields a polyhedral subdivision of the type space; the cells of maximal dimension have been called market segments by MV in the context of the monopoly problem.39 For the next definition, we define the set of facet-defining hyperplanes of an extended menu M ∈MFin, which requires some care when M is not d-dimensional. For each facet Fi of M, there is a unique outer normal vector ni ∈(aff M −a), where a ∈M is arbitrary, and a constant ci ∈R such that Fi = M ∩Hi and M ⊆Hi,−, where Hi = {z ∈Rd : z · ni = c} is the facet-defining hyperplane and Hi,−= {z ∈Rd : z · ni ≤ci} is the facet-defining halfspace.
39Subdivisions obtained from normal fans of polyhedra have appeared elsewhere in economic design as power diagrams (Frongillo and Kash, 2021; Kleiner et al., 2024) and as regular polyhedral complexes (Baldwin and Klemperer, 2019; Tran and Yu, 2019; Bedard and Goeree, 2023). They are also relevant in the context of the random expected utility model (Gul and Pesendorfer, 2006). See Doval et al. (2024) for another recent application.
35 Let HM be the union of the set of facet-defining hyperplanes of M with an arbitrary finite set of hyperplanes with corresponding halfspaces whose intersection is aff M. For brevity, we refer to HM as the set of facet-defining hyperplanes of M (although some of these define the improper face M).
Definition B.2. An extended menu M ′ ∈M is a deformation of M ∈MFin with HM = {H1, . . . , Hk} if there exist a deformation vector c′ = (c′ 1, . . . , c′ k) ∈Rk such that the following two conditions are satisfied: (1) M ′ = ∩k i=1H′ i,−, where H′ i,−= {z ∈Rd : z · ni ≤c′ i}.
(2) If ∩i∈IHi = {a} for I ⊆{1, . . . , k} and a ∈ext M, then there exists a′ ∈ext M ′ such that ∩i∈IH′ i = {a′}.
Let Def(M) ⊂M denote the set of deformations of M.
That is, (1) M ′ can be defined by translates of the facet-defining halfspaces of M, not all of which necessarily remain facet-defining, and (2) if some subset of the facet-defining hyperplanes of M defines a vertex of M, then the translated hyperplanes also define a vertex of M ′. See Figure 1 for an illustration, where the right-most facet-defining hyperplane of the menu is translated horizontally, yielding two deformations. (The left panel of Figure 7 in Appendix D.3 is another illustration). This definition of deformations is due to Castillo and Liu (2022, Definition 2.2), except here adapted to polyhedra rather than polytopes.
Remark. There is a bijection between deformations M ′ ∈Def(M) and deformation vectors c′ given by c′ i = max a∈M′ ni · a, ∀i = 1, . . . , k since every hyperplane H′ i in the definition of M ′ must support M ′ by condition (2). By condition (1), this bijection commutes with convex combinations.
Lemma B.3. Let x, x′ ∈X be finite menu mechanisms with associated extended menus M, M ′ ∈MFin. The following are equivalent: (1) IC(x) ⊆IC(x′).
(2) NM′ ≼NM.
(3) M ′ is a deformation of M.
(4) There exists a surjective map φ : ext M →ext M ′ such that for every edge40 ab of M there exists λab ∈R+ such that λab(a −b) = φ(a) −φ(b).
The lemma says that coarsening the normal fan is the geometric analogue of making inclusion-wise more incentive constraints binding. Deformations are exactly the operations on extended menus that coarsen the normal fan. The fourth condition is an equivalent formulation of deformations in terms of parallel edges and more readily reveals the algebraic nature of deformations.
Proof. For the proof, we will need the following basic observation about normal cones. For an extended menu M ∈MFin with HM = {H1, . . . , Hk} and a face f of M, let If = {1 ≤ i ≤k | f ⊆Hi} denote the set of facet-defining hyperplanes of M containing the face f. For every face f of M, we have: NCf = cone{ni}i∈If.
(10) In particular, dim NCf = d −dim f.
40One-dimensional face. ab = conv{a, b}.
36 We define NCθ := NCarg maxa∈M a·θ NC′ θ := NCarg maxa∈M′ a·θ.
NCθ and NC′ θ are the inclusion-wise smallest normal cones of M and M ′, respectively, to which θ belongs. By definition, NM = {NCθ}θ∈Θ and NM′ = {NC′ θ}θ∈Θ.
We also make the following preliminary observation: for all θ, ˜ θ ∈int cone Θ, (θ, ˜ θ) ∈IC(x) ⇐ ⇒arg max a∈M a · θ ⊇arg max a∈M′ a · θ ⇐ ⇒NCθ ⊆NC˜ θ (11) because menu(x) = ext M and menu(x′) = ext M ′ (Corollary A.3), a bounded face of polyhedron is the convex hull of some set of its extreme points, every type θ ∈int cone Θ is normal to a bounded face of M and M ′, and normal cones are dual to faces and, therefore, reverse the inclusion.
(1) = ⇒(2). By (10), if IC(x) ⊆IC(x′) and θ, ˜ θ ∈int cone Θ, then NCθ ⊆NC˜ θ implies NC′ θ ⊆NC′ ˜ θ. In particular, θ ∈NC˜ θ implies θ ∈NC′ ˜ θ. Thus, every cone in NM that meets int cone Θ is a subset of a cone in NM′. Every cone in NM that is contained in the boundary of cone Θ is also a subset of a cone in NM′ because it is a subset of a full-dimensional cone in NM, which meets int cone Θ. That the cones in NM are subsets of the cones in NM′ implies NM′ ≼NM (Lu and Robinson, 2008, Proposition 2).
(2) = ⇒(3). We first show condition (1) in the definition of a deformation, i.e., M ′ can be defined using translates of the facet-defining halfspaces of M. First, since every cone in NM contains the orthogonal complement of aff M −a, where a ∈M is arbitrary, the same must be true for the cones in NM′. Thus, aff M ′ must be contained in a translate of aff M and therefore the same normal vectors used to define aff M can be used to define aff M ′. Second, NM′ ≼NM implies that the cones in NM′ corresponding to the facets of M ′ are also cones in NM because these cones can only be written as the trivial union of themselves. Every such cone contains a unique normal vector in aff M −a, for arbitrary a ∈M. Thus, the same normal vectors used to define M can be used to define M ′, as desired.
We now show condition (2) in the definition of a deformation. Suppose ∩i∈IHi = {a} for I ⊆{1, . . . , k} and a ∈ext M. By (10), cone{ni}i∈I ⊆NC{a}. Since NC{a} is full-dimensional and NM′ ≼NM, there exists a′ ∈ext M ′ such that NC{a} ⊆NC{a′}. Thus, cone{ni}i∈I ⊆NC{a′}. Consequently, for all i ∈I, there exists c′ i = maxa∈M′ ni · a such that the hyperplane H′ i with normal ni and constant c′ i supports M ′ at a′. In particular, ∩i∈IH′ i = {a′}, as desired.
(3) = ⇒(4). It is immediate from condition (2) in the definition of deformations that there is a surjective map φ : ext M →ext M ′. Moreover, by condition (2), φ must map each edge e of M either to an edge e′ of M ′ that is parallel to e or to a vertex of M ′. This is because the hyperplanes of M defining e must intersect for M ′ in a translate of the line containing e.
(4) = ⇒ (1). Suppose M, M ′ ∈MFin have the properties stated in (4). Recall that menu(x) = ext M and menu(x′) = ext M ′ by Corollary A.3. To show that (θ, ˜ θ) ∈IC(x) implies (θ, ˜ θ) ∈IC(x′), it suffices to show that arg max a∈ext M′ a · θ = φ arg max a∈ext M a · θ (12) 37 for all θ ∈Θ.
Suppose ˜ a ∈arg maxa∈ext M a · θ. Fix any ˆ a ∈ext M. By the simplex algorithm, there exists a sequence (a0, a1, . . . , an−1, an) such that a0 = ˆ a, an = ˜ a, aiai−1 is an edge of M for all i = 1, . . . , n, and (ai −ai−1) · θ ≥0 for all i = 1, . . . , n. Condition (4) implies that (φ(ai) −φ(ai−1)) · θ ≥0 for all i = 1, . . . , n. Since ˆ a ∈ext M was arbitrary, φ(˜ a) · θ ≥φ(a) · θ for all a ∈A. That is, φ(˜ a) ∈arg maxa∈ext M′ a · θ.
Suppose ˜ a / ∈arg maxa∈ext M a · θ.
By the simplex algorithm, there exists a sequence (a0, a1, . . . , an−1, an) such that a0 = ˜ a, an ∈arg maxa∈ext M a · θ, aiai−1 is an edge of M for all i = 1, . . . , n, and (ai −ai−1) · θ > 0 for all i = 1, . . . , n. Condition (4) implies that either φ(˜ a) · θ < φ(an) · θ or φ(˜ a) = φ(an). In the first case, φ(˜ a) / ∈arg maxa∈ext M′ a · θ. In the second case, repeat the argument with an in place of ˜ a. Since | ext M| < ∞, either the procedure terminates and φ(˜ a) / ∈arg maxa∈ext M′ a · θ or | ext M ′| = 1, in which case (12) holds trivially.
□ We note (12) as a separate corollary for later use.
Corollary B.4. Suppose M ′ ∈Def(M). Then, there exists a surjective function φ : ext M → ext M ′ such that arg max a∈ext M′ a · θ = φ arg max a∈ext M a · θ (13) for all θ ∈Θ.
We can translate the definition of deformations into a polyhedral characterization of Def(M). For each vertex a ∈ext M, let Ia = {1 ≤i ≤l | a ∈Hi} denote the set of indices of facet-defining hyperplanes in HM = {H1, . . . , Hk} intersecting a. Under any feasible deformation and for each a ∈ext M, the hyperplanes in Ia still need to intersect in a single point φa ∈A. Thus, we have the following linear system with variables (φa)a∈ext M ∈Rd×| ext M| corresponding to the points in ext M ′ and variables c′ ∈Rk corresponding to the deformation vector of M ′: φa · ni = c′ i ∀a ∈ext M, ∀i ∈Ia (14) φa · ni ≤c′ i ∀a ∈ext M, ∀i ∈{1, . . . , k} \ Ia (15) φa · nH ≤cH ∀a ∈ext M, ∀H ∈F.
(16) Let us parse these (in)equalities. The inequalities in (16) capture the requirement that M ′ is a feasible extended menu, i.e., ext M ′ ⊂A. (Recall that F is the set of facet-defining hyperplanes of A.) The (in)equalities in (14) and (15) are jointly equivalent to condition (2) in the definition of a deformation. (Condition (1) is satisfied by construction: we use the facet-defining hyperplanes of M to define M ′.) (14) ensures that the facet-defining hyperplanes of M intersecting a ∈ext M still intersect in a single point φa ∈ext M under the deformation vector c′. (15) ensures that φa ∈ext M ′, i.e, the facet-defining halfspaces of M still contain φa under the deformation vector c′. In economic terms, recalling the equivalence between M ′ ∈Def(M) and the corresponding mechanisms x and x′ satisfying IC(x) ⊆IC(x′) (Lemma B.3), (14) and (15) are tantamount to IC(x) ⊆IC(x′). If none of the constraints in (15) are binding for M ′ (which is the case for M by definition of the index sets Ia), then IC(x) = IC(x′).
Lemma B.5. Def(M) is a polytope and a face of M. In particular, M ∈ext M if and only if M ∈ext Def(M).
38 Proof. Note that (14) to (16) define a polytope in Rd×| ext M| × Rk: (14) to (16) is a linear system with bounded solutions since A is bounded. The projection onto the second factor c′ ∈Rk is also a polytope. By construction, there is an affine bijection between the projected polytope and Def(M) given by the deformation vectors c′ ∈Rk. Thus, Def(M) is a polytope, i.e., the convex hull of finitely many extended menus.
To show that Def(M) is a face of M, first observe that if M ∈MFin, M ′, M ′′ ∈M and M = λM ′ + (1 −λ)M ′′ for some λ ∈(0, 1), then M ′, M ′′ ∈Def(M). This is because the normal fan of the Minkowski sum of polyhedra is finer than the normal fans of each summand.41 It is immediate that M ∈ext M if and only if M ∈ext Def(M).
To complete the proof that Def(M) is a face of M, consider any ˜ M ∈Def(M).
If ˜ M = λM ′ + (1 −λ)M ′′ for M ′, M ′′ ∈M, then M ′, M ′′ ∈Def( ˜ M) by the previous paragraph.
Observe that “deformation of” is a transitive relation, hence M ′, M ′′ ∈Def(M), as desired.
□ The polyhedral characterization of Def(M) immediately translates into an algebraic char-acterization of finite-menu extreme points: by Lemma B.5, M ∈ext M if and only if there is a non-zero direction (t, s) ∈Rd×| ext M| × Rk such that the two candidate solutions ((a ± ta)a∈ext M, (c ± s)) solve the linear system (14) to (16).
Using condition (4) in Lemma B.3, we can state an equivalent algebraic characterization of finite-menu extreme points that needs only minimal information about the underlying mechanism. For a mechanism x ∈X, let E = ( (a, b) ∈menu(x) × menu(x) ∃θ ∈Θ : {a, b} = arg max ˜ a∈menu(x) ˜ a · θ ) (17) denote the set of pairs (a, b) of menu items for which there exists a type whose favorite allocations are {a, b}. These are exactly the edges of the extended menu associated with x.
For an allocation a ∈menu(x), also define F(a) = {H ∈F | a ∈H}.
(18) Theorem B.6. Let x ∈X have finite menu size. Then x ∈ext X if and only if all solutions ((φa)a∈menu(x), (λab)(a,b)∈E) ∈Rd×| menu(x)| × R|E| + to λab(a −b) = φa −φb ∀(a, b) ∈E (19) φa · nH = cH ∀a ∈menu(x), H ∈F(a) (20) φa · nH ≤cH ∀a ∈menu(x), H / ∈F(a) (21) are the trivial solutions where {φa}a∈menu(x) = ext M and λab = 1 for all (a, b) ∈E.
Remark. If ab and a′b′ are not parallel for all (a, b), (a′, b′) ∈E, then the trivial solution is unique.
Proof. Let M ∈MFin be the extended menu associated with the finite-menu extreme point x ∈ext X. By Corollary A.3, menu(x) = ext M. By Lemma B.5, M ∈ext Def(M). An extreme point of a polytope in Euclidean space is uniquely determined from its incident facets, i.e., binding constraints. If φa = a for all a ∈ext M and c′ = c, then the constraints (15) are all slack by the definition of the index sets Ia. Thus, ((a)a∈ext M, c) must be the unique solution to (14), (15), (20) and (21). By Lemma B.3, there exist (λab)(a,b)∈E ∈R|E| such that 41An explicit reference for polyhedra is Maclagan and Sturmfels (2021, Equation 2.3.1).
39 ((φa)a∈ext M, (λab)(a,b)∈E) solve (19) if and only if there exists a permutation ξ : ext M →ext M and c′ ∈Rk such that ((φξ(a))a∈ext M, c′) solve (14) and (15).
□ B.1. Relation to MV. We summarize here, for readers of MV, how our results generalize their findings about the facial structure of IC mechanisms and their algebraic characterization of finite-menu extreme points to arbitrary linear screening problems.
Our Lemmas B.3 and B.5 generalize Theorems 17, 19, and 20 in MV. MV show for the multi-good monopoly problem that the decomposing summands of a finite-menu IC mechanism must have a coarser market segmentation (in our language: normal fan) than the mechanism itself (Theorem 17). In their Definition 18, MV then define the set of all IC mechanisms with a coarser market segmentation than a given IC mechanism that also satisfy an analogue of (20), i.e., have at least the same binding feasibility constraints as the given mechanism. This set is the analogue of our deformation polytope Def(M), modulo (20). MV show that the set is a face of the set of IC mechanisms (Theorem 19). We further show that the set is a polytope, which immediately gives us MV’s key technical result (Theorem 20): a finite-menu IC mechanism is an extreme point if and only if it is the singleton element of their set (i.e., Def(M) plus (20)).
Our polyhedral characterization (14) to (16) of Def(M) immediately translates into the algebraic characterization of finite-menu extreme points given in Theorem B.6, generalizing Theorem 24 in MV. In our result, (19) generalizes condition (13) in MV; (20) generalizes condition (14) in MV; (21) generalizes the condition “0 ≤za ≤1” in MV (which is feasibility for the monopoly problem); in our model, za = φa. Theorem B.6 amends a minor oversight in MV in that multiple solutions of (19) to (21) can correspond to the same extreme point x ∈ext X because there need not be a unique assignment of the variables (φa)a∈menu(x) to the menu items whenever the menu has parallel edges.
Appendix C. Extreme Points for One-Dimensional Type Spaces We deduce Theorem 6.1 from a general characterization of the extreme points for one-dimensional linear screening problems. We state the characterization in terms of extended menus. By Theorem A.2, we could equivalently state it in terms of menus of mechanisms.
The key concept in the characterization—a flexible chain—requires some notation to be defined. Let us first state the result, then define the concept, and then give the proof.
Theorem C.1. Let d = 2 and M ∈M. Then M ∈ext M if and only if (1) | ext M| ≤2 and M is exhaustive, or (2) 3 ≤| ext M| < ∞and M has no flexible chain.
Remark. The theorem is an extension of a result due to Mielczarek (1998, Theorem 3.1).
The result characterizes extremal convex bodies (ext M) contained in a given convex body in the plane (A). If cone Θ = R2 is unrestricted, then we can use Mielczarek’s Theorem.42 Otherwise, if cone Θ ̸= R2 is restricted, we have to make a minor modification to the result because we consider closed convex sets M ∈M with extreme points in A. In any case, the 42Specifically, in Mielczarek’s theorem, • condition 1◦is equivalent to condition (1) above; • if V (M) ̸= ∅, then conditions 2◦and (i) are equivalent to the absence of a flexible chain; • if V (M) = ∅, then conditions 2◦and (ii) are equivalent to the absence of a flexible chain.
Condition (iii) in Mielczarek’s theorem never applies if A (Q in the statement) is a polytope.
40 v1 v2 v3 v4 v5 v2 v3 v4 v1 β2 α2 β1 α1 β4 α4 α3 β3 Figure 7. An illustration of flexible chains and their connection to extreme points. Left: an extended menu M, depicted as the shaded area, with a flexible chain S = (v2, v3, v4, v5) and two deformations of M, depicted with dotted lines, that decompose M. Right: an extended menu M, depicted as the shaded area, with a chain S = (v1, v2, v3, v4) that is not flexible because it violates the symmetry condition (23) on the angles (αk)k and (βk)k. Intuitively, the start-and endpoints of a candidate deformation coincide only under the symmetry condition.
original presentation of the result and its proof are notationally tedious, so we have restated and shall reprove most of the result for the reader’s convenience.
To get a first sense of a flexible chain, recall Figure 1 (Section 4). This figure illustrates a non-extreme point that can be deformed by horizontally translating the right-most vertical edge in its menu. The two vertices of this edge form a flexible chain in the sense of Theorem C.1.
However, a menu may lack an edge that can be flexibly translated in both normal directions, yet the corresponding mechanism may still not be an extreme point. This is because multiple edges could potentially be translated jointly, which is the idea captured by a flexible chain.
The formal definition of a flexible chain requires some new notation.
For the following definitions, let d = 2 and fix an extended menu M ∈M of finite menu size | ext M| < ∞. Recall that M is a polyhedron that satisfies M = conv ext M + Θ◦. The vertices of any polyhedron in the plane can be ordered clockwise and adjacent vertices in the ordering are connected by an edge. If M ∈M is unbounded, i.e., cone Θ ̸= R2, we designate a placeholder ∗as the first and last vertex in the ordering (which can be thought of as a vertex at infinity).
We define four disjoint subsets V (M), I(M), B1(M), B2(M) ⊆ext M such that ext M = V (M) ∪I(M) ∪B1(M) ∪B2(M).
(22) V (M) = ext M ∩ext A. I(M) = ext M ∩int A. B1(M) is the set of vertices a ∈ext M ∩ (bndr A \ ext A) such that there is no other vertex b ∈ext M for which ab ⊂bndr A.
B2(M) = (bndr A \ ext A) \ B1(M) is the set of vertices a ∈ext M ∩(bndr A \ ext M) for which such a vertex b ∈ext M does exist.
41 Example C.2. We illustrate the definition of these subsets with several examples. In the left panel of Figure 7, v1 ∈V (M), v2 ∈B2(M), v3, v4 ∈B1(M), and v5 ∈I(M). In the right panel of Figure 7, all vertices are in B1(M). In Figure 5, v1 ∈V (M), v5 ∈I(M), and v4 ∈B1(M).
We define the following angles formed by the edges of M with the edges of the allocation polytope A. Let v ∈B1(M). Let u and w be the vertices preceding and succeeding v in the clock-wise ordering, respectively. Let ab be the edge of A on which v lies, where a preceeds b in the clock-wise ordering. Let αk be the measure of the angle ∠uva, and let βk be the measure of the angle ∠wvb; see the right panel in Figure 7 for an illustration.43 Definition C.3. A sequence S = (v1, . . . , vn) of vertices of M that are adjacent in the clock-wise ordering is a flexible chain if S ∩V (M) = ∅and one of the following holds: (1) v1, vn ∈I(M) ∪B2(M) ∪{∗}, and if n = 2, then v1vn ̸⊂bndr A; (2) S = ext M = B1(M), cone Θ = R2, n is even, and n Y k=1 sin αk = n Y k=1 sin βk.
(23) Example C.4. We illustrate the definition of a flexible chain with several examples. In the left panel of Figure 7, S = (v2, v3, v4, v5) forms a flexible chain. In the right panel of Figure 7, S = (v1, v2, v3, v4) does not form a flexible chain because the symmetry condition (23) is violated. In contrast, the vertices of a 45◦rotation of the allocation square would form a flexible chain. In Figure 5, (∗, v4, v5) forms a flexible chain. Indeed, the extended menu depicted there for the one-good monopoly problem has menu size 3. It is well-known that the corresponding mechanism cannot be an extreme point, i.e., the extended menu must have deformations that decompose it. This observation is generalized in Theorem C.1.
Proof of Theorem C.1. Suppose | ext M| ≤2. If ext M is a singleton, then M ∈ext M if and only if M is exhaustive. Suppose conv ext M is a line segment and M = λM ′ + (1 −λ)M ′′ for M ′, M ′′ ∈M and λ ∈(0, 1). Then, M ′ and M ′′ are homothetic to M because they must be deformations of M by Lemma B.5. Using Theorem 5.2, M ∈ext M if and only if M is exhaustive.
Thus, suppose | ext M| ≥3. We first show that if M ∈ext M, then | ext M| < ∞. Let U ∈U be the indirect utility function associated with M (i.e., the support function of M).
For the sake of contradiction, suppose | ext M| = ∞. Since A has only finitely many edges and on each edge of A there can be at most two vertices of M, | ext M ∩int A| = ∞. In particular, there must exist an open cone C ⊆cone Θ such that ext M ∩C ⊂int A and | ext M ∩C| = ∞. Let L be an open line segment such that cone L = C. Let γ : (0, 1) →L be a bijective isometry. Consider the convex function U ◦γ : (0, 1) →R, which completely determines U on C by 1-homogeneity.
Suppose there exist convex functions U1, U2 : (0, 1) →R such that U ◦γ = 1 2U1 + 1 2U2, U1(t) = U2(t) = U(t) for all t / ∈[ε, 1 −ε], where ε > 0 is sufficiently small, and the right-derivatives of U ◦γ, U1, and U2, respectively, are the same at ε, and the left-derivatives of U ◦γ, U1, and U2, respectively, are the same at 1 −ε. Then, U1 and U2 can be extended to sublinear functions on cone Θ such that U = 1 2U1 + 1 2U2 by first extending the functions to C 43That is, αk = cos−1 (u−v)·(a−v) ||u−v||||a−v|| and βk = cos−1 (w−v)·(b−v) ||w−v||||b−v|| .
42 by 1-homogeneity and then to cone Θ by setting U|cone Θ\C = U1|cone Θ\C = U2|cone Θ\C. If U1 and U2 are sufficiently close to U ◦γ, then their extensions are in U because ∂U(L) ⊂int A.
We now show that the convex functions U1, U2 : (0, 1) →R from the previous paragraph exist, contradicting that U ∈ext U. Consider the set G of convex functions g : [ε, 1 −ε] →R such that (1) g(ε) = (U ◦γ)(ε), (2) g(1 −ε) = (U ◦γ)(1 −ε), (3) g′ +(ε) = (U ◦γ)′ +(ε), where g′ + is the right-derivative, and (4) g′ −(1 −ε) = (U ◦γ)′ −(1 −ε), where g′ −is the left-derivative.
By combining a well-known result due to Blaschke and Pick (1916) about extremal convex functions on R and a result due to Winkler (1988) about the extreme points of convex sets obtained from a given convex set by imposing finitely many affine restrictions, one can show the the extreme points of G are piecewise-affine with at most three pieces. Thus, U / ∈ext U and M / ∈ext M since | ext M ∩cone(γ([ε, 1 −ε]))| can be made arbitrarily large by choosing ε > 0 small enough.
Suppose cone Θ ̸= R2 or ext M ̸= B1. By Theorem 4.1 and Lemma B.5, M / ∈ext M and | ext M| < ∞if and only if M has a deformation M ′ ∈Def(M) such that F(M) = F(M ′).
Thus, it suffices to show that M has a deformation M ′ ∈Def(M) such that F(M) = F(M ′) if and only if M has a flexible chain.
By Lemma B.5, M ′ ∈Def(M) if and only if the facet-defining hyperplanes (lines) of M ′ are parallel translates of the facet-defining hyperplanes of M and there is a surjective map φ : ext M →ext M ′. By taking a convex combination εM ′ + (1 −ε)M for ε > 0 sufficently small, we may assume that φ : ext M →ext M ′ is bijective.
For F(M) = F(M ′) to hold, (v, φ(v)) must lie on the same face of A. In particular, φ(a) = a for all a ∈V (M), φ(I(M)) = I(M ′), φ(B1(M)) = B1(M ′), and φ(B2(M)) = B2(M ′).
We observe that if v ∈B1(M) and v ̸= φ(v), then the two facet-defining hyperplanes of M intersecting in v must both be translated in M ′ for otherwise (v, φ(v)) cannot lie on a common edge of A.
Consider a deformation M ′ ∈Def(M) such that F(M) = F(M ′); we construct a flexible chain of M. Find a sequence S = (v1, . . . , vn) of vertices in ext M \ φ(ext M) that are adjacent in the clock-wise ordering and such that no other vertex in ext M \ φ(ext M) is adjacent to a vertex in S. S ∩V (M) = ∅follows since φ(a) = a for all a ∈V (M). If n = 2, then v1vn ̸⊂bndr A for otherwise the edge φ(v1)φ(vn) is not in bndr A, contradicting F(M) = F(M ′). If v1, vn ∈B1(M), then, by the previous paragraph, v1 and vn cannot be the first or last vertex in the sequence, contradicting the construction of S.
Conversely, suppose M ∈M has a flexible chain (v1, . . . , vn). We carry out the construction illustrated in Figure 7. Without loss of generality, we may assume v2, . . . , vn−1 ∈B1(M) for otherwise, (v1, . . . , vn) has a subsequence of adjacent vertices that is a flexible chain with the desired property. Let (H1, . . . , Hn−1) be the hyperplanes such that Hi defines the facet vivi+1 for all i = 1, . . . , n −1.
Suppose v1 ̸= ∗. Let H0 be the other hyperplane of M intersecting v1. Translate H1 by a sufficiently small amount, and let φ1 be the intersection of H′ 1 and H0. Since v1 ∈int A ∪B2, φ1 lies on the same face of A as v1. If v1 = ∗, translate H1 by a sufficiently small amount to obtain H′ 1.
Let φ2 be the intersection of H′ 1 with the edge of A on which v2 lies. (This intersection is non-empty as long as all translations are sufficiently small.) Let H′ 2 be the translate of H2 that intersects φ2.
43 Iterate the construction in the previous paragraphs to obtain a sequence of points (φ1, . . . , φn) and hyperplanes (H′ 1, . . . , H′ n−1). Since vn ∈int A ∪B2 ∪{∗}, the hyperplane Hn ̸= Hn−1 intersecting vn need not be translated to meet φn. Define M ′ as the polyhe-dron whose edges are defined by (H′ 1, . . . , H′ n−1) and by the facet-defining hyperplanes of M different from (H1, . . . , Hn−1). By construction, M ′ ∈Def(M) and F(M) = F(M ′).
It remains to consider the case where ext M = B1(M) and cone Θ = R2. We refer the reader to Lemmas 9, 17, and 18 in Mielczarek (1998) for the formal proof that M / ∈ext M if and only if the symmetry condition (23) holds. We illustrate the idea in the right panel of Figure 7: if (23) were to hold, then the dotted chain of line segments would have the same start- and endpoints, i.e., would become a deformation of the depicted extended menu.
□ Appendix D. Proofs This appendix gathers the proofs for the results in the main text in the order of appearance.
By Theorem A.2, we may prove all results either for the set of IC and IR mechanisms X, the set of extended menus M, or the set of indirect utility functions U.
D.1. Proofs for Section 4. We note the following observation.
Lemma D.1. Suppose x = λx′ +(1−λ)x′′ for x, x′, x′′ ∈X of finite menu size and λ ∈(0, 1).
Then, IC(x) = IC(x′) ∩IC(x′′) and F(x) = F(x′) ∩F(x′′).
Proof. Let M, M ′, M ′′ ∈M be the extended menus associated with x, x′, and x′′, respectively.
M = λM ′ + (1 −λ)M ′′ by Theorem A.2. For Z ⊂Rd, let Top(Z, θ) = arg maxa∈Z a · θ.
By Corollary A.3, menu(x) = ext M. Thus, (θ, θ′) ∈IC(x), i.e., Top(menu(x), θ′) ⊆ Top(menu(x), θ), if and only if Top(ext M, θ′) ⊆Top(ext M, θ).
We first show IC(x) ⊇IC(x′) ∩IC(x′′). Suppose (θ, θ′) ∈IC(x′) ∩IC(x′′). Then, Top(ext M ′, θ′) ⊆Top(ext M ′, θ) Top(ext M ′′, θ′) ⊆Top(ext M ′′, θ).
Thus, Top(λ ext M ′ + (1 −λ) ext M ′′, θ′) ⊆Top(λ ext M ′ + (1 −λ) ext M ′′, θ).
Since ext M ⊆λ ext M ′ + (1 −λ) ext M ′′, we conclude Top(ext M, θ′) ⊆Top(ext M, θ) or, equivalently, (θ, θ′) ∈IC(x).
We next show IC(x) ⊆IC(x′)∩IC(x′′). By interchanging the roles of x′ and x′′, it suffices to show that (θ, θ′) / ∈IC(x′) implies (θ, θ′) / ∈IC(x). Assume Top(ext M ′, θ′)\Top(ext M ′, θ) ̸= ∅, i.e., (θ, θ′) / ∈IC(x′). Then, Top(λ ext M ′ + (1 −λ) ext M ′′, θ′) \ Top(λ ext M ′ + (1 −λ) ext M ′′, θ) ̸= ∅.
Since conv ext M = conv(λ ext M ′ + (1 −λ) ext M ′′) and utility is linear, we conclude Top(ext M, θ′) \ Top(ext M, θ) ̸= ∅ or, equivalently, (θ, θ′) / ∈IC(x).
F(x) = F(x′) ∩F(x′′) is immediate. If one summand is bounded way from a hyperplane, then the the convex combination must also be bounded away from the hyperplane. Conversely, if both summands make allocations on the same hyperplane, then so does their convex combination.
□ 44 The following proof uses the polyhedral characterization of Def(M) given by (14), (15), and (16) in Appendix B. The proof idea is described right after the statement in the main text.
Proof of Theorem 4.1. The remark following Theorem 4.1 is immediate from Lemma D.1: for x ∈X, if another x′ ∈X satisfies IC(x) ⊆IC(x′) and F(x) ⊆F(x′), then x′′ = εx′ +(1−ε)x for 0 < ε < 1 satisfies IC(x) = IC(x′′) and F(x) = F(x′′).
Necessity is also immediate from Lemma D.1: if x / ∈ext X, then the summands in the decomposition make weakly more constraints binding.
For sufficiency, suppose IC(x) = IC(x′) for x, x′ ∈X of finite menu size. Let M, M ′ ∈M be the associated extended menus. By Lemma B.3, M and M ′ are mutual deformations. In particular, there is a bijection φ : ext M →ext M ′.
As an intermediate observation, we claim that F(x) = F(x′) implies, for all a ∈menu(x) and H ∈F, a ∈H if and only if φ(a) ∈H.
In words, each menu item of x makes the same feasibility constraints binding as the corresponding menu item in x′. We have H ∈F(M) = F(x) if and only if maxa∈ext M a · nH = cH, where nH is the normal vector and cH the right-hand side constant of the hyperplane H ∈F. Analogously, H ∈F(M ′) = F(x′) if and only if max a∈ext M′ a · nH = cH. The proof of the claim is completed using Corollary B.4, which gives arg max a∈ext M′ a · nH = φ(arg max a∈ext M a · nH).
(24) We complete the proof of Theorem 4.1 using the polyhedral characterization of Def(M) given by (14), (15), and (16). Let c and c′ denote the deformation vectors associated with M and M ′, respectively. By the previous paragraph, F(x) = F(x′) if and only if the variables (φa = φ(a))a∈ext A make the same constraints in (16) binding as the variables (a)a∈ext A.
IC(x) = IC(x′) if and only if the variables (c, (a)a∈ext M) and (c′, (φa)a∈ext M) both satisfy the constraints in (14) and make none of the constraints in (15) binding. (See the explanation of the constraints in Appendix B.) Thus, F(x) = F(x′) and IC(x) = IC(x′) if and only if (c, (a)a∈ext M) and (c′, (φa)a∈ext M) make the same constraints of Def(M) binding. The latter is equivalent to M, M ′ / ∈ext Def(M) because Def(M) is a polytope by Lemma B.5. Finally, M, M ′ / ∈ext Def(M) if and only if x, x′ / ∈ext X.
□ D.2. Proofs for Section 5. Recall that by Theorem A.2, the definitions of homothety and exhaustiveness translate straightforwardly to extended menus M ∈M, where F(M) ⊆F was defined to be the set of facet-defining hyperplanes of A intersected by ext M. Also recall that M = λM ′ + (1 −λ)M ′′ is a homothetic decomposition of M if λ ∈(0, 1) and M ′, M ′′ ∈M are homothetic to but distinct from M.
Lemma D.2. M ∈M is exhaustive if and only if M has no homothetic decomposition.
Proof. Suppose ext M = {a} is a singleton. If a / ∈ext A, then there exists a′ ∈A on the same faces of A as a. Thus, F(a′ + Θ◦) = F(M) and a′ + Θ◦is homothetic to M. Thus, M is not exhaustive. Conversely, if M is not exhaustive, then there exists M ′ homothetic to M, i.e., M ′ = t + λ(M + Θ◦) = t + λa + Θ◦, such that t + λa meets an inclusion-wise larger set of hyperplanes in F than a, which implies a / ∈ext A.
Suppose ext M = {a} is not a singleton. As an intermediate step, we will show that the set HC(M) = {(λ, t) ∈R+ × Rd | ext(λM + t) ⊂A} 45 of all (parameters of) homotheties of M is a polytope. HC(M) is bounded because A is bounded. Therefore, we show that HC(M) is the intersection of finitely many halfspaces. For this, let HC−(M, H) = {(λ, t) ∈R+ × Rd | ext(λM + t) ⊂H−}, where H−= {z ∈Rd : z · nH ≤cH} is the halfspace that contains A and is bounded by the facet-defining hyperplane H ∈F of A. Let HC(M, H) denote the associated hyperplane.
Equivalently, HC−(M, H) = (λ, t) ∈R × Rd λ max a∈ext M a · nH + t · nH ≤cH .
That is, HC−(M, F) is a halfspace in Rd+1 with normal (maxa∈ext M a · nH, nH). Thus, HC(M) = HC+ ∩ \ H∈F HC−(M, H) is a polytope, where HC+ = R+ × Rd.
We complete the proof by showing that M is exhaustive if and only if (λ, t) = (1, 0) ∈ ext HC(M). Note that (1, 0) does not lie on the boundary of HC+. Every other halfspace HC−(M, H) of HC(M) corresponds to a facet-defining hyperplane H of A. Thus, M is determined by its binding feasibility constraints F(M) up to homothety, i.e., exhaustive, if and only if (1, 0) lies on an inclusion-wise maximal set of facet-defining hyperplanes of HC(M).
The latter condition is what it means for a point to be an extreme point of a polytope.
□ Proof of Theorem 5.2. Immediate from Lemma D.2.
□ Proof of Theorem 5.3. By Lemma D.2, M ∈M is not exhaustive if and only if M has a homothetic decomposition, i.e., there exist M ′, M ′′ ∈M homothetic to M such that M = 1 2M ′ + 1 2M ′′.
Suppose ext M = {a} is a singleton. Then M has a homothetic decomposition if and only if a / ∈ext A. Thus, for the remainder of the proof, assume that ext M is not a singleton.
M has a homothetic decomposition if and only if one of the following holds: (1) There exists a point z ∈Rn and ε > 0 such that z + (1 + ε)(ext M −z) and z + (1 −ε)(ext M −z) are both subsets of A (dilation with center z).
(2) There exists a direction t ∈Rn \ {0} such that ext M + t and ext M −t are both subsets of A (translation).
The reason is that any homothety is itself either a dilation or translation.44 If (1) is true and a ∈H ∩ext M for some H ∈F(M), then z ∈H, for otherwise z + (1 + ε)(a −z) or z + (1 −ε)(a −z) is not in A. Thus, if (1) is true, T H∈F(M) H ̸= ∅.
Conversely, if T H∈F(M) H ̸= ∅, choose any z ∈T H∈F(M) H. For ε > 0 sufficiently small, z + (1 + ε)(ext M −z) and z + (1 −ε)(ext M −z) are both subsets of A. This is because ext M is uniformly bounded away from facet-defining hyperplanes H / ∈F(M) and because a ∈H ∈F(M) implies (z + (1 ± ε)(a −z)) ∈H by the definition of z, i.e., all facet-defining inequalities of A remain satisfied.
If (2) is true, then t is orthogonal to all the normals of the hyperplanes in F(M) for otherwise there is a point a ∈ext M ∩H, for some H ∈F(M), such that a + t / ∈H or 44Specifically, suppose M = 1 2M ′ + 1 2M ′′ and M ′ = z + (1 + ε)(M −z). Plugging in and rearranging for M ′′ yields M ′′ = z + (1 −ε)(M −z).
46 a−t / ∈H, which contradicts that ext M +t and ext M −t are subsets of A. Hence the spanning condition span{nH}H∈F(M) = Rd is violated. Conversely, if the spanning condition is violated, there is a direction t ∈Rd \ {0} such that t is orthogonal to all the facet normals in F(M).
As in the previous paragraph, ext M + t and ext M −t will still satisfy the facet-defining inequalities of A for ||t|| sufficiently small, i.e. ext M + t, ext M −t ⊂A.
The statement of the of Theorem 5.3 is the contraposition of what we have shown.
□ D.3. Proofs for Section 6. We use Theorem C.1 in Appendix C and the notation introduced for this result in the following proof.
Proof of Theorem 6.1. Let M ∈ext M be the extended menu associated with a mechanism x ∈ext X. Recall that menu(x) = ext M by Corollary A.3 (since d = 2); thus, we show | ext M| ≤|F|.
If cone Θ ̸= R2, then V (M) ̸= ∅for otherwise M ∈ext M has a flexible chain.
If cone Θ = R2 and V (M) = ∅, then M can only not have a flexible chain if ext M = B1(M).
In this case, | ext M| ≤|F|.Thus, we assume V (M) ̸= ∅going forward.
Consider any vertex v ∈V (M) such that the sequence of subsequent vertices S = (v1, . . . , vn) in the clockwise ordering of ext M satisfies S ∩V (M) = ∅and such that vn is adjacent to a vertex v′ ∈V (M). Since v, v′ ∈ext A, let (e1, . . . , ek) be the sequence of edges traversed when moving from v to v′ clockwise on the boundary of A. (If v = v′, then all edges are traversed.) We show that n ≤k −1. Since M ∈ext M, S does not contain a flexible chain. Thus, |(B2(M) ∪I(M)) ∩S| = 1. On every edge i = 2, . . . k −1, there lies at most one vertex in ext M, for otherwise |B2(M)| ≥2. Moreover, since v and v′ lie on e1 and ek, respectively, there can be at most one vertex in ext M \ {v, v′} on e1 ∪en. (This vertex would have to be in B2(M)). Thus, n ≤k −1.
By applying the previous argument to every v ∈V (M), we conclude that | ext M| ≤|F|.
□ Proof of Theorem 6.2. Immediate from Theorem 6.6 below.
□ Proof of Theorem 6.3. Let x ∈X be exhaustive and such that menu(x) is finite and in general position. Let M ∈M be the associated extended menu. By Corollary A.3, ext M = menu(x).
M = conv ext M + Θ◦is a polyhedron because ext M is finite and Θ◦is a polyhedral cone.
Since ext M is in general position, all proper bounded faces of M are simplices. Smilansky (1987, Theorem 5.1) shows that a polyhedron M of which every bounded face is a simplex cannot be represented as a convex combination of polyhedra with the same recession cone as M that are not homothetic to M. Therefore, M has no non-homothetic decomposition. By Lemma D.2, M has no homothetic decomposition because M is exhaustive. Thus M ∈ext M, and x ∈ext X by Theorem A.2.
□ We may define exhaustiveness for arbitrary subsets S of A: F(S) ⊆F are the facets of A intersected by S, and S is exhaustive if there is no S′ ⊂A positively homothetic to S such that F(S) ⊆F(S′). Theorem 5.3 applies as before. Recall that an extended menu M ∈M is exhaustive if ext M is exhaustive.
We use the following simple consequence of Theorem 5.3 in the proof of Theorem 6.4.
Corollary D.3. If S ⊂A is exhaustive, then there exists an exhaustive S′ ⊂S such that |F(S′)| ≤d + 1.
47 Proof. By Theorem 5.3, span{nH}H∈F(S) = Rd. Thus, there exists a subset F′ ⊂F(S) with |F′| = d such that span{nH}H∈F′ = Rd. Moreover, by Theorem 5.3, there must exist a hyperplane H′ ∈F(S) \ F′ such that T H∈F(S) H ∩H′ ̸= ∅. Select S′ ⊂S such that F(S′) = F′ ∪{F ′}. (Clearly, at most d + 1 points in S suffice.) Theorem 5.3 completes the proof.
□ Proof of Theorem 6.4. Let x ∈X be exhaustive with associated extended menu M ∈M and such that menu(x) = ext M is finite. We first construct a menu ˜ M ∈ext M of finite menu size that is arbitrarily close to M in the Hausdorff distance and satisfies | ext M| = | ext ˜ M|.
This suffices to show the denseness claim in the statement of Theorem 6.4 by Lemma A.7.
Select an inclusion-wise minimal subset V ⊆ext M such that V is exhaustive and ¯ a ∈V .
By Corollary D.3, |V | ≤d + 1. (If ¯ a ∈V , then |V | = 2 suffices.) If |V | ≤d, then V is trivially in general position (i.e., no more than d points lie on any hyperplane in Rd). Suppose |V | = d + 1. Then every vertex in V touches exactly one of the d + 1 facets in F(V ) by construction of V . Select an arbitrary vertex v ∈V and move v to a nearby point v′ in the same facet of A touched by v that is not in the affine hull of V \ {v} (which meets the facet of A touched by v in a (d −2)-dimensional convex set). Let W = V \ {v} ∪{v′}.
Now consider one-by-one v ∈ext M \ V . Perturb v to a point v′ ∈A arbitrarily close to v such that v does not lie in any hyperplane spanned by any subset of d points in W.
(This is possible since there are only finitely many such hyperplanes.) Update W = W ∪{v′} and V = V ∪{v}. Proceed iteratively until V = ext M. The resulting set of points W is in general position and exhaustive by construction.
Define ˜ M = conv W + Θ◦. By construction, ˜ M is a polyhedron in M. As long as all of the finitely many perturbations carried out are sufficiently small, W is in convex position, i.e., no point in W is in the convex hull of the other points, because ext M was in convex position. Moreover, for all v, v′ ∈W, v / ∈v′ + Θ◦because Θ◦is closed and the same holds for all v, v′ ∈ext M. Thus, ext ˜ M = W and ˜ M is exhaustive because W is exhaustive.
˜ M ∈ext M by Theorem 6.3 and | ext ˜ M| = |W| = | ext M| by construction, proving denseness.
For openness, every polytope in a sufficiently small Hausdorff-ball around a simplicial polytope conv ext M is simplicial since the vertices remain in general position (see e.g.
Gr¨ unbaum et al., 1967, Theorems 5.3.1 and 10.1.1). By Lemmas A.6 and A.7, the claim follows.
□ Remark. An alternative statement of Theorem 6.4 is that the set of extreme points of menu size k is relatively open and dense in the set of exhaustive mechanisms of menu size ≤k.
This is because the set of exhaustive extended menus of menu size k is relatively open and dense in the set of exhaustive extended menus of menu size ≤k.
Proof of Corollary 6.5. Take an arbitrary exhaustive extended menu M ∈M. Select a finite set of vertices V ⊆ext M, including ¯ a (if ¯ a exists) as well as points on the same facets of A as ext M, such that for every point of ext M there is a selected point in V at most ε > 0 away. By construction, V is an exhaustive set and ¯ a ∈V . Thus, ˜ M = conv V + Θ◦∈M is exhaustive, has finite menu size, and is arbitrarily close to M for ε sufficiently small. By Theorem 6.4, ˜ M is arbitrarily close to an element of ext M with finite menu size, which completes the proof.
□ 48 The proof of Theorem 6.6 proceeds with Baire-category type arguments, for which we need a few definitions: • exh M ⊂M is the set of exhaustive extended menus; • Ak ⊂exh M is the set of exhaustive extended menus M such that M = 1 2M ′ + 1 2M ′′ for M ′, M ′′ ∈M with d(M ′, M ′′) ≥1 k; • Bk ⊂exh M is the set of exhaustive extended menus M that have a bounded face f with diam(f) ≥1 k and outer unit normal vector nf ∈Θ such that d(nf, bndr cone Θ) ≥ 1/k (which is satisfied by convention if bndr cone Θ = ∅, i.e., cone Θ = Rd).45 We note that ext M = exh M \ S∞ k=1 Ak. Moreover, define exh Msc := exh M \ S∞ k=1 Bk.
Lemma D.4. Let x ∈X be a mechanism associated with an extended menu M ∈exh Msc.
Then, x : Θ →A is continuous on int cone Θ. In particular, menu(x) is uncountable whenever it is not a singleton.
Proof. For any M ∈exh Msc and θ ∈int cone Θ, arg maxa∈M θ · a is a singleton for otherwise the boundary of M would contain a line segment connecting two extreme points of M. In particular, x(θ) is uniquely determined by M on int cone Θ. Therefore, the associated indirect utility function U is differentiable on int cone Θ. By Rockafellar (1997, Corollary 25.5.1), U is continuously differentiable and therefore x = ∇U is continuous on int cone Θ.
□ Lemma D.5. Ak and Bk are closed subsets of exh M for all k ∈N.
Proof. Consider any convergent sequence {Mi}i∈N ⊂Ak with limit M ∈exh M. We show M ∈Ak. Selecting a subsequence, if necessary, we may assume that the associated sequences {M ′ i}i∈N ⊂M and {M ′′ i }i∈N ⊂M, where Mi = 1 2M ′ + 1 2M ′′, converge in M by Blaschke’s selection theorem and Lemma A.6. Let M ′ and M ′′ denote the respective limits. We have d(M ′, M ′′) ≥ 1 k since d(M ′ i, M ′′ i ) ≥ 1 k for all i ∈N. Moreover, M = 1 2M ′ + 1 2M ′′ since Mi = 1 2M ′ i + 1 2M ′′ i for all i ∈N and M is convex, so 1 2M ′ + 1 2M ′′ ∈M. Thus, M ∈Ak.
Consider any convergent sequence {Mi}i∈N ⊂Bk with limit M ∈exh M. We show M ∈Bk.
By definition, for each i ∈N, there exists a line segment Li ⊆bndr Mi of length ≥1 k with normal vector ni ∈Θ such that d(ni, bndr cone Θ) ≥1 k. Selecting a subsequence, if necessary, we may assume that the line segments {Li}i∈N and the normal vectors {ni}i∈N converge to limits L∗⊂A and n∗∈Θ, respectively, because Θ ⊆Sd−1 and A are compact. It is routine to verify that L∗⊆bndr M, L∗has length ≥1 k, n∗is normal to L∗on bndr M, and d(n∗, bndr cone Θ) ≥1 k. Thus, M ∈Bk.
□ Proof of Theorem 6.6. We show that ext M ∩exh Msc is a dense Gδ in exh M. This implies the statement by Lemma D.4.
The proof uses the Baire category theorem. For this, note that exh M is a compact metric space, hence a Baire space, because exh M is a closed subset of the compact metric space M (Lemmas A.6 and A.7). The set exh M is closed because every extended menu in a sufficiently small neighborhood of a non-exhaustive extended menu M ∈M intersects a weakly smaller set of facets of A than M and is hence also non-exhaustive by Theorem 5.3. Thus, it suffices 45The diameter of a set S ⊆Rn, denoted diam(S), is defined as: diam(S) = sup{∥a −b∥: a, b ∈S}.
49 to show that ext M and exh Msc are each a dense Gδ in exh M. For ext M, this follows immediately from Corollary 6.5 and Lemma D.5.
We complete the proof by showing that exh Msc is a dense Gδ in exh M. By Lemma D.5, exh Msc is a Gδ in exh M. To show denseness, consider the set exh M\Bk for some arbitrary k ∈N. By Lemma D.5, exh M\Bk is relatively open in exh M. Moreover, exh M\Bk is dense in exh M because every extended menu M ∈exh M can be approximated by a polyhedron in exh M whose bounded faces have diameter < 1 k. We have that exh Msc = T∞ k=1(exh M \ Bk) is a countable intersection of relatively open and dense sets in a Baire space. Thus, by the Baire category theorem, exh Msc is dense in exh M.
□ Proof of Corollary 6.7. Corollary 6.5 shows that the extreme points of X are dense in the set of exhaustive mechanisms. The Straszewicz-Klee theorem (Klee Jr, 1957, Theorem 2.1) implies that the exposed points of X are also dense in the set of exhaustive mechanisms. The Riesz representation theorem (Diestel and Uhl, 1977, Theorem IV.1) implies that, for every exposed point x ∈exp X, there exists an objective v and prior µ such that x is uniquely optimal.
□ D.4. Proofs for Section 7.
Proof of Lemma 7.1. It remains to show that in the linear delegation problem, the indecom-posability of an extended menu M ∈M is necessary for the non-existence of a non-homothetic decomposition. We show the converse. Assume that there exists an extended menu M ∈M that is decomposable; that is, there exist convex bodies K′, K′′ ⊂Rd, not homothetic to M, such that M = K′ + K′′.
We aim to construct from these summands K′ and K′′ a non-homothetic decomposition of M into extended menus. To achieve this, we will identify λ ∈(0, 1) and t ∈Rd such that the scaled and translated sets M ′ = 1 λ(K′ + t) and M ′′ = 1 1−λ(K′′ −t) are extended menus, i.e., subsets of the unit simplex A = ∆. This will complete the proof since M = λM ′ + (1 −λ)M ′′.
Since M ⊆A = ∆, M satisfies the following constraints: (1) Positivity: mina∈M ai ≥0 for all i ∈{1, . . . , d}; (2) Size: maxa∈M Pd i=1 ai ≤1.
We will now define t and λ such that the above constraints are binding for M ′. This ensures that the constraints are satisfied by M ′′ because they are satisfied by M and M is a convex combination of M ′ and M ′′. We set ti = −min a∈K′ ai.
This ensures min a∈K′+t ai = 0 for all i ∈{1, . . . , d}; hence M ′ satisfies the positivity constraint with equality, irrespective of our choice of λ.
Next, for any convex body K ⊂Rd, define: |K|∆= max a∈K d X i=1 ai − d X i=1 min a∈K ai.
Note that | · |∆commutes with positive scalar multiplication and Minkowski addition; that is, |αK|∆= α|K|∆and |K1 + K2|∆= |K1|∆+ |K2|∆.
Set λ = |K′|∆.
50 Since K′ and K′′ are not singletons (otherwise, the decomposition would be homothetic), we have |K′|∆, |K′′|∆> 0, hence λ > 0. Since |M|∆≤1, we have |K′|∆< 1 and |K′′|∆< 1, hence λ ∈(0, 1).
We can now compute max a∈M′ d X i=1 ai = |M ′|∆= 1 λ|K′ + t|∆= 1 λ|K′|∆= 1 and hence M ′ satisfies the size constraint with equality. Hence, M ′′ ⊆A by the earlier argument, which completes the proof.
□ D.5. Proofs for Section 8.
D.5.1. Undominated Mechanisms. We begin by establishing an important result for the proofs of Theorems 8.5 and 8.7, namely that uniquely optimal mechanisms are dense in the undominated extreme points when considering two mechanisms as being “close” when they are “close” with respect to the induced principal’s utility functions. Theorem 8.2 will be proved along the way.
To state the result, let V = {θ 7→x(θ) · v(θ) | x ∈X} denote the set of the principal’s utility functions induced by the set of (IC) and (IR) mechanisms.
This set of functions Θ →R is convex and L1-compact because it is a continuous image of the compact convex set X.
We say that a principal utility function V ∈V is undominated if there exists an undominated mechanism x ∈X such that V (θ) = x(θ) · v(θ).
We also define the following subsets of V: • und V ⊂V is the set of undominated principal utility functions; • undV ⊂und V is the set of undominated principal utility functions that are strictly suboptimal for every probability density f ∈L∞(Θ) that is uniformly bounded away from 0; • exp+ V ⊂V is the set of principal utility functions that are uniquely optimal for some probability density f ∈L∞(Θ) that is uniformly bounded away from 0. Note exp+ V ⊂und V ∩ext V.
As usual, we write ⟨V, f⟩= R Θ V (θ)f(θ) dθ.
Proposition D.6. exp+ V is dense in ext V ∩und V.
We proof the result in three steps. The argument for the first Lemma is inspired by the argument for Theorem 9 in Manelli and Vincent (2007); note the correction in Manelli and Vincent (2012).
Lemma D.7. undV ⊆cl conv(exp+ V).
Proof. Fix any V ∈undV. We show the claim by constructing a convergent sequence of points in V that are convex combinations of points in exp+ V with limit V .
For ε ≥0, let Fε = {f ∈L∞(Θ) | ε ≤f ≤1}.
51 Up to renormalization, these functions are essentially bounded probability densities that are uniformly bounded away from zero. By the Banach-Alaoglu theorem, Fε is weak-compact because it is a weak-closed subset of the dual unit ball.46 Recall that V ∈undV, i.e., V is strictly suboptimal for every density f ∈L∞(Θ) that is uniformly bounded away from 0. Thus, for every f ∈Fε, there exists Vf ∈V such that ⟨Vf, f⟩> ⟨V, f⟩.
By the continuity of the evaluation (see e.g. Aliprantis and Border, 2007, Corollary 6.40), for every f ∈Fε, there exists a weak-open neighborhood Of of f such that for all f ′ ∈Of, ⟨Vf, f ′⟩> ⟨V, f ′⟩.
Thus, {Of : f ∈Fε} is a weak-open cover of Fε.
By compactness, the open cover {Of : f ∈Fε} has a finite subcover {Om : m = 1, . . . , M}.
The functionals f ∈Fε that expose a point in V are norm-dense in Fε (see e.g. Lau (1976) and note that Fε has non-empty interior). Thus, for every m = 1, . . . , M, there exists f ′ ∈Om such that Vm := Vf′ ∈exp V.
Let G = {(⟨V1 −V, f⟩, . . . , ⟨Vm −V, f⟩) | f ∈Fε} ⊂RM.
The set G is • convex (because Fε is convex); • compact (because it is the continuous image of a weak-compact set); • and satisfies G ∩RM −= ∅(by construction of the open cover {Om : m = 1, . . . , M}), where RM −is the negative orthant.
By the Separating Hyperplane Theorem, there exists a vector α ∈RM + \ {0}, such that α · y > 0 for all y ∈G. Renormalize PM i=1 αi = 1.
Define ˜ Vε = M X i=1 αiVi.
Note ˜ Vε ∈V since V is convex. For all f ∈Fε, ⟨˜ Vε, f⟩−⟨V, f⟩= α · (⟨V1 −V, f⟩, . . . , ⟨Vm −V, f⟩) > 0.
Now consider a sequence εn →0 and the corresponding sequence of ˜ Vεn constructed above.
Since V is norm-compact, a subsequence of ( ˜ Vεn) converges to some ˜ V ∈V.
We show ˜ V = V , which proves the claim. Recall that V is undominated and suppose ˜ V ̸= V . Then there exists a set ˜ Θ ⊂Θ of non-zero (spherical) measure such that V (θ) > ˜ V (θ) for all θ ∈˜ Θ. Thus, any density f concentrated on ˜ Θ is such that ⟨V, f⟩> ⟨˜ V, f⟩.
By norm-norm continuity of the evaluation, there exists a strictly positive density f ′ and some ˜ Vεn for n large enough such that ⟨V, f⟩> ⟨˜ Vεn, f ′⟩, a contradiction.
□ 46Recall that by the Riesz representation theorem, every continuous linear functional on L1(Θ) can be represented by a function in L∞(Θ).
52 Proof of Theorem 8.2. Follows from the proof for Lemma D.7 with ε = 0.
□ We now extend Lemma D.7 to cover all undominated mechanisms.
Lemma D.8. und V ⊆cl conv(exp+ V).
Proof. Suppose not, i.e., V ∈und V \ cl conv(exp+ V). By Lemma D.7, V / ∈undV, i.e., V is optimal for some density f ∗∈L∞(Θ) that is uniformly bounded away from 0.
Since cl conv(exp+ V) is a closed subset of the norm-compact set V, it is norm-compact.
By the Hahn-Banach Separation Theorem, there exists f ∈L∞(Θ) such that ⟨V, f⟩> max V ′∈cl conv(exp+ V)⟨V ′, f⟩.
˜ f = εf + (1 −ε)f ∗is still uniformly bounded away from 0 for ε ∈(0, 1) small enough and, moreover, for ε ∈(0, 1) small enough ⟨V, ˜ f⟩> max V ′∈cl conv(exp+ V)⟨V ′, ˜ f⟩.
by norm-norm continuity of the evaluation and Berge’s maximum theorem (for the RHS). By the result of Lau (1976) used in Lemma D.7, there is another density ˆ f arbitrarily close to ˜ f and therefore also uniformly bounded away from ε that exposes a point ˆ V ∈V. Again by continuity and Berge’s maximum theorem, ⟨V, ˆ f⟩> max V ′∈cl conv(exp+ V)⟨V ′, ˆ f⟩.
By definition, ⟨ˆ V, ˆ f⟩> ⟨V, ˆ f⟩. Thus, the point ˆ V exposed by ˆ f cannot be in exp+ V, a contradiction.
□ We complete the proof of Proposition D.6.
Proof of Proposition D.6. The claim is a consequence of Milman’s theorem (see e.g. Klee Jr, 1957, Theorem 1.1.). The theorem implies that ext cl conv(exp+ V) ⊆cl exp+ V since cl conv(exp+ V) is compact and convex. In particular, by Lemma D.8, every undominated extreme point of V must be in cl conv(exp+ V). But since cl conv(exp+ V) is a convex subset of V, every undominated extreme point of V must also be in ext cl conv(exp+ V) and therefore arbitrarily close to a point in exp+ V.
□ D.5.2. Multi-Good Monopoly. We proceed with the multi-good monopoly problem.
To follow the standard terminology in mechanism design with transfers, we abuse language and refer to elements of [0, 1]m as types and allocations, and consider mechanisms and indirect utility functions as functions defined on [0, 1]m. In line with standard notation, we also write (x, t) ∈X to separate the “allocation component” of a mechanism from the “transfer component.” We use the following lemma about undominated mechanisms in the upcoming arguments.
Lemma D.9 (Manelli and Vincent, 2007, Lemma 11). Suppose (x′, t′) ∈X and (x, t) ∈X with indirect utility functions U ′ and U, respectively, are such that t′ ≥t almost everywhere.
Then, for all θ ∈[0, 1]m and λθ ∈[0, 1]m with λ > 1, (1) U ′(θ) > U(θ) = ⇒U ′(λθ) > U(λθ); (2) U ′(θ) ≥U(θ) = ⇒U ′(λθ) ≥U(λθ).
53 Proof of Lemma 8.3. Is is without loss of generality to consider only pricing functions with marginal prices in [0, 1] because types are in [0, 1]m.
Now consider a pricing function p with marginal prices in [δ, 1 −δ] for some δ > 0. Let (x, t) ∈X be the mechanism obtained from p, and let U be the associated indirect utility function. For the sake of contradiction, suppose (x′, t′) ∈X dominates (x, t), and let U ′ be the associated indirect utility function.
Since marginal prices are in [δ, 1 −δ], we have U ′(θ) ≥U(θ) = 0 for all θ ≤(δ, . . . , δ).
By Lemma D.9, we have U ′ ≥U. Thus, there is a type θ ∈[0, 1]m such that U ′(θ) > U(θ) (otherwise x and x′ are payoff-equivalent). By the continuity of indirect utility functions, we may assume θ ∈(0, 1)m.
Let λ1 > 1 be the largest scalar such that θ1 = λ1θ ∈[0, 1 −δ 2]m, and let i = 1, . . . , m be such that θ1 i = 1 −δ 2. Without loss of generality, suppose i = 1. By Lemma D.9, we have U ′(θ1) > U(θ1).
Consider the subspace H1 = {θ ∈[0, 1]m | θ1 = 1 −δ 2}. Up to an arbitrarily small translation of H1 in coordinate direction ±e1, we may assume by Fubini’s Theorem that x′(θ) · θ ≥x(θ) · θ for almost every θ ∈H1 (with respect to m −1-dimensional Lebesgue measure) because U ′ ≥U and t′ ≥t almost everywhere. For all θ ∈H1, we have xi(θ) = 1 since θi > 1 −δ and marginal prices are in [δ, 1 −δ]. Together with dominance, we have x′(θ) · (0, θ−1) ≥x(θ) · (0, θ−1) for almost every θ ∈H1.
Now let λ2 > 1 be the largest scalar such that θ2 = (θ1 1, λ2θ1 −1) ∈[0, 1 −δ 2]m, and let i = 2, . . . , m be such that θ2 i = 1 −δ 2. Without loss of generality, suppose i = 2. By the same arguments as for the proof of Lemma D.9, we have U ′(θ2) > U(θ2).
Iteratively proceed with this argument, constructing a sequence of affine subspaces (H1, . . . , Hm) and types (θ1, . . . , θm), where θi k = 1 −δ 2 for all i = 1, . . . , m and k ≤i, such that U ′(θi) > U(θi) for all i = 1, . . . , m and xi k = 1 for all i = 1, . . . , m and k ≤i.
Finally, U ′(θm) > U(θ)m implies U ′(θ) = x′(θ) · θ −t′(θ) > x(θ) · θ −t(θ) = U(θ) for all θ ∈B(θm) by continuity, where B(θm) is a sufficiently small ball around θm. We also have x(θ) = (1, . . . , 1) for all θ > (1 −δ, . . . , 1 −δ) since marginal prices are in [δ, 1 −δ]. Thus, t′(θ) < t(θ) for all θ ∈B(θm), a contradiction with dominance.
□ Proof of Corollary 8.4 . Take any pricing function p with marginal prices in [0, 1]. Then, for ε > δ > 0 small enough, the pricing function p′(a) = (1 −ε)p(a) + δa has marginal prices uniformly bounded away from 0 and 1. Moreover, the epigraphs epi p and epi p′ of p and p′, respectively, are arbitrarily close in the Hausdorff distance. Thus, the extended menus M = epi p + Θ◦and M ′ = epi p′ + Θ◦are arbitrarily close in the Hausdorff distance (Lemma A.6). By Lemma A.7, the associated mechanisms x and x′, respectively, are arbitrarily close in L1. By Lemma 8.3, x′ is undominated.
□ Proof of Theorem 8.5. The argument for why the undominated extreme points are dense in the set of (IC) and (IR) mechanisms when m ≥2 is analogous to the arguments in the proofs for Section 6 in Appendix D.3. By Corollary 8.4, for every (x, t) ∈X, find (x′, t′) ∈X arbitrarily close to (x, t) with marginal prices bounded away from 0 and 1. Then follow the construction for Corollary 6.5 and then the construction for Theorem 6.4. As long as all perturbations are small enough, the constructed extreme point still has marginal prices bounded away from 0 and 1 and is hence undominated.
54 We complete the the proof by showing that the mechanisms that are uniquely optimal for some type distribution are dense in the undominated mechanisms.
We first show that if (x, t), (x′, t′) ∈X are undominated and t = t′ almost everywhere, then x = x′ almost everywhere. It is easy to show (e.g., using Euler’s homogenous function theorem) that x −x′ is constant on almost every ray from the origin, i.e., HD0 up to tie-breaking. It therefore suffices to show that for every undominated mechanism (x, t) ∈X, limθ→0 x(θ) = 0 (independently of the choice of sequence). Let p denote the pricing function associated with x and assume, for the sake of contradiction, that limθ→0 x(θ) = a∗̸= 0. Then, p(a) = 0 for all a ≤a∗. Define a mechanism (x′, t′) ∈X by letting the agent buy from another pricing schedule p′(a) = ( p(a + a∗) if a + a∗∈[0, 1]m; maxa∈menu(x) p(a) + ε otherwise, where ε > 0. By construction, p′(0) = 0; thus, (x′, t′) is IC and IR. Since p′ is obtained from p by translation of the graph of p in direction −a∗with a new price maxa∈[0,1]m p(a) + ε for the grand bundle a = 1, almost every type either buys the same allocation as under p translated by −a∗or the grand bundle. Thus, t′ ≥t almost everywhere. For all sufficiently small ε > 0, a positive measure of types will buy the grand bundle. Thus, (x′, t′) dominates (x, t), a contradiction.
We next claim that ext V∩und V is dense in und V, where V is the set of IC transfer functions.
If (x, t) ∈ext X is undominated, then t ∈ext V. To see this, suppose t = 1 2t′ + 1 2t′′ / ∈ext V for t′, t′′ ∈V. Define ˜ x = 1 2x′ + 1 2x′′ ∈X, where x′, x′′ ∈X induce transfer t′ and t′′, respectively.
By definition, x and ˜ x both induce t. Thus, x = ˜ x by the previous paragraph, so x / ∈ext X.
The claim now follows because the undominated mechanisms in ext X are dense in the undominated mechanisms in X.
Fix any undominated mechanism (x, t) ∈X.
By Proposition D.6 and the previous paragraph, there exists a sequence of transfer functions (tn)n∈N ⊂exp+ V, each uniquely optimal for some type distribution µ, converging to t in L1. We have shown above that the associated sequence of allocation rules (xn)n∈N is uniquely determined. Since X is compact, up to taking a subsequence, (xn, tn)n∈N converges in L1 to some (x′, t) ∈X. But (x, t) is undominated, hence x = x′, as desired.
□ D.5.3. Linear Veto Bargaining. We proceed with the linear veto bargaining problem.
Proof of Lemma 8.6. We first show that the conditions given in the statement are necessary.
For this, fix any mechanism x ∈X. It is clear that ¯ a ∈menu(x) for otherwise x does not satisfy (IR) since there is a type ¯ θ for which ¯ a is their (unique) most preferred alternative in A. Next suppose menu(x) does not contain the principal’s (unique) favorite alternative a∗∈ext A. Obtain a new mechanism x′ ∈X by letting the agent choose from menu(x)∪{a∗}.
Thus, for all θ ∈Θ, either x′(θ) = x(θ) or x′(θ) = a∗. Since a∗∈ext A, there is a positive measure of types for which a∗is their most preferred allocation in A. Thus, x′ dominates x.
For sufficiency, let a∗∈ext A be the principal’s favorite alternative and suppose x, x′ ∈X are such that a∗∈menu(x), menu(x′) and x(θ) · ¯ v ≥x′(θ) · ¯ v for almost all θ ∈Θ. We show that x = x′ almost everywhere. We extend both mechanisms to cone Θ = Rd by letting each type chose their favorite allocation in menu(x) and menu(x′) (x and x′ are constant along 55 almost every ray from the origin). Let U and U ′ be the agent’s indirect utility functions associated with x and x′, respectively.
We claim that for all θ ∈cone Θ and λ ∈R, ϕθ(λ) := U(θ + λ¯ v) = U(θ) + Z λ 0 ⟨x(θ + z¯ v), ¯ v⟩dz and analogously for ϕ′ θ, x′, and U ′. Recall that x(θ) = ∂U(θ) and x′(θ) = ∂U ′(θ) (Theo-rem A.2). ϕθ(λ) is the restriction of a continuous convex function to a line, hence continuous and convex. It is easy to verify that ⟨x(θ + λ¯ v), ¯ v⟩as a function of λ is a subgradient of ϕθ(λ).
Hence the envelope formula follows (Rockafellar, 1997, Theorem 24.2).
By Fubini’s theorem, ϕθ(λ) −ϕ′ θ(λ) is non-decreasing for almost all θ ∈cone Θ because x(θ + λ¯ v) · ¯ v ≥x′(θ + λ¯ v) · ¯ v for almost all θ ∈cone Θ and all λ ∈R.
For all sufficiently large λ > 0, we have x(θ + λ¯ v) = x′(θ + λ¯ v) = a∗since a∗∈ext A is the principal’s, i.e., type ¯ v’s, (unique) favorite alternative in A and thus the favorite alternative of type θ + λ¯ v. Thus, ϕθ(λ) −ϕ′ θ(λ) = 0 for all sufficiently large λ > 0.
Similarly, for all sufficiently small λ < 0, we have x(θ + λ¯ v) = x′(θ + λ¯ v) = ¯ a since ¯ a is, by assumption, the principal’s (unique) least preferred alternative and the principal’s and agent’s preferences are sufficiently aligned.
Thus, for almost every θ ∈cone Θ and every λ ∈R, we have ϕθ(λ) = ϕ′ θ(λ) since ϕθ(λ) −ϕ′ θ(λ) is non-decreasing. Put differently, U = U ′ almost everywhere. By continuity, U = U ′. Consequently, x = x′ almost everywhere.
□ Proof of Theorem 8.7. The argument for statement (1) is immediate from Theorem 7.2 and Lemma 8.6. We proceed with statement (2).
The argument for why the undominated extreme points are dense in the undominated mechanisms when m ≥4 is completely analogous to the proofs of Corollary 6.5 and Theo-rem 6.4 when making sure that ¯ a, a∗∈V , where a∗is the principal’s favorite alternative and V is the set of vertices constructed in the proof of Theorem 6.4.
We complete the the proof by showing that the mechanisms that are uniquely optimal for some type distribution are dense in the undominated mechanisms.
The proof of Lemma 8.6 shows that if x, x′ ∈X are undominated and such that x(θ) · ¯ v = x(θ) · ¯ v for almost every θ ∈Θ, then x = x′ almost everywhere. Thus, an undominated principal utility function uniquely determines an undominated mechanism.
We claim that ext V∩und V is dense in und V. If x ∈ext X is undominated, then the induced principal utility function V ∈V is in ext V. To see this, suppose V = 1 2V ′ + 1 2V ′′ / ∈ext V for V ′, V ′′ ∈V. Define ˜ x = 1 2x′ + 1 2x′′ ∈X, where x′, x′′ ∈X induce V ′ and V ′′, respectively. By definition, x and ˜ x both induce v. Thus, x = ˜ x by the previous paragraph, so x / ∈ext X.
The claim now follows because the undominated mechanisms in ext X are dense in the undominated mechanisms in X.
Fix any undominated mechanism x ∈X. Let V ∈V be the associated principal utility function. By Proposition D.6 and the previous paragraph, there is a sequence (Vn)n∈N ⊂ exp+ V of uniquely optimal undominated principal utility functions converging to V . Let (xn)n∈N ⊂ext X be the sequence of mechanisms that is uniquely determined by (Vn)n∈N. (By definition, each mechanism in the sequence is uniquely optimal for some type distribution.) By compactness of X, up to taking a subsequence, xn →x′ ∈X. By continuity of the map that assigns to each mechanism in X a principal utility function in V, x′ must induce V since 56 Vn →V . Therefore, x′ is undominated. Thus, x = x′ almost everywhere, which completes the proof.
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7532 | https://opencurriculum.org/5409/balancing-chemical-equations/ | Balancing Chemical Equations
Joshua Siktar's files Science Chemistry The Basics of Chemistry
Article objectives
The objective of this article is to learn how to balance chemical equations and why it is important.
Introduction
We begin by introducing the Law of Conservation Matter, very important when studying chemical reactions.
The Law of Conservation of Matter states that no matter can be introduced nor removed from some medium under any circumstances.
In other words, all matter that is present at one time is present at all other times, and no additional matter will be found. In the context of chemical reactions, it is important to keep that in mind, because a Carbon atom cannot suddenly disappear during a reaction, even though some chemical equations imply this. For example, consider this single-replacement reaction where Titanium (II) Perchlorate and solid Chromium react to produce Chromium (III) Perchlorate and solid Titanium:
$$Ti(ClO_4)_2 + Cr \rightarrow Cr(ClO_4)_3 + Ti$$
If you look at the coefficients of the perchlorate ions ((ClO_4^{-})) on the two sides of the equation, you will notice that the reaction implies that an additional perchlorate ion appears out of nowhere during the reaction. However, this violates the Law of Conservation of Matter. When an equation does this, it is known to be unbalanced, because the number of each atom on the two sides of the equation are not the same. An equation that does not violate the Law of Conservation of Matter is called balanced.
Having a balanced equation is extremely important in many other topics in chemistry, and is one of its fundamental topics that is part of any first-year high school chemistry course.
Balancing Synthesis and Decomposition Reactions
The process of balancing synthesis and decomposition reactions is relatively straightforward since one side of the equation will usually only have one substance present. However, the process itself is best illustrated with examples. Pay careful attention to the rationale for each step; it will help you when practicing problems on your own.
Example 1: Balance this chemical equation:
$$Mn(CO_3)_2 \rightarrow MnO_2 + CO_2$$
Solution: The Manganese is already balanced. However, the left side has two Carbon and six Oxygen, while the right side has one Carbon and four Oxygen. The Carbon only appears once on each side of the equation, make those coefficients match by placing a "2" in front of Carbon Dioxide:
$$Mn(CO_3)_2 \rightarrow MnO_2 + 2CO_2$$
Both sides have two Carbon atoms and six Oxygen atoms; the number of Manganese atoms remained unchanged. Therefore the equation is now balanced.
As shown in the previous example, changing a single coefficient can change the quantity of atoms of multiple types on one side of the equation. Therefore, it is strongly advised to double-check the equations by doing an atom count. This is self-explanatory, and it means to check and make sure both sides of the equation have the same number of each atom, guaranteeing that the Law of Conservation of Matter is satisfied.
Also, remember that a decomposition reaction is a synthesis reaction in reverse, and so the two types of equations are balanced in the same way. The next example illustrates this.
Example 2: The reverse of the reaction in Example 1 is balanced in the same way. Therefore the equation
$$MnO_2 + CO_2 \rightarrow Mn(CO_3)_2$$
can be balanced this way, with the same procedure as in Example 1:
$$MnO_2 + 2CO_2 \rightarrow Mn(CO_3)_2$$
A new example of balancing a synthesis reaction's chemical equation will now be shown so the full process can be reemphasized.
Example 3: Balance this [very common] reaction:
$$H_3O^+ + OH^- \rightarrow H_2O$$
Solution: The left side of the equation has four Hydrogen atoms and two Oxygen atoms, while the right side of the equation has two Hydrogen atoms and one Oxygen atom. Therefore, the difference is a common factor of 2, which can easily be fixed by placing a “2” in front of (H_2O):
$$H_3O^+ + OH^- \rightarrow 2H_2O$$
Example 4: Balance the chemical equation for the reaction where Nitrogen trioxide decomposes into Nitrogen dioxide and Oxygen gas.
Solution: The equation itself is not given to us. Here are the chemical formulas for the chemicals involved:
Nitrogen trioxide: (NO_3)
Nitrogen dioxide: (NO_2)
Oxygen gas: (O_2)
Therefore, this is the chemical equation:
$$NO_3 \rightarrow NO_2 + O_2$$
Even though the Nitrogen atoms are already balanced, the oxygens cannot be balanced in the case where the coefficients of (NO_3) and (NO_2) are 1 because the number of Oxygen atoms cannot equal 3. Depending on the coefficient of (O_2), the number of Oxygen atoms on the right side of the equation could be 4, 6, 8, and so on. To fix this, multiply the entire equation by 2 in order to work with even numbers:
$$2NO_3 \rightarrow 2NO_2 + 2O_2$$
The left side of the equation has six Oxygen atoms, while the right side has eight. However, that can be fixed easily by changing the coefficient of Oxygen gas from "2" to "1":
$$2NO_3 \rightarrow 2NO_2 + O_2$$
Single Replacement Reactions
Single replacement reaction equations are typically relatively easy to balance because not many chemicals are involved. It is often helpful to treat a polyatomic ion like a regular element since it will not likely be broken up during the reaction.
Example 5: Balance this chemical equation:
$$Al(NO_3)_3 + Mg \rightarrow Mg(NO_3)_2 + Al$$
Solution: Initially the only elements that are not balanced are Nitrogen and Oxygen, and the ratio of these two atoms is consistent because they are found together in the Nitrate ion and nowhere else. However, it is not valid to balance it with
$$Al(NO_3)_3 + Mg \rightarrow \frac{3}{2}Mg(NO_3)_2 + Al$$
because atoms cannot be found in pieces. Instead, both coefficients must be changed to ensure the same number of Nitrate ions are on both sides of the equation. 3 is not evenly divisible by 2, so we need a least common multiple (to ensure integer coefficients), which is 6. Remember that this is not the coefficient on both sides of the equation, but the number of Nitrate ions on each side of the equation. Therefore, we add the following coefficients:
$$2Al(NO_3)_3 + Mg \rightarrow 3Mg(NO_3)_2 + Al$$
Now the Aluminum and Magnesium are not balanced, but that can be fixed easily by adding two more coefficients:
$$2Al(NO_3)_3 + 3Mg \rightarrow 3Mg(NO_3)_2 + 2Al$$
Here is one additional example that looks slightly more complicated, but is still a single replacement reaction.
Example 6: Balance this chemical equation:
$$MgTi(OH)_4 + O^{2-} \rightarrow MgTiO_2 + OH^-$$
Solution: The Magnesium and Titanium are already balanced, but the Oxygen and Hydrogen are not. Notice that there are 4 hydroxide ions on the left side of the equation and only one on the right side of the equation. Fix that by making the coefficient for the hydroxide ion a "4":
$$MgTi(OH)_4 + O^{2-} \rightarrow MgTiO_2 + 4OH^-$$
Similarly, if we ignore the Oxygen present in the hydroxide ions, there is one oxygen on the left side and two on the right side. Just make the coefficient for the oxide ion "2" so the number of oxygen atoms on both sides of the equation is the same:
$$MgTi(OH)_4 + 2O^{2-} \rightarrow MgTiO_2 + 4OH^-$$
Double Replacement Reactions
Double replacement reactions in general are more complicated than single replacement reactions, and that goes for balancing them as well. However, they are not much more difficult when you realize a double replacement reaction is very similar to a single replacement reaction.
Example 7a: In Example 6, the following equation was balanced:
$$MgTi(OH)_4 + O^{2-} \rightarrow MgTiO_2 + OH^-$$
However, the oxide and hydroxide ions are not generally found naturally without a cation. The cation, in theory, can be anything. For example, if the cation is Sodium, the chemical equation becomes
$$MgTi(OH)_4 + Na_2O \rightarrow MgTiO_2 + NaOH$$
Not only is the reaction more realistic, but it is a double replacement reaction instead of a single replacement reaction.
The next part of the example is a follow-up.
Example 7b: Balance the chemical equation written in Example 7a.
Solution: For reference, the equation will be written again here:
$$MgTi(OH)_4 + Na_2O \rightarrow MgTiO_2 + NaOH$$
Notice that Hydrogen only appears with Oxygen in the form of Hydroxide ions, so Hydroxide can be treated as an individual element when balancing, rather than a molecule. In fact, from here on out refer to the hydroxide ion as (X). Rewrite the chemical equation:
$$MgTiX_4 + Na_2O \rightarrow MgTiO_2 + NaX$$
Balance (X) by adding a "4" in front of (NaX):
$$MgTiX_4 + Na_2O \rightarrow MgTiO_2 + 4NaX$$
Now balance the sodium by adding a "2" in front of Sodium Oxide:
$$MgTiX_4 + 2Na_2O \rightarrow MgTiO_2 + 4NaX$$
The coefficient changes have also simultaneously balanced the Oxygen, so the equation is now fully balanced (do an atom count here). Lastly, substitute hydroxide back in for (X):
$$MgTi(OH)_4 + 2Na_2O \rightarrow MgTiO_2 + 4NaOH$$
The above chemical equation, once balanced, had the same coefficients as example 6's balanced equation. This is not a coincidence, because the addition of the extra cation is still in the same quantity before and after the reaction, by the Law of Conservation of Matter.
Here is one more example of balancing a double-replacement reaction.
Example 8: Balance this chemical equation:
$$AlPO_4 + Na_2CO_3 \rightarrow Al_2(CO_3)_3 + Na_3PO_4$$
Solution: Multiple types of atoms are not balanced. As usual, the strategy is to balance one at a time, beginning with the chemical that you will least likely need to adjust twice. Start with Aluminum Carbonate. Balance the Aluminum atoms on each side by changing the coefficient of the other chemical with Aluminum (least common multiples are not needed here):
$$2AlPO_4 + Na_2CO_3 \rightarrow Al_2(CO_3)_3 + Na_3PO_4$$
Now balance the other chemical with Carbonate:
$$2AlPO_4 + 3Na_2CO_3 \rightarrow Al_2(CO_3)_3 + Na_3PO_4$$
Lastly, both the Sodium and the Phosphate will be balanced by placing a "2" in front of Sodium Phosphate:
$$2AlPO_4 + 3Na_2CO_3 \rightarrow Al_2(CO_3)_3 + 2Na_3PO_4$$
As usual, do an atom count here to catch algebra mistakes.
The Special Case
Some chemical equations are balanced at the start. In other words, no coefficients need to be manipulated because they are all "1" in the balanced equation. Some instances of these will be illustrated here.
Example 9: When Nitric acid reacts with Lithium Hydroxide, Lithium Nitrate and water are produced. Here is the chemical equation:
$$HNO_3 + LiOH \rightarrow H_2O + LiNO_3$$
This equation is already balanced (to see this, it is helpful to treat (NO_3) as an individual atom). Doing an atom count verifies that all the coefficients should be "1".
Example 10: Copper (II) Fluoride and solid Manganese react to form solid Copper and Manganese (II) Fluoride. The chemical equation is below:
$$CuF_2 + Mn \rightarrow MnF_2 + Cu$$
Fluorine, Copper, and Manganese are all already balanced (do an atom count), so none of the coefficients need to be changed; they all remain "1".
Law of Conservation of Charge and Balancing Redox Reactions
Electrons are one form of matter on our Earth. Since electrons influence charge, the total charge of a group of substances will remain consistent even after a reaction because the number of protons and electrons must remain constant. That gives us the Law of Conservation of Charge:
The Law of Conservation of Charge states that before and after a reaction, the total charge of all substances involved must remain the same (even if it means lone electrons are left not attached to any atoms).
While this rule is very important in general, it can be very helpful in balancing chemical equations easier, particularly the chemical equations for redox reactions.
Example 11: Balance this redox reaction:
$$Mn^{2+} + Li \rightarrow Li^{+} + Mn$$
Solution: By the Law of Conservation of Charge, the charge on both sides of the equation must be the same. 2 is divisible by 1, so we can leave the Manganese alone and just change the coefficients of the Lithium. If the coefficients of (Li) and (Li^{+}) are both changed to “2,” the atoms remain balanced and both sides of the equation now have a total charge of +2.
$$Mn^{2+} + 2Li \rightarrow 2Li^{+} + Mn$$
Here is one other example:
Example 12: Consider this chemical equation that results from the products of a combination of two other reactions:
$$H^{+} + PO_4^{3-} \rightarrow H_3PO_4$$
Balance this equation.
Solution: Again the Law of Conservation of Charge is helpful. Once the equation is balanced, the charge of both sides must be 0 because there is only one chemical on the right side, and it has a neutral charge. Therefore the left side must also have 0 charge. If the Hydrogen and Phosphate ions are in a 1-1 ratio, (HPO_4^{2-}) will form, which has the wrong charge. However, lone Hydrogen ions are one of the reactants, so we can just add enough Hydrogen to get a neutral charge. In this case, we need two more, giving us (H_3PO_4), just as the right side of the equation says. Therefore, the coefficient of the Hydrogen ions is “3,” and everything else stays the same.
$$3H^{+} + PO_4^{3-} \rightarrow H_3PO_4$$
Reducing Coefficients
Sometimes missteps in balancing an equation can result in an equation with coefficients that all have a common multiple. If this is the case, the equation should be divided by the greatest common factor of all the coefficients in order to reduce them.
Example 13: Reduce the coefficients in this equation:
$$2H_2SO_4 + 4AgOH \rightarrow 4H_2O + 2Ag_2SO_4$$
Solution: There is a common factor of 2 in every coefficient. Therefore, we can divide the wjole equation by “2,” because no fractional coefficients will be introduced:
$$H_2SO_4 + 2AgOH \rightarrow 2H_2O + Ag_2SO_4$$
Doing an atom count verifies that this equation is now balanced.
Example 14: Do the coefficients need to be reduced in this equation?
$$3H_2Cr_2O_7 + 2Al(OH)_3 \rightarrow Al_2(Cr_2O_7)_3 + 6H_2O$$
Solution: The four coefficients (including the one for Aluminum Dichromate, (Al_2(Cr_2O_7)_3), which is "1") do not have a common factor, so the coefficients can not be reduced further.
Balancing Combustion Reactions
Although combustion reactions are a type of redox reaction, they are balanced quite differently from regular redox reactions.
Example 15: Balance this combustion reaction:
$$C_5H_{12} + O_2 \rightarrow CO_2 + H_2O$$
Solution: It is best to begin by balancing the Carbon, because it only appears once on each side of the equation. Add a "5" in front of Carbon Dioxide:
$$C_5H_{12} + O_2 \rightarrow 5CO_2 + H_2O$$
Now balance the Hydrogens similarly:
$$C_5H_{12} + O_2 \rightarrow 5CO_2 + 6H_2O$$
There are 16 oxygen on the right side, so adjust the (O_2) coefficient accordingly:
$$C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O$$
That was for a hydrocarbon. Let's do one for combusting a carbohydrate.
Example 16: Balance this [similar] combustion reaction:
$$C_5H_{10}O + O_2 \rightarrow CO_2 + H_2O$$
Solution: Begin by noting that this equation is slightly trickier to balance than the one in Example 14 because there is oxygen in all the reactants and products. Balance the Carbon and Hydrogen in the same way as before:
$$C_5H_{10}O + O_2 \rightarrow 5CO_2 + 5H_2O$$
There are currently 3 oxygen on the left side and 15 on the right side. Since (O_2) is present, and both sides having an odd number of oxygens (as opposed to one side having an odd number and one side having an even number, in which case we need to multiply the equation by 2), we can just adjust the coefficient of (O_2) and leave the others alone. If the coefficient is "7," there will be 14 oxygen atoms in elemental form, plus one from the hydrocarbon, making for a total of 15. Therefore the coefficient is "7," and the equation is balanced.
$$C_5H_{10}O + 7O_2 \rightarrow 5CO_2 + 5H_2O$$
In conclusion, there are many types of reactions you will come across, and many techniques to utilize when balancing them. This is a topic that requires practice to master, because there are no real formulas, only techniques that you must choose when to apply and then execute them.
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HomeSupportProduct FAQs
Xenon flash lamp
Q01. How are the maximum lamp input energy (per flash) and the maximum repetition rate of a lamp calculated?
Q02. What happens if a lamp is used at a main discharge voltage higher than its rating?
Q03. What happens if a lamp is used at a repetition rate that exceeds the maximum repetition rate?
Q04. How do the characteristics change depending on the arc length?
Q05. How do the characteristics change depending on the main discharge capacitance?
Q06. How do the characteristics change depending on the main discharge voltage?
Q07. What type of trigger signal input is needed from the signal source in order to operate a xenon flash lamp module or power supply for xenon flash lamp?
Q08. Is there any difference in life characteristics for each wavelength?
Q09. Is there any difference in life characteristics at each lamp input energy (energy per flash)?
Q10. What should be done to ensure the lamp is operated stably?
Q11. What device setup is used to measure the spectral distribution, emission pulse waveform, life characteristics and light output stability of xenon flash lamps?
Q12. What devices are needed to operate a lamp?
Q13. What is an important factor when selecting an optical fiber?
Q14. Are there any restrictions on the direction for installing a xenon flash lamp?
Q15. What should be checked before using a xenon flash lamp that has been stored for a long time?
Q16. Is it possible to change the cable length of the trigger socket?
Q01. How are the maximum lamp input energy (per flash) and the maximum repetition rate of a lamp calculated?
Refer to the following equations:
E ＝ 1/2 × Cm × Vm 2
P ＝ E × f
E: Lamp input energy (per flash) (J)
Cm: Main discharge capacitance (F)
Vm: Main discharge voltage (V)
P: Average lamp input (continuous) (W)
f: Repetition rate (Hz)
For example, when operating a 20 W xenon flash lamp at a main discharge voltage of 1000 V using a recommended power supply C13316-10 (main discharge capacitance: 1.0 μF (10-6 F) ), the maximum lamp input energy (per flash) is 0.5 J as calculated by the following equation:
E ＝ 1/2 × 10-6 (F) × 1000 (V)2 ＝ 0.5 (J)
In the above case, the maximum repetition rate of the 20 W xenon flash lamp is 40 Hz as calculated by the following equation:
f ＝ 20 (W) / 0.5 (J) ＝ 40 (Hz)
When selecting a lamp, the maximum lamp input energy and maximum repetition rate must be taken into account so that the maximum average lamp input (continuous) will not exceed the rating.
Q02. What happens if a lamp is used at a main discharge voltage higher than its rating?
The electrodes will wear down faster, and this will shorten the life of the xenon flash lamp.
While referring to the description in Q01, be sure to use the lamp under the operating conditions within the specified rating.
Q03. What happens if a lamp is used at a repetition rate that exceeds the maximum repetition rate?
The lamp will not emit light at the desired lamp input energy.
The electrodes will also be damaged by continuous lighting, and the life of the lamp will be shortened.
While referring to the description in Q01, be sure to use the lamp under the operating conditions within the specified rating.
Q04. How do the characteristics change depending on the arc length?
Xenon flash lamps with a long arc length provide higher light output with a wider flash pulse width (longer flash duration) and are ideal for applications that require a large irradiation area. On the other hand, xenon flash lamps with a short arc length emit higher brightness light and are used for applications that require higher accuracy.
Emission pulse waveform (Typ.)
Brightness characteristics (Typ.)
Q05. How do the characteristics change depending on the main discharge capacitance?
The larger the main discharge capacitance, the greater the maximum lamp input energy. This will produce a higher light output with a wider flash pulse width (longer flash duration).
Emission pulse waveform (Typ.)
Q06. How do the characteristics change depending on the main discharge voltage?
The higher the main discharge voltage, the greater the maximum lamp input energy and the higher the light output that can be obtained. Unlike the main discharge capacitance (Q05), the flash pulse width (flash duration) does not change.
Emission pulse waveform (Typ.)
Q07. What type of trigger signal input is needed from the signal source in order to operate a xenon flash lamp module or power supply for xenon flash lamp?
Input a rectangular wave signal, referring to the repetition rate and trigger signal on each specification page.
(Operation at 10 Hz or more is recommended for high stability)
In addition, use a signal source for trigger signal input that can output 15 to 30 mA.
Trigger signal waveform
Internal circuit diagram example of trigger signal input section
Q08. Is there any difference in life characteristics for each wavelength?
In general, the light output on the short wavelength side tends to decrease more rapidly than at longer wavelengths.
The lamp life is defined as the time when the light output at 190 nm to 1100 nm decreases to 50 % of the initial output level or the light output fluctuation exceeds the specified maximum value.
Life characteristics (2 W xenon flash lamp modules) (Typ.)
Q09. Is there any difference in life characteristics at each lamp input energy (energy per flash)?
In general, the larger the lamp input energy (energy per flash), the shorter the life.
Life characteristics (10 W xenon flash lamp) (Typ.)
NOTE:
Guaranteed lamp input energy for 10 W xenon flash lamps is 0.01 J to 0.1 J.
Q10. What should be done to ensure the lamp is operated stably?
The following solutions are recommended.
(1) Use the light in the center of the arc discharge.
The light output stability of a xenon flash lamp differs depending on the arc discharge measurement position. The closer to the center of the arc discharge, the more stable the light output.
Light output stability (Typ.)
(2) Do not use the light before warm-up.
Highly stable output light can be obtained from a xenon flash lamp by avoiding the warm-up time (time taken to reach stable operation) at the initial lighting.
Light output stability (Typ.)
Initial operation
After 5 seconds of operation
(3) Average the data.
Light output stability is improved by processing and averaging multiple acquired data.
Light output stability (Typ.)
Without data processing
With data processing
(average value of multiple acquired data)
Q11. What device setup is used to measure the spectral distribution, emission pulse waveform, life characteristics and light output stability of xenon flash lamps?
Typical measurement setups are as follows:
Measurement setup
○ Spectral distribution
○ Emission pulse waveform
○ Life characteristics and light output stability
Q12. What devices are needed to operate a lamp?
Prepare the following devices:
○ Operating a xenon flash lamp module
Required:
・DC power supply (input power)
・Pulse signal source such as a pulse generator (for external control of maximum repetition rate)
Optional:
・External control power supply (external control of main discharge voltage)
○ Operating a xenon flash lamp using a trigger socket and dedicated power supply
Required:
・DC power supply (input power)
Optional:
・Pulse signal source such as pulse generator (for external control of maximum repetition rate)
・External control power supply (external control of main discharge voltage)
Q13. What is an important factor when selecting an optical fiber?
Be sure to select an optical fiber that is resistant to UV light.
Q14. Are there any restrictions on the direction for installing a xenon flash lamp?
Installing a lamp with its light output window facing downward is not recommended.
Debris particles from the inside of the lamp may adhere to the light output window, causing a drop in the light output.
Q15. What should be checked before using a xenon flash lamp that has been stored for a long time?
Check the lead pins for any deterioration such as rust before checking the operation.
As long as the lamp is not stored in a harmful environment, there should be no problems with operating it unless deterioration such as rust is found on the lead pins.
Q16. Is it possible to change the cable length of the trigger socket?
The cable length of the trigger socket affects the flash pulse width (flash duration) and lamp input current.
When the cable length is increased, the flash pulse width becomes longer and the lamp input current tends to decrease, which might cause the lamp to fail to light up.
When the cable length is reduced, the flash pulse width becomes shorter and the lamp input current tends to increase, which might shorten the lamp life.
Therefore, changing the trigger socket cable length is not recommended.
Product FAQsSi photodiodePhotosensor amplifierSi strip detectorSi APDAPD moduleMPPC (SiPM)MPPC moduleInfrared sensorCCD image sensorCMOS image sensorInGaAs image sensorSi photodiode array with amplifierImage sensor driver circuit / Multichannel detector headCircuit for NMOS linear image sensorLight modulation photo ICPhoto IC for laser beam synchronous detectionPhoto IC for optical linkPhoto IC for optical switchPhoto IC for encoderIlluminance sensorColor sensorX-ray flat panel sensorPSDPSD signal processing circuitLCOS-SLMLEDMini-spectrometerXenon flash lampLIGHTNINGCURE LC-L5G: Linear irradiation type UV-LED unitsVUV IonizerFLAT EXCIMER
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7534 | https://activecalculus.org/single/sec-4-1-velocity-distance.html | Skip to main content
🔗
Section 4.1 Determining distance traveled from velocity
🔗
Motivating Questions
If we know the velocity of a moving body at every point in a given interval, can we determine the distance the object has traveled on the time interval?
How is the problem of finding distance traveled related to finding the area under a certain curve?
What does it mean to antidifferentiate a function and why is this process relevant to finding distance traveled?
If velocity is negative, how does this impact the problem of finding distance traveled?
🔗
🔗
In the first section of the text, we considered a moving object with known position at time , namely, a tennis ball tossed into the air with height (in feet) at time (in seconds) given by . We investigated the average velocity of the ball on an interval , computed by the difference quotient . We found that we could determine the instantaneous velocity of the ball at time by taking the derivative of the position function,
.
🔗
Thus, if its position function is differentiable, we can find the velocity of a moving object at any point in time.
🔗
From this study of position and velocity we have learned a great deal. We can use the derivative to find a function’s instantaneous rate of change at any point in the domain, to find where the function is increasing or decreasing, where it is concave up or concave down, and to locate relative extremes. The vast majority of the problems and applications we have considered have involved the situation where a particular function is known and we seek information that relies on knowing the function’s instantaneous rate of change. For all these tasks, we proceed from a function to its derivative, , and use the meaning of the derivative to help us answer important questions.
🔗
We have also encountered the reverse situation, where we know the derivative of a function, , and try to deduce information about . We will focus our attention in Chapter 4 on this problem: if we know the instantaneous rate of change of a function, can we find the function itself? We start with a more specific question: if we know the instantaneous velocity of an object moving along a straight line path, can we find its corresponding position function?
🔗
Preview Activity 4.1.1.
Suppose that a person is taking a walk along a long straight path and walks at a constant rate of 3 miles per hour.
On the left-hand axes provided in Figure 4.1.1, sketch a labeled graph of the velocity function .
Note that while the scale on the two sets of axes is the same, the units on the right-hand axes differ from those on the left. The right-hand axes will be used in question (d).
2. How far did the person travel during the two hours? How is this distance related to the area of a certain region under the graph of ?
3. Find an algebraic formula, , for the position of the person at time , assuming that . Explain your thinking.
4. On the right-hand axes provided in Figure 4.1.1, sketch a labeled graph of the position function .
5. For what values of is the position function increasing? Explain why this is the case using relevant information about the velocity function .
🔗
Subsection 4.1.1 Area under the graph of the velocity function
🔗
In Preview Activity 4.1.1, we learned that when the velocity of a moving object’s velocity is constant (and positive), the area under the velocity curve over an interval of time tells us the distance the object traveled.
🔗
🔗
The left-hand graph of Figure 4.1.2 shows the velocity of an object moving at 2 miles per hour over the time interval . The area of the shaded region under on is
mileshourhoursmile.
🔗
🔗
This result is simply the fact that distance equals rate times time, provided the rate is constant. Thus, if is constant on the interval , the distance traveled on is equal to the area given by
,
🔗
where is the change in over the interval. (Since the velocity is constant, we can use any value of on the interval , we simply chose , the value at the interval’s left endpoint.) For several examples where the velocity function is piecewise constant, see this applet1
gvsu.edu/s/9T
.
🔗
🔗
The situation is more complicated when the velocity function is not constant. But on relatively small intervals where does not vary much, we can use the area principle to estimate the distance traveled. The graph at right in Figure 4.1.2 shows a non-constant velocity function. On the interval , the velocity varies from down to . One estimate for the distance traveled is the area of the pictured rectangle,
mileshourhoursmiles.
🔗
Note that because is decreasing on , is an over-estimate of the actual distance traveled.
🔗
To estimate the area under this non-constant velocity function on a wider interval, say , one rectangle will not give a good approximation. Instead, we could use the six rectangles pictured in Figure 4.1.3, find the area of each rectangle, and add up the total. Obviously there are choices to make and issues to understand: How many rectangles should we use? Where should we evaluate the function to decide the rectangle’s height? What happens if the velocity is sometimes negative? Can we find the exact area under any non-constant curve?
🔗
We will study these questions and more in what follows; for now it suffices to observe that the simple idea of the area of a rectangle gives us a powerful tool for estimating distance traveled from a velocity function, as well as for estimating the area under an arbitrary curve. To explore the use of multiple rectangles to approximate area under a non-constant velocity function, see this applet2
gvsu.edu/s/9U
.
🔗
Activity 4.1.2.
Suppose that a person is walking in such a way that her velocity varies slightly according to the information given in Table 4.1.4 and graph given in Figure 4.1.5.
| | |
--- |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | |
Table 4.1.4. Velocity data for the person walking.
🔗
Using the grid, graph, and given data appropriately, estimate the distance traveled by the walker during the two hour interval from to . You should use time intervals of width , choosing a way to use the function consistently to determine the height of each rectangle in order to approximate distance traveled.
How could you get a better approximation of the distance traveled on ? Explain, and then find this new estimate.
Now suppose that you know that is given by . Remember that is the derivative of the walker’s position function, . Find a formula for so that .
Based on your work in (c), what is the value of ? What is the meaning of this quantity?
🔗
Subsection 4.1.2 Two approaches: area and antidifferentiation
🔗
When the velocity of a moving object is positive, the object’s position is always increasing. (We will soon consider situations where velocity is negative; for now, we focus on the situation where velocity is always positive.) We have established that whenever is constant on an interval, the exact distance traveled is the area under the velocity curve. When is not constant, we can estimate the total distance traveled by finding the areas of rectangles that approximate the area under the velocity curve.
🔗
Thus, we see that finding the area between a curve and the horizontal axis is an important exercise: besides being an interesting geometric question, if the curve gives the velocity of a moving object, the area under the curve tells us the exact distance traveled on an interval. We can estimate this area if we have a graph or a table of values for the velocity function.
🔗
In Activity 4.1.2, we encountered an alternate approach to finding the distance traveled. If is a formula for the instantaneous velocity of a moving object, then must be the derivative of the object’s position function, . If we can find a formula for from the formula for , we will know the position of the object at time , and the change in position over a particular time interval tells us the distance traveled on that interval.
🔗
For a simple example, consider the situation from Preview Activity 4.1.1, where a person is walking along a straight line with velocity function mph.
🔗
🔗
On the left-hand graph of the velocity function in Figure 4.1.6, we see the relationship between area and distance traveled,
mileshourhoursmiles.
🔗
In addition, we observe3
Here we are making the implicit assumption that ; we will discuss different possibilities for values of in subsequent study.
that if , then , so is the position function whose derivative is the given velocity function, . The respective locations of the person at times and are and , and therefore
miles.
🔗
This is the person’s change in position on , which is precisely the distance traveled. In this example there are profound ideas and connections that we will study throughout Chapter 4.
🔗
For now, observe that if we know a formula for a velocity function , it can be very helpful to find a function that satisfies . We say that is an antiderivative of . More generally, we have the following formal definition.
🔗
Definition 4.1.7.
If and are functions such that , we say that is an antiderivative of .
🔗
For example, if , is an antiderivative of , because . Note that we say “an” antiderivative of rather than “the” antiderivative of , because is also a function whose derivative is , and thus is another antiderivative of .
🔗
Activity 4.1.3.
A ball is tossed vertically in such a way that its velocity function is given by , where is measured in seconds and in feet per second. Assume that this function is valid for .
🔗
For what values of is the velocity of the ball positive? What does this tell you about the motion of the ball on this interval of time values?
Find an antiderivative, , of that satisfies .
Compute the value of . What is the meaning of the value you find?
Using the graph of provided in Figure 4.1.8, find the exact area of the region between the velocity curve and the -axis between and . What is the meaning of the value you find?
Answer the same questions as in (c) and (d) but instead using the interval .
What is the value of ? What does this result tell you about the flight of the ball? How is this value connected to the provided graph of ? Explain.
🔗
Subsection 4.1.3 When velocity is negative
🔗
The assumption that the velocity is positive on a given interval guarantees that the movement of an object is always in a single direction, and hence ensures that its change in position is the same as the distance it travels. As we saw in Activity 4.1.3, there are natural settings in which an object’s velocity is negative, and we would like to understand this scenario as well.
🔗
Consider a simple example where a woman goes for a walk on the beach along a stretch of very straight shoreline that runs east-west. We assume that her initial position is , and that her position function increases as she moves east from her starting location. For instance, mile represents one mile east of the start location, while tells us she is one mile west of where she began walking on the beach.
🔗
Now suppose she walks in the following manner. From the outset at , she walks due east at a constant rate of mph for 1.5 hours. After 1.5 hours, she stops abruptly and begins walking due west at a constant rate of mph and does so for 0.5 hours. Then, after another abrupt stop and start, she resumes walking at a constant rate of mph to the east for one more hour. What is the total distance she traveled on the time interval from to ? What is the total change in her position over that time?
🔗
🔗
These questions are possible to answer without calculus because the velocity is constant on each interval. From to , she traveled
miles per hour hours miles.
🔗
🔗
On to , the distance traveled is
miles per hour hours miles.
🔗
🔗
Finally, in the last hour she walked
miles per hour hour miles,
🔗
so the total distance she traveled is
miles.
🔗
🔗
Since the velocity for is , indicating motion in the westward direction, the woman first walked 4.5 miles east, then 2 miles west, followed by 3 more miles east. Thus, the total change in her position is
change in position miles.
🔗
We have been able to answer these questions fairly easily, and if we think about the problem graphically, we can generalize our solution to the more complicated setting when velocity is not constant, and possibly negative.
🔗
In Figure 4.1.9, we see how the distances we computed can be viewed as areas: comes from multiplying rate times time (), as do and . But while is an area (and is therefore positive), because the velocity function is negative for , this area has a negative sign associated with it. The negative area distinguishes between distance traveled and change in position.
🔗
🔗
The distance traveled is the sum of the areas,
miles.
🔗
🔗
But the change in position has to account for travel in the negative direction. An area above the -axis is considered positive because it represents distance traveled in the positive direction, while one below the -axis is viewed as negative because it represents travel in thenegative direction. Thus, the change in the woman’s position is
miles.
🔗
In other words, the woman walks 4.5 miles in the positive direction, followed by two miles in the negative direction, and then 3 more miles in the positive direction.
🔗
Negative velocity is also seen in the graph of the position function . Its slope is negative (specifically, ) on the interval because the velocity is on that interval. The negative slope shows the position function is decreasing because the woman is walking east, rather than west.
🔗
To summarize, we see that if velocity is sometimes negative, a moving object’s change in position is different from its distance traveled. If we compute separately the distance traveled on each interval where velocity is positive or negative, we can calculate either the total distance traveled or the total change in position. We assign a negative value to distances traveled in the negative direction when we calculate change in position, but a positive value when we calculate the total distance traveled.
🔗
Activity 4.1.4.
Suppose that an object moving along a straight line path has its velocity (in meters per second) at time (in seconds) given by the piecewise linear function whose graph is pictured at left in Figure 4.1.10. We view movement to the right as being in the positive direction (with positive velocity), while movement to the left is in the negative direction.
🔗
🔗
Suppose further that the object’s initial position at time is .
Determine the total distance traveled and the total change in position on the time interval . What is the object’s position at ?
On what time intervals is the moving object’s position function increasing? Why? On what intervals is the object’s position decreasing? Why?
What is the object’s position at ? How many total meters has it traveled to get to this point (including distance in both directions)? Is this different from the object’s total change in position on to ?
Find the exact position of the object at and use this data to sketch an accurate graph of on the axes provided at right in Figure 4.1.10. How can you use the provided information about to determine the concavity of on each relevant interval?
🔗
Subsection 4.1.4 Summary
🔗
If we know the velocity of a moving body at every point in a given interval and the velocity is positive throughout, we can estimate the object’s distance traveled and in some circumstances determine this value exactly.
In particular, when velocity is positive on an interval, we can find the total distance traveled by finding the area under the velocity curve and above the -axis on the given time interval. We may only be able to estimate this area, depending on the shape of the velocity curve.
An antiderivative of a function is a new function whose derivative is . That is, is an antiderivative of provided that . In the context of velocity and position, if we know a velocity function , an antiderivative of is a position function that satisfies . If is positive on a given interval, say , then the change in position, , measures the distance the moving object traveled on .
If its velocity is sometimes negative, a moving object is sometimes traveling in the opposite direction or backtracking. To determine distance traveled, we have to compute the distance separately on intervals where velocity is positive or negative, and account for the change in position on each such interval.
🔗
Exercises 4.1.5 Exercises
🔗
1. Estimating distance traveled from velocity data.
🔗
A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded:
| | | | | |
--- ---
| Time since brakes applied (sec) | 0 | 2 | 4 | 6 |
| Velocity (ft/s) | 90 | 46 | 17 | 0 |
🔗
Give lower and upper estimates (using all of the available data) for the distance the car traveled after the brakes were applied.
🔗
lower:
🔗
upper:
🔗
(for each, include help (units)4)
🔗
On a sketch of velocity against time, show the lower and upper estimates you found above..
🔗
2. Distance from a linear velocity function.
🔗
The velocity of a car is meters/second. Use a graph of to find the exact distance traveled by the car, in meters, from to seconds.
🔗
distance = (include help (units)5
/pg_files/helpFiles/Units.html
)
🔗
3. Change in position from a linear velocity function.
🔗
The velocity of a particle moving along the -axis is given by cm/sec. Use a graph of to find the exact change in position of the particle from time to seconds.
🔗
change in position = (include help (units)6
/pg_files/helpFiles/Units.html
)
🔗
4. Comparing distance traveled from velocity graphs.
🔗
Two cars start at the same time and travel in the same direction along a straight road. The figure below gives the velocity, (in km/hr), of each car as a function of time (in hr).
🔗
The velocity of car A is given by the solid, blue curve, and the velocity of car B by dashed, red curve.
🔗
(a)
🔗
🔗
Which car attains the larger maximum velocity?
🔗
(b)
🔗
🔗
Which stops first?
A
B
🔗
(c)
🔗
🔗
Which travels farther?
A
B
🔗
5. Finding average acceleration from velocity data.
🔗
Suppose that an accelerating car goes from 0 mph to 68.2 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: 1 mph is 22/15 feet per sec = 22/15 ft/s.) Find the average acceleration of the car over each of the first two seconds.
| | | | | | | |
--- --- ---
| | 0 | 1 | 2 | 3 | 4 | 5 |
| | 0.00 | 34.09 | 59.09 | 77.27 | 90.91 | 100.00 |
🔗
average acceleration over the first second = (include help (units)7
/pg_files/helpFiles/Units.html
)
🔗
average aceleration over the second second = (include help (units)8
/pg_files/helpFiles/Units.html
)
🔗
6. Change in position from a quadratic velocity function.
🔗
The velocity function is for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval .
🔗
displacement =
🔗
7.
Along the eastern shore of Lake Michigan from Lake Macatawa (near Holland) to Grand Haven, there is a bike path that runs almost directly north-south. For the purposes of this problem, assume the road is completely straight, and that the function tracks the position of the biker along this path in miles north of Pigeon Lake, which lies roughly halfway between the ends of the bike path.
🔗
Suppose that the biker’s velocity function is given by the graph in Figure 4.1.11 on the time interval (where is measured in hours), and that .
🔗
Approximately how far north of Pigeon Lake was the cyclist when she was the greatest distance away from Pigeon Lake? At what time did this occur?
What is the cyclist’s total change in position on the time interval ? At , was she north or south of Pigeon Lake?
What is the total distance the biker traveled on ? At the end of the ride, how close was she to the point at which she started?
Sketch an approximate graph of , the position function of the cyclist, on the interval . Label at least four important points on the graph of .
🔗
8.
A toy rocket is launched vertically from the ground on a day with no wind. The rocket’s vertical velocity at time (in seconds) is given by feet/sec.
At what time after the rocket is launched does the rocket’s velocity equal zero? Call this time value . What happens to the rocket at ?
Find the value of the total area enclosed by and the -axis on the interval . What does this area represent in terms of the physical setting of the problem?
Find an antiderivative of the function . That is, find a function such that .
Compute the value of . What does this number represent in terms of the physical setting of the problem?
Compute . What does this number tell you about the rocket’s flight?
🔗
9.
An object moving along a horizontal axis has its instantaneous velocity at time in seconds given by the function pictured in Figure 4.1.12, where is measured in feet/sec. Assume that the curves that make up the parts of the graph of are either portions of straight lines or portions of circles.
🔗
Determine the exact total distance the object traveled on .
What is the value and meaning of , where is the position function of the moving object?
On which time interval did the object travel the greatest distance: , , or ?
On which time interval(s) is the position function increasing? At which point(s) does achieve a relative maximum?
🔗
10.
Filters at a water treatment plant become dirtier over time and thus become less effective; they are replaced every 30 days. During one 30-day period, the rate at which pollution passes through the filters into a nearby lake (in units of particulate matter per day) is measured every 6 days and is given in the following table. The time is measured in days since the filters were replaced.
Table 4.1.13. Pollution data for the water filters.
| | | | | | | |
--- --- ---
| Day, | | | | | | |
| Rate of pollution in units per day, | | | | | | |
🔗
Plot the given data on a set of axes with time on the horizontal axis and the rate of pollution on the vertical axis.
Explain why the amount of pollution that entered the lake during this 30-day period would be given exactly by the area bounded by and the -axis on the time interval .
Estimate the total amount of pollution entering the lake during this 30-day period. Carefully explain how you determined your estimate.
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7535 | https://brainly.com/question/14844900 | [FREE] What is the role of the F-factor in conjugation? A. It allows auxotrophic bacterial cells to survive on - brainly.com
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What is the role of the F-factor in conjugation?
A. It allows auxotrophic bacterial cells to survive on minimal medium so that conjugation can occur.
B. It contains genes necessary for the formation of the pilus.
C. It contains genes that force recombination between the donor and recipient chromosomes.
D. It degrades the chromosome of the recipient cell after conjugation.
E. It contains genes necessary for replication of the donor's F plasmid.
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Final answer:
The F-factor contains genes important for the formation of the F pilus, necessary for bacterial conjugation, making option (b) the correct answer.
Explanation:
The role of the F-factor, or fertility factor, in conjugation is crucial for the process of DNA transfer between bacterial cells. Option (b) 'It contains genes necessary for the formation of the pilus' is correct. The F plasmid encodes the proteins that make up the F pilus (or sex pilus), which facilitates the contact between the donor (F+) and recipient (F-) cells. This contact is necessary for the conjugation process, during which a cytoplasmic bridge forms and allows the transfer of genetic material from one cell to another.
Importantly, the F factor is also responsible for the replication of the donor's F plasmid, which ensures that the genetic information can be passed on to the recipient cell, transforming it into a new F+ cell capable of future conjugation events.
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The F-factor, or F plasmid, is essential in bacterial conjugation as it contains the genes required for forming the conjugation pilus. This pilus facilitates the connection between donor (F+) and recipient (F-) cells, allowing DNA transfer. Therefore, option B is the correct answer.
Explanation
In bacterial conjugation, the F-factor, also known as the F plasmid or fertility factor, plays a crucial role in the transfer of genetic material between bacteria. The F-factor contains specific genes necessary for the formation of a structure called the conjugation pilus or F pilus.
The F pilus is a thin, hair-like projection that allows the donor cell (F+) to connect to the recipient cell (F-).
When the F pilus forms a connection between the two cells, a physical bridge is established, allowing for the transfer of DNA. The F-factor encodes the proteins that make up this pilus, making it essential for the initiation of the conjugation process.
During conjugation, a single strand of the F plasmid DNA is transferred from the donor cell to the recipient cell. Subsequently, both the donor and the recipient complete the DNA replication process, resulting in the recipient cell becoming F+ and capable of forming its own F pilus in the future.
Thus, the correct answer to the multiple-choice question is option B: 'It contains genes necessary for the formation of the pilus.' The F-factor is essential for establishing contact between bacterial cells, which is a crucial step in the exchange of genetic material.
Examples & Evidence
For instance, in E. coli, cells carrying the F plasmid (F+) can form a pilus to connect with F- cells, enabling the transfer of the plasmid and the potential acquisition of new traits, such as antibiotic resistance.
Scientific studies on bacterial conjugation reveal that the F plasmid harbors genes crucial for pilus formation, directly linking it to the process of genetic material transfer between bacteria.
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Community Answer Three pairs of bacterial cells with the given genotypes undergo conjugation. Place match the genotype of each cell after conjugation to its initial genotype. F+ × F− Hfr × F− F' × F− Answer Bank What is the role of the F‑factor in conjugation? It contains genes necessary for replication of the donor's F plasmid. It degrades the chromosome of the recipient cell after conjugation. It allows auxotrophic bacterial cells to survive on minimal medium so that conjugation can occur. It contains genes necessary for the formation of the pilus. It contains genes that force recombination between the donor and recipient chromosomes
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7537 | https://radiopaedia.org/articles/subsegmental-atelectasis?lang=us | Published Time: 2020-12-13T01:19:13.095Z
Subsegmental atelectasis | Radiology Reference Article | Radiopaedia.org
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Subsegmental atelectasis
Last revised by Andrew Murphy on 13 Dec 2020
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Citation, DOI, disclosures and article data
Citation:
Weerakkody Y, Murphy A, Iqbal S, et al. Subsegmental atelectasis. Reference article, Radiopaedia.org (Accessed on 24 Jul 2025)
DOI:
Permalink:
rID:
61933
Article created:
25 Jul 2018, Yuranga Weerakkody
Disclosures:
At the time the article was created Yuranga Weerakkody had no recorded disclosures.
View Yuranga Weerakkody's current disclosures
Last revised:
13 Dec 2020, Andrew Murphy
Disclosures:
At the time the article was last revised Andrew Murphy had no recorded disclosures.
View Andrew Murphy's current disclosures
Revisions:
15 times, by 10 contributors - see full revision history and disclosures
Systems:
Chest
Tags:
cases
Synonyms:
Subsegmental atelectatic changes
Subsegmental atelectases
More Cases Needed: This article has been tagged with "cases" because it needs some more cases to illustrate it. Read more...
Subsegmental atelectasis(plural: atelectases) is a descriptive term for the mildest form of lung atelectasis,involving less than one bronchopulmonary segment.
On this page:
Article:
Terminology
Pathology
Radiographic features
Differential diagnosis
Related articles
References
Terminology
The term subsegmental atelectasis includes any loss of lung volume so small that it does not cause indirect signs of volume loss (as might be seen with larger atelectases). A subtype of subsegmental atelectasis is linear atelectasis (also known as discoid or plate-like atelectasis, and historically as Fleischner lines on chest radiographs).
Subsegmental vs linear atelectasis
There is confusion about the use of the terms "subsegmental atelectasis" and "linear atelectasis" (and their synonyms). From an academic point of view, the term linear atelectasis is reserved for atelectasis which appears primarily in the lung bases and is secondary to hypoventilation. Conversely, subsegmental atelectasis includes both linear atelectases and all other forms of atelectasis that do not involve a whole bronchopulmonary segment.In other words, every linear atelectasis is subsegmental atelectasis, but not every subsegmental atelectasis is linear atelectasis.
Pathology
Etiology
Many subsegmental atelectases are secondary to airway obstruction of a small segment of the lung, either from benign (mucus plug, airway inflammation) or malignant causes (endobronchial tumor).
Linear atelectasis is usually due to a lack of adequate inspiration, and not due to any underlying airway obstruction. They are most commonly seen in post-surgical patients or those with abdominal pain, and may also be observed in the morbidly obese or immobile patients.
Radiographic features
Plain radiograph/CT
By definition, subsegmental atelectasis (regardless of its etiology)does not produce volume loss and subsequent shifting of mobile thoracic structures, and in most cases lacks clinical relevance and does not need to be reported.
Linear atelectases may result in minor linear densities of varying thickness usually parallel to the diaphragm, most commonly at the lung bases or less mobile regions of the lungs (e.g. lingula).
Other subsegmental atelectases present as linear or wedge-shaped densities and can affect any lung lobe. Unlike linear atelectases,subsegmental atelectases due to obstructive causes usually adopt a radial distribution rather than a horizontal one, as they are secondary to airway obstruction rather than parenchymal compression (as occurs in hypoventilation). Therefore, they follow the distribution of the bronchial and bronchiolar tree.
Differential diagnosis
small linear scars: can mimic this pattern and tends not to resolve on serial imaging
subpleural lines: commonly in asbestosis
References
Woodring JH, Reed JC. Types and mechanisms of pulmonary atelectasis. (1996) Journal of thoracic imaging. 11 (2): 92-108. doi:10.1097/00005382-199621000-00002 - Pubmed
Gurney JW. Atypical manifestations of pulmonary atelectasis. (1996) Journal of thoracic imaging. 11 (3): 165-75. Pubmed
Reed JC. Chest Radiology. (2017) ISBN: 9780323498319
W. Richard Webb, Charles B. Higgins. Thoracic Imaging. (2010) ISBN: 9781605479767
David M. Hansell, David A. Lynch, H. Page McAdams, Alexander A. Bankier. Imaging of Diseases of the Chest E-Book. (2009) ISBN: 9780723436676
Ozturk K, Soylu E, Topal U. Linear Atelectasis around the Hilum on Chest Radiography: A Novel Sign of Early Lung Cancer. (2018) Journal of clinical imaging science. 8: 27. doi:10.4103/jcis.JCIS_35_18 - Pubmed
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Articles:
Linear atelectasis
Small airways disease
Sickle cell disease (acute chest syndrome)
Lung atelectasis
Viral bronchiolitis
Cases:
Cocaine-induced pneumonitis
Neuroendocrine carcinoma of gallbladder - metastatic
Diffuse large B-cell lymphoma
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chronic unilateral airspace opacification (differential)
chronic bilateral airspace opacification (differential)
lobar consolidation[+][+]
right upper lobe consolidation
right middle lobe consolidation
right lower lobe consolidation
left upper lobe consolidation
left lower lobe consolidation
atelectasis
mechanism-based[+][+]
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morphology-based
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lobar consolidation[+][+]
right upper lobe consolidation
right middle lobe consolidation
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left upper lobe consolidation
left lower lobe consolidation
atelectasis
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passive (relaxation) atelectasis
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morphology-based
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lobar lung collapse[+][+]
right upper lobe collapse
right middle lobe collapse
right lower lobe collapse
left upper lobe collapse
left lower lobe collapse
chest x-ray in the exam setting[+][+]
adult chest x-ray in the exam setting
pediatric chest x-ray in the exam setting
neonatal chest x-ray in the exam setting
cardiomediastinal contour[+][+]
aortic knob
aortic nipple
cardiac silhouette
enlargement of the cardiac silhouette
right atrial enlargement
left atrial enlargement
right ventricular enlargement
left ventricular enlargement
cardiothoracic ratio
cavoatrial junction
moguls of the heart
vascular pedicle
chest radiograph zones[+][+]
apical zone
upper zone
mid zone
lower zone
tracheal air column[+][+]
tracheal bifurcation angle
fissures[+][+]
horizontal
oblique
accessory fissures
azygos fissure
superior accessory fissure
inferior accessory fissure
left horizontal fissure
vertical fissure line
normal chest x-ray appearance of the diaphragm[+][+]
flattening of the diaphragm
gastric bubble
nipple shadow[+][+]
nipple marker
lines and stripes[+][+]
anterior junction line
posterior junction line
right paratracheal stripe
left paratracheal stripe
posterior tracheal stripe/tracheo-esophageal stripe
posterior wall of bronchus intermedius
right paraspinal line
left paraspinal line
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retrosternal airspace
increased retrosternal airspace
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etiology
bird fancier's lung:pigeon fancier's lung
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talcosis
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non-small-cell lung cancer
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invasive mucinous adenocarcinoma (formerly mucinous BAC)
colloid adenocarcinoma
fetal adenocarcinoma
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pseudocavitation
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adenosquamous carcinoma
large cell carcinoma
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pleomorphic carcinoma of the lung
spindle cell carcinoma of the lung
giant cell carcinoma of the lung
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squamous cell carcinoma
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acinic cell carcinoma
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clear cell tumor of the lung
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7538 | https://community.ksde.gov/Portals/54/Documents/Standards/Standards_Review/Linked_Files/Numbers_and_Operations_Base_10_Progression_K-5.pdf | Progressions for the Common Core State Standards in Mathematics (draft) c ⃝The Common Core Standards Writing Team 6 March 2015 Suggested citation: Common Core Standards Writing Team. (2015, March 6).
Progressions for the Common Core State Stan-dards in Mathematics (draft).
Grades K–5, Number and Operations in Base Ten. Tucson, AZ: Institute for Mathematics and Education, University of Arizona.
For updates and more information about the Progres-sions, see
For discussion of the Progressions and related top-ics, see the Tools for the Common Core blog: http: //commoncoretools.me.
Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
Number and Operations in Base Ten, K–5 Overview Students’ work in the base-ten system is intertwined with their work on counting and cardinality, and with the meanings and properties of addition, subtraction, multiplication, and division.
Work in the base-ten system relies on these meanings and properties, but also contributes to deepening students’ understanding of them.
Position The base-ten system is a remarkably efficient and uni-form system for systematically representing all numbers. Using only the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, every number can be repre-sented as a string of digits, where each digit represents a value that depends on its place in the string. The relationship between values represented by the places in the base-ten system is the same for whole numbers and decimals: the value represented by each place is always 10 times the value represented by the place to its imme-diate right. In other words, moving one place to the left, the value of the place is multiplied by 10. In moving one place to the right, the value of the place is divided by 10. Because of this uniformity, standard algorithms for computations within the base-ten system for whole numbers extend to decimals.
Base-ten units Each place of a base-ten numeral represents a base-ten unit: ones, tens, tenths, hundreds, hundredths, etc. The digit in the place represents 0 to 9 of those units.
Because ten like units make a unit of the next highest value, only ten digits are needed to represent any quantity in base ten.
The basic unit is a one (represented by the rightmost place for whole numbers). In learning about whole numbers, children learn that ten ones com-pose a new kind of unit called a ten. They understand two-digit numbers as composed of tens and ones, and use this understanding in computations, decomposing 1 ten into 10 ones and composing a ten from 10 ones.
The power of the base-ten system is in repeated bundling by ten: 10 tens make a unit called a hundred. Repeating this process of creating new units by bundling in groups of ten creates units called Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 3 thousand, ten thousand, hundred thousand . . . In learning about decimals, children partition a one into 10 equal-sized smaller units, each of which is a tenth. Each base-ten unit can be understood in terms of any other base-ten unit. For example, one hundred can be viewed as a tenth of a thousand, 10 tens, 100 ones, or 1,000 tenths.
Algorithms• for operations in base ten draw on such relationships • From the Standards glossary: Computation algorithm.
A set of predefined steps applicable to a class of problems that gives the correct result in every case when the steps are carried out correctly. See also: computation strat-egy.
In mathematics, an algorithm is defined by its steps and not by the way those steps are recorded in writing. This progression gives examples of different recording methods and discusses their ad-vantages and disadvantages.
among the base-ten units.
Computations Standard algorithms• for base-ten computations with • The Standards do not specify a particular standard algorithm for each operation. This progression gives examples of algorithms that could serve as the standard algorithm and discusses their advantages and disadvantages.
the four operations rely on decomposing numbers written in base-ten notation into base-ten units. The properties of operations then allow any multi-digit computation to be reduced to a collection of single-digit computations. These single-digit computations some-times require the composition or decomposition of a base-ten unit.
Beginning in Kindergarten, the requisite abilities develop grad-ually over the grades.
Experience with addition and subtraction within 20 is a Grade 1 standard1.OA.6 and fluency is a Grade 2 1.OA.6Add and subtract within 20, demonstrating fluency for addi-tion and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 6 “ 8 2 4 “ 10 4 “ 14); decom-posing a number leading to a ten (e.g., 13 ´ 4 “ 13 ´ 3 ´ 1 “ 10´1 “ 9); using the relationship between addition and subtrac-tion (e.g., knowing that 8 4 “ 12, one knows 12 ´ 8 “ 4); and creating equivalent but easier or known sums (e.g., adding 6 7 by creating the known equivalent 6 6 1 “ 12 1 “ 13).
standard.2.OA.2 Computations within 20 that “cross 10,” such as 9 8 2.OA.2Fluently add and subtract within 20 using mental strate-gies.1 By end of Grade 2, know from memory all sums of two one-digit numbers.
or 13 ´ 6, are especially relevant to NBT because they afford the development of the Level 3 make-a-ten strategies for addition and subtraction described in the OA Progression. From the NBT per-spective, make-a-ten strategies are (implicitly) the first instances of composing or decomposing a base-ten unit. Such strategies are a foundation for understanding in Grade 1 that addition may require composing a ten1.NBT.4 and in Grade 2 that subtraction may involve 1.NBT.4Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a writ-ten method and explain the reasoning used.
Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
decomposing a ten.2.NBT.7 2.NBT.7Add and subtract within 1000, using concrete models or drawings and strategies based on place value, properties of op-erations, and/or the relationship between addition and subtrac-tion; relate the strategy to a written method.
Understand that in adding or subtracting three-digit numbers, one adds or sub-tracts hundreds and hundreds, tens and tens, ones and ones; and sometimes it is necessary to compose or decompose tens or hundreds.
Strategies and algorithms The Standards distinguish strategies• • From the Standards glossary: Computation strategy.
Purposeful manipula-tions that may be chosen for specific problems, may not have a fixed order, and may be aimed at converting one problem into another. See also: computation algorithm.
Examples of computation strategies are given in this progression and in the Operations and Algebraic Thinking Progression.
from algorithms. Work with computation begins with use of strate-gies and “efficient, accurate, and generalizable methods.” (See Grade 1 critical areas 1 and 2, Grade 2 critical area 2; Grade 4 critical area 1.) For each operation, the culmination of this work is signaled in the Standards by use of the term “standard algorithm.” Initially, students compute using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction (or multiplication and division). They relate their strategies to written methods and explain the reasoning used (for addition within 100 in Grade 1; for addition and subtraction within 1000 in Grade 2) or illustrate and explain their calculations with equations, rectangular arrays, and/or area models (for multiplication and division in Grade 4).
Students’ initial experiences with computation also include de-velopment, discussion, and use of “efficient, accurate, and general-izable methods.” So from the beginning, students see, discuss, and explain methods that can be generalized to all numbers represented in the base-ten system.
Initially, they may use written methods that include extra helping steps to record the underlying reasoning.
These helping step variations can be important initially for under-Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 4 standing. Over time, these methods can and should be abbreviated into shorter written methods compatible with fluent use of standard algorithms.
Students may also develop and discuss mental or written cal-culation methods that cannot be generalized to all numbers or are less efficient than other methods.
Mathematical practices The Standards for Mathematical Practice are central in supporting students’ progression from understanding and use of strategies to fluency with standard algorithms. The ini-tial focus in the Standards on understanding and explaining such calculations, with the support of visual models, affords opportunities for students to see mathematical structure as accessible, important, interesting, and useful.
Students learn to see a number as composed of its base-ten units (MP.7). They learn to use this structure and the properties of op-erations to reduce computing a multi-digit sum, difference, product, or quotient to a collection of single-digit computations in different base-ten units. (In some cases, the Standards refer to “multi-digit” operations rather than specifying numbers of digits. The intent is that sufficiently many digits should be used to reveal the standard algorithm for each operation in all its generality.) Repeated reason-ing (MP.8) that draws on the uniformity of the base-ten system is a part of this process. For example, in addition computations students generalize the strategy of making a ten to composing 1 base-ten unit of next-highest value from 10 like base-ten units.
Uniformity of the base-ten system ˜10 ˆ10 tens ˜10 ˆ10 ones ˜10 ˆ10 tenths hundredths For any base-ten unit, 10 copies compose 1 base-ten unit of next-highest value, e.g., 10 ones are 1 ten, 10 tens are 1 hundred, etc.
Students abstract quantities in a situation (MP.2) and use con-crete models, drawings, and diagrams (MP.4) to help conceptual-ize (MP.1), solve (MP.1, MP.3), and explain (MP.3) computational problems. They explain correspondences between different meth-ods (MP.1) and construct and critique arguments about why those methods work (MP.3). Drawings, diagrams, and numerical record-ings may raise questions related to precision (MP.6), e.g., does that 1 represent 1 one or 1 ten?, and to probe into the referents for sym-bols used (MP.2), e.g., does that 1 represent the number of apples in the problem?
Some methods may be advantageous in situations that require quick computation, but less so when uniformity is useful. Thus, com-paring methods offers opportunities to raise the topic of using ap-propriate tools strategically (MP.5). Comparing methods can help to illustrate the advantages of standard algorithms: standard al-gorithms are general methods that minimize the number of steps needed and, once, fluency is achieved, do not require new reason-ing.
Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 5 Kindergarten In Kindergarten, teachers help children lay the foundation for un-derstanding the base-ten system by drawing special attention to 10.
Children learn to view the whole numbers 11 through 19 as ten ones and some more ones. They decompose 10 into pairs such as 1 9, 28, 37 and find the number that makes 10 when added to a given number such as 3 (see the OA Progression for further discussion).
K.NBT.1Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (e.g., 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones.
Work with numbers from 11 to 19 to gain foundations for place valueK.NBT.1 Children use objects, math drawings,• and equations to • Math drawings are simple drawings that make essential math-ematical features and relationships salient while suppressing de-tails that are not relevant to the mathematical ideas.
describe, explore, and explain how the “teen numbers,” the counting numbers from 11 through 19, are ten ones and some more ones.
Children can count out a given teen number of objects, e.g., 12, and group the objects to see the ten ones and the two ones. It is also Number-bond diagram and equation 1 0 10 77 1 0 10 77 17 = 10 + 7 17 10 7 10 strip 5 strip Number Bond Drawing layered place value cards layered separated front: back: Figure 1: Decomposing 17 as 10 and 7 Equation Decompositions of teen numbers can be recorded with diagrams or equations.
helpful to structure the ten ones into patterns that can be seen as ten objects, such as two fives (see the OA Progression).
5- and 10-frames 1 0 10 77 1 0 10 77 17 = 10 + 7 17 10 7 10 strip 5 strip Number Bond Drawing layered place value cards layered separated front: Figure 1: Decomposing 17 as 10 and 7 Equation Children can place small objects into 10-frames to show the ten as two rows of five and the extra ones within the next 10-frame, or work with strips that show ten ones in a column.
A difficulty in the English-speaking world is that the words for teen numbers do not make their base-ten meanings evident. For example, “eleven” and “twelve” do not sound like “ten and one” and “ten and two.” The numbers “thirteen, fourteen, fifteen, . . . , nineteen” reverse the order of the ones and tens digits by saying the ones digit first. Also, “teen” must be interpreted as meaning “ten” and the prefixes “thir” and “fif” do not clearly say “three” and “five.” In contrast, the corresponding East Asian number words are “ten one, ten two, ten three,” and so on, fitting directly with the base-ten structure and drawing attention to the role of ten. Children could learn to say numbers in this East Asian way in addition to learning the standard English number names. Difficulties with number words beyond nineteen are discussed in the Grade 1 section.
The numerals 11, 12, 13, . . . , 19 need special attention for chil-dren to understand them. The first nine numerals 1, 2, 3, . . . , 9, and 0 are essentially arbitrary marks. These same marks are used again to represent larger numbers. Children need to learn the differences in the ways these marks are used. For example, initially, a numeral such as 16 looks like “one, six,” not “1 ten and 6 ones.” Layered place value cards can help children see the 0 “hiding” under the Place value cards 1 0 10 77 1 0 10 77 17 = 10 + 7 17 10 7 10 strip 5 strip Number Bond Drawing layered place value cards layered separated front: back: Figure 1: Decomposing 17 as 10 and 7 Equation Children can use layered place value cards to see the 10 “hiding” inside any teen number. Such decompositions can be connected to numbers represented with objects and math drawings. When any of the number arrangements is turned over, the one card is hidden under the tens card. Children can see this and that they need to move the ones dots above and on the right side of the tens card.
ones place and that the 1 in the tens place really is 10 (ten ones).
By working with teen numbers in this way in Kindergarten, stu-dents gain a foundation for viewing 10 ones as a new unit called a ten in Grade 1.
Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 6 Grade 1 In first grade, students learn to view ten ones as a unit called a ten.
The ability to compose and decompose this unit flexibly and to view the numbers 11 to 19 as composed of one ten and some ones allows development of efficient, general base-ten methods for addition and subtraction. Students see a two-digit numeral as representing some tens and they add and subtract using this understanding.
Extend the counting sequence and understand place value Through structured learning time, discussion, and practice students learn pat-terns in spoken number words and in written numerals, and how the two are related.
Part of a numeral list 91 101 111 92 102 112 93 103 113 94 104 114 95 105 115 96 106 116 97 107 117 98 108 118 99 109 119 100 110 120 In the classroom, a list of the numerals from 1 to 120 can be shown in columns of 10 to help highlight the base-ten structure, e.g., in the leftmost column, the 9s (indicating 9 tens) are lined up and the ones increase by 1 from 91 to 99. The numbers 101, . . . , 120 may be especially difficult for children to write.
Grade 1 students take the important step of viewing ten ones as a unit called a “ten.”1.NBT.2a They learn to view the numbers 11 1.NBT.2Understand that the two digits of a two-digit number rep-resent amounts of tens and ones. Understand the following as special cases: a 10 can be thought of as a bundle of ten ones—called a “ten.” through 19 as composed of 1 ten and some ones.1.NBT.2b They learn to b The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones.
view the decade numbers 10, . . . , 90, in written and in spoken form, as 1 ten, . . . , 9 tens.1.NBT.2c More generally, first graders learn that c The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
the two digits of a two-digit number represent amounts of tens and ones, e.g., 67 represents 6 tens and 7 ones. Saying 67 as “6 tens, 7 ones” as well as “sixty-seven” can help students focus on the tens and ones structure of written numerals.
The number words continue to require attention at first grade because of their irregularities. The decade words, “twenty,” “thirty,” “forty,” etc., must be understood as indicating 2 tens, 3 tens, 4 tens, etc.
Many decade number words sound much like teen number words. For example, “fourteen” and “forty” sound very similar, as do “fifteen” and “fifty,” and so on to “nineteen” and “ninety.” As discussed in the Kindergarten section, the number words from 13 to 19 give the number of ones before the number of tens. From 20 to 100, the number words switch to agreement with written numerals by giving the number of tens first. Because the decade words do not clearly indicate they mean a number of tens (“-ty” does mean tens but not clearly so) and because the number words “eleven” and “twelve” do not cue students that they mean “1 ten and 1” and “1 ten and 2,” children frequently make count errors such as “twenty-nine, twenty-ten, twenty-eleven, twenty-twelve.” Grade 1 students use their base-ten work to help them recognize that the digit in the tens place is more important for determining the size of a two-digit number.1.NBT.3 They use this understanding 1.NBT.3Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols ą, “, and ă.
to compare two two-digit numbers, indicating the result with the symbols ą, “, and ă. Correctly placing the ă and ą symbols is a challenge for early learners. Accuracy can improve if students think of putting the wide part of the symbol next to the larger number.
Use place value understanding and properties of operations to add and subtract First graders use their base-ten work to compute sums within 100 with understanding.1.NBT.4 Concrete objects, cards, or 1.NBT.4Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a writ-ten method and explain the reasoning used.
Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 7 drawings afford connections with written numerical work and dis-cussions and explanations in terms of tens and ones. In particular, showing composition of a ten with objects or drawings affords con-nection of the visual ten with the written numeral 1 that indicates 1 ten.
Combining tens and ones separately as illustrated in the margin Adding tens and ones separately 46 +37 46 +37 46 56 66 76 77 78 79 80 81 82 83 starting from 46 count on 3 tens then count on 7 ones combine ones view 6+7 as 1 ten and 3 ones combine 4 tens and 3 tens with the new group of 1 ten (shown below on the addition line) Using a sequence conception to add: Adding tens and ones separately: Figure 3: combine ones view 6 7 as 1 ten and 3 ones combine 4 tens and 3 tens with the newly composed ten (shown on the addition line) This method is an application of the commutative and associative properties. The diagrams can help children with understanding and explaining the steps (MP .1). Advantages of writing the 1 below the addends are discussed in the Grade 2 margin.
can be extended to the general method of combining like base-ten units. The margin illustrates combining ones, then tens. Like base-ten units can be combined in any order, but going from smaller to larger eliminates the need to go back to a given place to add in a new unit. For example, in computing 46 + 37 by combining tens, then ones (going left to right), one needs to go back to add in the new 1 ten: “4 tens and 3 tens is 7 tens, 6 ones and 7 ones is 13 ones which is 1 ten and 3 ones, 7 tens and 1 ten is 8 tens. The total is 8 tens and 3 ones: 83.” Students may also develop sequence methods that extend their Level 2 single-digit counting on strategies (see the OA Progression) to counting on by tens and ones, or mixtures of such strategies in which they add instead of count the tens or ones. Using objects or Counting on by tens 46 +37 46 +37 46 56 66 76 77 78 79 80 81 82 83 starting from 46 count on 3 tens then count on 7 ones combine ones view 6+7 as 1 ten and 3 ones combine 4 tens and 3 tens with the new group of 1 ten (shown below on the addition line) Using a sequence conception to add: Adding tens and ones separately: Figure 3: starting from 46 count on 3 tens then count on 7 ones Counting on by tens from 46, beginning 56, 66, 76, then counting on by ones. This method can be generalized, but the complexity of the counting on required and the lack of efficiency becomes apparent as the number of digits in the addends increases.
drawings of 5-groups can support students’ extension of the Level 3 make-a-ten methods discussed in the OA Progression for single-digit numbers.
First graders also engage in mental calculation, such as mentally finding 10 more or 10 less than a given two-digit number without having to count by ones.1.NBT.5 They may explain their reasoning by 1.NBT.5Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
saying that they have one more or one less ten than before. Draw-ings and layered cards can afford connections with place value and be used in explanations.
In Grade 1, children learn to compute differences of two-digit numbers for limited cases.1.NBT.6 Differences of multiples of 10, such 1.NBT.6Subtract multiples of 10 in the range 10–90 from multiples of 10 in the range 10–90 (positive or zero differences), using con-crete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addi-tion and subtraction; relate the strategy to a written method and explain the reasoning used.
as 70 ´ 40 can be viewed as 7 tens minus 4 tens and represented with concrete models such as objects bundled in tens or draw-ings. Children use the relationship between subtraction and addition when they view 80 ´ 70 as an unknown addend addition problem, 70 l “ 80, and reason that 1 ten must be added to 70 to make 80, so 80 ´ 70 “ 10.
First graders are not expected to compute differences of two-digit numbers other than multiples of ten. Deferring such work until Grade 2 allows two-digit subtraction with and without decompos-ing to occur in close succession, highlighting the similarity between these two cases. This helps students to avoid making the general-ization “in each column, subtract the larger digit from the smaller digit, independent of whether the larger digit is in the subtrahend or minuend,” e.g., making the error 82 - 45 = 43.
Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 8 Grade 2 At Grade 2, students extend their base-ten understanding to hun-dreds.
They now add and subtract within 1000, with composing and decomposing, and they understand and explain the reasoning of the processes they use. They become fluent with addition and subtraction within 100.
Understand place value In Grade 2, students extend their under-standing of the base-ten system by viewing 10 tens as forming a new unit called a “hundred.”2.NBT.1a This lays the groundwork for under-2.NBT.1a Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases: a 100 can be thought of as a bundle of ten tens—called a “hundred.” standing the structure of the base-ten system as based in repeated bundling in groups of 10 and understanding that the unit associated with each place is 10 of the unit associated with the place to its right.
Representations such as manipulative materials, math drawings and layered three-digit place value cards afford connections be-tween written three-digit numbers and hundreds, tens, and ones.
Drawings to support seeing 10 tens as 1 hundred drawings to support seeing 10 tens as 1 hundred 10 tens 1 hundred 1 hundred box (quick drawing to show 1 hundred) Number words and numbers written in base-ten numerals and as sums of their base-ten units can be connected with representations in drawings and place value cards, and by saying numbers aloud and in terms of their base-ten units, e.g., 456 is “Four hundred fifty six” and “four hundreds five tens six ones.”2.NBT.3 Unlayering three-digit 2.NBT.3Read and write numbers to 1000 using base-ten numerals, number names, and expanded form.
place value cards like the two-digit cards shown for Kindergarten and Grade 1 reveals the expanded form of the number.
Unlike the decade words, the hundred words indicate base-ten units. For example, it takes interpretation to understand that “fifty” means five tens, but “five hundred” means almost what it says (“five hundred” rather than “five hundreds”). Even so, this doesn’t mean that students automatically understand 500 as 5 hundreds; they may still only think of it as the number said after 499 or reached after 500 counts of 1.
A major task for Grade 2 is learning the counting sequence from 100 to 1,000. As part of learning and using the base-ten structure, students count by ones within various parts of this sequence, espe-cially the more difficult parts that “cross” tens or hundreds.
Building on their place value work, students continue to de-velop proficiency with mental computation.2.NBT.8 They extend this 2.NBT.8Mentally add 10 or 100 to a given number 100–900, and mentally subtract 10 or 100 from a given number 100–900.
to skip-counting by 5s, 10s, and 100s to emphasize and experi-ence the tens and hundreds within the sequence and to prepare for multiplication.2.NBT.2 2.NBT.2Count within 1000; skip-count by 5s, 10s, and 100s.
Comparing magnitudes of two-digit numbers uses the under-2.NBT.4Compare two three-digit numbers based on meanings of the hundreds, tens, and ones digits, using ą, =, and ă symbols to record the results of comparisons.
standing that 1 ten is greater than any amount of ones represented by a one-digit number. Comparing magnitudes of three-digit num-2.NBT.5Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the re-lationship between addition and subtraction.
2.NBT.6Add up to four two-digit numbers using strategies based on place value and properties of operations.
bers uses the understanding that 1 hundred (the smallest three-digit number) is greater than any amount of tens and ones represented by a two-digit number.
For this reason, three-digit numbers are compared by first inspecting the hundreds place (e.g. 845 ą 799; 849 ă 855).2.NBT.4 Drawings help support these understandings.
Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 9 Use place value understanding and properties of operations to add and subtract Students fluently add and subtract within 100.2.NBT.5 2.NBT.5Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the re-lationship between addition and subtraction.
They also add and subtract within 1000.2.NBT.7 They explain why ad-2.NBT.7Add and subtract within 1000, using concrete models or drawings and strategies based on place value, properties of op-erations, and/or the relationship between addition and subtrac-tion; relate the strategy to a written method.
Understand that in adding or subtracting three-digit numbers, one adds or sub-tracts hundreds and hundreds, tens and tens, ones and ones; and sometimes it is necessary to compose or decompose tens or hundreds.
dition and subtraction strategies work, using place value and the properties of operations, and may support their explanations with drawings or objects.2.NBT.9 Because adding and subtracting within 2.NBT.9Explain why addition and subtraction strategies work, us-ing place value and the properties of operations.2 100 is a special case of adding and subtracting within 1000, meth-ods within 1000 will be discussed before fluency within 100.
Two written methods for addition within 1000 are shown in the margins of this page and the next. The first explicitly shows the Addition: Recording newly composed units in separate rows 2 7 8 + 1 4 7 2 7 8 + 1 4 7 3 0 0 2 7 8 + 1 4 7 3 0 0 1 1 0 2 7 8 + 1 4 7 3 0 0 1 1 0 1 5 4 2 5 The computation shown proceeds from left to right, but could have gone from right to left. Working from left to right has two advantages: Many students prefer it because they read from left to right; working first with the largest units yields a closer approximation earlier.
Illustrating combining like units and composing new units The drawing shows the base-ten units of 278 and 147. Like units are shown together, with boundaries drawn around ten tens and ten ones to indicate the newly composed hundred and the newly composed ten. The newly composed units could also be indicated by crossing out grouped units and drawing a single next-highest unit, e.g., crossing out the group of ten ones and drawing a single ten. Drawings like this can be used to illustrate and explain both of the written computations below.
hundreds, tens, and ones that are being added; this can be helpful conceptually to students. The second method, shown on the next page, explicitly shows the adding of the single digits in each place and how this approach can continue on to places on the left.
Drawings can support students in explaining these and other methods. The drawing in the margin shows addends decomposed into their base-ten units (here, hundreds, tens, and ones), with the tens and hundreds represented by quick drawings.
These quick drawings show each hundred as a single unit rather than ten tens (see illustration on p. 8), generalizing the approach that students used in Grade 1 of showing a ten as a single unit rather than as 10 separate ones. The putting together of like quick drawings illus-trates adding like units as specified in 2.NBT.7: add ones to ones, tens to tens, and hundreds to hundreds. The drawing also shows newly composed units. Steps of adding like units and composing new units shown in the drawing can be connected with correspond-ing steps in other written methods. This also facilitates discussing how different written methods may show steps in different locations or different orders (MP.1 and MP.3). The associative and the com-mutative properties enable adding like units to occur.
The first written method is a helping step variation that gener-alizes to all numbers in base ten but becomes impractical because of writing so many zeros. Students can move from this method to the second method (or another compact method) by seeing how the steps of the two methods are related. Some students might make this transition in Grade 2, some in Grade 3, but all need to make it by Grade 4 where fluency requires a more compact method.
This first method can be seen as related to oral counting-on or written adding-on methods in which an addend is decomposed into hundreds, tens, and ones. These are successively added to the other addend, with the student saying or writing successive totals. These methods require keeping track of what parts of the decomposed addend have been added, and skills of mentally counting or adding hundreds, tens, and ones correctly.
For example, beginning with hundreds: 278 plus 100 is 378 (“I’ve used all of the hundreds”), 378 plus 30 is 408 and plus 10 (to add on all of the 40) is 418, and 418 plus 7 is 425. One way to keep track: draw the 147 and cross out parts as they are added on. Counting-on and adding-on methods become even more difficult with numbers over 1000. If they arise Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 10 from students, they should be discussed. But the major focus for addition within 1000 needs to be on methods such as those in the margin that are simpler for students and lead toward fluency (e.g., recording new units in separate rows shown) or are sufficient for fluency (e.g., recording new units in one row).
Addition: Recording newly composed units in the same row Add the ones, 8 7, and record these 15 ones with 1 on the line in the tens column and 5 below in the ones place.
Add the tens, 7 4 1, and record these 12 tens with 1 on the line in the hundreds column and 2 below in the tens place.
Add the hundreds, 2 1 1, and record these 4 hundreds below in the hundreds column.
Digits representing newly composed units are placed below the addends, on the line. This placement has several advantages.
Each two-digit partial sum (e.g., “15”) is written with the digits close to each other, suggesting their origin. In “adding from the top down,” usually sums of larger digits are computed first, and the easy-to-add “1” is added to that sum, freeing students from holding an altered digit in memory. The original numbers are not changed by adding numbers to the first addend; three multi-digit numbers (the addends and the total) can be seen clearly. It is easier to write teen numbers in their usual order (e.g., as 1 then 5) rather than “write the 5 and carry the 1” (write 5, then 1).
Drawings and steps for a generalizable method of subtracting within 1000 are shown in the margin. The total 425 does not have Subtraction: Decomposing where needed first decomposing left to right, 1 hundred, then 1 ten now subtract 425 - 278 425 - 278 Standard algorithm, ungroup where needed rst, then subtract: ungrouping left to right, 1 hundred, then 1 ten 425 - 278 now subtract [This is part 3 of Figure 5. Doug Clements drew the rst two parts.] All necessary decomposing is done first, then the subtractions are carried out. This highlights the two major steps involved and can help to inhibit the common error of subtracting a smaller digit on the top from a larger digit. Decomposing and subtracting can start from the left (as shown) or the right.
enough tens or ones to subtract the 7 tens or 8 ones in 278. There-fore one hundred is decomposed to make ten tens and one ten is decomposed to make ten ones. These decompositions can be done and written in either order; starting from the left is shown because many students prefer to operate in that order. In the middle step, one hundred has been decomposed (making 3 hundreds, 11 tens, 15 ones) so that the 2 hundreds 7 tens and 8 ones in 278 can be subtracted. These subtractions of like units can also be done in any order. When students alternate decomposing and subtracting like units, they may forget to decompose entirely or in a given column after they have just subtracted (e.g., after subtracting 8 from 15 to get 7, they move left to the tens column and see a 1 on the top and a 7 on the bottom and write 6 because they are in subtraction mode, having just subtracted the ones).
Students can also subtract within 1000 by viewing a subtraction as an unknown addend problem, e.g., 278 ? “ 425. Counting-on and adding-on methods such as those described above for addition can be used. But as with addition, the major focus needs to be on methods that lead toward fluency or are sufficient for fluency (e.g., recording as shown in the second row in the margin).
In Grade 1, students have added within 100 using concrete mod-els or drawings and used at least one method that is generalizable to larger numbers (such as between 101 and 1000).
In Grade 2, they can make that generalization, using drawings for explanation as discussed above. This extension could be done first for two-digit numbers (e.g., 78 47) so that students can see and discuss com-posing both ones and tens without the complexity of hundreds in the drawings or numbers (imagine the margin examples for 7847).
After computing totals that compose both ones and tens for two-digit numbers, then within 1000, the type of problems required for fluency in Grade 2 seem easy, e.g., 28 47 requires only composing a new ten from ones. This is now easier to do without drawings: one just records the new ten before it is added to the other tens or adds it to them mentally.
A similar approach can be taken for subtraction: first using con-crete models or drawings to solve subtractions within 100 that in-volve decomposing one ten, then rather quickly solving subtractions that require two decompositions. Spending a long time on subtrac-tion within 100 can stimulate students to count on or count down, which, as discussed above, are methods that are considerably more difficult with numbers above 100. Problems with different types of decompositions could be included so that students solve problems Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 11 requiring two, one, and no decompositions. Then students can spend time on subtractions that include multiple hundreds (totals from 201 to 1000). Relative to these experiences, the objectives for fluency at this grade are easy: focusing within 100 just on the two cases of one decomposition (e.g., 73 ´ 28) or no decomposition (e.g., 78 ´ 23) without drawings.
Students also add up to four two-digit numbers using strate-gies based on place value and properties of operations.2.NBT.6 This 2.NBT.6Add up to four two-digit numbers using strategies based on place value and properties of operations.
work affords opportunities for students to see that they may have to compose more than one ten, and as many as three new tens.
It is also an opportunity for students to reinforce what they have learned by informally using the commutative and associative prop-erties. They could mentally add all of the ones, then write the new tens in the tens column, and finish the computation in writing. They could successively add each addend or add the first two and last two addends and then add these totals. Carefully chosen problems could suggest strategies that depend on specific numbers. For ex-ample, 38479362 can be easily added by adding the first and last numbers to make 100, adding the middle two numbers to make 140, and increasing 140 by 100 to make 240.
Students also can develop special strategies for particularly easy computations such as 398 + 529, where the 529 gives 2 to the 398 to make 400, leaving 400 plus 527 is 927. But the major focus in Grade 2 needs to remain on the methods that work for all numbers and generalize readily to numbers beyond 1000.
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NBT, K–5 12 Grade 3 At Grade 3, the major focus is multiplication,• so students’ work • See the progression on Operations and Algebraic Thinking.
with addition and subtraction is limited to maintenance of fluency within 1000 for some students and building fluency to within 1000 for others.
Use place value understanding and properties of operations to perform multi-digit arithmetic Students fluently add and subtract within 1000 using methods based on place value, properties of oper-ations, and/or the relationship between addition and subtraction.3.NBT.2 3.NBT.2Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.
They focus on methods that generalize readily to larger numbers so that these methods can be extended to 1,000,000 in Grade 4 and fluency can be reached with such larger numbers. Fluency within 1000 implies that students use written methods without concrete models or drawings, though concrete models or drawings can be used with explanations to overcome errors and to continue to build understanding as needed.
Students use their place value understanding to round numbers to the nearest 10 or 100.3.NBT.1 They need to understand that when 3.NBT.1Use place value understanding to round whole numbers to the nearest 10 or 100.
moving to the right across the places in a number (e.g., 456), the dig-its represent smaller units. When rounding to the nearest 10 or 100, the goal is to approximate the number by the closest number with no ones or no tens and ones (e.g., so 456 to the nearest ten is 460; and to the nearest hundred is 500). Rounding to the unit represented by the leftmost place is typically the sort of estimate that is easiest for students and often is sufficient for practical purposes. Rounding to the unit represented by a place in the middle of a number may be more difficult for students (the surrounding digits are sometimes distracting). Rounding two numbers before computing can take as long as just computing their sum or difference.
The special role of 10 in the base-ten system is important in understanding multiplication of one-digit numbers with multiples of 10.3.NBT.3 For example, the product 3 ˆ 50 can be represented as 3 3.NBT.3Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., 9ˆ80, 5ˆ60) using strategies based on place value and properties of operations.
groups of 5 tens, which is 15 tens, which is 150. This reasoning relies on the associative property of multiplication: 3ˆ50 “ 3ˆp5ˆ10q “ p3 ˆ 5q ˆ 10 “ 15 ˆ 10 “ 150. It is an example of how to explain an instance of a calculation pattern for these products: calculate the product of the non-zero digits, then shift the product one place to the left to make the result ten times as large.
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NBT, K–5 13 Grade 4 At Grade 4, students extend their work in the base-ten system.
4.NBT.1Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right.
They use standard algorithms to fluently add and subtract. They use methods based on place value and properties of operations sup-ported by suitable representations to multiply and divide with multi-digit numbers.
Generalize place value understanding for multi-digit whole num-bers In the base-ten system, the value of each place is 10 times the 10 ˆ 30 represented as 3 tens each taken 10 times 10 × 30 10 groups of 30 10 10 10 ones tens hundreds 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 ones tens hundreds 10 × 30 = 300 100 100 100 ones tens hundreds 30 10 of each of the 3 tens 3 tens 10 times 3 tens is 3 hundreds Each of the 3 tens becomes a hundred and moves to the left. In the product, the 3 in the tens place of 30 is shifted one place to the left to represent 3 hundreds. In 300 divided by 10 the 3 is shifted one place to the right in the quotient to represent 3 tens.
value of the place to the immediate right.4.NBT.1 Because of this, multi-plying by 10 yields a product in which each digit of the multiplicand is shifted one place to the left.
To read numerals between 1,000 and 1,000,000, students need to understand the role of commas. Each sequence of three digits made by commas is read as hundreds, tens, and ones, followed by the name of the appropriate base-thousand unit (thousand, million, billion, trillion, etc.). Thus, 457,000 is read “four hundred fifty seven thousand.”4.NBT.2 The same methods students used for comparing and 4.NBT.2Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form.
Compare two multi-digit numbers based on meanings of the digits in each place, using ą, =, and ă symbols to record the results of com-parisons.
rounding numbers in previous grades apply to these numbers, be-cause of the uniformity of the base-ten system.4.NBT.3 4.NBT.3Use place value understanding to round multi-digit whole numbers to any place.
Decimal notation and fractions Students in Grade 4 work with fractions having denominators 10 and 100.4.NF.5 4.NF.5Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100.3 ones tens tenths hundredths ÷ 10 ÷ 10 ÷ 10 ÷ 10 .1 .1 .1 .1 .1 .1 .1 .1 .1 .1 1 100 1 10 each piece is .01 or each piece is .1 or ÷ 10 1 $1 dime penny dollar Because it involves partitioning into 10 equal parts and treating the parts as numbers called one tenth and one hundredth, work with these fractions can be used as prepa-ration to extend the base-ten system to non-whole numbers.
Using the unit fractions 1 10 and 1 100, non-whole numbers like 23 7 10 can be written in an expanded form that extends the form used with whole numbers: 2ˆ103ˆ17ˆ 1 10. As with whole-number expansions in the base-ten system, each unit in this decomposition is ten times the unit to its right, reflect-ing the uniformity of the base-ten system.
This can be connected with the use of base-ten notation to represent 2 ˆ 10 3 ˆ 1 7 ˆ 1 10 as 23.7. Using decimals allows stu-dents to apply familiar place value reasoning to fractional quantities.4.NF.6 The Number and 4.NF.6Use decimal notation for fractions with denominators 10 or 100.
Operations—Fractions Progression discusses decimals to hundredths and comparison of decimals4.NF.7 in more de-4.NF.7Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of com-parisons with the symbols ą, =, or ă, and justify the conclusions, e.g., by using a visual model.
tail.
The decimal point is used to signify the location of the ones place, but its location may suggest there should be a “oneths" place to its right in order to create symmetry with respect to the decimal point.
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NBT, K–5 14 However, because one is the basic unit from which the other base-ten units are derived, the symmetry occurs instead with respect to the ones place, as illustrated in the margin.
Symmetry with respect to the ones place 1 hundred hundredth ten tenth 4.NBT.4Fluently add and subtract multi-digit whole numbers using the standard algorithm.
4.NBT.5Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strate-gies based on place value and the properties of operations. Illus-trate and explain the calculation by using equations, rectangular arrays, and/or area models.
Ways of reading decimals aloud vary. Mathematicians and sci-entists often read 0.15 aloud as “zero point one five" or “point one five." (Decimals smaller than one may be written with or without a zero before the decimal point.) Decimals with many non-zero digits are more easily read aloud in this manner. (For example, the number π, which has infinitely many non-zero digits, begins 3.1415 . . .) Other ways to read 0.15 aloud are “1 tenth and 5 hundredths” and “15 hundredths,” just as 1,500 is sometimes read “15 hundred” or “1 thousand, 5 hundred.” Similarly, 150 is read “one hundred and fifty” or “a hundred fifty” and understood as 15 tens, as 10 tens and 5 tens, and as 100 50.
Multiplication: Illustrating partial products with an area model 549 × 8 8 549 = 500 + 40 + 9 Simplifed array/area drawing for 8 × 549 8 × 500 = 8 × 5 hundreds = 40 hundreds 8 × 40 = 8 × 4 tens = 32 tens 8 × 9 = 72 Three accessible ways to record the standard algorithm: Left to right showing the partial products Right to left showing the partial products Right to left recording the carries below 4000 320 72 4392 thinking: 8 × 5 hundreds 8 × 4 tens 8 × 9 549 × 8 549 × 8 72 320 4000 4392 thinking: 8 × 5 hundreds 8 × 4 tens 8 × 9 4022 4392 3 7 Each part of the region above corresponds to one of the terms in the computation below.
8 ˆ 549 “ 8 ˆ p500 40 9q “ 8 ˆ 500 8 ˆ 40 8 ˆ 9.
An area model can be used for any multiplication situation after students have discussed how to show an equal groups or a compare situation with an area model by making the length of the rectangle represent the size of the equal groups or the larger compared quantity imagining things inside the square units to make an array (but not drawing them), and understanding that the dimensions of the rectangle are the same as the dimensions of the imagined array, e.g., an array illustrating 8 x 549 would have 8 rows and 549 columns. (See the Operations and Algebraic Thinking Progression for discussion of “equal groups” and “compare” situations.) Just as 15 is understood as 15 ones and as 1 ten and 5 ones in computations with whole numbers, 0.15 is viewed as 15 hundredths and as 1 tenth and 5 hundredths in computations with decimals.
It takes time to develop understanding and fluency with the dif-ferent forms. Layered cards for decimals can help students under-stand how 2 tenths and 7 hundredths make 27 hundredths. Place value cards can be layered with the places farthest from the deci-mal point on the bottom (see illustration of the whole number cards on p. 5). These places are then covered by each place toward the decimal point: Tenths go on top of hundredth, and tens go on top of hundreds (for example, .2 goes on top of .07 to make .27, and 20 goes on top of 700 to make 720).
Use place value understanding and properties of operations to perform multi-digit arithmetic Students fluently add and subtract multi-digit numbers through 1,000,000 using the standard algorithm.4.NBT.4 Because students in Grade 2 and Grade 3 have been using at least one method that readily generalizes to 1,000,000, this extension does not have to take a long time. Thus, students will have time for the major NBT focus for this grade: multiplication and division.
Multiplication: Recording methods 549 × 8 8 549 = 500 + 40 + 9 Simplifed array/area drawing for 8 × 549 8 × 500 = 8 × 5 hundreds = 40 hundreds 8 × 40 = 8 × 4 tens = 32 tens 8 × 9 = 72 Three accessible ways to record the standard algorithm: Left to right showing the partial products Right to left showing the partial products Right to left recording the carries below 4000 320 72 4392 thinking: 8 × 5 hundreds 8 × 4 tens 8 × 9 549 × 8 549 × 8 72 320 4000 4392 thinking: 8 × 5 hundreds 8 × 4 tens 8 × 9 4022 4392 3 7 The first method proceeds from left to right, and the others from right to left. In the third method, the digits representing new units are written below the line rather than above 549, thus keeping the digits of the products close to each other, e.g., the 7 from 8 ˆ 9 “ 72 is written diagonally to the left of the 2 rather than above the 4 in 549. The colors indicate correspondences with the area model above.
In fourth grade, students compute products of one-digit num-bers and multi-digit numbers (up to four digits) and products of two two-digit numbers.4.NBT.5 They divide multi-digit numbers (up to four digits) by one-digit numbers. As with addition and subtraction, stu-dents should use methods they understand and can explain. Visual representations such as area and array diagrams that students draw and connect to equations and other written numerical work are use-ful for this purpose, which is why 4.NBT.5 explicitly states that they are to be used to illustrate and explain the calculation. By reason-ing repeatedly (MP.8) about the connection between math drawings and written numerical work, students can come to see multiplica-tion and division algorithms as abbreviations or summaries of their reasoning about quantities.
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NBT, K–5 15 One component of understanding general methods for multiplica-tion is understanding how to compute products of one-digit numbers and multiples of 10, 100, and 1000.
This extends work in Grade 3 on Illustrating partial products with an area model Simplifed array/area drawing for 36 × 94 30 + 6 90 + 4 30 × 90 = 3 tens × 9 tens = 27 hundreds = 2700 6 × 90 = 6 × 9 tens 54 tens = 540 30 × 4 = 3 tens × 4 = 12 tens = 120 6 × 4 = 24 Two accessible, right to left ways to record the standard algorithm: Showing the partial products Recording the carries below for correct place value placement 94 × 36 94 × 36 24 540 120 2700 thinking: 3 tens × 4 3 tens × 9 tens 6 × 9 tens 6 × 4 3384 5 2 2 1 1 1 3384 44 720 0 because we are multiplying by 3 tens in this row The products of base-ten units are shown as parts of a rectangular region. Such area models can support understanding and explaining of different ways to record multiplication. For students who struggle with the spatial demands of other methods, a useful helping step method is to make a quick sketch like this with the lengths labeled and just the partial products, then to add the partial products outside the rectangle.
Methods that compute partial products first Simplifed array/area drawing for 36 × 94 30 + 6 90 + 4 30 × 90 = 3 tens × 9 tens = 27 hundreds = 2700 6 × 90 = 6 × 9 tens 54 tens = 540 30 × 4 = 3 tens × 4 = 12 tens = 120 6 × 4 = 24 Two accessible, right to left ways to record the standard algorithm: Showing the partial products Recording the carries below for correct place value placement 94 × 36 94 × 36 24 540 120 2700 thinking: 3 tens × 4 3 tens × 9 tens 6 × 9 tens 6 × 4 3384 5 2 2 1 1 1 3384 44 720 0 because we are multiplying by 3 tens in this row These proceed from right to left, but could go left to right. On the right, digits that represent newly composed tens and hundreds are written below the line instead of above 94. The digits 2 and 1 are surrounded by a blue box. The 1 from 30 ˆ 4 “ 120 is placed correctly in the hundreds place and the digit 2 from 30 ˆ 90 “ 2700 is placed correctly in the thousands place. If these digits had been placed above 94, they would be in incorrect places. Note that the 0 (surrounded by a yellow box) in the ones place of the second row of the method on the right is there because the whole row of digits is produced by multiplying by 30 (not 3). Colors on the left correspond with the area model above.
Methods that alternate multiplying and adding These methods put the newly composed units from a partial product in the correct column, then they are added to the next partial product. These alternating methods are more difficult than the methods above that show the four partial products. The first method can be used in Grade 5 division when multiplying a partial quotient times a two-digit divisor.
Not shown is the recording method in which the newly composed units are written above the top factor (e.g., 94). This puts the hundreds digit of the tens times ones product in the tens column (e.g., the 1 hundred in 120 from 30 ˆ 4 above the 9 tens in 94).
This placement violates the convention that students have learned: a digit in the tens place represents tens, not hundreds.
products of one-digit numbers and multiples of 10. We can calculate 6 ˆ 700 by calculating 6 ˆ 7 and then shifting the result to the left two places (by placing two zeros at the end to show that these are hundreds) because 6 groups of 7 hundred is 6ˆ7 hundreds, which is 42 hundreds, or 4,200. Students can use this place value reasoning, which can also be supported with diagrams of arrays or areas, as they develop and practice using the patterns in relationships among products such as 6 ˆ 7, 6 ˆ 70, 6 ˆ 700, and 6 ˆ 7000. Products of 5 and even numbers, such as 5ˆ4, 5ˆ40, 5ˆ400, 5ˆ4000 and 4ˆ5, 4ˆ50, 4ˆ500, 4ˆ5000 might be discussed and practiced separately afterwards because they may seem at first to violate the patterns by having an “extra” 0 that comes from the one-digit product.
Another part of understanding general base-ten methods for multi-digit multiplication is understanding the role played by the distribu-tive property. This allows numbers to be decomposed into base-ten units, products of the units to be computed, then combined. By de-composing the factors into base-ten units and applying the distribu-tive property, multiplication computations are reduced to single-digit multiplications and products of numbers with multiples of 10, of 100, and of 1000. Students can connect diagrams of areas or arrays to numerical work to develop understanding of general base-ten mul-tiplication methods.
Computing products of two two-digit numbers requires using the distributive property several times when the factors are decomposed into base-ten units. For example, 36 ˆ 94 “ p30 6q ˆ p94q “ 30 ˆ 94 6 ˆ 94 “ 30 ˆ p90 4q 6 ˆ p90 4q “ 30 ˆ 90 30 ˆ 4 6 ˆ 90 6 ˆ 4.
The four products in the last line correspond to the four rectan-gles in the area model in the margin. Their factors correspond to the factors in written methods. When written methods are abbreviated, some students have trouble seeing how the single-digit factors are related to the two-digit numbers whose product is being computed (MP.2).
They may find it helpful initially to write each two-digit number as the sum of its base-ten units (e.g., writing next to the calculation 94 “ 90 4 and 36 “ 30 6) so that they see what the single digits are.
Some students also initially find it helpful to write what they are multiplying in front of the partial products (e.g., 6 ˆ 4 “ 24). These helping steps can be dropped when they are no longer needed. At any point before or after their acquisi-tion of fluency, some students may prefer to multiply from the left because they find it easier to align the subsequent products under this biggest product.
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NBT, K–5 16 General methods for computing quotients of multi-digit numbers and one-digit numbers rely on the same understandings as for mul-tiplication, but cast in terms of division.4.NBT.6 One component is quo-4.NBT.6Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relation-ship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
tients of multiples of 10, 100, or 1000 and one-digit numbers. For example, 42 ˜ 6 is related to 420 ˜ 6 and 4200 ˜ 6. Students can Division as finding group size 745 ÷ 3 = ? 3 groups 3 groups Thinking: Divide 7 hundreds, 4 tens, 5 ones equally among 3 groups, starting with hundreds.
7 hundreds ÷ 3 each group gets 2 hundreds; 1 hundred is left.
Unbundle 1 hundred.
Now I have 10 tens + 4 tens = 14 tens.
14 tens ÷ 3 each group gets 4 tens; 2 tens are left.
Unbundle 2 tens.
Now I have 20 + 5 = 25 left.
25 ÷ 3 each group gets 8; 1 is left.
3 )745 3 )745 - 6 1 2 hundr.
2 hundr.
2 hundr.
3 groups 2 hundr. + 4 tens 2 hundr. + 4 tens 2 hundr. + 4 tens 3 groups 2 hundr. + 4 tens + 8 2 hundr. + 4 tens + 8 2 hundr. + 4 tens + 8 2 3 )745 - 6 14 2 3 )745 - 6 14 - 12 2 24 3 )745 - 6 14 - 12 25 24 3 )745 - 6 14 - 12 25 - 24 1 248 Each group got 248 and 1 is left.
1 2 3 745 ˜ 3 can be viewed as allocating 745 objects bundled in 7 hundreds, 4 tens, and 3 ones equally among 3 groups. In Step 1, the 2 indicates that each group got 2 hundreds, the 6 is the number of hundreds allocated, and the 1 is the number of hundreds not allocated. After Step 1, the remaining hundred is decomposed as 10 tens and combined with the 4 tens (in 745) to make 14 tens.
draw on their work with multiplication and they can also reason that 4200 ˜ 6 means partitioning 42 hundreds into 6 equal groups, so there are 7 hundreds in each group.
Another component of understanding general methods for multi-digit division computation is the idea of decomposing the dividend into like base-ten units and finding the quotient unit by unit, start-ing with the largest unit and continuing on to smaller units. See the figure in the margin. As with multiplication, this relies on the distributive property. This can be viewed as finding the side length of a rectangle (the divisor is the length of the other side) or as al-locating objects (the divisor is the number of groups or the number of objects in each group). See the figure on the next page for an example.
Multi-digit division requires working with remainders. In prepa-ration for working with remainders, students can compute sums of a product and a number, such as 4 ˆ 8 3.• In multi-digit division, • A note on notation The result of division within the system of whole numbers is frequently written as: 84 ˜ 10 “ 8 R 4 and 44 ˜ 5 “ 8 R 4.
Because the two expressions on the right are the same, students should conclude that 84 ˜ 10 is equal to 44 ˜ 5, but this is not the case. (Because the equal sign is not used appropriately, this usage is a non-example of Standard for Mathematical Practice 6.) Moreover, the notation 8 R 4 does not indicate a number.
Rather than writing the result of division in terms of a whole-number quotient and remainder, the relationship of whole-number quotient and remainder can be written as: 84 “ 8 ˆ 10 4 and 44 “ 8 ˆ 5 4.
students will need to find the greatest multiple less than a given number. For example, when dividing by 6, the greatest multiple of 6 less than 50 is 6 ˆ 8 “ 48. Students can think of these “greatest multiples” in terms of putting objects into groups. For example, when 50 objects are shared among 6 groups, the largest whole number of objects that can be put in each group is 8, and 2 objects are left over. (Or when 50 objects are allocated into groups of 6, the largest whole number of groups that can be made is 8, and 2 objects are left over.) The equation 6 ˆ 8 2 “ 50 (or 8 ˆ 6 2 “ 50) corresponds with this situation.
Cases involving 0 in division may require special attention. See the figure below.
Cases involving 0 in division 2 ) 83 6 ) 901 - 6 3 4 1 - 8 0 12 ) 3714 -36 11 3 Stop now because of the 0?
No, there are still 3 ones left.
Stop now because 11 is less than 12?
No, it is 11 tens, so there are still 110 + 4 = 114 left.
Case 1 a 0 in the dividend: Case 2 a 0 in a remainder part way through: Case 3 a 0 in the quotient: What to do about the 0?
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NBT, K–5 17 Division as finding side length ? hundreds + ? tens + ? ones Find the unknown length of the rectangle; first find the hundreds, then the tens, then the ones. Method A Method B 100 + ?? The length has 1 hundred, making a rectangle with area 700. Method A records the difference of the areas as 966 – 700 = 266, showing the remaining area (266). Only hundreds are subtracted; the tens and ones digits do not change. Method B records only the hundreds digit (2) of the difference and “brings down” the unchanged tens digit (6). These digits represent: 2 hundreds + 6 tens = 26 tens. 100 + 30 + ? The length has 3 tens, making a rectangle with area 210. Method A records the difference of the areas as 266 – 210 = 56. Only hundreds and tens are subtracted; the ones digit does not change. Method B records only the tens digit (5) of the difference and “brings down” the ones digit (6). These digits represent: 5 tens + 6 ones = 56 ones. 100 + 30 + 8 The length has 8 ones, making an area of 56. The original rectangle can now be seen as composed of three smaller rectangles with areas of the amounts that were subtracted from 966. 966 ÷ 7 can be viewed as finding the unknown side length of a rectangular region with area 966 square units and a side of length 7 units. The divisor, partial quotients (100, 30, 8), and final quotient (138) represent quantities in length units and the dividend represents a quantity in area units. Method A shows each partial quotient and has the final step adding them (going from 100 + 30 + 8 to 138). Method B abbreviates these partial quotients. These can be said explicitly when explaining the method (e.g., 7 hundreds subtracted from the 9 hundreds is 2 hundreds). Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 18 Grade 5 In Grade 5, students extend their understanding of the base-ten sys-tem to decimals to the thousandths place, building on their Grade 4 work with tenths and hundredths. They become fluent with the stan-dard multiplication algorithm with multi-digit whole numbers. They reason about dividing whole numbers with two-digit divisors, and reason about adding, subtracting, multiplying, and dividing decimals to hundredths.
Understand the place value system Students extend their under-standing of the base-ten system to the relationship between adjacent places, how numbers compare, and how numbers round for decimals to thousandths.
New at Grade 5 is the use of whole number exponents to denote powers of 10.5.NBT.2 Students understand why multiplying by a power 5.NBT.2Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multi-plied or divided by a power of 10. Use whole-number exponents to denote powers of 10.
of 10 shifts the digits of a whole number or decimal that many places to the left. For example, multiplying by 104 is multiplying by 10 four times. Multiplying by 10 once shifts every digit of the multiplicand one place to the left in the product (the product is ten times as large) because in the base-ten system the value of each place is 10 times the value of the place to its right. So multiplying by 10 four times shifts every digit 4 places to the left. Patterns in the number of 0s in products of a whole number and a power of 10 and the location of the decimal point in products of decimals with powers of 10 can be explained in terms of place value. Because students have developed their understandings of and computations with decimals in terms of multiples (consistent with 4.OA.4) rather than powers, connecting the terminology of multiples with that of powers affords connections between understanding of multiplication and exponentiation.
Perform operations with multi-digit whole numbers and with dec-imals to hundredths At Grade 5, students fluently compute prod-ucts of whole numbers using the standard algorithm.5.NBT.5 Underly-5.NBT.5Fluently multiply multi-digit whole numbers using the stan-dard algorithm.
ing this algorithm are the properties of operations and the base-ten system (see the Grade 4 section).
Division in Grade 5 extends Grade 4 methods to two-digit divisors.5.NBT.6 5.NBT.6Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relation-ship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Students continue to decompose the dividend into base-ten units and find the quotient place by place, starting from the highest place.
They illustrate and explain their calculations using equations, rect-angular arrays, and/or area models. Estimating the quotients is a new aspect of dividing by a two-digit number. Even if students round the dividend appropriately, the resulting estimate may need to be adjusted up or down. Sometimes multiplying the ones of a two-digit divisor composes a new thousand, hundred, or ten. These newly composed units can be written as part of the division computation, added mentally, or as part of a separate multiplication computation.
Students who need to write decomposed units when subtracting need to remember to leave space to do so.
Recording division after an underestimate 27 ) 1655 -1350 305 -270 35 -27 8 1 10 50 61 1655 ÷ 27 Rounding 27 to 30 produces the underestimate 50 at the first step but this method allows the division process to be continued (30) Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 19 Because of the uniformity of the structure of the base-ten sys-tem, students use the same place value understanding for adding and subtracting decimals that they used for adding and subtracting whole numbers.5.NBT.7 Like base-ten units must be added and sub-5.NBT.7Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
tracted, so students need to attend to aligning the corresponding places correctly (this also aligns the decimal points). It can help to put 0s in places so that all numbers show the same number of places to the right of the decimal point.
A whole number is not usually written with a decimal point, but a decimal point followed by one or more 0s can be inserted on the right (e.g., 16 can also be written as 16.0 or 16.00). The process of composing and decompos-ing a base-ten unit is the same for decimals as for whole numbers and the same methods of recording numerical work can be used with decimals as with whole numbers. For example, students can write digits representing newly composed units on the addition line, and they can decompose units wherever needed before subtracting.
General methods used for computing products of whole numbers extend to products of decimals. Because the expectations for deci-mals are limited to thousandths and expectations for factors are lim-ited to hundredths at this grade level, students will multiply tenths with tenths and tenths with hundredths, but they need not multiply hundredths with hundredths. Before students consider decimal mul-tiplication more generally, they can study the effect of multiplying by 0.1 and by 0.01 to explain why the product is ten or a hundred times as small as the multiplicand (moves one or two places to the right). They can then extend their reasoning to multipliers that are single-digit multiples of 0.1 and 0.01 (e.g., 0.2 and 0.02, etc.).
There are several lines of reasoning that students can use to explain the placement of the decimal point in other products of dec-imals. Students can think about the product of the smallest base-ten units of each factor. For example, a tenth times a tenth is a hun-dredth, so 3.2ˆ7.1 will have an entry in the hundredth place. Note, however, that students might place the decimal point incorrectly for 3.2 ˆ 8.5 unless they take into account the 0 in the ones place of 32ˆ85. (Or they can think of 0.2ˆ0.5 as 10 hundredths.) They can also think of the decimals as fractions or as whole numbers divided by 10 or 100.5.NF.3 When they place the decimal point in the product, 5.NF.3Interpret a fraction as division of the numerator by the de-nominator (a{b “ a˜b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
they have to divide by a 10 from each factor or 100 from one factor.
For example, to see that 0.6 ˆ 0.8 “ 0.48, students can use frac-tions: 6 10 ˆ 8 10 “ 48 100.5.NF.4 Students can also reason that when they 5.NF.4Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
carry out the multiplication without the decimal point, they have multiplied each decimal factor by 10 or 100, so they will need to divide by those numbers in the end to get the correct answer. Also, students can use reasoning about the sizes of numbers to determine the placement of the decimal point. For example, 3.2 ˆ 8.5 should be close to 3 ˆ 9, so 27.2 is a more reasonable product for 3.2 ˆ 8.5 than 2.72 or 272. This estimation-based method is not reliable in all cases, however, especially in cases students will encounter in later grades. For example, it is not easy to decide where to place Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 20 the decimal point in 0.023 ˆ 0.0045 based on estimation. Students can summarize the results of their reasoning such as those above as specific numerical patterns and then as one general overall pattern such as “the number of decimal places in the product is the sum of the number of decimal places in each factor.” General methods used for computing quotients of whole numbers extend to decimals with the additional issue of placing the decimal point in the quotient. As with decimal multiplication, students can first examine the cases of dividing by 0.1 and 0.01 to see that the quotient becomes 10 times or 100 times as large as the dividend (see also the Number and Operations—Fractions Progression). For example, students can view 7 ˜ 0.1 “ l as asking how many tenths are in 7.5.NF.7b Because it takes 10 tenths to make 1, it takes 7 times 5.NF.7bApply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
b Interpret division of a whole number by a unit fraction, and compute such quotients.
as many tenths to make 7, so 7 ˜ 0.1 “ 7 ˆ 10 “ 70. Or students could note that 7 is 70 tenths, so asking how many tenths are in 7 is the same as asking how many tenths are in 70 tenths, which is 70. In other words, 7 ˜ 0.1 is the same as 70 ˜ 1. So dividing by 0.1 moves the number 7 one place to the left, the quotient is ten times as big as the dividend. As with decimal multiplication, students can then proceed to more general cases. For example, to calculate 7˜0.2, students can reason that 0.2 is 2 tenths and 7 is 70 tenths, so asking how many 2 tenths are in 7 is the same as asking how many 2 tenths are in 70 tenths. In other words, 7 ˜ 0.2 is the same as 70 ˜ 2; multiplying both the 7 and the 0.2 by 10 results in the same quotient. Or students could calculate 7 ˜ 0.2 by viewing 0.2 as 2 ˆ 0.1, so they can first divide 7 by 2, which is 3.5, and then divide that result by 0.1, which makes 3.5 ten times as large, namely 35. Dividing by a decimal less than 1 results in a quotient larger than the dividend5.NF.5 and moves the digits of the dividend one place 5.NF.5Interpret multiplication as scaling (resizing), by: a Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without per-forming the indicated multiplication.
b Explaining why multiplying a given number by a frac-tion greater than 1 results in a product greater than the given number (recognizing multiplication by whole num-bers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a{b “ pnˆaq{pnˆbq to the effect of multiplying a{b by 1.
to the left. Students can summarize the results of their reasoning as specific numerical patterns, then as one general overall pattern such as “when the decimal point in the divisor is moved to make a whole number, the decimal point in the dividend should be moved the same number of places.” Draft, 6 March 2015, comment at commoncoretools.wordpress.com.
NBT, K–5 21 Extending beyond Grade 5 At Grade 6, students extend their fluency with the standard algo-rithms, using these for all four operations with decimals and to com-pute quotients of multi-digit numbers. At Grade 6 and beyond, stu-dents may occasionally compute with numbers larger than those specified in earlier grades as required for solving problems, but the Standards do not specify mastery with such numbers.
In Grade 6, students extend the base-ten system to negative numbers. In Grade 7, they begin to do arithmetic with such numbers.
By reasoning about the standard division algorithm, students learn in Grade 7 that every fraction can be represented with a dec-imal that either terminates or repeats. In Grade 8, students learn informally that every number has a decimal expansion, and that those with a terminating or repeating decimal representation are rational numbers (i.e., can be represented as a quotient of integers).
There are numbers that are not rational (irrational numbers), such as the square root of 2. (It is not obvious that the square root of 2 is not rational, but this can be proved.) In fact, surprisingly, it turns out that most numbers are not rational. Irrational numbers can always be approximated by rational numbers.
In Grade 8, students build on their work with rounding and expo-nents when they begin working with scientific notation. This allows them to express approximations of very large and very small numbers compactly by using exponents and generally only approximately by showing only the most significant digits. For example, the Earth’s circumference is approximately 40,000,000 m. In scientific notation, this is 4 ˆ 107 m.
The Common Core Standards are designed so that ideas used in base-ten computation, as well as in other domains, can support later learning. For example, use of the distributive property occurs together with the idea of combining like units in the NBT and NF standards. Students use these ideas again when they calculate with polynomials in high school.
The distributive property and like units: Multiplication of whole numbers and polynomials 52 ¨ 73 “ p5 ¨ 10 2qp7 ¨ 10 3q “ 5 ¨ 10p7 ¨ 10 3q 2 ¨ p7 ¨ 10 3q “ 35 ¨ 102 15 ¨ 10 14 ¨ 10 2 ¨ 3 “ 35 ¨ 102 29 ¨ 10 6 p5x 2qp7x 3q “ p5x 2qp7x 3q “ 5xp7x 3q 2p7x 3q “ 35x2 15x 14x 2 ¨ 3 “ 35x2 29x ` 6 decomposing as like units (powers of 10 or powers of x) using the distributive property using the distributive property again combining like units (powers of 10 or powers of x) Draft, 6 March 2015, comment at commoncoretools.wordpress.com. |
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Keep Learning With Our AoPS Puzzle Library
At AoPS, we love a good challenge. And we know you do too! Our Keep Learning puzzles provide our community with problems and puzzles you can solve from anywhere — home, school, even on the sidewalk with chalk!
We post new problems every week, so check in often to find new complexities and perplexities.
Puzzles
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Puzzles
Test your mettle with our staff-curated puzzles.
Missing Lights
There are 8 lights arranged in a circle. Each of them is randomly lit either red, green, or blue. What is the probability that there are six consecutive lights which are lit up in at most two colors?
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Heads or Tails?
Winnie keeps flipping a coin until she gets either two heads in a row or three tails in a row, at which point she stops. What is the probability her final flip is heads?
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Matching Cards
Two shuffled standard 52-card decks are placed side by side.
What is the reciprocal of the probability that at least one card is in the same position in both decks, to nine decimal places?
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Ultimate End
Call a positive integer ultimate if it appears as the final digits of its square. For example, is ultimate because ends in However, is not ultimate, since does not end in How many positive integers less than million are ultimate?
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Base Bedlam
For which positive integers n is the following equation true in base n?
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Palindrome Pandemonium
A number is a palindrome in base n if its digits read the same forward and backward when written in that base.
How many numbers less than 1 million are palindromes in base 3? One such number is 23, written as 212 in base 3.
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Happy New Year!
Using just the four operations of arithmetic and three s, Grogg can make as follows:
How can Grogg make using the four operations of arithmetic, parentheses, and the fewest possible number of s?
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Infinite Reciprocals
Winnie likes positive integers whose only prime factors are and So, Winnie likes the numbers and but she does not like or
What is the sum of the reciprocals of all the numbers that Winnie likes?
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Like Twenty-four
Using exactly one multiplication, one addition, one division, and one subtraction, what is the smallest positive number you can create using the numbers 4, 5, 6, 7, and 8 once each? For instance, we can form the number ((8÷5) x 7) - (4+6)= 1.2 in this way.
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Cubic Chess
Cubic chess is played on a 6 x 6 x 6 chessboard, where pieces occupy one of 216 individual cubes.
A queen attacks all cubes that are in a straight line parallel to an edge of the cube, in a diagonal line parallel to a face diagonal of the cube, or in a diagonal line parallel to a space diagonal of the cube (i.e. a diagonal connecting two opposite corners).
How many queens can you place on this cubic chessboard so that no two are attacking each other?
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Unequal Equilateral Triangle
Given an equilateral triangle with ABC with side length 4, can you find points D, E, and F along its sides so that the areas x, y, and z are distinct positive integers.
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Consecutive Christmas
Seven gifts are placed in a circle around the Christmas tree, and three of these gifts are chosen at random. What is the probability no two chosen gifts are adjacent with each other?
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Age Conundrum
The sum of Dancer's, Dasher's, and Prancer's ages is 60. When Prancer was born, Dasher's age was double that of Dancer's. How old is Dancer today?
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Diabolical Digits
Can you find a four-digit number satisfying
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Polynomial Product
Suppose P(x) and Q(x) are two non-constant polynomials satisfying the equation P(Q(x)) = P(x) Q(x) for all x. What is Q(1)?
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Composite Composition
For an integer n, the function τ(n) is the number of positive divisors of n. Among all positive integers n<1000, what is the greatest possible value of τ(τ(n))?
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Geometric Sequence
Can you find a geometric sequence containing eight distinct 5-digit integers?
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Simon's Favorite Rectangle Problem
How many rectangles with integer side lengths have the property that the numerical value of their area is 2021 times the numerical value of their perimeter?
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Quasi-Fibonacci
Can you find integers and such that in the recursive sequence defined by and for all integers , we have
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Simple as a Square
What is the square root of the -digit number
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Easy as 1, 2, 3
A point is selected within square so that and as shown in the diagram. What is the measure of
![Image 69: [asy]
import markers;
size(250);
pen s = fontsize(8);
pair E = (1, 1);
pair F = (1, -1);
pair G = (-1, -1);
pair H = (-1, 1);
pair P = (.4, .2);
draw(E--F--G--H--E);
draw(P--E);
draw(P--F);
draw(P--G);
pair X = (P+E)/2;
pair Y = (P+F)/2;
pair Z = (P+G)/2;
dot(P);
label("A", E, NE);
label("B", F, SE);
label("C", G, SW);
label("D", H, NW);
label("P", P, N);
label("$1$", X, NW);
label("$2$", Y, SW);
label("$3$", Z, NW);
markangle(n=1,radius=22, F, P, E, heavyred);
[/asy]](
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A Peculiar Polynomial
Suppose is a polynomial with degree so that for all positive integers What is
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Infinite Roots
Which is greater: or
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Squaring the Hexagon
Suppose the plane is tessellated by congruent hexagons as shown here. Can you find four vertices of this tessellation which create a square?
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Listless Listing
Grogg lists five positive integers. Winnie then creates a new list by finding and recording the sum of each of the 10 pairs of Grogg's integers. If Winnie's list contains the integers 4, 40, 400, and 4000, then what is the largest integer that could possibly appear on Winnie's list?
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Absolute Minimum
Can you find the minimum possible value of this expression as ranges over all real numbers?
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Egyptian Fractions
Can you find four positive integers and so that
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Angle-Bisector Bisection
In triangle points and are chosen along segment so that is an altitude, is the interior angle bisector of angle , and is the midpoint of Suppose that also bisects the angle What is
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Like Limbo
Let be the smallest integer greater than or equal to Find given that
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Tough Value
What is the value of the following expression?
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Prime Time
Can you find all pairs of prime numbers such that is also a prime number? How do you know that you've found all of them?
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High Degree Difficulty
For which real numbers do there exist real numbers and such that
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Mathwalk Challenge: Mathscotch
Use the puzzle below for a Mathwalk challenge! After trying the puzzle yourself, chalk it out on the sidewalk, take a picture, and share it on social with #aopsmathwalk.
Mathwalk Challenge: Sum Shapes
Use the puzzle below for a Mathwalk challenge! After trying the puzzle yourself, chalk it out on the sidewalk, take a picture, and share it on social with #aopsmathwalk.
Count to N
Winnie and Alex play a game with a positive integer Winnie counts all pairs of positive integers such that Alex counts all pairs of positive integers such that Whoever has the higher count wins. For what numbers does Alex win?
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Decimal Digits
How many 's are in the first digits after the decimal point in the decimal expansion of
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Grazzling Grizzers
It takes Grogg 3 hours to fully grazzle a grizzer. It takes Winnie and Alex together 2 hours to grazzle a grizzer. One day, Alex grazzles a grizzer for 1 hour by himself, when Grogg and Winnie join him to finish grazzling the grizzer in 1 more hour. Who is the fastest grizzer-grazzler? You may assume beasts grazzle grizzers at fixed rates.
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Picking Probability
Grogg picks a real number between 0 and 1. Alex picks a real number between 1 and 2. What is the probability that Alex's number is more than double Grogg's number?
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What's the Point?
Grogg is trying to draw an equilateral triangle with a special point inside the triangle, so that and Can he do it? If so, what's the side-length of his triangle? If not, why not?
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Long Weekend!
Despising the short weekend, AoPS decided to invent Seconaturday: the Second Saturday, a day falling between Saturday and Sunday. The new calendar is completely the same with the typical 12 months spanning 365 days (except for leap years), although weeks now have 8 days total. If April 24th, 3000 is a Friday under the new calendar, how many Seconaturdays are in the year 3000?
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Some Sums
If and are positive real numbers that sum to what's the smallest possible value of
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Squares of Heptagons
A regular heptagon is divided as shown, where each variable or number denotes the length of the segment closest to it. What is
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Lovely Polygons
A polygon is lovely if all of its interior angles are equal and all of its side lengths are distinct. What is the smallest possible perimeter a lovely polygon could have, if all of its side lengths are integers?
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Does It Average Out?
Can you find seven distinct integers that satisfy all of the following? Can you find six distinct integers?
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Missing Digits
The six-digit number 4A8A2A is divisible by the two-digit number 8A for some digit A in base 10. What is A?
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Hat Trick
In the line for a hat store, 10 people are wearing hats, each of which is either blue, green, or red. You notice every person standing behind someone wearing a green hat is themselves wearing a blue hat. Additionally, no two people wearing red hats are standing next to each other. How many possible orders are there for the colors of the hats in the line?
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Find Four
What is the smallest n such that among any n integers, I can always find four integers whose sum is divisible by 4?
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Numbering Numbers
Lizzie writes the following sequence on a paper. Her sequence has one 1, two 2's, three 3's, and so on, ending with one thousand 1000's. Lizzie crosses out every other term in this sequence, beginning with the first term. What is the sum of the numbers that Lizzie does not cross out?
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Nice Triangles
A triangle is nice if the numerical values of all three of its side lengths and its area are distinct integers. If one side of a nice triangle has length 25, what is the smallest possible area it could have?
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Quadratic Quiz
The quadratic equation has roots and while the quadratic equation has roots and Given that all of and are non-zero real numbers, what is the greatest value any of them could be?
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Perfectly Imperfect
Winnie likes numbers that are one more or one less than a perfect cube. Lizzie likes numbers that are perfect squares. How many positive integers less than 1,000,000 do either Winnie or Lizzie (or both Winnie and Lizzie) like?
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Brain Breaker
This multiple choice question has at least one correct answer. If you guessed at random, what is the probability your answer would be incorrect?
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Black Holes
How many paths are there from the rocket to the moon (located at the red vertices) if you can only move up or to the right along the blue edges, without falling into any black holes? (You may walk along the edges of the black holes.)
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Multiples of the Year
What is the smallest positive multiple of 2021 whose digits are all distinct? What is the largest such number?
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Inscribed Inscription
What has the larger area: A square inscribed in a square inscribed in a circle inscribed in a unit square, or a regular octagon inscribed in a square inscribed in a unit circle?
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Divisibility Ability
What is the smallest positive integer with at least 25 positive divisors?
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Circle Challenge
The figure pictured here is a quarter circle. The two smaller quarter circles within have the same area and are also tangent. Which area is greater, green or purple?
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A Cute Time
What is the probability that at a random time of day, the smaller angle formed between the hour and minute hands of an analog clock is less than 90°?
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The Rose Gold Ratio
Let the rose gold ratio be . It turns out that is an integer. What number is it? How do you know?
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Cut Into Pieces
What is the largest number of enclosed regions you can create using any 3 lines and 3 circles? For example, this arrangement has 12 enclosed regions.
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A Quadrilateral Quandary
Grogg draws a quadrilateral and its two diagonals. Lizzie connects the midpoints of opposite sides of the quadrilateral, and the midpoints of the two diagonals. Miraculously, all three of these lines intersect at the same point! Did Grogg and Lizzie get lucky, or will this work every time? Why?
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Looking for a Match
Lizzie has three pairs of socks in different colors: one green, one blue, and one red. She grabs three socks randomly from her drawer. If they contain a pair of the same color, she wears it; if they are all different colors, she puts all three back. What is the probability she'll find a pair within her first three pulls?
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Tongue-Twister
If a trabble's nine trudds,
and a rabble's five rudds,
and a rudd and a trudd
are one half of a trabble,
then how many trudds
are in one hundred rabbles?
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Two Much Fibonacci
What is the value of the infinite sum where the numerators of each fraction form the famous Fibonacci sequence and the denominators of each fraction are successive powers of ?
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Putting the Pentagon in Pentadecagon
A pentadecagon is a 15-sided polygon. How many pentagons can you construct by connecting 5 non-adjacent vertices in a regular pentadecagon?
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I Can't Believe It's Not Irrational!
The number is an integer. Without using a calculator, can you tell which integer it is? How do you know?
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Counting Letters
Lizzie writes all the numbers from to on a chalkboard using words:
How many times does she write the letter ?
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Cube Folding
You can cut the following shape out of a piece of paper, and fold it into a cube. Can you similarly cut and fold a strip of paper (shown below) into a cube?
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Data Difficulties
Grogg was collecting some data on social networks and noticed an interesting phenomenon. Among the beasts he collected data from, there were pairs of beasts that were friends, and yet no three beasts were mutually friends with each other.
He explained his findings to Max, who thought about it for a minute before replying, "I think your data has an error!"
Is Max right? Why or why not?
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Three Squares
In the figure below, two non-overlapping squares are placed inside a larger square. Why must the combined perimeters of the smaller squares be less than the perimeter of the largest square?
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Loaded Dice
"Look!" Rosencrantz said. "I've got two dice that both have the numbers through on them. But they're loaded - on each die, some numbers have a higher probability of coming up than others. When I roll both of them, each of the sums from through has the same probability of coming up!"
"That's impossible!" Guildenstern exclaimed.
Who's right? Why?
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Tag
Winnie and Lizzie are playing tag in Beast Academy's circular playground, and Winnie is trying to catch Lizzie. Both beasts run at equal, constant speeds. Can Winnie always catch Lizzie in a finite amount of time?
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The Shaky Desk
"My new square teacher's desk is awful!" Ms. Q complained. "All four legs definitely have the same length, but the desk won't stop shaking!"
"Try rotating the desk about its center!" Professor Grok replied. "You'll definitely find a point where the desk stops shaking."
Is Professor Grok right? Why or why not?
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Tiling the AoPS Logo
The AoPS logo is a regular hexagon tiled with rhombi, each of which has two angles and two angles. There are three different rotational orientations for these tiles, and the tiles of each orientation are colored identically.
This is not the only way to tile the hexagon; you can see many others by looking at users' avatars on the AoPS Forums. However, any such tiling will always contain the same number of tiles of each orientation. Can you explain why?
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All But Two
Alex accidentally left his long division homework out in the rain. By the time he noticed, all but two of the numbers were unrecognizable! Can you help Alex figure out what this problem was supposed to read?
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The Perplexing Polynomial
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Squares in Squares
Can you fill in the 25 small squares below with the integers 1 to 25 once each, such that within each of the red and purple squares, the sum of the numbers in each row, column, and both main diagonals are the same?
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Polygon Perimeters
Could the perimeter of Lizzie's new polygon be greater than the perimeter of her original polygon? Why or why not?
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A Doubling Function
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A Scintillating Sequence
Prove that every number in the sequence below is composite.
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A Gnarly Grid
Consider this grid of 25 squares.
Can you write one integer in each square so that the sum of the integers in each row is greater than 300, but the sum of the integers in each column is less than 200? Why or why not?
Can you write one integer in each square so that the product of the integers in each row is greater than 300, but the product of the integers in each column is less than 200? Why or why not?
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An Eccentric Equation
Find all real solutions to the following equation. Why aren't there any other solutions?
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Five Quarter-Circles
Quarter-circles of the same color have equal areas. What's the ratio of the radius of the smallest quarter-circle to the radius of the largest quarter-circle?
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Grogg Tries to Skip School
Grogg said to his mom, "Why should I go to school when I don't have the time? I sleep hours a day, which adds up to about days per year. There's no school on weekends, which is days per year. We have days of summer vacation. I need hours a day to eat - that's days per year. And I need hours a day to have fun, or days per year. If I'm sick days per year, then there's no more days in the year, so I don't have time to go to school!"
Grogg's mom looked at Grogg's calculations and chuckled. "Your calculations aren't wrong," she said, "but that's no excuse not to go to school!"
What's wrong with Grogg's reasoning?
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Dizzying Divisors
For every positive integer let be the number of divisors of If and are positive integers, does the following equation have any solutions?
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The Tricky Triangle
The blue lines below divide an equilateral triangle with side length 7 into seven triangles with equal area. What is the combined length of the red segments?
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A Power Tower
What is the smallest positive integer such that
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A Round Robin Paradox
In a round robin tournament, every player plays one game against every other player, and there are no ties. Is it possible that for every pair of people in a round robin tournament, there is some other player that wins against both of them? If so, what's the fewest number of people this could happen with?
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Euclid's Game
Lizzie and Grogg are playing Euclid's Game. They start by writing two positive integers on a blackboard. Grogg goes first. On each player's turn, they must subtract some multiple of the smaller number from the larger number without making it negative, write that number on the board, and then erase the largest number from the board. The player that writes the integer wins.
For example, if it's Lizzie's turn and the numbers and are currently written on the blackboard, Lizzie could write down and erase or write down and erase
If Lizzie and Grogg start with the integers and who has a winning strategy?
What if they start with the integers and ?
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A Prime Problem
Is this number a prime or composite? How do you know?
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The Odd Octagon
Which area is greater...purple or orange?
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Drawing Doodles
Let's draw some doodles! Grab a blank piece of paper, and colored pencils or crayons in a few different colors. With a pencil, draw a doodle using the following rules:
Every stroke must be drawn using a continuous line or curve, without any retracing. A stroke may not intersect itself (though it may intersect other strokes).
Every stroke must either start and end at the same point, or start and end at different points along an edge or corner of the page.
Once you've drawn your doodle, color it in with as few colors as possible so that each region (bounded by pencil strokes) is colored in, and no two adjacent regions have the same color.
Try drawing a few doodles this way (or get someone else to contribute a doodle!). How are your doodles different? How are they similar? If you notice anything interesting, can you find a way to explain what's going on?
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Dice Trick
Here's a math trick! Roll a pair of standard six-sided dice, and write down the following four products:
The product of the top numbers of the dice;
The product of the bottom numbers of the dice;
The product of the top number of the first die and the bottom number of the second die;
The product of the bottom number of the first die and the top number of the second die.
Add the four products. You should get !
Why does this trick work?
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Dominono
Dominono is a two-player game played on a square grid, like the one below. Players take turns putting their mark on any vacant square, with one player using X's and the other using O's. The person who forms a domino - that is, who marks two squares that share an edge - loses.
Which player has a winning strategy on a board? How about a board? If you want a real challenge try thinking about a board.
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Blanks
Fill in each blank below with a digit so that the equation is true and no digits are unnecessarily included. (For example, the rightmost digit cannot be a as so the digit in the blank would be unnecessarily included.)
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The Word Store
At the word store, each vowel sells for a different price, but all consonants are free. The word "triangle" sells for $6, "square" sells for $9, "pentagon" sells for $7, "cube" sells for $7, and "tetrahedron" sells for $8. Try to find the longest word you can definitely buy for under $12!
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Cuckoo’s Egg
What number comes next in this famous sequence?
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WOW
How many whole numbers less than 10000 can be split into three numbers, each of which is a palindrome?
For example, you can split 113534 into 11 | 353 | 4 and each piece is a palindrome.
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Data Divide
Grogg has a data set of 2n distinct points in the plane. Can he always draw a separating line so that there are n data points on either side of his line?
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Queenpeace
Arrange n black queens and n white queens on a chessboard so that no black queen attacks a white queen. An example with n=4 is shown below.
Try to make n as big as you can!
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Marker Problem 3
Five rectangles. The tilted black rectangle has area 16. What's the shaded area?
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Polygon Symmetry
Start with a convex regular polygon Make a new convex polygon by using only some of the vertices of Is it possible to make a polygon in this way so that has rotational symmetry, but no reflectional symmetry?
An example and are shown below, with drawn in red.
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Marker Problem 2
Two copies of the same right triangle. What's the missing length?
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Candyland
Infinitely many math beasts stand in a line, all six feet apart, wearing masks, and with clean hands.
Since Grogg is generous, he decides to give away his pieces of candy. He gives one piece of candy to each of the next beasts in line and then leaves the line.
The other beasts repeat this process: the beast in the front, who has pieces of candy, passes one piece each to the next beasts in line and then leaves the line.
For some values of another beast (besides Grogg) temporarily holds all the candy. For which values of does this occur?
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Marker Problem
What fraction of this rectangle is shaded?
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A Flip Side
Every card in this deck has a number on one side and a letter on the other. The same number can appear on more than one card.
Euna places four of these cards in a row, then flips over some (maybe all) of the cards and mixes them up. The before and after state is pictured below. What number is on the other side of the A?
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Jewelry Design
10 silver beads and 10 gold beads are arranged randomly on a necklace. Is it always possible to make one straight line cut that divides the necklace into two pieces that each have 5 silver and 5 gold beads?
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Weird Equations [Pt. 2]
Let weird(n) be the number of ways to write n as a + a + b + c, where n, a, b, and c are distinct positive integers. Is weird(n) an increasing function?
Surprisingly, weird(n) is not increasing... it sure seems like it should be! See if you can find a pattern for when weird(n) goes down. Can you explain what is going on?
Hint: the author wrote a script to print the first few values of n where weird(n) goes down and a few other things before the pattern became apparent! This problem is a good example of how computers can help us find patterns!
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Last Year's Grid
Is it possible to fill in a 2020 x 2020 grid with the integers from 1 to 4,080,400 so that the sum of each row is 1 greater than the previous row? (USAMTS 2020)
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Resigned Resolutions
Since I always fail my New Year's resolutions, this year, my New Year's resolutions are:
Make a new year's resolution
Fail my new year's resolution
Am I guaranteed to succeed? Guaranteed to fail? Something else?
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Weird Equations
Let weird(n) be the number of ways to write n as a + a + b + c, where n, a, b, and c are distinct positive integers. Is weird(n) an increasing function?
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Naughty or Nice
Christmas is over, and Santa needs to get to work preparing for next year. Remember, he needs to check his list twice after all. In order to check his list twice, he first writes the list, then puts an "x" next to anyone on the naughty list and a "p" next to anyone on the nice list. About how long can Santa spend per row on his list if he wants to be ready for next year? (This is an estimation problem — you don't have enough information to answer exactly!)
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Grogg's Snowmen
Grogg is building snowmen. He is very particular with his snowmen. He is only happy if each snowman he builds uses twice as much snow as either the snowman before it, or the snowman before that. The first two snowmen each used 1 cubic foot of snow. How many different possibilities are there for how many cubic feet of snow Grogg uses for the 10th snowman? How about the 100th snowman?
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Happy New Year
Using the ice blocks below, construct a 3-digit number as follows: select a hundreds digit from the top block, a tens digit from the middle block, and a ones digit from the bottom block. Without reusing the same digit from a block, construct two more 3-digit numbers. For instance, you could create the numbers 967, 416, and 638. Find the sum of the three numbers. How do your choices affect the sum? Can you explain this?
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Exaggerating Elf
Santa checks on his favorite three elves: Holly, Noel, and Kris. They have made 12 toys in total.
Santa says: "Good work. Who made the most toys? ''
Holly says: "I made twice as many as Noel.''
Noel says: "I made twice as many as Holly.''
Kris says: "I made as many toys as Noel and Holly combined.''
Santa deduces that exactly one elf is lying. Can you figure out who the liar is? Can you figure out who made the most toys?
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Avoiding Fibonacci
Is it possible to split the positive integers into sets and so that no two elements in add up to a Fibonacci number, and no two elements in add up to a Fibonacci number?
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Santa's Flight Time
Santa needs to visit all of the houses below. The numbers between each pair of houses represent the time it takes for Santa to travel between those two houses (some houses have a lot of sky traffic between them!).
Try to find the shortest trip Santa can take that visits every house.
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Favorite Number
Binary Bat, Ternary Toad, and Decimal Dog all have the same favorite number.
Decimal Dog says: "When I write down our favorite number, it has two digits!"
Ternary Toad says: "When I write down our favorite number, it has no 2s in it!"
Binary Bat says: "When I write down our favorite number, it has no 0s in it!"
What is their favorite number?
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Grogg's Secret Numbers
Grogg randomly picks two integers from 1 to 10. Lizzie writes down 3 numbers. She wins if she can guess at least one of Grogg's numbers.
What are the chances that Lizzie wins, and what is the average number of correct guesses Lizzie makes?
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Student Puzzle: MaverickMan's Maze
(This problem was originally posted by one of our very own AoPS Community students, MaverickMan, and shared with their permission. Check out the message board for more of their puzzles.)
In the following maze, you start at the bottom green panel, and you have to get to one of the other 2 green panels. You move every one hour. However, if you touch a red panel, you move every 10 minutes, and if you touch a yellow panel, you move every 1 minute. What is the fastest path through the maze?
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Boxes and Squares
Can you rearrange the boxes so that the sum of the numbers on any two neighboring boxes is a perfect square?
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R & G's Chord Game
Rosencrans and Gildenstern play a game. They have a circle with 30 points on it. On their turn, each beast (starting with Rosencrans) draws a chord between a pair of points in such a way that any two chords have a shared point (they either intersect or have a common endpoint). For example, two potential legal moves for the second player are drawn in with dotted lines.
The game ends when someone cannot draw a chord. The last person to make a move wins. Assuming they play perfectly, who do you think will win this game?
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Baby Shark
The most-watched YouTube video of all time is now "Baby Shark", 2 minutes and 16 seconds of glorious singing. Doo doo doo doo doo doo. On November 2, 2020, the number of views of Baby Shark was 7.04 billion. On December 2, 2020, the number of views of Baby Shark was 7.37 billion. Given this information, how many people would you estimate are watching "Baby Shark" right now?
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Impostor(s)
In a popular video game, 10 players compete, and 2 are randomly selected to be impostors.
As the game begins, the green player, who is not an impostor, says "red orange sus" with no knowledge of who is or isn't an impostor. What is the probability that
both red and orange are impostors?
at least one of red and orange is an impostor?
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One Ninth
The fraction has the infinite repeating decimal . That means Use this to compute in two different ways. What does this tell you about the decimal representation of this number?
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Pinball Wizard
A pinball shot is worth 1 million points. Every additional shot scores 500,000 more points than the one before. So, the second shot is 1.5 million points, the third shot is 2 million points, etc.
Grogg makes just enough shots to score a total of 100 million points, then walks away. How many points was Grogg's final shot worth?
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The Elusive Matsuura
The Fermat point is a point inside most triangles that forms three angles with segments to the three vertices.
A Matsuura triangle is a triangle whose side lengths are all integers, and whose three interior segment lengths from the point are also integers.
Find some Matsuura triangles, or prove they do not exist.
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Leftovers
Grogg, Alex and Lizzie each bring a pie to their Thanksgiving dinner party. They each want to make one cut through every pie at the dinner table, which means the pies will have 3 cuts each.
Grogg suggested cutting his pie into 4 slices, like this. Alex said he would like to cut his pie into 6 slices. Lizzie found that there is a way to cut the pie into 7 slices!
It looks like there are many ways to cut these pies when they each made one cut!
Howard joined the party a bit later and bought another pie! So much pie! How many slices can they get if they each cut this pie?
Remember, this pie will be cut 4 times (since they will each be making one cut through the pie), but the cuts do not necessarily have to go through the center of the pie, and slices do not have to be the same size. The cuts do have to be straight and go all the way across the pie.
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Four Corners
Four cups are placed on the four corners of a rotating table. Each glass is randomly face-up or upside-down. Your task is to turn them all in the same direction. Sounds too easy, so here's the catch:
you have to play blindfolded
you have to grab exactly two glasses at a time and optionally flip either of them over.
after you put the two glasses back, the table is spun randomly
Can you find a way to get all the glasses facing in the same direction?
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Pepperoni Pizza Problem
The pizza below is covered with pepperoni. Can you make a straight cut to separate the pizza into two parts such that the ratio of the pepperoni on the two parts is 3:4? What other ratios can you get?
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Sense and Nonsense
Orangutoads tell the truth on Mondays, Wednesdays, Fridays, and Sundays. They lie on Tuesdays, Thursdays, and Saturdays.
Octapugs tell the truth on weekdays. They lie on weekends.
One day, an octapug and an orangutoad meet. The orangutoad says "I lied yesterday." The octapug replies, "so did I!" What day of the week is it?
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Eight Hungry Broccolemurs
Six broccolemurs can eat six ounces of Thanksgiving leftovers in six minutes. How long will it take eight broccolemurs to eat eight ounces of Thanksgiving leftovers?
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Pie Die Roll
Calamitous Clod and Winnie both want the last piece of pumpkin pie. Clod suggests a game to decide who gets it: they will each throw one of the dice shown below, and whoever gets the higher number gets the last piece.
"They all have different numbers on them," says Winnie. "Maybe you'll pick the best one."
"Okay," says Clod, "I'll let you pick first."
Should Winnie accept Clod's offer? If so, which die should she pick?
Here are the dice. They are drawn "unfolded" so you can see all the numbers.
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Pie Logic
Six logicians have just finished their Thanksgiving repast, and are ready for dessert. Their host comes around and asks "Would all of you like some pumpkin pie?"
First logician: "I don't know."
Second logician: "I don't know."
Third logician: "I don't know."
Fourth logician: "I don't know."
Fifth logician: "I don't know."
Sixth logician: "No."
Who wants pumpkin pie?
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Black Friday Strategy
Instead of going out to buy TVs on Black Friday, your friends choose to do a little geometry. You have the idea of the Shopper's Distance, where you measure how far away two aisle intersections are by counting the shortest number of aisles you must go through on a square grid.
For example, the two points below are a shopper's distance of 3 away from each other.
Can you place three friends in an equilateral triangle using shopper's distance? A square? Can you make a circle around an intersection point? What other shapes can you make, and how do they look different on Black Friday?
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Tryptophan
After eating too much turkey, Alex, Winnie, and Lizzie got very sleepy. They all fell asleep at exactly 8:00pm and have the following sleep patterns:
Lizzie sleeps for 5 hours at a time, wakes up, then immediately falls back asleep.
Winnie sleeps for 4 hours at a time, wakes up, then immediately falls back asleep.
Alex sleeps for 3 hours at a time, wakes up, then immediately falls back asleep.
The little monsters will all get up if they all happen to wake up at the same time. What time will it be when the three little monsters get up?
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Picture Proof Jeopardy
This handy picture provides a "proof" of an interesting fact in arithmetic. Can you figure out what it is proving?
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Clockwise
The numbers from to are placed clockwise on a circle. We move around the circle clockwise erasing every other number until only one number remains. If the first number we erase is what is the last remaining number on the circle?
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Creepy Coincidence?
Grogg noticed some strange things last week. Help him determine if each observation is a coincidence or mathematically guaranteed to happen.
In Grogg's woodshop class, which has 27 people in it, 3 people were born in the same month of the year!
Grogg picked 11 numbers at random from 1 to 22. Two of the numbers added to 23!
Grogg and his friends (14 beasts in total) split 100 pieces of Halloween candy. Two of them ate the same number of pieces of candy!
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Lo-Tech Baking
Alex and Lizzie are making brownies and need them to bake for exactly 15 minutes. Unfortunately, they only have an 11-minute hourglass and a 7-minute hourglass. Can you find a way for them to time exactly 15 minutes?
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Sideways Nim
Calamitous Clod has informed the little monsters that he will return Professor Grok if they can beat him in a game of Sideways Nim.
In a game of Sideways Nim, there are several piles of coins and each pile has some number of coins in it. On your turn, you can remove exactly one chip from any number of piles you want (with the caveat that you must remove at least one chip in total). The winner is the person who takes the final chip.
Since Calamitous Clod is very generous, he says he will let the little monsters choose how many piles of coins they want to play with and if they want to go first or second. Once they do this, a random number of coins will be placed in each pile and the game will begin.
Can you find a strategy to guarantee the little monsters have at least a chance of victory? How about
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The Vexing Hexagon
In this regular hexagon, which is greater, purple or orange?
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Two Groups of Three
Suppose that you're at a six-person party, where any two people are either acquaintances or strangers. Show that at this party, there must be two groups of three people who are either mutual acquaintances or mutual strangers. (These groups can overlap, and it's also possible to have one group of mutual acquaintances and one group of mutual strangers.) Must there be more than two such groups?
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Dissecting the Heart
The following shape is formed by putting together three congruent squares. How many values of can you find such that this shape can be divided into identical shapes? (For example, works, since we can divide this figure into three identical squares.)
What if these shapes had to be smaller versions of the original shape?
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Tiling the Plane
There are three ways to tile the plane using only one regular polygon, by using regular hexagons, squares, or equilateral triangles.
How many different tilings of the plane can you find which use two or more regular polygons, all of which have the same side length?
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Save Grogg
Oh no! Calamitous Clod has abducted Grogg and trapped him in the center of this maze.
Clod has given you the following rules for navigating the maze:
You must enter the maze from the arrow. The first tile you will step on is the tile in the bottom row labeled "".
Whenever you step on a tile, the next tile you step on must be that number of squares up, down, left, or right from the tile you're on. You cannot leave the maze. For example, the second tile you step on must either be one square up, left, or right from the bottom tile.
You can only step on a tile once.
If you can reach Grogg and have the tile numbers you step on along the way sum up to Calamitous Clod will set Grogg free!
Can you save Grogg?
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DIY Gerrymandering
In an upcoming election, each block votes for the yellow party or the green party, as shown below. The whole grid is to be divided into five districts, so that each district consists of five connected blocks. For each district, the party with the most number of votes wins the district. Then the party who wins the most number of districts wins the election. Divide the grid into five districts, so that the yellow party wins the election.
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Sidewalk Options
How many ways are there to tile a rectangle with squares and dominos?
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Tug of War Tourney
Beast Academy is going to run a single elimination tournament to determine the best tug-of-war player. The tournament will consist of 1 vs 1 tug-of-war matches. Starting with 1000 competitors, how many matches will they need to play to determine the winner?
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Knight Fight
How many knights can you place on an chess board so that no two attack each other? (It might be helpful to start with a board!)
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Breaking the Bar
R&G are playing a game where you start with an chocolate bar, and players take turns choosing any piece and breaking it along a line into two pieces. You win if the other player can't make a move i.e. the bar is completely broken down into pieces. Who has the winning strategy?
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3 out of 4
Wendell Willkie ran for President in 1940, but met only three of the four conditions for becoming President. Willkie was a natural-born citizen, at least 35 years old, and had been a resident of the US for at least 14 years. What was the fourth condition that Willkie was missing?
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Sum Equals Product
Howard bought three items at his local bodega, and when he added up the prices, it worked out to 9.96. Out of curiosity, Howard multiplied the prices, and to his surprise, the product was also 9.96! What were the prices of the three items that Howard bought?
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Two Semicircles
A semicircle is constructed on line segment Another semicircle is constructed on chord intersecting at and If and then find the length
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Friends and Friends of Friends
There are ten people at a party. Any two people are either friends, or do not know each other. If there are three people A, B, and C such that A and B are friends, and A and C are friends, but B and C do not know each other, then we say that these three people form a Friend-of-Friend triple. What is the maximum number of Friend-of-Friend triples at the party?
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Scheming
A rhyme scheme is A the first time,
Then is A again if it can rhyme,
But it switches to B,
If it changes, you see,
Then can go back to A (not A').
The rhyme scheme for the poem above is: AABBA.
Here are some other valid rhyme schemes for five-line poems:
AAAAA AABBB ABCDE ABBCA
Here are some things that are not valid rhyme schemes for five-line poems:
BABAB ABDAC ACBDE AACCA
How many possible rhyme schemes are there for five-line poems?
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Kindred Polygons
Three regular polygons share and completely surround a vertex. What are some different options for the three regular polygons? How many can you find?
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Area and Volume
A rectangular box has integer dimensions -by--by- with . Its surface area, in square units, equals its volume, in cubic units.
What are some different options for the dimensions of the box? How many can you find?
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Reciprocation
Three integers with have the property that their reciprocals add to exactly .
What are some different options for these integers? How many can you find?
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Chords
On a unit circle, points are equally spaced. One point is selected, and chords are drawn from that point to the other points.
In terms of , what is the product of the lengths of these chords.
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Algebeara's First Feast
Every bear attending the bear feast consumes exactly quarts of liquid, divided between porridge and honey. The newest bear, Algebeara, drank of the total amount of porridge and of the total amount of honey. At the end of the feast, all of the food has been consumed. How many bears are at the feast?
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Reflecting on the Next Term
What's the next term in this sequence? (Yes, we know there are lots of possibilities, but what do you think should be next?)
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Clod's Choice Claim
Calamitous Clod claims that for a certain choice of two numbers below, the product is nonnegative. Can you certify Clod's claim?
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Alex and His Primes
Alex found three numbers and such that and are all prime! What are and
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Just Add the Reciprocal
I generate an infinite sequence as follows: The first term in my sequence is Then, to get each following term in the sequence, I add the previous term in my sequence and its reciprocal. For example, the second term in my sequence is and the third term in my sequence is
How many terms of my sequence will be integers?
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Sticky Squares
Eight sticks fill in this large red rectangle to create 5 squares total. How can you rearrange these sticks to create 7 squares total?
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Lateral Quandary
The diagonals of this quadrilateral are perpendicular. The lengths of three of its sides, in some order, are 2, 3, and 4. What are the only two possible values for the fourth side of the quadrilateral?
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Sine
If A and B are both midpoints of this square, what is ?
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Grid-addition
How can you place the numbers 1, 2, 3, 4, 5, and 6 in the six empty boxes so that each row and column of the square adds up to the same number?
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Sticky Triangle
A stick is randomly broken into 3 pieces. What is the probability that a triangle can be made from the three segments?
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Orangutoad's Return
There are infinitely many lilypads in a single line. An orangutoad jumps 1 lilypad on its first jump, 2 lilypads on its second jump, 4 lilypads on its third jump, and so on, jumping a distance of lilypads on its t h jump. Can you find a way for the orangutoad to return to its starting point? (The orangutoad can choose to jump in either direction on each jump.)
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Sum 25 Product
A set of positive integers has sum 25. What is the biggest you can make the product of the numbers?
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Slow Motion
An object thrown into the air is called a projectile. The acceleration of a projectile on Earth's surface, ignoring air resistance, is
Suppose you took a video of a projectile and played it at half speed. What would be the acceleration of the object in the video?
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Triangles in a Pentagon
How many triangles can you find in the diagram below?
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Deven's Eleven
Is there a positive integer that ends in 11 and has all of its prime factors less than 11?
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The Four Cards
Shawn has dealt four cards in front of you. He claims that if a card has an even number on one side of it, then the other side of the card is blue. Which cards do you need to turn over, in order to confirm if Shawn is telling you the truth?
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100 Toys
Bebe spent 100 dollars to buy 100 toys. A spinning top costs 50 cents, a pet rock costs 3 dollars, and a slinky costs 10 dollars. If Bebe bought at least one toy of each kind, then how many toys of each kind did she buy?
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The Honest Coin Flip
You and your friend have agreed to a large wager, which is to be settled by a coin flip. The problem is, you and your friend live in different cities. Your friend could flip a coin and tell you the result by phone, but since the stakes are so high, you don't trust your friend to give a truthful result. (Similarly, you could flip a coin, but your friend doesn't trust you!) Is there a way to flip a coin (or something equivalent) that both of you could trust?
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Five Queens, Three Pawns
Place five queens and three pawns on a 5 x 5 chessboard, so that no queen attacks a pawn. Note: A queen attacks a pawn if they lie in the same row, column, or diagonal.
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Venn Diagrams: Four or More
It is easy to draw a Venn Diagram for three sets. Is it possible to draw a Venn Diagram for four sets? How about five sets? How high can you go?
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Every Day I'm Subtractin'
Pick a 3-digit number. Make the largest and smallest number you can from its digits. Then find the difference of these two numbers. Using this difference as the next 3-digit number, repeat this process.
For example, starting with 619, we first make 961 and 169, so we get the difference 961 - 169 = 792. Then, from 792, we make 972 and 279, recording the difference 972 - 279 = 693. We would then continue with 693, etc.
What happens in the end? Share your results!
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True or False?
Each statement is either true or false. How many of the following statements are true?
1. The answers to statements 2 and 3 are different.
2. The answers to statements 3 and 4 are different.
3. The answers to statements 4 and 5 are different.
4. The answers to statements 5 and 6 are different.
5. The answers to statements 6 and 7 are different.
6. The answers to statements 7 and 8 are different.
7. The answers to statements 8 and 1 are the same.
8. The answers to statements 1 and 2 are the same.
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Pooch Problem
You're walking home, a distance of 1.5 km. You walk at 1.5 m/s. With you is your dog, who excitedly runs from you all the way home at a blazing speed of 9 m/s.
Since you aren't home yet, your dog turns around and runs back to you, again at 9 m/s. (There is no delay for your dog to change direction.)
Then your dog runs home again, then back to you, etc.
What is the total distance of all the back and forth trips your dog makes while waiting for you to finish walking home?
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Shady Grid
Shade in some of the regions in the grid so that the shaded area is equal for each of the 11 rows and columns. Regions must be fully shaded or unshaded. At least one region must be shaded. The area of shaded regions must be at most half of the grid.
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Rote's Representative
Sergeant Rote needs to pick a Beast Academy representative for the international tug-of-war championships. Sergeant Rote is willing to send anyone from the Beast Academy team, as long as he doesn't choose the weakest person on the team.
There are a total of 20 beasts, who each pull with a different constant strength. Sergeant Rote wants to design a tournament, with each round planned ahead of time, which will allow him to pick a representative for Beast Academy. Each round of the tournament is a 10-on-10 tug-of-war match. Each round may end in one side winning, or in a tie if the strengths on each side are equally matched.
How many games should Sergeant Rote schedule in the tournament? Try to schedule as few games as you can and still allow Sergeant Rote to pick a valid representative.
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The Half Triangle Challenge
Calamitous Clod challenged Grogg to split the equilateral triangle below into two pieces of equal area using the shortest curve he could. Grogg's first idea was to draw a vertical line, like so.
Grogg's curve has length Can you help Grogg find a shorter curve? What is the shortest possible curve that splits this equilateral triangle into two pieces of equal area?
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An Unlikely Pattern
Winnie got an answer of on her probability homework. She typed it into a calculator just to see what the decimal form was. She saw:
What's going on there? Is that pattern a coincidence, or can you explain it?
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Don't Flip Out
Alex has discs that are black on one side and white on the other. He arranges them on a grid so that all the white sides are showing.
A move consists of taking any three consecutive discs in a row or column and flipping them over. You want the discs to make the checkerboard coloring shown below.
From the initial all-white position, what is the smallest number of moves needed to get to the checkerboard position?
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A Magic Star?
Can you place each number from 1 to 12 in the twelve circles below so that the sum along each of the 6 lines is equal?
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A Motley Crew
Snow White, Sirius Black, and Charlie Brown are wearing three solid-colored shirts, one each of white, black, and brown. "No one's shirt is the same color as their name," said the person in the brown shirt. "Wow, you're right!" replied Snow White.
Everyone is telling the truth. What color shirt is each person wearing?
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Welcome to Kentucky
Fill each box below with a digit from 1 to 9 to make a true equation. Each digit must be used exactly once. For an extra challenge, try putting in your area code instead of 606 and see if you can still make a true equation!
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Pace Yourself
In a 100-meter race, Amy, Brenda, and Calvin each run at a uniform pace throughout. If Amy beats Brenda by 10 meters and Brenda beats Calvin by 10 meters, by how many meters does Amy beat Calvin?
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On Brand
Can you divide the regular hexagon below into 12 congruent quadrilaterals? (A quadrilateral is a shape with 4 sides.)
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Triangles in Triangles
A circle is inscribed in an equilateral triangle and an equilateral triangle is inscribed in that circle. What is the ratio of the areas of the two triangles?
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Snail on a Cube
A snail starts in one corner of the unit cube and wishes to slide along the outside surface of the cube to the opposite corner. What is the shortest distance the snail would have to travel?
Bonus: What if instead the snail were traveling between opposite corners of a rectangular prism? Or an octahedron?
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Decimal Dilemma
Fill in each of the blanks with a different non-zero digit. What digit is ?
If we replace the and with boxes as well, how many more solutions do we have?
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Parentheses Puzzle
Can you add parentheses to the expression below to make a multiple of ? How about a perfect square?
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Jars of Chocolate
Sophia has access to an unlimited number of jars of chocolate sauce. The jars each either contain 360 ounces of chocolate or 193 ounces of chocolate. Can you find a way for Sophia to measure exactly 1 ounce?
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Infinite Fraction Fun
Can you write a simple expression to replace this infinite fraction?
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The Letter E
Each letter below is a variable. Compute the value of E.
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Sum Blob
Divide the square below into 4 different sections, each of which sums to the same value. For an added challenge, Calamitous Clod is covering three of the numbers that have the same value. What value is Calamitous Clod covering?
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Parallelograms in a Grid
Pick four points on the grid below to form a parallelogram. Make sure that none of the sides intersect any other points in the grid (also, there should be no points in the interior of your parallelogram). What is the area? Can you get a different value for area if you add more rows/columns to the grid?
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Birthday Seating Chart
To celebrate her birthday, Sophia throws a birthday party with Beast Academy characters. Grogg, Alex, Lizzie, Winnie, Calamitous Clod, and Professor Grok all attend. The cake is in the upper left corner of the table as pictured below. We know that:
Sophia sits closest to the cake;
Professor Grok and Calamitous Clod are as far away from each other as possible;
Grogg sits closer to the cake than Winnie does, and he sits further away from Clod;
Lizzie sits next to Sophia and across from Alex.
Where do each of the guests sit at the table?
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Great Maze Escape
Begin at the shaded 7. Follow a skip-counting pattern to escape by reaching the $75$. You can only move up, down, left, or right. Some of the numbers have been erased, but you can still use them if you figure out what they must be!
A skip-counting pattern is a sequence of numbers with a constant difference between each of the terms. For example, if we started by going from the 7 to the 9, then the next square would need to contain an 11 = 9+2.
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Transit Map of Beast Village
Below is a partial map of all the train stops in Beast Village. If the pattern continues, which train stop is directly west of train stop 100?
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Tip the Scale
A cup of water sits on a scale. You stick your finger into the cup of water, and the water does not overflow the cup. Your finger only touches the water, not the sides or bottom of the cup.
Does the reading on the scale go up when you put your finger in?
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Three Circles and a Chord
A circle of radius has a chord of length drawn. Two circles are inscribed above and below the chord as shown.
What is the area of the shaded region?
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Grid Town
Grid-town has a circular pond with radius and center There is also a horizontal river located on the -axis in Grid-town. Weven's tiny Grid-town home is located at Due to a recent drought, Weven needs to walk to the river and fill up a bucket, then walk to the pond and empty the bucket.
Of course, Weven is lazy, so he wants to walk as little as possible. Describe the shortest path that you can (draw a picture!) and find the length of your path.
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Invisible Tetris
Fames is playing tetris with 1 x 3 blocks, on a grid that is 7 units wide.
Unfortunately, Fames is playing in invisible mode. He can't see the blocks he has already dropped or his total number of points.
Can you find a strategy that will eventually allow Fames to earn at least one more point?
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Coin Flip Challenge
Two people each flip a coin. You have to guess the other person's result. If either person is right, you win. If both are wrong, you both lose. You can see the result of your own flip before guessing.
Before playing this game, you and your partner get to decide on a strategy you will use. Try to find a strategy that gives you as high of a probability of winning as you can!
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Floating Ball
A uniform trough of water sits on a table with of the trough on the table and the other overhanging the edge.
A ball which is less dense than water is placed in the trough on the part over the top of the table, where it floats. The ball is pushed gently down to the far end of the trough. Will moving the ball down the trough in this manner run a risk of making it topple off the table?
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Toothpick Puzzle
Grogg constructed the figure below out of toothpicks. Can you remove some toothpicks to leave exactly four rectangles in the picture? Try to do it by removing as few toothpicks as possible!
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Triangle in the Square
The square below has side length 1. There is a unique equilateral triangle that can be placed in this square with one vertex in the lower-left corner and the other two vertices on opposite sides, as shown. What is the area of this triangle?
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Cut the Net
A net for a polyhedron is cut along an edge to give two pieces. For example, we can cut a cube net along the red edge to form two pieces as shown.
Can you find two different polyhedra for which this process may result in the same two pairs of pieces?
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Eight Numbers, Four Equations
Fill in the boxes with the numbers 1 through 8, using each number once, so that all the equations are true.
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Half and Half
Kayla and Layla are twin sisters who have the same walking speed, and the same running speed. One day, they take a trip to the park. Kalya walks for half the distance, and runs for half the distance. Layla walks for half the time, and runs for half the time. Who arrives at the park first?
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The Folded Semicircle
A semicircle with a diameter of 6 is folded over a crease. The folded portion touches the diameter at a point that splits it into lengths of 4 and 2. Find the length of the crease.
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Attack of the Knights
What is the smallest number of knights that must be placed on an chessboard, so that all squares are under attack? Note: A knight also attacks the square it is on.
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Three Piles of Coins
You are playing a game with three piles of coins. You are allowed to make the following move: You can move coins from one pile (say pile A) to another pile (say pile B), as long as you double the number of coins in pile B. Using these moves, is it always possible to make one of the piles empty?
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Two Times Two
Solve the number puzzle below, where each letter stands for a different digit:
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Five Out of Six
Portia is thinking of four numbers. When she multiplies them in pairs, she gets six products. She tells you that five of these products are 2, 3, 4, 5, and 6. What is the sixth product?
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Triangle in a Grid
Do there exist three lattice points in the coordinate plane, which form the vertices of an equilateral triangle? (A lattice point is a point where both coordinates are integers.)
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Magic Multiplication
You want to fill each square in the grid below with a different positive integer, so that the product of the four numbers in each row, each column, and both diagonals is always the same constant. If is the largest number in the grid, how small can be?
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NEAR vs. EARN
In a dark corner at AoPS headquarters, a monkey is randomly pressing keys on a typewriter. If the monkey ever types the word NEAR, then it will be rewarded with a chocolate-covered banana. But if the monkey ever types the word EARN, then it will be immediately promoted and given a nice corner office. Which is more likely to occur first?
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Magically Latin
Fill each square in the grid below with a number from 1 to 5, so that every row contains all the numbers from 1 to 5, every column contains all the numbers from 1 to 5, and both main diagonals contain all the numbers from 1 to 5. Some of the squares have already been filled in.
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Points on a Pentagram
Ten points are marked in the pentagram below. Is it possible to color some of these ten points red, so that there are exactly three red points in each of the five lines of the pentagram?
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A Very Prime Product
A three-digit number and a two-digit number are multiplied. Each represents a prime digit and are not all necessarily the same. Determine the value of each digit.
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The Fuse Problem
You have two fuses that look like ropes, and each fuse burns for exactly one minute. However, the fuses don't burn uniformly, so for example, 90% of a fuse might burn in the first 10 seconds, and then the remaining 10% burns for 50 seconds.
(a) Using the two fuses, measure an interval of 30 seconds.
(b) Using the two fuses, measure an interval of 15 seconds.
Try to generalize. What time intervals can you measure with fuses?
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Two Rings in Space
Given a regular octahedron where all the edge lengths are 2, the incircle of one face is drawn (shown in red), and the circumcircle of an adjacent face is also drawn (shown in yellow). Find the shortest distance between a point on the red circle and a point on the yellow circle.
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A Six-Digit Equation
(a) Enter the digits 1 through 6 into the boxes below, to get a true equation.
(b) Enter the digits 2 through 7 into the boxes below, to get a true equation.
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Cutting a Ring
What is the maximum number of pieces a ring can be cut into by 3 straight lines? In the example below, the ring is cut into 6 pieces.
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The Egg Timers
I have two egg timers: One egg timer tells me when 7 minutes has passed, and the other egg timer tells me when 11 minutes has passed. Is it possible to use the egg timers to boil an egg for exactly 15 minutes?
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Tetromino Tiles
Find the largest number of tetrominos below that can be placed on a grid, without overlap. The tetromino can be rotated and/or reflected.
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Divisibility Graph
For a positive integer we mark points labelled in the plane. We join two points with an edge if the label of one point divides the label of the other point. Find the largest for which it is possible to draw all the edges, so that no two edges intersect.
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Twelve Circles
Place the numbers from 1 to 12 in the circles, so that the sum of the integers along each side of the square is 25.
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Angle in a Square
In square is the midpoint of side and is the midpoint of side Let be the intersection of and Find
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What is my Polynomial?
I am thinking of a polynomial with nonnegative integer coefficients. If you ask me, "What is the value of ", where is an integer, I will give you the answer. What is the minimum number of questions you need to ask me in order to determine
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Anyone's Game
Mr. X and Mrs. O are bored of playing regular tic-tac-toe, so they decide to come up with a variant: On any turn, a player can write either an X or an O. The first player to write three symbols in a row wins. In this variant, does either player have a winning strategy?
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Harmonic Integers
Prove that
is never an integer for
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A Twisted Square
Which is greater, teal or orange?
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Everything Adds Up
Is it possible to enter the numbers from 1 to 9 into the boxes below, so that all the equations work?
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A Domino Problem
Can you cover this grid with dominoes, without overlap?
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How to Tame Your Square Roots
Given positive real numbers a, b, and c, solve for x:
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Can You Trap the Monster?
A monster is moving in a straight line in the Cartesian plane at a constant speed so that every minute, it arrives at a lattice point (that is, a point with integer coordinates). Suppose that every minute you can throw a net at a lattice point that will trap the monster if it is there at that minute. You don't know where the monster started or where it is. Can you strategically throw nets in such a way that you are guaranteed to eventually trap the monster?
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A Star Stumper
In the star below, what is the sum of the angles marked in yellow?
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Five Circles
The large circle has twice the diameter of each of the small circles, and none of the four orange areas overlap. Which area is greater, the total orange area or the total red area?
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Cutting the AoPS Cube
Can you cut this cube into twenty-seven cubes using fewer than six straight cuts, each of which keeps individual cubes intact?
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The Magic Chessboard
I have a magic chessboard: I can choose any row or column and swap the colors of every square in that row or column (so that every white square turns blue, and every blue square turns white). Can I turn the coloring in the left picture to that of the right picture?
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Two Squares and a Semicircle
The diameter of the semicircle is 10. What's the total area of the two shaded squares?
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The Census Taker and the Children
A census taker walks up to a house, and knocks on the door. A woman answers.
Census taker: "Can you tell me about your children?"
Woman: "I have three children, and the product of their ages is 36. Also, the sum of their ages is the same as our house number."
The census taker observes the house number, and does some calculations.
Census taker: "I can't figure out the ages of your children."
Woman: "My oldest child plays the piano.
Census taker: "Now I know the ages of your children!"
How old are the children?
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Not Quite a Venn Diagram
Which is greater: red or teal?
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Revenge of the Decimals
Is the sum of all real numbers less than 1 with terminating decimal expansions finite or infinite?
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Warped Tic-Tac-Toe
The game of Warped Tic-Tac-Toe is played just like regular Tic-Tac-Toe, but with two differences: Firstly, the game is played on an grid instead of a grid. Secondly, a player can win by making a warped diagonal - a diagonal that wraps around the sides of the grid. For example, when X could win by placing five pieces into any of the following arrangements.
Can the player that goes second win Warped Tic-Tac-Toe?
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Hexagonal Tiles
A triangle is composed of regular hexagons, with n hexagons on each side. The triangle for n = 6 is shown below.
For which values of n can the triangle be tiled by the tiles shown below?
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3 in a Line
Use this puzzle for a MathWalk! Chalk out the puzzle on the sidewalk, solve it, and take a photo to share in the comments! Then erase your work and leave the puzzle for the next walker to try.
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Too Little Information?
The navy segment has length 100. Can you find the teal area?
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Teal and Yellow Squares
Enter a number from 1 to 5 in each of the squares below, so that:
Each row contains the numbers 1, 2, 3, 4, 5.
Each column contains the numbers 1, 2, 3, 4, 5.
The sum of the numbers in the yellow squares is 38.
The product of the numbers in the teal squares is 86400.
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Counters in a Circle
We divide a circle into sectors, and we place a counter in each sector. A move consists of choosing two counters, and moving each counter to an adjacent sector. (We can move the counters in the same direction, or in opposite directions.)
An example move is shown below. For which is it possible to move all the counters into the same sector?
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Odd Coefficients
Show that for any positive integer the number of odd coefficients in the expansion of is always a power of 2.
Can you determine which power of 2? Try to generalize your result, by finding the number of odd coefficients in the expansion of
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High-Low, High-Low, Round and Round We Go...
Pick your favorite three-digit number where not all the digits are the same. Put the digits in order from largest to smallest to form a number, and then in order from smallest to largest to form a second number (which might have leading zeros). Subtract the smaller number from the larger number. Then keep repeating this process with the number that you get.
For example, if my favorite three-digit number is 804, the number I'd get by putting the digits from largest to smallest is 840, and the number I'd get by putting the digits from smallest to largest is 048. Then, the number I'd get is 840-048=792, and I could repeat this process with the number 792.
What do you notice about the sequence of numbers you get? Can you explain your findings?
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Four Points, Two Distances
Arrange four points in the plane, so that they determine only two different distances. For example, four vertices of a square will work, because there are only two different distances (teal and yellow).
How many other arrangements can you find?
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A Stymying Grid
In the grid below, what are a, b, and c?
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A Geometric Irrationality Proof
In the following sequence of pictures, all segments of the same color have the same length. How can you use these pictures to show that is irrational?
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One Through Eight
Place the numbers 1-8 in the eight boxes below so that no two consecutive numbers are adjacent horizontally, vertically, or diagonally:
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The Liars Club and the Very Large Number
At a meeting of the Liars Club, the president writes a large number N on the blackboard.
The first member says "N is divisible by 1."
The second member says "N is divisible by 2."
The third member says "N is divisible by 3," and so on, until the 31st member who says "N is divisible by 31".
Exactly two members were lying, and they spoke consecutively. Who were the two members who were not telling the truth?
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Squares and Circles
Which is greater: teal or navy?
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Bella’s Nine Cards
Bella has a set of nine cards, and each card has a number written on it. Note that there are ways to choose a pair of cards. Bella claims if you choose two of her cards and add the two numbers, the probability of getting that sum is the same as if you rolled two dice. (For example, the probability of getting a sum of 4 with Bella's cards would be the same as the probability of getting 4 with a pair of dice, which is )
Can Bella be right? If so, what are the numbers on her cards?
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A Highly Divisible Cubic
Let where are integers and and let be a positive integer such that
What's the highest value of that you can get?
Note: The notation means for some integer
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MATHCOUNTS Week: 2019 State Team #5
Mr. Schwin has a large jar containing M&Ms, each with the letter “m” stamped on it. He removes 1000 candies from the jar and removes the letter “m” from each one. He then returns all of the M&Ms to the jar. After thoroughly mixing up the candies in the jar, he randomly removes 1000 candies from the jar and finds that 245 of them do not contain the letter “m”.
What is the expected number of M&Ms in the jar?
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MATHCOUNTS Week: 2018 State Sprint #23
Bryan visits a carnival booth where Carl shows him 10 boxes. Exactly one of the boxes contains a gold coin; the other boxes are empty. Bryan randomly takes one of the boxes, but he doesn’t open it. Carl then opens five other boxes that he knows are empty and shows Bryan that they are empty. Carl then tells Bryan he can either keep his initially chosen box or return it and choose one of the remaining closed boxes instead.
If Bryan chooses to return his box and choose another one instead, what is the probability Bryan will choose the box with the gold coin?
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MATHCOUNTS Week: 2017 State Target #4
The addition table shown has rows and columns labeled with integers a, b, c and d, in that order. A few of the sums in the table are already filled in; for example, the table shows that a+d = n – 2.
When all sixteen sums are filled in, what is the sum of the sixteen entries in the table, in terms of n?
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MATHCOUNTS Week: 2016 State Sprint #29
In the list of numbers 1, 2, …, 9999, the digits 0 through 9 are replaced with the letters A through J, respectively. For example, the number 501 is replaced by the string “FAB” and 8243 is replaced by the string “ICED”. The resulting list of 9999 strings is sorted alphabetically.
How many strings appear before “CHAI” in this list?
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MATHCOUNTS Week: 2015 State Team #2
Using the figure of 15 circles shown, how many sets of three distinct circles A, B, and C are there such that circle A encloses circle B encloses circle C?
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A Party Problem
Leif and their partner invited 5 couples over for a party. Some of the people at the party met each other for the first time (but, of course, no one met themselves or their partner for the first time). At the end of the party, Leif asked everyone else how many new people they met at the party, and got 11 different answers. How many people did Leif meet at the party?
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Sperry the Spinning Spider
Sperry the spider is practicing spinning a web, and wants to copy the spider's web below, which has 16 dots with 28 line segments connecting them. Sperry wants to start at one of the dots and spin a web from dot to dot along the given line segments, spinning across every line segment at least once. However, Sperry doesn't have a lot of thread, so they want to do this crossing the fewest number of line segments possible.
What's the shortest route you can find for Sperry? (Sperry can visit a dot or cross a line segment more than once.)
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#FlattenTheCurve
On the first day back at school, a math teacher gets their students to meet and greet each other by shaking hands. They specify that only one handshake is allowed between any two people, no one may shake hands with themselves, and all handshakes are between two people at a time. After a while, they say, "Stop - now write down on the board how many hands you shook. Then go wash your hands!" The students wrote down the following numbers on the board.
Can these students count?
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A Connected Network
Icarus Airlines is a new company vying for a place in the competitive airline market. Their CEO makes the following announcement: "We'll get you to your destination as fast as possible - we're only going to have direct flights that work in both directions. We're going to start flying from 25 cities, and fly to 12 cities from every city we serve. That may not sound like a lot, but we promise you can get between any of our 25 cities on Icarus Airlines flights."
Can the company uphold the CEO's promise?
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An Expanding Network
Icarus Airlines is expanding. Their CEO makes the following announcement: "Business has been good, and the time is ripe to expand the network of cities we serve. We will uphold our mission of only flying direct flight routes that work in both directions, but we're going to make sure we only run the flights that see the most demand. In particular, we won't run all three possible direct flight routes between any three cities we serve. We promise that by the end of our expansion, we'll be running a total of 650 different direct flight routes and serve 50 different cities."
Can the company uphold the CEO's promise?
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Student Puzzle: Liars and Truth-tellers
(This problem was originally posted by one of our very own AoPS Community students, LivelyQuotient, and has been featured here with their permission.)
You come across a fork in the road, in an unknown country. The only thing you know is that some people tell the truth, and some people lie. You are trying to get to the capitol, but you have no map. There are two people there, and when you ask them how to get to the capitol, here's what they tell you:
The girl on the left says, "If you asked her, she'd tell you to take the right path."
The girl on the right glares at her and says, "She's lying!"
You thank them, and continue on your way. Which path do you take?
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Three Semicircles
The area of the teal semicircle is 1. What's the navy area?
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The Star of David
Can you place the numbers 1 through 12 in the circles below so that the four numbers in each line add up to 26?
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Ribbit! Ribbit!
Hey, remember these frogs from a few weeks ago? Well, some purple-eyed frogs came to join the party.
The red-eyed frogs and blue-eyed frogs want to trade places so that the group of red-eyed frogs is sitting on the right two lily pads, the group of blue-eyed frogs is sitting on the left two lily pads, and the two purple-eyed frogs are back where they started. All the frogs can move to the left or right, but they're pretty restricted in how they move: Each frog can either slide from one lily pad to an empty lily pad next to it, or jump over a frog of the other group if there is an empty lily pad behind that frog.
What's the smallest number of moves you can do this in?
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The Bouncing Ball
A bouncy ball is fired at a 45 degree angle from one corner of a (frictionless) metal box measuring 209 cm by 161 cm. How many times will it rebound off the sides of the metal box before it reaches a corner? Can you generalize?
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AoPS is Not Responsible for Your Broken Picture Frames
If you hang a picture with a string looped around two nails, as shown below, removing either nail will not cause the picture to fall.
Can you find a way to hang a picture with a string looped around two nails so that the picture would fall if either one of the nails were removed?
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Triangle Sums
Place the numbers 1 through 9 in the circles below, so that the four numbers in each line add up to 20.
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A Rhombus Riddle
What fraction of the rhombus is shaded?
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We're Not in Hanoi Anymore
Four blocks are situated on one of three towers, as shown below. Holes are drilled through each block, so that each block fits perfectly on the tower at its current position. A move consists of moving the top block of a tower to another tower.
A possible first move is shown below in the left image. You can move a block as long as it can fit, but a block cannot exceed the height of the tower. For example, in the right image, block 4 can be moved to the middle tower, but not to the right tower.
Your goal is to move all four blocks to another tower. What is the smallest number of moves you can do this in? Can you generalize this to n blocks?
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The Confounding Clock
I have a weird clock: It tells me whether the current time is in A.M. or P.M., but the hour and minute hands are identical. Are there times during the day when I can't tell the time by looking at my clock? How many times does this occur in a 24-hour day?
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Not Quite Fermat's Last Theorem
If are positive integers, do there exist any solutions to the equation ?
Can you generalize?
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A Winding Road
Which color is greater: teal or navy?
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Ribbit!
There are three red-eyed frogs and three blue-eyed frogs sitting in a row of lily pads with a single empty lily pad separating the two groups. The two groups of frogs want to trade places so that the group of red-eyed frogs is sitting on the right three lily pads, and the group of blue-eyed frogs is sitting on the left three lily pads. Unfortunately, though, the frogs are pretty restricted in how they move:
Red-eyed frogs can only move to the right, and blue-eyed frogs can only move to the left.
Each frog can either slide from one lily pad to an empty lily pad next to it, or jump over a frog of the other group if there is an empty lily pad behind that frog.
What's the minimum number of moves required for the frogs to trade places? Why? What if more frogs come along?
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Increasing Distances
Fill the teal squares with the numbers 1 through 15, so that the distance between squares 1 and 2 is less than the distance between squares 2 and 3, which is less than the distance between squares 3 and 4, and so on, up to the distance between squares 14 and 15.
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The Points Don't Matter
1000 points are placed on the plane. Can you always draw a line l such that the two regions on either side of l (excluding l) contain exactly 500 points each?
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A Peculiar Product
The expression gets closer and closer to as gets larger. But is the infinite product
finite or infinite?
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Mushroom Puzzle
Which color is greater: teal or navy?
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Grid Problem
Each symbol below (circle, triangle, square) has a weight. The number to the right of a row or below a column represents the total weight of the symbols in that row or column. What total weight should replace the question mark (?) in the third row?
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Digit Sorting
Can you sort the digits 1 through 9 into the top or bottom row, using each digit exactly once, so that the product of the digits in the top row equals the sum of the digits in the bottom row?
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Numbercross
Every box contains a digit such that when the numbers in each row or column are read from left-to-right or top-to-bottom, they fit the clues above. Can you solve the puzzle? What's 5-Across?
Across
1
4 A multiple of
5 The smallest three-digit number in the gridDown
2 A multiple of
3 A power of
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Bot Golf
In the Bot Golf Challenge, we give you a starting number of golf balls, and a target number of golf balls. You need to make a combination of Bots that turns the starting number into the target number.
Below are three different Bots we can use. How few Bots can you use to get from 100 —> 99?
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7540 | https://www.omnicalculator.com/math/cube | Last updated:
Cube Calculator
Our cube calculator can help you find all the cube parameters. Whether you want to determine the volume of a box or check the area of a die, this flexible tool is the thing you're looking for. Type in one parameter out of five - cube volume, cube surface area, face diagonal, cube diagonal or cube side - and, in a blink of an eye, we will show you the rest. Give it a go! If you are still unsure how to find the volume of a cube formula, keep scrolling to the description part where we will explain everything in detail.
Cube and the others
Let's start from the beginning - what's a cube? It's a 3D solid object bounded by six square faces with three faces meeting at each vertex. It's a regular square prism in three orientations. The cube is the only regular hexahedron, and it has
6 faces;
12 edges; and
8 vertices.
Are you interested in a rectangular prism calculator? We have just the tool for you.
Cube surface area
The formula for the cube surface area is really easy - it's the area of one face times 6, as every cube has six identical square faces. The area of a square is equal to a² where a is the length of one edge, so the cube surface area equation is:
surface_area = 6 × a²
Cube volume formula
To calculate the cube volume, raise the edge length to the third power:
volume = a³
You can think about the volume of a cube formula as calculating any other prism volume - simply multiply the base area times height of the solid. Our base is a square, so its area is a², and our height also equals a as all edges are the same. So we got the same formula - a³ - as expected.
Cube diagonals - face and space
If you are wondering how to find the cube diagonals, think about the diagonal of a square for a while. The formula for square diagonal is the side length multiplied by square root of 2, and it comes from the Pythagorean theorem:
face diagonal = √(a² + a²) = √2a² = a√2 - it's our cube × face diagonal
For cube diagonal, all you need to do is to use the Pythagorean theorem once more:
cube diagonal = √((a√2)² + a²) = √3a² = a√3
So the diagonal of the cube is equal to the length of the side times the square root of 3.
Are you interested in learning more about the Pythagorean theorem? Visit our pythagorean theorem calculator.
How to find the volume of a cube?
Let's calculate the volume of a Rubik's cube:
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Wow, that was quick! All the other parameters appear. Thanks to the cube calculator, we have just learned that:
Volume is 11.4 cu in.
Surface area is 30.4 in².
Cube diagonal is 3.9 in.
Face diagonal is 3.18 in.
Now, try a small task to check the flexibility of our tool: imagine you have one gallon of water and you want to pour it into a cubic tank. You can use this cube calculator to check how big it needs to be!
If you are interested in learning more about calculating surface area you should check out our [surface area of a cube calculator].(calc:1934)
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7541 | https://math.stackexchange.com/questions/4774985/proving-two-triangles-congruent-given-two-congruent-sides-and-a-congruent-median | geometry - Proving two triangles congruent given two congruent sides and a congruent median - Mathematics Stack Exchange
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Proving two triangles congruent given two congruent sides and a congruent median
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The title was a bit too short for me to fit the full details, so here's the scenario I have. Prove that two triangles are congruent if in two triangles, the median from the common vertex and two sides is congruent with the corresponding median and the two sides in the second triangle.
Trying to figure this question out for my geometry class but unsure about exactly what I still need. We learned about SAS, ASA, and SSS in our last unit. The wording I'm a bit confused on, is it saying that all 4 sides are congruent or just that the corresponding side on each triangle is congruent with the other corresponding side? I can tell we'll have to use either SAS or SSS for this question as we know 2 of the sides are congruent on each triangle but I'm not fully sure of which congruency method I need to use yet. I'll include the most recent theorem we learned in our last lesson. Any help on how to solve this would be appreciated!
The median to the base of a triangle is also the angle bisector to the angle at the vertex opposite the base of this triangle in an isosceles triangle.
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edited Sep 25, 2023 at 3:02
dfnu
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asked Sep 24, 2023 at 23:29
nadelocknadelock
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Suppose you extend the median A D A D in triangle A B C A B C (with sides a a, b b, c c) beyond the point of intersection D D with the side B C B C by an amount, equal to the size of the median m m, giving the segment A E A E. Join E E with B B and C C. It is easy to see that A B E C A B E C is a parallelogram. His diagonals are d 1=A E=2 m d 1=A E=2 m and d 2=a d 2=a. By the parallelogram identity,
2(b 2+c 2)=d 2 1+d 2 2=(2 m)2+a 2 2(b 2+c 2)=d 1 2+d 2 2=(2 m)2+a 2
Therefore, sides b b, c c and d 1=2 m d 1=2 m uniquely determine side a a. As a consequence, your data completely determine the triangle.
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answered Sep 25, 2023 at 1:07
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First some clarification on the problem. It is not required that the four sides are all congruent. Let △A B C△A B C with median A M A M and △A′B′C′△A′B′C′ with median A′M′A′M′ be the triangles in hypothesis. Then we know that A B≅A′B′A B≅A′B′, A C≅A′C′A C≅A′C′, and A M≅A′M′A M≅A′M′.
(Similar to the other answer, but probably more closely related to the material you mention.)
Produce A M A M to D D so that A M≅M D A M≅M D and A′M′A′M′ to D′D′ so that A′M′≅M′D′A′M′≅M′D′.
By SAS criterion you have the following congruences: △A M D≅△C M B△A M D≅△C M B, △A C M≅△D B M△A C M≅△D B M, and analogously on the other triangle, △A′M′D′≅△C′M′B′△A′M′D′≅△C′M′B′, △A′C′M′≅△D′B′M′△A′C′M′≅△D′B′M′.
Given the hypotheses, and what we just said, by SSS criterion we have, e.g., △A C D≅△A′C′D′△A C D≅△A′C′D′.
Then the median of the above mentioned triangles must be congruent, that is A M≅A′M′A M≅A′M′. Hence A B≅A′B′A B≅A′B′, and finally the thesis, by SSS criterion.
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edited Sep 25, 2023 at 3:29
answered Sep 25, 2023 at 3:01
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7542 | https://www.youtube.com/watch?v=_LXOA8Lpq_g | Algebra I #3.11c, Writing equations for Consecutive integers
JoAnn's School
173000 subscribers
16 likes
Description
1482 views
Posted: 20 Oct 2016
An explanation of how to write equations for Consecutive integer phrases, including even or odd consecutive integers. #3.11c
Intro Algebra Word Problems #22 -
Write, solve equations for Consecutive integers
Intro Algebra Word Problems #23 -
Write, solve equations for Consecutive ODD integers
Intro Algebra Word Problems #24 -
Write, solve equations for Consecutive EVEN integers
Algebra I #3.11a,
Turn compound phrases into algebraic expressions
Algebra I #3.11b,
Consecutive integers - Odd or Even
Grade 8 Math #5.3a,
Bi-variate Data - Dependent variables vs Independent variables
Intro to Algebra word problems #1a,
Write an expression from clue words
Intro to Algebra word problems #1b,
Write more expressions from clue words
Grade 8 Math #1.2a,
Classify and sort Real Numbers
Grade 6 Math #9.1,
What are integers - Absolute value
If you like my work, think my videos are helpful,
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1 comments
Transcript:
Algebra 1 number 3.11c we're talking about compound algebraic expressions still and we're writing equations for consecutive integers so we learned in the last video that consecutive integers are found by a plus one or a minus 1 to the previous integer and consecutive odd integers are found by plus two or minus 2 to the previous odd integer and consecutive even integers are also found by that plus two or minus two but it would be to the previous even integer now the reason I'm saying plus 2 minus 2 or plus one minus 1 is because consecutive integers are just in order so it could be an order going smaller or going larger okay so there's going to be a link to the previous video in this description and other links to helpful videos okay so if x equals an integer then X Plus X Plus 1 plus X plus 2 would give us three consecutive integers so X would be the first One X Plus 1 would be the second one and X plus 2 would be the third one and we can compile combine these like terms to get an equivalent equation there's three x's and it's plus three the one and the two equal three so 3x plus 3 would be that new equivalent equation and if it said the sum of two consecutive integers well we know that X could be the first integer and X Plus 1 could be the second integer couldn't it that would give us 2x plus 1 because we would combine the like terms see if it said the sum of an integer and three times the next consecutive integer well that's the integer that's the next consecutive integer so the sum of that integer and three times the next consecutive one would be X plus three X plus one see if x were an integer and we were finding consecutive odd integers and we wanted three consecutive odd integers we'd have X as the integer X plus 2 as the next one and X plus 4 is the next one we can combine like terms to get an equivalent equation for these three consecutive odd integers and we'd get three X plus six one two three x and 2 and 4 is 6. see the sum of two consecutive odd integers would be X as the integer and then X plus 2 as the next one because they're odd we skip count by two remember so that would be combined as two X plus two that would be our equivalent equation and if it said the sum of an odd integer and two times the next odd integer well there's our odd integer that's the next odd integer because it goes up by two so if we had two times this one the next odd one it would be X plus 2X plus 2. see that would be our equation let's try it for an even one if x equals an integer and we want to find three even consecutive integers we would do X plus an X plus 2 plus an X plus 4 because just like these odd ones they go up by two remember the consecutive even in odd integers go up or down by two so the same equation works for both of them okay so we can do 3x plus 6 for either one to find the next largest uh consecutive integers as they're going up now if we wanted to find them going smaller then it would be X and then x minus 2 and then x minus 4 wouldn't it all right the sum of two consecutive even integers is just like the sum of two consecutive odd integers it's X Plus X Plus 2. these would be the second even integer so we have one integer the next consecutive even integer and that would give us 2x plus 2 the same as the odd ones see it's the same thing whether they're even or odd and look at this one what if it said the difference between two times an even integer and the next smallest even integer equals eight well two times an even integer would be the 2X and the next smallest even integer would be x minus two the difference tells us its subtractions and it equals eight see so that would be our equation see all right so in the next video 3.11 D we're going to actually do some word problems that involve consecutive integers okay I hope I'll see you there bye |
7543 | https://www.wiley.com/en-us/March%27s+Advanced+Organic+Chemistry%3A+Reactions%2C+Mechanisms%2C+and+Structure%2C+9th+Edition-p-00413743 | March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 9th Edition
Michael B. Smith
ISBN: 978-1-394-24302-0
July 2025
1680 pages
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## March's Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 9th Edition
Michael B. Smith
E-Book
978-1-394-24302-0
July 2025
$144.00
Hardcover
978-1-394-24299-3
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Description
Leading reference on the theories of organic chemistry, now updated to reflect the most recent literature from 2018 to 2023
Building on the success of the 8th Edition as winner of the Textbook & Academic Authors Association 2021 McGuffey Longevity Award, the revised and updated 9th Edition of March’s Advanced Organic Chemistryexplains the theories of organic chemistry, covers new advances in areas of organic chemistry published between 2018 and 2023, and guides readers to plan and execute multi-step synthetic reactions. Detailed examples and descriptions of all reactions are included throughout the text.
As in previous editions, the goal of this edition is to give equal weight to three fundamental aspects of the study of organic chemistry: reactions, mechanisms, and structure. Specific but specialized areas of organic chemistry, such as terpenes, polymerization, and steroids, have been incorporated into primary sections rather than segregated into their own sections.
The first nine chapters cover general organic chemistry with theoretical principles. The next 10 chapters address reactions and mechanistic discussion. Appendix A focuses on literature references and resources. More than 4,400 references are included throughout the text.
March’s Advanced Organic Chemistry provides information on:
Localized and delocalized chemical bonding and bonding weaker than covalent
Microwave chemistry, use of ultrasound, mechanochemistry, and reactions done under flow conditions
Acids and bases, irradiation processes, stereochemistry, structure of intermediates, and ordinary and photochemical reactions
Mechanisms and methods of determining carbocations, carbanions, free radicals, carbenes, and nitrenes
Aliphatic, alkenyl, and alkynyl substitution, additions to carbon-carbon and carbon-hetero bonds, eliminations, rearrangements, and oxidations and reductions
This 9th Edition of March’s Advanced Organic Chemistry continues to serve as a must-have reference for every student and professional working in organic chemistry or related fields.
About the Author
Michael B. Smith, PhD, is Professor Emeritus in the Department of Chemistry at the University of Connecticut. He is a coauthor of the fifth through eighth editions of March’s Advanced Organic Chemistry and the author of Volumes 6 - 13 of the Compendium of Organic Synthetic Methods, as well as several other monographs and textbooks.
To Purchase this product, please visit |
7544 | https://www.youtube.com/watch?v=rYSpO_1ZetI | Complex Numbers as Points (3 of 4: Geometric Meaning of Multiplication)
Eddie Woo
1940000 subscribers
184 likes
Description
16293 views
Posted: 9 Nov 2015
More resources available at www.misterwootube.com
11 comments
Transcript:
Here's what we've established so far. We established this mind-blowing idea that complex numbers are not just numbers, right? We can think of them, we can represent them geometrically on the complex plane. We can draw this fancy thing called an argan diagram. And um argan diagrams are going to become your very very good friends over the next couple weeks. Now, here's what we established first, right? We started to go back through the arithmetic that we did in our first lesson. We looked at addition. Uh I should put my numbers back on. Um actually no, I don't need them. Um zed 1 was like 3 plus i and zed 2 was 1 5 i, right? And so our conclusion was you add zed 1 and zed 2. The sum of zed 1 and zed 2 is really a translation of either from zed 1 you think about adding zed 2 to it you move over that much or vice versa because addition is cognitive. Right? Uh we looked at negation. What happens when you slap a minus sign on the front of a complex number? And what we concluded was you can think of it either as reflection like you you reflect across this way. Um or alternatively what I suggested is a better um way to think about it and the reason will become clear in the next 10 minutes or so is that negation really means you rotate it around the origin by pi radians. Okay. And that's what gets you from say the first quadrant around to the third. Okay. Now addition negation which gave us subtraction. What operation is next? Multiplication. Multiplication. Okay. Now, to demonstrate multiplication for you, I'm going to abandon that zed 1 and zed 2 I gave you before. Um, we can we can work with them, but um this is starting to get a bit trickier. So, I'm going to choose some very very simple examples for you. Okay. So, for instance, and I'm going to just pop this up in the right hand corner. We're going to fill this in like what is multiplication? We'll do that after we've gone through this process. Let's just say Z one is just the number one. Now every real number that you know about actually is like one actually is a complex number. It's just that it's got no imaginary component. Okay. So one is still a complex number. It's just I don't worry about the eyes. Where is Z1 on my diagram? It's going to be on the real axis, right? Because there's no imaginary part. Okay. Um, so I'm going to put him I guess I'll place him like there. Okay, that's that's um Z one. Okay, now here's what I want to do. Like I said, I'm choosing very simple example so I can see what's happening. I want to multiply by a number. Now, what's a nice simple complex number to choose for Z 2. I'm just going to go with I. I is the quintessential complex number, right? So multiplying one by I just gives me I. Now, where is I? on my complex plane. Okay. Being that I Zed 2 is purely imaginary, right? It belongs in line with zero on the real axis, but it belongs up here. There's Zed 2. Multiplying by I has taken me from there to there. Now, at the moment, I don't know enough. You know how we're talking about a series of sequences and you need three things for a pattern, right? I've got two things right now. One, two. But I don't know what's geometrically getting me from here to here. It could be translation. It could be reflection across the line, you know, y= x if I wanted to. I don't know enough yet. So, let's do it again. Okay, Z3. When I multiply by I again, I've got I squared now. But I know what I is by definition. It's 1. Where is1 on my complex plane? one is completely real. It's purely real. Okay, so I'm back on the real axis over here. Okay, we're going to do it one more time to make it bleedingly obvious. If I multiply by I one last time, Z 4 will be 1 I, which is I. And I won't ask it again. This time, let's just plot this guy. He belongs down here on the negative imaginary axis. Okay, four times I have done the same operation. Okay, the first time I went from here to here. The second time I went from here to here. This third time, wait, first, second, three, three times, of course, from one to four. Um, three times I've done this operation. What is the geometric thing that gets me repeatedly from here to here to here to here? And you can see even in the way that I've drawn like moving my hand, right? I am clearly rotating around the origin. Do you see that? Okay. So, what I've done is I've rotated this far. I've rotated in this case by pi on 2 radians. Right? That's 90°. When I multiply by I the second time, I rotate again another pi on 2 radians. And then the last time there's my final rotation. Okay. Now you can see here if I rotated one last time Z five where would zed five be back at he'd be back here and the reason why is because multiplying by i four times is the same as multiplying by -1 two times which is the same as multiplying by one and it lands you back where you started. Okay. What's something you can do four times and you end up back where you began? And the answer is you rotate pi on two radians and you'll get back there. Okay, so I'll just dot dot dot. Okay, so therefore my conclusion is multiplication in the complex world multiplication is rotation. And now can you see why I was insisting that negation really is actually a kind of rotation? What is negation? Negation is the multiplication by a number. Right? What what is my multiplying by? Negative1. Okay. Now, multiplying by I rotates me pi on 2 radians, right? Multiplying by1, which was what I said negation is, right? That's just multiplying by i twice, which is why you go pi on 2 and pi on 2 again gets you to pi. You see what's going on? So in fact that's why negation is not really reflection even though that that's what it ends up looking like. Okay, negation is just a particular example of multiplication. It's multiplication by negative one which is going to spin you around all the way to the opposite side. Okay. |
7545 | https://www.youtube.com/watch?v=HE6ywQxD7_s | Ramps and Inclines - Newton's 2nd Law
MrEScienceTheater
1550 subscribers
18 likes
Description
1655 views
Posted: 26 Oct 2020
This video is a tutorial about how to solve ramp problems with Newton's 2nd law. The key is to rotate your coordinate axes so that the direction of acceleration lines up with one of the axes.
(My hands don't look very clean in this video; I was working on some auto repair over the weekend, and they're as clean as I could get them.)
2 comments
Transcript:
in this lesson we're going to be learning about ramps and inclines so for example we have here a block of mass m on a ramp with an angle of theta as measured the horizontal so we we follow the steps here suppose we want to know the acceleration of the block along the ramp uh let's let's go ahead and do one without friction first and then we'll see how friction would affect the problem so we begin of course as we begin all of them with a force diagram so use a dot to represent my system in this case that's just my block i start with the force of gravity which as we know is always towards the center of the earth straight down and it's touching the surface so there's going to be a normal force perpendicular to the surface and there is no friction in this case so that's it so step number two would be to write my basic equation which is net force equals mass times acceleration but in this case there's one extra little thing that i want to do i want to ask myself what is the direction that it would accelerate in this case it's going to accelerate along the surface of the ramp parallel to the surface and downhill in this case so it's going to accelerate downhill what i want to do is i want to arrange my coordinate system so that it lines up with the acceleration now what that means is instead of having a standard x's horizontal wise vertical coordinate system i'm going to rotate this so that the surface of the ramp becomes my x direction see how that works surface of the ramp becomes my x direction i'm rotating my axes so on my diagram i can illustrate that by doing this that's my x direction and that's my y direction so as part of our force diagram if we're on a ramp we rotate our axes so that the direction of acceleration lines up with one of our axes for ramps we're usually talking about the x-axis always make the surface of the ramp your x-axis just because you know for example if if it were on a level surface that would be the x direction so the fact that it's on a ramp we're going to keep that surface being the x-direction okay then i'm going to write net force equals m a so step one is the force diagram including the rotating the axes so that it lines up with the acceleration then step two the basic equation net force equals ma step three i look at the direction it accelerates i'm going to identify the forces that make up this net force now it's accelerating in the x direction so i'm only looking at x direction forces notice that the y direction is perpendicular to the surface and the normal force is completely in the y direction so in the x direction notice that there is a component of gravity that is in the x direction i'm going to call that component f g x the x component of gravity and that's what will equal mass times acceleration in this case so f g x equals mass times acceleration so now step four is i need to plug in numbers and solve so how do i find f g x well let me take this diagram over here and let me look just at the force of gravity so here's the force of gravity which of course is equal to m times g right now it's got a component in the y direction that looks like this and a component in the x direction that looks like this where x and y meet there's always a right angle so this is f g x and this is f g y the x and y components of the force of gravity now as i'm looking at this angle theta right here this angle theta is going to appear in this triangle as one of these two corners so i need to figure out which corner it is well to do that let me just take my my axis here i'm going to lay it on top so here it is the force of gravity is in the old y direction the vertical direction now when i rotated right when i rotated the x-axis through this angle theta what happened to the y-axis it also rotated through the angle theta so the angle between force of gravity and fgy is going to be equal to the angle of the ramp as measured to the horizontal the angle between the horizontal and the surface of the ramp so that means up here in this corner is my angle theta once i recognize this i can see that f g y is the adjacent side and so that's going to equal hypotenuse times the cosine and f g x being the opposite side opposite that theta is going to equal the hypotenuse times the sine of theta so this is this is good to remember because as long as you set up the rank problem this way it will always be this this will always be the case so the angle theta is the angle of the ramp measured from the surface of the ramp to the horizontal then the x and y components of gravity will equal mg cosine theta for the y component and mg sine theta for the x component and by x we mean parallel to the ramp and by y we mean perpendicular to the ramp so the parallel component of gravity is going to be mg sine theta the perpendicular component will be mg cosine theta and this will always be true as long as you make the surface of the ramp your x-axis so when you rotate your axes right from horizontal rotate it through the angle theta so that the x-axis is parallel to the surface of the ramp and then you'll always have this case where fgx equals mg cosine theta sorry fgy fgy perpendicular component of gravity equals mg cosine theta and fgx or the parallel component of gravity will equal mg times the sine of theta so now if i take this and i plug this in for f g x mg times sine of theta equals m a notice that the mass will cancel out and i get the acceleration is equal to g sine theta it's important to note that the mass did cancel out and that's great so in other words the only the angle affects the acceleration and the acceleration of gravity not the mass of the object itself it's kind of like how if you drop two objects the mass will cancel out mg equals ma and the mass cancels out just similar thing here happens on the ramp okay so let's go look at another similar example what if the ramp is tilted the other way theta block mass m right our force diagram force of gravity the normal force is this way now and again there's no friction again i'm going to rotate my axes so that the x-axis is the surface of the ramp so instead of going horizontal the x-axis is going to go this way so i've got my x-axis here and my y-axis is going to be here so when i do net force equals m a again i've got just the x component it accelerates in the x direction so parallel to the ramp so i've got f g x equals m a f g x for the same reason will equal mg sine theta mass will cancel out g sine theta equals the acceleration there it is so it doesn't matter if the ramp is tilted to the right or tilted to the left you're going to get the same mathematical result because you're always tilting your axis so that the x-axis is parallel to the surface okay always tilt it so x-axis is parallel to the surface okay now let's look at what happens when there is friction [Music] so if there is friction mu not equal to zero okay and it's going to accelerate down the ramp again so i start with my force diagram i start with gravity always going to be straight down my normal force will be perpendicular to my surface and my friction is going to be parallel to my surface looks something like this then i'm going to rotate my axes make the x-axis the surface of the ramp okay like so so the x-axis is parallel to the surface which means friction is going to be all in the x-direction and my y-axis is going to be perpendicular to the ramp which means my normal force is completely in the y-direction so now when i do my net force equals ma and it is accelerating in the x direction i'm only looking at x direction forces it's going to accelerate downhill so f g x is going to be bigger than friction so i'm going to take f g x minus the friction equals m a f g x i know is equal to mg sine theta you can redrive that every time but you might as well just memorize it fgx can be mg sine theta minus the friction now friction is mu times the normal but the normal force because it's accelerating in the x direction there is no acceleration in the y direction which means that up equals down in the y direction so the normal force is going to be balanced by f g y so mu times the normal is mu times fgy and if we recall from before fgy is equal to mg cosine theta so i can replace normal force with mg cosine theta because that is fgy and the normal force in this case is equal to fgy so i replace friction with mu times mg cosine theta that will equal m a now i notice there is an m in every term so i can divide out the m so i just get g sine theta minus mu g cosine theta equals my acceleration now we can do this with numbers but base the basic principle is the same so i just fgx if i had if i had the m and the g and the theta i could plug in those numbers and get a numerical value here get a numerical value here and solve for the acceleration get a numerical value there and with a problem like a ramp you might end up with something you know how long does it take to slide two meters down the ramp so you would do this to find the acceleration and then use that acceleration in a kinematics part of the problem but that's essentially how you deal with ramp problems the key is identify the direction that the object accelerates and then rotate your axes so that one of your axes lines up with the direction of acceleration so typically for ramps that means make the surface of the ramp your x-axis and when you make the surface of the ramp your x-axis then gravity is going to have x and y-components and those components of gravity will be mg cosine theta for fgy the perpendicular component of gravity and mg sine theta or p for the parallel component or f gx keep that in mind and ramp problems will be a piece of cake for you |
7546 | https://link.springer.com/article/10.1023/A:1008767703006 | Journal of Algebraic Combinatorics 12 (2000), 95–99 c© 2000 Kluwer Academic Publishers. Manufactured in The Netherlands.
Singleton Bounds for Codes over Finite Rings ∗
KEISUKE SHIROMOTO keisuke@math.sci.kumamoto-u.ac.jp
Department of Mathematics, Kumamoto University, 2-39-1, Kurokami, Kumamoto 860-8555, Japan Received April 22, 1998; Revised May 6, 1999
Abstract. We introduce the Singleton bounds for codes over a finite commutative quasi-Frobenius ring.
Keywords: code, QF ring, module, bound, support, weight
Introduction
Let R be a finite commutative quasi-Frobenius (QF) ring (see ), and let V := R n be the free module of rank n consisting of all n-tuples of elements of R. A code C of length n
over R is an R-submodule of V . An element of C is called a codeword of C.In this paper, we will use a general notion of weight, abstracted from the Hamming, the Lee and the Euclidean weights. For every x = (x1, . . . , x n ) ∈ V and r ∈ R, the complete weight of x is defined by
n r (x) := |{ i | x i = r}| .
To define a general weight function w( x), let ar , ( 0 6 =)r ∈ R, be positive real numbers, and set a0 = 0. Set
w( x) := ∑
r∈R
ar n r (x). (1) If we set ar = 1, ( 0 6 =)∀r ∈ R, then w( x) is just the Hamming weight of x. For later use, we denote
A := max {ar | r ∈ R}. (2) For example, if R = Z4 = { 0, 1, 2, 3}, then setting a1 = a3 = 1 and a2 = 2 yields the Lee weight, while setting a1 = a3 = 1 and a2 = 4 yields the Euclidean weight. Put N := { 1, 2, . . . , n}. Define the support supp (x) of a vector x = (x1, . . . , x n ) ∈ V
by supp (x) := { i ∈ N | x i 6 = 0}.
∗
Supported in part by the Japan Society for the Promotion of Science. 96 SHIROMOTO The minimum weight of a code C, denoted by d, is
d := min {w( x) | (0 6 =)x ∈ C}.
We make the important (and elementary) observation that
w( x) ≤ A|supp (x)|. (3) The inner product of vectors x = (x1, . . . , x n ), y = (y1, . . . , yn ) ∈ V is defined by
〈x, y〉 = x1 y1 + · · · + x n yn .
The dual code of C is defined by
C⊥ := { y ∈ V | 〈 x, y〉 = 0 (∀x ∈ C)}.
The following proposition is well-known as the Singleton bound (see ).
Proposition 1 Let C be a linear [n, k, d]-code over GF (q), where d is the minimum Hamming weight of C. Then ,
d ≤ n − k + 1.
The main purpose of this paper is to find a similar bound for the minimum weight of a general weight function w( x) over R.
Singleton bound
For a submodule D of V and a subset M ⊆ N = { 1, 2, . . . , n}, let
D(M) := { x ∈ D | supp (x) ⊆ M},
D∗ := Hom R (D, R).
Clearly D(M) = D ∩ V (M) is a submodule of V , and |V (M)| = | R||M|. It is also the case that |D| = | D∗ | for any submodule of V . The following lemma is essential. (There is a similar result over GF (q) in ).
Lemma 1 Let C be a code of length n over R and M ⊆ N . Then there is an exact sequence of R-modules :0 → C⊥(M) inc
→ V (M) f
→ C∗ res
→ C(N − M)∗ → 0,
where the maps inc, res denote the inclusion map , restriction map , respectively , and the map f is defined by f : y 7 → ( ˆy : x 7 → 〈 x, y〉). SINGLETON BOUNDS FOR CODES OVER FINITE RINGS 97
Proof: The exactness of the sequence at C⊥(M) and at V (M) is clear. That the map res is surjective follows from R being an injective module over itself (the meaning of R
being QF). Clearly we note that Im f ⊆ ker (res ). Conversely, if we take any λ ∈ ker (res ), then
λ( x) = 0 (∀x ∈ C(N − M)).
Note that V → C∗ ; v 7 → ˆ v is surjective, so there exists y ∈ V with λ = ˆy. For any
x ∈ C(N − M), 〈x, y〉 = 0, so that,
y ∈ (C(N − M)) ⊥ = (C ∩ V (N − M)) ⊥= C⊥ + V (N − M)⊥ = C⊥ + V (M).
Since ˆz = 0 for any z ∈ C⊥, we have ker (res ) ⊆ Im f.
Thus the sequence is also exact at C∗, and the lemma follows. 2
We remark that we can prove the MacWilliams identity for codes over Z4 () by using Lemma 1 (there are similar results over GF (q) in and ). Using the above lemma, we establish the Singleton bound for a general weight function over R.
Theorem 1 Let C be a code of length n over a finite commutative QF ring R. Let w( x)
be a general weight function on C , as in (1), and with maximum a r -value A , as in (2).Suppose the minimum weight of w( x) on C is d. Then
[ d − 1
A
]
≤ n − log |R| |C|,
where [b] is the integer part of b.
Proof: By Lemma 1, we have
|C| · | C⊥(N − ˜M)| = | V (N − ˜M)| · | C( ˜M)|,
where ˜M = N − M. If we take a subset M of N with | ˜ M| = [ d−1
A
], then |C( ˜M)| = 1by (3). Since we always have |C⊥(N − ˜M)| ≥ 1, we see that
|C| ≤ | V (N − ˜M)| = | R| |N − ˜ M|.
Hence the theorem follows. 298 SHIROMOTO
An application to codes over Zl
The ring R = Zl is a good example of a finite commutative QF ring. Let k := [l/2], and regard Zl as the set {0, ±1, . . . , ±k} (with k = − k, when l = 2k is even). On codes over
Zl , there are three special weight functions: 1. the Hamming weight , where each ai = 1, i 6 = 0, 2. the Lee weight , where ai = | i|, and 3. the Euclidean weight , where ai = | i|2 .Denote the minimum weight of a code C with respect to these three weights by d H , d L and
d E , respectively. It is clear that the maximum ar -value A is 1 , k and k2, respectively. The next result follows immediately from Theorem 1.
Theorem 2 Using the above notation for a code C of length n over Zl , there are the following bounds on minimum weights :
d H ≤ n − log l |C| + 1,
[ d L − 1
k
]
≤ n − log l |C|,
[ d E − 1
k2
]
≤ n − log l |C|.
The Gray map φ : Z4 → Z22 is defined by φ( 0) = 00, φ( 1) = 01, φ( 2) = 11, and
φ( 3) = 10. It is well-known that φ is a weight-preserving map from ( Zn
4
, Lee weight) to ( Z2n
2
, Hamming weight) (see ). Using the above theorem, we have the Singleton bound for certain binary nonlinear codes.
Corollary 1 If a binary nonlinear (2n, M, d)-code B , where M := | B| and d is the minimum Hamming weight of B , is the Gray map image of a code C of length n over Z4,
then
[ d − 12
]
≤ n − log 4 M.
Proof: Since M = | C| and d is also the minimum Lee weight of C, the corollary follows from Theorem 2. 2
Acknowledgment
The author would like to thank the referee for his helpful advice on QF rings and general weight functions and other helpful suggestions. The author would like to thank adviser Professor Tomoyuki Yoshida for his helpful suggestions and Dr. Masaaki Harada for his helpful comments on codes over Z4 .SINGLETON BOUNDS FOR CODES OVER FINITE RINGS 99
References
C.W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras , Interscience Publishers, New York, 1962. 2. A.R. Hammons, P.V. Kumar, A.R. Calderbank, N.J.A. Sloane, and P. Sol´ e, “The Z4 -linearity of Kerdock, Preparata, Goethals, and related codes,” IEEE Trans. Inform. Theory 40 (1994), 301–319. 3. M. Klemm, “ ¨ Uber die Identit¨ at von MacWilliams f¨ ur die Gewichtsfunktion von Codes,” Arch. Math. 49 (1987), 400–406. 4. F.J. MacWilliams and N.J.A. Sloane, The Theory of Error-Correcting Codes , North Holland, Amsterdam, 1977. 5. K. Shiromoto, “A new MacWilliams type identity for linear codes,” Hokkaido Math. J . 25 (1996), 651–656. 6. T. Yoshida, “MacWilliams identities for linear codes with group action,” Kumamoto Math. J . 6 (1993), 29–45. |
7547 | https://ropercenter.cornell.edu/quick-answers/glossary-terms | Roper Center
Main Menu
Glossary of Terms
AAPOR Response Rate Definitions
| Response Rate 1 (RR1): the number of complete interviews divided by the number of interviews (complete and partial), the number of non-interviews (refusal, break-offs, non-contact and others), and number of cases of unknown eligibility. Response Rate 2 (RR2): the number of complete and partial interviews divided by the number of interviews (complete and partial), the number of non-interviews (refusal, break-offs, non-contact and others), and number of cases of unknown eligibility. Response Rate 3 (RR3): the number of complete interviews divided by the number of interviews (complete and partial), the number of non-interviews (refusal, break-offs, non-contact and others), and an estimate of the proportion of cases of unknown eligibility that are actually eligible. Response Rate 4 (RR4): the number of complete and partial interviews divided by the number of interviews (complete and partial), the number of non-interviews (refusal, break-offs, non-contact and others), and an estimate of the proportion of cases of unknown eligibility that are actually eligible. Response Rate 5 (RR5): a special case of RR3 in which either the proportion of eligible cases among the cases of unknown eligibility is assumed to be zero or there are no cases of unknown eligibility. Response Rate 6 (RR6): a special case of RR4 in which either the proportion of eligible cases among the cases of unknown eligibility is assumed to be zero or there are no cases of unknown eligibility. |
| A short description of the types of substantive questions in the questionnaire. Demographics and survey-organization defined variables, like date of interview, are not described. |
A group of persons (or any other units of analysis) that have some characteristics in common without necessarily having any other connection to each other. For example, Teachers in the U.S. , Internet-users, or Voters.
Archive catalog
| The Center’s online catalog of studies. |
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Banner book
| A presentation of survey results with many crosstab tables that indicate how different types of respondents responded to each survey question. |
Beginning and Ending Dates
The dates of interviewing.
Bivariate Analysis
The analysis of the relationship between an independent and dependent variable. (e.g. a cross tabulation showing presidential approval for men and women separately).
Used to search the Roper database, Boolean logic uses the terms "and," "or," and "not' as connectors between keywords or phrases to narrow the results of keyword searches.
Breakoff rate
| The percent of respondents who start the survey but do not finish it. |
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| This occurs when an interviewer attempts to reach a potential survey respondent by phone after failing to reach them previously. Most telephone surveys will set a maximum number of callbacks and continuing attempting to reach the respondent until that number has been reached. |
CAPI (computer-assisted personal interviewing)
| A mode of survey data collection in which an in-person interviewer uses a computer to administer the survey and record responses. |
CATI (computer-assisted telephone interviewing)
| A mode of survey data collection in which a telephone interviewer uses a computer to administer the survey and record responses. |
Cell weighting
| A technique of adjusting weights on respondents in a survey in which the weight applied to each subgroup in the dataset (e.g. men over the age of 55) is calculated based on the relevant distribution of the target population. |
A census is often similar to a survey, with the difference that the census collects data from all members of the population while the survey is limited to a sample.
| Redirecting an online respondent to another survey at the end of a completed survey |
| A reference intended to uniquely identify the study/dataset used in research. Mutiple citations formats are available on the Cite Study tab for each study in the archive, and citations are also available for individual questions. |
Codebook
A list of the variables and how they have been coded in a survey. Every polling organization has its own methods for coding, so every codebook will look a little different.
Coding (Responses)
The process of translating data (respondents' answers to questions) into a format that can be read and manipulated by a computer and later analyzed by the survey researcher. For example, a female respondent answers the following question:
Are you: a) Democrat b) Republican c) Independent d) Don't Know
When this question is coded later, it may look like this:
Sex: 1=Male 2=Female
Party: 1=Democrat 2= Republican 3= Independent 8=Don't Know
9=No Answer/Refused, and coded accordingly.
Collection Mode
| Method by which data were collected, such as telephone, in-person, online, etc. |
Continuous Variable
A continuous variable is a variable that can be expressed by an infinite number of measures. For survey purposes, they are usually measured on an interval or ratio scale. (i.e. time, speed, weight - since these may be broken down into an infinite number of smaller parts.)
Cross-Tabulation
A table which shows the influence of an independent variable (located in the column) on a dependent variable (located in the row.)(e.g. a table showing how income is related to the likelihood of voting for a certain candidate).
CSAQ (computerized self-administered questionnaires )
| A mode of survey data collection in which the respondent completes the survey using computer technology with little or no assistance from an administrator. CSAQ applications include online surveys. |
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D
Data Mining
This is a process where the researcher searches or "digs" through data-bases for information that may validate his/her own work or inspire ideas for new projects.
Dataset
The individual-level results of a survey, conceptualized as a table or "matrix" where the rows contain values for each individuals' coded responses. For example, a "1" on presidential approval might mean "approve" while a "2" might mean "disapprove." "Don'tknow" is often coded as "8" or "9". Datasets may be used for secondary analysis. The raw data. Each respondents responses to a question laid out in a table like form.
Datasets consist of all of the information gathered during a survey which needs to be analyzed. Learning how to interpret the results is a key component to the survey process.
The individual-level results of a survey, conceptualized as a table or "matrix." The rows contain values for each individual's coded responses to the questions asked (contained in the columns.)
Below you can see part of the SPSS downloaded dataset for the results of the USVNS National Election 2000. Column 9, for example, shows the variable "pres", which stands for "presidential vote." In the "Values" column, a "1" was coded for Gore, "2" for Bush, etc. (The pop-up box shows the other Value Labels in that category.)
Statistical software is necessary to define, manipulate, and extract variables and cases within data files.
Small data files may also be analyzed within spreadsheet software such as Excel.
Date of Source Document
This is a reference tool used by Roper Center staff {Link to page for Roper Center Staff} and is not necessarily the date of a survey's first public release. When no release date exists, the last day of the interviewing period is used.
Dependent Variable
In research, a dependent variable (also called the output variable) is the variable that is being measured in an experiment. It "depends" on other factors. For example, if the research question is "Does education level have an effect on annual income?" income is the dependent variable; the question is asking whether income "depends" on education level.
Dichotomous Question
A type of (close-ended) question which has two answer choices.
e.g. Are you: a) Female b) Male
Discrete Variable
A discrete (also known as categorical or nominal) variable is one that has two or more categories with no intrinsic order. For example, Sex, Hair color, and Favorite radio station. You can assign these variables to a category, but they have no order from highest to lowest. Therefore, you can't find "the average" of hair color. (See ordinal variable)
Disposition codes
A set of codes or categories used by survey researchers to document the ultimate outcome of contact attempts on individual cases in a survey sample.
DOI Link
| The DOI, or Digitial Object Identifier, a permanent identifier for the study |
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E
Exit polls
| A survey that is conducted on voters immediately after exiting a polling station, asking them how they voted. Exit polls are intended to allow for better understanding of voting behavior of different groups in the electorate and drivers of vote choice. |
External sponsor
| When applicable, the name of the organization that commissioned the survey. If the same organization funded, designed, and fielded a poll, no sponsor is listed. |
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F
Feeling Thermometer
A type of ratings scale where respondents are asked to gauge their attitudes about a particular topic or person. For example, ratings between 50 degrees and 100 degrees mean that you feel favorable and warm toward a person. Ratings between 0 degrees and 50 degrees mean that you Don'tfeel favorable toward the person. You would rate the person at the 50 degree mark if you Don'tfeel particularly warm or cold toward the person. The National Election Studies (NES) has often used this rating scale for questions about presidential candidates.
Filter Question
A type of question used on surveys in order to determine which subsequent (if any) questions to ask.
Focus Group
A small group selected from a wider population that is led by a moderator in an open discussion about the research topic. While there are many different functions of a focus group, there are typically three reasons why a focus group is used in qualitative research. The information gathered from the group interaction is used: 1) to gain insights that will help generate data for the development of a survey 2) as a supplementary source in a study along with other methods of data-collecting; used in either a preliminary or follow-up stage of the research 3) in combination with other methods of interviewing for the research project. Focus groups are often used in the areas of market research and political opinion analysis.
Form
| The version of a questionnaire used in a survey. While online or telephone surveys often utilize split samples to test different wordings or to included more questions than would be manageable in a single questionnaire, printed surveys require more than one form of the questionnaire to be printed and distributed if the questions are giong to vary. Multiple forms of questionnaires were often used in early polling. Information about the forms in Roper polls can be found in the survey documentation. |
Full Question ID
The unique identifier assigned by the Roper Center to every question in the iPOLL database.
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G
Geographic coverage
| The geographic area from which data were collected. |
Grant funder
| The funding agency that provided grant support for the study. |
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H
Hot deck imputation
| Replacing missing values for a respondent in a dataset with corresponding values from a similar respondent. For instance, if a respondent failed to answer a specific question, their non-response would be replaced with the response given by another respondent who was similar to them with regard to other characteristics. |
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I
Independent Variable
In research, an independent variable (also called experimental variable or predictor variable) is a variable that is measured for its effect on the dependent variable(s). As the independent variable changes, its effect on the dependent variable is observed by the researcher. For example, if the research question is "does education level have an effect on annual income?" education level is the independent variable.
Intercept sample
| A means of sampling or selecting people to participate in a survey that involves recruiting respondents at a particular public place or event. An intercept survey is usually administered by an in-person interviewer. |
Interval Variable
An interval variable is similar to an ordinal variable except that the intervals between the values of the categories are equidistant, or equally spaced. However, there is no meaningful "zero" point. For example, Age.
Interview
A data collection encounter in which one person (an interviewer) asks questions of another (a respondent). Interviews may be conducted face-to-face or by telephone.
Interview dates
| The date range during which data was collected from respondents. |
Interviewer Bias
Interviewers can intentionally or unintentionally prompt respondents to reply in a particular manner. Characteristics like sex, race, age, physical appearance and behavior can have subtle or sometimes, obvious affects on respondents during the interview process. Some respondents may answer in a manner that they believe would please the interviewer. It is for this reason that survey firms seeking to interview special samples of the population will carefully select interviewers with like characteristics to conduct the survey.
Interview Method/Mode
| The mode of interviewing: mail, telephone, in-person, online, etc. |
iPoll
| |
| The most comprehensive and up-to-date source for national public opinion data in the United States. Roper iPoll is a database and study catalog holding polling datasets, as well as a full-text question-level retrieval system. iPoll is designed so that users can locate, examine, and ultimately capture questions asked on national U.S. and state-level surveys on a variety of topics. |
IVR (interactive voice response)
| A technology that allows a computer to interact with humans through the use of voice and keypad input. IVR technology can replace a human interviewer in telephone surveys. IVR surveys are sometimes called "robocalls." |
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J
JPOLL
A special library collection of public opinion data from Japan. The archive includes over 1,000 reports containing full-question text and responses from surveys conducted in Japan from 1980 through 1995. These studies, done by the principal polling agencies in Japan, contain more than 20,000 survey questions and their corresponding responses.
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L
Likely voter screener
| A means of estimating whether a survey respondent is likely to vote in an election. Survey firms typically use questions about a respondent's past voting behavior and their intention to vote in future elections in order to identify respondents that are most likely to vote in an election, to increase the accuracy of electoral predictions based on surveys. Likely voter screens range widely in complexity, from single questions to multi-question indexes. Survey organizations often keep the elements of their likely voter model proprietary. |
Likert Scale
A widely used response format that allows for measuring typically qualitative attitudes into quantitative measures. For example, response options may include: "strongly agree", "agree", "disagree", "strongly disagree".
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M
Margin of error/margin of sampling error
| A statistic that captures the amount of random sampling error in a survey's results. Random sampling error is the difference between the values of the sample and the values of the population from which the sample is drawn. It is usually unknown but can be estimated. |
Mean
The average. To calculate this, simply add up the values for each case and divide by the total number of cases. (e.g. If you want to find out the average number of hours you spend on the computer each week, simply add up your daily hours and divide them by seven.)
Median
The middle score or measurement in a set of ranked scores or measurements.
Methodology
| The study of methods and research practices used in a field of study. In survey research, methodology refers to the study of different means of collecting and analyzing survey data. |
Mode
| The method or mode of interviewing: mail, telephone, in-person, online, etc. |
Most recent birthday
A technique for selecting members of a given household for a survey. Interviewers ask to speak to the member of the household with the most recent birthday. This selection method is called quasi-random.
Multi-stage (Cluster) Sampling
A sampling method using more than one stage in the process of gathering the sample. (e.g. You want to interview Missouri voters about their preferences in an upcoming election. However, you have limited resources in your ability to contact them. Therefore, you randomly select 30 voting districts in the state. From there, you randomly select towns. Within those towns, you randomly select neighborhoods. From the neighborhoods, you randomly select streets. and so on.)
Multivariate Analysis
The analysis of more than two variables simultaneously, for the purpose of determining the relationship between and/or among them. For example an issue by age and by sex.
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N
N
| The number of observed cases in a sample. In polling, N refers to the number of respondents. |
Next birthday
| A technique for selecting members of a given household for a survey. Interviewers ask to speak to the member of the household whose birthday is next. |
Noncontacts
| An instance in which a survey respondent selected to participate in a survey cannot be reached. For instance, if a respondent is selected for an in-person survey based on their address, but is not at home when the interviewer visits the address, this would be considered a noncontact. |
Non-probability sampling
| A type of sampling where samples are drawn utilizing non-random methods. In nonprobability sampling, all members of the universe do not have a known, non-zero probability of being sampled. Inferences about the target population should not be made based on such surveys unless methods, such as propensity-score weighting, have been put in place to adjust the results to better represent the total population, including those with a zero probability of being sampled. Non-probability sampling was standard in the United States before 1950, when U.S. polling organizations began to move to probability-based methods. In recent years, online polling utilizing non-probability sampling methods have become common. See also: quota sampling, river sampling. |
Non-probability Sampling
A type of sampling where samples are drawn arbitrarily, without regard to scientific methods; and, therefore, should not be used to make statistical inferences about the target population. (i.e. "person on the street" samples)
Nonresponse
| An instance in which a survey respondent fails to complete a survey or answer a survey question. Systematic or non-random nonresponses can result in nonresponse bias. |
Nonresponse bias
| A type of bias resulting from respondents failing to complete a survey or answer a specific survey question for systematic reasons. For instance, if younger respondents are less likely to complete a survey on political ideology than older ones, survey results may be biased in a particular direction. |
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O
Omnibus
| One survey that collects responses on a wide range of questions for multiple individuals or organizations. Pooling questions to field an omnibus survey may lower costs. |
Online panel
| A sample of respondents who have agreed to complete surveys online. Survey firms use online panels to quickly obtain responses from large groups of respondents. Online panels can utilize probability-based or non-probability-based sampling procedures. Probability-based online panels recruit panelists using traditional probability-based methods, like RDD telephone surveys, then provide internet access to those who require it. Non-probability-based online panels recruit their respondents through a variety of methods, including online ads. |
Ordinal Variable
An ordinal variable is similar to a discrete variable; however, the difference is that there is a clear ordering of the variables from low to high. For example, "Education." There is an order as well as a value from low to high when it comes to measuring years of education. We can code the variable as follows: "1"=Less than High School Graduate; "2"=H.S. Grad/Some College; "3" =College Graduate; "4" =Post Graduate or more
We know that there is an order from low to high in this case, but the size of the difference between each of the categories is not necessarily equidistant from one to the next. (If the difference between each of the categories were equally spaced, the variable would be an interval variable.)
Organization(s)
| Searches organizations associated with surveys, including both external sponsors and survey field organizations. |
Outlier Effect
An extreme value of a variable in a dataset. This extreme value can distort the results of your survey if you're dependent solely on the mean statistic for analysis. i.e. Here is a list of test scores: 90, 87, 99, 95, 85, 43, 91. The score 43 is an outlier and will distort the mean (84, in this case) would give a better sense of the overall scores.
Oversampling
| A means of selecting respondents that generates a sample in which some groups are over-represented compared to their share of the target population. This method allows for analysis of groups that would otherwise make up such a small percent of the sample that analysis would be difficult. For instance, oversampling Asian Americans in a survey of US citizens may allow for a more accurate analysis of this group. Oversamples are generally weighted down to their share of the population when results are aggregated to report overall results. |
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P
Population
The theoretical population from which the sample was drawn. For example, adults living in the contiguous U.S.; Some studies are based on sample sub-sets (such as national samples of women or African Americans).
Population Parameter
A characteristic of the target population described by a statistic. For example, if your target population is runners in the New York City Marathon, the average finishing time would be a population parameter. You calculate every runners' finishing time to get the parameter. (Not to be confused with sample statistic)
Population Size
In iPOLL, this is the total unweighted count of all completed interviews, also referred to as the sample size.
Poststratification
| A technique of adjusting weights on respondents in a survey so the adjusted weights add up to the known population sizes within each group, making the sample more closely resemble the target population. Poststratification is used when the grouping of like units is not possible during sampling. It is used in order to reduce bias and improve the precision of estimates. For instance, if respondents are selected for a survey and their gender is not known in advance, the gender distribution in the sample could be different from the gender distribution in the target population. Poststratification would be used in this case to adjust the weights of each gender to match the target population. |
Principal Investigator
| The lead researcher on a grant-funded study, often referred to as the PI. |
Probability Sampling
| A type of sampling which ensures that each member of the sampling frame has an equal, known chance of being selected. This kind of sampling allows researchers to make statistical inferences about the population at large. (see Non-probability Sampling) |
Propensity score weighting
| A technique used to adjust weights on different observations in an analysis based on their likelihood of receiving the treatment the analyst is interested in, given all other characteristics. Propensity score weighting is used in order to reduce selection bias and is commonly used by online polls to weight based on the likelihood of a respondent to have oneline access. |
Push poll
| A survey that is primarily intended to manipulate or influence respondents' opinion rather than to generate data for analysis. Push polls are especially used in political campaigning in order to influence potential voters under the guise of conducting a legitimate survey. Roper Center's acquisition policy excludes push polls. |
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Q
Question
The actual wording of the question. If additional information, such as text from a preceding question, is required for the question to be independently understandable, the added text will appear in parentheses. Similarly, the stem of multipart questions will be repeated and displayed in paratheses. For example: "(For each one, please tell me if you think it is a very serious problem, somewhat serious, not too serious, or not a problem at all.)...The large amount of American debt that is held by China." In iPOLL the question text is preceded by a number such as R18, Q08, or R02. This unique designation is assigned by the Roper Center and does not necessarily reflect the order in which the question appeared in the original study. Whenever possible the original survey instrument is used as the source document by Center staff. In many cases, though, the order of questions on the survey may have been altered in some way for publication in a final report or news release. Researchers requiring information on the original question order should contact the Roper Center.
Question-Level Text
The actual wording of the question used during the survey interview.
Quota sampling
| A means of sampling or selecting people to participate in surveys based on specific characteristics (such as age or gender). The objective is often to generate a sample that closely reflects the target population with regard to these characteristics, in order to reduce bias. For instance, if the target population consists of 50% men and 50% women, one might use quota sampling to recruit an equal number of men and women. In contrast to stratified sampling, quota sampling is a type of non-probability sampling. |
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R
Raking weighting
| A type of poststratification procedure that adjusts the sample weights in a survey in order to make the sample match the target population more closely with regard a number of different groups or post-strata (such as gender, race, age, etc). Raking adjusts the sample weights through repeated calculation of weights so they add up to the known population totals for the post-stratified classifications when only the marginal population totals are known (e.g. if the gender and age distribution of the population is known, but not the gender distribution for each age group). |
Random Digit Dialing (RDD)
A technique used to obtain a representative sample by using a device that randomly generates telephone numbers in order to contact eligible participants.
Ratio Variable
A ratio variable has all the properties of an interval variable. In addition, it has a zero point. For example, income.
Recontacts
| Survey respondents who either have to be contacted multiple times in order to be reached, or are contacted again after the initial survey has been conducted (e.g. to validate completed interviews or to measure behavioral changes over time). |
Refusals
| An instance in which a survey respondent selected to participate in a survey declines to do so or declines to answer one or more questions in the survey. This is a type of nonresponse that can bias survey results. |
Registration-Based Sampling (RBS)
| A means of sampling or selecting people to participate in election polls using a database of registered voters for a given geographical area. With RBS, the sampling frame from which the sample is drawn consists of the registered voters for the area. |
Reinterviews
| Interviews conducted with respondents who participated in a previous survey. Reinterviews are used in order to track changes in respondents' opinions over time or before and after an event, such as a presidential debate. |
Reliability
The quality of measurement that suggest that the same data would have been collected each time in repeated observations of the same phenomenon.
Research Sponsor
When applicable, the name of the organization that commissioned the survey.
Respondent selection
| The process of choosing the final respondent for a survey; for example, choosing the respondents for a particular survey from an online panel or choosing members of a household to participate in a survey once the household has already been selected (e.g. through random digit dialing). |
Response rate
| Proportion of contacted respondents who completed the survey. The American Association for Public Opinion Research (AAPOR) provides definitions for six measures of response rates (AAPOR response rate definitions). |
Responses (in iPoll)
The response categories and percentages of the sample answering each way. Generally, the percentages shown are weighted if the data were weighted better to reflect the population. Any special question-related information clarifying such things as multiple responses, partial responses, and the like will appear after the responses. These notes relate only to specific questions as opposed to the entire study and are referred to as question-level notes.
Rim weighting
| See raking. |
River sampling
| A means of sampling or selecting people to participate in surveys in which potential survey respondents are recruited through online ads or pop-ups on online platforms. It is a type of non-probability sampling in which the respondent's identity is unverifiable and the respondent cannot be recontacted. A respondent would typically click on an ad or offer, answer a number of pre-screening questions, and then be routed to a survey. |
Routers (breakout routers)
| Online survey routs screen respondents and directs them to open surveys for which they are qualified. |
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S
Sample
A description of the population from which the survey respondents were drawn.
Sample Error
One type of inaccuracy caused by making inferences about the target population based on the sample. The sampling error is an estimate of how a sample statistic is expected to differ from the population parameter.
Sample Frame
This is the list of eligible participants included in the target population. The sample is chosen from the sampling frame.
Sample Size
This is the total unweighted count of all completed interviews.
Sample Statistic
A statistic which describes the sample. (e.g. If you want to do a survey of New York City Marathon runners, including their finishing times, the average finishing time of the those surveyed would be an example of a sample statistic. Not to be confused with population parameter, which would calculate the average finishing time of all the runners, not just a sample of them.)
Sampling
A method of selecting elements (or units) from the target population in a way that is representative. Types of sampling include: Simple random sampling, stratified sampling, systematic sampling, and multi-stage cluster sampling.
Sampling procedure
| The method by which participants in a poll were selected. |
Secondary Data
This term refers to materials and information that has previously been documented. For example, a poll, a press release, a business report.
Simple Random Sample (SRS)
The most common sampling method where each element in the population has an equal chance of being selected.
Source Document
The document from which information was gathered. In iPOLL, the source document usually refers to the topline document released by a polling organization which was used as the source for questions and topline results.
Speeding
| Respondents answering at a rate too fast to allow for adequate comprehension of questions, particularly for paper or online questionnaires. |
Split sample
| A type of survey research design in which a sample is randomly split into different groups and assigned different treatments (e.g. questions or prompts) in order to determine the effect of the treatment on survey responses. A sample may also be split into groups and asked different questions in order to maximize the number of questions that can be asked in the survey. |
Standard Deviation
A statistic that shows the dispersion of scores in a distribution of scores. It is a measure of the average amount the scores in a distribution deviate from the mean. The more widely spread out the scores are, the larger the standard deviation will be.
Standard Error (of the Mean)
A statistic indicating how much the mean score of a single sample is likely to differ from the mean score of the population. It answers the question, "How good an estimate of the population mean is the sample mean?" (Not to be confused with sampling error)
Statistic
A number that describes some characteristic of a variable. (e.g. the mean, the standard deviation)
Straightlining
| A respondent providing identical answers across a range of questions (on a printed survey, literally marking off responses in a "straight line" through the instrument). |
Stratified Sampling
A method of sampling where groups that might not otherwise be equally represented are first divided proportionately into categories (“strata”); then, a sample is randomly selected from each of these categories. (e.g. If you wanted to do a study on hospitals, you’d separate them by size—small, medium-sized, and large hospitals. From there, you would draw samples from each category so that they’d all be equally represented.
Study Note
This note field on questions in the iPOLL database pertains to the entire release, report, or study from which the question was taken.
Subject
The topic classification(s) that best describe the question. The scheme for this categorization was developed by the Roper Center and contains over 100 subject categories.
Subpopulation
A subset of the population under study. In cases where responses are not based on the entire sample, question results in iPOLL will show the Subpopulation field with a description of the portion of the sample whose responses are being reported appears here (e.g. women, or those who favor a given policy).
Survey Organization
The organization that conducted the fieldwork for a survey.
Systematic Sampling
A method of sampling where units are selected from the sampling frame by every “nth” unit. (e.g. You have a directory of 100,000 names and you want a sample of 1,000 names. Divide 100,000 by 1,000 to get 100. You will select every 100th name from the directory. Randomly select a number between 1 and 100, say 42, and select every 42nd name in groups of 100 (42, 142, 242, 342, 442.) to complete your sample.
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Times-at-home weighting
| A technique of adjusting weights on respondents in a survey based on the estimated probability that the respondent would be found at home at the time the interview was completed. This type of weighting is used in order to reduce bias that would result from under-representing respondents who are difficult to reach at home. This method was used by Gallup to weight face to face, probability-based surveys from the 1960s to 1980s and is still used for weighting in-person interviews in some areas. |
| The subject classification(s) that best describe the question. The scheme for this categorization was developed by the Roper Center and contains over 100 subject categories. |
The topline is the result of how the aggregated sample answered a specific question.
Tracking poll
| A series of surveys repeated over time in order to measure changes in survey responses in a target population. Tracking polls are often used over the course of electoral campaigns to measure changes in support for political candidates. |
This term typically refers to the long-term patterns over time relating to topics of interest in public opinion that are measured by the repetition of the same question with unchanging wording over many years. An example of a survey that includes many important trends is the GSS, one of the nation's longest running surveys of social, cultural and political indicators.
Using more than one method to find meaning in a problem. i.e. If you want to interpret the President's Approval Rating, you could look at poll results, results of focus groups, and news stories of current events.
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Univariate Analysis
The analysis of a single variable, for purposes of description (e.g., averages, or the proportion of cases falling into a given category among the entire sample).
Universe/target population
| All entities that qualify for inclusion in the study or survey, from which the sample of respondents is drawn. The target population could consist of all adult American citizens, for instance, or all Fortune 500 companies. |
US National Adult
A common theoretical population for US “national” polls. Typically it means the age 18+, non-institutionalized (e.g. no prisons, nursing homes, or military bases) population in the 48 contiguous states, since Alaska and Hawaii are often omitted for practical reasons.
arrow_upward Back to Top
In survey research, a variable is an example of what is being measured. (i.e. income, age; presidential approval; support for a policy, etc.) There are different kinds of variables, including: categorical, continuous, interval, ratio, independent, dependent.
arrow_upward Back to Top
Also known as sample balancing, weighting is a technique used to reflect differences in the number of population units that each case in a dataset represents. Typically, for surveys designed to be representative of the population of the U.S., units are adjusted to reflect the U.S. Census on several demographic measures, including age, education, and sex. While polling organizations may have different methods for their weighting procedures, weighting generally involves the multiplication of survey observations by one or more factors in order to increase or decrease the emphasis that will be given to the observations when analyzing the data. See also propensity score weighting, raking weighting, times-at-home weighting.Also known as sample balancing, weighting is a technique used to reflect differences in the number of population units that each case in a dataset represents. Typically, for surveys designed to be representative of the population of the U.S., units are adjusted to reflect the U.S. Census. While polling organizations may have different methods for their weighting procedures, weighting generally involves the multiplication of survey observations by one or more factors in order to increase or decrease the emphasis that will be given to the observations when analyzing the data.
Weighting benchmark source
| Data source for benchmarks used to weight the sample. |
Within-household selection
| The process of choosing members of a household to participate in a survey once the household has already been selected (e.g. through random digit dialing). See also Respondent selection. Examples of methods of within household selection include most recent birthday and youngest man/oldest woman designs. |
arrow_upward Back to Top
Youngest man/oldest woman
| A systematic, non-random technique for selecting members of a given household for a survey. The member of the household who should be selected, e.g. the youngest man, the youngest woman, the oldest man or the oldest woman, can be randomly assigned across the sample, or interviewers can always start with the youngest man, then move to the oldest woman. |
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eBay item number:376286824794
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Subject Area
Mathematics, Education
Publication Name
Problem-Solving Strategies
Publisher
Springer New York
Item Length
9.3 in
Subject
Decision-Making & Problem Solving, General
Publication Year
1997
Series
Problem Books in Mathematics Ser.
Type
Textbook
Format
Trade Paperback
Language
English
Item Height
0.4 in
Author
Arthur Engel
Item Weight
45.2 Oz
Item Width
6.1 in
Number of Pages
403 Pages
ISBN
9780387982199
Category
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7549 | https://math.libretexts.org/Courses/Monroe_Community_College/MTH_098_Elementary_Algebra/5%3A_Systems_of_Linear_Equations/5.3%3A_Solve_Systems_of_Equations_by_Elimination | Published Time: 2020-01-06T03:30:26Z
5.3: Solve Systems of Equations by Elimination - Mathematics LibreTexts
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5: Systems of Linear Equations
MTH 098 Elementary Algebra
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Sun, 03 Mar 2024 18:38:21 GMT
5.3: Solve Systems of Equations by Elimination
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Mary Cameron
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Contents
1. Home
2. Campus Bookshelves
3. Monroe Community College
4. MTH 098 Elementary Algebra
5. 5: Systems of Linear Equations
6. 5.3: Solve Systems of Equations by Elimination
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MTH 098 Elementary Algebra
1: Foundations
2: Solving Linear Equations and Inequalities
3: Math Models
4: Graphs
5: Systems of Linear Equations
6: Polynomials
7: Factoring
5.3: Solve Systems of Equations by Elimination
Last updated Mar 3, 2024
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5.2E: Exercises
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Table of contents
1. Learning Objectives
2. Note
3. Solve a System of Equations by Elimination
1. Example 5.3.1 5.3.1: How to Solve a System of Equations by Elimination
2. Try It 5.3.2 5.3.2
3. Try It 5.3.3 5.3.3
4. HOW TO SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.
5. Example 5.3.4 5.3.4
6. Try It 5.3.5 5.3.5
7. Try It 5.3.6 5.3.6
8. Example 5.3.7 5.3.7
9. Try It 5.3.8 5.3.8
10. Try It 5.3.9 5.3.9
11. Example 5.3.10 5.3.10
12. Try It 5.3.11 5.3.11
13. Try It 5.3.12 5.3.12
14. Example 5.3.13 5.3.13
15. Try It 5.3.14 5.3.14
16. Try It 5.3.15 5.3.15
17. Example 5.3.16 5.3.16
18. Try It 5.3.17 5.3.17
19. Try It 5.3.18 5.3.18
20. Example 5.3.19 5.3.19
21. Try It 5.3.20 5.3.20
22. Try It 5.3.21 5.3.21
Solve Applications of Systems of Equations by Elimination
Example 5.3.22 5.3.22
Try It 5.3.23 5.3.23
Try It 5.3.24 5.3.24
Example 5.3.25 5.3.25
Try It 5.3.26 5.3.26
Try It 5.3.27 5.3.27
Choose the Most Convenient Method to Solve a System of Linear Equations
Example 5.3.28 5.3.28
Try It 5.3.29 5.3.29
Try It 5.3.30 5.3.30
Note
Key Concepts
Learning Objectives
By the end of this section, you will be able to:
Solve a system of equations by elimination
Solve applications of systems of equations by elimination
Choose the most convenient method to solve a system of linear equations
Note
Before you get started, take this readiness quiz.
Simplify −5(6−3a).
If you missed this problem, review Example 1.10.43.
2. Solve the equation 1 3 x+5 8=31 24 1 3 x+5 8=31 24.
If you missed this problem, reviewExample 2.5.1.
We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.
The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.
Solve a System of Equations by Elimination
The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.
For any expressions a, b, c, and d,
if and then a=b c=d a+c=b+d(5.3.1)(5.3.1)if a=b and c=d then a+c=b+d
To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.
Notice how that works when we add these two equations together:
3 x+y=5 2 x−y=0–––––––––––5 x=5(5.3.2)(5.3.2)3 x+y=5 2 x−y=0 _ 5 x=5
The y’s add to zero and we have one equation with one variable.
Let’s try another one:
{x+4 y=2 2 x+5 y=−2(5.3.3)(5.3.3){x+4 y=2 2 x+5 y=−2
This time we don’t see a variable that can be immediately eliminated if we add the equations.
But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.
Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.
Add the equations yourself—the result should be −3 y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.
We’ll do one more:
{4 x−3 y=10 3 x+5 y=−7(5.3.4)(5.3.4){4 x−3 y=10 3 x+5 y=−7
It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.
We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12 x and −12 x.
This gives us these two new equations:
{12 x−9 y−12 x−20 y=30=28{12 x−9 y=30−12 x−20 y=28
When we add these equations,
{12 x−9 y=30−12 x−20 y=28–––––––––––––––––−29 y=58(5.3.5)(5.3.5){12 x−9 y=30−12 x−20 y=28 _−29 y=58
the x’s are eliminated and we just have −29 y = 58.
Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.
Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.
Example 5.3.1 5.3.1: How to Solve a System of Equations by Elimination
Solve the system by elimination. {2 x+y=7 x−2 y=6{2 x+y=7 x−2 y=6
Solution
Try It 5.3.2 5.3.2
Solve the system by elimination. {3 x+y=5 2 x−3 y=7{3 x+y=5 2 x−3 y=7
Answer
(2,−1)
Try It 5.3.3 5.3.3
Solve the system by elimination. {4 x+y=−5−2 x−2 y=−2{4 x+y=−5−2 x−2 y=−2
Answer
(−2,3)
The steps are listed below for easy reference.
HOW TO SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.
Write both equations in standard form. If any coefficients are fractions, clear them.
Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites.
Add the equations resulting from Step 2 to eliminate one variable.
Solve for the remaining variable.
Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Write the solution as an ordered pair.
Check that the ordered pair is a solution to both original equations.
First we’ll do an example where we can eliminate one variable right away.
Example 5.3.4 5.3.4
Solve the system by elimination. {x+y=10 x−y=12{x+y=10 x−y=12
Solution
Both equations are in standard form.
The coefficients of y are already opposites.
Add the two equations to eliminate y.
The resulting equation has only 1 variable, x.
Solve for x, the remaining variable.
Substitute x = 11 into one of the original equations.
Solve for the other variable, y.
Write the solution as an ordered pair.The ordered pair is (11, −1).
Check that the ordered pair is a solution
to both original equations.
x+y 11+(−1)10==?=10 10 10✓x−y 11−(−1)12==?=12 12 12✓x+y=10 x−y=12 11+(−1)=?10 11−(−1)=?12 10=10✓12=12✓
The solution is (11, −1).
Try It 5.3.5 5.3.5
Solve the system by elimination. {2 x+y=5 x−y=4{2 x+y=5 x−y=4
Answer
(3,−1)
Try It 5.3.6 5.3.6
Solve the system by elimination.{x+y=3−2 x−y=−1{x+y=3−2 x−y=−1
Answer
(−2,5)
In Example 5.3.7 5.3.7, we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.
Example 5.3.7 5.3.7
Solve the system by elimination. {3 x−2 y=−2 5 x−6 y=10{3 x−2 y=−2 5 x−6 y=10
Solution
Both equations are in standard form.
None of the coefficients are opposites.
We can make the coefficients of y opposites by multiplying
the first equation by −3.
Simplify.
Add the two equations to eliminate y.
Solve for the remaining variable, x.
Substitute x = −4 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.The ordered pair is (−4, −5).
Check that the ordered pair is a solution to
both original equations.
3 x−2 y 3(−4)−2(−5)−12+10−2==?=?=−2−2−2−2✓5 x−6 y 5(−4)−6(−5)−20+30 10==?=?=10 10 10 10✓3 x−2 y=−2 5 x−6 y=10 3(−4)−2(−5)=?−2 5(−4)−6(−5)=?10−12+10=?−2−20+30=?10−2=−2✓10=10✓
The solution is (−4, −5).
Try It 5.3.8 5.3.8
Solve the system by elimination.{4 x−3 y=1 5 x−9 y=−4{4 x−3 y=1 5 x−9 y=−4
Answer
(1,1)
Try It 5.3.9 5.3.9
Solve the system by elimination.{3 x+2 y=2 6 x+5 y=8{3 x+2 y=2 6 x+5 y=8
Answer
(−2,4)
Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.
Example 5.3.10 5.3.10
Solve the system by elimination. {4 x−3 y=9 7 x+2 y=−6{4 x−3 y=9 7 x+2 y=−6
Solution
In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.
Both equations are in standard form. To get opposite
coefficients of y, we will multiply the first equation by 2
and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x.
Substitute x = 0 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.The ordered pair is (0, −3).
Check that the ordered pair is a solution to
both original equations.
4 x−3 y 4(0)−3(−3)9==?=9 9 9✓7 x+2 y 7(0)+2(−3)−6==?=−6−6−6✓4 x−3 y=9 7 x+2 y=−6 4(0)−3(−3)=?9 7(0)+2(−3)=?−6 9=9✓−6=−6✓
The solution is (0, −3).
Try It 5.3.11 5.3.11
Solve the system by elimination. {3 x−4 y=−9 5 x+3 y=14{3 x−4 y=−9 5 x+3 y=14
Answer
(1,3)
Try It 5.3.12 5.3.12
Solve the system by elimination. {7 x+8 y=4 3 x−5 y=27{7 x+8 y=4 3 x−5 y=27
Answer
(4,−3)
When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.
Example 5.3.13 5.3.13
Solve the system by elimination. {x+1 2 y=6 3 2 x+2 3 y=17 2{x+1 2 y=6 3 2 x+2 3 y=17 2
Solution
In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.
To clear the fractions, multiply each equation by its LCD.
Simplify.
Now we are ready to eliminate one of the variables. Notice that
both equations are in standard form.
We can eliminate y multiplying the top equation by −4.
Simplify and add.
Substitute x = 3 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.The ordered pair is (3, 6).
Check that the ordered pair is a solution
to both original equations.
x+1 2 y 3+1 2(6)3+3 6==?=?=6 6 6 6✓3 2 x+2 3 y 3 2(3)+2 3(6)9 2+4 9 2+8 2 17 2==?=?=?=17 2 17 2 17 2 17 2 17 2✓x+1 2 y=6 3 2 x+2 3 y=17 2 3+1 2(6)=?6 3 2(3)+2 3(6)=?17 2 3+3=?6 9 2+4=?17 2 6=6✓9 2+8 2=?17 2 17 2=17 2✓
The solution is (3, 6).
Try It 5.3.14 5.3.14
Solve the system by elimination. {1 3 x−1 2 y=1 3 4 x−y=5 2{1 3 x−1 2 y=1 3 4 x−y=5 2
Answer
(6,2)
Try It 5.3.15 5.3.15
Solve the system by elimination. {x+3 5 y=−1 5−1 2 x−2 3 y=5 6{x+3 5 y=−1 5−1 2 x−2 3 y=5 6
Answer
(1,−2)
In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.
Example 5.3.16 5.3.16
Solve the system by elimination.{3 x+4 y=12 y=3−3 4 x{3 x+4 y=12 y=3−3 4 x
Solution
Write the second equation in standard form.Clear the fractions by multiplying thesecond equation by 4.Simplify.To eliminate a variable, we multiply thesecond equation by −1.Simplify and add.⎧⎩⎨3 x+4 y y=12=3−3 4 x{3 x+4 y=12 3 4 x+y=3⎧⎩⎨⎪⎪3 x+4 y 4(3 4 x+y)=12=4(3){3 x+4 y=12 3 x+4 y=12{3 x+4 y=12−3 x−4 y=−12––––––––––––––––0=0{3 x+4 y=12 y=3−3 4 x Write the second equation in standard form.{3 x+4 y=12 3 4 x+y=3 Clear the fractions by multiplying thesecond equation by 4.{3 x+4 y=12 4(3 4 x+y)=4(3)Simplify.{3 x+4 y=12 3 x+4 y=12 To eliminate a variable, we multiply thesecond equation by −1.{3 x+4 y=12−3 x−4 y=−12 _ 0=0 Simplify and add.
This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.
After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.
Try It 5.3.17 5.3.17
Solve the system by elimination. {5 x−3 y=15 y=−5+5 3 x{5 x−3 y=15 y=−5+5 3 x
Answer
infinitely many solutions
Try It 5.3.18 5.3.18
Solve the system by elimination. {x+2 y=6 y=−1 2 x+3{x+2 y=6 y=−1 2 x+3
Answer
infinitely many solutions
Example 5.3.19 5.3.19
Solve the system by elimination. {−6 x+15 y=10 2 x−5 y=−5{−6 x+15 y=10 2 x−5 y=−5
Solution
The equations are in standard form.Multiply the second equation by 3 to eliminate a variable.Simplify and add.{−6 x+15 y 2 x−5 y=10=−5{−6 x+15 y=10 3(2 x−5 y)=3(−5){−6 x+15 y=10 6 x−15 y=−15––––––––––––––––0≠5 The equations are in standard form.{−6 x+15 y=10 2 x−5 y=−5 Multiply the second equation by 3 to eliminate a variable.{−6 x+15 y=10 3(2 x−5 y)=3(−5)Simplify and add.{−6 x+15 y=10 6 x−15 y=−15 _ 0≠5
This statement is false. The equations are inconsistent and so their graphs would be parallel lines.
The system does not have a solution.
Try It 5.3.20 5.3.20
Solve the system by elimination. {−3 x+2 y=8 9 x−6 y=13{−3 x+2 y=8 9 x−6 y=13
Answer
no solution
Try It 5.3.21 5.3.21
Solve the system by elimination. {7 x−3 y=−2−14 x+6 y=8{7 x−3 y=−2−14 x+6 y=8
Answer
no solution
Solve Applications of Systems of Equations by Elimination
Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.
Example 5.3.22 5.3.22
The sum of two numbers is 39. Their difference is 9. Find the numbers.
Solution
Step 1. Read the problem Step 2. Identify what we are looking for.Step 3. Name what we are looking for.Step 4. Translate into a system of equations.The system is:Step 5. Solve the system of equations.To solve the system of equations, use elimination. The equations are in standard form and the coefficients of m are opposites. Add.Solve for n.Substitute n=24 into one of the original equations and solve form.Step 6. Check the answer.Step 7. Answer the question.We are looking for two numbers.Let n = the first number.m = the second number The sum of two numbers is 39.n+m=39 Their difference is 9.n−m=9{n+m=39 n−m=9{n+m=39 n−m=9––––––––––2 n=48 n=24 n+m=39 24+m=39 m=15 Since 24+15=39 and 24−15=9, the answers check.The numbers are 24 and 15.Step 1. Read the problem Step 2. Identify what we are looking for.We are looking for two numbers.Step 3. Name what we are looking for.Let n = the first number.m = the second number Step 4. Translate into a system of equations.The sum of two numbers is 39.n+m=39 Their difference is 9.n−m=9 The system is:{n+m=39 n−m=9 Step 5. Solve the system of equations.To solve the system of equations, use elimination. The equations are in standard form and the coefficients of m are opposites. Add.{n+m=39 n−m=9 _ 2 n=48 Solve for n.n=24 Substitute n=24 into one of the original n+m=39 equations and solve form.24+m=39 m=15 Step 6. Check the answer.Since 24+15=39 and 24−15=9, the answers check.Step 7. Answer the question.The numbers are 24 and 15.
Try It 5.3.23 5.3.23
The sum of two numbers is 42. Their difference is 8. Find the numbers.
Answer
The numbers are 25 and 17.
Try It 5.3.24 5.3.24
The sum of two numbers is −15. Their difference is −35. Find the numbers.
Answer
The numbers are −25 and 10.
Example 5.3.25 5.3.25
Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?
Solution
Step 1. Read the problem.
Step 2. Identify what we are looking for.We are looking for the number of
calories in one order of medium fries
and in one small soda.
Step 3. Name what we are looking for.Let f = the number of calories in
1 order of medium fries.
s = the number of calories in
1 small soda.
Step 4. Translate into a system of equations:one medium fries and two small sodas had a
total of 620 calories
two medium fries and one small soda had a
total of 820 calories.
Our system is:
Step 5. Solve the system of equations.
To solve the system of equations, use
elimination. The equations are in standard
form. To get opposite coefficients of f,
multiply the top equation by −2.
Simplify and add.
Solve for s.
Substitute s = 140 into one of the original
equations and then solve for f.
Step 6. Check the answer.Verify that these numbers make sense
in the problem and that they are
solutions to both equations.
We leave this to you!
Step 7. Answer the question.The small soda has 140 calories and
the fries have 340 calories.
Try It 5.3.26 5.3.26
Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers cost? How much is one can of formula?
Answer
The bag of diapers costs $11 and the can of formula costs $13.
Try It 5.3.27 5.3.27
To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?
Answer
There are 105 calories in a banana and 5 calories in a strawberry.
Choose the Most Convenient Method to Solve a System of Linear Equations
When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.
Example 5.3.28 5.3.28
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
{3 x+8 y=40 7 x−4 y=−32{3 x+8 y=40 7 x−4 y=−32
{5 x+6 y=12 y=2 3 x−1{5 x+6 y=12 y=2 3 x−1
Solution
{3 x+8 y=40 7 x−4 y=−32{3 x+8 y=40 7 x−4 y=−32
Since both equations are in standard form, using elimination will be most convenient.
{5 x+6 y=12 y=2 3 x−1{5 x+6 y=12 y=2 3 x−1
Since one equation is already solved for y, using substitution will be most convenient.
Try It 5.3.29 5.3.29
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
{4 x−5 y=−32 3 x+2 y=−1{4 x−5 y=−32 3 x+2 y=−1
{x=2 y−1 3 x−5 y=−7{x=2 y−1 3 x−5 y=−7
Answer
1. Since both equations are in standard form, using elimination will be most convenient.
2. Since one equation is already solved for xx, using substitution will be most convenient.
Try It 5.3.30 5.3.30
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
{y=2 x−1 3 x−4 y=−6{y=2 x−1 3 x−4 y=−6
{6 x−2 y=12 3 x+7 y=−13{6 x−2 y=12 3 x+7 y=−13
Answer
1. Since one equation is already solved for y y, using substitution will be most convenient;
2. Since both equations are in standard form, using elimination will be most convenient.
Note
Access these online resources for additional instruction and practice with solving systems of linear equations by elimination.
Instructional Video-Solving Systems of Equations by Elimination
Instructional Video-Solving by Elimination
Instructional Video-Solving Systems by Elimination
Key Concepts
To Solve a System of Equations by Elimination
Write both equations in standard form. If any coefficients are fractions, clear them.
Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites.
Add the equations resulting from Step 2 to eliminate one variable.
Solve for the remaining variable.
Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Write the solution as an ordered pair.
Check that the ordered pair is a solution to both original equations.
This page titled 5.3: Solve Systems of Equations by Elimination is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.
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7550 | https://fishdb.sinica.edu.tw/eng/species.php?id=381037 | Taiwan Fish Database
Scientific Name Elops machnata Pronounce by: Lab of Fish Ecol. and Evo., BRCAS
Author(Forsskål, 1775)Depth 1 - 30 M
Chinese 海鰱 Poisonous Fish No
Family_Chinese 海鰱科 Economic Fish Yes
FamilyF069 ElopidaeEdible Fish Yes Chinese In Mainland China 海鰱
Max Length 118 cm Aquarium Fish No Common Name 夏威夷海鰱、瀾槽、四破、海鰱(臺東)、竹篙頭(台東)、瀾槽(澎湖)、竹梭(澎湖)、爛土梭(澎湖)
Distribution in World India Ocean to West Pacific Ocean Distribution in Taiwan East、West、South、North、North East、PonFu、ShaoLiuChew
Habitats Coral、Benthos、Estuary、Coastal、
Lagoon、Coral&Sand Holotype Locality Jeddah, Saudi Arabia, Red Sea
Synonyms Argentina machnata, Elops capensis, Elops hawaiensis, Elops indicus, Elops purpurascens, Elops saurus
Reference 臺灣魚類誌(沈等, 1993)；Jordan & Evermann, 1902: 327Forsskål, P. 1775Robert W. Hiatt 1947Fraser, T.H. 1973Tzeng, W. N. etc. 1986沈世傑 編 Shih-Chieh Shen ed. 1993Carpenter, K. E. etc. 1999
Specimen ListASIZP0060423. ASIZP0064777. ASIZP0070131. ASIZP0071151. ASIZP0072054. ASIZP0072837. ASIZP0072838. ASIZP0075511. ASIZP0075517. ASIZP0075523. ASIZP0075530. ASIZP0075549. ASIZP0075586. ASIZP0075590. ASIZP0075591. ASIZP0075601. ASIZP0075610. ASIZP0075801. ASIZP0075902. ASIZP0075914. ASIZP0076425. ASIZP0076471. ASIZP0076475. ASIZP0076487. ASIZP0076501. ASIZP0076534. ASIZP0076604. ASIZP0076640. ASIZP0076856. ASIZP0076876. ASIZP0076909. ASIZP0076935. ASIZP0076953. ASIZP0077014. ASIZP0077034. ASIZP0077052. ASIZP0077081. NTUM00091Holotype. ...List all 81 records...
Barcode2012-01-02,Chia-Hao Chang,CO1,98%2015-11-01,Yu-Ming Shih,CO1,100%2016-01-28,Chia-Hao Chang,CO1,100%
Common Name River skipjack; Tempounder; Tenpounder; Springer; Whole sardine; Ten pounder; Big-eye giant herring; Big-eye herring; Bony fish; Ladyfish
Redlist Status Not in IUCN Redlist
Characteristic Body elongate, fusiform, moderately compressed. Eye large. Mouth large, gape ending behind posterior margin of eye; mouth terminal, teeth on jaws villiform. All fins without spines; dorsal fin begins slightly behind midbody, with about 20-27 rays; anal fin short, with approximately 14 to 18 rays, begins well behind base of dorsal fin; caudal fin deeply forked. Scales very small, approximately 94-97 in lateral line. Blue or greenish grey above, silvery on sides; fins sometimes with a faint yellow tinge.
habitats Occurs in coastal waters, but enters lagoons and estuaries. Probably spawn at sea, but the transparent larvae migrate to inshore areas and are often found in brackish water. It has a leptocephalus larva. feeds on various fishes and crustaceans.
Distribution Distributed in Indo-West Pacific from Red Sea to Mossel Bay, South Africa and east to India and western Pacific. It is found around Taiwanese waters.
Utility Marketed fresh or frozen; in some places ground up as fish meal. A good sport fish. |
7551 | https://math.stackexchange.com/questions/323268/formula-to-reverse-digits | Skip to main content
Formula to reverse digits
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Is there a formula that can be used to reverse the digits in a number, given a certain base b? E.G.,
F10(32)=.23
F10(123.456)=654.321
If not, how can you write this out to show what you mean?
Also: what do you call a number that results from applying an irrational number to this formula/procedure? It is an integer (albeit infinitely large) which means it is no longer irrational (n/1 = n). It is also not ∞, unless I am mistaken - it is still it's own number, with its own properties, and you can compare it to other numbers and get back the original irrational number by applying the procedure again.
Thanks,
Brandon
P.S. I'm not really sure what tags to use, I apologize.
elementary-number-theory
irrational-numbers
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asked Mar 7, 2013 at 4:46
leviathanbadgerleviathanbadger
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The first and second operations there don't seem to be equivalent. 23, not .23 is the reverse of 32. There's no real formula that can do what you're asking, but one could write an _algorithm_—that is, a set of steps—to do so. It's also possible to write a formula for numbers of a specific length. For example, if x only has two digits, F10(x)=10(xmod10)+⌊x10⌋
– ahruss
Commented
Mar 7, 2013 at 4:59
As for the reverse of an irrational number, you're right that you do get an infinitely large number, but you're wrong that that's not infinity. By definition, that number is the same as infinity.
– ahruss
Commented
Mar 7, 2013 at 5:00
1
@ahruss That's...not right. You can't define the reversal of an irrational number precisely because there is no real number that does not terminate to the left. 'Infinity' is not a number.
– Potato
Commented
Mar 7, 2013 at 5:20
1
You can't sensibly write F(F(x))=x if you can't make sense of F(x). You can't use F(F(x))=x as evidence that F(x) makes sense.
– Gerry Myerson
Commented
Mar 7, 2013 at 5:32
1
That like saying why can't I use f(x)=1/x on x=0 because f(f(0))=0 is perfectly fine. Are you saying that dividing 1 by zero is okay? How can you do something twice if you can't even do it once?
– Fixed Point
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Mar 7, 2013 at 8:45
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So, you are looking for a way to flip the number about the decimal point in whatever base. Here is the function which will do it
f(x)=∑n=0∞10−n−1(⌊x10n⌋mod10)+∑n=1∞10n−1(⌊10nx⌋mod10)
which is just a compact form of an algorithm. This is basically starting from the decimal point, first going to the left the digits are flipped. Then the second sum starts from the decimal point and going to the right flips the digits.
For whatever base, just replace all of the 10's with whatever base you want. This will work fine when the number (in whatever base) has a terminated expansion on both sides of the decimal point meaning both sums are finite. For irrational numbers for example, the second sum won't converge and the function isn't defined at all so don't use this to flip the digits of 2–√ because it can't be done. Another way to say that is the algorithm won't terminate in finite time (and we don't have infinite memory) because we don't know the "last" digit of 2–√ in base 10 for example. And even for some rational numbers this function doesn't make sense. For example you can't flip the decimal expansion of 1/3 in base 10 because it doesn't terminate in base 10.
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edited Mar 7, 2013 at 6:33
answered Mar 7, 2013 at 5:58
Fixed PointFixed Point
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Thank you for the function, and thank you all for the explanations!
– leviathanbadger
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Mar 7, 2013 at 12:28
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Applying the procedure to a real irrational (even if that irrational is between zero and one) does not produce an integer. There is no such thing as an infinitely large integer. Applying the procedure to an irrational produces something that has no properties at all, until you can produce a coherent theory of strings of digits infinite to the left. But that first step is a doozy --- trying to produce a coherent theory of such strings.
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answered Mar 7, 2013 at 4:54
Gerry MyersonGerry Myerson
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Can't the coherent theory just be "These are the real numbers reflected across the decimal point"? Addition and multiplication follow the rules of F(a)+F(b)=F(a+b),F(a)F(b)=F(ab). Unless, what you mean, is that it's not coherent to the real numbers.
– Calvin Lin
Commented
Mar 7, 2013 at 4:57
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Strings of base-b digits infinite to the left and finite to the right can be considered as b-adic numbers. They work best when b is prime, but can still be defined if it isn't.
– Robert Israel
Commented
Mar 7, 2013 at 5:16
@RobertIsrael What do you mean?
– leviathanbadger
Commented
Mar 7, 2013 at 5:27
@Calvin, of course, you're right. But then you just have the real numbers, written backwards. People who propose strings infinite to the left generally think they are something new, not just a way to write the same old thing we already had.
– Gerry Myerson
Commented
Mar 7, 2013 at 5:28
aboveyouoo, if you do a websearch for "p-adic numbers" you'll see what Robert is talking about. But it doesn't interact in any nice way with the reversal that's infinite to the right.
– Gerry Myerson
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Mar 7, 2013 at 5:30
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Sorry I am not allowed to comment yet but the function provided by the first answer is not correct, it does perform digit reversal but is missing a composite function in it's denominator which i will post later for the non recurring fractions for which it does work in terms of digit reversal, but for recurring fractions such as 1/3,1/6,1/7 this formula does not work and requires the author to make some more alterations in addition to that composite function.
I can provide a formula that will work for all integers in all number bases, however I am yet to formulate an extension for this to Q∖Z.
The following will work for natural numbers in any base representation (b>1), but there are some pretty serious things i need to consider before moving to non integer fractions and eventually irrationals and resources on the subject are hard for me to find,but if somebody has one that works, please, do tell as it will take me a while on my own:
We start again with the Kronecker delta:
δ(x,y)={10x=yx≠y (1)
Which allows us to express the digits of a number 'a' in base 'b' in a computable integer sequence, in that we already know the exact length of the sequence which is of course the number of digits in total. The expression for this computation is:
dn(a,b)=∑k=1⌊ln(a)ln(b)⌋+1(δ(n,k)−bδ(n,k+1))⌊abk−⌊ln(a)ln(b)⌋−1⌋
For example,
a=12345
in base b=10: will, purely coincidentally of course, evaluate to the arithmetic progression with initial value of 1 and d=1 of length 5:
{d1(12345,10),d2(12345,10),d3(12345,10),d4(12345,10),d5(12345,10)}={1,2,3,4,5}
But this(2) will compute the nth digit for the number in any base b>1, and thus these values correspond to the coefficients of the b-adic expansion of the number thus we have as follows:
P(a,b)=∑n=0⌊ln(a)ln(b)⌋+1dn(a,b)bn (3)
And from this we can perform digit reversal as follows in any base as far as I can guess, although don't quote me on it because when I noticed this i thought it was too silly to share so i haven't checked other values of b:
P(N,b)b
P(12345,10)10=54321
P(13454345345,10)10=54354345431
P(842622684442,10)10=244486226248
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answered Apr 27, 2018 at 4:45
Adam LedgerAdam Ledger
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7552 | https://courses.cs.cornell.edu/cs6815/2022fa/lec3.pdf | CS 6815: Pseudorandomness and Combinatorial Constructions Fall 2022 Lecture 3: Aug 30, 2022 Lecturer: Eshan Chattopadhyay Scribe: Noam Ringach 1 Introduction We begin by recalling our definition of BPP: Definition 1.1 (BPP). A language L is in BPPif there exists a polynomial time probabilistic Turing Machine (PTM) A such that for all possible inputs x we have that Pr r∼{0,1}poly(|x|) [A(x, r) = L(x)] ≥2 3.
In the last lecture, we saw that to increase the accuracy of our algorithm from at least 2 3 to at least 1 −ε (i.e., decreasing our errors from at most 1 3 to at most ε), we can come up with a new PTM B defined as Algorithm 1 Definition of B based on A.
Require: Original PTM A, error rate ε > 0 t ←log 1 ε for i ∈[t] do Sample ri ∼{0, 1}poly(|x|) Yi ←A(x, ri) end for return Maj(Y1, . . . , Yt) Furthermore, B has the following guarantees on accuracy, runtime, and randomness, where T(n) = poly(n) is the runtime of A, R(n) = poly(|x|) is the randomness usage of A, and c is an overhead constant from B.
Algorithm Accuracy Runtime Randomness B 1 −ε c · T(n) · log 1 ε c · R(n) log 1 ε Table 1: Accuracy, runtime, and randomness usage of B.
In this lecture, we explore the following question.
Question 1. Can we construct another algorithm that uses less randomness than B to achieve the same 1 −ε error rate?
2 Pairwise Independence In the process of constructing such an algorithm, we will need to introduce a variation on the independence of random variables (RVs) called pairwise independence.
1 Lecture 3: Aug 30, 2022 2 Definition 2.1 (Pairwise Independence). A collection of RVs X1, X2, . . . , XN on the domain X is said to be pairwise independent if the following two conditions are satisfied: 1. For all i, j ∈[N] such that i ̸= j, we have that Xi and Xj are independent.
2. Each Xi is uniform on X.
We can easily construct a straightforward example of three pairwise independent RVs using the XOR function.
Example 2.2. Say that we uniformly sample x1, x2 ∼{0, 1} = F2 and define x3 = x1 ⊕x2. Then {x1, x2, x3} is a collection of pairwise independent RVs.
Proof. To see that x1, x2, x3 are pairwise independent, all we need to show is that {x1, x3} and {x2, x3} are independent, since {x1, x2} are independent by assumption. We will show that {x2, x3} are independent, and the logic for {x1, x3} being independent will be identical.
Take any a, b ∈F2. Then Pr x2,x3[x3 = a ∧x2 = b] = Pr x3 [x3 = a | x2 = b] Pr x2 [x2 = b].
Of course, Prx2[x2 = b] = 1 2 by definition. Additionally, we see that Prx3[x3 = a | x2 = b] = 1 2 because for any fixed x2 = b, there is exactly one value of x1 so that x1 ⊕x2 = a and one value so that x1 ⊕x2 ̸= a. Putting this together, we get that Pr x2,x3[x3 = a ∧x2 = b] = Pr x3 [x3 = a | x2 = b] Pr x2 [x2 = b] = 1 2 · 1 2 = Pr x3 [x3 = a] Pr x2 [x2 = b], meaning that x3 and x2 are indeed independent.
We can generalize this example to create even more pairwise independent RVs.
Example 2.3. For any uniformly sampled x1, . . . , xr ∼F2, take any non-empty subset S ⊆[r] and define ZS = M i∈S xi.
Then the collection Z = {ZS}S⊆[r] is a collection of 2r −1 pairwise independent RVs.
Proof. First, we easily see that |Z| = 2r −1 because there are 2r possible subsets of [r] and we are excluding the empty subset, so we have 2r −1 non-empty subsets over which we are indexing (i.e., there are only 2r −1 non-trivial ways we can take XOR over x1, . . . , xr).
Second, we will show that Z is a collection of pairwise independent RVs. Consider any S, U ⊆[r] such that S ̸= U and any a, b ∈F2. Then we have Pr x1,...,xr∼F2[ZS = a ∧ZU = b] = Pr x1,...,xr∼F2[ZS = a | ZU = b] Pr x1,...,xr∼F2[Zu = b], and all we need is for Prx1,...,xr∼F2[ZS = a | ZU = b] = Prx1,...,xr∼F2[ZS = a]. Similarly as in Example 2.2, since S ̸= U there are some xi’s (namely, {xi}i∈S\U) that change the value of ZS Lecture 3: Aug 30, 2022 3 regardless of the value of ZU. Hence, because the {xi}i∈S\U can change the value of ZS to 0 or 1 equally likely, we see that the value of ZU does not influence the result of ZS, giving us that Prx1,...,xr∼F2[ZS = a | ZU = b] = Prx1,...,xr∼F2[ZS = a], so Pr x1,...,xr∼F2[ZS = a ∧ZU = b] = Pr x1,...,xr∼F2[ZS = a | ZU = b] Pr x1,...,xr∼F2[Zu = b] = Pr x1,...,xr∼F2[ZS = a] Pr x1,...,xr∼F2[Zu = b], as desired.
While this example may seem like a forced generalization of the previous one, it amazingly allows us to efficiently1 generate n pairwise independent variables on F2 from r = log(1 + n) uniform bits. However, we aren’t creating randomness from nothing, since we quickly notice that if we consider (Z1, . . . , Zn) as an element of Fn 2, then the support generated by our example only has 2r = 2log(1+n) = n + 1 values in it (since this is how many possible inputs our algorithm takes), yet |Fn 2| = 2n, so our support is exponentially smaller than that of a uniform distribution over Fn 2.
This sparsity of our pairwise independent distribution on Fn 2 illustrates how much stronger regular independence is.
Eventually, we will use this technique of blowing up randomness to help us de-randomize algo-rithms, but first we will generalize this construction from RVs on F2 to RVs on larger domains. To do this, we will introduce the useful notion of universal hash functions.
Definition 2.4 (Family of Universal Hash Functions). The collection H = {h : [N] →[M]} is a family of universal hash functions (UHFs) if for all x, y ∈[N] and all a, b ∈[M] we have that 1. Prh∈H[h(x) = a] = 1 M 2. Prh∈H[h(x) = a ∧h(y) = b] = 1 M2 .
Immediately, we see that we can construct a collection of pairwise independent RVs simply by sampling h ∼H and taking Xi = h(i) for i ∈[N], and that doing this sampling only requires log(|H|) random bits (since this is how many bits it takes to enumerate the elements of H).
Now, of course, the question that remains is how to construct a family of UHFs efficiently.
Example 2.5 (Construction of a family of UHFs). Let N = 2n for some n ∈N and fix our base field as F2n (which, we recall, is isomorphic to Fn 2), so N = |F2n|. To remain consistent with our previous notation, it will be useful to view F2n as [N]. We then sample c, d ∼F2n and define a hash function by the linear expression hc,d : [N] →[N] x 7→cx + d The collection H = {hc,d}c,d∈F2n is a family of UHFs that can be sampled with O(log N) bits.
Proof. Take any x, y, a, b ∈F2n such that x ̸= y. Note that to sample from H, all we need to do is to sample c and d from F2n, which requires n + n = 2n = 2 log N = O(log N) bits. Thus, using the 1Here, by efficient we mean in terms of n.
I.e., we have to compute n XOR values, each of which looks at r = log(1 + n) bits, which in total takes n log(1 + n) = O(poly(n)) time.
Lecture 3: Aug 30, 2022 4 definition of hc,d, we can compute Pr h∈H[h(x) = a ∧h(y) = b] = Pr c,d∼F2n[hc,d(x) = a ∧hc,d(y) = b] = Pr c,d∼F2n[cx + d = a ∧cy + d = b] = Pr c,d∼F2n c = b −a y −x ∧d = bx −ay x −y .
But both b−a y−x and bx−ay x−y are just arbitrary, fixed elements of F2n, so the probability of c and d hitting them is exactly 1 N and are independent events, giving us that Pr c,d∼F2n[hc,d(x) = a ∧hc,d(y) = b] = Pr c,d∼F2n c = b −a y −x ∧d = bx −ay x −y = Pr c,d∼F2n c = b −a y −x Pr c,d∼F2n d = bx −ay x −y , giving us that H is indeed a family of UHFs.
With this construction, we have shown the following claim.
Claim 2.6. There exists an efficient algorithm that takes 2 log N random bits and produces a pairwise independent distribution on [N].
Although we will not show it here, this can be generalized to hash functions with domain [N] and range [M] while requiring log N + log M random bits. We will use this result shortly, so we state it here as a lemma.
Lemma 2.7. There exists an efficient algorithm that takes log N +log M random bits and produces a pairwise independent distribution on [M].2 In comparison, if we attempted the na¨ ıve approach of randomly picking a hash functions, one would require N log N bits. Again, this shows how far pairwise independence is from (complete) independence.
3 Pairwise Randomness and BPP Now we circle back to the question we asked at the beginning of lecture, namely Question 1: Can we construct another algorithm that uses less randomness than B to achieve the same 1−ε error rate?
Recall our definition of B in Algorithm 1, where we sample r1, . . . , rt i.i.d from {0, 1}poly(|x|) and then take Maj(Y1, . . . , Yt) as our result where Yi = A(x, ri). After introducing pairwise independent RVs, a reasonable question would be what occurs when we take r1, . . . , rt to be pairwise independent instead of independent. Call this modification of B with pairwise independent samples B2.
For consistency, since we can consider {0, 1}poly(|x|) as F2poly(|x|), let R = 2poly(|x|) so that we may again view F2poly(|x|) as [R]. With regular independence, to sample r1, . . . , rt from [R] would require t log R bits (since sampling each ri takes log R bits). On the other hand, applying Lemma 2.7 with N = t and M = R gives us that we can sample pairwise independent r1, . . . , rt from [R] only using log t + log R random bits, many fewer than t log R.
2For those wondering why we cannot just make N small and use few random bits to get a pairwise independent distribution on [M], recall that N is the number of elements that we have to be able to hash (i.e., the number of pairwise independent variables that we want in the end).
Lecture 3: Aug 30, 2022 5 However, now that we have changed our sampling strategy for r1, . . . , rt ∼[R] from independent to pairwise independent, it is reasonable to expect the value of t (the number of pairwise independent variables that we wish to sample) to differ from that of Algorithm 1, where we had t = log 1 ε. To find a new value for t in B2, we will perform a similar analysis as the one we originally did for B and use a tail bound.
Take any L ∈BPP and, without loss of generality, assume x ∈L (the proof in the other case is symmetrical. Recall from Algorithm 1 that Yi = A(x, ri) and define Y = Pt i=1 Yi. Then, the probability that B2 is correct on x is Pr r1,...,rt∼[R][B2(x) = L(x)] = Pr r1,...,rt∼[R] Y ≥t 2 .
Furthermore, by the linearity of expectation we see that E r1,...,rt∼[R][Y ] = E r1,...,rt∼[R] " t X i=1 Yi # = t X i=1 E r1,...,rt∼[R][Yi] ≥2 3t, where the last line follows from our definition of BPP. We now know that we want to avoid the case that Y < 1 2 and that Er1,...,rt∼[R][Y ] ≥2 3t, so we could reasonably ask for the probability of Y being within t 10 of its mean. In other words, we would like to upper bound3 Pr r1,...,rt∼[R] Y −2 3t > t 10 .
(1) Previously, for B we used the Chernoff bound to (1), but in this case that is no longer an option as the r1, . . . , rt ∼[R] are only pairwise independent. Instead, we will use Chebyshev’s inequality, which requires us to compute Var(Y ). Thankfully, since the Yi’s are pairwise independent (because the ri’s are), we have that Cov(Yi, Yj) = 0 for i ̸= j. Then sing the fact that each Yi is a Bernoulli RV with p = 2 3, meaning that Var(Yi) = 2 3 1 −2 3 = 2 9, we can compute Var(Y ) = Var t X i=1 Yi !
= t X i=1 Var(Yi) + X i̸=j Cov(Yi, Yj) = t X i=1 Var(Yi) ≤2t 9 We can now apply Chebyshev’s inequality to (1) as Pr r1,...,rt∼[R] Y −2 3t > t 10 ≤100 t2 · 2t 9 = c t 3Upper bounding this value does also limit the probability that we are in the upper end of the tail, but this is an acceptable relaxation as being near the mean of Y is sufficient in our case.
Lecture 3: Aug 30, 2022 6 Thus, to upper bound this value by ε we need c t ≤ε, so we choose t ≥c ε = Ω 1 ε , meaning that B2 repeats A on the order of Ω 1 ε times and uses log t + log R = log 1 ε + log R + O(1) random bits.
Putting this together with the facts about B in Table 1 gives us Algorithm Accuracy Runtime Randomness B 1 −ε c · T(n) · log 1 ε c · R(n) log 1 ε B2 1 −ε c′ · T(n) · 1 ε R(n) + log 1 ε + O(1) Table 2: Accuracy, runtime, and randomness usage of B and B2.
In conclusion, we have answered a definitive YES to Question 1, showing that we can indeed use fewer random bits than B while achieving the same 1 −ε error rate. Nevertheless, this was at the cost of an exponential increase in runtime. We can further explore this trade-off between runtime and randomness by defining Bk to be analogous to the algorithm that we have been using, but with the r1, . . . , rt used in it drawn k-wise independently (where we define k-wise independence as any collection of k RVs being independent, so pairwise independence is simply 2-wise independence).
These Bk’s will essentially interpolate between B and B2, and their runtime and randomness usage can be analyzed similarly as to what we have done so far, although higher moment methods are required as k increases. |
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Guide to margin of error (with examples)
Last updated
4 March 2023
Reviewed by
Hugh Good
No matter the number of survey respondents, generating the actual views of every member of the represented population is impossible (unless you survey every member of a population!). This is why, for every survey you conduct, you need to consider the margin of error (MOE).
Let's dive deeper to understand the margin of error and how to handle it. Get ready to crunch some numbers.
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What is the margin of error in a survey?
The margin of error is the range within which the actual value of a survey parameter falls within a certain confidence level. In other words, it quantifies how the results from a sample might differ from the actual value you would have obtained if you had studied the whole population.
The confidence interval (CI) is the range between the upper and lower bounds of an estimate. The narrower the interval, the better and more accurate your results.
For instance, suppose a random sample of people has a +/−2% margin of error at a 95% confidence level. That implies that if you repeated the same survey 100 times, you expect the percentage of people who gave a particular answer to fall within the range of 2% of the reported results, 95 times.
The margin of error usually decreases with an increase in the random sample size or the number of responses. That means you can be more confident your results are reliable.
When can you use a margin of error?
You use a margin of error when you have a random sample—a set of randomly selected respondents from the whole population you’re studying. A random sample is also known as a probability sample since every member of your population has a known probability of being part of your sample.
For example, your company wants to know whether workers would prefer an extra leave day or bonus pay. You can randomly select a few employees from your workforce and ask them to choose their preferences. A margin of error will help the decision-makers understand the accuracy of the results.
What factors affect the margin of error?
The size of the margin of error depends on various factors:
Sample size: the more respondents who complete your study, the smaller the margin of error
Confidence level (CL): increasing the confidence level leads to a wider margin of error
Population variance: the higher the variance, the larger the margin of error
Poll design: your questions' exact wording can influence how people answer them, affecting the MOE
Response rate: a lower response rate increases the MOE. It can also increase if the respondents don't closely resemble the larger population.
Non-sampling errors: errors from sources such as coding and measurement can affect the margin of error
What is an acceptable margin of error for a quantitative survey?
The MOE size varies with the percentage of the target population sampled. Knowing the MOE for a particular study is crucial since the value for all percentages will be assessed through this lens when reviewing data.
If you were surveying 1,000 people using a confidence level of 90, surveying at least 250 people would give you an MOE of 4%. With this margin, you could be reasonably confident that the results were reflective of the audience you were surveying.
How to calculate the margin of error
Whenever you perform a statistical survey, you should calculate the MOE. Before you do so, you need to define your population.
A population is a group of elements from which you want to survey and gather data. Properly defining the population and selecting an appropriate sample size can help reduce the MOE.
Margin of error calculator
Optimize your research’s impact when you improve the margin of error.
0%
Margin of error
The total number of people whose opinion or behavior your sample will represent.
The probability that your sample accurately reflects the attitudes of your population. The industry standard is 95%.
The number of people who took your survey.
0%
Margin of error
Here's the formula:
MOE= z-value √ (p(1-p)/n)
Where:
n is the sample size
p is the sample percentage
z-value is the critical value corresponding to your confidence level
Here are standard confidence levels with their corresponding z-values:
90%: 1.64
95%: 1.96
99%: 2.58
Finding the maximum margin of error formula
For the maximum MOE, you'll use 0.5 as p and the z-value for 95% (the standard confidence level), which is 1.96. Thus, the formula is a direct transformation of the sample size:
Maximum 95% MOE = 1.96√0.5 (1 - 0.5)/n =0.98/√n
Plug in your sample size to the formula, and voila! You'll get the maximum margin of error for whatever survey you've done. That means the figure is unique for a given size.
For example, the maximum MOE for a sample size of 100 would be 0.98/√100 = 0.098
How to calculate the margin of error with your survey data
Now you have your formula, crunching those figures won't be a headache.
Take n, the sample size, and p, the sample percentage.
Calculate p(1 - p) and divide the result by n.
Find the square root of the value.
Multiply the figure you got in Step 3 by the z-value.
An example of calculating the margin of error
Suppose you surveyed 1,000 respondents to find out their views on volunteering. 500 of them agreed that volunteering is an excellent part of life. What's the margin of error given a 50% CL?
MOE= z√(p (1 - p))/n
P = 500/1,000 = 50%
z-score = 1.96 (at 95% confidence level)
Therefore, MOE= 1.96√ (0.5(1 - 0.5))/1,000
=0.031=3.1%
So the sample's MOE is +/−3%. That implies if you repeated the survey several times, the number of people who love volunteering would be within 3% of the sample percentage (50%) 95% of the time.
How sample size affects the margin of error
The size of the sample is a key factor that affects your margin of error. Looking to increase your study’s precision? Interview more respondents and ensure they complete your survey.
Increasing sample size reduces the standard deviation (a measure of your estimate's variability). In other words, it narrows the range of possible values, leading to more precise estimates and lower MOE.
For example, if you're looking to estimate the average income of your existing customers, a sample of two people will lead to a wide range of figures, so the margin of error will be high. But if you increase the sample size to 1,000 people, the MOE will be significantly narrower.
That means MOE is critical to understanding whether your sample size is appropriate. Is it too large? You'll need to survey more people to capture your population's attitudes accurately.
#### Calculate your ideal sample size
How to increase your data's reliability
If you're looking to boost your survey's reliability, the trick is to minimize the MOE. Here are three tried-and-true ways you can generate more accurate results:
1. Minimize the variables
A high number of variables can introduce more errors to your survey. Variables cause your standard variation to shoot up, increasing the MOE. Here's a trick: change how you collect data. For example, ensure the process is rigorous and measure your factors or variables accurately.
2. Increase the sample size
This hack is often the easiest on our list. Statistically, if more people complete your study, your chances of getting a representative response increase because the confidence interval decreases. The result is a lower MOE. But ensure you have enough resources and time to generate a larger sample size.
3. Lower your confidence level
Using lower confidence is another trick that can lead to a narrower MOE. But be careful since a lower level means decision-makers will be less confident in the results.
Therefore, only reduce the CL if the cons of a smaller MOE outweigh the cons of a lower CL. For example, if it’s unfeasible to increase the sample size due to costs, you can reduce the CL to achieve a narrower interval.
FAQs
How can you calculate the margin of error when you have a confidence level?
You can calculate the margin of error using the following formula:
MOE= z √(p (1-p))/n
Where z is the z-value for your desired confidence level, p is the sample percentage, and n is the sample size.
What is a large margin of error?
A large MOE implies a high chance of the actual value being very different from the value you've estimated. It often occurs when dealing with small sample sizes and high data variability, combined with a high CL.
What if the margin of error is very high?
If the margin of error is very high, the survey's sample results won't accurately represent your population, so decision-makers can't rely on it. The solution is to survey more people, reduce variables, lower the CL, or use a one-sided CI.
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7554 | https://www.frontiersin.org/journals/physiology/articles/10.3389/fphys.2020.00260/full | Frontiers | Morphology of the Amazonian Teleost Genus Arapaima Using Advanced 3D Imaging
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ORIGINAL RESEARCH article
Front. Physiol., 27 April 2020
Sec. Integrative Physiology
Volume 11 - 2020 |
This article is part of the Research Topic Time Domains of Hypoxia Adaptation: Evolutionary Insights and ApplicationsView all 24 articles
Morphology of the Amazonian Teleost Genus Arapaima Using Advanced 3D Imaging
Miriam Scadeng1,2Christina McKenzie3Weston He1,4†Hauke Bartsch 1,5David J. Dubowitz 1,2Dominik Stec1Judy St. Leger 6
1 Department of Radiology, University of California, San Diego, San Diego, CA, United States
2 Department of Anatomy and Medical Imaging, University of Auckland, Auckland, New Zealand
3 Department of Pathobiology, University of Guelph, Guelph, ON, Canada
4 NOVA Southeastern University, Fort Lauderdale, FL, United States
5 Mohn Medical Imaging and Visualization Centre, Haukeland University Hospital, Bergen, Norway
6 Department of Biomedical Sciences, College of Veterinary Medicine, Cornell University, Ithaca, NY, United States
The arapaima is the largest of the extant air-breathing freshwater fishes. Their respiratory gas bladder is arguably the most striking of all the adaptations to living in the hypoxic waters of the Amazon basin, in which dissolved oxygen can reach 0 ppm (0 mg/l) at night. As obligatory air-breathers, arapaima have undergone extensive anatomical and physiological adaptations in almost every organ system. These changes were evaluated using magnetic resonance and computed tomography imaging, gross necropsy, and histology to create a comprehensive morphological assessment of this unique fish. Segmentation of advanced imaging data allowed for creation of anatomically accurate and quantitative 3D models of organs and their spatial relationships. The deflated gas bladder [1.96% body volume (BV)] runs the length of the coelomic cavity, and encompasses the kidneys (0.35% BV). It is compartmentalized by a highly vascularized webbing comprising of ediculae and inter-edicular septa lined with epithelium acting as a gas exchange surface analogous to a lung. Gills have reduced surface area, with severe blunting and broadening of the lamellae. The kidneys are not divided into separate regions, and have hematopoietic and excretory tissue interspersed throughout. The heart (0.21% BV) is encased in a thick layer of lipid rich tissue. Arapaima have an unusually large telencephalon (28.3% brain volume) for teleosts. The characteristics that allow arapaima to perfectly exploit their native environment also make them easy targets for overfishing. In addition, their habitat is at high risk from climate change and anthropogenic activities which are likely to result is fewer specimens living in the wild, or achieving their growth potential of up to 4.5 m in length.
Introduction
The arapaima, or pirarucu, is an air-breathing osteoglossid teleost native to the Amazonian floodplains, which can reach sizes of up to 4.5 m in length and 200 kg in weight (Graham, 1997; Castello, 2008) (Figure 1). As a genus, Arapaima are morphologically, biologically, taxonomically and commercially important.
FIGURE 1
Figure 1. Overview of specimen one. In all figures the traditional radiological convention is adopted with rostral end of fish to the left of the page for sagittal and coronal images and 3D reconstructions. For axial images, left side of image is right side of fish. Right (R), left (L), dorsal (D), ventral (V) marked for clarity. (A) 190 cm long female arapaima weighing 70.5 kg, volume of 68.85lt. The average density of the whole specimen is 1.02 g/cm 3. Whole fish surface reconstructed from CT and MRI data (CT cranial 85% MRI caudal 15% as demarcated by blue arrows). (B) Semi transparent surface outline showing relative position of internal organs and skeleton reconstructed from CT data. Heart: purple, liver: brown, spleens: pink, ovary: magenta, esophagus and stomach: green, gut: turquoise, pyloric caeca: orange, respiratory tissue: blue, air in gas bladder: yellow, esophageal sphincter: red. Axial skeleton head and body surface: semi-transparent.
Arapaima exhibit bimodal breathing, using both gills and a respiratory swim bladder to exchange gases with its environment (Gonzalez et al., 2010; Fernandes et al., 2012). If denied access to air, adult arapaima will drown within 10 min, making them obligate air-breathers (Farrell and Randall, 1978). Until they are approximately 9 days old (18 mm), arapaima exclusively breathe water (Graham, 1997). As they mature, arapaima undergo extensive anatomical and physiological changes, including atrophy of the gills, to become air-breathers (Brauner et al., 2004).
Sometimes referred to incorrectly as “lungs,” the highly specialized respiratory gas bladder of the adult arapaima runs along the dorsal surface of the coelomic cavity (Brauner et al., 2004; Daniels et al., 2004; Fernandes et al., 2012). Many fish possess gas bladders, which can be used for a variety of purposes including buoyancy, sound production and respiration. Many of the more derived teleosts, such as the rainbow trout (Oncorhynchus mykiss), a common fish model, have gas bladders that are a smooth bag-like structure (Hall, 1924). Even bichirs (Cladistia), which use a gas bladder for respiration like an arapaima, have a simple gas bladder (Perry et al., 2001). In arapaima, the respiratory gas bladder is complex and subdivided into many irregular ediculae to hugely increase surface area. The lining epithelium is similar to that seen in the gills of non-air-breathing fish (Fernandes et al., 2012). Similar to the majority of the actinopterygii, arapaima use a four-stroke buccal pump to move gas in and out of the respiratory gas bladder, rising to the surface to take a breath approximately every 4 min (Farrell and Randall, 1978; Brainerd, 1994). The respiratory gas bladder is not a true lung, as seen in sarcopterygians such as lungfish (Dipnoi) and tetrapods, because it is dorsal, unpaired and has a different ontological origin (Perry et al., 2001).
Taxonomic classification of arapaima species is a contentious issue. Until recently, it was believed Arapaima gigas was the only species (Stewart, 2013b). There are now five recognized species: A. mapae, A. gigas, A. leptosoma, A. agasizii, and A. arapaima, with the last being discovered as recently as 2013 (Stewart, 2013a).
Although all arapaima were thought to be A. gigas, that species is now only known from the museum specimen holotype in Paris (MNHN a-8837). The Amazon is a hotbed of biodiversity due to the variety of microenvironments, allowing for evolution in isolation, and there are likely still more arapaima species to be discovered (Castello et al., 2011; Stewart, 2013a). Aside from presenting an interesting taxonomical conundrum, this poses an important conservation issue as only A. gigas is protected under the International Union for Conservation (ICUN) Red List and the Convention on International Trade of Endangered Species (CITES) Appendix 2 (Castello and Stewart, 2010).
Advanced imaging, including computed tomography (CT) and magnetic resonance imaging (MRI), is becoming an increasingly popular means of collecting detailed morphological data from threatened and difficult to acquire species, that can be easily disseminated amongst scientists (Corfield et al., 2008; Berquist et al., 2012; Ponganis et al., 2015; Wright et al., 2016, 2017; York et al., 2018). These techniques allow for excellent imaging of tissue as it exists in situ, without disruption from autopsy. By combining the results of advanced imaging and 3D modeling, with detailed examinations using histologic tissue review, we can evaluate the whole animal from a cellular to a gross morphological scale.
Materials and Methods
Specimens
A 190cm long female arapaima weighing 70.5 kg was found dead in an aquarium enclosure. The primary findings on necropsy were fibrosing cardiomyopathy, 80 ml of serous pericardial fluid, and 400 ml of sero-hemorrhagic abdominal fluid. The effusions were likely related to the cardiac condition, the cause of which is unknown. This animal, the primary specimen or Specimen One, was examined using multiple modalities for anatomical investigation.
To enhance the histological investigation, three additional arapaima were examined for more detailed histologic information. Specimen Two was an 80 cm long juvenile male, which died in 2008 under anaesthesia. It was used for gill, air bladder and heart histology. Specimen Three was a 140 cm long sub-adult male that jumped from an enclosure in 2007, and was used for kidney histology. Specimen Four was a female arapaima, which died under anaesthesia while under-going enucleation, and was used for brain and digestive system histology.
Species Identification
Given the many uncertainties surrounding arapaima taxonomy we do not make definitive species identifications for any of our specimens, therefore our findings are representative of the genus Arapaima. The morphological indicators that differentiate between species include measurements such as tooth numbers, fin ray numbers and orbit diameter, vary based on body size; there is little interspecies difference in internal anatomy, which is the focus of this paper (Stewart, 2013b).
Data Collection and Quantitative Analysis
Imaging was performed on the Specimen One within 24 h of death. Necropsy was performed the following day at the facilities of SeaWorld San Diego. Histological slides were prepared using hematoxylin and eosin stain by San Diego Pathologists Medical Group, Inc.
T1 and T2 MR data was collected of the entire fish using a 3 Tesla MRI scanner (GE Sigma HDx MRI scanner – GE Medical Systems, Milwaukee, WI, United States) at the University of California, San Diego Center for Functional MRI. Due to the size of the specimen all imaging was acquired using the machine body coil, and in several sections, limited by the useable length of the imaging coil. T2-weighted images were acquired using a 2D fast spin echo imaging sequence with the following protocol parameters: echo time = 116 ms; repetition time = 3500 ms; averages; in-plane matrix = 512 × 512; in plane resolution = 0.55 × 0.55 mm; slice thickness = 4 mm. T1-weighted images were acquired using a 3D fast spoiled gradient echo sequence with the following protocol parameters: echo time = 4.3 ms; repetition time = 10.2 ms; averages; inversion time 450 ms, in-plane matrix = 512 × 512, in plane resolution = 0.43 × 0.43 mm; slice thickness = 1 mm. Distortion correction is performed automatically using GE software, and the length of the acquisition was limited to the linear portion of the gradient coil, with care taken to overlap the acquired segments. This ensured the whole length of the fish was acquired while avoiding either missing segments or distorted data.
CT data was collected using a GE lightspeed Discovery 750HD CT scanner (GE Medical Systems, Milwaukee, WI, United States). Scanning parameters were 120 kV, 226MA. In plane resolution 0.98 × 0.97 mm; slice thickness 1.25 mm. As the length of the specimen was longer than the CT machine table, the tip of the caudal fin was not imaged, though the surface model of the whole fish was reconstructed using MRI data of this region for accurate total body volumes (Figure 1).
Segmentation was manually performed using AMIRA software (FEI Visualization Sciences Group, Burlington, MA, United States). Segmentation was based on image grayscale intensity and a priori knowledge of anatomical characteristics of ostoglossid fish, spatial relationships and external landmarks. 3D models of the anatomy were constructed from the segmented imaging data.
Morphometric measurements were taken as follows: heart and vessel wall and lumen measurements were taken from MR data at three representative points and averaged. Tooth counts are an approximation. They were obtained from CT images, and as the teeth are similar in size to the image resolution this made some individual teeth difficult to define separate from neighboring teeth. The tooth counts for the dentary include only a single ramus. Orbit diameter was measured from CT images at the widest point. Interorbital width was measured from CT images from the most medial points of each orbit. Caudal peduncle length was measured from necropsy photos as the distance between the caudal insertion of the anal fin and the cranial insertion of the caudal fin. Anal fin rays were counted using a combination of CT and MR images. Total density of Specimen 1 was calculated using total body weight at necropsy divided by the total body volume measured using segmented imaging data.
Hematoxylin and eosin staining is done per Stevens (1982): tissues are fixed, dehydrated and embedded in melted paraffin wax, which facilitates microtoming into thin slices. After rehydrating, hematoxylin and a metallic salt is applied to the tissue, which is then rinsed in a weak acid solution, followed by rinsing in mildly alkaline water. This stains cell nuclei blue. The tissue is then stained with eosin, which stains the extracellular matrix and cytoplasm pink.
Results
Gill Morphology
There are four pairs of gill arches with accompanying filaments (Figure 2d). The filaments are short, squat and have only rudimentary lamellae. The lamellae are extremely blunt and broad compared to those of the gills of water-breathing fish and the pillar system is absent (Figure 2b). There is a slight curl to the distal portions of the filaments on necropsy, but this appears to be artifact as it not present on histology and may have been due to post mortem necrosis (Figure 2a). The lamellar epithelium is comprised of mucocytes, ionocytes, and pavement cells (Figure 2c). The gills subjectively appear to be less highly vascularized than most water-breathing fish.
FIGURE 2
Figure 2. Respiratory system: The gill arches and filaments. (a) On each side there are four gill arches (GA) with associated cartilaginous filaments (F). (b) Photomicrograph of the boney gill arch, cartilaginous filaments. Note extensive blunting and broadening of the lamellae. (c) A higher magnification light micrograph of the lamellae of a gill filament. The epithelium of the lamellae is made up of ionocytes (filled arrowheads), mucocytes (empty arrowhead) and pavement cells (notched arrowhead). Note the collapse of the pillar cell system. (d) Coronal CT slice at level of the base of mouth demonstrating the gill arches and filaments. (e) 3D reconstruction of gill arches (blue) from CT as seen from caudal end of fish looking cranially into the fish.
Esophageal Sphincter and Respiratory Gas Bladder
The respiratory gas bladder is unpaired and runs the dorsal length and width of the coelomic cavity, encompassing the kidneys (Figures 3, 4, 5). The gas bladder is connected to the esophagus cranially via a substantial dorsal muscular sphincter (Figures 3a–f). The sphincter is largely a triangular, disc-like shaped muscle measuring 5cm in length. In the midline is a longitudinal slit-like opening that measures 24 mm in length. It is through this opening that air moves between the esophagus and the rostral aspect of the gas bladder (GB), which was not inflated at this level. A tiny longitudinal strip of possible tissue is seen on CT and MR imaging overlying the opening (yellow arrow in 3c), and this may act as a valve regulating the flow of air, or preventing water/food from entering the GB. In addition a localized thickening of tissue (red arrow Figure 3c) sited on the opposing side of the esophagus, possibly occludes the slit opening when the surfaces of the esophagus are opposed.
FIGURE 3
Figure 3. Respiratory system: Esophageal Sphincter. (a) Midline sagittal, and (b,c) axial MRI image at level of upper esophagus through the muscular sphincter (MS), positioned in the dorsal aspect of the esophagus. White box on (b) indicates enlarged section (c). The sphincter is a triangular, disc-like shaped muscle through which air moves between the esophagus and the rostral aspect of the gas bladder (GB), which is not inflated at this level. A tiny longitudinal strip of tissue seen overlying the opening (yellow arrow in c), may act as a valve regulating the flow of air, or preventing water/food from entering the GB. In addition a localized thickening of tissue (red arrow) sited on the opposing side of the esophagus, possibly occludes the slit opening when the surfaces of the esophagus are opposed. (d,e) 3D reconstruction of muscular sphincter from CT data demonstrating relationship between the esophagus and the GB. Red: muscular sphincter, blue: rostral end of GB, green: esophagus, yellow: strip of tissue in valve opening. (d) From above sphincter, (e) from below sphincter, (f) lateral view.
FIGURE 4
Figure 4. Respiratory system: Respiratory gas bladder. (a) The gas bladder (GB) and respiratory tissue (RT) runs the length of the dorsal coelomic cavity and encompasses the kidneys (K). (b) The dorsal surface is covered by a mat of highly vascular tissue comprising of ediculae and inter-edicular septa. The inter-edicular septa are created by trabeculae made of smooth muscle and connective tissue. The interior surfaces of the gas bladder that are in contact with air are all covered with respiratory epithelium consisting of pavement and columnar epithelial cells. (c,d) A light micrograph stained with Masson’s Trichrome showing a blood vessel (BV), inter-edicular septa (IES) and ediculae (E). Collagen is stained blue. (e,f) Axial and zoomed axial MRI of respiratory tissue at level of mid of the GB demonstrating respiratory tissue (RT), gas in air bladder (A), Gas in stomach (G). The kidneys (K) are seen suspended in the GB, vertebrae (V) and gut (GU). Vertebral bodies appear to have spicules emanating from them (red arrow) – Similar appearance seen on CT in Figure 10. This feature is presumably to increase surface area for gas exchange and have not previously been described in this species. (g) Coronal CT scan at level of GB demonstrating relationships of contents of GB. (h–j) 3D reconstructions from CT data. (h) Air (A) fraction in GB (yellow), (i) Vascular respiratory tissue fraction (blue). (j) Reconstructions of GB showing the relationship of A to RT.
FIGURE 5
Figure 5. Ovary and kidneys. (a) Gross anatomy of ovary (O) in the peritoneal cavity. (b) Kidney- light micrograph showing hematopoietic tissue (H), renal tubules (T), capillary tuft (arrowhead) and melanomacrophages (empty arrowhead). (c) 3D reconstruction from CT data of kidneys (green) and ovary (magenta) as seen from left side and (d) from below looking up. Other organs as described above but semitransparent for orientation.
The dorsal aspect of the gas bladder is highly vascularized and appears dark red on necropsy (Figures 4a,b). Grossly, the surface of the gas bladder appear as a disorganized, dense web of tissue with air compartments, or ediculae, of varying size. The inter edicular septa are comprised of muscular supportive structures and blood vessels (trabeculae) with a layer of respiratory epithelium covering areas in contact with gas (Figures 4b–d). This epithelium is comprised of pavement cells and columnar cells. The respiratory tissue measured 60 mm at its thickest part. The total volume of the respiratory tissue is 785 ml excluding the air component.
At the time of CT and MR data collection, the gas bladder was almost completely collapsed, with much of the air residing within the mass of respiratory tissues. On the CT data of Specimen One the right side of the gas bladder was more inflated than the left side (total gas volume 564 ml, Table 1). The asymmetry in air volume is likely due to the left side being dependant during transportation of the specimen. The T2 weighted MR image (water has high signal intensity on T2-; Figures 4e,f) demonstrates that there is some fluid within the gas bladder, which highlights the visualization of the inter-edicular septa. The respiratory gas bladder makes up 1.96% of the total volume of this arapaima (Table 1).
TABLE 1
Table 1. Body structure volumes in milliliters (ml) and as percent of whole fish.
Kidneys and Ovary
The kidneys are paired and run medially along the spine within the respiratory gas bladder (Figures 4e–g, 5c,d) and occupy approximately the caudal three quarters of this space with the respiratory tissue on each side. The kidneys are in contact with each other along their length. They comprise 0.35% of the total volume of the arapaima (Table 1). There is no differentiation into a hematopoietic “head” and excretory “trunk,” as hematopoietic cells and nephrons are interspersed throughout the parenchyma. The nephron consists of a renal corpuscle and renal tubule, emptying into a collecting duct.
Specimen One is an adult female with an inactive left ovary and no right ovary. The ovary is long, thin, and flat and tapers to a point cranially. The left ovary comprises 0.08% of the total volume of the arapaima (Figures 5a,c,d).
Heart and Vasculature
The heart is found encased in a thick layer of lipid rich connective tissue just caudal to the gill arches. The heart consists of the sinus venosus, atrium, muscular ventricle, conus arteriosus, and bulbus arteriosus (Figures 6a–f). Blood enters the heart via the ductus Cuvier, which is fed by the hepatic and renal portal systems. Blood exits the heart to the dorsal aorta toward the gills. Only the heart parenchyma was included in segmentation; the lumen was excluded. The heart musculature constitutes 0.21% of the total volume of the arapaima, with the majority of the heart volume being the ventricle (Table 2). The ductus Cuvier has a diameter of 6.3 mm where it meets the sinus venosus. The wall of the sinus venosus is 1.1 mm thick and the atrium is 2.0 mm thick. The thickness of the ventricular wall varies between 7.8 and 14.8 mm, depending on the musculature. The lumen of the ventricle is very narrow in MR images, indicating a contracted state at the time of death. The conus arteriosus is a distinct entity between the ventricle and the bulbus arteriosus and supports the conus valve (Figure 5a). The bulbus arteriosus has a thick fibroelastic wall, measuring 1.9 mm thick, with tissue folds protruding into the lumen, which is 19.9 mm at the thickest point. Specimen one has moderate pericardial effusion, which can be seen in the MR images (Figure 6d).
FIGURE 6
Figure 6. The heart. (a,b,d) The heart consists of the sinus venosus (SV), the atrium (At), the ventricle (V) and the bulbus arteriosis (BA). Some of the substantial lipid rich connective tissue that surrounds the heart can be seen (L). (c) Photomicrograph of the ventricular myocardium. (d) Sagittal MRI of the heart showing the ductus of Cuvier (arrowhead) connecting to the sinus venosus. A small amount of pericardial effusion (PE) can be seen at the apex of the heart. (e–i) e-3D reconstruction of heart superimposed on axial MRI for orientation of heart. (f) Ventral (seen from below), (g) Dorsal (from above), (h) Left lateral, (i) Right Lateral. Hepatic vein: yellow, sinus venosus: light blue), atrium: dark blue, ventricle: red, bulbus arteriosus: purple.
TABLE 2
Table 2. Cardiac structure volumes in milliliters (ml) and as percent of whole heart volume.
Gastrointestinal System
The teeth of the dentary, premaxillary and maxillary bones are simple, conical, approximately 2 mm long and present in single rows. There are 36 dentary teeth and 34 maxillary teeth. The basibranchial tooth plate, or boney tongue, is covered in fine villiform lingual teeth (Figures 7d–f).
FIGURE 7
Figure 7. The digestive system. (a) Abdominal cavity with liver (L) and gut (G). (b) Several spleens. (c) Light micrograph of liver showing a bile duct (BD), blood vessel (BV), sinusoids (arrowhead) and hepatocytes (top half of image). (d,e) Sagittal midline and axial CT slice through head demonstrating how effective crushing of prey occurs between boney tongue and roof of mouth. (f) 3D reconstruction of head looking into mouth highlighting boney tongue (green). (g,h) 3D model of gut and liver reconstructed from CT data. The gut of the osteoglossomorpha is distinct from that of other fishes in that the intestine passes posteriorly to the left of the esophagus and stomach (Nelson, 1972) rather than to the right. (g) left lateral, (h) Right lateral. Esophagus and stomach: green, intestinal loops: turquoise, blind ending pyloric caeca originating from proximal gut: orange, liver: brown, spleens pink.
The esophagus is approximately 7 cm long and has a large dorsal muscular sphincter connecting it to the respiratory gas bladder as described above (Figure 3). The esophagus connects to a muscular stomach. The stomach has a thin tunica serosa, a thick tunica muscularis with perpendicular and circular muscle fibers, a tunica submucosa and a tunica mucosa with both gastric glands and goblet cells. There are two blind-ended pyloric ceca. The intestine passes posteriorly to the left of the esophagus and stomach rather than to the right (Figures 7g,h). The intestines loop back and forth four to five times within the coelom before terminating at the rectum. The gastrointestinal tract – esophagus –rectum, constitutes 2.56% of the total body volume (excluding gas volume).
The liver has one trough shaped lobe that lies ventral to the gastrointestinal tract in the cranial aspect of the coelomic cavity and comprises 0.57% of the total body volume (Figure 7a and Table 1). It is dark reddish-brown in color. The hepatocytes are hexagonal in shape, vacuolated, and have a round centrally located nucleus with a darkly staining nucleolus (Figure 7c). A network of sinusoids runs between the hepatocytes. There are many larger blood vessels within the liver parenchyma. The bile ducts are lined with a single layer of columnar epithelial cells. There are several spleens (Figure 7b).
Brain
The rostral-most structures are the paired olfactory bulbs, which are attached caudally to the cerebral hemispheres (Figures 8a–i). The telencephalon is large and well developed compared to the majority of fish, especially other osteoglossids; it has well differentiated right and left lobes. The diencephalon is primarily composed of the preoptic area. The mesencephalon is predominately optic tectum. Caudal to this is the cerebellum and the medulla. The brain makes up only 0.01% of the total body volume (Table 3) and the largest subdivision is the telencephalon at 28.3% of brain volume. The brain case is significantly larger than the brain itself (Figures 8b–i). There is a large and complex cerebro-spinal fluid (CSF) space around the brain which measures more than 3 times (323%) the volume of the brain itself (Figure 8b), and the space around and anterior to the CSF space is filled with fatty connective tissue.
FIGURE 8
Figure 8.(a) Photomicrograph showing the cerebellum (Ce), medulla (M), optic tectum (OT), valvula cerebelli (VC) and telencephalon (Te). (b) Sagittal midline T2 MR image through brain showing the cerebellum, optic tectum, telencephalon, olfactory bulb (OB), pituitary (P), spinal cord (SC) and medulla. Surrounding high signal is expansive cerebro-spinal fluid (CSF) space in which the brain is suspended. (c,d) Dorsal and lateral view 3D reconstructions of the head from CT data showing the extent of the CSF space encasing the brain (green), optic nerves (blue) and olfactory nerves (orange). Eyes are red, and lining of olfactory pits yellow. Other bony head structures are semi transparent head. (e–i) 3D reconstructions of the brain with large and complex CSF space surrounding brain (semitransparent surface). (e) Dorsal, (f) Ventral, (g) lateral, (h) anterior, and (i) Posterior view. Olfactory bulbs: green, telencephalon: dark yellow, hypothalamus: brown, optic tectum: blue, cerebellum: light yellow, medulla and proximal spinal cord: red, and proximal optic tract: turquoise.
TABLE 3
Table 3. Brain structure volumes in milliliters (ml) and as percent of whole brain volume.
Musculoskeletal and Integument
The cranium is heavily ossified and well protected consisting of 2 main layers of overlapping bones. (Figures 9A–C) The mouth opens upwards toward the surface. There are indentations in the cranium around pores to the acousticolateralis system (Figure 9A). The orbit diameter is 1.83% of total length and the interorbital width is 6.3% of total length.
FIGURE 9
Figure 9. Musculoskeletal -Head. (A–C) 3D reconstructions from CT data, of the external bony armor plates of the head with integument overlain as transparent surface. Abbtp, anterior basibranchial toothplate; ang, angular; ant, antorbital; br, branchiostegal rays; chy, Ceratohyal; cl; d, dentary; dsp, dermosphenotic; fr, frontal; io 1–4, infraorbital 1–4; iop, interopercular; iop, infraopercular; mx, maxilla; n, nasal; op, opercle; pa, parietal; pmx, premaxilla; pop, preopercle; q, quadrate; s, suture; sop, subopercle; spop, suprapreopercle; u, urohyal; v, vomer. (Bones referenced as per Steward 2013).
Bone (skull bones and axial skeleton) comprise 5.45% of the total volume, with skull bones constituting more than half of that, and the integument comprises 6.12% of the total volume. The 79 vertebrae are composed of a centrum, a neural spine and transverse processes. The ventral aspect of the vertebral bodies in the region of the gas bladder are spiculated and interdigitate with the respiratory tissue (Figures 10a–d). Many of the transverse processes of the vertebrae in the region of the gas bladder appear to partially encase air. There are 34 pairs of ribs (see Figure 10h). Arapaima have pectoral fins, pelvic fins, an adipose fin, an anal fin and a caudal fin. The adipose and anal fins are the most substantial, squaring off the peduncle before the caudal fin in a lateral view (Figure 1). Thirty-four anal fin rays support the anal fin and 36 support the adipose fin. The caudal fin is small and triangular with 27 boney rays.
FIGURE 10
Figure 10. Musculoskeletal and Integument. (a) Axial CT at level of GB demonstrating air partially encased between the bony cortices of the transvers processes of the vertebral bodies, and (b,c) shows spiculation of the vertebral body, possibly to increase surface area of respiratory epithelium exposed to air. (d) 3D rendering of vertebrae on background of coronal CT slice showing air density within part of the in the vertebral body between spicules. (e) Axial CT section demonstrating flexible dermal armor of elasmoid scales made from overlapping layers of type 1 collagen and a highly mineralized hydroxyapatite outer layer (Yang et al., 2014). Scales overlap with a thickness up to 9.7 mm. Single scale thickness is approximately 3.4 mm. (f,g) The muscle is divided in to left and right epaxial and hypaxial muscle by the vertical and horizontal septa. (h) Bony axial skeleton reconstructed from CT data (excluding distal tail).
The integument covering the head is green dorsally and white ventrally. There are pores on the mandible and ventral cranium that open to the acousticolateralis system. The scales of the body are thick and gray in color and lighter ventrally, creating heavy armor. Where the scales overlap the thickness is up to 9.7 mm. Single scale thickness is approximately 3.4 mm (Figure 10e). On the caudal trunk, some scales have bright red areas posteriorly which form diagonal lines along the sides of the fish. The amount of red pigment increases caudally. The body and tail are heavily muscled with mostly white muscle. Muscle makes up the majority of the fish (muscle and minor organ/soft tissue constitutes 82.6% of total). The muscle is divided in to left and right epaxial and hypaxial muscle by the vertical and horizontal septa (Figures 10f,g). The muscle is segmented into myomeres.
Discussion
Arapaima are highly specialized for life in the intermittently hypoxic environment of the Amazon, giving them a competitive edge during times of low water levels (Castello et al., 2011) in which dissolved oxygen can reach 0 ppm (0 mg/l) at night, though anoxic environments only occur in deeper waters (Junk et al., 2015). Many Amazonian fish have developed methods of compensating for hypoxia, such as the pharyngeal diverticula of the swamp eel (Snybranchus marmoratus) for air breathing, or the expansion of the lip in the pacu (Piaractus brachypomum) to exploit the more oxygen rich surface waters (Val et al., 1998). However, none are as striking or as comprehensive as the changes associated with the respiratory gas bladder in arapaima. The development of obligate air-breathing required the compensatory evolution of several additional morphological and physiological adaptations. In addition to changes in the respiratory organs, such as the gas bladder and gills, there are alterations in renal function and the vascular supply. Together, these morphological specializations allow the arapaima to fill a unique ecological niche in the Amazonian basin and make the arapaima a fascinating anatomical study.
Respiratory System
Although air-breathing has evolved multiple times in teleosts, no fish has freed itself entirely from using gills while remaining in an aquatic habitat, except the obligate air-breather adult arapaima (Graham, 1997). Arapaima are highly specialized to withstand hypoxia, with 50–100% of their oxygen coming from the air, depending on the oxygenation of the water (Stevens and Holeton, 1978) and size or age. They have higher metabolic rates than most fish and increased aerobic capacities, existing in a state of compensated respiratory acidosis with lower blood pH and higher pCO 2 levels than most fish (Hochachka et al., 1978b; Brauner et al., 2004; Gonzalez et al., 2010). Despite their reliance on air, arapaima ventilate their gills 16–24 times/minute and only stop during inhalation at the surface (Farrell and Randall, 1978; Stevens and Holeton, 1978) and this may be related to acid base balance and CO 2 excretion, though they may also maintain use of their gills in order to minimize time spent at the surface, where they are more prone to predation (Kramer et al., 1982).
Mechanics of Breathing
When arapaima ventilate their respiratory swim bladder, they only break the surface of the water for one second (Farrell and Randall, 1978). The four-stroke buccal pump breathing mechanism of the arapaima results in exchange the contents of their gas bladder (Greenwood and Liem, 1984; Liem, 1989). Farrell and Randall (1978) proposed that effective gas bladder ventilation is due to a diaphragm-like septum that stretches between the body flanks such that, when there is an outward movement of the flanks, the septum is pulled downwards creating negative pressure to fill the gas bladder. But, similar to Greenwood and Liem (1984), we found no evidence of such a structure. As the fish rises toward the surface, gas is moved from the bladder into the buccopharyngeal cavity and out through the opercular slit (Greenwood and Liem, 1984; Liem, 1989). When the mouth comes out of the water, the buccopharyngeal cavity is filled with air and the arapaima sinks below the surface while the buccopharyngeal floor rises and air moves to the gas bladder (Greenwood and Liem, 1984; Liem, 1989).
Gill Morphology
Like most teleosts, arapaima have four pairs of gill arches with two rows of filaments and associated lamellae which are used for gas exchange, ionoregulation, acid-base balance and nitrogenous waste excretion (Figure 2). Unlike most fish, arapaima gills are drastically decreased in surface area, have shorter lamellae, thicker blood-water barriers and a smaller role in homeostatic functions (Hulbert and Moon, 1978; Brauner et al., 2004; Gonzalez et al., 2010; Ramos et al., 2013). As arapaima mature and become increasingly reliant on air-breathing, the intralamellar spaces fill with a proliferation of chloride cells (more recently known as ionocytes), increasing the thickness of the blood-water barrier, and there is atrophy of the pillar system, decreasing blood supply (Brauner et al., 2004; Da Costa et al., 2007; Ramos et al., 2013). These changes are thought to help decrease the loss of oxygen from the blood into hypoxic waters while still allowing for CO 2 exchange (Graham, 1997; Brauner et al., 2004; Gonzalez et al., 2010). Ionocytes are usually only found on the edge of the filamental epithelium and function in the reuptake of Cl– and Ca 2+ ions, and their proliferation may indicate that ion and acid-base regulation is a more important role for arapaima gills than respiration (Brauner et al., 2004; Ramos et al., 2013). In addition to ionocytes, the epithelium of the gills also contains mucocytes (Figure 2C; Fernandes et al., 2012Ramos et al., 2018).
Respiratory Gas Bladder Morphology
As previously described, we found the gas bladder runs the length of the coelomic cavity, encompassing the kidneys. The dorsal surface is covered by a mat of highly vascular tissue comprising of ediculae and inter-edicular septa. The inter-edicular septa are created by trabeculae made of smooth muscle and connective tissue (Figure 4; Hochachka et al., 1978b; Greenwood and Liem, 1984; Fernandes et al., 2012). The interior surfaces of the gas bladder in contact with air are all covered with respiratory epithelium consisting of pavement and columnar epithelial cells (Figure 4c; Fernandes et al., 2012). Electron microscopy shows pavement cells have fewer mitochondria and are closely associated with capillaries on the basal surface, while the apical surface has short microvilli (Fernandes et al., 2012). The columnar cells have more mitochondria and microvilli (Fernandes et al., 2012). Columnar and pavement cells are connected by tight junctions (Fernandes et al., 2012). The blood-air barrier in arapaima is made up of capillary endothelial cells, the basal lamina and the pavement cells and is so thin it is comparable to avian lungs (Fernandes et al., 2012).
Our respiratory gas bladder volume of 2.56% total volume is substantially less than the 7.86% volume reported in Fernandes et al. (2012) and the 10% mass volume reported in Randall and Farrell (1978). This is due to the fact that the gas bladder in Specimen One were almost completely deflated at the time of advanced imaging (Table 1) and perhaps some variation in measuring techniques. Our measured value of 1.39% volume for the respiratory parenchyma (excluding air within it) is similar to the 2.14% volume reported in Fernandes et al. (2012), lending credence to the under-inflation hypothesis. Even this may be an overestimation as the respiratory tissue could have been congested due to the fibrosing cardiomyopathy. In Fernandes et al. (2012), volume was calculated using stereological point and intersection counting methods using a microscope and computer software in a juvenile arapaima. In Randall and Farrell (1978), gas bladder volume was measured as the volume of water required to fill it, which may have involved abnormal stretching of the organ. The gas bladder parenchyma has a surface to volume ratio of 221 cm–1, which is 5–33 times greater than the ratio in other air-breathing fish (Fernandes et al., 2012). For instance, the reported volume of the fully inflated gas bladder in a live O. mykiss is 2.65% of total volume (Martinez et al., 2014). The swim bladder of O. mykiss is a physostomus membranous sac with a rete mirabile and is used mainly for buoyancy rather than respiration (Hall, 1924). Fully inflated, the gas bladder of the arapaima is likely to appropriate a significant amount of space in the coelomic cavity, reflective of its change in function.
While the respiratory gas bladder of arapaima is highly complex, it is considered primitive because it lacks esophageal specialization and maintains a very short pneumatic duct, much like a bowfin (Amia calva), though the highly compartmentalized nature is more reminiscent of a lungfish (Figure 1; Greenwood and Liem, 1984; Liem, 1989). Although the muscular sphincter controlling the pneumatic duct is not very impressive in juvenile arapaima, we found it to be substantial in size in a mature adult (Figure 3; Greenwood and Liem, 1984; Liem, 1989). Anatomical features of the slit-like aperture in the sphincter that in vivo could act as a valve to control the movement of air or limit water movement through the sphincter were seen in the adult specimen (Figures 3a–f).
Renal System
Since the gills play a smaller role in homeostasis, air-breathing requires that the kidneys play a larger role in nitrogenous waste excretion, acid-base and ion regulation (Hochachka et al., 1978b; Brauner et al., 2004). Despite this, the gills are still the main site for urea and nitrogenous waste excretion (Gonzalez et al., 2010), and CO 2. Hochachka et al. (1978b) reported the arapaima kidney is 3.5 times larger in relation to body mass than the closely related water-breathing aruana based on personal observation. We found the kidney to be 0.35% of total body volume, which is only slightly larger than the 0.26% of body volume seen in O. mykiss, a water-breathing fish (Table 1; Martinez et al., 2014). It is unclear if the aruana has unusually small kidneys in comparison to other teleosts, such as O. mykiss. Arapaima kidneys have no differentiation into head and trunk segments; nephrons are distributed throughout the parenchyma, along with hematopoietic cells, intrarenal chromaffin cells, melanomacrophage centers, lymphatic cells and corpuscles of Stannius (Figure 5b; Hochachka et al., 1978b). The nephron tubule has the same three segments as other fish, but the third segment is much longer and more developed (Hochachka et al., 1978b). The neck has tall cuboidal cells with large basal nuclei and the transition segment has an apical tubular system and a less-developed brush border (Hochachka et al., 1978b). The first segment and second segment have a complex basilar membrane network with oblong mitochondria, like in the water-breathing aruana (Hochachka et al., 1978b). The third segment is the longest in the arapaima and is made up of columnar cells with large basal nuclei and small oblong mitochondria, with an appearance very similar to the ionocytes prevalent in the epithelium of the gills (Hochachka et al., 1978b).
Reproductive System
Fry production is the major restriction holding back arapaima aquaculture as reproduction is not as stable in captivity as in the wild and our understanding of arapaima reproduction is limited (Queiroz, 2000). Arapaima only have one developed ovary, the left, and they ovulate directly into the coelomic cavity as there is no oviduct (Godhino et al., 2005; Figures 5a,c,d). Eggs exit the coelomic cavity via a genital papilla that is approximately 6 mm in diameter in the caudal trunk (Godhino et al., 2005). There is considerable diversity in size at first maturity and nesting features for arapaima. This suggests that there may be multiple evolutionarily significant units or species (Godhino et al., 2005). The reported length at sexual maturity for females ranges from 137 to 207 cm depending on geographical site (Gurdak et al., 2019). At this length the females are approximately 5 years old, (Godhino et al., 2005), so our Specimen One was a mature adult. Arapaima are iteroparous but may not spawn every year. In years when they spawn, they often have multiple batches which reduce the risk of clutch failure (Queiroz, 2000). However as arapaima are partial spawners, they never have an empty ovary (Gurdak et al., 2019). With the start of the rising water season, reproductive activity begins and males and females pair off (Queiroz, 2000). The males build nests which are different in different regions. In the Lower Amazon, 90% of nests were found under woody vegetation (Gurdak et al., 2019). The males are responsible for caring for the young for approximately 4 months after breeding (Queiroz, 2000; Castello et al., 2011). Brood size per mating is currently unclear (Queiroz, 2000).
Circulatory System
In order to facilitate these changes in gas bladder function, the vasculature pattern adapted in a variety of ways, most noticeably in regard to the gas bladder, the kidney, and the gills. The afferent blood supply to the gas bladder is via the dorsal aorta to the renal portal vein (Greenwood and Liem, 1984). The efferent blood vessels from the respiratory gas bladder are well developed, especially the left posterior cardinal veins which drain into the ductus Cuvier (Figure 6d; Hulbert and Moon, 1978; Greenwood and Liem, 1984; Graham, 1997). Pulmonary veins that return oxygenated blood directly to the sinus venosus are lacking.
All fish hearts are comprised of two compartments, the atrium and the ventricle (Figure 6; Icardo et al., 2005; Grimes and Kirby, 2009; Ishimatsu, 2012). Blood enters the heart via the sinus venosus, a thin walled compartment with minimal myocardium (Icardo, 2006). As in most air-breathing fish, there is mixing of deoxygenated systemic blood and oxygenated blood from the gas bladder as both the hepatic vein and the ductus Cuvier empty into the heart at the sinus venosus, eroding the efficiency of gas exchange (Farrell, 1978; Greenwood and Liem, 1984; Burggren and Johansen, 1986; Graham, 1997). Some highly specialized air-breathing fish have specific cardiac modifications to decrease this mixing, such as the partial interventricular septum seen in lungfish (Fishman et al., 1985; Icardo et al., 2005; Grimes and Kirby, 2009). Mixing is limited in the arapaima because most blood returning from the posterior systemic circulation comes from the renal portal system and goes through the respiratory gas bladder before emptying into the heart (Johansen et al., 1978; Greenwood and Liem, 1984). The oxygen partial pressure in the dorsal aorta of arapaima is 45–50 mmHg, which is an oxygen saturation of 80–90% (Johansen et al., 1978). Schaller and Dorn (1973) made note of a balloon valve at the entry point of the hepatic vein into the sinus venosus, which could limit mixing of saturated and unsaturated blood. We did not find evidence of this valve.
Arapaima have a type 1 heart, as outlined in Grimes and Kirby (2009), with a completely trabeculated ventricle and no coronary vasculature (Figure 6b). Like most fish with an entirely trabeculated ventricle, arapaima have a distinct muscular conus arteriosus which supports the conus valve (Figures 6a,b,d; Icardo, 2006). In some air-breathing fish, the conus is well developed (e.g., gars (Lepisosteriformes), bichers and lungfish), while others [e.g., snakehead (Channa argus) and climbing perch (Anabas testudineus)] have a regressed conus arteriosus (Ishimatsu and Itazawa, 1983; Grimes and Kirby, 2009). However, prior to Icardo (2006), the outflow tract of fish was poorly defined so it is possible that the conus arteriosus in the snakehead and climbing perch was simply identified using a different paradigm. The thick fibroelastic wall of the bulbus arteriosus allows it to stretch and dampen systolic blood pressure, functioning as a “windkessel” vessel as in other species (Figures 6a–d; Grimes and Kirby, 2009). The bulbus arteriosus has elaborate tissue folds filling the lumen (Figure 6b). Specialization of the bulbus arteriosus and ventral aorta have been seen in other air-breathing fish and can allow for more control of blood flow. Lungfish have spiral and ventrolateral folds inside the bulbus arteriosus that aid in directing blood flow either systemically or to the gills (Parsons, 1929; Fishman et al., 1985; Graham, 1997; Ishimatsu, 2012). The air-breathing snakehead, has smooth muscle ridges in the bulbus arteriosus and a bifurcated ventral aorta, one half of which directs blood to the gills and air-breathing organ and the other systemically (Ishimatsu et al., 1979; Ishimatsu and Itazawa, 1983; Grimes and Kirby, 2009). The climbing perch, African knifefish (Gymnarchus niloticus), swamp eel (Monopterus cuchia), and Indian catfish (Heteropneustes fossilis) are also all air-breathing fish with ridges in the bulbus arteriosus (Parsons, 1929; Munshi et al., 1986; Olson et al., 1994; Grimes and Kirby, 2009). At this point, it is unclear if the luminal folds in the bulbus arteriosus of the arapaima play a role in maximizing oxygen delivery efficacy, although studies of the oxygen and carbon dioxide partial pressures in the dorsal aorta showed no evidence of separation of blood flow based on oxygenation (Johansen et al., 1978; Randall and Farrell, 1978).
The heart of the arapaima is larger than that of the closely-related aruana, and is surrounded by a thick layer of lipid stores (Hochachka et al., 1978a). In the aruana, the whole heart makes up 0.025% of body weight, while in the arapaima the ventricle alone accounts for 0.1% of body weight (Hochachka et al., 1978a). We found the volume of the heart to be 0.21% of total volume (Table 1). The presence of lipid-rich connective tissue around the heart could act as a supplementary energy source that the majority of fish would be prohibited from using due to hypoxic conditions (Hochachka et al., 1978a).
The microvascularization of the gills also reflects changes associated with air-breathing. The branchial arteries have thick layers of smooth muscle, and each branch of the arterioles serves a larger area and some of the vessels are deeper within the tissue, reducing the amount of gas exchange (Hulbert and Moon, 1978). Together these changes indicate that there is a cardiac bypass system that allows the blood to skip the lamellae, controlling the amount of cardiac output devoted to the gills and decreasing oxygen loss to hypoxic water (Hulbert and Moon, 1978). Similar changes to the gill vascularization can be seen in the air-breathing snakehead and swamp eel (Monopterus cuchia) (Olson et al., 1986).
Gastrointestinal System
Arapaima are an omnivorous species that mainly subsists on detritivorous and omnivorous species, such as catfishes and knifefishes, in the high-water season and macroinvertebrates in times of lower water (Watson et al., 2013). Grossly, the gastrointestinal tract can be divided in to four sections: the head gut is made up of the mouth and pharynx, the foregut is made up of the esophagus and stomach, the midgut is the longest section of the intestines, and the hindgut is the most terminal section including the rectum (Khojasteh, 2012).
The parasphenoid, vomer, premaxilla, maxilla, and dentary bones all bear teeth (Ridewood, 1905; Kershaw, 1976). The boney tongue, from which Osteoglossids derive their name, is made up of fine villiform lingual teeth from the medial hyobranchial bones to basibranchial toothplate (Figure 6; Ridewood, 1905; Stewart, 2013b; Watson et al., 2013). The boney tongue allows arapaima to crush the boney armor that protects catfish from most predators, enabling them to exploit an abundant food source (Watson et al., 2013).
The esophagus connects dorsally to the respiratory air bladder via a muscular sphincter. The stomach is where chemical digestion begins (Khojasteh, 2012). The intestine is 145% the total length of the fish, which is longer than would be expected for a carnivore (20% total length) and shorter than seen in herbivores (up to 2000% total length) (Khojasteh, 2012; Watson et al., 2013). In arapaima, the intestines loop back and forth within the coelom, in contrast to the carnivorous O. mykiss, where the intestine is a short straight tube (Khojasteh, 2012). Arapaima have two pyloric ceca, like most osteoglossid fish (Khojasteh, 2012; Watson et al., 2013). The gut of the osteoglossomorpha is distinct from that of other fishes in that the intestine passes posteriorly to the left of the esophagus and stomach (Nelson, 1972) rather than to the right (Figures 7G,H).
The volume of the liver of the arapaima was 0.57% of total body volume, compared to 0.43% in O. mykiss (Table 1). It is difficult to draw conclusions based on liver size in fish, as there is substantial interspecies variability as well as intraspecies variability based on sex, age, season and body condition (Datta-Mushi and Dutta, 1996).
Nervous System
There are relatively few studies looking at teleost brains compared to other clades and using MR is uncommon (Ullmann et al., 2010). As a result of their taxonomic breadth and the variety of ecological niches they fill, teleost brains are hugely morphologically variable to reflect the unique needs of each species (Huber et al., 1997; Salva et al., 2014). Most teleosts follow the standard configuration, where the spinal cord attaches to the brainstem, mesencephalon and diencephalon, with the mesencephalon made up primarily of the cerebellum and optic lobes (Figure 8). The telencephalon is made up of the cerebral hemispheres with the olfactory bulbs attached rostrally (Kotrschal et al., 1998). The braincase is often much larger than the brain, so brain size is not restricted (Kotrschal et al., 1998). The excess space is often filled with lymphatic fatty tissue (Kotrschal et al., 1998). In terms of brain-to-body weight, arapaima and other osteoglossids have a high degree of encephalization compared to many fish, comparable to that of a tuna or a jackfish (Bauchot et al., 1994). We found the volume of the brain to be 0.01% of total body volume (Table 1). Bauchot et al. (1994) reported that osteoglossids have very small olfactory bulbs compared to other freshwater fish making them microsmatic, but we found the olfactory bulbs to be twice the reported values for O. mykiss and 2.5 times the value reported for arapaima (Bauchot et al., 1994; Martinez et al., 2014). This is interesting because Bauchot et al. (1994) used juvenile arapaima in their study, so this difference may reflect an increase in the importance of olfaction as arapaima mature to become more migratory predators. In addition as seen in Bauchot et al. (1994), the size of the brain does not increase proportionately with the muscle volume of the fish as they age, resulting in younger fish brains constituting a larger percentage of total body volume. The cerebral hemispheres are large in the arapaima and may have some olfactory function as well as in coordinating motor centers (Bauchot et al., 1994). The telencephalon, is large, making up 28.3% (Table 3) of brain volume in our specimen, compared to other teleost species that have a mean telencephalon volume of 15.5% total brain volume. It is unclear why arapaima have such well-developed telencephalons, but it may be related to their migratory nature, high degree of parental care and piscivorous hunting. From our imaging data, the demarcation between optic tectum and the tectum mesencephalicum was unclear. In total they measured 27.89% of total volume with is similar to values (28.78%) for arapaima (Bauchot et al., 1994). The valvula cerebelli, a component of the tectum mesencephalicum along with the optic tectum, has gustatory and motor activity control functions (Bauchot et al., 1994). Like many basal fish, the arapaima has a relatively small cerebellum comprising 16.37% of brain volume in our specimen (Table 3). This relates to their low activity lifestyles in shallow water, combined with bursts of movement for hunting (Bauchot et al., 1994; Kotrschal et al., 1998).
Endocrine System
In our specimen, the pituitary was 1.77% of the volume of the brain (Table 3). The pituitary of arapaima is highly vascular, with the neurohypophysis and pars distalis being very well connected by what could be considered a hypophyseal-portal system (Borella et al., 2009). The neurohypophysis is dorsal to the pars distalis (Borella et al., 2009). The pars distalis contains ACTH, gonadotropin, prolactin, and growth hormone secreting cells, while the pars intermedia contains cells that can produce melanotropin, adrenocorticotropin and somatolactin (Borella et al., 2009).
Musculoskeletal System and Integument
When preparing an arapaima for human consumption, the scales, skin and fins are removed (Queiroz, 2000). Then a longitudinal cut is made along the spine (Queiroz, 2000). The ribs and remaining fin rays can then be loosened from the flesh leaving a triangular boneless piece of meat called the “manta” (Queiroz, 2000). The middle of this slab, the “ventrecha,” is a highly prized delicacy (Queiroz, 2000). As total muscle makes up near to 82% of the volume of the arapaima, fishing for arapaima is a lucrative pursuit (Table 1).
As with most teleosts, arapaima swim via undulation, which is achieved through conical formation of the myomeres (Müller and Van Leeuwen, 2006). Myomeres contract in waves causing the body to bend (Wardle et al., 1995). This form of swimming allows the entire body to be used to create thrust (Müller and Van Leeuwen, 2006).
The muscle tissue of the arapaima is uniquely modified to have increased aerobic and decreased anaerobic capabilities compared to water-breathing fish (Hochachka and Guppy, 1978). Like the closely-related jeju (Hopletrythrinus unitaeniatus), arapaima have mostly white muscle fibers but, unlike the jeju, arapaima have retained some red muscle fibers in small dorsal and lateral superficial bands which have a higher metabolism and require more oxygen, which the arapaima provides through air-breathing (Hochachka and Guppy, 1978). Red muscle is used to power continuous swimming, while white muscle is for burst swimming (Müller and Van Leeuwen, 2006). The white muscle in the arapaima has a low metabolism, which allows the arapaima to maintain a low overall metabolism (Hochachka and Guppy, 1978). This drive to conserve oxygen between breaths is evidenced by their sluggish movements between breaths, interspersed with bursts of predatory action (Almeida-Val and Hochachka, 1995). Their decreased reliance on anaerobic metabolism creates less lactic acid build up, allowing them to make each breath last longer underwater in a manner similar to diving mammals (Almeida-Val and Hochachka, 1995).
Arapaima have elongated skulls that retain many of the basal characteristics of actinopterygians but are still highly modified for their environment (Hilton et al., 2007; Figures 9A–C). The head is very flat dorsally with the eyes positioned in line with the top of the top of the head, which aids in breathing while minimally emerging from the surface of the water. The nasal bones are large and, like the frontal, parietal, pterotic and preopercular bones, have shallow depressions which surround the pores of the acousticolateralis system (Kershaw, 1976). These depressions are most evident in the preopercular bone, where there are three distinct elliptical hollows (Kershaw, 1976). The dermethmoid bone is rhomboidal, similar to other osteoglossids, but larger (Kershaw, 1976). The occipito-vertebral region of the arapaima is unique to the genus and can be used to identify specimens from the fossil record (Hilton et al., 2007). The parapophyses of the first vertebral centrum are enlarged and fused to the centrum, where they extend to the parasphenoid (Hilton et al., 2007). The bones of the cranium, as well as the pectoral girdle, play a role in creating the bucco-pharyngeal suction that allows arapaima to catch prey that are often faster and more agile (Kershaw, 1976).
Arapaima also have a flexible dermal armor of elasmoid scales made from overlapping layers of type 1 collagen and a highly mineralized hydroxyapatite outer layer (Yang et al., 2014). As in the heavily scaled gar, the flexural stiffness of the scales in the arapaima likely play an important role in the mechanics of undulatory swimming (Long and Nipper, 1996). Some of these scales are modified with an opening in the posterior third to accommodate the lateral line system (Stewart, 2013b). Scales near the caudal portion of the cranium at the nape insert into the parietal bone (Figure 9; Stewart, 2013b).
Summary
There are some acknowledged limitations of the current study. As adult arapaima are rare, only one specimen was available for CT and MR analysis. Although the primary specimen displayed some pathological changes such as pericardial, gas bladder and abdominal effusion and fibrosing cardiomyopathy, these changes did not drastically alter the morphometric analysis. Ideally, future quantitative research will be done on arapaima of varying sizes, sex, age, and species in order to gain more representative data.
Morphometric analysis from CT and MR data is subject to error associated with post-mortem processes and segmentation. Post-mortem changes were minimized by imaging within 24 h of death and by imaging fresh tissues (Berquist et al., 2012). Errors in the manual segmentation of imaging data into structures was minimized through reference to necropsy photographs and various arapaima specific literature.
As in many species, the major threats facing wild arapaima today are overfishing, climate change, and environmental degradation (Castello and Stewart, 2010; Freitas et al., 2013). Wild arapaima are especially prone to overfishing due to both their desirability as prey and their life history. Arapaima are easy to hunt with harpoons and gillnets as they periodically rise to the surface, and they are viewed as trophy species due to their extreme size and the fact they produce high quality meat (Castello and Stewart, 2010; Castello et al., 2011). In some areas of the Amazon basin, arapaima make up 47% of income from fishing and 70% of arapaima caught are below the minimum allowed size (Queiroz, 2000). Their populations are slow to recover from overfishing due to their size, high parental input to young, small clutch sizes, and high age at reproductive maturity (Castello et al., 2011). Despite some protection under ICUN and CITES, there is little enforcement of size and season regulations, or moratoriums (Castello and Stewart, 2010). Fortunately arapaima are increasingly be raised in aquaculture as both commercially and also as part of conservation (Schaefer et al., 2012).
Climate change is an increasing concern in the Amazon basin, where there is the highest fish diversity in the world, and it is expected to cause a 7–12% loss of fish species over the next 50 years (Freitas et al., 2013). The arapaima life cycle is heavily influenced by the annual hydrologic cycle of flooding and receding waters when they migrate between lakes, rivers and flooded forests (Queiroz, 2000; Castello, 2008). During times of low water, arapaima can become stranded in smaller lakes with deteriorating water quality where they are easy prey for fishermen (Queiroz, 2000). During the extreme drought brought about by a warming Atlantic in 2005, the population of arapaima’s closest relation in the Amazon, the aruana (Osteoglossum bicirrhosum), was decreased by over 50% with only a partial recovery by 2007 (Freitas et al., 2013). Arapaima are likely to be more heavily impacted by climate change as they are carnivores rather than planktivores or detritovores, but their annual migration between the lakes and rivers may be protective (Freitas et al., 2013). Although the impact of climate change is dire, direct human impact from activities such as agriculture, mining, urbanization, pollution, and hydroelectric dams are expected to have larger consequences in the Amazonian wetland ecosystem in the coming years (Junk, 2013).
Data Availability Statement
The raw data supporting the conclusions of this article will be made available by the authors, without undue reservation, to any qualified researcher.
Ethics Statement
All specimens were owned by SeaWorld and provided by Dr. St. Leger who was the director of research at SeaWorld San Diego.
Author Contributions
JS and MS conceived the project. MS acquired the MRI data. MS, CM, WH, DS, DD, and HB contributed to image data segmentations and designed and produced the figures. CM, MS, and JS, drafted the manuscript. All authors contributed to the final manuscript.
Funding
CM was supported by the Pitts Aquatic Veterinary Education Awards Program from the World Aquatic Veterinary Medical Association.
Conflict of Interest
MS and DD are partners and work at the same institutions. JS worked at SeaWorld when project was started.
The remaining authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Acknowledgments
All specimens were owned by SeaWorld and provided by JS who was the director of research at SeaWorld San Diego. We sincerely thank Erika Nilson for her help with the gross necropsy.
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Keywords: air-breathing, arapaima, morphology, imaging, pirarucu, osteoglossid, MRI
Citation: Scadeng M, McKenzie C, He W, Bartsch H, Dubowitz DJ, Stec D and St. Leger J (2020) Morphology of the Amazonian Teleost Genus Arapaima Using Advanced 3D Imaging. Front. Physiol. 11:260. doi: 10.3389/fphys.2020.00260
Received: 01 October 2019; Accepted: 06 March 2020;
Published: 27 April 2020.
Edited by:
James Todd Pearson, National Cerebral and Cardiovascular Center, Japan
Reviewed by:
Wallice Paxiuba Duncan, Federal University of Amazonas, Brazil
Steven F. Perry, University of Bonn, Germany
Copyright © 2020 Scadeng, McKenzie, He, Bartsch, Dubowitz, Stec and St. Leger. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Miriam Scadeng, m.scadeng@auckland.ac.nz; Judy St. Leger, judy.st.leger@gmail.com
†present address: Weston He, Indiana University School of Medicine, Evansville, IN, United States
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
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James Todd Pearson National Cerebral and Cardiovascular Center (Japan), Japan
Reviewed by
Wallice Paxiúba Duncan Federal University of Amazonas, Brazil Steven F. Perry University of Bonn, Germany
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7555 | https://support.khanacademy.org/hc/en-us/community/posts/360006621371-Calculus-Finding-the-Area-of-a-Surface-of-Revolution | Calculus : Finding the Area of a Surface of Revolution? – Khan Academy Help Center
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Calculus : Finding the Area of a Surface of Revolution?
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Renwick Preston
7 years ago
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I am reviewing calculus because I took it many years ago. Anyway, I am fairly sure that we calculated the area of a surface of revolution a lot more simply than I am finding from watching youtube videos and reading current textbooks.
Having said that, let me give an example. Given the function 1/x revolved around the x axis from 1 to infinity, why can't you just take the integral of the function 1/x (but mutliply the whole thing by 2pi because the circumference of the circle is 2pir)?
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Evan Lewis
7 years ago
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Just to be clear, you are meaning the surface area of the revolved region, not the volume?
I'm not very familiar with finding surface areas of revolved regions, I've mainly only dealt with volumes of solids, but I think I can offer some information.
I don't think this problem is quite as simple as you are remembering. You suggested that the surface area would be the integral of 1/x times 2pi. Try to think about this graphically: as we integrate with respect to x (assuming that's how you're going to integrate this function) the radii of the cross-sections of the solid are not all the same (2pi). The radii of the cross-sections are getting smaller as we move from left to right.
Renwick Preston
7 years ago
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Here is how I look at it. The changing circumference is the fact that 1/x is always decreasing from 1 to infinity. If I take an infinitesimal slice at any point it will be a tiny band.
The reason I remember it so vividly is that when the teacher wrote the problem on the board, I was immediately taken aback that the area was infinity because LN(infinity) is infinity.
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Chapter 23 Solutions 23.1 (a) N = 10.0 grams 107.87 grams mol 6.02 × 1023 atoms mol 47.0 electrons atom = 2.62 × 1024 (b) # electrons added = Q e = 1.00 × 10−3 C 1.60 × 10-19 C electron = 6.25 × 1015 or 2.38 electrons for every 109 already present 23.2 (a) Fe = ke q1q2 r2 = 8.99 × 109 N ⋅m2/C2 ( ) 1.60 × 10−19 C ( ) 2 (3.80 × 10−10 m)2 = 1.59 × 10−9 N (repulsion) (b) Fg = Gm1m2 r2 = 6.67 × 10−11 N ⋅m2 kg2 ( )(1.67 × 10−27 kg)2 (3.80 × 10−10 m)2 = 1.29 × 10−45 N The electric force is larger by 1.24 × 1036 times (c) If ke q1q2 r2 = G m1m2 r2 with q1 = q2 = q and m1 = m2 = m, then q m = G ke = 6.67 × 10−11 N ⋅m2 /kg2 8.99 × 109 N ⋅m2 /C2 = 8.61 × 10−11 C/kg 23.3 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains N ≈ 70,000 grams 18 grams mol 6.02 × 1023 molecules mol 10 protons molecule ≈2.3 × 1028 protons With an excess of 1% electrons over protons, each person has a charge q = (0.01)(1.6 × 10−19 C)(2.3 × 1028) = 3.7 × 107 C So F = ke q1q2 r2 = (9 × 109)(3.7 × 107)2 0.62 N = 4 × 1025 N ~ 1026 N This force is almost enough to lift a "weight" equal to that of the Earth: Mg = (6 × 1024 kg)(9.8 m s2) = 6 × 1025 N~ 1026 N 2 Chapter 23 Solutions 23.4 We find the equal-magnitude charges on both spheres: F = ke q1q2 r 2 = ke q2 r 2 so q = r F ke = 1.00 m ( ) 1.00 × 104 N 8.99 × 109 N ⋅m2/C2 = 1.05 × 10−3 C The number of electron transferred is then Nxfer = 1.05 × 10−3 C ( ) 1.60 × 10−19 C/e− ( ) = 6.59 × 1015 electrons The whole number of electrons in each sphere is Ntot = 10.0 g 107.87 g /mol 6.02 × 1023 atoms/mol ( ) 47 e−/atom ( ) = 2.62 × 1024 e− The fraction transferred is then f = Nxfer Ntot = 6.59 × 1015 2.62 × 1024 = 2.51 × 10–9 = 2.51 charges in every billion 23.5 F = ke q1q2 r2 = 8.99 × 109 N ⋅m2 C2 ( ) 1.60 × 10−19 C ( ) 2 6.02 × 1023 ( ) 2 2(6.37 × 106 m) [ ] 2 = 514 kN 23.6 (a) The force is one of attraction. The distance r in Coulomb's law is the distance between centers. The magnitude of the force is F = keq1q2 r2 = 8.99 × 109 N ⋅m2 C2 12.0 × 10−9 C ( ) 18.0 × 10−9 C ( ) (0.300 m)2 = 2.16 × 10−5 N (b) The net charge of − × − 6 00 10 9 .
C will be equally split between the two spheres, or −3.00 × 10−9 C on each. The force is one of repulsion, and its magnitude is F = keq1q2 r2 = 8.99 × 109 N ⋅m2 C2 3.00 × 10−9 C ( ) 3.00 × 10−9 C ( ) (0.300 m)2 = 8.99 × 10−7 N Chapter 23 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
23.7 F1 = ke q1q2 r2 = (8.99 × 109 N ⋅m2/C2)(7.00 × 10−6 C)(2.00 × 10−6 C) (0.500 m)2 = 0.503 N F2 = ke q1q2 r2 = (8.99 × 109 N ⋅m2 /C2)(7.00 × 10−6 C)(4.00 × 10−6 C) (0.500 m)2 = 1.01 N Fx = (0.503 + 1.01) cos 60.0°= 0.755 N Fy = (0.503 −1.01) sin 60.0°= −0.436 N F = (0.755 N)i −(0.436 N)j = 0.872 N at an angle of 330° Goal Solution Three point charges are located at the corners of an equilateral triangle as shown in Figure P23.7.
Calculate the net electric force on the 7.00−µC charge.
G : Gather Information: The 7.00−µC charge experiences a repulsive force F1 due to the 2.00−µC charge, and an attractive force F2 due to the −4.00−µC charge, where F2 = 2F1. If we sketch these force vectors, we find that the resultant appears to be about the same magnitude as F2 and is directed to the right about 30.0° below the horizontal.
O : Organize : We can find the net electric force by adding the two separate forces acting on the 7.00−µC charge. These individual forces can be found by applying Coulomb’s law to each pair of charges.
A : Analyze: The force on the 7.00−µC charge by the 2.00−µC charge is F1 = 8.99 × 109 N ⋅m2/C2 ( ) 7.00 × 10−6 C ( ) 2.00 × 10−6 C ( ) 0.500 m ( )2 cos60°i + sin60°j ( ) = F1 = 0.252i + 0.436j ( ) N Similarly, the force on the 7.00−µC by the −4.00−µC charge is F2 = −8.99 × 109 N ⋅m2 C2 7.00 × 10−6 C ( ) −4.00 × 10−6 C ( ) 0.500 m ( )2 cos60°i −sin60°j ( ) = 0.503i −0.872j ( ) N Thus, the total force on the 7.00−µC, expressed as a set of components, is F = F1 + F2 = 0.755 i −0.436 j ( ) N = 0.872 N at 30.0° below the +x axis L : Learn: Our calculated answer agrees with our initial estimate. An equivalent approach to this problem would be to find the net electric field due to the two lower charges and apply F=qE to find the force on the upper charge in this electric field.
4 Chapter 23 Solutions 23.8 Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by F = ke 3q ( )Q x2 i + ke q ( )Q d −x ( )2 −i ( ) The net force will be zero if 3 x2 = 1 d −x ( )2 , or d −x = x 3 This gives an equilibrium position of the third bead of x = 0.634d The equilibrium is stable if the third bead has positive charge .
23.9 (a) F = kee2 r 2 = (8.99 × 109 N ⋅ m2/C 2) (1.60 × 10–19 C)2 (0.529 × 10–10 m)2 = 8.22 × 10–8 N (b) We have F = mv2 r from which v = Fr m = 8.22 × 10−8 N ( ) 0.529 × 10−10 m ( ) 9.11× 10−31 kg = 2.19 × 106 m/s 23.10 The top charge exerts a force on the negative charge keqQ d 2 ( ) 2 + x2 which is directed upward and to the left, at an angle of tan−1 d/ 2x ( ) to the x-axis. The two positive charges together exert force 2keqQ d2 4 + x2 ( ) (−x)i d2 4 + x2 ( ) 1/2 = ma or for x << d 2, a ≈−2keqQ md 3 /8 x (a) The acceleration is equal to a negative constant times the excursion from equilibrium, as in a = −ω 2x, so we have Simple Harmonic Motion with ω 2 = 16keqQ md 3 . T = 2π ω = π 2 md3 keqQ , where m is the mass of the object with charge −Q.
(b) vmax = ωA = 4a keqQ md 3 Chapter 23 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
23.11 For equilibrium, Fe = −Fg, or qE = −mg −j ( ). Thus, E = mg q j.
(a) E = mg q j = (9.11× 10−31 kg)(9.80 m s2) −1.60 × 10−19 C ( ) j = −5.58 × 10−11 N C ( )j (b) E = mg q j = 1.67 × 10−27 kg ( ) 9.80 m s2 ( ) 1.60 × 10−19 C ( ) j = 1.02 × 10−7 N C ( )j 23.12 Fy = 0: ∑ QEj + mg(−j) = 0 ∴ m = QE g = (24.0 × 10-6 C)(610 N/C) 9.80 m /s2 = 1.49 grams 23.13 The point is designated in the sketch. The magnitudes of the electric fields, E1, (due to the –2.50 × 10–6 C charge) and E2 (due to the 6.00 × 10–6 C charge) are E1 = keq r 2 = (8.99 × 109 N · m2/C 2)(2.50 × 10–6 C) d2 (1) E2 = keq r 2 = (8.99 × 109 N · m2/C 2)(6.00 × 10–6 C) (d + 1.00 m)2 (2) Equate the right sides of (1) and (2) to get (d + 1.00 m)2 = 2.40d 2 or d + 1.00 m = ±1.55d which yields d = 1.82 m or d = – 0.392 m The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus, d = 1.82 m to the left of the -2.50 µC charge. 23.14 If we treat the concentrations as point charges, E+ = ke q r2 = 8.99 × 109 N ⋅m2 C2 40.0 C ( ) 1000 m ( ) 2 −j ( ) = 3.60 × 105 N/C −j ( ) (downward) E−= ke q r2 = 8.99 × 109 N ⋅m2 C2 40.0 C ( ) 1000 m ( ) 2 −j ( ) = 3.60 × 105 N/C −j ( ) (downward) E = E+ + E−= 7.20 × 105 N/C downward 6 Chapter 23 Solutions 23.15 (a) E1 = keq r2 = 8.99 × 109 ( ) 7.00 × 10−6 ( ) 0.500 ( )2 = 2.52 × 105 N C E2 = keq r2 = 8.99 × 109 ( ) 4.00 × 10−6 ( ) 0.500 ( )2 = 1.44 × 105 N C Ex = E2 −E1 cos 60°= 1.44 × 105 −2.52 × 105 cos 60.0°= 18.0 × 103 N C Ey = −E1 sin 60.0°= −2.52 × 105 sin 60.0°= −218 × 103 N C E = [18.0i −218j] × 103 N C = [18.0i −218j] kN C (b) F = qE = 2.00 × 10−6 C ( ) 18.0i −218j ( ) × 103 N C = 36.0i −436j ( ) × 10−3 N = 36.0i −436j ( )mN 23.16 (a) E1 = ke q1 r1 2 −j ( ) = 8.99 × 109 ( ) 3.00 × 10−9 ( ) 0.100 ( )2 −j ( ) = −2.70 × 103 N C ( )j E2 = ke q2 r2 2 −i ( ) = 8.99 × 109 ( ) 6.00 × 10−9 ( ) 0.300 ( )2 −i ( ) = −5.99 × 102 N C ( )i E = E2 + E1 = −5.99 × 102 N C ( )i −2.70 × 103 N C ( )j (b) F = qE = 5.00 × 10−9 C ( ) −599i −2700j ( )N C F = −3.00 × 10−6 i −13.5 × 10−6 j ( )N = − − ( ) 3 00 13 5 .
.
i j µN 23.17 (a) The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. (b) You may need to review vector addition in Chapter Three.
The magnitude of the field at point P due to each of the charges along the base of the triangle is E = ke q a2 . The direction of the field in each case is along the line joining the charge in question to point P as shown in the diagram at the right. The x components add to zero, leaving E = keq a2 sin60.0° ( )j + keq a2 sin60.0° ( )j = 3 keq a2 j Chapter 23 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Three equal positive charges q are at the corners of an equilateral triangle of side a, as shown in Figure P23.17. (a) Assume that the three charges together create an electric field. Find the location of a point (other than ∞) where the electric field is zero. (Hint: Sketch the field lines in the plane of the charges.) (b) What are the magnitude and direction of the electric field at P due to the two charges at the base?
G : The electric field has the general appearance shown by the black arrows in the figure to the right.
This drawing indicates that E = 0 at the center of the triangle, since a small positive charge placed at the center of this triangle will be pushed away from each corner equally strongly. This fact could be verified by vector addition as in part (b) below.
The electric field at point P should be directed upwards and about twice the magnitude of the electric field due to just one of the lower charges as shown in Figure P23.17. For part (b), we must ignore the effect of the charge at point P, because a charge cannot exert a force on itself.
O : The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E1 + E2 A : (b) The electric field from a point charge is As shown in the solution figure above, E1 = ke q a2 to the right and upward at 60° E2 = ke q a2 to the left and upward at 60° E = E1 + E2 = ke q a2 cos60°i + sin60°j ( ) + −cos60°i + sin60°j ( ) [ ] = ke q a2 2 sin60°j ( ) [ ] = 1.73ke q a2 j L : The net electric field at point P is indeed nearly twice the magnitude due to a single charge and is entirely vertical as expected from the symmetry of the configuration. In addition to the center of the triangle, the electric field lines in the figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically. 23.18 (a) E = ke q r2 = (8.99 × 109)(2.00 × 10−6) (1.12)2 = 14,400 N C Ex = 0 and Ey = 2(14,400) sin 26.6°= 1.29 × 104 N C so E = = 1.29 × 104 j N C (b) F = Eq = (1.29 × 104 j)(−3.00 × 10−6) = −3.86 × 10−2 j N 8 Chapter 23 Solutions 23.19 (a) E = ke q1 r1 2 ~ 1 + ke q2 r 2 2 ~ 2 + ke q3 r3 2 ~ 3 = ke 2q ( ) a2 i + ke 3q ( ) 2a2 (i cos 45.0° + j sin 45.0°) + ke 4q ( ) a2 j E = 3.06 keq a2 i + 5.06 keq a2 j = 5.91 keq a2 at 58.8° (b) F E = = q 5.91 keq2 a2 at 58.8° 23.20 The magnitude of the field at (x,y) due to charge q at (x0,y0) is given by E = keq r2 where r is the distance from (x0,y0) to (x,y). Observe the geometry in the diagram at the right.
From triangle ABC, r2 = (x −x0)2 + (y −y0)2, or r = (x −x0)2 +(y −y0)2 , sinθ = (y −y0) r , and cosθ = (x −x0) r Thus, Ex = Ecosθ = ke q r2 (x −x0) r = ke q(x −x0) [(x −x0)2 +(y −y0)2]3/2 and Ey = Esinθ = ke q r2 (y −y0) r = ke q(y −y0) [(x −x0)2 +(y −y0)2]3/2 23.21 The electric field at any point x is E = keq (x – a)2 – keq (x – (–a))2 = keq(4ax) (x2 – a2)2 When x is much, much greater than a, we find E ≈ (4a)(keq) x 3 23.22 (a) One of the charges creates at P a field E = ke Q/n R2 + x2 at an angle θ to the x-axis as shown.
When all the charges produce field, for n > 1, the components perpendicular to the x-axis add to zero.
The total field is nke (Q/n)i R2 + x2 cos θ = keQxi (R2 + x2)3/2 (b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field. .
Chapter 23 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
23.23 E = keq r2 ~ ∑ = keq a2 (−i)+ keq (2a)2 (−i) + keq (3a)2 (−i) + . . . = −keqi a2 1 + 1 22 + 1 33 + . . .
= −π 2keq 6a2 i 23.24 E = keλl d l+ d ( ) = ke Q/l ( )l d l+ d ( ) = keQ d l+ d ( ) = (8.99 × 109)(22.0 × 10–6) (0.290)(0.140 + 0.290) E = 1.59 × 106 N/C , directed toward the rod .
23.25 E = ∫ ke dq x 2 where dq = λ0 dx E = ke λ0 ⌡ ⌠ x0 ∞ dx x 2 = ke – 1 x ∞ x0 = keλ0 x0 The direction is –i or left for λ0 > 0 23.26 E = dE = keλ0x0 dx −i ( ) x3 x0 ∞ ∫ ∫ = −keλ0x0 i x−3 dx x0 ∞ ∫ = −keλ0x0 i −1 2x2 x0 ∞ = keλ0 2x0 −i ( ) 23.27 E = kexQ (x2 + a2)3/2 = (8.99 × 109)(75.0 × 10–6)x (x 2 + 0.1002)3/2 = 6.74 × 105 x (x 2 + 0.0100)3/2 (a) At x = 0.0100 m, E = 6.64 × 106 i N/C = 6.64 i MN/C (b) At x = 0.0500 m, E = 2.41 × 107 i N/C = 24.1 i MN/C (c) At x = 0.300 m, E = 6.40 × 106 i N/C = 6.40 i MN/C (d) At x = 1.00 m, E = 6.64 × 105 i N/C = 0.664 i MN/C 10 Chapter 23 Solutions 23.28 E = ke Qx (x 2 + a2)3/2 For a maximum, dE dx = Qke 1 (x2 + a2)3/2 − 3x2 (x2 + a2)5/2 = 0 x2 + a2 −3x2 = 0 or x = a 2 Substituting into the expression for E gives E = ke Qa 2( 3 2 a2)3/2 = ke Q 3 3 2 a2 = 2ke Q 3 3 a2 = Q 6 3π e0a2 23.29 E = 2π ke σ 1− x x2 + R2 E = 2π 8.99 × 109 ( ) 7.90 × 10−3 ( ) 1− x x2 + 0.350 ( )2 = 4.46 × 108 1− x x2 + 0.123 (a) At x = 0.0500 m, E = 3.83 × 108 N C = 383 MN C (b) At x = 0.100 m, E = 3.24 × 108 N C = 324 MN C (c) At x = 0.500 m, E = 8.07 × 107 N C = 80.7 MN C (d) At x = 2.00 m, E = 6.68 × 108 N C = 6.68 MN C 23.30 (a) From Example 23.9: E = 2π ke σ 1− x x2 + R2 σ = Q πR2 = 1.84 × 10−3 C m2 E = (1.04 × 108 N C)(0.900) = 9.36 × 107 N C = 93.6 MN/C appx: E = 2πke σ = 104 MN/C (about 11% high) (b) E = (1.04 × 108 N/C) 1− 30.0 cm 30.02 + 3.002 cm = (1.04 × 108 N C)(0.00496) = 0.516 MN/C appx: E = ke Q r2 = (8.99 × 109) 5.20 × 10−6 (0.30)2 = 0.519 MN/C (about 0.6% high) Chapter 23 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
23.31 The electric field at a distance x is Ex = 2πkeσ 1− x x2 + R 2 This is equivalent to Ex = 2πkeσ 1− 1 1+ R 2 x2 For large x, R 2 x2 << 1 and 1+ R 2 x2 ≈1+ R 2 2x2 so Ex = 2πkeσ 1− 1 1+ R2 (2x2) [ ] = 2πkeσ 1+ R 2 (2x2) −1 ( ) 1+ R 2 (2x2) [ ] Substitute σ = Q/π R2, Ex = keQ 1 x 2 ( ) 1+ R 2 (2x 2) [ ] = keQ x 2 + R 2 2 But for x > > R, 1 x2 + R2 2 ≈ 1 x2 , so Ex ≈keQ x2 for a disk at large distances 23.32 The sheet must have negative charge to repel the negative charge on the Styrofoam. The magnitude of the upward electric force must equal the magnitude of the downward gravitational force for the Styrofoam to "float" (i.e., Fe = Fg).
Thus, −qE = mg, or −q σ 2e0 = mg which gives σ = −2e0mg q 23.33 Due to symmetry Ey = ∫ dEy = 0, and Ex = ∫ dE sin θ = ke ∫ dq sin θ r 2 where dq = λ ds = λr dθ, so that, Ex = keλ r sinθ dθ 0 π ∫ = keλ r (−cosθ) 0 π = 2keλ r where λ = q L and r = L π . Thus, Ex = 2ke qπ L2 = 2(8.99 × 109 N · m2/C 2)(7.50 × 10–6 C)π (0.140 m)2 Solving, E = Ex = 2.16 × 107 N/C Since the rod has a negative charge, E = (–2.16 × 107 i) N/C = –21.6 i MN/C 12 Chapter 23 Solutions 23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has charge Qdx/h and produces, at the chosen point, a field dE = ke x (x2 + R 2)3/2 Qdx h i The total field is E = dE all charge ∫ = keQxdx h(x2 + R 2)3/2 i d d + h ∫ = keQi 2h (x2 + R 2)−3/2 2xdx x = d d + h ∫ E = keQi 2h (x2 + R 2)−1/2 (−1/ 2) x = d d + h = keQi h 1 (d2 + R 2)1/2 − 1 (d + h)2 + R 2 ( ) 1/2 (b) Think of the cylinder as a stack of disks, each with thickness dx, charge Q dx/h, and charge-per-area σ = Qdx / πR 2h. One disk produces a field dE = 2π keQdx πR2h 1 − x (x2 + R2)1/2 i So, E = dE all charge ∫ = x = d d + h ∫ 2keQdx R2 h 1 − x (x2 + R2)1/2 i E = 2keQi R 2h dx d d + h ∫ −1 2 (x2 + R 2)−1/2 2xdx x = d d + h ∫ = 2keQi R2h x d d+ h −1 2 (x2 + R2)1/2 1/ 2 d d + h E = 2keQi R 2h d + h −d −(d + h)2 + R 2 ( ) 1/2 + (d2 + R 2)1/2 E= 2keQi R 2h h + (d2 + R 2)1/2 −(d + h)2 + R 2 ( ) 1/2 23.35 (a) The electric field at point P due to each element of length dx, is dE = kedq (x2 + y2) and is directed along the line joining the element of length to point P. By symmetry, Ex = dEx = 0 ∫ and since dq = λ dx, E = Ey = dEy = dE cos θ ∫ ∫ where cos θ = y (x2 + y2)1 2 Therefore, dx (x2 + y2)3 2 = 2keλ sinθ0 y (b) For a bar of infinite length, θ → 90° and Ey = 2keλ y Chapter 23 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
23.36 (a) The whole surface area of the cylinder is A = 2π r2 + 2π rL = 2π r r + L ( ). Q = σA = 15.0 × 10−9 C m2 ( )2π 0.0250 m ( ) 0.0250 m + 0.0600 m [ ] = 2.00 × 10−10 C (b) For the curved lateral surface only, A = 2πrL. Q = σA = 15.0 × 10−9 C m2 ( )2π 0.0250 m ( ) 0.0600 m ( ) = 1.41× 10−10 C (c) Q = ρV = ρ π r2 L = 500 × 10−9 C m3 ( )π 0.0250 m ( )2 0.0600 m ( ) = 5.89 × 10−11 C 23.37 (a) Every object has the same volume, V = 8 0.0300 m ( )3 = 2.16 × 10−4 m3.
For each, Q = ρV = 400 × 10−9 C m3 ( ) 2.16 × 10−4 m3 ( ) = 8.64 × 10−11 C (b) We must count the 9.00 cm2 squares painted with charge: (i) 6 × 4 = 24 squares Q = σA = 15.0 × 10−9 C m2 ( )24.0 9.00 × 10−4 m2 ( ) = 3.24 × 10−10 C (ii) 34 squares exposed Q = σA = 15.0 × 10−9 C m2 ( )34.0 9.00 × 10−4 m2 ( ) = 4.59 × 10−10 C (iii) 34 squares Q = σA = 15.0 × 10−9 C m2 ( )34.0 9.00 × 10−4 m2 ( ) = 4.59 × 10−10 C (iv) 32 squares Q = σA = 15.0 × 10−9 C m2 ( )32.0 9.00 × 10−4 m2 ( ) = 4.32 × 10−10 C (c) (i) total edge length: = 24 × 0.0300 m ( ) Q = λ = 80.0 × 10−12 C m ( )24 × 0.0300 m ( ) = 5.76 × 10−11 C (ii) Q = λ = 80.0 × 10−12 C m ( )44 × 0.0300 m ( ) = 1.06 × 10−10 C (iii) Q = λ = 80.0 × 10−12 C m ( )64 × 0.0300 m ( ) = 1.54 × 10−10 C 14 Chapter 23 Solutions (iv) Q = λ = 80.0 × 10−12 C m ( )40 × 0.0300 m ( ) = 0.960 × 10−10 C Chapter 23 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
22.38 22.39 23.40 (a) q1 q2 = −6 18 = −1 3 (b) q1 is negative, q2 is positive 23.41 F = qE = ma a = qE m v = vi + at v = qEt m electron: ve = (1.602 × 10−19)(520)(48.0 × 10−9) 9.11× 10−31 = 4.39 × 106 m/s in a direction opposite to the field proton: vp = (1.602 × 10−19)(520)(48.0 × 10−9) 1.67 × 10−27 = 2.39 × 103 m/s in the same direction as the field 23.42 (a) a = qE m = (1.602 × 10−19)(6.00 × 105) (1.67 × 10−27) = 5.76 × 1013 m s so a = −5.76 × 1013 i m s2 (b) v = vi + 2a(x −xi) 0 = vi 2 + 2(−5.76 × 1013)(0.0700) vi = 2.84 × 106 i m s (c) v = vi + at 0 = 2.84 × 106 + (−5.76 × 1013)t t = 4.93 × 10−8 s 16 Chapter 23 Solutions 23.43 (a) a = qE m = 1.602 × 10−19 ( ) 640 ( ) 1.67 × 10−27 ( ) = 6.14 × 1010 m/s2 (b) v = vi + at 1.20 × 106 = (6.14 × 1010)t t = 1.95 × 10-5 s (c) x −xi = 1 2 vi + v ( )t x = 1 2 1.20 × 106 ( ) 1.95 × 10−5 ( ) = 11.7 m (d) K = 1 2 mv 2 = 1 2 (1.67 × 10−27 kg)(1.20 × 106 m /s)2 = 1.20 × 10-15 J 23.44 The required electric field will be in the direction of motion . We know that Work = ∆K So, –Fd = – 1 2 mv 2 i (since the final velocity = 0) This becomes Eed = 1 2 mvi 2 or E = 1 2 mv 2 i e d E = 1.60 × 10–17 J (1.60 × 10–19 C)(0.100 m) = 1.00 × 103 N/C (in direction of electron's motion) 23.45 The required electric field will be in the direction of motion .
Work done = ∆K so, –Fd = – 1 2 mv 2 i (since the final velocity = 0) which becomes eEd = K and E = K e d Chapter 23 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution The electrons in a particle beam each have a kinetic energy K . What are the magnitude and direction of the electric field that stops these electrons in a distance of d?
G : We should expect that a larger electric field would be required to stop electrons with greater kinetic energy. Likewise, E must be greater for a shorter stopping distance, d. The electric field should be in the same direction as the motion of the negatively charged electrons in order to exert an opposing force that will slow them down.
O : The electrons will experience an electrostatic force F = qE. Therefore, the work done by the electric field can be equated with the initial kinetic energy since energy should be conserved.
A : The work done on the charge is W = F ⋅d = qE⋅d and Ki +W = Kf = 0 Assuming v is in the + x direction, K + −e ( )E⋅di = 0 eE⋅di ( ) = K E is therefore in the direction of the electron beam: E = K ed i L : As expected, the electric field is proportional to K , and inversely proportional to d. The direction of the electric field is important; if it were otherwise the electron would speed up instead of slowing down! If the particles were protons instead of electrons, the electric field would need to be directed opposite to v in order for the particles to slow down.
23.46 The acceleration is given by v2 = v2 i + 2a(x – xi)or v2 = 0 + 2a(–h) Solving, a = – v2 2h Now ∑ F = ma: –mgj + qE = – mv2 j 2h Therefore qE = – mv 2 2h + mg j (a) Gravity alone would give the bead downward impact velocity 2 9.80 m /s2 ( ) 5.00 m ( ) = 9.90 m /s To change this to 21.0 m/s down, a downward electric field must exert a downward electric force.
(b) q = m E v 2 2h – g = 1.00 × 10–3 kg 1.00 × 104 N/C N · s2 kg · m (21.0 m/s)2 2(5.00 m) – 9.80 m/s2 = 3.43 µC 18 Chapter 23 Solutions 23.47 (a) t = x v = 0.0500 4.50 × 105 = 1.11 × 10-7 s = 111 ns (b) ay = qE m = (1.602 × 10−19)(9.60 × 103) (1.67 × 10−27) = 9.21× 1011 m /s2 y −yi = vyit + 1 2 ayt2 y = 1 2 (9.21× 1011)(1.11× 10−7)2 = 5.67 × 10-3 m = 5.67 mm (c) vx = 4.50 × 105 m/s vy = vy i + ay = (9.21 × 1011)(1.11 × 10-7) = 1.02 × 105 m/s 23.48 ay = qE m = (1.602 × 10−19)(390) (9.11× 10−31) = 6.86 × 1013 m /s2 (a) t = 2vi sinθ ay from projectile motion equations t = 2(8.20 × 105)sin 30.0° 6.86 × 1013 = 1.20 × 10-8 s) = 12.0 ns (b) h = vi 2 sin2 θ 2ay = (8.20 × 105)2 sin2 30.0° 2(6.86 × 1013) = 1.23 mm (c) R = vi 2 sin2θ 2ay = (8.20 × 105)2 sin 60.0° 2(6.86 × 1013) = 4.24 mm 23.49 vi = 9.55 × 103 m/s (a) ay = eE m = (1.60 × 10−19)(720) (1.67 × 10−27) = 6.90 × 1010 m s2 R = vi 2 sin2θ ay = 1.27 × 10-3 m so that (9.55 × 103)2sin2θ 6.90 × 1010 = 1.27 × 10−3 sin 2θ = 0.961 θ = 36.9° 90.0° – θ = 53.1° (b) t = R vix = R vi cosθ If θ = 36.9°, t = 167 ns If θ = 53.1°, t = 221 ns Chapter 23 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
23.50 (a) The field, E1, due to the 4.00 × 10–9 C charge is in the –x direction.
E1 = ke q r 2 = (8.99 × 109 N · m2/C 2)(− 4.00 × 10–9 C) (2.50 m)2 i = −5.75i N/C Likewise, E2 and E3, due to the 5.00 × 10–9 C charge and the 3.00 × 10–9 C charge are E2 = ke q r 2 = (8.99 × 109 N · m2/C 2)(5.00 × 10–9 C) (2.00 m)2 i = 11.2 N/C E3 = (8.99 × 109 N · m2/C 2)(3.00 × 10–9 C) (1.20 m)2 i = 18.7 N/C ER = E1 + E2 + E3 = 24.2 N/C in +x direction.
(b) E1 = ke q r 2 = −8.46 N/C ( ) 0.243i + 0.970j ( ) E2 = ke q r 2 = 11.2 N/C ( ) +j ( ) E3 = ke q r 2 = 5.81 N/C ( ) −0.371i + 0.928j ( ) Ex = E1x + E3x = – 4.21i N/C Ey = E1y + E2y + E3y = 8.43j N/C ER = 9.42 N/C θ = 63.4° above –x axis 23.51 The proton moves with acceleration ap = qE m = 1.60 × 10−19 C ( ) 640 N/C ( ) 1.67 × 10−27 kg = 6.13 × 1010 m s2 while the e− has acceleration ae = 1.60 × 10−19 C ( ) 640 N/C ( ) 9.11× 10−31 kg = 1.12 × 1014 m s2 = 1836ap (a) We want to find the distance traveled by the proton (i.e., d = 1 2 apt2), knowing: 4.00 cm = 1 2 apt2 + 1 2 aet2 = 1837 1 2 apt2 ( ) Thus, d = 1 2 apt2 = 4.00 cm 1837 = 21.8 µm 20 Chapter 23 Solutions (b) The distance from the positive plate to where the meeting occurs equals the distance the sodium ion travels (i.e., dNa = 1 2 aNat2). This is found from: 4.00 cm = 1 2 aNat2 + 1 2 aClt2: 4.00 cm = 1 2 eE 22.99 u t2 + 1 2 eE 35.45 u t2 This may be written as 4.00 cm = 1 2 aNat2 + 1 2 0.649aNa ( )t2 = 1.65 1 2 aNat2 ( ) so dNa = 1 2 aNat2 = 4.00 cm 1.65 = 2.43 cm 23.52 From the free-body diagram shown, ∑Fy = 0 and T cos 15.0° = 1.96 × 10–2 N So T = 2.03 × 10–2 N From ∑Fx = 0, we have qE = T sin 15.0° or q = T sin 15.0° E = (2.03 × 10–2 N) sin 15.0° 1.00 × 103 N/C = 5.25 × 10–6 C = 5.25 µC 23.53 (a) Let us sum force components to find ∑Fx = qEx – T sin θ = 0, and ∑Fy = qEy + T cos θ – mg = 0 Combining these two equations, we get q = mg (Ex cot θ + Ey) = (1.00 × 10-3)(9.80) (3.00 cot 37.0° + 5.00) × 105 = 1.09 × 10–8 C = 10.9 nC (b) From the two equations for ∑Fx and ∑Fy we also find T = qEx sin 37.0° = 5.44 × 10–3 N = 5.44 mN Free Body Diagram for Goal Solution Chapter 23 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A charged cork ball of mass 1.00 g is suspended on a light string in the presence of a uniform electric field, as shown in Fig. P23.53. When E = 3.00i + 5.00j ( ) × 105 N/C, the ball is in equilibrium at θ = 37.0°. Find (a) the charge on the ball and (b) the tension in the string.
G : (a) Since the electric force must be in the same direction as E, the ball must be positively charged. If we examine the free body diagram that shows the three forces acting on the ball, the sum of which must be zero, we can see that the tension is about half the magnitude of the weight.
O : The tension can be found from applying Newton's second law to this statics problem (electrostatics, in this case!). Since the force vectors are in two dimensions, we must apply ΣF = ma to both the x and y directions.
A : Applying Newton's Second Law in the x and y directions, and noting that ΣF = T + qE + Fg = 0, ΣFx = qEx −T sin 37.0°= 0 (1) ΣFy = qEy + T cos 37.0° −mg = 0 (2) We are given Ex = 3.00 × 105 N/C and Ey = 5.00 × 105 N/C; substituting T from (1) into (2): q = mg Ey + Ex tan 37.0° = (1.00 × 10−3 kg)(9.80 m /s2) 5.00 + 3.00 tan 37.0° × 105 N/C = 1.09 × 10−8 C (b) Using this result for q in Equation (1), we find that the tension is T = qEx sin 37.0° = 5.44 × 10−3 N L : The tension is slightly more than half the weight of the ball ( Fg = 9.80 × 10−3 N) so our result seems reasonable based on our initial prediction.
23.54 (a) Applying the first condition of equilibrium to the ball gives: ΣFx = qEx −T sinθ = 0 or T = qEx sinθ = qA sinθ and ΣFy = qEy + T cosθ −mg = 0 or qB + T cosθ = mg Substituting from the first equation into the second gives: q Acotθ + B ( ) = mg , or q = mg Acotθ + B ( ) (b) Substituting the charge into the equation obtained from ΣFx yields T = mg Acotθ + B ( ) A sinθ = mgA Acosθ + Bsinθ 22 Chapter 23 Solutions Goal Solution A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field, as shown in Figure P23.53. When E = Ai + Bj ( ) N/C, where A and B are positive numbers, the ball is in equilibrium at the angle θ . Find (a) the charge on the ball and (b) the tension in the string. G : This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ .
The unknowns are q and T. O : The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 53.
A : Again, Newton's second law: −T sin θ + qA = 0 (1) and + T cosθ + qB −mg = 0 (2) (a) Substituting T = qA sinθ , into Eq. (2), qAcosθ sinθ + qB = mg Isolating q on the left, q = mg Acot θ + B ( ) (b) Substituting this value into Eq. (1), T = mgA Acos θ + Bsinθ ( ) L : If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 53. If you find this problem more difficult than problem 53, the little list at the Gather step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the Analysis step, and for recognizing when we have an answer.
23.55 F = ke q1q2 r 2 tan θ = 15.0 60.0 θ = 14.0° F1 = (8.99 × 109)(10.0 × 10–6)2 (0.150)2 = 40.0 N F3 = (8.99 × 109)(10.0 × 10–6)2 (0.600)2 = 2.50 N F2 = (8.99 × 109)(10.0 × 10–6)2 (0.619)2 = 2.35 N Fx = –F3 – F2 cos 14.0° = –2.50 – 2.35 cos 14.0° = – 4.78 N Fy = –F1 – F2 sin 14.0° = – 40.0 – 2.35 sin 14.0° = – 40.6 N Fnet = F 2 x + F 2 y = (– 4.78)2 + (– 40.6)2 = 40.9 N tan φ = Fy Fx = – 40.6 – 4.78 φ = 263° Chapter 23 Solutions 23 © 2000 by Harcourt, Inc. All rights reserved.
23.56 From Fig. A: dcos .
.
30 0 15 0 ° = cm, or d = ° 15 0 30 0 .
cos . cm From Fig. B: θ = sin−1 d 50.0 cm = sin−1 15.0 cm 50.0 cm cos 30.0° ( ) = 20.3° Fq mg = tanθ or Fq = mgtan 20.3° (1) From Fig. C: Fq = 2Fcos 30.0° = 2 keq2 0.300 m ( )2 cos 30.0° (2) Equating equations (1) and (2), 2 keq2 0.300 m ( )2 cos 30.0° = mgtan 20.3° q2 = mg 0.300 m ( )2 tan 20.3° 2ke cos 30.0° q2 = 2.00 × 10−3 kg ( ) 9.80 m s2 ( ) 0.300 m ( )2 tan20.3° 2 8.99 × 109 N ⋅m2 C2 ( )cos 30.0° q = 4.20 × 10−14 C2 = 2.05 × 10−7 C= 0.205 µC Figure A Figure B Figure C 23.57 Charge Q/2 resides on each block, which repel as point charges: F = ke(Q/2)(Q/2) L 2 = k(L – L i) Q = 2L k L −Li ( ) ke = 2 0.400 m ( ) 100 N/m ( ) 0.100 m ( ) 8.99 × 109 N ⋅m2 /C2 ( ) = 26.7 µC 23.58 Charge Q/2 resides on each block, which repel as point charges: F = ke Q 2 ( ) Q 2 ( ) L2 = k L −Li ( ) Solving for Q, Q = 2L k L −Li ( ) ke 24 Chapter 23 Solutions 23.59 According to the result of Example 23.7, the lefthand rod creates this field at a distance d from its righthand end: E = keQ d(2a + d) dF = keQQ 2a dx d(d + 2a) F = keQ 2 2a ∫ b x = b – 2a dx x(x + 2a) = keQ 2 2a – 1 2a ln 2a + x x b b – 2a F = +keQ 2 4a2 – ln 2a + b b + ln b b – 2a = keQ 2 4a2 ln b2 (b – 2a)(b + 2a) = keQ 2 4a2 ln b2 b2 – 4a2 23.60 The charge moves with acceleration of magnitude a given by ∑F = ma = q E (a) a = q E m = 1.60 × 10–19 C (1.00 N/C) 9.11 × 10–31 kg = 1.76 × 1011 m/s2 Then v = vi + at = 0 + at gives t = v a = 3.00 × 107 m/s 1.76 × 1011 m/s2 = 171 µs (b) t = v a = vm qE = (3.00 × 107 m/s)(1.67 × 10–27 kg) (1.60 × 10–19 C)(1.00 N/C) = 0.313 s (c) From t = vm qE , as E increases, t gets shorter in inverse proportion.
23.61 Q = ∫ λdl = ∫ 90.0° –90.0° λ 0 cos θ Rdθ = λ 0 R sin θ 90.0° –90.0° = λ 0 R [1 – (–1)] = 2λ 0R Q = 12.0 µC = (2λ 0 )(0.600) m = 12.0 µC so λ 0 = 10.0 µC/m dFy = 1 4π e0 3.00 µC ( ) λdl ( ) R2 cosθ = 1 4π e0 3.00 µC ( ) λ0 cos2 θRdθ ( ) R2 Fy = ∫ 90.0° –90.0° 8.99 × 109 N · m2 C 2 (3.00 × 10–6 C)(10.0 × 10–6 C/m) (0.600 m) cos2 θdθ Fy = 8.99 30.0 ( ) 0.600 10−3 N ( ) 1 2 + 1 2 cos2θ ( )dθ −π/2 π/2 ∫ Fy = 0.450 N ( ) 1 2 π + 1 4 sin2θ −π/2 π/2 ( ) = 0.707 N Downward. Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0.
Chapter 23 Solutions 25 © 2000 by Harcourt, Inc. All rights reserved.
23.62 At equilibrium, the distance between the charges is r = 2 0.100 m ( )sin10.0°= 3.47 × 10−2 m Now consider the forces on the sphere with charge +q , and use ΣFy = 0: ΣFy = 0: T cos 10.0°= mg, or T = mg cos 10.0° (1) ΣFx = 0: Fnet = F2 −F1 = T sin10.0° (2) Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1). Fnet = mgsin 10.0° cos 10.0° = mgtan 10.0° = 2.00 × 10−3 kg ( ) 9.80 m /s2 ( )tan 10.0°= 3.46 × 10−3 N Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exerted by –q, and F2 is the force exerted on +q by the external electric field.
Fnet = F2 – F1 or F2 = Fnet + F1 F1 = 8.99 × 109 N ⋅m2 /C2 ( ) 5.00 × 10−8 C ( ) 5.00 × 10−8 C ( ) 3.47 × 10−3 m ( ) 2 = 1.87 × 10−2 N Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10−3 N + 1.87 × 10−2 N = 2.21× 10−2 N and F2 = qE, or E = F2 q = 2.21× 10−2 N 5.00 × 10−8 C = 4.43 × 105 N/C = 443 kN/C 23.63 (a) From the 2Q charge we have Fe −T2 sinθ2 = 0 and mg −T2 cosθ2 = 0 Combining these we find Fe mg = T2 sinθ2 T2 cosθ2 = tanθ2 From the Q charge we have Fe −T1sinθ1 = 0 and mg −T1cosθ1 = 0 Combining these we find Fe mg = T1sinθ1 T1cosθ1 = tanθ1 or θ2 = θ1 (b) Fe = ke2QQ r2 = 2keQ2 r2 If we assume θ is small then . Substitute expressions for Fe and tan θ into either equation found in part (a) and solve for r. Fe mg = tanθ then and solving for r we find 26 Chapter 23 Solutions 23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle θ corresponding to equilibrium. In terms of lengths s, 1 2 a 3, and r, shown in Figure P23.64, the charge at the origin exerts an attractive force keQq (s + 1 2 a 3)2 . The other two charges exert equal repulsive forces of magnitude keQq r2. The horizontal components of the two repulsive forces add, balancing the attractive force, Fnet = keQq 2 cos θ r2 − 1 (s + 1 2 a 3)2 = 0 From Figure P23.64, r = 1 2 a sin θ s = 1 2 a cot θ The equilibrium condition, in terms of θ, is Fnet = 4 a2 keQq 2 cos θ sin2θ − 1 ( 3 + cot θ)2 = 0 Thus the equilibrium value of θ is 2 cos θ sin2 θ( 3 + cot θ)2 = 1.
One method for solving for θ is to tabulate the left side. To three significant figures the value of θ corresponding to equilibrium is 81.7°. The distance from the origin to the equilibrium position is x = 1 2 a( 3 + cot 81.7°) = 0.939a θ 2 cosθ sin2θ( 3 + cotθ)2 60° 70° 80° 90° 81° 81.5° 81.7° 4 2.654 1.226 0 1.091 1.024 0.997 23.65 (a) The distance from each corner to the center of the square is L 2 ( ) 2 + L 2 ( ) 2 = L 2 The distance from each positive charge to −Q is then z2 + L2 2 .
Each positive charge exerts a force directed along the line joining q and −Q, of magnitude keQq z2 + L2 2 The line of force makes an angle with the z-axis whose cosine is z z2 + L2 2 The four charges together exert forces whose x and y components add to zero, while the z-components add to F = − 4keQqz z2 + L2 2 ( ) 3 2 k Chapter 23 Solutions 27 © 2000 by Harcourt, Inc. All rights reserved.
(b) For z << L, the magnitude of this force is Fz ≈−4keQqz L2 2 ( ) 3 2 = −4 2 ( )3 2keQq L3 z = maz Therefore, the object’s vertical acceleration is of the form az = −ω 2z with ω 2 = 4 2 ( )3 2keQq mL3 = keQq 128 mL3 Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by T = 2π ω = 2π 128 ( )1 4 mL3 keQq = π 8 ( )1 4 mL3 keQq 23.66 (a) The total non-contact force on the cork ball is: F = qE + mg = m g + qE m , which is constant and directed downward. Therefore, it behaves like a simple pendulum in the presence of a modified uniform gravitational field with a period given by: T = 2π L g + qE m = 2π 0.500 m 9.80 m /s2 + 2.00 × 10−6 C ( ) 1.00 × 105 N/C ( ) 1.00 × 10−3 kg = 0.307 s (b) Yes . Without gravity in part (a), we get T = 2π L qE m T = 2π 0.500 m 2.00 × 10−6 C ( ) 1.00 × 105 N/C ( ) 1.00 × 10−3 kg = 0.314 s (a 2.28% difference).
23.67 (a) Due to symmetry the field contribution from each negative charge is equal and opposite to each other. Therefore, their contribution to the net field is zero. The field contribution of the +q charge is E = keq r2 = keq 3a2 4 ( ) = 4keq 3a2 in the negative y direction, i.e., E = −4ke q 3a2 j x y 28 Chapter 23 Solutions (b) If Fe = 0, then E at P must equal zero. In order for the field to cancel at P, the −4q must be above + q on the y-axis.
Then, E = 0 = − keq 1.00 m ( )2 + ke(4q) y2 , which reduces to y2 = 4.00 m2.
Thus, y = ±2.00 m. Only the positive answer is acceptable since the −4q must be located above + q. Therefore, the −4q must be placed 2.00 meters above point P along the + y −axis .
23.68 The bowl exerts a normal force on each bead, directed along the radius line or at 60.0° above the horizontal. Consider the free-body diagram of the bead on the left: ΣFy = nsin 60.0°−mg = 0, or n = mg sin 60.0° Also, ΣFx = −Fe + ncos 60.0° = 0, or ke q2 R2 = ncos 60.0° = mg tan 60.0° = mg 3 Thus, q = R mg ke 3 1 2 n 60.0˚ mg Fe 23.69 (a) There are 7 terms which contribute: 3 are s away (along sides) 3 are 2 s away (face diagonals) and sin θ = 1 2 = cos θ 1 is 3 s away (body diagonal) and sinφ = 1 3 The component in each direction is the same by symmetry. F = keq2 s2 1+ 2 2 2 + 1 3 3 (i + j + k) = keq2 s2 (1.90)(i + j + k) (b) F = F x 2 + F y 2 + F z 2 = 3.29 ke q2 s2 away from the origin Chapter 23 Solutions 29 © 2000 by Harcourt, Inc. All rights reserved.
23.70 (a) Zero contribution from the same face due to symmetry, opposite face contributes 4 keq r2 sin φ where r = s 2 2 + s 2 2 + s2 = 1.5 s = 1.22 s E = 4 keqs r3 = 4 (1.22)3 keq s2 = 2.18 keq s2 sin φ = s/r (b) The direction is the k direction. 23.71 dE = ke dq x2 + 0.150 m ( )2 −xi + 0.150 mj x2 + 0.150 m ( )2 = keλ −xi + 0.150 mj ( )dx x2 + 0.150 m ( )2 [ ] 3 2 E = dE all charge ∫ = keλ −xi + 0.150 mj ( )dx x2 + 0.150 m ( )2 [ ] 3 2 x=0 0.400 m ∫ x y dq x 0.150 m dE E = keλ +i x2 + 0.150 m ( )2 0 0.400 m + 0.150 m ( )j x 0.150 m ( )2 x2 + 0.150 m ( )2 0 0.400 m E = 8.99 × 109 N ⋅m2 C2 35.0 × 10−9 C m i 2.34 −6.67 ( ) m + j 6.24 −0 ( ) m [ ] E = −1.36i + 1.96j ( ) × 103 N C = −1.36i + 1.96j ( ) kN C 23.72 By symmetry Ex ∑ = 0. Using the distances as labeled, Ey = ke q (a2 + y2) sin θ + q (a2 + y2) sin θ −2q y2 ∑ But sin θ = y (a2+y2) , so E = Ey = 2keq y (a2 + y2)3 2 −1 y2 ∑ Expand (a2 +y2)−3 2 as (a2 + y2)−3 2 = y−3 −(3 2)a2y−5 + . . .
Therefore, for a << y, we can ignore terms in powers higher than 2, and we have E = 2keq 1 y2 −3 2 a2 y4 −1 y2 or E = −ke3qa2 y4 j 30 Chapter 23 Solutions 23.73 The field on the axis of the ring is calculated in Example 23.8, E = Ex = kexQ (x2 + a2)3/2 The force experienced by a charge –q placed along the axis of the ring is F = −keQq x (x2 + a2)3/2 and when x << a, this becomes F = keQq a3 x This expression for the force is in the form of Hooke's law, with an effective spring constant of k = keQq a3 Since ω = 2π f = k m , we have f = 1 2π keQq ma3 23.74 The electrostatic forces exerted on the two charges result in a net torque τ = −2Fa sin θ = −2Eqa sin θ.
For small θ, sin θ ≈ θ and using p = 2qa, we have τ = –Epθ. The torque produces an angular acceleration given by τ = Iα = I d2θ dt2 Combining these two expressions for torque, we have d2θ dt2 + Ep I θ = 0 This equation can be written in the form d2θ dt2 = −ω 2θ where ω 2 = Ep I This is the same form as Equation 13.17 and the frequency of oscillation is found by comparison with Equation 13.19, or f = 1 2π pE I = 1 2π 2qaE I © 2000 by Harcourt, Inc. All rights reserved.
Chapter 24 Solutions 24.1 (a) ΦE = EA cos θ = (3.50 × 103)(0.350 × 0.700) cos 0° = 858 N · m2/C (b) θ = 90.0° ΦE = 0 (c) ΦE = (3.50 × 103)(0.350 × 0.700) cos 40.0° = 657 N · m2/C 24.2 ΦE = EA cos θ = (2.00 × 104 N/C)(18.0 m2)cos 10.0° = 355 kN · m2/C 24.3 ΦE = EA cos θ A = π r 2 = π (0.200)2 = 0.126 m2 5.20 × 105 = E (0.126) cos 0° E = 4.14 × 106 N/C = 4.14 MN/C 24.4 The uniform field enters the shell on one side and exits on the other so the total flux is zero .
24.5 (a) ′ = ( )( ) A 10 0 30 0 .
. cm cm ′ A = 300 cm2 = 0.0300 m2 ΦE, ′ A = E ′ A cosθ ΦE, ′ A = 7.80 × 104 ( ) 0.0300 ( )cos 180° ΦE, ′ A = −2.34 kN ⋅m2 C 0.0 cm 30.0 cm 0.0˚ (b) ΦE, A = EA cosθ = 7.80 × 104 ( ) A ( )cos 60.0° A = 30.0 cm ( ) w ( ) = 30.0 cm ( ) 10.0 cm cos 60.0° = 600 cm2 = 0.0600 m2 ΦE, A = 7.80 × 104 ( ) 0.0600 ( )cos 60°= + 2.34 kN ⋅m2 C (c) The bottom and the two triangular sides all lie parallel to E, so ΦE = 0 for each of these. Thus, Chapter 24 Solutions 33 © 2000 by Harcourt, Inc. All rights reserved. ΦE, total = −2.34 kN ⋅m2 C + 2.34 kN ⋅m2 C + 0 + 0 + 0 = 0 34 Chapter 24 Solutions 24.6 (a) ΦE = E⋅A = (ai + bj)⋅Ai = aA (b) ΦE = (ai + bj) ⋅Aj = bA (c) ΦE = (ai+ bj) ⋅Ak = 0 24.7 Only the charge inside radius R contributes to the total flux. ΦE = q/e0 24.8 ΦE = EA cosθ through the base ΦE = 52.0 ( ) 36.0 ( )cos 180°= –1.87 kN · m2/C Note the same number of electric field lines go through the base as go through the pyramid's surface (not counting the base).
For the slanting surfaces, ΦE = +1.87 kN ⋅m2 /C 24.9 The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side of the cone is ΦE = E⋅dA = ∫ ERh . This is the same as the flux that exits the right side of the cone. Note that for a uniform field only the cross sectional area matters, not shape.
24.10 (a) E = keQ r 2 8.90 × 102 = (8.99 × 109)Q (0.750)2 , But Q is negative since E points inward.
Q = – 5.56 × 10–8 C = – 55.6 nC (b) The negative charge has a spherically symmetric charge distribution.
24.11 (a) ΦE = qin e0 = +5.00 µC −9.00 µC + 27.0 µC −84.0 µC ( ) 8.85 × 10−12 C2 / N ⋅m2 = – 6.89 × 106 N · m2/C = – 6.89 MN · m2/C (b) Since the net electric flux is negative, more lines enter than leave the surface.
Chapter 24 Solutions 35 © 2000 by Harcourt, Inc. All rights reserved.
24.12 ΦE = qin e0 Through S1 ΦE = −2Q + Q e0 = −Q e0 Through S2 ΦE = + Q −Q e0 = 0 Through S3 ΦE = −2Q + Q −Q e0 = −2Q e0 Through S4 ΦE = 0 24.13 (a) One-half of the total flux created by the charge q goes through the plane. Thus, ΦE, plane = 1 2 ΦE, total = 1 2 q e0 = q 2e0 (b) The square looks like an infinite plane to a charge very close to the surface. Hence, ΦE, square ≈ΦE, plane = q 2e0 (c) The plane and the square look the same to the charge. 24.14 The flux through the curved surface is equal to the flux through the flat circle, E0 πr 2 .
24.15 (a) +Q 2 e0 Simply consider half of a closed sphere.
(b) –Q 2 e0 (from ΦΕ, total = ΦΕ, dome + ΦΕ, flat = 0) 36 Chapter 24 Solutions Goal Solution A point charge Q is located just above the center of the flat face of a hemisphere of radius R, as shown in Figure P24.15. What is the electric flux (a) through the curved surface and (b) through the flat face? G : From Gauss’s law, the flux through a sphere with a point charge in it should be Q e0 , so we should expect the electric flux through a hemisphere to be half this value: Φcurved = Q 2e0 . Since the flat section appears like an infinite plane to a point just above its surface so that half of all the field lines from the point charge are intercepted by the flat surface, the flux through this section should also equal Q 2e0 .
O : We can apply the definition of electric flux directly for part (a) and then use Gauss’s law to find the flux for part (b).
A : (a) With δ very small, all points on the hemisphere are nearly at distance R from the charge, so the field everywhere on the curved surface is keQ/ R2 radially outward (normal to the surface).
Therefore, the flux is this field strength times the area of half a sphere: Φcurved = E⋅dA ∫ = ElocalAhemisphere = ke Q R2 1 2 ( ) 4πR2 ( ) = 1 4π e0 Q 2π ( ) = Q 2e0 (b) The closed surface encloses zero charge so Gauss's law gives Φcurved + Φflat = 0 or Φflat = −Φcurved = −Q 2e0 L : The direct calculations of the electric flux agree with our predictions, except for the negative sign in part (b), which comes from the fact that the area unit vector is defined as pointing outward from an enclosed surface, and in this case, the electric field has a component in the opposite direction (down).
24.16 (a) ΦE, shell = qin e0 = 12.0 × 10−6 8.85 × 10−12 = 1.36 × 106 N ⋅m2 /C = 1.36 MN · m2/C (b) ΦE, half shell = 1 2 (1.36 × 106 N ⋅m2 /C) = 6.78 × 105 N ⋅m2 /C = 678 kN · m2/C (c) No, the same number of field lines will pass through each surface, no matter how the radius changes.
24.17 From Gauss's Law, ΦE = E⋅dA ∫ = qin e0 .
Thus, ΦE = Q e0 = 0.0462 × 10−6 C 8.85 × 10-12 C2 N ⋅m2 = 5.22 kN ⋅m2 C Chapter 24 Solutions 37 © 2000 by Harcourt, Inc. All rights reserved.
24.18 If R ≤ d, the sphere encloses no charge and ΦE = qin /e0 = 0 If R > d, the length of line falling within the sphere is 2 R 2 −d2 so ΦΕ = 2λ R2 −d2 e0 24.19 The total charge is Q −6 q . The total outward flux from the cube is Q −6 q ( )/e0, of which one-sixth goes through each face: ΦE ( )one face = Q −6 q 6e0 ΦE ( )one face = Q −6 q 6e0 = (5.00 −6.00) × 10−6 C⋅N ⋅m2 6 × 8.85 × 10−12 C2 = − ⋅ 18.8 kN m /C 2 24.20 The total charge is Q −6 q . The total outward flux from the cube is Q −6 q ( )/e0, of which one-sixth goes through each face: ΦE ( )one face = Q −6 q 6e0 24.21 When R < d, the cylinder contains no charge and ΦΕ = 0 . When R > d, ΦE = qin e0 = λL e0 24.22 ΦE, hole = E⋅Ahole = keQ R2 π r2 ( ) = 8.99 × 109 N ⋅m2 C2 ( ) 10.0 × 10−6 C ( ) 0.100 m ( )2 π 1.00 × 10−3 m ( ) 2 ΦE, hole = 28.2 N ⋅m2 C 38 Chapter 24 Solutions Chapter 24 Solutions 39 © 2000 by Harcourt, Inc. All rights reserved.
24.23 ΦE = qin e0 = 170 × 10−6 C 8.85 × 10-12 C2 N ⋅m2 = 1.92 × 107 N ⋅m2 C (a) ΦE ( )one face = 1 6 ΦE = 1.92 × 107 N ⋅m2 C 6 ΦE ( )one face = 3 20 . MN m C 2 ⋅ (b) ΦE = 19.2 MN ⋅m2 C (c) The answer to (a) would change because the flux through each face of the cube would not be equal with an unsymmetrical charge distribution. The sides of the cube nearer the charge would have more flux and the ones farther away would have less. The answer to (b) would remain the same, since the overall flux would remain the same.
24.24 (a) ΦE = qin e0 8.60 × 104 = qin 8.85 × 10−12 qin = 7.61× 10−7 C = 761 nC (b) Since the net flux is positive, the net charge must be positive . It can have any distribution.
(c) The net charge would have the same magnitude but be negative. 24.25 No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g. These three faces together fill solid angle equal to one-eighth of a sphere as seen from q, and together pass flux 1 8 q e0 ( ). Each face containing a intercepts equal flux going into the cube: 0 = ΦE, net = 3ΦE, abcd + q/8e0 ΦE, abcd = −q/ 24e0 40 Chapter 24 Solutions 24.26 The charge distributed through the nucleus creates a field at the surface equal to that of a point charge at its center: E = keq r2 E = (8.99 × 109 Nm2/C 2)(82 × 1.60 × 10–19 C) [(208)1/3 1.20 × 10–15 m] 2 E = 2.33 × 1021 N/C away from the nucleus 24.27 (a) E = ke Qr a3 = 0 (b) E = ke Qr a3 = (8.99 × 109)(26.0 × 10–6)(0.100) (0.400)3 = 365 kN/C (c) E = ke Q r 2 = (8.99 × 109)(26.0 × 10–6) (0.400)2 = 1.46 MN/C (d) E = ke Q r 2 = (8.99 × 109)(26.0 × 10–6) (0.600)2 = 649 kN/C The direction for each electric field is radially outward.
24.28 (a) E = 2ke λ r 3.60 × 104 = 2(8.99 × 109)(Q/2.40) (0.190) Q = + 9.13 × 10–7 C = +913 nC (b) E = 0 24.29 ∫ ο E · dA = qin e0 = ∫ ρ dV e0 = ρ e0 l π r 2 E2π rl = ρ e0 l π r 2 Chapter 24 Solutions 41 © 2000 by Harcourt, Inc. All rights reserved.
E = ρ 2 e0 r away from the axis Goal Solution Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ . Find the electric field at distance r from the axis where r < R.
G : According to Gauss’s law, only the charge enclosed within the gaussian surface of radius r needs to be considered. The amount of charge within the gaussian surface will certainly increase as ρ and r increase, but the area of this gaussian surface will also increase, so it is difficult to predict which of these two competing factors will more strongly affect the electric field strength. O : We can find the general equation for E from Gauss’s law.
A : If ρ is positive, the field must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is π r2L and it encloses charge ρπ r2L.
The circular end caps have no electric flux through them; there E⋅dA = EdAcos90.0°= 0. The curved surface has E⋅dA = EdAcos0°, and E must be the same strength everywhere over the curved surface.
Gauss’s law, E⋅dA ∫ = q e0 , becomes E dA ∫ Curved Surface = ρπ r2L e0 Now the lateral surface area of the cylinder is 2π rL: E 2π r ( )L = ρπ r2L e0 Thus, E = ρr 2e0 radially away from the cylinder axis L : As we expected, the electric field will increase as ρ increases, and we can now see that E is also proportional to r. For the region outside the cylinder ( r > R), we should expect the electric field to decrease as r increases, just like for a line of charge.
24.30 σ = 8.60 × 10−6 C/cm2 ( ) 100 cm m 2 = 8.60 × 10−2 C/m2 E = σ 2e0 = 8.60 × 10−2 2 8.85 × 10−12 ( ) = 4.86 × 109 N/C The field is essentially uniform as long as the distance from the center of the wall to the field point is much less than the dimensions of the wall.
24.31 (a) E = 0 42 Chapter 24 Solutions (b) E = keQ r2 = (8.99 × 109)(32.0 × 10−6) (0.200)2 = 7.19 MN/C Chapter 24 Solutions 43 © 2000 by Harcourt, Inc. All rights reserved.
24.32 The distance between centers is 2 × 5.90 × 10–15 m. Each produces a field as if it were a point charge at its center, and each feels a force as if all its charge were a point at its center.
F = keq1q2 r 2 = 8.99 × 109 N · m2 C 2 (46)2 (1.60 × 10–19 C)2 (2 × 5.90 × 10–15 m)2 = 3.50 × 103 N = 3.50 kN 24.33 Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a 0.05 m separation on the ends of strings making angles of 10˚ with the vertical.
(a) ΣFy = T cos 10° −mg = 0 ⇒ T = mg cos 10° ΣFx = T sin 10° −Fe = 0 ⇒ Fe = T sin 10°, so Fe = mg cos 10° sin 10° = mgtan 10°= 0.001 kg ( ) 9.8 m s2 ( )tan 10° Fe ≈2 × 10−3 N ~10-3 N or 1 mN (b) Fe = keq2 r2 2 × 10−3 N ≈ 8.99 × 109 N ⋅m2 C2 ( )q2 0.25 m ( )2 q ≈1.2 × 10−7 C ~10−7 C or 100 nC (c) E = keq r2 ≈ 8.99 × 109 N ⋅m2 C2 ( ) 1.2 × 10−7 C ( ) 0.25 m ( )2 ≈1.7 × 104 N C ~10 kN C (d) ΦE = q e0 ≈ 1.2 × 10−7 C 8.85 × 10−12 C2 N ⋅m2 = 1.4 × 104 N ⋅m2 C ~ 10 kN ⋅m2 C 24.34 (a) ρ = Q 4 3 πa3 = 5.70 × 10−6 4 3 π(0.0400)3 = 2.13 × 10−2 C/m3 qin = ρ 4 3 πr3 ( ) = 2.13 × 10−2 ( ) 4 3 π ( ) 0.0200 ( )3 = 7.13 × 10−7 C = 713 nC (b) qin = ρ 4 3 πr3 ( ) = 2.13 × 10−2 ( ) 4 3 π ( ) 0.0400 ( )3 = 5.70 µC 44 Chapter 24 Solutions 24.35 (a) E = 2keλ r = 2 8.99 × 109 N ⋅m2 C2 ( ) 2.00 × 10−6 C ( ) 7.00 m [ ] 0.100 m E = 51.4 kN/C, radially outward (b) ΦE = EAcosθ = E(2πr )cos 0˚ ΦE = 5.14 × 104 N C ( )2π 0.100 m ( ) 0.0200 m ( ) 1.00 ( ) = 646 N ⋅m2 C 24.36 Note that the electric field in each case is directed radially inward, toward the filament.
(a) E = 2keλ r = 2 8.99 × 109 N ⋅m2 C2 ( ) 90.0 × 10−6 C ( ) 0.100 m = 16.2 MN C (b) E = 2keλ r = 2 8.99 × 109 N ⋅m2 C2 ( ) 90.0 × 10−6 C ( ) 0.200 m = 8.09 MN C (c) E = 2keλ r = 2 8.99 × 109 N ⋅m2 C2 ( ) 90.0 × 10−6 C ( ) 1.00 m = 1.62 MN C 24.37 E = σ 2e0 = 9.00 × 10−6 C/m2 2(8.85 × 10−12 C2 / N ⋅m2) = 508 kN/C, upward 24.38 From Gauss's Law, EA = Q e0 σ = Q A = e0E = (8.85 × 10-12)(130)= 1.15 × 10-9 C/m2 = 1.15 nC/m2 24.39 ∫ ο E dA = E(2π rl ) = qin e0 E = qin/l 2π e0r = λ 2π e0r (a) r = 3.00 cm E = 0 inside the conductor (b) r = 10.0 cm E = 30.0 × 10–9 2π (8.85 × 10–12)(0.100) = 5400 N/C, outward (c) r = 100 cm E = 30.0 × 10–9 2π (8.85 × 10–12)(1.00) = 540 N/C, outward Chapter 24 Solutions 45 © 2000 by Harcourt, Inc. All rights reserved.
24.40 Just above the aluminum plate (a conductor), the electric field is E = ′ σ e0 where the charge Q is divided equally between the upper and lower surfaces of the plate: Thus ′ = ( ) = σ Q A Q A 2 2 and E = Q 2e0A For the glass plate (an insulator), E = σ / 2e0 where σ = Q/ A since the entire charge Q is on the upper surface. Therefore, E = Q 2e0A The electric field at a point just above the center of the upper surface is the same for each of the plates. E = Q 2e0A , vertically upward in each case (assuming Q > 0) 24.41 (a) E = σ e0 σ = (8.00 × 104)(8.85 × 10–12) = 7.08 × 10-7 C/m2 σ = 708 nC/m2 , positive on one face and negative on the other.
(b) σ = Q A Q = σA = (7.08 × 10–7) (0.500)2 C Q = 1.77 × 10–7 C = 177 nC , positive on one face and negative on the other.
24.42 Use Gauss's Law to evaluate the electric field in each region, recalling that the electric field is zero everywhere within conducting materials. The results are: E = 0 inside the sphere and inside the shell E = ke Q r2 between sphere and shell, directed radially inward E = ke 2Q r2 outside the shell, directed radially inward Charge –Q is on the outer surface of the sphere .
Charge +Q is on the inner surface of the shell , 46 Chapter 24 Solutions and +2Q is on the outer surface of the shell. Chapter 24 Solutions 47 © 2000 by Harcourt, Inc. All rights reserved.
24.43 The charge divides equally between the identical spheres, with charge Q/2 on each. Then they repel like point charges at their centers: F = ke(Q/2)(Q/2) (L + R + R)2 = ke Q 2 4(L + 2R)2 = 8.99 × 109 N · m2(60.0 × 10-6 C)2 4 C 2(2.01 m)2 = 2.00 N 24.44 The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is most intense where the radius of curvature is smallest and vise-versa. The local charge density and the electric field intensity are related by E = σ e0 or σ = e0E (a) Where the radius of curvature is the greatest, σ =e0Emin = 8.85 × 10−12 C2 N ⋅m2 ( ) 2.80 × 104 N C ( ) = 248 nC m2 (b) Where the radius of curvature is the smallest, σ =e0Emax = 8.85 × 10−12 C2 N ⋅m2 ( ) 5.60 × 104 N C ( ) = 496 nC m2 24.45 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = – λ. 0 = λ + qin ⇒ qin = –λ Outside surface: The total charge on the metal cylinder is 2λl = qin + qout. qout = 2λl+ λl so the outside charge/length = 3λ (b) E = 2ke (3λ) r = 6ke λ r = 3λ 2πe0r 48 Chapter 24 Solutions 24.46 (a) E = keQ r2 = 8.99 × 109 ( ) 6.40 × 10−6 ( ) 0.150 ( )2 = 2.56 MN/C, radially inward (b) E = 0 Chapter 24 Solutions 49 © 2000 by Harcourt, Inc. All rights reserved.
24.47 (a) The charge density on each of the surfaces (upper and lower) of the plate is: σ = 1 2 q A = 1 2 (4.00 × 10−8 C) (0.500 m)2 = 8.00 × 10−8 C/m2 = 80.0 nC/m2 (b) E = σ e0 k = 8.00 × 10−8 C/m2 8.85 × 10−12 C2 / N ⋅m2 k = 9.04 kN/C ( )k (c) E = −9.04 kN/C ( )k 24. 48 (a) The charge +q at the center induces charge −q on the inner surface of the conductor, where its surface density is: σa = −q 4π a2 (b) The outer surface carries charge Q + q with density σb = Q + q 4π b2 24.49 (a) E = 0 (b) E = keQ r2 = 8.99 × 109 ( ) 8.00 × 10−6 ( ) 0.0300 ( )2 = 7.99 × 107 N/C = 79.9 MN/C (c) E = 0 (d) E = keQ r2 = 8.99 × 109 ( ) 4.00 × 10−6 ( ) 0.0700 ( )2 = 7.34 × 106 N/C = 7.34 MN/C 24.50 An approximate sketch is given at the right. Note that the electric field lines should be perpendicular to the conductor both inside and outside.
50 Chapter 24 Solutions 24.51 (a) Uniform E, pointing radially outward, so ΦE = EA. The arc length is ds = Rd θ , and the circumference is 2π r = 2π R sin θ A = 2πrds = (2πRsinθ)Rdθ = 2πR2 sinθdθ 0 θ ∫ 0 θ ∫ ∫ = 2πR2(−cosθ) 0 θ = 2πR2(1−cosθ) ΦE = 1 4πe0 Q R2 ⋅2πR2(1−cosθ) = Q 2e0 (1−cosθ) [independent of R!] (b) For θ = 90.0° (hemisphere): ΦE = Q 2e0 (1−cos 90°) = Q 2e0 (c) For θ = 180° (entire sphere): ΦE = Q 2e0 (1−cos 180°) = Q e0 [Gauss's Law] 24.52 In general, E = ayi + bzj + cxk In the xy plane, z = 0 and E = ayi + cxk ΦE = E⋅dA = ayi + cxk ( ) ∫ ∫ ⋅kdA ΦE = ch xdx x=0 w ∫ = chx2 2 x=0 w = chw2 2 x y z x = 0 x = w y = 0 y = h dA = hdx 24.53 (a) qin = +3Q −Q = +2Q (b) The charge distribution is spherically symmetric and qin > 0. Thus, the field is directed radially outward .
(c) E = keqin r2 = 2keQ r2 for r ≥c (d) Since all points within this region are located inside conducting material, E = 0 for b < r < c.
(e) ΦE = E⋅dA ∫ = 0 ⇒ qin =e0ΦE = 0 (f) qin = +3Q (g) E = keqin r2 = 3keQ r2 (radially outward) for a ≤r < b Chapter 24 Solutions 51 © 2000 by Harcourt, Inc. All rights reserved.
(h) qin = ρV = +3Q 4 3 π a3 4 3 π r3 = +3Q r3 a3 (i) E = keqin r2 = ke r2 +3Q r3 a3 = 3keQ r a3 (radially outward) for 0 ≤r ≤a (j) From part (d), E = 0 for b < r < c. Thus, for a spherical gaussian surface with b < r < c, qin = +3Q + qinner = 0 where qinner is the charge on the inner surface of the conducting shell.
This yields qinner = −3Q (k) Since the total charge on the conducting shell is qnet = qouter + qinner = −Q, we have qouter = −Q −qinner = −Q −−3Q ( ) = +2Q (l) This is shown in the figure to the right.
E r a b c 24.54 The sphere with large charge creates a strong field to polarize the other sphere. That means it pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side. This patch of opposite charge is smaller in amount but located in a stronger external field, so it can feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker field on the other side.
24.55 (a) E⋅dA = E 4π r2 ( ) ∫ = qin e0 For r < a, qin = ρ 4 3 π r3 ( ) so E = pr 3e0 For a < r < b and c < r, qin = Q so that E = Q 4π r2e0 For b ≤ r ≤ c, E = 0, since E = 0 inside a conductor.
(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero. Therefore, q1 + Q = 0 and σ1 = q1 4π b 2 = – Q 4π b 2 Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require q1 + q2 = 0 and σ2 = q1 4π c2 = Q 4π c2 52 Chapter 24 Solutions 24.56 E⋅dA ∫ = E 4π r2 ( ) = qin e0 (a) −3.60 × 103 N C ( )4π 0.100 m ( )2 = Q 8.85 × 10−12 C2 N ⋅m2 ( a < r < b) Q = −4.00 × 10−9 C = −4.00 nC (b) We take ′ Q to be the net charge on the hollow sphere. Outside c, +2.00 × 102 N C ( )4π 0.500 m ( )2 = Q + ′ Q 8.85 × 10−12 C2 N ⋅m2 ( r > c) Q + ′ Q = +5.56 × 10−9 C, so ′ Q = +9.56 × 10−9 C = +9.56 nC (c) For b < r < c: E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface of the hollow sphere. Thus, Q1 = −Q = + 4.00 nC Then, if Q2 is the total charge on the outer surface of the hollow sphere, Q2 = ′ Q −Q1 = 9.56 nC −4.00 nC = +5.56 nC 24.57 The field direction is radially outward perpendicular to the axis. The field strength depends on r but not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius r and length L. If r < a, ΦE = qin e0 and E 2π rL ( ) = λL e0 E = λ 2π re0 or E = λ 2π re0 (r < a) If a < r < b, E 2π rL ( ) = λL + ρπ r2 −a2 ( )L e0 E = λ + ρπ r2 −a2 ( ) 2π re0 a < r < b ( ) If r > b, E 2π rL ( ) = λL + ρπ b2 −a2 ( )L e0 Chapter 24 Solutions 53 © 2000 by Harcourt, Inc. All rights reserved. E = λ + ρπ b2 −a2 ( ) 2π re0 ( r > b) 54 Chapter 24 Solutions 24.58 Consider the field due to a single sheet and let E+ and E– represent the fields due to the positive and negative sheets. The field at any distance from each sheet has a magnitude given by Equation 24.8: E+ = E– = σ 2e0 (a) To the left of the positive sheet, E+ is directed toward the left and E– toward the right and the net field over this region is E = 0 . (b) In the region between the sheets, E+ and E– are both directed toward the right and the net field is E = σ e 0 toward the right (c) To the right of the negative sheet, E+ and E– are again oppositely directed and E = 0 .
24.59 The magnitude of the field due to each sheet given by Equation 24.8 is E = σ 2e0 directed perpendicular to the sheet.
(a) In the region to the left of the pair of sheets, both fields are directed toward the left and the net field is E = σ e0 to the left (b) In the region between the sheets, the fields due to the individual sheets are oppositely directed and the net field is E = 0 (c) In the region to the right of the pair of sheets, both fields are directed toward the right and the net field is E = σ e0 to the right Chapter 24 Solutions 55 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Repeat the calculations for Problem 58 when both sheets have positive uniform charge densities of value σ. Note: The new problem statement would be as follows: Two infinite, nonconducting sheets of charge are parallel to each other, as shown in Figure P24.58. Both sheets have positive uniform charge densities σ. Calculate the value of the electric field at points (a) to the left of, (b) in between, and (c) to the right of the two sheets. G : When both sheets have the same charge density, a positive test charge at a point midway between them will experience the same force in opposite directions from each sheet. Therefore, the electric field here will be zero. (We should ask: can we also conclude that the electron will experience equal and oppositely directed forces everywhere in the region between the plates?) Outside the sheets the electric field will point away and should be twice the strength due to one sheet of charge, so E = σ /e0 in these regions.
O : The principle of superposition can be applied to add the electric field vectors due to each sheet of charge.
A : For each sheet, the electric field at any point is E = σ (2e0) directed away from the sheet.
(a) At a point to the left of the two parallel sheets E i i i = − + − = − E E E 1 2 2 ( ) ( ) ( ) = −σ e0 i (b) At a point between the two sheets E i i = + − = E E 1 2 0 ( ) (c) At a point to the right of the two parallel sheets E i i i = + = E E E 1 2 2 = σ e0 i L : We essentially solved this problem in the Gather information step, so it is no surprise that these results are what we expected. A better check is to confirm that the results are complementary to the case where the plates are oppositely charged (Problem 58).
24.60 The resultant field within the cavity is the superposition of two fields, one E+ due to a uniform sphere of positive charge of radius 2a, and the other E− due to a sphere of negative charge of radius a centered within the cavity. 4 3 π r3ρ e0 = 4π r2E+ so E+ = ρr 3e0 = ρr 3e0 – 4 3 π r1 3ρ e0 = 4π r1 2E− so E−= ρr1 3e0 (− 1 ) = −ρ 3e0 r1 Since r = a + r1, E−= −ρ(r −a) 3e0 E = E+ + E−= ρr 3e0 −ρr 3e0 + ρa 3e0 = ρa 3e0 = 0i + ρa 3e0 j Thus, Ex = 0 and Ey = ρa 3e0 at all points within the cavity.
56 Chapter 24 Solutions 24.61 First, consider the field at distance r < R from the center of a uniform sphere of positive charge Q = +e ( ) with radius R. 4π r2 ( )E = qin e0 = ρV e0 = +e 4 3 π R3 4 3 π r3 e0 so E = e 4π e0R3 r directed outward (a) The force exerted on a point charge q = −e located at distance r from the center is then F = qE = −e e 4π e0R3 r = − e2 4π e0R3 r = −Kr (b) K = e2 4π e0R3 = kee2 R3 (c) Fr = mear = −kee2 R3 r, so ar = − kee2 meR3 r = −ω 2r Thus, the motion is simple harmonic with frequency f = ω 2π = 1 2π kee2 meR3 (d) f = 2.47 × 1015 Hz = 1 2π 8.99 × 109 N ⋅m2 C2 ( ) 1.60 × 10−19 C ( ) 2 9.11× 10−31 kg ( )R3 which yields R3 = 1.05 × 10−30 m3, or R = 1.02 × 10−10 m = 102 pm 24.62 The electric field throughout the region is directed along x; therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore, ΦE = −Ex x=a ( )A + Ex x=a+c ( )A = −3 + 2a2 ( )ab + 3 + 2(a + c)2 ( )ab = 2abc(2a + c) Substituting the given values for a, b, and c, we find ΦE = 0.269 N · m2/C Q = ∈0ΦE = 2.38 × 10-12 C = 2.38 pC 24.63 E⋅dA ∫ = E(4π r2) = qin e0 (a) For r > R, qin = Ar2 0 R ∫ (4πr2)dr = 4π AR5 5 and E = AR5 5e0r2 (b) For r < R, qin = Ar2 0 r ∫ (4πr2)dr = 4πAr5 5 and E = Ar3 5e0 Chapter 24 Solutions 57 © 2000 by Harcourt, Inc. All rights reserved.
24.64 The total flux through a surface enclosing the charge Q is Q/ e0. The flux through the disk is Φdisk = E⋅dA ∫ where the integration covers the area of the disk. We must evaluate this integral and set it equal to 1 4 Q/ e0 to find how b and R are related. In the figure, take dA to be the area of an annular ring of radius s and width ds. The flux through dA is E · dA = E dA cos θ = E (2π sds) cos θ The magnitude of the electric field has the same value at all points within the annular ring, E = 1 4πe0 Q r2 = 1 4πe0 Q s2 + b2 and cosθ = b r = b (s2 + b2)1/2 Integrate from s = 0 to s = R to get the flux through the entire disk. ΦE, disk = Qb 2e0 s ds (s2 + b2)3/2 0 R ∫ = Qb 2e0 −(s2 + b2)1/2 [ ] 0 R = Q 2e0 1− b (R2 + b2)1/2 The flux through the disk equals Q/4 e0 provided that b (R2 + b2)1/2 = 1 2 .
This is satisfied if R = 3 b .
24.65 E⋅dA ∫ = qin e0 = 1 e0 a r 4πr2dr 0 r ∫ E4πr2 = 4πa e0 rdr 0 r ∫ = 4πa e0 r2 2 E = a 2e0 = constant magnitude (The direction is radially outward from center for positive a; radially inward for negative a.) 58 Chapter 24 Solutions 24.66 In this case the charge density is not uniform, and Gauss's law is written as E⋅dA = 1 e0 ρ dV ∫ ∫ . We use a gaussian surface which is a cylinder of radius r, length , and is coaxial with the charge distribution.
(a) When r < R, this becomes E( 2πrl) = ρ0 e0 a −r b 0 r ∫ dV. The element of volume is a cylindrical shell of radius r, length l, and thickness dr so that dV = 2πrl dr.
` E 2π rl ( ) = 2π r2lρ0 e0 a 2 −r 3b so inside the cylinder, E = ρ0r 2e0 a −2r 3b (b) When r > R, Gauss's law becomes E 2π rl ( ) = ρ0 e0 a −r b 0 R ∫ 2π rldr ( ) or outside the cylinder, E = ρ0R2 2e0r a −2R 3b 24.67 (a) Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x : Use Gauss's law: E⋅dA ∫ = qin e0 By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall of the gaussian cylinder.
Therefore, the only contribution to the integral is over the end cap containing the point x : E⋅dA ∫ = qin e0 or EA = ρ Ax ( ) e0 so that at distance x from the mid-line of the slab, E = ρx e0 x y z x gaussian surface (b) a = F me = −e ( )E me = − ρe mee0 x The acceleration of the electron is of the form a = −ω 2x with ω = ρe mee0 Thus, the motion is simple harmonic with frequency f = ω 2π = 1 2π ρe mee0 Chapter 24 Solutions 59 © 2000 by Harcourt, Inc. All rights reserved.
24.68 Consider the gaussian surface described in the solution to problem 67.
(a) For x > d 2 , dq = ρ dV = ρA dx = C Ax2 dx E⋅dA = 1 e0 dq ∫ ∫ EA = CA e0 x2 dx 0 d/2 ∫ = 1 3 CA e0 d3 8 E = Cd3 24e0 or E = Cd3 24e0 i for x d > 2 ; E = −Cd3 24e0 i for x d < −2 (b) For − < < d x d 2 2 E⋅dA = 1 e0 dq ∫ ∫ = C A e0 x2 dx = C Ax3 3e0 0 x ∫ E = Cx3 3e0 i for x > 0; E = −Cx3 3e0 i for x < 0 24.69 (a) A point mass m creates a gravitational acceleration g = −Gm r2 at a distance r.
The flux of this field through a sphere is g ∫ ⋅dA = −Gm r2 4πr2 ( ) = −4πGm Since the r has divided out, we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to g ⋅dA = −4πGmin ∫ (b) Take a spherical gaussian surface of radius r. The field is inward so g ∫ ⋅dA = g 4πr2 cos 180° = −g 4πr 2 and −4πGmin = −4G 4 3 πr3ρ Then, −g 4πr2 = −4πG 4 3 πr3ρ and g = 4 3 πrρG Or, since ρ = ME / 4 3 πRE 3, g = MEGr RE 3 or g = MEGr RE 3 inward © 2000 by Harcourt, Inc. All rights reserved.
Chapter 25 Solutions 25.1 ∆V = –14.0 V and Q = –NA e = – (6.02 × 1023)(1.60 × 10–19 C) = – 9.63 × 104 C ∆V = W Q , so W = Q(∆V) = (– 9.63 × 104 C)(–14.0 J/C) = 1.35 ΜJ 25.2 ∆K = q ∆V 7.37 × 10-17 = q(115) q = 6.41 × 10-19 C 25.3 W = ∆K = q ∆V 1 2 mv2 = e(120 V) = 1.92 × 10–17 J Thus, v = 3.84 × 10−17 J m (a) For a proton, this becomes v = 3.84 × 10−17 J 1.67 × 10−27 kg = 1.52 × 105 m/s = 152 km/s (b) If an electron, v = 3.84 × 10−17 J 9.11× 10−31 kg = 6.49 × 106 m/s = 6.49 Mm/s Goal Solution (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V.
(b) Calculate the speed of an electron that is accelerated through the same potential difference.
G : Since 120 V is only a modest potential difference, we might expect that the final speed of the particles will be substantially less than the speed of light. We should also expect the speed of the electron to be significantly greater than the proton because, with me << mp, an equal force on both particles will result in a much greater acceleration for the electron. O : Conservation of energy can be applied to this problem to find the final speed from the kinetic energy of the particles. (Review this work-energy theory of motion from Chapter 8 if necessary.) Chapter 25 Solutions 57 © 2000 by Harcourt, Inc. All rights reserved.
A : (a) Energy is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V: Ki + Ui + ∆Enc = Kf + Uf 0 + qV + 0 = 1 2 mvp 2 + 0 (1.60 × 10−19 C)(120 V) 1 J 1 V ⋅C = 1 2 (1.67 × 10−27 kg)vp 2 vp = 1.52 × 105 m /s (b) The electron will gain speed in moving the other way, from Vi = 0 to Vf = 120 V: Ki + Ui + ∆Enc = Kf + Uf 0 + 0 + 0 = 1 2 mve 2 + qV 0 = 1 2 (9.11× 10−31 kg)ve 2 + (−1.60 × 10−19 C)(120 J/C) ve = 6.49 × 106 m /s L : Both of these speeds are significantly less than the speed of light as expected, which also means that we were justified in not using the relativistic kinetic energy formula. (For precision to three significant digits, the relativistic formula is only needed if v is greater than about 0.1 c.) 25.4 For speeds larger than one-tenth the speed of light, 1 2 mv2 gives noticeably wrong answers for kinetic energy, so we use K = mc2 1 1−v2 /c2 −1 = 9.11× 10−31 kg ( ) 3.00 × 108 m /s ( ) 2 1 1−0.4002 −1 = 7.47 × 10–15 J Energy is conserved during acceleration: Ki + Ui + ∆E = Kf + Uf 0 + qVi + 0 = 7.47 × 10–15 J + qVf The change in potential is Vf – Vi : Vf – Vi = –7.47 × 10–15 J q = – 7.47 × 10–15 J –1.60 × 10–19 C = + 46.7 kV The positive answer means that the electron speeds up in moving toward higher potential.
25.5 W = ∆K = −q∆V 0 −1 2 9.11× 10−31 kg ( ) 4.20 × 105 m /s ( ) 2 = −−1.60 × 10−19 C ( )∆V From which, ∆V = – 0.502 V 58 Chapter 25 Solutions 25.6 (a) We follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm).
∆U = – (work done) ∆U = −work from origin to (20.0 cm,0) ( ) −work from (20.0 cm,0) to (20.0 cm,50.0 cm) ( ) Note that the last term is equal to 0 because the force is perpendicular to the displacement. ∆U = – (qEx)(∆x) = – (12.0 × 10–6 C)(250 V/m)(0.200 m) = – 6.00 × 10–4 J (b) ∆V = ∆U q = – 6.00 × 10–4 J 12.0 × 10–6 C = – 50.0 J/C = – 50.0 V 25.7 E = ∆V d = 25.0 × 103 J/C 1.50 × 10–2 m = 1.67 × 106 N/C = 1.67 MN/C 25.8 (a) ∆V = Ed = (5.90 × 103 V/m)(0.0100 m) = 59.0 V (b) 1 2 mvf 2 = q(∆V); 1 2 (9.11 × 10–31) vf2 = (1.60 × 10–19)(59.0) vf = 4.55 × 106 m/s 25.9 ∆U = −1 2 m vf 2 −vi 2 ( ) = −1 2 9.11× 10−31 kg 1.40 × 105 m /s ( ) 2 −3.70 × 106 m /s ( ) 2 = 6.23 × 10−18 J ∆U = q ∆V: + 6.23 × 10–18 = (–1.60 × 10–19)∆V ∆V = – 38.9 V The origin is at higher potential.
Chapter 25 Solutions 59 © 2000 by Harcourt, Inc. All rights reserved.
25.10 VB −VA = − A B ∫E⋅ds = − A C ∫E⋅ds − C B ∫E⋅ds VB −VA = (−Ecos180°) −0.300 0.500 ∫ dy −(Ecos90.0°) −0.200 0.400 ∫ dx VB – VA = (325)(0.800) = + 260 V 25.11 (a) Arbitrarily choose V = 0 at x = 0. Then at other points, V = −Ex and Ue = QV = −QEx. Between the endpoints of the motion, (K + Us + Ue)i = (K + Us + Ue)f 0 + 0 + 0 = 0 + 1 2 kxmax 2 −QExmax so the block comes to rest when the spring is stretched by an amount xmax = 2QE k = 2 50.0 × 10−6 C ( ) 5.00 × 105 V m ( ) 100 N m = 0.500 m (b) At equilibrium, ΣFx = −Fs + Fe = 0 or kx = QE. Thus, the equilibrium position is at x = QE k = 50.0 × 10−6 C ( ) 5.00 × 105 N C ( ) 100 N m = 0.250 m (c) The equation of motion for the block is ΣFx = −kx + QE = m d2x dt2 . Let ′ x = x −QE k , or x = ′ x + QE k so the equation of motion becomes: −k ′ x + QE k + QE = m d2 ′ x + QE k ( ) dt2 , or d2 ′ x dt2 = − k m ′ x This is the equation for simple harmonic motion a ′ x = −ω 2 ′ x ( ), with ω = k m. The period of the motion is then T = 2π ω = 2π m k = 2π 4.00 kg 100 N m = 1.26 s (d) (K + Us + Ue)i + ∆E = (K + Us + Ue)f 0 + 0 + 0 −µkmgxmax = 0 + 1 2 kxmax 2 −QExmax 60 Chapter 25 Solutions xmax = 2(QE −µkmg) k = 2 50.0 × 10−6 C ( ) 5.00 × 105 N C ( ) −0.200 4.00 kg ( ) 9.80 m s2 ( ) [ ] 100 N m = 0.343 m 25.12 (a) Arbitrarily choose V = 0 at 0. Then at other points V = −Ex and Ue = QV = −QEx. Between the endpoints of the motion, (K + Us + Ue)i = (K + Us + Ue)f 0 + 0 + 0 = 0 + 1 2 kxmax 2 −QExmax so xmax = 2QE k (b) At equilibrium, ΣFx = −Fs + Fe = 0 or kx = QE. So the equilibrium position is at x = QE k (c) The block's equation of motion is ΣFx = −kx + QE = m d2x dt2 . Let ′ x = x −QE k , or x = ′ x + QE k , so the equation of motion becomes: −k ′ x + QE k + QE = m d2 ′ x + QE k ( ) dt2 , or d2 ′ x dt2 = −k m ′ x This is the equation for simple harmonic motion a ′ x = −ω 2 ′ x ( ), with ω = k m The period of the motion is then T = 2π ω = 2π m k (d) (K + Us + Ue)i + ∆E = (K + Us + Ue)f 0 + 0 + 0 −µkmgxmax = 0 + 1 2 kxmax 2 −QExmax xmax = 2(QE −µkmg) k 25.13 For the entire motion, y −yi = vyit + 1 2 ayt2 0 −0 = vit + 1 2 ayt2 so ay = −2vi t ∑Fy = may: −mg −qE = −2mvi t E = m q 2vi t −g and E =−m q 2vi t −g j For the upward flight: vyf 2 = vyi 2 + 2ay(y −yi) 0 = vi 2 + 2 −2vi t (ymax −0) and ymax = 1 4 vit Chapter 25 Solutions 61 © 2000 by Harcourt, Inc. All rights reserved. ∆V = E⋅dy = + m q 2vi t −g y 0 ymax ∫ 0 ymax = m q 2vi t −g 1 4 vit ( ) ∆V = 2.00 kg 5.00 × 10−6 C 2(20.1 m s) 4.10 s −9.80 m s2 1 4 (20.1 m s) 4.10 s ( ) [ ] = 40.2 kV 62 Chapter 25 Solutions 25.14 Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length, V = −Ed and Ue = −λLEd (a) (K + U)i = (K + U)f 0 + 0 = 1 2 µLv 2 −λLEd v = 2λEd µ = 2(40.0 × 10−6 C/m)(100 N/C)(2.00 m) (0.100 kg /m) = 0.400 m/s (b) The same. 25.15 Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position of the charge is – E · s = –EL cos θ. At the final point a, V = –EL. Suppose the table is frictionless: (K + U)i = (K + U)f 0 −qEL cosθ = 1 2 mv 2 −qEL v = = 2qEL(1−cosθ) m = 2(2.00 × 10−6 C)(300 N/C)(1.50 m)(1−cos 60.0°) 0.0100 kg = 0.300 m/s 25.16 (a) The potential at 1.00 cm is V1 = ke q r = (8.99 × 109 N · m2/C 2)(1.60 × 10–19 C) 1.00 × 10–2 m = 1.44 × 10–7 V (b) The potential at 2.00 cm is V2 = ke q r = (8.99 × 109 N · m2/C 2)(1.60 × 10–19 C) 2.00 × 10–2 m = 0.719 × 10–7 V Thus, the difference in potential between the two points is ∆V = V2 – V1 = –7.19 × 10–8 V (c) The approach is the same as above except the charge is – 1.60 × 10–19 C. This changes the sign of all the answers, with the magnitudes remaining the same.
That is, the potential at 1.00 cm is –1.44 × 10–7 V The potential at 2.00 cm is – 0.719 × 10–7 V, so ∆V = V2 – V1 = 7.19 × 10–8 V .
Chapter 25 Solutions 63 © 2000 by Harcourt, Inc. All rights reserved.
25.17 (a) Since the charges are equal and placed symmetrically, F = 0 (b) Since F = qE = 0, E = 0 (c) V = 2ke q r = 2 8.99 × 109 N · m2 C 2 2.00 × 10–6 C 0.800 m V = 4.50 × 104 V = 45.0 kV 25.18 (a) Ex = ke q1 x2 + ke q2 (x – 2.00)2 = 0 becomes Ex = ke +q x2 + –2q (x – 2.00)2 = 0 Dividing by ke, 2qx2 = q(x – 2.00)2 x2 + 4.00x – 4.00 = 0 Therefore E = 0 when x = −4.00 ± 16.0 + 16.0 2 = – 4.83 m (Note that the positive root does not correspond to a physically valid situation.) (b) V = ke q1 x + ke q2 (2.00 – x) = 0 or V = ke +q x – 2q (2.00 – x) = 0 Again solving for x, 2qx = q(2.00 – x) For 0 ≤ x ≤ 2.00 V = 0 when x = 0.667 m and q x = –2q 2 – x For x < 0 x = –2.00 m 25.19 (a) U = keq1q2 r = −(8.99 × 109)(1.60 × 10−19)2 0.0529 × 10−9 = – 4.35 × 10–18 J = –27.2 eV (b) U = ke q1q2 r = – (8.99 × 109)(1.60 × 10–19)2 22(0.0529 × 10–9) = – 6.80 eV 64 Chapter 25 Solutions (c) U = ke q1q2 r = – ke e 2 ∞ = 0 Goal Solution The Bohr model of the hydrogen atom states that the single electron can exist only in certain allowed orbits around the proton. The radius of each Bohr orbit is r = n2(0.0529 nm) where n = 1, 2, 3, . . . .
Calculate the electric potential energy of a hydrogen atom when the electron is in the (a) first allowed orbit, n = 1; (b) second allowed orbit, n = 2; and (c) when the electron has escaped from the atom ( r = ∞).
Express your answers in electron volts. G : We may remember from chemistry that the lowest energy level for hydrogen is E1 = −13.6 eV, and higher energy levels can be found from En = E1 /n2, so that E2 = −3.40 eV and E∞= 0 eV. (see section 42.2) Since these are the total energies (potential plus kinetic), the electric potential energy alone should be lower (more negative) because the kinetic energy of the electron must be positive.
O : The electric potential energy is given by U = ke q1q2 r A : (a) For the first allowed Bohr orbit, U = 8.99 × 109 N ⋅m2 C2 (−1.60 × 10−19 C)(1.60 × 10−19 C) (0.0529 × 10−9 m) = −4.35 × 10−18 J = −4.35 × 10−18 J 1.60 × 10−19 J/eV = −27.2 eV (b) For the second allowed orbit, U = (8.99 × 109 N ⋅m2 /C2)(−1.60 × 10−19 C)(1.60 × 10−19 C) 22(0.0529 × 10−9 m) = −1.088 × 10−18 J = −6.80 eV (c) When the electron is at r = ∞, U = 8.99 × 109 N ⋅m2/C2 ( ) −1.60 × 10−19 C ( ) 1.60 × 10−19 C ( ) ∞ = 0 J L : The potential energies appear to be twice the magnitude of the total energy values, so apparently the kinetic energy of the electron has the same absolute magnitude as the total energy.
25.20 (a) U = qQ 4π e0r = (5.00 × 10–9 C)(– 3.00 × 10–9 C)(8.99 × 109 V · m/C) (0.350 m) = – 3.86 × 10–7 J The minus sign means it takes 3.86 × 10–7 J to pull the two charges apart from 35 cm to a much larger separation.
(b) V = Q1 4π e0r1 + Q2 4π e0r2 = (5.00 × 10−9 C)(8.99 × 109 V ⋅m/C) 0.175 m + (−3.00 × 10−9 C)(8.99 × 109 V ⋅m/C) 0.175 m V = 103 V Chapter 25 Solutions 65 © 2000 by Harcourt, Inc. All rights reserved.
25.21 V = ∑ i k qi ri V = (8.99 × 109)(7.00 × 10–6) –1 0.0100 – 1 0.0100 + 1 0.0387 V = –1.10 × 107 C = –11.0 MV 66 Chapter 25 Solutions 25.22 Ue = q4V1 + q4V2 + q4V3 = q4 1 4π ∈0 q1 r1 + q2 r2 + q3 r3 Ue = 10.0 × 10−6 C ( ) 2 8.99 × 109 N ⋅m2/C2 ( ) 1 0.600 m + 1 0.150 m + 1 0.600 m ( )2 + 0.150 m ( )2 Ue= 8.95 J 25.23 U = U1 + U2 + U3 + U4 U = 0 + U12 + (U13 + U23) + (U14 + U24 + U34) U = 0 + keQ2 s + keQ2 s 1 2 + 1 + keQ2 s 1+ 1 2 + 1 U = keQ2 s 4 + 2 2 = 5.41 ke Q2 s An alternate way to get the term 4 + 2 2 ( ) is to recognize that there are 4 side pairs and 2 face diagonal pairs.
25.24 (a) V = keq1 r1 + keq2 r2 = 2 keq r = 2 8.99 × 109 N ⋅m2 C2 ( ) 2.00 × 10−6 C ( ) 1.00 m ( )2 + 0.500 m ( )2 V = 3.22 × 104 V = 32.2 kV (b) U = qV = −3.00 × 10−6 C ( ) 3.22 × 104 J C = −9.65 × 10−2 J 2.00 µC 2.00 µC P (0, 0.500 m) (1.00 m, 0) (-1.00 m, 0) y x 25.25 Each charge creates equal potential at the center. The total potential is: V = 5 ke −q ( ) R = −5keq R Chapter 25 Solutions 67 © 2000 by Harcourt, Inc. All rights reserved.
25.26 (a) Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point located at a finite distance from the charges, where this total potential is zero.
(b) V = ke q a + ke q a = 2keq a 25.27 (a) Conservation of momentum: 0 = m1v1 i + m2v2 (–i) or v2 = m1v1 m 2 By conservation of energy, 0 + ke (–q1)q2 d = 1 2 m 1v2 1 + 1 2 m 2v2 2 + ke(–q1)q2 (r1 + r2) and ke q1q2 r1 + r2 – ke q1q2 d = 1 2 m 1v2 1 + 1 2 m 2 1v2 1 m 2 v1 = 2m2keq1q2 m1 m1 + m2 ( ) 1 r1 + r2 −1 d v1 = 2 0.700 kg ( ) 8.99 × 109 N ⋅m2/C2 ( ) 2 × 10−6 C ( ) 3 × 10−6 C ( ) 0.100 kg ( ) 0.800 kg ( ) 1 8 × 10−3 m − 1 1.00 m = 10.8 m/s v2 = m 1v1 m 2 = (0.100 kg)(10.8 m/s) 0.700 kg = 1.55 m/s (b) If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .
25.28 (a) Conservation of momentum: 0 = m1v1i + m2v2 (−i) or v2 = m1v1 /m2 By conservation of energy, 0 + ke(−q1)q2 d = 1 2 m1v 1 2 + 1 2 m2v 2 2 + ke(−q1)q2 (r1 + r2) and keq1q2 r1 + r2 −keq1q2 d = 1 2 m1v 1 2 + 1 2 m1 2v1 2 m2 v1 = 2m2keq1q2 m1(m1 + m2) 1 r1 + r2 −1 d v2 = m1 m2 v1 = 2m1keq1q2 m2(m1 + m2) 1 r1 + r2 −1 d (b) If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than calculated in (a) .
68 Chapter 25 Solutions 25.29 V = keQ r so r = keQ V = 8.99 × 109 N ⋅m2 C2 ( ) 8.00 × 10−9 C ( ) V = 72.0 V ⋅m V For V = 100 V, 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m The radii are inversely proportional to the potential.
25.30 (a) V x ( ) = keQ1 r1 + keQ2 r2 = ke +Q ( ) x2 + a2 + ke +Q ( ) x2 + −a ( )2 V x ( ) = 2keQ x2 + a2 = keQ a 2 x a ( ) 2 + 1 V x ( ) keQ a ( ) = 2 x a ( ) 2 + 1 (b) V y ( ) = keQ1 r1 + keQ2 r2 = ke +Q ( ) y −a + ke −Q ( ) y + a V y ( ) = keQ a 1 y a −1 − 1 y a + 1 V y ( ) keQ a ( ) = 1 y a −1 − 1 y a + 1 25.31 Using conservation of energy, we have Kf + Uf = Ki + Ui.
But Ui = keqαqgold ri , and ri ≈∞. Thus, Ui = 0.
Also Kf = 0 ( vf = 0 at turning point), so Uf = Ki, or keqαqgold rmin = 1 2 mαvα 2 rmin = 2keqαqgold mαvα 2 = 2(8.99 × 109 N ⋅m2 /C2)(2)(79) 1.60 × 10-19 C ( ) 2 (6.64 × 10−27 kg)(2.00 × 107 m /s)2 = 2.74 × 10−14 m = 27.4 fm Chapter 25 Solutions 69 © 2000 by Harcourt, Inc. All rights reserved.
25.32 Using conservation of energy we have: keeQ r1 = keeQ r2 + 1 2 mv2 which gives: v = 2keeQ m 1 r1 − 1 r2 or v = (2)(8.99 × 109 N ⋅m2 /C2)(−1.60 × 10−19 C)(10−9 C) 9.11× 10-31 kg 1 0.0300 m − 1 0.0200 m Thus, v = 7.26 × 106 m /s 25.33 U = keqiqj ri j ∑ , summed over all pairs of i, j ( ) where i ≠j U = ke q −2q ( ) b + −2q ( ) 3q ( ) a + 2q ( ) 3q ( ) b + q 2q ( ) a + q 3q ( ) a2 + b2 + 2q −2q ( ) a2 + b2 U = keq2 −2 0.400 − 6 0.200 + 6 0.400 + 2 0.200 + 3 0.447 − 4 0.447 U = 8.99 × 109 ( ) 6.00 × 10−6 ( ) 2 4 0.400 − 4 0.200 − 1 0.447 = – 3.96 J 25.34 Each charge moves off on its diagonal line. All charges have equal speeds. (K + U)i ∑ = (K + U)f ∑ 0 + 4keq2 L + 2keq2 2 L = 4 1 2 mv 2 ( ) + 4keq2 2L + 2keq2 2 2 L 2 + 1 2 ke q2 L = 2mv 2 v = 1 + 1 8 keq2 mL 70 Chapter 25 Solutions 25.35 A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 × 6 = 12 face diagonal pairs separated by 2 s, and 4 interior diagonal pairs separated 3 s. U = keq2 s 12 + 12 2 + 4 3 = 22.8 keq2 s 25.36 V = a + bx = 10.0 V + −7.00 V m ( )x (a) At x = 0, V = 10.0 V At x = 3.00 m, V = −11.0 V At x = 6.00 m, V = −32.0 V (b) E = −dV dx = −b = −−7.00 V m ( ) = 7.00 N C in + x direction 25.37 V = 5x – 3x2y + 2yz2 Evaluate E at (1, 0 – 2) Ex = – ∂V ∂x = – 5 + 6xy = – 5 + 6(1)(0) = – 5 Ey = – ∂V ∂y = +3x2 – 2z2 = 3(1)2 – 2(–2)2 = – 5 Ez = – ∂V ∂ z = – 4yz = – 4(0)(–2) = 0 E = Ex 2 + Ey 2 + Ez 2 = −5 ( )2 + −5 ( )2 + 02 = 7.07 N/C 25.38 (a) For r < R V = keQ R Er = −dV dr = 0 (b) For r ≥R V = keQ r Er = −dV dr = −−keQ r2 = keQ r2 Chapter 25 Solutions 71 © 2000 by Harcourt, Inc. All rights reserved.
25.39 Ey = −∂V ∂y = −∂ ∂y keQ l ln l+ l2 + y2 y Ey = keQ ly 1− y2 l2 + y2 + l l2 + y2 = keQ y l2 + y2 25.40 Inside the sphere, Ex = Ey = Ez = 0. Outside, Ex = −∂V ∂x = −∂ ∂x V0 −E0z + E0 a3z(x 2 + y2 + z 2)−3/2 ( ) So Ex = −0 + 0 + E0 a3z(−3/ 2)(x 2 + y2 + z2)−5/2 (2x) [ ] = 3E0 a3xz(x 2 + y2 + z2)−5/2 Ey = −∂V ∂y = −∂ ∂y V0 −E0 z + E0a3z(x 2 + y2 + z2)−3/2 ( ) Ey= −E0a3z(−3/ 2)(x 2 + y2 + z2)−5/2 2y = 3E0a3yz(x 2 + y2 + z2)−5/2 Ez = −∂V ∂z = E0 −E0a3z(−3/ 2)(x 2 +y 2 +z 2)−5/2 (2z) −E0a3(x 2 +y2 +z 2)−3/2 Ez = E0 + E0a3(2z 2 −x 2 −y 2)(x 2 +y 2 +z 2)−5/2 25.41 ∆V = V2R −V0 = keQ R2 + 2R ( )2 −keQ R = keQ R 1 5 −1 = – 0.553 keQ R 25.42 V = dV ∫ = 1 4π e0 dq r ∫ All bits of charge are at the same distance from O, so V = 1 4π e0 Q R = 8.99 × 109 N ⋅m2 C2 −7.50 × 10−6 C ( ) 0.140 m / π ( ) = –1.51 MV 72 Chapter 25 Solutions 25.43 (a) [α] = λ x = C m ⋅ 1 m = C m2 (b) V = ke ⌡ ⌠ dq r = ke ⌡ ⌠ λ dx r = keα xdx (d + x) 0 L ∫ = keα L −dln 1+ L d 25.44 V = ke dq r ∫ = ke α xdx b2 + L 2 −x ( ) 2 ∫ Let z = L 2 −x. Then x = L 2 −z, and dx = −dz V = keα L 2 −z ( )(−dz) b2 + z2 ∫ = −keα L 2 dz b2 + z2 ∫ + keα zdz b2 + z2 ∫ = −keαL 2 ln(z + z2 + b2 ) + keα z2 + b2 V = −keαL 2 ln L 2 −x ( ) + L 2 −x ( ) 2 + b2 0 L + keα L 2 −x ( ) 2 + b2 0 L V = −keα L 2 ln L 2 −L + L 2 ( ) 2 + b2 L 2 + L 2 ( ) 2 + b2 + keα L 2 −L ( ) 2 + b2 − L 2 ( ) 2 + b2 V = −keα L 2 ln b2 + (L2 4) −L 2 b2 + (L2 4) + L 2 25.45 dV = ke dq r2 + x2 where dq = σ dA = σ2πrdr V = 2πσke rdr r2 + x2 a b ∫ = 2πkeσ x2 + b2 − x2 + a2 25.46 V = ke dq r all charge ∫ = ke λ dx −x + ke λ ds R + ke λ dx x R 3R ∫ semicircle ∫ −3R −R ∫ V = −ke λ ln(−x) −3R −R + ke λ R πR + keλ lnx R 3R V = ke λ ln 3R R + ke λπ + ke λ ln 3 = ke λ (π + 2 ln 3) Chapter 25 Solutions 73 © 2000 by Harcourt, Inc. All rights reserved.
25.47 Substituting given values into V = ke q r , 7.50 × 103 V = (8.99 × 109 N·m2/C2) q (0.300 m) Substituting q = 2.50 × 10−7 C, N = 2.50 × 10-7 C 1.60 × 10-19 C/e− = 1.56 × 1012 electrons 25.48 q1 + q2 = 20.0 µC so q1 = 20.0 µC – q2 q1 q2 = r1 r2 so 20.0 µC – q2 q2 = 4.00 cm 6.00 cm Therefore 6.00(20.0 µC – q2) = 4.00q2 ; Solving, q2 = 12.0 µC and q1 = 20.0 µC – 12.0 µC = 8.00 µC (a) E1 = keq1 r1 2 = 8.99 × 109 ( ) 8.00 × 10−6 ( ) 0.0400 ( )2 = 4.50 × 107 V /m = 45.0 MV/m E2 = keq2 r2 2 = 8.99 × 109 ( ) 12.0 × 10−6 ( ) 0.0600 ( )2 = 3.00 × 107 V /m = 30.0 MV/m (b) V1 = V2 = keq2 r2 = 1.80 MV 25.49 (a) E = 0 ; V = keq R = (8.99 × 109)(26.0 × 10−6) 0.140 = 1.67 MV (b) E = keq r2 = (8.99 × 109)(26.0 × 10−6) (0.200)2 = 5.84 MN/C away V = keq r = (8.99 × 109)(26.0 × 10−6) (0.200) = 1.17 MV (c) E = keq R2 = (8.99 × 109)(26.0 × 10−6) (0.140)2 = 11.9 MN/C away V = keq R = 1.67 MV 25.50 No charge stays on the inner sphere in equilibrium. If there were any, it would create an electric field in the wire to push more charge to the outer sphere. Charge Q is on the outer sphere. Therefore, zero charge is on the inner sphere and 10.0 µC is on the outer sphere .
74 Chapter 25 Solutions 25.51 (a) Emax = 3.00 × 106 V /m = keQ r2 = keQ r 1 r = Vmax 1 r Vmax = Emaxr = 3.00 × 106(0.150) = 450 kV (b) keQmax r2 = Emax or keQmax r = Vmax Qmax = Emaxr2 ke = 3.00 × 106(0.150)2 8.99 × 109 = 7.51 µC Goal Solution Consider a Van de Graaff generator with a 30.0-cm-diameter dome operating in dry air. (a) What is the maximum potential of the dome? (b) What is the maximum charge on the dome?
G : Van de Graaff generators produce voltages that can make your hair stand on end, somewhere on the order of about 100 kV (see the Puzzler at beginning of Chapter 25). With these high voltages, the maximum charge on the dome is probably more than typical point charge values of about 1 µC.
The maximum potential and charge will be limited by the electric field strength at which the air surrounding the dome will ionize. This critical value is determined by the dielectric strength of air which, from page 789 or from Table 26.1, is Ecritical = 3 × 106 V /m. An electric field stronger than this will cause the air to act like a conductor instead of an insulator. This process is called dielectric breakdown and may be seen as a spark. O : From the maximum allowed electric field, we can find the charge and potential that would create this situation. Since we are only given the diameter of the dome, we will assume that the conductor is spherical, which allows us to use the electric field and potential equations for a spherical conductor.
With these equations, it will be easier to do part (b) first and use the result for part (a).
A : (b) For a spherical conductor with total charge Q, E = keQ r2 Q = Er2 ke = 3.00 × 106 V /m ( ) 0.150 m ( )2 8.99 × 109 N ⋅m2 /C2 1 N ⋅m / V ⋅C ( ) = 7.51 µC (a) V = keQ r = (8.99 × 109 N ⋅m2 /C2)(7.51× 10−6 C) 0.150 m = 450 kV L : These calculated results seem reasonable based on our predictions. The voltage is about 4000 times larger than the 120 V found from common electrical outlets, but the charge is similar in magnitude to many of the static charge problems we have solved earlier. This implies that most of these charge configurations would have to be in a vacuum because the electric field near these point charges would be strong enough to cause sparking in air. (Example: A charged ball with Q = 1 µC and r = 1 mm would have an electric field near its surface of E = keQ r2 = 9 × 109 N ⋅m2/C2 ( ) 1× 10−6 C ( ) 0.001 m ( )2 = 9 × 109 V /m which is well beyond the dielectric breakdown of air!) Chapter 25 Solutions 75 © 2000 by Harcourt, Inc. All rights reserved.
25.52 V = ke q r and E = ke q r2 Since E = V r , (b) r = V E = 6.00 × 105 V 3.00 × 106 V/m = 0.200 m and (a) q = Vr ke = 13.3 µC 25.53 U = qV = ke q1q2 r12 = (8.99 × 109) (38)(54)(1.60 × 10–19)2 (5.50 + 6.20) × 10–15 = 4.04 × 10–11 J = 253 MeV 25.54 (a) To make a spark 5 mm long in dry air between flat metal plates requires potential difference V = Ed = 3.0 × 106 V m ( ) 5.0 × 10−3 m ( ) = 1.5 × 104 V ~104 V (b) Suppose your surface area is like that of a 70-kg cylinder with the density of water and radius 12 cm. Its length would be given by 70 × 103 cm3 = π 12 cm ( )2l l = 1.6 m The lateral surface area is A = 2π r l = 2π 0.12 m ( ) 1.6 m ( ) = 1.2 m2 The electric field close to your skin is described by E = σ e0 = Q Ae0 , so Q = EA ∈0= 3.0 × 106 N C 1.2 m2 ( ) 8.85 × 10−12 C2 N ⋅m2 ~10−5 C 25.55 (a) V = ke Q 1 x + a – 2 x + 1 x – a V = ke Q x(x – a) – 2(x + a)(x – a) + x(x + a) x(x + a)(x – a) = 2ke Qa2 x3 – xa2 (b) V = 2ke Qa2 x3 for a x << 1 76 Chapter 25 Solutions 25.56 (a) Ex = −dV dx = −d dx 2keQa2 x3 −xa2 = (2keQa2)(3x2 −a2) (x3 −xa2)2 and Ey = Ez = 0 (b) Ex = 2(8.99 × 109 N ⋅m2/C2)(3 × 10−6 C)(2 × 10−3 m)2 3(6 × 10−3 m)2 −(2 × 10−3 m)2 [ ] (6 × 10−3 m)3 −(6 × 10−3 m)(2 × 10−3 m)2 [ ] 2 Ex = 609 × 106 N/C = 609 MN/C 25.57 (a) E = Q 4π∈0 r2 V = Q 4π∈0 r r = V E = 3000 V 500 V /m = 6.00 m (b) V = −3000 V = Q 4π∈0 (6.00 m) Q = −3000 V (8.99 × 109 V ⋅m/C) (6.00 m) = – 2.00 µC 25.58 From Example 25.5, the potential created by the ring at the electron's starting point is Vi = keQ xi 2 + a2 = ke 2π λ a ( ) xi 2 + a2 while at the center, it is Vf = 2π keλ . From conservation of energy, 0 + −eVi ( ) = 1 2 mevf 2 + −eVf ( ) vf 2 = 2e me Vf −Vi ( ) = 4πekeλ me 1− a xi 2 + a2 vf 2 = 4π 1.60 × 10−19 ( ) 8.99 × 109 ( ) 1.00 × 10−7 ( ) 9.11× 10−31 1− 0.200 0.100 ( )2 + 0.200 ( )2 vf = 1.45 × 107 m/s Chapter 25 Solutions 77 © 2000 by Harcourt, Inc. All rights reserved.
25.59 (a) Take the origin at the point where we will find the potential. One ring, of width dx, has charge Q dx/h and, according to Example 25.5, creates potential dV = keQ dx h x 2 + R 2 The whole stack of rings creates potential V = dV all charge ∫ = keQdx h x 2 + R 2 d d + h ∫ = ke Q h ln x + x 2 + R2 d d + h = ke Q h ln d + h + (d + h)2 + R2 d + d2 + R2 (b) A disk of thickness dx has charge Q dx/h and charge-per-area Q dx/π R 2h. According to Example 25.6, it creates potential dV = 2π ke Qdx π R2h x 2 + R2 −x Integrating, V = 2keQ R 2h d d+h ∫ x 2 + R 2 dx −xdx = 2ke Q R2 h 1 2 x x 2 + R2 + R2 2 ln x + x 2 + R2 −x 2 2 d d + h V = keQ R 2h (d + h) (d + h)2 + R2 −d d2 + R2 −2dh −h2 + R2 ln d + h + (d + h)2 + R2 d + d2 + R2 25.60 The positive plate by itself creates a field E = σ 2 ∈0 = 36.0 × 10–9 C/m2 2(8.85 × 10–12 C2/N · m2) = 2.03 kN C away from the + plate. The negative plate by itself creates the same size field and between the plates it is in the same direction. Together the plates create a uniform field 4.07 kN/C in the space between.
(a) Take V = 0 at the negative plate. The potential at the positive plate is then V −0 = − −4.07 kN/C ( )dx 0 12.0 cm ∫ The potential difference between the plates is V = (4.07 × 103 N/C)(0.120 m) = 488 V (b) 1 2 mv2 + q V i = 1 2 mv2 + q V f qV = (1.60 × 10–19 C)(488 V) = 1 2 mv 2 f = 7.81 × 10–17 J (c) vf = 306 km/s 78 Chapter 25 Solutions (d) vf 2 = v 2 i + 2a(x – xi) (3.06 × 105 m/s)2 = 0 + 2a(0.120 m) a = 3.90 × 1011 m/s2 (e) ∑F = ma = (1.67 × 10–27 kg)(3.90 × 1011 m/s2) = 6.51 × 10–16 N (f) E = F q = 6.51 × 10–16 N 1.60 × 10–19 C = 4.07 kN/C 25.61 W = V dq 0 Q ∫ where V = ke q R ; Therefore, W = keQ 2 2R 25.62 (a) VB −VA = − E⋅ds A B ∫ and the field at distance r from a uniformly charged rod (where r > radius of charged rod) is E = λ 2π e0r = 2keλ r In this case, the field between the central wire and the coaxial cylinder is directed perpendicular to the line of charge so that VB −VA = − 2keλ r dr ra rb ∫ = 2keλ ln ra rb , or ∆V = 2keλ ln ra rb rb ra λ − λ (b) From part (a), when the outer cylinder is considered to be at zero potential, the potential at a distance r from the axis is V = 2keλ ln ra r The field at r is given by E = −∂V ∂r = −2keλ r ra −ra r2 = 2keλ r But, from part (a), 2keλ = ∆V ln ra rb ( ) . Therefore, E = ∆V ln ra rb ( ) 1 r Chapter 25 Solutions 79 © 2000 by Harcourt, Inc. All rights reserved.
25.63 V2 −V1 = − E⋅dr r1 r2 ∫ = − λ 2πε0r r1 r2 ∫ dr V2 – V1 = −λ 2π∈0 ln r2 r1 25.64 For the given charge distribution, V x,y,z ( ) = ke q ( ) r1 + ke −2q ( ) r2 where r1 = x + R ( )2 + y2 + z2 and r2 = x2 + y2 + z2 The surface on which V x,y,z ( ) = 0 is given by keq 1 r1 −2 r2 = 0, or 2r1 = r2 This gives: 4 x + R ( )2 + 4y2 + 4z2 = x2 + y2 + z2 which may be written in the form: x2 + y2 + z2 + 8 3 R x + 0 ( )y + 0 ( )z + 4 3 R2 = 0 The general equation for a sphere of radius a centered at x0,y0,z0 ( ) is: x −x0 ( ) 2 + y −y0 ( ) 2 + z −z0 ( ) 2 −a2 = 0 or x2 + y2 + z2 + −2x0 ( )x + −2y0 ( )y + −2z0 ( )z + x0 2 + y0 2 + z0 2 −a2 ( ) = 0 Comparing equations and , it is seen that the equipotential surface for which V = 0 is indeed a sphere and that: −2x0 = 8 3 R; −2y0 = 0; −2z0 = 0; x0 2 + y0 2 + z0 2 −a2 = 4 3 R2 Thus, x0 = −4 3 R, y0 = z0 = 0, and a2 = 16 9 −4 3 R2 = 4 9 R2.
The equipotential surface is therefore a sphere centered at −4 3 R,0,0 , having a radius 2 3 R 80 Chapter 25 Solutions 25.65 (a) From Gauss's law, EA = 0 (no charge within) EB = ke qA r 2 = (8.99 × 109)(1.00 × 10−8) r 2 = 89.9 r 2 V /m EC = ke (qA + qB) r2 = (8.99 × 109) (−5.00 × 10−9) r2 = −45.0 r2 V /m (b) VC = ke (qA + qB) r = (8.99 × 109) (−5.00 × 10−9) r = −45.0 r V ∴At r2, V = −45.0 0.300 = −150V Inside r2, VB = −150 V + 89.9 r2 dr r2 r ∫ = −150 + 89.9 1 r − 1 0.300 = −450 + 89.9 r V ∴At r1, V = −450 + 89.9 0.150 = + 150V so VA = + 150 V 25.66 From Example 25.5, the potential at the center of the ring is Vi = keQ R and the potential at an infinite distance from the ring is Vf = 0. Thus, the initial and final potential energies of the point charge are: Ui = QVi = keQ2 R and Uf = QVf = 0 From conservation of energy, Kf + Uf = Ki + Ui or 1 2 Mvf 2 + 0 = 0 + keQ2 R giving vf = 2keQ2 MR 25.67 The sheet creates a field E1 = σ 2 ∈0 i for x > 0. Along the x −axis, the line of charge creates a field E2 = λ 2π r∈0 away = λ 2π ∈0 (3.00 m −x)(−i) for x < 3.00 m The total field along the x −axis in the region 0 < x < 3.00 m is then E = E1 + E2 = σ 2 ∈0 − λ 2π ∈0 3.00 −x ( ) i Chapter 25 Solutions 81 © 2000 by Harcourt, Inc. All rights reserved.
(a) The potential at point x follows from V −V0 = − E⋅idx = 0 x ∫ − σ 2 ∈0 − λ 2π ∈0 3.00 −x ( ) dx 0 x ∫ V = V0 −σ x 2∈0 − λ 2π ∈0 ln 1− x 3.00 V = 1.00 kV − (25.0 × 10−9 C/m2)x 2(8.85 × 10−12 C2 / N ⋅m2) − 80.0 × 10−9 C/m 2π(8.85 × 10−12 C2 / N ⋅m2) ln 1− x 3.00 V = 1.00 kV −1.41 kV m x −(1.44 kV) ln 1.00 − x 3.00 m (b) At x = 0.800 m, V = 316 V and U = QV = 2.00 × 10−9 C ( ) 316 J C ( ) = 6.33 × 10−7 J = 633 nJ 25.68 V = ke λ dx x2 + b2 = keλ ln a a+L ∫ x + (x2 + b2) a a+L = keλ ln a + L + a + L ( )2 + b2 a + a2 + b2 25.69 (a) Er = −∂V ∂r = 2kepcosθ r3 In spherical coordinates, the θ component of the gradient is 1 r ∂ ∂θ .
Therefore, Eθ = −1 r ∂V ∂θ = kepsinθ r3 For r >> a, Er(0°) = 2kep r3 and Er(90°) = 0, Eθ(0°) = 0 and Eθ(90°) = kep r3 These results are reasonable for r >> a .
However, for r → 0, E(0) → ∞ .
(b) V = kepy (x2 + y2)3/2 and Ex = −∂V ∂x = 3kepxy (x2 + y2)5/2 Ey = −∂V ∂y = kep(2y2 −x2) (x2 + y2)5/2 82 Chapter 25 Solutions 25.70 (a) EA > EB since E = ∆ ∆ V s (b) EB = – ∆V ∆s = – (6 – 2) V 2 cm = 200 N/C down (c) The figure is shown to the right, with sample field lines sketched in.
25.71 For an element of area which is a ring of radius r and width dr, dV = ke dq r2 + x2 dq = σ dA = Cr (2π r dr) and V = C(2πke) r2 dr r2 + x2 0 R ∫ = C(πke) R R2 + x2 + x2 ln x R + R2 + x2 25.72 dU = V dq where the potential V = ke q r . The element of charge in a shell is dq = ρ (volume element) or dq = ρ (4π r 2 dr) and the charge q in a sphere of radius r is q = 4πρ r2 dr = ρ 4πr3 3 0 r ∫ Substituting this into the expression for dU, we have dU = keq r dq = keρ 4πr3 3 1 r ρ(4πr2 dr) = ke 16π 2 3 ρ2r4 dr U = dU ∫ = ke 16π 2 3 ρ2 r4 dr 0 R ∫ = ke 16π 2 15 ρ2R5 But the total charge, Q = ρ 4 3 πR3. Therefore, U = 3 5 keQ2 R Chapter 25 Solutions 83 © 2000 by Harcourt, Inc. All rights reserved.
25.73 (a) From Problem 62, E = ∆V ln ra rb ( ) 1 r We require just outside the central wire 5.50 × 106 V m = 50.0 × 103 V ln 0.850 m rb 1 rb or 110 m-1 ( )rbln 0.850 m rb = 1 We solve by homing in on the required value rb (m) 0.0100 0.00100 0.00150 0.00145 0.00143 0.00142 110 m-1 ( )rb ln 0.850 m rb 4.89 0.740 1.05 1.017 1.005 0.999 Thus, to three significant figures, rb = 1.42 mm (b) At ra, E = 50.0 kV ln 0.850 m 0.00142 m ( ) 1 0.850 m = 9.20 kV m © 2000 by Harcourt, Inc. All rights reserved.
Chapter 26 Solutions 26.1 (a) Q = C (∆V) = (4.00 × 10–6 F)(12.0 V) = 4.80 × 10–5 C = 48.0 µC (b) Q = C (∆V) = (4.00 × 10–6 F)(1.50 V) = 6.00 × 10–6 C = 6.00 µC 26.2 (a) C = Q ∆V = 10.0 × 10−6 C 10.0 V = 1.00 × 10−6 F = 1.00 µF (b) ∆V = Q C = 100 × 10−6 C 1.00 × 10−6 F = 100 V 26.3 E = keq r2 : q = (4.90 × 104 N/C)(0.210 m)2 8.99 × 109 N ⋅m2 /C2 = 0.240 µC (a) σ = q A = 0.240 × 10−6 4π(0.120)2 = 1.33 µC/m2 (b) C = 4πe0r = 4π(8.85 × 10−12)(0.120) = 13.3 pF 26.4 (a) C = 4π e0R R = C 4π e0 = ke C = (8.99 × 109 N · m2/C 2)(1.00 × 10–12 F) = 8.99 mm (b) C = 4π e0R = 4π (8.85 × 10–12 C 2)(2.00 × 10–3 m) N · m2 = 0.222 pF (c) Q = CV = (2.22 × 10–13 F)(100 V) = 2.22 × 10–11 C 26.5 (a) Q1 Q2 = R1 R2 Q1 + Q2 = 1+ R1 R2 Q2 = 3.50Q2 = 7.00 µC Q2 = 2.00 µC Q1= 5.00 µC (b) V1 = V2 = Q1 C1 = Q2 C2 = 5.00 µC 8.99 × 109 m F ( ) −1(0.500 m) = 8.99 × 104 V = 89.9 kV Chapter 26 Solutions 85 © 2000 by Harcourt, Inc. All rights reserved.
26.6 C = κ e0A d = (1.00)(8.85 × 10–12 C 2)(1.00 × 103 m)2 N · m2(800 m) = 11.1 nF The potential between ground and cloud is ∆V = Ed = (3.00 × 106 N/C)(800 m) = 2.40 × 109 V Q = C (∆V) = (11.1 × 10-9 C/V)(2.40 × 109 V) = 26.6 C 26.7 (a) ∆V = Ed E = 20.0 V 1.80 × 10–3 m = 11.1 kV/m (b) E = σ e0 σ = (1.11 × 104 N/C)(8.85 × 10–12 C 2/N · m2) = 98.3 nC/m2 (c) C = e0A d = 8.85 × 10−12 C2 / N ⋅m2 ( ) 7.60 cm2 ( ) 1.00 m /100 cm ( )2 1.80 × 10−3 m = 3.74 pF (d) ∆V = Q C Q = (20.0 V)(3.74 × 10–12 F) = 74.7 pC 26.8 C = κe0A d = 60.0 × 10−15 F d = κe0A C = 1 ( ) 8.85 × 10−12 ( ) 21.0 × 10−12 ( ) 60.0 × 10−15 d = 3.10 × 10−9 m = 3.10 nm 26.9 Q = e0A d ∆V ( ) Q A = σ = e0 ∆V ( ) d d = e0 ∆V ( ) σ = 8.85 × 10−12 C 2 N ⋅m2 ( ) 150 V ( ) 30.0 × 10−9 C cm2 ( ) 1.00 × 104 cm2 m2 ( ) = 4.42 µm 86 Chapter 26 Solutions 26.10 With θ = π , the plates are out of mesh and the overlap area is zero.
With θ = 0, the overlap area is that of a semi-circle, πR2 2. By proportion, the effective area of a single sheet of charge is π −θ ( )R2 2. When there are two plates in each comb, the number of adjoining sheets of positive and negative charge is 3, as shown in the sketch.
When there are N plates on each comb, the number of parallel capacitors is 2N −1 and the total capacitance is + + + + + ---+ + + -----+ + + + d ---- C = 2N −1 ( ) e0Aeffective distance = 2N −1 ( )e0 π −θ ( )R2 2 d 2 = 2N −1 ( ) ∈0 π −θ ( )R2 d 26.11 (a) C = l 2ke ln b a = 50.0 2(8.99 × 109) ln 7.27 2.58 = 2.68 nF (b) Method 1: ∆V = 2ke λ ln b a λ = q / l = 8.10 × 10–6 C 50.0 m = 1.62 × 10–7 C/m ∆V = 2(8.99 × 10 9)(1.62 × 10–7) ln 7.27 2.58 = 3.02 kV Method 2: ∆V = Q C = 8.10 × 10–6 2.68 × 10–9 = 3.02 kV 26.12 Let the radii be b and a with b = 2a. Put charge Q on the inner conductor and – Q on the outer.
Electric field exists only in the volume between them. The potential of the inner sphere is Va = keQ a; that of the outer is Vb = keQ b. Then Va −Vb = ke Q a −ke Q b = Q 4π e0 b −a ab and C = Q Va −Vb = 4πe0ab b −a Here C = 4πe0 2a2 a = 8π e0 a a = C 8πe0 The intervening volume is Volume = 4 3 π b 3 −4 3 π a3 = 7 4 3 πa3 ( ) = 7 4 3 π ( ) C 3 83π 3e0 3 = 7C 3 384π 2e0 3 Volume = 7(20.0 × 10−6 C2 / N⋅m)3 384π 2(8.85 × 10−12 C2 / N⋅m2)3 = 2.13 × 1016 m3 The outer sphere is 360 km in diameter.
Chapter 26 Solutions 87 © 2000 by Harcourt, Inc. All rights reserved.
26.13 ΣFy = 0: T cosθ −mg = 0 ΣFx = 0: T sinθ −Eq = 0 Dividing, tanθ = Eq mg , so E = mg q tanθ ∆V = Ed = mgdtanθ q = (350 × 10−6 kg)(9.80 m /s2)(4.00 × 10−2 m) tan 15.0° 30.0 × 10−9 C = 1.23 kV 26.14 ΣFy = 0: T cosθ −mg = 0 ΣFx = 0: T sinθ −Eq = 0 Dividing, tanθ = Eq mg , so E mg q = tanθ and ∆V = Ed = mgdtanθ q 26.15 (a) C = ab ke(b −a) = (0.0700)(0.140) (8.99 × 109)(0.140 −0.0700) = 15.6 pF (b) C Q V = ∆ ∆V = Q C = 4.00 × 10−6 C 15.6 × 10−12 F = 256 kV Goal Solution An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and 14.0 cm, respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of 4.00 µ C on the capacitor? G : Since the separation between the inner and outer shells is much larger than a typical electronic capacitor with d ~ 0.1 mm and capacitance in the microfarad range, we might expect the capacitance of this spherical configuration to be on the order of picofarads, (based on a factor of about 700 times larger spacing between the conductors). The potential difference should be sufficiently low to prevent sparking through the air that separates the shells. O : The capacitance can be found from the equation for spherical shells, and the voltage can be found from Q = C∆V .
A : (a) For a spherical capacitor with inner radius a and outer radius b, C = ab k(b −a) = (0.0700 m)(0.140 m) 8.99 × 109 N ⋅m2 C2 ( )(0.140 −0.0700) m = 1.56 × 10−11 F = 15.6 pF (b) ∆V = Q C = (4.00 × 10−6 C) 1.56 × 10-11 F = 2.56 × 105 V = 256 kV L : The capacitance agrees with our prediction, but the voltage seems rather high. We can check this voltage by approximating the configuration as the electric field between two charged parallel plates separated by d = 7.00 cm, so E ~ ∆V d = 2.56 × 105 V 0.0700 m = 3.66 × 106 V /m This electric field barely exceeds the dielectric breakdown strength of air 3 × 106 V /m ( ), so it may not even be possible to place 4.00 µ C of charge on this capacitor!
88 Chapter 26 Solutions 26.16 C = 4π e0R = 4π 8.85 × 10−12 C N ⋅m2 ( ) 6.37 × 106 m ( ) = 7.08 × 10−4 F 26.17 (a) Capacitors in parallel add. Thus, the equivalent capacitor has a value of Ceq = C1 + C2 = 5.00 µF + 12.0 µF = 17.0 µF (b) The potential difference across each branch is the same and equal to the voltage of the battery.
∆V = 9.00 V (c) Q 5 = C (∆V) = (5.00 µF)(9.00 V) = 45.0 µC and Q12 = C (∆V) = (12.0 µF)(9.00 V) = 108 µC 26.18 (a) In series capacitors add as 1 C eq = 1 C 1 + 1 C 2 = 1 5.00 µF + 1 12.0 µF and C eq = 3.53 µF (c) The charge on the equivalent capacitor is Q eq = C eq (∆V) = (3.53 µF)(9.00 V) = 31.8 µC Each of the series capacitors has this same charge on it. So Q 1 = Q 2 = 31.8 µC (b) The voltage across each is ∆V1 = Q 1 C 1 = 31.8 µC 5.00 µF = 6.35 V and ∆V2 = Q 2 C 2 = 31.8 µC 12.0 µF = 2.65 V 26.19 C p = C 1 + C 2 1 C s = 1 C 1 + 1 C 2 Substitute C 2 = C p – C 1 1 C s = 1 C 1 + 1 C p – C 1 = C p – C 1 + C 1 C 1(C p – C 1) Simplifying, C 2 1 – C 1C p + C pC s = 0 C1 = Cp ± Cp 2 −4CpCs 2 = 1 2 Cp ± 1 4 Cp 2 −CpCs We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign, we would get the same two answers with their names interchanged.) C1 = 1 2 Cp + 1 4 Cp 2 −CpCs = 1 2 9.00 pF ( ) + 1 4 9.00 pF ( ) 2 −9.00 pF ( ) 2.00 pF ( ) = 6.00 pF C2 = Cp −C1 = 1 2 Cp − 1 4 Cp 2 −CpCs = 1 2 (9.00 pF) – 1.50 pF = 3.00 pF Chapter 26 Solutions 89 © 2000 by Harcourt, Inc. All rights reserved.
26.20 Cp = C1 + C2 and 1 Cs = 1 C1 + 1 C2 Substitute C2 = Cp −C1: 1 Cs = 1 C1 + 1 Cp −C1 = Cp −C1 + C1 C1 Cp− C1 Simplifying, C1 2 −C1Cp + CpCs = 0 and C1 = Cp ± Cp 2 −4CpCs 2 = 1 2 Cp + 1 4 Cp 2 −CpCs where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for the capacitances, with the names reversed). Then, from C2 = Cp −C1 C2 = 1 2 Cp − 1 4 Cp 2 −CpCs 26.21 (a) 1 Cs = 1 15.0 + 1 3.00 Cs = 2.50 µF Cp = 2.50 + 6.00 = 8.50 µF Ceq = 1 8.50 µF + 1 20.0 µF −1 = 5.96 µF (b) Q = ∆V ( )C = (15.0 V)(5.96 µF) = 89.5 µC on 20.0 µF ∆V = Q C = 89.5 µC 20.0 µF = 4.47 V 15.0 – 4.47 = 10.53 V Q = ∆V ( )C= (10.53)(6.00 µF) = 63.2 µC on 6.00 µF 89.5 – 63.2 = 26.3 µC on 15.0 µF and 3.00 µF 26.22 The circuit reduces first according to the rule for capacitors in series, as shown in the figure, then according to the rule for capacitors in parallel, shown below.
Ceq = C 1+ 1 2 + 1 3 = 11 6 C = 1.83C ⇒ 90 Chapter 26 Solutions 26.23 C = Q ∆V so 6.00 × 10–6 = Q 20.0 and Q = 120 µC Q1 = 120 µC – Q2 and ∆V = Q C 120 – Q 2 C 1 = Q 2 C 2 or 120 – Q2 6.00 = Q 2 3.00 (3.00)(120 – Q2) = (6.00)Q2 Q 2 = 360 9.00 = 40.0 µC Q1 = 120 µC – 40.0 µC = 80.0 µC 26.24 (a) In series , to reduce the effective capacitance: 1 32.0 µF = 1 34.8 µF + 1 Cs Cs = 1 2.51× 10−3 µ F = 398 µF (b) In parallel , to increase the total capacitance: 29.8 µF + Cp = 32.0 µF Cp = 2.20 µF 26.25 With switch closed, distance d' = 0.500d and capacitance ′ C = e0A ′ d = 2e0A d = 2C (a) Q = ′ C (∆V) = 2C(∆V) = 2(2.00 × 10−6 F)(100 V) = 400 µC (b) The force stretching out one spring is F = Q 2 2e0A = 4C 2(∆V)2 2e0A = 2C 2 (∆V)2 (e0A/ d)d = 2C(∆V)2 d One spring stretches by distance x = d /4, so k = F x = 2C(∆V)2 d 4 d = 8C(∆V)2 d 2 = 8(2.00 × 10−6 F)(100 V)2 (8.00 × 10−3 m)2 = 2.50 kN/m Chapter 26 Solutions 91 © 2000 by Harcourt, Inc. All rights reserved.
26.26 Positive charge on A will induce equal negative charges on B, D, and F, and equal positive charges on C and E. The nesting spheres form three capacitors in series. From Example 26.3, CAB = ab ke(b −a) = R(2R) keR = 2R ke CCD = 3R ( ) 4R ( ) keR = 12R ke CEF = 5R ( ) 6R ( ) keR = 30R ke Ceq = 1 ke / 2R + ke /12R + ke / 30R = 60R 37 ke 26.27 nC = = 100 n/C nC = 100C n so n2 = 100 and n = 10 Goal Solution A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group? G : Since capacitors in parallel add and ones in series add as inverses, 2 capacitors in parallel would have a capacitance 4 times greater than if they were in series, and 3 capacitors would give a ratio Cp /Cs = 9, so maybe n = Cp /Cs = 100 = 10. O : The ratio reasoning above seems like an efficient way to solve this problem, but we should check the answer with a more careful analysis based on the general relationships for series and parallel combinations of capacitors. A : Call C the capacitance of one capacitor and n the number of capacitors. The equivalent capacitance for n capacitors in parallel is Cp = C1 + C2 + . . . + Cn = nC The relationship for n capacitors in series is 1 Cs = 1 C1 + 1 C2 + . . . + 1 Cn = n C Therefore Cp Cs = nC C /n = n2 or n = Cp Cs = 100 = 10 L : Our prediction appears to be correct. A qualitative reason that Cp /Cs = n2 is because the amount of charge that can be stored on the capacitors increases according to the area of the plates for a parallel combination, but the total charge remains the same for a series combination.
92 Chapter 26 Solutions 26.28 Cs = 1 5.00 + 1 10.0 −1 = 3.33 µF Cp1 = 2(3.33) + 2.00 = 8.66 µF Cp2 = 2(10.0) = 20.0 µF Ceq = 1 8.66 + 1 20.0 −1 = 6.04 µF 26.29 Qeq = Ceq ∆V ( ) = 6.04 × 10−6 F ( ) 60.0 V ( ) = 3.62 × 10−4 C Qp1 = Qeq, so ∆Vp1 = Qeq Cp1 = 3.62 × 10−4 C 8.66 × 10−6 F = 41.8 V Q3 = C3 ∆Vp1 ( ) = 2.00 × 10−6 F ( ) 41.8 V ( ) = 83.6 µC 26.30 Cs = 1 5.00 + 1 7.00 −1 = 2.92 µF Cp = 2.92 + 4.00 + 6.00 = 12.9 µF 26.31 (a) U = 1 2 C (∆V)2 = 1 2 (3.00 µF)(12.0 V) 2 = 216 µ J (b) U = 1 2 C (∆V)2 = 1 2 (3.00 µF)(6.00 V) 2 = 54.0 µ J 26.32 U = 1 2 C (∆V)2 The circuit diagram is shown at the right.
(a) Cp = C1 + C2 = 25.0 µF + 5.00 µF = 30.0 µF U = 1 2 (30.0 × 10–6)(100) 2 = 0.150 J (b) Cs = 1 C1 + 1 C2 –1 = 1 25.0 µF + 1 5.00 µF –1 = 4.17 µF U = 1 2 C (∆V)2 ∆V = 2U C = 0.150 ( ) 2 ( ) 4.17 × 10−6 = 268 V Chapter 26 Solutions 93 © 2000 by Harcourt, Inc. All rights reserved.
26.33 Use U = 1 2 Q 2 C and C = e0 A d If d2 = 2d1, C2 = 1 2 C1. Therefore, the stored energy doubles .
26.34 u = U V = 1 2 e0E2 1.00 × 10−7 V = 1 2 (8.85 × 10−12)(3000)2 V = 2.51× 10−3 m3 = 2.51× 10−3 m3 ( ) 1000 L m3 = 2.51 L 26.35 W = U = Fdx ∫ so F = dU dx = d dx Q2 2c = d dx Q2x 2e0A = Q 2 2 e0A 26.36 Plate a experiences force – kx i from the spring and force QE i due to the electric field created by plate b according to E=σ / 2e0 = Q/ 2Ae0. Then, kx = Q 2 2A e0 x = Q2 2Ae0 k where A is the area of one plate.
26.37 The energy transferred is W Q V = = 1 2 ( ) ∆ 1 2 (50.0 C)(1.00 × 108 V) = 2.50 × 109 J and 1% of this (or W' = 2.50 × 107 J) is absorbed by the tree. If m is the amount of water boiled away, then W' = m(4186 J/kg °C)(100 °C – 30.0 °C) + m(2.26 × 106 J/kg) = 2.50 × 107 J giving m = 9.79 kg 94 Chapter 26 Solutions 26.38 U = 1 2 C ∆V ( )2 where C = 4π e0R = R ke and ∆V = keQ R −0 = keQ R U = 1 2 R ke keQ R 2 = keQ2 2R 26.39 keQ2 2R = mc2 R = ke e2 2mc2 = (8.99 × 109 N ⋅m2 /C)(1.60 × 10−19 C)2 2(9.11× 10−31 kg)(3.00 × 108 m /s)2 = 1.40 fm 26.40 C = κ ∈0 A d = 4.90(8.85 × 10–12 F/m)(5.00 × 10–4 m2) 2.00 × 10–3 m = 1.08 × 10–11 F = 10.8 pF 26.41 (a) C = κ ∈0 A d = 2.10(8.85 × 10–12 F/m)(1.75 × 10–4 m2) 4.00 × 10–5 m = 8.13 × 10–11 F = 81.3 pF (b) ∆Vmax = Emax d = (60.0 × 106 V/m)(4.00 × 10–5 m) = 2.40 kV 26.42 Qmax = C (∆Vmax), but ∆Vmax = Emax d Also, C = κ ∈0A d Thus, Qmax = κ ∈0A d (Emax d) = κ ∈0 AEmax (a) With air between the plates, κ = 1.00 and Emax = 3.00 × 106 V/m. Therefore, Qmax = κ ∈0 AEmax = (8.85 × 10–12 F/m)(5.00 × 10–4 m2)(3.00 × 106 V/m) = 13.3 nC (b) With polystyrene between the plates, κ = 2.56 and Emax = 24.0 × 106 V/m.
Qmax = κ ∈0 AEmax = 2.56(8.85 × 10–12 F/m)(5.00 × 10–4 m2)(24.0 × 106 V/m) = 272 nC 26.43 C = κ ∈0A d or Chapter 26 Solutions 95 © 2000 by Harcourt, Inc. All rights reserved. = 1.04 m 96 Chapter 26 Solutions 26.44 Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between them. Suppose the plastic has κ ≅ 3, Emax ~ 107 V/m and thickness 1 mil = 2.54 cm/1000. Then, C = κ ∈0 A d ~ 3(8.85 × 10–12 C2/N · m2)(0.4 m2) 2.54 × 10–5 m ~ 10–6 F ∆Vmax = Emaxd ~ 107 V m 2.54 × 10−5 m ( ) ~ 10 2 V 26.45 (a) With air between the plates, we find C0 = Q ∆V = 48.0 µC 12.0 V = 4.00 µF (b) When Teflon is inserted, the charge remains the same (48.0 µC) because the plates are isolated.
However, the capacitance, and hence the voltage, changes. The new capacitance is C' = κ C 0 = 2.10(4.00 µF) = 8.40 µF (c) The voltage on the capacitor now is ∆V' = Q C' = 48.0 µC 8.40 µF = 5.71 V and the charge is 48.0 µC 26.46 Originally, C =∈0A/d = Q/(∆V)i (a) The charge is the same before and after immersion, with value Q =∈0A(∆V)i / d . Q = (8.85 × 10−12 C2)(25.0 × 10−4 m2)(250 V) N ⋅m2 (1.50 × 10−2 m) = 369 pC (b) Finally, Cf = κ ∈0 A/ d = Q /(∆V)f Cf = 80.0(8.85 × 10−12 C2)(25.0 × 10−4 m2) N ⋅m2 (1.50 × 10−2 m) = 118 pF (∆V)f = Qd κ∈0A = ∈0A(∆V)i d κ∈0Ad = (∆V)i κ = 250 V 80.0 = 3.12 V (c) Originally, U = 1 2 C(∆V)i 2 = ∈0A(∆V)i 2 2d Finally, Uf = 1 2 Cf (∆V)f 2 = κ∈0A(∆V) i 2 2d κ 2 = ∈0A(∆V) i 2 2dκ So, ∆U = Uf −U = −∈0 A(∆V)i 2 (κ −1) 2d κ ∆U = (−8.85 × 10−12 C 2)(25.0 × 10−4 m2)(250 V)2 (79.0) N ⋅m2 2(1.50 × 10−2 m)80 = – 45.5 nJ Chapter 26 Solutions 97 © 2000 by Harcourt, Inc. All rights reserved.
26.47 1 C = 1 κ1ab ke(b −a) + 1 κ2bc ke(c −b) = ke(b −a) κ1ab + ke (c −b) κ2bc C = 1 ke (b −a) κ1 ab + ke (c −b) κ2bc = κ1κ2 abc ke κ2 (bc −ac) + ke κ1(ac −ab) = 4π κ1κ2 abc∈0 κ2bc −κ1ab + (κ1 −κ2)ac 26.48 (a) C = κC0 = κ∈0A d = (173)(8.85 × 10−12)(1.00 × 10−4 m2) 0.100 × 10−3 m = 1.53 nF (b) The battery delivers the free charge Q = C (∆V) = (1.53 × 10-9 F)(12.0 V) = 18.4 nC (c) The surface density of free charge is σ = Q A = 18.4 × 10−9 C 1.00 × 10−4 m2 = 1.84 × 10-4 C/m2 The surface density of polarization charge is σ σ κ σ p = − = − = 1 1 1 1 173 1.83 × 10-4 C/m2 (d) We have E = E0/κ and E0 = ∆V/d ; hence, E = ∆V κd = 12.0 V (173)(1.00 × 10−4 m) = 694 V/m 26.49 The given combination of capacitors is equivalent to the circuit diagram shown to the right.
Put charge Q on point A. Then, Q = (40.0 µF)∆VAB = (10.0 µF)∆VBC = (40.0 µF)∆VCD So, ∆VBC = 4∆VAB = 4∆VCD, and the center capacitor will break down first, at ∆VBC = 15.0 V. When this occurs, ∆ ∆ ∆ V V V AB CD BC = = ( ) = 1 4 3 75 . V and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V 98 Chapter 26 Solutions 26.50 (a) The displacement from negative to positive charge is 2 1 20 1 10 1 40 1 30 a i j i j = − + ( ) − − ( ) .
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mm mm = −2.60i + 2.40j ( ) × 10−3 m The electric dipole moment is p = 2aq = 3.50 × 10−9 C ( ) −2.60i + 2.40j ( ) × 10−3 m = −9.10i + 8.40j ( ) × 10−12 C⋅m (b) τ = p × E = −9.10i + 8.40j ( ) × 10−12 C⋅m [ ]× 7.80i −4.90j ( ) × 103 N C [ ] τ = + 44.6k −65.5k ( ) × 10−9 N ⋅m = −2.09 × 10−8 N ⋅mk (c) U = −p⋅E = −−9.10i + 8.40j ( ) × 10−12 C⋅m [ ]⋅7.80i −4.90j ( ) × 103 N C [ ] U = 71.0 + 41.2 ( ) × 10−9 J = 112 nJ (d) p = 9.10 ( )2 + 8.40 ( )2 × 10−12 C⋅m = 12.4 × 10−12 C⋅m E = 7.80 ( )2 + 4.90 ( )2 × 103 N C = 9.21× 103 N C Umax = p E = 114 nJ, Umin = −114 nJ Umax −Umin = 228 nJ 26.51 (a) Let x represent the coordinate of the negative charge.
Then x + 2acosθ is the coordinate of the positive charge.
The force on the negative charge is F−= −qE x ( )i. The force on the positive charge is F+ = +qE x + 2acosθ ( )i ≅qE x ( )i + q dE dx 2acosθ ( )i E θ p F+ F-The force on the dipole is altogether F = F−+ F+ = q dE dx 2acosθ ( )i = p dE dx cosθ i (b) The balloon creates field along the x −axis of keq x2 i. Thus, dE dx = −2 ( )keq x3 At x = 16.0 cm, dE dx = −2 ( ) 8.99 × 109 ( ) 2.00 × 10−6 ( ) 0.160 ( )3 = −8.78 MN C⋅m F = 6.30 × 10−9 C⋅m ( ) −8.78 × 106 N C⋅m cos0˚ i = – 55.3 i mN Chapter 26 Solutions 99 © 2000 by Harcourt, Inc. All rights reserved.
26.52 2πr E = qin ∈0 so E = λ 2πr∈0 ∆V = − E⋅dr = λ 2πr ∈0 r1 r2 ∫ r1 r2 ∫ dr = λ 2π ∈0 ln r1 r2 λmax 2π∈0 = Emaxrinner ∆V = 1.20 × 106 V m (0.100 × 10−3 m) ln 25.0 0.200 ∆Vmax = 579 V 26.53 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area ′ A << A parallel to the sheet. The side wall of the cylinder passes no flux of electric field since this surface is everywhere parallel to the field. Gauss’s law becomes E ′ A + E ′ A = Q ∈A ′ A , so E = Q 2 ∈A directed away from the positive sheet.
(b) In the space between the sheets, each creates field Q 2 ∈A away from the positive and toward the negative sheet. Together, they create a field of E = Q ∈A (c) Assume that the field is in the positive x −direction. Then, the potential of the positive plate relative to the negative plate is ∆V = − E⋅ds −plate +plate ∫ =− Q ∈A i ⋅−idx ( ) −plate +plate ∫ = + Qd ∈A (d) Capacitance is defined by: C = Q ∆V = Q Qd ∈A = ∈A d = κ∈0 A d 26.54 (a) C = 1 3.00 + 1 6.00 –1 + 1 2.00 + 1 4.00 –1 = 3.33 µF (c) Qac = Cac (∆Vac) = (2.00 µF)(90.0 V) = 180 µC Therefore, Q3 = Q6 = 180 µC Qdf = Cdf ∆Vdf ( ) = 1.33 µF ( ) 90.0 V ( ) = 120 µC 100 Chapter 26 Solutions (b) ∆V3 = Q 3 C 3 = 180 µC 3.00 µF = 60.0 V ∆V6 = Q 6 C 6 = 180 µC 6.00 µF = 30.0 V ∆V2 = Q 2 C 2 = 120 µC 2.00 µF = 60.0 V ∆V4 = Q 4 C 4 = 120 µC 4.00 µF = 30.0 V (d) UT = 1 2 C eq (∆V)2 = 1 2 (3.33 × 10–6)(90.0 V) 2 = 13.4 mJ 26.55 The electric field due to the charge on the positive wire is perpendicular to the wire, radial, and of magnitude E+ = λ 2π ∈0 r The potential difference between wires due to the presence of this charge is ∆V1 = − E⋅dr −wire +wire ∫ = − λ 2π ∈0 dr r D−d d ∫ = λ 2π ∈0 ln D −d d The presence of the linear charge density −λ on the negative wire makes an identical contribution to the potential difference between the wires. Therefore, the total potential difference is ∆V = 2 ∆V1 ( ) = λ π∈0 ln D −d d and the capacitance of this system of two wires, each of length , is The capacitance per unit length is: Chapter 26 Solutions 101 © 2000 by Harcourt, Inc. All rights reserved.
26.56 (a) We use Equation 26.11 to find the potential energy. As we will see, the potential difference ∆V changes as the dielectric is withdrawn. The initial and final energies are Ui = 1 2 Q 2 Ci and Uf = 1 2 Q 2 C f But the initial capacitance (with the dielectric) is Ci = κ C f. Therefore, Uf = 1 2 κ Q 2 Ci Since the work done by the external force in removing the dielectric equals the change in potential energy, we have W = Uf – Ui = 1 2 κ Q 2 Ci – 1 2 Q 2 Ci = 1 2 Q 2 C i (κ – 1) To express this relation in terms of potential difference ∆Vi , we substitute Q = Ci (∆Vi), and evaluate: W = 1 2 C i (∆Vi)2(κ – 1) = 1 2 (2.00 × 10–9 F)(100 V) 2(5.00 – 1.00) = 4.00 × 10–5 J The positive result confirms that the final energy of the capacitor is greater than the initial energy. The extra energy comes from the work done on the system by the external force that pulled out the dielectric.
(b) The final potential difference across the capacitor is ∆Vf = Q C f Substituting C f = Ci κ and Q = Ci (∆Vi) gives ∆Vf = κ ∆Vi = (5.00)(100 V) = 500 V Even though the capacitor is isolated and its charge remains constant, the potential difference across the plates does increase in this case.
26.57 κ = 3.00, Emax = 2.00 × 108 V/m = ∆Vmax /d For C = κ e0 A d = 0.250 × 10–6 F, A = Cd κ e0 = C ∆Vmax ( ) κ e0Emax = (0.250 × 10–6)(4000) (3.00)(8.85 × 10–12)(2.00 × 108) = 0.188 m2 26.58 (a) C 1 = κ1 e0 A/2 d ; C 2 = κ2 e0 A/2 d/2 ; C 3 = κ3 e0 A/2 d/2 1 C 2 + 1 C 3 –1 = C 2C 3 C 2 + C 3 = e0 A d κ2κ3 κ2 + κ3 C = C 1 + 1 C 2 + 1 C 3 –1 = e0 A d κ1 2 + κ2κ3 κ2 + κ3 102 Chapter 26 Solutions (b) Using the given values we find: Ctotal = 1.76 × 10–12 F = 1.76 pF Chapter 26 Solutions 103 © 2000 by Harcourt, Inc. All rights reserved.
26.59 The system may be considered to be two capacitors in series: C1 = e0A t1 and C2 = e0A t2 1 C = 1 C1 + 1 C2 = t1 + t2 e0A C = e0A t1 + t2 = e0A s – d Goal Solution A conducting slab of a thickness d and area A is inserted into the space between the plates of a parallel-plate capacitor with spacing s and surface area A, as shown in Figure P26.59. The slab is not necessarily halfway between the capacitor plates. What is the capacitance of the system?
G : It is difficult to predict an exact relationship for the capacitance of this system, but we can reason that C should increase if the distance between the slab and plates were decreased (until they touched and formed a short circuit). So maybe C ∝1/ s −d ( ). Moving the metal slab does not change the amount of charge the system can store, so the capacitance should therefore be independent of the slab position.
The slab must have zero net charge, with each face of the plate holding the same magnitude of charge as the outside plates, regardless of where the slab is between the plates.
O : If the capacitor is charged with + Q on the top plate and −Q on the bottom plate, then free charges will move across the conducting slab to neutralize the electric field inside it, with the top face of the slab carrying charge −Q and the bottom face carrying charge + Q. Then the capacitor and slab combination is electrically equivalent to two capacitors in series. (We are neglecting the slight fringing effect of the electric field near the edges of the capacitor.) Call x the upper gap, so that s −d −x is the distance between the lower two surfaces.
A : For the upper capacitor, C1 = ∈0 A x and the lower has C2 = ∈0 A s −d −x So the combination has C = 1 1 C1 + 1 C2 = 1 x ∈0 A + s −d −x ∈0 A = ∈0 A s −d L : The equivalent capacitance is inversely proportional to s −d ( ) as expected, and is also proportional to A. This result is the same as for the special case in Example 26.9 when the slab is just halfway between the plates; the only critical factor is the thickness of the slab relative to the plate spacing. 104 Chapter 26 Solutions 26.60 (a) Put charge Q on the sphere of radius a and – Q on the other sphere. Relative to V = 0 at infinity, the potential at the surface of a is Va = keQ a – keQ d and the potential of b is Vb = –keQ b + keQ d The difference in potential is Va – Vb = keQ a + keQ b – keQ d – keQ d and C = Q Va – Vb = 4π∈0 1 a ( ) + 1 b ( ) −2 d ( ) (b) As d → ∞, 1/d becomes negligible compared to 1/a. Then, C = 4π ∈0 1 a + 1 b and 1 C = 1 4π ∈0 a + 1 4π ∈0 b as for two spheres in series.
26.61 Note that the potential difference between the plates is held constant at ∆Vi by the battery. Ci = q0 ∆Vi and Cf = qf ∆Vi = q0 + q ∆Vi But Cf = κCi, so q0 + q ∆Vi = κ q0 ∆Vi Thus, κ = q0 + q q0 or κ = 1+ q q0 26.62 (a) (b) U C V = = 1 2 ( ) ∆ 2 (c) F = −dU dx = to the left Chapter 26 Solutions 105 © 2000 by Harcourt, Inc. All rights reserved.
(d) F = (2000)2(8.85 × 10−12) 0.0500 ( )(4.50 −1) 2(2.00 × 10−3) = 1.55 × 10−3 N 106 Chapter 26 Solutions 26.63 The portion of the capacitor nearly filled by metal has capacitance κ ∈0 ( x) d →∞ and stored energy Q2 2C →0. The unfilled portion has capacitance ∈0 ( −x) d. The charge on this portion is Q = ( −x)Q0 / .
(a) The stored energy is (b) F = −dU dx = F = to the right (c) (d) 26.64 Gasoline: 126000 Btu gal 1054 J Btu 1.00 gal 3.786 × 10-3 m3 1.00 m3 670 kg = 5.25 × 107 J kg Battery: 12.0 J C ( ) 100 C s ( ) 3600 s ( ) 16.0 kg = 2.70 × 105 J kg Capacitor: 1 2 0.100 F ( ) 12.0 V ( ) 2 0.100 kg = 72.0 J kg Gasoline has 194 times the specific energy content of the battery and 727 000 times that of the capacitor Chapter 26 Solutions 107 © 2000 by Harcourt, Inc. All rights reserved.
26.65 Call the unknown capacitance Cu Q = Cu(∆Vi) = (Cu + C)(∆Vf ) Cu = C(∆Vf ) (∆Vi) −(∆Vf ) = (10.0 µF)(30.0 V) (100 V −30.0 V) = 4.29 µF Goal Solution An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 V. When the charged capacitor is then connected in parallel to an uncharged 10.0-µ F capacitor, the voltage across the combination is 30.0 V. Calculate the unknown capacitance.
G : The voltage of the combination will be reduced according to the size of the added capacitance.
(Example: If the unknown capacitance were C = 10.0 µF, then ∆V1 = 50.0 V because the charge is now distributed evenly between the two capacitors.) Since the final voltage is less than half the original, we might guess that the unknown capacitor is about 5.00 µF. O : We can use the relationships for capacitors in parallel to find the unknown capacitance, along with the requirement that the charge on the unknown capacitor must be the same as the total charge on the two capacitors in parallel.
A : We name our ignorance and call the unknown capacitance Cu. The charge originally deposited on each plate, + on one, − on the other, is Q = Cu∆V = Cu 100 V ( ) Now in the new connection this same conserved charge redistributes itself between the two capacitors according to Q = Q1 + Q2. Q1 = Cu 30.0 V ( ) and Q2 = 10.0 µF ( ) 30.0 V ( ) = 300 µC We can eliminate Q and Q1 by substitution: Cu 100 V ( ) = Cu 30.0 V ( ) + 300 µC so Cu = 300 µC 70.0 V = 4.29 µF L : The calculated capacitance is close to what we expected, so our result seems reasonable. In this and other capacitance combination problems, it is important not to confuse the charge and voltage of the system with those of the individual components, especially if they have different values. Careful attention must be given to the subscripts to avoid this confusion. It is also important to not confuse the variable “ C” for capacitance with the unit of charge, “ C” for coulombs.
108 Chapter 26 Solutions 26.66 Put five 6.00 pF capacitors in series. The potential difference across any one of the capacitors will be: ∆V = ∆Vmax 5 = 1000 V 5 = 200 V and the equivalent capacitance is: 1 Ceq = 5 1 6.00 pF or Ceq = 6.00 pF 5 = 1.20 pF 26.67 When ∆Vdb = 0, ∆Vbc = ∆Vdc , and Q2 C2 = Q3 C3 Also, ∆Vba = ∆Vda or Q C Q C 1 1 4 4 = From these equations we have C C C Q Q Q Q C 2 3 4 2 1 4 3 1 = However, from the properties of capacitors in series, we have Q 1 = Q 2 and Q 3 = Q 4 Therefore, C2 = C3 C4 C1 = 9.00 12.0 (4.00 µF) = 3.00 µF 26.68 Let C = the capacitance of an individual capacitor, and Cs represent the equivalent capacitance of the group in series. While being charged in parallel, each capacitor receives charge Q = C ∆Vchg = (5.00 × 10-4 F)(800 V) = 0.400 C While being discharged in series, ∆Vdisch = Q Cs = Q C /10 = 0.400 C 5.00 × 10−5 F = 8.00 kV or 10 times the original voltage. 26.69 (a) C0 = ∈0A d = Q0 ∆V0 When the dielectric is inserted at constant voltage, Chapter 26 Solutions 109 © 2000 by Harcourt, Inc. All rights reserved. C = κC0 = Q ∆V0 ; U0 = C0(∆V0)2 2 U = C(∆V0)2 2 = κC0(∆V0)2 2 and U U0 = κ The extra energy comes from (part of the) electrical work done by the battery in separating the extra charge.
(b) Q0 = C0 ∆V0 and Q = C∆V0 = κC0 ∆V0 so Q /Q 0 = κ 110 Chapter 26 Solutions 26.70 (a) A slice of width (dx) at coordinate x in 0 ≤ x ≤ L has thickness x d / L filled with dielectric κ2, and d – x d / L is filled with the material having constant κ1. This slice has a capacitance given by 1 dC = 1 κ2 e0 (dx)W xd/L + 1 κ1e0 (dx)W d −xd/L = xd κ2 e0WL(dx) + dL −xd κ1e0WL(dx) = κ1xd + κ2dL −κ2xd κ1κ2 e0WL dx ( ) dC = κ1κ2 e0WL(dx) κ2 dL + (κ1 −κ2)xd The whole capacitor is all the slices in parallel: C = dC ∫ = κ1κ2 e0WL(dx) κ2Ld + (κ1 −κ2)xd x=0 L ∫ = κ1κ2 e0WL (κ1 −κ2)d κ2Ld + (κ1 −κ2)xd ( ) −1(κ1 −κ2)d(dx) x=0 L ∫ C= κ1κ2 e0WL (κ1 −κ2)d ln κ2Ld + (κ1 −κ2)xd [ ] 0 L = κ1κ2 e0WL (κ1 −κ2)d lnκ1Ld −ln κ2Ld [ ] = κ1κ2 e0WL (κ1 −κ2)d ln κ1 κ2 (b) To take the limit κ κ 1 2 → , write κ κ 1 2 = (1 + ) and let 0.
x x → Then C = κ2 2 (1+ x)e0WL (κ2 + κ2x −κ2)d ln (1+ x) Use the expansion of ln(1 + x) from Appendix B.5. C = κ2 2 (1 + x) ∈0WL κ2xd (x −1 2 x 2 + 1 3 x 3 . . . ) = κ2(1 + x) ∈0 WL d (1 −1 2 x + . . . ) lim x →0C = κ2 ∈0 WL d = κ∈0 A d 26.71 The vertical orientation sets up two capacitors in parallel, with equivalent capacitance Cp = ∈0 A 2 ( ) d + κ∈0 A 2 ( ) d = κ + 1 2 ∈0A d where A is the area of either plate and d is the separation of the plates. The horizontal orientation produces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric, the equivalent capacitance is 1 Cs = fd κ∈0A + 1−f ( )d ∈0A = f + κ 1−f ( ) κ d ∈0A , or Cs = κ f + κ 1−f ( ) ∈0A d Requiring that Cp = Cs gives κ + 1 2 = κ f + κ 1−f ( ) , or κ κ κ + ( ) + − [ ] = 1 1 2 f f ( ) For κ = 2.00, this yields 3 00 2 00 1 00 4 00 .
.
( .
) .
− [ ] = f , with the solution f = 2/ 3 .
Chapter 26 Solutions 111 © 2000 by Harcourt, Inc. All rights reserved.
26.72 Initially (capacitors charged in parallel), q1 = C1(∆V) = (6.00 µF)(250 V) = 1500 µC q2 = C2(∆V) = (2.00 µF)(250 V) = 500 µC After reconnection (positive plate to negative plate), ′ qtotal = q1 – q2 = 1000 µC and ∆′ V = ′ qtotal Ctotal = 1000 µC 8.00 µF = 125 V Therefore, ′ q1 = C1(∆′ V ) = (6.00 µF)(125 V) = 750 µC ′ q2 = C2(∆′ V ) = (2.00 µF)(125 V) = 250 µC 26.73 Emax occurs at the inner conductor's surface. Emax = 2keλ a from Equation 24.7. ∆V = 2keλ ln b a from Example 26.2 Emax = ∆V aln(b/ a) ∆Vmax = Emaxa ln b a = (18.0 × 106 V /m)(0.800 × 10−3 m) ln 3.00 0.800 = 19.0 kV 26.74 E = 2κλ a ; ∆V = 2κλ ln b a ∆Vmax = Emaxa ln b a dVmax da = Emax ln b a + a 1 b/ a −b a2 = 0 ln b a = 1 or b a = e1 so a = b e 112 Chapter 26 Solutions 26.75 Assume a potential difference across a and b, and notice that the potential difference across the 8.00 µF capacitor must be zero by symmetry. Then the equivalent capacitance can be determined from the following circuit: ⇒ ⇒ Cab = 3.00 µF 26.76 By symmetry, the potential difference across 3C is zero, so the circuit reduces to Ceq = (2C)(4C) 2C + 4C = 8 6 C = 4 3 C ⇒ ⇒ © 2000 by Harcourt, Inc. All rights reserved.
Chapter 27 Solutions 27.1 I = ∆Q ∆t ∆Q = I ∆t = (30.0 × 10–6 A)(40.0 s) = 1.20 × 10–3 C N = Q e = 1.20 × 10–3 C 1.60 × 10–19 C/electron = 7.50 × 1015 electrons 27.2 The atomic weight of silver = 107.9, and the volume V is V = (area)(thickness) = (700 × 10-4 m2)(0.133 × 10-3 m) = 9.31 × 10-6 m3 The mass of silver deposited is mAg = ρV = 10.5 × 103 kg /m3 ( ) 9.31× 10−6 m3 ( ) = 9.78 × 10−2 kg.
and the number of silver atoms deposited is N = (9.78 × 10-2 kg) 6.02 × 1026 atoms 107.9 kg = 5.45 × 1023 I = V R = 12.0 V 1.80 Ω = 6.67 A = 6.67 C/s ∆t = ∆Q I = Ne I = (5.45 × 1023)(1.60 × 10-19 C) 6.67 C/s = 1.31 × 104 s = 3.64 h 27.3 Q(t) = ∫ t 0 Idt = I0τ (1 – e–t/τ ) (a) Q(τ) = I0τ (1 – e–1) = (0.632)I0τ (b) Q(10τ ) = I0τ (1 – e–10) = (0.99995)I0τ (c) Q(∞) = I0τ (1 – e– ∞) = I0τ 27.4 (a) Using kee2 r2 = mv2 r , we get: v = kee2 mr = 2.19 × 106 m /s .
(b) The time for the electron to revolve around the proton once is: t = 2π r v = 2π(5.29 × 10−11 m) (2.19 × 106 m /s) = 1.52 × 10−16 s The total charge flow in this time is 1.60 × 10−19 C, so the current is 112 Chapter 27 Solutions I = 1.60 × 10−19 C 1.52 × 10−16 s = 1.05 × 10−3 A = 1.05 mA Chapter 27 Solutions 113 © 2000 by Harcourt, Inc. All rights reserved.
27.5 ω = 2π T where T is the period. I = q T = qω 2π = (8.00 × 10−9 C)(100π rad/s) 2π = 4.00 × 10−7 A = 400 nA 27.6 The period of revolution for the sphere is T = 2π ω , and the average current represented by this revolving charge is I = q T = qω 2π .
27.7 q = 4t 3 + 5t + 6 A = (2.00 cm2) 1.00 m 100 cm 2 = 2.00 × 10−4 m2 (a) I 1.00 s ( ) = dq dt t=1.00 s = 12t2 + 5 ( ) t=1.00 s = 17.0 A (b) J = I A = 17.0 A 2.00 × 10−4 m2 = 85.0 kA/m2 27.8 I = dq dt q = dq ∫ = I dt ∫ = (100 A) sin 0 1/240 s ∫ (120π t/s)dt q = −100 C 120π cos(π / 2) −cos 0 [ ] = +100 C 120π = 0.265 C 27.9 (a) J = I A = 5.00 A π(4.00 × 10−3 m)2 = 99.5 kA/m2 (b) J2 = 1 4 J1 ; I A2 = 1 4 I A1 A1 = 1 4 A2 so π(4.00 × 10−3)2 = 1 4 πr2 2 r2 = 2(4.00 × 10−3) = 8.00 × 10−3 m = 8.00 mm 114 Chapter 27 Solutions 27.10 (a) The speed of each deuteron is given by K = 1 2 mv 2 (2.00 × 106)(1.60 × 10–19 J) = 1 2 (2 × 1.67 × 10–27 kg) v2 and v = 1.38 × 107 m/s The time between deuterons passing a stationary point is t in I = q /t 10.0 × 10–6 C/s = 1.60 × 10–19 C/t or t = 1.60 × 10–14 s So the distance between them is vt = 1.38 × 107 m /s ( ) 1.60 × 10−14 s ( ) = 2.21 × 10–7 m (b) One nucleus will put its nearest neighbor at potential V = keq r = (8.99 × 109 N · m2/C 2)(1.60 × 10-19 C) 2 .21 × 10–7 m = 6.49 × 10–3 V This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect.
27.11 (a) J = I A = 8.00 × 10−6 A π 1.00 × 10−3 m ( ) 2 = 2.55 A/m2 (b) From J = nevd, we have n = J evd = 2.55 A/m2 1.60 × 10-19 C ( ) 3.00 × 108 m /s ( ) = 5.31× 1010 m−3 (c) From I = ∆Q/ ∆t, we have ∆t = ∆Q I = NAe I = 6.02 × 1023 ( ) 1.60 × 10−19 C ( ) 8.00 × 10−6 A = 1.20 × 1010 s (This is about 381 years!) 27.12 We use I = nqAvd where n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume). We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molecular weight of 27, we know that Avogadro's number of atoms, NA, has a mass of 27.0 g. Thus, the mass per atom is 27.0 g NA = 27.0 g 6.02 × 1023 = 4.49 × 10–23 g/atom Thus, n = density of aluminum mass per atom = 2.70 g/cm3 4.49 × 10–23 g/atom n = 6.02 × 1022 atoms cm3 = 6.02 × 1028 atoms m3 Therefore, vd = I nqA = 5.00 A (6.02 × 1028 m–3)(1.60 × 10–19 C)(4.00 × 10–6 m2) = 1.30 × 10–4 m/s or, vd = 0.130 mm/s Chapter 27 Solutions 115 © 2000 by Harcourt, Inc. All rights reserved.
27.13 I = ∆V R = 120 V 240 Ω= 0.500 A = 500 mA 27.14 (a) Applying its definition, we find the resistance of the rod, R = ∆V I = 15.0 V 4.00 × 10−3 A = 3750 Ω = 3.75 kΩ (b) The length of the rod is determined from Equation 27.11: R = ρ / A. Solving for and substituting numerical values for R, A, and the values of ρ given for carbon in Table 27.1, we obtain = RA ρ = (3.75 × 103 Ω)(5.00 × 10−6 m2) (3.50 × 10−5 Ω⋅m) = 536 m 27.15 ∆V = IR and R = ρl A : A = 0.600 mm2 1.00 m 1000 mm 2 = 6.00 × 10–7 m2 ∆V = Iρl A : I = ∆VA ρl = (0.900 V)(6.00 × 10–7 m2) (5.60 × 10–8 Ω · m)(1.50 m) I = 6.43 A 27.16 J = I π r2 = σE = 3.00 A π (0.0120 m)2 = σ (120 N/C) σ = 55.3(Ω · m)-1 ρ = 1 σ = 0.0181 Ω · m 27.17 (a) Given M = ρdV = ρdAl where ρd ≡mass density, we obtain: A = M ρdl Taking ρr ≡ resistivity, R = ρrl A = ρrl M ρdl = ρrρdl2 M Thus, l = MR ρrρd = 1.00 × 10−3 ( )(0.500) (1.70 × 10−8)(8.92 × 103) = 1.82 m 116 Chapter 27 Solutions (b) V = M ρd , or πr2l = M ρd Thus, r = M πρdl = 1.00 × 10−3 π(8.92 × 103)(1.82) = 1.40 × 10−4 m The diameter is twice this distance: diameter = 280 µm 27.18 (a) Suppose the rubber is 10 cm long and 1 mm in diameter. R = ρl A = 4ρl π d2 ~ 4 1013 Ω⋅m ( ) 10-1 m ( ) π 10−3 m ( ) 2 = ~1018 Ω (b) R = 4ρl π d2 ~ 4(1.7 × 10–8 Ω · m)(10–3 m) π (2 × 10–2 m)2 ~ 10–7 Ω (c) I = ∆V R ~ 10 2 V 1018 Ω ~ 10–16 A I ~ 10 2 V 10–7 Ω ~ 10 9 A 27.19 The distance between opposite faces of the cube is l = 90.0 g 10.5 g cm3 1 3 = 2.05 cm (a) R = ρl A = ρl l2 = ρ l = 1.59 × 10−8 Ω⋅m 2.05 × 10-2 m = 7.77 × 10−7 Ω= 777 nΩ (b) I = ∆V R = 1.00 × 10−5 V 7.77 × 10−7 Ω = 12.9 A n = 10.5 g cm3 107.87 g mol 6.02 × 1023 electrons mol n = 5.86 × 1022 electrons cm3 1.00 × 106 cm3 1.00 m3 = 5.86 × 1028 m3 I = nqvA and v = I nqA = 12.9 C s 5.86 × 1028 m3 ( ) 1.60 × 10−19 C ( ) 0.0205 m ( )2 = 3.28 µm/s Chapter 27 Solutions 117 © 2000 by Harcourt, Inc. All rights reserved.
27.20 Originally, R = ρl A Finally, Rf = ρ(l/ 3) 3A = ρl 9A = R 9 27.21 The total volume of material present does not change, only its shape. Thus, Aflf = Af 1.25li ( ) = Aili giving Af = Ai 1.25 The final resistance is then: Rf = ρlf Af = ρ 1.25li ( ) Ai 1.25 = 1.56 ρli Ai = 1.56R 27.22 ρAll π rAl ( ) 2 = ρCul π rCu ( ) 2 rAl rCu = ρAl ρCu = 2.82 × 10−8 1.70 × 10−8 = 1.29 27.23 J = σE so σ = J E = 6.00 × 10−13 A/m2 100 V /m = 6.00 × 10-15 (Ω · m)-1 27.24 R = ρ1l1 A1 + ρ2l2 A2 = (ρ1l1 + ρ2l2)/ d2 R = (4.00 × 10−3 Ω⋅m)(0.250 m)+(6.00 × 10−3 Ω⋅m)(0.400 m) (3.00 × 10−3 m)2 = 378 Ω 27.25 ρ = m nq2τ so τ = m ρnq2 = 9.11 × 10–31 (1.70 × 10–8)(8.49 × 1028)(1.60 × 1019)2 = 2.47 × 10–14 s vd = qE m τ so 7.84 × 10–4 = (1.60 × 10–19)E(2.47 × 10–14) 9.11 × 10–31 Therefore E = 0.181 V/m 118 Chapter 27 Solutions Goal Solution If the drift velocity of free electrons in a copper wire is 7.84 × 10−4 m /s, what is the electric field in the conductor? G : For electrostatic cases, we learned that the electric field inside a conductor is always zero. On the other hand, if there is a current, a non-zero electric field must be maintained by a battery or other source to make the charges flow. Therefore, we might expect the electric field to be small, but definitely not zero.
O : The drift velocity of the electrons can be used to find the current density, which can be used with Ohm’s law to find the electric field inside the conductor.
A : We first need the electron density in copper, which from Example 27.1 is n = 8.49 × 1028 e- /m3. The current density in this wire is then J = nqvd = (8.49 × 1028 e−/m3)(1.60 × 10−19 C/e-)(7.84 × 10−4 m /s) = 1.06 × 107 A/m2 Ohm’s law can be stated as J = σE = E/ρ where ρ = 1.7 × 10−8 Ω⋅m for copper, so then E = ρJ = (1.70 × 10−8 Ω⋅m)(1.06 × 107 A/m2) = 0.181 V /m L : This electric field is certainly smaller than typical static values outside charged objects. The direction of the electric field should be along the length of the conductor, otherwise the electrons would be forced to leave the wire! The reality is that excess charges arrange themselves on the surface of the wire to create an electric field that “steers” the free electrons to flow along the length of the wire from low to high potential (opposite the direction of a positive test charge). It is also interesting to note that when the electric field is being established it travels at the speed of light; but the drift velocity of the electrons is literally at a “snail’s pace”! 27.26 (a) n is unaffected (b) J = I A ∝I so it doubles (c) J = nevd so vd doubles (d) τ = mσ nq2 is unchanged as long as σ does not change due to heating.
27.27 From Equation 27.17, τ = me nq2ρ = 9.11× 10−31 8.49 × 1028 ( ) 1.60 × 10−19 ( ) 2 1.70 × 10−8 ( ) = 2.47 × 10−14 s l = vτ = 8.60 × 105 m /s ( ) 2.47 × 10−14 s ( ) = 2.12 × 10−8 m = 21.2 nm Chapter 27 Solutions 119 © 2000 by Harcourt, Inc. All rights reserved.
27.28 At the low temperature TC we write RC = ∆V IC = R0 1+ α(TC −T0) [ ] where T0 = 20.0°C At the high temperature Th , Rh = ∆V Ih = ∆V 1 A = R0 1+ α(Th −T0) [ ] Then (∆V)/(1.00 A) (∆V)/IC = 1 + (3.90 × 10–3)(38.0) 1 + (3.90 × 10–3)(–108) and IC = (1.00 A)(1.15/0.579) = 1.98 A 27.29 R = R0 1+ α(∆T) [ ] gives 140 Ω= (19.0 Ω) 1+ (4.50 × 10−3/°C)∆T [ ] Solving, ∆T = 1.42 × 103 °C = T – 20.0 °C And, the final temperature is T = 1.44 × 103 °C 27.30 R = Rc + Rn = Rc 1+ αc(T −T0) [ ] + Rn 1+ αn(T −T0) [ ] 0 = Rcα c (T – T0) + Rnα n (T – T0) so Rc = –Rn α n α c R = –Rn α n α c + Rn Rn = R(1 – α n /α c)–1 Rc = R(1 – α c /α n)–1 Rn = 10.0 k Ω 1 – (0.400 × 10–3/C°) (– 0.500 × 10–3/C°) –1 Rn = 5.56 k Ω and Rc = 4.44 k Ω 27.31 (a) ρ = ρ0 1+ α(T −T0) [ ] = (2.82 × 10−8 Ω⋅m) 1+ 3.90 × 10−3(30.0°) [ ] = 3.15 × 10-8 Ω · m (b) J = E ρ = 0.200 V /m 3.15 × 10−8 Ω⋅m = 6.35 × 106 A/m2 (c) I = JA = π d2 4 J = π(1.00 × 10−4 m)2 4 (6.35 × 106 A/m2) = 49.9 mA (d) n = 6.02 × 1023 electrons 26.98 g 2.70 × 106 g /m3 = 6.02 × 1028 electrons/m3 vd = J ne = (6.35 × 106 A/m2) (6.02 × 1028 electrons/m3)(1.60 × 10−19 C) = 659 µm/s (e) ∆V = E = (0.200 V /m)(2.00 m) = 0.400 V 120 Chapter 27 Solutions 27.32 For aluminum, αE = 3.90 × 10–3/°C (Table 27.1) α = 24.0 × 10–6/°C (Table 19.2) R = ρl A = ρ0 1+ αE∆T ( ) 1+ α∆T ( ) A 1+ α∆T ( )2 = R0 (1 + αE ∆T) (1 + α ∆T) = (1.234 Ω) (1.39) (1.0024) = 1.71 Ω 27.33 R = R0[1 + α ∆T] R – R0 = R0α ∆T R – R0 R0 = α ∆T = (5.00 × 10-3)25.0 = 0.125 27.34 Assuming linear change of resistance with temperature, R = R0(1 + α ∆T) R 77 K= 1.00 Ω ( ) 1+ 3.92 × 10−3 ( ) −216°C ( ) [ ] = 0.153 Ω 27.35 ρ = ρ0 1+ α ∆T ( ) or ∆TW = 1 αW ρW ρ0W −1 Require that ρW = 4ρ0Cu so that ∆TW = 1 4.50 × 10−3 /°C 4 1.70 × 10−8 ( ) 5.60 × 10−8 −1 = 47.6 °C Therefore, TW = 47.6 °C +T0 = 67.6°C 27.36 α = 1 R0 ∆R ∆T = 1 R0 2R0 −R0 T −T0 = 1 T −T0 so, T = 1 α + T0 and T = 1 0.400 × 10−3 C°−1 +20.0 °C so T = 2.52 × 103 °C 27.37 I = P ∆V = 600 W 120 V = 5.00 A and R = ∆V I = 120 V 5.00 A = 24.0 Ω Chapter 27 Solutions 121 © 2000 by Harcourt, Inc. All rights reserved.
27.38 P = 0.800(1500 hp)(746 W/hp) = 8.95 × 105 W P = I (∆V) 8.95 × 105 = I(2000) I = 448 A 27.39 The heat that must be added to the water is Q = mc ∆T = (1.50 kg)(4186 J/kg°C)(40.0°C) = 2.51 × 105 J Thus, the power supplied by the heater is P = W t = Q t = 2.51 × 105 J 600 s = 419 W and the resistance is R = ∆V ( )2 P = (110 V)2 419 W = 28.9 Ω 27.40 The heat that must be added to the water is Q = mc(T2 – T1) Thus, the power supplied by the heat is P = W ∆t = Q ∆t = mc T2 −T1 ( ) t and the resistance is R = ∆V ( )2 P = ∆V ( )2t mc T2 −T1 ( ) 27.41 P P0 = (∆V)2 / R (∆V0)2 / R = ∆V ∆V0 2 = 140 120 2 = 1.361 ∆% = P −P0 P0 100% ( ) = P P0 −1 100% ( ) = (1.361−1)100 = 36.1% 122 Chapter 27 Solutions Goal Solution Suppose that a voltage surge produces 140 V for a moment. By what percentage does the power output of a 120-V, 100-W light bulb increase? (Assume that its resistance does not change.) G : The voltage increases by about 20%, but since , the power will increase as the square of the voltage: or a 36.1% increase.
O : We have already found an answer to this problem by reasoning in terms of ratios, but we can also calculate the power explicitly for the bulb and compare with the original power by using Ohm’s law and the equation for electrical power. To find the power, we must first find the resistance of the bulb, which should remain relatively constant during the power surge (we can check the validity of this assumption later).
A : From , we find that The final current is, If = ∆Vf R = 140 V 144 Ω= 0.972 A The power during the surge is So the percentage increase is 136 W −100 W 100 W = 0.361 = 36.1% L : Our result tells us that this 100 - W light bulb momentarily acts like a 136 - W light bulb, which explains why it would suddenly get brighter. Some electronic devices (like computers) are sensitive to voltage surges like this, which is the reason that surge protectors are recommended to protect these devices from being damaged.
In solving this problem, we assumed that the resistance of the bulb did not change during the voltage surge, but we should check this assumption. Let us assume that the filament is made of tungsten and that its resistance will change linearly with temperature according to equation 27.21.
Let us further assume that the increased voltage lasts for a time long enough so that the filament comes to a new equilibrium temperature. The temperature change can be estimated from the power surge according to Stefan’s law (equation 20.18), assuming that all the power loss is due to radiation.
By this law, so that a 36% change in power should correspond to only about a 8% increase in temperature. A typical operating temperature of a white light bulb is about 3000 °C, so ∆T ≈0.08 3273 °C ( ) = 260 °C. Then the increased resistance would be roughly R = R0 1+ α T −T0 ( ) ( ) = 144 Ω ( ) 1+ 4.5 × 10−3 260 ( ) ( ) ≅310 Ω It appears that the resistance could change double from 144 Ω. On the other hand, if the voltage surge lasts only a very short time, the 136 W we calculated originally accurately describes the conversion of electrical into internal energy in the filament.
Chapter 27 Solutions 123 © 2000 by Harcourt, Inc. All rights reserved.
27.42 P = I (∆V) = (∆V)2 R = 500 W R = (110 V)2 (500 W) = 24.2 Ω (a) R = ρ Al so l = RA ρ = (24.2 Ω)π(2.50 × 10−4 m)2 1.50 × 10−6 Ω⋅m = 3.17 m (b) R = R0 [1 + α ∆T] = 24.2 Ω1+ (0.400 × 10−3)(1180) [ ] = 35.6 Ω P = (∆V)2 R = (110)2 35.6 = 340 W 27.43 R = ρl A = (1.50 × 10–6 Ω · m)25.0 m π (0.200 × 10–3 m)2 = 298 Ω ∆V = IR = (0.500 A)(298 Ω) = 149 V (a) E = ∆V l = 149 V 25.0 m = 5.97 V/m (b) P = (∆V)I = (149 V)(0.500 A) = 74.6 W (c) R = R0 1+ α(T −T0) [ ] = 298 Ω1+ (0.400 × 10−3 /C°)320 C° [ ] = 337 Ω I = ∆V R = (149 V) (337 Ω) = 0.443 A P = (∆V)I = (149 V)(0.443 A) = 66.1 W 27.44 (a) ∆U = q (∆V) = It (∆V) = (55.0 A · h)(12.0 V) 1 C 1 A · s 1 J 1 V · C 1 W · s 1 J = 660 W · h = 0.660 kWh (b) Cost = 0.660 kWh $0.0600 1 kWh = 3.96¢ 27.45 P = I (∆V) ∆V = IR P = (∆V)2 R = (10.0)2 120 = 0.833 W 124 Chapter 27 Solutions 27.46 The total clock power is 270 × 106 clocks ( ) 2.50 J s clock 3600 s 1 h = 2.43 × 1012 J h From e = Wout Qin , the power input to the generating plants must be: Qin t = Wout t e = 2.43 × 1012 J h 0.250 = 9.72 × 1012 J h and the rate of coal consumption is Rate = 9.72 × 1012 J h ( ) 1.00 kg coal 33.0 × 106 J = 2.95 × 105 kg coal h = 295 metric ton h 27.47 P = I ∆V ( ) = 1.70 A ( ) 110 V ( ) = 187 W Energy used in a 24-hour day = (0.187 kW)(24.0 h) = 4.49 kWh ∴ cost = 4.49 kWh $0.0600 kWh = $0.269 = 26.9¢ 27.48 P = I (∆V) = (2.00 A)(120 V) = 240 W ∆U = (0.500 kg)(4186 J/kg°C)(77.0°C) = 161 kJ = 1.61 × 105 J 240 W = 672 s 27.49 At operating temperature, (a) P = I (∆V) = (1.53 A)(120 V) = 184 W (b) Use the change in resistance to find the final operating temperature of the toaster.
R = R0(1 + α ∆T) 120 1.53 = 120 1.80 1+ (0.400 × 10−3)∆T [ ] ∆T = 441°C T = 20.0°C + 441°C = 461°C Chapter 27 Solutions 125 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A certain toaster has a heating element made of Nichrome resistance wire. When the toaster is first connected to a 120-V source of potential difference (and the wire is at a temperature of 20.0 °C), the initial current is 1.80 A. However, the current begins to decrease as the resistive element warms up. When the toaster has reached its final operating temperature, the current has dropped to 1.53 A. (a) Find the power the toaster consumes when it is at its operating temperature. (b) What is the final temperature of the heating element?
G : Most toasters are rated at about 1000 W (usually stamped on the bottom of the unit), so we might expect this one to have a similar power rating. The temperature of the heating element should be hot enough to toast bread but low enough that the nickel-chromium alloy element does not melt. (The melting point of nickel is 1455 °C, and chromium melts at 1907 °C.) O : The power can be calculated directly by multiplying the current and the voltage. The temperature can be found from the linear conductivity equation for Nichrome, with α = 0.4 × 10−3 °C-1 from Table 27.1.
A : (a) P = ∆V ( )I = 120 V ( ) 1.53 A ( ) = 184 W (b) The resistance at 20.0 °C is R0 = ∆V I = 120 V 1.80 A = 66.7 Ω At operating temperature, R = 120 V 1.53 A = 78.4 Ω Neglecting thermal expansion, R = ρl A = ρ0 1+ α T −T0 ( ) ( )l A = R0 1+ α T −T0 ( ) ( ) T = T0 + R R0 −1 α = 20.0 °C + 78.4 Ω66.7 Ω−1 0.4 × 10−3 °C-1 = 461 °C L : Although this toaster appears to use significantly less power than most, the temperature seems high enough to toast a piece of bread in a reasonable amount of time. In fact, the temperature of a typical 1000-W toaster would only be slightly higher because Stefan’s radiation law (Eq. 20.18) tells us that (assuming all power is lost through radiation) , so that the temperature might be about 700 °C.
In either case, the operating temperature is well below the melting point of the heating element.
27.50 P = (10.0 W /ft2)(10.0 ft)(15.0 ft) = 1.50 kW Energy = P t = (1.50 kW)(24.0 h) = 36.0 kWh Cost = (36.0 kWh)($0.0800/kWh) = $2.88 27.51 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy converted is P t = 400 J s ( ) 600 s d ( ) 365 d ( ) ≅9 × 107 J 1 kWh 3.6 × 106 J ≅20 kWh We suppose that electrical energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of Cost ≅20 kWh ( ) $0.100 kWh ( ) = $2 ~$1 126 Chapter 27 Solutions 27.52 (a) I = ∆V R so P = (∆V)I = (∆V)2 R = (120 V)2 25.0 W = 576 Ω and = (120 V)2 100 W = 144 Ω (b) I = = 25.0 W 120 V = 0.208 A = Q t = 1.00 C t t = 1.00 C 0.208 A = 4.80 s The charge has lower potential energy .
(c) P = 25.0 W = ∆U t = 1.00 J t t = 1.00 J 25.0 W = 0.0400 s The energy changes from electrical to heat and light .
(d) ∆U = Pt = (25.0 J/s)(86400 s/d)(30.0 d) = 64.8 × 106 J The energy company sells energy .
Cost = 64.8 × 106 J $0.0700 kWh k 1000 W · s J h 3600 s = $1.26 Cost per joule = $0.0700 kWh kWh 3.60 × 106 J = $1.94 × 10–8/J 27.53 We find the drift velocity from I = nqvd A = nqvd π r 2 vd = I nqπ r 2 = 1000 A 8.00 × 1028 m–3 (1.60 × 10–19 C)π (10–2 m)2 = 2.49 × 10–4 m/s v = x t t = x v = 200 × 103 m 2.49 × 10–4 m/s = 8.04 × 108 s = 25.5 yr 27.54 The resistance of one wire is 0.500 Ω m i (100 mi) = 50.0 Ω The whole wire is at nominal 700 kV away from ground potential, but the potential difference between its two ends is IR = (1000 A)(50.0 Ω) = 50.0 kV Then it radiates as heat power P = (∆V)I = (50.0 × 103 V)(1000 A) = 50.0 MW Chapter 27 Solutions 127 © 2000 by Harcourt, Inc. All rights reserved.
27.55 We begin with the differential equation α = 1 ρ dρ dT (a) Separating variables, ⌡ ⌠ ρ ρ0 dρ ρ = ⌡ ⌠ T T0 α dT ln ρ ρ0 = α (T – T0) and ρ = ρ0eα(T – T0) (b) From the series expansion e x ≅ 1 + x, (x << 1), ρ ≅ρ0 1+ α(T −T0) [ ] 27.56 Consider a 1.00-m length of cable. The potential difference between its ends is = 6.67 mV The resistance is R = ∆V I = 6.67 × 10–3 V 300 A = 22.2 µ Ω Then gives 1.56 cm 27.57 (m) R(Ω) ρ (Ω · m) 0.540 10.4 1.41 × 10–6 1.028 21.1 1.50 × 10–6 1.543 31.8 1.50 × 10–6 ρ – = 1.47 × 10–6 Ω · m (in agreement with tabulated value) 1.50 × 10–6 Ω · m (Table 27.1) 128 Chapter 27 Solutions 27.58 2 wires → = 100 m R = 0.108 Ω 300 m (100 m) = 0.0360 Ω (a) (∆V)home = (∆V)line – IR = 120 – (110)(0.0360) = 116 V (b) P = I (∆V) = (110 A)(116 V) = 12.8 kW (c) P wires = I 2R = (110 A)2(0.0360 Ω) = 436 W 27.59 (a) E = −dV dx i = − 0 −4.00 ( )V 0.500 −0 ( ) m = 8 00 .
i V m (b) R = ρl A = 4.00 × 10−8 Ω⋅m ( ) 0.500 m ( ) π 1.00 × 10−4 m ( ) 2 = 0.637 Ω (c) I = ∆V R = 4.00 V 0.637 Ω= 6.28 A (d) J = I A i = 6.28 A π 1.00 × 10−4 m ( ) 2 = 2.00 × 108 i A m2 = 200i MA m2 (e) ρJ = 4.00 × 10−8 Ω⋅m ( ) 2.00 × 108 i A m2 ( ) = 8.00i V m = E 27.60 (a) E = – dV(x) dx i = V L i (b) R = ρl A = 4ρL π d 2 (c) I = ∆V R = Vπ d 2 4ρL (d) J = I A i = V ρL i (e) ρ J = V L i = E Chapter 27 Solutions 129 © 2000 by Harcourt, Inc. All rights reserved.
27.61 R = R0 1+ α(T −T0) [ ] so T = T0 + 1 α R R0 −1 = T0 + 1 α I0 I −1 In this case, I = I0 10 , so T = T0 + 1 α (9) = 20° + 9 0.00450/°C = 2020 °C 27.62 R = ∆V I = 12.0 I = 6.00 (I – 3.00) thus 12.0I – 36.0 = 6.00I and I = 6.00 A Therefore, R = 12.0 V 6.00 A = 2.00 Ω 27.63 (a) = I ∆V ( ) so 667 A (b) and d = vt = 20.0 m /s ( ) 2.50 × 103 s ( ) = 50.0 km 27.64 (a) We begin with R = ρl A = ρ0 1 + α(T −T0) [ ] l0 1 + ′ α (T −T0) [ ] A0 1 + 2 ′ α (T −T0) [ ] , which reduces to R = R0 1 + α(T −T0) [ ] 1 + ′ α (T −T0) [ ] 1 + 2 ′ α (T −T0) [ ] (b) For copper: ρ0 = 1.70 × 10−8 Ω⋅m, α = 3.90 × 10−3 °C −1, and ′ α = 17.0 × 10−6 °C−1 R0 = ρ0l0 A0 = (1.70 × 10−8)(2.00) π(0.100 × 10−3)2 = 1.08 Ω The simple formula for R gives: R = 1.08 Ω ( ) 1+ (3.90 × 10−3°C−1)(100°C −20.0°C) [ ] = 1.420 Ω while the more complicated formula gives: R = 1.08 Ω ( ) 1+(3.90 × 10−3°C−1)(80.0°C) [ ] 1+(17.0 × 10−6 °C−1)(80.0°C) [ ] 1 + 2(17.0 × 10−6 °C−1)(80.0°C) [ ] = 1.418 Ω 130 Chapter 27 Solutions 27.65 Let α be the temperature coefficient at 20.0°C, and α′ be the temperature coefficient at 0 °C.
Then ρ = ρ0 1+ α T −20.0°C ( ) [ ], and ρ = ′ ρ 1+ ′ α T −0°C ( ) [ ] must both give the correct resistivity at any temperature T. That is, we must have: ρ0 1+ α(T −20.0°C) [ ] = ρ′ 1+ α′ T −0°C ( ) [ ] (1) Setting T = 0 in equation (1) yields: ′ ρ = ρ0 1−α(20.0°C) [ ], and setting T = 20.0°C in equation (1) gives: ρ0 = ′ ρ 1+ ′ α (20.0°C) [ ] Put ′ ρ from the first of these results into the second to obtain: ρ0 = ρ0 1−α(20.0°C) [ ] 1+ α′(20.0°C) [ ] Therefore 1+ α′ 20.0°C ( ) = 1 1−α 20.0°C ( ) which simplifies to α′= α 1−α (20.0 °C) [ ] From this, the temperature coefficient, based on a reference temperature of 0°C, may be computed for any material. For example, using this, Table 27.1 becomes at 0°C : Material Temp Coefficients at 0°C Silver 4.1 × 10− 3/˚C Copper 4.2 × 10− 3/˚C Gold 3.6 × 10− 3/˚C Aluminum 4.2 × 10− 3/˚C Tungsten 4.9 × 10− 3/˚C Iron 5.6 × 10− 3/˚C Platinum 4.25 × 10− 3/˚C Lead 4.2 × 10− 3/˚C Nichrome 0.4 × 10− 3/˚C Carbon −0.5 × 10− 3/˚C Germanium −24 × 10− 3/˚C Silicon −30 × 10− 3/˚C 27.66 (a) R = ρl A = ρL π rb 2 −ra 2 ( ) (b) R = 3.50 × 105 Ω⋅m ( ) 0.0400 m ( ) π 0.0120 m ( )2 −0.00500 m ( )2 [ ] = 3.74 × 107 Ω= 37.4 MΩ (c) , so R = ρ 2π L dr r ra rb ∫ = ρ 2π L ln rb ra (d) R = 3.50 × 105 Ω⋅m ( ) 2π 0.0400 m ( ) ln 1.20 0.500 = 1.22 × 106 Ω= 1.22 MΩ Chapter 27 Solutions 131 © 2000 by Harcourt, Inc. All rights reserved.
27.67 Each speaker receives 60.0 W of power. Using = I 2 R, we then have I = = 60.0 W 4.00 Ω = 3.87 A The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less .
27.68 ∆V = –E · l or dV = –E · dx ∆V = –IR = –E · l I = dq dt = E⋅l R = A ρlE⋅l = A ρ E = −σA dV dx = σA dV dx Current flows in the direction of decreasing voltage. Energy flows as heat in the direction of decreasing temperature.
27.69 R = ρ dx A = ρ dx wy ∫ ∫ where y = y1 + y2 −y1 L x R = ρ w dx y1 + y2 −y1 L x = ρL w(y2 −y1) ln y1 + y2 −y1 L x 0 L ∫ 0 L R = ρL w(y2 −y1) ln y2 y1 27.70 From the geometry of the longitudinal section of the resistor shown in the figure, we see that (b −r) y = (b −a) h From this, the radius at a distance y from the base is r = (a −b) y h + b For a disk-shaped element of volume dR = ρ dy πr 2 : R = ρ π dy (a −b)(y / h) + b [ ] 2 0 h ∫ . Using the integral formula du (au + b)2 = − 1 a(au + b) ∫ , R = ρ π h ab 132 Chapter 27 Solutions 27.71 I = I0 exp e∆V / kBT ( ) −1 [ ] and R = ∆V I with I0 = 1.00 × 10−9 A, e = 1.60 × 10−19 C, and kB = 1.38 × 10−23 J K.
The following includes a partial table of calculated values and a graph for each of the specified temperatures.
(i) For T = 280 K: ∆V V ( ) I A ( ) R Ω ( ) 0.400 0.0156 25.6 0.440 0.0818 5.38 0.480 0.429 1.12 0.520 2.25 0.232 0.560 11.8 0.0476 0.600 61.6 0.0097 (ii) For T = 300 K: ∆V V ( ) I A ( ) R Ω ( ) 0.400 0.005 77.3 0.440 0.024 18.1 0.480 0.114 4.22 0.520 0.534 0.973 0.560 2.51 0.223 0.600 11.8 0.051 (iii) For T = 320 K: ∆V V ( ) I A ( ) R Ω ( ) 0.400 0.0020 203 0.440 0.0084 52.5 0.480 0.0357 13.4 0.520 0.152 3.42 0.560 0.648 0.864 0.600 2.76 0.217 © 2000 by Harcourt, Inc. All rights reserved.
Chapter 28 Solutions 28.1 (a) P = ∆V ( )2 R becomes 20.0 W = (11.6 V)2 R so R = 6.73 Ω (b) ∆V = IR so 11.6 V = I (6.73 Ω) and I = 1.72 A ε = IR + Ir so 15.0 V = 11.6 V + (1.72 A)r r = 1.97 Ω Figure for Goal Solution Goal Solution A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is delivering 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance of the battery?
G : The internal resistance of a battery usually is less than 1 Ω, with physically larger batteries having less resistance due to the larger anode and cathode areas. The voltage of this battery drops significantly (23%), when the load resistance is added, so a sizable amount of current must be drawn from the battery. If we assume that the internal resistance is about 1 Ω, then the current must be about 3 A to give the 3.4 V drop across the battery’s internal resistance. If this is true, then the load resistance must be about R ≈12 V / 3 A = 4 Ω. O : We can find R exactly by using Joule’s law for the power delivered to the load resistor when the voltage is 11.6 V. Then we can find the internal resistance of the battery by summing the electric potential differences around the circuit.
A : (a) Combining Joule's law, P = ∆VI , and the definition of resistance, ∆V = IR, gives R = ∆V2 P = 11.6 V ( )2 20.
0 W = 6.73 Ω (b) The electromotive force of the battery must equal the voltage drops across the resistances: ε = IR + Ir , where I = ∆V R. r = ε −IR I = ε −∆V ( )R ∆V = (15.0 V −11.6 V)(6.73 Ω) 11.6 V = 1.97 Ω L : The resistance of the battery is larger than 1 Ω, but it is reasonable for an old battery or for a battery consisting of several small electric cells in series. The load resistance agrees reasonably well with our prediction, despite the fact that the battery’s internal resistance was about twice as large as we assumed. Note that in our initial guess we did not consider the power of the load resistance; however, there is not sufficient information to accurately solve this problem without this data.
Chapter 28 Solutions 135 © 2000 by Harcourt, Inc. All rights reserved.
28.2 (a) ∆Vterm = IR becomes 10.0 V = I (5.60 Ω) so I = 1.79 A (b) ∆Vterm = ε – Ir becomes 10.0 V = ε – (1.79 A)(0.200 Ω) so ε = 10.4 V 28.3 The total resistance is R = 3.00 V 0.600 A = 5.00 Ω (a) Rlamp = R – rbatteries = 5.00 Ω – 0.408 Ω = 4.59 Ω (b) Pbatteries Ptotal = (0.408 Ω)I 2 (5.00 Ω)I 2 = 0.0816 = 8.16% 28.4 (a) Here ε = I(R + r), so I = ε R + r = 12.6 V (5.00 Ω + 0.0800 Ω) = 2.48 A Then, ∆V = IR = 2.48 A ( ) 5.00 Ω ( ) = 12.4 V (b) Let I1 and I2 be the currents flowing through the battery and the headlights, respectively.
Then, I1 = I2 + 35.0 A, and ε −I1r −I2R = 0 so ε = (I2 + 35.0 A)(0.0800 Ω) + I2(5.00 Ω) = 12.6 V giving I2 = 1.93 A Thus, ∆V2 = (1.93 A)(5.00 Ω) = 9.65 V 28.5 ∆V = I 1R1 = (2.00 A)R 1 and ∆V = I 2(R 1 + R 2) = (1.60 A)(R 1 + 3.00 Ω) Therefore, (2.00 A)R 1 = (1.60 A)(R 1 + 3.00 Ω) or R 1 = 12.0 Ω 136 Chapter 28 Solutions 28.6 (a) Rp = 1 1 7.00 Ω ( ) + 1 10.0 Ω ( ) = 4.12 Ω Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω (b) ∆V = IR 34.0 V = I (17.1 Ω) I = 1.99 A for 4.00 Ω, 9.00 Ω resistors Applying ∆V = IR, (1.99 A)(4.12 Ω) = 8.18 V 8.18 V = I (7.00 Ω) so I = 1.17 A for 7.00 Ω resistor 8.18 V = I (10.0 Ω) so I = 0.818 A for 10.0 Ω resistor 28.7 If all 3 resistors are placed in parallel , 1 R = 1 500 + 2 250 = 5 500 and R = 100 Ω 28.8 For the bulb in use as intended, I = P ∆V = 75.0 W 120 V = 0.625 A and R = ∆V I = 120 V 0.625 A = 192 Ω Now, presuming the bulb resistance is unchanged, I = 120 V 193.6 Ω = 0.620 A Across the bulb is ∆V = IR = 192 Ω(0.620 A) = 119 V so its power is P = (∆V)I = 119 V(0.620 A) = 73.8 W Chapter 28 Solutions 137 © 2000 by Harcourt, Inc. All rights reserved.
28.9 If we turn the given diagram on its side, we find that it is the same as Figure (a). The 20.0-Ω and 5.00-Ω resistors are in series, so the first reduction is as shown in (b). In addition, since the 10.0-Ω, 5.00-Ω, and 25.0-Ω resistors are then in parallel, we can solve for their equivalent resistance as: Req = 1 1 10.0 Ω+ 1 5.00 Ω+ 1 25.0 Ω ( ) = 2.94 Ω This is shown in Figure (c), which in turn reduces to the circuit shown in (d).
Next, we work backwards through the diagrams applying I = ∆V/R and ∆V = IR. The 12.94-Ω resistor is connected across 25.0-V, so the current through the battery in every diagram is I = ∆V R = 25.0 V 12.94 Ω = 1.93 A In Figure (c), this 1.93 A goes through the 2.94-Ω equivalent resistor to give a potential difference of: ∆V = IR = (1.93 A)(2.94 Ω) = 5.68 V From Figure (b), we see that this potential difference is the same across Vab, the 10-Ω resistor, and the 5.00-Ω resistor.
(b) Therefore, Vab = 5.68 V (a) Since the current through the 20.0-Ω resistor is also the current through the 25.0-Ω line ab, I = Vab R ab = 5.68 V 25.0 Ω = 0.227 A = 227 mA (a) (b) (c) (d) 28.10 120 V = IReq = I ρl A1 + ρl A2 + ρl A3 + ρl A4 , or Iρl = 120 V ( ) 1 A1 + 1 A2 + 1 A3 + 1 A4 ∆V2 = Iρl A2 = 120 V ( ) A2 1 A1 + 1 A2 + 1 A3 + 1 A4 = 29.5 V 138 Chapter 28 Solutions 28.11 (a) Since all the current flowing in the circuit must pass through the series 100-Ω resistor, P = RI2 Pmax = RImax 2 so Imax = P R = 25.0 W 100 Ω= 0.500 A R eq = 100 Ω + 1 100 + 1 100 –1 Ω = 150 Ω ∆Vmax = R eq I max = 75.0 V (b) P = (∆V)I = (75.0 V)(0.500 A) = 37.5 W total power P 1 = 25.0 W P 2 = P 3 = RI 2 = (100 Ω)(0.250 A)2 = 6.25 W 28.12 Using 2.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series Parallel Mixed The resistors may be arranged in patterns: 2.00 Ω 6.00 Ω 3.00 Ω 7.00 Ω 4.00 Ω 9.00 Ω 5.00 Ω 0.923 Ω1.56 Ω 1.20 Ω 2.00 Ω 1.33 Ω 2.22 Ω 1.71 Ω 3.71 Ω 4.33 Ω 5.20 Ω 28.13 The potential difference is the same across either combination. ∆V = IR = 3I 1 1 R + 1 500 ( ) so R 1 R + 1 500 = 3 1 + R 500 = 3 and R = 1000 Ω = 1.00 kΩ 28.14 If the switch is open, I = ε /( ′ R + R) and P = ε 2 ′ R /( ′ R + R)2 If the switch is closed, I = ε /(R + ′ R / 2) and P = ε 2 ( ′ R / 2)/(R + ′ R / 2)2 Then, ε 2 ′ R ( ′ R + R)2 = ε 2 ′ R 2(R + ′ R / 2)2 2R2 + 2R ′ R + ′ R 2 / 2 = ′ R 2 + 2R ′ R + R2 The condition becomes R2 = ′ R 2 / 2 so ′ R = 2 R = 2 (1.00 ) = Ω 1.41 Ω Chapter 28 Solutions 139 © 2000 by Harcourt, Inc. All rights reserved.
28.15 Rp = 1 3.00 + 1 1.00 −1 = 0.750 Ω Rs = 2.00 + 0.750 + 4.00 ( ) Ω= 6.75 Ω Ibattery = ∆V Rs = 18.0 V 6.75 Ω= 2.67 A P = I2R: P 2 = 2.67 A ( ) 2 2.00 Ω ( ) P 2 = 14.2 W in 2.00 Ω P 4 = 2.67 A ( ) 2 4.00 Ω ( ) = 28.4 W in 4.00 Ω ∆V2 = 2.67 A ( ) 2.00 Ω ( ) = 5.33 V, ∆V4 = 2.67 A ( ) 4.00 Ω ( ) = 10.67 V ∆Vp = 18.0 V −∆V2 −∆V4 = 2.00 V = ∆V3 = ∆V1 ( ) P 3 = ∆V3 ( ) 2 R3 = 2.00 V ( ) 2 3.00 Ω = 1.33 W in 3.00 Ω P 1 = ∆V1 ( ) 2 R1 = 2.00 V ( ) 2 1.00 Ω = 4.00 W in 1.00 Ω 28.16 Denoting the two resistors as x and y, x + y = 690, and 1 150 = 1 x + 1 y 1 150 = 1 x + 1 690 – x = (690 – x) + x x(690 – x) x 2 – 690x + 103,500 = 0 x = 690 ± (690)2 – 414,000 2 x = 470 Ω y = 220 Ω 140 Chapter 28 Solutions 28.17 (a) ∆V = IR: 33.0 V = I1 11.0 Ω ( ) 33.0 V = I2 22.0 Ω ( ) I1 = 3.00 A I2 = 1.50 A P = I2R: P 1 = 3.00 A ( )2 11.0 Ω ( ) P 2 = 1.50 A ( )2 22.0 Ω ( ) P 1 = 99.0 W P 2 = 49.5 W The 11.0-Ω resistor uses more power. (b) P 1 + P 2 = 148 W P = I ∆V ( ) = 4.50 ( ) 33.0 ( ) = 148 W (c) Rs = R1 + R2 = 11.0 Ω+ 22.0 Ω= 33.0 Ω ∆V = IR: 33.0 V = I 33.0 Ω ( ), so I = 1.00 A P = I2R: P 1 = 1.00 A ( )2 11.0 Ω ( ) P 2 = 1.00 A ( )2 22.0 Ω ( ) P 1 = 11.0 W P 2 = 22.0 W The 22.0-Ω resistor uses more power. (d) P 1 + P 2 = I2 R1 + R2 ( ) = 1.00 A ( )2 33.0 Ω ( ) = 33.0 W P = I ∆V ( ) = 1.00 A ( ) 33.0 V ( ) = 33.0 W (e) The parallel configuration uses more power. 28.18 +15.0 – (7.00)I1 – (2.00)(5.00) = 0 5.00 = 7.00I1 so I1 = 0.714 A I3 = I1 + I2 = 2.00 A 0.714 + I2 = 2.00 so I2 = 1.29 A +ε – 2.00(1.29) – (5.00)(2.00) = 0 ε = 12.6 V Chapter 28 Solutions 141 © 2000 by Harcourt, Inc. All rights reserved.
28.19 We name the currents I1, I2, and I3 as shown.
From Kirchhoff's current rule, I3 = I1 + I2 Applying Kirchhoff's voltage rule to the loop containing I2 and I3 , 12.0 V – (4.00)I3 – (6.00)I2 – 4.00 V = 0 8.00 = (4.00)I3 + (6.00)I2 Applying Kirchhoff's voltage rule to the loop containing I1 and I2 , – (6.00)I2 – 4.00 V + (8.00)I1 = 0 (8.00)I1 = 4.00 + (6.00)I2 Solving the above linear systems, I1 = 846 mA, I2 = 462 mA, I3 = 1.31 A All currents flow in the directions indicated by the arrows in the circuit diagram.
28.20 The solution figure is shown to the right.
28.21 We use the results of Problem 19.
(a) By the 4.00-V battery: ∆U = (∆V)It = 4.00 V(– 0.462 A)120 s = – 222 J By the 12.0-V battery: 12.0 V (1.31 A) 120 s = 1.88 kJ (b) By the 8.00 Ω resistor: I 2 Rt = (0.846 A)2(8.00 Ω ) 120 s = 687 J By the 5.00 Ω resistor: (0.462 A)2(5.00 Ω ) 120 s = 128 J By the 1.00 Ω resistor: (0.462 A)2(1.00 Ω ) 120 s = 25.6 J By the 3.00 Ω resistor: (1.31 A)2(3.00 Ω ) 120 s = 616 J By the 1.00 Ω r e s i s t o r : (1.31 A)2(1.00 Ω ) 120 s = 205 J (c) –222 J + 1.88 kJ = 1.66 kJ from chemical to electrical. 687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from electrical to heat.
142 Chapter 28 Solutions 28.22 We name the currents I1, I2, and I3 as shown.
70.0 – 60.0 – I2 (3.00 kΩ) – I1 (2.00 kΩ) = 0 80.0 – I3 (4.00 kΩ) – 60.0 – I2 (3.00 kΩ) = 0 I2 = I1 + I3 (a) Substituting for I2 and solving the resulting simultaneous equations yields I1 = 0.385 mA (through R1) I3 = 2.69 mA (through R3) I2 = 3.08 mA (through R2) (b) ∆Vcf = – 60.0 V – (3.08 mA)(3.00 kΩ) = – 69.2 V Point c is at higher potential. 28.23 Label the currents in the branches as shown in the first figure.
Reduce the circuit by combining the two parallel resistors as shown in the second figure.
Apply Kirchhoff’s loop rule to both loops in Figure (b) to obtain: 2.71R ( )I1 + 1.71R ( )I2 = 250 and 1.71R ( )I1 + 3.71R ( )I2 = 500 With R = 1000 Ω, simultaneous solution of these equations yields: I1 = 10.0 mA and I2 = 130.0 mA From Figure (b), Vc −Va = I1 + I2 ( ) 1.71R ( ) = 240 V Thus, from Figure (a), I4 = Vc −Va 4R = 240 V 4000 Ω= 60.0 mA (a) (b) Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives: I = I4 −I1 = 60.0 mA −10.0 mA = +50.0 mA, or I = 50.0 mA flowing from point a to point e .
Chapter 28 Solutions 143 © 2000 by Harcourt, Inc. All rights reserved.
28.24 Name the currents as shown in the figure to the right. Then w + x + z = y. Loop equations are – 200w – 40.0 + 80.0x = 0 – 80.0x + 40.0 + 360 – 20.0y = 0 + 360 – 20.0y – 70.0z + 80.0 = 0 Eliminate y by substitution. x w x w z w x z = 2.50 + 0.500 400 100 20.0 20.0 = 0 440 20.0 20.0 90.0 = 0 − − − − − − Eliminate x : 350 270 20.0 = 0 430 70.0 90.0 = 0 − − − − w z w z Eliminate z = 17.5 – 13.5w to obtain 430 70.0 1575 +1215 = 0 − − w w w = 70.0/70.0 = 1.00 A upward in 200 Ω Now z = 4.00 A upward in 70.0 Ω x = 3.00 A upward in 80.0 Ω y = 8.00 A downward in 20.0 Ω and for the 200 Ω, ∆V = IR = (1.00 A)(200 Ω) = 200 V 28.25 Using Kirchhoff’s rules, 12.0 −0.0100 ( )I1 −0.0600 ( )I3 = 0 10.0 + 1.00 ( )I2 −0.0600 ( )I3 = 0 and I1 = I2 + I3 12.0 −0.0100 ( )I2 −0.0700 ( )I3 = 0 10.0 + 1.00 ( )I2 −0.0600 ( )I3 = 0 Solving simultaneously, I2 = 0.283 A downward in the dead battery, and I3 = 171 A downward in the starter.
144 Chapter 28 Solutions 28.26 Vab = 1.00 ( )I1 + 1.00 ( ) I1 −I2 ( ) Vab = 1.00 ( )I1 + 1.00 ( )I2 + 5.00 ( ) I −I1 + I2 ( ) Vab = 3.00 ( ) I −I1 ( ) + 5.00 ( ) I −I1 + I2 ( ) Let I = 1.00 A, I1 = x, and I2 = y Then, the three equations become: Vab = 2.00x −y, or y = 2.00x −Vab Vab = −4.00x + 6.00y + 5.00 and Vab = 8.00 −8.00x + 5.00y Substituting the first into the last two gives: 7.00Vab = 8.00x + 5.00 and 6.00Vab = 2.00x + 8.00 Solving these simultaneously yields Vab = 27 17 V Then, Rab = Vab I = 27 17 V 1.00 A or Rab = 27 17 Ω 28.27 We name the currents I1, I2, and I3 as shown.
(a) I1 = I2 + I3 Counterclockwise around the top loop, 12.0 V – (2.00 Ω)I3 – (4.00 Ω)I1 = 0 Traversing the bottom loop, 8.00 V – (6.00 Ω)I2 + (2.00 Ω)I3 = 0 I1 = 3.00 – 1 2 I3 I2 = 4 3 + 1 3 I3 and I3 = 909 mA (b) Va – (0.909 A)(2.00 Ω) = Vb Vb – Va = –1.82 V Chapter 28 Solutions 145 © 2000 by Harcourt, Inc. All rights reserved.
28.28 We apply Kirchhoff's rules to the second diagram.
50.0 – 2.00I1 – 2.00I2 = 0 (1) 20.0 – 2.00I3 + 2.00I2 = 0 (2) I1 = I2 + I3 (3) Substitute (3) into (1), and solve for I1, I2, and I3 I1 = 20.0 A; I2 = 5.00 A; I3 = 15.0 A Then apply P = I 2R to each resistor: (2.00 Ω)1 : P = I1 2(2.00 Ω) = (20.0 A)2 (2.00 Ω) = 800 W (4.00 Ω) : P = 5.00 2 A 2 (4.00 Ω) = 25.0 W (Half of I2 goes through each) (2.00 Ω)3 : P = I3 2(2.00 Ω) = (15.0 A)2(2.00 Ω) = 450 W 28.29 (a) RC = (1.00 × 106 Ω)(5.00 × 10–6 F) = 5.00 s (b) Q = Cε = (5.00 × 10–6 C)(30.0 V) = 150 µC (c) I(t) = ε R e−t/RC = 30.0 1.00 × 106 exp −10.0 (1.00 × 106)(5.00 × 10−6) = 4.06 µA 28.30 (a) I(t) = –I0e–t/RC I0 = Q RC = 5.10 × 10–6 C (1300 Ω)(2.00 × 10–9 F) = 1.96 A I(t) = – (1.96 A) exp –9.00 × 10–6 s (1300 Ω)(2.00 × 10–9 F) = – 61.6 mA (b) q(t) = Qe–t/RC = (5.10 µC) exp – 8.00 × 10–6 s (1300 Ω)(2.00 × 10–9 F) = 0.235 µC (c) The magnitude of the current is I0 = 1.96 A 146 Chapter 28 Solutions 28.31 U = 1 2 C ∆V ( )2 and ∆V = Q C Therefore, U = Q2 2C and when the charge decreases to half its original value, the stored energy is one-quarter its original value: Uf = 1 4U0 28.32 (a) τ = RC = (1.50 × 105 Ω)(10.0 × 10–6 F) = 1.50 s (b) τ = (1.00 × 105 Ω)(10.0 × 10–6 F) = 1.00 s (c) The battery carries current 10.0 V 50.0 × 103 Ω = 200 µA The 100 kΩ carries current of magnitude I = I0e–t/RC = 10.0 V 100 × 103 Ω e–t/ 1.00 s So the switch carries downward current 200 µA + (100 µA)e–t / 1.00 s 28.33 (a) Call the potential at the left junction VL and at the right VR. After a "long" time, the capacitor is fully charged.
VL = 8.00 V because of voltage divider: IL = 10.0 V 5.00 Ω = 2.00 A VL = 10.0 V – (2.00 A)(1.00 Ω) = 8.00 V Likewise, VR = 2.00 Ω 2.00 Ω + 8.00 Ω 10.0 V = 2.00 V or IR = 10.0 V 10.0 Ω = 1.00 A VR = (10.0 V) – (8.00 Ω)(1.00 A) = 2.00 V Therefore, ∆V = VL – VR = 8.00 – 2.00 = 6.00 V (b) Redraw the circuit R = 1 1/9.00 Ω ( ) + 1/6.00 Ω ( ) = 3.60 Ω RC = 3.60 × 10–6 s and e–t/RC = 1 10 so t = RC ln 10 = 8.29 µs Chapter 28 Solutions 147 © 2000 by Harcourt, Inc. All rights reserved.
28.34 (a) τ = RC = 4.00 × 106 Ω ( ) 3.00 × 10−6 F ( ) = 12.0 s (b) I = ε R e−t/RC = 12.0 4.00 × 106 e−t/12.0 s q = Cε 1−e−t/RC [ ] = 3.00 × 10−6 12.0 ( ) 1 −e−t/12.0 [ ) q = 36.0 µC 1 −e−t/12.0 [ ] I = 3.00 µAe−t/12.0 28.35 ∆V0 = Q C Then, if q(t) = Qe−t/RC ∆V(t) = ∆V0e−t/RC ∆V t ( ) ∆V0 = e−t/RC Therefore 1 2 = exp − 4.00 R 3.60 × 10−6 ( ) ln 1 2 = − 4.00 R 3.60 × 10−6 ( ) R = 1.60 MΩ 28.36 ∆V0 = Q C Then, if q(t) = Qe−t RC ∆V(t) = ∆V0 ( )e−t RC and ∆V(t) ∆V0 ( ) = e−t RC When ∆V(t) = 1 2 ∆V0 ( ), then e−t RC = 1 2 −t RC = ln 1 2 = −ln2 Thus, R = t C ln2 ( ) 148 Chapter 28 Solutions 28.37 q(t) = Q 1−e−t/RC [ ] so q(t) Q = 1−e−t/RC 0.600 = 1−e−0.900/RC or e−0.900/RC = 1−0.600 = 0.400 − = 0 900 0 400 .
ln( .
) RC thus RC = − = 0 900 0 400 .
ln( .
) 0.982 s 28.38 Applying Kirchhoff’s loop rule, −Ig 75.0 Ω ( ) + I −Ig ( )Rp = 0 Therefore, if I = 1.00 A when Ig = 1.50 mA, Rp = Ig 75.0 Ω ( ) I −Ig ( ) = 1.50 × 10−3 A ( ) 75.0 Ω ( ) 1.00 A −1.50 × 10−3 A = 0.113 Ω 28.39 Series Resistor → Voltmeter ∆V = IR: 25.0 = 1.50 × 10-3(Rs + 75.0) Solving, Rs = 16.6 kΩ Figure for Goal Solution Goal Solution The galvanometer described in the preceding problem can be used to measure voltages. In this case a large resistor is wired in series with the galvanometer in a way similar to that shown in Figure P28.24b This arrangement, in effect, limits the current that flows through the galvanometer when large voltages are applied. Most of the potential drop occurs across the resistor placed in series. Calculate the value of the resistor that enables the galvanometer to measure an applied voltage of 25.0 V at full-scale deflection.
G : The problem states that the value of the resistor must be “large” in order to limit the current through the galvanometer, so we should expect a resistance of kΩ to MΩ. O : The unknown resistance can be found by applying the definition of resistance to the portion of the circuit shown in Figure 28.24b.
A : ∆Vab = 25.0 V; From Problem 38, I = 1.50 mA and Rg = 75.0 Ω. For the two resistors in series, Req = Rs + Rg so the definition of resistance gives us: ∆Vab = I(Rs + Rg) Therefore, Rs = ∆Vab I −Rg = 25.0 V 1.50 × 10−3A −75.0 Ω= 16.6 kΩ L : The resistance is relatively large, as expected. It is important to note that some caution would be necessary if this arrangement were used to measure the voltage across a circuit with a comparable resistance. For example, if the circuit resistance was 17 kΩ, the voltmeter in this problem would cause a measurement inaccuracy of about 50%, because the meter would divert about half the current that normally would go through the resistor being measured. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.
Chapter 28 Solutions 149 © 2000 by Harcourt, Inc. All rights reserved.
28.40 We will use the values required for the 1.00-V voltmeter to obtain the internal resistance of the galvanometer.
∆V = Ig (R + rg) Solve for rg : rg = ∆V Ig – R = 1.00 V 1.00 × 10-3 A – 900 Ω = 100 Ω We then obtain the series resistance required for the 50.0-V voltmeter: R = V Ig – rg = 50.0 V 1.00 × 10-3 A – 100 Ω = 49.9 kΩ 28.41 ∆V = Igrg = I −Ig ( )Rp, or Rp = Igrg I −Ig ( ) = Ig 60.0 Ω ( ) I −Ig ( ) Therefore, to have I = 0.100 A = 100 mA when Ig = 0.500 mA: Rp = 0.500 mA ( ) 60.0 Ω ( ) 99.5 mA = 0.302 Ω Figure for Goal Solution Goal Solution Assume that a galvanometer has an internal resistance of 60.0 Ω and requires a current of 0.500 mA to produce full-scale deflection. What resistance must be connected in parallel with the galvanometer if the combination is to serve as an ammeter that has a full-scale deflection for a current of 0.100 A? G : An ammeter reads the flow of current in a portion of a circuit; therefore it must have a low resistance so that it does not significantly alter the current that would exist without the meter. Therefore, the resistance required is probably less than 1 Ω.
O : From the values given for a full-scale reading, we can find the voltage across and the current through the shunt (parallel) resistor, and the resistance value can then be found from the definition of resistance.
A : The voltage across the galvanometer must be the same as the voltage across the shunt resistor in parallel, so when the ammeter reads full scale, ∆V = 0.500 mA ( ) 60.0 Ω ( ) = 30.0 mV Through the shunt resistor, I = 100 mA −0.500 mA = 99.5 mA Therefore, R = ∆V I = 30.0 mV 99.5 mA = 0.302 Ω L : The shunt resistance is less than 1 Ω as expected. It is important to note that some caution would be necessary if this meter were used in a circuit that had a low resistance. For example, if the circuit resistance was 3 Ω, adding the ammeter to the circuit would reduce the current by about 10%, so the current displayed by the meter would be lower than without the meter. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.
150 Chapter 28 Solutions 28.42 Rx = R2R3 R1 = R2R3 2.50R2 = 1000 Ω 2.50 = 400 Ω 28.43 Using Kirchhoff’s rules with Rg << 1, −21.0 Ω ( )I1 + 14.0 Ω ( )I2 = 0, so I1 = 2 3 I2 70.0 −21.0I1 −7.00 I1 + Ig ( ) = 0, and 70.0 −14.0I2 −7.00 I2 −Ig ( ) = 0 The last two equations simplify to 10.0 −4.00 2 3 I2 ( ) = Ig, and 10.0 −3.00I2 = −Ig Solving simultaneously yields: Ig = 0.588 A + I1 21.0 Ω I2 14.0 Ω 7.00 Ω 7.00 Ω I2 - Ig Ig I1 + Ig G 70.0 V 28.44 R = ρL A and Ri = ρLi Ai But, V = AL = AiLi , so R = ρL2 V and Ri = ρLi 2 V Therefore, R = ρ Li + ∆L ( ) 2 V = ρLi 1+ ∆L Li ( ) [ ] 2 V = Ri 1+ α [ ]2 where α ≡∆L L This may be written as: R = Ri (1 + 2α + α 2) 28.45 εx Rs = εs Rs ; εx = εsRx Rs = 48.0 Ω 36.0 Ω (1.0186 V) = 1.36 V Chapter 28 Solutions 151 © 2000 by Harcourt, Inc. All rights reserved.
28.46 (a) In Figure (a), the emf sees an equivalent resistance of 200.00 Ω. I = 6.000 0 V 200.00 Ω= 0.030 000 A 6.0000 V 20.000 Ω 180.00 Ω 180.00 Ω 180.00 Ω A V A V (a) (b) (c) 20.000 Ω 20.000 Ω The terminal potential difference is ∆V = IR = 0.030 000 A ( ) 180.00 Ω ( ) = 5.400 0 V (b) In Figure (b), Req = 1 180.00 Ω+ 1 20 000 Ω −1 = 178.39 Ω The equivalent resistance across the emf is 178.39 Ω+ 0.500 00 Ω+ 20.000 Ω= 198.89 Ω The ammeter reads I = ε R = 6.000 0 V 198.89 Ω= 0.030 167 A and the voltmeter reads ∆V = IR = 0.030 167 A ( ) 178.39 Ω ( ) = 5.381 6 V (c) In Figure (c), 1 180.50 Ω+ 1 20 000 Ω −1 = 178.89 Ω Therefore, the emf sends current through Rtot = 178.89 Ω+ 20.000 Ω= 198.89 Ω The current through the battery is I = 6.000 0 V 198.89 Ω= 0.030 168 A but not all of this goes through the ammeter.
The voltmeter reads ∆V = IR = 0.030 168 A ( ) 178.89 Ω ( ) = 5.396 6 V The ammeter measures current I = ∆V R = 5.396 6 V 180.50 Ω= 0.029 898 A The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings.
28.47 (a) P = I(∆V) So for the Heater, I = P ∆V = 1500 W 120 V = 12.5 A For the Toaster, I = 750 W 120 W = 6.25 A And for the Grill, I = 1000 W 120 V = 8.33 A (Grill) (b) 12.5 + 6.25 + 8.33 = 27.1 A The current draw is greater than 25.0 amps, so this would not be sufficient.
152 Chapter 28 Solutions 28.48 (a) P = I2R = I2 ρl A = (1.00 A)2(1.70 × 10−8 Ω⋅m)(16.0 ft)(0.3048 m /ft) π(0.512 × 10−3 m)2 = 0.101 W (b) P = I 2 R = 100(0.101 Ω) = 10.1 W 28.49 IAl 2 RAl = ICu 2 RCu so IAl = RCu RAl ICu = ρCu ρAl ICu = 1.70 2.82 20.0 ( ) = 0.776(20.0) = 15.5 A 28.50 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is a chunk of rubber with contact area 4 mm2 and thickness 1 mm. Its resistance is R = ρl A ≅ 1013 Ω⋅m ( ) 10−3 m ( ) 4 × 10−6 m2 ≅2 × 1015 Ω The current will be driven by 120 V through total resistance (series) 2 × 1015 Ω+ 104 Ω+ 2 × 1015 Ω≅5 × 1015 Ω It is: I = ∆V R ~ 120 V 5 × 1015 Ω ~ 10−14 A (b) The resistors form a voltage divider, with the center of your hand at potential Vh 2, where Vh is the potential of the "hot" wire. The potential difference between your finger and thumb is ∆V = IR ~ 10−14 A ( ) 104 Ω ( ) ~ 10−10 V. So the points where the rubber meets your fingers are at potentials of ~ Vh 2 + 10−10 V and ~ Vh 2 −10−10 V 28.51 The set of four batteries boosts the electric potential of each bit of charge that goes through them by 4 × 1.50 V = 6.00 V. The chemical energy they store is ∆U = q∆V = (240 C)(6.00 J/C) = 1440 J The radio draws current I = ∆V R = 6.00 V 200 Ω = 0.0300 A So, its power is P = (∆V)I = (6.00 V)(0.0300 A) = 0.180 W = 0.180 J/s Then for the time the energy lasts, we have P = E t: t = E P = 1440 J 0.180 J/s = 8.00 × 103 s We could also compute this from I = Q/t: t = Q I = 240 C 0.0300 A = 8.00 × 103 s = 2.22 h Chapter 28 Solutions 153 © 2000 by Harcourt, Inc. All rights reserved.
28.52 I = ε R + r , so P = I2R = ε 2R R + r ( )2 or R + r ( )2 = ε 2 P R Let x ≡ε 2 P , then R + r ( )2 = xR or R2 + 2r −x ( )R −r2 = 0 With r = 1.20 Ω, this becomes R2 + 2.40 −x ( )R −1.44 = 0, which has solutions of R = −2.40 −x ( ) ± 2.40 −x ( )2 −5.76 2 (a) With ε = 9 20 . V and P = 12.8 W, x = 6.61: R = + 4.21± 4.21 ( )2 −5.76 2 = 3.84 Ω or 0.375 Ω (b) For ε = 9 20 . V and P = 21.2 W, x ≡ε 2 P = 3.99 R = +1.59 ± 1.59 ( )2 −5.76 2 = 1.59 ± −3.22 2 The equation for the load resistance yields a complex number, so there is no resistance that will extract 21.2 W from this battery. The maximum power output occurs when R = r = 1.20 Ω, and that maximum is: Pmax = ε 2 4r = 17.6 W 28.53 Using Kirchhoff’s loop rule for the closed loop, +12.0 −2.00I −4.00I = 0, so I = 2.00 A Vb −Va = + 4.00 V −2.00 A ( ) 4.00 Ω ( ) −0 ( ) 10.0 Ω ( ) = −4.00 V Thus, ∆Vab = 4.00 V and point a is at the higher potential .
28.54 The potential difference across the capacitor ∆V t ( ) = ∆Vmax 1−e−t RC [ ] Using 1 Farad = 1 s Ω, 4.00 V = 10.0 V ( ) 1−e −3.00 s ( ) R 10.0 × 10−6 s Ω ( ) Therefore, 0.400 = 1.00 −e −3.00 × 105 Ω ( ) R or e −3.00 × 105 Ω ( ) R = 0.600 Taking the natural logarithm of both sides, −3.00 × 105 Ω R = ln 0.600 ( ) and R = −3.00 × 105 Ω ln 0.600 ( ) = + 5.87 × 105 Ω= 587 kΩ 154 Chapter 28 Solutions 28.55 Let the two resistances be x and y.
Then, Rs = x + y = Ps I2 = 225 W 5.00 A ( )2 = 9.00 Ω y = 9.00 Ω – x and Rp = xy x + y = Pp I2 = 50.0 W 5.00 A ( )2 = 2.00 Ω x y x y so x 9.00 Ω−x ( ) x + 9.00 Ω−x ( ) = 2.00 Ω x2 −9.00x + 18.0 = 0 Factoring the second equation, x −6.00 ( ) x −3.00 ( ) = 0 so x = 6.00 Ω or x = 3.00 Ω Then, y = 9.00 Ω−x gives y = 3.00 Ω or y = 6.00 Ω The two resistances are found to be 6.00 Ω and 3.00 Ω .
28.56 Let the two resistances be x and y.
Then, Rs = x + y = Ps I2 and Rp = xy x + y = Pp I2 .
From the first equation, y = Ps I2 −x, and the second becomes x Ps I2 −x ( ) x + Ps I2 −x ( ) = Pp I2 or x2 −Ps I2 x + Ps Pp I4 = 0.
Using the quadratic formula, x = Ps ± Ps 2 −4Ps Pp 2I2 .
Then, y = Ps I2 −x gives y = Ps > Ps 2 −4Ps Pp 2I2 .
The two resistances are Ps + Ps 2 −4Ps Pp 2I2 and Ps − Ps 2 −4Ps Pp 2I2 Chapter 28 Solutions 155 © 2000 by Harcourt, Inc. All rights reserved.
28.57 The current in the simple loop circuit will be I = ε R + r (a) ∆Vter = ε – Ir = εR R + r and ∆Vter → ε as R → ∞ (b) I = ε R + r and I → ε r as R → 0 (c) P = I 2R = ε 2 R (R + r)2 dP dR = ε 2R(−2)(R + r)−3 +ε 2(R + r)−2 = −2ε 2R (R + r)3 + ε 2 (R + r)2 = 0 Then 2R = R + r and R = r Figure for Goal Solution Goal Solution A battery has an emf ε and internal resistance r. A variable resistor R is connected across the terminals of the battery. Determine the value of R such that (a) the potential difference across the terminals is a maximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is a maximum. G : If we consider the limiting cases, we can imagine that the potential across the battery will be a maximum when R = ∞ (open circuit), the current will be a maximum when R = 0 (short circuit), and the power will be a maximum when R is somewhere between these two extremes, perhaps when R = r.
O : We can use the definition of resistance to find the voltage and current as functions of R, and the power equation can be differentiated with respect to R.
A : (a) The battery has a voltage ∆Vterminal = ε −Ir = εR R + r or as R →∞, ∆Vterminal →ε (b) The circuit's current is I = ε R + r or as R →0, I →ε r (c) The power delivered is P = I2R = ε 2R (R + r)2 To maximize the power P as a function of R, we differentiate with respect to R, and require that dP/dR = 0 dP dR = ε 2R(−2)(R + r)−3 +ε 2(R + r)−2 = −2ε 2R (R + r)3 + ε 2 (R + r)2 = 0 Then 2R = R + r and R = r L : The results agree with our predictions. Making load resistance equal to the source resistance to maximize power transfer is called impedance matching.
156 Chapter 28 Solutions 28.58 (a) ε −I(ΣR) −(ε1 +ε2) = 0 40.0 V −(4.00 A) (2.00 + 0.300 + 0.300 + R)Ω [ ] −(6.00 + 6.00) V = 0; so R = 4.40 Ω (b) Inside the supply, P = I2R = 4.00 A ( )2 2.00 Ω ( ) = 32.0 W Inside both batteries together, P = I2R = 4.00 A ( )2 0.600 Ω ( ) = 9.60 W For the limiting resistor, P = 4.00 A ( )2 4.40 Ω ( ) = 70.4 W (c) P = I(ε1 +ε2) = (4.00 A) (6.00 + 6.00)V [ ] = 48.0 W 28.59 Let Rm = measured value, R = actual value, IR = current through the resistor R I = current measured by the ammeter.
(a) When using circuit (a), IRR = ∆V = 20 000(I – IR) or R = 20 000 I IR −1 But since I = ∆V Rm and IR = ∆V R , we have I IR = R Rm (a) (b) Figure for Goal solution and R = 20 000 (R – Rm) Rm (1) When R > Rm, we require (R – Rm) R ≤ 0.0500 Therefore, Rm ≥ R (1 – 0.0500) and from (1) we find R ≤ 1050 Ω (b) When using circuit (b), IRR = ∆V – IR (0.5 Ω). But since IR = ∆V Rm , Rm = (0.500 + R) (2) When Rm > R, we require (Rm – R) R ≤ 0.0500 From (2) we find R ≥ 10.0 Ω Chapter 28 Solutions 157 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in Figure P28.59.
The ammeter has a resistance of 0.500 Ω, and the voltmeter has a resistance of 20000 Ω. Within what range of actual values of R will the measured values be correct to within 5.00% if the measurement is made using (a) the circuit shown in Figure P28.59a (b) the circuit shown in Figure P28.59b?
G : An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance, so that adding the meter does not alter the current or voltage of the existing circuit. For the non-ideal meters in this problem, a low values of R will give a large voltage measurement error in circuit (b), while a large value of R will give significant current measurement error in circuit (a). We could hope that these meters yield accurate measurements in either circuit for typical resistance values of 1 Ω to 1 MΩ. O : The definition of resistance can be applied to each circuit to find the minimum and maximum current and voltage allowed within the 5.00% tolerance range.
A : (a) In Figure P28.59a, at least a little current goes through the voltmeter, so less current flows through the resistor than the ammeter reports, and the resistance computed by dividing the voltage by the inflated ammeter reading will be too small. Thus, we require that ∆V/ I = 0.950R where I is the current through the ammeter. Call IR the current through the resistor; then I −IR is the current in the voltmeter. Since the resistor and the voltmeter are in parallel, the voltage across the meter equals the voltage across the resistor. Applying the definition of resistance: ∆V = IRR = I −IR ( ) 20000 Ω ( ) so I = IR(R + 20000 Ω) 20000 Ω Our requirement is IRR IR(R + 20000 Ω) 20000 Ω ≥0.95R Solving, 20000 Ω≥0.95(R + 20000 Ω) = 0.95R +19000 Ω and R ≤1000 Ω 0.95 or R ≤1.05 kΩ (b) If R is too small, the resistance of an ammeter in series will significantly reduce the current that would otherwise flow through R. In Figure 28.59b, the voltmeter reading is I 0.500 Ω ( ) + IR, at least a little larger than the voltage across the resistor. So the resistance computed by dividing the inflated voltmeter reading by the ammeter reading will be too large. We require V I ≤1.05R so that I (0.500 Ω) + IR I ≤1.05R Thus, 0.500 Ω≤0.0500R and R ≥10.0 Ω L : The range of R values seems correct since the ammeter’s resistance should be less than 5% of the smallest R value ( 0.500 Ω≤0.05R means that R should be greater than 10 Ω), and R should be less than 5% of the voltmeter’s internal resistance ( R ≤0.05 × 20 kΩ= 1 kΩ). Only for the restricted range between 10 ohms and 1000 ohms can we indifferently use either of the connections (a) and (b) for a reasonably accurate resistance measurement. For low values of the resistance R, circuit (a) must be used. Only circuit (b) can accurately measure a large value of R.
158 Chapter 28 Solutions 28.60 The battery supplies energy at a changing rate dE dt = P = EI = E E R e−1/RC Then the total energy put out by the battery is dE ∫ = ε 2 R t=0 ∞ ∫ exp −t RC dt dE ∫ = ε 2 R (−RC) 0 ∞ ∫ exp −t RC −dt RC = −ε 2C exp −t RC 0 ∞ = −ε 2C[0 −1] = ε 2C The heating power of the resistor is dE dt = P = ∆VRI = I2R = Rε 2 R2 exp −2t RC So the total heat is dE ∫ = ε 2 R 0 ∞ ∫ exp −2t RC dt dE ∫ = ε 2 R −RC 2 exp 0 ∞ ∫ −2t RC −2dt RC = −ε 2C 2 exp −2t RC 0 ∞ = −ε 2C 2 [0 −1] = ε 2C 2 The energy finally stored in the capacitor is U = 1 2 C (∆V)2 = 1 2 C ε 2. Thus, energy is conserved: ε 2C = 1 2 ε 2C + 1 2 ε 2C and resistor and capacitor share equally in the energy from the battery.
28.61 (a) q = C ∆V ( ) 1−e−t RC [ ] q = 1.00 × 10−6 F ( ) 10.0 V ( ) = 9.93 µC (b) I = dq dt = ∆V R e−t RC I = 10.0 V 2.00 × 106 Ω e−5.00 = 3.37 × 10−8 A = 33.7 nA (c) dU dt = d dt 1 2 q2 C = q C dq dt = q C I dU dt = 9.93 × 10−6 C 1.00 × 10−6 C V 3.37 × 10−8 A ( ) = 3.34 × 10−7 W = 334 nW (d) Pbattery = IE = 3.37 × 10−8 A ( ) 10.0 V ( ) = 3.37 × 10−7 W = 337 nW Chapter 28 Solutions 159 © 2000 by Harcourt, Inc. All rights reserved.
28.62 Start at the point when the voltage has just reached 2 3 V and the switch has just closed. The voltage is 2 3 V and is decaying towards 0 V with a time constant RBC. VC(t) = 2 3 V e−t/RBC We want to know when VC(t) will reach 1 3 V . Therefore, 1 3 V = 2 3 V e−t/RBC or e−t/RBC = 1 2 or t1 = RBC ln 2 After the switch opens, the voltage is 1 3 V , increasing toward V with time constant RA + RB ( )C: VC (t) = V – 2 3 V e−t/(RA +RB )C When VC (t) = 2 3 V , 2 3 V = V −2 3 Ve−t/(RA +RB )C or e−t/(RA +RB )C = 1 2 so t2 = (RA + RB)C ln 2 and T = t1 + t2 = (RA + 2RB)C ln 2 28.63 (a) First determine the resistance of each light bulb: P = (∆V)2 R R = (∆V)2 P = (120 V)2 60.0 W = 240 Ω We obtain the equivalent resistance Req of the network of light bulbs by applying Equations 28.6 and 28.7: Req = R1 + 1 1/ R2 ( ) + 1/ R3 ( ) = 240 Ω+ 120 Ω= 360 Ω The total power dissipated in the 360 Ω is P = (∆V)2 Req = (120 V)2 360 Ω = 40.0 W (b) The current through the network is given by P = I2Req: I = P Req = 40.0 W 360 Ω= 1 3 A The potential difference across R1 is ∆V1 = IR1 = 1 3 A (240 Ω) = 80.0 V The potential difference ∆V23 across the parallel combination of R2 and R3 is ∆V23 = IR23 = 1 3 A 1 1/ 240 Ω ( ) + 1/ 240 Ω ( ) = 40.0 V 160 Chapter 28 Solutions 28.64 ∆V = IR (a) 20.0 V = (1.00 × 10-3 A)(R1 + 60.0 Ω) R1 = 1.994 × 104 Ω = 19.94 kΩ (b) 50.0 V = (1.00 × 10-3 A)(R2 + R1 + 60.0 Ω) R2 = 30.0 kΩ (c) 100 V = (1.00 × 10-3 A)(R3 + R1 + 60.0 Ω) R3 = 50.0 kΩ 28.65 Consider the circuit diagram shown, realizing that Ig = 1.00 mA. For the 25.0 mA scale: 24.0 mA ( ) R1 + R2 + R3 ( ) = 1.00 mA ( ) 25.0 Ω ( ) or R1 + R2 + R3 = 25.0 24.0 Ω (1) For the 50.0 mA scale: 49.0 mA ( ) R1 + R2 ( ) = 1.00 mA ( ) 25.0 Ω+ R3 ( ) or 49.0 R1 + R2 ( ) = 25.0 Ω+ R3 (2) For the 100 mA scale: 99.0 mA ( )R1 = 1.00 mA ( ) 25.0 Ω+ R2 + R3 ( ) or 99.0R1 = 25.0 Ω+ R2 + R3 (3) Solving (1), (2), and (3) simultaneously yields R1 = 0.260 Ω, R2 = 0.261 Ω, R3 = 0.521 Ω 28.66 Ammeter: Igr = 0.500 A −Ig ( ) 0.220 Ω ( ) or Ig r + 0.220 Ω ( ) = 0.110 V (1) Voltmeter: 2.00 V = Ig r + 2500 Ω ( ) (2) Solve (1) and (2) simultaneously to find: Ig = 0.756 mA and r = 145 Ω Chapter 28 Solutions 161 © 2000 by Harcourt, Inc. All rights reserved.
28.67 (a) After steady-state conditions have been reached, there is no DC current through the capacitor. Thus, for R3: IR3 = 0 (steady-state) For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 12-k Ω and 15-k Ω resistors in series: For R1 and R2: I(R1+R2 ) = ε R1 + R2 = 9.00 V (12.0 kΩ+ 15.0 kΩ)= 333 µA (steady-state) (b) After the transient currents have ceased, the potential difference across C is the same as the potential difference across R2(= IR2) because there is no voltage drop across R3. Therefore, the charge Q on C is Q = C (∆V)R2 = C (IR2) = (10.0 µF)(333 µA)(15.0 k Ω) = 50.0 µC (c) When the switch is opened, the branch containing R1 is no longer part of the circuit. The capacitor discharges through (R2 + R3) with a time constant of (R2 + R3)C = (15.0 k Ω + 3.00 k Ω)(10.0 µF) = 0.180 s. The initial current Ii in this discharge circuit is determined by the initial potential difference across the capacitor applied to (R2 + R3) in series: Ii = (∆V)C (R2 + R3) = IR2 (R2 + R3) = (333 µA)(15.0 kΩ) (15.0 kΩ+ 3.00 kΩ) = 278 µA Thus, when the switch is opened, the current through R2 changes instantaneously from 333 µA (downward) to 278 µA (downward) as shown in the graph. Thereafter, it decays according to IR2 = Iie−t/(R2 +R3 )C = (278 µA)e−t/(0.180 s) (for t > 0) (d) The charge q on the capacitor decays from Q i to Q i/5 according to q = Qie−t/(R2 +R3 )C Qi 5 = Qie(−t/0.180 s) 5 = et/0.180 s ln 5 = t 180 ms t = (0.180 s)(ln 5) = 290 ms (a) 162 Chapter 28 Solutions 28.68 ∆V = ε e−t RC so ln ε ∆V = 1 RC t A plot of ln ε ∆V versus t should be a straight line with slope = 1 RC . Using the given data values: t s ( ) ∆V (V) ln ε ∆V ( ) 0 6.19 0 4.87 5.55 0.109 11.1 4.93 0.228 19.4 4.34 0.355 30.8 3.72 0.509 46.6 3.09 0.695 67.3 2.47 0.919 102.2 1.83 1.219 (a) A least-square fit to this data yields the graph to the right. Σxi = 282, Σxi 2 = 1.86 × 104, Σxiyi = 244, Σyi = 4.03, N = 8 Slope = N Σxiyi ( ) −Σxi ( ) Σyi ( ) N Σxi 2 ( ) −Σxi ( ) 2 = 0.0118 Intercept = Σxi 2 ( ) Σyi ( ) −Σxi ( ) Σxiyi ( ) N Σxi 2 ( ) −Σxi ( ) 2 = 0.0882 The equation of the best fit line is: ln ε ∆V = 0.0118 ( )t + 0.0882 (b) Thus, the time constant is τ = = = = RC 1 1 0 0118 slope . 84.7 s and the capacitance is C = τ R = 84.7 s 10.0 × 106 Ω = 8.47 µF Chapter 28 Solutions 163 © 2000 by Harcourt, Inc. All rights reserved.
28.69 r r r a r r r b r r r r r r 3 junctions at the same potential another set of 3 junctions at the same potential a b i/3 i/3 i/3 i/3 i/3 i/3 i/6 i/6 i/6 i/6 i/6 i/6 28.70 (a) For the first measurement, the equivalent circuit is as shown in Figure 1. Rab = R1 = Ry + Ry = 2Ry so Ry = 1 2 R1 (1) For the second measurement, the equivalent circuit is shown in Figure 2. Thus, Rac = R2 = 1 2 Ry + Rx (2) a b c Ry Ry Rx R1 Figure 1 a c Rx Ry Ry R2 Figure 2 Substitute (1) into (2) to obtain: R2 = 1 2 1 2 R1 + Rx, or Rx = R2 −1 4 R1 (b) If R1 = 13.0 Ω and R2 = 6.00 Ω , then Rx = 2.75 Ω The antenna is inadequately grounded since this exceeds the limit of 2.00 Ω.
28.71 Since the total current passes through R3, that resistor will dissipate the most power. When that resistor is operating at its power limit of 32.0 W, the current through it is Itotal 2 = P R = 32.0 W 2.00 Ω= 16.0 A2, or Itotal = 4.00 A Half of this total current (2.00 A) flows through each of the other two resistors, so the power dissipated in each of them is: R1 R2 R3 R1 = R2 = R3 = 2.00 Ω P = 1 2 Itotal ( ) 2 R = (2.00 A)2(2.00 Ω) = 8.00 W Thus, the total power dissipated in the entire circuit is: 164 Chapter 28 Solutions Ptotal = 32.0 W + 8.00 W + 8.00 W = 48.0 W 28.72 The total resistance between points b and c is: R = ( )( ) + = 2 00 3 00 2 00 3 00 1 20 .
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. k k k k k Ω Ω Ω Ω Ω The total capacitance between points d and e is: C = 2.00 µF + 3.00 µF = 5.00 µF The potential difference between point d and e in this series RC circuit at any time is: ∆V = ε 1−e−t RC [ ] = 120.0 V ( ) 1−e−1000t 6 [ ] 20 V .00 kΩ .00 kΩ 1 = 2.00 µF 2 = 3.00 µF Therefore, the charge on each capacitor between points d and e is: q1 = C1 ∆V ( ) = 2.00 µF ( ) 120.0 V ( ) 1−e−1000t 6 [ ] = 240 µC ( ) 1−e−1000t 6 [ ] and q2 = C2 ∆V ( ) = 3.00 µF ( ) 120.0 V ( ) 1−e−1000t 6 [ ] = 360 µC ( ) 1−e−1000t 6 [ ] 28.73 (a) Req = 3R I = ε 3R Pseries = ε I = ε 2 3R (b) Req = 1 1/ R ( ) + 1/ R ( ) + 1/ R ( ) = R 3 I = 3ε R Pparallel = ε I = 3ε 2 R (c) Nine times more power is converted in the parallel connection.
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 29 Solutions 29.1 (a) up (b) out of the page, since the charge is negative.
(c) no deflection (d) into the page 29.2 At the equator, the Earth's magnetic field is horizontally north. Because an electron has negative charge, F = q v × B is opposite in direction to v × B. Figures are drawn looking down.
(a) Down × North = East, so the force is directed West (b) North × North = sin 0° = 0: Zero deflection (c) West × North = Down, so the force is directed Up (d) Southeast × North = Up, so the force is Down (a) (c) (d) 29.3 FB = q v × B; FB (–j) = –e v i × B Therefore, B = B (–k) which indicates the negative z direction 29.4 (a) FB = qvB sin θ = (1.60 × 10–19 C)(3.00 × 106 m/s)(3.00 × 10–1 T) sin 37.0° FB = 8.67 × 10–14 N (b) a = F m = 8.67 × 10–14 N 1.67 × 10–27 kg = 5.19 × 1013 m/s2 29.5 F = ma = (1.67 × 10–27 kg)(2.00 × 1013 m/s2) = 3.34 × 10–14 N = qvB sin 90° B = F qv = 3.34 × 10–14 N (1.60 × 10–19 C)(1.00 × 107 m/s) = 2.09 × 10–2 T The right-hand rule shows that B must be in the –y direction to yield a force in the +x direction when v is in the z direction.
2 Chapter 29 Solutions 29.6 First find the speed of the electron: ∆K = 1 2 mv2 = e(∆V) = ∆U v = 2e ∆V ( ) m = 2 1.60 × 10−19 C ( ) 2400 J/C ( ) 9.11× 10-31 kg ( ) = 2.90 × 107 m /s (a) FB, max = qvB = (1.60 × 10–19 C)(2.90 × 107 m/s)(1.70 T) = 7.90 × 10–12 N (b) FB, min = 0 occurs when v is either parallel to or anti-parallel to B 29.7 Gravitational force: Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10–30 N down Electric force: Fe = qE = (–1.60 × 10–19 C)100 N/C down = 1.60 × 10–17 N up Magnetic force: FB = qv × B = −1.60 × 10−19 C ( ) 6.00 × 106 m s E × 50.0 × 10−6 N ⋅s C⋅m N FB = – 4.80 × 10–17 N up = 4.80 × 10–17 N down 29.8 We suppose the magnetic force is small compared to gravity. Then its horizontal velocity component stays nearly constant. We call it v i.
From vy 2 = vyi 2 + 2ay y −yi ( ), the vertical component at impact is −2gh j. Then, FB = qv × B = Q vi − 2gh j ( ) × Bk = QvB −j ( ) −Q 2gh Bi FB = QvB vertical + Q 2gh B horizontal FB = 5.00 × 10–6 C(20.0 m/s)(0.0100 T) j + 5.00 × 10–6 C 2(9.80 m/s2)(20.0 m) (0.0100 T) i FB = (1.00 × 10–6 N) vertical + (0.990 × 10–6 N) horizontal 29.9 FB = qvB sin θ so 8.20 × 10-13 N = (1.60 × 10-19 C)(4.00 × 106 m/s)(1.70 T) sin θ sin θ = 0.754 and θ = sin-1(0.754) = 48.9° or 131° Chapter 29 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
29.10 qE = −1.60 × 10−19 C ( ) 20.0 N/C ( )k = −3.20 × 10−18 N ( )k ΣF = qE + qv × B = ma −3.20 × 10−18 N ( )k – 1.60 × 10-19 C (1.20 × 104 m/s i) × B = (9.11 × 10-31)(2.00 × 1012 m/s2)k – (3.20 × 10-18 N)k – (1.92 × 10-15 C · m/s)i × B = (1.82 × 10-18 N)k (1.92 × 10-15 C · m/s)i × B = – (5.02 × 10-18 N)k The magnetic field may have any x-component . Bz = 0 and By = –2.62 mT 29.11 FB = qv × B v × B = i j k +2 −4 +1 +1 +2 −3 = 12 −2 ( )i + 1+ 6 ( )j + 4 + 4 ( )k = 10i + 7j + 8k v × B = 102 + 72 + 82 = 14.6 T ⋅m /s FB = q v × B = 1.60 × 10−19 C ( ) 14.6 T ⋅m s ( ) = 2.34 × 10−18 N 29.12 FB = qv × B = −1.60 × 10−19 ( ) i j k 0 3.70 × 105 0 1.40 2.10 0 FB = −1.60 × 10−19 C ( ) 0 −0 ( )i + 0 −0 ( )j + 0 −1.40 T ( ) 3.70 × 105 m s ( ) ( ) k [ ] = 8.29 × 10−14 k ( ) N 29.13 FB = ILB sin θ with FB = Fg = mg mg = ILB sin θ so m L g = IB sin θ I = 2.00 A and m L = 0.500 g /cm ( ) 100 cm/m 1000 g /kg = 5.00 × 10−2 kg /m Thus (5.00 × 10–2)(9.80) = (2.00)B sin 90.0° B = 0.245 Tesla with the direction given by right-hand rule: eastward 4 Chapter 29 Solutions Goal Solution A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south.
What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?
G : Since I = 2.00 A south, B must be to the east to make F upward according to the right-hand rule for currents in a magnetic field.
The magnitude of B should be significantly greater than the earth’s magnetic field (~ 50 µT), since we do not typically see wires levitating when current flows through them.
O : The force on a current-carrying wire in a magnetic field is FB = I1× B, from which we can find B.
A : With I to the south and B to the east, the force on the wire is simply FB = I lBsin90°, which must oppose the weight of the wire, mg. So, B = FB Il = mg Il = g I m l = 9.80 m /s2 2.00 A 0.500 g cm 102 cm/m 103 g /kg = 0.245 T L : The required magnetic field is about 5000 times stronger than the earth’s magnetic field. Thus it was reasonable to ignore the earth’s magnetic field in this problem. In other situations the earth’s field can have a significant effect.
29.14 FB = I L × B = (2.40 A)(0.750 m)i × (1.60 T)k = (–2.88 j) N 29.15 (a) FB = ILB sin θ = (5.00 A)(2.80 m)(0.390 T) sin 60.0° = 4.73 N (b) FB = (5.00 A)(2.80 m)(0.390 T) sin 90.0° = 5.46 N (c) FB = (5.00 A)(2.80 m)(0.390 T) sin 120° = 4.73 N 29.16 FB L = mg L = I L × B L I = mg BL = (0.0400 kg/m)(9.80 m/s2) 3.60 T = 0.109 A The direction of I in the bar is to the right .
Chapter 29 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
29.17 The magnetic and gravitational forces must balance. Therefore, it is necessary to have FB = BIL = mg, or I = (mg/BL) = (λ g/B) [λ is the mass per unit length of the wire].
Thus, I = (1.00 × 10-3 kg/m)(9.80 m/s2) (5.00 × 10-5 T) = 196 A (if B = 50.0 µT) The required direction of the current is eastward , since East × North = Up.
29.18 For each segment, I = 5.00 A and B = 0.0200 N/ A ⋅m j Segment L FB = I L × B ( ) ab bc cd da −0.400 m j 0.400 m k – 0.400 m i + 0.400 m j 0.400 m i – 0.400 m k 0 (40.0 mN)(– i) (40.0 mN)(– k) (40.0 mN)(k + i) 29.19 The rod feels force FB = I d × B ( ) = Id k ( ) × B −j ( ) = IdB i ( ) The work-energy theorem is Ktrans + Krot ( )i + ∆E = Ktrans + Krot ( )f 0 + 0 + Fscosθ = 1 2 mv2 + 1 2 Iω 2 IdBLcos 0˚ = 1 2 mv2 + 1 2 1 2 mR2 ( ) v R 2 and IdBL = 3 4 mv2 v = 4IdBL 3m = 4(48.0 A)(0.120 m)(0.240 T)(0.450 m) 3(0.720 kg) = 1.07 m/s 29.20 The rod feels force FB = I d × B ( ) = Id k ( ) × B −j ( ) = IdB i ( ) The work-energy theorem is Ktrans + Krot ( )i + ∆E = Ktrans + Krot ( )f 0 + 0 + Fscosθ = 1 2 mv2 + 1 2 Iω 2 IdBLcos 0˚ = 1 2 mv2 + 1 2 1 2 mR2 ( ) v R 2 and v = 4IdBL 3m 6 Chapter 29 Solutions 29.21 The magnetic force on each bit of ring is I ds × B = I ds B radially inward and upward, at angle θ above the radial line. The radially inward components tend to squeeze the ring but all cancel out as forces. The upward components I ds B sin θ all add to I 2π rB sin θ up .
29.22 Take the x-axis east, the y-axis up, and the z-axis south. The field is B = 52.0 µT ( ) cos 60.0˚ −k ( ) + 52.0 µT ( ) sin 60.0˚ −j ( ) The current then has equivalent length: ′ L = 1.40 m −k ( ) + 0.850 m j ( ) FB = I ′ L × B = 0.0350 A ( ) 0.850j −1.40k ( )m × −45.0j −26.0k ( )10−6 T FB = 3.50 × 10−8 N −22.1i −63.0i ( ) = 2.98 × 10−6 N −i ( ) = 2.98 µN west 29.23 (a) 2π r = 2.00 m so r = 0.318 m µ = IA = (17.0 × 10-3 A) π(0.318)2 m2 [ ] = 5.41 mA · m2 (b) τ = µ × B so τ = (5.41 × 10-3 A · m2)(0.800 T) = 4.33 mN · m 29.24 τ = µB sin θ so 4.60 × 10-3 N · m = µ(0.250) sin 90.0° µ = 1.84 × 10–2 A · m2 = 18.4 mA · m2 29.25 τ = NBAI sin θ τ = 100 0.800 T ( ) 0.400 × 0.300 m2 ( ) 1.20 A ( )sin60° τ = 9.98 N · m Note that θ is the angle between the magnetic moment and the B field. The loop will rotate so as to align the magnetic moment with the B field.
Looking down along the y-axis, the loop will rotate in a clockwise direction. (a) (b) Chapter 29 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
29.26 (a) Let θ represent the unknown angle; L, the total length of the wire; and d, the length of one side of the square coil. Then, use the right-hand rule to find µ = NAI = L 4d d2I at angle θ with the horizontal.
At equilibrium, Στ = (µ × B) – (r × mg) = 0 ILBd 4 sin(90.0°−θ) −mgd 2 sinθ = 0 and mgd 2 sinθ = ILBd 4 cosθ θ = tan−1 ILB 2mg = tan−1 (3.40 A)(4.00 m)(0.0100 T) 2(0.100 kg)(9.80 m /s2) = 3.97° (b) τm = ILBd 4 cosθ = 1 4 (3.40 A)(4.00 m)(0.0100 T)(0.100 m) cos 3.97° = 3.39 mN · m 29.27 From τ = µ × B = IA × B, the magnitude of the torque is IAB sin 90.0° (a) Each side of the triangle is 40.0 cm/3.
Its altitude is 13.3 2 – 6.67 2 cm = 11.5 cm and its area is A = 1 2 (11.5 cm)(13.3 cm) = 7.70 × 10-3 m2 Then τ = (20.0 A)(7.70 × 10-3 m2)(0.520 N · s/C · m) = 80.1 mN · m (b) Each side of the square is 10.0 cm and its area is 100 cm2 = 10-2 m2.
τ = (20.0 A)(10-2 m2)(0.520 T) = 0.104 N· m (c) r = 0.400 m/2π = 0.0637 m A = π r 2 = 1.27 × 10-2 m2 τ = (20.0 A)(1.27 × 10-2 m2)(0.520) = 0.132 N · m (d) The circular loop experiences the largest torque.
29.28 Choose U = 0 when the dipole moment is at θ = 90.0° to the field. The field exerts torque of magnitude µBsinθ on the dipole, tending to turn the dipole moment in the direction of decreasing θ . Its energy is given by U −0 = µBsinθ dθ 90.0° θ ∫ = µB −cosθ ( ) 90.0° θ = −µBcosθ + 0 or U = – µ · B 8 Chapter 29 Solutions 29.29 (a) The field exerts torque on the needle tending to align it with the field, so the minimum energy orientation of the needle is: pointing north at 48.0 below the horizontal ° where its energy is Umin = −µBcos0˚ = −9.70 × 10−3 A ⋅m2 ( ) 55.0 × 10−6 T ( ) = −5.34 × 10−7 J It has maximum energy when pointing in the opposite direction, south at 48.0 above the horizontal ° where its energy is Umax = −µB cos 180˚ = + 9.70 × 10−3 A ⋅m2 ( ) 55.0 × 10−6 T ( ) = + 5.34 × 10−7 J (b) Umin +W = Umax: W = Umax −Umin = + 5.34 × 10−7 J −−5.34 × 10−7 J ( ) = 1.07 µJ 29.30 (a) τ = µ × B, so τ = µ × B= µB sin θ = NIAB sin θ τmax = NIABsin90.0˚ = 1 5.00 A ( ) π 0.0500 m ( )2 [ ] 3.00 × 10−3 T ( ) = 118 µN · m (b) U = −µ ⋅B, so −µB ≤U ≤+µB Since µB = NIA ( )B = 1 5.00 A ( ) π 0.0500 m ( )2 [ ] 3.00 × 10−3 T ( ) = 118 µJ, the range of the potential energy is: −118 µJ ≤U ≤+118 µJ 29.31 (a) B = 50.0 × 10-6 T; v = 6.20 × 106 m/s Direction is given by the right-hand-rule: southward FB = qvB sin θ FB = (1.60 × 10-19 C)(6.20 × 106 m/s)(50.0 × 10-6 T) sin 90.0° = 4.96 × 10-17 N (b) F = mv2 r so r = mv2 F = (1.67 × 10-27 kg)(6.20 × 106 m/s)2 4.96 × 10-17 N = 1.29 km 29.32 (a) 1 2 mv2 = q(∆V) 1 2 (3.20 × 10-26 kg) v2 = (1.60 × 10-19 C)(833 V) v = 91.3 km/s The magnetic force provides the centripetal force: qvB sin θ = mv2 r Chapter 29 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
r = mv qB sin 90.0° = (3.20 × 10-26 kg)(9.13 × 104 m/s) (1.60 × 10-19 C)(0.920 N · s/C · m) = 1.98 cm 10 Chapter 29 Solutions 29.33 For each electron, q vB sin 90.0° = mv2 r and v = eBr m The electrons have no internal structure to absorb energy, so the collision must be perfectly elastic: K = 1 2 mv1i 2 + 0 = 1 2 mv1f 2 + 1 2 mv2f 2 K = 1 2 m e2B2R1 2 m2 + 1 2 m e2B2R2 2 m2 = e2B2 2m R1 2 + R2 2 ( ) K = e(1.60 × 10−19 C)(0.0440 N ⋅s/C⋅m)2 2(9.11× 10−31 kg) (0.0100 m)2 + (0.0240 m)2 [ ] = 115 keV 29.34 We begin with qvB = mv2 R , so v = qRB m The time to complete one revolution is T = 2πR v = 2πR qRB m = 2πm qB Solving for B, B = 2πm qT = 6.56 × 10-2 T 29.35 q ∆V ( ) = 1 2 mv2 or v = 2q ∆V ( ) m Also, qvB = mv2 r so r = mv qB = m qB 2q ∆V ( ) m = 2m ∆V ( ) qB2 Therefore, rp 2 = 2mp ∆V ( ) eB2 rd 2 = 2md ∆V ( ) qdB2 = 2 2mp ( ) ∆V ( ) eB2 = 2 2mp ∆V ( ) eB2 = 2rp 2 and rα 2 = 2mα ∆V ( ) qαB2 = 2 4mp ( ) ∆V ( ) 2e ( )B2 = 2 2mp ∆V ( ) eB2 = 2rp 2 The conclusion is: rα = rd = 2 rp Chapter 29 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution 29.35 A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle, (charge + 2e, mass 4mp) are accelerated through a common potential difference ∆V . The particles enter a uniform magnetic field B with a velocity in a direction perpendicular to B. The proton moves in a circular path of radius rp. Determine the values of the radii of the circular orbits for the deuteron rd and the alpha particle rα in terms of rp.
G : In general, particles with greater speed, more mass, and less charge will have larger radii as they move in a circular path due to a constant magnetic force. Since the effects of mass and charge have opposite influences on the path radius, it is somewhat difficult to predict which particle will have the larger radius. However, since the mass and charge ratios of the three particles are all similar in magnitude within a factor of four, we should expect that the radii also fall within a similar range.
O : The radius of each particle’s path can be found by applying Newton’s second law, where the force causing the centripetal acceleration is the magnetic force: F = qv × B. The speed of the particles can be found from the kinetic energy resulting from the change in electric potential given.
A : An electric field changes the speed of each particle according to K + U ( )i = K + U ( )f . Therefore, assuming that the particles start from rest, we can write q∆V = 1 2 mv2.
The magnetic field changes their direction as described by ΣF = ma: qvBsin 90°= mv2 r thus r = mv qB = m qB 2q∆V m = 1 B 2m∆V q For the protons, rp = 1 B 2mp∆V e For the deuterons, rd = 1 B 2(2mp)∆V e = 2rp For the alpha particles, rα = 1 B 2(4mp)∆V 2e = 2rp L : Somewhat surprisingly, the radii of the deuterons and alpha particles are the same and are only 41% greater than for the protons.
29.36 (a) We begin with qvB = mv2 R , or qRB = mv. But, L = mvR = qR2B.
Therefore, R = L qB = 4.00 × 10−25 J ⋅s 1.60 × 10−19 C ( ) 1.00 × 10−3 T ( ) = 0.0500 m = 5.00 cm (b) Thus, v = L mR = 4.00 × 10−25 J ⋅s 9.11× 10−31 kg ( ) 0.0500 m ( ) = 8.78 × 106 m s 12 Chapter 29 Solutions 29.37 ω = qB m = (1.60 × 10−19 C)(5.20 T) 1.67 × 10−27 kg = 4.98 × 108 rad/s 29.38 1 2 mv2 = q(∆V) so v = 2q ∆V ( ) m r = mv qB so r = m 2q(∆V)/m qB r2 = m q ⋅2(∆V) B2 and ′ r ( )2 = ′ m ′ q ⋅2(∆V) B2 m = qB2r2 2(∆V) and ′ m ( ) = ′ q ( )B2 ′ r ( )2 2(∆V) so ′ m m = ′ q q ⋅ ′ r ( )2 r2 = 2e e 2R R 2 = 8 29.39 E = 1 2 mv2 = e(∆V) and evBsin90°= mv2 R B = mv eR = m eR 2e(∆V) m = 1 R 2m(∆V) e B = 1 5.80 × 1010 m 2(1.67 × 10−27 kg)(10.0 × 106 V) 1.60 × 10−19 C = 7.88 × 10-12 T 29.40 r = mv qB so m = rqB v = 7.94 × 10−3 m ( ) 1.60 × 10−19 C ( ) 1.80 T ( ) 4.60 × 105 m s m = 4.97 × 10−27 kg 1 u 1.66 × 10−27 kg = 2.99 u The particle is singly ionized: either a tritium ion, 1 3H+ , or a helium ion, 2 3He+ .
29.41 FB = Fe so qvB = qE where v = 2K /m . K is kinetic energy of the electrons. E = vB = 2K m B = 2 750 ( ) 1.60 × 10−19 ( ) 9.11× 10−31 1/2 0.0150 ( ) = 244 kV/m Chapter 29 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
29.42 K = 1 2 mv2 = q(∆V) so v = 2q ∆V ( ) m FB = qv × B = mv2 r r = mv qB = m q 2q(∆V)/m B = 1 B 2m(∆V) q (a) r238 = 2(238 × 1.66 × 10−27)2000 1.60 × 10−19 1 1.20 = 8.28 × 10−2 m = 8.28 cm (b) r235 = 8.23 cm r238 r235 = m238 m235 = 238.05 235.04 = 1.0064 The ratios of the orbit radius for different ions are independent of ∆V and B.
29.43 In the velocity selector: v = E B = 2500 V m 0.0350 T = 7.14 × 104 m s In the deflection chamber: r = mv qB = 2.18 × 10−26 kg ( ) 7.14 × 104 m s ( ) 1.60 × 10−19 C ( ) 0.0350 T ( ) = 0.278 m 29.44 K = 1 2 mv 2: 34.0 × 106 eV ( ) 1.60 × 10−19 J/eV ( ) = 1 2 1.67 × 10−27 kg ( )v2 v = 8.07 × 107 m/s r = mv qB = (1.67 × 10-27 kg)(8.07 × 107 m/s) (1.60 × 10-19 C)(5.20 T) = 0.162 m 29.45 (a) FB = qvB = mv2 R ω = v R = qBR mR = qB m = (1.60 × 10-19 C)(0.450 T) 1.67 × 10-27 kg = 4.31 × 107 rad/s (b) v = qBR m = (1.60 × 10-19 C)(0.450 T)(1.20 m) 1.67 × 10-27 kg = 5.17 × 107 m/s 29.46 FB = qvB = mv2 r B = m v qr = 4.80 × 10–16 kg · m/s (1.60 × 10–19 C)(1000 m) = 3.00 T 14 Chapter 29 Solutions 29.47 θ = tan-1 25.0 10.0 = 68.2° and R = 1.00 cm sin 68.2° = 1.08 cm Ignoring relativistic correction, the kinetic energy of the electrons is 1 2 mv2 = q(∆V) so v = 2q ∆V ( ) m = 1.33 × 108 m /s From the centripetal force m v 2 R = qvB, we find the magnetic field B = mv q R = (9.11× 10−31 kg)(1.33 × 108 m /s) (1.60 × 10−19 C)(1.08 × 10−2 m) = 70.1 mT 29.48 (a) RH ≡1 nq so n = 1 qRH = 1 1.60 × 10−19 C ( ) 0.840 × 10−10 m3 C ( ) = 7.44 × 1028 m−3 (b) ∆VH = IB nqt B = nqt ∆VH ( ) I = 7.44 × 1028 m−3 ( ) 1.60 × 10−19 C ( ) 0.200 × 10−3 m ( ) 15.0 × 10−6 V ( ) 20.0 A = 1.79 T 29.49 1 nq = t ∆VH ( ) IB = (35.0 × 10−6 V)(0.400 × 10-2 m) (21.0 A)(1.80 T) = 3.70 × 10-9 m3/C 29.50 Since ∆VH = IB nqt , and given that I = 50.0 A, B = 1.30 T, and t = 0.330 mm, the number of charge carriers per unit volume is n = IB e(∆VH)t = 1.28 × 1029 m-3 The number density of atoms we compute from the density: n0 = 8.92 g cm3 1 mole 63.5 g 6.02 × 1023 atoms mole 106 cm3 1 m3 = 8.46 × 1028 atom/m3 So the number of conduction electrons per atom is n n0 = 1.28 × 1029 8.46 × 1028 = 1.52 Chapter 29 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
29.51 B = nqt ∆VH ( ) I = 8.48 × 1028 m−3 ( ) 1.60 × 10−19 C ( ) 5.00 × 10−3 m ( ) 5.10 × 10−12 V ( ) 8.00 A B = 4.32 × 10−5 T = 43.2 µT Goal Solution In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.500 cm thick is positioned along an east-west direction. If a current of 8.00 A in the conductor results in a Hall voltage of 5.10 pV, what is the magnitude of the Earth's magnetic field? (Assume that n = 8.48 × 1028 electrons/m3 and that the plane of the bar is rotated to be perpendicular to the direction of B.) G : The Earth’s magnetic field is about 50 µT (see Table 29.1), so we should expect a result of that order of magnitude.
O : The magnetic field can be found from the Hall effect voltage: ∆VH = IB nqt or B = nqt∆VH I A : From the Hall voltage, B = 8.48 × 1028 e- m3 ( ) 1.60 × 10−19 C e-( ) 0.00500 m ( ) 5.10 × 10-12 V ( ) 8.00 A = 4.32 × 10−5 T = 43.2 µT L : The calculated magnetic field is slightly less than we expected but is reasonable considering that the Earth’s local magnetic field varies in both magnitude and direction.
29.52 (a) ∆VH = IB nqt so nqt I = B ∆VH = 0.0800 T 0.700 × 10−6 V = 1.14 × 105 T V Then, the unknown field is B = nqt I ∆VH ( ) B = 1.14 × 105 T V ( ) 0.330 × 10−6 V ( ) = 0.0377 T = 37.7 mT (b) nqt I = 1.14 × 105 T V so n = 1.14 × 105 T V I qt n = 1.14 × 105 T V 0.120 A 1.60 × 10−19 C ( ) 2.00 × 10−3 m ( ) = 4.29 × 1025 m−3 16 Chapter 29 Solutions 29.53 q vB sin 90° = mv2 r ∴ ω = v r = eB m = θ t (a) The time it takes the electron to complete π radians is t = θ ω = θm e B = (π rad)(9.11 × 10–31 kg) (1.60 × 10–19 C)(0.100 N · s/C · m) = 1.79 × 10–10 s (b) Since v = q Br m , Ke = 1 2 mv2 = q2B2r2 2m = e(1.60 × 10−19 C)(0.100 N ⋅s/Cm)2(2.00 × 10−2 m)2 2(9.11× 10−31 kg) = 351 keV 29.54 ∑ Fy = 0: +n – mg = 0 ∑ Fx = 0: –µkn + IBd sin 90.0° = 0 B = µkmg Id = 0.100(0.200 kg)(9.80 m/s2) (10.0 A)(0.500 m) = 39.2 mT 29.55 (a) The electric current experiences a magnetic force .
I(h × B) in the direction of L.
(b) The sodium, consisting of ions and electrons, flows along the pipe transporting no net charge. But inside the section of length L, electrons drift upward to constitute downward electric current J × (area) = JLw.
The current then feels a magnetic force I h × B = JLwhB sin 90° This force along the pipe axis will make the fluid move, exerting pressure F area = JLwhB hw = JLB 29.56 The magnetic force on each proton, FB = qv × B = qvB sin 90° downward perpendicular to velocity, supplies centripetal force, guiding it into a circular path of radius r, with qvB = mv2 r and r = mv qB We compute this radius by first finding the proton's speed: K = 1 2 mv 2 v = 2K m = 2 5.00 × 106 eV ( ) 1.60 × 10−19 J/eV ( ) 1.67 × 10−27 kg = 3.10 × 107 m /s Chapter 29 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
Now, r = mv qB = (1.67 × 10–27 kg)(3.10 × 107 m/s)(C · m) (1.60 × 10–19 C)(0.0500 N · s) = 6.46 m (b) From the figure, observe that sin ∝ = 1.00 m r = 1 m 6.46 m α = 8.90° (a) The magnitude of the proton momentum stays constant, and its final y component is – (1.67 × 10–27 kg)(3.10 × 107 m/s) sin(8.90°) = – 8.00 × 10–21 kg · m/s 29.57 (a) If B = Bx i + By j + Bz k, FB = qv × B = e vi i ( ) × Bx i + By j + Bz k ( ) = 0 + eviBy k −eviBz j Since the force actually experienced is FB = Fi j, observe that Bx could have any value , By = 0 , and Bz = −Fi evi (b) If v = −vi i, then FB = qv × B = e −vi i ( ) × Bx i + 0j −Fi evi k ( ) = –Fi j (c) If q = −e and v = vi i, then FB = qv × B = −e vi i ( ) × Bx i + 0j −Fi evi k ( ) = –Fi j Reversing either the velocity or the sign of the charge reverses the force.
29.58 A key to solving this problem is that reducing the normal force will reduce the friction force: FB = BIL or B = FB IL When the wire is just able to move, ΣFy = n + FB cosθ −mg = 0 so n = mg −FB cosθ and f = µ mg −FB cosθ ( ) Also, ΣFx = FB sinθ −f = 0 so FB sinθ = f : FB sinθ = µ mg −FB cosθ ( ) and FB = µmg sinθ + µ cosθ We minimize B by minimizing FB: dFB dθ = µmg ( ) cosθ −µ sinθ sinθ + µ cosθ ( )2 = 0 ⇒ µ sinθ = cosθ Thus, θ = tan−1 1 µ = tan−1 5.00 ( ) = 78.7° for the smallest field, and B = FB IL = µg I m L ( ) sinθ + µ cosθ Bmin = 0.200 ( ) 9.80 m s2 ( ) 1.50 A 0.100 kg m sin 78.7° + 0.200 ( ) cos 78.7° = 0.128 T Bmin = 0.128 T pointing north at an angle of 78.7˚ below the horizontal 18 Chapter 29 Solutions 29.59 (a) The net force is the Lorentz force given by F = qE + qv × B = q E + v × B ( ) F = 3.20 × 10−19 ( ) 4i −1j −2k ( ) + 2i + 3j −1k ( ) × 2i + 4j + 1k ( ) [ ]N Carrying out the indicated operations, we find: F = 3.52i −1.60j ( ) × 10−18 N (b) θ = cos−1 Fx F = cos−1 3.52 3.52 ( )2 + 1.60 ( )2 = 24.4° 29.60 r = mv qB = (1.67 × 10−27)(1.50 × 108) (1.60 × 10−19)(5.00 × 10−5) m = 3.13 × 104 m = 31.3 km No, the proton will not hit the Earth .
29.61 Let ∆x1 be the elongation due to the weight of the wire and let ∆x2 be the additional elongation of the springs when the magnetic field is turned on. Then Fmagnetic = 2k ∆x2 where k is the force constant of the spring and can be determined from k = mg/2∆x1. (The factor 2 is included in the two previous equations since there are 2 springs in parallel.) Combining these two equations, we find Fmagnetic = 2 mg 2∆x1 ∆x2 = mg∆x2 ∆x1 ; but FB = I L × B = ILB Therefore, where I = 24.0 V 12.0 Ω= 2.00 A, B = mg∆x2 IL∆x1 = (0.0100)(9.80)(3.00 × 10−3) (2.00)(0.0500)(5.00 × 10−3) = 0.588 T 29.62 Suppose the input power is 120 W = 120 V ( )I: I~ 1 A = 100 A Suppose ω = 2000 rev min 1 min 60 s 2π rad 1 rev ~200 rad s and the output power is 20 W = τω = τ 200 rad s τ ~10−1 N ⋅m Suppose the area is about 3 cm 4 cm ( ) × ( ), or A~10−3 m2 From Table 29.1, suppose that the field is B~10−1 T Then, the number of turns in the coil may be found from τ ≅NIAB: 0.1 N ⋅m ~ N 1 C s 10−3 m2 ( ) 10−1 N ⋅s Cm giving N ~103 Chapter 29 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
29.63 Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium: ΣFx = T sinθ −ILB sin 90.00 = 0 or T sinθ = ILB ΣFy = T cosθ −mg = 0, or T cosθ = mg Therefore, tanθ = ILB mg = IB m L ( )g or B = m L ( )g I tanθ B = 0.0100 kg m ( ) 9.80 m s2 ( ) 5.00 A tan 45.0˚ ( ) = 19.6 mT 29.64 Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibrium: ΣFx = T sinθ −ILBsin90.00 = 0 or T sinθ = ILB ΣFy = T cosθ −mg = 0, or T cosθ = mg tanθ = ILB mg = IB m L ( )g or B = m L ( )g I tanθ = µg I tanθ 29.65 ΣF = ma or qvB sin 90.0° = mv2 r ∴ the angular frequency for each ion is v r = ω = qB m = 2π f and ∆f = f12 −f14 = qB 2π 1 m12 − 1 m14 = (1.60 × 10−19 C)(2.40 T) 2π(1.66 × 10−27 kg / u) 1 12.0 u − 1 14.0 u ∆f = f12 – f14 = 4.38 × 105 s-1 = 438 kHz 29.66 Let vx and v⊥ be the components of the velocity of the positron parallel to and perpendicular to the direction of the magnetic field.
(a) The pitch of trajectory is the distance moved along x by the positron during each period, T (see Equation 29.15). p = vxT = (vcos 85.0°) 2πm Bq p = (5.00 × 106)(cos 85.0°)(2π)(9.11× 10−31) (0.150)(1.60 × 10−19) = 1.04 × 10-4 m (b) From Equation 29.13, r = mv⊥ Bq = mv sin 85.0° Bq r = (9.11× 10−31)(5.00 × 106)(sin 85.0°) (0.150)(1.60 × 10−19) = 1.89 × 10-4 m 20 Chapter 29 Solutions 29.67 τ = IAB where the effective current due to the orbiting electrons is I = ∆q ∆t = q T and the period of the motion is T = 2πR v The electron's speed in its orbit is found by requiring keq2 R2 = mv2 R or v = q ke mR Substituting this expression for v into the equation for T, we find T = 2π mR3 q2ke T = 2π (9.11× 10−31)(5.29 × 10−11)3 (1.60 × 10−19)2(8.99 × 109) = 1.52 × 10−16 s Therefore, τ = q T AB = 1.60 × 10−19 1.52 × 10−16 π(5.29 × 10−11)2(0.400) = 3.70 × 10-24 N · m Goal Solution Consider an electron orbiting a proton and maintained in a fixed circular path of radius R = 5.29 × 10-11 m by the Coulomb force. Treating the orbiting charge as a current loop, calculate the resulting torque when the system is in a magnetic field of 0.400 T directed perpendicular to the magnetic moment of the electron.
G : Since the mass of the electron is very small (~10-30 kg), we should expect that the torque on the orbiting charge will be very small as well, perhaps ~10-30 N⋅m.
O : The torque on a current loop that is perpendicular to a magnetic field can be found from τ = IAB sinθ . The magnetic field is given, θ = 90°, the area of the loop can be found from the radius of the circular path, and the current can be found from the centripetal acceleration that results from the Coulomb force that attracts the electron to proton.
A : The area of the loop is A = π r2 = π 5.29 × 10−11 m ( ) 2 = 8.79 × 10−21 m2.
If v is the speed of the electron, then the period of its circular motion will be T = 2πR v, and the effective current due to the orbiting electron is I = ∆Q/ ∆t = e T . Applying Newton’s second law with the Coulomb force acting as the central force gives ΣF = keq2 R2 = mv2 R so that v = q ke mR and T = 2π mR3 q2ke T = 2π (9.10 × 10−31 kg)(5.29 × 10-11 m)3 1.60 × 10-19 C ( ) 2 8.99 × 109 N ⋅m2 C2 ( ) = 1.52 × 10−16 s The torque is τ = q T AB: τ = 1.60 × 1019 C 1.52 × 10-16 s (π)(5.29 × 10−11 m)2(0.400 T) = 3.70 × 10−24 N ⋅m L : The torque is certainly small, but a million times larger than we guessed. This torque will cause the atom to precess with a frequency proportional to the applied magnetic field. A similar process on the nuclear, rather than the atomic, level leads to nuclear magnetic resonance (NMR), which is used for magnetic resonance imaging (MRI) scans employed for medical diagnostic testing (see Section 44.2).
Chapter 29 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
29.68 Use the equation for cyclotron frequency ω = qB m or m = qB ω = qB 2π f m = (1.60 × 10−19 C)(5.00 × 10−2 T) (2π)(5.00 rev/1.50 × 10−3 s) = 3.82 × 10-25 kg 29.69 (a) K = 1 2 mv2 = 6.00 MeV = 6.00 × 106 eV ( ) 1.60 × 10−19 J eV K = 9.60 × 10-13 J v = 2 9.60 × 10−13 J ( ) 1.67 × 10−27 kg = 3.39 × 107 m s x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 45˚ R 45.0˚ 45˚ v Bin = 1.00 T x θ’ FB = qvB = mv2 R so R = mv qB = 1.67 × 10−27 kg ( ) 3.39 × 107 m s ( ) 1.60 × 10−19 C ( ) 1.00 T ( ) = 0.354 m Then, from the diagram, x = 2R sin 45.0˚ = 2 0.354 m ( )sin 45.0˚ = 0.501 m (b) From the diagram, observe that θ' = 45.0° .
29.70 (a) See graph to the right. The Hall voltage is directly proportional to the magnetic field. A least-square fit to the data gives the equation of the best fitting line as: ∆VH = 1.00 × 10−4 V T ( )B (b) Comparing the equation of the line which fits the data best to ∆VH = I nqt B 0.00 20.00 40.00 60.00 80.00 100.00 120.00 0.00 0.20 0.40 0.60 0.80 1.00 1.20 B (T) ∆V H (µV) observe that: I nqt = 1.00 × 10−4 V T, or t = I nq 1.00 × 10−4 V T ( ) Then, if I = 0.200 A, q = 1.60 × 10−19 C, and n = 1.00 × 1026 m−3, the thickness of the sample is t = 0.200 A 1.00 × 1026 m−3 ( ) 1.60 × 10−19 C ( ) 1.00 × 10−4 V T ( ) = 1.25 × 10−4 m = 0.125 mm 22 Chapter 29 Solutions 29.71 (a) The magnetic force acting on ions in the blood stream will deflect positive charges toward point A and negative charges toward point B. This separation of charges produces an electric field directed from A toward B. At equilibrium, the electric force caused by this field must balance the magnetic force, so qvB = qE = q ∆V d or v = ∆V Bd = 160 × 10−6 V 0.040 0 T ( ) 3.00 × 10−3 m ( ) = 1.33 m/s (b) No . Negative ions moving in the direction of v would be deflected toward point B, giving A a higher potential than B. Positive ions moving in the direction of v would be deflected toward A, again giving A a higher potential than B. Therefore, the sign of the potential difference does not depend on whether the ions in the blood are positively or negatively charged.
29.72 When in the field, the particles follow a circular path according to qvB = mv2 r, so the radius of the path is: r = mv/ qB (a) When r = h = mv qB , that is, when v = qBh m , the particle will cross the band of field. It will move in a full semicircle of radius h, leaving the field at 2h, 0, 0 ( ) with velocity v f = −vj .
(b) When v < qBh m , the particle will move in a smaller semicircle of radius r = mv qB < h. It will leave the field at 2r, 0, 0 ( ) with velocity v f = −vj .
(c) When v > qBh m , the particle moves in a circular arc of radius r = mv qB > h, centered at r, 0, 0 ( ).
The arc subtends an angle given by θ = sin−1 h r ( ). It will leave the field at the point with coordinates r 1−cosθ ( ), h, 0 [ ] with velocity v f = vsinθ i + vcosθ j .
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 30 Solutions 30.1 B = µ0I 2R = µ0q(v/2π R) 2R = 12.5 T 30.2 We use the Biot-Savart law. For bits of wire along the straight-line sections, ds is at 0° or 180° to ~, so ds × ~= 0. Thus, only the curved section of wire contributes to B at P. Hence, ds is tangent to the arc and ~ is radially inward; so ds × ~= ds l sin 90° = ds ⊗. All points along the curve are the same distance r = 0.600 m from the field point, so B = ⌡ ⌠ all current dB = ⌡ ⌠ µ0 4π I ds × ~ r 2 = µ0 4π I r 2 ⌡ ⌠ ds = µ0 4π I r 2 s where s is the arclength of the curved wire, s = rθ = (0.600 m)30.0° 2π 360° = 0.314 m Then, B = 10–7 T · m A (3.00 A) (0.600 m)2 (0.314 m) B = 261 nT into the page 30.3 (a) B = 4µ0I 4π a cos π 4 – cos 3π 4 where a = l 2 is the distance from any side to the center.
B = 4.00 × 10–6 0.200 2 2 + 2 2 = 2 2 × 10–5 T = 28.3 µT into the paper (b) For a single circular turn with 4 l = 2π R, B = µ0I 2R = µ0π I 4 l = (4π 2 × 10–7)(10.0) 4(0.400) = 24.7 µT into the paper Figure for Goal Solution Chapter 30 Solutions 191 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution (a) A conductor in the shape of a square of edge length l= 0.400 m carries a current I = 10.0 A (Fig. P30.3). Calculate the magnitude and direction of the magnetic field at the center of the square. (b) If this conductor is formed into a single circular turn and carries the same current, what is the value of the magnetic field at the center?
G : As shown in the diagram above, the magnetic field at the center is directed into the page from the clockwise current. If we consider the sides of the square to be sections of four infinite wires, then we could expect the magnetic field at the center of the square to be a little less than four times the strength of the field at a point l/2 away from an infinite wire with current I. B < 4 µ0I 2π a = 4 4π × 10−7 T ⋅m / A ( ) 10.0 A ( ) 2π 0.200 m ( ) = 40.0 µT Forming the wire into a circle should not significantly change the magnetic field at the center since the average distance of the wire from the center will not be much different.
O : Each side of the square is simply a section of a thin, straight conductor, so the solution derived from the Biot-Savart law in Example 30.1 can be applied to part (a) of this problem. For part (b), the Biot-Savart law can also be used to derive the equation for the magnetic field at the center of a circular current loop as shown in Example 30.3.
A : (a) We use Equation 30.4 for the field created by each side of the square. Each side contributes a field away from you at the center, so together they produce a magnetic field: B = 4µ0I 4π a cos π 4 −cos 3π 4 = 4 4π × 10−6 T ⋅m / A ( ) 10.0 A ( ) 4π 0.200 m ( ) 2 2 + 2 2 so at the center of the square, B = 2.00 2 × 10−5 T = 28.3 µT perpendicularly into the page (b) As in the first part of the problem, the direction of the magnetic field will be into the page. The new radius is found from the length of wire: 4 = 2π R, so R = 2 /π = 0.255 m. Equation 30.8 gives the magnetic field at the center of a circular current loop: B = µ0I 2R = (4π × 10−7 T ⋅m / A) 10.0 A ( ) 2(0.255 m) = 2.47 × 10−5 T = 24.7 µT Caution! If you use your calculator, it may not understand the keystrokes: To get the right answer, you may need to use .
L : The magnetic field in part (a) is less than 40µT as we predicted. Also, the magnetic fields from the square and circular loops are similar in magnitude, with the field from the circular loop being about 15% less than from the square loop. Quick tip: A simple way to use your right hand to find the magnetic field due to a current loop is to curl the fingers of your right hand in the direction of the current. Your extended thumb will then point in the direction of the magnetic field within the loop or solenoid.
192 Chapter 30 Solutions 30.4 B = µ0I 2πr = 4π × 10−7 (1.00 A) 2π(1.00 m) = 2.00 × 10-7 T 30.5 For leg 1, ds × ~= 0, so there is no contribution to the field from this segment. For leg 2, the wire is only semi-infinite; thus, B = 1 2 µ0I 2π x = µ0I 4π x into the paper 30.6 B = µ0I 2R R = µ0I 2B = 20.0π × 10−7 2.00 × 10−5 = 31.4 cm 30.7 We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude µ0I 2πR and directed into the page) and the field due to the circular loop (having magnitude µ0I 2R and directed into the page). The resultant magnetic field is: B = 1 + 1 π µ0I 2R = 1 + 1 π 4π × 10−7 T ⋅m / A ( ) 7.00 A ( ) 2 0.100 m ( ) = 5.80 × 10−5 T or B = 58.0 µT directed into the page ( ) 30.8 We can think of the total magnetic field as the superposition of the field due to the long straight wire (having magnitude µ0I 2πR and directed into the page) and the field due to the circular loop (having magnitude µ0I 2R and directed into the page). The resultant magnetic field is: B = 1 + 1 π µ0I 2R directed into the page ( ) 30.9 For the straight sections ds × ~= 0. The quarter circle makes one-fourth the field of a full loop: B = 1 4 µ0I 2R = µ0I 8R into the paper B = (4π × 10−7 T ⋅m / A)(5.00 A) 8(0.0300 m) = 26.2 µT into the paper Chapter 30 Solutions 193 © 2000 by Harcourt, Inc. All rights reserved.
30.10 Along the axis of a circular loop of radius R, B = µ0IR2 2 x2 + R2 ( ) 3 2 or B B0 = 1 x R ( ) 2 + 1 3 2 where B0 ≡µ0I 2R.
B Along Axis of Circular Loop 0.00 0.20 0.40 0.60 0.80 1.00 0.00 1.00 2.00 3.00 4.00 5.00 x/R B/B 0 x R B B0 0.00 1.00 1.00 0.354 2.00 0.0894 3.00 0.0316 4.00 0.0143 5.00 0.00754 30.11 dB = µ0I 4π d1×~ r2 B = µ0I 4π 1 6 2πa a2 − 1 6 2πb b2 B = µ0I 12 1 a −1 b directed out of the paper 30.12 Apply Equation 30.4 three times: B = µ 0I 4πa cos 0 − d d2 + a2 toward you + µ0I 4πd a d2 + a2 + a d2 + a2 away from you + µ0I 4π a −d d2 + a2 −cos 180° toward you B = µ 0I a2 + d2 −d a2 + d2 2πad a2 + d2 away from you 194 Chapter 30 Solutions 30.13 The picture requires L = 2R B = 1 2 µ0I 2R + µ0I 4π R (cos 90.0° −cos 135°) + µ0I 4π R (cos 45.0° −cos 135°) + µ0I 4π R (cos 45.0° −cos 90.0°) into the page B = µ0I R 1 4 + 1 π 2 = 0.475 µ0I R (into the page) 30.14 Label the wires 1, 2, and 3 as shown in Figure (a) and let the magnetic field created by the currents in these wires be B1, B2, and B3 respectively.
(a) At Point A : B1 = B2 = µ0I 2π a 2 ( ) and B3 = µ0I 2π 3a ( ) . The directions of these fields are shown in Figure (b).
Observe that the horizontal components of B1 and B2 cancel while their vertical components both add to B3.
Figure (a) Therefore, the net field at point A is: BA = B1 cos 45.0˚ + B2 cos 45.0˚ + B3 = µ0I 2π a 2 2 cos 45.0˚ + 1 3 BA = 4π × 10−7 T ⋅m A ( ) 2.00 A ( ) 2π 1.00 × 10−2 m ( ) 2 2 cos 45˚ + 1 3 = 53.3 µT (b) At point B : B1 and B2 cancel, leaving BB = B3 = µ0I 2π 2a ( ) . BB = 4π × 10−7 T ⋅m A ( ) 2.00 A ( ) 2π 2 ( ) 1.00 × 10−2 m ( ) = 20.0 µT Figure (b) Figure (c) (c) At point C : B1 = B2 = µ0I 2π a 2 ( ) and B3 = µ0I 2πa with the directions shown in Figure (c). Again, the horizontal components of B1 and B2 cancel. The vertical components both oppose B3 giving BC = 2 µ0I 2π a 2 ( ) cos 45.0˚ −µ0I 2πa = µ0I 2πa 2 cos 45.0˚ 2 −1 = 0 Chapter 30 Solutions 195 © 2000 by Harcourt, Inc. All rights reserved.
30.15 Take the x-direction to the right and the y-direction up in the plane of the paper. Current 1 creates at P a field B1 = µ0I 2πa = 2.00 × 10−7 T ⋅m ( ) 3.00 A ( ) A 0.0500 m ( ) B1 = 12.0 µT downward and leftward, at angle 67.4° below the –x axis.
Current 2 contributes B2 = 2.00 × 10−7 T ⋅m ( ) 3.00 A ( ) A 0.120 m ( ) clockwise perpendicular to 12.0 cm B2 = 5.00 µT to the right and down, at angle –22.6° Then, B = B1 + B2 = 12.0 µT ( ) −i cos 67.4°−j sin 67.4° ( ) + 5.00 µT ( ) i cos 22.6°−j sin 22.6° ( ) B = −11.1 µT ( )j −1.92 µT ( )j = (–13.0 µT)j 30.16 Let both wires carry current in the x direction, the first at y = 0 and the second at y = 10.0 cm.
(a) B = µ0I 2πr k = 4π × 10−7 T ⋅m A ( ) 5.00 A ( ) 2π 0.100 m ( ) k B = 1.00 × 10−5 T out of the page (b) FB = I2L × B = 8.00 A ( ) 1.00 m ( )i × 1.00 × 10−5 T ( )k [ ] = 8.00 × 10−5 N ( ) −j ( ) FB = 8.00 × 10−5 N toward the first wire (c) B = µ0I 2πr −k ( ) = 4π × 10−7 T ⋅m A ( ) 8.00 A ( ) 2π 0.100 m ( ) −k ( ) = (1.60 × 10−5 T) −k ( ) B = 1.60 × 10−5 T into the page (d) FB = I1L × B = 5.00 A ( ) 1.00 m ( )i × 1.60 × 10−5 T ( ) −k ( ) [ ] = 8.00 × 10−5 N ( ) +j ( ) FB = 8.00 × 10−5 N toward the second wire 30.17 By symmetry, we note that the magnetic forces on the top and bottom segments of the rectangle cancel. The net force on the vertical segments of the rectangle is (using Equation 30.12) FB = µ0 I1I2 l 2π 1 c + a – 1 c i Substituting given values FB = –2.70 × 10–5 i N = – 27.0 µN i 196 Chapter 30 Solutions Goal Solution In Figure P30.17, the current in the long, straight wire is I 1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and l = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. G : Even though there are forces in opposite directions on the loop, we must remember that the magnetic field is stronger near the wire than it is farther away. By symmetry the forces exerted on sides 2 and 4 (the horizontal segments of length a) are equal and opposite, and therefore cancel. The magnetic field in the plane of the loop is directed into the page to the right of I1. By the right-hand rule, F = I1× B is directed toward the left for side 1 of the loop and a smaller force is directed toward the right for side 3. Therefore, we should expect the net force to be to the left, possibly in the µN range for the currents and distances given. O : The magnetic force between two parallel wires can be found from Equation 30.11, which can be applied to sides 1 and 3 of the loop to find the net force resulting from these opposing force vectors.
A : F = F1 + F2 = µ0I1I2l 2π 1 c + a −1 c i = µ0I1I2l 2π −a c c + a ( ) i F = 4π × 10−7 N/ A2 ( ) 5.00 A ( )(10.0 A)(0.450 m) 2π −0.150 m (0.100 m)(0.250 m) i F i = − × − ( .
2 70 10 5 ) N or F = × − 2 70 10 5 . N toward the left L : The net force is to the left and in the µN range as we expected. The symbolic representation of the net force on the loop shows that the net force would be zero if either current disappeared, if either dimension of the loop became very small ( a → 0 or l → 0), or if the magnetic field were uniform ( c → ∞) .
30.18 The separation between the wires is a = 2(6.00 cm) sin 8.00° = 1.67 cm.
(a) Because the wires repel, the currents are in opposite directions .
(b) Because the magnetic force acts horizontally, FB Fg = µ0I 2 l 2π a mg = tan 8.00° I 2 = mg 2π a l µ0 tan 8.00° so I = 67.8 A Chapter 30 Solutions 197 © 2000 by Harcourt, Inc. All rights reserved.
30.19 Each wire is distant from P by (0.200 m) cos 45.0° = 0.141 m Each wire produces a field at P of equal magnitude: BA = µ0I 2π a = (2.00 × 10–7 T · m)(5.00 A) A(0.141 m) = 7.07 µT Carrying currents into the page, A produces at P a field of 7.07 µT to the left and down at –135°, while B creates a field to the right and down at – 45°. Carrying currents toward you, C produces a field downward and to the right at – 45°, while D 's contribution is downward and to the left. The total field is then 4 (7.07 µT) sin 45.0° = 20.0 µT toward the page's bottom 30.20 Let the current I flow to the right. It creates a field B = µ0I 2π d at the proton's location.
And we have a balance between the weight of the proton and the magnetic force mg(−j) + qv(−i) × µ0I 2πd (k) = 0 at a distance d from the wire d = qvµ0I 2πmg = (1.60 × 10−19 C)(2.30 × 104 m /s)(4π × 10−7 T ⋅m / A)(1.20 × 10−6 A) 2π(1.67 × 10−27 kg) (9.80 m /s2) = 5.40 cm 30.21 From Ampère's law, the magnetic field at point a is given by Ba = µ0Ia 2π ra , where Ia is the net current flowing through the area of the circle of radius ra. In this case, Ia = 1.00 A out of the page (the current in the inner conductor), so Ba = 4π × 10−7 T ⋅m / A ( )(1.00 A) 2π(1.00 × 10−3 m) = 200 µT toward top of page Similarly at point b : Bb = µ0 Ib 2π rb , where Ib is the net current flowing through the area of the circle having radius rb. Taking out of the page as positive, Ib = 1.00 A −3.00 A = −2.00 A , or Ib = 2.00 A into the page.
Therefore, Bb = (4π × 10−7 T ⋅m / A)(2.00 A) 2π(3.00 × 10−3 m) = 133 µT toward bottom of page 198 Chapter 30 Solutions 30.22 (a) In B = µ0I 2π r , the field will be one-tenth as large at a ten-times larger distance: 400 cm (b) B = µ0I 2π r1 k + µ0I 2π r2 (–k) so B = 4π × 10–7 T · m (2.00 A) 2π A 1 0.3985 m – 1 0.4015 m = 7.50 nT (c) Call r the distance from cord center to field point and 2d = 3.00 mm the distance between conductors.
B = µ0I 2π 1 r – d – 1 r + d = µ0I 2π 2d r 2 – d 2 7.50 × 10–10 T = 2.00 × 10–7 T · m A (2.00 A) (3.00 × 10–3 m) r 2 – 2.25 × 10–6 m2 so r = 1.26 m The field of the two-conductor cord is weak to start with and falls off rapidly with distance.
(d) The cable creates zero field at exterior points, since a loop in Ampère's law encloses zero total current. Shall we sell coaxial-cable power cords to people who worry about biological damage from weak magnetic fields?
30.23 (a) Binner = µ0NI 2π r = 3.60 T (b) Bouter = µ0NI 2π r = 1.94 T 30.24 (a) B = µ0I 2π a2 r for r ≤ a so B = µ0(2.50 A) 2π (0.0250 m)2 (0.0125 m) = 10.0 µT (b) r = µ0I 2π B = µ0(2.50 A) 2π (10.0 × 10–6 T) = 0.0500 m = 2.50 cm beyond the conductor's surface 30.25 (a) One wire feels force due to the field of the other ninety-nine. Within the bundle, B = µ0I 2π R2 r = 3.17 × 10−3 T. The force, acting inward, is FB = I lB, and the force per unit length is FB l = 6.34 × 10–3 N/m inward (b) B ∝ r, so B is greatest at the outside of the bundle. Since each wire carries the same current, F is greatest at the outer surface .
Figures for Goal Solution Chapter 30 Solutions 199 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R = 0.500 cm. (a) If each wire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on a wire located 0.200 cm from the center of the bundle? (b) Would a wire on the outer edge of the bundle experience a force greater or less than the value calculated in part (a)?
G : The force on one wire comes from its interaction with the magnetic field created by the other ninety-nine wires. According to Ampere’s law, at a distance r from the center, only the wires enclosed within a radius r contribute to this net magnetic field; the other wires outside the radius produce magnetic field vectors in opposite directions that cancel out at r. Therefore, the magnetic field (and also the force on a given wire at radius r) will be greater for larger radii within the bundle, and will decrease for distances beyond the radius of the bundle, as shown in the graph to the right. Applying F = I1× B, the magnetic force on a single wire will be directed toward the center of the bundle, so that all the wires tend to attract each other.
O : Using Ampere’s law, we can find the magnetic field at any radius, so that the magnetic force F = I1× B on a single wire can then be calculated.
A : (a) Ampere’s law is used to derive Equation 30.15, which we can use to find the magnetic field at r = 0.200 cm from the center of the cable: B = µoIor 2π R2 = 4π × 10−7 T ⋅m / A ( ) 99 ( ) 2.00 A ( ) 0.200 × 10−2 m ( ) 2π (0.500 × 10−2 m)2 = 3.17 × 10−3 T This field points tangent to a circle of radius 0.200 cm and exerts a force F = I1× B toward the center of the bundle, on the single hundredth wire: F l= IBsinθ = 2.00 A ( ) 3.17 × 10−3 T ( ) sin 90° ( ) = 6.34 mN/m (b) As is shown above in Figure 30.12 from the text, the magnetic field increases linearly as a function of r until it reaches a maximum at the outer surface of the cable. Therefore, the force on a single wire at the outer radius r = 5.00 cm would be greater than at r = 2.00 cm by a factor of 5/2.
L : We did not estimate the expected magnitude of the force, but 200 amperes is a lot of current. It would be interesting to see if the magnetic force that pulls together the individual wires in the bundle is enough to hold them against their own weight: If we assume that the insulation accounts for about half the volume of the bundle, then a single copper wire in this bundle would have a cross sectional area of about 1 2 ( ) 0.01 ( )π 0.500 cm ( )2 = 4 × 10−7 m2 with a weight per unit length of ρ gA = 8920 kg /m3 ( ) 9.8 N/kg ( ) 4 × 10−7 m2 ( ) = 0.03 N/m Therefore, the outer wires experience an inward magnetic force that is about half the magnitude of their own weight. If placed on a table, this bundle of wires would form a loosely held mound without the outer sheathing to hold them together.
30.26 From B⋅d1 = µ0I, ∫ I = 2π rB µ0 = (2π)(1.00 × 10-3)(0.100) 4π × 10–7 = 500 A 200 Chapter 30 Solutions 30.27 Use Ampère’s law, B⋅ds = µ0I ∫ . For current density J, this becomes B⋅ds = µ0 J ∫ ∫ ⋅dA (a) For r1 < R, this gives 2πr1 = µ0 br ( ) 2πr dr ( ) 0 r1 ∫ and B = µ0br1 2 3 for r1 < R or inside the cylinder ( ) (b) When r2 > R, Ampère’s law yields 2πr2 ( )B = µ0 br ( ) 2πr dr ( ) 0 R ∫ = 2π µ0bR3 3, or B = µ0bR3 3r2 for r2 > R or outside the cylinder ( ) 30.28 (a) See Figure (a) to the right.
(b) At a point on the z axis, the contribution from each wire has magnitude B = µ0I 2π a2 + z2 and is perpendicular to the line from this point to the wire as shown in Figure (b). Combining fields, the vertical components cancel while the horizontal components add, yielding By = 2 µ0I 2π a2 + z2 sinθ = µ0I π a2 + z2 z a2 + z2 = µ0I z π a2 + z2 ( ) The condition for a maximum is: dBy dz = −µ0I z 2z ( ) π a2 + z2 ( ) 2 + µ0I π a2 + z2 ( ) = 0, or µ0I π a2 −z2 ( ) a2 + z2 ( ) 2 = 0 Thus, along the z axis, the field is a maximum at d = a .
(Currents are into the paper) Figure (a) Figure (b) Chapter 30 Solutions 201 © 2000 by Harcourt, Inc. All rights reserved.
30.29 B = µ0 N l I so I = B µ0n = 31.8 mA 30.30 (a) I = 10.0 (4π × 10–7)(2000) = 3.98 kA (b) FB l = IB = 39.8 kN/m radially outward This is the force the windings will have to resist when the magnetic field in the solenoid is 10.0 T.
30.31 The resistance of the wire is Re = ρl π r2 , so it carries current I = ε Re = επr2 ρl .
If there is a single layer of windings, the number of turns per length is the reciprocal of the wire diameter: n = 1/ 2r .
So, B = nµ0I = µ0ε π r 2 ρ l 2r = µ0ε π r 2ρ l = (4π × 10–7 T · m/A)(20.0 V)π (2.00 × 10–3 m) 2(1.70 × 10–8 Ω · m)(10.0 m) = 464 mT 30.32 The field produced by the solenoid in its interior is given by B = µ0nI −i ( ) = 4π × 10−7 T ⋅m A 30.0 10-2 m 15.0 A ( ) −i ( ) B = −5.65 × 10−2 T ( )i The force exerted on side AB of the square current loop is FB ( )AB = IL × B = 0.200 A ( ) 2.00 × 10−2 m ( )j × 5.65 × 10−2 T ( ) −i ( ) [ ] FB ( )AB = 2.26 × 10−4 N ( )k Similarly, each side of the square loop experiences a force, lying in the plane of the loop, of 226 µN directed away from the center . From the above result, it is seen that the net torque exerted on the square loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of the square loop is given by µ = IA = 0.200 A ( ) 2.00 × 10−2 m ( ) 2 −i ( ) = −80.0 µA ⋅m2 i The torque exerted on the loop is then τ = µ × B = −80.0 µA ⋅m2 i ( ) × −5.65 × 10−2 Ti ( ) = 0 202 Chapter 30 Solutions 30.33 (a) ΦB = B⋅dA ∫ = B⋅A = 5i + 4j + 3k ( )T ⋅2.50 × 10−2 m ( ) 2 i ΦB = 3.13 × 10−3 T ⋅m2 = 3.13 × 10−3 Wb = 3.13 mWb (b) ΦB ( )total = B⋅dA ∫ = 0 for any closed surface (Gauss’s law for magnetism) 30.34 (a) ΦB = B⋅A = BA where A is the cross-sectional area of the solenoid. ΦB = µ0NI l π r2 ( ) = 7.40 µWb (b) ΦB = B⋅A = BA = µ0NI l π r2 2 −r1 2 ( ) [ ] ΦB = 4π × 10−7 T ⋅m A ( ) 300 ( ) 12.0 A ( ) 0.300 m ( ) π 8.00 ( )2 −4.00 ( )2 [ ] 10−3 m ( ) 2= 2.27 µWb 30.35 (a) ΦB ( )flat = B⋅A = BπR2 cos 180 −θ ( )= –Bπ R 2 cos θ (b) The net flux out of the closed surface is zero: ΦB ( )flat + ΦB ( )curved = 0 ΦB ( )curved = Bπ R 2 cos θ 30.36 dΦE dt = d dt (EA) = dQ/ dt e0 = I e0 (a) dE dt = I e0A = 7.19 × 1011 V/m · s (b) B⋅ds = e0µ0 ΦE dt ∫ so 2πrB = e0µ0 d dt Q e0A ⋅πr2 B = µ0Ir 2A = µ0(0.200)(5.00 × 10−2) 2π(0.100)2 = 2.00 × 10-7 T 30.37 (a) dΦE dt = dQ/ dt e0 = I e0 = (0.100 A) 8.85 × 10−12 C2 / N ⋅m2 = 11.3 × 109 V ⋅m /s (b) Id = e0 dΦE dt = I = 0.100 A Chapter 30 Solutions 203 © 2000 by Harcourt, Inc. All rights reserved.
30.38 (a) I = ev 2π r µ = IA = ev 2π r π r 2 = 9.27 × 10–24 A · m2 The Bohr model predicts the correct magnetic moment.
However, the "planetary model" is seriously deficient in other regards.
(b) Because the electron is (–), its [conventional] current is clockwise, as seen from above, and µ points downward .
30.39 Assuming a uniform B inside the toroid is equivalent to assuming r << R, then B0 ≅ µ0 NI 2π R and a tightly wound solenoid. B0 = µ0 (630)(3.00) 2π(0.200) = 0.00189 T With the steel, B = κmB0 = (1 + χ)B0 = (101)(0.00189 T) B = 0.191 T 30.40 B = µnI = µ N 2πr I so I = 2πr ( )B µN = 2π 0.100 m ( ) 1.30 T ( ) 5000 4π × 10−7 Wb A ⋅m ( ) 470 ( ) = 277 mA 30.41 ΦB = µnIA B = µnI = (750 × 4π × 10−7) 500 2π(0.200) (0.500) = 0.188 T A = 8.00 × 10-4 m2 and ΦB = (0.188 T)(8.00 × 10-4 m2) = 1.50 × 10-4 T · m2) = 150 µT · m2 30.42 The period is T = 2π /ω. The spinning constitutes a current I = Q T = Qω 2π . µ = IA = Qω 2π πR2 = QωR2 2 in the direction of ω µ = (6.00 × 10−6 C)(4.00/s)(0.0200 m)2 2 = 4.80 × 10-9 A · m2 204 Chapter 30 Solutions 30.43 B = µ0(H + M) so H = B µ0 −M = 2.62 × 106 A/m 30.44 B = µ0 H + M ( ) If µ0M = 2.00 T, then the magnetization of the iron is M = 2.00 T µ0 .
But M = xnµB where µB is the Bohr magneton, n is the number of atoms per unit volume, and x is the number of electrons that contribute per atom. Thus, x = M nµB = 2.00 T nµBµ0 = 2.00 T 8.50 × 1028 m−3 ( ) 9.27 × 10−24 N ⋅m T ( ) 4π × 10−7 T ⋅m A ( ) = 2.02 30.45 (a) Comparing Equations 30.29 and 30.30, we see that the applied field is described by B0 = µ0H.
Then Eq. 30.35 becomes M = C B0 T = C T µ0H , and the definition of susceptibility (Eq. 30.32) is χ = M H = C T µ0 (b) C = χT µ0 = 2.70 × 10−4 ( ) 300 K ( ) 4π × 10−7 T ⋅m A = 6.45 × 104 K ⋅A T ⋅m 30.46 (a) Bh = Bcoil = µ0NI 2R = (4π × 10−7)(5.00)(0.600) 0.300 = 12.6 µT (b) Bh = Bsinφ →B = Bh sinφ = 12.6 µT sin 13.0° = 56.0 µT 30.47 (a) Number of unpaired electrons = 8.00 × 1022 A · m2 9.27 × 10–24 A · m2 = 8.63 × 1045 Each iron atom has two unpaired electrons, so the number of iron atoms required is 1 2 8.63 × 1045 ( ).
(b) Mass = (4.31 × 1045 atoms)(7900 kg/m3) 8.50 × 1028 atoms/m3 = 4.01 × 1020 kg Chapter 30 Solutions 205 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution The magnetic moment of the Earth is approximately 8.00 × 1022 A·m2. (a) If this were caused by the complete magnetization of a huge iron deposit, how many unpaired electrons would this correspond to?
(b) At two unpaired electrons per iron atom, how many kilograms of iron would this correspond to?
(Iron has a density of 7 900 kg/m3, and approximately 8.50 × 1028 atoms/m3.) G : We know that most of the Earth is not iron, so if the situation described provides an accurate model, then the iron deposit must certainly be less than the mass of the Earth ( MEarth = 5.98 × 1024 kg). One mole of iron has a mass of 55.8 g and contributes 2(6.02 × 10 23 ) unpaired electrons, so we should expect the total unpaired electrons to be less than 1050.
O : The Bohr magneton µB is the measured value for the magnetic moment of a single unpaired electron. Therefore, we can find the number of unpaired electrons by dividing the magnetic moment of the Earth by µB. We can then use the density of iron to find the mass of the iron atoms that each contribute two electrons.
A : (a) µB = 9.27 × 10−24 J T 1 N ⋅m J 1 T N ⋅s C⋅m 1 A C/s = 9.27 × 10−24 A ⋅m2 The number of unpaired electrons is N = 8.00 × 1022 A ⋅m2 9.27 × 10−24 A ⋅m2 = 8.63 × 1045 e-(b) Each iron atom has two unpaired electrons, so the number of iron atoms required is 1 2 N = 1 2 (8.63 × 1045) = 4.31× 1045 iron atoms.
Thus, MFe = 4.31× 1045 atoms ( ) 7900 kg /m3 ( ) 8.50 × 1028 atoms/m3 = 4.01× 1020 kg L : The calculated answers seem reasonable based on the limits we expected. From the data in this problem, the iron deposit required to produce the magnetic moment would only be about 1/15 000 the mass of the Earth and would form a sphere 500 km in diameter. Although this is certainly a large amount of iron, it is much smaller than the inner core of the Earth, which is estimated to have a diameter of about 3000 km.
30.48 B = µ0I 2π R = 2.00 × 10–5 T = 20.0 µT 30.49 B = µ0IR 2 2(R 2 + R 2)3/2 so I = 2.00 × 109 A flowing west 30.50 (a) BC = µ0I 2π (0.270) – µ0(10.0) 2π (0.0900) = 0 so I = 30.0 A (b) BA = 4µ0(10.0) 2π (0.0900) = 88.9 µT out of paper 206 Chapter 30 Solutions 30.51 Suppose you have two 100-W headlights running from a 12-V battery, with the whole 200 W 12 V = 17 A current going through the switch 60 cm from the compass. Suppose the dashboard contains little iron, so µ ≅ µ0. Model the current as straight. Then, B = µ0I 2π r = (4π × 10–7)17 2π (0.6) ~ 10– 5 T If the local geomagnetic field is 5 × 10–5 T, this is ~10–1 times as large, enough to affect the compass noticeably.
30.52 A ring of radius r and width dr has area dA = 2π r dr. The current inside radius r is I = 2π J rdr 0 r ∫ = 2π J0 rdr −2π J0 R2 ( ) 0 r ∫ r3 dr 0 r ∫ = 2π J0 r2 2 −2π J0 R2 ( ) r4 4 ( ) (a) Ampère's law says B 2π r ( ) = µ0I = µ0π J0 r2 −r4 2R2 ( ), or B = µ0J0R 1 2 r R −1 4 r R 3 for r ≤R and B 2π r ( ) = µ0Itotal = µ0 π J0R2 −π J0R2 2 [ ] = µ0 π J0R2 2 or B = µ0J0R2 4r = µ0J0R 4 r R ( ) for r ≥R (b) 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0 2 4 6 r/R B /µ 0J 0R (c) To locate the maximum in the region r ≤R, require that dB dr = µ0J0 2 −3 µ0J0r2 4R2 = 0 This gives the position of the maximum as r = 2/ 3 R .
Here B = µ0J0R 1 2 2 3 1 2 −1 4 2 3 3 2 = 0.272µ0J0R Chapter 30 Solutions 207 © 2000 by Harcourt, Inc. All rights reserved.
30.53 Consider a longitudinal filament of the strip of width dr as shown in the sketch. The contribution to the field at point P due to the current dI in the element dr is dB = µ0dI 2π r where dI = I dr w ( ) B = ∫ d B = ⌡ ⌠ b b + w µ0I dr 2π wr k = µ0I 2π w ln 1 + w b k 30.54 We find the total number of turns: B = µ0NI l N = B l µ0I = (0.0300 T)(0.100 m)A (4π × 10–7 T · m)(1.00 A) = 2.39 × 103 Each layer contains (10.0 cm/0.0500 cm) = 200 closely wound turns so she needs (2.39 × 103/200) = 12 layers .
The inner diameter of the innermost layer is 10.0 mm. The outer diameter of the outermost layer is 10.0 mm + 2 × 12 × 0.500 mm = 22.0 mm. The average diameter is 16.0 mm, so the total length of wire is (2.39 × 103)π (16.0 × 10–3 m) = 120 m 30.55 On the axis of a current loop, the magnetic field is given by B = µ0IR 2 2(x 2 + R 2)3/2 where in this case I = q (2π /ω) . The magnetic field is directed away from the center, with a strength of B = µ0ωR 2q 4π (x 2 + R 2)3/2 = µ0(20.0)(0.100)2(10.0 × 10−6) 4π (0.0500)2 + (0.100)2 [ ] 3/2 = 1.43 × 10–10 T 30.56 On the axis of a current loop, the magnetic field is given by B = µ0IR 2 2(x 2 + R 2)3/2 where in this case I = q (2π /ω) . Therefore, B = µ0ωR 2q 4π (x 2 + R 2)3/2 when x = R 2 , then B = µ0ωqR2 4π 5 4 R2 ( ) 3/2 = µ0qω 2.5 5 π R 208 Chapter 30 Solutions 30.57 (a) Use Equation 30.7 twice: Bx = µ0IR 2 2(x 2 + R 2)3/2 B = Bx1 + Bx2 = µ0IR 2 2 1 (x 2 + R 2)3/2 + 1 ((R – x)2 + R 2)3/2 B = µ0IR 2 2 1 (x 2 + R 2)3/2 + 1 (2R 2 + x 2 – 2xR)3/2 (b) dB dx = µ0IR2 2 −3 2 2x ( ) x2 + R2 ( ) −5 2 −3 2 2R2 + x2 −2xR ( ) −5 2 2x −2R ( ) Substituting x = R 2 and cancelling terms, dB dx = 0 d2B dx2 = −3µ0IR2 2 (x2+ R2)−5 2−5x2(x2+ R2)−7 2+ (2R2 + x2−2xR)−5 2−5(x −R)2(2R2 + x2−2xR)−7 2 [ ] Again substituting x = R 2 and cancelling terms, d2B dx2 = 0 30.58 "Helmholtz pair" → separation distance = radius B = 2µ0IR 2 2 R/ 2 ( )2 + R 2 [ ] 3/2 = µ0IR 2 1 4 + 1 3/2 R 3 = µ0I 1.40R for 1 turn For N turns in each coil, B = µ0NI 1.40R = 4π × 10−7 ( )100 10.0 ( ) 1.40 0.500 ( ) = 1.80 × 10- 3 T Chapter 30 Solutions 209 © 2000 by Harcourt, Inc. All rights reserved.
30.59 Model the two wires as straight parallel wires (!) (a) FB = µ0I 2L 2π a (Equation 30.12) FB = (4π × 10–7)(140)22π(0.100) 2π (1.00 × 10–3) = 2.46 N upward (b) aloop = 2.46 N – mloop g mloop = 107 m/s2 upward 30.60 (a) In dB = µ0 4π r2 Ids × ~, the moving charge constitutes a bit of current as in I = nqvA. For a positive charge the direction of ds is the direction of v, so dB = µ0 4πr2 nqA ds ( )v × ~. Next, A ds ( ) is the volume occupied by the moving charge, and nA ds ( ) = 1 for just one charge. Then, B = µ0 4πr2 qv × ~ (b) B = 4π × 10−7 T ⋅m A ( ) 1.60 × 10−19 C ( ) 2.00 × 107 m s ( ) 4π 1.00 × 10−3 ( ) 2 sin 90.0˚= 3.20 × 10−13 T (c) FB = q v × B = 1.60 × 10−19 C ( ) 2.00 × 107 m s ( ) 3.20 × 10−13 T ( ) sin 90.0˚ FB = 1.02 × 10−24 N directed away from the first proton (d) Fe = qE = keq1q2 r2 = 8.99 × 109 N ⋅m2 C2 ( ) 1.60 × 10−19 C ( ) 2 1.00 × 10−3 ( ) 2 Fe = 2.30 × 10−22 N directed away from the first proton Both forces act together. The electrical force is stronger by two orders of magnitude. It is productive to think about how it would look to an observer in a reference frame moving along with one proton or the other.
30.61 (a) B = µ0I 2π r = 4π × 10−7 T ⋅m A ( ) 24.0 A ( ) 2π 0.0175 m ( ) = 2.74 × 10−4 T (b) At point C, conductor AB produces a field 1 2 2.74 × 10−4 T ( ) −j ( ), conductor DE produces a field of 1 2 2.74 × 10−4 T ( ) −j ( ), BD produces no field, and AE produces negligible field. The total field at C is 2.74 × 10−4 T −j ( ) .
210 Chapter 30 Solutions (c) FB = IL × B = 24.0 A ( ) 0.0350 mk ( ) × 5 2.74 × 10−4 T ( ) −j ( ) [ ] = 1.15 × 10−3 N ( )i (d) a = ΣF m = 1.15 × 10−3 N ( )i 3.00 × 10−3 kg = 0.384 m s2 i (e) The bar is already so far from AE that it moves through nearly constant magnetic field. The force acting on the bar is constant, and therefore the bar’s acceleration is constant .
(f) vf 2 = vi 2 + 2ax = 0 + 2 0.384 m s2 ( ) 1.30 m ( ), so v f = 0 999 . m s ( )i Chapter 30 Solutions 211 © 2000 by Harcourt, Inc. All rights reserved.
30.62 At equilibrium, FB l = µ0IAIB 2π a = mg l or IB = 2π a m l ( )g µ0IA IB = 2π 0.0250 m ( ) 0.0100 kg m ( ) 9.80 m s2 ( ) 4π × 10−7 T ⋅m A ( ) 150 A ( ) = 81.7 A 30.63 (a) The magnetic field due to an infinite sheet of charge (or the magnetic field at points near a large sheet of charge) is given by B = µ0Js 2 .
The current density Js = I l and in this case the equivalent current of the moving charged belt is I = dq dt = d dt (σ lx) = σ lv; v = dx dt Therefore, Js = σ v and B = µ0σ v 2 (b) If the sheet is positively charged and moving in the direction shown, the magnetic field is out of the page, parallel to the roller axes. 30.64 C = TM B = (4.00 K)(10.0%)(8.00 × 1027 atoms/m3)(5.00)(9.27 × 10−24 J/ T2) 5.00 T = 2.97 × 104 K ⋅J T2 ⋅m3 30.65 At equilibrium, Στ = + µ × B −mg L 2 cos 5.00˚ = 0, or µB sin 5.00˚ = mgL 2 cos 5.00˚ Therefore, B = mgL 2µ tan 5.00˚ = 0.0394 kg ( ) 9.80 m s2 ( ) 0.100 m ( ) 2 7.65 J T ( )tan 5.00˚ B = 28.8 mT 30.66 The central wire creates field B = µ0I1 2πR counterclockwise. The curved portions of the loop feels no force since 1 × B = 0 there. The straight portions both feel I 1 × B forces to the right, amounting to FB = I2 2L µ0 I1 2πR = µ0 I1 I2 L πR to the right 212 Chapter 30 Solutions 30.67 When the conductor is in the rectangular shape shown in figure (a), the segments carrying current straight toward or away from point P1 do not contribute to the magnetic field at P1. Each of the other four setions of length l makes an equal contribution to the total field into the page at P1. To find the contribution of the horizontal section of current in the upper right, we use B = µ0I 4πa(cos θ1 – cos θ2) with a = l, θ1 = 90°, and θ2 = 135° So B1 = 4µ0I 4πl 0 – 1 2 = µ0I 2 πl When the conductor is in the shape of a circular arc, the magnitude or the field at the center is given by Equation 30.6, B = µ0I 4πR θ . From the geometry in this case, we find R = 4l π and θ = π. Therefore, B2 = µ0Iπ 4π(4l/ π) = µ0Iπ 16l ; so that B1 B2 = 8 2 π 2 30.68 I = 2πrB µ0 = 2π 9.00 × 103 ( ) 1.50 × 10−8 ( ) 4π × 10−7 = 675 A Flow of positive current is downward or negative charge flows upward .
30.69 By symmetry of the arrangement, the magnitude of the net magnetic field at point P is B = 8B0x where B0 is the contribution to the field due to current in an edge length equal to L/2. In order to calculate B0, we use the Biot-Savart law and consider the plane of the square to be the yz-plane with point P on the x-axis.
The contribution to the magnetic field at point P due to a current element of length dz and located a distance z along the axis is given by Equation 30.3. B0 = µ0I 4π d1×~ r2 ∫ From the figure we see that r = x2 + (L2 / 4) + z2 and d1×~ = dzsinθ = dz L2 / 4 + x2 L2 / 4 + x2 + z2 By symmetry all components of the field B at P cancel except the components along x (perpendicular to the plane of the square); and Chapter 30 Solutions 213 © 2000 by Harcourt, Inc. All rights reserved. B0x = B0 cosφ where cosφ = L/ 2 L2 / 4 + x2 . Therefore, B0x = µ0I 4π sinθ cosφ dz r2 0 L/2 ∫ and B = 8B0x.
Using the expressions given above for sin θ cos φ, and r, we find B = µ0IL2 2π x2 + L2 4 x2 + L2 2 30.70 (a) From Equation 30.10, the magnetic field produced by one loop at the center of the second loop is given by B = µ0IR2 2x3 = µ0I π R2 ( ) 2π x3 = µ0µ 2π x3 where the magnetic moment of either loop is µ = I π R2 ( ). Therefore, Fx = µ dB dx = µ µ0µ 2π 3 x4 = 3µ0 πR2 I ( ) 2 2π x4 = 3π 2 µ0I2R4 x4 (b) Fx = 3π 2 µ0I2R4 x4 = 3π 2 4π × 10−7 T ⋅m A ( ) 10.0 A ( )2 5.00 × 10−3 m ( ) 4 5.00 × 10−2 m ( ) 4 = 5.92 × 10−8 N 30.71 There is no contribution from the straight portion of the wire since ds×~= 0. For the field of the spiral, dB = µ0I (4π) (ds×~ ) r2 B = µ0I 4π ds sinθ ~ r2 θ =0 2π ∫ = µ0I 4π 2 dr ( ) sin 3π 4 1 r2 θ =0 2π ∫ B = µ0I 4π r−2 dr = − θ =0 2π ∫ µ0I 4π r−1 ( ) θ =0 2π Substitute r = eθ: B = −µ0I 4π e−θ [ ]0 2π = −µ0I 4π e−2π −e0 [ ] = µ0I 4π 1−e−2π ( ) (out of the page) 214 Chapter 30 Solutions 30.72 (a) B = B0 + µ0M M = B −B0 µ0 and M = B −B0 µ0 Assuming that B and B0 are parallel, this becomes M = B −B0 ( ) µ0 The magnetization curve gives a plot of M versus B0.
(b) The second graph is a plot of the relative permeability B B0 ( ) as a function of the applied field B0.
30.73 Consider the sphere as being built up of little rings of radius r, centered on the rotation axis. The contribution to the field from each ring is dB = µ0 r2 dI 2 x2 + r2 ( ) 3 2 where dI = dQ t = ω dQ 2π dQ = ρ dV = ρ 2πrdr ( ) dx ( ) dB = µ0ρωr3 drdx 2 x2 + r2 ( ) 3 2 where ρ = Q 4 3 π R3 ( ) B = µ0ρω 2 r=0 R2 −x2 ∫ x=−R +R ∫ r3 drdx x2 + r2 ( ) 3 2 r x dr dx ω R Let v = r2 + x2, dv = 2rdr , and r2 = v −x2 B = µ0ρω 2 v=x2 R2 ∫ x=−R +R ∫ v −x2 ( )dv 2v3 2 dx = µ0ρω 4 v−1 2 dv v=x2 R2 ∫ −x2 v−3 2 dv v=x2 R2 ∫ dx x=−R R ∫ B = µ0ρω 4 2v1 2 x2 R2 + 2x2 ( )v−1 2 x2 R2 dx x=−R R ∫ = µ0ρω 4 2 R −x ( ) + 2x2 1 R −1 x dx x=−R R ∫ B = µ0ρω 4 2 x2 R −4 x + 2R dx −R R ∫ = 2µ0ρω 4 2 x2 R −4x + 2R dx 0 R ∫ B = 2µ0ρω 4 2R3 3R −4R2 2 + 2R2 = µ0ρωR2 3 Chapter 30 Solutions 215 © 2000 by Harcourt, Inc. All rights reserved.
30.74 Consider the sphere as being built up of little rings of radius r, centered on the rotation axis. The current associated with each rotating ring of charge is dI = dQ t = ω 2π ρ 2πrdr ( ) dx ( ) [ ] The magnetic moment contributed by this ring is dµ = A dI ( ) = π r2 ω 2π ρ 2πrdr ( ) dx ( ) [ ] = πωρr3 drdx r x dr dx ω R µ = πωρ r3 dr r=0 R2 −x2 ∫ x=−R +R ∫ dx = πωρ R2 −x2 4 4 x=−R +R ∫ dx = πωρ R2 −x2 ( ) 2 4 x=−R +R ∫ dx µ = πωρ 4 R4 −2R2x2 + x4 ( ) x=−R +R ∫ dx = πωρ 4 R4 2R ( ) −2R2 2R2 3 + 2R5 5 µ = πωρ 4 R5 2 −4 3 + 2 5 = πωρR5 4 16 15 = 4πωρR5 15 up 30.75 Note that the current I exists in the conductor with a current density J = I A, where A = π a2 −a2 4 −a2 4 [ ] = π a2 2 Therefore, J = 2I π a2 .
To find the field at either point P1 or P2, find Bs which would exist if the conductor were solid, using Ampère’s law. Next, find B1 and B2 that would be due to the conductors of radius a 2 that could occupy the void where the holes exist. Then use the superposition principle and subtract the field that would be due to the part of the conductor where the holes exist from the field of the solid conductor.
θ θ θ Bs -B1 -B2 P1 r r r2 + a 2 ( ) 2 a/2 a/2 Bs −′ B1 −′ B2 P2 (a) At point P1, Bs = µ0J π a2 ( ) 2π r , B1 = µ0Jπ a 2 ( ) 2 2π r −a 2 ( ) , and B2 = µ0Jπ a 2 ( ) 2 2π r + a 2 ( ) . B = Bs −B1 −B2 = µ0Jπ a2 2π 1 r − 1 4 r −a 2 ( ) − 1 4 r + a 2 ( ) B = µ0 2I ( ) 2π 4r2 −a2 −2r2 4r r2 −a2 4 ( ) = µ0I π r 2r2 −a2 4r2 −a2 directed to the left 216 Chapter 30 Solutions (b) At point P2, Bs = µ0J π a2 ( ) 2π r and ′ B1 = ′ B2 = µ0Jπ a 2 ( ) 2 2π r2 + a 2 ( ) 2 . The horizontal components of ′ B1 and ′ B2 cancel while their vertical components add. B = Bs − ′ B1cosθ − ′ B2 cosθ = µ0J π a2 ( ) 2π r −2 µ0Jπ a2 4 2π r2 + a2 4 r r2 + a2 4 B = µ0J π a2 ( ) 2π r 1− r2 2 r 2 + a2 4 ( ) = µ0 2I ( ) 2π r 1− 2r2 4r2 + a2 = µ0I π r 2r2 + a2 4r2 + a2 directed toward the top of the page © 2000 by Harcourt, Inc. All rights reserved.
Chapter 31 Solutions 31.1 ε = ∆ΦB ∆t = ∆NBA ( ) ∆t = 500 mV 31.2 ε = ∆ΦB ∆t = ∆B⋅A ( ) ∆t = 1.60 mV and Iloop = ε R = 1.60 mV 2.00 Ω= 0.800 mA 31.3 ε = –N ∆BA cos θ ∆t = –NB π r 2 cos θf – cos θi ∆t = −25.0 50.0 × 10−6 T ( )π 0.500m ( ) 2 cos 180°−cos 0 0.200 s E = + 9.82 mV 31.4 (a) ε = – dΦB dt = –A dB dt = ABmax τ e–t/τ (b) ε = (0.160 m2)(0.350 T) 2.00 s e– 4.00/2.00 = 3.79 mV (c) At t = 0, ε = 28.0 mV 31.5 ε = N dΦB dt = ∆(NBA) ∆t = 3.20 kV so I = ε R = 160 A Chapter 31 Solutions 219 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A strong electromagnet produces a uniform field of 1.60 T over a cross-sectional area of 0.200 m2. A coil having 200 turns and a total resistance of 20.0 Ω is placed around the electromagnet. The current in the electromagnet is then smoothly decreased until it reaches zero in 20.0 ms. What is the current induced in the coil? G : A strong magnetic field turned off in a short time ( 20.0 ms) will produce a large emf, maybe on the order of 1 kV. With only 20.0 Ω of resistance in the coil, the induced current produced by this emf will probably be larger than 10 A but less than 1000 A. O : According to Faraday’s law, if the magnetic field is reduced uniformly, then a constant emf will be produced. The definition of resistance can be applied to find the induced current from the emf.
A : Noting unit conversions from F = qv × B and U = qV , the induced voltage is ε = −N d(B⋅A) dt = −N 0 −BiAcosθ ∆t = +200 1.60 T ( ) 0.200 m2 ( ) cos 0° ( ) 20.0 × 10−3 s 1 N ⋅s/C⋅m T 1 V ⋅C N ⋅m = 3200 V I = ε R = 3200 V 20.0 Ω= 160 A L : This is a large current, as we expected. The positive sign is indicative that the induced electric field is in the positive direction around the loop (as defined by the area vector for the loop).
31.6 ε = –N dΦB dt = – N(BA – 0) ∆t ∆t = NBA ε = NB(πr 2) ε = 500(0.200)π (5.00 × 10-2)2 10.0 × 103 = 7.85 × 10–5 s 31.7 ε = d(BA) dt = 0.500 µ0nA dI dt = 0.480 × 10–3 V (a) Iring = ε R = 4.80 × 10-4 3.00 × 10-4 = 1.60 A (b) Bring = µ0I 2rring = 20.1 µT (c) Coil's field points downward, and is increasing, so Bring points upward 31.8 ε = d(BA) dt = 0.500 µ0nA dI dt = 0.500 µ0nπ r 2 2 ∆I ∆t 220 Chapter 31 Solutions (a) Iring = ε R = µ0nπ r 2 2 2R ∆I ∆t (b) B = µ0I 2r1 = µ2 0nπ r2 2 4r1R ∆I ∆t (c) The coil's field points downward, and is increasing, so Bring points upward .
31.9 (a) dΦB = B⋅dA = µ0I 2πx Ldx: ΦB = µ0IL 2π x=h h+w ∫ dx x = µ0IL 2π ln h + w h (b) ε = −dΦB dt = −d dt µ0IL 2π ln h + w h = −µ0L 2π ln h + w h dI dt ε = − 4π × 10−7 T ⋅m A ( ) 1.00 m ( ) 2π ln 1.00 + 10.0 1.00 10.0 A s = −4.80 µV The long wire produces magnetic flux into the page through the rectangle (first figure, above).
As it increases, the rectangle wants to produce its own magnetic field out of the page, which it does by carrying counterclockwise current (second figure, above).
Chapter 31 Solutions 221 © 2000 by Harcourt, Inc. All rights reserved.
31. 10 ΦB = (µ0nI)Asolenoid ε = −N dΦB dt = −Nµ0n πrsolenoid 2 ( ) dI dt = −Nµ0n πrsolenoid 2 ( ) 600 A s ( ) cos(120t) ε = −15.0 4π × 10−7 T ⋅m A ( ) 1.00 × 103 m ( )π 0.0200 m ( )2 600 A s ( ) cos(120t) E = −14.2 cos(120t) mV 31.11 For a counterclockwise trip around the left-hand loop, with B = At d dt At(2a2)cos 0° [ ] −I1(5R) −IPQR = 0 and for the right-hand loop, d dt Ata2 [ ] + IPQ R −I2(3R) = 0 where IPQ = I1 −I2 is the upward current in QP Thus, 2Aa2 −5R(IPQ + I2) −IPQR = 0 and Aa2 + IPQR = I2(3R) 2Aa2 −6RIPQ −5 3 (Aa2 + IPQR) = 0 IPQ = Aa2 23R upward, and since R = (0.1 0 /m)(0.65 m) = 0.065 0 0 0 Ω Ω IPQ = (1.00 × 10−3 T/s)(0.650 m)2 23(0.0650 Ω) = 283 µA upward 31.12 ε = ∆ΦB ∆t = N dB dt A = N 0.0100 + 0.0800t ( )A At t = 5.00 s, ε = 30.0(0.410 T) π 0.0400 m ( )2 [ ] = 61.8 mV 222 Chapter 31 Solutions 31.13 B = µ0nI = µ0n 30.0 A ( ) 1−e−1.60t ( ) ΦB = BdA ∫ = µ0n 30.0 A ( ) 1−e−1.60t ( ) dA ∫ ΦB = µ0n 30.0 A ( ) 1−e−1.60t ( )πR2 ε = −N dΦB dt = −Nµ0n 30.0 A ( )πR2 1.60 ( )e−1.60t ε = −(250)(4π × 10−7 N A2)(400 m−1)(30.0 A) π(0.0600 m)2 [ ]1.60 s−1 e−1.60t ε = 68.2 mV ( )e−1.60t counterclockwise 31.14 B = µ0nI = µ0nImax(1−e−αt) ΦB = BdA ∫ = µ0nImax(1 −e−αt) dA ∫ ΦB = µ0nImax(1 −e−αt)πR2 ε = −N dΦB dt = −Nµ0nImaxπR2αe−αt ε = Nµ0nImaxπR2αe−αt counterclockwise 31.15 ε = d dt (NBl2 cos θ) = Nl2 ∆B cos θ ∆t l= ε ∆t N ∆B cosθ = (80.0 × 10−3 V)(0.400 s) (50)(600 × 10−6 T −200 × 10−6 T)cos(30.0°) = 1.36 m Length = 4 lN = 4(1.36 m)(50) = 272 m Chapter 31 Solutions 223 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A coil formed by wrapping 50.0 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased uniformly from 200 µ T to 600 µ T in 0.400 s, an emf of 80.0 mV is induced in the coil. What is the total length of the wire? G : If we assume that this square coil is some reasonable size between 1 cm and 1 m across, then the total length of wire would be between 2 m and 200 m. O : The changing magnetic field will produce an emf in the coil according to Faraday’s law of induction.
The constant area of the coil can be found from the change in flux required to produce the emf.
A : By Faraday’s law, ε = −N dΦB dt = −N d dt (BAcosθ) = −NAcosθ dB dt For magnitudes, ε = NA cosθ ∆B ∆t and the area is A = ε N cosθ ∆B ∆t = 80.0 × 10−3 V 50(cos 30.0°) 600 × 10−6 T −200 × 10−6 T 0.400 s = 1.85 m2 Each side of the coil has length d = A , so the total length of the wire is L = N 4d ( ) = 4N A = (4)(50) 1.85 m2 = 272 m L : The total length of wire is slightly longer than we predicted. With d = 1.36 m, a normal person could easily step through this large coil! As a bit of foreshadowing to a future chapter on AC circuits, an even bigger coil with more turns could be hidden in the ground below high-power transmission lines so that a significant amount of power could be “stolen” from the electric utility. There is a story of one man who did this and was arrested when investigators finally found the reason for a large power loss in the transmission lines!
31.16 The average induced emf is given by ε = – N ∆ΦB ∆t Here N = 1, and ∆ΦB = B(Asquare −Acircle) with Acircle = πr2 = π(0.500 m)2 = 0.785 m2 Also, the circumference of the circle is 2π r = 2π (0.500 m) = 3.14 m Thus, each side of the square has a length L = 3.14 m 4 = 0.785 m, and Asquare = L2 = 0.617 m2 So ∆ΦB = (0.400 T)(0.617 m2 −0.785 m2) = −0.0672 T ⋅ m2 The average induced emf is therefore: ε = – −0.0672 T ⋅ m2 0.100 s = 0.672 V 224 Chapter 31 Solutions 31.17 In a toroid, all the flux is confined to the inside of the toroid. B = µ0NI 2πr = 500 µ0I 2πr ΦB = BdA ∫ = 500 µ0Imax 2π sinωt dzdr r ∫ ΦB = 500 µ0Imax 2π asinωtln b + R R ε = ′ N dΦB dt = 20 500 µ0Imax 2π ωaln b + R R cosωt ε = 104 2π 4π × 10−7 N A2 50.0 A ( ) 377 rad s 0.0200 m ( )ln 3.00 + 4.00 ( ) cm 4.00 cm cosωt = (0.422 V) cos ω t 31.18 The field inside the solenoid is: B = µ0nI = µ0 N l I Thus, through the single-turn loop ΦB = BAsolenoid = µ0 N l πr2 ( )I and the induced emf in the loop is ε = −∆ΦB ∆t = −µ0 N l πr2 ( ) ∆I ∆t = −µ0Nπr2 l I2 – I1 ∆t 31.19 ε = −N dΦB dt IR = −N dΦB dt Idt = −N R dΦB Idt ∫ = −N R dΦB ∫ Q = −N R ∆ΦB = −N R A Bf −Bi ( ) Q = − 200 5.00 Ω (100 × 10−4 m2)(−1.10 −1.10) T = 0.880 C 31.20 I = ε R = Blv R v = 1.00 m/s Chapter 31 Solutions 225 © 2000 by Harcourt, Inc. All rights reserved.
31.21 (a) FB = I 1× B = I lB. When I = E / R and ε = Blv, we get FB = Blv R (lB) = B2l2v R = (2.50)2(1.20)2(2.00) 6.00 = 3.00 N The applied force is 3.00 N to the right (b) P = I2R = B2l2v2 R = 6.00 W or P = Fv = 6.00 W 31.22 FB = IlB and E = Blv I = E R = Blv R so B = IR lv (a) FB = I2lR lv and I = FBv R = 0.500 A (b) I 2R = 2.00 W (c) For constant force, P = F ⋅v = 1.00 N ( ) 2.00 m /s ( ) = 2.00 W 31.23 The downward component of B, perpendicular to v, is (50.0 × 10–6 T) sin 58.0° = 4.24 × 10–5 T E = Blv = 4.24 × 10−5 T ( ) 60.0 m ( ) 300 m /s ( ) = 0.763 V The left wing tip is positive relative to the right.
31.24 ε = –N d dt BA cos θ = –NB cos θ ∆A ∆t ε = –1(0.100 T) cos 0° (3.00 m × 3.00 m sin 60.0°) – (3.00 m)2 0.100 s = 1.21 V I = 1.21 V 10.0 Ω = 0.121 A The flux is into the page and decreasing. The loop makes its own magnetic field into the page by carrying clockwise current.
31.25 ω = (2.00 rev/s)(2π rad/rev) = (4.00)π rad/s E = 1 2 Bω l2 = 2.83 mV 226 Chapter 31 Solutions 31.26 (a) Bext = Bext i and Bext decreases; therefore, the induced field is B0 = B0 i (to the right). Therefore, the current is to the right in the resistor.
(b) Bext = Bext (–i) increases; therefore, the induced field B0 = B0 (+ i) is to the right, and the current is to the right in the resistor.
(c) Bext = Bext (–k) into the paper and Bext decreases; therefore, the induced field is B0 = B0 (–k) into the paper. Therefore, the current is to the right in the resistor.
(d) By the Lorentz force law, FB = q (v × B). Therefore, a positive charge will move to the top of the bar if B is into the paper . (a) (b) (c) (d) 31.27 (a) The force on the side of the coil entering the field (consisting of N wires) is F = N ILB ( ) = N IwB ( ) The induced emf in the coil is ε = N dΦB dt = N d Bwx ( ) dt = NBwv, so the current is I = ε R = NBwv R counterclockwise.
The force on the leading side of the coil is then: F = N NBwv R wB = N2B2w2v R to the left (b) Once the coil is entirely inside the field, ΦB = NBA = constant, so ε = 0, I = 0, and F = 0 .
(c) As the coil starts to leave the field, the flux decreases at the rate Bwv, so the magnitude of the current is the same as in part (a), but now the current flows clockwise. Thus, the force exerted on the trailing side of the coil is: Chapter 31 Solutions 227 © 2000 by Harcourt, Inc. All rights reserved. F = N2B2w2v R to the left again 31.28 (a) Motional emf ε = Bwv appears in the conducting water. Its resistance, if the plates are submerged, is ρL A = ρw ab Kirchhoff's loop theorem says Bwv – IR – Iρw ab = 0 I = B w v R + ρw ab = abvB ρ + abR w (b) Isc = (100 m)(5.00 m)(3.00 m/s)(50.0 × 10– 6 T) 100 Ω · m = 0.750 mA 31.29 Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The loop will create a field out of the page by carrying a counterclockwise current.
Therefore, current must flow from b to a through the resistor. Hence, Va – Vb will be negative .
31.30 E = 1 2 Bω l2 = 0.259 mV 31.31 Name the currents as shown in the diagram: Left loop: + Bdv2 −I2R2 −I1R1 = 0 Right loop: + Bdv3 −I3R3 + I1R1 = 0 At the junction: I2 = I1 + I3 Then, Bdv2 −I1R2 −I3R2 −I1R1 = 0 I3 = Bdv3 R3 + I1R1 R3 So, Bdv2 −I1(R1 + R2) − Bdv3 R2 R3 − I1R1R2 R3 = 0 228 Chapter 31 Solutions I1 = Bd v2R3 −v3R2 R1R2 + R1R3 + R2R3 upward I1 = (0.0100 T)(0.100 m) (4.00 m /s)(15.0 Ω) −(2.00 m /s)(10.0 Ω) (5.00 Ω)(10.0 Ω)+(5.00 Ω)(15.0 Ω)+(10.0 Ω)(15.0 Ω) = 145 µA upward 31.32 (a) dB dt = 6.00t 2 – 8.00t ε = dΦB dt At t = 2.00 s, E = π R2(dB/dt) 2π r2 = 8.00π (0.0250)2 2π (0.0500) F = qE = 8.00 × 10–21 N clockwise for electron (b) When 6.00t2 – 8.00t = 0, t = 1.33 s 31.33 dB dt = 0.0600t ε = dΦB dt At t = 3.00 s, E = π r2 1 dB 2π r1 dt = 1.80 × 10–3 N/C perpendicular to r1 and counterclockwise 31.34 ε = dΦB dt = πr2 dB dt = E⋅d1 ∫ E(2πR) = πr2 dB dt , or E = πr2 2πR dB dt B = µ0nI dB dt = µ0n dI dt I = 3.00e0.200t dI dt = 0.600e0.200t At t = 10.0 s, E = πr2 2πR µ0n ( )(0.600e0.200t) becomes E = (0.0200 m)2 2(0.0500 m)(4π × 10−7 N/ A2)(1000 turns/m)(0.600)e2.00= 2.23 × 10−5 N/C Chapter 31 Solutions 229 © 2000 by Harcourt, Inc. All rights reserved.
31.35 (a) E⋅d1 = dΦB dt ∫ 2πrE = (πr2) dB dt so E = (9.87 mV/m) cos (100 π t) (b) The E field is always opposite to increasing B. ∴ clockwise 230 Chapter 31 Solutions 31.36 For the alternator, ω = 3000 rev min 2π rad 1 rev 1 min 60 s = 314 rad/s ε = −N dΦB dt = −250 d dt (2.50 × 10−4 T ⋅m2)cos(314 t/s) [ ] = +250(2.50 × 10–4 T · m2)(314/s) sin(314t) (a) ε = (19.6 V) sin(314t) (b) ε max = 19.6 V 31.37 (a) εmax = NABω = (1000)(0.100)(0.200)(120π) = 7.54 kV (b) ε(t) = –NBAω · sin ωt = –NBAω sin θ ε is maximal when sin θ = 1, or θ = ± π 2 , so the plane of coil is parallel to B 31.38 Let θ represent the angle through which the coil turns, starting from θ = 0 at an instant when the horizontal component of the Earth's field is perpendicular to the area. Then, ε = − N d dt BA cos θ = − NBA d dt cos ωt = + NBAω sin ωt Here sin ω t oscillates between +1 and –1, so the spinning coil generates an alternating voltage with amplitude εmax = NBAω = NBA2π f = 100(2.00 × 10−5 T)(0.200 m)2(1500) 2π rad 60.0 s = 12.6 mV 31.39 B = µ0nI = 4π × 10−7 T ⋅m A ( ) 200 m−1 ( ) 15.0 A ( ) = 3.77 × 10−3 T For the small coil, ΦB = NB⋅A = NBA cosωt = NB πr2 ( )cosωt Thus, ε = −dΦB dt = NBπr2ω sinωt ε = 30.0 ( ) 3.77 × 10−3 T ( )π 0.0800 m ( )2 4.00π s−1 ( )sin 4.00πt ( )= 28 6 4 00 .
sin . mV ( ) ( ) πt Chapter 31 Solutions 231 © 2000 by Harcourt, Inc. All rights reserved.
31.40 As the magnet rotates, the flux through the coil varies sinusoidally in time with ΦB = 0 at t = 0. Choosing the flux as positive when the field passes from left to right through the area of the coil, the flux at any time may be written as ΦB = −Φmax sinωt so the induced emf is given by ε = −dΦB dt = ωΦmax cosωt. - 1 -0.5 0 0.5 1 0 0.5 1 1.5 2 t/T = (ω t/2 π ) I / I max The current in the coil is then I = ε R = ωΦmax R cosωt = Imax cosωt 31.41 (a) F = NI lB τmax = 2Fr = NI lwB = 0.640 N · m (b) P = τω = (0.640 N · m)(120π rad/s) Pmax = 241 W (about 1 3 hp) 31.42 (a) εmax = BAω = B 1 2 πR2 ( )ω εmax = 1.30 T ( ) π 2 0.250 m ( )2 4.00π rad s εmax = 1.60 V (b) ε = ε 2π 0 2π ∫ dθ = BAω 2π sinθ dθ = 0 2π ∫ 0 (c) The maximum and average ε would remain unchanged.
(d) See Figure 1 at the right.
(e) See Figure 2 at the right.
Figure 2 ε ε Figure 1 31.43 (a) ΦB = BA cos θ = BA cos ωt = (0.800 T)(0.0100 m2) cos 2π (60.0)t = (8.00 mT · m2) cos (377t) 232 Chapter 31 Solutions (b) ε = – dΦB dt = (3.02 V) sin (377t) (c) I = ε R = (3.02 A) sin (377t) (d) P = I 2R = (9.10 W) sin2 (377t) (e) P = Fv = τ ω so τ = P ω = (24.1 mN · m) sin2 (377t) 31.44 At terminal speed, the upward magnetic force exerted on the lower edge of the loop must equal the weight of the loop. That is, Mg = FB = IwB = ε R wB = Bwvt R wB = B2w2vt R Thus, B = MgR w2vt = 0.150 kg ( ) 9.80 m s2 ( ) 0.750 Ω ( ) 1.00 m ( )2 2.00 m s ( ) = 0.742 T 31.45 See the figure above with Problem 31.44.
(a) At terminal speed, Mg = FB = IwB = ε R wB = Bwvt R wB = B2w2vt R or vt = MgR B2w2 (b) The emf is directly proportional to vt, but the current is inversely proportional to R. A large R means a small current at a given speed, so the loop must travel faster to get Fm = mg.
(c) At given speed, the current is directly proportional to the magnetic field. But the force is proportional to the product of the current and the field. For a small B, the speed must increase to compensate for both the small B and also the current, so vt ∝B2.
31.46 The current in the magnet creates an upward magnetic field, so the N and S poles on the solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as the coil approaches, so a current is induced here to create a downward magnetic field. This is clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic field by being counterclockwise as the picture correctly shows.
Chapter 31 Solutions 233 © 2000 by Harcourt, Inc. All rights reserved.
31.47 F = ma = qE + qv × B a = e m[E + v × B] where v × B = i j k 200 0 0 0.200 0.300 0.400 = −200(0.400)j + 200(0.300)k a = 1.60 × 10−19 1.67 × 10−27 [50.0j −80.0j + 60.0k] = 9.58 × 107[−30.0j + 60.0k] a = 2.87 × 109[−j + 2k] m s2 = (−2.87 × 109 j + 5.75 × 109 k) m s2 31.48 F = ma = qE + qv × B so a = −e m [E + v × B] where v × B = i j k 10.0 0 0 0 0 0.400 = −4.00j a = −1.60 × 10−19 ( ) 9.11× 10−31 [2.50i + 5.00j −4.00j] = −1.76 × 1011 ( )[2.50i + 1.00j] a = −4.39 × 1011i −1.76 × 1011 j ( ) m s2 31.49 ε = −N d dt BAcosθ ( ) = −N πr2 ( ) cos0˚ dB dt ε = −30.0 ( )π 2.70 × 10−3 m ( ) 2 1 ( ) d dt 50.0 mT + 3.20 mT ( ) sin 2π 523t s ( ) [ ] ε = −30.0 ( )π 2.70 × 10−3 m ( ) 2 3.20 × 10−3 T ( ) 2π ( ) 523 s ( ) cos 2π 523t s ( ) ε = −7.22 × 10−3 V ( ) cos 2π 523t s ( ) 31.50 (a) Doubling the number of turns.
Amplitude doubles: period unchanged (b) Doubling the angular velocity.
doubles the amplitude: cuts the period in half (c) Doubling the angular velocity while reducing the number of turns to one half the original value.
Amplitude unchanged: cuts the period in half 234 Chapter 31 Solutions 31.51 ε = −N ∆ ∆t BAcosθ ( ) = −N πr2 ( )cos0˚ ∆B ∆t = −1 0.00500 m2 ( ) 1 ( ) 1.50 T −5.00 T 20.0 × 10−3 s = 0.875 V (a) I = ε R = 0.875 V 0.0200 Ω= 43.8 A (b) P = EI = 0.875 V ( ) 43.8 A ( ) = 38.3 W 31.52 In the loop on the left, the induced emf is ε = dΦB dt = A dB dt = π 0.100 m ( )2 100 T s ( ) = π V and it attempts to produce a counterclockwise current in this loop.
In the loop on the right, the induced emf is ε = dΦB dt = π 0.150 m ( )2 100 T s ( ) = 2.25π V and it attempts to produce a clockwise current. Assume that I1 flows down through the 6.00-Ω resistor, I2 flows down through the 5.00-Ω resistor, and that I3 flows up through the 3.00-Ω resistor.
From Kirchhoff’s point rule: I3 = I1 + I2 (1) Using the loop rule on the left loop: 6.00I1 + 3.00I3 = π (2) Using the loop rule on the right loop: 5.00I2 + 3.00I3 = 2.25π (3) Solving these three equations simultaneously, I1 = 0.0623 A , I2 = 0.860 A , and I3 = 0.923 A 31.53 The emf induced between the ends of the moving bar is ε = Blv = 2.50 T ( ) 0.350 m ( ) 8.00 m s ( ) = 7.00 V The left-hand loop contains decreasing flux away from you, so the induced current in it will be clockwise, to produce its own field directed away from you. Let I1 represent the current flowing upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwise current. Let I3 be the upward current in the 5.00-Ω resistor.
Chapter 31 Solutions 235 © 2000 by Harcourt, Inc. All rights reserved.
(a) Kirchhoff’s loop rule then gives: +7.00 V −I1 2.00 Ω ( ) = 0 I1 = 3.50 A and +7.00 V −I3 5.00 Ω ( ) = 0 I3 = 1.40 A (b) The total power dissipated in the resistors of the circuit is P = EI1 + EI3 = E I1 + I3 ( ) = 7.00 V ( ) 3.50 A + 1.40 A ( ) = 34.3 W (c) Method 1 : The current in the sliding conductor is downward with value I2 = 3.50 A + 1.40 A = 4.90 A. The magnetic field exerts a force of Fm = IlB = 4.90 A ( ) 0.350 m ( ) 2.50 T ( ) = 4.29 N directed toward the right on this conductor. An outside agent must then exert a force of 4.29 N to the left to keep the bar moving. Method 2 : The agent moving the bar must supply the power according to P = F ⋅v = Fvcos0˚ . The force required is then: F = P v = 34.3 W 8.00 m s = 4.29 N 31.54 Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm.
Suppose we use a bar magnet to produce field 10−3 T through the coil in one direction along its axis. Suppose we then flip the magnet to reverse the flux in 10−1 s. The average induced emf is then ε = −N ∆ΦB ∆t = −N ∆BAcosθ [ ] ∆t = −NB πr2 ( ) cos180˚ −cos0˚ ∆t ε = −20 ( ) 10−3 T ( )π 0.0150 m ( )2 −2 10−1 s ~10−4 V 31.55 I = ε + ε Induced R and ε Induced = – d dt (BA) F = m dv dt = IBd d v dt = IBd m = Bd mR (ε + ε Induced) = Bd mR (ε – Bvd) To solve the differential equation, let u = (ε – Bvd), d u dt = –Bd dv dt .
– 1 Bd du dt = Bd mR u so u0 u ∫ du u = − t=0 t ∫ (Bd)2 mR dt Integrating from t = 0 to t = t, ln u u0 = – (Bd)2 mR t or u u0 = e−B2d2t/mR Since v = 0 when t = 0, u0 = ε and u = ε – Bvd 236 Chapter 31 Solutions ε – Bvd = εe−B2d2t/mR and v = ε Bd (1−e−B2d2t/mR) Chapter 31 Solutions 237 © 2000 by Harcourt, Inc. All rights reserved.
31.56 (a) For maximum induced emf, with positive charge at the top of the antenna, F+ = q+ (v × B), so the auto must move east (b) ε = Blv = (5.00 × 10–5 T)(1.20 m) 65.0 × 103 m 3600 s cos 65.0° = 4.58 × 10– 4 V 31.57 I = ε R = B R ∆A ∆t so q = I ∆t = (15.0 µT)(0.200 m)2 0.500 Ω = 1.20 µC Goal Solution The plane of a square loop of wire with edge length a = 0.200 m is perpendicular to the Earth's magnetic field at a point where B = 15.0 µT, as shown in Figure P31.57. The total resistance of the loop and the wires connecting it to the galvanometer is 0.500 Ω. If the loop is suddenly collapsed by horizontal forces as shown, what total charge passes through the galvanometer?
G : For the situation described, the maximum current is probably less than 1 mA. So if the loop is closed in 0.1 s, then the total charge would be Q = I∆t = 1 mA ( ) 0.1 s ( ) = 100 µC O : We do not know how quickly the loop is collapsed, but we can find the total charge by integrating the change in magnetic flux due to the change in area of the loop ( a2 → 0).
A : Q = Idt ∫ = ε dt R ∫ = 1 R −dΦB dt dt ∫ = −1 R dΦB ∫ = −1 R d(BA) = −B R ∫ dA A1=a2 A2 =0 ∫ Q = −B R A A1=a2 A2 =0 = Ba2 R = (15.0 × 10−6 T)(0.200 m)2 0.500 Ω = 1.20 × 10−6 C L : The total charge is less than the maximum charge we predicted, so the answer seems reasonable. It is interesting that this charge can be calculated without knowing either the current or the time to collapse the loop. Note: We ignored the internal resistance of the galvanometer. D’Arsonval galvanometers typically have an internal resistance of 50 to 100 Ω, significantly more than the resistance of the wires given in the problem. A proper solution that includes RG would reduce the total charge by about 2 orders of magnitude ( Q ~ 0.01 µC). 238 Chapter 31 Solutions 31.58 (a) I = dq dt = ε R where E = −N dΦB dt so dq ∫ = N R dΦB Φ1 Φ2 ∫ and the charge through the circuit will be Q = N R (Φ2 – Φ1) (b) Q = N R BAcos0 −BAcos π 2 = BAN R so B = RQ NA = (200 Ω)(5.00 × 10– 4 C) (100)(40.0 × 10– 4 m2) = 0.250 T 31.59 (a) ε = B lv = 0.360 V I = ε R = 0.900 A (b) FB = IlB = 0.108 N (c) Since the magnetic flux B · A is in effect decreasing, the induced current flow through R is from b to a. Point b is at higher potential.
(d) No . Magnetic flux will increase through a loop to the left of ab. Here counterclockwise current will flow to produce upward magnetic field. The in R is still from b to a.
31.60 ε = Blv at a distance r from wire ε = µ0I 2πr lv 31.61 (a) At time t , the flux through the loop is ΦB = BAcosθ = a + bt ( ) πr2 ( )cos0˚ = π a + bt ( )r2 At t = 0, ΦB = πar2 (b) ε = −dΦB dt = −πr2 d a + bt ( ) dt = −πbr2 (c) I = ε R = −πbr2 R (d) P = ε I = −πbr2 R −πbr2 ( ) = π 2b2r4 R Chapter 31 Solutions 239 © 2000 by Harcourt, Inc. All rights reserved.
31.62 ε = −d dt (NBA) = −1 dB dt πa2 = πa2K (a) Q = C ε = C π a2K (b) B into the paper is decreasing; therefore, current will attempt to counteract this. Positive charge will go to upper plate .
(c) The changing magnetic field through the enclosed area induces an electric field , surrounding the B-field, and this pushes on charges in the wire.
31.63 The flux through the coil is ΦB = B⋅A = BAcosθ = BAcosωt. The induced emf is ε = −N dΦB dt = −NBA d cosωt ( ) dt = NBAω sinωt.
(a) εmax = NBAω = 60.0 1.00 T ( ) 0.100 × 0.200 m2 ( ) 30.0 rad s ( ) = 36.0 V (b) dΦB dt = ε N , thus dΦB dt max = εmax N = 36.0 V 60.0 = 0.600 V = 0.600 Wb/s (c) At t t = = 0 0500 1 50 .
. s, rad ω and ε = εmax sin 1.50 rad ( ) = 36.0 V ( )sin 1.50 rad ( ) = 35.9 V (d) The torque on the coil at any time is τ = µ × B = NIA × B = NAB ( )I sinωt = εmax ω ε R sinωt When ε = εmax, sin .
ωt = 1 00 and τ = εmax 2 ωR = 36.0 V ( )2 30.0 rad s ( ) 10.0 Ω ( ) = 4.32 N · m 31.64 (a) We use ε = – N ∆ΦB ∆t , with N = 1.
Taking a = 5.00 × 10- 3 m to be the radius of the washer, and h = 0.500 m, ∆ΦB = B2A −B1A = A(B2 −B1) = πa2 µ0I 2π(h + a) −µ0I 2πa = a2µ0I 2 1 + 1 h a a − = −µ0ahI 2(h + a) The time for the washer to drop a distance h (from rest) is: ∆t = 2h g Therefore, ε = µ0ahI 2(h + a)∆t = µ0ahI 2(h + a) g 2h = µ0aI 2(h + a) gh 2 and ε = T m/A) m) A m m) m/s m) ( ( .
( .
) ( .
.
( .
)( .
4 10 5 00 10 10 0 2 0 500 0 00500 9 80 0 500 2 7 3 2 π × ⋅ × + − − = 97.4 nV (b) Since the magnetic flux going through the washer (into the plane of the paper) is decreasing in time, a current will form in the washer so as to oppose that decrease. Therefore, the current will flow in a clockwise direction .
240 Chapter 31 Solutions 31.65 ε = −N dΦB dt = −N d dt BAcosθ ( ) ε = −NBcosθ ∆A ∆t = −200 50.0 × 10−6 T ( ) cos 62.0˚ ( ) 39.0 × 10−4 m2 1.80 s = –10.2 µV 31.66 Find an expression for the flux through a rectangular area "swept out" by the bar in time t. The magnetic field at a distance x from wire is B = µ0I 2πx and ΦB = BdA.
∫ Therefore, ΦB = µ0Ivt 2π dx x r r+l ∫ where vt is the distance the bar has moved in time t. Then, ε = dΦB dt = µ0Iv 2π ln 1+ l r 31.67 The magnetic field at a distance x from a long wire is B = µ0I 2πx . Find an expression for the flux through the loop. dΦB = µ0I 2πx (ldx) so ΦB = µ0Il 2π dx x r r+w ∫ = µ0Il 2π ln 1+ w r Therefore, ε = −dΦB dt = µ0Ilv 2πr w (r + w) and I = ε R = µ0Ilv 2πRr w (r + w) 31.68 As the wire falls through the magnetic field, a motional emf ε = Blv is induced in it. Thus, a counterclockwise induced current of I = ε R = Blv R flows in the circuit.
The falling wire is carrying a current toward the left through the magnetic field. Therefore, it experiences an upward magnetic force given by FB = IlB = B2l2v R. The wire will have attained terminal speed when the magnitude of this magnetic force equals the weight of the wire.
Thus, B2l2vt R = mg, or the terminal speed is vt = mgR B2l2 31.69 ΦB = (6.00t 3 – 18.0t 2) T · m2 and ε = – dΦB dt = –18.0t 2 + 36.0t Maximum ε occurs when dε dt = – 36.0t + 36.0 = 0, which gives t = 1.00 s.
Therefore, the maximum current (at t = 1.00 s) is I = ε R = (–18.0 + 36.0)V 3.00 Ω = 6.00 A Chapter 31 Solutions 241 © 2000 by Harcourt, Inc. All rights reserved.
31.70 For the suspended mass, M: ΣF = Mg – T = Ma For the sliding bar, m: ΣF = T – I lB = ma, where I = ε R = Blv R Mg −B2l2v R = (m + M)a or a = dv dt = Mg m + M − B2l2v R(M + m) dv (α −βv) = dt 0 t ∫ 0 v ∫ where α = Mg M + m and β = B2l2 R(M + m) Therefore, the velocity varies with time as v = α β (1−e−βt) = MgR B2l2 1−e−B2l2t/R(M+m) 31.71 (a) ε = –N dΦB dt = –NA dB dt = –NA d dt (µ0nI) where A = area of coil, N = number of turns in coil, and n = number of turns per unit length in solenoid. Therefore, ε = Nµ0An d dt 4 120 sin( ) πt [ ] = Nµ0An(480π) cos (120π t) ε = 40(4π × 10–7 ) π( .
) 0 0500 2 m (480π) cos(120π t) = (1.19 V) cos(120π t) (b) I = ∆V R and P = ∆VI = (1.19 V)2 cos2(120π t) (8.00 Ω) From cos2 θ = 1 2 + 1 2 cos 2θ, the average value of cos2θ is 1 2 , so P = 1 2 1.19 V ( )2 8.00 Ω ( ) = 88.5 mW 31.72 The induced emf is ε = Blv where B = µ0I 2π y , v = vi + gt = 9.80 m s2 ( )t, and y = yi −1 2 gt2 = 0.800 m −4.90 m s2 ( )t2. ε = 4π × 10−7 T ⋅m A ( ) 200 A ( ) 2π 0.800 m −4.90 m s2 ( )t2 [ ] 0.300 m ( ) 9.80 m s2 ( )t = 1.18 × 10−4 ( )t 0.800 −4.90t2 [ ] V At t = 0.300 s , ε = 1.18 × 10−4 ( ) 0.300 ( ) 0.800 −4.90 0.300 ( )2 [ ] V = 98.3 µV 242 Chapter 31 Solutions 31.73 The magnetic field produced by the current in the straight wire is perpendicular to the plane of the coil at all points within the coil.
The magnitude of the field is B = µ0I 2πr. Thus, the flux linkage is NΦB = µ0NIL 2π dr r h h+w ∫ = µ0NImaxL 2π ln h + w h sin(ωt + φ) Finally, the induced emf is ε = −µ0NImaxLω 2π ln 1+ w h cos(ωt + φ) ε = − 4π × 10−7 ( )(100)(50.0)(0.200 m)(200π s−1) 2π ln 1+ 5.00 cm 5.00 cm cos(ωt + φ) ε = −87.1 mV ( ) cos(200πt + φ) The term sin(ωt + φ)in the expression for the current in the straight wire does not change appreciably when ωt changes by 0.100 rad or less. Thus, the current does not change appreciably during a time interval t < 0.100 (200π s−1) = 1.60 × 10−4 s.
We define a critical length, ct = (3.00 × 108 m /s)(1.60 × 10 −4 s) = 4.80 × 104 m equal to the distance to which field changes could be propagated during an interval of 1.60 × 10 −4 s. This length is so much larger than any dimension of the coilor its distance from the wire that, although we consider the straight wire to be infinitely long, we can also safely ignore the field propagation effects in the vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the vicinity of the coil.
If the frequency ω were much larger, say, 200π × 105 s−1, the corresponding critical length would be only 48.0 cm. In this situation propagation effects would be important and the above expression for ε would require modification. As a "rule of thumb" we can consider field propagation effects for circuits of laboratory size to be negligible for frequencies, f = ω 2π , that are less than about 106 Hz.
31.74 ΦB = BA cos θ dΦB dt = –ωBA sinθ ; I ∝ – sin θ τ ∝ IB sin θ ∝ – sin2 θ 31.75 The area of the tent that is effective in intercepting magnetic field lines is the area perpendicular to the direction of the magnetic field. This is the same as the base of the tent.
In the initial configuration, this is A1 = L(2L cos θ) = 2(1.50 m)2 cos 60.0˚ = 2.25 m2 After the tent is flattened, A2 = L(2L) = 2L2 = 2(1.50 m)2 = 4.50 m2 The average induced emf is: ε = −∆ΦB ∆t = −B ∆A ( ) ∆t = −0.300 T ( ) 4.50 −2.25 ( ) m2 0.100 s = – 6.75 V © 2000 by Harcourt, Inc. All rights reserved.
Chapter 32 Solutions 32.1 ε = L ∆I ∆t = (3.00 × 10– 3 H) 1.50 A −0.200 A 0.200 s = 1.95 × 10–2 V = 19.5 mV 32.2 Treating the telephone cord as a solenoid, we have: L = µ0N2A l = (4π × 10−7 T ⋅ m / A)(70.0)2(π)(6.50 × 10−3 m)2 0.600 m = 1.36 H µ 32.3 ε = + L ∆I ∆t = (2.00 H) 0.500 A 0.0100 s = 100 V 32.4 L = µ0n2Al so n = L µ0Al = 7.80 × 103 turns/m 32.5 L = N ΦB I →ΦB = LI N = 240 nT · m2 (through each turn) 32.6 ε = L dI dt where L = µ0N2A l Thus, ε = µ0N2A l dI dt = 4π × 10−7 T ⋅m A ( ) 300 ( )2 π × 10−4 m2 ( ) 0.150 m 10.0 A s ( ) = 2.37 mV 32.7 ε back = –ε = L dI dt = L d dt (Imax sin ω t) = Lω Imax cos ω t = (10.0 × 10-3)(120π )(5.00) cos ω t ε back = (6.00π ) cos (120π t) = (18.8 V) cos (377t) Chapter 32 Solutions 243 © 2000 by Harcourt, Inc. All rights reserved.
32.8 From ε = L ∆I ∆t , we have L = ε ∆Ι ∆t = 24.0 × 10– 3 V 10.0 A/s = 2.40 × 10– 3 H From L = NΦB I , we have ΦB = LI N = (2.40 × 10– 3 H)(4.00 A) 500 = 19.2 µT · m2 32.9 L = µ0N 2A l = µ0(420)2(3.00 × 10– 4) 0.160 = 4.16 × 10– 4 H ε = –L dI dt → dI dt = −ε L = –175 × 10– 6 V 4.16 × 10– 4 H = – 0.421 A/s 32.10 The induced emf is ε = −L dI dt , where the self-inductance of a solenoid is given by L = µ0N2A l . Thus, dI dt = −ε L = − εl µ0N2A 32.11 ε = L dI dt = (90.0 × 10-3) d dt (t 2 – 6t) V (a) At t = 1.00 s, ε = 360 mV (b) At t = 4.00 s, ε = 180 mV (c) ε = (90.0 × 10-3)(2t – 6) = 0 when t = 3.00 s 32.12 (a) B = µ0nI = µ0 450 0.120 (0.0400 mA) = 188 µT (b) ΦB = BA = 3.33 × 10-8 T · m2 (c) L = NΦB I = 0.375 mH 244 Chapter 32 Solutions (d) B and ΦB are proportional to current; L is independent of current Chapter 32 Solutions 245 © 2000 by Harcourt, Inc. All rights reserved.
32.13 (a) L = µ0N 2A l = µ0(120)2π ( 5.00 × 10–3)2 0.0900 = 15.8 µH (b) ′ ΦB = µm µ0 ΦB → L = µmN 2A l = 800(1.58 × 10– 5 H) = 12.6 mH 32.14 L = NΦB I = NBA I ≈ NA I · µ0NI 2π R = µ0N 2A 2π R 32.15 ε = ε0e−kt = −L dI dt dI = −ε0 L e−kt dt If we require I → 0 as t → ∞, the solution is I = ε0 kL e−kt = dq dt Q = I dt ∫ = ε0 kL e−kt 0 ∞ ∫ dt = −ε0 k2L Q = ε0 k2L 32.16 I = ε R (1−e−Rt/L) 0.900 ε R = ε R 1−e−R(3.00 s)/2.50 H [ ] exp −R(3.00 s) 2.50 H = 0.100 R = 2.50 H 3.00 s ln 10.0 = 1.92 Ω 246 Chapter 32 Solutions 32.17 τ = L R = 0.200 s: I Imax = 1 – e–t/τ (a) 0.500 = 1 – e–t/0.200 → t = τ ln 2.00 = 0.139 s (b) 0.900 = 1 – e–t/0.200 → t = τ ln 10.0 = 0.461 s Figure for Goal Solution Goal Solution A 12.0-V battery is about to be connected to a series circuit containing a 10.0-Ω resistor and a 2.00-H inductor. How long will it take the current to reach (a) 50.0% and (b) 90.0% of its final value? G : The time constant for this circuit is τ = L R = 0.2 s, which means that in 0.2 s, the current will reach 1/e = 63% of its final value, as shown in the graph to the right. We can see from this graph that the time to reach 50% of Imax should be slightly less than the time constant, perhaps about 0.15 s, and the time to reach 0.9Imax should be about 2.5τ = 0.5 s. O : The exact times can be found from the equation that describes the rising current in the above graph and gives the current as a function of time for a known emf, resistance, and time constant. We set time t = 0 to be the moment the circuit is first connected.
A : At time t, I t ( ) = ε(1−e−t/τ ) R where, after a long time, Imax = ε(1−e−∞) R = ε R At I t ( ) = 0.500Imax, 0.500 ( )ε R = ε(1−e−t/0.200 s) R so 0.500 = 1−e−t/0.200 s Isolating the constants on the right, ln e−t/2.00 s ( ) = ln 0.500 ( ) and solving for t, − t 0.200 s = −0.693 or t = 0.139 s (b) Similarly, to reach 90% of Imax, 0.900 = 1−e−t/τ and t = −τ ln 1−0.900 ( ) Thus, t = −0.200 s ( )ln 0.100 ( ) = 0.461 s L : The calculated times agree reasonably well with our predictions. We must be careful to avoid confusing the equation for the rising current with the similar equation for the falling current.
Checking our answers against predictions is a safe way to prevent such mistakes.
Chapter 32 Solutions 247 © 2000 by Harcourt, Inc. All rights reserved.
32.18 Taking τ = L R, I = I0e−t/τ : dI dt = I0e−t/τ −1 τ IR + L dI dt = 0 will be true if I0Re−t/τ + L I0 e−t/τ ( ) −1 τ = 0 Because τ = L R, we have agreement with 0 = 0 32.19 (a) τ = L R = 2.00 × 10– 3 s = 2.00 ms (b) I = Imax 1−e−t/τ ( ) = 6.00 V 4.00 Ω 1−e−0.250/2.00 ( ) = 0.176 A (c) Imax = ε R = 6.00 V 4.00 Ω = 1.50 A (d) 0.800 = 1 – e–t/2.00 ms → t = – (2.00 ms) ln(0.200) = 3.22 ms 32.20 I = ε R (1 – e–t/τ ) = 120 9.00 (1 – e–1.80/7.00) = 3.02 A ∆VR = IR = (3.02)(9.00) = 27.2 V ∆VL = ε – ∆VR = 120 – 27.2 = 92.8 V 32.21 (a) ∆VR = IR = (8.00 Ω)(2.00 A) = 16.0 V and ∆VL = ε −∆VR = 36.0 V −16.0 V = 20.0 V Therefore, ∆VR ∆VL = 16.0 V 20.0 V = 0.800 (b) ∆VR = IR = (4.50 A)(8.00 Ω) = 36.0 V ∆VL = ε −∆VR = 0 Figure for Goal Solution 248 Chapter 32 Solutions Goal Solution For the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 Ω, and ε = 36.0 V. (a) Calculate the ratio of the potential difference across the resistor to that across the inductor when I = 2.00 A. (b) Calculate the voltage across the inductor when I = 4.50 A.
G : The voltage across the resistor is proportional to the current, ∆ IR VR = , while the voltage across the inductor is proportional to the rate of change in the current, εL = −LdI dt. When the switch is first closed, the voltage across the inductor will be large as it opposes the sudden change in current. As the current approaches its steady state value, the voltage across the resistor increases and the inductor’s emf decreases. The maximum current will be ε /R = 4.50 A, so when I = 2.00 A, the resistor and inductor will share similar voltages at this mid-range current, but when I = 4.50 A, the entire circuit voltage will be across the resistor, and the voltage across the inductor will be zero. O : We can use the definition of resistance to calculate the voltage across the resistor for each current.
We will find the voltage across the inductor by using Kirchhoff's loop rule.
A : (a) When I = 2.00 A, the voltage across the resistor is ∆VR = IR = 2.00 A ( ) 8.00 Ω ( ) = 16.0 V Kirchhoff's loop rule tells us that the sum of the changes in potential around the loop must be zero: ε −∆VR −εL = 36.0 V −16.0 V −εL = 0 so εL = 20.0 V and ∆VR εL = 16.0 V 20.0 V = 0.800 (b) Similarly, for I = 4.50 A, ∆VR = IR = 4.50 A ( ) 8.00 Ω ( ) = 36.0 V ε −∆VR −εL = 36.0 V −36.0 V −εL = 0 so εL = 0 L : We see that when I = 2.00 A, ∆VR < εL, but they are similar in magnitude as expected. Also as predicted, the voltage across the inductor goes to zero when the current reaches its maximum value.
A worthwhile exercise would be to consider the ratio of these voltages for several different times after the switch is reopened.
32.22 After a long time, 12.0 V = (0.200 A)R Thus, R = 60.0 Ω. Now, τ = L R gives L = τ R = (5.00 × 10– 4 s)(60.0 V/A) = 30.0 mH 32.23 I = Imax 1−e−t/τ ( ): dI dt = −Imax e−t/τ ( ) −1 τ τ = L R = 15.0 H 30.0 Ω = 0.500 s : dI dt = R L Imax e–t/τ and Imax = ε R (a) t = 0: dI dt = R L Imax e0 = ε L = 100 V 15.0 H = 6.67 A/s (b) t = 1.50 s: dI dt = ε L e–t/τ = (6.67 A/s)e– 1.50/(0.500) = (6.67 A/s)e–3.00 = 0.332 A/s Chapter 32 Solutions 249 © 2000 by Harcourt, Inc. All rights reserved.
32.24 I = Imax 1−e−t/τ ( ) 0.980 = 1−e−3.00×10−3/τ 0.0200 = e−3.00×10−3/τ τ = −3.00 × 10−3 ln(0.0200) = 7.67 × 10−4 s τ = L R, so L = τR = (7.67 × 10−4)(10.0) = 7.67 mH 32.25 Name the currents as shown. By Kirchhoff’s laws: I1 = I2 + I3 (1) +10.0 V −4.00I1 −4.00I2 = 0 (2) +10.0 V −4.00I1 −8.00I3 −1.00 ( ) dI3 dt = 0 (3) From (1) and (2), +10.0 −4.00I1 −4.00I1 + 4.00I3 = 0 and I1 = 0.500I3 + 1.25 A Then (3) becomes 10.0 V −4.00 0.500I3 + 1.25 A ( ) −8.00I3 −1.00 ( ) dI3 dt = 0 1.00 H ( ) dI3 dt ( ) + 10.0 Ω ( )I3 = 5.00 V We solve the differential equation using Equations 32.6 and 32.7: I3 t ( ) = 5.00 V 10.0 Ω1−e−10.0 Ω ( )t 1.00 H [ ] = 0.500 A ( ) 1−e−10t/s [ ] I1 = 1.25 + 0.500I3 = 1.50 A −0.250 A ( )e−10t/s 32.26 (a) Using τ = RC = L R , we get R = L C = 3.00 H 3.00 × 10−6 F = 1.00 × 103 Ω= 1.00 kΩ (b) τ = RC = 1.00 × 103 Ω ( ) 3.00 × 10−6 F ( ) = 3.00 × 10−3 s = 3.00 ms 250 Chapter 32 Solutions 32.27 For t ≤0, the current in the inductor is zero. At t = 0, it starts to grow from zero toward 10.0 A with time constant τ = L R = 10.0 mH ( ) 100 Ω ( ) = 1.00 × 10−4 s.
For 0 ≤t ≤200 µs, I = Imax 1−e −t/τ = 10.00 A ( ) 1−e−10000t/s ( ) At t = 200 µs, I = 10.00 A ( ) 1−e−2.00 ( ) = 8.65 A Thereafter, it decays exponentially as I = I0e−′ t τ , so for t ≥200 µs, I = 8.65 A ( )e−10000 t−200 µs ( ) s = 8.65 A ( )e−10000t s +2.00 = 8.65e2.00 A ( )e−10000t s = 63.9 A ( )e−10000t s 32.28 (a) I = ε R = 12.0 V 12.0 Ω = 1.00 A (b) Initial current is 1.00 A, : ∆V12 = (1.00 A)(12.00 Ω) = 12.0 V ∆V1200 = (1.00 A)(1200 Ω) = 1.20 kV ∆VL = 1.21 kV (c) I = Imax e–Rt/L: dI dt = – Imax R L e–Rt/L and –L dI dt = ∆VL = Imax Re–Rt/L Solving 12.0 V = (1212 V)e–1212t/2.00 so 9.90 × 10– 3 = e– 606t Thus, t = 7.62 ms 32.29 τ = L R = 0.140 4.90 = 28.6 ms; Imax = ε R = 6.00 V 4.90 Ω= 1.22 A (a) I = Imax 1−e−t/τ ( ) so 0.220 = 1.22 1−e−t/τ ( ) e−t/τ = 0.820 t = −τ ln(0.820) = 5.66 ms (b) I = Imax 1−e − 10.0 0.0286 = (1.22 A) 1−e−350 ( ) = 1.22 A (c) I = Imaxe−t/τ and 0.160 = 1.22e−t/τ so t = –τ ln(0.131) = 58.1 ms Chapter 32 Solutions 251 © 2000 by Harcourt, Inc. All rights reserved.
32.30 (a) For a series connection, both inductors carry equal currents at every instant, so dI/dt is the same for both. The voltage across the pair is Leq dI dt = L1 dI dt + L 2 dI dt so Leq = L1 + L 2 (b) Leq dI dt = L1 dI1 dt = L 2 dI2 dt = ∆VL where I = I1 + I2 and dI dt = dI1 dt + dI2 dt Thus, ∆VL Leq = ∆VL L1 + ∆VL L 2 and 1 Leq = 1 L1 + 1 L 2 (c) Leq dI dt + R eq I = L1 dI dt + IR1 + L 2 dI dt + IR 2 Now I and dI/dt are separate quantities under our control, so functional equality requires both Leq = L1 + L 2 and R eq = R 1 + R 2 (d) ∆V = Leq dI dt + R eqI = L1 dI1 dt + R1I1 = L 2 dI2 dt + R 2I2 where I = I1 + I2 and dI dt = dI1 dt + dI2 dt We may choose to keep the currents constant in time. Then, 1 R eq = 1 R 1 + 1 R 2 We may choose to make the current swing through 0. Then, 1 Leq = 1 L1 + 1 L 2 This equivalent coil with resistance will be equivalent to the pair of real inductors for all other currents as well. 32.31 L = N ΦB I = 200(3.70 × 10– 4) 1.75 = 42.3 mH so U = 1 2 LI 2 = 1 2 (0.423 H)(1.75 A) 2 = 0.0648 J 32.32 (a) The magnetic energy density is given by u = B 2 2µ0 = (4.50 T)2 2(1.26 × 10– 6 T · m/A) = 8.06 × 106 J/m3 (b) The magnetic energy stored in the field equals u times the volume of the solenoid (the volume in which B is non-zero).
U = uV = (8.06 × 106 J/m3) (0.260 m)π(0.0310 m)2 [ ] = 6.32 kJ 252 Chapter 32 Solutions 32.33 L = µ0 N 2A l = µ0 (68.0)2 π(0.600 × 10−2)2 0.0800 = 8.21 µH U = 1 2 LI 2 = 1 2 (8.21× 10−6 H)(0.770 A)2 = 2.44 µJ 32.34 (a) U = 1 2 LI 2 = 1 2 L ε 2R 2 = Lε 2 8R2 = (0.800)(500)2 8(30.0)2 = 27.8 J (b) I = ε R 1−e−(R/L)t [ ] so ε 2R = ε R 1−e−(R/L)t [ ] →e−(R/L)t = 1 2 R L t = ln2 so t = L R ln2 = 0.800 30.0 ln2 = 18.5 ms 32.35 u = ε 0 E 2 2 = 44.2 nJ/m3 u = B 2 2µ0 = 995 µ J/m3 32.36 (a) U = 1 2 LI 2 = 1 2 (4.00 H)(0.500 A) 2 = 0.500 J (b) dU dt = LI = (4.00 H)(1.00 A) = 4.00 J/s = 4.00 W (c) P = (∆V)I = (22.0 V)(0.500 A) = 11.0 W 32.37 From Equation 32.7, I = ε R 1−e−Rt L ( ) (a) The maximum current, after a long time t , is I = ε R = 2.00 A.
At that time, the inductor is fully energized and P = I(∆V) = (2.00 A)(10.0 V) = 20.0 W (b) Plost = I 2R = (2.00 A)2(5.00 Ω) = 20.0 W (c) Pinductor = I(∆Vdrop) = 0 (d) U = LI 2 2 = (10.0 H)(2.00 A)2 2 = 20.0 J Chapter 32 Solutions 253 © 2000 by Harcourt, Inc. All rights reserved.
32.38 We have u = e0 E2 2 and u = B2 2µ0 Therefore e0 E2 2 = B2 2µ0 so B2 = e0µ0E2 B = E e0µ0 = 6.80 × 105 V /m 3.00 × 108 m /s = 2.27 × 10– 3 T 32.39 The total magnetic energy is the volume integral of the energy density, u = B2 2µ0 Because B changes with position, u is not constant. For B = B0 R/r ( )2, u = B0 2 2µ0 R r 4 Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness dr. Such a shell has a volume 4π r 2 dr, so the energy stored in it is dU = u 4π r2dr ( ) = 2πB0 2R4 µ0 dr r2 We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the sphere. This gives U = 2π B2 0 R 3 µ0 = 2π (5.00 × 10–5 T)2(6.00 × 106 m)3 (1.26 × 10– 6 T · m/A) = 2.70 × 1018 J 32.40 I1(t) = Imaxe−αt sinωt with Imax = 5.00 A, α = 0.0250 s−1, and ω = 377 rad s. dI1 dt = Imaxe−αt −α sinωt + ω cosωt ( ) At t = 0.800 s, dI1 dt = 5.00 A s ( )e−0.0200 −0.0250 ( )sin 0.800 377 ( ) ( ) + 377cos 0.800 377 ( ) ( ) [ ] dI1 dt = 1.85 × 103 A s Thus, ε2 = −M dI1 dt : M = −ε2 dI1 dt = +3.20 V 1.85 × 103 A s = 1.73 mH 254 Chapter 32 Solutions 32.41 ε2 = −M dI1 dt = −(1.00 × 10−4 H)(1.00 × 104 A/s) cos(1000t) ε2 ( )max = 1.00 V 32.42 M = ε2 dI1 dt = 96.0 mV 1.20 A/s = 80.0 mH 32.43 (a) M = NBΦBA IA = 700(90.0 × 10−6) 3.50 = 18.0 mH (b) LA = ΦA IA = 400(300 × 10−6) 3.50 = 34.3 mH (c) εB = −M dIA dt = −(18.0 mH)(0.500 A/s) = – 9.00 mV 32.44 M = N2Φ12 I1 = N2 B1A1 ( ) I1 = N2 µ0n1I1 ( )A1 [ ] I1 = N2 µ 0n1A1 M = (1.00) 4π × 10−7 T ⋅m A ( ) 70.0 0.0500 m π 5.00 × 10−3 m ( ) 2 = 138 nH 32.45 B at center of (larger) loop: B1 = µ0I1 2R (a) M = Φ2 I1 = B1A 2 I1 = (µ0I1 / 2R)(πr2) I1 = µ π 0 2 2 r R (b) M = µ0π(0.0200)2 2(0.200) = 3.95 nH Chapter 32 Solutions 255 © 2000 by Harcourt, Inc. All rights reserved.
32.46 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at distance x from the wire is B = µ0I 2πx, and this field passes perpendicularly through the plane of the loop. The flux through the loop is ΦB = B⋅dA = BdA ∫ = B ldx ( ) ∫ = ∫ µ0I l 2π dx x 0.400 mm 1.70 mm ∫ = µ0Il 2π ln 1.70 0.400 The mutual inductance between the wire and the loop is then M = N2Φ12 I1 = N2µ0Il 2πI ln 1.70 0.400 = N2µ0l 2π 1.45 ( ) = 1(4π × 10−7 T ⋅m A)(2.70 × 10−3 m) 2π 1.45 ( ) M = 7.81× 10−10 H = 781 pH 32.47 With I = I1 + I2, the voltage across the pair is: ∆V = −L1 dI1 dt −M dI2 dt = −L 2 dI2 dt −M dI1 dt = −Leq dI dt So, −dI1 dt = ∆V L1 + M L1 dI2 dt and −L 2 dI2 dt + M ∆V ( ) L1 + M2 L1 dI2 dt = ∆V (a) (b) (−L1L2 + M2) dI2 dt = ∆V(L1 −M) By substitution, −dI2 dt = ∆V L 2 + M L 2 dI1 dt leads to (−L1L 2 + M 2) dI1 dt = ∆V (L2 −M) Adding to , (−L1L 2 + M 2) dI dt = ∆V(L1 + L 2 −2M) So, Leq = − ∆V dI / dt = L1L 2 −M 2 L1 + L 2 −2M 32.48 At different times, UC ( )max = UL ( )max so 1 2 C ∆V ( )2 [ ]max = 1 2 LI2 ( )max Imax = C L ∆V ( )max = 1.00 × 10−6 F 10.0 × 10−3 H 40.0 V ( ) = 0.400 A 256 Chapter 32 Solutions 32.49 1 2 C ∆V ( )2 [ ]max = 1 2 LI2 ( )max so ∆VC ( )max = L C Imax = 20.0 × 10−3 H 0.500 × 10−6 F 0.100 A ( ) = 20.0 V 32.50 When the switch has been closed for a long time, battery, resistor, and coil carry constant current Imax = ε / R. When the switch is opened, current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the LC loop.
We interpret the problem to mean that the voltage amplitude of these oscillations is ∆V, in 1 2 C ∆V ( )2 = 1 2 LImax 2 .
Then, L = C ∆V ( ) 2 Imax 2 = C ∆V ( )2R2 ε 2 = 0.500 × 10−6 F ( ) 150 V ( )2 250 Ω ( )2 50.0 V ( )2 = 0.281 H 32.51 C = 1 (2π f)2L = 1 (2π ⋅6.30 × 106)2 (1.05 × 106) = 608 pF Goal Solution A fixed inductance L = 1.05 µ H is used in series with a variable capacitor in the tuning section of a radio.
What capacitance tunes the circuit to the signal from a station broadcasting at 6.30 MHz?
G : It is difficult to predict a value for the capacitance without doing the calculations, but we might expect a typical value in the µF or pF range.
O : We want the resonance frequency of the circuit to match the broadcasting frequency, and for a simple RLC circuit, the resonance frequency only depends on the magnitudes of the inductance and capacitance.
A : The resonance frequency is f0 = 1 2π LC Thus, C = 1 (2π f0)2L = 1 (2π)(6.30 × 106 Hz) [ ] 2(1.05 × 10−6 H) = 608 pF L : This is indeed a typical capacitance, so our calculation appears reasonable. However, you probably would not hear any familiar music on this broadcast frequency. The frequency range for FM radio broadcasting is 88.0 – 108.0 MHz, and AM radio is 535 – 1605 kHz. The 6.30 MHz frequency falls in the Maritime Mobile SSB Radiotelephone range, so you might hear a ship captain instead of Top 40 tunes! This and other information about the radio frequency spectrum can be found on the National Telecommunications and Information Administration (NTIA) website, which at the time of this printing was at Chapter 32 Solutions 257 © 2000 by Harcourt, Inc. All rights reserved.
32.52 f = 1 2π LC : L = 1 (2π f)2C = 1 (2π ⋅120)2(8.00 × 10−6) = 0.220 H 32.53 (a) f = 1 2π LC = 1 2π (0.0820 H)(17.0 × 10−6 F) = 135 Hz (b) Q = Qmax cosωt = (180 µC) cos(847 × 0.00100) = 119 µC (c) I = dQ dt = −ωQmax sinωt = −(847)(180) sin(0.847) = – 114 mA 32.54 (a) f = 1 2π LC = 1 2π (0.100 H)(1.00 × 10−6 F) = 503 Hz (b) Q = Cε = (1.00 × 10−6 F)(12.0 V) = 12.0 µC (c) 1 2 Cε 2 = 1 2 LImax 2 Imax = ε C L = 12 V 1.00 × 10−6 F 0.100 H = 37.9 mA (d) At all times U = 1 2 Cε 2 = 1 2 (1.00 × 10−6 F)(12.0 V)2 = 72.0 µ J 32.55 ω = 1 LC = 1 3.30 H ( ) 840 × 10−12 F ( ) = 1.899 × 104 rad s Q = Qmax cosωt, I = dQ dt = −ωQmax sinωt (a) UC = Q2 2C = 105 × 10−6 [ ]cos 1.899 × 104 rad s ( ) 2.00 × 10−3 s ( ) [ ] ( ) 2 2 840 × 10−12 ( ) = 6.03 J (b) UL = 1 2 LI2 = 1 2 Lω 2Qmax 2 sin2 ωt ( ) = Qmax 2 sin2 ωt ( ) 2C UL = 105 × 10−6 C ( ) 2 sin2 1.899 × 104 rad s ( ) 2.00 × 10−3 s ( ) [ ] 2 840 × 10−12 F ( ) = 0.529 J (c) Utotal = UC + UL = 6.56 J 258 Chapter 32 Solutions 32.56 (a) ωd = 1 LC − R 2L 2 = 1 2.20 × 10−3 ( ) 1.80 × 10−6 ( ) − 7.60 2 2.20 × 10−3 ( ) 2 = 1.58 × 104 rad/s Therefore, fd = ωd 2π = 2.51 kHz (b) Rc = 4L C = 69.9 Ω 32.57 (a) ω0 = 1 LC = 1 (0.500)(0.100 × 10−6) = 4.47 krad/s (b) ωd = 1 LC − R 2L 2 = 4.36 krad/s (c) ∆ω ω0 = 2.53% lower 32.58 Choose to call positive current clockwise in Figure 32.19. It drains charge from the capacitor according to I = – dQ/dt. A clockwise trip around the circuit then gives + Q C −IR −L dI dt = 0 + Q C + dQ dt R + L d dt dQ dt = 0, identical with Equation 32.29.
32.59 (a) Q = Qmaxe −Rt 2L cos ωdt so Imax ∝e −Rt 2L 0.500 = e −Rt 2L and Rt 2L = −ln(0.500) t = −2L R ln 0.500 ( ) = 0.693 2L R (b) U0 ∝Qmax 2 and U = 0.500U0 so Q = 0.500 Qmax = 0.707Qmax t = −2L R ln(0.707) = 0.347 2L R (half as long) Chapter 32 Solutions 259 © 2000 by Harcourt, Inc. All rights reserved.
32.60 With Q = Qmax at t = 0, the charge on the capacitor at any time is Q = Qmax cosωt where ω = 1 LC . The energy stored in the capacitor at time t is then U = Q2 2C = Qmax 2 2C cos2 ωt = U0 cos2 ωt.
When U = 1 4U0, cos ω t = 1 2 and ω t = 1 3 π rad Therefore, t LC = π 3 or t2 LC = π 2 9 The inductance is then: L = 9t 2 π 2C 32.61 (a) εL = −L dI dt = −1.00 mH ( ) d 20.0t ( ) dt = – 20.0 mV (b) Q = I dt 0 t ∫ = 20.0t ( )dt 0 t ∫ = 10.0t 2 ∆VC = −Q C = −10.0t 2 1.00 × 10−6 F = −10.0 MV s2 ( )t 2 (c) When Q 2 2C ≥1 2 LI 2, or −10.0t 2 ( ) 2 2 1.00 × 10−6 ( ) ≥1 2 1.00 × 10−3 ( ) 20.0t ( ) 2, then 100t 4 ≥400 × 10−9 ( )t 2. The earliest time this is true is at t = 4.00 × 10−9 s = 63.2 µs 32.62 (a) ε L = −L dI dt = −L d dt (Kt) = –LK (b) I = dQ dt , so Q = I dt 0 t ∫ = Ktdt 0 t ∫ = 1 2 Kt 2 and ∆VC = −Q C = −Kt 2 2C (c) When 1 2 C ∆VC ( ) 2 = 1 2 LI 2, 1 2 C K2 t 4 4C 2 = 1 2 L K 2t 2 ( ) Thus t = 2 LC 260 Chapter 32 Solutions 32.63 1 2 Q2 C = 1 2C Q 2 2 + 1 2 LI2 so I = 3Q 2 4CL The flux through each turn of the coil is ΦB = LI N = Q 2N 3L C where N is the number of turns.
32.64 Equation 30.16: B = µ0NI 2πr (a) ΦB = BdA ∫ = µ0NI 2πr hdr a b ∫ = µ0NIh 2π dr r a b ∫ = µ0NIh 2π ln b a L = NΦB I = µ0N2h 2π ln b a (b) L = µ0(500)2(0.0100) 2π ln 12.0 10.0 = 91.2 µH (c) Lappx = µ0N2 2π A R = µ0(500)2 2π 2.00 × 10−4 m2 0.110 = 90.9 µH 32.65 (a) At the center, B = Nµ0IR2 2(R2 + 02)3/2 = Nµ0I 2R So the coil creates flux through itself ΦB ≈BAcosθ = Nµ0I 2R πR2 cos0°= π 2 Nµ0IR When the current it carries changes, εL = −N dΦB dt ≈−N π 2 Nµ0R dI dt = −L dI dt so L ≈π 2 N2µ0R (b) 2π r ≈ 3(0.3 m), so r ≈ 0.14 m; L ≈ π 2 12 4π × 10–7 T · m A 0.14 m = 2.8 × 10–7 H ~ 100 nH (c) L R ≈ 2.8 × 10–7 V · s/A 270 V/A = 1.0 × 10– 9 s ~ 1 ns Chapter 32 Solutions 261 © 2000 by Harcourt, Inc. All rights reserved.
32.66 (a) If unrolled, the wire forms the diagonal of a 0.100 m (10.0 cm) rectangle as shown. The length of this rectangle is ′ L = 9.80 m ( )2 −0.100 m ( )2 ′ L 0.100 m 9.80 m The mean circumference of each turn is C r = ′ 2π , where ′ r = 24.0 + 0.644 2 mm is the mean radius of each turn. The number of turns is then: N = ′ L C = 9.80 m ( )2 −0.100 m ( )2 2π 24.0 + 0.644 2 × 10−3 m = 127 (b) R = ρl A = 1.70 × 10−8 Ω⋅m ( ) 10.0 m ( ) π 0.322 × 10−3 m ( ) 2 = 0.522 Ω (c) L = µN2A ′ l = 800µ0 ′ l ′ L C 2 π ′ r ( )2 L = 800 4π × 10−7 ( ) 0.100 m 9.80 m ( )2 −0.100 m ( )2 π 24.0 + 0.644 ( ) × 10−3 m 2 π 24.0 + 0.644 2 × 10−3 m 2 L = 7.68 × 10−2 H = 76.8 mH 32.67 From Ampere’s law, the magnetic field at distance r ≤R is found as: B 2πr ( ) = µ0J πr 2 ( ) = µ0 I πR2 πr 2 ( ), or B = µ0Ir 2πR2 The magnetic energy per unit length within the wire is then U l = B2 2µ0 2πrdr ( ) 0 R ∫ = µ0I2 4πR4 r 3 dr 0 R ∫ = µ0I2 4πR4 R4 4 = µ0I 2 16π This is independent of the radius of the wire.
262 Chapter 32 Solutions 32.68 The primary circuit (containing the battery and solenoid) is an RL circuit with R = 14.0 Ω, and L = µ0N2A l = 4π × 10−7 ( ) 12 500 ( ) 2 1.00 × 10−4 ( ) 0.0700 = 0.280 H (a) The time for the current to reach 63.2% of the maximum value is the time constant of the circuit: τ = L R = 0.280 H 14.0 Ω= 0.0200 s = 20.0 ms (b) The solenoid's average back emf is ε L = L ∆I ∆t = L If −0 ∆t where If = 0.632Imax = 0.632 ∆V R = 0.632 60.0 V 14.0 Ω = 2.71 A Thus, εL = 0.280 H ( ) 2.71 A 0.0200 s = 37.9 V (c) The average rate of change of flux through each turn of the overwrapped concentric coil is the same as that through a turn on the solenoid: ∆ΦB ∆t = µ0n ∆I ( )A ∆t = 4π × 10−7 T ⋅m A ( ) 12500 0.0700 m ( ) 2.71 A ( ) 1.00 × 10−4 m2 ( ) 0.0200 s = 3.04 mV (d) The magnitude of the average induced emf in the coil is εL = N ∆ΦB ∆t ( ) and magnitude of the average induced current is I = ε L R = N R ∆ΦB ∆t = 820 24.0 Ω3.04 × 10−3 V ( ) = 0.104 A = 104 mA 32.69 Left-hand loop: E −(I + I2)R 1 −I2R2 = 0 Outside loop: E −(I + I2)R 1 −L dI dt = 0 Eliminating I 2 gives ′ E −I ′ R −L dI dt = 0 This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: I t ( ) = ′ E ′ R (1−e− ′ R t L) But ′ R = R1R2 / R1 + R2 ( ) and ′ E = R2E / R1 + R2 ( ), so ′ E ′ R = ER 2 /(R 1 + R2) R 1R2 /(R1 + R2) = E R 1 Thus I(t) = E R 1 (1−e− ′ R t L) Chapter 32 Solutions 263 © 2000 by Harcourt, Inc. All rights reserved.
32.70 When switch is closed, steady current I0 = 1.20 A. When the switch is opened after being closed a long time, the current in the right loop is I = I0e−R2 t L so e Rt L = I0 I and Rt L = ln I0 I Therefore, L = R2 t ln I0 I ( ) = 1.00 Ω ( ) 0.150 s ( ) ln 1.20 A 0.250 A ( ) = 0.0956 H = 95.6 mH 32.71 (a) While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives: +ε0 − 2.00 + 6.00 ( ) × 103 Ω [ ] 9.00 × 10−3 A ( ) = 0 +ε0 = 72 0 . V with end at the higher potential b (b) (c) After the switch is opened, the current around the outer loop decays as I = Imaxe−Rt L with Imax = 9.00 mA, R = 8.00 kΩ, and L = 0.400 H Thus, when the current has reached a value I = 2.00 mA, the elapsed time is: t = L R ln Imax I = 0.400 H 8.00 × 103 Ω ln 9.00 2.00 = 7.52 × 10−5 s = 75.2 µs 264 Chapter 32 Solutions 32.72 (a) The instant after the switch is closed, the situation is as shown in the circuit diagram of Figure (a). The requested quantities are: IL = 0, IC = ε0 R, IR = ε0 R ∆VL = ε0, ∆VC = 0, ∆VR = ε0 (b) After the switch has been closed a long time, the steady-state conditions shown in Figure (b) will exist. The currents and voltages are: IL = 0, IC = 0, IR = 0 ∆VL = 0, ∆VC = ε0, ∆VR = 0 IR = 0 + -ε0 Q = 0 ∆VC = 0 IR = ε0/R + -IL = 0 ∆VL = ε0 ∆VR = ε0 Figure (a) + -ε0 Q = Cε0 ∆VC = ε0 + -IL = 0 ∆VL = 0 ∆VR = 0 Figure (b) IC = ε0/R 32.73 When the switch is closed, as shown in Figure (a), the current in the inductor is I : 12.0 – 7.50I – 10.0 = 0 → I = 0.267 A When the switch is opened, the initial current in the inductor remains at 0.267 A.
IR = ∆V: (0.267 A)R ≤ 80.0 V R ≤ 300 Ω (a) (b) Goal Solution To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallel with the armature. If the motor is suddenly unplugged while running, this resistor limits the voltage that appears across the armature coils. Consider a 12.0-V dc motor with an armature that has a resistance of 7.50 Ω and an inductance of 450 mH. Assume that the back emf in the armature coils is 10.0 V when the motor is running at normal speed. (The equivalent circuit for the armature is shown in Figure P32.73.) Calculate the maximum resistance R that limits the voltage across the armature to 80.0 V when the motor is unplugged.
Chapter 32 Solutions 265 © 2000 by Harcourt, Inc. All rights reserved.
G : We should expect R to be significantly greater than the resistance of the armature coil, for otherwise a large portion of the source current would be diverted through R and much of the total power would be wasted on heating this discharge resistor. O : When the motor is unplugged, the 10-V back emf will still exist for a short while because the motor’s inertia will tend to keep it spinning. Now the circuit is reduced to a simple series loop with an emf, inductor, and two resistors. The current that was flowing through the armature coil must now flow through the discharge resistor, which will create a voltage across R that we wish to limit to 80 V. As time passes, the current will be reduced by the opposing back emf, and as the motor slows down, the back emf will be reduced to zero, and the current will stop.
A : The steady-state coil current when the switch is closed is found from applying Kirchhoff's loop rule to the outer loop: + 12.0 V −I 7.50 Ω ( ) −10.0 V = 0 so I = 2.00 V 7.50 Ω= 0.267 A We then require that ∆VR = 80.0 V = 0.267 A ( )R so R = ∆VR I = 80.0 V 0.267 A = 300 Ω L : As we expected, this discharge resistance is considerably greater than the coil’s resistance. Note that while the motor is running, the discharge resistor turns P = (12 V)2 300 Ω= 0.48 W of power into heat (or wastes 0.48 W). The source delivers power at the rate of about P = IV = 0.267 A + 12 V / 300 Ω ( ) [ ] 12 V ( ) = 3.68 W, so the discharge resistor wastes about 13% of the total power. For a sense of perspective, this 4-W motor could lift a 40-N weight at a rate of 0.1 m/s. 32.74 (a) L1 = µ0N1 2A l 1 = 4π × 10−7 T ⋅m A ( ) 1000 ( )2 1.00 × 10−4 m2 ( ) 0.500 m = 2.51× 10−4 H = 251 µH (b) M = N2Φ2 I1 = N2Φ1 I1 = N2BA I1 = N2 µ0 N1 l1 ( )I1 [ ]A I1 = µ0N1N2A l1 M = 4π × 10−7 T ⋅m A ( ) 1000 ( ) 100 ( ) 1.00 × 10−4 m2 ( ) 0.500 m = 2.51× 10−5 H = 25.1 µH (c) ε 1 = −M dI2 dt , or I1R1 = −M dI2 dt and I1 = dQ1 dt = −M R1 dI2 dt Q1 = −M R1 dI2 0 tf ∫ = −M R1 I2f −I2i ( ) = −M R1 0 −I2i ( ) = M I2i R1 Q1 = 2.51× 10−5 H ( ) 1.00 A ( ) 1000 Ω = 2.51× 10−8 C = 25.1 nC 266 Chapter 32 Solutions 32.75 (a) It has a magnetic field, and it stores energy, so L = 2U I 2 is non-zero.
(b) Every field line goes through the rectangle between the conductors.
(c) Φ = LI so L = Φ I = 1 I Bda y=a w−a ∫ L = 1 I xdy a w−a ∫ µ0I 2π y + µ0I 2π w −y ( ) = 2 I µ0Ix 2π y dy = ∫ 2µ0x 2π ln y a w−a Thus L = µ0x π ln w −a a 32.76 For an RL circuit, I(t) = Imaxe −R L t : I(t) Imax = 1−10−9 = e −R L t ≅1−R L t R L t = 10−9 so Rmax = (3.14 × 10−8)(10−9) (2.50 yr)(3.16 × 107 s/ yr) = 3.97 × 10−25 Ω (If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm2, its resistance would be at least 10– 6 Ω).
32.77 (a) UB = 1 2 LI 2 = 1 2 (50.0 H)(50.0 × 10 3 A) 2 = 6.25 × 1010 J (b) Two adjacent turns are parallel wires carrying current in the same direction. Since the loops have such large radius, a one-meter section can be regarded as straight. Then one wire creates a field of B = µ0I 2π r This causes a force on the next wire of F = IlB sin θ giving F = Il µ0I 2π r sin90°= µ0lI2 2π r Solving for the force, F = (4π × 10–7 N/A2) (1.00 m)(50.0 × 10 3 A) 2 (2π)(0.250 m) = 2000 N Chapter 32 Solutions 267 © 2000 by Harcourt, Inc. All rights reserved.
32.78 P = I ∆V ( ) I = P ∆V = 1.00 × 109 W 200 × 103 V = 5.00 × 103 A From Ampere’s law, B 2πr ( ) = µ0Ienclosed or B = µ0Ienclosed 2πr (a) At r = a = 0.0200 m, Ienclosed = 5.00 × 103 A and B = 4π × 10−7 T ⋅m A ( ) 5.00 × 103 A ( ) 2π 0.0200 m ( ) = 0.0500 T = 50.0 mT (b) At r = b = 0.0500 m, Ienclosed = I = 5.00 × 103 A and B = 4π × 10−7 T ⋅m A ( ) 5.00 × 103 A ( ) 2π 0.0500 m ( ) = 0.0200 T = 20.0 mT (c) U = udV ∫ = B r ( ) [ ] 2 2π rldr ( ) 2µ 0 r=a r=b ∫ = µ0I2l 4π dr r a b ∫ = µ0I2l 4π ln b a U = 4π × 10−7 T ⋅m A ( ) 5.00 × 103 A ( ) 2 1000 × 103 m ( ) 4π ln 5.00 cm 2.00 cm = 2.29 × 106 J = 2.29 MJ (d) The magnetic field created by the inner conductor exerts a force of repulsion on the current in the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small rectangular section of the outer cylinder of length l and width w. It carries a current of 5.00 × 103 A ( ) w 2π 0.0500 m ( ) and experiences an outward force F = IlBsinθ = 5.00 × 103 A ( )w 2π 0.0500 m ( ) l 20.0 × 10−3 T ( ) sin 90.0˚ The pressure on it is P = F A = F wl = 5.00 × 103 A ( ) 20.0 × 10−3 T ( ) 2π 0.0500 m ( ) = 318 Pa 268 Chapter 32 Solutions 32.79 (a) B = µ0NI l = 4π × 10−7 T ⋅m A ( ) 1400 ( ) 2.00 A ( ) 1.20 m = 2.93 × 10−3 T (upward) (b) u = B2 2µ0 = 2.93 × 10−3 T ( ) 2 2 4π × 10−7 T ⋅m A ( ) = 3.42 J m3 1 N ⋅m 1 J = 3.42 N m2 = 3.42 Pa (c) To produce a downward magnetic field, the surface of the super conductor must carry a clockwise current.
(d) The vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges and has a radially outward horizontal component. This component exerts upward force on the clockwise superconductor current. The total force on the core is upward . You can think of it as a force of repulsion between the solenoid with its north end pointing up, and the core, with its north end pointing down.
(e) F = PA = 3.42 Pa ( ) π 1.10 × 10−2 m ( ) 2 = 1.30 × 10−3 N Note that we have not proven that energy density is pressure. In fact, it is not in some cases; see problem 12 in Chapter 21.
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Chapter 33 Solutions 33.1 ∆v(t) = ∆Vmax sin(ωt) = 2 ∆Vrms sin(ωt) = 200 2 sin[2π(100t)] = (283 V) sin (628 t) 33.2 ∆Vrms = 170 V 2 = 120 V (a) P = (∆Vrms)2 R →R = (120 V)2 75.0 W = 193 Ω (b) R = (120 V)2 100 W = 144 Ω 33.3 Each meter reads the rms value. ∆Vrms = 100 V 2 = 70.7 V Irms = ∆Vrms R = 70.7 V 24.0 Ω= 2.95 A 33.4 (a) ∆vR = ∆Vmax sinωt ∆vR = 0.250 ∆Vmax ( ), so sinωt = 0.250, or ωt = sin−1 0.250 ( ) The smallest angle for which this is true is ωt = 0 253 . rad. Thus, if t = 0.0100 s, ω = 0.253 rad 0.0100 s = 25.3 rad/s (b) The second time when ∆vR = 0.250 ∆Vmax ( ), ωt = sin−1 0.250 ( ) again. For this occurrence, ω π t = − = 0 253 2 89 .
. rad rad (to understand why this is true, recall the identity sin π −θ ( ) = sinθ from trigonometry). Thus, t = 2.89 rad 25.3 rad s = 0.114 s 33.5 iR = Imax sinωt becomes 0.600 = sin(ω 0.00700) Thus, (0.00700)ω = sin–1(0.600) = 0.644 and ω = 91.9 rad/s = 2π f so f = 14.6 Hz Chapter 33 Solutions 271 © 2000 by Harcourt, Inc. All rights reserved.
33.6 P = Irms ∆Vrms ( ) and ∆Vrms = 120 V for each bulb (parallel circuit), so: I1 = I2 = P1 ∆Vrms = 150 W 120 V = 1.25 A , and R1 = ∆Vrms I1 = 120 V 1.25 A = 96.0 Ω = R2 I3 = P3 ∆Vrms = 100 W 120 V = 0.833 A , and R3 = ∆Vrms I3 = 120 V 0.833 A = 144 Ω 33.7 ∆Vmax = 15.0 V and Rtotal = 8.20 Ω+ 10.4 Ω= 18.6 Ω Imax = ∆Vmax Rtotal = 15.0 V 18.6 Ω= 0.806 A = 2 Irms Pspeaker = Irms 2 Rspeaker = 0.806 A 2 2 10.4 Ω ( ) = 3.38 W 33.8 For Imax = 80.0 mA, Irms = 80.0 mA 2 = 56.6 mA (XL)min = Vrms Irms = 50.0 V 0.0566 A = 884 Ω XL = 2π f L → L = XL 2π f ≥ 884 Ω 2π (20.0) ≥ 7.03 H 33.9 (a) XL = ∆Vmax Imax = 100 7.50 = 13.3 Ω L = XL ω = 13.3 2π(50.0) = 0.0424 H = 42.4 mH (b) XL = ∆Vmax Imax = 100 2.50 = 40.0 Ω ω = XL L = 40.0 42.4 × 10−3 = 942 rad/s 33.10 At 50.0 Hz, XL = 2π 50.0 Hz ( )L = 2π 50.0 Hz ( ) XL 60.0 Hz 2π 60.0 Hz ( ) = 50.0 60.0 54.0 Ω ( ) = 45.0 Ω Imax = ∆Vmax XL = 2 ∆Vrms ( ) XL = 2 100 V ( ) 45.0 Ω = 3.14 A 272 Chapter 33 Solutions 33.11 iL t ( ) = ∆Vmax ωL sin ωt −π 2 ( ) = 80.0 V ( )sin 65.0π ( ) 0.0155 ( ) −π 2 [ ] 65.0π rad s ( ) 70.0 × 10−3 H ( ) iL t ( ) = 5.60 A ( )sin 1.59 rad ( ) = 5.60 A 33.12 ω = 2π f = 2π(60.0/s) = 377 rad/s XL = ωL = (377 /s)(0.0200 V ⋅s/ A) = 7.54 Ω Irms = ∆Vrms XL = 120 V 7.54 Ω= 15.9 A Imax = 2 Irms = 2 (15.9 A) = 22.5 A i(t) = Imax sinωt = (22.5 A)sin 2π 60.0 ( ) s ⋅1 s 180 = (22.5 A) sin 120° = 19.5 A U = 1 2 Li 2 = 1 2 0.0200 V ⋅s A (19.5 A)2 = 3.80 J 33.13 L = N ΦB I where ΦB is the flux through each turn. N ΦB,max = LIB, max = XL ω ∆VL, max ( ) XL N ΦB,max = 2 ∆VL,rms ( ) 2π f = 120 V ⋅s 2 π(60.0) T ⋅C⋅m N ⋅s N ⋅m J J V ⋅C = 0.450 T · m2 33.14 (a) XC = 1 2π fC : 1 2π f(22.0 × 10−6) < 175 Ω 1 2π(22.0 × 10−6)(175) < f f > 41.3 Hz (b) XC ∝1 C , so X(44) = 1 2 X(22): XC < 87.5 Ω 33.15 Imax = 2 Irms = 2 ∆Vrms ( ) XC = 2 ∆Vrms ( )2π fC (a) Imax = 2 (120 V)2π(60.0/s)(2.20 × 10−6 C/ V) = 141 mA (b) Imax = 2 (240 V)2π(50.0/s)(2.20 × 10−6 F) = 235 mA Chapter 33 Solutions 273 © 2000 by Harcourt, Inc. All rights reserved.
33.16 Qmax = C ∆Vmax ( ) = C 2 ∆Vrms ( ) [ ] = 2 C ∆Vrms ( ) 33.17 Imax = ∆Vmax ( )ωC = (48.0 V)(2π)(90.0 s−1)(3.70 × 10−6 F) = 100 mA 33.18 XC = 1 ωC = 1 2π(60.0/s)(1.00 × 10−3 C/ V) = 2.65 Ω vC(t) = ∆Vmax sin ωt, to be zero at t = 0 iC = ∆Vmax XC sin(ωt + φ) = 2 (120 V) 2.65 Ω sin 2π 60 s-1 180 s-1 + 90.0° = (64.0 A)sin(120°+ 90.0°) = – 32.0 A 33.19 (a) XL = ω L = 2π (50.0)(400 × 10- 3) = 126 Ω XC = 1 ωC = 1 2π(50.0)(4.43 × 10−6) = 719 Ω Z = R2 + (XL −XC)2 = 5002 + (126 −719)2 = 776 Ω ∆Vmax = Imax Z = (250 × 10- 3)(776) = 194 V (b) φ = tan−1 XL −XC R = tan−1 126 −719 500 = – 49.9° Thus, the Current leads the voltage. 33.20 ωL = 1 ωC →ω = 1 LC = 1 (57.0 × 10−6)(57.0 × 10−6) = 1.75 × 104 rad/s f = ω 2π = 2.79 kHz 33.21 (a) XL = ω L = 2π (50.0 s-1)(250 × 10-3 H) = 78.5 Ω (b) XC = 1 ωC = 2π(50.0 s−1)(2.00 × 10−6 F) [ ] −1 = 1.59 kΩ (c) Z = R2 + (XL −XC)2 = 1.52 kΩ (d) Imax = ∆Vmax Z = 210 V 1.52 × 103 Ω = 138 mA (e) φ = tan−1 XL −XC R = tan−1(−10.1) = – 84.3° 274 Chapter 33 Solutions 33.22 (a) Z = R2 + XL −XC ( ) = 68.02 + 16.0 −101 ( )2 = 109 Ω XL = ωL = 100 ( ) 0.160 ( ) = 16.0 Ω XC = 1 ωC = 1 100 ( ) 99.0 × 10−6 ( ) = 101 Ω (b) Imax = ∆Vmax Z = 40.0 V 109 Ω= 0.367 A (c) tanφ = XL −XC R = 16.0 −101 68.0 = −1.25: φ = −0.896 rad = −51.3° Imax = 0.367 A ω = 100 rad/s φ = – 0.896 rad = – 51.3° 33.23 XL = 2π fL = 2π 60.0 ( ) 0.460 ( ) = 173 Ω XC = 1 2π fC = 1 2π 60.0 ( ) 21.0 × 10−6 ( ) = 126 Ω (a) tanφ = XL −XC R = 173 Ω−126 Ω 150 Ω = 0.314 φ = 0.304 rad = 17.4° (b) Since XL > XC , φ is positive; so voltage leads the current .
33.24 XC = 1 2π fC = 1 2π(60.0 Hz)(20.0 × 10−12 F) = 1.33 × 108 Ω Z = (50.0 × 103 Ω)2 + (1.33 × 108 Ω)2 ≈1.33 × 108 Ω Irms = ∆Vrms Z = 5000 V 1.33 × 108 Ω = 3.77 × 10–5 A ∆Vrms ( )body = IrmsRbody = (3.77 × 10−5 A)(50.0 × 10 3 Ω) = 1.88 V Chapter 33 Solutions 275 © 2000 by Harcourt, Inc. All rights reserved.
33.25 XC = 1 ωC = 1 2π(50.0)(65.0 × 10−6) = 49.0 Ω XL = ωL = 2π(50.0)(185 × 10−3) = 58.1 Ω Z = R2 + (XL −XC)2 = (40.0)2 + (58.1−49.0)2 = 41.0 Ω Imax = ∆Vmax Z = 150 41.0 = 3.66 A (a) ∆VR = Imax R = (3.66)(40) = 146 V (b) ∆VL = Imax XL = (3.66)(58.1) = 212.5 = 212 V (c) ∆VC = Imax XC = (3.66)(49.0) = 179.1 V = 179 V (d) ∆VL – ∆VC = 212.5 – 179.1 = 33.4 V 33.26 R = 300 Ω XL = ωL = 2π 500 π s−1 0.200 H ( ) = 200 Ω XC = 1 ωC = 2π 500 π s−1 11.0 × 10−6 F ( ) −1 = 90.9 Ω XL = 200 Ω XC = 90.9 Ω R = 300 Ω φ Z { XL - XC = 109 Ω Z = R2 + XL −XC ( ) 2 = 319 Ω and φ = tan−1 XL −XC R = 20.0° 33.27 (a) XL = 2π 100 Hz ( ) 20.5 H ( ) = 1.29 × 104 Ω Z = ∆Vrms Irms = 200 V 4.00 A = 50.0 Ω XL −XC ( ) 2 = Z2 −R2 = 50.0 Ω ( )2 −35.0 Ω ( )2 XL −XC = 1.29 × 104 Ω − 1 2π 100 Hz ( )C = ±35.7 Ω C = 123 nF or 124 nF (b) ∆VL,rms = Irms XL = 4.00 A ( ) 1.29 × 104 Ω ( ) = 51.5 kV Notice that this is a very large voltage!
276 Chapter 33 Solutions 33.28 XL = ωL = (1000/s)(0.0500 H) [ ] = 50.0 Ω XC = 1/ωC = (1000/s)(50.0 × 10−6 F) [ ] −1 = 20.0 Ω Z = R2 +(XL −XC)2 Z = (40.0)2 +(50.0 −20.0)2 = 50.0 Ω (a) Irms = ∆Vrms ( )/Z = 100 V / 50.0 Ω Irms = 2.00 A φ = Arctan XL −XC R φ = Arctan 30.0 Ω 40.0 Ω= 36.9° (b) P = ∆Vrms ( )Irms cos φ = 100 V(2.00 A) cos 36.9° = 160 W (c) PR = Irms 2 R = (2.00 A)2 40.0 Ω= 160 W 33.29 ω = 1000 rad/s, R = 400 Ω, C = 5.00 × 10– 6 F, L = 0.500 H ∆Vmax = 100 V, ω L = 500 Ω, 1 ω C = 200 Ω Z = R2 + ωL −1 ωC 2 = 4002 + 3002 = 500 Ω Imax = ∆Vmax Z = 100 500 = 0.200 A The average power dissipated in the circuit is P = Irms 2 R = Imax 2 2 R P = (0.200 A)2 2 (400 Ω) = 8.00 W Chapter 33 Solutions 277 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution An ac voltage of the form ∆v = 100 V ( )sin 1000 t ( ) is applied to a series RLC circuit. If R = 400 Ω, C = 5.00 µ F, and L = 0.500 H, what is the average power delivered to the circuit?
G : Comparing ∆v = 100 V ( )sin 1000 t ( ) with ∆v = ∆Vmax sin ω t, we see that ∆Vmax = 100 V and ω = 1000 s-1 Only the resistor takes electric energy out of the circuit, but the capacitor and inductor will impede the current flow and therefore reduce the voltage across the resistor. Because of this impedance, the average power dissipated by the resistor must be less than the maximum power from the source: Pmax = ∆Vmax ( ) 2 2R = 100 V ( )2 2 400 Ω ( ) = 12.5 W O : The actual power dissipated by the resistor can be found from P = Irms 2 R, where Irms = ∆Vrms /Z.
A : ∆Vrms = 100 2 = 70.7 V In order to calculate the impedance, we first need the capacitive and inductive reactances: XC = 1 ω C = 1 (1000 s-1)(5.00 × 10−6 F) = 200 Ω and XL = ω L = 1000 s-1 ( ) 0.500 H ( ) = 500 Ω Then, Z = R2 + (XL −XC)2 = (400 Ω)2 + (500 Ω−200 Ω)2 = 500 Ω Irms = ∆Vrms Z = 70.7 V 500 Ω= 0.141 A and P = Irms 2 R = 0.141 A ( )2 400 Ω ( ) = 8.00 W L : The power dissipated by the resistor is less than 12.5 W, so our answer appears to be reasonable. As with other RLC circuits, the power will be maximized at the resonance frequency where XL = XC so that Z = R. Then the average power dissipated will simply be the 12.5 W we calculated first.
33.30 Z = R2 + XL −XC ( ) 2 or XL −XC ( ) = Z2 −R2 XL −XC ( ) = 75.0 Ω ( )2 −45.0 Ω ( )2 = 60.0 Ω φ = tan−1 XL −XC R = tan−1 60.0 Ω 45.0 Ω = 53.1° Irms = ∆Vrms Z = 210 V 75.0 Ω= 2.80 A P = ∆Vrms ( )Irms cosφ = 210 V ( ) 2.80 A ( )cos 53.1˚ ( ) = 353 W 278 Chapter 33 Solutions 33.31 (a) P = Irms(∆Vrms)cosφ = (9.00)(180) cos(– 37.0°) = 1.29 × 103 W P = Irms 2 R so 1.29 × 103 = (9.00)2R and R = 16.0 Ω (b) tanφ = XL −XC R becomes tan( .
) − ° = − 37 0 16 X X L C : so XL – XC = – 12.0 Ω 33.32 XL = ωL = 2π(60.0/s)(0.0250 H) = 9.42 Ω Z = R2 + (XL −XC)2 = 20.0 ( )2 + 9.42 ( )2 Ω= 22.1 Ω (a) Irms = ∆Vrms Z = 120 V 22.1 Ω= 5.43 A (b) φ = tan−1 9.42/ 20.0 ( ) = 25.2° so power factor = cos φ = 0.905 (c) We require φ = 0. Thus, XL = XC: 9.42 Ω= 1 2π(60.0 s−1)C and C = 281 µF (d) Pb = Pd or ∆Vrms ( )b Irms ( )b cos φb = ∆Vrms ( )d 2 R ∆Vrms ( )d = R ∆Vrms ( )b Irms ( )b cos φb = 20.0 Ω ( ) 120 V ( ) 5.43 A ( ) 0.905 ( ) = 109 V 33.33 Consider a two-wire transmission line: Irms = P ∆Vrms = 100 × 106 W 50.0 × 103 V = 2.00 × 103 A loss = 0.0100 ( )P = Irms 2 Rline = Irms 2 2R1 ( ) R1 R1 RL ∆Vrms Thus, R1 = 0.0100 ( )P 2Imax 2 = 0.0100 ( ) 100 × 106 W ( ) 2 2.00 × 103 A ( ) 2 = 0.125 Ω But R1 = ρl A or A = π d2 4 = ρl R1 Therefore d = 4ρl π R1 = 4 1.70 × 10−8 Ω⋅m ( ) 100 × 103 m ( ) π 0.125 Ω ( ) = 0.132 m = 132 mm Chapter 33 Solutions 279 © 2000 by Harcourt, Inc. All rights reserved.
33.34 Consider a two-wire transmission line: Irms = P ∆Vrms and power loss = Irms 2 Rline = P 100 Thus, P ∆Vrms 2 2R1 ( ) = P 100 or R1 = ∆Vrms ( ) 2 200P R1 = ρ d A = ∆Vrms ( ) 2 200P or A = π 2r ( )2 4 = 200ρ P d ∆Vrms ( ) 2 and the diameter is 2r = 800ρ P d π ∆Vrms ( ) 2 R1 R1 RL ∆Vrms 33.35 One-half the time, the left side of the generator is positive, the top diode conducts, and the bottom diode switches off. The power supply sees resistance 1 2R + 1 2R −1 = R and the power is ∆Vrms ( ) 2 R The other half of the time the right side of the generator is positive, the upper diode is an open circuit, and the lower diode has zero resistance. The equivalent resistance is then R1 R1 RL ∆Vrms Req = R + 1 3R + 1 R −1 = 7R 4 and P = ∆Vrms ( ) 2 Req = 4 ∆Vrms ( ) 2 7R The overall time average power is: ∆Vrms ( ) 2 R [ ] + 4 ∆Vrms ( ) 2 7R [ ] 2 = 11 ∆Vrms ( ) 2 14R 33.36 At resonance, 1 2π fC = 2π fL and 1 2π f ( ) 2L = C The range of values for C is 46.5 pF to 419 pF 33.37 ω0 = 2π(99.7 × 106) = 6.26 × 108 rad/s = 1 LC C = 1 ω0 2L = 1 (6.26 × 108)2(1.40 × 10−6) = 1.82 pF 280 Chapter 33 Solutions 33.38 L = 20.0 mH, C = 1.00 × 10–7, R = 20.0 Ω, ∆Vmax = 100 V (a) The resonant frequency for a series –RLC circuit is f = 1 2π 1 LC = 3.56 kHz (b) At resonance, Imax = ∆Vmax R = 5.00 A (c) From Equation 33.36, Q = ω0L R = 22.4 (d) ∆VL,max = XLImax = ω0LImax = 2.24 kV 33.39 The resonance frequency is ω0 = 1 LC . Thus, if ω = 2ω0, XL = ω L = 2 LC L = 2 L C and XC = 1 ω C = LC 2C = 1 2 L C Z = R2 + XL −XC ( ) 2 = R2 + 2.25 L C ( ) so Irms = ∆Vrms Z = ∆Vrms R2 + 2.25 L C ( ) and the energy dissipated in one period is Q = P∆t: Q = ∆Vrms ( ) 2R R2 + 2.25 L C ( ) 2π ω = ∆Vrms ( ) 2RC R2C + 2.25L π LC ( ) = 4π ∆Vrms ( ) 2RC LC 4R2C + 9.00L With the values specified for this circuit, this gives: Q = 4π 50.0 V ( )2 10.0 Ω ( ) 100 × 10−6 F ( ) 3 2 10.0 × 10−3 H ( ) 1 2 4 10.0 Ω ( )2 100 × 10−6 F ( ) + 9.00 10.0 × 10−3 H ( ) = 242 mJ 33.40 The resonance frequency is ω0 = 1 LC . Thus, if ω = 2ω0, XL = ω L = 2 LC L = 2 L C and XC = 1 ω C = LC 2C = 1 2 L C Then Z = R2 + XL −XC ( ) 2 = R2 + 2.25 L C ( ) so Irms = ∆Vrms Z = ∆Vrms R2 + 2.25 L C ( ) and the energy dissipated in one period is Q = P∆t = ∆Vrms ( ) 2R R2 + 2.25 L C ( ) 2π ω = ∆Vrms ( ) 2RC R2C + 2.25L π LC ( ) = 4π ∆Vrms ( ) 2RC LC 4R2C + 9.00L Chapter 33 Solutions 281 © 2000 by Harcourt, Inc. All rights reserved.
33.41 For the circuit of problem 22, ω0 = 1 LC = 1 160 × 10−3 H ( ) 99.0 × 10−6 F ( ) = 251 rad s Q = ω0L R = 251 rad s ( ) 160 × 10−3 H ( ) 68.0 Ω = 0.591 For the circuit of problem 23, Q = ω0L R = L R LC = 1 R L C = 1 150 Ω 460 × 10−3 H 21.0 × 10−6 F = 0.987 The circuit of problem 23 has a sharper resonance.
33.42 (a) ∆V2,rms = 1 13 120 V ( ) = 9.23 V (b) ∆V1,rms I1,rms = ∆V2,rms I2,rms (120 V)(0.350 A) = (9.23 V)I2,rms I2,rms = 42.0 W 9.23 V = 4.55 A for a transformer with no energy loss (c) P = 42.0 W from (b) 33.43 ∆Vout ( )max = N2 N1 ∆Vin ( )max = 2000 350 (170 V) = 971 V ∆Vout ( )rms = (971 V) 2 = 687 V 33.44 (a) ∆V2, rms ( ) = N2 N1 ∆V1, rms ( ) N2 = (2200)(80) 110 = 1600 windings (b) I1,rms ∆V1,rms ( ) = I2,rms ∆V2,rms ( ) I1,rms = (1.50)(2200) 110 = 30.0 A (c) 0.950I1,rms ∆V1,rms ( ) = I2,rms ∆V2,rms ( ) I1,rms = (1.20)(2200) 110(0.950) = 25.3 A 282 Chapter 33 Solutions 33.45 The rms voltage across the transformer primary is N1 N2 ∆V2,rms ( ) so the source voltage is ∆Vs,rms = I1,rms Rs + N1 N2 ∆V2,rms ( ) The secondary current is ∆V2,rms ( ) RL , so the primary current is N2 N1 ∆V2,rms ( ) RL = I1,rms Then ∆Vs,rms = N2 ∆V2,rms ( )Rs N1RL + N1 ∆V2,rms ( ) N2 and Rs = N1RL N2 ∆V2,rms ( ) ∆Vs,rms − N1 ∆V2,rms ( ) N2 = 5(50.0 Ω) 2(25.0 V) 80.0 V −5(25.0 V) 2 = 87.5 Ω 33.46 (a) ∆V2,rms = N2 N1 ∆V1,rms ( ) N2 N1 = ∆V2,rms ∆V1,rms = 10.0 × 103 V 120 V = 83.3 (b) I2,rms ∆V2,rms ( ) = 0.900I1,rms ∆V1,rms ( ) I2,rms 10.0 × 103 V ( ) = 0.900 120 V 24.0 Ω 120 V ( ) I2,rms = 54.0 mA (c) Z2 = ∆V2,rms I2,rms = 10.0 × 103 V 0.054 A = 185 kΩ 33.47 (a) R = (4.50 × 10−4 Ω/m)(6.44 × 105 m) = 290 Ω and Irms = P ∆Vrms = 5.00 × 106 W 5.00 × 105 V = 10.0 A Ploss = Irms 2 R = (10.0 A)2(290 Ω) = 29.0 kW (b) Ploss P = 2.90 × 104 5.00 × 106 = 5.80 × 10−3 (c) It is impossible to transmit so much power at such low voltage. Maximum power transfer occurs when load resistance equals the line resistance of 290 Ω, and is (4.50 × 103 V)2 2 ⋅2(290 Ω) = 17.5 kW, far below the required 5 000 kW Chapter 33 Solutions 283 © 2000 by Harcourt, Inc. All rights reserved.
33.48 For the filter circuit, ∆Vout ∆Vin = XC R2 + XC 2 (a) At f = 600 Hz, XC = 1 2π fC = 1 2π 600 Hz ( ) 8.00 × 10−9 F ( ) = 3.32 × 104 Ω and ∆Vout ∆Vin = 3.32 × 104 Ω 90.0 Ω ( )2 + 3.32 × 104 Ω ( ) 2 ≈ 1.00 (b) At f = 600 kHz, XC = 1 2π fC = 1 2π 600 × 103 Hz ( ) 8.00 × 10−9 F ( ) = 33.2 Ω and ∆Vout ∆Vin = 33.2 Ω 90.0 Ω ( )2 + 33.2 Ω ( )2 = 0.346 33.49 For this RC high-pass filter, ∆Vout ∆Vin = R R2 + XC 2 (a) When ∆Vout ∆Vin = 0.500, then 0.500 Ω 0.500 Ω ( )2 + XC 2 = 0.500 or XC = 0.866 Ω If this occurs at f = 300 Hz, the capacitance is C = 1 2π fXC = 1 2π 300 Hz ( ) 0.866 Ω ( ) = 6.13 × 10−4 F = 613 µF (b) With this capacitance and a frequency of 600 Hz, XC = 1 2π 600 Hz ( ) 6.13 × 10−4 F ( ) = 0.433 Ω ∆Vout ∆Vin = R R2 + XC 2 = 0.500 Ω 0.500 Ω ( )2 + 0.433 Ω ( )2 = 0.756 (a) (b) (c) Figures for Goal Solution 284 Chapter 33 Solutions Goal Solution The RC high-pass filter shown in Figure 33.22 has a resistance R = 0.500 Ω. (a) What capacitance gives an output signal that has one-half the amplitude of a 300-Hz input signal? (b) What is the gain ( ∆Vout / ∆Vin) for a 600-Hz signal?
G : It is difficult to estimate the capacitance required without actually calculating it, but we might expect a typical value in the µF to pF range. The nature of a high-pass filter is to yield a larger gain at higher frequencies, so if this circuit is designed to have a gain of 0.5 at 300 Hz, then it should have a higher gain at 600 Hz. We might guess it is near 1.0 based on Figure (b) above. O : The output voltage of this circuit is taken across the resistor, but the input sees the impedance of the resistor and the capacitor. Therefore, the gain will be the ratio of the resistance to the impedance. A : ∆Vout ∆Vin = R R2 + 1 ω C ( ) 2 (a) When ∆Vout / ∆Vin = 0.500 solving for C gives C = 1 ωR ∆Vin ∆Vout 2 −1 = 1 (2π)(300 Hz)(0.500 Ω) (2.00)2 −1 = 613 µF (b) At 600 Hz, we have ω = 2π rad ( ) 600 s-1 ( ) so ∆Vout ∆Vin = 0.500 Ω 0.500 Ω ( )2 + 1 1200π rad/s ( ) 613 µF ( ) 2 = 0.756 L : The capacitance value seems reasonable, but the gain is considerably less than we expected. Based on our calculation, we can modify the graph in Figure (b) to more transparently represent the characteristics of this high-pass filter, now shown in Figure (c). If this were an audio filter, it would reduce low frequency “humming” sounds while allowing high pitch sounds to pass through. A low pass filter would be needed to reduce high frequency “static” noise.
33.50 ∆V1 = I r + R ( )2 + XL 2 , and ∆V2 = I R2 + XL 2 Thus, when ∆V1 = 2∆V2 r + R ( )2 + XL 2 = 4 R2 + XL 2 ( ) or 25.0 Ω ( )2 + XL 2 = 4 5.00 Ω ( )2 + 4XL 2 ∆V1 ∆V2 r = 20.0 Ω R =5.00 Ω L = 250 mH which gives XL = 2π f 0.250 H ( ) = 625 −100 3 Ω and f = 8.42 Hz Chapter 33 Solutions 285 © 2000 by Harcourt, Inc. All rights reserved.
33.51 ∆Vout ∆Vin = R R2 + XL −XC ( ) 2 (a) At 200 Hz: 1 4 = 8.00 Ω ( )2 8.00 Ω ( )2 + 400πL − 1 400πC 2 At 4000 Hz: 8.00 Ω ( )2 + 8000π L − 1 8000π C 2 = 4 8.00 Ω ( )2 At the low frequency, XL −XC < 0. This reduces to 400 1 400 13 9 π π L C − = − . Ω For the high frequency half-voltage point, 8000 1 8000 13 9 π π L C − = + . Ω Solving Equations (1) and (2) simultaneously gives C = 54.6 µF and L = 580 µH (b) When XL = XC, ∆Vout ∆Vin = ∆Vout ∆Vin max = 1.00 (c) XL = XC requires f0 = 1 2π LC = 1 2π 5.80 × 10−4 H ( ) 5.46 × 10−5 F ( ) = 894 Hz (d) At 200 Hz, ∆Vout ∆Vin = R Z = 1 2 and XC > XL, so the phasor diagram is as shown: R Z XL - XC φ or φ ∆Vout ∆Vin φ = −cos−1 R Z = −cos−1 1 2 so ∆Vout leads ∆Vin by 60.0° At f0, XL = XC so ∆Vout and ∆Vin have a phase difference of 0° At 4000 Hz, ∆Vout ∆Vin = R Z = 1 2 and XL −XC > 0 Thus, φ = cos−1 1 2 = 60.0° R Z XL - XC φ or φ ∆Vout ∆Vin or ∆Vout lags ∆Vin by 60.0° (e) At 200 Hz and at 4 kHz, P = ∆Vout,rms ( ) 2 R = 1 2 ∆Vin,rms ( ) 2 R = 1 2 1 2 ∆Vin,max ( ) 2 R = 10.0 V ( )2 8 8.00 Ω ( ) = 1.56 W At f0, P = ∆Vout,rms ( ) 2 R = ∆Vin,rms ( ) 2 R = 1 2 ∆Vin,max ( ) 2 R = 10.0 V ( )2 2 8.00 Ω ( ) = 6.25 W (f) We take: Q = ω 0L R = 2π f0 L R = 2π 894 Hz ( ) 5.80 × 10−4 H ( ) 8.00 Ω = 0.408 286 Chapter 33 Solutions 33.52 For a high-pass filter, ∆Vout ∆Vin = R R2 + 1 ωC 2 ∆Vout ( )1 ∆Vin ( )1 = R R2 + 1 ωC 2 and ∆Vout ( )2 ∆Vin ( )2 = R R2 + 1 ωC 2 Now ∆Vin ( )2 = ∆Vout ( )1 so ∆Vout ( )2 ∆Vin ( )1 = R2 R2 + 1 ωC 2 = 1 1+ 1 ωRC 2 33.53 Rewrite the circuit in terms of impedance as shown in Fig. (b).
Find: ∆Vout = ZR ZR + ZC ∆Vab From Figure (c), ∆Vab = ZC | | ZR + ZC ( ) ZR + ZC | | ZR + ZC ( ) ∆Vin So Eq. becomes ∆Vout = ZR ZC | | ZR + ZC ( ) [ ] ZR + ZC ( ) ZR + ZC| | ZR + ZC ( ) [ ] ∆Vin or ∆Vout ∆Vin = ZR 1 ZC + 1 ZR + ZC −1 ZR + ZC ( ) ZR + 1 ZC + 1 ZR + ZC −1 ∆Vout ∆Vin = ZRZC ZC ZC + ZR ( ) + ZR ZR + 2ZC ( ) = ZR 3ZR + ZC + ZR ( ) 2 ZC Now, ZR = R and ZC = −j ω C where j= −1 ∆Vout ∆Vin = R 3R − 1 ω C j+ R2ω Cj where we used 1 j= −j.
ZR ZR ZC ZC ∆Vin ∆Vout a b Figure (a) a b ∆Vout ∆Vab ZC ZR Figure (b) ZR ZR ZC ZC ∆Vin ∆Vab a b Figure (c) R Z −ˆ j ω C ∆Vout ∆Vin = R 3R − 1 ω C −R2ω C j = R 3R ( )2 + 1 ω C −R2ω C 2 = 1.00 × 103 3.00 × 103 ( ) 2 + 1592 −628 ( )2 = 0.317 Chapter 33 Solutions 287 © 2000 by Harcourt, Inc. All rights reserved.
33.54 The equation for ∆v t ( ) during the first period (using y = mx + b) is: ∆v t ( ) = 2 ∆Vmax ( )t T −∆Vmax ∆v ( )2 [ ]ave = 1 T ∆v t ( ) [ ] 0 T ∫ 2 dt = ∆Vmax ( ) 2 T 2 T t −1 0 T ∫ 2 dt ∆v ( )2 [ ]ave = ∆Vmax ( ) 2 T T 2 2t T −1 [ ] 3 3 t=0 t=T = ∆Vmax ( ) 2 6 +1 ( )3 −−1 ( )3 [ ] = ∆Vmax ( ) 2 3 ∆Vrms = ∆v ( )2 [ ]ave = ∆Vmax ( ) 2 3 = ∆Vmax 3 33.55 ω0 = 1 LC = 1 (0.0500 H)(5.00 × 10−6 F) = 2000 s−1 so the operating frequency of the circuit is ω = ω0 2 = 1000 s−1 Using Equation 33.35, P = ∆Vrms ( ) 2Rω 2 R2ω 2 + L2 ω 2 −ω0 2 ( ) 2 P = (400)2(8.00)(1000)2 (8.00)2(1000)2 + (0.0500)2 (1.00 −4.00) × 106 [ ] 2 = 56.7 W Q ≈12.5 ( ) Figure for Goal Solution Goal Solution A series RLC circuit consists of an 8.00-Ω resistor, a 5.00-µ F capacitor, and a 50.0-mH inductor. A variable frequency source applies an emf of 400 V (rms) across the combination. Determine the power delivered to the circuit when the frequency is equal to one half the resonance frequency. G : Maximum power is delivered at the resonance frequency, and the power delivered at other frequencies depends on the quality factor, Q. For the relatively small resistance in this circuit, we could expect a high Q = ω0L R. So at half the resonant frequency, the power should be a small fraction of the maximum power, Pav, max = ∆Vrms 2 R = 400 V ( )2 8 Ω= 20 kW. O : We must first calculate the resonance frequency in order to find half this frequency. Then the power delivered by the source must equal the power taken out by the resistor. This power can be found from Pav = Irms 2 R where Irms = ∆Vrms /Z.
288 Chapter 33 Solutions A : The resonance frequency is f0 = 1 2π LC = 1 2π 0.0500 H ( ) 5.00 × 10−6 F ( ) = 318 Hz The operating frequency is f = f0 / 2 = 159 Hz. We can calculate the impedance at this frequency: XL = 2π f L = 2π(159 Hz) 0.0500 H ( ) = 50.0 Ω and XC = 1 2π f C = 1 2π(159 Hz) 5.00 × 10-6 F ( ) = 200 Ω Z = R2 + (XL −XC)2 = 8.002 + (50.0 −200)2 Ω= 150 Ω So, Irms = ∆Vrms Z = 400 V 150 Ω= 2.66 A The power delivered by the source is the power dissipated by the resistor: Pav = Irms 2R = (2.66 A)2 8.00 Ω ( ) = 56.7 W L : This power is only about 0.3% of the 20 kW peak power delivered at the resonance frequency. The significant reduction in power for frequencies away from resonance is a consequence of the relatively high Q-factor of about 12.5 for this circuit. A high Q is beneficial if, for example, you want to listen to your favorite radio station that broadcasts at 101.5 MHz, and you do not want to receive the signal from another local station that broadcasts at 101.9 MHz.
33.56 The resistance of the circuit is R = ∆V I = 12.0 V 0.630 A = 19.0 Ω The impedance of the circuit is Z = ∆Vrms Irms = 24.0 V 0.570 A = 42.1 Ω Z2 = R2 + ω 2L2 L = 1 ω Z2 −R2 = 1 377 (42.1)2 −(19.0)2 = 99.6 mH 33.57 (a) When ω L is very large, the bottom branch carries negligible current. Also, 1/ω C will be negligible compared to 200 Ω and 45 0 V/200 = .
Ω 225 mA flows in the power supply and the top branch.
(b) Now 1/ω C → ∞ and ω L → 0 so the generator and bottom branch carry 450 mA Chapter 33 Solutions 289 © 2000 by Harcourt, Inc. All rights reserved.
33.58 (a) With both switches closed, the current goes only through generator and resistor. i(t) = ∆Vmax R cosωt (b) P = 1 2 ∆Vmax ( ) 2 R (c) i(t) = ∆Vmax R2 + ω 2L2 cos ωt + Arctan(ωL/ R) [ ] (d) For 0 = φ = Arctan ω0L − 1 ω0C R We require ω0L = 1 ω0 C , so C = 1 ω0 2 L (e) At this resonance frequency, Z = R (f) U = 1 2 C ∆VC ( ) 2 = 1 2 C I2XC 2 Umax = 1 2 CImax 2 XC 2 = 1 2 C ∆Vmax ( ) 2 R2 1 ω0 2C2 = ∆Vmax ( ) 2L 2R2 (g) Umax = 1 2 LImax 2 = 1 2 L ∆Vmax ( ) 2 R2 (h) Now ω = 2ω0 = 2 LC So φ = Arctan ωL −1 ωC R = Arctan 2 L C −1 2 L C R = Arctan 3 2R L C (i) Now ωL = 1 2 1 ωC ω = 1 2LC = ω0 2 290 Chapter 33 Solutions 33.59 (a) As shown in part (b), circuit (a) is a high-pass filter and circuit (b) is a low-pass filter .
(b) For circuit (a), ∆Vout ∆Vin = RL 2 + XL 2 RL 2 + XL −XC ( ) 2 = RL 2 + ω L ( ) 2 RL 2 + ω L −1 ω C ( ) 2 As ω →0, ∆Vout ∆Vin ≈ω RLC ≈0 As ω →∞, ∆Vout ∆Vin ≈1 (high-pass filter) For circuit (b), ∆Vout ∆Vin = XC RL 2 + XL −XC ( ) 2 = 1 ω C RL 2 + ω L −1 ω C ( ) 2 As ω →0, ∆Vout ∆Vin ≈1 As ω →∞, ∆Vout ∆Vin ≈ 1 ω 2LC ≈0 (low-pass filter) ∆Vout ∆Vin Circuit (b) ∆Vout ∆Vin Circuit (a) 33.60 (a) IR,rms = ∆Vrms R = 100 V 80.0 Ω= 1.25 A (b) The total current will lag the applied voltage as seen in the phasor diagram at the right. IL,rms = ∆Vrms XL = 100 V 2π 60.0 s−1 ( ) 0.200 H ( ) = 1.33 A φ IR IL ∆V I Thus, the phase angle is: φ = tan−1 IL,rms IR,rms = tan−1 1.33 A 1.25 A = 46.7° 33.61 Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or $36 per 30 days at 10¢ per kWh). Suppose the transmission line is at 20 kV. Then Irms = P ∆Vrms = 20 000 ( ) 500 W ( ) 20 000 V ~103 A If the transmission line had been at 200 kV, the current would be only ~102 A .
Chapter 33 Solutions 291 © 2000 by Harcourt, Inc. All rights reserved.
33.62 L = 2.00 H, C = 10.0 × 10– 6 F, R = 10.0 Ω, ∆v(t) = (100 sin ω t) (a) The resonant frequency ω0 produces the maximum current and thus the maximum power dissipation in the resistor. ω0 = 1 LC = 1 (2.00)(10.0 × 10−6) = 224 rad/s (b) P = ∆Vmax ( ) 2 2R = (100)2 2(10.0) = 500 W (c) Irms = ∆Vrms Z = ∆Vrms R2 + ωL −1 ωC 2 and Irms ( )max = ∆Vrms R Irms 2 R = 1 2 Irms 2 ( )maxR or ∆Vrms ( ) 2 Z2 R = 1 2 ∆Vrms ( ) 2 R2 R This occurs where Z 2 = 2R 2: R2 + ωL −1 ωC 2 = 2R2 ω 4L2C 2 −2Lω 2C −R2ω 2C 2 + 1 = 0 or L2C 2ω 4 −(2LC + R2C 2)ω 2 + 1 = 0 (2.00)2(10.0 × 10−6)2 [ ]ω 4 −2(2.00)(10.0 × 10−6) + (10.0)2(10.0 × 10−6)2 [ ]ω 2 + 1 = 0 Solving this quadratic equation, we find that ω 2 = 51130, 48 894 ω1 = 48 894 = 221 rad/s and ω2 = 51130 = 226 rad/s 33.63 R = 200 Ω, L = 663 mH, C = 26.5 µF, ω = 377 s−1, ∆Vmax = 50.0 V ωL = 250 Ω, 1 ωC = 100 Ω, Z = R2 + XL −XC ( ) 2 = 250 Ω (a) Imax = ∆Vmax Z = 50.0 V 250 Ω= 0.200 A φ = tan−1 XL −XC R = 36.8° (∆V leads I) (b) ∆VR,max = Imax R = 40.0 V at φ = 0° (c) ∆VC,max = Imax ωC = 20.0 V at φ = – 90.0° (I leads ∆V) (d) ∆VL,max = ImaxωL = 50.0 V at φ = + 90.0° (∆V leads I) 292 Chapter 33 Solutions 33.64 P = Irms 2 R = ∆Vrms Z 2 R, so 250 W = 120 V ( )2 Z2 40.0 Ω ( ): Z = R2 + ωL −1 ωC ( ) 2 250 = 120 ( )2 40.0 ( ) 40.0 ( )2 + 2π f 0.185 ( ) − 1 2π f 65.0 × 10−6 ( ) 2 and 250 = 576 000 f 2 1600 f 2 + 1.1624 f 2 −2448.5 ( ) 2 1 = 2304 f 2 1600 f 2 + 1.3511f 4 −5692.3 f 2 + 5 995 300 so 1.3511f 4 −6396.3 f 2 + 5995300 = 0 f 2 = 6396.3 ± 6396.3 ( )2 −4 1.3511 ( ) 5 995 300 ( ) 2 1.3511 ( ) = 3446.5 or 1287.4 f = 58.7 Hz or 35.9 Hz 33.65 (a) From Equation 33.39, N1 N2 = ∆V1 ∆V2 Let output impedance Z1 = ∆V1 I1 and the input impedance Z2 = ∆V2 I2 so that N1 N2 = Z1I1 Z2I2 But from Eq. 33.40, I1 I2 = ∆V2 ∆V1 = N2 N1 So, combining with the previous result we have N1 N2 = Z1 Z2 (b) N1 N2 = Z1 Z2 = 8000 8.00 = 31.6 33.66 IR = ∆Vrms R ; IL = ∆Vrms ωL ; IC = ∆Vrms (ωC)−1 (a) Irms = IR 2 + (IC −IL)2 = ∆Vrms 1 R2 + ωC −1 ωL 2 (b) tanφ = IC −IL IR = ∆Vrms 1 XC −1 XL 1 ∆Vrms / R tanφ = R 1 XC −1 XL Chapter 33 Solutions 293 © 2000 by Harcourt, Inc. All rights reserved.
33.67 (a) Irms = ∆Vrms 1 R2 + ωC −1 ωL 2 ∆Vrms →∆Vrms ( )max when ω ω C L = 1 f = 1 2π LC f = 1 2π 200 × 10−3 H)(0.150 × 10−6 F) = 919 Hz (b) IR = ∆Vrms R = 120 V 80.0 Ω= 1.50 A IL = ∆Vrms ωL = 120 V (374 s−1)(0.200 H) = 1.60 A IC = ∆Vrms(ωC) = (120 V)(374 s−1)(0.150 × 10−6 F) = 6.73 mA (c) Irms = IR 2 + (IC −IL)2 = (1.50)2 + (0.00673 −1.60)2 = 2.19 A (d) φ = tan−1 IC −IL IR = tan−1 0.00673 −1.60 1.50 = – 46.7° The current is lagging the voltage .
33.68 (a) tanφ = ∆VL ∆VR = I ωL ( ) IR = ωL R Thus, R = ωL tanφ = 200 s−1 ( ) 0.500 H ( ) tan 30.0° ( ) = 173 Ω (b) ∆Vout ∆Vin = ∆VR ∆Vin = cosφ ∆Vout = ∆Vin ( )cosφ = 10.0 V ( ) cos 30.0°= 8.66 V ∆VR = IR ∆VL = IXL ∆V = IZ φ ∆Vin L R ∆Vout 294 Chapter 33 Solutions 33.69 (a) XL = XC = 1884 Ω when f = 2000 Hz L = XL 2π f = 1884 Ω 4000π rad s = 0.150 H and C = 1 2π f ( )XC = 1 4000π rad s ( ) 1884 Ω ( ) = 42.2 nF XL = 2π f 0.150 H ( ) XC = 1 2π f ( ) 4.22 × 10−8 F ( ) Z = 40.0 Ω ( )2 + XL −XC ( ) 2 f (Hz) XL (Ω) XC (Ω) Z (Ω) 300 283 12600 12300 600 565 6280 5720 800 754 4710 3960 1000 942 3770 2830 1500 1410 2510 1100 2000 1880 1880 40 3000 2830 1260 1570 4000 3770 942 2830 6000 5650 628 5020 10000 9420 377 9040 (b) Impedence, Ω 33.70 ω0 = 1 LC = 1.00 × 106 rad s ω ω0 ωL Ω ( ) 1 ω C Ω ( ) Z Ω ( ) P = I2R W ( ) For each angular frequency, we find Z = R2 + ω L −1/ω C ( ) 2 then I = 1.00 V ( )/ Z and P = I2 1.00 Ω ( ) 0.9990 0.9991 0.9993 0.9995 0.9997 0.9999 1.0000 1.0001 1.0003 1.0005 1.0007 1.0009 1.0010 999.0 999.1 999.3 999.5 999.7 999.9 1000 1000.1 1000.3 1000.5 1000.7 1000.9 1001 1001.0 1000.9 1000.7 1000.5 1000.3 1000.1 1000.0 999.9 999.7 999.5 999.3 999.1 999.0 2.24 2.06 1.72 1.41 1.17 1.02 1.00 1.02 1.17 1.41 1.72 2.06 2.24 0.19984 0.23569 0.33768 0.49987 0.73524 0.96153 1.00000 0.96154 0.73535 0.50012 0.33799 0.23601 0.20016 The full width at half maximum is: ∆f = ∆ω 2π = 1.0005 −0.9995 ( )ω0 2π ∆f = 1.00 × 10 3 s−1 2π = 159 Hz while R 2πL = 1.00 Ω 2π 1.00 × 10−3 H ( ) = 159 Hz Chapter 33 Solutions 295 © 2000 by Harcourt, Inc. All rights reserved.
33.71 ∆Vout ∆Vin = R R2 + 1 ωC ( ) 2 = R R2 + 1 2π f C ( ) 2 (a) ∆Vout ∆Vin = 1 2 when 1 ω C = R 3 Hence, f = ω 2π = 1 2π RC 3 = 1.84 kHz ∆Vin R C ∆Vout (b) Log Gain versus Log Frequency - 4 - 3 - 2 - 1 0 0 1 2 3 4 5 6 Log f Log∆V out/∆V in © 2000 by Harcourt, Inc. All rights reserved.
Chapter 34 Solutions 34.1 Since the light from this star travels at 3.00 × 108 m/s, the last bit of light will hit the Earth in 6.44 × 1018 m 3.00 × 108 m /s = 2.15 × 1010 s = 680 years. Therefore, it will disappear from the sky in the year 1999 + 680 = 2.68 × 10 3 A.D. 34.2 v = 1 kµ0e0 = 1 1.78 c = 0.750c = 2.25 × 108 m/s 34.3 E B = c or 220 B = 3.00 × 108; so B = 7.33 × 10–7 T = 733 nT 34.4 Emax Bmax = v is the generalized version of Equation 34.13. Bmax = Emax v = 7.60 × 10−3 V /m (2/ 3)(3.00 × 108 m /s) N ⋅m V ⋅C T ⋅C⋅m N ⋅s = 3.80 × 10–11 T = 38.0 pT 34.5 (a) f λ = c or f (50.0 m) = 3.00 × 108 m/s so f = 6.00 × 106 Hz = 6.00 MHz (b) E B = c or 22.0 Bmax = 3.00 × 108 so Bmax = (73.3 nT)(–k) (c) k = 2π λ = 2π 50.0 = 0.126 m–1 and ω = 2π f = 2π (6.00 × 106 s–1) = 3.77 × 107 rad/s B = Bmax cos(kx – ω t) = (73.3 nT) cos (0.126x – 3.77 × 107 t)(–k) 34.6 ω = 2π f = 6.00π × 109 s–1 = 1.88 × 1010 s-1 k = 2π λ = ω c = 6.00π × 109 s–1 3.00 × 108 m/s = 20.0π = 62.8 m–1 Bmax = E c = 300 V/m 3.00 × 108 m/s = 1.00 µT E = 300 V m cos 62.8x −1.88 × 1010t ( ) B = (1.00 µT) cos (62.8x – 1.88 × 1010 t) Chapter 34 Solutions 299 © 2000 by Harcourt, Inc. All rights reserved.
34.7 (a) B = E c = 100 V/m 3.00 × 10 8 m/s = 3.33 × 10–7 T = 0.333 µT (b) λ = 2π k = 2π 1.00 × 10 7 m–1 = 0.628 µm (c) f = c λ = 3.00 × 108 m/s 6.28 × 10–7 m = 4.77 × 1014 Hz 34.8 E = Emax cos(kx – ω t) ∂E ∂ x = –Emax sin(kx – ω t)(k) ∂E ∂ t = –Emax sin(kx – ω t)(–ω) ∂ 2E ∂ x 2 = –Emax cos(kx – ω t)(k 2) ∂ 2E ∂ t 2 = –Emax cos(kx – ω t)(–ω)2 We must show: ∂E ∂x2 = µ0e0 ∂2E ∂t2 That is, −k2 ( )Emax cos kx −ω t ( ) = −µ0e0 −ω ( )2Emax cos kx −ω t ( ) But this is true, because k2 ω 2 = 1 fλ 2 = 1 c2 = µ0e0 The proof for the wave of magnetic field is precisely similar.
34.9 In the fundamental mode, there is a single loop in the standing wave between the plates.
Therefore, the distance between the plates is equal to half a wavelength.
λ = 2L = 2(2.00 m) = 4.00 m Thus, f = c λ = 3.00 × 108 m/s 4.00 m = 7.50 × 10 7 Hz = 75.0 MHz 34.10 dA to A = 6 cm ± 5% = λ 2 λ = ± 12 cm 5% 300 Chapter 34 Solutions v = λ f = 0.12 m ± 5% ( ) 2.45 × 109 s−1 ( ) = 2.9 × 108 m s ± 5% 34.11 S = I = U At = Uc V = uc Energy Unit Volume = u = I c = 1000 W/m2 3.00 × 108 m/s = 3.33 µ J/m3 34.12 Sav = P 4πr2 = 4.00 × 10 3 W 4π (4.00 × 1609 m)2 = 7.68 µW/m2 Emax = 2µ0cSav = 0.0761 V /m ∆Vmax = Emax · L = (76.1 mV/m)(0.650 m) = 49.5 mV (amplitude) or 35.0 mV (rms) 34.13 r = 5.00 mi ( ) 1609 m /mi ( ) = 8.04 × 103 m S = P 4πr 2 = 250 × 10 3 W 4π (8.04 × 10 3 m)2 = 307 µW/m2 Chapter 34 Solutions 301 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically with an average power of 250 kW? G : As the distance from the source is increased, the power per unit area will decrease, so at a distance of 5 miles from the source, the power per unit area will be a small fraction of the Poynting vector near the source. O : The Poynting vector is the power per unit area, where A is the surface area of a sphere with a 5-mile radius.
A : The Poynting vector is In meters, r = 5.00 mi ( ) 1609 m /mi ( ) = 8045 m and the magnitude is S = 250 × 103 W (4π)(8045)2 = 3.07 × 10−4 W/m2 L : The magnitude of the Poynting vector ten meters from the source is 199 W/m2, on the order of a million times larger than it is 5 miles away! It is surprising to realize how little power is actually received by a radio (at the 5-mile distance, the signal would only be about 30 nW, assuming a receiving area of about 1 cm2).
34.14 I = 100 W 4π (1.00 m)2 = 7.96 W/m2 u = I c = 2.65 × 10 – 8 J/m3 = 26.5 n J/m3 (a) uE = 1 2 u = 13.3 n J/m3 (b) uB = 1 2 u = 13.3 n J/m3 (c) I = 7.96 W/m2 34.15 Power output = (power input)(efficiency) Thus, Power input = power out eff = 1.00 × 106 W 0.300 = 3.33 × 106 W and A = P I = 3.33 × 10 6 W 1.00 × 10 3 W/m2 = 3.33 × 10 3 m2 302 Chapter 34 Solutions 34.16 I = Bmax 2 c 2µ0 = P 4π r 2 Bmax = P 4π r2 2µ0 c = 10.0 × 103 ( ) 2 ( ) 4π × 10−7 ( ) 4π 5.00 × 103 ( ) 2 3.00 × 108 ( ) = 5.16 × 10–10 T Since the magnetic field of the Earth is approximately 5 × 10 –5 T, the Earth's field is some 100,000 times stronger.
34.17 (a) P = I 2R = 150 W; A = 2π rL = 2π (0.900 × 10–3 m)(0.0800 m) = 4.52 × 10 –4 m2 S = P A = 332 kW/m2 (points radially inward) (b) B = µ0 I 2π r = µ0(1.00) 2π (0.900 × 10 –3) = 222 µT E = ∆V ∆x = IR L = 150 V 0.0800 m = 1.88 kV/m Note: S = EB µ0 = 332 kW/m2 Chapter 34 Solutions 303 © 2000 by Harcourt, Inc. All rights reserved.
34.18 (a) E · B = (80.0 i + 32.0 j – 64.0 k)(N/C) · (0.200 i + 0.0800 j + 0.290 k)µT E · B= (16.0 + 2.56 – 18.56)N 2 · s/C 2 · m = 0 (b) S = 1 µ0 E × B= (80.0 i + 32.0 j – 64.0 k)(N/C) × (0.200 i + 0.0800 j + 0.290 k)µT 4π × 10 –7 T · m/A S = (6.40 k – 23.2 j – 6.40 k + 9.28 i – 12.8 j + 5.12 i)10 – 6 W/m2 4π × 10–7 S = (11.5 i – 28.6 j) W/m2 = 30.9 W/m2 at – 68.2° from the +x axis 34.19 We call the current Irms and the intensity I . The power radiated at this frequency is P = (0.0100)(∆Vrms)Irms = 0.0100(∆Vrms)2 R = 1.31 W If it is isotropic, the intensity one meter away is I = P A = 1.31 W 4π (1.00 m)2 = 0.104 W/m2 = Sav = c 2µ0 B2 max Bmax = 2µ0I c = 2 4π × 10−7 T ⋅m / A ( ) 0.104 W /m2 ( ) 3.00 × 108 m /s = 29.5 nΤ 34.20 (a) efficiency = useful power output total power input × 100% = 700 W 1400 W × 100% = 50.0% (b) Sav = P A = 700 W 0.0683 m ( ) 0.0381 m ( ) = 2.69 × 105 W m2 Sav = 269 kW m2 toward the oven chamber (c) Sav = Emax 2 2µ0c Emax = 2 4π × 10−7 T ⋅m A 3.00 × 108 m s 2.69 × 105 W m2 = 1.42 × 104 V m = 14.2 kV m 304 Chapter 34 Solutions 34.21 (a) Bmax = Emax c = 7.00 × 105 N/C 3.00 × 108 m/s = 2.33 mT (b) I = E2 max 2µ0c = (7.00 × 10 5)2 2(4π × 10 –7 )(3.00 × 108) = 650 MW/m2 (c) I = P A : P = I A = (6.50 × 108 W/m2) π 4 (1.00 × 10 – 3 m) 2 = 510 W 34.22 Power = SA = Emax 2 2µ0c (4πr2); solving for r , r = Pµ0c Emax 2 2π = (100 W)µ0c 2π(15.0 V /m)2 = 5.16 m 34.23 (a) I = ( .
) ( .
10 0 10 0 800 10 3 3 × × = − − W m)2 π 4.97 kW/m2 (b) uav = I c = 4.97 × 103 J/m2 ⋅s 3.00 × 108 m /s = 16.6 µ J/m3 34.24 (a) E cB = = × × − ( .
3 00 10 0 8 m/s)(1.8 10 T) = 6 540 V/m (b) uav = B2 µ0 = (1.80 × 10−6)2 4π × 10−7 = 2.58 µ J/m3 (c) Sav = cuav = (3.00 × 108)(2.58 × 10−6) = 773 W/m2 (d) This is 77.3% of the flux in Example 34.5 . It may be cloudy, or the Sun may be setting.
34.25 For complete absorption, P = S c = 25.0 3.00 × 108 = 83.3 nPa 34.26 (a) P = Sav ( ) A ( ) = 6.00 W /m2 ( ) 40.0 × 10−4 m2 ( ) = 2.40 × 10−2 J/s In one second, the total energy U impinging on the mirror is 2.40 × 10–2 J. The momentum p transferred each second for total reflection is p = 2U c = 2(2.40 × 10–2 J) 3.00 × 108 m/s = 1.60 × 10 –10 kg · m s (each second) (b) F = dp dt = 1.60 × 10–10 kg · m/s 1 s = 1.60 × 10–10 N Chapter 34 Solutions 305 © 2000 by Harcourt, Inc. All rights reserved.
34.27 (a) The radiation pressure is (2)(1340 W /m2) 3.00 × 108 m /s2 = 8.93 × 10−6 N/m2 Multiplying by the total area, A = 6.00 × 105 m2 gives: F = 5.36 N (b) The acceleration is: a = F m = 5.36 N 6000 kg = 8.93 × 10– 4 m/s2 (c) It will take a time t where: d = 1 2 at 2 or t = 2d a = 2 3.84 × 108 m ( ) 8.93 × 10−4 m /s2 ( ) = 9.27 × 105 s = 10.7 days 34.28 The pressure P upon the mirror is P = 2Sav c where A is the cross-sectional area of the beam and Sav = P A The force on the mirror is then F = PA = 2 c P A A = 2P c Therefore, F = 2(100 × 10−3) (3 × 108) = 6.67 × 10–10 N 34.29 I = P π r2 = Emax 2 2µ0c (a) Emax = P 2µ0c ( ) π r2 = 1.90 kN/C (b) 15 × 10–3 J/s 3.00 × 108 m/s (1.00 m) = 50.0 pJ (c) p = U c = 5 × 10–11 3.00 × 108 = 1.67 × 10–19 kg · m/s 306 Chapter 34 Solutions 34.30 (a) If PS is the total power radiated by the Sun, and rE and rM are the radii of the orbits of the planets Earth and Mars, then the intensities of the solar radiation at these planets are: IE = PS 4πrE 2 and IM = PS 4πrM 2 Thus, IM = IE rE rM 2 = 1340 W m2 ( ) 1.496 × 1011 m 2.28 × 1011 m 2 = 577 W/m2 (b) Mars intercepts the power falling on its circular face: PM = IM π RM 2 ( ) = 577 W m2 ( )π 3.37 × 106 m ( ) 2 = 2.06 × 1016 W (c) If Mars behaves as a perfect absorber, it feels pressure P = SM c = IM c and force F = PA = IM c π RM 2 ( ) = PM c = 2.06 × 1016 W 3.00 × 108 m s = 6.87 × 107 N (d) The attractive gravitational force exerted on Mars by the Sun is Fg = GMSMM rM 2 = 6.67 × 10−11 N ⋅m2 kg2 ( ) 1.991× 1030 kg ( ) 6.42 × 1023 kg ( ) 2.28 × 1011 m ( ) 2 = 1.64 × 1021 N which is ~1013 times stronger than the repulsive force of (c).
34.31 (a) The total energy absorbed by the surface is U = 1 2 I ( )At = 1 2 750 W m2 0.500 × 1.00 m2 ( ) 60.0 s ( ) = 11.3 kJ (b) The total energy incident on the surface in this time is 2U = 22.5 kJ, with U = 11.3 kJ being absorbed and U = 11.3 kJ being reflected. The total momentum transferred to the surface is p = momentum from absorption ( ) + momentum from reflection ( ) p = U c + 2U c = 3U c = 3 11.3 × 103 J ( ) 3.00 × 108 m s = 1.13 × 10−4 kg ⋅m s 34.32 Sav = µ0Jmax 2 c 8 or 570 = (4π × 10−7)Jmax 2 (3.00 × 108) 8 so Jmax = 3.48 A/m2 Chapter 34 Solutions 307 © 2000 by Harcourt, Inc. All rights reserved.
34.33 (a) P = SavA = µ0Jmax 2 c 8 A P = 4π × 10−7(10.0)2(3.00 × 108) 8 (1.20 × 0.400) = 2.26 kW (b) Sav = µ0Jmax 2 c 8 = (4π × 10−7(10.0)2(3.00 × 108) 8 = 4.71 kW/m2 34.34 P = ∆V ( )2 R or P ∝∆V ( )2 ∆V = − ( )Ey ⋅∆y = Ey ⋅lcosθ ∆V ∝cosθ so P ∝cos2 θ (a) θ = 15.0°: P = Pmax cos2 15.0° ( ) = 0.933Pmax= 93.3% (b) θ = 45.0°: P = Pmax cos2 45.0° ( ) = 0.500Pmax= 50.0% (c) θ = 90.0°: P = Pmax cos2 90.0° ( ) = 0 θ l eceiving ntenna ∆y 34.35 (a) Constructive interference occurs when d cos θ = nλ for some integer n. cosθ = n λ d = n λ λ / 2 = 2n n = 0, ± 1, ± 2, . . . ∴ strong signal @ θ = cos–1 0 = 90°, 270° (b) Destructive interference occurs when dcosθ = 2n + 1 2 λ : cos θ = 2n + 1 ∴ weak signal @ θ = cos–1 (±1) = 0°, 180° 308 Chapter 34 Solutions Goal Solution Two radio-transmitting antennas are separated by half the broadcast wavelength and are driven in phase with each other. In which directions are (a) the strongest and (b) the weakest signals radiated?
G : The strength of the radiated signal will be a function of the location around the two antennas and will depend on the interference of the waves.
O : A diagram helps to visualize this situation. The two antennas are driven in phase, which means that they both create maximum electric field strength at the same time, as shown in the diagram. The radio EM waves travel radially outwards from the antennas, and the received signal will be the vector sum of the two waves. A : (a) Along the perpendicular bisector of the line joining the antennas, the distance is the same to both transmitting antennas. The transmitters oscillate in phase, so along this line the two signals will be received in phase, constructively interfering to produce a maximum signal strength that is twice the amplitude of one transmitter.
(b) Along the extended line joining the sources, the wave from the more distant antenna must travel one-half wavelength farther, so the waves are received 180° out of phase. They interfere destructively to produce the weakest signal with zero amplitude.
L : Radio stations may use an antenna array to direct the radiated signal toward a highly-populated region and reduce the signal strength delivered to a sparsely-populated area.
34.36 λ = c f = 536 m so h = λ 4 = 134 m λ = c f = 188 m so h = λ 4 = 46.9 m 34.37 For the proton: ΣF = ma ⇒ qvB sin 90.0° = mv2 R The period and frequency of the proton’s circular motion are therefore: T = 2πR v = 2π m qB = 2π 1.67 × 10−27 kg ( ) 1.60 × 10 −19 C ( ) 0.350 T ( ) = 1.87 × 10−7 s f = 5.34 × 106 Hz. The charge will radiate at this same frequency, with λ = c f = 3.00 × 108 m s 5.34 × 106 Hz = 56.2 m 34.38 For the proton, ΣF = ma yields qvB sin 90.0° = mv2 R The period of the proton’s circular motion is therefore: T = 2πR v = 2π m qB The frequency of the proton's motion is f = 1/T The charge will radiate electromagnetic waves at this frequency, with λ = = = c f cT 2π mc qB Chapter 34 Solutions 309 © 2000 by Harcourt, Inc. All rights reserved.
34.39 From the electromagnetic spectrum chart and accompanying text discussion, the following identifications are made: Frequency f Wavelength, λ = c f Classification 2 Hz = 2 × 100 Hz 150 Mm Radio 2 kHz = 2 × 103 Hz 150 km Radio 2 MHz = 2 × 106 Hz 150 m Radio 2 GHz = 2 × 109 Hz 15 cm Microwave 2 THz = 2 × 1012 Hz 150 µm Infrared 2 PHz = 2 × 1015 Hz 150 nm Ultraviolet 2 EHz = 2 × 1018 Hz 150 pm x-ray 2 ZHz = 2 × 1021 Hz 150 fm Gamma ray 2 YHz = 2 × 1024 Hz 150 am Gamma Ray Wavelength, λ Frequency f c = λ Classification 2 km = 2 × 103 m 1.5 × 105 Hz Radio 2 m = 2 × 100 m 1.5 × 108 Hz Radio 2 mm = 2 × 10−3 m 1.5 × 1011 Hz Microwave 2 µm = 2 × 10−6 m 1.5 × 1014 Hz Infrared 2 nm = 2 × 10−9 m 1.5 × 1017 Hz Ultraviolet/x-ray 2 pm = 2 × 10−12 m 1.5 × 1020 Hz x-ray/Gamma ray 2 fm = 2 × 10−15 m 1.5 × 1023 Hz Gamma ray 2 am = 2 × 10−18 m 1.5 × 1026 Hz Gamma ray 34.40 (a) f = c λ = 3 × 108 m/s 1.7 m ~ 108 Hz radio wave (b) 1000 pages, 500 sheets, is about 3 cm thick so one sheet is about 6 × 10 – 5 m thick f = 3 × 108 m/s 6 × 10–5 m ~ 1013 Hz infrared 34.41 f = c λ = 3.00 × 108 m/s 5.50 × 10–7 m = 5.45 × 1014 Hz 34.42 (a) λ = c f = 3.00 × 108 m/s 1150 × 103/s = 261 m so 180 m 261 m = 0.690 wavelengths (b) λ = c f = 3.00 × 108 m/s 98.1 × 106/s = 3.06 m so 180 m 3.06 m = 58.9 wavelengths 310 Chapter 34 Solutions 34.43 (a) f λ = c gives (5.00 × 1019 Hz)λ = 3.00 × 108 m/s: λ = 6.00 × 10 –12 m = 6.00 pm (b) f λ = c gives (4.00 × 109 Hz)λ = 3.00 × 108 m/s: λ = 0.075 m = 7.50 cm 34.44 Time to reach object = 1 2 (total time of flight) = 1 2 (4.00 × 10 – 4 s) = 2.00 × 10 – 4 s Thus, d = vt = (3.00 × 10 8 m/s)(2.00 × 10 – 4 s) = 6.00 × 10 4 m = 60.0 km 34.45 The time for the radio signal to travel 100 km is: tr = 100 × 103 m 3.00 × 108 m /s = 3.33 × 10 – 4 s The sound wave to travel 3.00 m across the room in: ts = 3.00 m 343 m/s = 8.75 × 10– 3 s Therefore, listeners 100 km away will receive the news before the people in the newsroom by a total time difference of ∆t = 8.75 × 10– 3 s – 3.33 × 10– 4 s = 8.41 × 10– 3 s 34.46 The wavelength of an ELF wave of frequency 75.0 Hz is λ = c f = 3.00 × 108 m s 75.0 Hz = 4.00 × 106 m The length of a quarter-wavelength antenna would be L = 1.00 × 106 m = 1.00 × 103 km or L = 1000 km ( ) 0.621 mi 1.00 km = 621 mi Thus, while the project may be theoretically possible, it is not very practical.
34.47 (a) For the AM band, λmax = c fmin = 3.00 × 108 m /s 540 × 103 Hz = 556 m λmin = c fmax = 3.00 × 108 m /s 1600 × 103 Hz = 187 m (b) For the FM band, λmax = c fmin = 3.00 × 108 m /s 88.0 × 106 Hz = 3.41 m λmin = c fmax = 3.00 × 108 m /s 108 × 106 Hz = 2.78 m Chapter 34 Solutions 311 © 2000 by Harcourt, Inc. All rights reserved.
34.48 CH 4 : fmin = 66 MHz λ max = 4.55 m fmax = 72 MHz λ min = 4.17 m CH 6 : fmin = 82 MHz λ max = 3.66 m fmax = 88 MHz λ min = 3.41 m CH 8 : fmin = 180 MHz λ max = 1.67 m fmax = 186 MHz λ min = 1.61 m 34.49 (a) P = SA = (1340 W/m2)4π (1.496 × 1011 m)2 = 3.77 × 1026 W (b) S = cBmax 2 2µ0 so Bmax = 2µ0S c = 2(4π × 10−7 N/ A2)(1340 W /m2) 3.00 × 108 m /s = 3.35 µT S = Emax 2 2µ0c so Emax = 2µ0cS = 2 4π × 10−7 ( ) 3.00 × 108 ( ) 1340 ( ) = 1.01 kV/m 34.50 Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60°. Then the target area you fill in the Sun's field of view is (1.7 m )( 0.3 m )( cos 30° ) = 0.4 m2 Now I = P A = E At: E = IAt = 1340 W m2 (0.6)(0.5)(0.4 m2) 3600 s ~ 106 J 34.51 (a) ε = −dΦB dt = −d dt (BA cos θ) = −A d dt (Bmax cos ωt cos θ) = ABmax ω(sin ωt cos θ) ε(t) = 2π fBmax A sin 2π ft cos θ = 2π 2r 2fBmax cos θ sin 2π ft Thus, εmax = 2π 2r 2fBmax cos θ , where θ is the angle between the magnetic field and the normal to the loop.
(b) If E is vertical, then B is horizontal, so the plane of the loop should be vertical and the plane should contain the line of sight to the transmitter .
312 Chapter 34 Solutions 34.52 (a) Fgrav = GMsm R2 = GMs R2 ρ 4/ 3 ( )π r3 where Ms = mass of Sun, r = radius of particle and R = distance from Sun to particle. Since Frad = Sπ r 2 c , Frad Fgrav = 1 r 3SR2 4cGMsρ ∝1 r (b) From the result found in part (a), when Fgrav = Frad, we have r = 3SR2 4cGMs ρ r = 3 214 W /m2 ( ) 3.75 × 1011 m ( ) 2 4 6.67 × 10−11 N ⋅m2 kg2 ( ) 1.991× 1030 kg ( ) 1500 kg m3 ( ) 3.00 × 108 m s ( ) = 3.78 × 10–7 m 34.53 (a) Bmax = Emax c = 6.67 × 10–16 T (b) Sav = E c max 2 0 2µ = 5.31 × 10–17 W/m2 (c) P = Sav A = 1.67 × 10–14 W (d) F = PA = Sav c A = 5.56 × 10–23 N (≈ weight of 3000 H atoms!) 34.54 (a) The electric field between the plates is E = ∆V l, directed downward in the figure. The magnetic field between the plate's edges is B = µ0 i 2π r counterclockwise. The Poynting vector is: S = 1 µ0 E × B = ∆V ( )i 2π rl (radially outward) (b) The lateral surface area surrounding the electric field volume is A = 2π rl, so the power output is P = SA = ∆V ( )i 2π rl 2π rl ( ) = (∆V)i ---- - -- -- ----- - -+ + + + + + + + + + + + + + E S B i i (c) As the capacitor charges, the polarity of the plates and hence the direction of the electric field is unchanged. Reversing the current reverses the direction of the magnetic field, and therefore the Poynting vector. The Poynting vector is now directed radially inward. Chapter 34 Solutions 313 © 2000 by Harcourt, Inc. All rights reserved.
34.55 (a) The magnetic field in the enclosed volume is directed upward, with magnitude B = µ0ni and increasing at the rate dB dt = µ0n di dt .
The changing magnetic field induces an electric field around any circle of radius r, according to Faraday’s Law: E 2π r ( ) = −µ0n di dt π r2 ( ) E = −µ0nr 2 di dt or E = µ0nr 2 di dt (clockwise) B i S E Then, S = 1 µ0 E × B = 1 µ0 µ0nr 2 di dt µ0ni ( ) inward, or the Poynting vector is S = µ0n2ri 2 di dt (radially inward) (b) The power flowing into the volume is P = SAlat where Alat is the lateral area perpendicular to S. Therefore, P = µ0n2ri 2 di dt 2π rl ( ) = µ0π n2r2li di dt (c) Taking Across to be the cross-sectional area perpendicular to B, the induced voltage between the ends of the inductor, which has N = nl turns, is ∆V = ε = N dB dt Across = nl µ0n di dt π r2 ( ) = µ0π n2r2l di dt and it is observed that P = ∆V ( )i 34.56 (a) The power incident on the mirror is: P I= IA = 1340 W m2 π 100 m ( )2 [ ] = 4.21× 107 W The power reflected through the atmosphere is PR = 0.746 4.21× 107 W ( ) = 3.14 × 107 W (b) S = PR A = 3.14 × 107 W π 4.00 × 103 m ( ) 2 = 0.625 W/m2 (c) Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface: PN = 0.746 1340 W /m2 ( )sin 7.00°= 122 W /m2 The radiation intensity received from the mirror is 0.625 W m2 122 W m2 100% = 0.513% of that from the noon Sun in January.
314 Chapter 34 Solutions 34.57 u = 1 2 e0Emax 2 (Equation 34.21) Emax = 2u e0 = 95.1 mV/m 34.58 The area over which we model the antenna as radiating is the lateral surface of a cylinder, A = 2π rl= 2π 4.00 × 10−2 m ( ) 0.100 m ( ) = 2.51× 10−2 m2 (a) The intensity is then: S = P A = 0.600 W 2.51× 10−2 m2 = 23.9 W m2 (b) The standard is: 0.570 mW cm2 = 0.570 mW cm2 1.00 × 10−3 W 1.00 mW 1.00 × 104 cm2 1.00 m2 = 5.70 W m2 While it is on, the telephone is over the standard by 23.9 W m2 5.70 W m2 = 4.19 times 34.59 (a) Bmax = Emax c = 175 V/m 3.00 × 108 m/s = 5.83 × 10–7 T k = 2π λ = 2π (0.0150 m) = 419 rad/m ω = kc = 1.26 × 1011 rad/s Since S is along x, and E is along y, B must be in the z direction . (That is S ∝ E × B.) (b) Sav = EmaxBmax 2µ0 = 40.6 W/m2 (c) Pr = 2S c = 2.71 × 10–7 N/m2 (d) a = ΣF m = PA m = 2.71× 10−7 N/m2 ( ) 0.750 m2 ( ) 0.500 kg = 4.06 × 10−7 m /s2 Chapter 34 Solutions 315 © 2000 by Harcourt, Inc. All rights reserved.
34.60 (a) At steady-state, Pin = Pout and the power radiated out is Pout = eσAT4.
Thus, 0.900 1000 W m2 A = 0.700 ( ) 5.67 × 10−8 W m2 ⋅K4 AT4 or T = 900 W m2 0.700 5.67 × 10−8 W m2 ⋅K4 ( ) 1 4 = 388 K = 115°C (b) The box of horizontal area A, presents projected area A sin .
50 0° perpendicular to the sunlight. Then by the same reasoning, 0.900 1000 W m2 A sin 50.0° = 0.700 ( ) 5.67 × 10−8 W m2 ⋅K4 AT4 or T = 900 W m2 ( )sin 50.0° 0.700 5.67 × 10−8 W m2 ⋅K4 ( ) 1 4 = 363 K = 90.0 °C 34.61 (a) P = F A = I c F = IA c = P c = 100 J/s 3.00 × 108 m /s = 3.33 × 10−7 N = 110 kg ( )a a = 3.03 × 10–9 m/s2 and x = 1 2 at 2 t = 2x a = 8.12 × 104 s = 22.6 h (b) 0 = (107 kg)v – (3.00 kg)(12.0 m/s – v) = (107 kg)v – 36.0 kg · m/s + (3.00 kg)v v = 36.0 110 = 0.327 m/s t = 30.6 s 316 Chapter 34 Solutions Goal Solution An astronaut, stranded in space 10.0 m from his spacecraft and at rest relative to it, has a mass (including equipment) of 110 kg. Since he has a 100-W light source that forms a directed beam, he decides to use the beam as a photon rocket to propel himself continuously toward the spacecraft. (a) Calculate how long it takes him to reach the spacecraft by this method. (b) Suppose, instead, he decides to throw the light source away in a direction opposite the spacecraft. If the mass of the light source has a mass of 3.00 kg and, after being thrown, moves at 12.0 m/s relative to the recoiling astronaut , how long does it take for the astronaut to reach the spacecraft?
G : Based on our everyday experience, the force exerted by photons is too small to feel, so it may take a very long time (maybe days!) for the astronaut to travel 10 m with his “photon rocket.” Using the momentum of the thrown light seems like a better solution, but it will still take a while (maybe a few minutes) for the astronaut to reach the spacecraft because his mass is so much larger than the mass of the light source.
O : In part (a), the radiation pressure can be used to find the force that accelerates the astronaut toward the spacecraft. In part (b), the principle of conservation of momentum can be applied to find the time required to travel the 10 m.
A : (a) Light exerts on the astronaut a pressure P = F A = S c, and a force of F = SA c = = 100 J/s 3.00 × 108 m /s = 3.33 × 10−7 N By Newton’s 2nd law, a = F m = 3.33 × 10−7 N 110 kg = 3.03 × 10−9 m /s2 This acceleration is constant, so the distance traveled is x = 1 2 at2, and the amount of time it travels is t = 2x a = 2 10.0 m ( ) 3.03 × 10−9 m /s2 = 8.12 × 104 s = 22.6 h (b) Because there are no external forces, the momentum of the astronaut before throwing the light is the same as afterwards when the now 107-kg astronaut is moving at speed v towards the spacecraft and the light is moving away from the spacecraft at 12.0 m /s −v ( ). Thus, pi = pf gives 0 = 107 kg ( )v −3.00 kg ( ) 12.0 m /s −v ( ) 0 = 107 kg ( )v −36.0 kg ⋅m /s ( ) + 3.00 kg ( )v v = 36.0 110 = 0.327 m /s t = x v = 10.0 m 0.327 m /s = 30.6 s L : Throwing the light away is certainly a more expedient way to reach the spacecraft, but there is not much chance of retrieving the lamp unless it has a very long cord. How long would the cord need to be, and does its length depend on how hard the astronaut throws the lamp? (You should verify that the minimum cord length is 367 m, independent of the speed that the lamp is thrown.) Chapter 34 Solutions 317 © 2000 by Harcourt, Inc. All rights reserved.
34.62 The 38.0% of the intensity S = 1340 W m2 that is reflected exerts a pressure P1 = 2Sr c = 2 0.380 ( )S c The absorbed light exerts pressure P2 = Sa c = 0.620 ( )S c Altogether the pressure at the subsolar point on Earth is (a) Ptot = P1 + P2 = 1.38S c = 1.38(1340 W/m2) 3.00 × 108 m/s = 6.16 × 10– 6 Pa (b) Pa Ptot = 1.01× 105 N/m2 6.16 × 10−6 N/m2 = 1.64 × 1010 times smaller than atmospheric pressure 34.63 Think of light going up and being absorbed by the bead which presents a face area π r 2 b . The light pressure is P = S c = I c .
(a) Fl = Iπ r 2 b c = mg = ρ 4 3 π r 3 b g and I = 4ρ gc 3 3m 4πρ 1/3 = 8.32 × 10 7 W/m2 (b) P = IA = (8.32 × 10 7 W/m2)π (2.00 × 10–3 m)2 = 1.05 kW 34.64 Think of light going up and being absorbed by the bead which presents face area π r 2 b .
If we take the bead to be perfectly absorbing, the light pressure is P = Sav c = I c = Fl A (a) Fl = Fg so I = Flc A = Fgc A = mgc π rb 2 From the definition of density, ρ = m V = m 4 3 π rb 3 so 1 rb = 4 3 π ρ m ( ) 1/3 Substituting for rb, I = mgc π 4πρ 3m 2/3 = gc 4ρ 3 2/3 m π 1/3 = 4ρgc 3 3m 4π ρ 1/3 (b) P = IA = π r 24ρgc 3 3m 4π ρ 1/3 318 Chapter 34 Solutions 34.65 The mirror intercepts power P = I1A1 = (1.00 × 10 3 W/m2)π (0.500 m)2 = 785 W In the image, (a) I2 = P A2 = 785 W π 0.0200 m ( )2 = 625 kW/m2 (b) I2 = E 2 max 2µ0c so Emax = (2µ0c I2)1/2 = [2(4π × 10–7)(3.00 × 108)(6.25 × 105)]1/2= 21.7 kN/C Bmax = Emax c = 72.4 µT (c) 0.400 P t = mc ∆T 0.400(785 W)t = (1.00 kg) 4186 J kg · C° (100°C – 20.0°C) t = 3.35 × 105 J 314 W = 1.07 × 103 s = 17.8 min 34.66 (a) λ = c f = 3.00 × 108 m s 20.0 × 109 s−1 = 1.50 cm (b) U = P ∆t ( ) = 25.0 × 103 J s 1.00 × 10−9 s ( ) = 25.0 × 10−6 J = 25.0 µJ (c) uav = U V = U π r2 ( )l = U π r2 ( )c ∆t ( ) = 25.0 × 10−6 J π 0.0600 m ( )2 3.00 × 108 m s ( ) 1.00 × 10−9 s ( ) uav = 7.37 × 10−3 J m3 = 7.37 mJ m3 (d) Emax = 2uav e0 = 2 7.37 × 10−3 J m3 ( ) 8.85 × 10−12 C2 N ⋅m2 = 4.08 × 104 V m = 40.8 kV/m Bmax = Emax c = 4.08 × 104 V m 3.00 × 108 m s = 1.36 × 10−4 T = 136 µT (e) F = PA = S c A = cuav c A = uavA = 7.37 × 10−3 J m3 π 0.0600 m ( )2 = 8.33 × 10−5 N = 83.3 µN Chapter 34 Solutions 319 © 2000 by Harcourt, Inc. All rights reserved.
34.67 (a) On the right side of the equation, C2 m s2 ( ) 2 C2 N ⋅m2 ( ) m s ( ) 3 = N ⋅m2 ⋅C2 ⋅m2 ⋅s3 C2 ⋅s4 ⋅m3 = N ⋅m s = J s = W (b) F = ma = qE or a = qE m = 1.60 × 10−19 C ( ) 100 N C ( ) 9.11× 10−31 kg = 1.76 × 1013 m s2 The radiated power is then: P = q2a2 6π e0c3 = 1.60 × 10−19 ( ) 2 1.76 × 1013 ( ) 2 6π 8.85 × 10−12 ( ) 3.00 × 108 ( ) 3 = 1 75 10 27 .
× − W (c) F = mar = m v2 r = qvB so v = qBr m The proton accelerates at a = v2 r = q2B2r m2 = 1.60 × 10−19 ( ) 2 0.350 ( )2 0.500 ( ) 1.67 × 10−27 ( ) 2 = 5.62 × 1014 m s2 The proton then radiates P = q2a2 6π e0c3 = 1.60 × 10−19 ( ) 2 5.62 × 1014 ( ) 2 6π 8.85 × 10−12 ( ) 3.00 × 108 ( ) 3 = 1.80 × 10−24 W 34.68 P = S c = Power Ac = P 2π rlc = 60.0 W 2π(0.0500 m)(1.00 m)(3.00 × 108 m \s) = 6.37 × 10–7 Pa 34.69 F = PA = SA c = P / A ( )A c = P c , τ = F l 2 = Pl 2c , and τ = κ θ Therefore, θ = Pl 2cκ = 3.00 × 10−3 ( ) 0.0600 ( ) 2 3.00 × 108 ( ) 1.00 × 10−11 ( ) = 3.00 × 10– 2 deg 34.70 We take R to be the planet’s distance from its star. The planet, of radius r, presents a projected area π r2 perpendicular to the starlight. It radiates over area 4π r2. At steady-state, Pin = Pout: e Iin π r2 ( ) = eσ 4π r2 ( )T4 e 6.00 × 1023 W 4π R2 π r2 ( ) = eσ 4π r2 ( )T4 so that 6.00 × 1023 W = 16π σ R2 T4 R = 6.00 × 1023 W 16π σ T4 = 6.00 × 1023 W 16π 5.67 × 10−8 W m2 ⋅K4 ( ) 310 K ( )4 = 4.77 × 109 m = 4.77 Gm 320 Chapter 34 Solutions 34.71 The light intensity is I = Sav = E2 2µ0c The light pressure is P = S c = E2 2µ0c2 = 1 2 e0 E2 For the asteroid, PA = ma and a = e0E2A 2m 34.72 f = 90.0 MHz, Emax = 2.00 × 10−3 V /m = 200 mV/m (a) λ = c f = 3.33 m T = 1 f = 1.11× 10−8 s = 11.1 ns Bmax = Emax c = 6.67 × 10−12 T = 6.67 pT (b) E = (2.00 mV /m) cos 2π x 3.33 m − t 11.1 ns j B = 6.67 pT ( )k cos2π x 3.33 m − t 11.1 ns (c) I = Emax 2 2µ0c = (2.00 × 10−3)2 2(4π × 10−7)(3.00 × 108) = 5.31× 10−9 W/m2 (d) I =cuav so uav = 1.77 × 10−17 J/m3 (e) P = 2I c = (2)(5.31× 10−9) 3.00 × 108 = 3.54 × 10−17 Pa © 2000 by Harcourt, Inc. All rights reserved.
Chapter 35 Solutons 35.1 The Moon's radius is 1.74 × 10 6 m and the Earth's radius is 6.37 × 10 6 m. The total distance traveled by the light is: d = 2(3.84 × 10 8 m – 1.74 × 10 6 m – 6.37 × 10 6 m) = 7.52 × 10 8 m This takes 2.51 s, so v = 7.52 × 10 8 m 2.51 s = 2.995 × 10 8 m/s = 299.5 Mm/s 35.2 ∆x = ct ; c = ∆x t = 2(1.50 × 10 8 km)(1000 m/km) (22.0 min)(60.0 s/min) = 2.27 × 10 8 m/s = 227 Mm/s 35.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing light through one notch and returning light through the next: t = 2l c θ = ω t = ω 2l c so ω = cθ 2l = (2.998 × 108)[2π /(720)] 2(11.45 × 10 3) = 114 rad/s The returning light would be blocked by a tooth at one-half the angular speed, giving another data point.
35.4 (a) For the light beam to make it through both slots, the time for the light to travel the distance d must equal the time for the disk to rotate through the angle θ, if c is the speed of light, d c = θ ω , so c = dω θ (b) We are given that d = 2.50 m, θ π = ° ° = × − 1 00 60 0 180 2 91 10 4 .
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. rad rad, ω = 5555 rev s 2π rad 1.00 rev = 3.49 × 104 rad s c = dω θ = 2.50 m ( ) 3.49 × 10 4 rad s ( ) 2.91× 10−4 rad = 3.00 × 108 m s = 300 Mm/s 35.5 Using Snell's law, sin sin θ θ 2 1 2 1 = n n θ 2 = 25.5° λ λ 2 1 1 = = n 442 nm Chapter 35 Solutions 323 © 2000 by Harcourt, Inc. All rights reserved.
35.6 (a) f c = = × × = − λ 3 00 108 7 . m/s 6.328 10 m 4.74 × 10 14 Hz (b) λ λ glass air nm 1.50 = = = n 632 8 . 422 nm (c) v c n glass air m/s 1.50 m/s = = = × = × 3 00 10 2 00 10 8 8 .
. 200 Mm/s 35.7 n n 1 1 2 2 sin sin θ θ = sin θ1 = 1.333 sin 45.0° sin θ1 = (1.33)(0.707) = 0.943 θ1 = 70.5° → 19.5° above the horizon Figure for Goal Solution Goal Solution An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the vertical. What is the actual direction of the Sun?
G : The sunlight refracts as it enters the water from the air. Because the water has a higher index of refraction, the light slows down and bends toward the vertical line that is normal to the interface.
Therefore, the elevation angle of the Sun above the water will be less than 45° as shown in the diagram to the right, even though it appears to the diver that the sun is 45° above the horizon.
O : We can use Snell’s law of refraction to find the precise angle of incidence.
A : Snell’s law is: n n 1 1 2 2 sin sin θ θ = which gives sin θ1 = 1.333 sin 45.0° sinθ1 = (1.333)(0.707) = 0.943 The sunlight is at θ1 = 70.5° to the vertical, so the Sun is 19.5° above the horizon.
L : The calculated result agrees with our prediction. When applying Snell’s law, it is easy to mix up the index values and to confuse angles-with-the-normal and angles-with-the-surface. Making a sketch and a prediction as we did here helps avoid careless mistakes.
324 Chapter 35 Solutions 35.8 (a) n n 1 1 2 2 sin sin θ θ = 1.00 sin 30.0° = n sin 19.24° n = 1.52 (c) f c = = × × = − λ 3 00 108 7 . m/s 6.328 10 m 4.74 × 10 14 Hz in air and in syrup.
(d) v c n = = × = × 3 00 10 1 98 10 8 8 .
. m/s 1.52 m/s = 198 Mm/s (b) λ = = × × = v f 1 98 10 4 74 10 8 14 .
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/ m/s s 417 nm 35.9 (a) Flint Glass: v c n = = × = × = 3 00 10 1 81 10 8 8 .
. m s 1.66 m s 181 Mm/s (b) Water: v c n = = × = × = 3 00 10 2 25 10 8 8 .
. m s 1.333 m s 225 Mm/s (c) Cubic Zirconia: v c n = = × = × = 3 00 10 1 36 10 8 8 .
. m s 2.20 m s 136 Mm/s 35.10 n n 1 1 2 2 sin sin θ θ = ; 1.333 sin 37.0° = n2 sin 25.0° n2 = 1.90 = c v ; v = c 1.90 = 1.58 × 10 8 m/s = 158 Mm/s 35.11 n n 1 1 2 2 sin sin θ θ = ; θ 2 = sin–1 n1 sin θ 1 n2 θ2 1 1 00 30 1 50 = ° = − sin ( .
)(sin ) . 19.5° θ 2 and θ 3 are alternate interior angles formed by the ray cutting parallel normals. So, θ 3 = θ 2 = 19.5° .
1.50 sin θ 3 = (1.00) sin θ4 θ4 = 30.0° Chapter 35 Solutions 325 © 2000 by Harcourt, Inc. All rights reserved.
35.12 (a) Water λ λ = = = 0 436 n nm 1.333 327 nm (b) Glass λ λ = = = 0 436 n nm 1.52 287 nm 35.13 sin sin θ θ 1 2 = nw sin .
sin .
sin( .
.
) .
θ θ 2 1 1 1 333 1 1 333 90 0 28 0 0 662 = = ° − ° = θ2 1 0 662 41 5 = = ° − sin .
. h d = = ° = tan .
θ2 3 00 m tan 41.5 3.39 m 35.14 (a) From geometry, 1.25 m = d sin 40.0° so d = 1.94 m (b) 50.0° above horizontal , or parallel to the incident ray 35.15 The incident light reaches the left-hand mirror at distance (1.00 m) tan 5.00° = 0.0875 m above its bottom edge. The reflected light first reaches the right-hand mirror at height 2(0.0875 m) = 0.175 m It bounces between the mirrors with this distance between points of contact with either. Since 1.00 m 0.175 m = 5.72, the light reflects five times from the right-hand mirror and six times from the left. 326 Chapter 35 Solutions 35.16 At entry, n n 1 1 2 2 sin sin θ θ = or 1.00 sin 30.0° = 1.50 sin θ 2 θ 2 = 19.5° The distance h the light travels in the medium is given by cos θ 2 = (2.00 cm) h or h = ° = ( .
.
2 00 2 12 cm) cos 19.5 cm The angle of deviation upon entry is α θ θ = − = ° − ° = ° 1 2 30 0 19 5 10 5 .
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The offset distance comes from sin α = d h : d = (2.21 cm) sin 10.5° = 0.388 cm 35.17 The distance, h, traveled by the light is h = ° = 2 00 2 12 .
. cm cos 19.5 cm The speed of light in the material is v c n = = × = × 3 00 10 2 00 10 8 8 .
. m/s 1.50 m/s Therefore, t h v = = × × = × − − 2 12 10 2 00 10 1 06 10 2 8 10 .
.
. m m/s s = 106 ps 35.18 Applying Snell's law at the air-oil interface, n n air oil sin sin .
θ = ° 20 0 yields θ = 30.4° Applying Snell's law at the oil-water interface n n w sin sin .
′ = ° θ oil 20 0 yields ′ = ° θ 22 3 . 35.19 time difference = (time for light to travel 6.20 m in ice) – (time to travel 6.20 m in air) ∆t = 6.20 m vice −6.20 m c but v = c n ∆t = (6.20 m) 1.309 c −1 c = (6.20 m) c (0.309) = 6.39 × 10−9 s = 6.39 ns Chapter 35 Solutions 327 © 2000 by Harcourt, Inc. All rights reserved.
35.20 Consider glass with an index of refraction of 1.5, which is 3 mm thick The speed of light in the glass is 3 10 1 5 2 10 8 8 × = × m/s m/s .
The extra travel time is 3 × 10−3 m 2 × 108 m /s −3 × 10−3 m 3 × 108 m /s ~ 10–11 s For light of wavelength 600 nm in vacuum and wavelength 600 400 nm 1.5 nm = in glass, the extra optical path, in wavelengths, is 3 10 4 10 3 10 6 10 3 7 3 7 × × − × × − − − − m m m m ~ 103 wavelengths 35.21 Refraction proceeds according to 1 00 1 66 1 2 .
sin .
sin ( ) = ( ) θ θ (1) (a) For the normal component of velocity to be constant, v1 cos θ1 = v2 cos θ2 or c c ( ) = ( ) cos .
cos θ θ 1 2 1 66 (2) We multiply Equations (1) and (2), obtaining: sin cos sin cos θ θ θ θ 1 1 2 2 = or sin sin 2 2 1 2 θ θ = The solution θ θ 1 2 0 = = does not satisfy Equation (2) and must be rejected. The physical solution is 2 180 2 1 2 θ θ = ° − or θ θ 2 1 90 0 = ° − .
. Then Equation (1) becomes: sin .
cos θ θ 1 1 1 66 = , or tan .
θ1 1 66 = which yields θ1 = 58.9° (b) Light entering the glass slows down and makes a smaller angle with the normal. Both effects reduce the velocity component parallel to the surface of the glass, so that component cannot remain constant, or will remain constant only in the trivial case θ1 = θ2 = 0 35.22 See the sketch showing the path of the light ray. α and γ are angles of incidence at mirrors 1 and 2.
For triangle abca, 2α + 2γ + β = 180° or β α γ = ° − + ( ) 180 2 (1) Now for triangle bcdb, 90.0° −α ( ) + 90.0° −γ ( ) + θ = 180° or θ = α + γ (2) 328 Chapter 35 Solutions Substituting Equation (2) into Equation (1) gives β = 180° −2θ Note: From Equation (2), γ = θ −α . Thus, the ray will follow a path like that shown only if α < θ . For α > θ , γ is negative and multiple reflections from each mirror will occur before the incident and reflected rays intersect.
35.23 Let n(x) be the index of refraction at distance x below the top of the atmosphere and n x = h ( ) = n be its value at the planet surface. Then, n x ( ) = 1.000 + n −1.000 h x (a) The total time required to traverse the atmosphere is t = dx v 0 h ∫ = n x ( ) c dx 0 h ∫ = 1 c 1.000 + n −1.000 h x 0 h ∫ dx = h c + n −1.000 ( ) ch h2 2 = h c n + 1.000 2 t = × × + = 20 0 10 1 005 1 000 2 3 .
.
. m 3.00 10 m s 8 66.8 µs (b) The travel time in the absence of an atmosphere would be h/c. Thus, the time in the presence of an atmosphere is n + 1.000 2 = 1.0025 times larger or 0.250% longer .
35.24 Let n(x) be the index of refraction at distance x below the top of the atmosphere and n x = h ( ) = n be its value at the planet surface. Then, n x ( ) = 1.000 + n −1.000 h x (a) The total time required to traverse the atmosphere is t = dx v 0 h ∫ = n x ( ) c dx 0 h ∫ = 1 c 1.000 + n −1.000 h x 0 h ∫ dx = h c + n −1.000 ( ) ch h2 2 = h c n + 1.000 2 (b) The travel time in the absence of an atmosphere would be h/c. Thus, the time in the presence of an atmosphere is n + 1 000 2 . times larger Chapter 35 Solutions 329 © 2000 by Harcourt, Inc. All rights reserved.
35.25 From Fig. 35.20 nv = 1.470 at 400 nm and nr = 1.458 at 700 nm Then ( .
)sin .
sin 1 00 1 470 θ θ = v and ( .
)sin .
sin 1 00 1 458 θ θ = r δr −δv = θr −θv = sin−1 sin θ 1.458 −sin−1 sin θ 1.470 ∆δ = sin−1 sin 30.0° 1.458 −sin−1 sin 30.0° 1.470 = 0.171° 35.26 n1 sin θ1 = n2 sin θ2 so θ2 = sin−1 n1 sin θ1 n2 θ2 = sin−1 (1.00)(sin 30.0°) 1.50 = 19.5° θ3 = (90.0° −19.5°) + 60.0° [ ] −180° ( ) + 90.0° = 40.5° n3 sin θ3 = n4 sin θ4 so θ4 = sin−1 n3 sin θ3 n4 = sin−1 (1.50)(sin 40.5°) 1.00 = 77.1° 35.27 Taking Φ to be the apex angle and δmin to be the angle of minimum deviation, from Equation 35.9, the index of refraction of the prism material is n = sin Φ + δmin 2 sin Φ 2 ( ) Solving for δmin, δmin = 2 sin−1 n sin Φ 2 −Φ = 2 sin−1 2.20 ( ) sin 25.0° ( ) [ ] −50.0° = 86.8° 35.28 n(700 nm) = 1.458 (a) (1.00) sin 75.0° = 1.458 sin θ 2; θ 2 = 41.5° (b) Let θ3 + β = 90.0°, θ2 + α = 90.0°; then α + β + 60.0° = 180° So 60.0° −θ2 −θ3 = 0 ⇒60.0° −41.5° = θ3 = 18.5° (c) 1.458 sin 18.5° = 1.00 sin θ4 θ4 = 27.6° (d) γ = (θ1 −θ2) + [β −(90.0° −θ4)] γ = 75.0° −41.5° + (90.0° −18.5°) −(90.0° −27.6°) = 42.6° 330 Chapter 35 Solutions 35.29 For the incoming ray, sin sin θ θ 2 1 = n Using the figure to the right, (θ2)violet = sin−1 sin 50.0° 1.66 = 27.48° (θ2)red = sin−1 sin 50.0° 1.62 = 28.22° For the outgoing ray, ′ θ3 = 60.0° – θ2 and sinθ4 = nsinθ3 (θ4)violet = sin−1[1.66 sin 32.52°] = 63.17° (θ4)red = sin−1[1.62 sin 31.78°] = 58.56° The dispersion is the difference ∆θ 4 = (θ4)violet −(θ4)red= 63.17° – 58.56° = 4.61° 35.30 n = sin Φ + δmin 2 sin Φ 2 ( ) For small Φ, δmin ≈Φ so Φ + δmin 2 is also a small angle. Then, using the small angle approximation ( sinθ ≈θ when θ << 1 rad), we have: n ≈Φ + δmin ( ) 2 Φ 2 = Φ + δmin Φ or δmin ≈n −1 ( )Φ where Φ is in radians.
35.31 At the first refraction, 1 00 1 2 .
sin sin ( ) = θ θ n The critical angle at the second surface is given by n sin θ3 = 1.00, or θ3 = sin−1 1.00 1.50 = 41.8°.
But, θ2 = 60.0°−θ3. Thus, to avoid total internal reflection at the second surface (i.e., have θ3 < 41.8°), it is necessary that θ2 > 18.2°.
Since sin θ1 = n sin θ2, this requirement becomes sin θ1 > 1.50 ( )sin 18.2° ( ) = 0.468, or θ 1 > 27.9° Chapter 35 Solutions 331 © 2000 by Harcourt, Inc. All rights reserved.
35.32 At the first refraction, 1.00 ( )sin θ1 = n sin θ2. The critical angle at the second surface is given by n sin θ3 = 1.00, or θ3 = sin−1 1.00 n ( ) But 90.0° −θ2 ( ) + 90.0° −θ3 ( ) + Φ = 180°, which gives θ2 = Φ −θ3.
Thus, to have θ3 < sin−1 1.00 n ( ) and avoid total internal reflection at the second surface, it is necessary that θ2 > Φ −sin−1 1.00 n ( ). Since sin sin θ θ 1 2 = n , this requirement becomes sin θ1 > n sin Φ −sin−1 1.00 n or θ1 > sin−1 n sin Φ −sin−1 1.00 n Through the application of trigonometric identities, θ1 > sin−1 n2 −1 sin Φ −cos Φ 35.33 n = sin δ + φ ( ) sin φ / 2 ( ) so 1.544sin 1 2 φ ( ) = sin 5°+ 1 2 φ ( ) = cos 1 2 φ ( )sin5°+sin 1 2 φ ( )cos5° tan 1 2 φ ( ) = sin5° 1.544 −cos5° and φ = 18.1° 35.34 Note for use in every part: Φ + 90.0° −θ2 ( ) + 90.0° −θ3 ( ) = 180° so θ3 = Φ −θ2 At the first surface is α = θ1 −θ2 At exit, the deviation is β = θ4 −θ3 The total deviation is therefore δ = α + β = θ1 + θ4 −θ2 −θ3 = θ1 + θ4 −Φ (a) At entry: n1 sin θ1 = n2 sin θ2 or θ2 = sin−1 sin 48.6° 1.50 = 30.0° Thus, θ3 60 0 30 0 30 0 = ° − ° = ° .
.
.
At exit: 1.50 sin 30.0°= 1.00 sin θ4 or θ4 = sin−1 1.50 sin 30.0° ( ) [ ] = 48.6° so the path through the prism is symmetric when θ1 48 6 = ° .
.
(b) δ = ° + ° − ° = 48 6 48 6 60 0 .
.
. 37.2° (c) At entry: sin sin .
.
.
θ θ 2 2 45 6 1 50 28 4 = ° ⇒ = ° θ3 60 0 28 4 31 6 = ° − ° = ° .
.
.
At exit: sin .
sin .
.
θ θ 4 4 1 50 31 6 51 7 = ° ( ) ⇒ = ° δ = ° + ° − ° = 45 6 51 7 60 0 .
.
. 37.3° (d) At entry: sin sin .
.
.
θ θ 2 2 51 6 1 50 31 5 = ° ⇒ = ° θ3 60 0 31 5 28 5 = ° − ° = ° .
.
.
At exit: sin .
sin .
.
θ θ 4 4 1 50 28 5 45 7 = ° ( ) ⇒ = ° δ = ° + ° − ° = 51 6 45 7 60 0 .
.
. 37.3° 332 Chapter 35 Solutions 35.35 n sin θ = 1. From Table 35.1, (a) θ = = − sin .
1 1 2 419 24.4° (b) θ = = − sin .
1 1 1 66 37.0° (c) θ = = − sin .
1 1 1 309 49.8° 35.36 sin ; sin θ θ c c n n n n = = − 2 1 1 2 1 (a) Diamond: θc = = − sin .
.
1 1 333 2 419 33.4° (b) Flint glass: θc = = − sin .
.
1 1 333 1 66 53.4° (c) Ice: Since n2 > n1, there is no critical angle .
35.37 sin θc n n = 2 1 (Equation 35.10) n2 = n1 sin 88.8° = (1.0003)(0.9998) = 1.000 08 35.38 sin .
.
.
θc n n = = = air pipe 1 00 1 36 0 735 θ c = 47.3° Geometry shows that the angle of refraction at the end is θ r = 90.0° – θ c = 90.0° – 47.3° = 42.7° Then, Snell's law at the end, 1.00 sin θ = 1.36 sin 42.7° gives θ = 67.2° 35.39 For total internal reflection, n n 1 1 2 90 0 sin sin .
θ = ° (1.50) sin θ1 = (1.33)(1.00) or θ1 = 62.4° Chapter 35 Solutions 333 © 2000 by Harcourt, Inc. All rights reserved.
35.40 To avoid internal reflection and come out through the vertical face, light inside the cube must have θ3 1 1 < − sin ( / ) n So θ2 1 90 0 1 > ° − − .
sin ( / ) n But θ θ 1 2 90 0 1 < ° < .
sin and n In the critical case, sin ( / ) .
sin ( / ) − − = ° − 1 1 1 90 0 1 n n 1/n = sin 45.0° n = 1.41 35.41 From Snell's law, n n 1 1 2 2 sin sin θ θ = At the extreme angle of viewing, θ 2 = 90.0° (1.59)(sin θ1) = (1.00) · sin 90.0° So θ1 = 39.0° Therefore, the depth of the air bubble is rd tan θ1 < d < rp tan θ1 or 1.08 cm < d < 1.17 cm 35.42 (a) sin sin θ θ 2 1 2 1 = v v and θ2 90 0 = ° . at the critical angle sin .
sin 90 0 1850 343 ° = θc m s m s so θc = = − sin .
1 0 185 10.7° (b) Sound can be totally reflected if it is traveling in the medium where it travels slower: air (c) Sound in air falling on the wall from most directions is 100% reflected , so the wall is a good mirror.
334 Chapter 35 Solutions 35.43 For plastic with index of refraction n ≥1.42 surrounded by air, the critical angle for total internal reflection is given by θc = sin−1 1 n ≤sin−1 1 1.42 = 44.8° In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical are totally reflected from both the sides of the slab and from facets at the lower end of the plastic, where it is not immersed in gasoline. This light returns up inside the plastic and makes it look bright. Where the plastic is immersed in gasoline, with index of refraction about 1.50, total internal reflection should not happen. The light passes out of the lower end of the plastic with little reflected, making this part of the gauge look dark. To frustrate total internal reflection in the gasoline, the index of refraction of the plastic should be n < 2.12 , since θc = sin−1 1.50 2.12 ( ) = 45.0°.
35.44 Assume the lifeguard’s path makes angle θ1 with the north-south normal to the shoreline, and angle θ2 with this normal in the water. By Fermat’s principle, his path should follow the law of refraction: sin θ1 sin θ2 = v1 v2 = 7.00 m s 1.40 m s = 5.00 or θ2 = sin−1 sin θ1 5 The lifeguard on land travels eastward a distance x = ( ) 16 0 1 .
tan m θ . Then in the water, he travels 26 0 20 0 2 .
.
tan m m − = ( ) x θ further east. Thus, 26 0 16 0 20 0 1 2 .
.
tan .
tan m m m = ( ) + ( ) θ θ or 26.0 m = 16.0 m ( )tan θ1 + 20.0 m ( )tan sin−1 sin θ1 5 We home in on the solution as follows: θ1 (deg) 50.0 60.0 54.0 54.8 54.81 right-hand side 22.2 m 31.2 m 25.3 m 25.99 m 26.003 m The lifeguard should start running at 54.8° east of north .
35.45 Let the air and glass be medium 1 and 2, respectively. By Snell's law, n2 sin θ2 = n1 sin θ1 or 1.56 sin θ2 = sin θ1 But the conditions of the problem are such that θ1 = 2θ2. 1.56 sin θ2 = sin 2θ2 We now use the double-angle trig identity suggested. 1.56 sin θ2 = 2 sin θ2 cos θ2 or cos θ2 = 1.56 2 = 0.780 Thus, θ2 = 38.7° and θ1 = 2θ2 = 77.5° Chapter 35 Solutions 335 © 2000 by Harcourt, Inc. All rights reserved.
35.46 (a) ′ = = θ θ 1 1 30.0° n n 1 1 2 2 sin sin θ θ = (1.00) sin 30.0° = 1.55 sinθ2 θ2 = 18.8° (b) ′ θ1 = θ1 = 30.0° θ2 = sin−1 n1 sin θ1 n2 = ° = − sin .
sin .
1 1 55 30 0 1 50.8° (c) and (d) The other entries are computed similarly, and are shown in the table below.
(c) air into glass, angles in degrees (d) glass into air, angles in degrees incidence reflection refraction incidence reflection refraction 0 0 0 0 0 0 10.0 10.0 6.43 10.0 10.0 15.6 20.0 20.0 12.7 20.0 20.0 32.0 30.0 30.0 18.8 30.0 30.0 50.8 40.0 40.0 24.5 40.0 40.0 85.1 50.0 50.0 29.6 50.0 50.0 none 60.0 60.0 34.0 60.0 60.0 none 70.0 70.0 37.3 70.0 70.0 none 80.0 80.0 39.4 80.0 80.0 none 90.0 90.0 40.2 90.0 90.0 none total internal reflection 35.47 For water, sin θc = 1 4/ 3 = 3 4 Thus θc = sin−1(0.750) = 48.6° and d = 2 (1.00 m)tan θc [ ] d = (2.00 m)tan 48.6° = 2.27 m Figure for Goal Solution 336 Chapter 35 Solutions Goal Solution A small underwater pool light is 1.00 m below the surface. The light emerging from the water forms a circle on the water's surface. What is the diameter of this circle?
G : Only the light that is directed upwards and hits the water’s surface at less than the critical angle will be transmitted to the air so that someone outside can see it. The light that hits the surface farther from the center at an angle greater than θc will be totally reflected within the water, unable to be seen from the outside. From the diagram above, the diameter of this circle of light appears to be about 2 m.
O : We can apply Snell’s law to find the critical angle, and the diameter can then be found from the geometry.
A : The critical angle is found when the refracted ray just grazes the surface (θ2 = 90°). The index of refraction of water is n2 = 1.33, and n1 = 1.00 for air, so n n c 1 2 sin sin θ = ° 90 gives θc = sin−1 1 1.333 = sin−1 0.750 ( ) = 48.6° The radius then satisfies tanθc = r (1.00 m) So the diameter is d = 2r = 2 1.00 m ( )tan48.6°= 2.27 m L : Only the light rays within a 97.2° cone above the lamp escape the water and can be seen by an outside observer (Note: this angle does not depend on the depth of the light source). The path of a light ray is always reversible, so if a person were located beneath the water, they could see the whole hemisphere above the water surface within this cone; this is a good experiment to try the next time you go swimming!
35.48 Call θ1 the angle of incidence and of reflection on the left face and θ2 those angles on the right face. Let α represent the complement of θ1 and β be the complement of θ2. Now α = γ and β = δ because they are pairs of alternate interior angles. We have A = γ + δ = α + β and B = α + A + β = α + β + A = 2A Chapter 35 Solutions 337 © 2000 by Harcourt, Inc. All rights reserved.
35.49 (a) We see the Sun swinging around a circle in the extended plane of our parallel of latitude. Its angular speed is ω = ∆θ ∆t = 2π rad 86 400 s = 7.27 × 10−5 rad s The direction of sunlight crossing the cell from the window changes at this rate, moving on the opposite wall at speed v = rω = 2.37 m ( ) 7.27 × 10−5 rad s ( ) = 1.72 × 10−4 m s = 0.172 mm s (b) The mirror folds into the cell the motion that would occur in a room twice as wide: v = rω = 2 0.174 mm s ( ) = 0.345 mm s (c) and (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the window as fixed, and both patches of light move northward and downward at 50.0° .
35.50 By Snell's law, n1 sin θ1 = n2 sin θ2 With v = c n , c v1 sin θ1 = c v2 sin θ2 or sin sin θ θ 1 1 2 2 v v = This is also true for sound. Here, sin .
sin 12 0 340 1 2 ° = m/s 510 m/s θ θ2 = arcsin (4.44 sin 12.0°) = 67.4° 35.51 (a) n = c v = 2.998 × 108 m s 61.15 km hr 1.00 h 3600 s 1.00 × 103 m 1.00 km = 1.76 × 107 (b) n1 sin θ1 = n2 sin θ2 so 1 76 10 1 00 90 0 7 1 .
sin .
sin .
× ( ) = ( ) ° θ θ1 = 3 25 10 6 .
× − degree This problem is misleading. The speed of energy transport is slow, but the speed of the wavefront advance is normally fast. The condensate's index of refraction is not far from unity.
338 Chapter 35 Solutions 35.52 Violet light: 1 00 25 0 1 689 14 490 2 2 .
sin .
.
sin .
( ) ° = ⇒ = ° θ θ yv = ( ) = ( ) ° 5 00 5 00 14 490 2 .
tan .
tan . cm cm θ = 1.2622 cm Red Light: 1 00 25 0 1 642 14 915 2 2 .
sin .
.
sin .
( ) ° = ⇒ = ° θ θ yR = ( ) ° = 5 00 14 915 1 3318 .
tan .
. cm cm The emergent beams are both at 25.0° from the normal. Thus, w y = ° ∆cos .
25 0 where ∆y = 1.3318 cm −1.2622 cm = 0.0396 cm w = ( ) ° = 0 396 25 0 .
cos . mm 0.359 mm 35.53 Horizontal light rays from the setting Sun pass above the hiker. The light rays are twice refracted and once reflected, as in Figure (b) below, by just the certain special raindrops at 40.0° to 42.0° from the hiker's shadow, and reach the hiker as the rainbow.
The hiker sees a greater percentage of the violet inner edge, so we consider the red outer edge. The radius R of the circle of droplets is Figure (a) R = (8.00 km)(sin 42.0°) = 5.35 km Then the angle φ, between the vertical and the radius where the bow touches the ground, is given by cos .
.
.
φ = = = 2 00 5 3 0 374 km 2.00 km 5 km R or φ = 68.1° The angle filled by the visible bow is 360° – (2 × 68.1°) = 224°, so the visible bow is 224° 360° = 62.2% of a circle Figure (b) Chapter 35 Solutions 339 © 2000 by Harcourt, Inc. All rights reserved.
35.54 From Snell’s law, 1 00 4 3 1 2 .
sin sin ( ) = θ θ x R r = = sin sin θ θ 2 1 so r R = sin θ2 sin θ1 = 3 4 θ 2 θ 1 R r z d x eye Fish at depth d Image at depth z Fish apparent depth actual depth = z d = r cos θ1 R cos θ2 = 3 4 cos θ1 1−sin2 θ2 But sin2 θ2 = 3 4 sin θ1 2 = 9 16 1−cos2 θ1 ( ) So z d = 3 4 cos θ1 1−9 16 + 9 16 cos2 θ1 = 3 4 cos θ1 7 + 9 cos2 θ1 16 or z = 3d cos θ1 7 + 9 cos2 θ1 35.55 As the beam enters the slab, 1 00 50 0 1 48 2 .
sin .
.
sin ( ) ° = ( ) θ giving θ2 31 2 = ° .
. The beam then strikes the top of the slab at x1 1 55 = ° ( ) . mm tan 31.2 from the left end. Thereafter, the beam strikes a face each time it has traveled a distance of 2 1 x along the length of the slab. Since the slab is 420 mm long, the beam has an additional 420 1 mm −x to travel after the first reflection.
The number of additional reflections is 420 mm −x1 2x1 = 420 mm −1.55 mm tan 31.2° ( ) 3.10 mm tan 31.2° ( ) = 81.5 or 81 reflections since the answer must be an integer. The total number of reflections made in the slab is then 82 .
35.56 (a) ′ S1 S1 = n2 −n1 n2 + n1 2 = 1.52 −1.00 1.52 + 1.00 2 = 0.0426 (b) If medium 1 is glass and medium 2 is air, ′ S1 S1 = n2 −n1 n2 + n1 2 = 1.00 −1.52 1.00 + 1.52 2 = 0.0426; There is no difference (c) ′ S1 S1 = 1.76 × 107 −1.00 1.76 × 107 + 1.00 2 = 1.76 × 107 + 1.00 −2.00 1.76 × 107 + 1.00 2 ′ S1 S1 = − × + ≈ − × + 1 00 2 00 1 76 10 1 00 1 00 2 2 00 1 76 10 1 00 7 2 7 .
.
.
.
.
.
.
. = 1.00 −2.27 × 10−7 or 100% This suggests he appearance would be very shiny, reflecting practically all incident light .
See, however, the note concluding the solution to problem 35.51.
340 Chapter 35 Solutions 35.57 (a) With n1 = 1 and n2 = n, the reflected fractional intensity is ′ S1 S1 = n −1 n + 1 2 . The remaining intensity must be transmitted: S2 S1 = 1−n −1 n + 1 2 = n + 1 ( )2 −n −1 ( )2 n + 1 ( )2 = n2 + 2n + 1−n2 + 2n −1 n + 1 ( )2 = 4n n + 1 ( )2 (b) At entry, S2 S1 = 1−n −1 n + 1 2 = 4 2.419 ( ) 2.419 + 1 ( )2 = 0.828 At exit, S3 S2 = 0.828 Overall, S3 S1 = S3 S2 S2 S1 = 0.828 ( )2 = 0.685 or 68.5% 35.58 Define T = 4n n + 1 ( )2 as the transmission coefficient for one encounter with an interface. For diamond and air, it is 0.828, as in problem 57.
As shown in the figure, the total amount transmitted is T2 + T2 1−T ( )2 + T2 1−T ( )4 + T2 1−T ( )6 + . . . + T2 1−T ( )2n + . . . We have 1−T = 1−0.828 = 0.172 so the total transmission is 0.828 ( )2 1+ 0.172 ( )2 + 0.172 ( )4 + 0.172 ( )6 + . . .
[ ] To sum this series, define F = 1+ 0.172 ( )2 + 0.172 ( )4 + 0.172 ( )6 + . . . .
Note that 0.172 ( )2F = 0.172 ( )2 + 0.172 ( )4 + 0.172 ( )6 + . . ., and 1+ 0.172 ( )2F = 1+ 0.172 ( )2 + 0.172 ( )4 + 0.172 ( )6 + . . . = F.
Then, 1 = F −0.172 ( )2F or F = 1 1−0.172 ( )2 The overall transmission is then 0.828 2 ( ) −( ) = 1 0 172 0 706 2 .
. or 70.6% Chapter 35 Solutions 341 © 2000 by Harcourt, Inc. All rights reserved.
35.59 n sin .
sin .
42 0 90 0 ° = ° so n = ° = 1 42 0 1 49 sin .
. sin sin .
θ1 18 0 = ° n and sin sin .
sin .
θ1 18 0 42 0 = ° ° θ1 = 27.5° Figure for Goal Solution Goal Solution The light beam shown in Figure P35.59 strikes surface 2 at the critical angle. Determine the angle of incidence θ1. G : From the diagram it appears that the angle of incidence is about 40°.
O : We can find θ1 by applying Snell’s law at the first interface where the light is refracted. At surface 2, knowing that the 42.0° angle of reflection is the critical angle, we can work backwards to find θ1.
A : Define n1 to be the index of refraction of the surrounding medium and n2 to be that for the prism material. We can use the critical angle of 42.0° to find the ratio n2 n1: n2 sin 42.0°= n1sin 90.0° So, n2 n1 = 1 sin 42.0° = 1.49 Call the angle of refraction θ2 at the surface 1. The ray inside the prism forms a triangle with surfaces 1 and 2, so the sum of the interior angles of this triangle must be 180°. Thus, 90.0°−θ2 ( ) + 60.0°+ 90.0°−42.0° ( ) = 180° Therefore, θ2 = 18.0° Applying Snell’s law at surface 1, n1sinθ1 = n2 sin 18.0° sinθ1 = n2 n1 ( )sin θ2 = 1.49 ( )sin 18.0° θ1 = 27.5° L : The result is a bit less than the 40.0° we expected, but this is probably because the figure is not drawn to scale. This problem was a bit tricky because it required four key concepts (refraction, reflection, critical angle, and geometry) in order to find the solution. One practical extension of this problem is to consider what would happen to the exiting light if the angle of incidence were varied slightly.
Would all the light still be reflected off surface 2, or would some light be refracted and pass through this second surface?
342 Chapter 35 Solutions 35.60 Light passing the top of the pole makes an angle of incidence φ1 = 90.0° −θ . It falls on the water surface at distance s1 = (L −d) tan θ from the pole, and has an angle of refraction φ2 from (1.00)sin φ1 = n sin φ2.
Then s d 2 2 = tan φ and the whole shadow length is s1 + s2 = L −d tan θ + dtan sin−1 sin φ 1 n s1 + s2 = L −d tan θ + dtan sin−1 cos θ n = 2.00 m tan 40.0° + 2.00 m ( )tan sin−1 cos 40.0° 1.33 = 3.79 m 35.61 (a) For polystyrene surrounded by air, internal reflection requires θ3 = sin−1 1.00 1.49 = 42.2° Then from the geometry, θ 2 = 90.0° – θ 3 = 47.8° From Snell's law, sin θ1 = 1.49 ( )sin 47.8°= 1.10 This has no solution. Therefore, total internal reflection always happens .
(b) For polystyrene surrounded by water, θ3 = sin−1 1.33 1.49 = 63.2° and θ2 = 26.8° From Snell's law, θ1 = 30.3° (c) No internal refraction is possible since the beam is initially traveling in a medium of lower index of refraction. 35.62 δ θ θ = − = ° 1 2 10 0 .
and n1 sin θ1 = n2 sin θ2 with n1 = 1, n2 = 4 3 Thus, θ1 = sin−1(n2 sin θ2) = sin−1 n2 sin(θ1 −10.0°) [ ] Chapter 35 Solutions 343 © 2000 by Harcourt, Inc. All rights reserved.
(You can use a calculator to home in on an approximate solution to this equation, testing different values of θ1 until you find that θ1 = 36.5° . Alternatively, you can solve for θ1 exactly, as shown below.) We are given that sin θ1 = 4 3 sin(θ1 −10.0°) This is the sine of a difference, so 3 4 sin θ1 = sin θ1 cos 10.0° −cos θ1 sin 10.0° Rearranging, sin .
cos cos .
sin 10 0 10 0 3 4 1 1 ° = ° − θ θ sin .
cos .
.
tan 10 0 10 0 0 750 1 ° ° − = θ and θ1 1 0 740 = = − tan . 36.5° 35.63 tan .
θ1 4 00 = cm h and tan .
θ2 2 00 = cm h tan tan tan 2 1 2 2 2 2 θ θ θ = ( ) 2.00 = 4.00 sin ( sin ) .
sin ( sin ) 2 1 2 1 2 2 2 2 1 4 00 1 θ θ θ θ − = − (1) Snell's law in this case is: n n 1 1 2 2 sin sin θ θ = sin .
sin θ θ 1 2 1 333 = Squaring both sides, sin2 θ1 = 1.777 sin2 θ2 (2) Substituting (2) into (1), 1.777 sin2 θ2 1−1.777 sin2 θ2 = 4.00 sin2 θ2 1−sin2 θ2 Defining x = sin2 θ , 0.444 1−1.777x ( ) = 1 1−x ( ) Solving for x, 0.444 −0.444x = 1−1.777x and x = 0.417 From x we can solve for θ2: θ2 = sin−1 0.417 = 40.2° Thus, the height is h = = ° = ( ) tan ( ) tan( .
) 2.00 cm 2.00 cm θ2 40 2 2.37 cm 344 Chapter 35 Solutions 35.64 Observe in the sketch that the angle of incidence at point P is γ , and using triangle OPQ: sin γ = L/ R.
Also, cos γ = 1−sin2 γ = R2 −L2 R Applying Snell’s law at point P, 1.00 ( )sin γ = n sin φ Thus, sin φ = sin γ n = L nR and cos φ = 1−sin2 φ = n2R2 −L2 nR From triangle OPS, φ + α + 90.0° ( ) + 90.0° −γ ( ) = 180° or the angle of incidence at point S is α = γ −φ . Then, applying Snell’s law at point S gives 1.00 ( )sin θ = n sin α = n sin γ −φ ( ), or sin θ = n sin γ cos φ −cos γ sin φ [ ] = n L R n2R2 −L2 nR − R2 −L2 R L nR sinθ = L R2 n2R2 −L2 − R2 −L2 and θ = sin−1 L R2 n2R2 −L2 − R2 −L2 35.65 To derive the law of reflection, locate point O so that the time of travel from point A to point B will be minimum.
The total light path is L a b = + sec sec θ θ 1 2 The time of travel is t = 1 v (a sec θ1 + b sec θ2) If point O is displaced by dx, then dt = 1 v (a sec θ1tan θ1 dθ1 + b sec θ2 tan θ2 dθ2) = 0 (1) (since for minimum time dt = 0).
Also, c + d = a tan θ1 + b tan θ2 = constant so, a sec2 θ1 dθ1 + b sec2 θ2 dθ2 = 0 (2) Divide equations (1) and (2) to find θ 1 = θ 2 Chapter 35 Solutions 345 © 2000 by Harcourt, Inc. All rights reserved.
35.66 As shown in the sketch, the angle of incidence at point A is: θ = sin−1 d 2 ( ) R = sin−1 1.00 m 2.00 m = 30.0° If the emerging ray is to be parallel to the incident ray, the path must be symmetric about the center line CB of the cylinder . In the isosceles triangle ABC, γ = α and β θ = ° − 180 . Therefore, α + β + γ = 180° becomes 2 180 180 α θ + ° − = ° or α = θ 2 = 15.0° Then, applying Snell’s law at point A, n sin .
sin α θ = ( ) 1 00 or n = = ° ° = sin sin sin .
sin .
θ α 30 0 15 0 1.93 35.67 (a) At the boundary of the air and glass, the critical angle is given by sin θc = 1 n Consider the critical ray PB ′ B : tan θc = d 4 t or sin θc cos θc = d 4t Squaring the last equation gives: sin2 θc cos2 θc = sin2 θc 1−sin2 θc = d 4t 2 Since sin θc = 1 n , this becomes 1 n2 −1 = d 4t 2 or n = 1+ 4t d ( ) 2 (b) Solving for d, d = 4t n2 −1 Thus, if n = 1.52 and t = 0.600 cm, d = 4 0.600 cm ( ) 1.52 ( )2 −1 = 2.10 cm (c) Since violet light has a larger index of refraction, it will lead to a smaller critical angle and the inner edge of the white halo will be tinged with violet light.
346 Chapter 35 Solutions 35.68 From the sketch, observe that the angle of incidence at point A is the same as the prism angle θ at point O. Given that θ = 60.0°, application of Snell’s law at point A gives 1 50 1 00 60 0 .
sin .
sin .
β = ° or β = 35.3° From triangle AOB , we calculate the angle of incidence (and reflection) at point B. θ β γ + ° − ( ) + ° − ( ) = ° 90 0 90 0 180 .
.
so γ θ β = − = ° − ° = ° 60 0 35 3 24 7 .
.
.
Now, using triangle BCQ : 90 0 90 0 90 0 180 .
.
.
° − ( ) + ° − ( ) + ° − ( ) = ° γ δ θ Thus the angle of incidence at point C is δ θ γ = ° − ( ) − = ° − ° = ° 90 0 30 0 24 7 5 30 .
.
.
.
Finally, Snell’s law applied at point C gives 1 00 1 50 5 30 .
sin .
sin .
φ = ° or φ = ° ( ) = − sin .
sin .
1 1 50 5 30 7.96° 35.69 (a) Given that θ θ 1 2 45 0 76 0 = ° = ° .
. and , Snell’s law at the first surface gives n sin .
sin .
α = ( ) ° 1 00 45 0 (1) Observe that the angle of incidence at the second surface is β α = ° − 90 0 .
. Thus, Snell’s law at the second surface yields n n sin sin .
.
sin .
β α = ° − ( ) = ( ) ° 90 0 1 00 76 0 , or n cos sin .
α = ° 76 0 (2) Dividing Equation (1) by Equation (2), tan sin .
sin .
.
α = ° ° = 45 0 76 0 0 729 or α = 36.1° Then, from Equation (1), n = ° = ° ° = sin .
sin sin .
sin .
45 0 45 0 36 1 α 1.20 (b) From the sketch, observe that the distance the light travels in the plastic is d = L sin α . Also, the speed of light in the plastic is v = c n, so the time required to travel through the plastic is t = d v = nL c sin α = 1.20 ( ) 0.500 m ( ) 3.00 × 108 m s ( )sin 36.1° = 3.40 × 10−9 s = 3.40 ns Chapter 35 Solutions 347 © 2000 by Harcourt, Inc. All rights reserved.
35.70 sin θ1 sin θ2 sin θ1/ sin θ2 0.174 0.131 1.3304 0.342 0.261 1.3129 0.500 0.379 1.3177 0.643 0.480 1.3385 0.766 0.576 1.3289 0.866 0.647 1.3390 0.940 0.711 1.3220 0.985 0.740 1.3315 The straightness of the graph line demonstrates Snell's proportionality.
The slope of the line is n = ± 1 3276 0 01 .
.
and n = 1.328 ± 0.8% © 2000 by Harcourt, Inc. All rights reserved.
Chapter 36 Solutions 36.1 I stand 40 cm from my bathroom mirror. I scatter light which travels to the mirror and back to me in time 0.8 m 3 × 108 m/s ~ 10–9 s showing me a view of myself as I was at that look-back time. I'm no Dorian Gray!
36.2 The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror.
The image of the choir is 0.800 m + 5.30 m = 6.10 m from the organist. Using similar triangles: ′ h 0.600 m = 6.10 m 0.800 m or ′ h = 0.600 m ( ) 6.10 m 0.800 m = 4.58 m h’ 0.600 m 5.30 m 0.800 m Organist mirror image of choir View Looking Down South 36.3 The flatness of the mirror is described by R = ∞, f = ∞, and 1/ f = 0. By our general mirror equation, 1 p + 1 q = 1 f , or q = −p Thus, the image is as far behind the mirror as the person is in front. The magnification is then Figure for Goal Solution M = – q p = 1 = h' h so h' = h = 70.0" The required height of the mirror is defined by the triangle from the person's eyes to the top and bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height must be: h' p p – q = h' p 2p = h' 2 Thus, the mirror must be at least 35.0" high .
Chapter 36 Solutions 351 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Determine the minimum height of a vertical flat mirror in which a person 5'10" in height can see his or her full image. (A ray diagram would be helpful.) G : A diagram with the optical rays that create the image of the person is shown above. From this diagram, it appears that the mirror only needs to be about half the height of the person.
O : The required height of the mirror can be found from the mirror equation, where this flat mirror is described by R = ∞, f = ∞, and 1/ f = 0.
A : The general mirror equation is 1 p + 1 q = 1 f , so with f = ∞, q = −p Thus, the image is as far behind the mirror as the person is in front. The magnification is then M = −q p = 1 = ′ h h so ′ h = h = 70.0 in.
The required height of the mirror is defined by the triangle from the person's eyes to the top and bottom of the image, as shown. From the geometry of the similar triangles, we see that the length of the mirror must be: L = ′ h p p −q = ′ h p 2p = ′ h 2 = 70.0 in 2 = 35.0 in.
Thus, the mirror must be at least 35.0 in high. L : Our result agrees with our prediction from the ray diagram. Evidently, a full-length mirror only needs to be a half-height mirror! On a practical note, the vertical positioning of such a mirror is also important for the person to be able to view his or her full image. To allow for some variation in positioning and viewing by persons of different heights, most full-length mirrors are about 5’ in length.
36.4 A graphical construction produces 5 images, with images I1 and I2 directly into the mirrors from the object O, and (O, I3, I4) and (I1, I2, I5) forming the vertices of equilateral triangles.
352 Chapter 36 Solutions 36.5 (1) The first image in the left mirror is 5.00 ft behind the mirror, or 10.0 ft from the position of the person.
(2) The first image in the right mirror is located 10.0 ft behind the right mirror, but this location is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the mirror, or 30.0 ft from the person.
(3) The first image in the left mirror forms an image in the right mirror. This first image is 20.0 ft from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed. This image in the right mirror also forms an image in the left mirror. The distance from this image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is, thus, 35.0 ft behind the mirror, or 40.0 ft from the person.
36.6 For a concave mirror, both R and f are positive. We also know that f = R 2 = 10.0 cm (a) 1 q = 1 f – 1 p = 1 10.0 cm – 1 40.0 cm = 3 40.0 cm , and q = 13.3 cm M = q p = – 13.3 cm 40.0 cm = – 0.333 The image is 13.3 cm in front of the mirror, is real, and inverted .
(b) 1 q = 1 f – 1 p = 1 10.0 cm – 1 20.0 cm = 1 20.0 cm , and q = 20.0 cm M = q p = – 20.0 cm 20.0 cm = –1.00 The image is 20.0 cm in front of the mirror, is real, and inverted .
(c) 1 q = 1 f – 1 p = 1 10.0 cm – 1 10.0 cm = 0 Thus, q = infinity.
No image is formed. The rays are reflected parallel to each other.
36.7 1 q = 1 f – 1 p = – 1 0.275 m – 1 10.0 m gives q = – 0.267 m Thus, the image is virtual .
M = – q p = – – 0.267 10.0 m = 0.0267 Thus, the image is upright (+M) and diminished ( M < 1 ( ) Chapter 36 Solutions 353 © 2000 by Harcourt, Inc. All rights reserved.
36.8 With radius 2.50 m, the cylindrical wall is a highly efficient mirror for sound, with focal length f = R 2 = 1.25 m In a vertical plane the sound disperses as usual, but that radiated in a horizontal plane is concentrated in a sound image at distance q from the back of the niche, where 1 p + 1 q = 1 f so 1 2.00 m + 1 q = 1 1.25 m q = 3.33 m 36.9 (a) 1 p + 1 q = 2 R gives 1 (30.0 cm) + 1 q = 2 (– 40.0 cm) 1 q = – 2 (40.0 cm) – 1 (30.0 cm) = – 0.0833 cm–1 so q = –12.0 cm M = – q p = – (–12.0 cm) (30.0 cm) = 0.400 (b) 1 p + 1 q = 2 R gives 1 (60.0 cm) + 1 q = – 2 (40.0 cm) 1 q = – 2 (40.0 cm) – 1 (60.0 cm) = – 0.0666 cm–1 so q = –15.0 cm M = – q p = – (–15.0 cm) (60.0 cm) = 0.250 (c) Since M > 0, the images are upright .
36.10 (a) M = −q p . For a real image, q > 0 so in this case M = −4.00 q pM cm = − = 120 and from 1 p + 1 q = 2 R R = 2pq (p + q) = 2(30.0 cm)(120 cm) (150 cm) = 48.0 cm 354 Chapter 36 Solutions (b) 36.11 (a) 1 p + 1 q = 2 R becomes 1 q = 2 (60.0 cm) − 1 90.0 cm ( ) q = 45.0 cm and M = −q p = −(45.0 cm) (90.0 cm) = – 0.500 (b) 1 p + 1 q = 2 R becomes 1 q = 2 (60.0 cm) − 1 (20.0 cm) , q = – 60.0 cm and M = −q p = −(−60.0 cm) (20.0 cm) = 3.00 (c) The image in (a) is real, inverted and diminished. That of (b) is virtual, upright, and enlarged. The ray diagrams are similar to Figures 36.15(a) and 36.15(b), respectively. (a) (b) Figures for Goal Solution Chapter 36 Solutions 355 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A concave mirror has a radius of curvature of 60.0 cm. Calculate the image position and magnification of an object placed in front of the mirror (a) at a distance of 90.0 cm and (b) at a distance of 20.0 cm. (c) In each case, draw ray diagrams to obtain the image characteristics. G : It is always a good idea to first draw a ray diagram for any optics problem. This gives a qualitative sense of how the image appears relative to the object. From the ray diagrams above, we see that when the object is 90 cm from the mirror, the image will be real, inverted, diminished, and located about 45 cm in front of the mirror, midway between the center of curvature and the focal point.
When the object is 20 cm from the mirror, the image is be virtual, upright, magnified, and located about 50 cm behind the mirror.
O : The mirror equation can be used to find precise quantitative values.
A : (a) The mirror equation is applied using the sign conventions listed in the text. 1 p + 1 q = 2 R or 1 90.0 cm + 1 q = 2 60.0 cm so q = 45.0 cm (real, in front of the mirror) M = −q p = −45.0 cm 90.0 cm = −0.500 (inverted) (b) 1 p + 1 q = 2 R or 1 20.0 cm + 1 q = 2 60.0 cm so q = − 60.0 cm (virtual, behind the mirror) M = −q p = −−60.0 cm 20.0 cm = 3.00 (upright) L : The calculated image characteristics agree well with our predictions. It is easy to miss a minus sign or to make a computational mistake when using the mirror-lens equation, so the qualitative values obtained from the ray diagrams are useful for a check on the reasonableness of the calculated values.
36.12 For a concave mirror, R and f are positive. Also, for an erect image, M is positive. Therefore, M = −q p = 4 and q = – 4p.
1 f = 1 p + 1 q becomes 1 40.0 cm = 1 p – 1 4p = 3 4p ; from which, p = 30.0 cm 36.13 (a) q = (p + 5.00 m) and, since the image must be real, M = – q p = – 5 or q = 5p. Therefore, p + 5.00 = 5p or p = 1.25 m and q = 6.25 m.
From 1 p + 1 q = 2 R , R = 2pq (q + p) = 2(1.25)(6.25) (6.25 + 1.25) = 2.08 m (concave) (b) From part (a), p = 1.25 m; the mirror should be 1.25 m in front of the object.
356 Chapter 36 Solutions 36.14 (a) The image is the trapezoid ′ a ′ b ′ d ′ e as shown in the ray diagram.
C F a b d e ′ a ′ b ′ d ′ e h ′ hL ′ hR qR - qL pR = 40.0 cm pL = 60.0 cm (b) To find the area of the trapezoid, the image distances, qR and qL, along with the heights ′ hR and ′ hL, must be determined. The mirror equation, 1 p + 1 q = 2 R becomes 1 40.0 cm + 1 qR = 2 20.0 cm or qR = 13.3 cm hR = hMR = h −qR pR = 10.0 cm ( ) −13.3 cm 40.0 cm = −3.33 cm Also 1 60.0 cm + 1 qL = 2 20.0 cm or qL = 12 0 . cm hL = hML = 10.0 cm ( ) −12.0 cm 60.0 cm = −2.00 cm The area of the trapezoid is the sum of the area of a square plus the area of a triangle: At = A1 + A2 = qR −qL ( )hL + 1 2 qR −qL ( ) hR −hL ( ) = 3.56 cm2 36.15 Assume that the object distance is the same in both cases (i.e., her face is the same distance from the hubcap regardless of which way it is turned). Also realize that the near image (q = – 10.0 cm) occurs when using the convex side of the hubcap. Applying the mirror equation to both cases gives: (concave side: R = R , q = −30.0 cm) 1 p − 1 30.0 = 2 R , or 2 R = 30.0 cm −p (30.0 cm)p (convex side: R = −R , q = −10.0 cm) 1 p − 1 10.0 = −2 R , or 2 R = p −10.0 cm 10.0 cm ( )p (a) Equating Equations (1) and (2) gives: 30.0 cm−p 3.00 = p −10.0 cm or p = 15.0 cm Thus, her face is 15.0 cm from the hubcap.
(b) Using the above result ( p = 15.0 cm) in Equation gives: 2 R = 30.0 cm −15.0 cm (30.0 cm)(15.0 cm) or 2 R = 1 30.0 cm , and R = 60.0 cm Chapter 36 Solutions 357 © 2000 by Harcourt, Inc. All rights reserved.
The radius of the hubcap is 60.0 cm .
36.16 1 p + 1 q = 1 f f = R 2 = –1.50 cm q = – 15.0 11.0 cm (behind mirror) M = –q p = 1 11.0 36.17 (a) The image starts from a point whose height above the mirror vertex is given by 1 1 1 2 p q f R + = = 1 3.00 m + 1 q = 1 0.500 m Therefore, q = 0.600 m As the ball falls, p decreases and q increases. Ball and image pass when q p 1 1 = . When this is true, 1 1 1 0 500 2 1 1 1 p p p + = = . m or p1 1 00 = . m.
As the ball passes the focal point, the image switches from infinitely far above the mirror to infinitely far below the mirror. As the ball approaches the mirror from above, the virtual image approaches the mirror from below, reaching it together when p2 = q2 = 0.
(b) The falling ball passes its real image when it has fallen 3.00 m −1.00 m = 2.00 m = 1 2 gt2, or when t = ( ) = 2 2 00 9 80 .
. m m s2 0.639 s .
The ball reaches its virtual image when it has traversed 3.00 m −0 = 3.00 m = 1 2 gt2, or at t = ( ) = 2 3 00 9 80 .
. m m s2 0.782 s .
36.18 When R → ∞, Equation 36.8 for a spherical surface becomes q = −p n2 n1 ( ). We use this to locate the final images of the two surfaces of the glass plate.
First, find the image the glass forms of the bottom of the plate: qB1 = −1.33 1.66 (8.00 cm) = −6.41 cm 358 Chapter 36 Solutions This virtual image is 6.41 cm below the top surface of the glass or 18.41 cm below the water surface. Next, use this image as an object and locate the image the water forms of the bottom of the plate. qB2 = −1.00 1.33 (18.41 cm) = −13.84 cm or 13.84 cm below the water surface. Now find image the water forms of the top surface of the glass. q3 = − 1 1.33 (12.0 cm) = −9.02 cm, or 9.02 cm below the water surface.
Therefore, the apparent thickness of the glass is ∆t = 13.84 cm – 9.02 cm = 4.82 cm 36.19 n1 p + n2 q = n2 – n1 R = 0 and R → ∞ q = – n2 n1 p = – 1 1.309 (50.0 cm) = – 38.2 cm Thus, the virtual image of the dust speck is 38.2 cm below the top surface of the ice.
36.20 n1 p + n2 q = n2 – n1 R so 1.00 ∞ + 1.40 21.0 mm = 1.40 – 1.00 6.00 mm and 0.0667 = 0.0667 They agree.
The image is inverted, real and diminished. Chapter 36 Solutions 359 © 2000 by Harcourt, Inc. All rights reserved.
36.21 From Equation 36.8, n1 p + n2 q = (n2 – n1) R Solve for q to find q = n2 R p p(n2 – n1) – n1 R In this case, n1 = 1.50, n2 = 1.00, R = –15.0 cm, and p = 10.0 cm, So q = (1.00)(–15.0 cm)(10.0 cm) (10.0 cm)(1.00 – 1.50) – (1.50)(–15.0 cm) = – 8.57 cm Therefore, the apparent depth is 8.57 cm .
36.22 p = ∞and q = +2R 1.00 p + n2 q = n2 – 1.00 R 0 + n2 2R = n2 – 1.00 R so n2 = 2.00 36.23 n1 p + n2 q = (n2 −n1) R because 1.00 p + 1.50 q = 1.50 −1.00 6.00 cm = 1.00 12.0 cm (a) 1 20.0 cm + 1.50 q = 1.00 12.0 cm or q = 1.50 1.00 12.0 cm − 1.00 20.0 cm = 45.0 cm (b) 1.00 10.0 cm + 1.50 q = 1.00 12.0 cm or q = 1.50 1.00 12.0 cm − 1.00 10.0 cm = – 90.0 cm (c) 1.00 3.00 cm + 1.50 q = 1.00 12.0 cm or q = 1.50 1.00 12.0 cm − 1.00 3.00 cm = – 6.00 cm 36.24 For a plane surface, n1 p + n2 q = n2 −n1 R becomes q = −n2p n1 . Thus, the magnitudes of the rate of change in the image and object positions are related by dq dt = n2 n1 dp dt If the fish swims toward the wall with a speed of 2.00 cm s, the speed of the image is given by vimage = dq dt = 1.00 1.33 2.00 cm s ( ) = 1.50 cm/s 360 Chapter 36 Solutions 36.25 n1 p + n2 q = n2 −n1 R n1 = 1.33 n2 = 1.00 p = +10.0 cm R = −15.0 cm q = −9.01 cm, or the fish appears to be 9.01 cm inside the bowl 36.26 Let R1 = outer radius and R2 = inner radius 1 f = (n – 1) 1 R1 – 1 R2 = (1.50 – 1) 1 2.00 m – 1 2.50 cm = 0.0500 cm so f = 20.0 cm 36.27 (a) 1 f = (n – 1) 1 R1 – 1 R2 = (0.440) 1 (12.0 cm) – 1 (–18.0 cm) : f = 16.4 cm (b) 1 f = (0.440) 1 (18.0 cm) – 1 (–12.0 cm) : f = 16.4 cm Figure for Goal Solution Goal Solution The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens. (b) Calculate the focal length if the radii of curvature of the two faces are interchanged.
G : Since this is a biconvex lens, the center is thicker than the edges, and the lens will tend to converge incident light rays. Therefore it has a positive focal length. Exchanging the radii of curvature amounts to turning the lens around so the light enters the opposite side first. However, this does not change the fact that the center of the lens is still thicker than the edges, so we should not expect the focal length of the lens to be different (assuming the thin-lens approximation is valid).
O : The lens makers’ equation can be used to find the focal length of this lens.
A : The centers of curvature of the lens surfaces are on opposite sides, so the second surface has a negative radius: (a) 1 f = n −1 ( ) 1 R1 −1 R2 = 1.44 −1.00 ( ) 1 12.0 cm − 1 −18.0 cm so f = 16.4 cm (b) 1 f = 0.440 ( ) 1 18.0 cm − 1 −12.0 cm so f = 16.4 cm L : As expected, reversing the orientation of the lens does not change what it does to the light, as long as the lens is relatively thin (variations may be noticed with a thick lens). The fact that light rays can be traced forward or backward through an optical system is sometimes referred to as the principle of reversibility. We can see that the focal length of this biconvex lens is about the same magnitude as the average radius of curvature. A few approximations, useful as checks, are that a symmetric biconvex lens with radii of magnitude R will have focal length f ≈R; a plano-convex lens with radius R will have f ≈R/ 2; and a symmetric biconcave lens has f ≈−R. These approximations apply when the lens has n ≈1.5, which is typical of many types of clear glass and plastic.
36.28 For a converging lens, f is positive. We use 1 p + 1 q = 1 f .
Chapter 36 Solutions 361 © 2000 by Harcourt, Inc. All rights reserved.
(a) 1 q = 1 f – 1 p = 1 20.0 cm – 1 40.0 cm = 1 40.0 cm q = 40.0 cm M = – q p = – 40.0 40.0 = –1.00 The image is real, inverted , and located 40.0 cm past the lens.
(b) 1 q = 1 f – 1 p = 1 20.0 cm – 1 20.0 cm = 0 q = infinity No image is formed. The rays emerging from the lens are parallel to each other.
(c) 1 q = 1 f – 1 p = 1 20.0 cm – 1 10.0 cm = – 1 20.0 cm q = – 20.0 cm M = – q p = – –20.0 10.0 = 2.00 The image is upright, virtual , and 20.0 cm in front of the lens.
36.29 (a) 1 q = 1 f – 1 p = 1 25.0 cm – 1 26.0 cm q = 650 cm The image is real, inverted, and enlarged .
(b) 1 q = 1 f – 1 p = 1 25.0 cm – 1 24.0 cm q = – 600 cm The image is virtual, upright, and enlarged .
36.30 (a) 1 p + 1 q = 1 f ⇒ 1 (32.0 cm) + 1 (8.00 cm) = 1 f so f = 6.40 cm (b) M = – q p = – (8.00 cm) (32.00 cm) = – 0.250 (c) Since f > 0, the lens is converging .
362 Chapter 36 Solutions 36.31 We are looking at an enlarged, upright, virtual image: M = h' h = 2 = – q p so p = – q 2 = – – 2.84 cm 2 = +1.42 cm 1 p + 1 q = 1 f gives 1 1.42 cm + 1 – 2.84 cm = 1 f f = 2.84 cm 36.32 To use the lens as a magnifying glass, we form an upright, virtual image: M = +2.00 = – q p or 1 p + 1 q = 1 f We eliminate q = – 2.00p:: 1 p + 1 – 2.00p = 1 15.0 cm or – 2.00 + 1.00 – 2.00p = 1 15.0 cm Solving, p = 7.50 cm 36.33 (a) Note that q = 12.9 cm – p so 1 p + 1 12.9 – p = 1 2.44 which yields a quadratic in p: – p2 + 12.9p = 31.5 which has solutions p = 9.63 cm or p = 3.27 cm Both solutions are valid.
(b) For a virtual image, – q = p + 12.9 cm 1 p – 1 12.9 + p = 1 2.44 or p2 + 12.9p = 31.8 from which p = 2.10 cm or p = – 15.0 cm.
We must have a real object so the – 15.0 cm solution must be rejected.
Chapter 36 Solutions 363 © 2000 by Harcourt, Inc. All rights reserved.
36.34 (a) 1 p + 1 q = 1 f : 1 p + 1 – 30.0 cm = 1 12.5 cm p = 8.82 cm M = – q p = – (– 30.0) 8.82 = 3.40, upright (b) See the figure to the right.
36.35 1 p + 1 q = 1 f : p–1 + q–1 = constant We may differentiate through with respect to p: –1p–2 – 1q–2 dq d p = 0 dq d p = – q2 p2 = – M2 36.36 The image is inverted: M = h' h = –1.80 m 0.0240 m = – 75.0 = – q p q = 75.0p (b) q + p = 3.00 m = 75.0p + p p = 39.5 mm (a) q = 2.96 m 1 f = 1 p + 1 q = 1 0.0395 m + 1 2.96 m f = 39.0 mm 36.37 (a) 1 p + 1 q = 1 f 1 (20.0 cm) 1 1 ( 32.0 cm) + = − q so q = − 1 20.0 + 1 32.0 −1 = –12.3 cm The image is 12.3 cm to the left of the lens. (b) M = −q p = −(−12.3 cm) (20.0 cm) = 0.615 364 Chapter 36 Solutions (c) See the ray diagram above.
36.38 (a) 1 f = (n −1) 1 R1 −1 R2 = (1.50 −1) 1 15.0 cm − 1 −12.0 cm ( ) , or f = 13.3 cm (b) Ray Diagram: F F h = 10.0 cm pR = 20.0 cm pL = 30.0 cm a b c d a’ b’ c’ d’ h’R h’L qR qL 10.0 cm (c) To find the area, first find qR and qL , along with the heights ′ hR and ′ hL, using the thin lens equation. 1 pR + 1 qR = 1 f becomes: 1 20.0 cm + 1 qR = 1 13.3 cm or qR = 40 0 . cm ′ hR = hMR = h − qR pR = (10.0 cm)(−2.00) = −20.0 cm 1 30.0 cm + 1 qL = 1 13.3 cm : qL = 24.0 cm ′ hL = hML = (10.0 cm)(−0.800) = −8.00 cm Thus, the area of the image is: Area = qR −qL ′ hL + 1 2 qR −qL ′ hR −′ hL = 224 cm2 36.39 (a) The distance from the object to the lens is p, so the image distance is q = 5.00 m −p.
Thus, 1 p + 1 q = 1 f becomes: 1 p + 1 5.00 m −p = 1 0.800 m This reduces to a quadratic equation: p2 −5.00 m ( )p + 4.00 m ( ) = 0 which yields p = 4.00 m, or p = 1.00 m .
Thus, there are two possible object distances, both corresponding to real objects.
(b) For p = 4.00 m: q = 5.00 m −4.00 m = 1.00 m: M = −1.00 m 4.00 m = – 0.250 .
For p = 1.00 m: q = 5.00 m −1.00 m = 4.00 m: M = −4.00 m 1.00 m = – 4.00 .
Chapter 36 Solutions 365 © 2000 by Harcourt, Inc. All rights reserved.
Both images are real and inverted , but the magnifications are different, with one being larger than the object and the other smaller.
36.40 (a) The image distance is: q = d −p. Thus, 1 p + 1 q = 1 f becomes 1 p + 1 d −p = 1 f This reduces to a quadratic equation: p2 + −d ( )p + f d ( ) = 0 which yields: p = d ± d2 −4f d 2 = d 2 ± d2 4 −f d Since f < d 4, both solutions are meaningful and the two solutions are not equal to each other. Thus, there are two distinct lens positions that form an image on the screen. (b) The smaller solution for p gives a larger value for q, with a real, enlarged, inverted image .
The larger solution for p describes a real, diminished, inverted image .
36.41 To properly focus the image of a distant object, the lens must be at a distance equal to the focal length from the film ( q1 = 65.0 mm). For the closer object: 1 p2 + 1 q2 = 1 f becomes 1 2000 mm + 1 q2 = 1 65.0 mm and q2 = 65.0 mm ( ) 2000 2000 −65.0 The lens must be moved away from the film by a distance D = q2 −q1 = 65.0 mm ( ) 2000 2000 −65.0 −65.0 mm = 2.18 mm 36.42 (a) The focal length of the lens is given by 1 f = n −1 ( ) 1 R1 −1 R2 = 1.53 −1.00 ( ) 1 −32.5 cm − 1 42.5 cm f = −34.7 cm Note that R1 is negative because the center of curvature of the first surface is on the virtual image side. R1 R2 366 Chapter 36 Solutions When p = ∞, the thin lens equation gives q = f . Thus, the violet image of a very distant object is formed at q = −34 7 . cm . The image is virtual, upright, and diminished .
(b) The same ray diagram and image characteristics apply for red light.
Again, q = f , and now 1 f = 1.51−1.00 ( ) 1 −32.5 cm − 1 42.5 cm giving f = −36.1 cm .
⇑ F F I 36.43 Ray h1 is undeviated at the plane surface and strikes the second surface at angle of incidence given by θ1 = sin−1 h1 R = sin−1 0.500 cm 20.0 cm = 1.43° Then, (1.00)sinθ2 = (1.60)sinθ1 = (1.60) 0.500 20.0 cm so θ 2 = 2.29° The angle this emerging ray makes with the horizontal is θ 2 – θ1 = 0.860° It crosses the axis at a point farther out by f1 where f1 = h 1 tan(θ 2 – θ1) = 0.500 cm tan(0.860°) = 33.3 cm The point of exit for this ray is distant axially from the lens vertex by 20.0 cm – (20.0 cm)2 – (0.500 cm)2 = 0.00625 cm so ray h1 crosses the axis at this distance from the vertex: x1 = 33.3 cm – 0.00625 cm = 33.3 cm Now we repeat this calculation for ray h2: θ1 = sin−1 12.0 cm 20.0 cm = 36.9° (1.00)sinθ2 = (1.60)sinθ1 = (1.60) 12.00 20.0 θ 2 = 73.7° f2 = h 2 tan(θ 2 – θ1) = 12.0 cm tan(36.8°) = 16.0 cm x2 = 16.0 cm ( ) 20.0 cm − (20.0 cm)2 −(12.0 cm)2 = 12.0 cm Now ∆x = 33.3 cm – 12.0 cm = 21.3 cm Chapter 36 Solutions 367 © 2000 by Harcourt, Inc. All rights reserved.
36.44 For starlight going through Nick's glasses, 1 p + 1 q = 1 f 1 ∞+ 1 −0.800 m ( ) = 1 f = −1.25 diopters For a nearby object, 1 p + 1 −0.180 m ( ) = −1.25 m-1, so p = 23.2 cm 36.45 P = 1 f = 1 p + 1 q = 1 ∞− 1 0.250 m = −4.00 diopters = −4.00 diopters, a diverging lens 36.46 Consider an object at infinity, imaged at the person's far point: 1 p + 1 q = 1 f 1 ∞+ 1 q = −4.00 m-1 q = −25.0 cm The person's far point is 25.0 cm + 2.00 cm = 27.0 cm from his eyes. For the contact lenses we want 1 ∞+ 1 −0.270 m ( ) = 1 f = −3.70 diopters 36. 47 First, we use the thin lens equation to find the object distance: 1 p + 1 −25.0 cm ( ) = 1 10.0 cm Then, p = 7.14 cm and Then, M = −q p = −−25.0 cm ( ) 7.14 cm = 3.50 36.48 (a) From the thin lens equation: 1 p + 1 −25.0 cm ( ) = 1 5.00 cm or p = 4.17 cm (b) M = −q p = 1 + 25.0 cm f = 1 + 25.0 cm 5.00 cm = 6.00 36.49 Using Equation 36.20, M = − L fo 25.0 cm fe = − 23.0 cm 0.400 cm 25.0 cm 2.50 cm = – 575 368 Chapter 36 Solutions 36.50 M = M1me = M1 25.0 cm fe ⇒fe = M1 M 25.0 cm ( ) = −12.0 −140 25.0 cm ( ) = 2.14 cm 36.51 fo = 20.0 m fe = 0.0250 m (a) The angular magnification produced by this telescope is: m = −fo fe = – 800 (b) Since m < 0, the image is inverted .
Chapter 36 Solutions 369 © 2000 by Harcourt, Inc. All rights reserved.
36.52 (a) The lensmaker's equation 1 p + 1 q = 1 f gives q = 1 1 f −1 p = 1 p −f f p = f p p −f Then, M = ′ h h = −q p = − f p −f gives ′ h = hf f −p (b) For p >> f , f −p ≈−p. Then, ′ h = −hf p .
(c) Suppose the telescope observes the space station at the zenith: ′ h = −hf p = −108.6 m ( ) 4.00 m ( ) 407 × 103 m = −1.07 mm 36.53 (b) Call the focal length of the objective fo and that of the eyepiece −fe . The distance between the lenses is fo −fe . The objective forms a real diminished inverted image of a very distant object at q1 = fo. This image is a virtual object for the eyepiece at p2 = −fe . For it 1 p + 1 q = 1 f becomes 1 −fe + 1 q = 1 −fe , 1 q2 = 0 and q2 = ∞ (a) The user views the image as virtual . Letting ′ h represent the height of the first image, θo = ′ h fo and θ = ′ h fe . The angular magnification is m = θ θo = ′ h fe ′ h fo = fo fe (c) Here, fo −fe = 10.0 cm and fo fe = 3.00.
Thus, fe = fo 3.00 and 2 3 fo = 10.0 cm.
F0 I θ 0 h’ F0 θ 0 L1 Fe O Fe θ L2 fo = 15.0 cm fe = 5.00 cm and fe = −5 00 . cm 370 Chapter 36 Solutions 36.54 Let I0 represent the intensity of the light from the nebula and θo its angular diameter. With the first telescope, the image diameter ′ h on the film is given by θo = −′ h fo as ′ h = −θo 2000 mm ( ).
The light power captured by the telescope aperture is P1 = I0A1 = I0 π 200 mm ( )2 4 [ ], and the light energy focused on the film during the exposure is E1 = P1t1 = I0 π 200 mm ( )2 4 [ ] 1.50 min ( ).
Likewise, the light power captured by the aperture of the second telescope is P2 = I0A2 = I0 π 60.0 mm ( )2 4 [ ] and the light energy is E I t 2 0 2 2 60 0 4 = ( ) [ ] π . mm . Therefore, to have the same light energy per unit area, it is necessary that I0 π 60.0 mm ( )2 4 [ ]t2 π θo 900 mm ( )2 4 [ ] = I0 π 200 mm ( )2 4 [ ] 1.50 min ( ) π θo 2000 mm ( )2 4 [ ] The required exposure time with the second telescope is t2 = 200 mm ( )2 900 mm ( )2 60.0 mm ( )2 2000 mm ( )2 1.50 min ( ) = 3.38 min 36.55 Only a diverging lens gives an upright diminished image. The image is virtual and d = p −q = p + q: M = – q p so q = – Mp and d = p – Mp p = d 1 – M : 1 p + 1 q = 1 f = 1 p + 1 – Mp = – M + 1 – Mp = (1 – M)2 – Md f = – Md (1 – M)2 = – (0.500)(20.0 cm) (1 – 0.500)2 = – 40.0 cm 36.56 If M < 1, the lens is diverging and the image is virtual.
d = p – q = p + q M = – q p so q = –Mp and d = p – Mp p = d 1 – M : 1 p + 1 q = 1 f = 1 p + 1 – Mp = – M + 1 – Mp = (1 – M)2 – Md f = – Md (1 – M)2 If M > 1, the lens is converging and the image is still virtual. Now d = – q – p. We obtain in this case f = Md (M – 1)2 Chapter 36 Solutions 371 © 2000 by Harcourt, Inc. All rights reserved.
36.57 Start with the first pass through the lens.
1 q1 = 1 f1 – 1 p1 = 1 80.0 cm – 1 100 cm q1 = 400 cm to right of lens For the mirror, p2 = – 300 cm 1 q2 = 1 f2 – 1 p2 = 1 – 50.0 cm – 1 – 300 cm q2 = – 60.0 cm For the second pass through the lens, p3 = 160 cm 1 q3 = 1 f1 – 1 p3 = 1 80.0 cm – 1 160 cm q3 = 160 cm to left of lens M1 = – q1 p1 = – 400 cm 100 cm = – 4.00 M2 = – q2 p2 = – – 60.0 cm – 300 cm = – 1 5 M3 = – q3 p3 = – 160 cm 160 cm = –1 M = M1M2M3 = – 0.800 Since M < 0 the final image is inverted .
36.58 (a) 1 f = n −1 ( ) 1 R1 −1 R2 1 −65.0 cm = 1.66 −1 ( ) 1 50.0 cm −1 R2 1 R2 = 1 50.0 cm + 1 42.9 cm so R2 = 23.1 cm D B R1 R2 2.00 cm A C (b) The distance along the axis from B to A is R1 − R1 2 −2.00 cm ( ) 2 = 50.0 cm − 50.0 cm ( )2 −2.00 cm ( ) 2 = 0.0400 cm Similarly, the axial distance from C to D is 23 1 23 1 2 00 0 0868 2 2 .
.
.
. cm cm cm cm −( ) −( ) = Then, AD = 0.100 cm −0.0400 cm + 0.0868 cm = 0.147 cm .
372 Chapter 36 Solutions 36.59 1 q1 = 1 f1 – 1 p1 = 1 10.0 cm – 1 12.5 cm so q1 = 50.0 cm (to left of mirror) This serves as an object for the lens (a virtual object), so 1 q2 = 1 f2 – 1 p2 = 1 – 16.7 cm – 1 – 25.0 cm q2 = – 50.3 cm (to right of lens) Thus, the final image is located 25.3 cm to right of mirror .
M1 = – q1 p1 = – 50.0 cm 12.5 cm = – 4.00 M2 = – q2 p2 = – – 50.3 cm – 25.0 cm = – 2.01 M = M1M2 = 8.05 Thus, the final image is virtual, upright , 8.05 times the size of object, and 25.3 cm to right of the mirror.
36.60 We first find the focal length of the mirror.
1 f = 1 p + 1 q = 1 10.0 cm + 1 8.00 cm = 9 40.0 cm and f = 4.44 cm Hence, if p = 20.0 cm, 1 q = 1 f – 1 p = 1 4.44 cm – 1 20.0 cm = 15.56 88.8 cm Thus, q = 5.71 cm , real 36.61 A hemisphere is too thick to be described as a thin lens.
The light is undeviated on entry into the flat face. W e next consider the light's exit from the second surface, for which R = – 6.00 cm The incident rays are parallel, so p = ∞.
Then, n1 p + n2 q = n2 – n1 R becomes 0 + 1 q = (1.00 – 1.56) – 6.00 cm and q = 10.7 cm Chapter 36 Solutions 373 © 2000 by Harcourt, Inc. All rights reserved.
36.62 (a) I = P 4π r2 = 4.50 W 4π 1.60 × 10-2 m ( ) 2 = 1 40 . kW m2 (b) I = P 4π r2 = 4.50 W 4π 7.20 m ( )2 = 6 91 . mW m2 (c) 1 p + 1 q = 1 f ⇒ 1 7.20 m + 1 q = 1 0.350 m so q = 0.368 m and M = ′ h 3.20 cm = −q p = −0.368 m 7.20 m ′ h = 0.164 cm (d) The lens intercepts power given by P = IA = 6.91× 10−3 W m2 ( ) 1 4 π 0.150 m ( )2 [ ] and puts it all onto the image where I = P A = 6.91× 10−3 W m2 ( ) π 15.0 cm ( )2 4 [ ] π 0.164 cm ( )2 4 = 58.1 W m2 36.63 From the thin lens equation, q1 = f1p1 p1 – f1 = (– 6.00 cm)(12.0 cm) 12.0 cm – (– 6.00 cm) = – 4.00 cm When we require that q2 → ∞, the thin lens equation becomes p2 = f2; In this case, p2 = d – (– 4.00 cm) Therefore, d + 4.00 cm = f2 = 12.0 cm and d = 8.00 cm 36.64 (a) For the light the mirror intercepts, P = I0A = I0π Ra 2 350 W = 1000 W m2 ( )π Ra 2 and Ra = 0.334 m or larger (b) In 1 p + 1 q = 1 f = 2 R we have p →∞ so q = R 2 . M = ′ h h = −q p , so ′ h = −q h p ( ) = −R 2 0.533° π rad 180° = −R 2 9.30 m rad ( ) where h p is the angle the Sun subtends. The intensity at the image is then I = P π ′ h 2 4 = 4I0π Ra 2 π ′ h 2 = 4I0Ra 2 R 2 ( ) 2 9.30 × 10−3 rad ( ) 2 120 × 103 W m2 = 16 1000 W m2 ( )Ra 2 R2 9.30 × 10−3 rad ( ) 2 so Ra R = 0.0255 or larger 374 Chapter 36 Solutions 36.65 For the mirror, f = R/2 = +1.50 m. In addition, because the distance to the Sun is so much larger than any other figures, we can take p = ∞. The mirror equation, 1 p + 1 q = 1 f , then gives q = f = 1.50 m .
Now, in M = – q p = h' h , the magnification is nearly zero, but we can be more precise: h p is the angular diameter of the object. Thus, the image diameter is h' = – hq p = (– 0.533°) π 180 rad/deg (1.50 m) = – 0.140 m = –1.40 cm 36.66 (a) The lens makers' equation, 1 f = (n −1) 1 R1 −1 R2 , becomes: 1 5.00 cm = (n −1) 1 9.00 cm − 1 −11.0 cm ( ) giving n = 1.99 .
(b) As the light passes through the lens for the first time, the thin lens equation 1 p1 + 1 q1 = 1 f becomes: 1 8.00 cm + 1 q1 = 1 5.00 cm or q1 = 13.3 cm, and M1 = −q1 p1 = −13.3 cm 8.00 cm = −1.67 This image becomes the object for the concave mirror with: pm = 20.0 cm −q1 = 20.0 cm −13.3 cm = 6.67 cm, and f = R 2 = + 4.00 cm. The mirror equation becomes: 1 6.67 cm + 1 qm = 1 4.00 cm giving qm = 10.0 cm and M2 = −qm pm = −10.0 cm 6.67 cm = −1.50 The image formed by the mirror serves as a real object for the lens on the second pass of the light through the lens with: p3 = 20.0 cm −qm = +10.0 cm The thin lens equation yields: 1 10.0 cm + 1 q3 = 1 5.00 cm or q3 = 10.0 cm, and M3 = −q3 p3 = −10.0 cm 10.0 cm = −1.00.
The final image is a real image located 10.0 cm to the left of the lens . The overall magnification is Mtotal = M1M2M3 = −2 50 . .
(c) Since the total magnification is negative, this final image is inverted .
Chapter 36 Solutions 375 © 2000 by Harcourt, Inc. All rights reserved.
36.67 In the original situation, p1 + q1 = 1.50 m In the final situation, p2 = p1 + 0.900 m and q2 = q1 – 0.900 m.
Our lens equation is 1 p1 + 1 q1 = 1 f = 1 p2 + 1 q2 Substituting, we have 1 p1 + 1 1.50 m – p1 = 1 p1 + 0.900 + 1 0.600 – p1 Adding the fractions, 1.50 m – p1 + p1 p1(1.50 m – p1) = 0.600 – p1 + p1 + 0.900 (p1 + 0.900)(0.600 – p1) Simplified, this becomes p1(1.50 m – p1) = (p1 + 0.900)(0.600 – p1) (a) Thus, p1 = 0.540 1.80 m = 0.300 m p2 = p1 + 0.900 = 1.20 m (b) 1 f = 1 0.300 m + 1 1.50 m – 0.300 m and f = 0.240 m (c) The second image is real, inverted, and diminished , with M = – q2 p2 = – 0.250 36.68 As the light passes through, the lens attempts to form an image at distance q1 where 1 q1 = 1 f −1 p1 or q1 = f p1 p1 −f This image serves as a virtual object for the mirror with p2 = −q1. The plane mirror then forms an image located at q2 = −p2 = +q1 above the mirror and lens.
This second image serves as a virtual object ( p3 = −q2 = −q1) for the lens as the light makes a return passage through the lens. The final image formed by the lens is located at distance q3 above the lens where 1 q3 = 1 f −1 p3 = 1 f + 1 q1 = 1 f + p1 −f f p1 = 2p1 −f f p1 or q3 = f p1 2p1 −f If the final image coincides with the object, it is necessary to require q p 3 1 = , or f p1 2p1 −f = p1. This yields the solution p1 = f or the object must be located at the focal point of the lens .
376 Chapter 36 Solutions 36.69 For the objective: 1 1 1 p q f + = becomes 1 3.40 mm + 1 q = 1 3.00 mm so q = 25.5 mm The objective produces magnification M1 = −q/ p = −25.5 mm 3.40 mm = −7.50 For the eyepiece as a simple magnifier, me = 25.0 cm f = 25.0 cm 2.50 cm = 10.0 and overall M = M1me = −75.0 36.70 (a) Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of it (i.e., q2 19 0 = − . cm). The required object distance for this lens is then p2 = q2 f2 q2 −f2 = −19.0 cm ( ) 20.0 cm ( ) −19.0 cm −20.0 cm = 380 cm 39.0 The image formed by the first lens serves as the object for the second lens. Therefore, the image distance for the first lens is q p 1 2 50 0 50 0 380 1570 = − = − = .
. cm cm cm 39.0 cm 39.0 The distance the original object must be located to the left of the first lens is then given by 1 p1 = 1 f1 −1 q1 = 1 10.0 cm − 39.0 1570 cm = 157 −39.0 1570 cm = 118 1570 cm or p1 1570 118 = = cm 13.3 cm (b) M = M1M2 = −q1 p1 −q2 p2 = 1570 cm 39.0 118 1570 cm −19.0 cm ( ) 39.0 ( ) 380 cm = −5 90 . (c) Since M < 0, the final image is inverted .
36.71 (a) P f p q = = + = + ∞= 1 1 1 1 (0.0224 m) 1 44.6 diopters (b) P = 1 f = 1 p + 1 q = 1 (0.330 m) + 1 ∞= 3.03 diopters Chapter 36 Solutions 377 © 2000 by Harcourt, Inc. All rights reserved.
36.72 The object is located at the focal point of the upper mirror.
Thus, the upper mirror creates an image at infinity (i.e., parallel rays leave this mirror).
The lower mirror focuses these parallel rays at its focal point, located at the hole in the upper mirror. Thus, the image is real, inverted, and actual size .
For the upper mirror: 1 p + 1 q = 1 f ⇒ 1 7.50 cm + 1 q1 = 1 7.50 cm : q1 = ∞ For the lower mirror: 1 ∞+ 1 q2 = 1 7.50 cm : q2 = 7.50 cm Light directed into the hole in the upper mirror reflects as shown, to behave as if it were reflecting from the hole.
36.73 (a) Lens one: 1 40.0 cm + 1 q1 = 1 30.0 cm : q1 = 120 cm M q p 1 1 1 120 40 0 = − = − cm cm . = −3.00 This real image is a virtual object for the second lens, at p2 = 110 cm −120 cm = −10.0 cm 1 10 0 1 1 20 0 2 − + = − .
. cm cm q : q2 = 20.0 cm M2 = −q2 p2 = − 20.0 cm −10.0 cm ( ) = +2.00 Moverall = M1M2 = −6.00 (b) Moverall < 0, so final image is inverted .
(c) Lens two converging: 1 10 0 1 1 20 0 2 − + = .
. cm cm q q2 = 6.67 cm M2 = − 6.67 cm (−10.0 cm) = +0.667 Moverall = M1M2 = −2 00 . Again, Moverall < 0 and the final image is inverted .
378 Chapter 36 Solutions © 2000 by Harcourt, Inc. All rights reserved.
Chapter 37 Solutions 37.1 ∆ybright = λL d = (632.8 × 10- 9)(5.00) 2.00 × 10- 4 m = 1.58 cm 37.2 y L d m bright = λ For m = 1, λ = = × ( ) × ( ) = − − yd L 3 40 10 5 00 10 3 30 3 4 .
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. m m m 515 nm 37.3 Note, with the conditions given, the small angle approximation does not work well. That is, sin θ, tan θ, and θ are significantly different. The approach to be used is outlined below.
(a) At the m = 2 maximum, tan θ = 400 m 1000 m = 0.400 θ = 21.8° so λ = d sin θ m = (300 m) sin 21.8° 2 = 55.7 m (b) The next minimum encountered is the m = 2 minimum; and at that point, d sin θ = m + 1 2 λ which becomes d sin θ = 5 2 λ or sin θ = 5λ 2d = 5(55.7 m) 2(300 m) = 0.464 and θ = 27.7° so y = (1000 m) tan 27.7° = 524 m Therefore, the car must travel an additional 124 m .
37.4 λ = v f = 354 m/s 2000/s = 0.177 m (a) d sin θ = mλ so (0.300 m) sin θ = 1(0.177 m) and θ = 36.2° (b) d sin θ = mλ so d sin 36.2° = 1(0.0300 m) and d = 5.08 cm (c) (1.00 × 10- 6 m) sin 36.2° = 1λ so λ = 590 nm f = c λ = 3.00 × 108 m/s 5.90 × 10–7 m = 508 THz 2 Chapter 37 Solutions 37.5 For the tenth minimum, m = 9. Using Equation 37.3, sin θ = λ d 9 + 1 2 Also, tan θ = y L . For small θ, sin θ ≈ tan θ. Thus, d = 9.5λ sin θ = 9.5λ L y = 9.5(589 × 10–9 m)(2.00 m) 7.26 × 10–3 m = 1.54 × 10–3 m = 1.54 mm Goal Solution Young's double-slit experiment is performed with 589-nm light and a slit-to-screen distance of 2.00 m.
The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. G : For the situation described, the observed interference pattern is very narrow, (the minima are less than 1 mm apart when the screen is 2 m away). In fact, the minima and maxima are so close together that it would probably be difficult to resolve adjacent maxima, so the pattern might look like a solid blur to the naked eye. Since the angular spacing of the pattern is inversely proportional to the slit width, we should expect that for this narrow pattern, the space between the slits will be larger than the typical fraction of a millimeter, and certainly much greater than the wavelength of the light (d >> λ = 589 nm).
O : Since we are given the location of the tenth minimum for this interference pattern, we should use the equation for destructive interference from a double slit. The figure for Problem 7 shows the critical variables for this problem. A : In the equation dsinθ = m + 1 2 ( )λ , The first minimum is described by m = 0 and the tenth by m = 9: sinθ = λ d 9 + 1 2 ( ) Also, tanθ = y L, but for small θ , sinθ ≈tanθ . Thus, d = 9.5λ sinθ = 9.5λL y d = 9.5(5890·10-10 m)(2.00 m) 7.26·10-3 m = 1.54·10-3 m = 1.54 mm = 1.54 mm L : The spacing between the slits is relatively large, as we expected (about 3 000 times greater than the wavelength of the light). In order to more clearly distinguish between maxima and minima, the pattern could be expanded by increasing the distance to the screen. However, as L is increased, the overall pattern would be less bright as the light expands over a larger area, so that beyond some distance, the light would be too dim to see.
Chapter 37 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
37.6 λ = 340 m/s 2000 Hz = 0.170 m Maxima are at d sin θ = mλ: m = 0 gives θ = 0° m = 1 gives sin θ = λ d = 0.170 m 0.350 m θ = 29.1° m = 2 gives sin θ = 2λ d = 0.971 θ = 76.3° m = 3 gives sin θ = 1.46 No solution.
Minima at d sin θ = m + 1 2 λ: m = 0 gives sin θ = λ 2d = 0.243 θ = 14.1° m = 1 gives sin θ = 3λ 2d = 0.729 θ = 46.8° m = 2 gives No solution.
So we have maxima at 0°, 29.1°, and 76.3° and minima at 14.1° and 46.8°. 37.7 (a) For the bright fringe, ybright = mλ L d where m = 1 y = (546.1× 10−9 m)(1.20 m) 0.250 × 10−3 m = 2.62 × 10−3 m = 2.62 mm (b) For the dark bands, ydark = λ L d m + 1 2 ; m = 0, 1, 2, 3, . . . y2 −y1 = λ L d 1+ 1 2 −0 + 1 2 = λ L d 1 ( ) = (546.1× 10−9 m)(1.20 m) 0.250 × 10−3 m ∆y = 2.62 mm Figures for Goal Solution 4 Chapter 37 Solutions Goal Solution A pair of narrow, parallel slits separated by 0.250 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.20 m away from the plane of the slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands.
G : The spacing between adjacent maxima and minima should be fairly uniform across the pattern as long as the width of the pattern is much less than the distance to the screen (so that the small angle approximation is valid). The separation between fringes should be at least a millimeter if the pattern can be easily observed with a naked eye. O : The bright regions are areas of constructive interference and the dark bands are destructive interference, so the corresponding double-slit equations will be used to find the y distances.
It can be confusing to keep track of four different symbols for distances. Three are shown in the drawing to the right. Note that: y is the unknown distance from the bright central maximum (m = 0) to another maximum or minimum on either side of the center of the interference pattern.
λ is the wavelength of the light, determined by the source.
A : (a) For very small θ sin θ ≈tan θ and tan θ = y /L and the equation for constructive interference sinθ = mλ d (Eq. 37.2) becomes ybright ≈λL d ( )m (Eq. 37.5) Substituting values, ybright = (546 × 10−9 m)(1.20 m) 0.250 × 10−3 m 1 ( ) = 2.62 mm (b) If you have trouble remembering whether Equation 37.5 or Eq. 37.6 applies to a given situation, you can instead remember that the first bright band is in the center, and dark bands are halfway between bright bands. Thus, Eq. 37.5 describes them all, with m = 0, 1, 2 . . . for bright bands, and with m = 0.5, 1.5, 2.5 . . . for dark bands. The dark band version of Eq. 37.5 is simply Eq. 37.6: ydark = λL d m + 1 2 ( ) ∆ydark = 1+ 1 2 ( ) λL d −0 + 1 2 ( ) λL d = λL d = 2.62 mm L : This spacing is large enough for easy resolution of adjacent fringes. The distance between minima is the same as the distance between maxima. We expected this equality since the angles are small: θ = 2.62 mm ( ) 1.20 m ( ) = 0.00218 rad = 0.125° When the angular spacing exceeds about 3°, then sin θ differs from tan θ when written to three significant figures.
Chapter 37 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
37.8 Taking m = 0 and y = 0.200 mm in Equation 37.6 gives L dy ≈2 λ = × × × = − − − 2 0 400 10 10 442 10 0 362 3 3 9 ( .
.
m)(0.200 m) m m L ≈ 36.2 cm Geometric optics incorrectly predicts bright regions opposite the slits and darkness in between. But, as this example shows, interference can produce just the opposite.
37.9 Location of A = central maximum, Location of B = first minimum. So, ∆y y y L d L d = − = + − = = [ ] .
min max λ λ 0 1 2 0 1 2 20 0 m Thus, d L = ( ) = ( )( ) = λ 2 20 0 3 00 150 40 0 .
.
. m m m m 11.3 m 37.10 At 30.0°, d m sin θ λ = (3.20 × 10−4 m) sin 30.0°= m(500 × 10−9 m) so m = 320 There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead. There are 641 maxima .
37.11 φ = 2π λ dsin θ ≈2π λ d y L (a) φ π = × × ( ) = − − ° 2 5 00 10 1 20 10 0 500 7 4 ( .
( .
sin . m) m) 13.2 rad (b) φ = 2π (5.00 × 10−7 m) (1.20 × 10−4 m) 5.00 × 10−3 m 1.20 m = 6.28 rad 6 Chapter 37 Solutions (c) If φ = 0.333 rad = 2π dsin θ λ , θ = sin−1 λ φ 2π d = sin−1 (5.00 × 10−7 m)(0.333 rad) 2π(1.20 × 10−4 m) θ = 1 27 10 2 .
× − deg (d) If dsin θ λ = 4 , θ = sin−1 λ 4d = sin−1 5 × 10−7 m 4(1.20 × 10−4 m) θ = 5 97 10 2 .
× − deg 37.12 The path difference between rays 1 and 2 is: δ = dsin θ1 −dsin θ2 For constructive interference, this path difference must be equal to an integral number of wavelengths: dsin θ1 −dsin θ2 = mλ , or d(sin θ1 −sin θ2) = mλ 37.13 (a) The path difference δ = dsin θ and when L >> y δ ≈ = × × = × = − − − yd L ( .
)( .
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1 80 10 1 50 10 1 40 1 93 10 2 4 6 m m) m m 1.93 µm (b) δ λ = × × = − − 1 93 10 6 43 10 3 00 6 7 .
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. m m , or δ λ = 3 00 . (c) Point P will be a maximum since the path difference is an integer multiple of the wavelength.
37.14 (a) I Imax = cos2 φ 2 (Equation 37.11) Therefore, φ = 2 cos–1 I Imax 1/2 = 2 cos–1 (0.640)1/2 = 1.29 rad (b) δ = λφ 2π = (486 nm)(1.29 rad) 2π = 99.8 nm Chapter 37 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
37.15 Iav = Imax cos2 π d sin θ λ For small θ, sin θ = y L and Iav = 0.750 Imax y = λ L π d cos–1 Iav Imax 1/2 y = (6.00 × 10–7)(1.20 m) π (2.50 × 10–3 m) cos–1 0.750 Imax Imax 1/2 = 48.0 µm 37.16 I I yd L = max cos2 π λ I Imax cos .
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.
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= × ( ) × ( ) × ( )( ) = − − − 2 3 4 9 6 00 10 1 80 10 656 3 10 0 800 π m m m m 0.987 37.17 (a) From Equation 37.8, φ = 2π d λ sin θ = 2π d λ ⋅ y y2 + D2 φ ≈2π yd λD = 2π 0.850 × 10−3 m ( ) 2.50 × 10−3 m ( ) 600 × 10−9 m ( ) 2.80 m ( ) = 7.95 rad (b) I Imax = cos2 π d λ sin θ cos2 π d λ sin θmax = cos2 φ 2 cos2 mπ I Imax = cos2 φ 2 = cos2 7.95 rad 2 = 0.453 8 Chapter 37 Solutions Goal Solution Two narrow parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe? (b) What is the ratio of the intensity at this point to the intensity at the center of a bright fringe?
G : It is difficult to accurately predict the relative intensity at the point of interest without actually doing the calculation. The waves from each slit could meet in phase ( φ = 0) to produce a bright spot of constructive interference, out of phase ( φ = 180°) to produce a dark region of destructive interference, or most likely the phase difference will be somewhere between these extremes, 0 < φ < 180°, so that the relative intensity will be 0 < I Imax < 1. O : The phase angle depends on the path difference of the waves according to Equation 37.8. This phase difference is used to find the average intensity at the point of interest. Then the relative intensity is simply this intensity divided by the maximum intensity.
A : (a) Using the variables shown in the diagram for problem 7 we have, φ = 2π d λ sinθ = 2π d λ y y2 + L2 ≅2π yd λL = 2π 0.850·10-3 m ( )0.00250 m ( ) 600·10-9 m ( )2.80 m ( ) = 7.95 rad = 2π +1.66 rad = 95.5 (b) I Imax = cos2 π d λ sinθ cos2 π d λ sinθmax = cos2 φ 2 cos2 mπ ( ) = cos2 φ 2 = cos2 95.5 2 Ł ł= 0.452 L : It appears that at this point, the waves show partial interference so that the combination is about half the brightness found at the central maximum. We should remember that the equations used in this solution do not account for the diffraction caused by the finite width of each slit. This diffraction effect creates an “envelope” that diminishes in intensity away from the central maximum as shown by the dotted line in Figures 37.13 and P37.60. Therefore, the relative intensity at y = 2.50 mm will actually be slightly less than 0.452. 37.18 (a) The resultant amplitude is Er = E0 sin ωt + E0 sin ωt + φ ( ) + E0 sin ωt + 2φ ( ), where φ = 2π λ dsin θ . Er = E0 sin ωt + sin ωt cos φ + cos ωt sin φ + sin ωt cos 2φ + cos ωt sin 2φ ( ) Er = E0(sin ωt) 1+ cos φ + 2 cos2 φ −1 ( ) + E0(cos ωt)(sin φ + 2 sin φ cos φ) Er = E0(1+ 2 cos φ)(sin ωt cos φ + cos ωt sin φ) = + + E t 0 (1 2 cos ) sin ( ) φ ω φ Chapter 37 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
Then the intensity is I ∝Er 2 = E0 2 (1+ 2 cos φ)2 1 2 where the time average of sin2 (ωt + φ) is 1 2. From one slit alone we would get intensity Imax ∝E0 2 1 2 so I = Imax 1+ 2 cos 2πdsin θ λ 2 (b) Look at the N = 3 graph in Figure 37.13. Minimum intensity is zero, attained where cos 1/2 φ = − . One relative maximum occurs at cos 1.00 φ = − , where I I = max.
The larger local maximum happens where cos φ = +1.00, giving I = 9.00 I0.
The ratio of intensities at primary versus secondary maxima is 9.00 .
37.19 (a) We can use sin A + sin B = 2 sin A 2 + B 2 ( ) cos A 2 −B 2 ( ) to find the sum of the two sine functions to be E1 + E2 = 24.0 kN C ( ) sin 15x −4.5t + 35.0° ( ) cos 35.0° E1 + E2 = 19.7 kN C ( ) sin 15x −4.5t + 35.0° ( ) Thus, the total wave has amplitude 19.7 kN/C and has a constant phase difference of 35.0° from the first wave.
(b) In units of kN/C, the resultant phasor is ER = E1 + E2 = ( ) + ° + ° ( ) 12 0 12 0 70 0 12 0 70 0 .
. cos .
. sin .
i i j = + 16 1 11 3 .
.
i j ER = 16.1 ( )2 + 11.3 ( )2 at tan−1 11.3 16.1 ( ) = 19.7 kN/C at 35.0° E1 E2 70.0° ER α x y kx - ωt (c) E i j R = ° + ° 12 0 70 0 12 0 70 0 . cos .
. sin . +15.5cos80.0°i −15.5sin80.0°j +17.0 cos 160°i + 17.0 sin 160°j E i j R = − + = 9 18 1 83 .
. 9.36 kN/C at 169° E1 E2 ER x y kx - ωt E3 The wave function of the total wave is EP = 9.36 kN C ( ) sin 15x −4.5t + 169° ( ) 10 Chapter 37 Solutions 37.20 (a) E i i j R E = + ° + ° ( ) [ 0 20 0 20 0 cos .
sin . + i cos 40.0°+j sin 40.0° ( )] E i j R E = + [ ] 0 2 71 0 985 .
. = ° = 2 88 0 .
E at 20.0 2.88E0 at 0.349 rad EP = 2.88E0 sin ωt + 0.349 ( ) E1 E2 E3 ER y x (b) E i i j R E = + ° + ° ( ) [ 0 60 0 60 0 cos .
sin . + i cos120°+j sin 120° ( )] ER = E0 1.00i + 1.73j [ ] = 2.00E0 at 60.0°= 2.00E0 at π 3 rad EP = 2.00E0 sin ωt + π 3 ( ) E1 E2 E3 ER y x (c) E i i j R E = + ° + ° ( ) [ 0 120 120 cos sin + i cos 240°+j sin 240° ( )] E i j R E = + [ ] = 0 0 0 0 EP = 0 E1 E2 E3 y x (d) E i i j R E = + + ( ) [ 0 3 2 3 2 cos sin π π + i cos 3π +j sin 3π ( )] E i j R E E = − [ ] = ° = 0 0 0 1 00 . at 270 E0 3 at 2 rad π EP = E0 sin ωt + 3π 2 ( ) E1 y x E2 E3 ER 37.21 E i j R = + = ( ) + ( ) ( ) − 6 00 8 00 6 00 8 00 8 00 6 00 2 2 1 .
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tan .
. at ER = 10.0 at 53.1°= 10.0 at 0.927 rad EP = 10.0 sin 100πt + 0.927 ( ) 6.00 8.00 y x ER π 2 α 37.22 If E E t 1 10 = sin ω and E E t 2 20 = + ( ) sin ω φ , then by phasor addition, the amplitude of E is E E E E 0 10 20 2 20 2 = + ( ) + ( ) cos sin φ φ = E10 2 + 2E10E20 cos φ + E20 2 and the phase angle is found from sinθ = E20 sin φ E0 E10 E20 E0 x y θ φ Chapter 37 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
37.23 E i i j R = + ° + ° ( ) 12 0 18 0 60 0 18 0 60 0 .
. cos .
. sin . E i j R = + = ° 21 0 15 6 26 2 .
.
. at 36.6 EP = 26 2 36 6 . sin .
ω t + ° ( ) 12.0 18.0 ER x y θ 60.0° 37.24 Constructive interference occurs where m = 0, 1, 2, 3, . . ., for 2π x1 λ −2π f t + π 6 −2π x2 λ −2π f t + π 8 = 2π m 2π(x1 −x2) λ + π 6 −π 8 = 2π m (x1 −x2) λ + 1 12 −1 16 = m x1 −x2 = m −1 48 λ m = 0, 1, 2, 3, . . . 37.25 See the figure to the right: φ = π /2 37.26 ER 2 = E1 2 + E2 2 −2E1E2 cos β, where β = 180 −φ .
Since I ∝E2, IR = I1 + I2 + 2 I1I2 cos φ 37.27 Take φ = 360° N where N defines the number of coherent sources. Then, ER = E0 sin ω t + mφ ( ) m=1 N ∑ = 0 In essence, the set of N electric field components complete a full circle and return to zero.
12.0 x y The N = 6 case φ = 360° N = 360° 6 = 60.0° 60.0° 12 Chapter 37 Solutions 37.28 Light reflecting from the first surface suffers phase reversal. Light reflecting from the second surface does not, but passes twice through the thickness t of the film. So, for constructive interference, we require λ n 2 + 2t = λ n where λ n = λ n is the wavelength in the material. Then 2t = λ n 2 = λ 2n λ = 4nt = 4 × 1.33 × 115 nm = 612 nm 37.29 (a) The light reflected from the top of the oil film undergoes phase reversal. Since 1.45 > 1.33, the light reflected from the bottom undergoes no reversal. For constructive interference of reflected light, we then have 2nt = m + 1 2 λ or λ m = 2nt m + 1 2 = 2(1.45)(280 nm) m + 1 2 Substituting for m, we have m = 0: λ 0 = 1620 nm (infrared) m = 1: λ 1 = 541 nm (green) m = 2: λ 2 = 325 nm (ultraviolet) Both infrared and ultraviolet light are invisible to the human eye, so the dominant color in the reflected light is green .
(b) The dominant wavelengths in the transmitted light are those that produce destructive interference in the reflected light. The condition for destructive interference upon reflection is 2nt m = λ or λ m nt m m = = 2 812 nm Substituting for m gives: m = 1, λ 1 812 = nm (near infrared) m = 2, λ2 406 = nm (violet) m = 3, λ 3 271 = nm (ultraviolet) Of these. the only wavelength visible to the human eye (and hence the dominate wavelength observed in the transmitted light) is 406 nm. Thus, the dominant color in the transmitted light is violet .
Chapter 37 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
37.30 Since 1 < 1.25 < 1.33, light reflected both from the top and from the bottom surface of the oil suffers phase reversal. For constructive interference we require 2t = mλ cons n and for destructive interference, 2t = m + 1 2 λ des n Then λ cons λ dest = 1 + 1 2m = 640 nm 512 nm = 1.25 and m = 2 Therefore, t = 2(640 nm) 2(1.25) = 512 nm 37.31 Treating the anti-reflectance coating like a camera-lens coating, 2t = m + 1 2 λ n Let m = 0: t = λ 4n = 3.00 cm 4(1.50) = 0.500 cm This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to 1.50 cm—now creating maximum reflection!
37.32 2nt = m + 1 2 λ so t = m + 1 2 λ 2n Minimum t = 1 2 (500 nm) 2(1.30) = 96.2 nm 37.33 Since the light undergoes a 180° phase change at each surface of the film, the condition for constructive interference is 2nt = mλ , or λ = 2nt m. The film thickness is t = × = × = − − 1 00 10 1 00 10 100 5 7 .
. cm m nm. Therefore, the wavelengths intensified in the reflected light are λ = 2 1.38 ( ) 100 nm ( ) m = 276 nm m where m = 1, 2, 3, . . . or λ 1 276 = nm, λ2 138 = nm, . . . . All reflection maxima are in the ultraviolet and beyond.
No visible wavelengths are intensified. 14 Chapter 37 Solutions 37.34 (a) For maximum transmission, we want destructive interference in the light reflected from the front and back surfaces of the film. If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface suffers no phase reversal and light reflected from the back does undergo phase reversal. This effect by itself would produce destructive interference, so we want the distance down and back to be one whole wavelength in the film: 2t = λ n. t n = = ( ) = λ 2 656 3 . nm 2 1.378 238 nm (b) The filter will expand. As t increases in 2nt = λ , so does λ increase .
(c) Destructive interference for reflected light happens also for λ in 2 2 nt = λ, or λ = ( ) = 1 378 238 . nm 328 nm (near ultraviolet) .
37.35 If the path length ∆= λ , the transmitted light will be bright. Since ∆= = 2d λ, dmin nm 2 = = = λ 2 580 290 nm 37.36 The condition for bright fringes is 2t + λ 2n = m λ n m = 1, 2, 3, . . .
From the sketch, observe that t = R 1−cos θ ( ) ≈R 1−1+ θ 2 2 = R 2 r R 2 = r2 2R R θ r t The condition for a bright fringe becomes r2 R = m −1 2 λ n . Thus, for fixed m and λ , nr2 = constant.
Therefore, nliquidrf 2 = nairri 2 and nliquid cm cm = ( )( ) ( ) = 1 00 1 50 1 31 2 2 .
.
. 1.31 Chapter 37 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
37.37 For destructive interference in the air, 2t = mλ.
For 30 dark fringes, including the one where the plates meet, t = 29(600 nm) 2 = 8.70 × 10– 6 m Therefore, the radius of the wire is r = d 2 = 8.70 µm 2 = 4.35 µm Goal Solution An air wedge is formed between two glass plates separated at one edge by a very fine wire as shown in Figure P37.37. When the wedge is illuminated from above by 600-nm light, 30 dark fringes are observed.
Calculate the radius of the wire.
G : The radius of the wire is probably less than 0.1 mm since it is described as a “very fine wire.” O : Light reflecting from the bottom surface of the top plate undergoes no phase shift, while light reflecting from the top surface of the bottom plate is shifted by π, and also has to travel an extra distance 2t, where t is the thickness of the air wedge.
For destructive interference, 2t = mλ m = 0, 1, 2, 3, . . .
( ) The first dark fringe appears where m = 0 at the line of contact between the plates. The 30th dark fringe gives for the diameter of the wire 2t = 29λ , and t = 14.5λ .
A : r = t 2 = 7.25λ = 7.25 600 × 10−9 m ( ) = 4.35 µm L : This wire is not only less than 0.1 mm; it is even thinner than a typical human hair ( ~ 50 µm).
37.38 For destructive interference, 2t = λm n . At the position of the maximum thickness of the air film, m = 2tn λ = 2(4.00 × 10−5 m)(1.00) 5.461× 10−7 m = 146.5 The greatest integer value is m = 146.
Therefore, including the dark band at zero thickness, there are 147 dark fringes .
16 Chapter 37 Solutions 37.39 For total darkness, we want destructive interference for reflected light for both 400 nm and 600 nm. With phase reversal at just one reflecting surface, the condition for destructive interference is 2nairt = mλ m = 0, 1, 2, . . .
The least common multiple of these two wavelengths is 1200 nm, so we get no reflected light at 2 1 00 3 400 2 600 1200 .
( ) = ( ) = ( ) = t nm nm nm, so t = 600 nm at this second dark fringe.
By similar triangles, 600 0 0500 nm mm 10.0 cm x = .
, or the distance from the contact point is x = × ( ) × = − − 600 10 9 m 0.100 m 5.00 10 m 5 1.20 mm 37.40 2 2 2 1 80 10 550 5 10 4 9 t m m t = = = ( .
m) m = λ λ ⇒ × × − − . 654 dark fringes 37.41 When the mirror on one arm is displaced by ∆ l, the path difference increases by 2∆ l. A shift resulting in the formation of successive dark (or bright) fringes requires a path length change of one-half wavelength. Therefore, 2∆ l = mλ/2, where in this case, m = 250. ∆l = m λ 4 = 250 ( ) 6.328 × 10−7 m ( ) 4 = 39.6 µm 37.42 Distance = 2(3.82 × 10– 4 m) = 1700λ λ = 4.49 × 10– 7 m = 449 nm The light is blue 37.43 Counting light going both directions, the number of wavelengths originally in the cylinder is m1 = 2L λ . It changes to m2 = 2L λ n = 2nL λ as the cylinder is filled with gas. If N is the number of bright fringes passing, N = m2 −m1 = 2L λ n −1 ( ), or the index of refraction of the gas is n = 1+ Nλ 2L = 1+ 35 633 × 10-9 m ( ) 2 0.0300 m ( ) = 1.000 369 Chapter 37 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
37.44 Counting light going both directions, the number of wavelengths originally in the cylinder is m1 = 2L λ . It changes to m2 = 2L λ n = 2nL λ as the cylinder is filled with gas. If N is the number of bright fringes passing, N = m2 −m1 = 2L λ n −1 ( ), or the index of refraction of the gas is n = 1+ Nλ 2L 37.45 The wavelength is λ = c f = 3.00 × 108 m s 60.0 × 106 s−1 = 5.00 m.
Along the line AB the two traveling waves going in opposite directions add to give a standing wave. The two transmitters are exactly 2.00 wavelengths apart and the signal from B, when it arrives at A, will always be in phase with transmitter B. Since B is 180° out of phase with A, the two signals always interfere destructively at the position of A.
The first antinode (point of constructive interference) is located at distance λ 4 5 00 = = . m 4 1.25 m from the node at A.
37.46 My middle finger has width d = 2 cm.
(a) Two adjacent directions of constructive interference for 600-nm light are described by d sin θ = mλ θ0 = 0 (2 × 10– 2 m) sin θ1 = 1 ( 6 × 10– 7 m) Thus, θ1 = 2 × 10– 3 degree and θ1 – θ0 ~ 10– 3 degree (b) Choose θ1 = 20° 2 × 10– 2 m sin 20° = 1λ λ = 7 mm Millimeter waves are microwaves f = c λ = 3 × 108 m/s 7 × 10– 3 m ~ 1011 Hz 37.47 If the center point on the screen is to be a dark spot rather than bright, passage through the plastic must delay the light by one-half wavelength. Calling the thickness of the plastic t.
18 Chapter 37 Solutions t λ + 1 2 = t λ n ( ) = nt λ or t = λ 2( 1) n − where n is the index of refraction for the plastic.
Chapter 37 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
37.48 No phase shift upon reflection from the upper surface (glass to air) of the film, but there will be a shift of λ 2 due to the reflection at the lower surface of the film (air to metal). The total phase difference in the two reflected beams is then δ λ = + 2 2 nt . For constructive interference, δ λ = m , or 2 1 00 2 ( .
)t m + = λ λ. Thus, the film thickness for the mth order bright fringe is: tm = m −1 2 λ 2 = m λ 2 −λ 4 , and the thickness for the m – 1 bright fringe is: tm−1 = (m −1) λ 2 −λ 4 .
Therefore, the change in thickness required to go from one bright fringe to the next is ∆t = tm −tm−1 = λ 2. To go through 200 bright fringes, the change in thickness of the air film must be: 200 λ 2 ( ) = 100λ . Thus, the increase in the length of the rod is ∆L = 100λ = 100 5.00 × 10−7 m ( ) = 5.00 × 10−5 m, From ∆ ∆ L L T i = ( ) α , we have: α = ∆L Li ∆T ( ) = 5.00 × 10−5 m 0.100 m ( ) 25.0°C ( ) = 20 0 10 6 1 . × ° − − C 37.49 Since 1 < 1.25 < 1.34, light reflected from top and bottom surfaces of the oil undergoes phase reversal. The path difference is then 2t, which must be equal to mλ n = mλ n for maximum reflection, with m = 1 for the given first-order condition and n = 1.25. So t = mλ 2n = 1(500 nm) 2(1.25) = 200 nm The volume we assume to be constant: 1.00 m3 = (200 nm)A A = 1.00 m3 200(10– 9 m) = 5.00 × 106 m2 = 5.00 km2 37.50 For destructive interference, the path length must differ by mλ. We may treat this problem as a double slit experiment if we remember the light undergoes a π/2-phase shift at the mirror.
The second slit is the mirror image of the source, 1.00 cm below the mirror plane. Using Equation 37.5, ydark = mλ L d = 1(5.00 × 10– 7 m)(100 m) (2.00 × 10– 2 m) = 2.50 mm 20 Chapter 37 Solutions 37.51 One radio wave reaches the receiver R directly from the distant source at an angle θ above the horizontal. The other wave undergoes phase reversal as it reflects from the water at P.
Constructive interference first occurs for a path difference of d = λ 2 (1) The angles θ in the figure are equal because they each form part of a right triangle with a shared angle at R'.
It is equally far from P to R as from P to R', the mirror image of the telescope.
So the path difference is d = 2(20.0 m) sin θ = (40.0 m) sin θ The wavelength is λ = c f = 3.00 × 108 m/s 60.0 × 106 Hz = 5.00 m Substituting for d and λ in Equation (1), (40.0 m) sin θ = 5.00 m 2 Solving for the angle θ, sin θ = 5.00 m 80.0 m and θ = 3.58° 37.52 2 (15.0 km)2 + h2 = 30.175 km (15.0 km)2 + h2 = 227.63 h = 1.62 km 37.53 From Equation 37.13, I Imax = cos2 π yd λL Let λ 2 equal the wavelength for which I Imax → I2 Imax = 0.640 Then λ 2= π yd/L cos– 1 (I2/Imax)1/2 But π yd L = λ 1 cos– 1 I1 Imax 1/2 = (600 nm) cos– 1 (0.900) = 271 nm Substituting this value into the expression for λ 2, λ 2 = 271 nm cos– 1 (0.6401/2) = 421 nm Note that in this problem, cos– 1 I Imax 1/2 must be expressed in radians.
Chapter 37 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
37.54 For Young's experiment, use δ θ λ = = d m sin . Then, at the point where the two bright lines coincide, d m m sin θ λ λ = = 1 1 2 2 so λ λ 1 2 2 1 540 450 6 5 = = = m m sin θ = 6λ2 d = 6(450 nm) 0.150 mm = 0.0180 Since sin θ θ ≈ and L = 1 40 . m, x L = = ( )( ) = θ 0 0180 1 40 .
. m 2.52 cm 37.55 For dark fringes, 2nt = mλ and at the edge of the wedge, t = 84(500 nm) 2 .
When submerged in water, 2nt = mλ m = 2(1.33)(42)(500 nm) 500 nm so m + 1 = 113 dark fringes 37.56 At entrance, 1.00 sin 30.0° = 1.38 sin θ2 θ2 = 21.2° Call t the unknown thickness. Then cos 21.2° = t a a = t cos 21.2° tan 21.2° = c t c = t tan 21.2° sin θ1 = b 2c b = 2t tan 21.2° sin 30.0° The net shift for the second ray, including the phase reversal on reflection of the first, is 2an – b – λ 2 where the factor n accounts for the shorter wavelength in the film. For constructive interference, we require 2an – b – λ 2 = mλ The minimum thickness will be given by 2an – b – λ 2 = 0.
λ 2 = 2an – b = 2 nt cos 21.2° – 2t(tan 21.2°) sin 30.0° 590 nm 2 = 2 × 1.38 cos 21.2° – 2 tan 21.2° sin 30.0° t = 2.57t t = 115 nm 22 Chapter 37 Solutions 37.57 The shift between the two reflected waves is δ = 2na −b −λ 2 where a and b are as shown in the ray diagram, n is the index of refraction, and the factor of λ 2 is due to phase reversal at the top surface. For constructive interference, δ λ = m where m has integer values. This condition becomes φ 1 φ 1 b 2c c a θ 2 t n 2na −b = m + 1 2 λ (1) From the figure's geometry, a t = cos θ2 , c = asin θ2 = tsin θ2 cos θ2 , b c t = = 2 2 1 2 2 1 sin sin cos sin φ θ θ φ Also, from Snell’s law, sin sin φ θ 1 2 = n . Thus, b = 2nt sin2 θ2 cos θ2 With these results, the condition for constructive interference given in Equation (1) becomes: 2n t cos θ2 −2nt sin2 θ2 cos θ2 = 2nt cos θ2 1−sin2 θ2 ( ) = m + 1 2 λ or 2nt cos θ2 = m + 1 2 λ 37.58 (a) Minimum: 2nt = mλ2 m = 0, 1, 2, . . .
Maximum: 2nt = m' + 1 2 λ 1 m' = 0, 1, 2, . . .
for λ 1 > λ 2, m' + 1 2 < m so m' = m – 1 Then 2nt = mλ 2 = m – 1 2 λ 1 2mλ 2 = 2mλ 1 – λ 1 so m = λ 1 2(λ 1 – λ 2) (b) m = 500 2(500 – 370) = 1.92 → 2 (wavelengths measured to ± 5 nm) [Minimum]: 2nt = mλ 2 2(1.40)t = 2(370 nm) t = 264 nm [Maximum]: 2nt = m – 1 + 1 2 λ = 1.5λ 2(1.40)t = (1.5)500 nm t = 268 nm Film thickness = 266 nm Chapter 37 Solutions 23 © 2000 by Harcourt, Inc. All rights reserved.
37.59 From the sketch, observe that x = h2 + d 2 ( ) 2 = 4h2 + d2 2 Including the phase reversal due to reflection from the ground, the total shift between the two waves is δ = 2x −d −λ 2. d d/2 x x h (a) For constructive interference, the total shift must be an integral number of wavelengths, or δ = mλ where m = 0, 1, 2, 3, . . . .
Thus, 2x −d = m + 1 2 λ or λ = 4x −2d 2m + 1 For the longest wavelength, m = 0, giving λ = − = 4 2 x d 2 4h2 + d2 −2d (b) For destructive interference, δ = m −1 2 λ where m = 1, 2, 3, . . . .
Thus, 2x −d = mλ or λ = 2x −d m . For the longest wavelength, m = 1 giving λ = − = 2x d 4h2 + d2 −d 37.60 Call t the thickness of the sheet. The central maximum corresponds to zero phase difference. Thus, the added distance ∆r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. As light advances through distance t in air, the number of cycles it goes through is t/λa.
The number of cycles in the sheet is t (λa /n) = nt λa Thus, the sheet introduces phase difference φ = 2π nt λa – t λa The corresponding difference in path length is ∆r = φ λa 2π = 2π λa (nt – t) λa 2π = (n – 1)t Note that the wavelength of the light does not appear in this equation. In the figure, the two rays from the slits are essentially parallel, so the angle θ may be expressed as tanθ = ∆r d = ′ y L.
Substituting for ∆r and solving for y' gives y' = ∆r L d = t(n – 1)L d = (5.00 × 10– 5 m)(1.50 – 1)(1.00 m) (3.00 × 10– 4 m) = 0.0833 m = 8.33 cm 24 Chapter 37 Solutions 37.61 Call t the thickness of the film. The central maximum corresponds to zero phase difference. Thus, the added distance ∆r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. The phase difference φ is φ = 2π t λa (n – 1) The corresponding difference in path length ∆r is ∆r = φ λa 2π = 2π t λa (n – 1) λa 2π = t(n – 1) Note that the wavelength of the light does not appear in this equation. In the figure, the two rays from the slits are essentially parallel.
Thus the angle θ may be expressed as tan θ = ∆r d = y' L Eliminating ∆r by substitution, y' L = t(n – 1) d gives y' = t(n – 1)L d Goal Solution Consider the double-slit arrangement shown in Figure P37.60, where the slit separation is d and the slit to screen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y'. Find y'.
G : Since the film shifts the pattern upward, we should expect ′ y to be proportional to n, t, and L.
O : The film increases the optical path length of the light passing through the upper slit, so the physical distance of this path must be shorter for the waves to meet in phase ( φ = 0) to produce the central maximum. Thus, the added distance ∆r traveled by the light from the lower slit must introduce a phase difference equal to that introduced by the plastic film. A : First calculate the additional phase difference due to the plastic. Recall that the relation between phase difference and path difference is φ = 2π δ λ . The presence of plastic affects this by changing the wavelength of the light, so that the phase change of the light in air and plastic, as it travels over the thickness t is φair = 2π t λair and φplastic = 2π t λair /n Thus, plastic causes an additional phase change of ∆φ = 2π t λair n −1 ( ) Next, in order to interfere constructively, we must calculate the additional distance that the light from the bottom slit must travel. ∆r = ∆φ λair 2π = t n −1 ( ) In the small angle approximation we can write ∆r = ′ y d L, so ¢ y = t n -1 ( ) L dL L : As expected, ′ y is proportional to t and L. It increases with increasing n, being proportional to n −1 ( ). It is also inversely proportional to the slit separation d, which makes sense since slits that are closer together make a wider interference pattern.
Chapter 37 Solutions 25 © 2000 by Harcourt, Inc. All rights reserved.
37.62 λ = = × × = c f 3 00 10 1 50 10 200 8 6 .
. m s Hz m For destructive interference, the path difference is one-half wavelength. Thus, λ 2 = 100 m = x + x2 + 2.00 × 104 m ( ) 2 −2.00 × 104 m, or 2.01× 104 m −x = x2 + 2.00 × 104 m ( ) 2 Squaring and solving, x = 99.8 m 37.63 (a) Constructive interference in the reflected light requires 2t = m + 1 2 ( )λ . The first bright ring has m = 0 and the 55th has m = 54, so at the edge of the lens t = 54.5(650 10 m) 2 = 17.7 m 9 × − µ Now from the geometry in Figure 37.18, the distance from the center of curvature down to the flat side of the lens is R 2 −r 2 = R −t or R2 −r2 = R2 −2Rt + t2 R = r 2 + t 2 2t = (5.00 × 10−2 m)2 +(1.77 × 10−5 m)2 2(1.77 × 10−5 m) = 70.6 m (b) 1 = (n 1) 1 1 = 0.520 1 1 70.6 m 2 2 f R R − − ∞−− so f = 136 m 37.64 Bright fringes occur when 2t = λ n m + 1 2 and dark fringes occur when 2t = λ n m The thickness of the film at x is t = h l x.
Therefore, xbright = λl 2hn m + 1 2 and xdark = λlm 2hn 26 Chapter 37 Solutions 37.65 E E E E R = + + 1 2 3 = + + cos .
cos .
cos π π π 6 3 00 7 2 6 00 4 3 i + + + sin .
sin .
sin π π π 6 3 00 7 2 6 00 4 3 j E i j R = − − 2 13 7 70 .
. ER = − ( ) + − ( ) − − = 2 13 7 70 7 70 2 13 7 99 2 2 .
.
.
.
. at tan at 4.44 rad -1 Thus, EP = 7 99 4 44 .
sin .
ω t + ( ) rad x y E1 E2 E3 ER 37.66 For bright rings the gap t between surfaces is given by 2t = m + 1 2 ( )λ The first bright ring has m = 0 and the hundredth has m = 99.
So, t = ( ) × ( ) = − 1 2 99 5 500 10 24 9 9 . m . m µ .
Call r b the ring radius. From the geometry of the figure at the right, t = r − r2 −rb 2 −R − R2 −rb 2 Since r r b << , we can expand in series: t = r −r 1−1 2 rb 2 r 2 −R + R 1−1 2 rb 2 R2 = 1 2 rb 2 r −1 2 rb 2 R rb = 2t 1/r −1/ R 1/2 = 2 24.9 × 10−6 m ( ) 1/ 4.00 m −1/12.0 m 1/2 = 1.73 cm 37.67 The shift between the waves reflecting from the top and bottom surfaces of the film at the point where the film has thickness t is δ = 2tnfilm + λ 2 ( ), with the factor of λ 2 being due to a phase reversal at one of the surfaces.
For the dark rings (destructive interference), the total shift should be δ = m + 1 2 ( )λ with m = 0, 1, 2, 3, . . . . This requires that t = mλ 2nfilm .
R θ r t nfilm To find t in terms of r and R, R2 = r2 + (R −t)2 so r2 = 2Rt + t2 Since t is much smaller than R, t2 << 2Rt and r2 ≈2Rt = 2R mλ 2nfilm . Thus, where m is an integer, r ≈ mλR nfilm Chapter 37 Solutions 27 © 2000 by Harcourt, Inc. All rights reserved.
37.68 (a) Bright bands are observed when 2nt = m + 1 2 λ Hence, the first bright band (m = 0) corresponds to nt = λ 4.
Since x x t t 1 2 1 2 = , we have x x t t x 2 1 2 1 1 2 1 3 00 = = = ( ) = λ λ . cm 680 nm 420 nm 4.86 cm (b) t n 1 1 4 420 = = = λ nm 4(1.33) 78.9 nm t n 2 2 4 680 = = = λ nm 4(1.33) 128 nm (c) θ θ ≈ = = = tan .
t x 1 1 78 9 3 00 nm . cm 2 63 10 6 .
× − rad 37.69 2h sin θ = m + 1 2 λ bright 2h ∆y 2L = 1 2 λ so h = Lλ 2∆y = 2.00 m ( ) 606 × 10−9 m ( ) 2 1.2 × 10−3 m ( ) = 0.505 mm 37.70 Superposing the two vectors, ER = E1 + E2 ER = E1 + E2 = E0 + E0 3 cos φ 2 + E0 3 sin φ 2 = E0 2 + 2 3 E0 2 cos φ + E0 2 9 cos2 φ + E0 2 9 sin2 φ ER = 10 9 E0 2 + 2 3 E0 2 cos φ Since intensity is proportional to the square of the amplitude, I = 10 9 Imax + 2 3 Imax cos φ Using the trigonometric identity cos cos φ φ = − 2 2 1 2 , this becomes I = 10 9 Imax + 2 3 Imax 2 cos2 φ 2 −1 = 4 9 Imax + 4 3 Imax cos2 φ 2 , or I = 4 9 Imax 1+ 3 cos2 φ 2 © 2000 by Harcourt, Inc. All rights reserved.
CHAPTER 38 38.1 sin θ = λ a = 6.328 × 10– 7 3.00 × 10– 4 = 2.11 × 10– 3 y 1.00 m = tan θ ≈ sin θ = θ (for small θ) 2y = 4.22 mm 38.2 The positions of the first-order minima are y L ≈sin θ = ±λ a. Thus, the spacing between these two minima is ∆y = 2 λ a ( )L and the wavelength is λ = ∆y 2 a L = 4.10 × 10−3 m 2 0.550 × 10−3 m 2.06 m = 547 nm 38.3 y L = sin θ = mλ a ∆y = 3.00 × 10– 3 m ∆m = 3 – 1 = 2 and a = ∆mλ L ∆y a = (2)(690 × 10– 9 m)(0.500 m) 3.00 × 10– 3 m = 2.30 × 10– 4 m 38.4 For destructive interference, sin θ = m λ a = λ a = 5.00 cm 36.0 cm = 0.139 and θ = 7.98° d L = tan θ gives d = L tan θ = (6.50 m) tan 7.98° = 0.912 m d = 91.2 cm Chapter 38 Solutions 407 © 2000 by Harcourt, Inc. All rights reserved.
38.5 If the speed of sound is 340 m/s, λ = v f = 340 m/s 650 /s = 0.523 m Diffraction minima occur at angles described by a sin θ = mλ 1.10 m sin θ 1 = 1(0.523 m) θ 1 = 28.4° 1.10 m sin θ 2 = 2(0.523 m) θ 2 = 72.0° 1.10 m sin θ 3 = 3(0.523 m) θ 3 nonexistent Maxima appear straight ahead at 0° and left and right at an angle given approximately by (1.10 m) sin θ x = 1.5(0.523 m) θ x ≈ 46° There is no solution to a sin θ = 2.5λ, so our answer is already complete, with three sound maxima.
38.6 (a) sin θ λ = = y L m a Therefore, for first minimum, m = 1 and L ay m = = × ( ) × ( ) ( ) × ( ) = − − − λ 7 50 10 8 50 10 1 587 5 10 4 4 9 . m . m . m 1.09 m (b) w = 2y1 yields y1 = 0.850 mm w = 2 0.850 × 10−3 m ( ) = 1.70 mm 38.7 sin θ ≈ = × − y L 4.1 10 m 1.20 m 3 0 β θ λ 2 4 00 10 546 1 10 4 10 10 1 20 7 86 4 9 3 = = × ( ) × × = − − − π π asin . m . m . m . m . rad I Imax sin / / sin .
.
= ( ) = ( ) = β β 2 2 7 86 7 86 2 2 1 62 10 2 .
× − 408 Chapter 38 Solutions 38.8 Bright fringes will be located approximately midway between adjacent dark fringes.
Therefore, for the second bright fringe, let m = 2.5 and use sin θ = mλ a ≈y L .
The wavelength will be λ ≈ay mL = (0.800 × 10−3 m)(1.40 × 10−3 m) 2.5(0.800 m) = 5.60 × 10−7 m = 560 nm 38.9 Equation 38.1 states that sin θ = mλ a , where m = ± 1, ± 2, ± 3, . . . .
The requirement for m = 1 is from an analysis of the extra path distance traveled by ray 1 compared to ray 3 in Figure 38.5. This extra distance must be equal to λ / 2 for destructive interference. When the source rays approach the slit at an angle β , there is a distance added to the path difference (of ray 1 compared to ray 3) of a/ 2 ( )sinβ Then, for destructive interference, a 2 sin β + a 2 sin θ = λ 2 so sin θ = λ a − sin β. Dividing the slit into 4 parts leads to the 2nd order minimum: sin θ = 2λ a − sin β Dividing the slit into 6 parts gives the third order minimum: sin θ = 3λ a − sin β Generalizing, we obtain the condition for the mth order minimum: sin θ = m a λ − sin β 38.10 (a) Double-slit interference maxima are at angles given by d m sin θ λ = . For m = 0, θ0 = 0° For m = ( ) = ( ) 1 2 1 0 5015 , sin . .80 m m µ θ µ : θ1 1 0 179 = ( ) = − sin . 10.3° Similarly, for m = 2, 3, 4, 5 and 6, θ2 = 21.0° , θ3 = 32.5° , θ4 = 45.8° , θ5 = 63.6° , and θ6 = sin−1 1.07 ( ) = nonexistent.
Thus, there are 5 + 5 + 1 = 11 directions for interference maxima .
(b) We check for missing orders by looking for single-slit diffraction minima, at a m sin θ λ = . For m = 1, 0 700 1 0 5015 .
sin . m m µ θ µ ( ) = ( ) and θ1 45 8 = ° .
. Thus, there is no bright fringe at this angle. There are only nine bright fringes , at θ = ° ± ° ± ° ± ° ± ° 0 32 5 , , .
, 10.3 , 21.0 and 63.6 .
Chapter 38 Solutions 409 © 2000 by Harcourt, Inc. All rights reserved.
(c) I = Imax sin π asin θ λ ( ) π asin θ λ 2 At θ = 0°, sin θ θ →1 and I Imax → 1.00 At θ = 10.3°, π asin θ λ = π 0.700 µm ( ) sin 10.3° 0.5015 µm = 0.785 rad = 45.0° I Imax = sin 45.0° 0.785 2 = 0.811 Similarly, at θ = 21.0°, π asin θ λ = 1.57 rad = 90.0° and I Imax = 0.405 At θ = 32.5°, π asin θ λ = 2.36 rad = 135° and I Imax = 0.0901 At θ = 63.6°, π asin θ λ = 3.93 rad = 225° and I Imax = 0.0324 38.11 sin θ = λ a = 5.00 × 10– 7 m 5.00 × 10– 4 = 1.00 × 10– 3 rad 38.12 θ min = y L = 1.22 λ D y = (1.22)(5.00 × 10– 7)(0.0300) 7.00 × 10– 3 = 2.61 µm y = radius of star-image L = length of eye λ = 500 nm D = pupil diameter θ = half angle 38.13 Following Equation 38.9 for diffraction from a circular opening, the beam spreads into a cone of half-angle θ min = 1.22 λ D = 1.22 (632.8 × 10– 9 m) (0.00500 m) = 1.54 × 10– 4 rad The radius of the beam ten kilometers away is, from the definition of radian measure, r beam = θ min (1.00 × 104 m) = 1.544 m and its diameter is d beam = 2r beam = 3.09 m 410 Chapter 38 Solutions Goal Solution A helium-neon laser emits light that has a wavelength of 632.8 nm. The circular aperture through which the beam emerges has a diameter of 0.500 cm. Estimate the diameter of the beam 10.0 km from the laser.
G : A typical laser pointer makes a spot about 5 cm in diameter at 100 m, so the spot size at 10 km would be about 100 times bigger, or about 5 m across. Assuming that this HeNe laser is similar, we could expect a comparable beam diameter.
O : We assume that the light is parallel and not diverging as it passes through and fills the circular aperture. However, as the light passes through the circular aperture, it will spread from diffraction according to Equation 38.9.
A : The beam spreads into a cone of half-angle θmin = 1.22 λ D = 1.22 632.8 × 10−9 m ( ) 0.00500 m ( ) = 1.54 × 10−4 rad The radius of the beam ten kilometers away is, from the definition of radian measure, rbeam = θmin 1.00 × 104 m ( ) = 1.54 m and its diameter is dbeam = 2rbeam = 3.09 m L : The beam is several meters across as expected, and is about 600 times larger than the laser aperture.
Since most HeNe lasers are low power units in the mW range, the beam at this range would be so spread out that it would be too dim to see on a screen. 38.14 θ min = 1.22 λ D = d L 1.22 5.80 × 10– 7 m 4.00 × 10– 3 m = d 1.80 mi 1 mi 1609 m d = 0.512 m The shortening of the wavelength inside the patriot's eye does not change the answer.
38.15 By Rayleigh's criterion, two dots separated center-to-center by 2.00 mm would overlap when θmin = d L = 1.22 λ D Thus, L = dD 1.22λ = 2.00 × 10−3 m ( ) 4.00 × 10−3 m ( ) 1.22 500 × 10−9 nm ( ) = 13.1 m 38.16 D = 1.22 λ θ min = 1.22(5.00 × 10– 7) 1.00 × 10– 5 m = 6.10 cm Chapter 38 Solutions 411 © 2000 by Harcourt, Inc. All rights reserved.
38.17 θmin = 1.22 wavelength pupil diameter (distance between sources) = L Thus, 1.22λ d = w vt , or w = 1.22λ vt ( ) d Taillights are red. Take λ ≈650 nm: w ≈ 1.22 650 × 10−9 m ( ) 20.0 m s ( ) 600 s ( ) 5.00 × 10−3 m = 1.90 m 38.18 θmin = 1.22 wavelength pupil diameter = (distance between sources) L so 1.22λ d = w vt w = 1.22λ vt ( ) d where λ ≈650 nm is the average wavelength radiated by the red taillights.
38.19 1.22λ D = d L λ = c f = 0.0200 m D = 2.10 m L = 9000 m d = 1.22 (0.0200 m)(9000 m) 2.10 m = 105 m 38.20 Apply Rayleigh's criterion, θmin = x D = 1.22 λ d where θmin = half-angle of light cone, x = radius of spot, λ =wavelength of light, d = diameter of telescope, D = distance to Moon.
Then, the diameter of the spot on the Moon is 2x = 2 1.22 λD d = 2 1.22 ( ) 694.3 × 10−9 m ( ) 3.84 × 108 m ( ) 2.70 m = 241 m 38.21 For 0.100° angular resolution, 1.22 3.00 × 10−3 m ( ) D = 0.100° ( ) π 180° D = 2.10 m 38.22 L = 88.6 × 109 m, D = 0.300 m, λ = 590 × 10−9 m (a) 1.22 λ D = θmin = 2 40 10 6 .
× −rad (b) d = θminL = 213 km 412 Chapter 38 Solutions 38.23 d = 1.00 cm 2000 = 1.00 × 10−2 m 2000 = 5.00 µm sin θ = mλ d = 1 640 × 10−9 m ( ) 5.00 × 10−6 m = 0.128 θ = 7.35° 38.24 The principal maxima are defined by d sin θ = mλ m = 0, 1, 2, . . .
For m = 1, λ = d sin θ where θ is the angle between the central (m = 0) and the first order (m = 1) maxima. The value of θ can be determined from the information given about the distance between maxima and the grating-to-screen distance. From the figure, tan θ = 0.488 m 1.72 m = 0.284 so θ = 15.8° and sin θ = 0.273 The distance between grating "slits" equals the reciprocal of the number of grating lines per centimeter d = 1 5310 cm– 1 = 1.88 × 10– 4 cm = 1.88 × 103 nm The wavelength is λ = d sin θ = (1.88 × 103 nm)(0.273) = 514 nm 38.25 The grating spacing is d = (1.00 × 10– 2 m) 4500 = 2.22 × 10– 6 m In the 1st-order spectrum, diffraction angles are given by sin θ = λ d : sin θ 1 = 656 × 10– 9 m 2.22 × 10– 6 m = 0.295 so that for red θ1 = 17.17° and for violet sin θ 2 = 434 × 10– 9 m 2.22 × 10–6 m = 0.195 so that θ 2 = 11.26° Figure for Goal Solution Chapter 38 Solutions 413 © 2000 by Harcourt, Inc. All rights reserved.
The angular separation is in first-order, ∆θ = 17.17° – 11.26° = 5.91° In the second-order spectrum, ∆θ = sin– 1 2λ 1 d – sin– 1 2λ 2 d = 13.2° Again, in the third order, ∆θ = sin– 1 3λ 1 d – sin– 1 3λ 2 d = 26.5° Since the red line does not appear in the fourth-order spectrum, the answer is complete.
Goal Solution The hydrogen spectrum has a red line at 656 nm and a violet line at 434 nm. What is the angular separation between two spectral lines obtained with a diffraction grating that has 4500 lines/cm?
G : Most diffraction gratings yield several spectral orders within the 180° viewing range, which means that the angle between red and violet lines is probably 10° to 30°. O : The angular separation is the difference between the angles corresponding to the red and violet wavelengths for each visible spectral order according to the diffraction grating equation, dsinθ = mλ .
A : The grating spacing is d = 1.00 × 10−2 m ( ) 4500 lines = 2.22 × 10−6 m In the first-order spectrum (m = 1), the angles of diffraction are given by sin θ = λ d: sinθ1r = 656 × 10−9 m 2.22 × 10−6 m = 0.295 so θ1r = 17.17° sinθ1v = 434 × 10−9 m 2.22 × 10−6 m = 0.195 so θ1v = 11.26° The angular separation is ∆θ1 = θ1r -θ1v = 17.17 -11.26 = 5.91 In the 2nd-order ( m = 2) ∆θ2 = sin−1 2λr d −sin−1 2λv d = 13.2° In the third order ( m = 3), ∆θ3 = sin−1 3λr d −sin−1 3λv d = 26.5° In the fourth order, the red line is not visible: θ4r = sin−1 4λr / d ( ) = sin−1 1.18 ( ) does not exist L : The full spectrum is visible in the first 3 orders with this diffraction grating, and the fourth is partially visible. We can also see that the pattern is dispersed more for higher spectral orders so that the angular separation between the red and blue lines increases as m increases. It is also worth noting that the spectral orders can overlap (as is the case for the second and third order spectra above), which makes the pattern look confusing if you do not know what you are looking for.
414 Chapter 38 Solutions 38.26 sin θ = 0.350: d = λ sin θ = 632.8 nm 0.350 = 1.81 × 103 nm Line spacing = 1.81 µm 38.27 (a) d = 1 3660 lines/cm = 2.732 × 10– 4 cm = 2.732 × 10– 6 m = 2732 nm λ = d sin θ m : At θ = 10.09° λ = 478.7 nm At θ = 13.71°, λ = 647.6 nm At θ = 14.77°, λ = 696.6 nm (b) d = λ sin θ 1 and λ = d sin θ 2 so sin θ 2 = 2λ d = 2λ λ sin θ 1 = 2 sin θ 1 Therefore, if θ 1 = 10.09° then sin θ 2 = 2 sin (10.09°) gives θ 2 = 20.51° Similarly, for θ 1 = 13.71°, θ 2 = 28.30° and for θ 1 = 14.77°, θ 2 = 30.66° 38.28 d = 1 800/mm = 1.25 × 10– 6 m The blue light goes off at angles sin θm = mλ d : θ 1 = sin– 1 1 × 5.00 × 10– 7 m 1.25 × 10– 6 m = 23.6° θ 2 = sin– 1 (2 × 0.400) = 53.1° θ 3 = sin– 1 (3 × 0.400) = nonexistent The red end of the spectrum is at θ 1 = sin– 1 1 × 7.00 × 10– 7 m 1.25 × 10– 6 m = 34.1° θ 2 = sin– 1 (2 × 0.560) = nonexistent So only the first-order spectrum is complete, and it does not overlap the second-order spectrum.
Chapter 38 Solutions 415 © 2000 by Harcourt, Inc. All rights reserved.
38.29 (a) From Equation 38.12, R = Nm where N = 3000 lines cm ( ) 4.00 cm ( ) = 1.20 × 104 lines.
In the 1st order, R = (1)(1.20 × 104 lines) = 1 20 104 .
× In the 2nd order, R = (2)(1.20 × 104 lines) = 2 40 104 .
× In the 3rd order, R = (3)(1.20 × 104 lines) = 3 60 104 .
× (b) From Equation 38.11, R = λ ∆λ : In the 3rd order, ∆λ = λ R = 400 nm 3.60 × 104 = 0.0111 nm = 11.1 pm 38.30 sin θ = mλ d Therefore, taking the ends of the visible spectrum to be λ v = 400 nm and λ r = 750 nm, the ends the different order spectra are: End of second order: sin θ2r = 2λ r d = 1500 nm d Start of third order: sin θ3v = 2λ v d = 1200 nm d Thus, it is seen that θ2r > θ3v and these orders must overlap regardless of the value of the grating spacing d.
38.31 (a) Nm = λ ∆λ N(1) = 531.7 nm 0.19 nm = 2800 (b) 1.32 × 10– 2 m 2800 = 4.72 µm 38.32 dsin θ = mλ and, differentiating, d(cos θ)dθ = mdλ or d 1−sin2 θ ∆θ ≈m∆λ d 1−m2λ2 / d2 ∆θ ≈m∆λ so ∆θ ≈ ∆λ d2 /m2 −λ2 416 Chapter 38 Solutions 38.33 d = 1.00 × 10−3 m mm 250 lines mm = 4.00 × 10−6 m = 4000 nm dsin θ = mλ ⇒ m = dsin θ λ (a) The number of times a complete order is seen is the same as the number of orders in which the long wavelength limit is visible. mmax = dsin θmax λ = 4000 nm ( )sin 90.0° 700 nm = 5.71 or 5 orders is the maximum .
(b) The highest order in which the violet end of the spectrum can be seen is: mmax = dsin θmax λ = 4000 nm ( )sin 90.0° 400 nm = 10.0 or 10 orders in the short - wavelength region 38.34 d = = × = − 1 4200 cm 2.38 10 m nm 6 2380 dsin θ = mλ or θ = sin−1 mλ d and y = Ltan θ = Ltan sin−1 mλ d Thus, ∆y = L tan sin−1 mλ 2 d −tan sin−1 mλ 1 d For m = 1, ∆y = 2.00 m ( ) tan sin−1 589.6 2380 −tan sin−1 589 2380 = 0.554 mm For m = 2, ∆y = 2.00 m ( ) tan sin−1 2 589.6 ( ) 2380 −tan sin−1 2 589 ( ) 2380 = 1.54 mm For m = 3, ∆y = 2.00 m ( ) tan sin−1 3 589.6 ( ) 2380 −tan sin−1 3 589 ( ) 2380 = 5.04 mm Thus, the observed order must be m = 2 .
38.35 2d sin θ = mλ: λ = 2d sin θ m = 2(0.353 × 10– 9 m) sin (7.60°) (1) = 9.34 × 10– 11 m = 0.0934 nm 38.36 2d sin θ = mλ ⇒ d = m λ 2 sin θ = (1)(0.129 nm) 2 sin (8.15°) = 0.455 nm Chapter 38 Solutions 417 © 2000 by Harcourt, Inc. All rights reserved.
38.37 2d sin θ = mλ so sin θ = mλ 2d = 1(0.140 × 10– 9 m) 2(0.281 × 10– 9 m) = 0.249 and θ = 14.4° 38.38 sin θm = mλ 2d : sin 12.6° = 1λ 2d = 0.218 sin θ 2 = 2λ 2d = 2(0.218) so θ 2 = 25.9° Three other orders appear: θ 3 = sin–1 (3 × 0.218) = 40.9° θ 4 = sin–1 (4 × 0.218) = 60.8° θ 5 = sin–1 (5 × 0.218) = nonexistent 38.39 2dsin θ = mλ θ = sin−1 mλ 2d = sin−1 2 × 0.166 2 × 0.314 = 31.9° 38.40 Figure 38.25 of the text shows the situation. 2dsin θ = mλ or λ = 2dsin θ m m = ⇒ = ( ) ° = 1 2 2 80 80 0 1 m 1 λ .
sin . 5.51 m m = ⇒ = ( ) ° = 2 2 2 80 80 0 2 m 2 λ .
sin . 2.76 m m = ⇒ = ( ) ° = 3 2 2 80 80 0 3 m 3 λ .
sin . 1.84 m 38.41 The average value of the cosine-squared function is one-half, so the first polarizer transmits 1 2 the light.
The second transmits cos2 30.0° = 3 4 .
If = 1 2 × 3 4 Ii = 3 8 Ii 418 Chapter 38 Solutions 38.42 (a) θ 1 = 20.0°, θ 2 = 40.0°, θ 3 = 60.0° If = Ii cos2(θ 1 – 0°) cos2(θ 2 – θ 1) cos2(θ 3 – θ 2) If = (10.0 units) cos2(20.0°) cos2(20.0°) cos2(20.0°) = 6.89 units (b) θ 1 = 0°, θ 2 = 30.0°, θ 3 = 60.0° If = (10.0 units) cos2(0°) cos2(30.0°) cos2(30.0°) = 5.63 units 38.43 I = Imax cos2 θ ⇒ θ = cos– 1 I Imax 1/2 (a) I Imax = 1 3.00 ⇒ θ = cos– 1 1 3.00 1/2 = 54.7° (b) I Imax = 1 5.00 ⇒ θ = cos– 1 1 5.00 1/2 = 63.4° (c) I Imax = 1 10.0 ⇒ θ = cos– 1 1 10.0 1/2 = 71.6° 38.44 By Brewster's law, n p = tan = tan(48.0 ) = θ ° 1.11 38.45 sin θc n = 1 or n c = = ° = 1 1 34 4 1 77 sin sin .
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θ Also, tan θp = n. Thus, θp = tan−1 n ( ) = tan−1 1.77 ( ) = 60.5° 38.46 sin θc = 1 n and tan θp = n Thus, sin θc = 1 tan θp or cot sin θ θ p c = Chapter 38 Solutions 419 © 2000 by Harcourt, Inc. All rights reserved.
38.47 Complete polarization occurs at Brewster's angle tan θp = 1.33 θp = 53.1° Thus, the Moon is 36.9° above the horizon.
38.48 For incident unpolarized light of intensity Imax: After transmitting 1st disk: I = 1 2 Imax After transmitting 2nd disk: I = 1 2 Imax cos2 θ After transmitting 3rd disk: I = 1 2 Imax cos2 θ cos2 90°−θ ( ) where the angle between the first and second disk is θ = ω t.
Using trigonometric identities cos2 θ = 1 2 (1+ cos 2θ) and cos2 90 −θ ( ) = sin2 θ = 1 2 1−cos 2θ ( ) we have I = 1 2 Imax (1+ cos 2θ) 2 (1−cos 2θ) 2 = 1 8 Imax(1−cos2 2θ) = 1 8 Imax 1 2 (1−cos 4θ) Since θ = ω t, the intensity of the emerging beam is given by I = 1 16 Imax(1−cos 4ω t) 38.49 Let the first sheet have its axis at angle θ to the original plane of polarization, and let each further sheet have its axis turned by the same angle. The first sheet passes intensity I max cos2 θ.
The second sheet passes I max cos4 θ, nd the nth sheet lets through I max cos2n θ ≥ 0.90I max where θ = 45° n Try different integers to find cos2 × 5 45° 5 = 0.885, cos2 × 6 45° 6 = 0.902, (a) So n = 6 (b) θ = 7.50° 420 Chapter 38 Solutions 38.50 Consider vocal sound moving at 340 m s and of frequency 3000 Hz. Its wavelength is λ = v f = 340 m s 3000 Hz = 0.113 m If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then a m sin θ λ = predicts no diffraction minima. You are a nearly isotropic source of this sound. It spreads out from you nearly equally in all directions. On the other hand, if you use a megaphone with width 60.0 cm at its wide end, then a m sin θ λ = predicts the first diffraction minimum at θ = sin−1 mλ a = sin−1 0.113 m 0.600 m = 10.9° This suggests that the sound is radiated mostly toward the front into a diverging beam of angular diameter only about 20°. With less sound energy wasted in other directions, more is available for your intended auditors. We could check that a distant observer to the side or behind you receives less sound when a megaphone is used.
38.51 The first minimum is at a sin θ = 1λ. This has no solution if λ a > 1 or if a < λ = 632.8 nm 38.52 x = 1.22 λ d D = 1.22 5.00 × 10−7 m 5.00 × 10−3 m 250 × 103 m ( ) = 30.5 m D = 250 × 103 m λ = 5.00 × 10– 7 m d = 5.00 × 10– 3 m 38.53 d = 1 400/mm = 2.50 × 10– 6 m (a) d sin θ = mλ θa = sin– 1 2 × 541 × 10– 9 m 2.50 × 10– 6 m = 25.6° (b) λ = 541 × 10– 9 m 1.33 = 4.07 × 10– 7 m θb = sin– 1 2 × 4.07 × 10– 7 m 2.50 × 10– 6 m = 19.0° (c) d sin θa = 2λ d sin θb = 2λ n n sin θb = 1 sin θa Chapter 38 Solutions 421 © 2000 by Harcourt, Inc. All rights reserved.
38.54 (a) λ = v f = 3.00 × 108 m s 1.40 × 109 s−1 = 0.214 m θmin = 1.22 λ D = 1.22 0.214 m 3.60 × 104 m = 7 26 . rad µ = 7.26 µ rad 180 × 60 × 60 s π = 1.50 arc seconds (b) θmin = d L : d = θminL = 7.26 × 10−6 rad ( ) 26 000 ly ( ) = 0.189 ly (c) θ λ min .
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= = × × = − − 1 22 1 22 500 10 12 0 10 9 3 D m m 50 8 . rad µ 10.5 seconds of arc ( ) (d) d = θminL = 50.8 × 10−6 rad ( ) 30.0 m ( ) = 1.52 × 10−3 m = 1.52 mm 38.55 θ λ min 2.00 m 10.0 m = = ( ) ( ) = 1 22 1 22 .
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D 0.244 rad = 14.0° 38.56 With a grazing angle of 36.0°, the angle of incidence is 54.0° tan θp = n = tan 54.0° = 1.38 In the liquid, λn = λ /n = 750 nm/1.38 = 545 nm 38.57 (a) dsin θ = mλ , or d = mλ sin θ = 3 500 × 10−9 m ( ) sin 32.0° = 2.83 µm Therefore, lines per unit length = 1 m d = × − 1 2 83 10 6 .
or lines per unit length = × = 3 53 105 .
m 3 53 103 .
× cm .
(b) sin θ = mλ d = m 500 × 10−9 m ( ) 2.83 × 10−6 m = m 0.177 ( ) For sin .
θ ≤1 00, we must have m 0.177 ( ) ≤1.00 or m ≤5.65 Therefore, the highest order observed is m = 5 Total number primary maxima observed is 2m + 1 = 11 422 Chapter 38 Solutions Goal Solution Light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at 32.0°, (a) what is the number of rulings per centimeter for the grating? (b) Determine the total number of primary maxima that can be observed in this situation.
G : The diffraction pattern described in this problem seems to be similar to previous problems that have diffraction gratings with 2 000 to 5 000 lines/mm. With the third-order maximum at 32°, there are probably 5 or 6 maxima on each side of the central bright fringe, for a total of 11 or 13 primary maxima.
O : The diffraction grating equation can be used to find the grating spacing and the angles of the other maxima that should be visible within the 180° viewing range.
A : (a) Use Equation 38.10, d m sinθ λ = d = mλ sinθ = (3)(5.00 × 10−7 m) sin(32.0°) = 2.83 × 10−6 m Thus, the grating gauge is 1 d = 3.534 × 105 lines/m = 3530 lines/cm ◊ (b) sinθ = m λ d = m(5.00 × 10−7 m) 2.83 × 10-6 m = m(0.177) For sin θ ≤1, we require that m 1.77 ( ) ≤1 or m ≤5.65. Since m must be an integer, its maximum value is really 5. Therefore, the total number of maxima is 2m + 1 = 11 L : The results agree with our predictions, and apparently there are 5 maxima on either side of the central maximum. If more maxima were desired, a grating with fewer lines/cm would be required; however, this would reduce the ability to resolve the difference between lines that appear close together.
38.58 For the air-to-water interface, tan θp = nwater nair = 1.33 1.00 θp = 53.1° and 1.00 sin sin 2 ( ) = ( ) θ θ p 1 33 . θ2 1 53 1 1 33 36 9 = ° = ° − sin sin .
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θ3 θ2 For the water-to-glass interface, tan θp = tan θ3 = nglass nwater = 1.50 1.33 so θ3 48 4 = ° .
The angle between surfaces is θ θ θ = − = 3 2 11.5° Chapter 38 Solutions 423 © 2000 by Harcourt, Inc. All rights reserved.
38.59 The limiting resolution between lines θ λ min .
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= = × ( ) × ( ) = × − − − 1 22 1 22 1 34 10 4 D 550 10 m 5 00 10 m rad 9 3 Assuming a picture screen with vertical dimension l, the minimum viewing distance for no visible lines is found from θmin = l 485 ( ) L. The desired ratio is then L l = 1 485θmin = 1 485 1.34 × 10−4 rad ( ) = 15.4 38.60 (a) Applying Snell's law gives n2 sin φ = n1sin θ . From the sketch, we also see that: θ + φ +β = π , or φ = π -(θ +β) Using the given identity: sin φ = sin π cos(θ + β) −cos π sin(θ + β), which reduces to: sinφ = sin(θ + β). Applying the identity again: sin φ = sin θ cos β + cos θ sin β Snell's law then becomes: n2 sin θ cos β + cos θ sin β ( ) = n1sin θ or (after dividing by cos θ ): n2(tan θ cos β + sin β) = n1tan θ .
Solving for tan θ gives: tan θ = n2 sin β n1 −n2 cos β (b) If β = 90.0°, n1 = 1.00, and n2 = n, the above result becomes: tan θ = n(1.00) 1.00 −0 , or n = tan θ , which is Brewster's law.
38.61 (a) From Equation 38.1, θ = sin−1 mλ a In this case m = 1 and λ = = × × = × − c f 3.00 10 m/s 7.50 10 Hz 4.00 10 m 8 9 2 Thus, θ = × × = − − − sin .
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1 2 2 4 00 10 6 00 10 m m 41.8° (b) From Equation 38.4, I Imax = sin β 2 ( ) β 2 2 where β = 2π asin θ λ 424 Chapter 38 Solutions When θ = 15.0°, β = 2π 0.0600 m ( )sin 15.0° 0.0400 m = 2.44 rad and I Imax = sin 1.22 rad ( ) 1.22 rad 2 = 0.593 (c) sin θ = λ a so θ = 41.8°: This is the minimum angle subtended by the two sources at the slit. Let α be the half angle between the sources, each a distance l= 0.100 m from the center line and a distance L from the slit plane. Then, L = l cot α = 0.100 m ( )cot 41.8 2 ( )= 0.262 m 38.62 I Imax = 1 2 (cos2 45.0°)(cos2 45.0°) = 1 8 38.63 (a) The E and O rays, in phase at the surface of the plate, will have a phase difference θ = 2π λ ( )δ after traveling distance d through the plate. Here δ is the difference in the optical path lengths of these rays. The optical path length between two points is the product of the actual path length d and the index of refraction. Therefore, δ = dnO −dnE The absolute value is used since nO nE may be more or less than unity. Therefore, θ = 2π λ dnO −dnE = 2π λ d nO −nE (b) d = λ θ 2π nO −nE = 550 × 10−9 m ( ) π 2 ( ) 2π 1.544 −1.553 = 1.53 × 10−5 m = 15.3 µm 38.64 For a diffraction grating, the locations of the principal maxima for wavelength λ are given by sin θ = mλ d ≈y L. The grating spacing may be expressed as d = a N where a is the width of the grating and N is the number of slits. Thus, the screen locations of the maxima become Chapter 38 Solutions 425 © 2000 by Harcourt, Inc. All rights reserved. y = NLmλ / a. If two nearly equal wavelengths are present, the difference in the screen locations of corresponding maxima is ∆y = NLm ∆λ ( ) a For a single slit of width a, the location of the first diffraction minimum is sinθ = λ a ≈y L, or y = L/ a ( )λ . If the two wavelengths are to be just resolved by Rayleigh’s criterion, y = ∆y from above. Therefore, L a λ = NLm ∆λ ( ) a or the resolving power of the grating is R” λ ∆λ = Nm .
38.65 The first minimum in the single-slit diffraction pattern occurs at sin θ = λ a » ymin L Thus, the slit width is given by a = λL ymin For a minimum located at ymin .
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= ± 6 36 0 08 mm mm, the width is a = 632.8·10-9 m ( )1.00 m ( ) 6.36·10-3 m = 99.5 µm ± 1% 38.66 (a) From Equation 38.4, I Imax = sin β 2 ( ) β 2 ( ) 2 If we define φ ≡β 2 this becomes I Imax sin = φ φ 2 Therefore, when I Imax = 1 2 we must have sin φ φ = 1 2 , or sin φ φ = 2 426 Chapter 38 Solutions (b) Let y1 = sin φ and y2 = φ 2 . A plot of y y 1 2 and in the range 1.00 ≤φ ≤π 2 is shown to the right.
The solution to the transcendental equation is found to be φ = 1.39 rad . (c) β = 2π asin θ λ = 2φ gives sin θ = φ π Ł ł λ a = 0.443 λ a .
If λ a is small, then θ ≈0.443 λ a . This gives the half-width, measured away from the maximum at θ = 0. The pattern is symmetric, so the full width is given by ∆θ = 0.443 λ a −−0.443 λ a = 0.886λ a 38.67 φ 2 sin φ 1 1.19 bigger than φ 2 1.29 smaller than φ 1.5 1.41 smaller 1.4 1.394 1.39 1.391 bigger 1.395 1.392 1.392 1.3917 smaller 1.3915 1.39154 bigger 1.39152 1.39155 bigger 1.3916 1.391568 smaller 1.39158 1.391563 1.39157 1.391560 1.39156 1.391558 1.391559 1.3915578 1.391558 1.3915575 1.391557 1.3915573 1.3915574 1.3915574 We get the answer to seven digits after 17 steps. Clever guessing, like using the value of 2 sin φ as the next guess for φ, could reduce this to around 13 steps.
Chapter 38 Solutions 427 © 2000 by Harcourt, Inc. All rights reserved.
38.68 In I = Imax sin β 2 ( ) β 2 ( ) Ø º Œ ø ß œ 2 find dI dβ = Imax 2sin β 2 ( ) β 2 ( ) β 2 ( )cos β 2 ( ) 1 2 ( ) −sin β 2 ( ) 1 2 ( ) β 2 ( ) 2 and require that it be zero. The possibility sin β 2 ( ) = 0 locates all of the minima and the central maximum, according to β 2 = 0, π, 2π, . . . ; β = 2π asin θ λ = 0, 2π, 4π, . . . ; asin θ = 0, λ , 2λ , . . . .
The side maxima are found from β 2 cos β 2 Ł ł- sin β 2 Ł ł= 0, or tan β 2 Ł ł= β 2 . This has solutions β 2 = 4.4934 , β 2 = 7.7253 , and others, giving (a) π asin θ = 4.4934λ asin θ = 1.4303λ (b) π asin θ = 7.7253λ asin θ = 2.4590λ 38.69 (a) We require θmin = 1.22 λ D = radius of diffraction disk L = D 2L.
Then D2 = 2.44λ L (b) D = × ( )( ) = − 2 44 500 10 0 150 9 .
. m m 428 µm © 2000 by Harcourt, Inc. All rights reserved.
Chapter 39 Solutions 39.1 In the rest frame, pi = m1v1i + m2v2i = (2000 kg)(20.0 m/s) + (1500 kg)(0 m/s) = 4.00 × 104 kg · m/s pf = (m1 + m2)vf = (2000 kg + 1500 kg)vf Since pi = pf, vf = 4.00 × 104 kg · m/s 2000 kg + 1500 kg = 11.429 m/s In the moving frame, these velocities are all reduced by +10.0 m/s. ′ v1i = v1i −′ v = 20.0 m/s – (+10.0 m/s) = 10.0 m/s ′ v2i = v2i −′ v = 0 m/s – (+10.0 m/s) = –10.0 m/s ′ vf = 11.429 m/s – (+10.0 m/s) = 1.429 m/s Our initial momentum is then ′ pi = m1 ′ v1i + m2 ′ v2i = (2000 kg)(10.0 m/s) + (1500 kg)(–10.0 m/s) = 5000 kg · m/s and our final momentum is ′ pf = (2000 kg + 1500 kg) ′ vf = (3500 kg)(1.429 m/s) = 5000 kg · m/s 39.2 (a) v = vT + vB = 60.0 m/s (b) v = vT – vB = 20.0 m/s (c) v = vT 2 + vB 2 = 202 + 402 = 44.7 m/s 39.3 The first observer watches some object accelerate under applied forces. Call the instantaneous velocity of the object v1. The second observer has constant velocity v21 relative to the first, and measures the object to have velocity v2 = v1 −v21.
The second observer measures an acceleration of a2 = dv2 dt = dv1 dt This is the same as that measured by the first observer. In this nonrelativistic case, they measure the same forces as well. Thus, the second observer also confirms that ΣF = ma.
2 Chapter 39 Solutions 39.4 The laboratory observer notes Newton's second law to hold: F1 = ma1 (where the subscript 1 implies the measurement was made in the laboratory frame of reference). The observer in the accelerating frame measures the acceleration of the mass as a2 = a1 – ′ a (where the subscript 2 implies the measurement was made in the accelerating frame of reference, and the primed acceleration term is the acceleration of the accelerated frame with respect to the laboratory frame of reference). If Newton's second law held for the accelerating frame, that observer would then find valid the relation F2 = ma2 or F1 = ma2 (since F1 = F2 and the mass is unchanged in each). But, instead, the accelerating frame observer will find that F2 = ma2 – m ′ a which is not Newton's second law.
39.5 L = Lp 1−v2 c2 ⇒v = c 1−L Lp ( ) 2 Taking L = Lp / 2 where Lp = 1.00 m gives v = c 1− Lp 2 Lp 2 = c 1−1 4 = 0.866 c 39.6 ∆t = ∆tp 1−v c ( ) 2 [ ] 1 2 so v = c 1− ∆tp ∆t 2 1 2 For ∆t = 2∆tp ⇒ v = c 1− ∆tp 2∆tp 2 1/2 = c 1−1 4 1/2 = 0.866 c 39.7 (a) γ = 1 1−v c ( ) 2 = 1 1−0.500 ( )2 = 2 3 The time interval between pulses as measured by the Earth observer is ∆t = γ ∆tp = 2 3 60.0 s 75.0 = 0.924 s Thus, the Earth observer records a pulse rate of 60 0 0 924 .
. s min s = 64.9/min (b) At a relative speed v = 0.990c, the relativistic factor γ increases to 7 09 . and the pulse rate recorded by the Earth observer decreases to 10 6 .
min . That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is much longer as measured by an Earth clock than by a clock aboard the space vehicle.
Chapter 39 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
39.8 The observed length of an object moving at speed v is L = Lp 1−v2 /c2 with Lp as the proper length. For the two ships, we know L2 = L1, L2p = 3L1p, and v1 = 0.350c Thus, L 2 2 = L1 2 and 9L1p 2 1−v2 2 c2 = L1p 2 1−0.350 ( )2 [ ] giving 9 −9 v 2 2 c2 = 0.878, or v2 = 0 950 .
c 39.9 ∆t = γ ∆tp = ∆tp 1−v2 /c2 so ∆tp = 1−v2 /c2 ∆t ≈1−v2 2c2 ∆t and ∆t −∆tp = v2 2c2 ∆t If v = 1000 km h = 1.00 × 106 m 3600 s = 277.8 m s, then v c = 9.26 × 10−7 and ∆t −∆tp ( ) = (4.28 × 10−13)(3600 s) = 1.54 × 10−9 s = 1.54 ns 39.10 γ −1 = 1−v2 c2 = 1−0.950 ( )2 = 0.312 (a) astronauts' time: ∆tp = γ −1 ∆t = 0.312 ( ) 4.42 yr ( ) = 1.38 yr (b) astronauts' distance: L = γ −1 ∆Lp = 0.312 ( ) 4.20 ly ( ) = 1.31 ly 39.11 The spaceship appears length-contracted to the Earth observer as given by L = Lp 1−v2 c2 or L2 = Lp 2 1−v2 c2 ( ) Also, the contracted length is related to the time required to pass overhead by: L = vt or L2 = v2t2 = v2 c2 ct ( ) 2 Equating these two expressions gives Lp 2 −Lp 2 v2 c2 = (ct)2 v2 c2 or Lp 2 + (ct)2 [ ] v2 c2 = Lp 2 Using the given values: Lp = 300 m and t = 7.50 × 10– 7 s this becomes (1.41 × 105 m2) v2 c 2 = 9.00 × 104 m2 giving v = 0.800 c 4 Chapter 39 Solutions Goal Solution A spaceship with a proper length of 300 m takes 0.750 µs seconds to pass an Earth observer. Determine its speed as measured by the Earth observer.
G : We should first determine if the spaceship is traveling at a relativistic speed: classically, v = (300m)/(0.750 µs) = 4.00 × 108 m/s, which is faster than the speed of light (impossible)! Quite clearly, the relativistic correction must be used to find the correct speed of the spaceship, which we can guess will be close to the speed of light. O : We can use the contracted length equation to find the speed of the spaceship in terms of the proper length and the time. The time of 0.750 µs is the proper time measured by the Earth observer, because it is the time interval between two events that she sees as happening at the same point in space. The two events are the passage of the front end of the spaceship over her stopwatch, and the passage of the back end of the ship.
A : L = Lp / γ , with L = v∆t: v∆t = Lp 1−v2 /c2 ( ) 1/2 Squaring both sides, v2∆t2 = Lp 2 1−v2 /c2 ( ) v2c2 = Lp 2c2 / ∆t2 −v2Lp 2 / ∆t2 Solving for the velocity, v = cLp / ∆t c2 + Lp 2 / ∆t2 So v = 3.00 × 108 ( ) 300 m ( ) 0.750 × 10−6 s ( ) 3.00 × 108 ( ) 2 + 300 m ( )2 0.750 × 10−6 s ( ) 2 = 2.40 × 108 m /s L : The spaceship is traveling at 0.8c. We can also verify that the general equation for the speed reduces to the classical relation v = Lp / ∆t when the time is relatively large.
39.12 The spaceship appears to be of length L to Earth observers, where L = Lp 1−v2 c2 1/2 and L = vt vt = Lp 1−v2 c2 1/2 so v2t2 = Lp 2 1−v2 c2 Solving for v, v2 t2 + Lp 2 c2 = Lp 2 v c = Lp c2t2 + Lp2 −1/2 Chapter 39 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
39.13 For v c = 0.990, γ = 7.09 (a) The muon’s lifetime as measured in the Earth’s rest frame is ∆t = 4.60 km 0.990c and the lifetime measured in the muon’s rest frame is ∆tp = ∆t γ = 1 7.09 4.60 × 103 m 0.990 3.00 × 108 m s ( ) = 2.18 µs (b) L = Lp 1−v c ( ) 2 = Lp γ = 4.60 × 103 m 7.09 = 649 m 39.14 We find Carpenter's speed: GMm r2 = mv2 r v = GM (R + h) 1/2 = (6.67 × 10−11)(5.98 × 1024) (6.37 × 106 + 0.160 × 106) 1/2 = 7.82 km/s Then the time period of one orbit, T = 2π(R + h) v = 2π(6.53 × 106) 7.82 × 103 = 5.25 × 103 s (a) The time difference for 22 orbits is ∆t −∆tp = (γ −1)∆tp = 1−v2 c2 ( ) −1/2 −1 22T ( ) ∆t −∆tp ≈1+ 1 2 v2 c2 −1 22T ( ) = 1 2 7.82 × 103 m /s 3 × 108 m /s 2 22 5.25 × 103 s ( )= 39.2 µs (b) For one orbit, ∆t −∆tp = 39.2 µs 22 = 1.78 µs.
The press report is accurate to one digit .
39.15 For pion to travel 10.0 m in ∆t in our frame, 10.0 m = v∆t = v(γ ∆tp) = v(26.0 × 10−9 s) 1−v 2 /c2 Solving for the velocity, (3.85 × 108 m /s)2(1−v2 /c2) = v 2 1.48 × 1017 m2 /s2 = v2(1+1.64) v = 2.37 × 108 m /s = 0.789 c 39.16 γ = 1 1−v2 c2 = 1.01 so v = 0.140 c 6 Chapter 39 Solutions 39.17 (a) Since your ship is identical to his, and you are at rest with respect to your own ship, its length is 20.0 m .
(b) His ship is in motion relative to you, so you see its length contracted to 19.0 m .
(c) We have L = Lp 1−v2 c2 from which L Lp = 19.0 m 20.0 m = 0.950 = 1−v2 c2 and v = 0.312 c 39.18 (a) ∆t = γ ∆tp = ∆tp 1−v c ( ) 2 = 15.0 yr 1−0.700 ( )2 = 21.0 yr (b) d = v ∆t ( ) = 0.700c [ ] 21.0 yr ( ) = 0.700 ( ) 1.00 ly yr ( ) [ ] 21.0 yr ( ) = 14.7 ly (c) The astronauts see Earth flying out the back window at 0 700 .
c: d = v ∆tp ( ) = 0.700c [ ] 15.0 yr ( ) = 0.700 ( ) 1.00 ly yr ( ) [ ] 15.0 yr ( ) = 10.5 ly (d) Mission control gets signals for 21.0 yr while the battery is operating, and then for 14.7 years after the battery stops powering the transmitter, 14.7 ly away: 21.0 yr + 14.7 yr = 35.7 yr 39.19 The orbital speed of the Earth is as described by ΣF = ma: GmSmE r2 = mEv2 r v = GmS r = 6.67 × 10−11 N ⋅m2 kg2 ( ) 1.99 × 1030 kg ( ) 1.496 × 1011 m = 2.98 × 104 m s The maximum frequency received by the extraterrestrials is fobs = fsource 1+ v c 1−v c = 57.0 × 106 Hz ( ) 1+ 2.98 × 104 m s ( ) 3.00 × 108 m s ( ) 1−2.98 × 104 m s ( ) 3.00 × 108 m s ( ) = 57.005 66 × 106 Hz The minimum frequency received is fobs = fsource 1−v c 1+ v c = 57.0 × 106 Hz ( ) 1−2.98 × 104 m s ( ) 3.00 × 108 m s ( ) 1+ 2.98 × 104 m s ( ) 3.00 × 108 m s ( ) = 56.994 34 × 106 Hz The difference, which lets them figure out the speed of our planet, is 57 005 66 56 994 34 106 .
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− ( ) × = Hz 1 13 104 .
× Hz Chapter 39 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
39.20 (a) Let fc be the frequency as seen by the car. Thus, fc = fsource c + v c −v and, if f is the frequency of the reflected wave, f = fc c + v c −v Combining gives f = fsource (c + v) (c −v) (b) Using the above result, f c −v ( ) = fsource(c + v) which gives (f −fsource)c = (f + fsource)v ≈2fsourcev The beat frequency is then fb = f −fsource = 2fsourcev c = 2v λ (c) fb = (2)(30.0 m s)(10.0 × 109 Hz) 3.00 × 108 m s = (2)(30.0 m s) (0.0300 m) = 2000 Hz = 2.00 kHz λ = c fsource = 3.00 × 108 m s 10.0 × 109 Hz = 3.00 cm (d) v = fb λ 2 so ∆v = ∆fb λ 2 = 5 Hz ( )(0.0300 m) 2 = 0.0750 m s ≈0.2 mi/h 39.21 (a) When the source moves away from an observer, the observed frequency is fobs = fsource c −vs c + vs 1/2 where vs = vsource When vs << c, the binomial expansion gives c −vs c + vs 1 2 = 1−vs c 1 2 1+ vs c −1 2 ≈1−vs 2c 1−vs 2c ≈1−vs c So, fobs ≈fsource 1−vs c The observed wavelength is found from c = λobs fobs = λ fsource: λobs = λ fsource fobs ≈ λ fsource fsource 1−vs c ( ) = λ 1−vs c ∆λ = λobs −λ = λ 1 1−vs c −1 = λ 1 1−vs c −1 = λ vs c 1−vs c Since 1−vs c ≈1, ∆λ λ ≈vsource c (b) vsource = c ∆λ λ = c 20.0 nm 397 nm = 0.050 4 c 8 Chapter 39 Solutions 39.22 ′ ux = ux −v 1−uxv/c2 = 0.950c −0.750c 1−0.950 × 0.750 = 0.696 c 39.23 ′ ux = ux −v 1−uxv c2 = −0.750c −0.750c 1−(−0.750)(0.750) = – 0.960 c 39.24 γ = 10.0 We are also given: L1 = 2.00 m, and θ1 = 30.0° (both measured in a reference frame moving relative to the rod).
Thus, L1x = L1 cos θ1 = (2.00 m)(0.867) = 1.73 m and L1y = L1 sin θ1 = (2.00 m)(0.500) = 1.00 m L2x = a "proper length" is related to L1x by L1x = L2x γ Therefore, L2x = 10.0L1x = 17.3 m and L2y = L1y = 1.00 m (Lengths perpendicular to the motion are unchanged).
(a) L2 = (L2x)2 + (L2y)2 gives L2 = 17.4 m (b) θ2 = tan−1 L2y L2x gives θ 2 = 3.30° 39.25 ux = Enterprise velocity v = Klingon velocity From Equation 39.16, ′ ux = ux −v 1−ux v c2 = 0.900c −0.800c 1−0.900 ( ) 0.800 ( ) = 0.357c Chapter 39 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
39.26 (a) From Equation 39.13, ∆′ x = γ ∆x −v∆t ( ), 0 = γ 2.00 m −v 8.00 × 10−9 s ( ) [ ] v = 2.00 m 8.00 × 10−9 s = 2.50 × 108 m s γ = 1 1−2.50 × 108 m s ( ) 2 3.00 × 108 m s ( ) 2 = 1.81 (b) From Equation 39.11, ′ x = γ x −vt ( ) = 1.81 3.00 m −2.50 × 108 m s ( ) 1.00 × 10−9 s ( ) [ ] = 4.97 m (c) ′ t = γ t −v c2 x = 1.81 1.00 × 10−9 s − 2.50 × 108 m s ( ) 3.00 × 108 m s ( ) 2 3.00 m ( ) ′ t = −1.33 × 10−8 s 39.27 p = γ mu (a) For an electron moving at 0.0100 c, γ = 1 1−u c ( ) 2 = 1 1−(0.0100)2 = 1.00005 ≈1.00 Thus, p = 1.00 9.11× 10−31 kg ( ) 0.0100 ( ) 3.00 × 108 m /s ( ) = 2.73 × 10−24 kg ⋅m s (b) Following the same steps as used in part (a), we find at 0.500 c γ = 1.15 and p = 1.58 × 10−22 kg ⋅m s (c) At 0.900 c, γ = 2.29 and p = 5.64 × 10−22 kg ⋅m s 39.28 Using the relativistic form, p = mu 1−u c ( ) 2 = γ mu, we find the difference ∆p from the classical momentum, mu: ∆p = γ mu −mu = (γ −1)mu (a) The difference is 1.00% when (γ −1)mu = 0.0100γ mu: γ = 1 0.990 = 1 1−u c ( ) 2 ⇒ 1−u c ( ) 2 = 0.990 ( )2 or u = 0.141 c (b) The difference is 10.0% when (γ −1)mu = 0.100γ mu: γ = 1 0.900 = 1 1−u c ( ) 2 ⇒ 1−u c ( ) 2 = 0.900 ( )2 or u = 0.436 c 10 Chapter 39 Solutions 39.29 p −mu mu = γ mu −mu mu = γ −1 γ −1 = 1 1−u c ( ) 2 −1 ≈1+ 1 2 u c 2 −1 = 1 2 u c 2 p −mu mu = 1 2 90.0 m s 3.00 × 108 m s 2 = 4.50 × 10−14 39.30 p = mu 1−u c ( ) 2 becomes 1−u2 c2 = m2u2 p2 which gives: 1 = u2 m2 p2 + 1 c2 or c2 = u2 m2c2 p2 + 1 and u = c m2c2 p2 + 1 39.31 Relativistic momentum must be conserved: For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter, p2 = p1 or γ 2m2u2 = γ 1m1u1 = 2.50 × 10−28 kg 1−(0.893)2 × (0.893c) or (1.67 × 10−27 kg)u2 1−u2 c ( ) 2 = (4.960 × 10−28 kg)c and u2 = 0.285 c Chapter 39 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution An unstable particle at rest breaks into two fragments of unequal mass. The rest mass of the lighter fragment is 2.50 × 10−28 kg, and that of the heavier fragment is 1.67 × 10−27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment? G : The heavier fragment should have a speed less than that of the lighter piece since the momentum of the system must be conserved. However, due to the relativistic factor, the ratio of the speeds will not equal the simple ratio of the particle masses, which would give a speed of 0.134c for the heavier particle. O : Relativistic momentum of the system must be conserved. For the total momentum to be zero after the fission, as it was before, p1 + p2 = 0, where we will refer to the lighter particle with the subscript '1', and to the heavier particle with the subscript '2.' A : γ 2m2v2 + γ 1m1v1 = 0 so γ 2m2v2 + 2.50 × 10−28 kg 1- 0.8932 0.893c ( ) = 0 Rearranging, 1.67 × 10−27 kg 1−v2 2 c2 v2 c = −4.96 × 10−28 kg Squaring both sides, 2.79 × 10−54 ( ) v2 c 2 = 2.46 × 10−55 ( ) 1−v2 2 c2 and v2 = −0.285c We choose the negative sign only to mean that the two particles must move in opposite directions.
The speed, then, is v2 = 0.285c L : The speed of the heavier particle is less than the lighter particle, as expected. We can also see that for this situation, the relativistic speed of the heavier particle is about twice as great as was predicted by a simple non-relativistic calculation.
39.32 ∆E = (γ 1 −γ 2)mc2.
For an electron, mc2 = 0.511 MeV.
(a) ∆E = 1 (1−0.810) − 1 (1−0.250) mc2 = 0.582 MeV (b) ∆E = 1 1−(0.990)2 − 1 1−0.810 mc2 = 2.45 MeV 39.33 E = γ mc2 = 2mc2, or γ = 2 Thus, u c = 1−1 γ ( ) 2 = 3 2 , or u = c 3 2 .
The momentum is then p = γ mu = 2m c 3 2 = mc2 c 3 = 938.3 MeV c 3 = 1.63 × 103 MeV c 12 Chapter 39 Solutions 39.34 The relativistic kinetic energy of an object of mass m and speed u is Kr = 1 1−u2 /c2 −1 mc2 For u = 0.100 c, Kr = 1 1−0.0100 −1 mc2 = 0.005038mc2 The classical equation Kc = 1 2 mu2 gives Kc = 1 2 m(0.100c)2 = 0.005000mc2 different by 0.005038 – 0.005000 0.005038 = 0.751% For still smaller speeds the agreement will be still better.
39.35 (a) ER = mc2 = (1.67 × 10−27kg)(2.998 × 108 m /s)2 = 1.50 × 10−10 J = 938 MeV (b) E = γ mc2 = 1.50 × 10−10 J [1−(0.95c/c)2]1/2 = 4.81× 10−10 J = 3.00 × 103 MeV (c) K = E −mc2 = 4.81× 10−10 J −1.50 × 10−10 J = 3.31× 10−10 J = 2 07 103 .
× MeV 39.36 (a) KE = E – ER = 5ER E = 6ER = 6(9.11× 10−31 kg)(3.00 × 108 m /s)2 = 4.92 × 10−13 J = 3.07 MeV (b) E = γ mc2 = γ ER Thus, γ = E Er = 6 = 1 1−u2 c2 which yields u = 0.986 c 39.37 The relativistic density is ER c2 V = mc2 c2 V = m V = m Lp ( ) Lp ( ) Lp 1−u c ( ) 2 = 8.00 g 1.00 cm ( )3 1−0.900 ( )2 = 18.4 g/cm3 Chapter 39 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
39.38 We must conserve both mass-energy and relativistic momentum. With subscript 1 referring to the 0.868c particle and subscript 2 to the 0.987c particle, γ 1 = 1 1−0.868 ( )2 = 2.01 and γ 2 = 1 1−0.987 ( )2 = 6.22 Conservation of mass-energy gives E1 + E2 = Etotal which is γ 1m1c2 + γ 2m2c2 = mtotalc2 or 2.01m1 + 6.22m2 = 3.34 × 10−27 kg This reduces to: m1 + 3.09m2 = 1.66 × 10−27 kg Since the momentum after must equal zero, p1 = p2 gives γ 1m1u1 = γ 2m2u2 or (2.01)(0.868c)m1 = (6.22)(0.987c)m2 which becomes m1 = 3.52m2 Solving and simultaneously, m1 = 8.84 × 10−28 kg and m2 = 2.51× 10−28 kg 39.39 E = γ mc2, p = γ mu; E2 = (γ mc2)2; p2 = (γ mu)2; E2 −p2c2 = (γ mc2)2 −(γ mu)2c2 = γ 2 (mc2)2 −(mc)2u2 = (mc2)2 1−u2 c2 1−u2 c2 −1 = (mc2)2 Q.E.D.
39.40 (a) K = 50.0 GeV mc2 = 1.67 × 10−27 kg ( ) 2.998 × 108 m s ( ) 2 1 1.60 × 10−10 J GeV = 0.938 GeV E = K + mc2 = 50.0 GeV + 0.938 GeV = 50.938 GeV E2 = p2c2 + mc2 ( ) 2 ⇒ p = E2 −mc2 ( ) 2 c2 = 50.938 GeV ( )2 −0.938 GeV ( )2 c2 p = 50.9 GeV c = 50.9 GeV 3.00 × 108 m s 1.60 × 10−10 J 1 GeV = 2.72 × 10−17 kg ⋅m s (b) E = γ mc2 = mc2 1−u c ( ) 2 ⇒ u = c 1−mc2 E ( ) 2 v = 3.00 × 108 m s ( ) 1− 0.938 GeV 50.938 GeV 2 = 2 9995 108 .
× m s 39.41 (a) q ∆V ( ) = K = γ −1 ( )mec2 14 Chapter 39 Solutions Thus, γ = 1 1−u c ( ) 2 = 1+ q ∆V ( ) mec2 from which u = 0.302 c (b) K = γ −1 ( )mec2 = q ∆V ( ) = 1.60 × 10−19 C ( ) 2.50 × 104 J C ( ) = 4.00 × 10−15 J 39.42 (a) E = γ mc2 = 20.0 GeV with mc2 = 0.511 MeV for electrons. Thus, γ = 20.0 × 109 eV 0.511× 106 eV = 3.91× 104 (b) γ = 1 1−u c ( ) 2 = 3.91× 104 from which u =0.999 999 999 7c (c) L = Lp 1−u c ( ) 2 = Lp γ = 3.00 × 103 m 3.91× 104 = 7.67 × 10−2 m = 7.67 cm 39.43 Conserving total momentum, pBefore decay = pafter decay = 0: pν = pµ = γ mµu = γ 206me ( )u Conservation of mass-energy gives: Eµ + Eν = Eπ γ mµc2 + pνc = mπc2 γ 206me ( ) + pν c = 270me Substituting from the momentum equation above, γ 206me ( ) + γ 206me ( ) u c = 270me or γ 1+ u c = 270 206 = 1.31 ⇒ u c = 0.264 Then, Kµ = γ −1 ( )mµc2 = γ −1 ( )206 mec2 ( ) = 1 1−0.264 ( )2 −1 206 0.511 MeV ( ) = 3.88 MeV Also, Eν = Eπ −Eµ = mπc2 −γ mµc2 = 270 −206γ ( )mec2 Eν = 270 − 206 1−0.264 ( )2 0.511 MeV ( ) = 28.8 MeV 39.44 Let a 0.3-kg flag be run up a flagpole 7 m high.
We put into it energy mgh = 0.3 kg(9.8 m/s2) 7 m ≈ 20 J Chapter 39 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
So we put into it extra mass ∆m = E c 2 = 20 J (3 × 108 m/s)2 = 2 × 10–16 kg for a fractional increase of 2 × 1016 kg 0.3 kg ~10–15 39.45 E = 2.86 × 105 J. Also, the mass-energy relation says that E = mc2.
Therefore, m = E c2 = 2.86 × 105 J (3.00 × 108 m /s)2 = 3.18 × 10–12 kg No, a mass loss of this magnitude (out of a total of 9.00 g) could not be detected .
39.46 (a) K = (γ −1)mc2 = 1 1−u2 /c2 −1 mc2 = 0.25mc2 = 2.25 × 1022 J (b) E = mfuel c2 so mfuel = 2.25 × 1022 9.00 × 1016 = 2.50 × 105 kg 39.47 ∆m = E c2 = P t c2 = 0.800 1.00 × 109 J s ( ) 3.00 yr ( ) 3.16 × 107 s yr ( ) 3.00 × 108 m s ( ) 2 = 0.842 kg 39.48 Since the total momentum is zero before decay, it is necessary that after the decay pnucleus = pphoton = Eγ c = 14.0 keV c Also, for the recoiling nucleus, E2 = p2c2 + mc2 ( ) 2 with mc2 = 8.60 × 10−9 J = 53.8 GeV Thus, mc2 + K ( ) 2 = 14.0 keV ( )2 + mc2 ( ) 2 or 1+ K mc2 2 = 14.0 keV mc2 2 + 1 So 1+ K mc2 = 1+ 14.0 keV mc2 2 ≈1+ 1 2 14.0 keV mc2 2 Binomial Theorem ( ) and K ≈14.0 keV ( )2 2mc2 = 14.0 × 103 eV ( ) 2 2 53.8 × 109 eV ( ) = 1 82 10 3 .
× − eV 39.49 P = dE dt = d mc2 ( ) dt = c2 dm dt = 3.77 × 1026 W 16 Chapter 39 Solutions Thus, dm dt = 3.77 × 1026 J s 3.00 × 108 m s ( ) 2 = 4 19 109 .
× kg s 39.50 2mec2 = 1.02 MeV: Eγ ≥ 1.02 MeV 39.51 The moving observer sees the charge as stationary, so she says it feels no magnetic force. q(E + v × B) = q( ′ E + 0) and ′ E = E + v × B 39.52 (a) When Ke = Kp, mec2 γ e −1 ( ) = mpc2 γ p −1 ( ) In this case, mec2 = 0.511 MeV, mpc2 = 938 MeV and γ e = 1−0.750 ( )2 [ ] −1/2 = 1.5119 Substituting, γ p = 1+ mec2(γ e −1) mpc2 = 1+ 0.511 MeV ( ) 1.5119 −1 ( ) 938 MeV = 1.000279 but γ p = 1 1−up c 2 1/2 . Therefore, up = c 1−γ p −2 = 0.0236 c (b) When pe = pp, γ pmpup = γ emeue or γ pup = γ emeue mp .
Thus, γ pup = 1.5119 ( ) 0.511 MeV c2 ( ) 0.750c ( ) 938 MeV c2 = 6.1772 × 10−4 c and up c = 6.1772 × 10−4 1− up c 2 which yields up = 6.18 × 10−4 c = 185 km s 39.53 (a) 1013 MeV = (γ – 1)mpc 2 so γ ≈ 1010 vp ≈ c ′ t = t γ = 105 yr 1010 = 10−5 yr ~ 10 2 s (b) ′ d = c ′ t ~1011 m Chapter 39 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution The cosmic rays of highest energy are protons, which have kinetic energy on the order of 1013 MeV.
(a) How long would it take a proton of this energy to travel across the Milky Way galaxy, having a diameter on the order of ~105 light-years, as measured in the proton's frame? (b) From the point of view of the proton, how many kilometers across is the galaxy?
G : We can guess that the energetic cosmic rays will be traveling close to the speed of light, so the time it takes a proton to traverse the Milky Way will be much less in the proton’s frame than 105 years. The galaxy will also appear smaller to the high-speed protons than the galaxy’s proper diameter of 105 light-years.
O : The kinetic energy of the protons can be used to determine the relativistic γ-factor, which can then be applied to the time dilation and length contraction equations to find the time and distance in the proton’s frame of reference.
A : The relativistic kinetic energy of a proton is K = γ −1 ( )mc2 = 1013 MeV Its rest energy is mc2 = 1.67 × 10−27 kg ( ) 2.998 × 108 m s 2 1 eV 1.60 × 10−19 kg ⋅m2/s2 = 938 MeV So 1013 MeV = γ −1 ( ) 938 MeV ( ), and therefore γ = 1.07 × 1010 The proton's speed in the galaxy’s reference frame can be found from γ = 1 1−v2 /c2 : 1−v2 c2 = 8.80 × 10−21 and v = c 1−8.80 × 10−21 = 1−4.40 × 10−21 ( )c ≈3.00 × 108 m /s The proton’s speed is nearly as large as the speed of light. In the galaxy frame, the traversal time is ∆t = x / v = 105 light - years/c = 105 years (a) This is dilated from the proper time measured in the proton's frame. The proper time is found from ∆t = γ∆tp: ∆tp = ∆t/ γ = 105 yr 1.07 × 1010 = 9.38 × 10−6 years = 296 s ~ a few hundred seconds (b) The proton sees the galaxy moving by at a speed nearly equal to c, passing in 296 s: ∆Lp = v∆tp = 3.00 × 108 ( ) 296 s ( ) = 8.88 × 107 km ~ 108 km ∆Lp = 8.88 × 1010 m ( ) 9.46 × 1015 m /ly ( ) = 9.39 × 10−6 ly ~ 10-5 ly L : The results agree with our predictions, although we may not have guessed that the protons would be traveling so close to the speed of light! The calculated results should be rounded to zero significant figures since we were given order of magnitude data. We should also note that the relative speed of motion v and the value of γ are the same in both the proton and galaxy reference frames. 18 Chapter 39 Solutions 39.54 Take the primed frame as: (a) The mother ship: ux = ′ u ′ x + v 1+ ′ u ′ x v c2 = v + v 1+ v2 c2 = 2v 1+ v2 c2 = 2(0.500c) 1+ (0.500)2 = 0.800 c (b) The shuttle: ux = v + 2v 1+ v2 /c2 1+ v c2 2v 1+ v2 /c2 = 3v + v3 /c2 1+ 3v2 /c2 = 3(0.500c) + (0.500c)3 c2 1+ 3(0.500)2 = 0.929 c 39.55 ∆mc2 mc2 = 4 938.78 MeV ( ) −3728.4 MeV 4 938.78 MeV ( ) × 100% = 0.712% 39.56 dearth = vtearth = vγ tastro so 2.00 × 106 yr ⋅c = v 1 1−v2 /c2 30.0 yr 1−v2 /c2 = v/c ( )(1.50 × 10−5) 1−v2 c2 = v2 c2 (2.25 × 10−10) 1 = v2 c2 (1+ 2.25 × 10−10) so v c = 1+ 2.25 × 10−10 ( ) −1/2 = 1−1 2 (2.25 × 10−10) v c = 1−1.12 × 10−10 39.57 (a) Take the spaceship as the primed frame, moving toward the right at v c = +0 600 .
. Then ′ = + u c x 0 800 .
, and ux = ′ ux + v 1+ ′ ux v ( ) c2 = 0.800c +0.600c 1+ 0.800 ( ) 0.600 ( ) = 0.946 c (b) L = Lp γ = 0.200 ly ( ) 1−0.600 ( )2 = 0.160 ly (c) The aliens observe the 0.160-ly distance closing because the probe nibbles into it from one end at 0.800c and the Earth reduces it at the other end at 0.600c. Thus, time ly = + = 0 160 0 800 0 600 .
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c c 0.114 yr (d) K = 1 1−u2 c2 −1 mc2 = 1 1−0.946 ( )2 −1 4.00 × 105 kg ( ) 3.00 × 108 m s ( ) 2 = 7 50 1022 .
× J Chapter 39 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
39.58 In this case, the proper time is T0 (the time measured by the students on a clock at rest relative to them). The dilated time measured by the professor is: ∆t = γ T0 where ∆t = T + t. Here T is the time she waits before sending a signal and t is the time required for the signal to reach the students.
Thus, we have: T + t = γ T0 (1) To determine the travel time t, realize that the distance the students will have moved beyond the professor before the signal reaches them is: d = v(T + t) The time required for the signal to travel this distance is: t = d c = v c T + t ( ) Solving for t gives: t = v c ( )T 1−v c ( ) Substituting this into equation (1) yields: T + v c ( )T 1−v c ( ) = γT0 or T = 1−v c ( ) −1 = γT0 Then T = T0 1−v c ( ) 1−v2 c2 ( ) = T0 1−v/c ( ) 1+ v/c ( ) [ ] 1−v/c ( ) [ ] = T0 1−v c ( ) 1+ v c ( ) 39.59 Look at the situation from the instructor's viewpoint since they are at rest relative to the clock, and hence measure the proper time. The Earth moves with velocity v = – 0.280 c relative to the instructors while the students move with a velocity ′ u = – 0.600 c relative to Earth. Using the velocity addition equation, the velocity of the students relative to the instructors (and hence the clock) is: u = v + ′ u 1+ v ′ u c2 = (−0.280c) −(0.600c) 1+ (−0.280c)(−0.600c) c2 = −0.753c (students relative to clock) (a) With a proper time interval of ∆ tp = 50.0 min, the time interval measured by the students is: ∆t = γ∆tp with γ = 1 1−0.753c ( )2 /c2 = 1.52 Thus, the students measure the exam to last T = 1.52(50.0 min) = 76.0 minutes (b) The duration of the exam as measured by observers on Earth is: ∆t = γ∆tp with γ = 1 1−0.280c ( )2 c2 so T = 1.04(50.0 min) = 52.1 minutes 20 Chapter 39 Solutions 39.60 The energy which arrives in one year is E = Pt = 1.79 × 1017 J/s ( ) 3.16 × 107 s ( ) = 5.66 × 1024 J Thus, m = E c2 = 5.66 × 1024 J 3.00 × 108 m /s ( ) 2 = 6.28 × 107 kg 39.61 The observer sees the proper length of the tunnel, 50.0 m, but sees the train contracted to length L = Lp 1−v2 c2 = 100 m 1−(0.950)2 = 31.2 m shorter than the tunnel by 50.0 – 31.2 = 18.8 m so it is completely within the tunnel.
39.62 If the energy required to remove a mass m from the surface is equal to its mass energy mc2, then GMsm Rg = mc2 and Rg = GMs c2 = (6.67 × 10−11 N ⋅m2 /kg2)(1.99 × 1030 kg) (3.00 × 108 m /s)2 = 1.47 × 103 m = 1.47 km 39.63 (a) At any speed, the momentum of the particle is given by p = γ mu = mu 1−u c ( ) 2 Since F = qE = dp dt qE = d dt mu 1−u2 c2 ( ) −1/2 qE = m 1−u2 c2 ( ) −1/2 du dt + 1 2 mu 1−u2 c2 ( ) −3/2 2u c2 ( ) du dt So qE m = du dt 1−u2 c2 + u2 c2 1−u2 c2 ( ) 3/2 and a = du dt = qE m 1−u2 c2 3/2 (b) As u →c, a →0 (c) du 1−u2 /c2 3/2 0 v ∫ = qE m dt t=0 t ∫ so u = qEct m2c2 + q2E2t2 x = udt 0 t ∫ = qEc tdt m2c2 + q2E2t2 0 t ∫ = c qE m2c2 + q2E2t2 −mc Chapter 39 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
39.64 (a) fobserved = fsource 1+ v c 1−v c implies c λ + ∆λ = c λ 1+ v c 1−v c , or 1−v c 1+ v c = λ + ∆λ λ and 1+ ∆λ λ = 1−v c 1+ v c (b) 1+ 550 nm −650 nm 650 nm = 1−v c 1+ v c = 0.846 1−v c = 0.846 ( )2 1+ v c = 0.716 + 0.716 v c v = 0.166c = 4 97 107 .
× m s 39.65 (a) An observer at rest relative to the mirror sees the light travel a distance D = 2d −x = 2 1.80 × 1012 m ( ) −0.800c ( )t where x = 0.800c ( )t is the distance the ship moves toward the mirror in time t. Since this observer agrees that the speed of light is c, the time for it to travel distance D is: t = D c = 2(1.80 × 1012 m) 3.00 × 108 m /s −0.800t = 6 67 103 .
× s (b) The observer in the rocket sees a length-contracted initial distance to the mirror of: L = d 1−v2 c2 = 1.80 × 1012 m ( ) 1−(0.800c)2 c2 = 1.08 × 1012 m, and the mirror moving toward the ship at speed v = 0.800c. Thus, he measures the distance the light travels as: D = 2 1.08 × 1012 m −y ( ) where y = (0.800c) t/ 2 ( ) is the distance the mirror moves toward the ship before the light reflects off it. This observer also measures the speed of light to be c, so the time for it to travel distance D is: t = D c = 2 c 1.08 × 1012 m −0.800c ( ) t 2 , which gives t = 4 00 103 .
× s 22 Chapter 39 Solutions 39.66 (a) An observer at rest relative to the mirror sees the light travel a distance D = 2d −x, where x = vt is the distance the ship moves toward the mirror in time t. Since this observer agrees that the speed of light is c, the time for it to travel distance D is t = D c = 2d −vt c = 2d c + v (b) The observer in the rocket sees a length-contracted initial distance to the mirror of L = d 1−v2 c2 and the mirror moving toward the ship at speed v. Thus, he measures the distance the light travels as D = 2 L −y ( ) where y = vt 2 is the distance the mirror moves toward the ship before the light reflects off it.
This observer also measures the speed of light to be c, so the time for it to travel distance D is: t = D c = 2 c d 1−v2 c2 −vt 2 so c + v ( )t = 2d c c + v ( ) c −v ( ) or t = 2d c c −v c + v 39.67 (a) Since Mary is in the same reference frame, ′ S , as Ted, she observes the ball to have the same speed Ted observes, namely ′ ux = 0.800c .
(b) ∆′ t = Lp ′ ux = 1.80 × 1012 m 0.800 3.00 × 108 m s ( ) = 7.5 10 s 3 0 × (c) L = Lp 1−v2 c2 = 1.80 × 1012 m ( ) 1−(0.600c)2 c2 = 1.44 10 m 12 × Since v = 0.600c and ′ ux = −0.800c, the velocity Jim measures for the ball is ux = ′ ux + v 1+ ′ uxv c2 = −0.800c ( ) + 0.600c ( ) 1+ −0.800 ( ) 0.600 ( ) = −0.385c (d) Jim observes the ball and Mary to be initially separated by 1.44 × 1012 m. Mary's motion at 0.600c and the ball's motion at 0.385c nibble into thi distance from both ends. The gap closes at the rate 0.600c + 0.385c = 0.985c, so the ball and catcher meet after a time ∆t = 1.44 × 1012 m 0.985 3.00 × 1018 m /s ( ) = 4.88 × 103 s Chapter 39 Solutions 23 © 2000 by Harcourt, Inc. All rights reserved.
39.68 (a) L0 2 = L0x 2 + L0y 2 and L2 = L x 2 + L y 2 The motion is in the x direction: Ly = L0y = L0 sin θ0 Lx = L0x 1−v c ( ) 2 = L0 cos θ0 ( ) 1−v c ( ) 2 Thus, L2 = L0 2 cos2 θ0 1−v c 2 + L0 2 sin2 θ0 = L0 2 1−v c 2 cos2 θ0 or L = L0 1−v c ( ) 2 cos2 θ0 [ ] 1 2 (b) tan θ = Ly Lx = L0y L0x 1−v c ( ) 2 = γ tanθ0 39.69 (a) First, we find the velocity of the stick relative to ′ S using L = Lp 1− ′ ux ( ) 2 c2 Thus ′ ux = ± c 1−L Lp ( ) 2 Selecting the negative sign because the stick moves in the negative x direction in ′ S gives: ′ ux = −c 1−0.500 m 1.00 m 2 = −0.866c so the speed is ′ ux = 0.866 c Now determine the velocity of the stick relative to S, using the measured velocity of the stick relative to ′ S and the velocity of ′ S relative to S. From the velocity addition equation, we have: ux = ′ ux + v 1+ v ′ ux c2 = −0.866c ( ) + 0.600c ( ) 1+ 0.600c ( ) −0.866c ( ) = −0.554c and the speed is ux = 0.554 c (b) Therefore, the contracted length of the stick as measured in S is: L = Lp 1−ux c ( ) 2 = 1.00 m ( ) 1−0.554 ( )2 = 0.833 m 24 Chapter 39 Solutions 39.70 (b) Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti, stationary with respect to both. Just as our spaceship is passing him, he also sees the blast waves from both explosions. Judging both stars to be stationary, this observer concludes that the two stars blew up simultaneously .
(a) We in the spaceship moving past the hermit do not calculate the explosions to be simultaneous. We see the distance we have traveled from the Sun as L = Lp 1−v c ( ) 2 = 6.00 ly ( ) 1−0.800 ( )2 = 3.60 ly We see the Sun flying away from us at 0.800c while the light from the Sun approaches at 1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time we calculate to have elapsed since the Sun exploded is 3.60 ly 1.80c = 2.00 yr.
We see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only 0.200c faster. We see the gap between that star and its blast wave as 3.60 ly and growing at 0.200c. We calculate that it must have been opening for 3.60 ly 0.200c = 18.0 yr and conclude that Tau Ceti exploded 16.0 years before the Sun .
39.71 The unshifted frequency is fsource = c λ = 3.00 × 108 m s 394 × 10−9 m = 7.61× 1014 Hz We observe frequency f = 3.00 × 108 m s 475 × 10−9 m = 6.32 × 1014 Hz Then f = fsource 1+ v c 1−v c gives: 6.32 = 7.61 1+ v c 1−v c or 1+ v c 1−v c = 0.829 ( )2 Solving for v yields: v = −0.185c = 0.185c away ( ) Chapter 39 Solutions 25 © 2000 by Harcourt, Inc. All rights reserved.
39.72 Take m = 1.00 kg. The classical kinetic energy is Kc = 1 2 mu2 = 1 2 mc2 u c 2 = 4.50 × 1016 J ( ) u c 2 and the actual kinetic energy is Kr = 1 1−u c ( ) 2 −1 mc2 = 9.00 × 1016 J ( ) 1 1−u c ( ) 2 −1 u c Kc J ( ) Kr J ( ) 0.000 0.000 0.000 0.100 0.045 × 1016 0.0453 × 1016 0.200 0.180 × 1016 0.186 × 1016 0.300 0.405 × 1016 0.435 × 1016 0.400 0.720 × 1016 0.820 × 1016 0.500 1.13 × 1016 1.39 × 1016 0.600 1.62 × 1016 2.25 × 1016 0.700 2.21 × 1016 3.60 × 1016 0.800 2.88 × 1016 6.00 × 1016 0.900 3.65 × 1016 11.6 × 1016 0.990 4.41 × 1016 54.8 × 1016 Kc = 0.990Kr when 1 2 u c ( ) 2 = 0.990 1 1−u c ( ) 2 −1 , yielding u = 0.115 c Similarly, Kc = 0.950Kr when u = 0.257 c and Kc = 0.500Kr when u = 0.786 c 39.73 ∆m = E c2 = mc ∆T ( ) c2 = ρVc ∆T ( ) c2 = 1030 kg /m3 ( ) 1.40 × 109 ( ) 103 m ( ) 3 4186 J/kg⋅°C ( ) 10.0 °C ( ) 3.00 × 108 m /s ( ) 2 ∆m = 6.71 10 kg 8 × © 2000 by Harcourt, Inc. All rights reserved.
Chapter 40 Solutions 40.1 T = 2.898 × 10−3 m ⋅K 560 × 10−9 m = 5 18 103 .
× K 40.2 (a) λ max = 2.898 × 10−3 m ⋅K T ~ 2.898 × 10−3 m ⋅K 104 K ~ 10 7 − m ultraviolet .
(b) λ max ~ 2.898 × 10−3 m ⋅K 107 K ~ 10 10 − m . γ −ray 40.3 (a) Using λ maxT = 2.898 10 m K 3 × ⋅ − we getλ max = 2.898 10 m 2900 K × −3 = 9.99 × 10– 7 m = 999 nm (b) The peak wavelength is in the infrared region of the electromagnetic spectrum, which is much wider than the visible region of the spectrum.
40.4 Planck's radiation law gives intensity-per-wavelength. Taking E to be the photon energy and n to be the number of photons emitted each second, we multiply by area and wavelength range to have energy-per-time leaving the hole: P = 2π hc2 (λ 2 −λ 1)π(d/ 2)2 λ 1 + λ 2 2 5 e 2hc (λ 1+λ 2 )kBT −1 = En = nhf where E = hf = 2hc λ 1 + λ 2 n = P E = 8π 2cd2(λ 2 −λ 1) (λ 1 + λ 2)4 e2hc (λ 1+λ 2 )kBT −1 ( ) = 8π 2 3.00 × 108 m s ( ) 5.00 × 10−5 m ( ) 2 1.00 × 10−9 m ( ) 1001× 10−9 m ( ) 4 e 2 6.626 × 10−34 J⋅s ( ) 3.00×108 m s ( ) 1001×10−9 m ( ) 1.38×10−23 J K ( ) 7.50×103 K ( ) −1 n e = × − ( ) = 5.90 10 s 1 16/ .
3 84 1.30 × 1015 /s 2 Chapter 40 Solutions 40.5 (a) P = eAσT4 = 1 20.0 × 10−4 m2 ( ) 5.67 × 10−8 W m2 ⋅K4 ( ) 5000 K ( )4 = 7 09 104 .
× W (b) λ λ λ max max max .
T = ( ) = × ⋅ ⇒ = − 5000 2 898 10 3 K m K 580 nm (c) We compute: hc kBT = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 1.38 × 10−23 J K ( ) 5000 K ( ) = 2.88 × 10−6 m The power per wavelength interval is P λ ( ) = AI λ ( ) = 2π hc2A λ5 exp hc λ kBT ( ) −1 [ ] , and 2π hc2A = 2π 6.626 × 10−34 ( ) 3.00 × 108 ( ) 2 20.0 × 10−4 ( ) = 7.50 × 10−19 J ⋅m4 s P 580 nm ( ) = 7.50 × 10−19 J ⋅m4 s 580 × 10−9 m ( ) 5 exp 2.88 µm 0.580 µm ( ) −1 [ ] = 1.15 × 1013 J m ⋅s e4.973 −1 = 7 99 1010 .
× W m (d) - (i) The other values are computed similarly: λ hc/kBT ehc/λkBT –1 2π hc2A/λ5 P (λ), W/m (d) 1.00 nm 2882.6 7.96 × 101251 7.50 × 1026 9.42 × 10–1226 (e) 5.00 nm 576.5 2.40 × 10250 2.40 × 1023 1.00 × 10–227 (f) 400 nm 7.21 1347 7.32 × 1013 5.44 × 1010 (c) 580 nm 4.97 143.5 1.15 × 1013 7.99 × 1010 (g) 700 nm 4.12 60.4 4.46 × 1012 7.38 × 1010 (h) 1.00 mm 0.00288 0.00289 7.50 × 10– 4 0.260 (i) 10.0 cm 2.88 × 10–5 2.88 × 10–5 7.50 × 10–14 2.60 × 10–9 (j) We approximate the area under the P λ ( ) versus λ curve, between 400 nm and 700 nm, as two trapezoids: P ≈ 5.44 + 7.99 ( ) × 1010 W m 580 −400 ( ) × 10−9 m [ ] 2 + 7.99 + 7.38 ( ) × 1010 W m 700 −580 ( ) × 10−9 m [ ] 2 P = 2.13 × 104 W so the power radiated as visible light is approximately 20 kW .
Chapter 40 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
40.6 (a) P = eAσT4, so T = P eAσ 1 4 = 3.77 × 1026 W 1 4π 6.96 × 108 m ( ) 2 5.67 × 10−8 W m2 ⋅K4 1 4 = 5 75 103 .
× K (b) λmax .
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= × ⋅ = × ⋅ × = × = − − − 2 898 10 2 898 10 5 75 10 5 04 10 3 3 3 7 m K m K K m T 504 nm 40.7 (a) E = hf = 6.626 × 10−34 J ⋅s ( ) 620 × 1012 s−1 ( ) 1.00 eV 1.60 × 10−19 J = 2.57 eV (b) E hf = = × ⋅ ( ) × ( ) × = − − − 6 626 10 3 10 10 1 00 1 60 10 34 9 19 .
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. J s s eV J 1 1 28 10 5 .
× − eV (c) E = hf = 6.626 × 10−34 J ⋅s ( ) 46.0 × 106 s−1 ( ) 1.00 eV 1.60 × 10−19 J = 1 91 10 7 .
× − eV (d) λ = c f = 3.00 × 108 m s 620 × 1012 Hz = 4.84 × 10−7 m = 484 nm, visible light (blue) λ = c f = 3.00 × 108 m s 3.10 × 109 Hz = 9.68 × 10−2 m = 9.68 cm, radio wave λ = c f = 3.00 × 108 m s 46.0 × 106 Hz = 6.52 m, radio wave 40.8 E = hf = hc λ = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 589.3 × 10−9 m = 3.37 × 10−19 J photon n = P E = 10.0 J s 3.37 × 10−19 J photon = 2 96 1019 .
× photons s 40.9 Each photon has an energy E = hf = (6.626 × 10– 34)(99.7 × 106) = 6.61 × 10– 26 J This implies that there are 150 × 103 J/s 6.61 × 10– 26 J/photons = 2.27 × 1030 photons/s 4 Chapter 40 Solutions 40.10 Energy of a single 500-nm photon: Eγ = hf = hc λ = (6.626 × 10 – 34 J · s)(3.00 × 108 m/s) 500 × 10 – 9 m = 3.98 × 10 – 19 J The energy entering the eye each second E = Pt= (IA)t = (4.00 × 10 – 11 W/m2) π 4 (8.50 × 10 – 3 m) 2(1.00 s) = 2.27 × 10 – 15 J The number of photons required to yield this energy n = E Eγ = 2.27 × 10 – 15 J 3.98 × 10 – 19 J/photon = 5.71 × 103 photons 40.11 We take θ = 0.0300 radians. Then the pendulum's total energy is E = mgh = mg(L – L cos θ) E = (1.00 kg)(9.80 m/s2)(1.00 – 0.9995) = 4.41 × 10–3 J The frequency of oscillation is f = ω 2π = 1 2π g L = 0.498 Hz The energy is quantized, E = nhf Therefore, n = E h f = 4.41 × 10–3 J (6.626 × 10–34 J · s)(0.498 s–1) = 1.34 × 1031 40.12 The radiation wavelength of λ ′= 500 nm that is observed by observers on Earth is not the true wavelength, λ , emitted by the star because of the Doppler effect. The true wavelength is related to the observed wavelength using: c ′ λ = c λ 1−v c ( ) 1+ v c ( ) λ = λ ′ 1−v c ( ) 1+ v c ( ) = 500 nm ( ) 1−0.280 ( ) 1+ 0.280 ( ) = 375 nm The temperature of the star is given by λ maxT = 2.898 × 10−3 m ⋅K: T = 2.898 × 10−3 m ⋅K λ max = 2.898 × 10−3 m ⋅K 375 × 10−9 = 7.73 10 K 3 × Chapter 40 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
40.13 This follows from the fact that at low T or long λ , the exponential factor in the denominator of Planck's radiation law is large compared to 1, so the factor of 1 in the denominator can be neglected. In this approximation, one arrives at Wien's radiation law.
40.14 Planck’s radiation law is I λ ,T ( ) = 2π hc2 λ5 ehc λ kBT −1 ( ) Using the series expansion ex = 1+ x + x2 2! + x3 3! + . . .
Planck’s law reduces to I λ ,T ( ) = 2π hc2 λ5 1+ hc λ kBT + . . .
( ) −1 [ ] ≈ 2π hc2 λ5 hc λ kBT ( ) = 2π ckBT λ4 which is the Rayleigh-Jeans law, for very long wavelengths.
40.15 (a) λ c = hc φ = (6.626 × 10– 34 J · s)(3.00 × 108 m/s) (4.20 eV)(1.60 × 10–19 J/eV) = 296 nm fc = c λ c = 3.00 × 108 m/s 296 × 10– 9 m = 1.01 × 1015 Hz (b) h c λ = φ + e(∆VS): (6.626 × 10– 34)(3.00 × 108) 180 × 10– 9 = (4.20 eV)(1.60 × 10– 19 J/eV) + (1.60 × 10–19)(∆VS) Therefore, ∆VS = 2.71 V 40.16 K mv max max = 1 2 2 = 1 2 (9.11 × 10– 31)(4.60 × 105) 2 = 9.64 × 10 – 20 J = 0.602 eV (a) φ = E – Kmax = 1240 eV · nm 625 nm – 0.602 eV = 1.38 eV (b) fc = φ h = 1.38 eV 6.626 × 10– 34 J · s 1.60 × 10– 19 J 1 eV = 3.34 × 1014 Hz 6 Chapter 40 Solutions 40.17 (a) λ c = hc φ Li: λ c = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( ) 2.30 eV ( ) 1.60 × 10−19 J/eV ( ) = 540 nm Be: λ c = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( ) 3.90 eV ( ) 1.60 × 10−19 J/eV ( ) = 318 nm Hg: λ c = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( ) 4.50 eV ( ) 1.60 × 10−19 J/eV ( ) = 276 nm λ < λ c for photo current. Thus, only lithium will exhibit the photoelectric effect. (b) For lithium, hc λ = φ + Kmax 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( ) 400 × 10−9 m = 2.30 eV ( ) 1.60 × 10−19 ( ) + Kmax Kmax = 1.29 × 10−19 J = 0.808 eV 40.18 From condition (i), hf = e(∆VS 1) + φ1 and hf = e(∆VS 2) + φ2 (∆VS 1) = (∆VS 2) + 1.48 V Then φ2 – φ1 = 1.48 eV From condition (ii), h fc 1 = φ1 = 0.600hfc 2 = 0.600φ2 φ2 – 0.600φ2 = 1.48 eV φ2 = 3.70 eV φ1 = 2.22 eV 40.19 (a) e ∆VS ( ) = hc λ −φ → φ = 1240 nm ⋅eV 546.1 nm −0.376 eV = 1.90 eV (b) e ∆VS ( ) = hc λ −φ = 1240 nm ⋅eV 587.5 nm −1.90 eV → ∆VS = 0.216 V Chapter 40 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp (λ = 546.1 nm) is used, a retarding potential of 0.376 V reduces the photocurrent to zero. (a) Based on this measurement, what is the work function for this metal? (b) What stopping potential would be observed when using the yellow light from a helium discharge tube (λ = 587.5 nm)?
G : According to Table 40.1, the work function for most metals is on the order of a few eV, so this metal is probably similar. We can expect the stopping potential for the yellow light to be slightly lower than 0.376 V since the yellow light has a longer wavelength (lower frequency) and therefore less energy than the green light. O : In this photoelectric experiment, the green light has sufficient energy hf to overcome the work function of the metal φ so that the ejected electrons have a maximum kinetic energy of 0.376 eV.
With this information, we can use the photoelectric effect equation to find the work function, which can then be used to find the stopping potential for the less energetic yellow light.
A : (a) Einstein’s photoelectric effect equation is Kmax = hf −φ , and the energy required to raise an electron through a 1 V potential is 1 eV, so that Kmax = eVs = 0.376 eV.
A photon from the mercury lamp has energy: hf = hc λ = 4.14 × 10−15 eV ⋅s ( ) 3.00 × 108 m s ( ) 546.1× 10−9 m E = hf = 2.27 eV Therefore, the work function for this metal is: φ = hf −Kmax = 2.27 eV −0.376 eV ( ) = 1.90 eV (b) For the yellow light, λ = 587.5 nm, and hf = hc λ = 4.14 × 10−15 eV ⋅s ( ) 3.00 × 108 m /s ( ) 587.5 × 10−9 m E = 2.11 eV Therefore, Kmax = hf −φ = 2.11 eV −1.90 eV = 0.216 eV, so Vs = 0.216 V L : The work function for this metal is lower than we expected, and does not correspond with any of the values in Table 40.1. Further examination in the CRC Handbook of Chemistry and Physics reveals that all of the metal elements have work functions between 2 and 6 eV. However, a single metal’s work function may vary by about 1 eV depending on impurities in the metal, so it is just barely possible that a metal might have a work function of 1.90 eV. The stopping potential for the yellow light is indeed lower than for the green light as we expected.
An interesting calculation is to find the wavelength for the lowest energy light that will eject electrons from this metal. That threshold wavelength for Kmax = 0 is 658 nm, which is red light in the visible portion of the electromagnetic spectrum.) 8 Chapter 40 Solutions 40.20 From the photoelectric equation, we have: e ∆VS1 ( ) = Eγ 1 −φ and e ∆VS2 ( ) = Eγ 2 −φ Since ∆VS2 = 0.700 ∆VS1 ( ), then e ∆VS2 ( ) = 0.700(Eγ 1 −φ) = Eγ 2 −φ or (1−0.700)φ = Eγ 2 −0.700Eγ 1 and the work function is: φ = Eγ 2 −0.700Eγ 1 0.300 The photon energies are: Eγ 1 = hc λ 1 = 1240 nm ⋅eV 410 eV = 3.03 eV and E hc γ λ 2 2 1240 445 2 79 = = ⋅ = nm eV eV eV .
Thus, the work function is φ = 2.79 eV −0.700 3.03 eV ( ) 0.300 = 2.23 eV and we recognize this as characteristic of potassium .
40.21 The energy needed is E = 1.00 eV = 1.60 × 10– 19 J The energy absorbed in time t is E = Pt = (IA)t so t = E IA = 1.60 × 10– 19 J (500 J/s · m2)[π (2.82 × 10– 15 m)2] = 1.28 × 107 s = 148 days The gross failure of the classical theory of the photoelectric effect contrasts with the success of quantum mechanics.
40.22 Ultraviolet photons will be absorbed to knock electrons out of the sphere with maximum kinetic energy Kmax = hf −φ , or Kmax = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 200 × 10−9 m 1.00 eV 1.60 × 10−19 J −4.70 eV = 1.51 eV The sphere is left with positive charge and so with positive potential relative to V = 0 at r = ∞.
As its potential approaches 1.51 V, no further electrons will be able to escape, but will fall back onto the sphere. Its charge is then given by V = keQ r or Q = rV ke = 5.00 × 10−2 m ( ) 1.51 N ⋅m C ( ) 8.99 × 109 N ⋅m2 C2 = 8 41 10 12 .
× − C Chapter 40 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
40.23 (a) By having the photon source move toward the metal, the incident photons are Doppler shifted to higher frequencies, and hence, higher energy.
(b) If v = 0.280c, ′ f = f 1+ v/c 1−v/c = 7.00 × 1014 ( ) 1.28 0.720 = 9.33 × 1014 Hz Therefore, φ = 6.626 × 10−34 J ⋅s ( ) 9.33 × 1014 Hz ( ) = 6.18 × 10−19 J = 3.87 eV (c) At v = 0.900c, f = 3.05 × 1015 Hz and Kmax = hf −φ = 6.626 × 10−34 J ⋅s ( ) 3.05 × 1015 Hz ( ) 1.00 eV 1.60 × 10−19 J −3.87 eV = 8.78 eV 40.24 E = hc λ = (6.626 × 10– 34 J · s)(3.00 × 108 m/s) 700 × 10– 9 m = 2.84 × 10– 19 J = 1.78 eV p = h λ = 6.626 × 10– 34 J · s 700 × 10– 9 m = 9.47 × 10– 28 kg · m/s 40.25 (a) ∆λ = h mec (1−cos θ) = 6.626 × 10 – 34 (9.11 × 10 – 31)(3.00 × 108) (1 – cos 37.0°) = 4.88 × 10 – 13 m (b) E0 = hc/λ0: 300 × 103 eV ( ) 1.60 × 10−19 J/eV ( ) = 6.626 × 10−34 ( ) 3.00 × 108 m /s ( ) λ0 λ 0 = 4.14 × 10 – 12 m and ′ λ = λ0 + ∆λ = 4.63 × 10−12 m ′ E = hc ′ λ = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( ) 4.63 × 10−12 m = 4.30 × 1014 J = 268 keV (c) Ke = E0 − ′ E = 300 keV −268.5 keV = 31.5 keV 40.26 This is Compton scattering through 180°: E0 = hc λ 0 = (6.626 × 10 – 34 J · s)(3.00 × 108 m/s) (0.110 × 10 – 9 m)(1.60 × 10–19 J/eV) = 11.3 keV ∆λ = h mec (1−cos θ) = (2.43 × 10 – 12 m)(1 – cos 180°) = 4.86 × 10 – 12 m ′ λ = λ 0 + ∆λ = 0.115 nm so ′ E = hc ′ λ = 10.8 keV Momentum conservation: h λ 0 i = h ′ λ (–i) + pe (i) and pe = h 1 λ0 −1 ′ λ pe = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( )/c 1.60 × 10−19 J/eV 1 0.110 × 10−9 m + 1 0.115 × 10−9 m = 22.1 keV/c 10 Chapter 40 Solutions Energy conservation: 11.3 keV = 10.8 keV + Ke so that Ke = 478 eV Check: E 2 = p 2 c 2 + m c e 2 4 or (mec2 + Ke)2 = (pc)2 + (mec2)2 (511 keV + 0.478 keV)2 = (22.1 keV)2 + (511 keV)2 2.62 × 1011 = 2.62 × 1011 40.27 Ke = E0 – ′ E With Ke = ′ E , ′ E = E0 – ′ E : ′ E = E0 2 ′ λ = hc ′ E = h c 1 2 E0 = 2 hc E0 = 2λ 0 ′ λ = λ 0 + λ C (1 – cos θ) 2λ 0 = λ 0 + λ C (1 – cos θ) 1 – cos θ = λ 0 λ C = 0.00160 0.00243 → θ = 70.0° 40.28 We may write down four equations, not independent, in the three unknowns λ0, ′ λ , and v using the conservation laws: hc λ 0 = hc ′ λ + γ mec2 −mec2 (Energy conservation) h λ 0 = γ mev cos 20.0°(momentum in x-direction) 0 = h ′ λ −γ mev sin 20.0° (momentum in y-direction) and Compton's equation ′ λ −λ 0 = h mec 1−cos 90.0° ( ).
It is easiest to ignore the energy equation and, using the two momentum equations, write h/λ 0 h/ ′ λ = γ mev cos 20.0° γ mev sin 20.0° or λ 0 = ′ λ tan 20.0° Then, the Compton equation becomes ′ λ − ′ λ tan 20.0°= 0.00243 nm, or ′ λ = 0.00243 nm 1−tan 20.0° = 0.00382 nm = 3.82 pm Chapter 40 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
40.29 (a) Conservation of momentum in the x direction gives: pγ = ′ pγ cos θ + pe cos φ or since θ = φ , h λ 0 = pe + h ′ λ cos θ Conservation of momentum in the y direction gives: 0 = ′ pγ sin θ −pe sin θ, which (neglecting the trivial solution θ = 0) gives: pe = ′ pγ = h ′ λ Substituting into gives: h λ 0 = 2h ′ λ cos θ , or ′ λ = 2λ 0 cos θ Then the Compton equation is ′ λ −λ 0 = h mec (1−cos θ) giving 2λ 0 cos θ −λ 0 = h mec (1−cos θ) or 2 cos θ −1 = hc λ 0 1 mec2 (1−cos θ) Since Eγ = hc λ 0 , this may be written as: 2 cos θ −1 = Eγ mec2 (1−cos θ) which reduces to: 2 + Eγ mec2 cos θ = 1+ Eγ mec2 or cos θ = mec2 + Eγ 2mec2 + Eγ = 0.511 MeV + 0.880 MeV 1.02 MeV + 0.880 MeV = 0.732 so that θ = φ = 43.0° (b) Using Equation (3): ′ Eγ = hc ′ λ = hc λ 0 2 cos θ ( ) = Eγ 2 cos θ = 0.880 MeV 2 cos 43.0° = 0.602 MeV = 602 keV Then, ′ pγ = ′ Eγ c = 0.602 MeV c = 3.21× 10−22 kg ⋅m s (c) From Equation (2), pe = ′ pγ = 0.602 MeV c = 3.21× 10−22 kg ⋅m s From energy conservation: Ke = Eγ − ′ Eγ = 0.880 MeV −0.602 MeV = 0.278 MeV = 278 keV 12 Chapter 40 Solutions 40.30 The energy of the incident photon is E0 = pγ c = hc λ 0 .
(a) Conserving momentum in the x direction gives pγ = pe cos φ + ′ pγ cos θ , or since φ = θ , E0 c = pe + ′ pγ ( )cos θ Conserving momentum in the y direction (with φ = θ ) yields 0 = ′ pγ sin θ −pe sin θ, or pe = ′ pγ = h ′ λ Substituting Equation into Equation gives E0 c = h ′ λ + h ′ λ cos θ , or ′ λ = 2hc E0 cos θ By the Compton equation, ′ λ −λ 0 = h mec 1−cos θ ( ), 2hc E0 cos θ −2hc E0 = h mec 1−cos θ ( ) which reduces to 2mec2 + E0 ( )cos θ = mec2 + E0 Thus, φ = θ = cos−1 mec2 + E0 2mec2 + E0 (b) From Equation , ′ λ = 2hc E0 cos θ = 2hc E0 mec2 + E0 2mec2 + E0 Therefore, ′ Eγ = hc ′ λ = hc 2hc E0 ( ) mec2 + E0 ( ) 2mec2 + E0 ( ) = E0 2 2mec2 + E0 mec2 + E0 , and ′ pγ = ′ Eγ c = E0 2c 2mec2 + E0 mec2 + E0 (c) From conservation of energy, Ke = E0 − ′ Eγ = E0 −E0 2 2mec2 + E0 mec2 + E0 or Ke = E0 2 2mec2 + 2E0 −2mec2 −E0 mec2 + E0 = E0 2 2 mec2 + E0 ( ) Finally, from Equation (2), pe = ′ pγ = E0 2c 2mec2 + E0 mec2 + E0 Chapter 40 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
40.31 (a) Thanks to Compton we have four equations in the unknowns φ, v, and ′ λ : hc λ 0 = hc ′ λ + γ mec2 −mec2 (energy conservation) h λ 0 = h λ ' cos 2φ + γ mev cos φ (momentum in x direction) 0 = h ′ λ sin 2φ −γ mev sin φ (momentum in y direction) ′ λ −λ 0 = h mec (1−cos 2φ) (Compton equation) Using sin 2φ = 2 sin φ cos φ in Equation gives γ mev = 2h ′ λ cos φ .
Substituting this into Equation and using cos 2φ = 2 cos2 φ −1 yields h λ 0 = h ′ λ (2 cos2φ −1) + 2h ′ λ cos2φ = h ′ λ (4 cos2φ −1), or λ ' = 4λ 0 cos2φ −λ 0 Substituting the last result into the Compton equation gives 4λ 0 cos2φ −2λ 0 = h mec 1−2 cos2φ −1 ( ) [ ] = 2 hc mec2 1−cos2 φ ( ).
With the substitution λ 0 = hc E0 , this reduces to cos2 φ = mec2 + E0 2mec2 + E0 = 1+ x 2 + x where x ≡ E0 mec2 .
For x = 0.700 MeV 0.511 MeV = 1.37, this gives φ = cos−1 1+ x 2 + x = 33.0° (b) From Equation , ′ λ = λ 0 4 cos2φ −1 ( ) = λ 0 4 1+ x 2 + x −1 = λ 0 2 + 3x 2 + x . Then, Equation becomes hc λ 0 = hc λ 0 2 + x 2 + 3x + γ mec2 −mec2 or E0 mec2 −E0 mec2 2 + x 2 + 3x + 1 = γ .
Thus, γ = 1+ x −x 2 + x 2 + 3x , and with x = 1.37 we get γ = 1.614.
Therefore, v c = 1−γ −2 = 1−0.384 = 0.785 or v = 0.785 c .
14 Chapter 40 Solutions 40.32 ′ λ −λ = h mec (1 −cos θ) ′′ λ − ′ λ = h mec 1 −cos(π −θ) [ ] ′′ λ −λ = h mec − h mec cos(π −θ) + h mec − h mec cos θ Now cos (π – θ) = – cos θ, so ′′ λ −λ = 2 h mec = 0.00486 nm 40.33 (a) K = 1 2 mev2 = 1 2 9.11× 10−31 kg ( ) 1.40 × 106 m s ( ) 2 = 8.93 × 10−19 J = 5.58 eV E0 = hc λ 0 = 1240 eV ⋅nm 0.800 nm = 1550 eV ′ E = E0 −K , and ′ λ = hc ′ E = 1240 eV ⋅nm 1550 eV −5.58 eV = 0.803 nm ∆λ = ′ λ −λ 0 = 0.00288 nm = 2.88 pm (b) ∆λ = λ C 1−cos θ ( ) ⇒cos θ = 1−∆λ λ C = 1−0.00288 nm 0.00243 nm = −0.189, so θ = 101° 40.34 Maximum energy loss appears as maximum increase in wavelength, which occurs for scattering angle 180°. Then ∆λ = 1−cos 180° ( ) h/mc ( ) = 2h/mc where m is the mass of the target particle. The fractional energy loss is E0 − ′ E E0 = hc λ 0 −hc ′ λ hc λ 0 = ′ λ −λ 0 ′ λ = ∆λ λ 0 + ∆λ = 2h mc λ 0 + 2h mc Further, λ 0 = hc E0 , so E0 − ′ E E0 = 2h mc hc E0 + 2h mc = 2E0 mc2 + 2E0 .
(a) For scattering from a free electron, mc2 = 0.511 MeV, so E0 − ′ E E0 = 2 0.511 MeV ( ) 0.511 MeV + 2 0.511 MeV ( ) = 0.667 (b) For scattering from a free proton, mc2 = 938 MeV, and E0 − ′ E E0 = 2 0.511 MeV ( ) 938 MeV + 2 0.511 MeV ( ) = 0.00109 Chapter 40 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
40.35 Start with Balmer's equation, 1 λ = RH 1 22 −1 n 2 , or λ = (4n 2 / RH) (n 2 −4) .
Substituting RH = 1.0973732 × 107 m−1, we obtain λ = (3.645 × 10−7 m)n 2 n 2 −4 = 364.5n 2 n 2 −4 nm, where n = 3, 4, 5, . . .
40.36 (a) Using 1 λ = RH 1 nf 2 −1 ni 2 , for nf = 2, and ni ≥3, we get: λ = 4n2 RH n2 −4 ( ) = 4n2 2.00 × 107 m−1 ( ) n2 −4 ( ) = 200.0 ( )n2 n2 −4 nm This says that 200 nm ≤λ ≤360 nm, which is ultraviolet .
(b) Using n ≥3, λ = 4n2 RH n2 −4 ( ) = 4n2 0.500 × 107 m−1 ( ) n2 −4 ( ) = (800.0)n2 n2 −4 nm This says that 800 nm ≤λ ≤1440 nm, which is in the infrared .
40.37 (a) Lyman series: 1 λ = R 1−1 n2 n = 2, 3, 4, . . . 1 λ = 1 94.96 × 10−9 = (1.097 × 107) 1−1 n2 n = 5 (b) Paschen series: 1 λ = R 1 32 −1 n2 n = 4, 5, 6, . . .
The shortest wavelength for this series corresponds to n = ∞ for ionization 1 λ = 1.097 × 107 1 9 −1 n2 For n = ∞, this gives λ = 820 nm This is larger than 94.96 nm, so this wave length cannot be associated with the Paschen series Brackett series: 1 λ = R 1 42 −1 n2 n = 5, 6, 7, . . . 1 λ = 1.097 × 107 1 16 −1 n2 n = ∞ for ionization λ min = 1458 nm Once again this wavelength cannot be associated with the Brackett series 16 Chapter 40 Solutions 40.38 (a) λ min = hc Emax Lyman (nf = 1): λ min = hc E1 = 1240 eV ⋅nm 13.6 eV = 91.2 nm (Ultraviolet) Balmer (nf = 2): λ min = hc E2 = 1240 eV ⋅nm 1 4 ( )13.6 eV = 365 nm (UV) Paschen (nf = 3): λ min = . . . = 32(91.2 nm) = 821 nm (Infrared) Bracket (nf = 4): λ min = . . . = 42(91.2 nm) = 1460 nm (IR) (b) Emax = hc λ min Lyman: Emax = 13.6 eV = E1 ( ) Balmer: Emax = 3.40 eV = E2 ( ) Paschen: Emax = 1.51 eV = E3 ( ) Brackett: Emax = 0.850 eV = E4 ( ) 40.39 Liquid O2 λ abs = 1269 nm E = hc λ = 1.2398 × 10−6 1.269 × 10−6 = 0.977 eV for each molecule.
For two molecules, λ = hc 2E = 634 nm, red By absorbing the red photons, the liquid O2 appears to be blue.
40.40 (a) v1 = kee2 mer1 where r1 = 1 ( )2a0 = 0.00529 nm = 5.29 × 10−11 m v1 = 8.99 × 109 N ⋅m2 C2 ( ) 1.60 × 10−19 C ( ) 2 9.11× 10−31 kg ( ) 5.29 × 10−11 m ( ) = 2.19 × 106 m s (b) K1 = 1 2 mev1 2 = 1 2 9.11× 10−31 kg ( ) 2.19 × 106 m s ( ) 2 = 2.18 × 10−18 J = 13.6 eV (c) U1 = −kee2 r1 = − 8.99 × 109 N ⋅m2 C2 ( ) 1.60 × 10−19 C ( ) 2 5.29 × 10−11 m = −4.35 × 10−18 J = –27.2 eV Chapter 40 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
40.41 (a) r2 2 = 0.0529 nm ( ) 2 ( )2 = 0.212 nm (b) mev2 = mekee2 r2 = 9.11× 10−31 kg ( ) 8.99 × 109 N ⋅m2 C 2 ( ) 1.60 × 10−19 C ( ) 2 0.212 × 10−9 m = 9.95 × 10−25 kg ⋅m s (c) L2 = mev2r2 = 9.95 × 10−25 kg ⋅m s ( ) 0.212 × 10−9 m ( ) = 2.11× 10−34 kg ⋅m2 s (d) K2 = 1 2 mev2 2 = mev2 ( ) 2 2me = 9.95 × 10−25 kg ⋅m s ( ) 2 2 9.11× 10−31 kg ( ) = 5.43 × 10−19 J = 3.40 eV (e) U2 = −kee2 r2 = − 8.99 × 109 N ⋅m2 C 2 ( ) 1.60 × 10−19 C ( ) 2 0.212 × 10−9 m = −1.09 × 10−18 J = – 6.80 eV (f) E2 = K2 + U2 = 3.40 eV −6.80 eV = – 3.40 eV 40.42 ∆E = (13.6 eV) 1 ni 2 −1 n f 2 Where for ∆E > 0 we have absorption and for ∆E < 0 we have emission.
(A) for ni = 2 and nf = 5 ∆E = 2.86 eV (absorption) (B) for ni = 5 and nf = 3 ∆E = – 0.967 eV (emission) (C) for ni = 7 and nf = 4 ∆E = – 0.572 eV (emission) (D) for ni = 4 and nf = 7 ∆E = 0.572 eV (absorption) (a) E = hc λ so the shortest wavelength is emitted in transition B .
(b) The atom gains most energy in transition A .
(c) The atom loses energy in transitions B and C .
40.43 (b) 1 λ = R 1 n f 2 −1 ni 2 = (1.097 × 107 m−1) 1 22 −1 62 so λ = 410 nm (a) E = hc λ = (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) 410 × 10−9 m = 4.85 × 10−19 J = 3.03 eV (c) f = c λ = 3.00 × 108 410 × 10−9 = 7.32 × 1014 Hz 18 Chapter 40 Solutions 40.44 We use En = –13.6 eV n2 To ionize the atom when the electron is in the nth level, it is necessary to add an amount of energy given by E = – En = 13.6 eV n2 (a) Thus, in the ground state where n = 1, we have E = 13.6 eV (b) In the n = 3 level, E = 13.6 eV 9 = 1.51 eV 40.45 Starting with 1 2 m ev2 = kee2 2r , we have v2 = kee2 m er and using rn = n2h2 mekee2 gives vn 2 = kee2 me n2h2 mekee2 or vn = kee2 nh 40.46 (a) The velocity of the moon in its orbit is v = 2π r T = 2π (3.84 × 108 m) 2.36 × 106 s = 1.02 × 103 m/s So, L = mvr = (7.36 × 1022 kg)(1.02 × 103 m/s)(3.84 × 108 m) = 2.89 × 1034 kg · m2/s (b) We have L = n h or n = L h = 2.89 × 1034 kg · m2/s 1.055 × 10 – 34 J · s = 2.74 × 1068 (c) We have n h = L = mvr = m(GMe /r)1/2 r, so r = h2 m2 GMe n2 = Rn2 and ∆r r = (n + 1)2 R – n2R n2R = 2n + 1 n2 which is approximately equal to 2 n = 7.30 × 10– 69 Chapter 40 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
40.47 The batch of excited atoms must make these six transitions to get back to state one: 2 → 1, and also 3 → 2 and 3 → 1, and also 4 → 3 and 4 → 2 and 4 → 1. Thus, the incoming light must have just enough energy to produce the 1 → 4 transition. It must be the third line of the Lyman series in the absorption spectrum of hydrogen. The absorbing atom changes from energy Ei = −13.6 eV 12 = −13.6 eV to Ef = −13.6 eV 42 = −0.850 eV, so the incoming photons have wavelength λ = hc Ef −Ei = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) −0.850 eV −−13.6 eV ( ) 1.00 eV 1.60 × 10−19 J = 9.75 × 10−8 m = 97.5 nm 40.48 Each atom gives up its kinetic energy in emitting a photon, so 1 2 mv2 = hc λ = (6.626 × 10– 34 J · s)(3.00 × 108 m/s) (1.216 × 10– 7 m) = 1.63 × 10–18 J v = 4.42 × 104 m/s 40.49 (a) The energy levels of a hydrogen-like ion whose charge number is Z are given by En = −13.6 eV ( ) Z2 n2 Thus for He lium Z = 2 ( ), the energy levels are En = −54.4 eV n2 n = 1, 2, 3, . . . (b) For He+, Z = 2, so we see that the ionization energy (the energy required to take the electron from the n = 1 to the n = ∞ state is E = E∞−E1 = 0 −−13.6 eV ( ) 2 ( )2 1 ( )2 = 54.4 eV 20 Chapter 40 Solutions 40.50 r = n2h2 Zmekee2 = n2 Z h2 mekee2 ; n = 1 r = 1 Z (1.055 × 10−34 J ⋅s)2 (9.11× 10−31 kg)(8.99 × 109 N ⋅m2 /C 2)(1.602 × 10−19 C)2 = 5.29 × 10−11 m Z (a) For He+, Z = 2 r = 5.29 × 10−11 m 2 = 2.65 × 10−11 m = 0.0265 nm (b) For Li2+, Z = 3 r = 5.29 × 10−11 m 3 = 1.77 × 10−11 m = 0.0177 nm (c) For Be3+, Z = 4 r = 5.29 × 10−11 m 4 = 1.32 × 10−11 m = 0.0132 nm 40.51 Since F = qvB = mv2 r we have qrB = mv, or qr2B = mvr = nh so rn = nh qB 40.52 (a) The time for one complete orbit is: T = 2πr v From Bohr's quantization postulate, L = mevr = nh, we see that v = nh mer Thus, the orbital period becomes: T = 2πmer 2 nh = 2πme(a0n 2)2 nh = 2πmea0 2 h n3 or T = t0n3 where t0 = 2πmea0 2 h = 2π(9.11× 10−31 kg)(0.0529 × 10−9 m)2 (1.055 × 10−34 J ⋅s) = 1.52 10 s 16 × − (b) With n = 2, we have T = 8t0 = 8(1.52 × 10−16 s) = 1.21× 10−15 s Thus, if the electrons stay in the n = 2 state for 10 µs, it will make 10.0 × 10−6 s 1.21× 10−15 s/rev = 8.23 × 109 revolutions of the nucleus (c) Yes, for 8.23 × 109 "electron years" Chapter 40 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
40.53 λ = h p = h mv = 6.626 × 10–34 J · s (1.67 × 10–27 kg)(1.00 × 106 m/s) = 3.97 × 10–13 m 40.54 (a) p2 2m = (50.0)(1.60 × 10–19 J) p = 3.81 × 10–24 kg · m/s λ = h p = 0.174 nm (b) p2 2m = (50.0 × 103)(1.60 × 10–19 J) p = 1.20 × 10– 22 kg · m/s λ = h p = 5.49 × 10 –12 m The relativistic answer is slightly more precise: λ = h p = hc (mc2 + K)2 −m2c4 [ ] 1/2 = 5.37 × 10−12 m 40.55 (a) Electron: λ = h p and K = 1 2 mev2 = me 2v2 2me = p2 2me so p = 2meK and λ = h 2meK = 6.626 × 10−34 J ⋅s 2 9.11× 10-31 kg ( ) 3.00 ( ) 1.60 × 10−19 J ( ) λ = 7.09 × 10−10 m = 0.709 nm (b) Photon: λ = c/ f and E = hf so f = E h and λ = hc E = (6.626 × 10– 34 J · s)(3.00 × 108 m/s) (3.00)(1.60 × 10– 19 J) = 4.14 × 10–7 m = 414 nm 22 Chapter 40 Solutions 40.56 From the Bragg condition (Eq. 38.13), mλ = 2d sin θ = 2d cos φ 2 ( ) But, d = a sin φ 2 ( ) where a is the lattice spacing.
Thus, with m = 1, λ = 2a sin φ 2 ( ) cos φ 2 ( ) = a sin φ λ = h p = h 2meK = 6.626 × 10−34 J ⋅s 2 9.11× 10−31 kg ( ) 54.0 × 1.60 × 10−19 J ( ) = 1.67 × 10−10 m Therefore, the lattice spacing is a = λ sin φ = 1.67 × 10−10 m sin 50.0° = 2.18 × 10−10 m = 0.218 nm 40.57 (a) λ ~ 10−14 m or less. p = h λ ~ 6.6 × 10−34 J ⋅s 10−14 m = 10−19 kg ⋅m s or more.
The energy of the electron is E = p2c2 + me 2c4 ~ 10−19 ( ) 2 3 × 108 ( ) 2 + 9 × 10−31 ( ) 2 3 × 108 ( ) 4 1 2 ~ 10−11 J~ 108 eV or more, so that K = E −mec2 ~ 108 eV −0.5 × 106 eV ( ) ~ 108 eV or more.
(b) The electric potential energy of the electron would be Ue = keq1q2 r ~ 9 × 109 N ⋅m2 C2 ( ) 10−19 C ( ) −e ( ) 10−14 m ~ −105 eV With its kinetic energy much larger than its negative potential energy, the electron would immediately escape the nucleus .
Chapter 40 Solutions 23 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution The nucleus of an atom is on the order of 10−14 m in diameter. For an electron to be confined to a nucleus, its de Broglie wavelength would have to be of this order of magnitude or smaller. (a) What would be the kinetic energy of an electron confined to this region? (b) On the basis of this result, would you expect to find an electron in a nucleus? Explain. G : The de Broglie wavelength of a normal ground-state orbiting electron is on the order 10−10 m (the diameter of a hydrogen atom), so with a shorter wavelength, the electron would have more kinetic energy if confined inside the nucleus. If the kinetic energy is much greater than the potential energy from its attraction with the positive nucleus, then the electron will escape from its electrostatic potential well.
O : If we try to calculate the velocity of the electron from the de Broglie wavelength, we find that v = h meλ = 6.63 × 10−34 J ⋅s 9.11× 10-31 kg ( ) 10−14 m ( ) = 7.27 × 1010 m /s which is not possible since it exceeds the speed of light. Therefore, we must use the relativistic energy expression to find the kinetic energy of this fast-moving electron. A : (a) The relativistic kinetic energy of a particle is K = E −mc2, where E2 = pc ( ) 2 + mc2 ( ) 2, and the momentum is p = h λ : p = 6.63 × 10−34 J ⋅s 10-14 m = 6.63 × 10−20 N ⋅s E = 1.99 × 10−11 J ( ) 2 + 8.19 × 10−14 J ( ) 2 = 1.99 × 10−11 J K = E −mc2 = 1.99 × 10−11 J −8.19 × 10−14 J 1.60 × 10−19 J/eV = 124 MeV ~ 100 MeV (b) The electrostatic potential energy of the electron 10−14 m away from a positive proton is : U = −kee2 r = − 8.99 × 109 N ⋅m2 C2 1.60 × 10−19 C ( ) 2 10−14 m = −2.30 × 10−14 J ~ −0.1 MeV L : Since the kinetic energy is nearly 1000 times greater than the potential energy, the electron would immediately escape the proton’s attraction and would not be confined to the nucleus. It is also interesting to notice in the above calculations that the rest energy of the electron is negligible compared to the momentum contribution to the total energy.
24 Chapter 40 Solutions 40.58 (a) From E = γ me c2 γ = 20.0 × 103 MeV 0.511 MeV = 3.91 × 104 (b) p ≈ E c (for me c2 << pc) p = (2.00 × 104 MeV)(1.60 × 10–13 J/MeV) 3.00 × 108 m/s = 1.07 × 10–17 kg · m/s (c) λ = h p = 6.626 × 10–34 J · s 1.07 × 10–17 kg · m/s = 6.22 × 10–17 m Since the size of a nucleus is on the order of 10–14 m, the 20-GeV electrons would be small enough to go through the nucleus.
40.59 (a) E 2 = p 2 c 2 + m2c4 with E = hf , p = h λ , and mc = h λ C so h2f 2 = h2c2 λ2 + h c 2 2 2 λ C and f c 2 = 1 λ2 + 1 λ C 2 (Eq. 1) (b) For a photon f /c = 1/λ . The third term 1/λC in Equation 1 for electrons and other massive particles shows that they will always have a different frequency from photons of the same wavelength 40.60 (a) The wavelength of the student is λ = h p = h mv. If w is the width of the diffraction aperture, then we need w ≤10.0λ = 10.0 h mv ( ), so that v ≤10.0 h mw = 10.0 6.626 × 10−34 J ⋅s (80.0 kg)(0.750 m) = 1.10 × 10−34 m s (b) Using t = d v we get: t ≥ 0.150 m 1.10 × 10−34 m /s = 1.36 × 1033 s (c) No . The minimum time to pass through the door is over 1015 times the age of the Universe.
Chapter 40 Solutions 25 © 2000 by Harcourt, Inc. All rights reserved.
40.61 The de Broglie wavelength is: λ = h γ mev The Compton wavelength is: λ C = h mec Therefore, we see that to have λ = λ C, it is necessary that γ v = c.
This gives: v 1−v2 /c2 = c, or v c 2 = 1−v c 2 , yielding v = c 2 .
40.62 ∆VS = h e f −φ e From two points on the graph 0 = h e 4.1× 1014 Hz ( ) −φ e and 3.3 V = h e 12 × 1014 Hz ( ) −φ e Combining these two expressions we find: (a) φ = 1.7 eV (b) h e = 4.2 × 10 – 15 V · s (c) At the cutoff wavelength hc λ c = φ = h e ec λ c λ c = (4.2 × 10 – 15 V · s)(1.6 × 10 – 19 C) (3.0 × 108 m/s) (1.7 eV)(1.6 × 10 – 19 J/eV) = 730 nm 40.63 Kmax = q2B2R2 2m e = (1.60 × 10 – 19 C)2(2.00 × 10 – 5 T)2(0.200 m)2 2(9.11 × 10 – 31 kg) = 2.25 × 10 – 19 J = 1.40 eV = hf – φ φ = hf – Kmax = hc λ – Kmax= (4.14 × 10 – 15 eV · s)(3.00 × 108 m/s) 450 × 10 – 9 m – 1.40 eV = 1.36 eV 26 Chapter 40 Solutions 40.64 From the path the electrons follow in the magnetic field, the maximum kinetic energy is seen to be: Kmax = e2B2R2 2me From the photoelectric equation, Kmax = hf −φ = hc λ −φ Thus, the work function is φ = hc λ −Kmax = hc λ −e2B2R2 2me 40.65 We want an Einstein plot of Kmax versus f λ, nm f, 1014 Hz Kmax, eV 588 5.10 0.67 505 5.94 0.98 445 6.74 1.35 399 7.52 1.63 (a) slope = 0.402 eV 1014 Hz ± 8% (b) e(∆VS) = hf – φ h = (0.402) 1.60 × 10– 19 J · s 1014 = 6.4 × 10– 34 J · s ± 8% (c) Kmax = 0 at f ≈ 344 × 1012 Hz φ = hf = 2.32 × 10– 19 J = 1.4 eV f THz ( ) 40.66 ∆λ = h mpc (1−cos θ) = (6.626 × 10−34 J ⋅s) (1.67 × 10−27 kg)(3.00 × 108 m /s) (0.234) = 3.09 × 10−16 m λ 0 = hc E0 = (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) (200 MeV)(1.60 × 10−13 J/ MeV) = 6.20 × 10−15 m ′ λ = λ 0 + ∆λ = 6.51× 10−15 m (a) Eγ = hc ′ λ = 191 MeV (b) Kp = 9.20 MeV Chapter 40 Solutions 27 © 2000 by Harcourt, Inc. All rights reserved.
40.67 M is the mass of the positron which equals me, the mass of the electron.
So µ ≡reduced mass = meM me + M = me 2 rpos = n2h2 Zµkee2 = n2h2 Z me / 2 ( )kee2 = 2n2h2 Zmekee2 or rpos = 2rHyd = 1.06 × 10−10 m ( )n2 This is the separation of the two particles. Epos = −µke 2e4 2h2 1 n2 = −meke 2e4 4h2 1 n2 ; n = 1, 2, 3, . . .
or Epos = EHyd 2 = −6.80 eV n2 Goal Solution Positronium is a hydrogen-like atom consisting of a positron (a positively charged electron) and an electron revolving around each other. Using the Bohr model, find the allowed radii (relative to the center of mass of the two particles) and the allowed energies of the system.
G : Since we are told that positronium is like hydrogen, we might expect the allowed radii and energy levels to be about the same as for hydrogen: r = a0n2 = 5.29 × 10−11 m ( )n2 and En = −13.6 eV ( )/n2.
O : Similar to the textbook calculations for hydrogen, we can use the quantization of angular momentum of positronium to find the allowed radii and energy levels.
A : Let r represent the distance between the electron and the positron. The two move in a circle of radius r/2 around their center of mass with opposite velocities. The total angular momentum is quantized according to Ln = For each particle, ΣF = ma expands to kee2 r2 = mv2 r / 2 We can eliminate to find So the separation distances are The orbital radii are r/2 = a0n2, the same as for the electron in hydrogen.
The energy can be calculated from E = K + U = 1 2 mv2 + 1 2 mv2 −kee2 r Since mv2 = kee2 2r , E = kee2 2r −kee2 r = −kee2 2r = −kee2 4a0n2 = −6.80 eV n2 L : It appears that the allowed radii for positronium are twice as large as for hydrogen, while the energy levels are half as big. One way to explain this is that in a hydrogen atom, the proton is much more massive than the electron, so the proton remains nearly stationary with essentially no kinetic energy. However, in positronium, the positron and electron have the same mass and therefore both have kinetic energy that separates them from each other and reduces their total energy compared with hydrogen.
28 Chapter 40 Solutions 40.68 Isolate the terms involving φ in Equations 40.12 and 40.13. Square and add to eliminate φ . h2 1 λ 0 2 + 1 ′ λ 2 −2 cos θ λ 0 ′ λ = γ 2me 2v2 Solve for v2 c2 = b b + c2 ( ) : b = h2 me 2 1 λ 0 2 + 1 ′ λ 2 −2 cos θ λ 0 ′ λ Substitute into Eq. 40.11: 1+ h mec 1 λ 0 −1 ′ λ = γ = 1− b b + c2 Square each side: c2+ 2hc me 1 λ 0 −1 ′ λ + h2 me 2 1 λ 0 −1 ′ λ 2 = c2+ h2 me 2 1 λ 0 2 + 1 ′ λ 2 −2 cos θ λ 0 ′ λ From this we get Eq. 40.10: ′ λ −λ 0 = h mec ( ) 1−cos θ [ ] 40.69 hf = ∆E = 4π 2meke 2e4 2h2 1 (n −1)2 −1 n2 so f = 2π 2meke 2e4 h3 2n −1 (n −1)2n2 As n approaches infinity, we have f approaching 2π 2meke 2e4 h3 2 n3 The classical frequency is f = v 2πr = 1 2π kee2 me 1 r3/2 where r = n2h2 4πmekee2 Using this equation to eliminate r from the expression for f, f = 2π 2meke 2e4 h3 2 n3 40.70 Show that if all of the energy of a photon is transmitted to an electron, momentum will not be conserved.
Energy: hc λ 0 = hc ′ λ + Ke = mec2(γ −1) if hc ′ λ = 0 (1) Momentum: h λ 0 = h ′ λ + γ mev = γ mev if ′ λ = ∞ (2) From (1), γ = h λ 0mec + 1 (3) v = c 1− λ 0mec h + λ 0mec 2 (4) Substitute (3) and (4) into (2) and show the inconsistency: h λ 0 = 1+ h λ 0mec mec 1− λ 0mec h + λ 0mec 2 = λ 0mec + h λ 0 h(h + 2λ 0mec) (h + λ 0mec)2 = h λ 0 h + 2λ 0mec h This is impossible, so all of the energy of a photon cannot be transmitted to an electron.
40.71 Begin with momentum expressions: p = h λ , and p = γ mv = γ mc v c . Chapter 40 Solutions 29 © 2000 by Harcourt, Inc. All rights reserved.
Equating these expressions, γ v c = h mc 1 λ = λ C λ Thus, v c ( ) 2 1−v c ( ) 2 = λ C λ 2 or v c 2 = λ C λ 2 − λ C λ 2 v c 2 = λ C λ ( ) 2 1+ λ C λ ( ) 2 = 1 λ λ C ( ) 2 + 1 giving v = c 1+ (λ /λ C)2 40.72 (a) The energy of the ground state is: E1 = − hc λ series limit = −1240 eV ⋅nm 152.0 nm = – 8.16 eV From the wavelength of the Lα line, we see: E2 −E1 = hc λ = 1240 nm ⋅eV 202.6 nm = 6.12 eV E2 = E1 + 6.12 eV = – 2.04 eV Using the wavelength of the Lβ line gives: E3 −E1 = 1240 nm ⋅eV 170.9 nm = 7.26 eV so E3 = – 0 . 9 0 2 e V Next, using the Lγ line gives: E4 −E1 = 1240 nm ⋅eV 162.1 nm = 7.65 eV and E4 = – 0 . 5 0 8 e V From the Lδ line, E5 −E1 = 1240 nm ⋅eV 158.3 nm = 7.83 eV so E5 = – 0 . 3 2 5 e V (b) For the Balmer series, hc λ = Ei −E2, or λ = 1240 nm ⋅eV Ei −E2 For the α line, Ei = E3 and so λα = 1240 nm ⋅eV −0.902 eV ( ) −(−2.04 eV) = 1090 nm Similarly, the wavelengths of the β line, γ line, and the short wavelength limit are found to be: 811 nm , 724 nm , and 609 nm .
30 Chapter 40 Solutions (c) Computing 60.0% of the wavelengths of the spectral lines shown on the energy-level diagram gives: 0.600(202.6 nm) = 122 nm , 0.600(170.9 nm) = 103 nm , 0.600(162.1 nm) = 97.3 nm , 0.600(158.3 nm) = 95.0 nm , and 0.600(152.0 nm) = 91.2 nm . These are seen to be the wavelengths of the α , β , γ , and δ lines as well as the short wavelength limit for the Lyman series in Hydrogen.
(d) The observed wavelengths could be the result of Doppler shift when the source moves away from the Earth. The required speed of the source is found from ′ f f = λ ′ λ = 1−v c ( ) 1+ v c ( ) = 0.600 yielding v = 0.471c 40.73 (a) Starting with Planck’s law, I λ ,T ( ) = 2π hc2 λ5 ehc λ kBT −1 [ ] the total power radiated per unit area I λ ,T ( ) 0 ∞ ∫ dλ = 2π hc2 λ5 ehc λ kBT −1 [ ] dλ 0 ∞ ∫ .
Change variables by letting x = hc λ kBT and dx = −hcdλ kBTλ2 Note that as λ varies from 0 →∞, x varies from ∞→0.
Then I λ ,T ( ) 0 ∞ ∫ dλ = −2π kB 4T4 h3c2 x3 ex −1 ( ) dx ∞ 0 ∫ = 2π kB 4T4 h3c2 π 4 15 Therefore, I λ ,T ( ) 0 ∞ ∫ dλ = 2π 5 kB 4 15h3c2 T4 = σ T 4 (b) From part (a), σ = 2π 5 kB 4 15h3c2 = 2π 5 1.38 × 10−23 J K ( ) 4 15 6.626 × 10−34 J ⋅s ( ) 3 3.00 × 108 m s ( ) 2 σ = 5.67 × 10−8 W m2 ⋅K4 Chapter 40 Solutions 31 © 2000 by Harcourt, Inc. All rights reserved.
40.74 Planck’s law states I λ ,T ( ) = 2π hc2 λ5 ehc λ kBT −1 [ ] = 2π hc2λ−5 ehc λ kBT −1 [ ] −1 To find the wavelength at which this distribution has a maximum, compute dI dλ = 2π hc2 −5λ−6 ehc λ kBT −1 [ ] −1 −λ−5 ehc λ kBT −1 [ ] −2ehc λ kBT − hc λ2kBT = 0 dI dλ = 2π hc2 λ6 ehc λ kBT −1 [ ] −5 + hc λ kBT ehc λ kBT ehc λ kBT −1 [ ] = 0 Letting x = hc λ kBT , the condition for a maximum becomes xex ex −1 = 5.
We zero in on the solution to this transcendental equation by iterations as shown in the table below. The solution is found to be x xex ex −1 ( ) 4.00000 4.0746294 4.50000 4.5505521 5.00000 5.0339183 4.90000 4.9367620 4.95000 4.9853130 4.97500 5.0096090 4.96300 4.9979452 4.96900 5.0037767 4.96600 5.0008609 4.96450 4.9994030 4.96550 5.0003749 4.96500 4.9998890 4.96525 5.0001320 4.96513 5.0000153 4.96507 4.9999570 4.96510 4.9999862 4.965115 5.0000008 x = hc λ max kBT = 4.965115 and λ maxT = hc 4.965115kB Thus, λ maxT = 6.626075 × 10−34 J ⋅s ( ) 2.997925 × 108 m s ( ) 4.965115 1.380658 × 10−23 J K ( ) = 2.897755 × 10−3 m ⋅K This result is very close to Wien’s experimental value of λ maxT = 2.898 × 10−3 m ⋅K for this constant.
32 Chapter 40 Solutions 40.75 ∆λ = h mec (1−cos θ) = ′ λ −λ 0 ′ E = hc ′ λ = hc λ 0+ ∆λ = hc λ 0 + h mec (1−cos θ) −1 ′ E = hc λ 0 1+ hc mec2λ 0 (1−cos θ) −1 ′ E = hc λ 0 1+ hc mec2λ 0 (1−cos θ) −1 = E0 1+ E0 mec2 (1−cos θ) −1 40.76 r1 = 1 ( )2h2 Zµkee2 = h2 82 ( ) 207me ( )kee2 = a0 82 ( ) 207 ( ) = 0.0529 nm 82 ( ) 207 ( ) = 3.12 fm E1 = −13.6 eV 1 ( )2 207 1 82 1 2 = – 18.9 MeV 40.77 This is a case of Compton scattering with a scattering angle of 180°. ∆λ = ′ λ −λ 0 = h mec 1−cos 180° ( ) = 2h mec E0 = hc λ 0 , so λ 0 = hc E0 and ′ λ = λ 0 + ∆λ = hc E0 + 2h mec = hc E0 1+ 2E0 mec2 The kinetic energy of the recoiling electron is then K = E0 −hc ′ λ = E0 − E0 1+ 2E0 mec2 ( ) = E0 1+ 2E0 mec2 −1 1+ 2E0 mec2 = 2E0 2 mec2 1+ 2E0 mec2 Defining a ≡E0 mec2 , the kinetic energy can be written as K = 2E0a 1+ 2a = 2 hf ( )a 1+ 2a = 2h f a 1+ 2a ( )−1 where f is the frequency of the incident photon.
Chapter 40 Solutions 33 © 2000 by Harcourt, Inc. All rights reserved.
40.78 (a) Planck's radiation law predicts maximum intensity at a wavelength λmax we find from dI dλ = 0 = d dλ 2π hc2 λ−5 e(hc/λ kBT) −1 [ ] −1 0 = 2π hc2λ−5(−1) e(hc/λ kBT) −1 [ ] −2 e(hc/λ kBT) −hc/λ2kBT ( ) +2π hc2(−5)λ−6 e(hc/λ kBT) −1 [ ] −1 or −hce(hc/λ kBT) λ7 kBT e(hc/λ kBT) −1 [ ] 2 + 5 λ6 e(hc/λ kBT) −1 [ ] = 0 which reduces to 5 λ kBT hc ( ) e(hc/λ kBT) −1 [ ] = e(hc/λ kBT) Define x = hc λ kBT. Then we require 5ex −5 = xex.
Numerical solution of this transcendental equation gives x = 4.965 to four digits. So λ max = hc 4.965kBT , in agreement with Wien's law.
The intensity radiated over all wavelengths is I(λ ,T)dλ = A + B = 0 ∞ ∫ 2π hc2 dλ λ5 e(hc/λ kBT) −1 [ ] 0 ∞ ∫ Again, define x = hc λ kBT so λ = hc xkBT and dλ = −hc x2kBT ( )dx Then, A + B = −2π hc2 x5kB 5T5 hc dx h5c5x2 kBT ex −1 ( ) x=∞ 0 ∫ = 2π kB 4T4 h3c2 x3dx ex −1 ( ) 0 ∞ ∫ The integral is tabulated as π 4 /15, so (in agreement with Stefan's law) A + B = 2π 5 kB 4 T4 15h3 c2 The intensity radiated over wavelengths shorter than λ max is I(λ ,T)dλ = A = 2π hc2 dλ λ5 e(hc/λ kBT) −1 [ ] 0 λ max ∫ 0 λ max ∫ With x = hc λ kBT, this similarly becomes A = 2π kB 4 T4 h3c2 x3 dx ex −1 4.965 ∞ ∫ So the fraction of power or of intensity radiated at wavelengths shorter than λ max is A A + B = 2π kB 4T4 h3c2 π 4 15 − x3 dx ex −1 0 4.965 ∫ 2π 5 kB 4T4 15h3 c2 = 1−15 π 4 x3 dx ex −1 0 4.965 ∫ 34 Chapter 40 Solutions (b) Here are some sample values of the integrand, along with a sketch of the curve: Approximating the integral by trapezoids gives A A + B ≈1−15 π 4 4.870 ( ) = 0.2501 40.79 λ C = h mec and λ = h p : λ C λ = h/mec h/ p = p mec ; E2 = c2p2 + (mec2)2: p = E2 c2 −(mec)2 λ C λ = 1 mec E2 c2 −(mec)2 = 1 (mec)2 E2 c2 −(mec)2 = E mec2 2 −1 40.80 p = mv = 2mE = 2 1.67 × 10−27 kg ( ) 0.0400 eV ( ) 1.60 × 10−19 J/eV ( ) λ = h mv = 1.43 × 10– 10 m = 0.143 nm This is of the same order of magnitude as the spacing between atoms in a crystal so diffraction should appear.
40.81 Let u′ represent the final speed of the electron and let ′ γ = 1−′ u 2 c2 ( ) −1/2. We must eliminate β and u′ from the three conservation equations: hc λ 0 + γ mec2 = hc ′ λ + ′ γ mec2 h λ 0 + γ meu −h ′ λ cos θ = ′ γ me ′ u cos β h ′ λ sin θ = ′ γ me ′ u sin β Chapter 40 Solutions 35 © 2000 by Harcourt, Inc. All rights reserved.
Square Equations and and add: h2 λ 0 2 + γ 2 me 2u2 + h2 ′ λ 2 + 2hγ meu λ 0 −2h2 cos θ λ 0 ′ λ −2hγ meu cos θ ′ λ = ′ γ 2 me 2 ′ u 2 h2 λ 0 2 + h2 ′ λ 2 + γ 2 me 2u2 + 2hγ meu λ 0 −2hγ meu cos θ ′ λ −2h2 cos θ λ0 ′ λ = me 2 ′ u 2 1 − ′ u 2 /c2 Call the left-hand side b. Then b −b ′ u 2 c2 = me 2 ′ u 2 and ′ u 2 = b me 2 + b c2 = c2b me 2c2 + b Now square Equation and substitute to eliminate ′ γ : h2 λ2 + γ 2me 2c2 + h2 ′ λ 2 + 2hγ mec λ 0 −2h2 λ 0 ′ λ −2hγ mec ′ λ = me 2c2 1−′ u 2 /c2 = me 2c2 + b So we have h2 λ 0 2 + h2 ′ λ 2 + γ 2me 2c2 + 2hγ mec λ 0 −2hγ mec ′ λ −2h2 λ 0 ′ λ = mec2 + h2 λ 0 2 + h2 ′ λ 2 + γ 2me 2u2 + 2hγ meu λ 0 −2hγ meucosθ ′ λ −2h2 cosθ λ0 ′ λ Multiply through by λ 0 ′ λ me 2c2 λ 0 ′ λ γ 2 + 2h ′ λ γ mec −2hλ0 γ mec −2h2 me 2c2 = λ 0 ′ λ + λ 0 ′ λ γ 2 u2 c2 + 2h ′ λ uγ mec2 − 2hγ λ 0 u cos θ mec2 −2h2 cos θ me 2c2 λ 0 ′ λ γ 2 −1−γ 2 u2 c2 + 2hγ ′ λ mec 1−u c = 2hγ λ 0 mec 1−u cos θ c + 2h2 me 2c2 1−cos θ ( ) The first term is zero. Then ′ λ = λ 0 1−u cos θ ( ) c 1−u c + hγ −1 mec 1 1−u c 1−cos θ ( ) Since γ −1 = 1−u c ( ) 2 = 1−u c ( ) 1+ u c ( ) this result may be written as ′ λ = λ 0 1−u cos θ ( ) c 1−u c + h mec 1+ u c 1−u c 1−cos θ ( ) Chapter 41 Solutions 41.1 (a) λ = = × ⋅ × ( )( ) = − h mv 6 626 10 0 400 34 .
. J s 1.67 10 kg m/s -27 9 92 10 7 .
× − m (b) For destructive interference in a multiple-slit experiment, d m sinθ λ = + 1 2 with m = 0 for the first minimum. Then, θ λ = = ° − sin .
1 2 0 0284 d y L = tanθ so y L = = ( ) ° ( ) = tan .
tan .
θ 10 0 0 0284 m 4 96 . mm (c) We cannot say the neutron passed through one slit. We can only say it passed through the slits.
41.2 Consider the first bright band away from the center: d m sinθ λ = 6 00 10 0 400 200 1 1 20 10 8 1 10 .
sin tan .
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× ( ) = = × − − − m m λ λ = h m v e so m v h e = λ and K m v m v m h m e V e e e e = = = = ( ) 1 2 2 2 2 2 2 2 2 λ ∆ ∆V h eme = = × ⋅ ( ) × ( ) × ( ) × ( ) = − − − − 2 2 34 2 19 31 10 2 2 6 626 10 2 1 60 10 9 11 10 1 20 10 λ .
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. J s C kg m 105 V 41.3 (a) The wavelength of a non-relativistic particle of mass m is given by λ = = h p h mK / 2 where the kinetic energy K is in joules. If the neutron kinetic energy Kn is given in electron volts, its kinetic energy in joules is K Kn = × ( ) − 1 60 10 19 . J/eV and the equation for the wavelength becomes λ = = × ⋅ × ( ) × ( ) = − − − h mK Kn 2 6 626 10 2 1 67 10 1 60 10 34 27 19 .
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. J s kg J/eV 2 87 10 11 .
× − Kn m where Kn is expressed in electron volts.
(b) If Kn = = 1 00 1000 . keV eV, then Chapter 41 Solutions 495 41.4 λ = = h p h mK 2 , so K h m = 2 2 2 λ If the particles are electrons and λ ~ .
0 1 10 10 nm m = − , the kinetic energy in electron volts is K = × ⋅ ( ) × ( )( ) × = − − − 6 626 10 2 9 11 10 10 1 34 2 31 10 2 .
. J s kg m eV 1.602 10 J -19 ~ 102 eV 41.5 λ = h p p h = = × ⋅ × = × ⋅ − − − λ 6 626 10 1 00 10 6 63 10 34 11 23 .
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. J s m kg m/s (a) electrons: K p m e e = = × ⋅ ( ) × ( ) = − − 2 34 2 31 2 6 63 10 2 9 11 10 .
. J s J 15 1 . keV The relativistic answer is more precisely correct: K p c m c m c e e e = + ( ) − = 2 2 2 4 1 2 2 14 9 / /. keV (b) photons: E pc γ = = × ( ) × ( ) = − 6 63 10 3 00 10 23 8 .
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124 keV 41.6 The theoretical limit of the electron microscope is the wavelength of the electrons. If Ke = 40 0 . keV , then E K m c e e = + = 2 551 keV and p c E m c e = − = ( ) −( ) × × = × ⋅ − − 1 551 511 3 00 10 1 60 10 1 00 1 10 10 2 2 4 2 2 8 16 22 keV keV m/s J keV kg m/s .
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The electron wavelength, and hence the theoretical limit of the microscope, is then λ = = × ⋅ × ⋅ = × = − − − h p 6 626 10 6 03 10 34 12 .
. J s 1.10 10 kg m/s m 22 6 03 . pm 41.7 E K m c e = + = + = 2 1 00 0 511 1 51 .
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. MeV MeV MeV p E m c e 2 2 2 4 2 2 1 51 0 511 c MeV MeV 2 = − = ( ) −( ) .
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so p c = 1 42 . MeV/ λ = = = × ⋅ ( ) × ( ) × ( ) × ( ) = × − − − h p hc 1 42 6 626 10 3 00 10 1 42 10 1 60 10 8 74 10 34 8 6 19 13 .
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. MeV J s m/s J m Suppose the array is like a flat diffraction grating with openings 0 250 . nm apart: d m sinθ λ = −13 5 8 74 10 Chapter 41 Solutions 496 41.8 (a) ∆∆ ∆∆ p x m v x = ≥h/2 so ∆ ∆ v h m x ≥ = ⋅ ( )( ) = 4 2 2 00 1 00 π π π J s 4 kg m .
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0 250 . m/s (b) The duck might move by 0 25 5 1 25 .
. m/s s m ( )( ) = . With original position uncertainty of 1 0 . 0 m, we can think of ∆x growing to 1 00 1 25 .
. m m + = 2 25 . m 41.9 For the electron, ∆ ∆ p m v e = = × ( )( ) × ( ) = × ⋅ − − − 9 11 10 500 1 00 10 4 56 10 31 4 32 .
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. kg m/s kg m/s ∆ ∆ x h p = = × ⋅ × ⋅ ( ) = − − 4 6 626 10 4 56 10 34 32 π π .
. J s 4 kg m/s 1 16 . mm For the bullet, ∆ ∆ p m v = = ( )( ) × ( ) = × ⋅ − − 0 0200 500 1 00 10 1 00 10 4 3 .
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. kg m/s kg m/s ∆ ∆ x h p = = 4π 5 28 10 32 .
× − m Goal Solution An electron (me = × − 9 11 10 31 . kg ) and a bullet ( .
m = 0 0200 kg) each have a speed of 500 m/s, accurate to within 0 0100 .
%. Within what limits could we determine the position of the objects?
G: It seems reasonable that a tiny particle like an electron could be located within a more narrow region than a bigger object like a bullet, but we often find that the realm of the very small does not obey common sense.
O: Heisenberg’s uncertainty principle can be used to find the uncertainty in position from the uncertainty in the momentum.
A: The uncertainty principle states: ∆∆ x px ≥h/2 where ∆ ∆ p m v x = and h = h/2π .
Both the electron and bullet have a velocity uncertainty, ∆v = ( )( ) = 0 000100 500 0 0500 .
. m/s m/s For the electron, the minimum uncertainty in position is ∆ ∆ x h m v = = × ⋅ × ( )( ) = − − 4 6 63 10 9 11 10 0 0500 1 16 34 31 π π .
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. J s 4 kg m/s mm For the bullet, ∆ ∆ x h m v = = × ⋅ ( )( ) = × − − 4 6 63 10 0 0200 0 0500 5 28 10 34 32 π π .
. J s 4 kg m/s m Chapter 41 Solutions 497 Chapter 41 Solutions 498 41.10 ∆ ∆ y x p p y x = and d p h y ∆ ≥ /4π Eliminate ∆py and solve for x.
x p y d h x = ( ) = × ( )( ) × ( ) × ( ) × ⋅ ( ) = − − − − 4 4 1 00 10 100 1 00 10 2 00 10 6 626 10 3 2 3 34 π π ∆ .
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. kg m/s m m J s 3 79 1028 .
× m This is 190 times greater than the diameter of the Universe!
41.11 ∆∆ p x ≥h 2 so ∆ ∆ ∆ p m v h x e = ≥4π ∆ ∆ v h m x e ≥ = × ⋅ × ( ) × ( ) = − − − 4 6 626 10 9 11 10 34 31 π π .
. J s 4 kg 5.00 10 m 11 1 16 106 .
× m/s 41.12 With ∆x = × − 2 10 15 m, the uncertainty principle requires ∆ ∆ p x x ≥ = × ⋅ − h 2 2 6 10 20 . kg m/s The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty in momentum measures the root-mean-square momentum, so we take prms = × ⋅ − 3 10 20 kg m/s .
For an electron, the non-relativistic approximation p m v e = would predict v = × 3 1010 m/s, while vcannot be greater than c.
Thus, a better solution would be E m c pc m c e e = ( ) + ( ) = = 2 2 2 1 2 2 56 / MeV γ γ ≈ = − 110 1 1 2 2 v c / so v c ≈0 99996 .
For a proton, v p m = / gives v = × 1 8 107 . m / s, less than one-tenth the speed of light.
41.13 (a) At the top of the ladder, the woman holds a pellet inside a small region ∆xi. Thus, the uncertainty principle requires her to release it with typical horizontal momentum ∆ ∆ ∆ p m v x x x i = = h/2 . It falls to the floor in time given by H gt = + 0 1 2 2 as t H g = 2 , so the total width of the impact points is ∆ ∆ ∆ ∆ ∆ ∆ ∆ x x v t x m x H g x A x f i x i i i i = + ( ) = + = + h 2 2 , where A H = h 2 Chapter 41 Solutions 499 so ∆x A i = , and the minimum width of the impact points is ∆ ∆ ∆ ∆ x x A x A f i i x A i ( ) = + = = = min 2 2 2 1 2 1 4 h m H g = / / (b) ∆xf ( ) = × ⋅ ( ) × ( ) = − − min / / .
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2 1 0546 10 5 00 10 2 2 00 9 80 34 4 1 2 1 4 J s kg m m/s2 5 19 10 16 .
× − m Chapter 41 Solutions 500 41.14 Probability P x a x a dx a a x a a a a a a a = ( ) = + ( ) = − − − − ∫ ∫ ψ π π 2 2 2 1 1 tan P = − − ( ) [ ] = −− = − − 1 1 1 1 4 4 1 1 π π π π tan tan 1 2 / 41.15 (a) ψ π λ x A x A x ( ) = = × ( ) sin sin .
2 5 00 1010 so 2 5 00 1010 π λ = × − . m 1 λ π = × ( ) = 2 5 00 1010 .
1 26 10 10 .
× − m (b) p h = = × ⋅ × = − − λ 6 626 10 34 10 . J s 1.26 10 m 5 27 10 24 .
× ⋅ − kg m/s (c) m = × − 9 11 10 31 . kg K p m = = × ⋅ ( ) × × ( ) = × = × × = − − − − − 2 24 2 31 17 17 19 2 5 27 10 2 9 11 10 1 52 10 1 52 10 1 602 10 .
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. kg m/s kg J J J/eV 95 5 . eV 41.16 For an electron to “fit” into an infinitely deep potential well, an integral number of half-wavelengths must equal the width of the well.
nλ 2 1 00 10 9 = × − . m so λ = × = − 2 00 10 9 .
n h p (a) Since K p m h m h m n n e e e = = ( ) = × ( ) = ( ) − 2 2 2 2 2 9 2 2 2 2 2 2 10 0 377 / .
λ eV For K ≈6 eV , n = 4 (b) With n = 4 , K = 6 03 . eV 41.17 (a) We can draw a diagram that parallels our treatment of standing mechanical waves. In each state, we measure the distance d from one node to another (N to N), and base our solution upon that: Since dN to N = λ 2 and λ h Chapter 41 Solutions 501 Next, K p m h m d d e e = = = × ⋅ ( ) × ( ) − − 2 2 2 34 2 31 2 8 1 6 626 10 8 9 11 10 .
. J s kg Evaluating, K d = × ⋅ − 6 02 10 38 2 . J m2 K d = × ⋅ − 3 77 10 19 2 . eV m2 In state 1, d = × − 1 00 10 10 . m K1 37 7 = . eV In state 2, d = × − 5 00 10 11 . m K2 151 = eV In state 3, d = × − 3 33 10 11 . m K3 339 = eV In state 4, d = × − 2 50 10 11 . m K4 603 = eV (b) When the electron falls from state 2 to state 1, it puts out energy E hf hc = − = = = 151 37 7 113 eV eV eV .
λ i n t o e m i t t i n g a p h o to n o f w a v e l e n g t h λ = = × ⋅ ( ) × ( ) ( ) × ( ) = − − hc E 6 626 10 3 00 10 113 1 60 10 11 0 34 8 19 .
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. J s m/s eV J/eV nm The wavelengths of the other spectral lines we find similarly: T r a n s i t i o n E eV ( ) λ nm ( ) Chapter 41 Solutions 502 41.18 E1 19 2 00 3 20 10 = = × − .
. eV J For the ground-state, E h m L e 1 2 2 8 = (a) L h m E e = = × = − 8 4 34 10 1 10 . m 0 434 . nm (b) ∆E E E h m L h m L e e = − = − = 2 1 2 2 2 2 4 8 8 6 00 . eV 41.19 ∆E hc h m L h m L e e = = − [ ] = λ 2 2 2 2 2 2 8 2 1 3 8 L h m c e = = × = − 3 8 7 93 10 10 λ . m 0 793 . nm 41.20 ∆E hc h m L h m L e e = = − [ ] = λ 2 2 2 2 2 2 8 2 1 3 8 so L = 3 8 h m c e λ 41.21 E n h mL n = 2 2 2 8 so ∆E E E h mL hc mc L = − = = ( ) 2 1 2 2 2 2 2 3 8 3 8 and ∆E hf hc = = λ Hence, λ = = × ( ) × ( ) ⋅ ( ) − 8 3 8 938 10 1 00 10 3 1240 2 2 6 5 2 mc L hc eV nm eV nm .
λ = 2 02 10 4 .
× − nm (gamma ray) E hc = = λ 6 15 . MeV Fi gu re fo r G oa l So lu ti Chapter 41 Solutions 503 Goal Solution The nuclear potential energy that binds protons and neutrons in a nucleus is often approximated by a square well. Imagine a proton confined in an infinitely high square well of width 10 0 . fm , a typical nuclear diameter. Calculate the wavelength and energy associated with the photon emitted when the proton moves from the n = 2 state to the ground state. In what region of the electromagnetic spectrum does this wavelength belong?
G: Nuclear radiation from nucleon transitions is usually in the form of high energy gamma rays with short wavelengths.
O: The energy of the particle can be obtained from the wavelengths of the standing waves corresponding to each level. The transition between energy levels will result in the emission of a photon with this energy difference.
A: At level 1, the node-to-node distance of the standing wave is 1 00 10 14 .
× − m, so the wavelength is twice this distance: h p / .
= × − 2 00 10 14 m. The proton’s kinetic energy is K mv p m h m = = = = × ⋅ ( ) × ( ) × ( ) = × × = − − − − − 1 2 2 2 2 2 34 2 27 14 2 13 19 2 2 6 63 10 2 1 67 10 2 00 10 3 29 10 1 60 10 2 06 λ .
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. J s kg m J J/eV MeV In the first excited state, level 2, the node-to-node distance is two times smaller than in state 1. The momentum is two times larger and the energy is four times larger: K = 8 23 . MeV .
The proton has mass, has charge, moves slowly compared to light in a standing-wave state, and stays inside the nucleus. When it falls from level 2 to level 1, its energy change is 2 06 8 23 6 17 .
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. MeV MeV MeV − = − Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the speed of light, and that it has an energy of +6 17 . MeV.
Its frequency is f E h = = × ( ) × ( ) × ⋅ = × − − 6 17 1 60 10 6 63 10 1 49 10 19 34 21 .
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10 eV J/eV J s Hz 6 and its wavelength is λ = = × × = × − − c f 3 00 10 1 49 10 2 02 10 8 21 1 13 .
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. m/s s m This is a gamma ray, according to Figure 34.17.
L:The radiated photons are energetic gamma rays as we expected for a nuclear transition. In the above calculations, we assumed that the proton was not relativistic ( v c < 0 1 .
), but we should check this assumption for the highest energy state we examined (n = 2): v K m c = = × ( ) × ( ) × = × = − − 2 2 8 23 10 1 60 10 1 67 10 3 97 10 0 133 6 19 27 7 .
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. eV J/eV kg m/s This appears to be a borderline case where we should probably use relativistic equations, but our classical treatment should give reasonable results, within ( .
) % 0 133 1 2 = accuracy.
Chapter 41 Solutions 504 41.22 λ = 2D for the lowest energy state K p m h m h mD = = = = × ⋅ × ( ) [ ] × ( ) = × = − − − − 2 2 2 2 34 27 14 2 14 2 2 8 6 626 10 8 4 1 66 10 1 00 10 8 27 10 λ ( .
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. J s) kg m J 2 0 517 . MeV p h h D = = = × ⋅ × ( ) = − − λ 2 6 626 10 2 1 00 10 34 14 .
. J s m 3 31 10 20 .
× ⋅ − kg m/s 41.23 E h mL n n = 2 2 2 8 E1 34 2 27 14 2 14 6 626 10 8 1 67 10 2 00 10 8 21 10 = × ⋅ ( ) × ( ) × ( ) = × − − − − .
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. J s kg m J E1 = 0 513 . MeV E E 2 1 4 = = 2 05 . MeV E E 3 1 9 = = 4 62 . MeV 41.24 (a) < >= = − ∫ ∫ x x L x L dx L x x L dx L L 2 2 2 1 2 1 2 4 2 0 0 sin cos π π < >= − + = x L x L L x L x L x L L L 1 2 1 16 4 4 4 2 0 2 2 0 π π π π sin cos L/2 (b) Probability = = − ∫ 2 2 1 1 4 4 2 0 490 0 510 0 490 0 510 L x L dx L x L L x L L L L L sin sin .
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π π π Probability = − − ( ) = 0 20 1 4 2 04 1 96 .
sin .
sin .
π π π 5 26 10 5 .
× − (c) Probability = − = x L x L L L 1 4 4 0 240 0 260 π π sin .
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3 99 10 2 .
× − (d) In the n = 2 graph in Figure 41.11 (b), it is more probable to find the particle either near x L = 4 or x L = 3 4 than at the center, where the probability density is zero.
Chapter 41 Solutions 505 A n x L dx A L x L 2 2 2 0 2 1 sin π = = = ∫ or A L = 2 41.26 The desired probability is P dx L x L dx x x L L = = = = ∫ ∫ ψ π 2 0 4 2 0 4 2 2 / / sin where sin cos 2 1 2 2 θ θ = − Thus, P x L x L L = − = − − + = 1 4 4 1 4 0 0 0 0 4 π π sin / 0 250 .
41.27 In 0 ≤ ≤ x L, the argument 2π x L / of the sine function ranges from 0 to 2π . The probability density 2 2 2 / sin / L x L ( ) ( ) π reaches maxima at sinθ = 1 and sinθ = −1 at 2 2 π π x L = and 2 3 2 π π x L = ∴ The most probable positions of the particle are at x L x L = = 4 3 4 and 41.28 (a) The probability is P dx L x L dx L x L dx L L L = = = − ∫ ∫ ∫ ψ π π 2 0 3 2 0 3 0 3 2 2 1 2 1 2 2 / / / sin cos P x L x L L = − = − = − = 1 2 2 1 3 1 2 2 3 1 3 3 4 0 3 π π π π π sin sin / 0 196 .
(b) The probability density is symmetric about x L = /2. Thus, the probability of finding the particle between x L = 2 3 / and x L = is the same 0 196 .
. Therefore, the probability of finding it in the range L x L / / 3 2 3 ≤ ≤ is P = −( ) = 1 00 2 0 196 0 609 .
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(c) Classically, the electron moves back and forth with constant speed between the walls, and the probability of finding the electron is the same for all points between the walls. Thus, the classical probability of finding the electron in any range equal to one-third of the available space is P classical = 1 3 / .
Chapter 41 Solutions 506 41.29 The ground state energy of a particle (mass m) in a 1-dimensional box of width L is E h mL 1 2 2 8 = .
(a) For a proton m = × ( ) − 1 67 10 27 . kg in a 0 200 .
- nm wide box: E1 34 2 27 10 2 22 6 626 10 8 1 67 10 2 00 10 8 22 10 = × ⋅ ( ) × ( ) × ( ) = × = − − − − .
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. J s kg m J 5 13 10 3 .
× − eV (b) For an electron m = × ( ) − 9 11 10 31 . kg in the same size box: E1 34 2 31 10 2 18 6 626 10 8 9 11 10 2 00 10 1 51 10 = × ⋅ ( ) × ( ) × ( ) = × = − − − − .
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. J s kg m J 9 41 . eV (c) The electron has a much higher energy because it is much less massive.
41.30 (a) ψ π 1 2 x L x L ( ) = cos P x x L x L 1 1 2 2 2 ( ) = ( ) = ψ π cos ψ π 2 2 2 x L x L ( ) = sin P x x L x L 2 2 2 2 2 2 ( ) = ( ) = ψ π sin ψ π 3 2 3 x L x L ( ) = cos P x x L x L 3 3 2 2 3 ( ) = ( ) = ψ π cos 41.31 We have ψ ω = − ( ) Aei kx t and ∂ψ ∂ ψ 2 2 2 x k = − Schrödinger’s equation: ∂ψ ∂ ψ ψ 2 2 2 2 2 x k m E U = − = − ( ) h Since k p p 2 2 2 2 2 2 2 2 2 = ( ) = ( ) = π λ π h h and E U p m − ( ) = 2 2 / Chapter 41 Solutions 507 41.32 ψ x A kx B kx ( ) = + cos sin ∂ψ ∂x kA kx kB kx = − + sin cos ∂ψ ∂ 2 2 2 2 x k A k x k B kx = − − cos sin − − ( ) = − + ( ) 2 2 2 2 m E U mE A kx B kx h h ψ cos sin Therefore the Schrödinger equation is satisfied if ∂ψ ∂ ψ 2 2 2 2 x m E U = − − ( ) h or − + ( ) = − + ( ) k A kx B kx mE A kx B kx 2 2 2 cos sin cos sin h This is true as an identity (functional equality) for all x if E k m = h2 2 2 41.33 Problem 45 in Ch. 16 helps students to understand how to draw conclusions from an identity.
(a) ψ x A x L ( ) = − 1 2 2 d dx Ax L ψ = −2 2 ∂ψ ∂ 2 2 2 2 x A L = − Schrödinger’s equation ∂ψ ∂ ψ 2 2 2 2 x m E U = − − ( ) h becomes − = − + − ( ) − − ( ) 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 A L m EA x L m x A x L mL L x h h h − = − + − 1 2 2 2 2 2 2 4 L mE mEx L x L h h This will be true for all x if both 1 2 2 L mE = h and mE L L h2 2 4 1 0 − = Both of these conditions are satisfied for a particle of energy E L m = h2 2 .
(b) For normalization, 1 1 1 2 2 2 2 2 2 2 2 4 4 = − = − + − − ∫ ∫ A x L dx A x L x L dx L L L L 1 2 3 5 2 3 5 2 3 5 16 15 2 3 2 5 4 2 2 = − + = − + + − + = − A x x L x L A L L L L L L A L L L A L = 15 16 (c) Chapter 41 Solutions 508 P = = 47 81 0 580 .
Chapter 41 Solutions 509 41.34 (a) Setting the total energy E equal to zero and rearranging the Schrödinger equation to isolate the potential energy function gives U x m d dx ( ) = h2 2 2 2 1 ψ ψ If ψ x Axe x L ( ) = − 2 2 / Then d dx Ax AxL e L x L 2 2 3 2 4 4 6 2 2 ψ = − ( ) − / or d dx x L L x 2 2 2 2 4 4 6 ψ ψ = − ( ) ( ) and U x mL x L ( ) = − h2 2 2 2 2 4 6 S ee figure to the right .
41.35 (a) S ee figure to the right .
(b) The wavelength of the transmitted wave traveling to the left is the same as the original wavelength, which equals 2L .
4 41.37 T e CL = −2 (Use Equation 41.17) 2 2 2 9 11 10 8 00 10 1 055 10 2 00 10 4 58 31 19 34 10 CL = × ( ) × ( ) × × ( ) = − − − − .
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F i Chapter 41 Solutions 510 o r G o a l S o l u t i o n Chapter 41 Solutions 511 Goal Solution An electron with kinetic energy E = 5 00 . eV is incident on a barrier with thickness L = 0 100 . nm and height U = 10 0 . eV (Fig. P41.37). What is the probability that the electron (a) will tunnel through the barrier and (b) will be reflected?
G: Since the barrier energy is higher than the kinetic energy of the electron, transmission is not likely, but should be possible since the barrier is not infinitely high or thick.
O: The probability of transmission is found from the transmission coefficient equation 41.18.
A: The transmission coefficient is C m U E = − ( ) = × ( ) − ( ) × ( ) × ⋅ = × − − − 2 2 9 11 10 10 0 5 00 1 60 10 6 63 10 1 14 10 31 19 34 10 h .
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. kg eV eV J/eV J s/2 m-1 π (a) The probability of transmission is T e e e CL = = = = − − × ( ) × ( ) − − − 2 2 1 14 10 2 00 10 4 58 10 1 10 0 0103 .
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. m m (b) If the electron does not tunnel, it is reflected, with probability 1 0 0103 0 990 − = .
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L:Our expectation was correct: there is only a 1% chance that the electron will penetrate the barrier. This tunneling probability would be greater if the barrier were thinner, shorter, or if the kinetic energy of the electron were greater.
41.38 C = × ( ) − ( ) × ( ) ⋅ × ⋅ − − 2 9 11 10 5 00 4 50 1 60 10 31 19 .
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. kg m/s 1.055 10 J s -34 T e CL = = − × ( ) × ( ) [ ] = − ( ) − − − 2 9 1 12 2 3 62 10 950 10 6 88 exp .
exp . m m T = 1 03 10 3 .
× − 41.39 From problem 38, C = × − 3 62 109 1 . m 10 2 3 62 10 6 9 1 − − = − × ( ) [ ] exp . m L Taking logarithms, − = − × ( ) − 13 816 2 3 62 109 1 .
. m L New L = 1 91 . nm Chapter 41 Solutions 512 41.40 With the wave function proportional to e CL − , the transmission coefficient and the tunneling current are proportional to ψ 2 , to e CL − .
Then, I I e e e 0 500 0 515 2 10 0 0 500 2 10 0 0 515 20 0 0 015 .
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. nm nm /nm nm /nm nm ( ) ( ) = = = −( )( ) −( )( ) ( ) 1 35 .
41.41 With transmission coefficient e CL − , the fractional change in transmission is e e e e L L L −( ) −( ) + ( ) −( ) ( ) − = − = = 2 10 0 2 10 0 0 00200 2 10 0 29 0 0 00200 1 0 0392 .
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. /nm /nm nm /nm 3 92 .
% 41.42 ψ ω = −( ) Be m x /2 2 h so d dx m x ψ ω ψ = − h and d dx m x m 2 2 2 2 ψ ω ψ ω ψ = + − h h Substituting into Equation 41.19 gives m x m mE m x ω ψ ω ψ ψ ω ψ h h h h + − = + 2 2 2 2 2 2 which is satisfied provided that E = hω 2 .
41.43 Problem 45 in Chapter 16 helps students to understand how to draw conclusions from an identity.
ψ = − Axe bx2 so d dx Ae bx Ae bx bx ψ = − − − 2 2 2 2 and d dx bxAe bxAe b x Ae b b x bx bx bx 2 2 2 3 2 2 2 4 4 6 4 2 2 2 ψ ψ ψ = − − + = − + − − − Substituting into Equation 41.19, − + = − + 6 4 2 2 2 2 2 b b x mE m x ψ ψ ψ ω ψ h h For this to be true as an identity, it must be true for all values of x.
So we must have both − = − 6 2 2 b mE h and 4 2 2 b m = ω h (a) Therefore b m = ω 2h (b) and E b m = = 3 2 h 3 2hω (c) The wave function is that of the f irst excited state .
Chapter 41 Solutions 513 41.44 The longest wavelength corresponds to minimum photon energy, which must be equal to the spacing between energy levels of the oscillator: hc k m λ ω = = h h so λ π π = = × ( ) × = − 2 2 3 00 10 9 11 10 8 99 8 31 1 2 c m k .
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/ m/s kg N/m 600 nm 41.45 (a) With ψ ω = −( ) Be m x /2 2 h , the normalization condition ψ 2 1 dx = ∫all becomes 1 2 2 1 2 2 2 2 2 2 2 0 2 2 2 = = = −( ) −∞ ∞ −( ) ∞ ∫ ∫ B e dx B e dx B m m x m x ω ω π ω / / / h h h where Table B.6 in Appendix B was used to evaluate the integral.
Thus, 1 2 = B m π ω h and B m = ω π h 1 4 / (b) For small δ , the probability of finding the particle in the range − ≤ ≤ δ δ / / 2 2 x is ψ δ ψ δ δ δ 2 2 2 2 2 0 0 − − ∫ = ( ) = = / / dx B e δ ω π m h 1 2 / 41.46 (a) With < >= x 0 and < >= px 0, the average value of x2 is ∆x ( )2 and the average value of px 2 is ∆px ( ) 2. Then ∆ ∆ x px ≥h/2 requires E p m k p x x ≥ + = 2 2 2 2 2 4 h p m k p x x 2 2 2 2 8 + h (b) To minimize this as a function of px 2, we require dE dp m k p x x 2 2 4 0 1 2 8 1 1 = = + − ( ) h Then k p m x h2 4 8 1 2 = p mk mk x 2 2 1 2 2 8 2 = = h h / and E mk m k mk k m k m ≥( ) + = + h h h h h 2 2 3 8 4 4 2 .
Chapter 41 Solutions 514 41.47 Suppose the marble has mass 20 g. Suppose the wall of the box is 12 cm high and 2 mm thick.
While it is inside the wall, U mgy = = ( )( )( ) = 0 02 9 8 0 12 0 0235 .
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. kg m/s m J 2 and E K mv = = = ( )( ) = 1 2 2 1 2 2 0 02 0 8 0 0064 .
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. kg m/s J Then C m U E = − ( ) = ( )( ) × ⋅ = × − − 2 2 0 02 0 0171 1 055 10 2 5 10 34 32 1 h .
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. kg J J s m and the transmission coefficient is e e e e CL − − × ( ) × ( ) − × − × ( ) − × = = = = = − 2 2 2 5 10 2 10 10 10 2 30 4 3 10 4 3 10 32 3 29 29 29 10 .
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~ 10 1030 − 41.48 (a) λ = = 2L 2 00 10 10 .
× − m (b) p h = = × ⋅ × = − − λ 6 626 10 34 10 . J s 2.00 10 m 3 31 10 24 .
× ⋅ − kg m/s (c) E p m = = 2 2 0 172 . eV 41.49 (a) S ee the first figure to the right .
(b) S ee the second figure to the right .
(c) ψ is continuous and ψ →0 as x →±∞ (d) Since ψ is symmetric, ψ ψ 2 2 0 2 1 dx dx = = ∞ −∞ ∞ ∫ ∫ or 2 2 2 1 2 2 0 2 0 A e dx A e e x − ∞ −∞ ∫ = − − ( ) = α α This gives A = α Chapter 41 Solutions 515 (a) Use Schrödinger’s equation ∂ψ ∂ ψ 2 2 2 2 x m E U = − − ( ) h with solutions ψ1 1 1 = + − Ae Be ik x ik x [region I] ψ 2 2 = Ceik x [region II] Where k mE 1 2 = h and k m E U 2 2 = − ( ) h Then, matching functions and derivatives at x = 0: ψ ψ 1 0 2 0 ( ) = ( ) ⇒ + = A B C and d dx d dx k A B k C ψ ψ 1 0 2 0 1 2 = ⇒ − ( ) = Then B k k k k A = − + 1 1 2 1 2 1 C k k A = + 2 1 2 1 / Incident wave Aeikx reflects Be ikx − , with probability R B A k k k k = = − ( ) + ( ) = 2 2 2 1 2 2 1 2 1 1 / / k k k k 1 2 2 1 2 2 − ( ) + ( ) (b) With E = 7 00 . eV and U = 5 00 . eV, k k E U E 2 1 2 00 7 00 0 535 = − = = .
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The reflection probability is R = − ( ) + ( ) = 1 0 535 1 0 535 2 2 .
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0 0920 .
The probability of transmission is T R = − = 1 0 908 .
41.51 R k k k k k k k k = − ( ) + ( ) = − ( ) + ( ) 1 2 2 1 2 2 2 1 2 2 1 2 1 1 / / h2 2 2 k m E U = − for constant U h k m E 2 1 2 2 = since U = 0 (1) Chapter 41 Solutions 516 Dividing (2) by (1), k k U E 2 2 1 2 1 1 1 2 1 2 = − = − = so k k 2 1 1 2 = and therefore, R = − ( ) + ( ) = − ( ) + ( ) = 1 1 2 1 1 2 2 1 2 1 2 2 2 2 / / 0 0294 .
Chapter 41 Solutions 517 41.52 (a) The wave functions and probability densities are the same as those shown in the two lower curves in Figure 41.11 of the textbook.
(b) P dx x dx x x 1 1 2 0 150 2 0 150 0 150 0 350 2 1 00 1 00 2 00 2 1 00 4 2 1 00 = = = − ∫ ∫ ψ π π π .
sin .
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sin .
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. nm nm nm nm nm nm 0.350 nm nm 0.350 nm nm nm In the above result we used sin / / sin 2 2 1 4 2 axdx x a ax = ( ) −( ) ( ) ∫ P x x 1 0 150 0 350 1 00 1 00 2 2 1 00 = − .
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sin .
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nm nm nm nm nm π π P 1 1 00 0 350 0 1 1 00 2 0 700 0 300 = − − ( ) − ( ) [ ] = .
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sin .
sin .
nm nm 50 nm nm π π π 0 200 .
(c) P x dx x x 2 2 0 150 0 350 0 150 0 350 2 1 00 2 1 00 2 00 2 1 00 8 4 1 00 = = − ∫ .
sin .
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sin .
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π π π P x x 2 0 150 0 350 1 00 1 00 4 4 1 00 1 00 0 350 0 150 1 00 4 1 40 0 600 = − = − ( ) − ( ) − ( ) [ ] = .
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sin .
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sin .
sin .
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π π π π π 0 351 .
(d) Using E n h mL n = 2 2 2 8 , we find that E1 = 0 377 . eV and E2 = 1 51 . eV 41.53 (a) mgy mv i f = 1 2 2 v gy f i = = ( )( ) = 2 2 9 80 50 0 31 3 .
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. m/s m m/s 2 λ = = × ⋅ ( )( ) = − h mv 6 626 10 31 3 34 .
. J s 75.0 kg m/s 2 82 10 37 .
× − m (not observable) (b) ∆∆ E t ≥h/2 so ∆E ≥ × ⋅ × ( ) = − − 6 626 10 5 00 10 34 3 .
. J s 4 s π 1 06 10 32 .
× − J (c) ∆E E = × ( )( )( ) = − 1 06 10 75 0 9 80 50 0 32 . J kg m/s m 2 2 87 10 35 .
% × − Chapter 41 Solutions 518 41.54 From the uncertainty principle ∆∆ E t −h/2 or ∆ ∆ mc t 2 2 ( ) = h/ . Therefore, ∆ ∆ ∆ m m h c t m h t ER = ( ) = ( ) = × ⋅ × ( )( ) × = − − 4 4 6 626 10 4 8 70 10 135 1 2 34 17 π π π .
. J s s MeV MeV 1.60 10 J -13 2 81 10 8 .
× − Chapter 41 Solutions 519 41.55 (a) f E h = = × ⋅ × = − − 180 1 60 10 1 00 19 eV 6.626 10 J s J eV 34 .
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4 34 1014 .
× Hz (b) λ = = × × = × = − c f 3 00 10 4 34 10 6 91 10 8 14 7 .
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. m/s Hz m 691 nm (c) ∆∆ E t ≥h 2 so ∆ ∆ ∆ E t h t ≥ = = × ⋅ × ( ) = × = − − − h 2 4 6 626 10 2 00 10 2 64 10 34 8 29 π π .
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. J s 4 s J 1 65 10 10 .
× − eV 41.56 (a) f = E h (b) λ = = c f hc E (c) ∆∆ E t ≥h 2 so ∆ ∆ E t ≥ = h 2 h T 4π 41.57 x x dx 2 2 2 = −∞ ∞ ∫ ψ For a one-dimensional box of width L , ψ π n L n x L = 2 sin Thus, x L x n x L dx L 2 2 2 0 2 = = ∫ sin π L L n 2 2 2 2 3 2 − π (from integral tables) 41.58 (a) ψ 2 1 dx −∞ ∞ ∫ = becomes A x L dx A L x L x L A L L L L L 2 2 4 4 2 4 4 2 2 2 1 4 4 2 2 1 cos sin / / / / π π π π π π = + = = − − ∫ or A L 2 4 = and A L = 2 (b) The probability of finding the particle between 0 and L/8 is ψ π π 0 8 2 2 2 0 8 2 1 4 1 2 L L dx A x L dx / / cos ∫ ∫ = = + = 0 409 .
Chapter 41 Solutions 520 41.59 For a particle with wave function ψ x ae x a ( ) = − 2 / for x > 0 and 0 for x < 0 (a) ψ x ( ) = 2 0 , x < 0 and ψ 2 2 2 x a e x a ( ) = − / , x > 0 (b) Prob x x dx dx < ( ) = ( ) = ( ) = −∞ −∞ ∫ ∫ 0 0 2 0 0 ψ 0 (c) Normalization ψ ψ ψ x dx dx dx ( ) = + = −∞ ∞ −∞ ∞ ∫ ∫ ∫ 2 2 0 2 0 1 0 2 0 1 1 0 2 2 0 0 dx a e dx e e x a x a −∞ − − ∞ ∞ −∞ ∫ ∫ + ( ) = − = − − ( ) = / / / Prob 0 2 1 2 0 2 0 2 0 2 < < ( ) = = ( ) = = − = ∫ ∫ − − − x a dx a e dx e e a x a a x a a ψ / / / 0 865 .
41.60 (a) λ = = − = + ( ) −( ) h p hc E m c hc m c K m c e e e 2 2 4 2 2 2 2 λ = × ⋅ ( ) × ( ) ( ) −( ) × = − − 6 626 10 3 00 10 576 511 1 34 8 2 2 .
. J s m/s keV keV keV 1.60 10 J 16 4 68 10 12 .
× − m (b) 50 0 . λ = 2 34 10 10 .
× − m 41.61 (a) ∆∆ x p ≥h 2 so if ∆x r = , ∆p ≥h 2r (b) Choosing ∆p r = h , K p m p m e e = = ( ) = 2 2 2 2 ∆ h2 2 2m r e U k e r e = − 2 , so E K U = + = h2 2 2 2m r k e r e e − (c) To minimize E, dE k e e h 2 h2 Chapter 41 Solutions 521 Then, E m m k e k e m k e m k e e e e e e e e e = − = − = h h h h 2 2 2 2 2 2 2 2 4 2 2 2 −13 6 . eV 41.62 (a) The requirement that n L λ 2 = so p h nh L = = λ 2 is still valid.
E pc mc E nhc L mc n = ( ) +( ) ⇒ = +( ) 2 2 2 2 2 2 2 K E mc n n = − = 2 nhc L mc mc 2 2 2 2 2 +( ) − (b) Taking L = × − 1 00 10 12 . m , m = × − 9 11 10 31 . kg , and n = 1, we find K1 = 4 69 10 14 .
× − J Nonrelativistic, E h mL 1 2 2 34 31 12 2 14 8 6 626 10 8 9 11 10 1 00 10 6 02 10 = = × ⋅ ( ) × ( ) × ( ) = × − − − − .
.
.
. J s kg m J 2 Comparing this to K1, we see that this value is too large by 28 6 . % .
41.63 (a) U e d e d = −+ − + −+ + − ( ) = − ( ) = 2 0 2 0 4 1 1 2 1 3 1 1 2 1 7 3 4 π π e e / −7 3 2 k e d e (b) From Equation 41.9, K E h m d e = = ( ) = 2 2 8 9 1 2 2 h m d e 2 2 36 (c) E U K = + and dE dd = 0 for a minimum: 7 3 18 0 2 2 2 3 k e d h m d e e − = d h k e m h m k e e e e e = ( )( ) = = × ( ) ( ) × ( ) × ( ) × ( ) = − − − 3 7 18 42 6 626 10 42 9 11 10 8 99 10 1 602 10 2 2 2 2 34 2 31 9 19 2 .
.
.
. C 0 0499 . nm (d) Since the lithium spacing is a, where Na V 3 = , and the density is Nm V / , where m is the mass of one atom, we get: a Vm Nm m = = = × × = × = − − 1 3 1 3 27 1 3 10 1 66 10 530 2 80 10 / / / .
.
density kg 7 kg m m 0 280 . nm (5.62 times larger than c).
Chapter 41 Solutions 522 41.64 (a) ψ ω = −( ) Bxe m x /2 2 h d dx Be Bx m xe Be B m x e m x m x m x m x ψ ω ω ω ω ω ω = + − = − −( ) −( ) −( ) −( ) / / 2 2 2 2 2 2 2 2 2 2 2 h h h h h h d dx Bx m xe B m xe B m x m xe m x m x m x 2 2 2 2 2 2 2 2 2 2 ψ ω ω ω ω ω ω ω = − − − − −( ) −( ) −( ) h h h h h h h / / d dx B m xe B m x e m x m x 2 2 2 3 2 3 2 2 ψ ω ω ω ω = − + −( ) −( ) h h h h / / Substituting into the Schrödinger Equation (41.19), we have − + = − + −( ) −( ) −( ) −( ) 3 2 2 3 2 2 2 2 2 2 2 2 2 2 B m xe B m x e mE Bxe m x Bxe m x m x m x m x ω ω ω ω ω ω ω h h h h h h h h / / / / This is true if − = − 3 2 ω E h ; it is true if E = 3 2hω (b) We never find the particle at x = 0 because ψ = 0 there.
(c) ψ is maximized if d dx x m ψ ω = = − 0 1 2 h , which is true at x m = ± h ω (d) We require ψ 2 1 dx = −∞ ∞ ∫ : 1 2 2 1 4 2 2 2 2 2 0 2 3 2 1 2 3 2 3 2 2 2 = = = ( ) = ( ) −( ) −( ) ∞ −∞ ∞ ∫ ∫ B x e dx B x e dx B m B m m x m x ω ω π ω π ω / / / / / / h h h h Then B m = = 21 2 1 4 3 4 / / / π ω h 4 3 3 3 1 4 m ω π h / (e) At x m = 2 h/ ω , the potential energy is 1 2 2 2 1 2 2 4 2 m x m m ω ω ω ω = ( ) = h h / . This is larger than the total energy 3 2 hω/ , so there is z ero classical probability of finding the particle here.
(f) Probability = = = −( ) −( ) ψ δ δ ω ω 2 2 2 2 2 2 2 dx Bxe B x e m x m x / / h h Probability = = −( ) ( ) δ π ω ω ω ω 2 4 1 2 3 2 4 / / / / m m e m m h h h h 8 1 2 4 δ ω π m e h − / Chapter 41 Solutions 523 41.65 (a) ψ 2 0 1 dx L = ∫ : A x L x L x L x L dx L 2 2 2 0 16 2 8 2 1 sin sin sin sin π π π π + + = ∫ A L L x L x L dx L 2 0 2 16 2 8 2 1 + + = ∫sin sin π π A L x L x L dx A L L x L L x x L 2 2 0 2 3 0 17 2 16 17 2 16 3 1 + = + = ∫ = = sin cos sin π π π π A L 2 2 17 = , so the normalization constant is A L = 2 17 / (b) ψ 2 1 dx a a = − ∫ : A x a B x a A B x a x a dx a a 2 2 2 2 2 2 2 1 cos sin cos sin π π π π + + = − ∫ The first two terms are A a 2 and B a 2 . The third term is: 2 2 2 2 2 4 2 2 8 3 2 0 2 3 A B x a x a x a dx A B x a x a dx a A B x a a a a a a a cos sin cos cos sin cos π π π π π π π = = = − − − ∫ ∫ so that a A B 2 2 1 + ( ) = , giving A B a 2 2 1 + = / .
41.66 With one slit open P 1 1 2 = ψ or P 2 2 2 = ψ With both slits open, P = + ψ ψ 1 2 2 At a maximum, the wave functions are in phase P max = + ( ) ψ ψ 1 2 2 At a minimum, the wave functions are out of phase P min = − ( ) ψ ψ 1 2 2 Now P P 1 1 2 2 25 0 = = ψ . , so ψ ψ 1 5 00 = .
Chapter 41 Solutions 524 41.67 (a) The light is unpolarized. It contains both horizontal and vertical field oscillations.
(b) The interference pattern appears, but with diminished overall intensity.
(c) The results are the same in each case.
(d) The interference pattern appears and disappears as the polarizer turns, with alternately increasing and decreasing contrast between the bright and dark fringes. The intensity on the screen is precisely zero at the center of a dark fringe four times in each revolution, when the filter axis has turned by 45°, 135°, 225°, and 315° from the vertical.
(e) Looking at the overall light energy arriving at the screen, we see a low-contrast interference pattern.
After we sort out the individual photon runs into those for trial 1, those for trial 2, and those for trial 3, we have the original results replicated: The runs for trials 1 and 2 form the two blue graphs in Figure 41.3, and the runs for trial 3 build up the red graph.
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 42 Solutions 42.1 (a) The point of closest approach is found when E = K + U = 0 + keqαqAu r or rmin = ke 2e ( ) 79e ( ) E rmin = 8.99 × 10 9 N ⋅m2 C 2 ( )158 1.60 × 10−19 C ( ) 2 4.00 MeV ( ) 1.60 × 10−13 J MeV ( ) = 5.68 × 10−14 m (b) The maximum force exerted on the alpha particle is Fmax = keqαqAu rmin 2 = 8.99 × 10 9 N ⋅m2 C 2 ( )158 1.60 × 10−19 C ( ) 2 5.68 × 10−14 m ( ) 2 = 11.3 N away from the nucleus 42.2 (a) The point of closest approach is found when E = K + U = 0 + keqαqT r or rmin = ke 2e ( ) Ze ( ) E = 2Zkee2 E (b) The maximum force exerted on the alpha particle is Fmax = keqαqT rmin 2 = 2Zkee2 E 2Zkee2 2 = E 2 2Zkee2 away from the target nucleus 42.3 (a) The photon has energy 2.28 eV.
And 13.6 eV ( )/ 22 = 3.40 eV is required to ionize a hydrogen atom from state n = 2. So while the photon cannot ionize a hydrogen atom pre-excited to n = 2, it can ionize a hydrogen atom in the n = 3 state, with energy – 13.6 eV 3 2 = – 1.51 eV (b) The electron thus freed can have kinetic energy Ke = 2.28 eV – 1.51 eV = 0.769 eV = 1 2 mev2 v = 2(0.769)(1.60 × 10−19)J (9.11× 10−31)kg = 520 km/s 2 Chapter 42 Solutions 42.4 (a) Longest wavelength implies lowest frequency and smallest energy: the electron falls from n = 3 to n = 2, losing energy – 13.6 eV 3 2 + 13.6 eV 2 2 = 1.89 eV The photon frequency is f = ∆E h and its wavelength is λ = c f = ch ∆E = (3.00 × 108 m /s)(6.626 × 10−34 J ⋅s) 1.89 eV eV 1.60 × 10−19 J λ = 656 nm Balmer Series (b) The biggest energy loss is for an electron to fall from ionization, n = ∞, to the n = 2 state. It loses energy – 13.6 eV ∞ + 13.6 eV 2 2 = 3.40 eV to emit light of wavelength λ = hc ∆E = 3.00 × 108 m /s ( ) 6.626 × 10−34 J ⋅s ( ) 3.40 eV 1.60 × 10-19 J/eV ( ) = 365 nm 42.5 (a) For positronium, µ = me 2 , so λ 32 = (656 nm)2 = 1312 nm = 1.31 µm (infrared region) .
(b) For He+, µ ≈ me, q1 = e, and q2 = 2e, so λ 32 = (656/4) nm = 164 nm (ultraviolet region) .
Goal Solution A general expression for the energy levels of one-electron atoms and ions is En = −µke 2q1 2q2 2 2h2n2 where ke is the Coulomb constant, q1 and q2 are the charges of the two particles, and µ is the reduced mass, given by µ = m1m2 / m1 + m2 ( ). In Problem 4 we found that the wavelength for the n = 3 to n = 2 transition of the hydrogen atom is 656.3 nm (visible red light). What are the wavelengths for this same transition in (a) positronium, which consists of an electron and a positron, and (b) singly ionized helium? (Note: A positron is a positively charged electron.) G : The reduced mass of positronium is less than hydrogen, so the photon energy will be less for positronium than for hydrogen. This means that the wavelength of the emitted photon will be longer than 656.3 nm. On the other hand, helium has about the same reduced mass but more charge than hydrogen, so its transition energy will be larger, corresponding to a wavelength shorter than 656.3 nm. O : All the factors in the above equation are constant for this problem except for the reduced mass and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the transition can be found simply from the ratio of mass and charge variables.
Chapter 42 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
A : For hydrogen, µ = mpme mp + me ≈me The photon energy is ∆E = E3 −E2 Its wavelength is λ = 656.3 nm, where λ = c f = hc ∆E (a) For positronium, µ = meme me + me = me 2 so the energy of each level is one half as large as in hydrogen, which we could call "protonium." The photon energy is inversely proportional to its wavelength, so for positronium, λ32 = 2(656.3 nm) = 1313 nm (in the infrared region) (b) For He+, µ ≈me, q1 = e, and q2 = 2e, so the transition energy is 22 = 4 times larger than hydrogen. Then, λ32 = 656 4 nm = 164 nm (in the ultraviolet region) L : As expected, the wavelengths for positronium and helium are respectively larger and smaller than for hydrogen. Other energy transitions should have wavelength shifts consistent with this pattern. It is important to remember that the reduced mass is not the total mass, but is generally close in magnitude to the smaller mass of the system (hence the name reduced mass).
42.6 (a) For a particular transition from ni to nf , ∆EH = −µHke 2e4 2h2 1 nf 2 −1 ni 2 = hc λH and ∆ED = −µDke 2e4 2h2 1 nf 2 −1 ni 2 = hc λD where µH = memp me + mp and µD = memD me + mD By division, ∆EH ∆ED = µH µD = λ D λ H or λ D = µH µD λ H Then, λ H −λ D = 1−µH µD λ H (b) µH µD = memp me + mp me + mD memD = 1.007 276 u ( ) 0.000 549 u + 2.013 553 u ( ) 0.000 549 u + 1.007 276 u ( ) 2.013 553 u ( ) = 0.999728 λ H −λ D = 1−0.999 728 ( ) 656.3 nm ( ) = 0.179 nm 4 Chapter 42 Solutions 42.7 (a) In the 3d subshell, n = 3 and l = 2, we have n 3 3 3 3 3 3 3 3 3 3 l 2 2 2 2 2 2 2 2 2 2 ml +2 +2 +1 +1 0 0 -1 -1 -2 -2 m s +1/2 -1/2 +1/2 -1/2 +1/2 -1/2 +1/2 -1/2 +1/2 -1/2 (A total of 10 states) (b) In the 3p subshell, n = 3 and l = 1, we have n 3 3 3 3 3 3 l 1 1 1 1 1 1 ml +1 +1 +0 +0 -1 -1 m s +1/2 -1/2 +1/2 -1/2 +1/2 -1/2 (A total of 6 states) 42.8 ψ1s(r) = 1 πa0 3 e−r/a0 (Eq. 42.3) P1s(r) = 4r 2 a0 3 e−2r/a0 (Eq. 42.7) 42.9 (a) ψ 2dV ∫ = 4π ψ 2r 2 dr 0 ∞ ∫ = 4π 1 π a0 3 r 2 e−2r a0 0 ∞ ∫ dr Using integral tables, ψ 2 dV ∫ = −2 a0 2 e−2r a0 r 2 + a0r + a0 2 2 r = 0 r = ∞ = −2 a0 2 −a0 2 2 = 1 so the wave function as given is normalized.
(b) Pa0 2 →3a0 2 = 4π ψ 2r 2 dr a0 2 3a0 2 ∫ = 4π 1 π a0 3 r 2 e−2r a0 a0 2 3a0 2 ∫ dr Again, using integral tables, Pa0 2 →3a0 2 = −2 a0 2 e−2r a0 r 2 + a0r + a0 2 2 a0 2 3a0 2 = −2 a0 2 e−3 17 a0 2 4 −e−1 5a0 2 4 = 0.497 Chapter 42 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
42.10 ψ = 1 3 1 (2a0)3/2 r a0 e−r/2a0 so Pr = 4πr 2 ψ 2 = 4πr 2 r 2 24a0 5 e−r/a0 Set dP dr = 4π 24a0 5 4r 3e−r/a0 + r 4 −1 a0 e−r/a0 = 0 Solving for r, this is a maximum at r = 4 a0 42.11 ψ = 1 πa0 3 e−r/a0 2 r dψ dr = −2 r πa0 5 e−r/a0 = 2 ra0 ψ d 2ψ dr 2 = 1 πa0 7 e−r/a0 = 1 a0 2 ψ −h2 2me 1 a0 2 −2 ra0 ψ − e 2 4π e0r ψ = Eψ But a0 = h24πe0 mee2 , so − e2 8πe0a0 = E or E = −kee2 2a0 This is true, so the Schrödinger equation is satisfied.
42.12 The hydrogen ground-state radial probability density is P(r) = 4πr 2 ψ1s 2 = 4r 2 a0 3 exp −2r a0 The number of observations at 2a0 is, by proportion N = 1000 P(2a0) P(a0 / 2) = 1000 (2a0)2 (a0 / 2)2 e−4a0/a0 e−a0/a0 = 1000(16)e−3 = 797 times 42.13 (a) For the d state, l = 2, L = 6h = 2.58 × 10– 34 J · s (b) For the f state, l = 3, L = l(l+ 1)h = 12h = 3.65 × 10– 34 J · s 6 Chapter 42 Solutions 42.14 L = l(l+ 1)h so 4.714 × 10−34 = l(l+ 1) 6.626 × 10−34 2π l(l+ 1) = 4.714 × 10−34 ( ) 2 2π ( )2 6.626 × 10−34 ( ) 2 = 1.998 × 101 ≈20 = 4(4 + 1) so l = 4 42.15 The 5th excited state has n = 6, energy – 13.6 eV 36 = – 0.378 eV The atom loses this much energy: hc λ = (6.626 × 10– 34 J · s)(3.00 × 108 m/s) (1090 × 10– 9 m)(1.60 × 10– 19 J/eV) = 1.14 eV to end up with energy – 0.378 eV – 1.14 eV = – 1.52 eV which is the energy in state 3: – 13.6 eV 3 3 = – 1.51 eV While n = 3, l can be as large as 2, giving angular momentum l(l+ 1)h = 6 h 42.16 For a 3d state, n = 3 and l = 2. Therefore, L = l(l+ 1)h = 2(2 + 1)h = 6h = 2.58 × 10– 34 J · s ml can have the values – 2, – 1, 0, 1, and 2, so Lz can have the values –2 h, – h, 0, and 2 h Using the relation cos θ = Lz /L, we find that the possible values of θ are equal to 145°, 114°, 90.0°, 65.9°, and 35.3° .
42.17 (a) n = 1: For n = 1, l = 0, ml = 0, ms = ± 1 2 , → 2 sets n l ml m s 1 0 0 –1/2 2 1 1 2 2 n = = ( ) 2 1 0 0 +1/2 (b) For n = 2, we have n l ml m s 2 0 0 ±1 2 2 1 –1 ±1 2 yields 8 sets; 2 2 2 2 2 n = ( ) = 8 2 1 0 ±1 2 2 1 1 ±1 2 Chapter 42 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
Note that the number is twice the number of ml values. Also, for each l there are (2l+ 1) different ml values. Finally, l can take on values ranging from 0 to n −1. So the general expression is s = 2 2l+ 1 ( ) 0 n −1 ∑ The series is an arithmetic progression: 2 + 6 + 10 + 14, the sum of which is s = n 2 2a + (n −1)d [ ] where a = 2, d = 4: s = n 2 4 + (n −1)4 [ ] = 2n2 (c) n = 3: 2(1) + 2(3) + 2(5) = 2 + 6 + 10 = 18 2n2 = 2(3)2 = 18 (d) n = 4: 2(1) + 2(3) + 2(5) + 2(7) = 32 2n2 = 2(4)2 = 32 (e) n = 5: 32 + 2(9) = 32 + 18 = 50 2n2 = 2(5)2 = 50 42.18 µB = eh 2me e = 1.60 × 10– 19 C h= 1.055 × 10– 34 J · s me = 9.11 × 10– 31 kg µB = 9.27 × 10– 24 J/T = 5.79 × 10– 5 eV/T 42.19 (a) Density of a proton: ρ = m V = 1.67 × 10−27 kg (4/ 3)π(1.00 × 10−15 m)3 = 3.99 × 1017 kg/m3 (b) Size of model electron: r = 3m 4πρ 1/3 = 3 × 9.11× 10−31 kg ⋅m3 4π × 3.99 × 1017 kg 1/3 = 8.17 × 10– 17 m (c) Moment of inertia: I = 2 5 mr2 = 2 5 (9.11× 10−31 kg)(8.17 × 10−17 m)2 = 2.43 × 10−63 kg ⋅m2 Lz = Iω = h 2 = Iv r Therefore, v = hr 2I = 6.626 × 10−34 J ⋅s ( )(8.17 × 10−17 m) 2π 2 ( ) 2.43 × 10−63 kg ⋅m2 ( ) = 1.77 × 1012 m/s (d) This is 5.91 × 103 times larger than the speed of light.
8 Chapter 42 Solutions 42.20 (a) L = mvr = m 2πr T r = l(l+ 1)h = (l2 + l)h ≈lh (5.98 × 1024 kg) 2π (1.496 × 1011 m)2 3.156 × 107 s = lh so 2.66 × 1040 1.055 × 10– 34 J · s = l = 2.52 × 1074 (b) E = −U + K = −K = 1 2 mv2 = 1 2 mr 2 mr 2 mv2 = 1 2 L2 mr 2 = 1 2 l(l+ 1)h2 mr 2 ≈1 2 l2h2 mr 2 dE dl = 1 2 2lh2 mr 2 l l = 2 E l so dE = 2 E l dl = 2 1 2 (5.98 × 1024 kg) 2π × 1.496 × 1011 m 3.156 × 107 s 2 1 2.52 × 1074 ∆E = × × = 5 30 10 2 52 10 33 74 .
. J 2.10 × 10– 41 J 42.21 µn = eh 2mp e = 1.60 × 10– 19 C h = 1.055 × 10– 34 J · s mp = 1.67 × 10– 27 kg (a) µn = 5.05 × 10– 27 J/T = 31.6 neV/T (b) µn µB = 1 1836 = me mp Apparently it is harder to "spin up" a nucleus than a electron, because of its greater mass.
42.22 In the N shell, n = 4. For n = 4, l can take on values of 0, 1, 2, and 3. For each value of l, ml can be −l to l in integral steps. Thus, the maximum value for ml is 3. Since Lz = mlh, the maximum value for Lz is Lz = 3h .
42.23 The 3d subshell has l = 2, and n = 3. Also, we have s = 1. Therefore, we can have n = 3; l = 2; ml = – 2, – 1, 0, 1, 2; s = 1; and ms = – 1, 0, 1 , leading to the following table: n 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 l 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ml –2 –2 –2 –1 –1 –1 0 0 0 1 1 1 2 2 2 s 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ms –1 0 1 –1 0 1 –1 0 1 –1 0 1 –1 0 1 Chapter 42 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
42.24 (a) 1s22s22p 4 (b) For the 1s electrons, n = 1, l = 0, ml = 0, ms = + 1/2 and – 1/2 For the two 2s electrons, n = 2, l = 0, ml = 0, ms = + 1/2 and – 1/2 For the four 2p electrons, n = 2; l = 1; ml = –1, 0, or 1; and ms = + 1/2 or – 1/2 42.25 The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for scandium through zinc. Thus, we would first suppose that Ar [ ]3d 4 4s2 would have lower energy than Ar [ ]3d 5 4s1. But the latter has more unpaired spins, six instead of four, and Hund’s rule suggests that this could give the latter configuration lower energy. In fact it must, for Ar [ ]3d 5 4s1 is the ground state for chromium.
42.26 (a) For electron one and also for electron two, n = 3 and l = 1. The possible states are listed here in columns giving the other quantum numbers: electron ml 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 one ms 1 2 1 2 1 2 1 2 1 2 −1 2 −1 2 −1 2 −1 2 −1 2 1 2 1 2 1 2 1 2 1 2 electron ml 1 0 0 –1 –1 1 0 0 –1 –1 1 1 0 –1 –1 two ms −1 2 1 2 −1 2 1 2 −1 2 1 2 1 2 −1 2 1 2 −1 2 1 2 −1 2 −1 2 1 2 −1 2 electron ml 0 0 0 0 0 –1 –1 –1 –1 –1 –1 –1 –1 –1 –1 one ms −1 2 −1 2 −1 2 −1 2 −1 2 1 2 1 2 1 2 1 2 1 2 −1 2 −1 2 −1 2 −1 2 −1 2 electron ml 1 1 0 –1 –1 1 1 0 0 –1 1 1 0 0 –1 two ms 1 2 −1 2 1 2 1 2 −1 2 1 2 −1 2 1 2 −1 2 −1 2 1 2 −1 2 1 2 −1 2 1 2 There are thirty allowed states, since electron one can have any of three possible values for ml for both spin up and spin down, amounting to six states, and the second electron can have any of the other five states.
(b) Were it not for the exclusion principle, there would be 36 possible states, six for each electron independently.
42.27 Shell K L M N n 1 2 3 4 l 0 0 1 0 1 2 0 ml 0 0 1 0 – 1 0 1 0 – 1 2 1 0 – 1 – 2 0 ms ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ count 1 2 3 4 10 12 18 21 30 20 He Be Ne Mg Ar Zn Ca (a) zinc or copper (b) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 or 1s2 2s2 2p6 3s2 3p6 4s1 3d10 42.28 Listing subshells in the order of filling, we have for element 110, 10 Chapter 42 Solutions 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f 14 5d10 6p6 7s2 5f 14 6d8 In order of increasing principal quantum number, this is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f 14 5s2 5p6 5d10 5f 14 6s2 6p6 6d8 7s2 42.29 (a) n + l 1 2 3 4 5 6 7 subshell 1s 2s 2p , 3s 3p , 4s 3d , 4p , 5s 4d , 5p , 6s 4f , 5d , 6p , 7s (b) Z = 15: Filled subshells: 1s, 2s, 2p, 3s (12 electrons) Valence subshell: 3 electrons in 3p subshell Prediction: Valance = + 3 or – 5 Element is phosphorus Valence + 3 or – 5 (Prediction correct) Z = 47: Filled subshells: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s (38 electrons) Outer subshell: 9 electrons in 4d subshell Prediction: Valence = – 1 Element is silver, (Prediction fails) Valence is + 1 Z = 86: Filled subshells: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p (86 electrons) Prediction Outer subshell is full: inert gas Element is radon, inert (Prediction correct) 42.30 Electronic configuration: Sodium to Argon [1s2 2s2 2p6] + 3s1 → Na11 +3s2 → Mg12 +3s2 3p1 → Al13 +3s2 3p2 → Si14 +3s2 3p3 → P15 +3s2 3p4 → S16 +3s2 3p5 → Cl17 +3s2 3p6 → Ar18 [1s2 2s2 2p6 3s2 3p6]4s1 → K19 Chapter 42 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
42.31 n = 3, l = 0, ml = 0 ψ 300 corresponds to E 300 = – Z2 E0 n2 = – 22(13.6) (3)2 = – 6.05 eV n = 3, l = 1, ml = –1, 0, 1 ψ 31 – 1, ψ 310, ψ 311 have the same energy since n is the same.
For n = 3, l = 2, ml = – 2, – 1, 0, 1, 2 ψ 32 – 2, ψ 32 – 1, ψ 320, ψ 321, ψ 322 have the same energy since n is the same. All states are degenerate.
42.32 E = hc λ = e(∆V) ⇒ (6.626 × 10– 34 J · s)(3.00 × 108 m/s) 10.0 × 10– 9 m = (1.60 × 10– 19)(∆V) ∆V = 124 V 42.33 Ephoton max = hc λ min = e ∆V ( ) = 40.0 keV λ min = hc Emax = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 40.0 × 103 eV 1.00 eV 1.60 × 10−19 J = 0.031 0 nm 42.34 Some electrons can give all their kinetic energy Ke = e ∆V ( ) to the creation of a single photon of x-radiation, with hf = hc λ = e ∆V ( ) λ = hc e ∆V ( ) = 6.6261× 10−34 J ⋅s ( ) 2.9979 × 108 m s ( ) 1.6022 × 10−19 C ( ) ∆V ( ) = 1240 nm ⋅V ∆V 42.35 Following Example 42.7, Eγ = 3 4 (42 – 1) 2(13.6 eV) = 1.71 × 104 eV = 2.74 × 10– 15 J f = 4.14 × 1018 Hz and λ = 0.072 5 nm 12 Chapter 42 Solutions 42.36 The Kβ x-rays are emitted when there is a vacancy in the (n = 1) K shell and an electron from the (n = 3) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and by one in its final state. hc λ = – 13.6(Z – 9)2 3 2 eV + 13.6(Z – 1)2 12 eV (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) (0.152 × 10−9 m)(1.60 × 10−19 J/eV) = 13.6 eV −Z2 9 + 18Z 9 −81 9 + Z2 −2Z + 1 8.17 × 103 eV = 13.6 eV 8Z2 9 – 8 so 601 = 8Z2 9 – 8 and Z = 26 Iron 42.37 (a) Suppose the electron in the M shell is shielded from the nucleus by two K plus seven L electrons. Then its energy is − − − 13.6 eV(83 9) 3 = 8.27 keV 2 2 Suppose, after it has fallen into the vacancy in the L shell, it is shielded by just two K-shell electrons. Then its energy is − − − 13.6 eV(83 2) 2 = 22.3 keV 2 2 Thus the electron's energy loss is the photon energy: (22 3 8.27) keV = . − 14.0 keV (b) ∆E = hc λ so λ = 6.626 × 10−34 J ⋅s (3.00 × 108 m /s) 14.0 × 103 × 1.60 × 10−19 J = 8. 5 10 m 11 8 × − 42.38 E = hc λ = 1240 eV · nm λ = 1.240 keV · nm λ for λ 1 = 0.0185 nm, E = 67.11 keV λ 2 = 0.0209 nm, E = 59.4 keV λ 3 = 0.0215 nm, E = 57.7 keV The ionization energy for K shell = 69.5 keV, so, the ionization energies for the other shells are: L shell = 11.8 keV : M shell = 10.1 keV : N shell = 2.39 keV Chapter 42 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
42.39 (a) The outermost electron in sodium has a 3s state for its ground state. The longest wavelength means minimum photon energy and smallest step on the energy level diagram. Since n = 3, ′ n must be 4. With l = 0, ′ l must be 1 , since l must change by 1 in a photon absorption process.
(b) 1 330 × 10−9 m = 1.097 × 107 1 m 1 3 −1.35 ( )2 − 1 4 −δ1 ( ) 2 0.276 = 1 1.65 ( )2 − 1 4 −δ1 ( ) 2 = 0.367 − 1 4 −δ1 ( ) 2 so 4 −δ1 ( ) 2 = 10.98 and δ1 = 0.686 42.40 λ = c f = hc hf = (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) (2.10 eV)(1.60 × 1019 J/eV) = 590 nm 42.41 We require A = ufB = 16π 2h λ3 B or uf = 16π 2h λ3 = 16π 2 1.055 × 10−34 J ⋅s ( ) 645 × 10−9 m ( ) 3 = 6.21× 10−14 J ⋅s m3 42.42 f = E h = 2 82 1013 .
× − s 1 λ = c f = 10.6 µm , infrared 42.43 E = Pt = (1.00 × 106 W)(1.00 × 10– 8 s) = 0.0100 J Eγ = h f = hc λ = (6.626 × 10– 34)(3.00 × 108) 694.3 × 10– 9 J = 2.86 × 10– 19 J N = E Eγ = 0.0100 2.86 × 10– 19 = 3.49 × 1016 photons 14 Chapter 42 Solutions Goal Solution A ruby laser delivers a 10.0-ns pulse of 1.00 MW average power. If the photons have a wavelength of 694.3 nm, how many are contained in the pulse?
G : Lasers generally produce concentrated beams that are bright (except for IR or UV lasers that produce invisible beams). Since our eyes can detect light levels as low as a few photons, there are probably at least 1 000 photons in each pulse.
O : From the pulse width and average power, we can find the energy delivered by each pulse. The number of photons can then be found by dividing the pulse energy by the energy of each photon, which is determined from the photon wavelength.
A : The energy in each pulse is E = Pt = (1.00 × 106 W)(1.00 × 10−8 s) = 1.00 × 10−2 J The energy of each photon is Eγ = hf = hc λ = (6.626 × 10−34)(3.00 × 108) 694.3 × 10−9 J = 2.86 × 10−19 J So N = E Eγ = 1.00 × 10−2 J 2.86 × 10−19 J/photon = 3.49 × 1016 photons L : With 1016 photons/pulse, this laser beam should produce a bright red spot when the light reflects from a surface, even though the time between pulses is generally much longer than the width of each pulse. For comparison, this laser produces more photons in a single ten-nanosecond pulse than a typical 5 mW helium-neon laser produces over a full second (about 1.6 × 1016 photons/second). 42.44 In G = eσ nu −nl ( )L we require 1.05 = e 1.00 × 10−18 m2 ( ) nu −nl ( ) 0.500 m ( ) Thus, ln 1.05 ( ) = 5.00 × 10−19 m3 ( ) nu −nl ( ) so nu −nl = ln 1.05 ( ) 5.00 × 10−19 m3 = 9 76 1016 .
× − m 3 42.45 (a) N3 N2 = Nge−E3 kB ⋅300 K ( ) Nge−E2 kB ⋅300 K ( ) = e−(E3 −E2 ) kB ⋅300 K ( ) = e−hc λ kB ⋅300 K ( ) where λ is the wavelength of light radiated in the 3 →2 transition: N3 N2 = e −6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 632.8 × 10 −9 m ( ) 1.38 × 10−23 J K ( ) 300 K ( ) = e−75.8 = 1 22 10 33 .
× − (b) N3 N2 = ehc/λ kBT = e −6.626 × 10−34 J ⋅s ( ) 3.00 × 10 8 m s ( ) 694.3 × 10 −9 m ( ) 1.38 × 10−23 J K ( ) 4.00 K ( ) = − e 5187 To avoid overflowing your calculator, note that 10 = eln 10. Take N3 N2 = eln 10 × (−5187/ln 10) = 10 2253 − Chapter 42 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
42.46 Nu Nl = e−Eu −El ( ) kBT where the subscript u refers to an upper energy state and the subscript l to a lower energy state.
(a) Since Eu −El = Ephoton = hc λ , Nu Nl = e−hc λkBT Thus, we require 1.02 = e−hc λkBT or ln 1.02 ( ) = − 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 694.3 × 10−9 m ( ) 1.38 × 10−23 J K ( )T T = −2.07 × 104 K ln 1.02 ( ) = − × 1 05 106 . K A negative-temperature state is not achieved by cooling the system below 0 K, but by heating it above T = ∞, for as T →∞ the populations of upper and lower states approach equality.
(b) Because Eu −El > 0, and in any real equilibrium state T > 0, e−Eu −El ( ) kBT < 1 and Nu < Nl.
Thus, a population inversion cannot happen in thermal equilibrium.
42.47 (a) I = (3.00 × 10−3 J) (1.00 × 10−9 s)π(15.0 × 10−6 m)2 = 4.24 × 1015 W/m2 (b) (3.00 × 10−3 J)(0.600 × 10−9 m)2 (30.0 × 10−6 m)2 = 1.20 × 10– 12 J = 7.50 MeV 42.48 (a) The energy difference between these two states is equal to the energy that is absorbed.
Thus, E = E2 – E1 = (– 13.6 eV) 4 – (– 13.6 eV) 1 = 10.2 eV = 1.63 × 10– 18 J (b) We have E = 3 2 kBT, or T = 2 3kB E = 2(1.63 × 10– 18 J) 3(1.38 × 10– 23 J/K) = 7.88 × 104 K 42.49 rav = 0 ∞ ∫rP(r)dr = 0 ∞ ∫ 4r 3 a0 3 (e−2r/a0 )dr Make a change of variables with 2r a0 = x and dr = a0 2 dx.
Then rav = a0 4 0 ∞ ∫ x3e−x dx = a0 4 −x3e−x + 3 −x2e−x + 2e−x(−x −1) ( ) [ ] 0 ∞ = 3 2 a0 16 Chapter 42 Solutions 42.50 1 r = 4r 2 a0 3 e−2r a0 1 r dr 0 ∞ ∫ = 4 a0 3 re−2 a0 ( )r dr 0 ∞ ∫ = 4 a0 3 1 2 a0 ( ) 2 = 1 a0 We compare this to 1 r = 1 3a0 2 = 2 3a0 , and find that the average reciprocal value is NOT the reciprocal of the average value.
42.51 The wave equation for the 2s state is given by Eq. 42.7: ψ2s r ( ) = 1 4 2π 1 a0 3 2 2 −r a0 e−r 2a0 (a) Taking r = a0 = 0.529 × 10−10 m, we find ψ2s a0 ( ) = 1 4 2π 1 0.529 × 10−10 m 3 2 2 −1 [ ]e−1 2 = 1 57 1014 .
× − m 3 2 (b) ψ2s a0 ( ) 2 = 1.57 × 1014 m−3 2 ( ) 2 = 2 47 1028 .
× − m 3 (c) Using Equation 42.5 and the results to (b) gives P2s a0 ( ) = 4π a0 2 ψ2s a0 ( ) 2 = 8 69 108 .
× − m 1 42.52 We define the reduced mass to be µ , and the ground state energy to be E1: µ = m1m2 m1 + m2 = 207memp 207me + mp = 207 9.11× 10−31 kg ( ) 1.67 × 10−27 kg ( ) 207 9.11× 10−31 kg ( ) + 1.67 × 10−27 kg ( ) = 1.69 × 10−28 kg E1 = −µke 2q1 2q2 2 2h2 1 ( )2 = − 1.69 × 10−28 kg ( ) 8.99 × 109 N ⋅m2 C2 ( ) 2 1.60 × 10−19 C ( ) 3 e 2 1.055 × 10−34 J ⋅s ( ) 2 = −2.52 × 103 eV To ionize the muonium "atom" one must supply energy +2 52 . keV .
42.53 (a) (3.00 × 108 m/s)(14.0 × 10– 12 s) = 4.20 mm (b) E = hc λ = 2.86 × 10– 19 J N = 3.00 J 2.86 × 10– 19 J = 1.05 × 1019 photons (c) V = (4.20 mm)π (3.00 mm)2 = 119 mm3 n = 1.05 × 1019 119 = 8.82 × 1016 mm– 3 Chapter 42 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
42.54 (a) The length of the pulse is ∆L = ct (b) The energy of each photon is Eγ = hc λ so N = E Eγ = Eλ h c (c) V = ∆Lπ d 2 4 n = N V = 4 ctπ d2 Eλ hc 42.55 We use ψ2s(r) = 1 4 (2πa0 3)−1/2 2 −r a0 e−r/2a0 By Equation 42.5, P(r) = 4πr 2ψ 2 = 1 8 r2 a0 3 2 −r a0 2 e−r/a0 (a) dP(r) dr = 1 8 2r a0 3 2 −r a0 2 −2r 2 a0 3 1 a0 2 −r a0 −r2 a0 3 2 −r a0 2 1 a0 e−r/a0 = 0 or 1 8 r a0 3 2 −r a0 2 2 −r a0 −2r a0 −r a0 2 −r a0 e−r/a0 = 0 Therefore we require the roots of dP dr = 0 at r = 0, r = 2a0, and r = ∞ to be minima with P r ( ) = 0. [ . . . . . ] = 4 −6r / a0 ( ) + r / a0 ( ) 2 = 0 with solutions r a = ± ( ) 3 5 0.
We substitute the last two roots into P(r) to determine the most probable value: When r a a = − ( ) = 3 5 0 7639 0 0 .
, then P r ( ) = 0.0519/ a0 When r = 3 + 5 ( )a0 = 5.236a0, then P r ( ) = 0.191/ a0 Therefore, the most probable value of r is 3 5 0 + ( )a = 5.236a0 (b) 0 ∞ ∫ P(r)dr = 0 ∞ ∫ 1 8 r2 a0 3 2 −r a0 2 e−r/a0 dr Let u = r a0 , dr = a0 du, 0 ∞ ∫ P(r)dr = 0 ∞ ∫ 1 8 u2(4 −4u + u2)e−u dr = 0 ∞ ∫ 1 8 (u4 −4u3 + 4u2)e−u du = −1 8 (u4 + 4u2 + 8u + 8)e−u 0 ∞ = 1 This is as desired .
42.56 ∆z = at2 2 = 1 2 Fz m0 t2 = µz dBz dz ( ) 2m0 ∆x v 2 and µz = eh 2me dBz dz = 2m0(∆z)v22me ∆x2eh = × × × × ⋅ − − − − − 2 108 1 66 10 10 2 9 11 10 10 1 00 1 60 10 1 05 10 27 4 2 2 31 3 2 19 34 ( )( .
)( / ) ( .
)( ) ( .
)( .
)( .
) kg m s kg m m C J s = 0.389 T/m 18 Chapter 42 Solutions 42.57 With one vacancy in the K shell, excess energy ∆E ≈−(Z −1)2(13.6 eV) 1 22 −1 12 = 5.40 keV We suppose the outermost 4s electron is shielded by 20 electrons inside its orbit: Eionization ≈22(13.6 eV) 42 = 3.40 eV Note the experimental ionization energy is 6.76 eV. K = ∆E −Eionization ≈ 5.39 keV 42.58 E = hc λ = 1240 eV · nm λ = ∆E λ 1 = 310 nm, so ∆E 1 = 4.00 eV λ 2 = 400 nm, ∆E 2 = 3.10 eV λ 3 = 1378 nm, ∆E 3 = 0.900 eV and the ionization energy = 4.10 eV The energy level diagram having the fewest levels and consistent with these energies is shown at the right.
42.59 (a) One molecule's share of volume Al: V = mass per molecule density = 27.0 g mol 6.02 × 1023 molecules mole 1.00 × 10−6 m3 2.70 g = × − 1 66 10 29 . m3 V 3 = 2 55 10 10 .
× − − m ~ 10 nm 1 U: V = 238 g 6.02 × 1023 molecules 1.00 × 10−6 m3 18.9 g = 2.09 × 10−29 m3 V 3 = 2 76 10 10 .
× − − m ~ 10 nm 1 (b) The outermost electron in any atom sees the nuclear charge screened by all the electrons below it. If we can visualize a single outermost electron, it moves in the electric field of net charge, +Ze −(Z −1)e = +e, the charge of a single proton, as felt by the electron in hydrogen. So the Bohr radius sets the scale for the outside diameter of every atom. An innermost electron, on the other hand, sees the nuclear charge unscreened, and the scale size of its (K-shell) orbit is a0 Z.
Chapter 42 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
42.60 (a) No orbital magnetic moment to consider: higher energy for N S N S parallel magnetic moments, for antiparallel spins of the electron and proton.
(b) E = hc λ = 9.42 × 10−25 J = 5 89 . eV µ (c) ∆E∆t ≈h 2 so ∆E ≈ 1.055 × 10−34 J ⋅s 2 107 yr ( ) 3.16 × 107 s yr ( ) 1.00 eV 1.60 × 10−19 J = 1.04 × 10−30 eV 42.61 P = 4r2 a0 3 e−2r a0 dr 2.50a0 ∞ ∫ = 1 2 z2 e−z dz 5.00 ∞ ∫ where z ≡2r a0 P = −1 2 (z2 + 2z + 2)e−z 5.00 ∞ = −1 2 + 1 2 25.0 + 10.0 + 2.00 ( )e−5 = 37 2 0.00674 ( ) = 0.125 Goal Solution For hydrogen in the 1s state, what is the probability of finding the electron farther than 2.50 ao from the nucleus?
G : From the graph shown in Figure 42.8, it appears that the probability of finding the electron beyond 2.5 a0 is about 20%. O : The precise probability can be found by integrating the 1s radial probability distribution function from r = 2.50 ao to ∞.
A : The general radial probability distribution function is P r ( ) = 4π r2 ψ 2 With ψ1s = π a0 3 ( ) −1/2e−r/a0 it is P r ( ) = 4r2a0 −3e−2r/a0 The required probability is then P = P r ( )dr 2.50a0 ∞ ∫ = 4r2 a0 3 e−2r/a0dr 2.50a0 ∞ ∫ Let z = 2r a0 and dz = 2dr a0 : P = 1 2 z2e−zdz 5.00 ∞ ∫ Performing this integration by parts, P = −1 2 z2 + 2z + 2 e−z]5.00 ∞ P = −1 2 0 ( ) + 1 2 25.0 + 10.0 + 2.00 ( )e−5.00 = 37 2 ( ) 0.00674 ( ) = 0.125 L : The probability of 12.5% is less than the 20% we estimated, but close enough to be a reasonable result.
In comparing the 1s probability density function with the others in Figure 42.8, it appears that the ground state is the most narrow, indicating that a 1s electron will probably be found in the narrow range of 0 to 4 Bohr radii, and most likely at r = a0.
20 Chapter 42 Solutions 42.62 The probability, P, of finding the electron within the Bohr radius is P = P1s r ( )dr r = 0 a0 ∫ = 4r2 a0 3 e−2r a0 dr r = 0 a0 ∫ Defining z ≡2r a0 , this becomes P = −1 2 (z2 + 2z + 2)e−z 0 2 = −1 2 4 + 4 + 2 ( )e−2 −0 + 0 + 2 ( )e0 [ ] = 1 2 2 −10 e2 = 0.323 The electron is likely to be within the Bohr radius about one-third of the time. The Bohr model indicates none of the time.
42.63 (a) For a classical atom, the centripetal acceleration is a = v2 r = 1 4πe0 e2 r 2me E = − e2 4πe0r + mev2 2 = − e2 8πe0r so dE dt = e2 8πe0r 2 dr dt = −1 6πe0 e2a2 c3 = −e2 6πe0c3 e2 4πe0r 2me 2 Therefore, dr dt = − e4 12π 2e0 2r 2me 2c3 (b) − r= 2.00 × 10−10 m r = 0 ∫ 12π 2e 0 2r 2me 2c3dr = e4 t = 0 T ∫ dt 12π 2e0 2me 2c3 e4 r 3 3 0 2.00 × 10−10 = T = 8.46 × 10– 10 s Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold for systems of atomic size.
42.64 (a) +3e – 0.85e – 0.85e = 1.30e (b) The valence electron is in an n = 2 state, with energy – 13.6 eV Z2 eff n2 = – 13.6 eV(1.30)2 2 2 = – 5.75 eV To ionize the atom you must put in + 5.75 eV This differs from the experimental value by 6%, so we could say the effective value of Z is accurate within 3%.
Chapter 42 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
42.65 ∆E = 2µBB = hf so 2 9.27 × 10−24 J/ T ( ) 0.350 T ( ) = 6.626 × 10−34 J ⋅s ( )f and f = 9.79 × 109 Hz 42.66 The photon energy is E4 −E3 = 20.66 −18.70 eV = 1.96 eV = hc λ λ = (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) 1.96 × 1.60 × 10−19 J = 633 nm 42.67 (a) 1 α = hc kee2 = 6.626 × 10−34 ( ) 3.00 × 108 ( ) 2π 8.99 × 109 ( ) 1.60 × 10−19 ( ) 2 = 137 (b) λ C re = h mc mc2 kee2 = hc ke2 = 2π α (c) a0 λ C = h2 mkee 2 mc h = 1 2π hc kee 2 = 137 2π = 1 2πα (d) 1 RH a0 = 1 RHa0 = 4πch3 mke 2e4 mkee2 h2 = 4π hc kee2 = 4π α 42.68 ψ = 1 4 (2π)−1/2 1 a0 3/2 2 −r a0 e−r/2a0 = A 2 −r a0 e−r/2a0 ∂2ψ ∂r 2 = Ae−r/2a0 a0 2 3 2 − r 4a0 Substituting into Schrödinger's equation and dividing by ψ , 1 a0 2 1 2 − r 4a0 = −2m h2 [E −U] 2 −r a0 Now E −U = h2 2m a0 2 1 4 − ke2 4a0 ( ) m h2 ( ) m h2 ( ) = −1 4 h2 2m a0 2 and 1 a0 2 1 2 − r 4a0 = 1 4a0 2 2 −r a0 ∴ψ is a solution.
22 Chapter 42 Solutions 42.69 The beam intensity is reduced by absorption of photons into atoms in the lower state. The number of transitions per time and per area is −BNl I x ( )ndx c. The beam intensity is increased by stimulating emission in atoms in the upper state, with transition rate +BNuI x ( )ndx c. The net rate of change in photon numbers per area is then −B Nl −Nu ( )I x ( )ndx c.
Each photon has energy hf , so the net change in intensity is dI x ( ) = −hf B Nl −Nu ( )I x ( )ndx c = −hf B∆N I x ( )ndx c Then, dI x ( ) I x ( ) = −hf B∆N n c dx so dI x ( ) I x ( ) I0 I L ( ) ∫ = −hf B∆N n c dx x = 0 L ∫ ln I L ( ) [ ] −ln I0 [ ] = ln I L ( ) I0 = −hf B∆N n c L −0 ( ) I L ( ) = I0 e−h f B∆N nL c = I0 e−α L This result is also expressed in problem 42.44 as I L ( ) I0 = G = e−σ nl−nu ( )L = e+σ nu −nl ( )L 42.70 (a) Suppose the atoms move in the +x direction. The absorption of a photon by an atom is a completely inelastic collision, described by mvi i + h λ −i ( ) = mvf i so vf −vi = −h mλ This happens promptly every time an atom has fallen back into the ground state, so it happens every 10−8 s = ∆t. Then, a = vf −vi ∆t = − h mλ ∆t ~ − 6.626 × 10−34 J ⋅s 10−25 kg ( ) 500 × 10−9 m ( ) 10−8 s ( ) −106 m s2 (b) With constant average acceleration, vf 2 = vi 2 + 2a ∆x ( ) 0 ~ 103 m s ( ) 2 + 2 −106 m s2 ( )∆x so ∆x ~ 103 m s ( ) 2 106 m s2 ~ 1 m © 2000 by Harcourt, Inc. All rights reserved.
Chapter 43 Solutions 43.1 (a) F = q2 4π e0r 2 = (1.60 × 10−19)2(8.99 × 109) (5.00 × 10−10)2 N = 0.921 10 N 9 × − toward the other ion.
(b) U = −q2 4πe0r = −(1.60 × 10−19)2(8.99 × 109) 5.00 × 10−10 J ≈ – 2.88 eV 43.2 We are told K + Cl + 0.7 eV → K+ + Cl− and Cl + e− → Cl−+ 3.6 eV or Cl− → Cl + e−−3.6 eV By substitution, K + Cl + 0.7 eV → K+ + Cl + e−−3.6 eV K + 4.3 eV → K+ + e− or the ionization energy of potassium is 4.3 eV 43.3 (a) Minimum energy of the molecule is found from dU dr = −12Ar−13 + 6Br−7 = 0, yielding r0 = 2A B 1 6 (b) E = U r = ∞−U r = r 0= 0 − A 4A2 B2 − B 2A B = − 1 4 −1 2 B2 A = B2 4A This is also the equal to the binding energy, the amount of energy given up by the two atoms as they come together to form a molecule.
(c) r0 = 2 0.124 × 10−120 eV ⋅m12 ( ) 1.488 × 10−60 eV ⋅m6 1 6 = 7.42 × 10−11 m = 74.2 pm E = 1.488 × 10−60 eV ⋅m6 ( ) 2 4 0.124 × 10−120 eV ⋅m12 ( ) = 4.46 eV 43.4 At the boiling or condensation temperature, kBT ≈10−3 eV = 10−3 1.6 × 10−19 J ( ) T ≈ 1.6 × 10−22 J 1.38 × 10−23 J K ~10 K 546 Chapter 43 Solutions 43.5 µ = m1m2 m1 + m2 = 132.9(126.9) 132.9 +126.9 1.66 × 10−27 kg ( ) = 1.08 × 10−25 kg I = µr 2 = (1.08 × 10−25 kg)(0.127 × 10−9 m)2 = 1.74 × 10−45 kg ⋅m2 (a) E = 1 2 I ω 2 = (I ω)2 2I = J(J +1)h2 2I J = 0 gives E = 0 J = 1 gives E = h2 I = (6.626 × 10−34 J ⋅s)2 4π 2(1.74 × 10−45 kg ⋅m2) = 6.41× 10−24 J = 40.0 µeV hf = 6.41× 10−24 J −0 to f = 9.66 × 109 Hz (b) f = E1 h = h2 hI = h 4π 2µr2 ∝r−2 If r is 10% too small, f is 20% too large. 43.6 hf = ∆E = h2 2I 2 2 + 1 ( ) [ ] −h2 2I 1 1+ 1 ( ) [ ] = h2 2I 4 ( ) I = 4 h 2π ( ) 2 2hf = h 2π 2 f = 6.626 × 10−34 J ⋅s 2π 2 2.30 × 1011 Hz ( ) = 1.46 × 10−46 kg ⋅m2 43.7 For the HCl molecule in the J = 1 rotational energy level, we are given r0 = 0.1275 nm. Erot = h2 2I J(J + 1) Taking J = 1, we have Erot = h2 I = 1 2 Iω 2 or ω = 2h2 I2 = 2 h I The moment of inertia of the molecule is given by Equation 43.3. I = µr0 2 = m1m2 m1 + m2 r0 2 I = (1 u)(35 u) 1 u + 35 u r0 2 = (0.972 u)(1.66 × 10−27 kg / u)(1.275 × 10−10 m)2 = 2.62 × 10−47 kg ⋅m2 Therefore, ω = 2 h I = 2 1.055 × 10−34 J ⋅s 2.62 × 10−47 kg ⋅m2 = 5.69 × 1012 rad/s Chapter 43 Solutions 547 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution An HCl molecule is excited to its first rotational-energy level, corresponding to J = 1. If the distance between its nuclei is 0.127 5 nm, what is the angular speed of the molecule about its center of mass?
G : For a system as small as a molecule, we can expect the angular speed to be much faster than the few rad/s typical of everyday objects we encounter. O : The rotational energy is given by the angular momentum quantum number, J . The angular speed can be calculated from this kinetic rotational energy and the moment of inertia of this one-dimensional molecule.
A : For the HCl molecule in the J = 1 rotational energy level, we are given r0 = 0.1275 nm. Erot = h 2I J J + 1 ( ) so with J = 1, Erot = h2 I = 1 2 Iω 2 and ω = 2h2 I2 = h 2 I The moment of inertia of the molecule is given by: I = µ r0 2 = m1m2 m1 + m2 r0 2 = (1 u)(35 u) 1 u + 35 u r0 2 I = (0.972 u)(1.66 × 10-27 kg / u)(1.275 × 10-10 m)2 = 2.62 × 10−47 kg ⋅m2 Therefore, 2 1.055 × 10−34 J ⋅s 2.62 × 10-47 kg ⋅m2 = 5.69 × 1012 rad/s L : This angular speed is more than a billion times faster than the spin rate of a music CD, which rotates at 200 to 500 revolutions per minute, or ω = 20 rad/s to 50 rad/s.
43.8 I = m1r1 2 + m2r2 2 where m1r1 = m2r2 and r1 + r2 = r Then r1 = m2r2 m1 so m2r2 m1 + r2 = r and r2 = m1r m1 + m2 Also, r2 = m1r1 m2 Thus, r1 + m1r1 m2 = r and r1 = m2r m1 + m2 I = m1 m2 2r 2 m1 + m2 ( ) 2 + m2m1 2r 2 m1 + m2 ( ) 2 = m1m2r 2 m2 + m1 ( ) m1 + m2 ( ) 2 = m1m2r 2 m1 + m2 = µ r2 43.9 (a) µ = 22.99 35.45 ( ) 22.99 + 35.45 ( ) 1.66 × 10−27 kg ( ) = 2.32 × 10−26 kg I = µr 2 = 2.32 × 10−26 kg ( ) 0.280 × 10−9 m ( ) 2 = 1.81× 10−45 kg ⋅m2 (b) hc λ = h2 2I 2 2 + 1 ( ) −h2 2I 1 1+ 1 ( ) = 3h2 I −h2 I = 2h2 I = 2h2 4π 2I λ = c4π 2I 2h = 3.00 × 108 m /s ( )4π 2 1.81× 10−45 kg ⋅m2 ( ) 2 6.626 × 10−34 J ⋅s ( ) = 1.62 cm 548 Chapter 43 Solutions 43.10 The energy of a rotational transition is ∆E = h2 I ( )J where J is the rotational quantum number of the higher energy state (see Equation 43.7). We do not know J from the data.
However, ∆E = hc λ = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) λ 1 eV 1.60 × 10−19 J For each observed wavelength, λ mm ( ) ∆E eV ( ) 0.1204 0.01032 0.0964 0.01288 0.0804 0.01544 0.0690 0.01800 0.0604 0.02056 The ∆′ E s consistently increase by 0.00256 eV. E1 = h2 I = 0.00256 eV and I = h2 E1 = 1.055 × 10−34 J ⋅s ( ) 2 0.00256 eV ( ) 1 eV 1.60 × 10−19 J = 2.72 × 10−47 kg ⋅m2 For the HCl molecule, the internuclear radius is r = I µ = 2.72 × 10−47 1.62 × 10−27 m = 0.130 nm 43.11 µ = m1m2 m1 + m2 = 35 36 × 1.66 × 10−27 kg = 1.61× 10−27 kg ∆Evib = h k µ = 1.055 × 10−34 ( ) 480 1.61× 10−27 = 5.74 × 10−20 J = 0.358 eV 43.12 (a) Minimum amplitude of vibration of HI is 1 2 kA2 = 1 2 hf : A = hf k = (6.626 × 10−34 J ⋅s)(6.69 × 1013 /s) 320 N/m = 1.18 × 10−11 m = 0.0118 nm (b) For HF, A = (6.626 × 10−34 J ⋅s)(8.72 × 1013 /s) 970 N/m = 7.72 × 10−12 m = 0.00772 nm Since HI has the smaller k, it is more weakly bound.
Chapter 43 Solutions 549 © 2000 by Harcourt, Inc. All rights reserved.
43.13 (a) The reduced mass of the O2 is µ = 16 u ( ) 16 u ( ) 16 u ( ) + 16 u ( ) = 8 u = 8 1.66 × 10−27 kg ( ) = 1.33 × 10−26 kg The moment of inertia is then I = µr 2 = 1.33 × 10−26 kg ( ) 1.20 × 10−10m ( ) 2 = 1.91× 10−46 kg ⋅m2 The rotational energies are Erot = h2 2I J J + 1 ( ) = 1.055 × 10−34 J ⋅s ( ) 2 2 1.91× 10−46 kg ⋅m2 ( ) J J + 1 ( ) Thus Erot = 2.91× 10−23 J ( )J(J + 1) And for J = 0, 1, 2, Erot = 0, 3.64 × 10−4 eV, 1.09 × 10−3 eV (b) Evib = v + 1 2 h k µ = v + 1 2 1.055 × 10−34 J ⋅s ( ) 1177 N/m 8(1.66 × 10−27kg) Evib = v + 1 2 3.14 × 10−20 J ( ) 1 eV 1.60 × 10−19 J = v + 1 2 0.196 eV ( ) For v = 0, 1, 2, Evib = 0.0982 eV, 0.295 eV, 0.491 eV 43.14 In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is 0.110 + 0.100 nm ( ) = 0.210 nm from the axis. Thus, I = Σmr 2: I = 6 1.99 × 10−26 kg ( ) 0.110 × 10−9 m ( ) 2 + 6 1.67 × 10−27 kg ( ) 0.210 × 10−9 m ( ) 2 = 1.89 × 10−45 kg ⋅m2 The allowed rotational energies are then Erot = h2 2I J J + 1 ( ) = 1.055 × 10−34 J ⋅s ( ) 2 2 1.89 × 10−45 kg ⋅m2 ( ) J J + 1 ( ) = 2.95 × 10−24 J ( )J J + 1 ( ) = 18.4 × 10−6 eV ( )J J + 1 ( ) Erot = 18.4 µeV ( )J J + 1 ( ) where J = 0, 1, 2, 3, . . . The first five of these allowed energies are: Erot eV, 111 eV, 221 eV, and 369 eV = 0 36 9 , . µ µ µ µ 43.15 hf = h2 4π 2I J where the rotational transition is from J – 1 to J , where f = 6.42 × 1013 Hz and I = 1.46 × 10−46 kg ⋅m2 from Example 43.1. J = 4π 2If h = 4π 2 (1.46 × 10−46 kg ⋅m2 )(6.42 × 1013 /s) 6.626 × 10−34 J ⋅s = 558 550 Chapter 43 Solutions 43.16 The emission energies are the same as the absorption energies, but the final state must be below v = 1, J = 0 ( ). The transition must satisfy ∆J = ±1, so it must end with J = 1. To be lower in energy, it must be v = 0, J = 1 ( ). The emitted photon energy is therefore hfphoton = Evib v = 1 + Erot J = 0 −Evib v = 0 + Erot J = 1 = Evib v = 1 −Evib v = 0 −Erot J = 1 −Erot J = 0 hfphoton = hfvib −hfrot Thus, fphoton = fvib −frot = 6.42 × 1013 Hz −1.15 × 1011 Hz = 6.41× 1013 Hz 43.17 The moment of inertia about the molecular axis is Ix = 2 5 mr2 + 2 5 mr2 = 4 5 m 2.00 × 10−15 m ( ) 2 The moment of inertia about a perpendicular axis is Iy = m R 2 2 + m R 2 2 = m 2 2.00 × 10−10 m ( ) 2 The allowed rotational energies are Erot = h2 2I ( )J J + 1 ( ), so the energy of the first excited state is E1 = h2 I . The ratio is therefore E1, x E1, y = h2 Ix ( ) h2 Iy ( ) = Iy Ix = 1 2 m 2.00 × 10−10 m ( ) 2 4 5 m 2.00 × 10−15 m ( ) = 5 8 105 ( ) 2 = 6.25 × 109 43.18 Consider a cubical salt crystal of edge length 0.1 mm. The number of atoms is 10−4 m 0.261× 10−9 m 3 ~ 1017 This number of salt crystals would have volume 10−4 m ( ) 3 10−4 m 0.261× 10−9 m 3 ~ 105 m3 If it is cubic, it has edge length 40 m.
43.19 U = −αkee2 r0 1−1 m = −(1.7476)(8.99 × 109)(1.60 × 10−19)2 (0.281× 10−9) 1−1 8 = −1.25 × 10−18 J = – 7.84 eV 43.20 Visualize a K+ ion at the center of each shaded cube, a C l– ion at the center of each white one. The distance ab is 2 0.314 nm) = ( 0.444 nm Distance ac is 2(0.314 nm) = 0.628 nm Distance ad is 2 +( 2) 0.314 nm) 2 2 ( = 0.769 nm Chapter 43 Solutions 551 © 2000 by Harcourt, Inc. All rights reserved.
43.21 U = −kee2 r −kee2 r + kee2 2r + kee2 2r −kee2 3r −kee2 3r + kee2 4r + kee2 4r − . . . = −2kee2 r 1−1 2 + 1 3 −1 4 + . . . But, ln 1 1 2 3 4 2 3 4 + ( ) = − + − + x x x x . . . so, U = −2kee2 r ln 2, or U = −keα e2 r where α = 2 ln 2 43.22 EF = h2 2m 3ne 8π 2 3 = (6.625 × 10−34 J ⋅s)2 2(9.11× 10−31 kg)(1.60 × 10−19 J/eV) (3/8π)2 3n2 3 EF = (3.65 × 10−19)n2 3 eV with n measured in electrons/m3 43.23 The density of conduction electrons n is given by EF = h 2 2m 3ne 8π 2 3 or ne = 8π 3 2mEF h 2 3/2 = 8π 3 2(9.11× 10−31 kg)(5.48)(1.60 × 10−19 J) [ ] 3/2 (6.626 × 10−34 J ⋅s)3 = 5.80 × 1028 m-3 The number-density of silver atoms is nAg = 10.6 × 103 kg /m3 ( ) 1 atom 108 u 1 u 1.66 × 10−27 kg = 5.91× 1028 m-3 So an average atom contributes 5.80 5.91 = 0.981 electron to the conduction band 43.24 (a) 1 2 mv2 = 7.05 eV v = 2(7.05 eV)(1.60 × 10−19 J/eV) 9.11× 10−31 kg = 1.57 10 m/s 6 × (b) Larger than 10– 4 m/s by ten orders of magnitude. However, the energy of an electron at room temperature is typically kBT = 1 40 eV.
552 Chapter 43 Solutions 43.25 For sodium, M = 23.0 g mol and ρ = 0 971 . g cm3.
(a) ne = NA ρ M = 6.02 × 1023 electrons mol ( ) 0.971 g cm3 ( ) 23.0 g mol ne = 2.54 × 1022 electrons cm3 = 2 54 1028 .
× electrons m3 (b) EF = h2 2m 3ne 8π 2 3 = 6.626 × 10−34 J ⋅s ( ) 2 2 9.11× 10−31 kg ( ) 3 2.54 × 1028 m−3 ( ) 8π 2 3 = 5.05 × 10−19 J = 3.15 eV (c) vF = 2EF m = 2 5.05 × 10−19 J ( ) 9.11× 10−31 kg = 1.05 × 106 m s 43.26 The melting point of silver is 1234 K. Its Fermi energy at 300 K is 5.48 eV. The approximate fraction of electrons excited is kBT EF = 1.38 × 10−23 J K ( ) 1234 K ( ) 5.48 eV ( ) 1.60 × 10−19 J eV ( ) ≈ 2% 43.27 Taking EF = 5.48 eV for sodium at 800 K, f = e E −EF ( ) kBT + 1 [ ] −1 = 0.950 e E −EF ( ) kBT = (1/0.950) −1 = 0.0526 E −EF kBT = ln 0.0526 ( ) = −2.94 E −EF = −2.94 (1.38 × 10−23)(800)J 1.60 × 10−19 J/eV = −0.203 eV or E = 5.28 eV Chapter 43 Solutions 553 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Calculate the energy of a conduction electron in silver at 800 K if the probability of finding an electron in that state is 0.950. The Fermi energy is 5.48 eV at this temperature.
G : Since there is a 95% probability of finding the electron in this state, its energy should be slightly less than the Fermi energy, as indicated by the graph in Figure 43.21. O : The electron energy can be found from the Fermi-Dirac distribution function.
A : Taking EF = 5.48 eV for silver at 800 K, and given f E ( ) = 0.950, we find f E ( ) = 1 e E−EF ( ) kBT + 1 = 0.950 or e E−EF ( ) kBT = 1 0.950 −1 = 0.05263 E −EF kBT = ln 0.05263 ( ) = −2.944 so E −EF = −2.944kBT = −2.944 1.38 × 10−23 J/K ( ) 800 K ( ) E = EF −3.25 × 10−20 J = 5.48 eV −0.203 eV = 5.28 eV L : As expected, the energy of the electron is slightly less than the Fermi energy, which is about 5 eV for most metals. There is very little probability of finding an electron significantly above the Fermi energy in a metal.
43.28 d = 1.00 mm, so V = 1.00 × 10−3 m ( ) 3 = 1.00 × 10−9 m3 The density of states is g E ( ) = CE 1 2 = 8 2 π m3 2 h3 E 1 2 or g E ( ) = 8 2 π 9.11× 10−31 kg ( ) 3 2 6.626 × 10−34 J ⋅s ( ) 3 4.00 eV ( ) 1.60 × 10−19 J eV ( ) g E ( ) = 8.50 × 1046 m−3 ⋅J −1 = 1.36 × 1028 m−3 ⋅ eV −1 So, the total number of electrons is N = g E ( ) [ ] ∆E ( )V = 1.36 × 1028 m−3 ⋅ eV −1 ( ) 0.0250 eV ( ) 1.00 × 10−9 m3 ( ) = 3 40 1017 .
× electrons 43.29 Eav = 1 ne EN E ( )dE 0 ∞ ∫ At T = 0, N E ( ) = 0 for E > EF; Since f E ( ) = 1 for E < EF and f E ( ) = 0 for E > EF, we can take N(E) = CE1/2 Eav = 1 ne 0 EF ∫ CE3/2 dE = C ne 0 EF ∫ E 3 2 dE = 2C 5ne E 5 2 But from Equation 43.24, C ne = 3 2 EF −3/2, so that Eav = 2 5 3 2 EF −3/2 EF 5/2 = 3 5 EF 554 Chapter 43 Solutions 43.30 Consider first the wave function in x. At x = 0 and x = L , ψ = 0.
Therefore, sin kxL = 0 and kxL = π , 2π , 3π , . . .
Similarly, sin kyL = 0 and kyL = π , 2π , 3π , . . .
sin kzL = 0 and kzL = π , 2π , 3π , . . . ψ = Asin nxπx L sin nyπy L sin nzπz L From d2ψ dx2 + d2ψ dy2 + d2ψ dz2 = 2me h2 (U −E)ψ , we have inside the box, where U = 0, −nx 2π 2 L2 − ny 2π 2 L2 −nz 2π 2 L2 ψ = 2me h2 (−E)ψ E = h2π 2 2meL2 (nx 2 + ny 2 + nz 2) nx, ny, nz = 1, 2, 3, . . . Outside the box we require ψ = 0.
The minimum energy state inside the box is nx = ny = nz = 1, with E = 3h2π 2 2meL2 43.31 (a) The density of states at energy E is g E ( ) = CE 1 2 Hence, the required ratio is g 8.50 eV ( ) g 7.00 eV ( ) = C 8.50 ( )1 2 C 7.00 ( )1 2 = 1.10 (b) From Eq. 43.22, the number of occupied states having energy E is N E ( ) = CE 1 2 e E−EF ( ) kBT + 1 Hence, the required ratio is N 8.50 eV ( ) N 7.00 eV ( ) = 8.50 ( )1 2 7.00 ( )1 2 e 7.00 −7.00 ( ) kBT + 1 e 8.50 −7.00 ( ) kBT + 1 At T = 300 K, kBT = 4.14 × 10−21 J = 0.0259 eV, N 8.50 eV ( ) N 7.00 eV ( ) = 8.50 ( )1 2 7.00 ( )1 2 2.00 e 1.50 ( ) 0.0259 + 1 And N 8.50 eV ( ) N 7.00 eV ( ) = 1 55 10 25 .
× − Comparing this result with that from part (a), we conclude that very few states with E > EF are occupied.
43.32 (a) Eg = 1.14 eV for Si h f = 1.14 eV = (1.14 eV)(1.60 × 10– 19 J/eV) = 1.82 × 10– 19 J so f ≥ 2.75 × 1014 Hz (b) c = λ f ; λ = c f = 3.00 × 108 m /s 2.75 × 1014 Hz = 1.09 × 10−6 m = 1.09 µm (in the infrared region) Chapter 43 Solutions 555 © 2000 by Harcourt, Inc. All rights reserved.
43.33 Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than λ = hc E = (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) 2.42 × 1.60 × 10−19 J = 514 nm All the hydrogen Balmer lines except for the red line at 656 nm will be absorbed.
43.34 Eg = hc λ = (6.626 × 10−34 J ⋅s)(3.00 × 108 m /s) 650 × 10−9 m J ≈ 1.91 eV 43.35 If λ ≤1.00 × 10−6 m, then photons of sunlight have energy E ≥ hc λ max = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 1.00 × 10−6 m 1 eV 1.60 × 10−19 J = 1.24 eV Thus, the energy gap for the collector material should be Eg ≤1.24 eV . Since Si has an energy gap Eg ≈1.14 eV, it will absorb radiation of this energy and greater. Therefore, Si is acceptable as a material for a solar collector.
Goal Solution Most solar radiation has a wavelength of 1 µm or less. What energy gap should the material in a solar cell have in order to absorb this radiation? Is silicon appropriate (see Table 43.5)?
G : Since most photovoltaic solar cells are made of silicon, this semiconductor seems to be an appropriate material for these devices.
O : To absorb the longest-wavelength photons, the energy gap should be no larger than the photon energy.
A : The minimum photon energy is hf = hc λ = 6.63 × 10−34 J ⋅s ( ) 3.00 × 108 m /s ( ) 10-6 m 1 eV 1.60 × 10-19 J = 1.24 eV Therefore, the energy gap in the absorbing material should be smaller than 1.24 eV.
L : So silicon, with gap of 1.14 eV < 1.24 eV, is an appropriate material for absorbing solar radiation.
43.36 If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here, hf ( )max = hc λ min = 5.50 eV: λ min = hc 5.5 eV = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 5.5 eV ( ) 1.60 × 10−19 J eV ( ) λ min = 2.26 × 10−7 m = 226 nm 556 Chapter 43 Solutions 43.37 I = I0 ee ∆V ( ) kBT −1 ( ) Thus, ee ∆V ( ) kBT = 1+ I I0 and ∆V = kBT e ln 1+ I I0 ( ) At T = 300 K, ∆V = 1.38 × 10−23 J K ( ) 300 K ( ) 1.60 × 10−19 C ln 1+ I I0 = 25.9 mV ( ) ln 1+ I I0 (a) If I = 9.00I0, ∆V = 25.9 mV ( ) ln 10.0 ( ) = 59.5 mV (b) If I = −0.900I0, ∆V = 25.9 mV ( ) ln 0.100 ( ) = – 59.5 mV The basic idea behind a semiconductor device is that a large current or charge can be controlled by a small control voltage.
43.38 The voltage across the diode is about 0.6 V. The voltage drop across the resistor is (0.025 A)(150 Ω) = 3.75 V. Thus, E −0.6 V −3.8 V = 0 and E = 4.4 V 43.39 First, we evaluate I0 in I = I0 ee ∆V ( ) kBT −1 ( ), given that I = 200 mA when ∆V = 100 mV and T = 300 K. e ∆V ( ) kBT = 1.60 × 10−19 C ( ) 0.100 V ( ) 1.38 × 10−23 J K ( ) 300 K ( ) = 3.86 so I0 = I ee ∆V ( ) kBT −1 = 200 mA e3.86 −1 = 4.28 mA If ∆V = −100 mV, e ∆V ( ) kBT = −3.86; and the current will be I = I0 ee ∆V ( ) kBT −1 ( ) = 4.28 mA ( ) e−3.86 −1 ( ) = – 4.19 mA 43.40 (a) See the figure at right. (b) For a surface current around the outside of the cylinder as shown, B = Nµ0I l or NI = Bl µ0 = (0.540 T)(2.50 × 10−2 m) (4π × 10−7) T ⋅m / A = 10.7 kA Chapter 43 Solutions 557 © 2000 by Harcourt, Inc. All rights reserved.
43.41 By Faraday’s law (Equation 32.1), ∆ΦB ∆t = L ∆I ∆t = A ∆B ∆t .
Thus, ∆I = A ∆B ( ) L = π 0.0100 m ( )2 0.0200 T ( ) 3.10 × 10−8 H = 203 A The direction of the induced current is such as to maintain the B – field through the ring.
Goal Solution Determine the current generated in a superconducting ring of niobium metal 2.00 cm in diameter if a 0.0200-T magnetic field in a direction perpendicular to the ring is suddenly decreased to zero. The inductance of the ring is 3.10 × 10-8 H.
G : The resistance of a superconductor is zero, so the current is limited only by the change in magnetic flux and self-inductance. Therefore, unusually large currents (greater than 100 A) are possible. O : The change in magnetic field through the ring will induce an emf according to Faraday’s law of induction. Since we do not know how fast the magnetic field is changing, we must use the ring’s inductance and the geometry of the ring to calculate the magnetic flux, which can then be used to find the current. A : From Faraday's law (Eq. 31.1), we have ε = ∆ΦB ∆t = A ∆B ∆t = L ∆I ∆t or ∆I = A∆B L = π 0.0100 m ( )2 0.0200 T ( ) 3.10 × 10−8 H = 203 A The current is directed so as to produce its own magnetic field in the direction of the original field.
L : This induced current should remain constant as long as the ring is superconducting. If the ring failed to be a superconductor (e.g. if it warmed above the critical temperature), the metal would have a non-zero resistance, and the current would quickly drop to zero. It is interesting to note that we were able to calculate the current in the ring without knowing the emf. In order to calculate the emf, we would need to know how quickly the magnetic field goes to zero.
43.42 (a) ∆V = IR If R = 0, then ∆V = 0, even when I ≠0.
(b) The graph shows a direct proportionality. Slope = 1 R = ∆I ∆V = − ( ) − ( ) 155 57 8 . mA 3.61 1.356 mV = − 43 1 . 1 Ω R = 0 0232 . Ω (c) Expulsion of magnetic flux and therefore fewer current-carrying paths could explain the decrease in current.
558 Chapter 43 Solutions 43.43 (a) Since the interatomic potential is the same for both molecules, the spring constant is the same. Then f = 1 2π k µ where µ 12 = 12 u ( ) 16 u ( ) 12 u + 16 u = 6.86 u and µ 14 = 14 u ( ) 16 u ( ) 14 u + 16 u = 7.47 u Therefore, f 14 = 1 2π k µ 14 = 1 2π k µ 12 µ 12 µ 14 = f 12 µ 12 µ 14 = 6.42 × 1013 Hz ( ) 6.86 u 7.47 u = 6 15 1013 .
× Hz (b) The equilibrium distance is the same for both molecules. I14 = µ 14r 2 = µ 14 µ 12 µ 12r 2 = µ 14 µ 12 I12 I14 = 7.47 u 6.86 u 1.46 × 10−46 kg ⋅m2 ( ) = 1.59 × 10−46 kg ⋅m2 (c) The molecule can move to the v = 1, J = 9 ( ) state or to the v = 1, J = 11 ( ) state. The energy it can absorb is either ∆E = hc λ = 1+ 1 2 ( )hf 14 + 9 9 + 1 ( ) h2 2I14 − 0 + 1 2 ( )hf 14 + 10 10 + 1 ( ) h2 2I14 , or ∆E = hc λ = 1+ 1 2 ( )hf 14 + 11 11+ 1 ( ) h2 2I14 − 0 + 1 2 ( )hf 14 + 10 10 + 1 ( ) h2 2I14 .
The wavelengths it can absorb are then λ = c f 14 −10h 2π I14 ( ) or λ = c f 14 + 11h 2π I14 ( ) These are: λ = 3.00 × 108 m s 6.15 × 1013 Hz − 10 1.055 × 10−34 J ⋅s ( ) 2π 1.59 × 10−46 kg ⋅m2 ( ) = 4.96 µm and λ = 3.00 × 108 m s 6.15 × 1013 Hz + 11 1.055 × 10−34 J ⋅s ( ) 2π 1.59 × 10−46 kg ⋅m2 ( ) = 4.79 µm Chapter 43 Solutions 559 © 2000 by Harcourt, Inc. All rights reserved.
43.44 For the N2 molecule, k = 2297 N/m, m = 2.32 × 10−26 kg, r = 1.20 × 10−10 m, µ = m/ 2 ω = k µ = 4.45 × 1014 rad/s, I = µr 2 = (1.16 × 10−26 kg)(1.20 × 10−10 m)2 = 1.67 × 10−46 kg ⋅m2 For a rotational state sufficient to allow a transition to the first exited vibrational state, h2 2I J(J + 1) = hω so J(J + 1) = 2Iω h = 2(1.67 × 10−46)(4.45 × 1014) 1.055 × 10−34 = 1410 Thus J = 37 43.45 ∆Emax = 4.5 eV = v + 1 2 hω so (4.5 eV)(1.6 × 10−19 J/eV) (1.055 × 10−34 J ⋅s)(8.28 × 1014 s−1) ≥v + 1 2 8.25 > 7.5 v = 7 43.46 With 4 van der Waal bonds per atom pair or 2 electrons per atom, the total energy of the solid is E = 2(1.74 × 10−23 J/atom) 6.02 × 1023 atoms 4.00 g = 5.23 J/g 43.47 The total potential energy is given by Equation 43.16: Utotal = −α kee2 r + B r m The total potential energy has its minimum value U0 at the equilibrium spacing, r = r0. At this point, dU dr r = r 0 = 0, or dU dr r = r 0 = d dr −α kee2 r + B r m r = r 0 = α kee2 r0 2 −mB r0 m + 1 = 0 Thus, B = α kee2 m r0 m −1 Substituting this value of B into Utotal, U0 = −α kee2 r0 + α kee2 m r0 m −1 1 r0 m = −α kee2 r0 1−1 m 43.48 Suppose it is a harmonic-oscillator potential well. Then, 1 2 hf + 4.48 eV = 3 2 hf + 3.96 eV is the depth of the well below the dissociation point. We see hf = 0.520 eV, so the depth of the well is 1 2 hf + 4.48 eV = 1 2 0.520 eV ( ) + 4.48 eV = 4.74 eV 560 Chapter 43 Solutions 43.49 (a) For equilibrium, dU dx = 0: d dx Ax−3 −Bx−1 ( ) = −3Ax−4 + Bx−2 = 0 x →∞ describes one equilibrium position, but the stable equilibrium position is at 3Ax0 −2 = B. x0 = 3A B = 3 0.150 eV ⋅nm3 ( ) 3.68 eV ⋅nm = 0.350 nm (b) The depth of the well is given by U0 = U x = x0 = A x0 3 −B x0 = AB 3 2 3 3 2A 3 2 − BB 1 2 3 1 2A 1 2 U0 = U x = x0 = − 2B 3 2 3 3 2A 1 2 = − 2 3.68 eV ⋅nm ( ) 3 2 3 3 2 0.150 eV ⋅nm3 ( ) 1 2 = – 7.02 eV (c) Fx = −dU dx = 3Ax−4 −Bx−2 To find the maximum force, we determine finite xm such that dFx dx x = xm = 0 Thus, −12Ax−5 + 2Bx−3 [ ] x = x0 = 0 so that xm = 6A B 1 2 Then Fmax = 3A B 6A 2 −B B 6A = −B2 12A = − 3.68 eV ⋅nm ( )2 12 0.150 eV ⋅nm3 ( ) or Fmax = −7.52 eV nm 1.60 × 10−19 J 1 eV 1 nm 10−9 m = −1.20 × 10−9 N = −1 20 . nN 43.50 (a) For equilibrium, dU dx = 0: d dx Ax−3 −Bx−1 ( ) = −3Ax−4 + Bx−2 = 0 x →∞ describes one equilibrium position, but the stable equilibrium position is at 3Ax0 −2 = B or x0 = 3A B (b) The depth of the well is given by U0 = U x = x0 = A x0 3 −B x0 = AB 3 2 3 3 2A 3 2 − BB 1 2 3 1 2A 1 2 = −2 B3 27A (c) Fx = −dU dx = 3Ax−4 −Bx−2 To find the maximum force, we determine finite xm such that dFx dx x = xm = −12Ax−5 + 2Bx−3 [ ] x = x0 = 0 then Fmax = 3A B 6A 2 −B B 6A = −B2 12A Chapter 43 Solutions 561 © 2000 by Harcourt, Inc. All rights reserved.
43.51 (a) At equilibrium separation, r = re, dU dr r = re = −2aB e −a re −r 0 ( ) −1 e −a re −r 0 ( ) = 0 We have neutral equilibrium as re →∞ and stable equilibrium at e −a re −r 0 ( ) = 1, or re = r0 (b) At r = r0, U = 0. As r →∞, U →B. The depth of the well is B .
(c) We expand the potential in a Taylor series about the equilibrium point: U r ( ) ≈U r0 ( ) + dU dr r = r 0 r −r0 ( ) + 1 2 d2U dr 2 r = r 0 r −r0 ( ) 2 U r ( ) ≈0 + 0 + 1 2 −2Ba ( ) −ae −2 r −r 0 ( ) −ae −r −r 0 ( ) e −2 r −r 0 ( ) −1 r = r 0 r −r0 ( ) 2 ≈Ba2 r −r0 ( ) 2 This is of the form 1 2 kx2 = 1 2 k r −r0 ( ) 2 for a simple harmonic oscillator with k = 2Ba2 Then the molecule vibrates with frequency f = 1 2π k µ = a 2π 2B µ = a π B 2µ (d) The zero-point energy is 1 2hω = 1 2 hf = ha π B 8µ Therefore, to dissociate the molecule in its ground state requires energy B −ha π B 8µ 562 Chapter 43 Solutions 43.52 T = 0 T = 0.1TF T = 0.2TF T = 0.5TF E/EF e E EF −1 TF T f E ( ) e E EF −1 TF T f E ( ) e E EF −1 TF T f E ( ) e E EF −1 TF T f E ( ) 0 e−∞ 1.00 e−10.0 1.000 e−5.00 0.993 e−2.00 0.881 0.500 e−∞ 1.00 e−5.00 0.993 e−2.50 0.924 e−1.00 0.731 0.600 e−∞ 1.00 e−4.00 0.982 e−2.00 0.881 e−0.800 0.690 0.700 e−∞ 1.00 e−3.00 0.953 e−1.50 0.818 e−0.600 0.646 0.800 e−∞ 1.00 e−2.00 0.881 e−1.00 0.731 e−0.400 0.599 0.900 e−∞ 1.00 e−1.00 0.731 e−0.500 0.622 e−0.200 0.550 1.00 e0 0.500 e0 0.500 e0 0.500 e0 0.500 1.10 e+∞ 0.00 e1.00 0.269 e0.500 0.378 e0.200 0.450 1.20 e+∞ 0.00 e2.00 0.119 e1.00 0.269 e0.400 0.401 1.30 e+∞ 0.00 e3.00 0.0474 e1.50 0.182 e0.600 0.354 1.40 e+∞ 0.00 e4.00 0.0180 e2.00 0.119 e0.800 0.310 1.50 e+∞ 0.00 e5.00 0.00669 e2.50 0.0759 e1.00 0.269 Chapter 43 Solutions 563 © 2000 by Harcourt, Inc. All rights reserved.
43.53 (a) There are 6 Cl− ions at distance r = r0. The contribution of these ions to the electrostatic potential energy is −6kee2 r0 .
There are 12 Na+ ions at distance r = 2 r0. Their contribution to the electrostatic potential energy is +12kee2 2 r0 . Next, there are 8 Cl− ions at distance r = 3 r0. These contribute a term of −8kee2 3 r0 to the electrostatic potential energy.
To three terms, the electrostatic potential energy is: U = −6 + 12 2 −8 3 kee2 r0 = −2.13 kee2 r0 or U = −α kee2 r0 with α = 2.13 (b) The fourth term consists of 6 Na+ at distance r r = 2 0. Thus, to four terms, U = −2.13 + 3 ( ) kee2 r0 = 0.866 kee2 r0 So we see that the electrostatic potential energy is not even attractive to 4 terms, and that the infinite series does not converge rapidly when groups of atoms corresponding to nearest neighbors, next-nearest neighbors, etc. are added together.
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 44 Solutions 44.1 An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing mass (say) 35 kg: 35 kg 1 nucleon 1.67 × 10–27 kg ~ 1028 protons and ~ 1028 neutrons The electron number is precisely equal to the proton number, ~ 1028 electrons 44.2 1 2 mv2 = q ∆V ( ) and mv2 r = qvB ⇒ 2m ∆V ( ) = qr 2B2 r = 2m ∆V ( ) qB2 = 2 1000 V ( ) 1.60 × 10−19 C ( ) 0.200 T ( )2 1 2 m r = 5.59 × 1011 m kg m (a) For 12C: m = 12 u and r = × × ( ) = = − 5 59 10 12 1 66 10 0 0789 11 27 .
.
. m kg kg m 7.89 cm For 13C: r = × × ( ) = = − 5 59 10 13 1 66 10 0 0821 11 27 .
.
. m kg kg m 8.21 cm (b) With r m V qB 1 1 2 2 = ( ) ∆ and r m V qB 2 2 2 2 = ( ) ∆ , the ratio gives r1 r2 = m1 m2 . r1 r2 = 7.89 cm 8.21 cm = 0.961 and m1 m2 = 12 u 13 u = 0.961 so they do agree.
566 Chapter 44 Solutions 44.3 (a) F = ke Q 1Q 2 r 2 = (8.99 × 109 N · m2/C 2) (2)(6)(1.60 × 10– 19 C)2 (1.00 × 10–14 m)2 = 27.6 N (b) a = F m = 27.6 N 6.64 × 10– 27 kg = 4.17 × 10 27 m/s2 away from the nucleus..
(c) U = ke Q 1Q 2 r = (8.99 × 109 N · m2/C 2) (2)(6)(1.60 × 10– 19 C)2 (1.00 × 10–14 m) = 2.76 × 10– 13 J = 1.73 MeV 44.4 Eα = 7.70 MeV (a) d k Ze mv k Ze E e e min ( .
)( )( .
) .
( .
) = = = × × × − − 4 2 2 8 99 10 79 1 60 10 7 70 1 60 10 2 2 2 9 19 2 13 α = 29.5 × 10– 15 m = 29.5 fm (b) The de Broglie wavelength of the α is λ = h mαvα = h 2mαEα = 6.626 × 10−34 2(6.64 × 10−27)7.70(1.60 × 10−13) = 5.18 × 10– 15 m = 5.18 fm (c) Since λ is much less than the distance of closest approach , the α may be considered a particle.
44.5 (a) The initial kinetic energy of the alpha particle must equal the electrostatic potential energy at the distance of closest approach. Ki = Uf = keqQ rmin rmin = keqQ Ki = 8.99 × 109 N ⋅m2 C 2 ( ) 2 ( ) 79 ( ) 1.60 × 10−19 C ( ) 2 0.500 MeV ( ) 1.60 × 10−13 J MeV ( ) = 4 55 10 13 .
× − m (b) Since Ki = 1 2 mα vi 2 = keqQ rmin , vi = 2keqQ mα rmin = 2 8.99 × 109 N ⋅m2 C 2 ( ) 2 ( ) 79 ( ) 1.60 × 10−19 C ( ) 2 4.00 u ( ) 1.66 × 10−27 kg u ( ) 3.00 × 10−13 m ( ) = 6 04 106 .
× m s Chapter 44 Solutions 567 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution (a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle having an initial energy of 0.500 MeV and a gold nucleus (197Au) at rest. (Assume the gold nucleus remains at rest during the collision.) (b) What minimum initial speed must the alpha particle have in order to get as close as 300 fm? G : The positively charged alpha particle (q = +2e) will be repelled by the positive gold nucleus (Q = +79e), so that the particles probably will not touch each other in this electrostatic “collision.” Therefore, the closest the alpha particle can get to the gold nucleus would be if the two nuclei did touch, in which case the distance between their centers would be about 6 fm (using r = r0A1/3 for the radius of each nucleus). To get this close, or even within 300 fm, the alpha particle must be traveling very fast, probably close to the speed of light (but of course v must be less than c).
O : At the distance of closest approach, rmin, the initial kinetic energy will equal the electrostatic potential energy between the alpha particle and gold nucleus.
A : (a) Kα = U = ke qQ rmin and rmin = ke qQ Kα = (8.99 × 109 N ⋅m2 /C2)(2)(79)(1.60 × 10−19 C)2 (0.500 MeV)(1.60 × 10-13 J/ MeV) = 455 fm (b) Since Kα = 1 2 mv2 = ke qQ rmin v = 2keqQ mrmin = 2(8.99 × 109 N ⋅m2/C2)(2)(79)(1.60 × 10−19 C)2 4(1.66 × 10−27 kg)(3.00 × 10-13 m) = 6.04 × 106 m /s L : The minimum distance in part (a) is about 100 times greater than the combined radii of the particles.
For part (b), the alpha particle must have more than 0.5 MeV of energy since it gets closer to the nucleus than the 455 fm found in part (a). Even so, the speed of the alpha particle in part (b) is only about 2% of the speed of light, so we are justified in not using a relativistic approach. In solving this problem, we ignored the effect of the electrons around the gold nucleus that tend to “screen” the nucleus so that the alpha particle sees a reduced positive charge. If this screening effect were considered, the potential energy would be slightly reduced and the alpha particle could get closer to the gold nucleus for the same initial energy. 44.6 It must start with kinetic energy equal to Ki = Uf = keqQ rf . Here rf stands for the sum of the radii of the 2 4He and 79 197Au nuclei, computed as rf = r0 A1 1 3 + r0 A2 1 3 = 1.20 × 10−15 m ( ) 41 3 + 197 1 3 ( ) = 8.89 × 10−15 m Thus, K U i f = = × ⋅ ( )( )( ) × ( ) × − − 8 99 10 2 79 1 60 10 8 89 10 9 19 2 15 .
.
. N m C C m 2 2 = 4.09 × 10−12 J = 25.6 MeV 568 Chapter 44 Solutions 44.7 (a) r = r A 0 1 3 / = (1.20 × 10– 15 m)(4)1/3 = 1.90 × 10– 15 m (b) r = r A 0 1 3 / = (1.20 × 10– 15 m)(238)1/3 = 7.44 × 10– 15 m 44.8 From r = r A 0 1 3 / , the radius of uranium is r U = r 0(238)1/3. Thus, if r = 1 2 r U then r 0 A1/3 = 1 2 r 0(238)1/3 from which A = 30 44.9 The number of nucleons in a star of two solar masses is A = 2 1.99 × 1030 kg ( ) 1.67 × 10−27 kg nucleon = 2.38 × 1057 nucleons Therefore r = r0A1/3 = 1.20 × 10−15 m ( ) 2.38 × 1057 ( ) 1 3 = 16.0 km 44.10 V = 4 3 πr 3 = 4.16 × 10−5 m3 m = ρV = (2.31× 1017 kg /m3)(4.16 × 10−5 m3) = 9.61× 1012 kg and F = G m1m2 r 2 = (6.67 × 10−11 Nm2 /kg2)(9.61× 1012 kg)2 (1.00 m)2 = 6.16 × 1015 N toward the other ball.
44.11 The stable nuclei that correspond to magic numbers are: Z magic: 2He 8O 20Ca 28Ni 50Sn 82Pb 126 N magic: T1 3, He 2 4, N7 15, O 8 16, Cl17 37, K 19 39, Ca 20 40, V 23 51, Cr 24 52, Sr 38 88, Y 39 89, Zr 40 90, Xe 54 136, Ba 56 138, La 57 139, Ce 58 140, Pr 59 141, Nd60 142, Pb82 208, Bi 83 209, Po84 210 Chapter 44 Solutions 569 © 2000 by Harcourt, Inc. All rights reserved.
44.12 Of the 102 stable nuclei listed in Table A.3, (a) Even Z, Even N 48 (b) Even Z, Odd N 6 (c) Odd Z, Even N 44 (d) Odd Z, Odd N 4 44.13 (a) (b) 44.14 (a) fn = 2µB h = 2(1.9135)(5.05 × 10−27 J/ T)(1.00 T) 6.626 × 10−34 J ⋅s = 29.2 MHz (b) fp = 2(2.7928)(5.05 × 10−27 J/ T)(1.00 T) 6.626 × 10−34 J ⋅s = 42.6 MHz (c) In the Earth's magnetic field, fp = 2(2.7928)(5.05 × 10−27)(50.0 × 10−6) 6.626 × 10−34 = 2.13 kHz 44.15 Using atomic masses as given in Table A.3, (a) For 2H1, – 2.014 102 + 1(1.008 665) + 1(1.007 825) 2 570 Chapter 44 Solutions Eb = (0.00119 4 u) 931.5 MeV u = 1.11 MeV/nucleon Chapter 44 Solutions 571 © 2000 by Harcourt, Inc. All rights reserved.
(b) For 4He, 2 (1.008 665) + 2 (1.007 825) – 4.002 602 4 Eb = 0.00759 u = 7.07 MeV/nucleon (c) For 56Fe26, 30(1.008 665) + 26(1.007 825) – 55.934 940 = 0.528 u Eb = 0.528 56 = 0.00944 u = 8.79 MeV/nucleon (d) For 238U92, 146(1.008 665) + 92(1.007 825) – 238.050 784 = 1.934 2 u Eb = 1.934 2 238 = 0.00813 u = 7.57 MeV/nucleon 44.16 ∆M = ZmH + Nmn – M BE A = ∆M(931.5) A Nuclei Z N M in u ∆M in u BE/A in MeV 55Mn 25 30 54.938048 0.517527 8.765 56Fe 26 30 55.934940 0.528460 8.786 59Co 27 32 58.933198 0.555357 8.768 ∴ 56Fe has a greater BE/A than its neighbors. This tells us finer detail than is shown in Figure 44.8.
44.17 (a) The neutron-to-proton ratio, A −Z ( )/Z is greatest for 55 139Cs and is equal to 1.53.
(b) 139La has the largest binding energy per nucleon of 8.378 MeV.
(c) 139C s with a mass of 138.913 u. We locate the nuclei carefully on Figure 44.3, the neutron-proton plot of stable nuclei. Cesium appears to be farther from the center of the zone of stability. Its instability means extra energy and extra mass.
44.18 Use Equation 44.4.
The 11 23Na, Eb A = 8.11 MeV/nucleon and for 12 23Mg, Eb A = 7.90 MeV/nucleon The binding energy per nucleon is greater for 11 23Na by 0.210 MeV . (There is less proton repulsion in Na23.) 572 Chapter 44 Solutions 44.19 The binding energy of a nucleus is Eb MeV ( ) = Z M H ( ) + Nmn −M Z AX ( ) [ ] 931.494 MeV u ( ) For 8 15O: Eb = 8 1.007 825 u ( ) + 7 1.008 665 u ( ) −15.003 065 u [ ] 931.494 MeV u ( )= 111.96 MeV For 7 15N: Eb = 7 1.007 825 u ( ) + 8 1.008 665 u ( ) −15.000 108 u [ ] 931.494 MeV u ( )= 115.49 MeV Therefore, the binding energy of 7 15N is larger by 3.54 MeV .
44.20 (a) The radius of the 40Ca nucleus is: R = r0A1 3 = 1.20 × 10−15 m ( ) 40 ( )1 3 = 4.10 × 10−15 m The energy required to overcome electrostatic repulsion is U = 3keQ2 5R = 3 8.99 × 109 N ⋅m2 C 2 ( ) 20 1.60 × 10−19 C ( ) [ ] 2 5 4.10 × 10−15 m ( ) = 1.35 × 10−11 J = 84.1 MeV (b) The binding energy of 20 40Ca is Eb = 20 1.007 825 u ( ) + 20 1.008 665 u ( ) −39.962 591 u [ ] 931.5 MeV u ( )= 342 MeV (c) The nuclear force is so strong that the binding energy greatly exceeds the minimum energy needed to overcome electrostatic repulsion.
44.21 Removal of a neutron from 20 43Ca would result in the residual nucleus, 20 42Ca. If the required separation energy is Sn, the overall process can be described by mass 20 43Ca mass Ca mass n 20 42 ( ) + = ( ) + Sn ( ) Sn = (41.958 618 + 1.008 665 – 42.958 767) u = (0.008 516 u)(931.5 MeV/u) = 7.93 MeV 44.22 (a) The first term overstates the importance of volume and the second term subtracts this overstatement.
(b) For spherical volume 4 3 πR3 4πR2 = R 3 For cubical volume R3 6R2 = R 6 The maximum binding energy or lowest state of energy is achieved by building "nearly" spherical nuclei.
Chapter 44 Solutions 573 © 2000 by Harcourt, Inc. All rights reserved.
44.23 ∆Eb = Ebf −Ebi For A = 200, Eb A = 7.4 MeV so Ebi = 200 7.4 MeV ( ) = 1480 MeV For A ≈100, Eb A ≈8.4 MeV so Ebf = 2 100 ( ) 8.4 MeV ( ) = 1680 MeV ∆Eb = Ebf −Ebi Eb = 1680 MeV −1480 MeV = 200 MeV 44.24 (a) "Volume" term: E 1 = C 1A = (15.7 MeV)(56) = 879 MeV "Surface" term: E 2 = – C 2A 2/3 = – (17.8 MeV)(56)2/3 = – 260 MeV "Coulomb" term: E 3 = – C 3 Z(Z – 1) A 1/3 = – (0.71 MeV) (26)(25) (56)1/3 = – 121 MeV "Asymmetry" term: E 4 = C 4 (A – 2Z)2 A = – (23.6 MeV) (56 – 52)2 56 = – 6.74 MeV Eb = 491 MeV (b) E Eb 1 179 = %; E Eb 2 53 0 = − . %, E Eb 3 24 6 = − . %; E Eb 4 1 37 = −.
% 44.25 dN dt = – λ N so λ = − = × × = × − − − 1 1 00 10 6 00 10 6 00 10 15 11 4 1 N dN dt ( .
)( .
) . s T 1 2 2 / ln = = λ 1 16 103 .
× s (= 19.3 min) 44.26 R R e e e t = = ( ) = ( )( ) − − ( ) − 0 2 8 04 40 2 2 5 6 40 6 40 λ .
.
ln .
.
ln mCi mCi d d = 6.40 mCi ( ) 1 25 = 0.200 mC i 574 Chapter 44 Solutions 44.27 (a) From R R e t = − 0 λ , λ = 1 0 t R R ln = 1 4.00 h ln 10.0 8.00 = 5.58 × 10−2 h−1 = 1 55 10 5 1 .
× − − s T 1 2 2 / ln = = λ 12.4 h (b) N R 0 0 3 5 10 10 0 10 1 55 10 3 70 10 = = × × × − − λ .
.
/ .
/ Ci s s 1 Ci = 2.39 × 1013 atoms (c) R R e t = = − × × = − − 0 2 10 5 58 10 30 0 λ ( .
) exp ( .
. ) 0 mCi 1.87 mC i Goal Solution A freshly prepared sample of a certain radioactive isotope has an activity of 10.0 mCi. After 4.00 h, its activity is 8.00 mCi. (a) Find the decay constant and half-life. (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30.0 h after it is prepared? G : Over the course of 4 hours, this isotope lost 20% of its activity, so its half-life appears to be around 10 hours, which means that its activity after 30 hours (~3 half-lives) will be about 1 mCi. The decay constant and number of atoms are not so easy to estimate.
O : From the rate equation, R = R0e−λt, we can find the decay constant λ , which can then be used to find the half life, the original number of atoms, and the activity at any other time, t.
A : (a) λ = 1 t ln Ro R = 1 4.00 h ( ) 60.0 s/h ( ) ln 10.0 mCi 8.00 mCi = 1.55 × 10−5 s−1 T1/2 = ln 2 λ = 0.693 0.0558 h-1 = 12.4 h (b) The number of original atoms can be found if we convert the initial activity from curies into becquerels (decays per second): 1 Ci ≡3.7 × 1010 Bq R0 = 10.0 mCi = 10.0 × 10−3 Ci ( ) 3.70 × 1010 Bq /Ci ( ) = 3.70 × 108 Bq Since R0 = λN0 , N0 = R0 λ = 3.70 × 108 decays/s 1.55 × 10−5 s = 2.39 × 1013 atoms (c) R = Roe−λt = (10.0 mCi)e−(5.58×10−2 h−1)(30.0 h) = 1.87 mCi L : Our estimate of the half life was about 20% short because we did not account for the non-linearity of the decay rate. Consequently, our estimate of the final activity also fell short, but both of these calculated results are close enough to be reasonable.
The number of atoms is much less than one mole, so this appears to be a very small sample. To get a sense of how small, we can assume that the molar mass is about 100 g/mol, so the sample has a mass of only m ≈2.4 × 1013 atoms ( ) 100 g /mol ( ) 6.02 × 1023 atoms/mol ( ) ≈0.004 µg This sample is so small it cannot be measured by a commercial mass balance! The problem states that this sample was “freshly prepared,” from which we assumed that all the atoms within the sample are initially radioactive. Generally this is not true, so that N0 only accounts for the formerly radioactive atoms, and does not include additional atoms in the sample that were not radioactive.
Realistically then, the sample mass should be significantly greater than our above estimate.
Chapter 44 Solutions 575 © 2000 by Harcourt, Inc. All rights reserved.
44.28 R R e t = − 0 λ where λ = ln 2 26.0 h = 0.0266/h R R e t 0 0 100 = = − .
λ so ln (0.100) = – λ t 2.30 = 0.0266 h t t = 86.4 h 44.29 The number of nuclei which decay during the interval will be N1 −N2 = N0 e−λ t1 −e−λ t 2 First we find λ : λ = ln 2 T1/2 = 0.693 64.8 h = 0.0107 h−1 = 2.97 × 10−6 s−1 and N0 = R0 λ = (40.0 µCi)(3.70 × 104 cps/µCi) 2.97 × 10−6 s−1 = 4.98 × 1011 nuclei Substituting these values, N1 −N2 = 4.98 × 1011 ( ) e−(0.0107 h−1)(10.0 h) −e−(0.0107 h−1)(12.0 h) Hence, the number of nuclei decaying during the interval is N1 −N2 = 9.47 × 109 nuclei 44.30 The number of nuclei which decay during the interval will be N1 −N2 = N0 e−λ t1 −e−λ t 2 First we find λ : λ = ln / 2 1 2 T so e−λ t = eln 2(−t/T1/2 ) = 2−t/T1/2 and N 0 = R0 λ = R0T1/2 ln 2 Substituting in these values N1 −N2 = R0T1/2 ln 2 e−λ t 1 −e−λ t 2 ( ) = R0T1/2 ln 2 2−t1/T1/2 −2−t2/T1/2 ( ) 44.31 R = λN = ln2 5.27 yr 1.00 g 59.93 g /mol 6.02 × 1023 ( ) R = 1.32 × 1021 decays yr 1 yr 3.16 × 107 s = 4 18 1013 .
× Bq 576 Chapter 44 Solutions 44.32 (a) 28 65Ni (b) 82 211Pb (c) 27 55Co (d) −1 0e (e) 1 1H or p ( ) 44.33 Q = M238U −M234Th −M4He ( ) 931.5 MeV u ( ) Q = 238.050 784 −234.043 593 −4.002 602 ( )u 931.5 MeV u ( ) = 4.27 MeV 44.34 NC = 0.0210 g 12.0 g mol 6.02 × 1023 molecules mol ( ) NC = 1.05 × 1021 carbon atoms ( ) of which 1 in 7.70 × 1011 is a 14C atom N0 9 14 1 37 10 ( ) = × C .
, λ 14C = ln 2 5730 yr = 1.21× 10−4 yr−1 = 3.83 × 10−12 s−1 R = λ N = λ N0e−λ t At t = 0, R0 = λ N0 = 3.83 × 10−12 s-1 ( ) 1.37 × 109 ( ) 7 86400 s ( ) 1 week = 3.17 × 103 decays week At time t , R = 837 0.88 = 951 decays week Taking logarithms, ln R R 0 = −λt so t = −1 λ ln R R0 t = −1 1.21× 10−4 yr−1 ln 951 3.17 × 103 = 9.96 × 103 yr 44.35 In the decay 1 3H He e → + + − 2 3 1 0 ν , the energy released is E m c M M c = ( ) = − ∆ 2 2 1 3 3 2 H He since the antineutrino is massless and the mass of the electron is accounted for in the masses of 1 3H and 2 3He.
Thus, E = 3.016 049u −3.016 029u [ ] 931.5 MeV u ( ) = 0.0186 MeV = 18.6 keV Chapter 44 Solutions 577 © 2000 by Harcourt, Inc. All rights reserved.
44.36 (a) For e+ decay, Q = MX −MY −2me ( )c 2 = 39.962 591 u −39.964 000 u −2 0.0000 549 u ( ) [ ] 931.5 MeV u ( ) Q = −2.34 MeV Since Q < 0, the decay cannot occur spontaneously.
(b) For alpha decay, Q = MX −Mα −MY ( )c 2 = 97.905 287 u −4.002 602 u −93.905 085 u [ ] 931.5 MeV u ( ) Q = −2.24 MeV Since Q < 0, the decay cannot occur spontaneously.
(c) For alpha decay, Q = MX −Mα −MY ( )c 2 = 143.910 082 u −4.002 602 u −139.905 434 u [ ] 931.5 MeV u ( ) Q = 1.91 MeV Since Q > 0, the decay can occur spontaneously.
44.37 (a) e p n −+ → + ν (b) For nuclei, 15O + e−→15N + ν . Add seven electrons to both sides to obtain 8 15 7 15 O atom N atom → + ν .
(c) From Table A.3, m 15O ( ) = m 15N ( ) + Q c 2 ∆m = 15.003 065 u – 15.000 108 u = 0.002 957 u Q = (931.5 MeV/u)(0.002 957 u) = 2.75 MeV 578 Chapter 44 Solutions 44.38 (a) Let N be the number of 238U nuclei and N' be 206Pb nuclei.
Then N = N0e−λ t and N0 = N + ′ N so N = N + ′ N ( )e−λ t or eλ t = 1+ ′ N N Taking logarithms, λt = ln 1+ ′ N N where λ = ln2 ( )/T1/2.
Thus, t = T1 2 ln 2 ln 1+ ′ N N If N ′ N = 1.164 for the 238 206 U Pb → chain with T1 2 = 4.47 × 109 yr, the age is: t = 4.47 × 109 yr ln 2 ln 1+ 1 1.164 = 4 00 109 .
× yr (b) From above, eλ t = 1+ ′ N N . Solving for N ′ N gives N ′ N = e−λ t 1−e−λ t With t = × 4 00 109 . yr and T 1 2 8 7 04 10 = × . yr for the 235 207 U Pb → chain, λt = ln 2 T1 2 t = ln 2 ( ) 4.00 × 109 yr ( ) 7.04 × 108 yr = 3.938 and N ′ N = 0.0199 With t = × 4 00 109 . yr and T 1 2 10 1 41 10 = × . yr for the 232 208 Th Pb → chain, λt = ln 2 ( ) 4.00 × 109 yr ( ) 1.41× 1010 yr = 0.1966 and N ′ N = 4.60 44.39 Chapter 44 Solutions 579 © 2000 by Harcourt, Inc. All rights reserved.
44.40 (a) 4 00 4 00 10 3 70 10 1 00 10 1 12 10 3 .
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. pCi L Ci 1 L Bq 1 Ci L m3 = × × × = − 148 Bq m3 (b) N = R λ = R T1 2 ln 2 = 148 Bq m3 3.82 d ln 2 86 400 s 1 d = 7 05 107 .
× atoms m3 (c) mass atoms m mol 6.02 10 atoms g 1 mol g m 3 23 3 = × × = × − 7 05 10 1 222 2 60 10 7 14 .
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Since air has a density of 1.20 kg m3 , the fraction consisting of radon is fraction g m 1.20 kg m 3 3 = × = − 2 60 10 14 . 2 17 10 17 .
× − 44.41 Number remaining: N = N0e−ln 2 ( )t T1 2 Fraction remaining: N N0 = e−λ t = e−ln 2 ( )t T1 2 (a) With T1 2 = 3.82 d and t = 7.00 d, N N0 = e−ln 2 ( ) 7.00 ( ) 3.82 ( ) = 0.281 (b) When t = 1.00 yr = 365.25 d, N N0 = e−ln 2 ( ) 365.25 ( ) 3.82 ( ) = 1 65 10 29 .
× − (c) Radon is continuously created as one daughter in the series of decays starting from the long-lived isotope 238U.
44.42 Q = M 27Al + Mα −M 30P −mn [ ]c2 Q = 26.981 538 + 4.002 602 −29.978 307 −1.008 665 [ ]u 931.5 MeV u ( ) = −2 64 . MeV 44.43 (a) 79 197Au n Au Hg e + → → + + − 0 1 79 197 80 198 1 0 ν (b) Consider adding 79 electrons: 79 197Au atom + 0 1n→ 80 198Hg atom + ν + Q Q = M197Au + mn −M198Hg c2 Q = 196.966 543 + 1.008 665 −197.966 743 [ ]u 931.5 MeV u ( ) = 7.89 MeV 580 Chapter 44 Solutions 44.44 (a) For X, A = 24 + 1−4 = 21 and Z = 12 + 0 −2 = 10, so X is 10 21Ne (b) A = 235 + 1−90 −2 = 144 and Z = 92 + 0 −38 −0 = 54, so X is 54 144Xe (c) A = 2 −2 = 0 and Z = − = + 2 1 1, so X must be a positron.
As it is ejected, so is a neutrino: X = 1 0e+ and ′ X = 0 0ν 44.45 Neglect recoil of product nucleus, (i.e., do not require momentum conservation). The energy balance gives Kemerging = Kincident + Q. To find Q: Q = MH + MAl ( ) −MSi + mn ( ) [ ]c2 Q = 1.007 825 + 26.981 528 ( ) −26.986 721+ 1.008 665 ( ) [ ]u 931.5 MeV u ( ) = −5 61 . MeV Thus, Kemerging MeV MeV = − = 6 61 5 61 .
. 1.00 MeV 44.46 (a) 5 10 2 4 6 13 1 1 B He C H + → + The product nucleus is 6 13C (b) 6 13 1 1 5 10 2 4 C H B He + → + The product nucleus is 5 10B 44.47 4 9Be MeV Be n + → + 1 666 4 8 0 1 .
, so M 4 8Be = M 4 9Be −Q c2 −mn M 4 8Be = 9.012 174 u −−1.666 MeV ( ) 931.5 MeV u −1.008 665 u = 8.005 3 u 4 9Be n Be MeV + → + 0 1 4 10 6 810 .
, so M 4 10Be = M 4 9Be + mn −Q c 2 M 4 10Be = 9.012 174 u + 1.008 665 u − 6.810 MeV 931.5 MeV u = 10.013 5 u Chapter 44 Solutions 581 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Using the Q values of appropriate reactions and from Table 44.5, calculate the masses of 8Be and 10Be in atomic mass units to four decimal places. G : The mass of each isotope in atomic mass units will be approximately the number of nucleons (8 or 10), also called the mass number. The electrons are much less massive and contribute only about 0.03% to the total mass. O : In addition to summing the mass of the subatomic particles, the net mass of the isotopes must account for the binding energy that holds the atom together. Table 44.5 includes the energy released for each nuclear reaction. Precise atomic masses values are found in Table A.3.
A : The notation 9Be γ , n ( ) 8Be with Q = −1.666 MeV means 9Be + γ → 8Be + n −1.666 MeV Therefore m 8Be ( ) = m 9Be ( ) −mn + 1.666 MeV 931.5 MeV/ u m 8Be ( ) = 9.012174 −1.008665 + 0.001789 = 8.0053 u The notation 9Be n, γ ( ) 10Be with Q = 6.810 MeV means 9Be + n → 10Be + γ + 6.810 MeV m 10Be ( ) = m 9Be ( ) + mn + 6.810 MeV 931.5 MeV/ u m 10Be ( ) = 9.012174 + 1.008665 −0.001789 = 10.0135 u L : As expected, both isotopes have masses slightly greater than their mass numbers. We were asked to calculate the masses to four decimal places, but with the available data, the results could be reported accurately to as many as six decimal places.
44.48 92 236U Rb Cs n → + + 37 90 55 143 0 1 3 , so Q = M 92 236U −M 37 90Rb −M 55 143Cs −3mn c 2 From Table A.3, Q = 236.045 562 −89.914 811−142.927 220 −3 1.008 665 ( ) [ ]u 931.5 MeV u ( ) = 165 MeV 44.49 N1 N2 = N0 −N0e−λ Th/2 N0e−λ Th/2 −N0e−λ Th = 1−e−ln 2/2 e−ln 2/2 −e−ln 2 = − − = − − − 1 2 2 2 1 2 1 2 1 / / 2 582 Chapter 44 Solutions 44.50 1 1 3 7 4 7 0 1 H Li Be n + → + Q = (MH + MLi) −(MBe + Mn) Q = (1.007 825 u + 7.016 003 u) −(7.016 928 u + 1.008 665 u) Q = (−1.765 × 10−3 u)(931.5 MeV/ u) = −1.644 MeV Thus, KEmin = 1+ mincident projectile mtarget nucleus Q = 1+ 1.007 825 7.016 003 (1.644 MeV) = 1.88 MeV 44.51 (a) N0 = mass mass per atom = 1.00 kg 239.05 u ( ) 1.66 × 10−27 kg u ( ) = 2.52 1024 × (b) λ = = × ( ) × ( ) = × − − ln ln .
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/ 2 2 2 412 10 3 156 10 9 106 10 1 2 4 7 13 T yr s yr s 1 R0 = λ N0 = 9.106 × 10−13 s−1 ( ) 2.52 × 1024 ( ) = 2 29 1012 .
× Bq (c) R = R0e−λ t, so t = −1 λ ln R R0 = 1 λ ln R0 R t = 1 9.106 × 10−13 s−1 ln 2.29 × 1012 Bq 0.100 Bq = 3.38 × 1013 s 1 yr 3.156 × 107 s = 1 07 106 .
× yr 44.52 (a) 27 57Co Fe e → + + + 26 57 1 0 0 0ν The Q −value for this positron emission is Q M M m c e = − − [ ] 57 57 2 2 Co Fe Q = − −( ) [ ] ( ) 56 936 294 56 935 396 2 0 000 549 .
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u 931.5 MeV u = −0 186 . MeV Since Q < 0, this reaction cannot spontaneously occur .
(b) 6 14C N e → + + − 7 14 1 0 0 0ν The Q −value for this e− decay is Q = M14C −M14N [ ]c 2. Q = 14.003 242 −14.003 074 [ ]u 931.5 MeV u ( ) = 0.156 MeV = 156 keV Since Q > 0, the decay can spontaneously occur .
(c) The energy released in the reaction of (b) is shared by the electron and neutrino. Thus, Ke can range from zero to 156 keV .
Chapter 44 Solutions 583 © 2000 by Harcourt, Inc. All rights reserved.
44.53 (a) r = r0A1/3 = 1.20 × 10−15A1/3 m. When A = 12, r = 2.75 × 10– 15 m (b) F = ke(Z −1)e2 r 2 = (8.99 × 109 N ⋅m2 /C 2)(Z −1)(1.60 × 10−19 C)2 r 2 When Z = 6 and r = 2.75 × 10– 15 m, F = 152 N (c) U = keq1q2 r = ke(Z −1)e2 r = (8.99 × 109)(1.6 × 10−19)2(Z −1) r When Z = 6 and r = 2.75 × 10– 15 m, U = 4.19 × 10– 13 J = 2.62 MeV (d) A = 238; Z = 92, r = 7.44 × 10– 15 m F = 379 N and U = 2.82 × 10– 12 J = 17.6 MeV 44.54 (a) ln(counts/min) vs time 0.00 3.00 6.00 9.00 0.00 4.00 8.00 12.00 t (h) ln(cpm) A least-square fit to the graph yields: λ = − = −− ( ) = − − slope h h 1 1 0 250 0 250 .
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, and ln .
cpm intercept ( ) = = = t 0 8 30 (b) λ = 0.250 h−1 1 h 60.0 min = 4 17 10 3 .
× − − min 1 T1 2 = ln 2 λ = ln 2 4.17 × 10−3 min−1 = 166 min = 2.77 h (c) From (a), intercept = ln (cpm)0 = 8.30.
Thus, cpm ( )0 = e8.30 counts min = 4 02 103 .
× counts min (d) N0 = R0 λ = 1 λ cpm ( )0 Eff = 4.02 × 103 counts min 4.17 × 10−3 min−1 ( ) 0.100 ( ) = 9 65 106 .
× atoms 584 Chapter 44 Solutions 44.55 (a) Because the reaction p → n + e+ + ν would violate the law of conservation of energy , mp = 1.007 276 u mn = 1.008 665 u me+ = 5.49 × 10−4 u Note that mn + me+ > mp (b) The required energy can come from the electrostatic repulsion of protons in the nucleus.
(c) Add seven electrons to both sides of the reaction for nuclei 7 13N→6 13C + e+ + ν to obtain the reaction for neutral atoms 7 13N atom→6 13C atom + e+ + e−+ ν Q = c 2 m 13N ( ) −m 13C ( ) −me+ −me−−mν [ ] Q = (931.5 MeV/ u) 13.005 738 −13.003 355 −2 5.49 × 10−4 ( ) −0 [ ]u Q = (931.5 MeV/u)(1.285 × 10– 3 u) = 1.20 MeV 44.56 (a) If we assume all the 87Sr came from 87Rb, then N = N0e−λ t yields t = −1 λ ln N N0 = T1 2 ln 2 ln N0 N , where N = N 87Rr and N0 = N 87Sr + N 87Rb. t = 4.75 × 1010 yr ( ) ln 2 ln 1.82 × 1010 + 1.07 × 109 1.82 × 1010 = 3 91 109 .
× yr (b) It could be no longer . The rock could be younger if some 87Sr were originally present.
44.57 (a) Let us assume that the parent nucleus (mass Mp) is initially at rest, and let us denote the masses of the daughter nucleus and alpha particle by Md and Mα, respectively. Applying the equations of conservation of momentum and energy for the alpha decay process gives Mdvd = Mαvα (1) Mpc 2 = Mdc 2 + Mαc 2 + 1 2 Mαvα 2 + 1 2 Mdvd 2 (2) The disintegration energy Q is given by Q M M M c M v M v p d d d = − − = + ( ) α α α 2 2 2 1 2 1 2 (3) Eliminating vd from Equations (1) and (3) gives Q = 1 2 Md Mα Md vα 2 + 1 2 Mαvα 2 = 1 2 Mα 2 Md vα 2 + 1 2 Mαvα 2 = 1 2 Mαvα 2 1+ Mα Md = Kα 1+ Mα Md (b) Kα = Q 1+ Mα Md ( ) = 4.87 MeV 1+ 4/ 222 ( ) = 4.78 MeV Chapter 44 Solutions 585 © 2000 by Harcourt, Inc. All rights reserved.
44.58 (a) The reaction is 61 145 59 141 Pm→ + Pr α (b) Q = (MPm −Mα −MPr)931.5 = (144.912 745 −4.002 602 −140.907 647)931.5 = 2.32 MeV (c) The alpha and daughter have equal and opposite momenta p pd α = E p m α α α = 2 2 E p m d d d = 2 2 Eα Etot = Eα Eα + Ed = pα 2 2mα pα 2 2ma + pα 2 2md = 1 2mα 1 2mα + 1 2md = md md + mα = 141 141+ 4 = 97.2% or 2.26 MeV This is carried away by the alpha 44.59 (a) If ∆E is the energy difference between the excited and ground states of the nucleus of mass M, and h f is the energy of the emitted photon, conservation of energy gives ∆E = h f + Er (1) Where Er is the recoil energy of the nucleus, which can be expressed as Er = Mv2 2 = (Mv)2 2M (2) Since momentum must also be conserved, we have Mv = h f c (3) Hence, Er can be expressed as Er = (hf)2 2Mc2 .
When h f << Mc 2, we can make the approximation that h f ≈ ∆E, so Er ≈ (∆E)2 2Mc 2 (b) Er = (∆E)2 2Mc 2 where ∆E = 0.0144 MeV and Mc 2 = (57 u)(931.5 MeV/u) = 5.31 × 10 4 MeV Therefore, Er = (1.44 × 10−2 MeV)2 (2)(5.31× 104 MeV) = 1.94 × 10– 3 eV 586 Chapter 44 Solutions 44.60 (a) One liter of milk contains this many 40K nuclei: N = 2.00 g ( ) 6.02 × 1023 nuclei mol 39.1 g mol 0.0117 100 = 3.60 × 1018 nuclei λ = ln 2 T1 2 = ln 2 1.28 × 109 yr 1 yr 3.156 × 107 s = 1.72 × 10−17 s−1 R = λ N = 1.72 × 10−17 s−1 ( ) 3.60 × 1018 ( ) = 61.8 Bq (b) For the iodine, R = R0e−λ t with λ = ln 2 d 8 04 .
. t = 1 λ ln R0 R = 8.04 d ln 2 ln 2000 61.8 = 40.3 d 44.61 (a) For cobalt-56, λ = = = − ln 2 7.1 d d yr 3.28 yr 1 T 1 2 2 7 365 25 1 ln .
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The elapsed time from July 1054 to July 2000 is 946 yr. R = R0e−λ t implies R R0 = e−λ t = e −3.28 yr−1 ( ) 946 yr ( ) = e−3106 = e−ln 10 ( )1349 = ~ 10 1349 − (b) For carbon-14, λ = ln 2 5730 yr = 1.21× 10−4 yr−1 R R0 = e−λ t = e −1.21 × 10−4 yr−1 ( ) 946 yr ( ) = e−0.114 = 0.892 44.62 We have N235 = N0, 235e−λ 235t and N238 = N0, 238e−λ 238t N235 N238 = 0.00725 = e −ln 2 ( )t Th, 235 + ln 2 ( )t Th, 238 ( ) Taking logarithms, −4.93 = − ln 2 0.704 × 109 yr + ln 2 4.47 × 109 yr t or − = − × + × ( ) 4 93 1 0 704 10 1 4 47 10 2 9 9 .
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ln yr yr t t = − − × ( ) = − − 4 93 1 20 10 2 9 .
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ln yr 1 5 94 109 .
× yr Chapter 44 Solutions 587 © 2000 by Harcourt, Inc. All rights reserved.
44.63 (a) Add two electrons to both sides of the reaction to have it in energy terms: 4 1 1H atom → 2 4He atom + Q Q = ∆mc2 = 4 M 1 1H −M 2 4He [ ]c2 Q = 4 1.007 825 u ( ) −4.002 602 u [ ] 931.5 MeV u ( ) 1.60 × 10−13 J 1 MeV = 4 28 10 12 .
× − J (b) N = 1.99 × 1030 kg 1.67 × 10−27 kg atom = 1 19 1057 .
× atoms = 1 19 1057 .
× protons (c) The energy that could be created by this many protons in this reaction is: 1 19 10 4 28 10 4 1 27 10 57 12 45 .
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× ( ) × = × − protons J protons J P = E t so t = E P = 1.27 × 1045 J 3.77 × 1026 W = 3.38 × 1018 s = 107 billion years 44.64 (a) Q = M9Be + M 4He −M12C −mn [ ]c2 Q = 9.012 174 u + 4.002 602 u −12.000 000 u −1.008 665 u [ ] 931.5 MeV u ( ) = 5.69 MeV (b) Q M M mn = − − [ ] 2 2 3 H He Q = ( ) − − [ ] ( ) 2 2 014 102 3 016 029 1 008 665 .
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u 931.5 MeV u = 3.27 MeV (exothermic) 44.65 E = – µ · B so the energies are E 1 = +µB and E 2 = – µB µ = 2.7928µn and µn = 5.05 × 10– 27 J/T ∆E = 2µB = 2 (2.7928)(5.05 × 10– 27 J/T)(12.5 T) = 3.53 × 10– 25 J = 2.20 × 10– 6 eV 588 Chapter 44 Solutions 44.66 (a) λ = ln 2 T1 2 = ln 2 5.27 yr 1 yr 3.156 × 107 s = 4.17 × 10−9 s−1 t = 30.0 months = 2.50 yr ( ) 3.156 × 107 s 1 yr = 7.89 × 107 s R = R0e−λ t = λ N0 ( )e−λ t so N0 = R λ eλ t = 10.0 Ci ( ) 3.70 × 1010 Bq Ci ( ) 4.17 × 10−9 s−1 e 4.17 × 10−9 s−1 ( ) 7.89 × 107 s ( ) N0 = × 1 23 1020 . nuclei Mass atoms g mol atoms mol = × ( ) × 1 23 10 59 93 6 02 10 20 23 .
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. = 1.23 × 10−2 g = 12.3 mg (b) We suppose that each decaying nucleus promptly puts out both a beta particle and two gamma rays, for Q = + + ( ) = 0 310 1 17 1 33 .
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. MeV 2.81 Mev P = QR = 2.81 Mev ( ) 1.6 × 10−13 J MeV ( ) 3.70 × 1011 s−1 ( ) = 0.166 W 44.67 For an electric charge density ρ = Ze 4 3 πR 3 Using Gauss's Law inside the sphere, E⋅4πr 2 = 4 3 πr 3 e0 Ze 4 3 πR 3 : E = 1 4π e0 Zer R 3 (r ≤ R) E = 1 4π e0 Ze r 2 (r ≥ R) We now find the electrostatic energy: U = 1 2 e0E 24πr 2 dr r = 0 ∞ ∫ U = 1 2 e0 0 R ∫ 1 4π e0 2 Z2e2r 2 R 6 4πr 2 dr + 1 2 e0 1 4π e0 2 Z2e2 r 4 R ∞ ∫ 4πr 2 dr = Z2e2 8π e0 R 5 5R 6 + 1 R = 3 20 Z2e2 π e0R Chapter 44 Solutions 589 © 2000 by Harcourt, Inc. All rights reserved.
44.68 (a) For the electron capture, 43 93Tc + −1 0e→42 93Mo + γ The disintegration energy is Q = M93Tc −M93Mo [ ]c2. Q = − [ ] ( ) = 92 910 2 92 906 8 3 17 .
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u 931.5 MeV u MeV > 2.44 MeV Electron capture is allowed to all specified excited states in 42 93Mo.
For positron emission, 43 93Tc→42 93Mo + +1 0e + γ The disintegration energy is ′ = − − [ ] Q M M m c e 93 93 2 2 Tc Mo . ′ = − −( ) [ ] ( ) = Q 92 910 2 92 906 8 2 0 000 549 2 14 .
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u 931.5 MeV u MeV Positron emission can reach the 1.35, 1.48, and 2.03 MeV states but there is insufficient energy to reach the 2.44 MeV state.
(b) The daughter nucleus in both forms of decay is 42 93Mo .
44.69 K mv = 1 2 2, so v = 2K m = 2 0.0400 eV ( ) 1.60 × 10−19 J eV ( ) 1.67 × 10−27 kg = 2.77 × 103 m s The time for the trip is t x v = = × × = 1 00 10 2 77 10 3 61 4 3 .
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. m m s s The number of neutrons finishing the trip is given by N = N0e−λ t.
The fraction decaying is 1 1 1 0 004 00 0 2 2 3 61 1 2 − = − = − = = −( ) −( )( ) N N e e t T ln ln .
. s 624 s 0.400% 590 Chapter 44 Solutions 44.70 (a) At threshold, the particles have no kinetic energy relative to each other. That is, they move like two particles which have suffered a perfectly inelastic collision. Therefore, in order to calculate the reaction threshold energy, we can use the results of a perfectly inelastic collision.
Initially, the projectile Ma moves with velocity va while the target MX is at rest. We have from momentum conservation: Mava = Ma + MX ( )vc The initial energy is: Ei = 1 2 Mava 2 The final kinetic energy is: Ef = 1 2 Ma + MX ( )vc 2 = 1 2 Ma + MX ( ) Mava Ma + MX 2 = Ma Ma + MX ( ) Ei From this, we see that Ef is always less than Ei and the loss in energy, Ei −Ef , is given by Ei −Ef = 1− Ma Ma + MX Ei = MX Ma + MX Ei In this problem, the energy loss is the disintegration energy −Q and the initial energy is the threshold energy Eth. Therefore, −Q = MX Ma + MX Eth or Eth = −Q MX + Ma MX = −Q 1+ Ma MX (b) First, calculate the Q −value for the reaction: Q = M14N + M 4He −M17O −M1H [ ]c2 Q = 14.003 074 + 4.002 602 −16.999 132 −1.007 825 [ ]u 931.5 MeV u ( ) = −1.19 MeV Then, Eth = −Q MX + Ma MX = −−1.19 MeV ( ) 1+ 4.002 602 14.003 074 = 1.53 MeV 44.71 R= R 0 exp (– λ t) ln R = ln R 0 – λ t (the equation of a straight line) slope = λ The logarithmic plot shown in Figure P44.71 is fitted by ln R = 8.44 – 0.262t.
If t is measured in minutes, then the decay constant λ is 0.262 per minute. The half-life is T 1 2 2 2 0 262 = = = ln ln .
/min λ 2.64 min The reported half-life of 137Ba is 2.55 min. The difference reflects experimental uncertainties.
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 45 Solutions 45.1 ∆m = (mn + MU) – (MZr + MTe + 3mn) ∆m = (1.008 665 u + 235.043 924 u) – (97.912 0 u + 134.908 7 u + 3(1.008 665 u)) ∆m = 0.205 89 u = 3.418 × 10–28 kg so Q = ∆mc2 = 3.076 × 10–11 J = 192 MeV 45.2 Three different fission reactions are possible: 0 1 92 235 38 90 54 144 0 1 2 n U Sr Xe n + → + + 54 144Xe 0 1 92 235 38 90 54 143 0 1 3 n U Sr Xe n + → + + 54 143Xe 0 1n U Sr Xe n + → + + 92 235 38 90 54 142 0 1 4 54 142Xe 45.3 0 1n + 90 232Th→ 90 233Th→ 91 233Pa + e−+ ν 91 233Pa→ 92 233U + e−+ ν 45.4 0 1n + 92 238U→ 92 239U→ 93 239Np + e−+ ν 93 239Np→ 94 239Pu + e−+ ν 45.5 (a) Q = ∆m ( )c2 = mn + MU235 −MBa141 −MKr92 −3mn [ ]c2 ∆m = 1.008 665 + 235.043 924 ( ) −140.913 9 + 91.897 3 + 3 × 1.008 665 ( ) [ ]u = 0.215 39 u Q = 0.215 39 u ( ) 931.5 MeV u ( ) = 201 MeV (b) f = ∆m mi = 0.215 39 u 236.052 59 u = 9.13 × 10−4 = 0.0913% 45.6 If the electrical power output of 1000 MW is 40.0% of the power derived from fission reactions, the power output of the fission process is 1000 MW 0.400 J s d J/d = × × = × 2 50 10 8 64 10 2 16 10 9 4 14 .
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The number of fissions per day is 2 16 10 1 6 74 10 14 24 1 .
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× × × = × − − J d fission 200 10 eV 1 eV 1.60 10 J d 6 19 This also is the number of 235U nuclei used, so the mass of 235U used per day is 6 74 10 235 2 63 10 24 3 .
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× × = × nuclei d g/mol 6.02 10 nuclei/mol g/d = 23 2.63 kg/d In contrast, a coal-burning steam plant producing the same electrical power uses more than 6 × 106 kg/d of coal.
Chapter 45 Solutions 593 © 2000 by Harcourt, Inc. All rights reserved.
45.7 The available energy to do work is 0.200 times the energy content of the fuel. 1 00 0 0340 1000 1 6 02 10 208 1 60 10 235 23 13 .
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( )( . kg fuel U fuel g 1 kg mol 235 g mol J) fission ( ) × × − 2.90 × 1012 J ( ) 0.200 ( ) = 5.80 × 1011 J = 1.00 × 105 N ( )⋅d d = 5.80 × 106 m = 5.80 Mm Goal Solution Suppose enriched uranium containing 3.40% of the fissionable isotope 92 235U is used as fuel for a ship. The water exerts an average frictional drag of 1.00 × 105 N on the ship. How far can the ship travel per kilogram of fuel? Assume that the energy released per fission event is 208 MeV and that the ship's engine has an efficiency of 20.0%.
G : Nuclear fission is much more efficient for converting mass to energy than burning fossil fuels.
However, without knowing the rate of diesel fuel consumption for a comparable ship, it is difficult to estimate the nuclear fuel rate. It seems plausible that a ship could cross the Atlantic ocean with only a few kilograms of nuclear fuel, so a reasonable range of uranium fuel consumption might be 10 km/kg to 10 000 km/kg. O : The fuel consumption rate can be found from the energy released by the nuclear fuel and the work required to push the ship through the water.
A : One kg of enriched uranium contains 3.40% 92 235U so m235 = 1000 g ( ) 0.0340 ( ) = 34.0 g In terms of number of nuclei, this is equivalent to N235 = 34.0 g ( ) 1 235 g /mol 6.02 × 1023 atoms/mol ( ) = 8.71× 1022 nuclei If all these nuclei fission, the thermal energy released is equal to 8.71× 1022 nuclei ( ) 208 MeV nucleus 1.602 × 10−19 J eV ( ) = 2.90 × 1012 J Now, for the engine, efficiency = work output heat input or e = fdcosθ Qh So the distance the ship can travel per kilogram of uranium fuel is d = eQh f cos 0 ( ) = 0.200 2.90 × 1012 J ( ) 1.00 × 105 N = 5.80 × 106 m L : The ship can travel 5 800 km/kg of uranium fuel, which is on the high end of our prediction range.
The distance between New York and Paris is 5 851 km, so this ship could cross the Atlantic ocean on just one kilogram of uranium fuel.
594 Chapter 45 Solutions 45.8 (a) For a sphere: V = 4 3 πr 3 and r = 3V 4π 1 3 so A V = 4πr 2 (4/ 3)πr 3 = 4.84V – 1/3 (b) For a cube: V = l3 and l = V1 3 so A V = 6l2 l3 = 6V – 1/3 (c) For a parallelepiped: V = 2a3 and a = V 2 1/3 so A V = 2a2 + 8a2 ( ) 2a3 = 6.30V – 1/3 (d) Therefore, the sphere has the least leakage and the parallelepiped has the greatest leakage for a given volume.
45.9 mass of 235U available ≈0.007 ( ) 109 metric tons ( ) 106 g 1 metric ton = 7 × 1012 g number of nuclei ~ 7 × 1012 g 235 g mol 6.02 × 1023 nuclei mol = 1.8 × 1034 nuclei The energy available from fission (at 208 MeV/event) is E ~ 1.8 × 1034 events ( ) 208 MeV/event ( ) 1.60 × 10−13 J/ MeV ( ) = 6.0 × 1023 J This would last for a time of t = E P ~ 6.0 × 1023 J 7.0 × 1012 J s = 8.6 × 1010 s ( ) 1 yr 3.16 × 107 s ~ 3000 yr 45.10 In one minute there are 60.0 s 1.20 ms = 5.00 × 104 fissions.
So the rate increases by a factor of (1.000 25)50000 = 2.68 × 105 45.11 P = 10.0 MW = 1.00 × 107 J/s If each decay delivers 1.00 MeV = 1.60 × 10–13 J, then the number of decays/s = 6.25 × 1019 Bq Chapter 45 Solutions 595 © 2000 by Harcourt, Inc. All rights reserved.
45.12 (a) The Q value for the D-T reaction is 17.59 MeV. Heat content in fuel for D-T reaction: ( .
)( .
) ( )( .
) .
17 59 1 60 10 1 66 10 3 39 10 13 27 14 MeV J/MeV 5 u kg/u J/kg × × = × − − r DT = (3.00 × 109 J/s)(3600 s/hr) (3.39 × 1014 J/kg)(10−3 kg /g) = 31.9 g/h burning of D and T (b) Heat content in fuel for D-D reaction: Q = 1 2 (3.27 + 4.03) = 3.65 MeV average of two Q values ( ( .
)( .
.
3 1 60 10 4 1 66 10 8 80 10 13 27 13 .65 MeV) J/MeV) ( u kg/u) J/kg × × = × − − r DD = (3.00 × 109 J/s)(3600 s/hr) (8.80 × 1012 J/kg)(10−3 kg /g) = 122 g/h burning of D 45.13 (a) At closest approach, the electrostatic potential energy equals the total energy E. Uf = ke Z1e ( ) Z2e ( ) rmin = E: E = 8.99 × 109 N ⋅m2 C2 ( ) 1.6 × 10−19 C ( ) 2Z1Z2 1.00 × 10−14 m = 2.30 × 10−14 J ( )Z1Z2 (b) For both the D-D and the D-T reactions, Z1 = Z2 = 1. Thus, the minimum energy required in both cases is E = 2.30 × 10−14 J ( ) 1 MeV 1.60 × 10−13 J = 0.144 MeV 45.14 (a) rf = rD + rT = 1.20 × 10−15 m ( ) 2 ( )1 3 + 3 ( )1 3 [ ] = 3 24 10 15 .
× − m (b) Uf = kee2 rf = 8.99 × 109 N ⋅m2 C2 ( ) 1.60 × 10−19 C ( ) 2 3.24 × 10−15 m = 7.10 × 10−14 J= 444 keV (c) Conserving momentum, mDvi = mD + mT ( )vf , or vf = mD mD + mT vi = 2 5 vi (d) Ki + Ui = Kf + Uf : Ki + 0 = 1 2 mD + mT ( )vf 2 + Uf = 1 2 mD + mT ( ) mD mD + mT 2 vi 2 + Uf Ki + 0 = mD mD + mT 1 2 mDvi 2 ( ) + Uf = mD mD + mT Ki + Uf 1− mD mD + mT Ki = Uf : Ki = Uf mD + mT mT = 5 3 444 keV ( ) = 740 keV (e) Possibly by tunneling. 596 Chapter 45 Solutions 45.15 (a) Average KE per particle is 3 2 kBT = 1 2 mv2.
Therefore, vrms = 3kBT m = 3 1.38 × 10−23 J/K ( ) 4.00 × 108 K ( ) 2 1.67 × 10−27 kg ( ) = 2.23 × 106 m/s (b) t = x v ~ 0.1 m 106 m /s ~ 10–7 s 45.16 (a) V = 317 × 106 mi3 ( ) 1609 m 1 mi 3 = 1.32 × 1018 m3 mwater = ρV = 103 kg m3 ( ) 1.32 × 1018 m3 ( ) = 1.32 × 1021 kg mH2 = MH2 MH2O mH2O = 2.016 18.015 1.32 × 1021 kg ( ) = 1.48 × 1020 kg mDeuterium = 0.0300% ( )mH2 = 0.0300 × 10−2 ( ) 1.48 × 1020 kg ( ) = 4.43 × 1016 kg The number of deuterium nuclei in this mass is N = mDeuterium mDeuteron = 4.43 × 1016 kg 2.014 u ( ) 1.66 × 10−27 kg u ( ) = 1.33 × 1043 Since two deuterium nuclei are used per fusion, 1 2H + 1 2H→2 4He + Q, the number of events is N 2 = 6.63 × 1042.
The energy released per event is Q = M 2H + M 2H −M 4He [ ]c2 = 2 2.014 102 ( ) −4.002 602 [ ]u 931.5 MeV u ( ) = 23.8 MeV The total energy available is then E = N 2 Q = 6.63 × 1042 ( ) 23.8 MeV ( ) 1.60 × 10−13 J 1 MeV = 2 52 1031 .
× J (b) The time this energy could possibly meet world requirements is t = E P = 2.52 × 1031 J 100 7.00 × 1012 J s ( ) = 3.61× 1016 s ( ) 1 yr 3.16 × 107 s = 1.14 × 109 yr ~ 1 billion years.
Chapter 45 Solutions 597 © 2000 by Harcourt, Inc. All rights reserved.
45.17 (a) Including both ions and electrons, the number of particles in the plasma is N = 2nV where n is the ion density and V is the volume of the container. Application of Equation 21.6 gives the total energy as E = 3 2 NkBT = 3nVkBT = 3 2.0 × 1013 cm−3 ( ) 50 m3 ( ) 106 cm3 1 m3 1.38 × 10−23 J K ( ) 4.0 × 108 K ( ) E = 1.7 × 107 J (b) From Table 20.2, the heat of vaporization of water is Lv = 2.26 × 106 J kg. The mass of water that could be boiled away is m = E Lv = 1.7 × 107 J 2.26 × 106 J kg = 7.3 kg 45.18 (a) Lawson’s criterion for the D-T reaction is nτ ≥1014 s cm3. For a confinement time of τ = 1.00 s, this requires a minimum ion density of n = 1014 cm 3 − (b) At the ignition temperature of T = 4.5 × 107 K and the ion density found above, the plasma pressure is P = 2nkBT = 2 1014 cm−3 ( ) 106cm3 1 m3 1.38 × 10−23 J K ( ) 4.5 × 107 K ( ) = 1 24 105 .
× J m3 (c) The required magnetic energy density is then uB = B2 2µ0 ≥10P = 10 1.24 × 105 J m3 ( ) = 1.24 × 106 J m3, B ≥ 2 4π × 10−7 N A2 ( ) 1.24 × 106 J m3 ( ) = 1.77 T 45.19 Let the number of 6Li atoms, each having mass 6.015 u, be N6 while the number of 7Li atoms, each with mass 7.016 u, is N7 . Then, N6 = 7.50% of Ntotal = 0.0750 N6 + N7 ( ), or N7 = 0.925 0.0750 N6 Also, total mass = N6 6.015 u ( ) + N7 7.016 u ( ) [ ] 1.66 × 10−27 kg u ( ) = 2.00 kg, or N6 6.015 u ( ) + 0.925 0.0750 7.016 u ( ) 1.66 × 10−27 kg u ( ) = 2.00 kg.
This yields N6 = 1 30 1025 .
× as the number of 6Li atoms and N7 = 0.925 0.0750 1.30 × 1025 ( ) = 1 61 1026 .
× as the number of 7Li atoms.
598 Chapter 45 Solutions 45.20 The number of nuclei in 1.00 metric ton of trash is N = 1000 kg 1000 g /kg ( ) 6.02 × 1023 nuclei/mol ( ) 56.0 g /mol ( ) = 1.08 × 1028 nuclei At an average charge of 26.0 e/nucleus, q = 1.08 × 1028 ( ) 26.0 ( ) 1.60 × 10−19 ( ) = 4.47 × 1010 C Therefore t = q I = 4.47 × 1010 1.00 × 106 = 4.47 × 104 s = 12.4 h 45.21 N0 = mass present mass of nucleus = 5.00 kg 89.9077 u ( ) 1.66 × 10−27 kg u ( ) = 3.35 × 1025 nuclei λ = ln 2 T1/2 = ln 2 29.1 yr = 2.38 × 10−2 yr−1 = 4.52 × 10−8 min−1 R0 = λ N0 = 4.52 × 10−8 min−1 ( ) 3.35 × 1025 ( ) = 1.52 × 1018 counts min R R0 = e−λ t = 10.0 counts min 1.52 × 1018 counts min = 6.60 × 10−18 and λt = −ln 6.60 × 10−18 ( ) = 39.6 giving t = 39.6 λ = 39.6 2.38 × 10−2 yr−1 = 1 66 103 .
× yr 45.22 Source: 100 mrad of 2-MeV γ - rays/h at a 1.00-m distance.
(a) For γ - rays, dose in rem = dose in rad.
Thus a person would have to stand 10.0 hours to receive 1.00 rem from a 100-mrad/h source.
(b) If the γ - radiation is emitted isotropically, the dosage rate falls off as 1/r2.
Thus a dosage 10.0 mrad/h would be received at a distance r = 10.0 m = 3.16 m .
45.23 (a) The number of x-rays taken per year is n = 8 x - ray d ( ) 5 d wk ( ) 50 wk yr ( ) = 2.0 × 103 x - ray yr The average dose per photograph is 5.0 rem yr 2.0 × 103 x - ray yr = 2.5 × 10−3 rem x - ray (b) The technician receives low-level background radiation at a rate of 0.13 rem yr. The dose of 5.0 rem yr received as a result of the job is Chapter 45 Solutions 599 © 2000 by Harcourt, Inc. All rights reserved. 5.0 rem yr 0.13 rem yr = 38 times background levels 600 Chapter 45 Solutions 45.24 (a) I = I0e−µ x, so x = 1 µ ln I0 I With µ = 1.59 cm−1, the thickness when I = I0 2 is x = 1 1.59 cm−1 ln 2 ( ) = 0.436 cm (b) When I0 I = 1.00 × 104, x = 1 1.59 cm−1 ln 1.00 × 104 ( ) = 5.79 cm 45.25 1 rad = 10−2 J/kg Q = mc∆T Pt = mc∆T t = mc∆T P = m 4186 J/kg⋅°C ( ) 50.0 °C ( ) 10 ( ) 10−2 J/kg ⋅s ( )(m) = 2.09 × 106 s ≈ 24 days!
Note that power is the product of dose rate and mass.
45.26 Q m = absorbed energy unit mass = (1000 rad) 10−2 J/kg 1 rad = 10.0 J/kg The rise in body temperature is calculated from Q = mc ∆T where c = 4186 J/kg · °C for water and the human body ∆T = Q mc = (10.0 J/kg) 1 4186 J/kg⋅°C = 2 39 10 3 .
× ° − C (Negligible) 45.27 If half of the 0.140-MeV gamma rays are absorbed by the patient, the total energy absorbed is E = 0.140 MeV ( ) 2 1.00 × 10−8 g 98.9 g mol 6.02 × 1023 nuclei 1 mol = 4.26 × 1012 MeV E = 4.26 × 1012 MeV ( ) 1.60 × 10−13 J MeV ( ) = 0.682 J Thus, the dose received is Dose = 0.682 J 60.0 kg 1 rad 10−2 J kg = 1.14 rad Chapter 45 Solutions 601 © 2000 by Harcourt, Inc. All rights reserved.
45.28 The nuclei initially absorbed are N0 = 1.00 × 10−9 g ( ) 6.02 × 1023 nuclei mol 89.9 g mol = 6.70 × 1012 The number of decays in time t is ∆N = N0 −N = N0 1−e−λ t ( ) = N0 1−e−ln 2 ( )t T1 2 At the end of 1 year, t T1 2 = 1.00 yr 29.1 yr = 0.0344 and ∆N = N0 −N = 6.70 × 1012 ( ) 1−e−0.0238 ( ) = 1.58 × 1011 The energy deposited is E = 1.58 × 1011 ( ) 1.10 MeV ( ) 1.60 × 10−13 J MeV ( ) = 0.0277 J Thus, the dose received is Dose .0277 J kg = = 0 70 0 . 3 96 10 4 .
× − J kg = 0.0396 rad 45.29 (a) E Eβ = 1 2 C(∆V)2 0.500 MeV = 1 2 5.00 × 10−12 F ( ) 1.00 × 103 V ( ) 2 0.500 MeV ( ) 1.60 × 10−13 J/ MeV ( ) = 3.12 × 107 (b) N = Q e = C(∆V) e = 5.00 × 10−12 F ( ) 1.00 × 103 V ( ) 1.60 × 10−19 C = 3.12 × 1010 electrons 45.30 (a) amplification = energy discharged E = 1 2 C ∆V ( )2 E = C ∆V ( )2 2E (b) N = charge released charge of electron = C ∆V ( ) e 45.31 (a) EI = 10.0 eV is the energy required to liberate an electron from a dynode. Let ni be the number of electrons incident upon a dynode, each having gained energy e ∆V ( ) as it was accelerated to this dynode. The number of electrons that will be freed from this dynode is Ni = nie ∆V ( ) EI : At the first dynode, ni = 1 and N1 = (1)e 100 V ( ) 10.0 eV = 101 electrons 602 Chapter 45 Solutions (b) For the second dynode, ni = N1 = 101, so N2 = (101)e 100 V ( ) 10.0 eV = 102. At the third dynode, ni = N2 = 102 and N3 = (102)e 100 V ( ) 10.0 eV = 103.
Observing the developing pattern, we see that the number of electrons incident on the seventh and last dynode is n7 = N6 = 106 .
(c) The number of electrons incident on the last dynode is n7 = 106. The total energy these electrons deliver to that dynode is given by E = nie ∆V ( ) = 106 e 700 V −600 V ( ) = 108 eV 45.32 (a) The average time between slams is 60 min 38 = 1.6 min. Sometimes, the actual interval is nearly zero. Perhaps about equally as often, it is 2 × 1.6 min. Perhaps about half as often, it is 4 × 1.6 min. Somewhere around 5 × 1.6 min = 8.0 min , the chances of randomness producing so long a wait get slim, so such a long wait might likely be due to mischief.
(b) The midpoints of the time intervals are separated by 5.00 minutes. We use R = R0e−λ t.
Subtracting the background counts, 337 5 15 372 5 15 2 5 00 1 2 −( ) = −( ) [ ] −( )( ) e T ln . min or ln 262 297 = ln 0.882 ( ) = −3.47 min T1 2 which yields T1 2 = 27.6 min .
(c) As in the random events in part (a), we imagine a ±5 count counting uncertainty. The smallest likely value for the half-life is then given by ln 262 −5 297 + 5 = −3.47 min T1 2 , or T1 2 ( )min = 21.1 min The largest credible value is found from ln 262 + 5 297 −5 = −3.47 min T1 2 , yielding T1 2 ( )max = 38.8 min Thus, T1 2 = 38.8 + 21.1 2 ± 38.8 −21.1 2 min = 30 ± 9 ( ) min = 30 min ± 30% Chapter 45 Solutions 603 © 2000 by Harcourt, Inc. All rights reserved.
45.33 The initial specific activity of 59Fe in the steel, (R/m)0 = 20.0 µCi 0.200 kg = 100 µCi kg 3.70 × 104 Bq 1 µCi = 3.70 × 106 Bq /kg After 1000 h, R m = (R/m)0e−λ t = (3.70 × 106 Bq /kg)e−(6.40 × 10−4 h−1)(1000 h) = 1.95 × 106 Bq /kg The activity of the oil, Roil = 800 60.0 Bq /liter (6.50 liters) = 86.7 Bq Therefore, min oil = Roil (R/m) = 86.7 Bq 1.95 × 106 Bq /kg = 4.45 × 10−5 kg So that wear rate is 4 45 10 100 5 .
× = − kg 0 h 4 45 10 8 .
× − kg/h 45.34 The half-life of 14O is 70.6 s, so the decay constant is λ = ln 2 T1 2 = ln 2 70.6 s = 0.009 82 s−1 The 14O nuclei remaining after five min is N = N0e−λ t = 1010 ( )e −0.009 82 s−1 ( ) 300 s ( ) = 5.26 × 108 The number of these in one cubic centimeter of blood is ′ N = N 1.00 cm3 total vol. of blood = 5.26 × 108 ( ) 1.00 cm3 2000 cm3 = 2.63 × 105 and their activity is R = λ ′ N = 0.009 82 s−1 ( ) 2.63 × 105 ( ) = 2.58 × 103 Bq ~ 103 Bq 45.35 (a) The number of photons is 104 MeV 1.04 MeV = 9.62 × 103. Since only 50% of the photons are detected, the number of 65Cu nuclei decaying is twice this value, or 1 92 104 .
× . In two half-lives, three-fourths of the original nuclei decay, so 3 4 N0 = 1.92 × 104 and N0 = 2.56 × 104. This is 1% of the 65Cu, so the number of 65Cu is 2 56 106 .
× ~ 106 .
(b) Natural copper is 69.17% 63Cu and 30.83% 65Cu. Thus, if the sample contains NCu copper atoms, the number of atoms of each isotope is N63 = 0.6917 NCu and N65 = 0.3083NCu.
Therefore, N63 N65 = 0.6917 0.3083 or N63 = 0.6917 0.3083 N65 = 0.6917 0.3083 2.56 × 106 ( ) = 5.75 × 106 The total mass of copper present is then mCu = 62.93 u ( )N63 + 64.93 u ( )N65: mCu = 62.93 ( ) 5.75 × 106 ( ) + 64.93 ( ) 2.56 × 106 ( ) [ ]u 1.66 × 10−24 g u ( ) = 8.77 × 10−16 g ~ 10 15 − g 45.36 (a) Starting with N = 0 radioactive atoms at t = 0, the rate of increase is (production – decay) 604 Chapter 45 Solutions dN dt = R – λ N so dN = (R – λ N)dt The variables are separable. N = 0 N ∫ dN R −λ N = t = 0 t ∫ dt −1 λ ln R −λ N R = t so ln R −λ N R = −λt R −λ N R = e−λ t and 1−λ R N = e−λ t Therefore, N = R λ 1−e−λ t ( ) (b) The maximum number of radioactive nuclei would be R/λ 45.37 (a) At 6 × 108 K, each carbon nucleus has thermal energy of 3 2 kBT = (1.5)(8.62 × 10−5 eV /K)(6 × 108 K) = 8 × 104 eV (b) The energy released is E = 2m C12 ( ) −m Ne20 ( ) −m He4 ( ) [ ]c2 E = (24.000 000 – 19.992 435 – 4.002 602)(931.5) MeV = 4.62 MeV In the second reaction, E = 2m C12 ( ) −m Mg24 ( ) [ ] 931.5 ( )MeV/ u E = (24.000 000 – 23.985 042)(931.5)MeV = 13.9 MeV (c) The energy released is the energy of reaction of the # of carbon nuclei in a 2.00-kg sample, which corresponds to ∆E = 2.00 × 103g ( ) 6.02 × 1023 atoms/mol 12.0 g /mol 4.62 MeV/fusion event 2 nuclei/fusion event 1 kWh 2.25 × 1019 MeV ∆E = 1.00 × 1026 ( ) 4.62 ( ) 2 2.25 × 1019 ( ) kWh = 1.03 × 107 kWh Chapter 45 Solutions 605 © 2000 by Harcourt, Inc. All rights reserved.
45.38 (a) Suppose each 235U fission releases 208 MeV of energy. Then, the number of nuclei that must have undergone fission is N = total release energy per nuclei = 5 × 1013 J 208 MeV ( ) 1.60 × 10−13 J MeV ( ) = 1 5 1024 . × nuclei (b) mass nuclei 6.02 nuclei mol g mol 23 = × × ( ) ≈ 1 5 10 10 235 24 . 0.6 kg 45.39 For a typical 235U, Q = 208 MeV; and the initial mass is 235 u. Thus, the fractional energy loss is Q mc2 = 208 MeV 235 u ( ) 931.5 MeV u ( ) = 9.50 × 10−4 = 0.0950% For the D-T fusion reaction, Q = 17.6 MeV The initial mass is m = ( ) + ( ) = 2 014 3 016 5 03 .
.
. u u u The fractional loss in this reaction is Q mc2 = 17.6 MeV 5.03 u ( ) 931.5 MeV u ( ) = 3.75 × 10−3 = 0.375% 0.375% 0.0950% = 3.95 or the fractional loss in D T fusion is about 4 times that in U fission − 235 .
45.40 To conserve momentum, the two fragments must move in opposite directions with speeds v1 and v2 such that m1v1 = m2v2 or v2 = m1 m2 v1 The kinetic energies after the break-up are then K1 = 1 2 m1v1 2 and K2 = 1 2 m2v2 2 = 1 2 m2 m1 m2 2 v1 2 = m1 m2 K1 The fraction of the total kinetic energy carried off by m1 is K1 K1 + K2 = K1 K1 + m1 m2 ( )K1 = m2 m1 + m2 and the fraction carried off by m2 is 1− m2 m1 + m2 = m1 m1 + m2 606 Chapter 45 Solutions 45.41 The decay constant is λ = ln 2 T1 2 = ln 2 12.3 yr ( ) 3.16 × 107 s yr ( ) = 1.78 × 10−9 s−1 The tritium in the plasma decays at a rate of R = λ N = 1.78 × 10−9 s−1 ( ) 2.00 × 1014 cm3 106 cm3 1 m3 50.0 m3 ( ) R = 1.78 × 1013 Bq = 1.78 × 1013 Bq ( ) 1 Ci 3.70 × 1010 Bq = 482 Ci The fission inventory is Ci Ci times greater 4 10 482 10 10 8 × ~ than this amount.
Goal Solution The half-life of tritium is 12.3 yr. If the TFTR fusion reactor contained 50.0 m3 of tritium at a density equal to 2.00 × 1014 ions/cm3, how many curies of tritium were in the plasma? Compare this value with a fission inventory (the estimated supply of fissionable material) of 4 × 1010 Ci. G : It is difficult to estimate the activity of the tritium in the fusion reactor without actually calculating it; however, we might expect it to be a small fraction of the fission (not fusion) inventory.
O : The decay rate (activity) can be found by multiplying the decay constant λ by the number of 1 3H particles. The decay constant can be found from the half-life of tritium, and the number of particles from the density and volume of the plasma. A : The number of Hydrogen-3 nuclei is N = 50.0 m3 ( ) 2.00 × 1014 particles m3 100 cm m 3 = 1.00 × 1022 particles λ = ln 2 T1/2 = 0.693 12.3 yr 1 yr 3.16 × 107 s = 1.78 × 10−9 s-1 The activity is then R = λN = 1.78 × 10−9 s-1 ( ) 1.00 × 1022 nuclei ( ) = 1.78 × 1013 Bq = 1.78 × 1013 Bq ( ) 1 Ci 3.70 × 1010 Bq = 482 Ci L : Even though 482 Ci is a large amount of radioactivity, it is smaller than 4.00 × 1010 Ci by about a hundred million. Therefore, loss of containment is a smaller hazard for a fusion power reactor than for a fission reactor.
Chapter 45 Solutions 607 © 2000 by Harcourt, Inc. All rights reserved.
45.42 Momentum conservation: 0 = mLivLi + mαvα, or, mLivLi = mαvα Thus, KLi = 1 2 mLivLi 2 = 1 2 mLivLi ( ) 2 mLi = mαvα ( ) 2 2mLi = mα 2 2mLi vα 2 KLi = 4.002 6 u ( ) 2 7.016 9 u ( ) 2 9.30 × 106 m s 2 = 1.14 u ( ) 9.30 × 106 m s 2 KLi = 1.14 1.66 × 10−27 kg ( ) 9.30 × 106 m s 2 = 1.64 × 10−13 J = 1.02 MeV 45.43 The complete fissioning of 1.00 gram of U235 releases ∆Q = (1.00 g) 235 grams/mol 6.02 × 1023 atoms mol 200 MeV fission 1.60 × 10−13 J MeV = 8.20 × 1010 J If all this energy could be utilized to convert m kilograms of 20.0°C water to 400°C steam (see Chapter 20 of text for values), then ∆Q = mcw ∆T + mLv + mcs ∆T ∆Q = m 4186 J/kg⋅°C ( ) 80.0 °C ( ) + 2.26 × 106 J/kg + 2010 J/kg⋅°C ( ) 300 °C ( ) [ ] Therefore m = 8.20 × 1010 J 3.20 × 106 J/kg = 2.56 × 104 kg 45.44 When mass m of 235U undergoes complete fission, releasing 200 MeV per fission event, the total energy released is: Q = m 235 g mol NA 200 MeV ( ) where NA is Avogadro’s number.
If all this energy could be utilized to convert a mass mw of liquid water at Tc into steam at Th, then, Q = mw cw 100°C −Tc ( ) + Lv + cs Th −100°C ( ) [ ] where cw is the specific heat of liquid water, Lv is the latent heat of vaporization, and cs is the specific heat of steam. Solving for the mass of water converted gives mw = Q cw 100°C −Tc ( ) + Lv + cs Th −100°C ( ) [ ] = mNA 200 MeV ( ) 235 g mol ( ) cw 100°C −Tc ( ) + Lv + cs Th −100°C ( ) [ ] 608 Chapter 45 Solutions 45.45 (a) The number of molecules in 1.00 liter of water (mass = 1000 g) is N = 1.00 × 103 g 18.0 g mol 6.02 × 1023 molecules mol ( ) = 3.34 × 1025 molecules The number of deuterium nuclei contained in these molecules is ′ N = 3.34 × 1025 molecules ( ) 1 deuteron 3300 molecules = 1.01× 1022 deuterons Since 2 deuterons are consumed per fusion event, the number of events possible is ′ N 2 = 5.07 × 1021 reactions, and the energy released is Efusion = 5.07 × 1021 reactions ( ) 3.27 MeV reaction ( ) = 1.66 × 1022 MeV Efusion = 1.66 × 1022 MeV ( ) 1.60 × 10−13 J MeV ( ) = 2 65 .
× 10 J 9 (b) In comparison to burning 1.00 liter of gasoline, the energy from the fusion of deuterium is Efusion Egasoline = 2.65 × 109 J 3.40 × 107 J = 78.0 times larger .
45.46 The number of nuclei in 0.155 kg of 210Po is N0 = 155 g 209.98 g mol 6.02 × 1023 nuclei g ( ) = 4.44 × 1023 nuclei The half-life of 210Po is 138.38 days, so the decay constant is given by λ = ln 2 T1 2 = ln 2 138.38 d ( ) 8.64 × 104 s d ( ) = 5.80 × 10−8 s−1 The initial activity is R0 = λ N0 = 5.80 × 10−8 s−1 ( ) 4.44 × 1023 nuclei ( ) = 2.58 × 1016 Bq The energy released in each 84 210Po Pb He → + 82 206 2 4 reaction is Q = M 84 210Po −M 82 206Pb −M 2 4He c2: Q = 209.982 848 −205.974 440 −4.002 602 [ ]u 931.5 MeV u = 5.41 MeV Thus, assuming a conversion efficiency of 1.00%, the initial power output of the battery is P = 0.0100 ( )R0Q = 0.0100 ( ) 2.58 × 1016 decays s 5.41 MeV decay 1.60 × 10−13 J MeV = 223 W Chapter 45 Solutions 609 © 2000 by Harcourt, Inc. All rights reserved.
45.47 (a) The thermal power transferred to the water is Pw = 0.970 waste heat ( ) Pw = 0.970 3065 −1000 ( )MW = 2.00 × 109 J s rw is the mass of heated per hour: rw = Pw c ∆T ( ) = 2.00 × 109 J s ( ) 3600 s h ( ) 4186 J kg⋅°C ( ) 3.50 °C ( ) = 4 91 108 .
× kg h The volume used per hour is 4 91 10 1 00 10 8 3 .
.
× × = kg h kg m3 4 91 105 .
× m h 3 (b) The 235U fuel is consumed at a rate rf = 3065 × 106 J s 7.80 × 1010 J g 1 kg 1000 g 3600 s 1 h = 0.141 kg/h 45.48 (a) ∆V = 4π r 2 ∆r ( ) = 4π 14.0 × 103 m ( ) 2 0.05 m ( ) = 1.23 × 108 m3 ~ 10 m 8 3 (b) The force on the next layer is determined by atmospheric pressure. W = P ∆V ( ) = 1.013 × 105 N m2 1.23 × 108 m3 ( ) = 1.25 × 1013 J ~ 10 J 13 (c) 1.25 × 1013 J = 1 10 yield ( ), so yield J = × 1 25 1014 . ~ 10 J 14 (d) 1 25 10 4 2 10 2 97 10 14 9 4 .
.
.
× × = × J J ton TNT ton TNT ~ 10 ton TNT 4 or ~ 10 kilotons 45.49 (a) V = l3 = m ρ , so l = m ρ 1 3 = 70.0 kg 18.7 × 103 kg m3 1 3 = 0.155 m (b) Add 92 electrons to both sides of the given nuclear reaction. Then it becomes 92 238U atom →8 2 4He atom + 82 206Pb atom + Qnet. Qnet = M 92 238U −8 M 2 4He −M 82 206Pb c2 = 238.050 784 −8 4.002 602 ( ) −205.974 440 [ ]u 931.5 MeV u ( ) Qnet = 51.7 MeV (c) If there is a single step of decay, the number of decays per time is the decay rate R and the energy released in each decay is Q. Then the energy released per time is P = QR . If there is a series of decays in steady state, the equation is still true, with Q representing the net decay energy.
610 Chapter 45 Solutions (d) The decay rate for all steps in the radioactive series in steady state is set by the parent uranium: N = 7.00 × 104 g 238 g mol 6.02 × 1023 nuclei mol ( ) = 1.77 × 1026 nuclei λ = ln2 T1 2 = ln2 4.47 × 109 yr = 1.55 × 10−10 1 yr R = λ N = 1.55 × 10−10 1 yr 1.77 × 1026 nuclei ( ) = 2.75 × 1016 decays yr, so P = QR = 51.7 MeV ( ) 2.75 × 1016 1 yr 1.60 × 10−13 J MeV ( ) = 2 27 105 .
× J yr (e) dose in rem = dose in rad x RBE 5 00 1 10 .
. rem yr dose in rad yr = , giving dose in rad yr rad yr = 4 55 .
The allowed whole-body dose is then 70 0 4 55 10 1 2 .
. kg rad yr J kg rad ( ) = − 3.18 J/yr 45.50 ET ≡E thermal ( ) = 3 2 kBT = 0.039 eV ET = 1 2 ( ) n E where n ≡ number of collisions, and 0.039 = 1 2 ( ) n 2.0 × 106 ( ) Therefore, n = 25.6 = 26 collisions 45.51 From conservation of energy: Kα + Kn = Q or 1 2 mαvα 2 + 1 2 mnvn 2 = 17.6 MeV Conservation of momentum: mαvα = mnvn or vα = mn mα vn.
The energy equation becomes: 1 2 mα mn mα 2 vn 2 + 1 2 mnvn 2 = mn + mα mα 1 2 mnvn 2 ( ) = 17.6 MeV Thus, Kn = mα mn + mα 17.6 MeV ( ) = 4.002 602 1.008 665 + 4.002 602 = 14.1 MeV Chapter 45 Solutions 611 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution Assuming that a deuteron and a triton are at rest when they fuse according to 2H + 3H →4He + n +17.6 MeV, determine the kinetic energy acquired by the neutron. G : The products of this nuclear reaction are an alpha particle and a neutron, with total kinetic energy of 17.6 MeV. In order to conserve momentum, the lighter neutron will have a larger velocity than the more massive alpha particle (which consists of two protons and two neutrons). Since the kinetic energy of the particles is proportional to the square of their velocities but only linearly proportional to their mass, the neutron should have the larger kinetic energy, somewhere between 8.8 and 17.6 MeV. O : Conservation of linear momentum and energy can be applied to find the kinetic energy of the neutron. We first suppose the particles are moving nonrelativistically. A : The momentum of the alpha particle and that of the neutron must add to zero, so their velocities must be in opposite directions with magnitudes related by mnvn + mαvα = 0 or 1.0087 u ( )vn = 4.0026 u ( )vα At the same time, their kinetic energies must add to 17.6 MeV E = 1 2 mnvn 2 + 1 2 mαvα 2 = 1 2 1.0087 u ( )vn 2 + 1 2 4.0026 u ( )vα 2 = 17.6 MeV Substitute vα = 0.2520vn: E = 0.50435 u ( )vn 2 + 0.12710 u ( )vn 2 = 17.6 MeV 1 u 931.494 MeV/c2 vn = 0.0189c2 0.63145 = 0.173c = 5.19 × 107 m /s Since this speed is not too much greater than 0.1c, we can get a reasonable estimate of the kinetic energy of the neutron from the classical equation, K = 1 2 mv2 = 1 2 1.0087 u ( ) 0.173c ( ) 931.494 MeV/c2 u = 14.1 MeV L : The kinetic energy of the neutron is within the range we predicted. For a more accurate calculation of the kinetic energy, we should use relativistic expressions. Conservation of momentum gives γ nmnvn + γ αmαvα = 0 1.0087 vn 1−vn 2/c2 = 4.0026 vα 1−vα 2/c2 yielding vα 2 c2 = vn 2 15.746c2 −14.746vn 2 Then γ n −1 ( )mnc2 + γ α −1 ( )mαc2 = 17.6 MeV and vn = 0.171c, implying that γ n −1 ( )mnc2 = 14.0 MeV 612 Chapter 45 Solutions 45.52 From Table A.3, the half-life of 32P is 14.26 d. Thus, the decay constant is λ = ln 2 T1 2 = ln 2 14.26 d = 0.0486 d−1 = 5.63 × 10−7 s−1. N0 = R0 λ = 5.22 × 106 decay s 5.63 × 10−7 s−1 = 9.28 × 1012 nuclei At t = 10.0 days, the number remaining is N = N0 e−λ t = 9.28 × 1012 nuclei ( )e −0.0486 d−1 ( ) 10.0 d ( ) = 5.71× 1012 nuclei so the number of decays has been N0 −N = 3.57 × 1012 and the energy released is E = 3.57 × 1012 ( ) 700 keV ( ) 1.60 × 10−16 J keV = 0.400 J If this energy is absorbed by 100 g of tissue, the absorbed dose is Dose J .100 kg rad J kg = = − 0 400 0 1 10 2 . 400 rad 45.53 (a) The number of Pu nuclei in 1.00 kg = 6.02 × 1023 nuclei/mol 239.05 g/mol (1000 g) The total energy = (25.2 × 1023 nuclei)(200 MeV) = 5.04 × 1026 MeV E = (5.04 × 1026 MeV)(4.44 × 10– 20 kWh/MeV) = 2.24 × 107 kWh or 22 million kWh (b) E = ∆mc2 = (3.016 049 u + 2.014 102 u – 4.002 602 u – 1.008 665 u) (931.5 MeV/u) E = 17.6 MeV for each D-T fusion (c) En = (Total number of D nuclei)(17.6)(4.44 × 10–20) En = (6.02 × 1023)(1000/2.014)(17.6)(4.44 × 10–20) = 2.34 × 108 kWh (d) En = the number of C atoms in 1.00 kg × 4.20 eV En = (6.02 × 1026/12.0)(4.20 × 10– 6 MeV)(4.44 × 10– 20) = 9.36 kWh (e) Coal is cheap at this moment in human history. We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels.
Chapter 45 Solutions 613 © 2000 by Harcourt, Inc. All rights reserved.
45.54 Add two electrons to both sides of the given reaction. Then 4 1 1H atom →2 4He atom + Q where Q = ∆m ( )c 2 = 4 1.007 825 ( ) −4.002 602 [ ]u 931.5 MeV u ( ) = 26.7 MeV or Q = 26.7 MeV ( ) 1.60 × 10−13 J MeV ( ) = 4.28 × 10−12 J The proton fusion rate is then rate power output energy per proton = = 3.77 × 1026 J s 4.28 × 10−12 J ( ) 4 protons ( ) = 3 53 1038 .
× protons s 45.55 (a) QI = MA + MB −MC −ME [ ]c2, and QII = MC + MD −MF −MG [ ]c2 Qnet = QI + QII = MA + MB −MC −ME + MC + MD −MF −MG [ ]c2 Qnet = QI + QII = MA + MB + MD −ME −MF −MG [ ]c2 Thus, reactions may be added. Any product like C used in a subsequent reaction does not contribute to the energy balance.
(b) Adding all five reactions gives 1 1H + 1 1H + −1 0e + 1 1H + 1 1H + −1 0e → 2 4He + 2ν + Qnet or 4 1 1H + 2 −1 0e → 2 4He + 2ν + Qnet Adding two electrons to each side 4 1 1H atom → 2 4He atom + Qnet Thus, Qnet = 4 M 1 1H −M 2 4He [ ]c2 = 4 1.007 825 ( ) −4.002 602 [ ]u 931.5 MeV u ( ) = 26.7 MeV 45.56 (a) The mass of the pellet is m = ρV = 0.200 g cm3 4π 3 1.50 × 10−2 cm 2 3 = 3.53 × 10−7 g The pellet consists of equal numbers of 2H and 3H atoms, so the average atomic weight is 2.50 and the total number of atoms is N = 3.53 × 10−7 g 2.50 g mol 6.02 × 1023 atoms mol ( ) = 8.51× 1016 atoms When the pellet is vaporized, the plasma will consist of 2N particles ( N nuclei and N electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The temperature of the plasma is found from E = 2N ( ) 3 2 kBT ( ) as T = E 3NkB = 2.00 × 103 J 3 8.51× 1016 ( ) 1.38 × 10−23 J K ( ) = 5 68 108 .
× K (b) Each fusion event uses 2 nuclei, so N 2 events will occur. The energy released will be E = N 2 Q = 8.51× 1016 2 17.59 MeV ( ) 1.60 × 10−13 J MeV = 1.20 × 105 J = 120 kJ 614 Chapter 45 Solutions 45.57 (a) The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the Coulomb-repulsion barrier to 1 1 2 3 2 4 H He He e + → + + + ν , estimated as ke e ( ) 2e ( ) r. The Coulomb barrier to Bethe’s fifth and eight reactions is like ke e ( ) 7e ( ) r, larger by 7 2 times, so the temperature should be like 7 2 15 106 × ( ) ≈ K 5 × 107 K .
(b) For 12C + 1H →13N + Q, Q1 = 12.000 000 + 1.007 825 −13.005 738 ( ) 931.5 MeV ( ) = 1.94 MeV For the second step, add seven electrons to both sides to have: 13N atom →13C atom + e−+ e+ + Q. Q2 = 13.005 738 −13.003 355 −2 0.000 549 ( ) [ ] 931.5 MeV ( ) = 1.20 MeV Q3 = Q7 = 2 0.000 549 ( ) 931.5 MeV ( ) = 1.02 MeV Q4 = 13.003 355 + 1.007 825 −14.003 074 [ ] 931.5 MeV ( ) = 7.55 MeV Q5 = 14.003 074 + 1.007 825 −15.003 065 [ ] 931.5 MeV ( ) = 7.30 MeV Q6 = 15.003 065 −15.000 108 −2 0.000 549 ( ) [ ] 931.5 MeV ( ) = 1.73 MeV Q8 = 15.000 108 + 1.007 825 −12 −4.002 602 [ ] 931.5 MeV ( ) = 4.97 MeV The sum is 26.7 MeV , the same as for the proton-proton cycle.
(c) Not all of the energy released heats the star. When a neutrino is created, it will likely fly directly out of the star without interacting with any other particle.
45.58 (a) I2 I1 = I0e−µ2x I0e−µ1x = e−µ2 −µ1 ( )x (b) I50 I100 = e−5.40 −41.0 ( ) 0.100 ( ) = e3.56 = 35.2 (c) I50 I100 = e−5.40 −41.0 ( ) 1.00 ( ) = e35.6 = 2.89 × 1015 Thus, a 1.00-cm aluminum plate has essentially removed the long-wavelength x-rays from the beam.
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Chapter 46 Solutions 46.1 Assuming that the proton and antiproton are left nearly at rest after they are produced, the energy of the photon E, must be E = 2E0 = 2(938.3 MeV) = 1876.6 MeV = 3.00 × 10– 10 J Thus, E = hf = 3.00 × 10– 10 J f = 3.00 × 10– 10 J 6.626 × 10– 34 J · s = 4.53 × 10 23 Hz λ = c f = 3.00 × 10 8 m/s 4.53 × 10 23 Hz = 6.62 × 10– 16 m 46.2 The minimum energy is released, and hence the minimum frequency photons are produced, when the proton and antiproton are at rest when they annihilate. That is, E = E0 and K = 0. To conserve momentum, each photon must carry away one-half the energy. Thus, Emin = hfmin = (2E0) 2 = E0 = 938.3 MeV Thus, fmin = (938.3 MeV)(1.60 × 10– 13 J/MeV) 6.626 × 10– 34 J · s = 2.27 × 10 23 Hz λ = c fmin = 3.00 × 10 8 m/s 2.27 × 10 23 Hz = 1.32 × 10– 15 m 46.3 In γ → p+ + p–, we start with energy 2.09 GeV we end with energy 938.3 MeV + 938.3 MeV + 95.0 MeV + K2 where K2 is the kinetic energy of the second proton.
Conservation of energy gives K2 = 118 MeV 2 Chapter 46 Solutions Goal Solution A photon with an energy Eγ = 2.09 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 95.0 MeV. What is the kinetic energy of the antiproton? ( mpc2 = 938.3 MeV). G : An antiproton has the same mass as a proton, so it seems reasonable to expect that both particles will have similar kinetic energies.
O : The total energy of each particle is the sum of its rest energy and its kinetic energy. Conservation of energy requires that the total energy before this pair production event equal the total energy after. A : Eγ = ERp + Kp ( ) + ERp + Kp ( ) The energy of the photon is given as Eγ = 2.09 GeV = 2.09 × 103 MeV. From Table 46.2, we see that the rest energy of both the proton and the antiproton is ERp = ERp = mpc2 = 938.3 MeV If the kinetic energy of the proton is observed to be 95.0 MeV, the kinetic energy of the antiproton is Kp = Eγ −ERp −ERp −Kp = 2.09 × 103 MeV −2 938.5 MeV ( ) −95.0 MeV = 118 MeV L : The kinetic energy of the antiproton is slightly (~20%) greater than the proton. The two particles most likely have different shares in momentum of the gamma ray, and therefore will not have equal energies, either. 46.4 The reaction is µ+ + e– → ν + ν muon-lepton number before reaction: (–1) + (0) = –1 electron-lepon number before reaction: (0) + (1) = 1 Therefore, after the reaction, the muon-lepton number must be –1. Thus, one of the neutrinos must be the anti-neutrino associated with muons, and one of the neutrinos must be the neutrino associated with electrons: ν µ and νe Then µ+ + e– → ν – µ + νe 46.5 The creation of a virtual Z0 boson is an energy fluctuation ∆E = 93 × 109 eV. It can last no longer than ∆t = h 2∆E and move no farther than c ∆t ( ) = hc 4π ∆E = 6.626 × 10−34 J ⋅s ( ) 3.00 × 108 m s ( ) 4π 93 × 109 eV ( ) 1 eV 1.60 × 10−19 J = 1.06 × 10−18 m = ~ 10−18 m Chapter 46 Solutions 3 © 2000 by Harcourt, Inc. All rights reserved.
46.6 (a) ∆E = (mn – mp – me)c 2 From Table A-3, ∆E = (1.008 665 – 1.007 825)931.5 = 0.782 MeV (b) Assuming the neutron at rest, momentum is conserved, pp = pe relativistic energy is conserved, mpc2 ( ) 2 + pp 2c 2 + mec2 ( ) 2 + pe 2c 2 = mnc2 Since pp = pe , 938.3 ( )2 + pc ( ) 2 + 0.511 ( )2 + pc ( ) 2 = 939.6 MeV Solving the algebra pc = 1.19 MeV If pec = γ mevec = 1.19 MeV, then γ v c e = 1.19 MeV 0.511 MeV = x 1 – x 2 = 2.33 where x = ve c Solving, x 2 = (1 – x 2)5.43 and x = ve/c = 0.919 ve = 0.919c Then mpvp = γem eve : vp = γ emevec mpc = (1.19 MeV)(1.60 × 10–13 J/MeV) (1.67 × 10 – 2 7 )(3.00 × 10 8) = 3.80 × 10 5 m/s = 380 km/s (c) The electron is relativistic, the proton is not. 46.7 The time for a particle traveling with the speed of light to travel a distance of 3 × 10–15 m is ∆t = d v = 3 × 10–15 m 3 × 108 m/s = ~ 10–23 s 46.8 With energy 938.3 MeV, the time that a virtual proton could last is at most ∆t in ∆E ∆t ~ h.
The distance it could move is at most c ∆t ~ hc ∆E = (1.055 × 10–34 J · s)(3 × 108 m/s) (938.3)(1.6 × 10–13 J) = ~ 10–16 m 4 Chapter 46 Solutions 46.9 By Table 46.2, Mπ 0 = 135 MeV/c 2 Therefore, Eγ = 67.5 MeV for each photon p = Eγ c = 67.5 MeV c and f = Eγ h = 1.63 × 1022 Hz 46.10 In ? + p+ → n + µ+, charge conservation requires the unknown particle to be neutral. Baryon number conservation requires baryon number = 0. The muon-lepton number of ? must be –1.
So the unknown particle must be ν – µ .
46.11 Ω+ → Λ 0 + K+ KS 0 →π + + π − (or π0 + π0) Λ 0 →p + π + n →p + e+ + νe 46.12 (a) p + p – → µ+ + e– Le 0 + 0 → 0 + 1 and Lµ 0 + 0 → –1 + 0 (b) π – + p → p + π + charge –1 + 1 → +1 + 1 (c) p + p → p + π + baryon number 1 + 1 → 1 + 0 (d) p + p → p + p + n baryon number 1 + 1 → 1 + 1 + 1 (e) γ + p → n + π 0 charge 0 + 1 → 0 + 0 46.13 (a) Baryon number and charge are conserved, with values of 0 + 1 = 0 + 1 and 1 + 1 = 1 + 1 in both reactions.
(b) Strangeness is not conserved in the second reaction.
Chapter 46 Solutions 5 © 2000 by Harcourt, Inc. All rights reserved.
6 Chapter 46 Solutions 46.14 Baryon number conservation allows the first and forbids the second .
46.15 (a) π µ − − → + νµ Lµ: 0 →1−1 (b) K+ → + + µ νµ Lµ: 0 →−1+ 1 (c) νe + → + + p n e + Le: −1+ 0 →0 −1 (d) νe + → + − n p e + Le: 1+ 0 →0 + 1 (e) νµ + → + − n p+ µ Lµ: 1+ 0 →0 + 1 (f) µ −→e−+ νe + νµ Lµ: 1→0 + 0 + 1 and Le: 0 →1−1+ 0 46.16 Momentum conservation requires the pions to have equal speeds. The total energy of each is 497.7 MeV 2 so E2 = p2c2 + (mc 2)2 gives (248.8 MeV)2 = (pc)2 + (139.6 MeV)2 Solving, pc = 206 MeV = γ mvc = mc2 1−v c ( ) 2 v c pc mc 2 = 206 MeV 139.6 MeV = 1 1−v c ( ) 2 v c = 1.48 v c ( ) = 1.48 1−v c ( ) 2 and v c ( ) 2 = 2.18 1−v c ( ) 2 [ ] = 2.18 −2.18 v c ( ) 2 3.18 v c ( ) 2 = 2.18 so v c = 2.18 3.18 = 0.828 and v = 0.828c 46.17 (a) p+ → π + + π 0 Baryon number is violated: 1 → 0 + 0 (b) p+ + p+ → p+ + p+ + π 0 This reaction can occur .
(c) p+ + p+ → p+ + π + Baryon number is violated: 1 + 1 → 1 + 0 (d) π + → µ+ + νµ This reaction can occur .
(e) n0 → p+ + e– + ν – e This reaction can occur .
(f) π + → µ+ + n Violates baryon number : 0 → 0 + 1 Chapter 46 Solutions 7 © 2000 by Harcourt, Inc. All rights reserved.
Violates muon-lepton number : 0 → –1 + 0 8 Chapter 46 Solutions 46.18 (a) p →e+ + γ Baryon number: +1→0 + 0 ∆B ≠0 , so baryon number is violated.
(b) From conservation of momentum: pe = pγ Then, for the positron, Ee 2 = pec ( ) 2 + E0, e 2 becomes Ee 2 = pγ c ( ) 2 + E0, e 2 = Eγ 2 + E0, e 2 From conservation of energy: E0, p = Ee + Eγ or Ee = E0, p −Eγ so Ee 2 = E0, p 2 −2E0, pEγ + Eγ 2. Equating this to the result from above gives Eγ 2 + E0, e 2 = E0, p 2 −2E0,pEγ + Eγ 2, or Eγ = E0, p 2 −E0, e 2 2E0, p = 938.3 MeV ( )2 −0.511 MeV ( )2 2 938.3 MeV ( ) = 469 MeV Thus, Ee = E0, p −Eγ = 938.3 MeV – 469 MeV = 469 MeV Also, pγ = Eγ c = 469 MeV/c and pe = pγ = 469 MeV/c (c) The total energy of the positron is Ee = 469 MeV.
But, Ee = γ E0, e = E0, e 1−(v/c)2 so 1−v c ( ) 2 = E0, e Ee = 0.511 MeV 469 MeV = 1.09 × 10−3 which yields: v c = 0 999 999 4 . 46.19 The relevant conservation laws are: ∆Le = 0, ∆Lµ = 0, and ∆Lτ = 0.
(a) π π + + → + + 0 e ? L L L e e e : 0 0 1 1 → − + = ⇒ and we have a νe (b) ? + → + + − + p p µ π L L L µ µ µ : + →+ + + = ⇒ 0 1 0 0 1 and we have a νµ (c) Λ0 → + + − p µ ? L L L µ µ µ : 0 0 1 1 → + + = − ⇒ and we have a νµ (d) τ µ + + → + + ?
? L L L µ µ µ : 0 1 1 →− + = ⇒ and we have a νµ L L L τ τ τ : + → + = ⇒ 1 0 1 and we have a ντ Conclusion for (d): Lµ = 1 for one particle, and Lτ = 1 for the other particle. We have νµ and ντ .
46.20 The ρ π π 0 → + + − decay must occur via the strong interaction. The KS 0 → + + − π π decay must occur via the weak interaction.
Chapter 46 Solutions 9 © 2000 by Harcourt, Inc. All rights reserved.
46.21 (a) Λ0 → p + π – Strangeness: –1 → 0 + 0 (strangeness is not conserved ) (b) π – + p → Λ0 + K 0 Strangeness: 0 + 0 → –1 + 1 (0 = 0 and strangeness is conserved ) (c) p – + p → Λ – 0 + Λ0 Strangeness: 0 + 0 → +1 – 1 (0 = 0 and strangeness is conserved ) (d) π – + p → π – + Σ + Strangeness: 0 + 0 → 0 – 1 (0 ≠ –1: strangeness is not conserved ) (e) Ξ – → Λ0 + π – Strangeness: –2 → –1 + 0 (–2 ≠ –1 so strangeness is not conserved ) (f) Ξ 0 → p + π – Strangeness: –2 → 0 + 0 (–2 ≠ 0 so strangeness is not conserved ) 46.22 (a) µ– → e– + γ Le : 0 → 1 + 0, and Lµ : 1 → 0 (b) n → p + e– + νe Le : 0 → 0 + 1 + 1 (c) Λ0 → p + π 0 Strangeness: –1 → 0 + 0, and charge: 0 → +1 + 0 (d) p → e+ + π 0 Baryon number: +1 → 0 + 0 (e) Ξ 0 → n + π 0 Strangeness: –2 → 0 + 0 46.23 (a) π −+ p →2η violates conservation of baryon number as 0 + 1→0. not allowed (b) K−+ n →Λ0 + π − Baryon number = 0 + 1 → 1 + 0 Charge = –1 + 0 → 0 – 1 Strangeness, – 1 + 0 → –1 + 0 Lepton number, 0 → 0 The interaction may occur via the strong interaction since all are conserved.
(c) K− − → + π π 0 Strangeness, –1 → 0 + 0 Baryon number, 0 → 0 Lepton number, 0 → 0 Charge, –1 → –1 + 0 Strangeness is violated by one unit, but everything else is conserved. Thus, the reaction can occur via the weak interaction , but not the strong or electromagnetic interaction.
(d) Ω Ξ − − → + π 0 Baryon number, 1 → 1 + 0 Lepton number, 0 → 0 Charge, –1 → –1 + 0 Strangeness, –3 → –2 + 0 May occur by weak interaction , but not by strong or electromagnetic.
(e) η →2γ Baryon number, 0 → 0 Lepton number, 0 → 0 Charge, 0 → 0 Strangeness, 0 → 0 10 Chapter 46 Solutions No conservation laws are violated, but photons are the mediators of the electromagnetic interaction. Also, the lifetime of the η is consistent with the electromagnetic interaction .
46.24 (a) Ξ Λ − − → + + 0 µ νµ Baryon number: + 1 → + 1 + 0 + 0 Charge: –1 → 0 – 1 + 0 Le: 0 → 0 + 0 + 0 Lµ: 0 → 0 + 1 + 1 Lτ : 0 → 0 + 0 + 0 Strangeness: –2 → –1 + 0 + 0 Conserved quantities are: B L L e , , , charge and τ (b) KS 0 →2 0 π Baryon number: 0 → 0 Charge: 0 → 0 Le: 0 → 0 Lµ: 0 → 0 Lτ : 0 → 0 Strangeness: +1 → 0 Conserved quantities are: B L L L e , , , charge , and µ τ (c) K−+ p →Σ0 + n Baryon number: 0 + 1 → 1 + 1 Charge: –1 + 1 → 0 + 0 Le: 0 + 0 → 0 + 0 Lµ: 0 + 0 → 0 + 0 Lτ : 0 + 0 → 0 + 0 Strangeness: –1 + 0 → –1 + 0 Conserved quantities are: S L L L e , , , charge , and µ τ (d) Σ0 →Λ0 + γ Baryon number: + 1 → 1 + 0 Charge: 0 → 0 Le: 0 → 0 + 0 Lµ: 0 → 0 + 0 Lτ : 0 → 0 + 0 Strangeness: –1 → –1 + 0 Conserved quantities are: B S L L L e , , , , charge , and µ τ (e) e e + − + − + → + µ µ Baryon number: 0 + 0 → 0 + 0 Charge: +1 – 1 → +1 – 1 Le: –1 + 1 → 0 + 0 Lµ: 0 + 0 → + 1 – 1 Lτ : 0 + 0 → 0 + 0 Strangeness: 0 + 0 → 0 + 0 Conserved quantities are: B S L L L e , , , , charge , and µ τ (f) p n + → + − Λ Σ 0 Baryon number: –1 + 1 → –1 + 1 Charge: –1 + 0 → 0 – 1 Le: 0 + 0 → 0 + 0 Lµ: 0 + 0 → 0 + 0 Lτ : 0 + 0 → 0 + 0 Strangeness: 0 + 0 → +1 – 1 Conserved quantities are: B S L L L e , , , , charge , and µ τ Chapter 46 Solutions 11 © 2000 by Harcourt, Inc. All rights reserved.
46.25 (a) K p ? p + + → + The strong interaction conserves everything.
Baryon number, 0 + 1→B + 1 so B = 0 Charge, +1+ 1→Q + 1 so Q = +1 Lepton numbers, 0 + 0 →L + 0 so L L L e = = = µ τ 0 Strangeness, +1+ 0 →S + 0 so S = 1 The conclusion is that the particle must be positively charged, a non-baryon, with strangeness of +1. Of particles in Table 46.2, it can only be the K+ . Thus, this is an elastic scattering process.
The weak interaction conserves all but strangeness, and ∆S = ±1.
(b) Ω− − → + ? π Baryon number, +1→B + 0 so B = 1 Charge, −1→Q −1 so Q = 0 Lepton numbers, 0 →L + 0 so Le = Lµ = Lτ = 0 Strangeness, −3 →S + 0 so ∆S = 1: S = −2 The particle must be a neutral baryon with strangeness of –2. Thus, it is the Ξ0 .
(c) K ? + + → + + µ νµ : Baryon number, 0 →B + 0 + 0 so B = 0 Charge, +1→Q + 1+ 0 so Q = 0 Lepton Numbers L L e e , 0 0 0 → + + so Le = 0 L L µ µ , 0 1 1 → − + so Lµ = 0 L L τ τ , 0 0 0 → + + so Lτ = 0 Strangeness: 1→S + 0 + 0 so ∆S = ±1 (for weak interaction): S = 0 The particle must be a neutral meson with strangeness = 0 ⇒ π 0 .
46.26 (a) proton u u d total strangeness 0 0 0 0 0 baryon number 1 1/3 1/3 1/3 1 charge e 2e/3 2e/3 –e/3 e (b) neutron u d d total strangeness 0 0 0 0 0 baryon number 1 1/3 1/3 1/3 1 charge 0 2e/3 –e/3 –e/3 0 46.27 (a) The number of protons Np = 1000 g 6.02 × 1023 molecules 18.0 g 10 protons molecule = 3.34 × 1026 protons 12 Chapter 46 Solutions and there are Nn = 1000 g ( ) 6.02 × 1023 molecules 18.0 g 8 neutrons molecule = 2.68 × 1026 neutrons So there are for electric neutrality 3.34 × 1026 electrons The up quarks have number 2 × 3.34 × 1026 + 2.68 × 1026 = 9.36 × 1026 up quarks and there are 2 × 2.68 × 1026 + 3.34 × 1026 = 8.70 × 1026 down quarks (b) Model yourself as 65 kg of water. Then you contain 65 × 3.34 × 1026 ~ 1028 electrons 65 × 9.36 × 1026 ~ 1029 up quarks 65 × 8.70 × 1026 ~ 1029 down quarks Only these fundamental particles form your body. You have no strangeness, charm, topness or bottomness.
46.28 (a) K0 d s total strangeness 1 0 1 1 baryon number 0 1/3 –1/3 0 charge 0 –e/3 e/3 0 (b) Λ0 u d s total strangeness –1 0 0 –1 –1 baryon number 1 1/3 1/3 1/3 1 charge 0 2e/3 –e/3 –e/3 0 46.29 Quark composition of proton = uud and of neutron = udd.
Thus, if we neglect binding energies, we may write mp = 2mu + md (1) and mn = mu + 2md (2) Solving simultaneously, we find mu = 1 3 (2mp – mn) = 1 3 2 938 939 2 2 ( .
/ ) .
/ 3 MeV 6 MeV c c − [ ] = 312 MeV/c2 and from either (1) or (2), md = 314 MeV/c 2 46.30 In the first reaction, π −+ → + p K0 Λ0, the quarks in the particles are: ud + uud → ds + uds.
There is a net of 1 up quark both before and after the reaction, a net of 2 down quarks both Chapter 46 Solutions 13 © 2000 by Harcourt, Inc. All rights reserved.
before and after, and a net of zero strange quarks both before and after. Thus, the reaction conserves the net number of each type of quark.
In the second reaction, π −+ → + p K n 0 , the quarks in the particles are: ud + uud → ds + udd. In this case, there is a net of 1 up and 2 down quarks before the reaction but a net of 1 up, 3 down, and 1 anti-strange quark after the reaction. Thus, the reaction does not conserve the net number of each type of quark.
46.31 (a) π −+ → + p Κ Λ 0 0 In terms of constituent quarks: ud + uud →ds + uds .
up quarks: – 1 + 2 → 0 + 1, or 1 → 1 down quarks: 1 + 1 → 1 + 1, or 2 → 2 strange quarks: 0 + 0 → –1 + 1, or 0 → 0 (b) π + + + ⇒ + → + p Κ Σ ud + uud →us + uus up quarks: 1 + 2 → 1 + 2, or 3 → 3 down quarks: –1 + 1 → 0 + 0, or 0 → 0 strange quarks: 0 + 0 → –1 + 1, or 0 → 0 (c) Κ Κ Κ Ω − + − ⇒ + → + + p 0 us + uud →us + ds + sss up quarks: –1 + 2 → 1 + 0 + 0, or 1 → 1 down quarks: 0 + 1 → 0 + 1 + 0, or 1 → 1 strange quarks: 1 + 0 → –1 – 1 + 3, or 1 → 1 (d) p + p →K0 + p + π + + ? ⇒ uud + uud →ds + uud + ud + ? The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks.
up quarks: 2 + 2 = 0 + 2 + 1 + ? (has 1 u quark) down quarks: 1 + 1 = 1 + 1 – 1 + ? (has 1 d quark) strange quarks: 0 + 0 = –1 + 0 + 0 + ? (has 1 s quark) quark composite = uds = Λ0 or Σ0 46.32 Σ Σ 0 + → + + + p γ X dds + uud → uds + 0 + ?
The left side has a net 3d, 2u and 1s. The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing.
The unknown particle is a neutron, udd. Baryon and strangeness numbers are conserved.
14 Chapter 46 Solutions 46.33 Compare the given quark states to the entries in Tables 46.4 and 46.5.
(a) suu = Σ+ (b) ud = π − (c) sd = Κ0 (d) ssd = Ξ − 46.34 (a) u ud : charge = −2 3 e ( ) + −2 3 e ( ) + 1 3 e ( ) = −e . This is the antiproton .
(b) u d d : charge = −2 3 e ( ) + 1 3 e ( ) + 1 3 e ( ) = 0 . This is the antineutron .
46.35 Section 39.4 says fobserver = fsource 1+ va c 1−va c The velocity of approach, va, is the negative of the velocity of mutual recession: va = −v. Then, c c v c v c ′ = − + λ λ 1 1 and ′ = + − λ λ 1 1 v c v c 46.36 v = HR (Equation 46.7) H = (1.7 × 10–2 m/s) ly (a) v (2.00 × 106 ly) = 3.4 × 104 m/s λ' = λ 1 1 + − v c v c / / = 590(1.0001133) = 590.07 nm (b) v (2.00 × 108 ly) = 3.4 × 106 m/s λ' = 590 1 0 01133 1 0 01133 + − .
. = 597 nm (c) v (2.00 × 109 ly) = 3.4 × 107 m/s λ' = 590 1 0 1133 1 0 1133 + − .
. = 661 nm 46.37 (a) λ' λ = 650 nm 434 nm = 1.50 = 1 1 + − v c v c / / 1 + v/c 1 – v/c = 2.24 v = 0.383c, 38.3% the speed of light (b) Equation 46.7, v = HR R = v H = (0.383)(3.00 × 108) (1.7 × 10–2) = 6.76 × 109 light years Chapter 46 Solutions 15 © 2000 by Harcourt, Inc. All rights reserved.
Goal Solution A distant quasar is moving away from Earth at such high speed that the blue 434-nm hydrogen line is observed at 650 nm, in the red portion of the spectrum. (a) How fast is the quasar receding? You may use the result of Problem 35. (b) Using Hubble's law, determine the distance from Earth to this quasar.
G : The problem states that the quasar is moving very fast, and since there is a significant red shift of the light, the quasar must be moving away from Earth at a relativistic speed (v > 0.1c). Quasars are very distant astronomical objects, and since our universe is estimated to be about 15 billion years old, we should expect this quasar to be ~109 light-years away.
O : As suggested, we can use the equation in Problem 35 to find the speed of the quasar from the Doppler red shift, and this speed can then be used to find the distance using Hubble’s law. A : (a) ′ λ λ = 650 nm 434 nm = 1.498 = 1+ v c 1−v c or squared, 1+ v c 1−v c = 2.243 Therefore, v = 0.383c or 38.3% the speed of light (b) Hubble’s law asserts that the universe is expanding at a constant rate so that the speeds of galaxies are proportional to their distance R from Earth, v = HR so, R = v H = 0.383 ( ) 3.00 × 108 m /s ( ) 1.70 × 10-2 m /s⋅ly ( ) = 6.76 × 109 ly L : The speed and distance of this quasar are consistent with our predictions. It appears that this quasar is quite far from Earth but not the most distant object in the visible universe. 46.38 (a) ′ λ n = λ n 1+ v/c 1−v/c = (Z + 1)λ n 1 + v/c 1 – v/c = ( Z + 1)2 1 + v c = ( Z + 1)2 – v c ( Z + 1) 2 v c ( Z2 + 2 Z + 2) = Z2 + 2 Z v = c Z2 + 2 Z Z2 + 2 Z + 2 (b) R = v H = c H Z2 + 2 Z Z2 + 2 Z + 2 16 Chapter 46 Solutions 46.39 The density of the Universe is ρ = 1.20ρc = 1.20 3H2 8π G ( ).
Consider a remote galaxy at distance r. The mass interior to the sphere below it is M = ρ 4π r3 3 = 1.20 3H2 8π G 4π r3 3 = 0.600H2r3 G both now and in the future when it has slowed to rest from its current speed v = Hr. The energy of this galaxy is constant as it moves to apogee distance R: 1 2 mv2 −GmM r = 0 −GmM R so 1 2 mH2r2 −Gm r 0.600H2r3 G = 0 −Gm R 0.600H2r3 G − = − 0 100 0 600 .
.
r R so R = 6.00r The Universe will expand by a factor of 6.00 from its current dimensions.
46.40 (a) kBT ≈2mpc2 so T m c k p ≈ = ( ) × ( ) × − − 2 2 938 3 1 38 10 1 60 10 2 23 13 B MeV J K J 1 MeV .
.
. ~ 1013 K (b) k T m c e B ≈2 2 T m c k e ≈ = ( ) × ( ) × − − 2 2 0 1 38 10 1 60 10 2 23 13 B .511 MeV J K J 1 MeV .
. ~ 1010 K 46.41 (a) Wien’s law: λmax .
T = × ⋅ − 2 898 10 3 m K Thus, λmax .
.
.
.
= × ⋅ = × ⋅ = × = − − − 2 898 10 2 898 10 2 73 1 06 10 3 3 3 m K m K K m T 1.06 mm (b) This is a microwave .
46.42 (a) L = hG c3 = 1.055 × 10−34 J ⋅s ( ) 6.67 × 10−11 N ⋅m2 kg2 ( ) 3.00 × 108 m s ( ) 3 = 1 61 10 35 .
× − m (b) This time is given as T = L c = 1.61× 10−35 m 3.00 × 108 m s = 5 38 10 44 .
× − s , which is approximately equal to the ultra-hot epoch.
Chapter 46 Solutions 17 © 2000 by Harcourt, Inc. All rights reserved.
46.43 (a) ∆E∆t ≈h, and ∆t ≈r c = 1.4 × 10−15 m 3.0 × 108 m s = 4.7 × 10−24 s ∆E ≈h ∆t = 1.055 × 10−34 J ⋅s 4.7 × 10−24 s = 2.3 × 10−11 J 1 MeV 1.60 × 10−13 J = 1.4 × 102 MeV m = ∆E c2 ≈1.4 × 102 MeV c2 ~ 102 2 MeV c (b) From Table 46.2, m c π 2 139 6 = . MeV, a pi-meson 46.44 (a) π – + p → Σ + + π 0 is forbidden by charge conservation (b) µ– → π – + νe is forbidden by energy conservation (c) p → π + + π + + π – is forbidden by baryon number conservation 46.45 The total energy in neutrinos emitted per second by the Sun is: (0.4)(4π)(1.5 × 1011)2 = 1.1 × 1023 W Over 109 years, the Sun emits 3.6 × 1039 J in neutrinos. This represents an annihilated mass mc2 = 3.6 × 1039 J m = 4.0 × 1022 kg About 1 part in 50,000,000 of the Sun's mass, over 109 years, has been lost to neutrinos.
18 Chapter 46 Solutions Goal Solution The energy flux carried by neutrinos from the Sun is estimated to be on the order of 0.4 W/m2 at Earth's surface. Estimate the fractional mass loss of the Sun over 109 years due to the radiation of neutrinos.
(The mass of the Sun is 2 × 1030 kg. The Earth-Sun distance is 1.5 × 1011 m.) G : Our Sun is estimated to have a life span of about 10 billion years, so in this problem, we are examining the radiation of neutrinos over a considerable fraction of the Sun’s life. However, the mass carried away by the neutrinos is a very small fraction of the total mass involved in the Sun’s nuclear fusion process, so even over this long time, the mass of the Sun may not change significantly (probably less than 1%).
O : The change in mass of the Sun can be found from the energy flux received by the Earth and Einstein’s famous equation, E = mc2.
A : Since the neutrino flux from the Sun reaching the Earth is 0.4 W/m2, the total energy emitted per second by the Sun in neutrinos in all directions is 0.4 W /m2 ( ) 4π r2 ( ) = 0.4 W /m2 ( ) 4π ( ) 1.5 × 1011 m ( ) 2 = 1.13 × 1023 W In a period of 109 yr, the Sun emits a total energy of 1.13 × 1023 J/s ( ) 109 yr ( ) 3.156 × 107 s/ yr ( ) = 3.57 × 1039 J in the form of neutrinos. This energy corresponds to an annihilated mass of E = mν c2 = 3.57 × 1039 J so mν = 3.97 × 1022 kg Since the Sun has a mass of about 2 × 103 kg, this corresponds to a loss of only about 1 part in 50 000 000 of the Sun's mass over 109 yr in the form of neutrinos. L : It appears that the neutrino flux changes the mass of the Sun by so little that it would be difficult to measure the difference in mass, even over its lifetime!
46.46 p p p + → + + + π X We suppose the protons each have 70.4 MeV of kinetic energy. From conservation of momentum, particle X has zero momentum and thus zero kinetic energy. Conservation of energy then requires Mpc2 + Mπc2 + MXc2 = Mpc2 + Kp ( ) + Mpc2 + Kp ( ) MXc2 = Mpc2 + 2Kp −Mπc2 = 938.3 MeV + 2 70.4 MeV ( ) −139.6 MeV = 939.5 MeV X must be a neutral baryon of rest energy 939.5 MeV. Thus X is a neutron .
Chapter 46 Solutions 19 © 2000 by Harcourt, Inc. All rights reserved.
46.47 We find the number N of neutrinos: 1046 J = N(6 MeV) = N(6 × 1.6 × 10–13 J) N = 1.0 × 1058 neutrinos The intensity at our location is N A = N 4πr 2 = 1.0 × 1058 4π(1.7 × 105 ly)2 1 ly (3.0 × 108 m /s)(3.16 × 107 s) 2 = 3.1× 1014 /m2 The number passing through a body presenting 5000 cm2 = 0.50 m2 is then 3 1 10 1 0 5 1 5 10 14 2 2 14 .
.
.
× ( ) = × m 0 m or ~ 1014 46.48 By relativistic energy conservation, Eγ + mec2 = 3mec2 1−v2 /c2 (1) By relativistic momentum conservation, Eγ c = 3mev 1−v2 /c2 (2) Dividing (2) by (1), X = Eγ Eγ + mec2 = v c Subtracting (2) from (1), mec2 = 3mec2 1−X2 −3mec2X 1−X2 Solving, 1 = 3 −3X 1−X2 and X = 4 5 so Eγ = 4mec2 = 2.04 MeV 46.49 mΛc2 = 1115.6 MeV Λ0 → p + π – m pc 2 = 938.3 MeV m c π 2 = 139.6 MeV The difference between starting mass-energy and final mass-energy is the kinetic energy of the products.
Kp + Kπ = 37.7 MeV and pp = pπ = p Applying conservation of relativistic energy, ( . ) .
( . ) .
.
938 3 938 3 139 6 139 6 37 2 2 2 2 2 2 + − + + − = p c p c 7 MeV Solving the algebra yields pπ c = ppc = 100.4 MeV Then, Kp = (m pc 2)2 + (100.4)2 – mpc 2 = 5.35 MeV Kπ = (139.6)2 + (100.4)2 – 139.6 = 32.3 MeV 20 Chapter 46 Solutions 46.50 Momentum of proton is qBr = 1.60 × 10−19 C ( ) 0.250 kg /C⋅s ( ) 1.33 m ( ) pp = 5.32 × 10–20 kg · m s cpp = 1.60 × 10–11 kg m2 s2 = 1.60 × 10–11 J = 99.8 MeV Therefore, pp = 99.8 MeV c The total energy of the proton is Ep = E0 2 + (cp)2 = (938.3)2 + (99.8)2 = 944 MeV For pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2-particle decay).
pπ = 99.8 MeV c E0π = 139.6 MeV Eπ = E0 2 + (cp)2 = (139.6)2 + (99.8)2 = 172 MeV Thus, ETotal after = ETotal before = Rest Energy Rest Energy of unknown particle = 944 MeV + 172 MeV = 1116 MeV (This is a Λ0 particle!) Mass = 1116 MeV/c 2 46.51 Σ Λ 0 0 → + γ From Table 46.2, mΣ = 1192.5 MeV/c 2 and m c Λ = 1115 6 2 . MeV Conservation of energy requires E E K E 0 0 , , Σ Λ Λ = + ( ) + γ , or 1192 5 1115 6 2 2 .
. MeV MeV = + + p m E Λ Λ γ Momentum conservation gives p p Λ = γ , so the last result may be written as 1192 5 1115 6 2 2 .
. MeV MeV = + + p m E γ γ Λ or 1192 5 1115 6 2 2 2 2 .
. MeV MeV = + + p c m c E γ γ Λ Recognizing that m c Λ 2 1115 6 = . MeV and p c E γ γ = , we now have 1192 5 1115 6 2 1115 6 2 .
.
. MeV MeV MeV = + ( ) + E E γ γ Solving this quadratic equation, Eγ ≈ 74.4 MeV Chapter 46 Solutions 21 © 2000 by Harcourt, Inc. All rights reserved.
46.52 p p p n + → + + + π The total momentum is zero before the reaction. Thus, all three particles present after the reaction may be at rest and still conserve momentum. This will be the case when the incident protons have minimum kinetic energy. Under these conditions, conservation of energy gives 2 2 2 2 2 m c K m c m c m c p p p n + ( ) = + + π so the kinetic energy of each of the incident protons is K m c m c m c p n p = + − = + − ( ) = 2 2 2 2 939 6 139 6 938 3 2 π .
.
. MeV 70.4 MeV 46.53 Time-dilated lifetime: T = γ T0 = 0.900 × 10–10 s 1 – v2/c2 = 0.900 × 10–10 s 1 – (0.960)2 = 3.214 × 10–10 s distance = (0.960)(3.00 × 108 m/s)(3.214 × 10–10 s) = 9.26 cm 46.54 π µ ν − − → + µ From the conservation laws, m c E E π µ ν 2 139 5 = = + . MeV and p p µ ν = , E p c ν ν = Thus, E p c p c µ µ ν 2 2 2 2 2 105 7 105 7 = ( ) + ( ) = ( ) + ( ) .
. MeV MeV or Eµ 2 −Eν 2 = 105.7 MeV ( )2 Since E E µ ν + = 139 5 . MeV and E E E E µ ν µ ν + ( )( ) = ( ) − 105 7 2 . MeV then E E µ ν − = ( ) = 105 7 139 5 80 1 2 .
.
. MeV MeV Subtracting from , 2 59 4 Eν = . MeV and Eν = 29 7 . MeV 22 Chapter 46 Solutions 46.55 The expression e−E kBT dE gives the fraction of the photons that have energy between E and E + dE. The fraction that have energy between E and infinity is e−E kBT dE E ∞ ∫ e−E kBT dE 0 ∞ ∫ = e−E kBT −dE kBT ( ) E ∞ ∫ e−E kBT −dE kBT ( ) 0 ∞ ∫ = e−E kBT E ∞ e−E kBT 0 ∞= e−E kBT We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = = × − 1 00 1 60 10 19 .
. eV J Thus, 0 0100 1 60 10 1 38 10 19 23 .
.
.
= − × ( ) × ( ) − − e T J J K or ln 0.0100 ( ) = − 1.60 × 10−19 J 1.38 × 10−23 J K ( )T = −1.16 × 104 K T giving T = 2 52 103 .
× K 46.56 (a) This diagram represents the annihilation of an electron and an antielectron. From charge and lepton-number conservation at either vertex, the exchanged particle must be an electron, e– .
(b) This is the tough one. A neutrino collides with a neutron, changing it into a proton with release of a muon. This is a weak interaction. The exchanged particle has charge +1e and is a W+ . (a) (b) 46.57 (a) The mediator of this weak interaction is a Z0 boson .
(b) The Feynman diagram shows a down quark and its antiparticle annihilating each other. They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge. In this case the product particle is a photon . For conservation of both energy and momentum, we would expect two photons; but (a) (b) momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matter-antimatter pair of particles, as shown in Figure P46.57. Depending on the color charges of the d and d quarks, Chapter 46 Solutions 23 © 2000 by Harcourt, Inc. All rights reserved.
the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.13(b). |
7558 | https://instituteofliving.org/health-wellness/news/newsroom-detail?articleId=29129&publicid=501 | Cardio-Oncology: When Heart Disease Meets Cancer | Institute of Living | Hartford HealthCare | CT
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Cardio-Oncology: When Heart Disease Meets Cancer
October 04, 2020
By Dr. Eric Oligino
Director, Cardio-Oncology Program
Hartford HealthCare Cancer Institute
Over recent decades, advances in cancer therapies have translated into a rising population of cancer survivors estimated to reach 19 million in the United States alone by 2024.
Similarly, novel treatment options continue to reduce the morbidity and mortality associated with cardiovascular disease. Despite great strides in both arenas, cardiovascular disease and cancer remain the two leading causes of death in the United States, accounting for approximately 46 percent of all incidences.
Not surprisingly, these trends have led to a greater number of patients with cardiovascular disease and risk factors being treated for cancer. Widely used cancer therapies have known cardiovascular toxicities that affect those even without pre-existing heart disease. Toxicities include, but are not limited to, cardiac dysfunction and heart failure, arrhythmias, valvular disease, accelerated coronary disease and pericardial disease.
Coexisting cancer and cardiovascular disease leads to complex management decisions that extend beyond the boundaries of traditional medicine specialties. A new discipline called “Cardio-Oncology” has formed in response to this growing clinical need.
The Cardio-Oncology team at the Hartford HealthCare Cancer Institute consists of dedicated professionals across multiple disciplines who provide specialized care to cancer patients throughout all treatment phases. Screening, preventing and treating cardiotoxicities of cancer therapies require specialized knowledge to balance risks and benefits.
Importantly, the scope of Cardio-Oncology extends beyond cardiotoxicity and encompasses comprehensive cardiovascular care throughout survivorship. Long after the active cancer treatment phase is complete, Cardio-Oncology aims to maximize cardiovascular health.
Traditional cardiac risk factors — namely hypertension, diabetes, hyperlipidemia, and smoking — need to be aggressively controlled in survivorship. In women with breast cancer, weight gain and decreased physical activity are recognized consequences of treatment and the disease itself. A recent meta-analysis found that prediagnosis and postdiagnosis physical activity were associated with a lower mortality and breast cancer recurrence. Cardio-Oncology programs emphasize the American Heart Association’s recommendation of at least 30 minutes of moderate-intense aerobic physical activity five days each week.
The European Society of Cardiology recently published a position paper detailing the rationale, organization and implementation of Cardio-Oncology programs. Multiple professional societies, including the American Heart Association, have published guidelines to minimize the heterogeneous approach of current clinical practices.
For example, anthracyclines are common anti-cancer agents used against multiple diseases, including breast cancer and hematologic malignancies. These agents are known to carry a risk of cardiac dysfunction and heart failure. In fact, the origins of Cardio-Oncology date to the 1960s, when these complications were first recognized.
As a result, a baseline assessment of cardiac function, most commonly by an echocardiogram, is required before initiating an anthracycline-based regimen. But follow-up assessments are now recommended immediately following and six months after completing treatment. Adoption of these new guidelines has been slow, particularly in settings without a dedicated Cardio-Oncology team.
Hartford HealthCare’s Cardio-Oncology Program incorporates the most cutting-edge approach to the cardiovascular health of cancer patients throughout all treatment phases. Not all patients require a formal assessment, but specialists are readily available to address the broad range of issues that may arise.
Transitioning from the treatment phase into survivorship is an opportune time to determine if a Cardio-Oncology consultation is right for you.
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7559 | https://virology.ws/2009/07/06/detecting-viruses-the-plaque-assay/ | Skip to content
Detecting viruses: the plaque assay
104 Comments
/ By Vincent Racaniello
/ 6 July 2009
One of the most important procedures in virology is measuring the virus titer – the concentration of viruses in a sample. A widely used approach for determining the quantity of infectious virus is the plaque assay. This technique was first developed to calculate the titers of bacteriophage stocks. Renato Dulbecco modified this procedure in 1952 for use in animal virology, and it has since been used for reliable determination of the titers of many different viruses.
To perform a plaque assay, 10-fold dilutions of a virus stock are prepared, and 0.1 ml aliquots are inoculated onto susceptible cell monolayers. After an incubation period, to allow virus to attach to cells, the monolayers are covered with a nutrient medium containing a substance, usually agar, that causes the formation of a gel. When the plates are incubated, the original infected cells release viral progeny. The spread of the new viruses is restricted to neighboring cells by the gel. Consequently, each infectious particle produces a circular zone of infected cells called a plaque. Eventually the plaque becomes large enough to be visible to the naked eye. Dyes that stain living cells are often used to enhance the contrast between the living cells and the plaques. Only viruses that cause visible damage of cells can be assayed in this way. An example of plaques formed by poliovirus on a monolayer of HeLa cells is shown at left. In this image, the cells have been stained with crystal violet, and the plaques are readily visible where the cells have been destroyed by viral infection.
The titer of a virus stock can be calculated in plaque-forming units (PFU) per milliliter. To determine the virus titer, the plaques are counted. To minimize error, only plates containing between 10 and 100 plaques are counted, depending on the size of the cell culture plate that is used. Statistical principles dictate that when 100 plaques are counted, the sample titer will vary by plus or minus 10%. Each dilution is plated in duplicate to enhance accuracy. In the example shown below, there are 17 plaques on the plate made from the 10-6 dilution. The titer of the virus stock is therefore 1.7 x 108 PFU/ml.
Next we’ll consider how the plaque assay can be used to prepare clonal virus stocks, a step that is essential for studying viral genetics.
Dulbecco, R., & Vogt, M. (1953). Some problems of animal virology as studied by the plaque technique. Cold Spring Harbor Symp. Quant. Biol., 18, 273-279
104 thoughts on “Detecting viruses: the plaque assay”
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Szote
Dear Sir , we are facing problem with viral plaque assay that always our cell lines die after poring the mixture of agrose .. can i hear recoomendations from you ?
3. RLG
Dear all, I am performing a plaque assay and looking for a recipe for the agarose/agar medium that I can make myself. Due to time constraints it is not feasible to order pre-made in. Any suggestions welcome, thanks.
4. UP
sir i am doing plaque assay for NDV on cef-df1 cell lines but the 2%agarose i use for overlay has a detrimental effect on my cell line and its dead overnight what can be done to prevent the same |
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Congruent Triangles
Proofs (mathematics)
Vectors (mathematics)
PLANE GEOMETRY
Geometry Class
Concept of Geometry
Mathematical Proof
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5
How can we prove that two sides of a triangle are equal using vectors?
Michael Paglia
Former Journeyman Wireman IBEW
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Author has 33.3K answers and 5.1M answer views
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1y
Vectors have magnitude and direction
No lengths in plane geometry
I think youd use polar for coordinates
Maybe Pythagoras
And definitely some algebra if he didnt work
Equations of lines and points
I was awesome in plane
Bombed polar
Turning point in my younger life
Algebra geometry trig all easy as pie
Hit Algebra 2 then polar then calculus
I wasnt a true mathematician
Just a real good amateur
Tjat sounds light
Believe me at 16 it was a hard pill to swallow
I.hepled football players pass in freshman year Algebra
I had it in 8th grade aced it
Loved it
I marked tests while they were taking them
It was a reverse 70
Vectors have magnitude and direction
No lengths in plane geometry
I think youd use polar for coordinates
Maybe Pythagoras
And definitely some algebra if he didnt work
Equations of lines and points
I was awesome in plane
Bombed polar
Turning point in my younger life
Algebra geometry trig all easy as pie
Hit Algebra 2 then polar then calculus
I wasnt a true mathematician
Just a real good amateur
Tjat sounds light
Believe me at 16 it was a hard pill to swallow
I.hepled football players pass in freshman year Algebra
I had it in 8th grade aced it
Loved it
I marked tests while they were taking them
It was a reverse 70s math nerd movie
I was a smart kid used my head
I helped and made friends for 4 years
Then junior year.
Pow
Started out EE with partial scholarship.
Retired making 60$ hour as a construction electrician
Learned digital microprocessors
Never let stuff deter you from.doing what you like
I just needed a different path
Id do it all.over again
2 seconds from death in 1997
Still do it all.over
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Related questions
How can we prove that if there are two equal sides in a triangle, the resulting two angles opposite to these sides will also have to be equal?
What is a four-sided shape with two equal sides and two unequal sides?
If two angles of a triangle are equal, are all three sides equal?
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Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
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Author has 6.2K answers and 6.9M answer views
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Jan 25
Related
How can you prove by vector method that a line segment joining the midpoint of any two sides of a triangle is parallel to the third side and is half of it?
Take a scalene △PQR as shown above. This is to make things easy and there won't be any loss of generality. Variables a, b and c are taken in multiples of 2 so that we don't have to deal with tough algebra!
−−→RQ=(2c,0)=2(c,0)
−→ST=(c,0)
There’s no need to take magnitudes, from the components itself, it's very clear that ST is half of RQ. The other thing is that both the vectors are scalar multiples of each other and their cross product is zero which proves parallelism!
Take a scalene △PQR as shown above. This is to make things easy and there won't be any loss of generality. Variables a, b and c are taken in multiples of 2 so that we don't have to deal with tough algebra!
−−→RQ=(2c,0)=2(c,0)
−→ST=(c,0)
There’s no need to take magnitudes, from the components itself, it's very clear that ST is half of RQ. The other thing is that both the vectors are scalar multiples of each other and their cross product is zero which proves parallelism!
John Pereira
Retired lecturer (Maths)
·
Author has 1.6K answers and 884K answer views
·
1y
Please check your question. (Exactly) two sides of a triangle are equal if any two angles are equal and every pair of two sides are equal if every angle is 60 deg.
Neil Morrison
I do all my own differentiation!
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Author has 9.4K answers and 27.4M answer views
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Jan 25
Related
How can you prove by vector method that a line segment joining the midpoint of any two sides of a triangle is parallel to the third side and is half of it?
−−→PR+−−→RT+−−→TP=0––
⇒−−→PR+−−→RT=−−→PT[∗∗]
−−→QS=−−→QP+−−→PT+−→TS
=−12−−→PR+−−→PT−12−−→RT
=−12(−−→PR+−−→RT)+−−→PT
=−12−−→PT+−−→PT
−−→PR+−−→RT+−−→TP=0––
⇒−−→PR+−−→RT=−−→PT[∗∗]
−−→QS=−−→QP+−−→PT+−→TS
=−12−−→PR+−−→PT−12−−→RT
=−12(−−→PR+−−→RT)+−−→PT
=−12−−→PT+−−→PT[Using ]
=12−−→PT
So, since −−→QS is a scalar multiple of −−→PT, they are parallel, andsince that scalar multiplier is12, the line joining Q to S is halfthe length of side PT.
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Related questions
What are two pairs of equal sides?
What is the name of a triangle with two equal sides and one unequal side?
Can a rhombus have two equal sides?
How do you prove that if two angles of one triangle are equal to two angles of another triangle, then all three sides will be equal?
How do you prove that two triangles have equal angles and sides?
David Joyce
Dave's Short Course in Trig,
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Author has 9.9K answers and 68.1M answer views
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5y
Related
How do you prove that the sum of the vectors directed from the vertices to the mid-points of opposite sides of a triangle is zero?
Here’s a visual, geometric proof. Start with a triangle and its three medians.
We’re considering the sum of the three vectors −−→AD, −−→BE, and −−→CF.
One thing we know about vectors going around a triangle is that their sum is zero: −−→AB+−−→BC+−−→CA=0.
So half that sum is also zero: −−→AF+−−→BD+−−→CE=0.
But −−→AF=−−→AG+−−→GF, −−→BD=−−→BG+−−→GD, and [math]\overrightarrow{CE}=\overrightarrow{CG}+[/math]
Here’s a visual, geometric proof. Start with a triangle and its three medians.
We’re considering the sum of the three vectors −−→AD, −−→BE, and −−→CF.
One thing we know about vectors going around a triangle is that their sum is zero: −−→AB+−−→BC+−−→CA=0.
So half that sum is also zero: −−→AF+−−→BD+−−→CE=0.
But −−→AF=−−→AG+−−→GF, −−→BD=−−→BG+−−→GD, and −−→CE=−−→CG+−−→GE.
Therefore, (−−→AG+−−→GF)+(−−→BG+−−→GD)+(−−→CG+−−→GE)=0.
Rearranging that sum, we get (−−→AG+−−→GD)+(−−→BG+−−→GE)+(−−→CG+−−→GF)=0.
But −−→AG+−−→GD=−−→AD, −−→BG+−−→GE=−−→BE, and −−→CG+−−→GF=−−→CF.
Therefore, −−→AD+−−→BE+−−→CF=0.
Richard Goldstone
Emeritus at Manhattan College
·
Author has 1.8K answers and 3.8M answer views
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4y
Related
How do you prove that three vectors form a triangle?
I assume the space can be for any [math]n \ge 2[/math]. If [math]{ a, b, c }[/math] is a solution, then math + (\pm b) + ( \pm c) = \vec{0}[/math] for some choices of the [math]\pm[/math] signs.
Given three vectors in [math]\mathbb{R}^n,[/math] choose a basis and write the vectors in coordinate form. Let [math]A[/math] be the [math]n \times 3[/math] matrix whose columns are the three coordinate vectors.
If the three vectors form a triangle, then they span the plane of the triangle and so the matrix A has to be of rank 2.
Moreover, in order to get a triangle, we need
[math]A \begin{bmatrix} \pm 1 \ \pm 1 \ \pm 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ \vdots \ 0 [/math]
I assume the space can be [math]\mathbb{R}^n[/math] for any [math]n \ge 2[/math]. If [math]{ a, b, c }[/math] is a solution, then math + (\pm b) + ( \pm c) = \vec{0}[/math] for some choices of the [math]\pm[/math] signs.
Given three vectors in [math]\mathbb{R}^n,[/math] choose a basis and write the vectors in coordinate form. Let [math]A[/math] be the [math]n \times 3[/math] matrix whose columns are the three coordinate vectors.
If the three vectors form a triangle, then they span the plane of the triangle and so the matrix A has to be of rank 2.
Moreover, in order to get a triangle, we need
[math]A \begin{bmatrix} \pm 1 \ \pm 1 \ \pm 1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ \vdots \ 0 \end{bmatrix}[/math]
for some choice of the [math]\pm[/math] signs. In other words, for some choice of the [math]\pm[/math] signs, we must have
[math]v= \begin{bmatrix} \pm 1 \ \pm 1 \ \pm 1 \end{bmatrix} \in \text{null}\,(A).[/math]
To determine which possible such [math]v[/math]’s are in [math]\text{null}\,(A)[/math], if any, row reduce [math]A[/math] (assuming rank 2 now) to get
[math]A_{\text{red}}=\begin{bmatrix} 1 & 0 & \alpha_1 \ 0 & 1 & \alpha_2 \ 0 & 0 & \alpha_3 \ \vdots & \vdots & \vdots \ 0 & 0 & \alpha_n \end{bmatrix}.[/math]
If any one of [math]\alpha_3, \alpha_4, \dots, \alpha_n[/math] is non-zero, then every vector [math]\begin{bmatrix} x & y & z \end{bmatrix}^\top[/math] in the one-dimensional null space of [math]A[/math] will have to have [math]z=0,[/math] and then no [math]v[/math] of the right form can be in the null space of [math]A[/math] and so the three vectors cannot form a triangle. This necessary condition says that all rows below the first two rows in the reduced matrix must be zero rows.
So suppose that after row reduction, the necessary condition of zero rows is satisfied. Then the null space of [math]A[/math] is spanned by the vector [math]n=\begin{bmatrix} -\alpha_1 & -\alpha_2 & 1 \end{bmatrix}^\top.[/math] In order for the null space to contain a vector [math]v[/math] of the proper form—and so for the three vectors to make a triangle—we must have [math]\alpha_1=\pm 1, \alpha_2=\pm 1[/math]. Any of those four possible outcomes signify that the original three vectors form a triangle.
For example, suppose
[math]a = \begin{bmatrix} 1 \ 0 \ 2 \end{bmatrix}, \quad b = \begin{bmatrix} 5 \ 1\ 0 \end{bmatrix}, \quad c = \begin{bmatrix} -4 \ -1 \ 2 \end{bmatrix}.[/math]
The matrix [math]A[/math] is
[math]A = \begin{bmatrix} 1 & 5 & -4 \ 0 & 1 & -1 \ 2 & 0 & 2 \end{bmatrix},[/math]
and its row-reduced form is
[math]A_{\text{red}} = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & -1 \ 0 & 0 & 0 \end{bmatrix}.[/math]
This meets all the conditions. The kernel is spanned by [math]\begin{bmatrix} -1 & 1 & 1 \end{bmatrix}^\top,[/math] which shows that[math] -a+b+c=0[/math] as required for triangularity.
If we change the last entry in [math]c[/math] and so have
[math]c' = \begin{bmatrix} -4 \ -1 \ 4 \end{bmatrix},[/math]
then the row-reduced matrix is
[math]A_{\text{red}}' = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & -1 \ 0 & 0 & 2 \end{bmatrix},[/math]
and the condition that all rows below the first two is violated and so the vectors do not form a triangle. This is obvious from the row-reduced form, which is of rank 3 so that the original vectors are linearly independent in [math]\mathbb{R}^3.[/math]
If we change the first entry in [math]c’[/math] and so have
[math]c'' = \begin{bmatrix} -3 \ -1 \ 4 \end{bmatrix},[/math]
then the row-reduced matrix is
[math]A_{\text{red}}'' = \begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & -1 \ 0 & 0 & 0 \end{bmatrix},[/math]
and now the 2 in position math[/math] violates the conditions and the vectors do not make a triangle. They’d like to make a triangle, but the vector [math]a[/math] is too short. The null space is spanned by
[math]\begin{bmatrix} -2 \ 1 \ 1 \end{bmatrix}[/math]
and this means that [math]-2a+b+c''=\vec{0},[/math] but we are only supplied with the vector [math]a.[/math]
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David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979)
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11y
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If 2 sides of a triangle are described by 2 vectors a and b, how does one work out what vector defines the third side?
You subtract a and b to get the third vector. Here's a picture to demonstrate what's really going on
In your example where a = 2i + 3j + 5k and b = i - 2j + 4k, the third side of the triangle would be i + 5j + k. But since what we call a and b are arbitrary, we can switch the definitions around in which case, the third side would be -i - 5j - k. It's just the negative of what we had first. In any case, the resulting triangle would be the same
You subtract a and b to get the third vector. Here's a picture to demonstrate what's really going on
In your example where a = 2i + 3j + 5k and b = i - 2j + 4k, the third side of the triangle would be i + 5j + k. But since what we call a and b are arbitrary, we can switch the definitions around in which case, the third side would be -i - 5j - k. It's just the negative of what we had first. In any case, the resulting triangle would be the same
Subramanya R
Lives in Mysuru, Karnataka, India
·
Author has 2K answers and 1.5M answer views
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2y
Related
How can we prove that if two sides of any right-angled triangle are equal, then the angle opposite of the two equal sides will also be equal?
How can we prove that if two sides of any right-angled triangle are equal, then the angle opposite of the two equal sides will also be equal?
[math] \text {When sides of the right triangle are equal or legs of the right triangle}[/math]
[math] \text {are equal, then it is half of the square halved along its diagonal} [/math]
[math] \text {Interio angles of a square are always equal to 90° }[/math]
[math] \text {Moreover, diagonal bisect the angles equally and are always} [/math]
[math] \text {equal to 45° }[/math]
[math] \text {This one of the proofs to prove that equal sides of a triangle subtend} [/math]
[math] \text {equal angle opposite to them} [/math]
[math] \text {Mathematically, it can al[/math]
How can we prove that if two sides of any right-angled triangle are equal, then the angle opposite of the two equal sides will also be equal?
[math] \text {When sides of the right triangle are equal or legs of the right triangle}[/math]
[math] \text {are equal, then it is half of the square halved along its diagonal} [/math]
[math] \text {Interio angles of a square are always equal to 90° }[/math]
[math] \text {Moreover, diagonal bisect the angles equally and are always} [/math]
[math] \text {equal to 45° }[/math]
[math] \text {This one of the proofs to prove that equal sides of a triangle subtend} [/math]
[math] \text {equal angle opposite to them} [/math]
[math] \text {Mathematically, it can also be proved as follows} [/math]
[math] \text {The hypotenuse of right triangle with equal legs will always be equal to }[/math]
[math] \sqrt{2}a \text { ; where a is the length of a leg} [/math]
[math] \text{Then } \sin B=\dfrac{a}{\sqrt{2}a}=\dfrac{1}{\sqrt{2}}[/math]
[math] ∠ B=\sin^{-1}\left (\dfrac{1}{\sqrt{2}}\right )=45° [/math]
[math] \text {Similarly, } \sin A=\dfrac{a}{\sqrt{2}a}=\dfrac{1}{\sqrt{2}}[/math]
[math] ∠ A=\sin^{-1}\left (\dfrac{1}{\sqrt{2}}\right )=45° [/math]
[math] \Rightarrow ∠ A=∠ B[/math]
[math] \text {It can also be proved using cosine law }[/math]
[math] \cos A=\dfrac{b^2+c^2-a^2}{2bc}[/math]
[math] \text {here, } b=a, c=\sqrt{2}a [/math]
[math] \cos A=\dfrac{a^2+(\sqrt{2}a)^2 –a^2}{2 a(\sqrt{2}a )} [/math]
[math] \cos A=\dfrac{2a^2 }{2 \sqrt{2}a^2 )} [/math]
[math] \cos A=\dfrac{1}{\sqrt{2}}[/math]
[math] \Rightarrow ∠ A=45° [/math]
[math] \text {Similarly, } \cos B=\dfrac{b^2+c^2-a^2}{2bc}[/math]
[math] \text {here, } b=a, c=\sqrt{2}a [/math]
[math] \cos B=\dfrac{a^2+(\sqrt{2}a)^2 –a^2}{2 a(\sqrt{2}a )} [/math]
[math] \cos B=\dfrac{2a^2 }{2 \sqrt{2}a^2 )} [/math]
[math] \cos B=\dfrac{1}{\sqrt{2}}[/math]
[math] \Rightarrow ∠ B=45° [/math]
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Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
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1y
Related
How can we prove that a triangle is equilateral using vectors?
Equal magnitudes, 60° angle between vectors.
Raghav Arora ( ਰਾਘਵ ਅਰੋੜਾ )
B.Sc. in Computer Science & Mathematics, University of Toronto St. George Campus (Graduated 2025)
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8y
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How do I prove that two triangles are congruent?
Since, angle SRT = angle SRT = 90 degrees
SR = TU
And the two vertically opposite angles formed by the intersection of SU and RT are also equal-:
(LET POINT of intersection of SU and RT be O.)
Triangle SRO is congruent to Triangle TUO by AAS criterion.
Now SO = TO and RO = UO by corresponding parts of congruent triangles axiom.
Therefore, SO + UO = TO + RO which gives us SU = RT.
Now, I've given you a really big hint and I hope you can solve your question by using the proof I've given.
Brian Overland
Tutor, author, computer programmer and tech writer
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Author has 31.3K answers and 102.8M answer views
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May 12
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If two sides of a right-angled triangle are equal, is the third side also equal?
No. Take a perfect square and divide it diagonally. You will get two triangles, each with the angle measurements as follows: 45–90–45.
Two of the angles of each triangle are equal to each other, but the third is not.
Moreover, if each side of the square measured 1 unit, then the Pythagorean Theorem tells us that the diagonal is equal to the square root of 2 (roughly 1.414 units).
Therefore, two of the sides of each triangle would measure 1 unit, and the third side would measure approx. 1.414 units.
Douglas Magowan
Private Pilot
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Author has 1.2K answers and 1.3M answer views
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8y
Related
How do I prove that two triangles are congruent?
There is a notable exception to the SSA does not prove congruence. It does if that angle is right.
But, if I am invoking a theorem that hasn’t yet been proven…. lets define the point X at the intersection of RT and SU.
angle UXT is congruent to angle RXS by vertical angles.
Triangle RSX is congruent to Triangle UXT by AAS.
SXT is iscocles.
angle RTS is congruent to angle UST.
Triangle RTS is congruent to triangle UST by AAS
Subramanian Tirunellayi Ramachandran
Author has 792 answers and 1.5M answer views
·
6y
Related
How do I prove that in a triangle, the smallest angle is the one opposite the smallest side?
By Law of Sines!!
if a is small sin A needs to be small too which means the angle A needs to be small too.
By Law of Sines!!
if a is small sin A needs to be small too which means the angle A needs to be small too.
Related questions
How can we prove that if there are two equal sides in a triangle, the resulting two angles opposite to these sides will also have to be equal?
What is a four-sided shape with two equal sides and two unequal sides?
If two angles of a triangle are equal, are all three sides equal?
How can we prove that two triangles are similar if three sides of one triangle are equal to three sides of the other?
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Understanding the Arithmetical Hierarchy
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I am trying to get acquainted with the arithmetical hierarchy, and as I wrote down some examples, I got a bit confused.
Consider the language L={+⋅,<,=,0,1} of PA.
For example, the sentence ϕ:=∀x∃y(y<x∧x=y+1) is a Π1-sentence because it is of the form ∀xθ, where θ=∃y(y<x∧x=y+1) is a formula having only bounded quantifiers.
On the other hand, in the sentence ψ:=∀x∃y(x=y+1), the formula (x=y+1) has no bounded quantifiers. Hence, what type of formula is ψ? How are formulas with no bounded quantifiers classified? In this case, θ and ∃y(x=y+1) are logically equivalent.
What about the sentence ∀x(x=x)? Is it Π0? If so, what about (x=x)?
A final example: Goodstein's Theorem (where the sequence is defined recursively starting with x in hereditary base 2, bumping the base by 1 and subtracting 1 from the result) can be expressed as the sentence G:=∀x∃y(gy(x)=0), where gy(x) denotes the yth term of the Goodstein sequence starting at x. I have read that G is a Π2-sentence. But the formula (gy(x)=0) does not have bounded quantifiers. Is it equivalent to a formula having only bounded quantifiers?
These examples describe the nature of my confusion. Any guidance will be appreciated.
Edit: As far as I can tell now, there are several approaches to defining the arithmetic hierarchy. At least three. This and this do a good job of exhibiting different approaches and hashing out the details and how they may be linked (e.g. Post's Theorem). Differences occur at the bottom level Σ0=Π0=Δ0. But crucially all approaches agree as one moves up the hierarchy.
Some references with various points of view:
Rogers, H. jun., Theory of recursive functions and effective computability, Maidenhead, Berksh.: McGraw-Hill Publishing Company, Ltd. 512 p. (1967). ZBL0183.01401. (See Ch. 14.1 & 14.7)
Avigad, Jeremy, Mathematical logic and computation, Cambridge: Cambridge University Press (ISBN 978-1-108-47875-5/hbk; 978-1-108-77875-6/ebook). xii, 513 p. (2023). ZBL1524.03001. (See Ch. 10.2)
Simpson, Stephen G., Subsystems of second order arithmetic, Perspectives in Mathematical Logic. Berlin: Springer. xiv, 444 p. (1999). ZBL0909.03048. (See Ch. I.7 - Definition I.7.8)
Schwichtenberg, Helmut; Wainer, Stanley S., Proofs and computations, Perspectives in Logic. Cambridge: Cambridge University Press; Urbana, IL: Association for Symbolic Logic (ASL) (ISBN 978-0-521-51769-0/hbk; 978-1-139-03190-5/ebook). xiii, 465 p. (2012). ZBL1294.03006. (See Ch. 2.6, 3.1 & the definition on p.151)
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edited Mar 27, 2024 at 15:34
John
asked Mar 26, 2024 at 16:10
JohnJohn
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gy(x)=0 is a computable relation, so is Σ1, which is all you need for G to be Π2.
– spaceisdarkgreen
Commented
Mar 26, 2024 at 20:29
I’m saying even if you use the bounded quantifier definition, computable relations (say, implemented by the beta function) are syntactically Σ1 so a ∀∃ of one is syntactically Π2.
– spaceisdarkgreen
Commented
Mar 26, 2024 at 21:23
@spaceisdarkgreen I see, thank you. Let me see if I understand you. For example, a sentence like ϕ:=∀x(0+x=x) is Π1 because θ:=(0+x=x) is Σ0 (=Π0=Δ0). And if I wanted to prove ϕ in IΣ0 I would use IΣ0-induction on θ. Am I correct? In any case, is there a reference you may recommend for learning these topics?
– John
Commented
Mar 26, 2024 at 21:40
you didn’t say anything incorrect there but I fail to see how it has anything to do with what I said. I don’t have anything in particular… intro books on computability theory tend to have treatments of arithmetical hierarchy, as do the “intro to godel’s theorem” variety
– spaceisdarkgreen
Commented
Mar 26, 2024 at 22:10
My point was just this: if ψ is Σ1, then ∀x∃yψ is Π2, by definition. In your post you're worried that you can't show ψ is Δ0... it isn't (under the bounded quantifiers only definition), but it doesn't need to be... all it needs to be is Σ1.
– spaceisdarkgreen
Commented
Mar 27, 2024 at 2:45
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"Only bounded quantifiers" should be understood as "all quantifiers are bounded", so a formula with no quantifiers at all would qualify vacuously as having "only bounded quantifiers".
So x=y+1 would have "only bounded quantifiers". It is Π0 (and Σ0).
Thus θ:=∃y(x=y+1) would be Σ1.
Now, in the logically equivalent formula θ′:=∃y(y<x∧x=y+1), the quantifier on y is bounded. Thus θ′ is Σ0 (and Π0).
Since θ and θ′ are logically equivalent, then θ is also Σ0 and Π0.
Usually we are interested in the least n which classifies a formula as either Σn or Πn.
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edited Mar 26, 2024 at 18:21
answered Mar 26, 2024 at 16:31
TedTed
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The way I learned that, but I'm aware that the notation differs between sources, is that Πn would be the form in which a formula is written, but Π0n would be the class "up to equivalence".
– Asaf Karagila
♦
Commented
Mar 26, 2024 at 17:39
@John Please read the answer again. I made some significant changes since I realized it doesn't really make sense to talk about classifying closed sentences.
– Ted
Commented
Mar 26, 2024 at 17:50
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"we don't usually classify closed sentences" I don't think so. Logicians discuss the complexity of sentences (like, the complexity of the continuum hypothesis is Σ21.)
– Hanul Jeon
Commented
Mar 26, 2024 at 18:12
@HanulJeon Ok, it seems like I may be somewhat confused, so I deleted the part about closed formulas.
– Ted
Commented
Mar 26, 2024 at 18:21
1
The possible reason for the confusion is that you reason semantically, but the formula complexity is a syntactic concept.
– Hanul Jeon
Commented
Mar 26, 2024 at 18:23
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7562 | https://artofproblemsolving.com/downloads/printable_post_collections/400595?srsltid=AfmBOoppP63x7vIx5dLUwKXfHMWyfgrI6VC5Fe5pnjavaaKhIpfu5bzN | AoPS Community 2017 CMIMC Number Theory
Problems from the 2017 CMIMC Number Theory Test
www.artofproblemsolving.com/community/c400595 by djmathman
1 There exist two distinct positive integers, both of which are divisors of 10 10 , with sum equal to
157 . What are they?
2 Determine all possible values of m + n, where m and n are positive integers satisfying
lcm( m, n ) − gcd( m, n ) = 103 .
3 For how many triples of positive integers (a, b, c ) with 1 ≤ a, b, c ≤ 5 is the quantity
(a + b)( a + c)( b + c)
not divisible by 4?
4 Let a1, a 2, a 3, a 4, a 5 be positive integers such that a1, a 2, a 3 and a3, a 4, a 5 are both geometric sequences and a1, a 3, a 5 is an arithmetic sequence. If a3 = 1575 , find all possible values of
|a4 − a2|.
5 One can define the greatest common divisor of two positive rational numbers as follows: for
a, b, c, and d positive integers with gcd( a, b ) = gcd( c, d ) = 1 , write
gcd
( ab , cd
)
= gcd( ad, bc )
bd .
For all positive integers K, let f (K) denote the number of ordered pairs of positive rational numbers (m, n ) with m < 1 and n < 1 such that
gcd( m, n ) = 1
K .
What is f (2017) − f (2016) ?
6 Find the largest positive integer N satisfying the following properties: -N is divisible by 7;-Swapping the ith and jth digits of N (for any i and j with i 6 = j) gives an integer which is not
divisible by 7.
©2019 AoPS Incorporated
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7 The arithmetic derivative D(n) of a positive integer n is defined via the following rules: - D(1) = 0 ;- D(p) = 1 for all primes p;- D(ab ) = D(a)b + aD (b) for all positive integers a and b.Find the sum of all positive integers n below 1000 satisfying D(n) = n.
8 Let N be the number of ordered triples (a, b, c ) ∈ { 1, . . . , 2016 }3 such that a2 + b2 + c2 ≡ 0(mod 2017) . What are the last three digits of N ?
9 Find the smallest prime p for which there exist positive integers a, b such that
a2 + p3 = b4.
10 For each positive integer n, define
g(n) = gcd {0! n!, 1!( n − 1)! , 2( n − 2)! , . . . , k !( n − k)! , . . . , n !0! } .
Find the sum of all n ≤ 25 for which g(n) = g(n + 1) .
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World population growth rate for 2022 was 0.87%, a 0.04% increase from 2021.
World population growth rate for 2021 was 0.83%, a 0.19% decline from 2020.
World population growth rate for 2020 was 1.02%, a 0.03% decline from 2019.
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【原】黄金分割在生活应用中熠熠生辉
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✅黄金分割（0.618）作为一种数学比例和美学法则，在生活、科学及人生规划中的确被广泛应用。以下是基于搜索结果的综合分析与案例佐证：
📍### 一、黄金分割的理论依据与生活优化原理
黄金分割的本质是将整体分为两部分，较长部分与整体的比值等于较短部分与较长部分的比值（约0.618）。这一比例被认为在自然界和人类感知中具有天然的和谐性。
数学与心理学基础：古希腊毕达哥拉斯学派通过观察铁匠打铁节奏首次总结出黄金分割规律，并将其视为宇宙和谐的数理表达。现代心理学研究表明，人类对接近0.618比例的事物（如矩形纸张、建筑结构）会感到更舒适愉悦。
优化方法论：在优选法中，黄金分割法（0.618法）被用于快速寻找最优解。例如，华罗庚推广的“0.618法”在工业生产中大幅减少试验次数，16次试验即可达到传统均分法2500次的精度。
📍### 二、黄金分割在生活中的实际应用案例
1. 人体健康与舒适度
体温与环境温度：人体最舒适的环境温度为23℃，接近正常体温37℃的0.618倍（37×0.618≈22.9℃），此时新陈代谢效率最高。
解剖结构：肚脐是身高的黄金分割点，喉部、膝关节、肘关节等位置均符合黄金比例，这些比例被用于医疗和运动康复设计（如针灸穴位定位）。
2. 艺术与设计
建筑美学：巴黎圣母院、埃及金字塔、埃菲尔铁塔等建筑的比例均严格遵循黄金分割，东方明珠电视塔的球体位置也位于塔高的0.618处。
摄影与绘画：达芬奇的《蒙娜丽莎》构图将人物面部和身体关键点置于黄金分割线上，现代摄影中主体常被放置在画面的0.618位置以增强视觉平衡。
3. 生产与决策优化
工业案例：某饮料公司通过黄金分割法优化化学成分添加量，仅需少量试验即可确定最佳配比，节约了90%的研发成本。
军事策略：成吉思汗的骑兵阵型中，重骑兵与轻骑兵的比例为2:3（≈0.618），这种配置在实战中实现了攻防效率最大化。
📍### 三、人生阶段的黄金分割规划与实证
1. 生命周期关键节点
以人均寿命为基准，黄金分割可划分人生重要阶段：
70岁模型：黄金分割点约43岁（70×0.618），对应事业巅峰期；进一步分割得到26岁（婚育黄金年龄）、16岁（学习关键期）等，与传统俗语“三岁看大，七岁看老”契合。
现代寿命模型（77.93岁）：黄金分割点约48岁（精力与事业顶峰期），次分割点为29岁（最佳婚龄）、18岁（高等教育期）等，与现代社会节奏相匹配。
2. 压力与调节
责任分配：43-44岁（传统模型）或48岁（现代模型）时，个体处于“上有老下有小”的高压期，通过黄金分割法可优化时间分配，例如将60%精力投入核心责任（如职业发展），40%用于家庭与自我调节。
📍### 四、科学研究的支持与局限性
1. 数据佐证
医学研究：情绪调节与黄金分割的关联性被用于中医情志疗法，例如通过释放负面情绪改善乳腺增生等疾病，相关成果获中国民族医药学会奖项。
社会统计：华罗庚推广优选法后，中国工业领域生产效率提升显著，案例覆盖化工、冶金等20余个行业。
2. 应用边界
黄金分割并非万能公式，其有效性需结合具体场景：
动态适应：如人均寿命延长后，黄金分割节点需重新计算；
个体差异：心理偏好、文化背景可能影响对黄金比例的感知。
📍### 五、总结与建议
黄金分割为生活优化提供了一种科学框架，但其核心价值在于启发人们通过比例思维平衡资源分配与目标优先级。实际应用中可结合以下策略：
关键决策：使用0.618法筛选最优选项（如职业选择、投资配比）；
时间管理：将60%时间投入高价值任务，剩余40%用于缓冲与创新；
健康调节：参考黄金比例安排运动、饮食与休息节奏。
需注意的是，黄金分割是工具而非教条，需灵活适配个人与社会变迁。
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7565 | https://stackoverflow.com/questions/54166546/formal-language-theory-regular-expressions-and-regular-languages-concept-of | Skip to main content
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Formal language theory (regular expressions and regular languages) - concept of "OR"
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Okay, so in programming the logical OR symbol (typically ||) when applied to operands a and b, that is, a || b, means that either a or b can be true, OR both can be true. If you want only one to be true, you use XOR (sometimes, the ^ symbol.)
However, in formal language theory, the concept of OR (typically the + symbol) seems to imply exclusive-or (xor) instead of regular OR. For example, if we describe a language L with a regular expression aa + bb + ab, a valid string (word) from the language would be one of those (aa, bb, or ab), not some concatenation of them. To do that, you must use the Kleene closure, as in (aa + bb + ab), right?
Perhaps I'm just thinking of + as being defined in a peculiar way, or perhaps it's that the operands are no longer Boolean?
I'm just looking for verification if I seem to be understanding that + (OR) has a seemingly different meaning in formal language / computational modeling than it does in programming languages. Thanks!
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asked Jan 13, 2019 at 6:19
GinzorfGinzorf
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The formal language OR is an inclusive ("regular") OR. E.g., the regular language ab + ab includes strings that are in both ab and ab (i.e., the string ab).
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answered Jan 13, 2019 at 23:59
Michael DyckMichael Dyck
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The problem is not with the operator - the + in regular expressions really does mean the same thing as union of sets - the problem is with your understanding of the operands. Specifically, in your regular expression, aa + bb + ab, aa does not represent a string over your alphabet, but a sub-regular expression. Regular expressions describe sets of strings; so the regular expression aa describes the set of strings {aa}. So, the regular expression aa + bb + ab describes the set of strings {aa} union {bb} union {ab} = {aa, bb, ab}. The exclusive-or of set theory, symmetric difference, does not have an operator in the regular expression syntax. We can recursively define the language of a regular expression, written L(r) for regular expression r, as follows:
L(r) = {r}, if r is a string over the alphabet;
L(r) = L(s)L(t) if r = st;
L(r) = L(s) if r = s;
L(r) = L(s) union L(t) if r = s + t.
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answered Feb 14, 2019 at 20:58
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7566 | https://k12.libretexts.org/Bookshelves/Mathematics/Trigonometry/03%3A_Trigonometric_Identities/3.04%3A_Double_and_Half_Angle_Identities/3.4.06%3A_Trigonometric_Equations_Using_Half_Angle_Formulas | Skip to main content
3.4.6: Trigonometric Equations Using Half Angle Formulas
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: Mar 27, 2022
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3.4.5: Half Angle Formulas
3.5: Sum to Product and Triple Angle Formulas
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( \newcommand{\kernel}{\mathrm{null}\,})
Simplifying all six trigonometric functions with half a given angle.
As you've seen many times, the ability to find the values of trig functions for a variety of angles is a critical component to a course in Trigonometry. If you were given an angle as the argument of a trig function that was half of an angle you were familiar with, could you solve the trig function?
For example, if you were asked to find
would you be able to do it? Keep reading, and in this section you'll learn how to do this.
using Half Angle Formulas on Trigonometric Equations
It is easy to remember the values of trigonometric functions for certain common values of . However, sometimes there will be fractional values of known trig functions, such as wanting to know the sine of half of the angle that you are familiar with. In situations like that, a half angle identity can prove valuable to help compute the value of the trig function.
In addition, half angle identities can be used to simplify problems to solve for certain angles that satisfy an expression. To do this, first remember the half angle identities for sine and cosine:
if is located in either the first or second quadrant.
if is located in the third or fourth quadrant.
if is located in either the first or fourth quadrant.
if is located in either the second or fourth quadrant.
When attempting to solve equations using a half angle identity, look for a place to substitute using one of the above identities. This can help simplify the equation to be solved.
Let's look at some problems that use the half angle formula.
Solve the trigonometric equation over the interval .
Then or , which is .
.
Solve for
To solve , first we need to isolate cosine, then use the half angle formula.
when
Solve for
To solve , first isolate tangent, then use the half angle formula.
using your graphing calculator, when
Example
Earlier, you were asked to solve sin .
Solution
Knowing the half angle formulas, you can compute easily:
Example
Find the exact value of
Solution
Example
Find the exact value of
Solution
Example
Find the exact value of .
Solution
Review
Use half angle identities to find the exact value of each expression.
Use half angle identities to help solve each of the following equations on the interval .
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.12.
Vocabulary
| Term | Definition |
--- |
| Half Angle Identity | A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument. |
3.4.5: Half Angle Formulas
3.5: Sum to Product and Triple Angle Formulas |
7567 | https://courses.lumenlearning.com/uvu-introductoryalgebra/chapter/8-1-definition-of-polynomials/ | 8.1: Polynomials in One Variable
Learning Outcomes
Define a polynomial
Determine the degree of a polynomial
Classify a polynomial as a monomial, binomial or trinomial
Determine the leading term and leading coefficient of a polynomial
Write a polynomial in standard form (descending order)
Key WORDS
Term: a number, variable, or product or quotient of numbers and variables
Coefficient: a number multiplied onto a variable; the number in a term
Algebraic expression: terms that are combined using operations like +,−,×,÷. etc.
Polynomial: an algebraic expression of the form anxn+an−1xn−1+...+a1x+a0
Monomial:a polynomial with one terms
Binomial: a polynomial with two terms
Trinomial: a polynomial with three terms
Degree: the largest exponent
Descending order: written from highest to lowest terms based on degree
Standard Form: a polynomial written in descending order
Leading term: the monomial with the highest degree
Leading coefficient: the coefficient of the leading term
Polynomials in One Variable
We previously defined a term to be a number, a variable, or the product or quotient of numbers and variables. For example, 6x,4x25y3,−4x3y2 are all terms. The coefficient of a term is the number in the term. So, for the list of examples just given, the coefficients are, 6,45,−4. When algebraic terms are combined using operations like +,−,×,÷. etc., they form an algebraic expression. For example, 5x3−34x2+6 is an algebraic expression containing three terms.
A polynomial in one variable is a special algebraic expression that consists of a term or a sum (or difference) of terms in which each term is a real number, a variable, or the product of a real number and a variable with a whole number exponent.
For example,
5x2+7x−3 is a polynomial with three terms: 5x2,7x,−3. The coefficient of x2 is 5; the coefficient of x is 7; and −3 is the constant term.
−43x9+8x4−35x2−9 is a polynomial with four terms: −43x9,8x4,−35x2,−9. The coefficient of x9 is −43; the coefficient of x4 is 8; the coefficient of x2 is 35; the constant term −9.
√5x3 and 4.563 are polynomials with just one term.
However, 3x,x12,4y2−4x3 are NOT polynomials because the exponents on the variables are not whole numbers. Whole number exponents exclude radicals (fractional exponents) and variables on the denominator of a fraction (negative exponents).
The following table is intended to help us tell the difference between what is a polynomial and what is not.
| | | |
---
| IS a Polynomial | Is NOT a Polynomial | Because |
| 2x2−12x−9 | 2x2+x | Polynomials only have variables in the numerator |
| y4−y3 | 2y+4 | Polynomials only have variables in the numerator |
| √12(a)+9 | √a+7 | Variables under a root are not allowed in polynomials |
The basic building block of a polynomial is a monomial. A monomial is a term of the form axn, where a is a constant and n is a whole number. A monomial can be a number, a variable, or the product of a number and variable with an exponent. The number part of the term is called the coefficient.
Examples of monomials:
constant: 6
variable with a coefficient of 1: x
product of a coefficient and a variable: 6x
product of a coefficient and a variable with a whole number exponent: 6x3
The coefficient can be any real number, including 0. The exponent of the variable must be a whole number (i.e., 0,1,2,3,...). The value of the exponent is the degree of the monomial. Remember that a variable, such as x, that appears to have no exponent really has an exponent of 1: x=x1. A monomial with no variable has a degree of 0. A number such as 3 can be written as 3x0, since x0=1 if [latex]x\neq0[/latex]. A monomial cannot have a variable in the denominator as that would have a negative exponent, nor can it have a variable under a radical as that would have a fractional exponent.
Example
Identify the coefficient, variable, and degree of the monomial:
1) 9
2) x
3) 35k8
Solution
1) 9 is a constant so there is no variable. Since 9=9x0, the degree is 0 and the coefficient is 9. 9 is called a constant term.
2) The variable is x.
The degree of x is 1, because x=x1.
The coefficient of x is 1, because x=1x.
3) The variable is k.
The degree of35k8 is 8.
The coefficient of k8 is 35.
Try It
Identify the coefficient, variable, and degree of the monomial:
6x7
−9x
47
Show Answer
Coefficient = 6; variable = x; degree = 7
2. Coefficient = −9; variable = x; degree = 1
3. Coefficient = 47; no variable; degree = 0
Classifying Polynomials
The word polynomial joins two diverse roots: the Greek poly, meaning “many”, and the Latin nomen, meaning “name.” Consequently, polynomial means many names, or in math, many terms. In this section, we will look at different ways that we classify polynomials. First, we will classify polynomials by the number of terms in the polynomial and then we will classify them by the largest exponent.
By Number of Terms
A polynomial is defined as a monomial or the sum (or difference) of monomials. This means that a polynomial that contains just one term is called a monomial. A polynomial containing two terms, such as 2x−9, is called a binomial. A polynomial containing three terms, such as −3x2+8x−7, is called a trinomial. After three terms we tend to simply refer to them as polynomials.
Polynomials
polynomial—A monomial; or two or more monomials, combined by addition (or subtraction)
monomial—A polynomial with exactly one term (“mono” means one)
binomial— A polynomial with exactly two terms (“bi” means two)
trinomial—A polynomial with exactly three terms (“tri” means three)
Here are some examples of polynomials classified by the number of terms.
| | | | |
--- --- |
| Polynomial | b+1 | 4y2−7y+2 | 5x5−4x4+x3+8x2−9x+1 |
| Monomial | 5 | 4b2 | −9x3 |
| Binomial | 3a−7 | y2−9 | 17x3+14x2 |
| Trinomial | x2−5x+6 | 4y2−7y+2 | 5a4−3a3+a |
Notice that every monomial, binomial, and trinomial is also a polynomial. in other words:
{monomials} ⊂ {binomials} ⊂ {trinomials} ⊂ {polynomials}
example
Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial:
8x2−7x−9
−5a4
x4−7x3−6x2+5x+2
11−4y3
n
Solution
| | Polynomial | Number of terms | Type |
--- --- |
| 1. | 8x2−7x−9 | 3 | Trinomial |
| 2. | −5a4 | 1 | Monomial |
| 3. | x4−7x3−6x2+5x+2 | 5 | Polynomial |
| 4. | 11−4y3 | 2 | Binomial |
| 5. | n | 1 | Monomial |
Example
For the following expressions, determine whether they are a polynomial. If so, categorize them as a monomial, binomial, or trinomial.
x−31−x+x2
t2+2t−3
x3+x8
√y2−y−1
Solution
x−31−x+x2 is not a polynomial because it violates the rule that polynomials cannot have variables in the denominator of a fraction (negative exponent).
t2+2t−3 is a polynomial because it is an expression whose monomial terms are connected by addition and subtraction. There are three terms in this polynomial so it is a trinomial.
x3+x8is a polynomial because it is an expression whose monomial terms are connected by addition and subtraction. There are two terms in this polynomial so it is a binomial.
√y2−y−1 is not a polynomial because it violates the rule that polynomials cannot have variables under a root (fractional exponent).
try it
The following video shows more examples of how to identify and categorize polynomials.
By Degree
We can find the degree of a one variable polynomial by identifying the highest power of the variable that occurs in the polynomial. Polynomials can be classified by the degree of the polynomial. The degree of a polynomial is the degree of its highest degree monomial term. So the degree of 2x3+3x2+8x+5 is 3.
A one variable polynomial is said to be written in standard form when the terms are arranged from the highest degree to the lowest degree. This is referred to as descending order. When it is written in standard form it is easy to determine the degree of the polynomial. The term with the highest degree is called the leading term because it is written first in standard form. The coefficient of the leading term is called the leading coefficient.
Degree of a ONE VARIABLE Polynomial
The degree of a term is the exponent of its variable.
The degree of a constant is 0.
The degree of a polynomial is the highest degree of all its terms.
When the coefficient of a polynomial term is 0, e.g., 0x2, we usually do not write the term at all because 0xn=0, and adding 0 does not change the value of the polynomial. However, we will see later, that occasionally, it is necessary to include all terms of the polynomial, including terms with zero coefficients.
A term without a variable is called a constant term, and the degree of that term is 0. This is because any constant a can be written as ax0, since x0=1. For example, we can write the polynomial 3x+13 as 3x1+13x0. Although this is not how we would normally write this, it allows us to see that 13 is the constant term because its degree is 0, and the degree of 3x is 1. Consequently, the degree of the binomial3x+13 is 1.
If a polynomial does not have a constant term, like the polynomial 14x3+3x we say that the constant term is 0.
Let’s see how this works by looking at several polynomials. We’ll take it step by step, starting with monomials, and then progressing to polynomials with more terms.
Remember: Any variable base written without an exponent has an implied exponent of 1, and any constant term has an implied exponent of zero.
| | | | | |
--- ---
| Monomials | 5 | 4b2 | −9x3 | −18 |
| Degree | 0 | 2 | 3 | 0 |
| | | | | |
| Binomials | b+17 | 3a−7 | y2−9 | 17x3+14x2 |
| Degree of each term | 1 0 | 1 0 | 2 0 | 3 2 |
| Degree of polynomial | 1 | 1 | 2 | 3 |
| | | | | |
| Trinomial | x2−5x+6 | 4y2−7y+2 | 5a12−3a9+a | 12x4+2x2−5 |
| Degree of each term | 2 1 0 | 2 1 0 | 12 9 1 | 4 2 0 |
| Degree of polynomial | 2 | 2 | 12 | 4 |
| | | | | |
| Polynomial | −7x8+9x6−7x4+x2−8 | 35x7−8x6+78x+9 | 5.4x17−8.1x13+2.8x10−6.5x5 | 7+6y−9y2+6y3−3y4+y5 |
| Degree of each term | 8 6 4 2 0 | 7 6 1 0 | 17 13 10 5 | 0 1 2 3 4 5 |
| Degree of polynomial | 8 | 7 | 17 | 5 |
example
Find the degree of the following polynomials:
4x
3x3−5x+7
−11
−6x2+9x−3
8x+2
Solution
| | |
--- |
| 1. | 4x |
| The exponent of x is one. x=x1 | The degree is 1. |
| 2. | 3x3−5x+7 |
| The highest degree of all the terms is 3. | The degree is 3 |
| 3. | −11 |
| The degree of a constant is 0. | The degree is 0. |
| 4. | −6x2+9x−3 |
| The highest degree of all the terms is 2. | The degree is 2. |
| 5. | 8x+2 |
| The highest degree of all the terms is 1. | The degree is 1. |
Working with polynomials is easier when we list the terms in descending order of degrees. When a polynomial is written this way, it is said to be in standard form. Look back at the polynomials in the previous example. Notice that they are all written in standard form. Get in the habit of writing polynomials in standard form
try it
Example
For the following polynomials, write them in standard form (descending order), identify the degree, the leading term, and the leading coefficient.
3+2x2−4x3
5t5+7t−2t3
6p−p3−2
Solution
Standard form: −4x32x2+3. The highest power of x is 3, so the degree is 3. The leading term is −4x3. The leading coefficient is the coefficient of that term, −4.
Standard form: 5t5−2t3+7t. The highest power of t is 5, so the degree is 5. The leading term is 5t5. The leading coefficient is the coefficient of that term, 5.
Standard form:−p3+6p−2. The highest power of p is 3, so the degree is 3. The leading term is −p3, The leading coefficient is the coefficient of that term, −1.
The following video shows how to identify the terms, leading coefficient, and degree of a polynomial.
Try It
For the following polynomials, write them in standard form (descending order), identify the degree, the leading term, and the leading coefficient.
9+2x2−5x5
−5t5+7+8t−4t3
4p4−p7−2p
Show Answer
Standard form: −5x5+2x2+9; Degree = 5; Leading term = −5x5; Leading coefficient = −5
Standard form: −5t5−4t3+8t+7; Degree = 5; Leading term = −5t5; Leading coefficient = −5
Standard form: −p7+4p4−2p; Degree = 7; Leading term = −p7; Leading coefficient = −1
Candela Citations
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Determine if an Expression is a Polynomial. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning.. Located at: License: CC BY: Attribution
Revision and Adaption. Authored by: Hazel McKenna and Roxanne Brinkerhoff. Provided by: Utah Valley University. License: CC BY: Attribution
Ex: Intro to Polynomials in One Variable. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Try It hjm334; hjm847. Authored by: Hazel McKenna. Provided by: Utah Valley University. License: CC BY: Attribution
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Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. Authored by: Monterey Institute of Technology and Education. License: CC BY: Attribution
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CC licensed content, Original
Determine if an Expression is a Polynomial. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning.. Located at: License: CC BY: Attribution
Revision and Adaption. Authored by: Hazel McKenna and Roxanne Brinkerhoff. Provided by: Utah Valley University. License: CC BY: Attribution
Ex: Intro to Polynomials in One Variable. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
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7568 | https://brainly.com/question/39146519 | [FREE] The solutions to the equation x^2 + px + q = 0 are x = 5 \pm \sqrt{3}. Find the values of p and q. - brainly.com
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The solutions to the equation x 2+p x+q=0 are x=5±3. Find the values of p and q.
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The values of p and q are 0 and −3 25, respectively.
Given the quadratic equation x 2+p x+q=0 with the roots x=35 and x=−35,[/tex] we need to determine the values of p and q.
Step 1: Use Vieta's Formulas
For a quadratic equation ax^2 + bx + c = 0 ) with roots ( \alpha ) and ( \beta \
The sum of the roots \alpha + \beta = -\frac{b}{a} \
The product of the roots \alpha \beta = \frac{c}{a} \
Here, [tex]α=35 and β=−35.
Step 2: Calculate the Sum and Product of the Roots
Sum of the roots:
α+β=35+(−35)=0
Since the sum is 0, -p = 0, so p = 0.
Product of the roots:
α β=(35)(−35)=−3 25
So, q=−3 25.[/tex]
Step 3: Values of p and q:
[tex]p=0
q=−3 25
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The values of p and q for the equation x 2+p x+q=0 with solutions x=5±3 are p=−10 and q=22.
Explanation
To find the values of p and q in the quadratic equation x 2+p x+q=0 with solutions x=5±3, we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots.
Step 1: Identify the Roots
The roots are given as:
r 1=5+3
r 2=5−3
Step 2: Calculate the Sum of the Roots
According to Vieta's formulas, the sum of the roots r 1+r 2 for the equation x 2+p x+q=0 is equal to −p:
r 1+r 2=(5+3)+(5−3)=10
This means that −p=10 or p=−10.
Step 3: Calculate the Product of the Roots
The product of the roots r 1⋅r 2 is given by Vieta's formulas and is equal to q:
r 1⋅r 2=(5+3)(5−3)=5 2−(3)2=25−3=22
Thus, we have q=22.
Final Values:
p=−10
q=22
Therefore, the values of p and q are p=−10 and q=22.
Examples & Evidence
For example, if the roots of a quadratic equation are 3 and 7, then using Vieta's formulas, we can find that the sum is 10 (which would equal -p) and the product is 21 (which would equal q). This method helps in solving various quadratic equations efficiently.
Vieta's formulas state that in a quadratic equation a x 2+b x+c=0, the sum of the roots α+β=−a b and the product of the roots α β=a c, which confirms the approach used to derive the values for p and q.
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7569 | http://www.freestudy.co.uk/thermodynamics/t15.pdf | (c) www.freestudy.co.uk Author D. J. Dunn 1 THERMODYNAMICS TUTORIAL 15 COMBUSTION This tutorial is set at QCF Levels 5 and 6 On completion of this tutorial you should be able to Analyse combustion processes in terms of stoichiometric, internal energy of reaction and enthalpy of reaction and formation. Apply First Law of thermodynamics to chemical reactions. Explain chemical dissociation and determine its effect in reactions involving perfect gases. Contents 1. Introduction 2. Combustion Chemistry 2.1 Solid and Liquid Fuels 2.2 Gaseous Fuels 3. Combustion by Mass 4. Combustion by Volume 5. Relationship between Product and Excess Air. 6. Energy Released by the Reaction 7. Dissociation Let's start by revising the basics. (c) www.freestudy.co.uk Author D. J. Dunn 2 1. Introduction Combustion is the process of chemical reaction between fuel and oxygen (reactants). The process releases heat and produces products of combustion. The main elements which burn are: Carbon Hydrogen Sulphur The heat released by 1 kg or m3 of fuel is called the calorific value. The oxygen used in combustion processes normally comes from the atmosphere and this brings nitrogen in with it which normally does nothing in the process but makes up the bulk of the gases remaining after combustion. The main elements in combustion are: Symbol Atomic Mass Molecular Mass Product Carbon C 12 CO2 Hydrogen H2 1 2 H2O Sulphur S 32 SO2 Oxygen O2 16 32 Nitrogen N2 14 28 If the water formed during combustion leaves as vapour, it takes with it the latent heat of evaporation and thus reduces the energy available from the process. In this case the calorific value is called the lower Calorific value (LCV). If the products cool down after combustion so that the vapour condenses, the latent heat is given up and the calorific value is then the higher calorific value (HCV). Solid and liquid fuels are normally analysed by mass to give the content of carbon, hydrogen, sulphur and any other elements present. Often there is silica, moisture and oxygen present in small quantities which have some effect on process. The silica leaves slaggy deposits on the heat transfer surfaces in boilers. Gaseous fuels are normally analysed by volumetric content and are in the main hydrocarbon fuels. For purposes of calculation, the content of air is considered to be Volumetric Gravimetric Oxygen 21% 23% Nitrogen 79% 77% The sulphur content of the fuel is considered to be a pollutant and so undesirable. The theoretically correct quantity of air or oxygen required to just exactly burn the fuel expressed as a ratio to the fuel burned, is called the Stoichiometric Ratio. In practice it is found that not all the oxygen in the reactant reaches the fuel elements and that excess air is required in order to ensure complete combustion. This results in oxygen appearing in the products. If too little air or oxygen is supplied, the result is incomplete combustion resulting in the formation of carbon monoxide CO instead of carbon dioxide CO2. The resulting products contain water H2O. Industrial equipment for measuring the contents of the products usually remove the water from the sample and the products are then called the dry products. (c) www.freestudy.co.uk Author D. J. Dunn 3 2. Combustion Chemistry 2.1 Solid and Liquid Fuels In the case of solid and liquid fuels, we do the combustion of each element separately. The important rule is that you must have the same number of atoms of each substance before and after the process. This may be obtained by juggling with the number of molecules. CARBON C + O2 = CO2 Mass ratio 12 + 32 = 44 Hence 1kg of C needs 32/12kg of O2 and makes 44/12kg of CO2 HYDROGEN 2H2 + O2 = 2H2O Mass ratio 4 + 32 = 36 Hence 1kg of H2 needs 8kg of O2 and makes 9 kg of H2O SULPHUR S + O2= SO2 32 + 32 = 64 Hence 1 kg of S needs 1kg of O2 and makes 2 kg of SO2. (c) www.freestudy.co.uk Author D. J. Dunn 4 2.2. Gaseous Fuels Typical hydrocarbons are : Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Pentane C5H12 Hexane C6H14 Heptane C7H16 Octane C8H18 Ethene C2H4(Ethylene) Propene C3H6 (Propylene) Ethyne C2H2 (Acetylene) Benzenol C6H6 (Benzene) Cyclohexane C6H12 The combustion equation follows the following rule If this results in fractional numbers of molecules, then the whole equation may be multiplied up. WORKED EXAMPLE No. 1 Write out the combustion equation for C8H18 SOLUTION 2C8H18 + 25O2 = 16CO2 +18H2O There are other gases which burn and the main one to know about is Carbon Monoxide (CO) which is partially burned carbon. The equation for the combustion of CO is 2CO + O2 = 2CO2 (c) www.freestudy.co.uk Author D. J. Dunn 5 3. Combustion by Mass The only rule to be observed in deducing the quantities of each substance is the law of conservation of mass. The proportions of the mass are that of the molecular masses. This is shown in the following example. WORKED EXAMPLE No. 2 A fuel contains by mass 88% C, 8%H2, 1%S and 3% ash (silica). Calculate the stoichiometric air. SOLUTION CARBON C + O2 = CO2 Mass ratio 12 + 32 = 44 Hence 0.88kg of C need (32/12) 0.88 = 2.347 kg of oxygen. It makes (44/12) 0.88 = 3.227 kg of carbon dioxide. HYDROGEN 2H2 + O2 = 2H2O Mass ratio 4 + 32 = 36 Hence 0.08kg of hydrogen needs (32/4) 0.08 = 0.64 kg of oxygen. SULPHUR S + O2= SO2 Mass ratio 32 + 32 = 64 Hence 0.01kg of sulphur needs 0.01kh of oxygen and makes 0.02kg of sulphur dioxide. Total Oxygen needed is 2.347 + 0.64 + 0.01 = 2.997kg Total Air needed is 2.997/23% = 13.03kg The Stoichiometric air/fuel ratio is 13.03/1 (c) www.freestudy.co.uk Author D. J. Dunn 6 WORKED EXAMPLE No. 3 If the air supplied is 20% more than the stoichiometric value, find the analysis of the dry products by mass. SOLUTION If 20% excess air is supplied then the air supplied is: 120% 13.03 = 15.637 kg Oxygen is also 20% excess so 0.2 2.997 = 0.599 kg is left over. Nitrogen in the air is 77% 15.637 = 12.04kg List of products Nitrogen 12.04 kg = 75.8% Carbon dioxide 3.227 kg = 20.3% Sulphur dioxide 0.02 kg = 0.1% Oxygen 0.599 kg = 3.8% Total dry product 15.886 kg = 100% It is of interest to note that for a given fuel, the % of any product is a direct indication of the excess air and in practice the carbon dioxide and/or oxygen is used to indicate this. This is important in obtaining optimal efficiency in a combustion process. (c) www.freestudy.co.uk Author D. J. Dunn 7 SELF ASSESSMENT EXERCISE No. 1 1. A boiler burns fuel oil with the following analysis by mass 80% C 18% H2 2% S 30% excess air is supplied to the process. Calculate the stoichiometric ratio by mass and the % Carbon Dioxide present in the dry products. (15.62/1 14.9% CO2) 2. A boiler burns coal with the following analysis by mass 75% C 15% H2 7%S remainder ash Calculate the % Carbon Dioxide present in the dry products if 20% excess air is supplied. (16.5% CO2) 3. Calculate the % of each dry product when coal is burned stoichiometrically in air. The analysis of the coal is 80% C 10% H2 5% S and 5% ash. (76.7%N, 22.5% CO2 0.8% SO2) (c) www.freestudy.co.uk Author D. J. Dunn 8 4. Combustion by Volume First we need to revise gas mixtures and understand the meaning of Volumetric Content. To do this we must understand Dalton's law of partial pressures and Avogadro's Law. First let us define the kmol. A kmol of substance is the number of kg numerically equal to the apparent molecular mass. For example 12 kg of Carbon is a kmol, so is 32 kg of O2 and 2 kg of H2 and 28 kg of N2. The molecular mass of a substance is expressed as kg/kmol so the molecular mass of O2 for example is 32 kg/kmol. Avogadro's Law states 1m3 of any gas at the same pressure and temperature contains the same number of molecules. It follows that the volume of a gas at the same p and T is directly proportional to the number of molecules. From this we find that the volume of a kmol of any gas is the same if p and T are the same. Dalton's law states: The total pressure of a mixture is the sum of the partial pressures. The partial pressure is the pressure each gas would exert if it alone occupied the same volume at the same temperature. Consider two gases A and B occupying a volume V at temperature T. Using the Universal gas law for each: Ñ is the relative molecular mass. pA and pB are the partial pressures. VA and VB are the partial volumes. These are the volumes each gas would occupy if they were separated and kept at the original p and T. This concept is very useful in problems involving the combustion of gases. It also follows that the partial volumes are directly related to the partial pressures so that Figure 1 (c) www.freestudy.co.uk Author D. J. Dunn 9 When not mixed the pressure is p and the volumes are VA and VB. Hence Since (1) = (2) then This shows that in a mixture, the partial volumes are in the same ratio as the kmol fractions which in turn are in proportion to the number of molecules of each gas. When mixed they both have volume V, hence: Consider the combustion of Methane. CH4 + 2O2= CO2 + 2H2O Since the volumetric content of each gas is in the same ratio as the kmol fractions then volumetric content is in the same proportion as the molecules. Hence it needs 2 volumes of oxygen to burn 1 volume of methane. The volume of air needed is 2 21% = 9.52 volumes. Hence it burn 1 m3 of methane we need 9.52 m3 of air for stoichiometric combustion. If the products are at the same p and T as the original reactants, we would obtain 1 m3 of carbon dioxide and 2 m3 of water vapour which would probably condense and cause a reduction in volume and/or pressure. (c) www.freestudy.co.uk Author D. J. Dunn 10 WORKED EXAMPLE No. 4 Calculate the % CO2 in the dry products when methane is burned with 15% excess air by volume. SOLUTION CH4 + 2O2= CO2 + 2H2O Volume ratio 1 2 1 2 The stoichiometric air is 2/21% = 9.524 m3 The actual air is 9.524 115% = 10.95 m3 Analysis of dry products Nitrogen 79% 10.95 8.65 m3 Carbon Dioxide 1.00 m3 Oxygen 15% 2 0.30 m3 Total 9.95 m3 The % Carbon Dioxide = (1/9.95) 100 = 10% When the fuel is a mixture of gases, the procedure outlined must be repeated for each combustible gas and the oxygen deduced for the volume of each in 1 m3 of total fuel. WORKED EXAMPLE No. 5 A fuel is a mixture of 60% Methane and 30% carbon monoxide and 10% oxygen by volume. Calculate the stoichiometric oxygen needed. SOLUTION As before, the volume of oxygen required to burn 1 m3 of methane is 2m3.To burn 0.6m3 needs 1.2m3 of oxygen. For carbon monoxide we use the combustion equation 2CO + O2 = 2CO2 Hence to burn 1 m3 of CO need 0.5 m3 of oxygen, so to burn 0.3 m3 needs 0.15 m3 of oxygen. The total oxygen needed is 1.2 + 0.15 = 1.35 m3 . However there is already 0.1 m3 in the fuel so the stoichiometric oxygen needed 1.25m3 (c) www.freestudy.co.uk Author D. J. Dunn 11 SELF ASSESSMENT EXERCISE No. 2 1. Find the air fuel ratio for stoichiometric combustion of Ethene by volume. (26.19/1) 2. Find the air fuel ratio for stoichiometric combustion of Butane by volume. (30.95/1). Calculate the % carbon dioxide present in the dry flue gas if 30% excess air is used. (10.6%) 3. Find the air fuel ratio for stoichiometric combustion of Propane by volume. (23.81/1) Calculate the % oxygen present in the dry flue gas if 20% excess air is used. (3.8%) 4. A gaseous fuel contains by volume : 5% CO2, 40% H2 , 40% CH4, 15% N2 Determine the stoichiometric air and the % content of each dry product. (4.76 m3, 89.7%,N2 10.3% CO2) (c) www.freestudy.co.uk Author D. J. Dunn 12 5. Relationship between Product and Excess Air. It follows that if we can deduce the % product then given the figure, we can work backwards to determine the air or oxygen that was used. WORKED EXAMPLE No. 6 Consider the combustion of methane again. CH4 + 2O2 = CO2 + 2H2O 1 vol 2 vol 1 vol 2 vols SOLUTION Let the excess air be x (as a decimal) The stoichiometric air is 9.52 vols. Actual air is 9.52(1 + x) Dry Products: Nitrogen 0.79 x 9.52(1 + x) = 7.524x + 7.524 Oxygen 2.000x Carbon Dioxide 1.000 Total 9.524x + 8.524 % Carbon monoxide = 100 {1/(9.524x + 8.524)} % Oxygen = 100{2/(9.524x + 8.524)} For example if the % CO2 is 10% then the excess air is found as follows 10% = 100 {1/(9.524x + 8.524)} 0.1 = 1/(9.524x + 8.524) (9.524x + 8.524) = 10 9.524x = 1.476 x = 0.155 or 15.5% Similarly if the O2 is 10% then the excess air is 81% (show this for yourself) If the analysis of the fuel is by mass, then a different approach is needed as in the following: (c) www.freestudy.co.uk Author D. J. Dunn 13 WORKED EXAMPLE No. 7 An analysis of the dry exhaust gas from an engine burning Benzole shows 15% Carbon Dioxide present by volume. The Benzole contains 90% C and 10% H2 by mass. Assuming complete combustion, determine the air/fuel ratio used. SOLUTION 1 kg of fuel contains 0.9kg of C and 0.1kg of H2. Converting these into kmol we have 0.9/12 kmol of C and 0.1/2 kmol of H2. For 1 kmol of dry exhaust gas we have 0.15 kmol of CO2 Y kmol of excess O2 1 - 0.15 - Y = 0.85 - Y kmol of N2 1 kmol of CO2 is 44 kg 1 kmol of N2 is 28 kg 1 kmol of O is 32 kg 0.15 kmol of CO2 is 0.15 x 44kg This contains (12/44) carbon so the carbon present is 0.15 12 kg The carbon in the fuel is 0.9 kmol per kmol of fuel. Hence the number of kmols of DEG must be 0.9/(0.15 12) = 0.5 There are 0.5 kmol of DEG for each kmol of fuel burned. The Nitrogen present in the DEG is 0.85 - Y kmol per kmol of DEG. This has a mass of 28(0.85 - Y) per kmol of DEG The oxygen supplied to the process must be (23.3/76.7) 28 (0.85 - Y) = 7.24 - 8.5Y kg per kmol of DEG. (Use precise proportions of air for accuracy). The oxygen contained within the carbon dioxide is: (32/44) 0.15 44 = 4.8 kg per kmol DEG 1 kmol of CO2 contains 44 kg and 32/44 of this is oxygen. The oxygen in the CO2 is hence 32 0.15 kg per kmol DEG. The excess oxygen is 32Y kg per kmol DEG Total oxygen in the products excluding that used to make H2O is 32 0.15 + 32Y The oxygen used to burn hydrogen is hence 7.24 - 8.5Y - 32 0.15 + 32Y O2 used to burn H2 is 2.44 - 40.5Y kg per kmol DEG (c) www.freestudy.co.uk Author D. J. Dunn 14 For 0.5 kmol this is 1.22 - 20.25Y kg To burn hydrogen requires oxygen in a ratio of 8/1. There is 0.1 kg of H2 in each kmol of fuel so 0.8 kg of O2 is needed. Hence 0.8 =1.22 - 20.25Y Y = (1.22 – 0.8)/20.25 = 0.0208kmol per kmol DEG The nitrogen in the DEG is 0.85 - Y = 0.829 kmol per kmol DEG The actual Nitrogen = 0.829 0.5 28 = 11.61 kg The air supplied must be 11.61/.767 = 15.14kg per kg of fuel. A simple calculation shows the stoichiometric mass of air is 13.73 so there is 10.3% excess air. SELF ASSESSMENT EXERCISE No. 3 1. C2H6 is burned in a boiler and the dry products are found to contain 8% CO2 by volume. Determine the excess air supplied. (59%) 2. The analysis of the dry exhaust gas from a boiler shows 10% carbon dioxide. Assuming the rest is oxygen and nitrogen; determine the excess air supplied to the process and the % excess air. The fuel contains 85% C and 15% H2 (21.5 kg , 44.5%) Now we will look at a complete example involving all the principles so far covered. (c) www.freestudy.co.uk Author D. J. Dunn 15 6. Energy Released by the Reaction The contents of the fuel and air or oxygen prior to combustion are called the reactants. The resulting material is called the products. The process releases energy but the amount of energy depends upon the temperature before and after the reaction. Consider a mixture of reactants at condition (1) which is burned and the resulting products are at condition (2). In order to solve problems we consider that the reactants are first cooled to a reference condition (0) by removing energy Q1. The reaction then takes place and energy is released. The products are then brought back to the same reference conditions (0) by removing energy Q2. The energy Q1 and Q2 are then returned so that the final condition of the products is reached (2). Figure 2 For constant volume combustion (closed system), we use Internal Energy. Balancing we have Up2 - UR1 = (URo - UR1) + (Upo - URo) + (Up2 - Upo) The energy released by combustion is in this case the Internal Energy of combustion and this occurs at standard conditions of 1 bar and 25oC. This pressure is designated p and the internal energy of combustion is designated U. When this is based on 1 kmol it is designated u Up2 - UR1 = (URo - UR1) + Uo + (Up2 - Upo) The standard conditions chosen for the combustion are 1 bar and 25oC. At this temperature the internal energy of all gases is the same (-2 479 kJ/kmol). The figure is negative because the zero value of internal energy arbitrarily occurs at a higher temperature. If the process is conducted in a steady flow system, enthalpy is used instead of internal energy. The reasoning is the same but U is replaced by H. Hp2 - HR1 = (HRo - HR1) + Ho + (Hp2 - Hpo) ho may be found in the thermodynamic tables for some fuels. The figures are quoted in kJ per kmol of substance. For the products In terms of kmol fractions hpo = upo + npRoTo For the reactants In terms of kmol fractions hRo = uRo + nRRoTo Where n is the kmols. (c) www.freestudy.co.uk Author D. J. Dunn 16 ho = (upo + npRoTo) - (uRo + nRRoTo ) ho = (upo - uRo ) - nRRoTo + npRoTo ho = (upo - uRo ) + (np - nR)RoTo uo = ho + (np - nR)RoTo If the combustion produces equal numbers of kmols before and after, the pressure would be constant (assuming constant volume and no condensation). np = nR so ho is the same as the internal energy of reaction o. For example consider the combustion of ethylene. C2H4 + 3O2 = 2CO2 + 2H2O In this case there are 4 kmols before and after . When this occurs, we may use the specific heat cp to solve the problems. uo= ho = cpT The specific heats are listed in the thermodynamic tables. Note that in order to make the method of solution conform to standard data, the combustion equations should always be based on 1 kmol of fuel. The heat transfer Q1 is found either by use of the mean specific heat or by looking up the enthalpy of the gas at the required temperatures (enthalpy of formation) and deducing the change. In general for a constant volume we should use uo and Cv to solve problems. For constant pressure with no work being done (e.g. a combustion chamber) we should use ho and Cp. Since tables only list ho and Cp we may find uo = ho + (np - nR)RoTo where is 298.1 K and Ro is 8.314 kJ/kmol K and n is the number of kmols of product. Cv = Cp – R = Cp – Ro/molecular mass (c) www.freestudy.co.uk Author D. J. Dunn 17 WORKED EXAMPLE No. 8 1. A vessel contains 0.2 m3 of C2H4O and oxygen in its stoichiometric ratio. The mixture is at 1 bar pressure and 25oC. The mixture is ignited and allowed to cool back to 25oC. Determine i. The final pressure. ii. The amount of condensate formed. iii. The heat transfer. iv. The enthalpy of reaction per kmol of C2H4O. The enthalpy of formation (ho ) for the gases involved is shown below for a temperature of 298 K. Molecular mass. Enthalpy of reaction kg/kmol (kJ/kmol) CO2(gas) 44 - 393 520 H2O(gas) 18 - 241 830 H2O(liquid) 18 -285 820 O2(gas) 32 0 C2H4O 44 -52 630 (c) www.freestudy.co.uk Author D. J. Dunn 18 SOLUTION C2H4O + 2½O2 = 2CO2 + 2H2O mass ratio 44 80 88 36 kmol ratio 1 2½ 2 1 p = 1bar V = 0.2 m3 T = 298 K V = 0.2m3 = Vf + Vox = 3½ Volumes p = 1 bar = pf + pox pf = 1 (1/3.5) = 0.2857 bar pox = 1 (2.5/3.5) =0.7143 bar Mass of fuel Mass of oxygen Total mass = 0.286 kg Mass of CO2 = (88/124) 0.286 = 0.203 kg Mass of H2O = (36/124) 0.286 = 0.083 kg Total mass = 0.286 kg Since condensate forms, the gas is saturated with water vapour. Hence p(vapour) = pg @ 28oC from the steam tables. p(vapour) = 0.03166 bar Total pressure = 0.5716 + 0.03166 = 0.603 bar (answer (i) Mass of vapour = V/vg Where vg = 43.4 m3/kg @ 28oC from tables. Mass of vapour = 0.2/43.4 = 0.004608 kg Mass of condensate formed = 0.083 - 0.0046 = 0.0784 kg Answer (ii) Now consider the reaction. Since it starts and finishes at 25oC there is no initial cooling required (Q1= 0). (c) www.freestudy.co.uk Author D. J. Dunn 19 REACTANT Fuel Mass = 0.1015 kg kmol = 0.1015/44 = 0.00231 kmol Oxygen Mass = 0.1845 kg kmol = 0.1845/32 = 0.00576 kmol Enthalpy of formation for oxygen = 0 Enthalpy of formation for C2H4O = -52 630 kJ/kmol Hf = 0.00231(-52 630) = -121.4 kJ (Minus relative to higher point of reference) PRODUCTS CO2 kmol = 0.203/44 = 0.00461 kmol Hf = 0.00461 x (-393 520) = - 1 815.6 kJ H2O (gas) kmol = 0.0408/18 = 0.00227 kmol Hf = 0.00227 x (-241 830) = - 548.1 kJ H2O (water) kmol = 0.0784/18 = 0.00436 kmol Hf = 0.00436 (-285 820) = - 1 244.9 kJ Total = -3 608 kJ The change in enthalpy = -3 608 - (-121.4) = -3 486.6 kJ = Ho Answer (iv) ho = -3 486.6/0.00231 =1.509 GJ/kmol Answer (v) (c) www.freestudy.co.uk Author D. J. Dunn 20 WORKED EXAMPLE No. 9 Air and Ethane (C2H6) are mixed with twice the stoichiometric ratio at 600K in a vessel at 12 bar pressure. Determine the temperature and pressure after combustion assuming no energy losses. The enthalpy of combustion at 25oC is Ho = - 1 427 860 kJ/kmol SOLUTION C2H6 + 3.5 O2 = 3H2O + 2CO2 kmol 1 3.5 3 2 The air required = 3.5/0.21 = 16.67 kmol Actual air = 33.33 kmol Nitrogen = 0.79 33.33 = 26.33 kmol Excess oxygen = 3.5 kmol The equation may be rewritten as C2H6 + 7O2 + 26.33 N2 = 3H2O + 2CO2 + 3.5 O2 + 26.33 N2 The process may be idealised as follows First find the enthalpy of the reactants. The mean temperature of the reactants relative to 25oC is {(25+273) + 600}/2 = 449K near enough 450K for the tables. We look up specific heats in the thermodynamic tables at 450 K. The temperature change from 25oC to 600 K is 302 K. We proceed to work out the heat transfer based on 1 kmol of fuel, Q1 as follows. Table of values C2H6 Cp = 2.402 kJ/kg K mass = 1 kmol 30 = 30 kg Q1 = 30 2.402 302 = 21 762 kJ O2 Cp = 0.956 kJ/kg K mass = 7 kmol 32 = 224 kg Q1 = 224 0.956 302 = 64 671.5 kJ N2 Cp = 1.049 kJ/kg K mass = 26.33 kmol 28 = 737.24 kg Q1 = 737.24 1.049 302 = 233 556 kJ Total Q1 =- 319 989.7 kJ per kmol of fuel (negative leaving system) Next we repeat the process for the products to find Q1 + Q2 In order to use a mean specific heat we must guess the approximate final temperature of the products. A good guess is always 2000 K so the mean of 25oC and 2000 K is near enough 1150 K. Using this we work out the heat transfer to the products with an unknown temperature change from 25oC to T2 of T. (c) www.freestudy.co.uk Author D. J. Dunn 21 H2O Cp = 2.392 kJ/kg K mass = 3 kmol x 16 = 48 kg Q = 48 2.392 T = 114.8 T O2 Cp = 1.109 kJ/kg K mass = 3.5 kmol x 32 = 112 kg Q1 = 112 1.109 T = 124.2 T N2 Cp = 1.196 kJ/kg K mass = 26.33 kmol 28 = 737.24 kg Q = 737.24 1.196 T = 881.7 T CO2 Cp = 1.270 kJ/kg K mass = 2 kmol 44 = 88 kg Q = 88 1.270 x T = 111.76 T Total Q1 + Q2 = 1 232.5 T kJ per kmol of fuel (positive entering system) Q2 = - 1 427 860 kJ/kmol of fuel (from question). Conserving energy we have 1 232.5 T = 1 427 860 + 319 989.7 Hence T = 1 232.5 K and T2 = 1 716 K which is different from our original guess of 2000 K but more accurate. Next we must repeat the last stage with a more accurate mean temperature. Mean temperature = (298 + 1716)/2 = 1007 K. Say 1 000 K. H2O Cp = 2.288 kJ/kg K Q = 48 2.288 T = 109.8T O2 Cp = 1.090 kJ/kg K Q1 = 112 1.09 T = 122.1T N2 Cp = 1.167 kJ/kg K Q = 737.24 1.167 x T = 860.4T CO2 Cp = 1.234 kJ/kg K Q = 88 1.234 T = 108.6T Total Q1 + Q2 = 1 201T kJ per kmol of fuel (positive entering system) Conserving energy we have 1 201T = 1 427 860 + 319 989.7 T = 1 455 K and T2 = 1 753 K which is different from our original guess of 1 716K but more accurate. The true answer is between 1 716 and 1 753 K and may be narrowed down by making more steps but two is usually sufficient. Finally the pressure. p1V1 / N1T1 =Ro = p2V2 / N2T2 and the volumes are equal. p1=12 bar T1 = 600 K N1= 4.5 kmol T2 = 1753 K N2= 5 kmol p2 = 12 5 1753/(4.5 600) = 38.9 bar (c) www.freestudy.co.uk Author D. J. Dunn 22 SELF ASSESSMENT EXERCISE No. 4 1. The gravimetric analysis of a fuel is Carbon 78%, hydrogen 12%, oxygen 5% and ash 5%. The fuel is burned with 20% excess air. Assuming complete combustion, determine i. the composition of the products. (72.6% N2, 17.3% CO2, 6.5%H2O and 3.6% O2) ii. the dew point of the products. (47oC) iii. the mass of water condensed when the products are cooled to 30oC. (0.67 kg) 2. Carbon monoxide is burned with 25% excess oxygen in a fixed volume of 0.2 m3. The initial and final temperature is 25oC. The initial pressure is 1 bar. Calculate i. the final pressure.(0.874 bar) ii. the heat transfer. (574.5 kJ) Use your thermodynamic tables for enthalpies of reaction. 3. Prove that the enthalpy and the internal energy of reaction are related by Ho= Uo + RoTo(nP - nR) where nP and nR are the kmols of products and reactants respectively. Ethylene (C2H4) and 20% excess air at 77oC are mixed in a vessel and burned at constant volume. Determine the temperature of the products. You should use your thermodynamic tables to find Uo or Ho and the table below. (Answer 2263 K) T(K) C2H4 O2 N2 CO2 H2O U (kJ/kmol) 298.15 - 2 479 -2 479 -2 479 -2 479 -2 479 300 - 2 415 -2 440 -2 440 -2 427 -2 432 400 -1 557 -297 -355 683 126 2 400 54 537 50 696 95 833 73 650 2 600 60 657 50 696 95 833 73 650 2 800 66 864 62 065 117 160 92 014 3 000 73 155 67 795 127 920 101 420 4. An engine burns hexane (C6H14) in air. At a particular running condition the volumetric analysis of the dry products are CO2 8.7% CO 7.8 % N2 83.5% Calculate the air-fuel ratio by mass and find the stoichiometric ratio. (Answer 12.9 and 15.17) (c) www.freestudy.co.uk Author D. J. Dunn 23 7. Dissociation At the high temperatures and pressures experienced in combustion, dissociation occurs. This results in some of the fuel not burning. CO is produced and in the case of hydrogen, some of it remains as hydrogen after the process even though oxygen is present. The reasons for this will not be covered here other than to say it is predicted by the 2nd law of thermodynamics and involves equilibrium in the chemical process. When dissociation occurs, the energy released is reduced accordingly and if the amount of unburned fuel is known the previous examples may easily be modified to take account of it. When hydrogen is burned, it can be shown that the partial pressures of the hydrogen, oxygen and water vapour present in the products are related by The properties tables list values of ln k. Similarly when dissociation occurs in the formation of carbon dioxide, the relationship between the partial pressures of CO2, CO and O2 is given by Other similar equations for other combinations of products may be found in the tables. (c) www.freestudy.co.uk Author D. J. Dunn 24 SELF ASSESSMENT EXERCISE No. 5 1. Hydrogen is mixed with stoichiometric air at 25oC and burned adiabatically at constant volume. After combustion 6% of the hydrogen remains unburned. Determine the temperature and pressure of the products. (Answer the temperature is 2344K after two approximations) You need to find K in the tables. Also find Ho = 241 800 kJ/kmol. Deduce the partial pressures of the products as a fraction of p and then use K to solve p. 2. A mixture of air and CO is burned adiabatically at constant volume. The air is 90% of the stoichiometric requirement. The mixture is initially at 5 bar and 400K. The only dissociation that occurs is CO2CO + ½O2. Show that the equilibrium constant at the final temperature T2 is a is the amount of CO2 kmol in the products per kmol of CO in the reactants. If it assumed that initially T2 = 2 900 K for which log Kp = 0.649, the solution of the above equation gives a = 0.784. Check that at the assumed value of T2 given that the internal energy of reaction at T0 = 298.15 K is -281 750 kJ/kmol. T (K) U kJ/kmol CO O2 N2 CO2 298.15 -2 479 -2 479 -2 479 -2 479 400 - 351 - 297 - 355 + 683 2900 +65 715 +69 999 +64 924 +122 530 |
7570 | https://opencw.aprende.org/courses/mechanical-engineering/2-71-optics-spring-2009/video-lectures/lecture-2-reflection-and-refraction-prisms-waveguides-and-dispersion/ | Lecture 2: Reflection and refraction; prisms, waveguides, and dispersion | Video Lectures | Optics | Mechanical Engineering | MIT OpenCourseWare
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Topics: Laws of reflection and refraction; prisms; dispersion; paraboloidal reflector.
Instructors: George Barbastathis, Colin Sheppard
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7571 | https://dn790008.ca.archive.org/0/items/chem-7-zumdahl/Zumdahl_Text.pdf | Chemistry Seventh Edition Steven S. Zumdahl University of Illinois Susan A. Zumdahl University of Illinois Houghton Mifflin Company Boston New York Executive Editor: Richard Stratton Developmental Editor: Rebecca Berardy Schwartz Senior Project Editor: Cathy Labresh Brooks Editorial Assistant: Susan Miscio Senior Art & Design Coordinator: Jill Haber Composition Buyer: Chuck Dutton Manufacturing Coordinator: Renee Ostrowski Senior Marketing Manager: Katherine Greig Marketing Assistant: Naveen Hariprasad Cover image: Masaaki Kazama/Photonica Photo credits: Page A39.
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Library of Congress Catalog Card Number: 2005929890 Student edition: ISBN 13: 978-0-618-52844-8 ISBN 10: 0-618-52844-X Instructor’s Annotated Edition: ISBN 13: 978-0-618-52845-5 ISBN 10: 0-618-52845-8 Advanced Placement edition: ISBN 13: 978-0-618-71370-7 ISBN 10: 0-618-71370-0 123456789-WEB-09 08 07 06 05 iii 3.3 The Mole 82 ■CHEMICAL IMPACT Elemental Analysis Catches Elephant Poachers 84 3.4 Molar Mass 86 ■CHEMICAL IMPACT Measuring the Masses of Large Molecules, or Making Elephants Fly 87 3.5 Percent Composition of Compounds 89 3.6 Determining the Formula of a Compound 91 3.7 Chemical Equations 96 3.8 Balancing Chemical Equations 98 3.9 Stoichiometric Calculations: Amounts of Reactants and Products 102 ■CHEMICAL IMPACT High Mountains—Low Octane 103 3.10 Calculations Involving a Limiting Reactant 106 For Review 113 • Key Terms 113 • Questions and Exercises 115 4 Types of Chemical Reactions and Solution Stoichiometry 126 4.1 Water, the Common Solvent 127 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 129 ■CHEMICAL IMPACT Arrhenius: A Man with Solutions 132 4.3 The Composition of Solutions 133 ■CHEMICAL IMPACT Tiny Laboratories 138 4.4 Types of Chemical Reactions 140 4.5 Precipitation Reactions 140 4.6 Describing Reactions in Solution 145 4.7 Stoichiometry of Precipitation Reactions 147 4.8 Acid–Base Reactions 149 4.9 Oxidation–Reduction Reactions 154 ■CHEMICAL IMPACT Iron Zeroes in on Pollution 156 ■CHEMICAL IMPACT Pearly Whites 159 ■CHEMICAL IMPACT Aging: Does It Involve Oxidation?
160 4.10 Balancing Oxidation–Reduction Equations 162 For Review 168 • Key Terms 168 • Questions and Exercises 170 Contents To the Professor ix To the Student xv 1 Chemical Foundations 1 1.1 Chemistry: An Overview 2 ■CHEMICAL IMPACT The Chemistry of Art 4 1.2 The Scientific Method 5 ■CHEMICAL IMPACT A Note-able Achievement 7 ■CHEMICAL IMPACT Critical Units!
8 1.3 Units of Measurement 8 1.4 Uncertainty in Measurement 10 1.5 Significant Figures and Calculations 13 1.6 Dimensional Analysis 16 1.7 Temperature 19 ■CHEMICAL IMPACT Faux Snow 22 1.8 Density 24 1.9 Classification of Matter 25 For Review 29 • Key Terms 29 • Questions and Exercises 30 2 Atoms, Molecules, and Ions 38 2.1 The Early History of Chemistry 39 ■CHEMICAL IMPACT There’s Gold in Them There Plants!
40 2.2 Fundamental Chemical Laws 41 2.3 Dalton’s Atomic Theory 43 2.4 Early Experiments to Characterize the Atom 45 ■CHEMICAL IMPACT Berzelius, Selenium, and Silicon 46 2.5 The Modern View of Atomic Structure: An Introduction 49 ■CHEMICAL IMPACT Reading the History of Bogs 51 2.6 Molecules and Ions 52 2.7 An Introduction to the Periodic Table 55 ■CHEMICAL IMPACT Hassium Fits Right in 57 2.8 Naming Simple Compounds 57 For Review 67 • Key Terms 67 • Question and Exercises 69 3 Stoichiometry 76 3.1 Counting by Weighing 77 3.2 Atomic Masses 78 ■CHEMICAL IMPACT Buckyballs Teach Some History 80 iv 5 Gases 178 5.1 Pressure 179 5.2 The Gas Laws of Boyle, Charles, and Avogadro 181 5.3 The Ideal Gas Law 186 5.4 Gas Stoichiometry 190 5.5 Dalton’s Law of Partial Pressures 194 ■CHEMICAL IMPACT Separating Gases 196 ■CHEMICAL IMPACT The Chemistry of Air Bags 197 5.6 The Kinetic Molecular Theory of Gases 199 5.7 Effusion and Diffusion 206 5.8 Real Gases 208 5.9 Characteristics of Several Real Gases 210 5.10 Chemistry in the Atmosphere 211 ■CHEMICAL IMPACT Acid Rain: A Growing Problem 212 For Review 215 • Key Terms 215 • Questions and Exercises 217 6 Thermochemistry 228 6.1 The Nature of Energy 229 6.2 Enthalpy and Calorimetry 235 ■CHEMICAL IMPACT Nature Has Hot Plants 238 ■CHEMICAL IMPACT Firewalking: Magic or Science?
241 6.3 Hess’s Law 242 6.4 Standard Enthalpies of Formation 246 6.5 Present Sources of Energy 252 6.6 New Energy Sources 256 ■CHEMICAL IMPACT Farming the Wind 258 ■CHEMICAL IMPACT Veggie Gasoline?
262 For Review 264 • Key Terms 264 • Questions and Exercises 265 7 Atomic Structure and Periodicity 274 7.1 Electromagnetic Radiation 275 ■CHEMICAL IMPACT Flies That Dye 277 7.2 The Nature of Matter 277 ■CHEMICAL IMPACT Chemistry That Doesn’t Leave You in the Dark 280 ■CHEMICAL IMPACT Thin Is In 282 7.3 The Atomic Spectrum of Hydrogen 284 7.4 The Bohr Model 285 ■CHEMICAL IMPACT Fireworks 288 7.5 The Quantum Mechanical Model of the Atom 290 7.6 Quantum Numbers 293 7.7 Orbital Shapes and Energies 295 7.8 Electron Spin and the Pauli Principle 296 7.9 Polyelectronic Atoms 298 7.10 The History of the Periodic Table 299 ■CHEMICAL IMPACT The Growing Periodic Table 302 7.11 The Aufbau Principle and the Periodic Table 302 7.12 Periodic Trends in Atomic Properties 309 7.13 The Properties of a Group: The Alkali Metals 314 ■CHEMICAL IMPACT Potassium—Too Much of a Good Thing Can Kill You 317 For Review 318 • Key Terms 318 • Questions and Exercises 320 8 Bonding: General Concepts 328 8.1 Types of Chemical Bonds 330 ■CHEMICAL IMPACT No Lead Pencils 332 8.2 Electronegativity 333 8.3 Bond Polarity and Dipole Moments 335 8.4 Ions: Electron Configurations and Sizes 338 8.5 Energy Effects in Binary Ionic Compounds 342 8.6 Partial Ionic Character of Covalent Bonds 346 8.7 The Covalent Chemical Bond: A Model 347 8.8 Covalent Bond Energies and Chemical Reactions 350 8.9 The Localized Electron Bonding Model 353 8.10 Lewis Structures 354 ■CHEMICAL IMPACT Nitrogen Under Pressure 358 8.11 Exceptions to the Octet Rule 358 8.12 Resonance 362 8.13 Molecular Structure: The VSEPR Model 367 ■CHEMICAL IMPACT Chemical Structure and Communication: Semiochemicals 378 For Review 380 • Key Terms 380 • Questions and Exercises 382 v 9 Covalent Bonding: Orbitals 390 9.1 Hybridization and the Localized Electron Model 391 9.2 The Molecular Orbital Model 403 9.3 Bonding in Homonuclear Diatomic Molecules 406 9.4 Bonding in Heteronuclear Diatomic Molecules 412 9.5 Combining the Localized Electron and Molecular Orbital Models 413 ■CHEMICAL IMPACT What’s Hot?
414 For Review 416 • Key Terms 416 • Questions and Exercises 417 10 Liquids and Solids 424 10.1 Intermolecular Forces 426 10.2 The Liquid State 429 10.3 An Introduction to Structures and Types of Solids 430 ■CHEMICAL IMPACT Smart Fluids 434 10.4 Structure and Bonding in Metals 436 ■CHEMICAL IMPACT Seething Surfaces 438 ■CHEMICAL IMPACT Closest Packing of M & Ms 441 ■CHEMICAL IMPACT What Sank the Titanic?
443 10.5 Carbon and Silicon: Network Atomic Solids 444 ■CHEMICAL IMPACT Golfing with Glass 449 ■CHEMICAL IMPACT Transistors and Printed Circuits 452 10.6 Molecular Solids 454 ■CHEMICAL IMPACT Explosive Sniffer 455 10.7 Ionic Solids 456 10.8 Vapor Pressure and Changes of State 459 10.9 Phase Diagrams 467 ■CHEMICAL IMPACT Making Diamonds at Low Pressures: Fooling Mother Nature 470 For Review 472 • Key Terms 472 • Questions and Exercises 474 11 Properties of Solutions 484 11.1 Solution Composition 485 ■CHEMICAL IMPACT Electronic Ink 488 11.2 The Energies of Solution Formation 488 11.3 Factors Affecting Solubility 492 ■CHEMICAL IMPACT Ionic Liquids?
494 ■CHEMICAL IMPACT The Lake Nyos Tragedy 497 11.4 The Vapor Pressures of Solutions 497 ■CHEMICAL IMPACT Spray Power 500 11.5 Boiling-Point Elevation and Freezing-Point Depression 504 11.6 Osmotic Pressure 508 11.7 Colligative Properties of Electrolyte Solutions 512 ■CHEMICAL IMPACT The Drink of Champions— Water 514 11.8 Colloids 514 ■CHEMICAL IMPACT Organisms and Ice Formation 516 For Review 516 • Key Terms 516 • Questions and Exercises 518 12 Chemical Kinetics 526 12.1 Reaction Rates 527 12.2 Rate Laws: An Introduction 532 12.3 Determining the Form of the Rate Law 534 12.4 The Integrated Rate Law 538 12.5 Rate Laws: A Summary 548 12.6 Reaction Mechanisms 549 12.7 A Model for Chemical Kinetics 552 12.8 Catalysis 557 ■CHEMICAL IMPACT Automobiles: Air Purifiers?
560 ■CHEMICAL IMPACT Enzymes: Nature’s Catalysts 562 For Review 564 • Key Terms 564 • Questions and Exercises 566 13 Chemical Equilibrium 578 13.1 The Equilibrium Condition 579 13.2 The Equilibrium Constant 582 13.3 Equilibrium Expressions Involving Pressures 586 13.4 Heterogeneous Equilibria 588 13.5 Applications of the Equilibrium Constant 591 vi 13.6 Solving Equilibrium Problems 600 13.7 Le Châtelier’s Principle 604 For Review 610 • Key Terms 610 • Questions and Exercises 613 14 Acids and Bases 622 14.1 The Nature of Acids and Bases 623 14.2 Acid Strength 626 14.3 The pH Scale 631 ■CHEMICAL IMPACT Arnold Beckman, Man of Science 632 14.4 Calculating the pH of Strong Acid Solutions 634 14.5 Calculating the pH of Weak Acid Solutions 635 ■CHEMICAL IMPACT Household Chemistry 643 14.6 Bases 644 ■CHEMICAL IMPACT Amines 648 14.7 Polyprotic Acids 650 14.8 Acid–Base Properties of Salts 655 14.9 The Effect of Structure on Acid–Base Properties 661 14.10 Acid–Base Properties of Oxides 662 14.11 The Lewis Acid–Base Model 663 ■CHEMICAL IMPACT Self-Destructing Paper 666 14.12 Strategy for Solving Acid–Base Problems: A Summary 666 For Review 668 • Key Terms 668 • Questions and Exercises 672 15 Applications of Aqueous Equilibria 680 Acid–Base Equilibria 681 15.1 Solutions of Acids or Bases Containing a Common Ion 681 15.2 Buffered Solutions 684 15.3 Buffering Capacity 693 15.4 Titrations and pH Curves 696 15.5 Acid–Base Indicators 711 Solubility Equilibria 717 15.6 Solubility Equilibria and the Solubility Product 771 ■CHEMICAL IMPACT The Chemistry of Teeth 720 15.7 Precipitation and Qualitative Analysis 724 Complex Ion Equilibria 731 15.8 Equilibria Involving Complex Ions 731 For Review 736 • Key Terms 736 • Questions and Exercises 739 16 Spontaneity, Entropy, and Free Energy 748 16.1 Spontaneous Processes and Entropy 749 16.2 Entropy and the Second Law of Thermodynamics 755 ■CHEMICAL IMPACT Entropy: An Organizing Force?
756 16.3 The Effect of Temperature on Spontaneity 756 16.4 Free Energy 759 16.5 Entropy Changes in Chemical Reactions 762 16.6 Free Energy and Chemical Reactions 766 16.7 The Dependence of Free Energy on Pressure 770 16.8 Free Energy and Equilibrium 774 16.9 Free Energy and Work 778 For Review 780 • Key Terms 780 • Questions and Exercises 782 17 Electrochemistry 790 17.1 Galvanic Cells 791 17.2 Standard Reduction Potentials 794 17.3 Cell Potential, Electrical Work, and Free Energy 800 17.4 Dependence of Cell Potential on Concentration 803 17.5 Batteries 808 ■CHEMICAL IMPACT Printed Batteries 809 ■CHEMICAL IMPACT Thermophotovoltaics: Electricity from Heat 810 ■CHEMICAL IMPACT Fuel Cells for Cars 812 vii 17.6 Corrosion 813 ■CHEMICAL IMPACT Paint that Stops Rust— Completely 814 17.7 Electrolysis 816 ■CHEMICAL IMPACT The Chemistry of Sunken Treasure 820 17.8 Commercial Electrolytic Processes 821 For Review 826 • Key Terms 826 • Questions and Exercises 829 18 The Nucleus: A Chemist’s View 840 18.1 Nuclear Stability and Radioactive Decay 841 18.2 The Kinetics of Radioactive Decay 846 18.3 Nuclear Transformations 849 ■CHEMICAL IMPACT Stellar Nucleosynthesis 850 18.4 Detection and Uses of Radioactivity 852 18.5 Thermodynamic Stability of the Nucleus 856 18.6 Nuclear Fission and Nuclear Fusion 859 18.7 Effects of Radiation 863 ■CHEMICAL IMPACT Nuclear Physics: An Introduction 864 For Review 867 • Key Terms 867 • Questions and Exercises 869 19 The Representative Elements: Groups 1A Through 4A 874 19.1 A Survey of the Representative Elements 875 19.2 The Group 1A Elements 880 19.3 Hydrogen 883 19.4 The Group 2A Elements 885 19.5 The Group 3A Elements 888 ■CHEMICAL IMPACT Boost Your Boron 889 19.6 The Group 4A Elements 890 ■CHEMICAL IMPACT Concrete Learning 892 ■CHEMICAL IMPACT Beethoven: Hair Is the Story 893 For Review 894 • Key Terms 894 • Questions and Exercises 895 20 The Representative Elements: Groups 5A Through 8A 900 20.1 The Group 5A Elements 901 20.2 The Chemistry of Nitrogen 903 ■CHEMICAL IMPACT Nitrous Oxide: Laughing Gas That Propels Whipped Cream and Cars 912 20.3 The Chemistry of Phosphorus 913 ■CHEMICAL IMPACT Phosphorus: An Illuminating Element 914 20.4 The Group 6A Elements 918 20.5 The Chemistry of Oxygen 919 20.6 The Chemistry of Sulfur 920 20.7 The Group 7A Elements 924 ■CHEMICAL IMPACT Photography 926 20.8 The Group 8A Elements 931 ■CHEMICAL IMPACT Automatic Sunglasses 931 For Review 933 • Key Terms 933 • Questions and Exercises 936 21 Transition Metals and Coordination Chemistry 942 21.1 The Transition Metals: A Survey 943 21.2 The First-Row Transition Metals 949 ■CHEMICAL IMPACT Titanium Dioxide—Miracle Coating 951 ■CHEMICAL IMPACT Titanium Makes Great Bicycles 952 21.3 Coordination Compounds 955 ■CHEMICAL IMPACT Alfred Werner: Coordination Chemist 960 21.4 Isomerism 960 ■CHEMICAL IMPACT The Importance of Being cis 963 viii 22.5 Polymers 1016 ■CHEMICAL IMPACT Heal Thyself 1018 ■CHEMICAL IMPACT Wallace Hume Carothers 1022 ■CHEMICAL IMPACT Plastic That Talks and Listens 1024 22.6 Natural Polymers 1025 ■CHEMICAL IMPACT Tanning in the Shade 1032 For Review 1040 • Key Terms 1040 • Questions and Exercises 1044 Appendix 1 Mathematical Procedures A1 A1.1 Exponential Notation A1 A1.2 Logarithms A4 A1.3 Graphing Functions A6 A1.4 Solving Quadratic Equations A7 A1.5 Uncertainties in Measurements A10 Appendix 2 The Quantitative Kinetic Molecular Model A13 Appendix 3 Spectral Analysis A16 Appendix 4 Selected Thermodynamic Data A19 Appendix 5 Equilibrium Constants and Reduction Potentials A22 A5.1 Values of Ka for Some Common Monoprotic Acids A22 A5.2 Stepwise Dissociation Constants for Several Common Polyprotic Acids A23 A5.3 Values of Kb for Some Common Weak Bases A23 A5.4 Ksp Values at 25C for Common Ionic Solids A24 A5.5 Standard Reduction Potentials at 25C (298K) for Many Common Half-Reactions A25 Appendix 6 SI Units and Conversion Factors A26 Glossary A27 Photo Credits A39 Answers to Selected Exercises A41 Index A70 21.5 Bonding in Complex Ions: The Localized Electron Model 965 21.6 The Crystal Field Model 967 ■CHEMICAL IMPACT Transition Metal Ions Lend Color to Gems 970 21.7 The Biologic Importance of Coordination Complexes 973 ■CHEMICAL IMPACT The Danger of Mercury 975 ■CHEMICAL IMPACT Supercharged Blood 978 21.8 Metallurgy and Iron and Steel Production 978 For Review 987 • Key Terms 987 • Questions and Exercises 989 22 Organic and Biological Molecules 996 22.1 Alkanes: Saturated Hydrocarbons 997 22.2 Alkenes and Alkynes 1005 22.3 Aromatic Hydrocarbons 1008 22.4 Hydrocarbon Derivatives 1010 ix With this edition of Chemistry, students and instructors alike will experience a truly integrated learning program. The textbook’s strong emphasis on conceptual learning and prob-lem solving is extended through the numerous online media as-signments and activities. It was our mission to create a media program that embodies the spirit of the textbook so that, when instructors and students look online for either study aids or on-line homework, that each resource supports the goals of the textbook—a strong emphasis on models, real-world applica-tions, and visual learning.
We have gone over every page in the sixth edition thor-oughly, fine-tuning in some cases and rewriting in others. In doing so, we have incorporated numerous constructive sugges-tions from instructors who used the previous edition. Based on this feedback new content has been added, such as the treat-ment of real gases in Chapter 5, which has been expanded to include a discussion of specific gases, and also coverage of pho-toelectric effect has been added to Chapter 7. In addition, the Sample Exercises in Chapter 2 have been revised to cover the naming of compounds given the formula and the opposite process of writing the formula from the name. To help students review key concepts, the For Review section of each chapter has been reorganized to provide an easy-to-read bulleted sum-mary; this section includes new review questions. The art pro-gram has been enhanced to include electrostatic potential maps to show a more accurate distribution of charge in molecules.
In the media program instructors will find a variety of re-sources to assign additional practice, study, and quiz mater-ial. ChemWork interactive assignments, end-of-chapter online homework, HM Testing, and classroom response system ap-plications allow you to assess students in multiple ways. The Online Study Center promotes self-study with animations, video demonstrations, and practice exercises.
Important Features of Chemistry ●Chemistry contains numerous discussions, illustrations, and exercises aimed at overcoming common misconcep-tions. It has become increasingly clear from our own teach-ing experience that students often struggle with chemistry because they misunderstand many of the fundamental con-cepts. In this text, we have gone to great lengths to pro-vide illustrations and explanations aimed at giving students more accurate pictures of the fundamental ideas of chem-istry. In particular, we have attempted to represent the microscopic world of chemistry so that students have a pic-ture in their minds of “what the atoms and molecules are doing.” The art program along with animations emphasize this goal. Also, we have placed a larger emphasis on the qualitative understanding of concepts before quantitative problems are considered. Because using an algorithm to correctly solve a problem often masks misunderstanding— students assume they understand the material because they got the right “answer”—it is important to probe their understanding in other ways. In this vein the text includes a number of Active Learning Questions (previously called In-Class Discussion Questions) at the end of each chapter that are intended for group discussion. It is our experience that students often learn the most when they teach each other. Students are forced to recognize their own lack of conceptual understanding when they try and fail to explain a concept to a colleague.
●With a strong problem-solving orientation, this text talks to the student about how to approach and solve chemical prob-lems. We have made a strong pitch to students for using a thoughtful and logical approach rather than simply memo-rizing procedures. In particular, an innovative method is given for dealing with acid–base equilibria, the material the typical student finds most difficult and frustrating. The key to this approach involves first deciding what species are pre-sent in solution, then thinking about the chemical properties of these species. This method provides a general framework for approaching all types of solution equilibria.
●The text contains almost 300 sample exercises, with many more examples given in the discussions leading to sample exercises or used to illustrate general strategies. When a spe-cific strategy is presented, it is summarized, and the sample exercise that follows it reinforces the step-by-step attack on the problem. In general, in approaching problem solving we emphasize understanding rather than an algorithm-based approach.
●We have presented a thorough treatment of reactions that occur in solution, including acid–base reactions. This mate-rial appears in Chapter 4, directly after the chapter on chem-ical stoichiometry, to emphasize the connection between solution reactions and chemical reactions in general. The early presentation of this material provides an opportunity to cover some interesting descriptive chemistry and also sup-ports the lab, which typically involves a great deal of aque-ous chemistry. Chapter 4 also includes oxidation–reduction reactions, because a large number of interesting and impor-tant chemical reactions involve redox processes. However, coverage of oxidation–reduction is optional at this point and depends on the needs of a specific course.
To the Professor ●Descriptive chemistry and chemical principles are thor-oughly integrated in this text. Chemical models may appear sterile and confusing without the observations that stimu-lated their invention. On the other hand, facts without orga-nizing principles may seem overwhelming. A combination of observations and models can make chemistry both inter-esting and understandable. In addition, in those chapters that deal with the chemistry of the elements systematically, we have made a continuous effort to show how properties and models correlate. Descriptive chemistry is presented in a va-riety of ways—as applications of the principles in separate sections, in Sample Exercises and exercise sets, in pho-tographs, and in Chemical Impact features.
●Throughout the book a strong emphasis on models prevails.
Coverage includes how they are constructed, how they are tested, and what we learn when they inevitably fail. Mod-els are developed naturally, with pertinent observations always presented first to show why a particular model was invented.
●Everyday-life applications of chemistry that should be of interest to students taking general chemistry appear throughout the text. For example, the Chemical Impact “Pearly Whites” illustrates the procedures for keeping teeth white, and “Thin is In” discusses the new technology being used to produce plasma flat-panel displays. Many industrial applications have also been incorporated into the text.
●A double-helix icon in the Instructor’s Annotated Edition highlights organic and biological examples of applications that are integrated throughout the text, in end-of-chapter problems, in exercises, or in-text discussions or examples.
This feature allows instructors to quickly locate material that will be of particular interest to students in pre-medicine, biology, or other health-related fields.
●Judging from the favorable comments of instructors and students who have used the sixth edition, the text seemed to work very well in a variety of courses. We were espe-cially pleased that readability was cited as a key strength when students were asked to assess the text. Thus, although the text has been fine-tuned in many areas, we have en-deavored to build on the basic descriptions, strategies, analogies, and explanations that were successful in the pre-vious editions.
New to the Seventh Edition The seventh edition of Chemistry incorporates many significant improvements and is accompanied by new and enhanced me-dia products and support services.
●Electrostatic potential maps have been added to Chapter 8 to show a more accurate distribution of charge in molecules.
These maps are based on ab initio molecular modeling calculations and provide a convenient method for better student understanding of bond and molecular polarity.
●Additional topics have been added to the text, which include a treatment of real gases in Chapter 5 and coverage of photoelectric effect to Chapter 7. In addition, the Sample Exercises in Chapter 2 have been revised to cover the nam-ing of compounds given the formula and the opposite process of writing the formula from the name.
●The end-of-chapter exercises and problems have been revised, providing approximately 20% new problems, in-cluding some that feature molecular art. End-of-chapter problems include: Active Learning Questions to test students’conceptual grasp of the material; Questions to help review important facts; Exercises that are paired and organized by topic; Additional Exercises, which are not keyed by topic; Challenge Problems, which require students to combine skills and problems; and Marathon Problems, which are the most comprehensive and challenging type of problem. New to the seventh edition are Integrative Problems that require students to understand multiple concepts across chapters.
●The For Review section, at the beginning of the end-of-chapter exercises, has been reorganized to help students more easily identify key concepts and test themselves on these concepts with review questions.
●A large number of new Chemical Impacts have been in-cluded in the seventh edition to continue the emphasis on up-to-date application of chemistry in the real world. These essays feature intriguing topics such as “Faux Snow,” and “Closest Packing of M&M’s®.” ●To support the use of active learning in chemical education, we have created new PowerPoint presentations—Active Learning PowerPoints with Lecture Outlines. These Power-Point presentations feature in-class discussion questions called Reacts, chemical demonstrations, animations, and fig-ures from the text. This material is designed to help in-structors present chemistry using an interactive teaching style, which we believe is most effective in promoting stu-dent learning. An Active Learning Guide includes the dis-cussion questions and supporting information in a workbook format. The questions are repeated in the workbook (with space to record answers) so that students can focus on par-ticipation in class sessions. This guide can then be used ef-fectively for independent student review outside of class.
●The Online Study Center has been enhanced to include a variety of tools to support visual learning and to give stu-dents extra practice. A For Review section summarizes the key topics of each chapter and helps students visualize the concepts with animations and video demonstrations. Visu-alization quiz questions allow students to test their knowl-edge of the concepts presented through the animations and video demonstrations. ACE practice tests allow students to practice problems on their own, and get immediate feed-back. Additional resources include a molecule library, in-teractive periodic table, and flashcards to help students study key terms.
x To the Professor To the Professor xi ●A very important feature accompanying the seventh edition is the online homework in the Eduspace® online learning tool. In addition to new algorithmic end-of-chapter ques-tions, Eduspace also includes ChemWork™interactive on-line homework. ChemWork is structured to help students learn chemistry in a conceptual way and is a series of text-based assignments. The system is modeled on a one-to-one teacher- student problem session. When a student cannot an-swer a given question, instead of giving him/her the correct answer, a system of interactive hints is available to help them think through each problem. Often the hints are in the form of a question on which the student receives feedback. Links to text material are also available for reference to key con-cepts at points of learning. The philosophy behind the home-work is to help students understand the material so that they can arrive at the correct answer by their own efforts, sup-ported by the kind of help an instructor would provide in a one-to-one tutoring session.
Another important feature of this homework system is that each student, even in a very large course, receives a unique set of tasks for each homework assignment, which is accomplished using random number–generation and sim-ilar versions of algorithmic problems. Each student’s work is assessed by the system, and the score for each task in the assignment is recorded in the electronic gradebook for im-mediate access by both student and instructor. The system also encourages increased student responsibility by setting firm deadlines for assignments. From the instructor’s per-spective, Eduspace encourages student study without the burden of tracking student efforts through grading. Our ex-perience with a similar system at the University of Illinois convinces us that this interactive homework represents an important breakthrough in helping students learn chemistry.
Flexibility of Topic Order The order of topics in the text was chosen because it is pre-ferred by the majority of instructors. However, we consciously constructed the book so that many other orders are possible.
During our tenure at the University of Illinois, for a two-chapter sequence, we used the chapters in this order: 1–6, 13–15, 7–9, 18, 21, 12, 10, 11, 16, 17, and parts of 22. Sections of Chap-ters 19, 20, and parts of 22 are used throughout the two se-mesters as appropriate. This order, chosen because of the way the laboratory is organized, is not necessarily recommended, but it illustrates the flexibility of order built into the text.
Some specific points about topic order: ●About half of chemistry courses present kinetics before equi-libria; the other half present equilibria first. This text is writ-ten to accommodate either order.
●The introductory aspects of thermodynamics are presented relatively early (in Chapter 6) because of the importance of energy in various chemical processes and models, but the more subtle thermodynamic concepts are left until later (Chapter 16). These two chapters may be used together if desired.
●To make the book more flexible, the derivation of the ideal gas law from the kinetic molecular theory and quantitative analysis using spectroscopy are presented in the appendixes.
Although mainstream general chemistry courses typically do not cover this material, some courses may find it appropriate.
By using the optional material in the appendixes and by as-signing the more difficult end-of-chapter exercises (from the additional exercises section), an instructor will find the level of the text appropriate for many majors courses or for other courses requiring a more extensive coverage of these topics.
●Because some courses cover bonding using only a Lewis structure approach, orbitals are not presented in the intro-ductory chapter on bonding (Chapter 8). In Chapter 9 both hybridization and the molecular orbital model are covered, but either or both of these topics may be omitted if desired.
●Chapter 4 can be tailored to fit the specific course involved.
Used in its entirety where it stands in the book, it provides interesting examples of descriptive chemistry and supports the laboratory program. Material in this chapter can also be skipped entirely or covered at some later point, whenever appropriate. For example, the sections on oxidation and reduction can be taught with electrochemistry. Although many instructors prefer early introduction of this concept, these sections can be omitted without complication since the next few chapters do not depend on this material.
Supplements An extensive teaching and learning package has been designed to make this book more useful to both instructors and students.
Technology: For Instructors Chemistry is accompanied by a complete suite of teaching and learning tools, including the customizable media resources be-low. Whether online or via CD, these integrated resources are designed to save you time and help make class preparation, pre-sentation, assessment, and course management more efficient and effective.
●Media Integration Guide for Instructors is your portal to the digital assets for this text. It includes the CDs described below as well as a user name and password to the Online Teaching Center, giving you instant access to text-related materials.
HM ClassPrep™CD includes everything an instructor needs to develop lectures: Active Learning PowerPoints with Lec-ture Outlines; virtually all text figures, tables, and photos in PowerPoint slides and as JPEGs; the Instructor’s Resource Guide in Word; Word files of the printed Test Bank; and Word files of the Complete Solutions Manual.
xii To the Professor HM Testing™(powered by Diploma®) is Houghton Mifflin’s new version of HM Testing. It significantly improves on functionality and ease of use by offering instructors all the tools they will need to create, author, deliver, and customize multiple types of tests—including authoring and editing algorithmic questions. New content includes 150 new Con-ceptual Questions, skill-level coding, and preprogrammed, algorithmic questions. HM Testing combines a flexible test-editing program with a comprehensive gradebook func-tion for easy administration and tracking. It enables in-structors to administer tests via print, network server, or the web. The HM Testing database contains a wealth of ques-tions and can produce multiple-choice, true/false, fill-in-the-blank, and essay tests. Questions can be customized based on the chapter being covered, the question format, level of difficulty, and specific topics. Available on the HM ClassPrep CD.
HM ClassPresent™2006: General Chemistry features new animations and video demonstrations. HM ClassPre-sent provides a library of high-quality, scaleable lab demon-strations and animations covering core chemistry concepts arranged by chapter and topic. The resources within it can be browsed by thumbnail and description or searched by chapter, title, or keyword. Instructors can export the anima-tions and videos into a variety of presentation formats or use for presentation directly from the CD. Full transcripts ac-company all audio commentary to reinforce visual presen-tations and to cater to different learning styles.
Online Teaching Center includes classroom presentation and preparation materials. Animations; videos; virtually all figures, tables, and photos from the text are available in JPEG and PowerPoint format; the Transition Guide from the sixth to seventh edition; Active Learning PowerPoints with Lecture Outlines; and classroom response system content are all available online.
Eduspace (powered by Blackboard™), Houghton Mifflin’s complete course-management solution, features algorith-mic, end-of-chapter questions along with ChemWork in-teractive online homework. Both types of homework prob-lems include links to relevant pages from the text. These integrated resources allow students to reference core con-cepts at the point of learning. ChemWork assignments help students learn the process of thinking like a chemist: as students work through unique, text-based assignments, a system of interactive hints is available to help them think through each problem. Eduspace includes all of Black-board’s powerful features for teaching and learning, and comes preloaded with course materials including videos and animations, and a link to SMARTHINKING™live online tutoring. Customized functions allow instructors to tailor these materials to their specific needs, select, create and post homework assignments and tests, communicate with students in a variety of different ways, track student progress, and manage their portfolio of course work in the gradebook. To help instructors best utilize the media that accompanies the textbook, lesson plans have been created based on the sections of the book. Each section correlates the relevant ChemWork assignments, Visualization (ani-mations and videos), and online end-of-chapter questions.
Please note: instructors who want their students to use Eduspace must request a Getting Started Guide for Stu-dents which will be bundled free with new copies of the text. Instructors who adopt Eduspace will receive a separate Getting Started Guide for Instructors for the program with a passkey to set up their course.
●Classroom Response System (CRS) compatible content on the Online Teaching Center, HM ClassPrep CD, and in Edu-space allows professors to perform “on-the-spot” assess-ments, deliver quick quizzes, gauge students’ understanding of a particular question or concept, and take their class ros-ter easily. Students get immediate feedback on how well they know the content and where they need to improve. Two sets of questions are available in PowerPoint slides: one based on Test Bank content and the other with unique, conceptual questions. Both question types are correlated to sections in the textbook. The conceptual questions are also correlated to relevant media and art from the book.
●TeamUP Integration Services Houghton Mifflin aims to provide customers with quality textbooks, technology, and superior training and implemen-tation services. TeamUP, our integration program, offers flexible, personalized training and consultative services by phone, online, or on campus. Experienced faculty advisors and media specialists will assist you and your department in using our products most effectively.
●Course-Management Software is available through WebCT and Blackboard. These two distributed learning systems al-low instructors to create a virtual classroom without any knowledge of HTML. Features include: assessment tools, a gradebook, online file exchange between instructors and stu-dents, online syllabi, and course descriptions. The customized Chemistry cartridges feature Test Bank questions, lecture materials, and study aids related to the text.
Print Supplements: For Instructors ●Complete Solutions Guide, by Thomas J. Hummel, Susan Arena Zumdahl, and Steven S. Zumdahl, presents detailed solutions for all of the end-of-chapter exercises in the text for the convenience of faculty and staff involved in instruc-tion and for instructors who wish their students to have so-lutions for all exercises. Departmental approval is required for the sale of the Complete Solutions Guide to students.
●Instructor’s Resource Guide, by Donald J. DeCoste, in-cludes suggestions for alternative orders of topics, suggested responses to the Active Learning Questions, amplification To the Professor xiii of strategies used in various chapters, lesson plans of media resources correlated to section, answers to Reacts, and a sec-tion on notes for teaching assistants.
●Lecture Demonstration Guide, by Fred Jurgens of the Uni-versity of Wisconsin—Madison, lists the sources for over 750 classroom demonstrations that can be used in general chemistry courses. Icons in the margins of the Instructor’s Annotated Edition of the text key the demonstrations to their corresponding text discussions.
●Instructor’s Resource Guide for Experimental Chemistry, Seventh Edition, by James F. Hall, contains tips including hints on running experiments, approximate times for each experiment, and answers to all prelab and postlab questions posed in the laboratory guide.
●Bibliobase (www.bibliobase.com) allows instructors to cre-ate a completely customized lab manual by mixing and matching from 88 general chemistry labs—including all the labs from Experimental Chemistry—and 56 labs for the course in general, organic, and biochemistry. At the On-line Teaching Center, instructors search through the data-base of labs, make their selections, organize the sequence of the manual, and submit their order via the Internet. Cus-tomized, printed, and bound lab manuals are delivered to the bookstore within weeks.
●Test Item File, by Steven S. Zumdahl, Susan Arena Zum-dahl, and Gretchen Adams (available to adopters), offers a printed version of more than 2000 exam questions, 10 per-cent of which are new to this edition, referenced to the appropriate text section. Questions are in multiple-choice, open-ended, and true-false formats.
●Transparencies, in a full-color set of 255, are available to adopters of the seventh edition of the text.
Technology: For Students Chemistry is supported by an array of learning tools designed to help students succeed in their chemistry course. It includes the following media resources: A passkey to the Online Study Center is bound into the front of the textbook. From the Online Study Center, students have access to practice, visualization, and self-study aids. Visu-alization animations and video demonstrations help students see key concepts, and each Visualization is accompanied by quiz questions for students’ review. A For Review section helps students review key topics at a glance and includes video demon-strations and animations for additional reinforcement. Flash-cards and ACE practice tests help students study key concepts and problem-solve. A molecule library, glossary, and interactive periodic table are also available for support. A Student CD, with many of these Online Study Center resources, is available upon request for students who do not have Internet access.
Eduspace (powered by Blackboard), Houghton Mifflin’s complete course-management solution, features algorithmic end-of-chapter questions along with ChemWork interactive online homework. Through Eduspace, students can also access the Online Study Center and SMARTHINKING live, online tutoring. Instructors who adopt Eduspace will receive a sepa-rate user guide for the program with a passkey to set up their course. Students using Eduspace will also receive a separate user guide and passkey.
SMARTHINKING live, online tutoring is also available free with new books upon instructor request. Students may also purchase stand-alone access to it. SMARTHINKING provides personalized, text-specific tutoring and is available during peak study hours when students need it most. Limits apply; terms and hours of SMARTHINKING service are subject to change.
Print Supplements: For Students ●Study Guide, by Paul B. Kelter of the University of Illinois— Urbana. Written to be a self-study aid for students, this guide includes alternate strategies for solving problems, supple-mental explanations for the most difficult material, and self-tests. There are approximately 500 worked examples and 1200 practice problems (with answers), designed to give stu-dents mastery and confidence.
●Student Solutions Manual, by Thomas J. Hummel, Susan Arena Zumdahl, and Steven S. Zumdahl, all of the Univer-sity of Illinois, Urbana, provides detailed solutions for half of the end-of-chapter exercises (designated by the blue ques-tion numbers) using the strategies emphasized in the text.
To ensure the accuracy of the solutions, this supplement and the Complete Solutions Guide were checked independently by several instructors.
●Active Learning Guide, by Donald J. DeCoste. This printed workbook can be used in lecture or recitation in conjunc-tion with the instructor PowerPoint slides. It provides a com-plete set of React questions with space for student answers.
Students can use the workbook as a self-study aid outside of class.
●Solving Equilibrium Problems with Applications to Qual-itative Analysis, by Steven S. Zumdahl. Successfully used by thousands of students, this book offers thorough, step-by-step procedures for solving problems related to equi-libria taking place both in the gas phase and in solution.
Containing hundreds of sample exercises, test exercises with complete solutions, and end-of-chapter exercises with answers, the text utilizes the same problem-solving methods found in Chemistry and is an excellent source of additional drill-type problems. The last chapter presents an exploratory qualitative analysis experiment with ex-planations based on the principles of aqueous equilibria.
●Experimental Chemistry, Seventh Edition, by James F. Hall of the University of Massachusetts—Lowell, provides an ex-tensively revised laboratory program compatible with the text. The 48 experiments present a wide variety of chem-istry, and many experiments offer choices of procedures.
Safety is strongly emphasized throughout the program.
Acknowledgments: This book represents the efforts of many talented and dedicated people. We particularly want to thank Richard Stratton, Executive Editor, for his vision and oversight of this project. Richard’s knowledge, judgment, and enthusiasm have contributed immeasurably to the success of this text. He is not only an outstanding editor but also one of the nicest people in the business.
We also want to thank Cathy Brooks, Senior Project Edi-tor, who did a miraculous job of coordinating the production of an incredibly complex project with grace and good humor.
We also especially appreciate the excellent work of Rebecca Berardy Schwartz, Developmental Editor, who managed the re-vision process in a very supportive and organized manner.
We are especially grateful to Tom Hummel, who managed the revision of the end-of-chapter problems and the solutions manuals. Tom’s extensive experience teaching general chem-istry and his high standards of accuracy and clarity have resulted in great improvements in the quality of the problems and the solutions in this edition. In addition, we very much ap-preciate the contributions of Don DeCoste, who has helped us comprehend more clearly the difficulties students have with conceptual understanding and who contributed the Challenge Problems.
We also extend our thanks to Jason Overby, who rendered the electrostatic potential maps and who contributed the Integrative Problems. Our thanks and love also go to Leslie, Steve, Whitney, Scott, Tyler, Sunshine, and Tony for their con-tinuing support.
Thanks to the others at Houghton Mifflin who supplied valuable assistance on this revision: Jill Haber, Senior Art/Design Coordinator; Sharon Donahue, Photo Researcher; Katherine Greig, Senior Marketing Manager; Naveen Hariprasad, Mar-keting Assistant; and Susan Miscio, Editorial Assistant.
Special thanks go to the following people who helped shape this edition by offering suggestions for its improvement: Dawood Afzal, Truman State (media reviewer); Carol Anderson, University of Connecticut—Avery Point (media reviewer); Jeffrey R. Appling, Clemson University (media reviewer); Dave Blackburn, University of Minnesota; Robert S. Boikess, Rutgers University; Ken Carter, Truman State (media reviewer); Bette Davidowitz, University of Cape Town; Natalie Foster, Lehigh University; Tracy A.
Halmi, Penn State Erie, The Behrend College; Carl A.
Hoeger, UC—San Diego; Ahmad Kabbani, Lebanese American University; Arthur Mar, University of Alberta; Jim McCormick, Truman State (media reviewer); Richard Orwell, Blue Ridge Community College (media reviewer); Jason S. Overby, College of Charleston; Robert D. Pike, The College of William and Mary; Daniel Raftery, Purdue University; Jimmy Rogers, University of Texas—Arlington (media reviewer); Raymond Scott, Mary Washington College; Alan Stolzenberg, West Virginia University; Rashmi Venkateswaran, University of Ottawa. AP reviewers: Annis Hapkiewicz, Okemos High School; Tina Ohn-Sabatello, Maine Township HS East. Interactive Course Guide Reviewers: Lynne C. Cary, Ph.D., Bethel College; Craig C. Martens, University of California— Irvine; Jeffrey P. Osborne, Manchester College; Donald W.
Shive, Muhlenberg College; Craig Sockwell, Northwest Shoals Community College; Richard Pennington, College of St. Mary. Accuracy reviewers: Linda Bush (textbook reviewer), Jon Booze (media reviewer) Reviewers of the sixth edition: Ramesh D. Arasasingham, University of California—Irvine; Stanley A. Bajue, Medgar Evans College, CUNY; V.G.
Berner, New Mexico Junior College; Dave Blackburn, University of Minnesota; Steven R. Boone, Central Missouri State University; Gary S. Buckley, Cameron University; Lara L. Chappell, SUNY College at Oswego; David Cramb, University of Calgary; Philip W. Crawford, Southeast Missouri State University; Philip Davis, University of Tennessee; Michael P. Garoutte, Missouri Southern State College; Daniel Graham, Loyola University; David R.
Hawkes, Lambuth University; Dale Hawley, Kansas State University; Thomas B. Higgins, Harold Washington College; John C. Hogan, Louisiana State University; Donald P. Land, University of California—Davis; Michael P. Masingale, LeMoyne College; Julie T. Millard, Colby College; Robert H.
Paine, Rochester Institute of Technology; Brenda Ross, Cottey College; Jay S. Shore, South Dakota State University; Richard T. Toomey, Northwest Missouri State University; Robert Zoellner, Humboldt State University.
xiv To the Professor xv The major purpose of this book, of course, is to help you learn chemistry. However, this main thrust is closely linked to two other goals: to show how important and how interesting the subject is, and to show how to think like a chemist. To solve complicated problems the chemist uses logic, trial and error, intuition, and, above all, patience. A chemist is used to being wrong. The important thing is to learn from a mistake, recheck assumptions, and try again. A chemist thrives on puzzles that seem to defy solutions.
Many of you using this text do not plan to be practicing chemists. However, the nonchemist can benefit from the chemist’s attitude. Problem solving is important in all profes-sions and in all walks of life. The techniques you will learn from this book will serve you well in any career you choose.
Thus, we believe that the study of chemistry has much to offer the nonmajor, including an understanding of many fascinating and important phenomena and a chance to hone problem-solving skills.
This book attempts to present chemistry in a manner that is sensible to the novice. Chemistry is not the result of an in-spired vision. It is the product of countless observations and many attempts, using logic and trial and error, to account for these observations. In this book the concepts are developed in a natural way: The observations come first and then models are constructed to explain the observed behavior.
Models are a major focus in this book. The uses and lim-itations of models are emphasized, and science is treated as a human activity, subject to all the normal human foibles. Mis-takes are discussed as well as successes.
A central theme of this book is a thoughtful, systematic approach to problem solving. Learning encompasses much more than simply memorizing facts. Truly educated people use their factual knowledge as a starting point—a base for creative approaches to solving problems.
Read through the material in the text carefully. For most concepts, illustrations or photos will help you visualize what is going on. To further help you visualize concepts by using animations and videos, we have included Visualization exer-cises on the Online Study Center or on an optional free CD.
Icons in the text margin signal that there is companion mater-ial available on the CD.
Often a given type of problem is “walked through” in the text before the corresponding Sample Exercises appear. Strate-gies for solving problems are given throughout the text.
Thoroughly examine the Sample Exercises and the prob-lem-solving strategies. The strategies summarize the approach taken in the text; the Sample Exercises follow the strategies step-by-step. Schematics in Chapter 15 also illustrate the logical pathways to solving aqueous equilibrium problems.
Throughout the text, we have used margin notes to high-light key points, to comment on an application of the text material, or to reference material in other parts of the book.
Chemical Impact, the boxed feature that appears frequently throughout the text, discusses especially interesting applica-tions of chemistry to the everyday world.
Each chapter has a summary and key terms list for review, and the glossary gives a quick reference for definitions.
Learning chemistry requires working the end-of-chapter exercises assigned by your professor. Answers to exercises denoted by blue question numbers are in the back of the book, and complete solutions to those exercises are in the Partial Solutions Guide. To help you assess your level of proficiency, the Online Study Center (college.hmco.com/PIC/ zumdahl7e) offers quizzes and electronic homework assign-ments that feature instant feedback.
The Study Guide contains extra practice problems and many worked examples. The supplement, Solving Equilibrium Problems with Applications to Qualitative Analysis, reinforces in great detail the text’s step-by-step approach to solving equilibrium problems and contains many worked examples and self-quiz questions.
It is very important to use the exercises and electronic homework assignments to your best advantage. Your main goal should not be to simply get the correct answer but to under-stand the process for getting the answer. Memorizing the so-lutions for specific problems is not a very good way to prepare for an exam. There are too many pigeonholes required to cover every possible problem type. Look within the problem for the solution. Use the concepts you have learned along with a sys-tematic, logical approach to find the solution. Learn to trust yourself to think it out. You will make mistakes, but the im-portant thing is to learn from these errors. The only way to gain confidence is to do lots of practice problems and use these to diagnose your weaknesses.
Be patient and thoughtful and work hard to understand rather than simply memorize. We wish you an interesting and satisfying year.
To the Student Features of Chemistry Seventh Edition The Contents gives students an overview of the topics to come.
Conceptual Understanding and Problem Solving 350 Chapter Eight Bonding: General Concepts Fundamental Properties of Models Models are human inventions, always based on an incomplete understanding of how nature works. A model does not equal reality.
Models are often wrong. This property derives from the first property. Models are based on speculation and are always oversimplifications.
Models tend to become more complicated as they age. As flaws are discovered in our models, we “patch” them and thus add more detail.
It is very important to understand the assumptions inherent in a particular model be-fore you use it to interpret observations or to make predictions. Simple models usu-ally involve very restrictive assumptions and can be expected to yield only qualitative information. Asking for a sophisticated explanation from a simple model is like expecting to get an accurate mass for a diamond using a bathroom scale.
For a model to be used effectively, we must understand its strengths and weak-nesses and ask only appropriate questions. An illustration of this point is the simple aufbau principle used to account for the electron configurations of the elements. Al-though this model correctly predicts the configuration for most atoms, chromium and copper, for example, do not agree with the predictions. Detailed studies show that the configurations of chromium and copper result from complex electron interactions that are not taken into account in the simple model. However, this does not mean that we should discard the simple model that is so useful for most atoms. Instead, we must apply it with caution and not expect it to be correct in every case.
When a model is wrong, we often learn much more than when it is right. If a model makes a wrong prediction, it usually means we do not understand some fundamen-tal characteristics of nature. We often learn by making mistakes. (Try to remember this when you get back your next chemistry test.) 8.8 Covalent Bond Energies and Chemical Reactions In this section we will consider the energies associated with various types of bonds and see how the bonding concept is useful in dealing with the energies of chemical reactions.
One important consideration is to establish the sensitivity of a particular type of bond to its molecular environment. For example, consider the stepwise decomposition of methane: Process Energy Required (kJ/mol) CH4(g) S CH3(g) H(g) 435 CH3(g) S CH2(g) H(g) 453 CH2(g) S CH(g) H(g) 425 CH(g) S C(g) H(g) 339 Total 1652 Average 413 l h h C b d i b k i h h i d i i 1652 4 8.13 Molecular Structure: The VSEPR Model The structures of molecules play a very important role in determining their chemical prop-erties. As we will see later, this is particularly important for biological molecules; a slight change in the structure of a large biomolecule can completely destroy its usefulness to a cell or may even change the cell from a normal one to a cancerous one.
Many accurate methods now exist for determining molecular structure, the three-dimensional arrangement of the atoms in a molecule. These methods must be used if precise information about structure is required. However, it is often useful to be able to predict the approximate molecular structure of a molecule. In this section we consider a simple model that allows us to do this. This model, called the valence shell electron-pair repulsion (VSEPR) model, is useful in predicting the geometries of molecules formed from nonmetals. The main postulate of this model is that the structure around a given atom is determined principally by minimizing electron-pair repulsions. The idea here is that the bonding and nonbonding pairs around a given atom will be positioned as far apart as possible. To see how this model works, we will first consider the molecule BeCl2, which has the Lewis structure Avogadro’s Law Suppose we have a 12.2-L sample containing 0.50 mol oxygen gas (O2) at a pressure of 1 atm and a temperature of 25C. If all this O2 were converted to ozone (O3) at the same temperature and pressure, what would be the volume of the ozone?
Solution The balanced equation for the reaction is To calculate the moles of O3 produced, we must use the appropriate mole ratio: Avogadro’s law states that V an, which can be rearranged to give Since a is a constant, an alternative representation is where V1 is the volume of n1 moles of O2 gas and V2 is the volume of n2 moles of O3 gas.
In this case we have Solving for V2 gives Reality Check: Note that the volume decreases, as it should, since fewer moles of gas molecules will be present after O2 is converted to O3.
See Exercises 5.35 and 5.36.
V2 an2 n1 b V1 a0.33 mol 0.50 molb 12.2 L 8.1 L V1 12.2 L V2 ?
n1 0.50 mol n2 0.33 mol V1 n1 a V2 n2 V n a 0.50 mol O2 2 mol O3 3 mol O2 0.33 mol O3 3O21g2 ¡ 2O31g2 N2 H2 Ar CH4 FIGURE 5.10 These balloons each hold 1.0 L of gas at 25C and 1 atm. Each balloon contains 0.041 mol of gas, or 2.5 1022 molecules.
Sample Exercise 5.5 Avogadro’s law also can be written as V1 n1 V2 n2 Sample Exercises model a step-by-step approach to solving problems. Cross-references to similar end-of-chapter exercises are provided at the end of each Sample Exercise. Reality Checks appear after the solutions in selected exercises, helping students evaluate their answers to ensure that they are reasonable.
By stressing the limitations and uses of scientific models, the authors show students how chemists think and work. The authors’ emphasis on modeling (or chemical theories) throughout the text addresses the problem of rote memorization by helping students better understand and appreciate the process of scientific thinking.
xvi Connections Much of the chemistry that affects each of us occurs among substances dissolved in water. For example, virtually all the chemistry that makes life possible occurs in an aqueous environment. Also, various medical tests involve aqueous reactions, depending heavily on analyses of blood and other body fluids. In addition to the common tests for sugar, cholesterol, and iron, analyses for specific chemical markers allow detection of many diseases before obvious symptoms occur.
Aqueous chemistry is also important in our environment. In recent years, contami-nation of the groundwater by substances such as chloroform and nitrates has been widely publicized. Water is essential for life, and the maintenance of an ample supply of clean water is crucial to all civilization.
To understand the chemistry that occurs in such diverse places as the human body, the atmosphere, the groundwater, the oceans, the local water treatment plant, your hair as you shampoo it, and so on, we must understand how substances dissolved in water react with each other.
However, before we can understand solution reactions, we need to discuss the nature of solutions in which water is the dissolving medium, or solvent. These solutions are called aqueous solutions. In this chapter we will study the nature of materials after they are dis-solved in water and various types of reactions that occur among these substances. You will see that the procedures developed in Chapter 3 to deal with chemical reactions work very well for reactions that take place in aqueous solutions. To understand the types of reactions that occur in aqueous solutions, we must first explore the types of species present.
This requires an understanding of the nature of water.
4.1 Water, the Common Solvent Water is one of the most important substances on earth. It is essential for sustaining the reactions that keep us alive, but it also affects our lives in many indirect ways. Water helps moderate the earth’s temperature; it cools automobile engines, nuclear power plants, and many industrial processes; it provides a means of transportation on the earth’s surface and a medium for the growth of a myriad of creatures we use as food; and much more.
One of the most valuable properties of water is its ability to dissolve many different substances. For example, salt “disappears” when you sprinkle it into the water used to cook vegetables, as does sugar when you add it to your iced tea. In each case the “dis-appearing” substance is obviously still present—you can taste it. What happens when a solid dissolves? To understand this process, we need to consider the nature of water. Liquid water consists of a collection of H2O molecules. An individual H2O molecule is “bent” or V-shaped, with an HOOOH angle of approximately 105 degrees: The OOH bonds in the water molecule are covalent bonds formed by electron shar-ing between the oxygen and hydrogen atoms. However, the electrons of the bond are not shared equally between these atoms. For reasons we will discuss in later chapters, oxy-gen has a greater attraction for electrons than does hydrogen. If the electrons were shared equally between the two atoms, both would be electrically neutral because, on average, the number of electrons around each would equal the number of protons in that nucleus.
H H O 105˚ 127 Each chapter begins with an engaging introduction that demonstrates how chemistry is related to everyday life.
CHEMICAL IMPACT The Chemistry of Air Bags M ost experts agree that air bags represent a very impor-tant advance in automobile safety. These bags, which are stored in the auto’s steering wheel or dash, are designed to inflate rapidly (within about 40 ms) in the event of a crash, cushioning the front-seat occupants against impact. The bags then deflate immediately to allow vision and movement af-ter the crash. Air bags are activated when a severe deceler-ation (an impact) causes a steel ball to compress a spring and electrically ignite a detonator cap, which, in turn, causes sodium azide (NaN3) to decompose explosively, forming sodium and nitrogen gas: This system works very well and requires a relatively small amount of sodium azide (100 g yields 56 L N2(g) at 25C and 1.0 atm).
When a vehicle containing air bags reaches the end of its useful life, the sodium azide present in the activators must be given proper disposal. Sodium azide, besides being ex-plosive, has a toxicity roughly equal to that of sodium 2NaN31s2 ¡ 2Na1s2 3N21g2 cyanide. It also forms hydrazoic acid (HN3), a toxic and explosive liquid, when treated with acid.
The air bag represents an application of chemistry that has already saved thousands of lives.
Inflated air bags.
Chemical Impact boxes describe current applications of chemistry. These special-interest boxes cover such topics as preserving works of art, molecules as a means of communication, and the heat of chili peppers. xvii Visualization 346 Chapter Eight Bonding: General Concepts more negative than that for combining gaseous Na and F ions to form NaF(s). Thus the energy released in forming a solid containing Mg2 and O2 ions rather than Mg and O ions more than compensates for the energies required for the processes that produce the Mg2 and O2 ions.
If there is so much lattice energy to be gained in going from singly charged to doubly charged ions in the case of magnesium oxide, why then does solid sodium fluoride contain Na and F ions rather than Na2 and F2 ions? We can answer this question by recognizing that both Na and F ions have the neon electron configura-tion. Removal of an electron from Na requires an extremely large quantity of energy (4560 kJ/mol) because a 2p electron must be removed. Conversely, the addition of an electron to F would require use of the relatively high-energy 3s orbital, which is also an unfavorable process. Thus we can say that for sodium fluoride the extra energy required to form the doubly charged ions is greater than the gain in lattice energy that would result.
This discussion of the energies involved in the formation of solid ionic compounds illustrates that a variety of factors operate to determine the composition and structure of these compounds. The most important of these factors involve the balancing of the ener-gies required to form highly charged ions and the energy released when highly charged ions combine to form the solid.
8.6 Partial Ionic Character of Covalent Bonds Recall that when atoms with different electronegativities react to form molecules, the elec-trons are not shared equally. The possible result is a polar covalent bond or, in the case of a large electronegativity difference, a complete transfer of one or more electrons to form ions. The cases are summarized in Fig. 8.12.
How well can we tell the difference between an ionic bond and a polar covalent bond?
The only honest answer to this question is that there are probably no totally ionic bonds between discrete pairs of atoms. The evidence for this statement comes from calculations of the percent ionic character for the bonds of various binary compounds in the gas phase.
These calculations are based on comparisons of the measured dipole moments for mole-cules of the type X—Y with the calculated dipole moments for the completely ionic case, XY. The percent ionic character of a bond can be defined as Application of this definition to various compounds (in the gas phase) gives the results shown in Fig. 8.13, where percent ionic character is plotted versus the difference in the electronegativity values of X and Y. Note from this plot that ionic character increases with electronegativity difference, as expected. However, none of the bonds reaches 100% ionic character, even though compounds with the maximum possible electronegativity differ-ences are considered. Thus, according to this definition, no individual bonds are com-pletely ionic. This conclusion is in contrast to the usual classification of many of these compounds (as ionic solids). All the compounds shown in Fig. 8.13 with more than 50% ionic character are normally considered to be ionic solids. Recall, however, the results in Fig. 8.13 are for the gas phase, where individual XY molecules exist. These results can-not necessarily be assumed to apply to the solid state, where the existence of ions is fa-vored by the multiple ion interactions.
Another complication in identifying ionic compounds is that many substances contain polyatomic ions. For example, NH4Cl contains NH4 and Cl ions, and Na2SO4 contains Na and SO4 2 ions. The ammonium and sulfate ions are held together by covalent bonds. Thus, calling NH4Cl and Na2SO4 ionic compounds is somewhat ambiguous.
Percent ionic character of a bond a measured dipole moment of X¬Y calculated dipole moment of XYb 100% FIGURE 8.12 The three possible types of bonds: (a) a covalent bond formed between identical F atoms; (b) the polar covalent bond of HF, with both ionic and covalent components; and (c) an ionic bond with no electron sharing.
(a) (b) (c) – + H F F F Since the equation for lattice energy con-tains the product Q1Q2, the lattice energy for a solid with 2 and 2 ions should be four times that for a solid with 1 and 1 ions. That is, For MgO and NaF, the observed ratio of lattice energies (see Fig. 8.11) is 3916 kJ 923 kJ 4.24 122122 112112 4 Electrostatic potential maps help students visualize the distribution of charge in molecules.
FIGURE 4.17 Photos and accompanying molecular-level representations illustrating the reaction of KCl(aq) with AgNO3(aq) to form AgCl(s). Note that it is not possible to have a photo of the mixed solution before the reaction occurs, because it is an imaginary step that we use to help visualize the reaction. Actually, the reaction occurs immediately when the two solutions are mixed.
K+ Cl– Ag+ Ag+ NO3– Solutions are mixed The art program emphasizes molecular-level interactions that help students visualize the “micro-macro” connection.
Visualization animations and video demonstrations help students further understand and visualize chemical concepts. Animations and videos (Visualizations) are found via the Online Study Center and Online Teaching Center, and HM ClassPresent instructor CD.
xviii Practice 4. As you increase the temperature of a gas in a sealed, rigid con-tainer, what happens to the density of the gas? Would the results be the same if you did the same experiment in a container with a piston at constant pressure? (See Figure 5.17.) 5. A diagram in a chemistry book shows a magnified view of a flask of air as follows: What do you suppose is between the dots (the dots represent air molecules)?
a. air b. dust c. pollutants d. oxygen e. nothing 6. If you put a drinking straw in water, place your finger over the opening, and lift the straw out of the water, some water stays in the straw. Explain.
7. A chemistry student relates the following story: I noticed my tires were a bit low and went to the gas station. As I was filling the tires, I thought about the kinetic molecular theory (KMT). I noticed the tires because the volume was low, and I realized that I was increasing both the pressure and volume of the tires.
“Hmmm,” I thought, “that goes against what I learned in chem-istry, where I was told pressure and volume are inversely pro-portional.” What is the fault in the logic of the chemistry student in this situation? Explain why we think pressure and volume to be inversely related (draw pictures and use the KMT).
8. Chemicals X and Y (both gases) react to form the gas XY, but it takes a bit of time for the reaction to occur. Both X and Y are placed in a container with a piston (free to move), and you note the volume. As the reaction occurs, what happens to the volume of the container? (See Fig. 5.18.) 9. Which statement best explains why a hot-air balloon rises when the air in the balloon is heated?
a. According to Charles’s law, the temperature of a gas is directly related to its volume. Thus the volume of the balloon increases, making the density smaller. This lifts the balloon.
b. Hot air rises inside the balloon, and this lifts the balloon.
c. The temperature of a gas is directly related to its pressure.
The pressure therefore increases, and this lifts the balloon.
d. Some of the gas escapes from the bottom of the balloon, thus decreasing the mass of gas in the balloon. This decreases the density of the gas in the balloon, which lifts the balloon.
e. Temperature is related to the root mean square velocity of the gas molecules. Thus the molecules are moving faster, hitting the balloon more, and thus lifting the balloon.
Justify your choice, and for the choices you did not pick, explain what is wrong with them.
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Consider the following apparatus: a test tube covered with a non-permeable elastic membrane inside a container that is closed with a cork. A syringe goes through the cork.
a. As you push down on the syringe, how does the membrane covering the test tube change?
b. You stop pushing the syringe but continue to hold it down.
In a few seconds, what happens to the membrane?
2. Figure 5.2 shows a picture of a barometer. Which of the following statements is the best explanation of how this barometer works?
a. Air pressure outside the tube causes the mercury to move in the tube until the air pressure inside and outside the tube is equal.
b. Air pressure inside the tube causes the mercury to move in the tube until the air pressure inside and outside the tube is equal.
c. Air pressure outside the tube counterbalances the weight of the mercury in the tube.
d. Capillary action of the mercury causes the mercury to go up the tube.
e. The vacuum that is formed at the top of the tube holds up the mercury.
Justify your choice, and for the choices you did not pick, explain what is wrong with them. Pictures help!
3. The barometer below shows the level of mercury at a given at-mospheric pressure. Fill all the other barometers with mercury for that same atmospheric pressure. Explain your answer.
Hg(l) Syringe Membrane Cork 217 For Review 215 Key Terms Section 5.1 barometer manometer mm Hg torr standard atmosphere pascal Section 5.2 Boyle’s law ideal gas Charles’s law absolute zero Avogadro’s law Section 5.3 universal gas constant ideal gas law Section 5.4 molar volume standard temperature and pressure (STP) Section 5.5 Dalton’s law of partial pressures partial pressure mole fraction Section 5.6 kinetic molecular theory (KMT) root mean square velocity joule Section 5.7 diffusion effusion Graham’s law of effusion Section 5.8 real gas van der Waals equation Section 5.10 atmosphere air pollution photochemical smog acid rain For Review State of a gas The state of a gas can be described completely by specifying its pressure (P), volume (V), temperature (T) and the amount (moles) of gas present (n) Pressure • Common units • SI unit: pascal Gas laws Discovered by observing the properties of gases Boyle’s law: PV k Charles’s law: V bT Avogadro’s law: V an Ideal gas law: PV nRT Dalton’s law of partial pressures: Ptotal P1 P2 P3 , where Pn represents the partial pressure of component n in a mixture of gases Kinetic molecular theory (KMT) Model that accounts for ideal gas behavior Postulates of the KMT: • Volume of gas particles is zero • No particle interactions • Particles are in constant motion, colliding with the container walls to produce pressure • The average kinetic energy of the gas particles is directly proportional to the temperature of the gas in kelvins Gas properties The particles in any gas sample have a range of velocities The root mean square (rms) velocity for a gas represents the average of the squares of the particle velocities Diffusion: the mixing of two or more gases Effusion: the process in which a gas passes through a small hole into an empty chamber Real gas behavior Real gases behave ideally only at high temperatures and low pressures Understanding how the ideal gas equation must be modified to account for real gas behavior helps us understand how gases behave on a molecular level Van der Waals found that to describe real gas behavior we must consider particle interactions and particle volumes REVIEW QUESTIONS 1. Explain how a barometer and a manometer work to measure the pressure of the atmosphere or the pressure of a gas in a container.
urms B 3RT M p 1 atm 101,325 Pa 1 atm 760 torr 1 torr 1 mm Hg Each chapter has a For Review section to reinforce key concepts, and includes review questions. Key Terms are printed in bold type and are defined where they first appear. They are also grouped at the end of the chapter and in the Glossary at the back of the text.
226 Chapter Five Gases If 2.55 102 mL of NO(g) is isolated at 29C and 1.5 atm, what amount (moles) of UO2 was used in the reaction?
128. Silane, SiH4, is the silicon analogue of methane, CH4. It is prepared industrially according to the following equations: a. If 156 mL of HSiCl3 (d 1.34 g/mL) is isolated when 15.0 L of HCl at 10.0 atm and 35C is used, what is the percent yield of HSiCl3?
b. When 156 mL of HSiCl3 is heated, what volume of SiH4 at 10.0 atm and 35C will be obtained if the percent yield of the reaction is 93.1%?
129. Solid thorium(IV) fluoride has a boiling point of 1680C. What is the density of a sample of gaseous thorium(IV) fluoride at its boiling point under a pressure of 2.5 atm in a 1.7-L container?
Which gas will effuse faster at 1680C, thorium(IV) fluoride or uranium(III) fluoride? How much faster?
130. Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6). A typical mixture might have methane 0.915 and ethane 0.085. What are the partial pres-sures of the two gases in a 15.00-L container of natural gas at 20.C and 1.44 atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
131. Use the following information to identify element A and com-pound B, then answer questions a and b.
An empty glass container has a mass of 658.572 g. It has a mass of 659.452 g after it has been filled with nitrogen gas at a pressure of 790. torr and a temperature of 15C. When the con-tainer is evacuated and refilled with a certain element (A) at a pressure of 745 torr and a temperature of 26C, it has a mass of 660.59 g.
Compound B, a gaseous organic compound that consists of 85.6% carbon and 14.4% hydrogen by mass, is placed in a stain-less steel vessel (10.68 L) with excess oxygen gas. The vessel is placed in a constant-temperature bath at 22C. The pressure in the vessel is 11.98 atm. In the bottom of the vessel is a container that is packed with Ascarite and a desiccant. Ascarite is asbestos impregnated with sodium hydroxide; it quantitatively absorbs carbon dioxide: 2NaOH1s2 CO21g2 ¡ Na2CO31s2 H2O1l2 4HSiCl31l2 ¡ SiH41g2 3SiCl41l2 Si1s2 3HCl1g2 ¡ HSiCl31l2 H21g2 25C. The air has a mole fraction of nitrogen of 0.790, the rest being oxygen.
a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. Be complete.
b. Above what temperature would you heat the balloon so that it would float?
123. You have a helium balloon at 1.00 atm and 25C. You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0% nitrogen, 21.0% oxygen by volume. The “lift” of a balloon is given by the difference be-tween the mass of air displaced by the balloon and the mass of gas inside the balloon.
a. Will the temperature in the hot-air balloon have to be higher or lower than 25C? Explain.
b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and 25C. Assume atmospheric conditions are 1.00 atm and 25C.
124. We state that the ideal gas law tends to hold best at low pres-sures and high temperatures. Show how the van der Waals equa-tion simplifies to the ideal gas law under these conditions.
125. Atmospheric scientists often use mixing ratios to express the con-centrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): On a recent autumn day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached 3.0 102 ppmv. The atmospheric pressure at that time was 628 torr, and the temperature was 0C.
a. What was the partial pressure of CO?
b. What was the concentration of CO in molecules per cubic centimeter?
126. Nitrogen gas (N2) reacts with hydrogen gas (H2) to form am-monia gas (NH3). You have nitrogen and hydrogen gases in a 15.0-L container fitted with a movable piston (the piston allows the container volume to change so as to keep the pressure con-stant inside the container). Initially the partial pressure of each reactant gas is 1.00 atm. Assume the temperature is constant and that the reaction goes to completion.
a. Calculate the partial pressure of ammonia in the container af-ter the reaction has reached completion.
b. Calculate the volume of the container after the reaction has reached completion.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
127. In the presence of nitric acid, UO2 undergoes a redox process.
It is converted to UO2 2 and nitric oxide (NO) gas is produced according to the following unbalanced equation: NO3 1aq2 UO21aq2 ¡ NO1g2 UO2 21aq2 ppmv of X vol. of X at STP total vol. of air at STP 106 Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
Questions give students an opportunity to review key concepts; Exercises (paired and organized by topic) reinforce students’ understanding of each section; Additional Exercises require students to identify and apply the appropriate concepts themselves; Challenge Problems take students one step further and challenge students more rigorously than Additional Exercises; Integrative Problems combine concepts from multiple chapters; Marathon Problems also combine concepts from multiple chapters, and they are the most challenging problems in the end-of-chapter material. Active Learning Questions are designed to promote discussion among groups of students in class.
xix Online Problem Solving and Practice Developed by the Zumdahls to reinforce the approach of the book, ChemWork interactive online homework offers problems accompanied by hints to help students as they think through each problem. ChemWork assignments are offered in Eduspace—Houghton Mifflin’s course-management system.
The Online Study Center features Visualization practice exercises.
Visualizations include animations and video demonstrations that help students to further understand chemical concepts.
Each Visualization is accompanied by quiz questions. xx Algorithmic, end-of-chapter exercises from the text also appear in Eduspace. Exercises also include helpful links to art, tables, and equations from the textbook.
HM ClassPrep with HM Testing (powered by Diploma) CD is a cross-platform CD that contains extensive text-specific resources for instructors to incorporate into their lecture presentations.
These customizable assets include PowerPoint slides, Word files of the printed Test Bank and Solutions Manual, figures from the text, the Instructor’s Resource Guide and more. HM Testing (powered by Diploma) is Houghton Mifflin’s new flexible test-editing program, which features algorithmically generated ques-tions, conceptual questions, and factual questions coded by level of difficulty to allow you to more easily choose appropriate test items. Select from 2400 test items designed to measure the concepts and principles covered in the seventh edition. HM ClassPresent includes animations and video demons-trations that can be used to illustrate concepts and ideas that will help students further understand and visualize chemical concepts. Animations and videos can be projected directly from the CD, exported to your computer, and also come embedded in PowerPoint files.
Online Teaching Center for Chemistry offers access to lecture preparation materials; PowerPoint presentation resources; JPEGs of virtually all text illustrations, tables, and photos; video demonstrations and animations; molecule library with CHIME; as well as service and support. Also included on the Online Teaching Center, you will find classroom response-system slides. These slides allow you to get on-the-spot feedback on how well your students are grasping key concepts.
Eduspace, featuring online homework, is Houghton Mifflin’s course-management system. Eduspace allows for online delivery of course materials, chat and discussion tools, and includes two types of algorithmic online homework: ChemWork and end-of-chapter exercises.
ChemWork helps students learn the process of problem solving with interactive hints that help students think through each problem. Media Resources for Instructors xxi The Online Study Center supports the goals of the seventh edition with visualization, practice, and study aids. The Visualizations use animations and video demonstrations to help students see the chemistry concepts, and each Visualization is accompanied by a set of quiz questions so that students can test their knowledge of the concept. The Online Study Center also includes an interactive review for each chapter, flashcards of key terms, and ACE practice tests, which help students prepare for quizzes and exams.
Many of the resources on the Online Study Center are also available on the optional free, student CD-ROM. Students get access to live, online help through SMARTHINKING™.
E-structors are available when students need it the most and help students problem-solve rather than supply answers. Available free with new books on instructor’s request. Also available via Eduspace.
Technological Resources for Students xxii This page intentionally left blank b 1Chemical Foundations Contents 1.1 Chemistry: An Overview • Science: A Process for Understanding Nature and Its Changes 1.2 The Scientific Method • Scientific Models 1.3 Units of Measurement 1.4 Uncertainty in Measurement • Precision and Accuracy 1.5 Significant Figures and Calculations 1.6 Dimensional Analysis 1.7 Temperature 1.8 Density 1.9 Classification of Matter Male Monarch butterflies use the pheromones produced by a gland on their wings to make themselves attractive to females.
When you start your car, do you think about chemistry? Probably not, but you should. The power to start your car is furnished by a lead storage battery. How does this battery work, and what does it contain? When a battery goes dead, what does that mean?
If you use a friend’s car to “jump start” your car, did you know that your battery could explode? How can you avoid such an unpleasant possibility? What is in the gasoline that you put in your tank, and how does it furnish the energy to drive to school? What is the vapor that comes out of the exhaust pipe, and why does it cause air pollution? Your car’s air conditioner might have a substance in it that is leading to the destruction of the ozone layer in the upper atmosphere. What are we doing about that? And why is the ozone layer important anyway?
All these questions can be answered by understanding some chemistry. In fact, we’ll consider the answers to all these questions in this text.
Chemistry is around you all the time. You are able to read and understand this sen-tence because chemical reactions are occurring in your brain. The food you ate for break-fast or lunch is now furnishing energy through chemical reactions. Trees and grass grow because of chemical changes.
Chemistry also crops up in some unexpected places. When archaeologist Luis Alvarez was studying in college, he probably didn’t realize that the chemical elements iridium and niobium would make him very famous when they helped him solve the problem of the disappearing dinosaurs. For decades scientists had wrestled with the mystery of why the dinosaurs, after ruling the earth for millions of years, suddenly became extinct 65 million years ago. In studying core samples of rocks dating back to that period, Alvarez and his coworkers recognized unusual levels of iridium and niobium in these samples—levels much more characteristic of extraterrestrial bodies than of the earth. Based on these observations, Alvarez hypothesized that a large meteor hit the earth 65 million years ago, changing atmospheric conditions so much that the dinosaurs’ food couldn’t grow, and they died—almost instantly in the geologic timeframe.
Chemistry is also important to historians. Did you realize that lead poisoning proba-bly was a significant contributing factor to the decline of the Roman Empire? The Romans had high exposure to lead from lead-glazed pottery, lead water pipes, and a sweetening syrup called sapa that was prepared by boiling down grape juice in lead-lined vessels. It turns out that one reason for sapa’s sweetness was lead acetate (“sugar of lead”) that formed as the juice was cooked down. Lead poisoning with its symptoms of lethargy and mental malfunctions certainly could have contributed to the demise of the Roman society.
Chemistry is also apparently very important in determining a person’s behavior.
Various studies have shown that many personality disorders can be linked directly to imbalances of trace elements in the body. For example, studies on the inmates at Stat-eville Prison in Illinois have linked low cobalt levels with violent behavior. Lithium salts have been shown to be very effective in controlling the effects of manic depressive dis-ease, and you’ve probably at some time in your life felt a special “chemistry” for another person. Studies suggest there is literally chemistry going on between two people who are attracted to each other. “Falling in love” apparently causes changes in the chemistry of the brain; chemicals are produced that give that “high” associated with a new relation-ship. Unfortunately, these chemical effects seem to wear off over time, even if the rela-tionship persists and grows.
The importance of chemistry in the interactions of people should not really surprise us, since we know that insects communicate by emitting and receiving chemical signals via molecules called pheromones. For example, ants have a very complicated set of chemical 1 2 Chapter One Chemical Foundations signals to signify food sources, danger, and so forth. Also, various female sex attractants have been isolated and used to lure males into traps to control insect populations. It would not be surprising if humans also emitted chemical signals that we were not aware of on a conscious level. Thus chemistry is pretty interesting and pretty important. The main goal of this text is to help you understand the concepts of chemistry so that you can better ap-preciate the world around you and can be more effective in whatever career you choose.
1.1 Chemistry: An Overview Since the time of the ancient Greeks, people have wondered about the answer to the ques-tion: What is matter made of? For a long time humans have believed that matter is com-posed of atoms, and in the previous three centuries we have collected much indirect evidence to support this belief. Very recently, something exciting has happened—for the first time we can “see” individual atoms. Of course, we cannot see atoms with the naked eye but must use a special microscope called a scanning tunneling microscope (STM).
Although we will not consider the details of its operation here, the STM uses an electron current from a tiny needle to probe the surface of a substance. The STM pictures of several substances are shown in Fig. 1.1. Notice how the atoms are connected to one another by “bridges,” which, as we will see, represent the electrons that interconnect atoms.
In addition to “seeing” the atoms in solids such as salt, we have learned how to iso-late and view a single atom. For example, the tiny white dot in the center of Fig. 1.2 is a single mercury atom that is held in a special trap.
So, at this point, we are fairly sure that matter consists of individual atoms. The na-ture of these atoms is quite complex, and the components of atoms don’t behave much like the objects we see in the world of our experience. We call this world the macroscopic world—the world of cars, tables, baseballs, rocks, oceans, and so forth. One of the main jobs of a scientist is to delve into the macroscopic world and discover its “parts.” For example, when you view a beach from a distance, it looks like a continuous solid substance.
As you get closer, you see that the beach is really made up of individual grains of sand.
(a) (b) (c) FIGURE 1.1 (a) The surface of a single grain of table salt. (b) An oxygen atom (indicated by arrow) on a gallium arsenide surface.
(c) Scanning tunneling microscope image showing rows of ring-shaped clusters of benzene molecules on a rhodium surface.
Each “doughnut”-shaped image represents a benzene molecule.
1.1 Chemistry: An Overview 3 As we examine these grains of sand, we find they are composed of silicon and oxygen atoms connected to each other to form intricate shapes (see Fig. 1.3). One of the main challenges of chemistry is to understand the connection between the macroscopic world that we experience and the microscopic world of atoms and molecules. To truly under-stand chemistry you must learn to think on the atomic level. We will spend much time in this text helping you learn to do that.
One of the amazing things about our universe is that the tremendous variety of sub-stances we find there results from only about 100 different kinds of atoms. You can think of these approximately 100 atoms as the letters in an alphabet out of which all the “words” in the universe are made. It is the way the atoms are organized in a given substance that determines the properties of that substance. For example, water, one of the most common and important substances on earth, is composed of two types of atoms: hydrogen and oxygen. There are two hydrogen atoms and one oxygen atom bound together to form the water molecule: When an electric current passes through it, water is decomposed to hydrogen and oxygen.
These chemical elements themselves exist naturally as diatomic (two-atom) molecules: We can represent the decomposition of water to its component elements, hydrogen and oxygen, as follows: Notice that it takes two molecules of water to furnish the right number of oxygen and hy-drogen atoms to allow for the formation of the two-atom molecules. This reaction explains one oxygen molecule written O2 two water molecules written 2H2O electric current two hydrogen molecules written 2H2 written O2 oxygen molecule written H2 hydrogen molecule hydrogen atom oxygen atom water molecule FIGURE 1.2 A charged mercury atom shows up as a tiny white dot (indicated by the arrow).
O Si FIGURE 1.3 Sand on a beach looks uniform from a distance, but up close the irregular sand grains are visible, and each grain is com-posed of tiny atoms.
4 Chapter One Chemical Foundations why the battery in your car can explode if you jump start it improperly. When you hook up the jumper cables, current flows through the dead battery, which contains water (and other things), and causes hydrogen and oxygen to form by decomposition of some of the water.
A spark can cause this accumulated hydrogen and oxygen to explode, forming water again.
This example illustrates two of the fundamental concepts of chemistry: (1) matter is com-posed of various types of atoms, and (2) one substance changes to another by reorganiz-ing the way the atoms are attached to each other.
These are core ideas of chemistry, and we will have much more to say about them.
O2 2H2O spark 2H2 T he importance of chemistry can show up in some unusual places. For example, a knowledge of chemistry is crucial to authenticating, preserving, and restoring art objects. The J. Paul Getty Museum in Los Angeles has a state-of-the-art chemical laboratory that costs many millions of dollars and employs many scientists. The National Gallery of Art (NGA) in Washington, D.C., also operates a highly sophisticated laboratory that employs 10 people: five chemists, a botanist, an art historian, a technician with a chemistry degree, and two fellows (interns).
One of the chemists at NGA is Barbara Berrie, who spe-cializes in identifying paint pigments. One of her duties is to analyze a painting to see whether the paint pigments are ap-propriate for the time the picture was supposedly painted and consistent with the pigments known to be used by the artist given credit for the painting. This analysis is one way in which paintings can be authenticated. One of Berrie’s recent projects was to analyze the 1617 oil painting St. Cecilia and an Angel.
Her results showed the painting was the work of two artists of the time, Orazio Gentileschi and Giovanni Lanfranco.
Originally the work was thought to be by Gentileschi alone.
Berrie is also working to define the range of colors used by water colorist Winslow Homer (the NGA has 30 Homer paintings in its collection) and to show how his color palette changed over his career. In addition, she is exploring how acidity affects the decomposition of a particular deep green transparent pigment (called copper resinate) used by Italian Renaissance artists so that paintings using this pigment can be better preserved.
Berrie says, “The chemistry I do is not hot-dog chem-istry, just good old-fashioned general chemistry.” CHEMICAL IMPACT The Chemistry of Art Dr. Barbara Berrie of the National Gallery of Art is shown analyzing the glue used in the wooden supports for a 14th century altar piece.
1.2 The Scientific Method 5 Science: A Process for Understanding Nature and Its Changes How do you tackle the problems that confront you in real life? Think about your trip to school. If you live in a city, traffic is undoubtedly a problem you confront daily. How do you decide the best way to drive to school? If you are new in town, you first get a map and look at the possible ways to make the trip. Then you might collect information from people who know the area about the advantages and disadvantages of various routes. Based on this information, you probably try to predict the best route. However, you can find the best route only by trying several of them and comparing the results. After a few experiments with the various possibilities, you probably will be able to select the best way. What you are doing in solving this everyday problem is applying the same process that scientists use to study nature. The first thing you did was collect relevant data. Then you made a prediction, and then you tested it by trying it out. This process contains the fundamental elements of science.
1. Making observations (collecting data) 2. Making a prediction (formulating a hypothesis) 3. Doing experiments to test the prediction (testing the hypothesis) Scientists call this process the scientific method. We will discuss it in more detail in the next section. One of life’s most important activities is solving problems—not “plug and chug” exercises, but real problems—problems that have new facets to them, that involve things you may have never confronted before. The more creative you are at solving these problems, the more effective you will be in your career and your per-sonal life. Part of the reason for learning chemistry, therefore, is to become a better problem solver. Chemists are usually excellent problem solvers, because to master chemistry, you have to master the scientific approach. Chemical problems are frequently very complicated—there is usually no neat and tidy solution. Often it is difficult to know where to begin.
1.2 The Scientific Method Science is a framework for gaining and organizing knowledge. Science is not simply a set of facts but also a plan of action—a procedure for processing and understanding cer-tain types of information. Scientific thinking is useful in all aspects of life, but in this text we will use it to understand how the chemical world operates. As we have said in our pre-vious discussion, the process that lies at the center of scientific inquiry is called the scientific method. There are actually many scientific methods, depending on the nature of the specific problem under study and on the particular investigator involved. However, it is useful to consider the following general framework for a generic scientific method (see Fig. 1.4): Steps in the Scientific Method ➥ 1 Making observations. Observations may be qualitative (the sky is blue; water is a liquid) or quantitative (water boils at 100C; a certain chemistry book weighs 2 kilograms). A qualitative observation does not involve a number.
A quantitative observation (called a measurement) involves both a number and a unit.
➥ 2 Formulating hypotheses. A hypothesis is a possible explanation for an observation.
➥ 3 Performing experiments. An experiment is carried out to test a hypothesis. This involves gathering new information that enables a scientist to decide whether Experiment Prediction Observation Hypothesis Experiment Theory (model) Theory modified as needed FIGURE 1.4 The fundamental steps of the scientific method.
6 Chapter One Chemical Foundations the hypothesis is valid—that is, whether it is supported by the new information learned from the experiment. Experiments always produce new observations, and this brings the process back to the beginning again.
To understand a given phenomenon, these steps are repeated many times, gradually ac-cumulating the knowledge necessary to provide a possible explanation of the phenomenon.
Scientific Models Once a set of hypotheses that agrees with the various observations is obtained, the hy-potheses are assembled into a theory. A theory, which is often called a model, is a set of tested hypotheses that gives an overall explanation of some natural phenomenon.
It is very important to distinguish between observations and theories. An observation is something that is witnessed and can be recorded. A theory is an interpretation—a possible explanation of why nature behaves in a particular way. Theories inevitably change as more information becomes available. For example, the motions of the sun and stars have remained virtually the same over the thousands of years during which humans have been observing them, but our explanations—our theories—for these motions have changed greatly since ancient times. (See the Chemical Impact on Observations, Theories, and the Planets on the Web site.) The point is that scientists do not stop asking questions just because a given the-ory seems to account satisfactorily for some aspect of natural behavior. They continue doing experiments to refine or replace the existing theories. This is generally done by using the currently accepted theory to make a prediction and then performing an experiment (making a new observation) to see whether the results bear out this prediction.
Always remember that theories (models) are human inventions. They represent at-tempts to explain observed natural behavior in terms of human experiences. A theory is actually an educated guess. We must continue to do experiments and to refine our theo-ries (making them consistent with new knowledge) if we hope to approach a more nearly complete understanding of nature.
As scientists observe nature, they often see that the same observation applies to many different systems. For example, studies of innumerable chemical changes have shown that the total observed mass of the materials involved is the same before and after the change.
Such generally observed behavior is formulated into a statement called a natural law. For example, the observation that the total mass of materials is not affected by a chemical change in those materials is called the law of conservation of mass.
Note the difference between a natural law and a theory. A natural law is a summary of observed (measurable) behavior, whereas a theory is an explanation of behavior. A law summarizes what happens; a theory (model) is an attempt to explain why it happens.
In this section we have described the scientific method as it might ideally be applied (see Fig. 1.5). However, it is important to remember that science does not always progress smoothly and efficiently. For one thing, hypotheses and observations are not totally inde-pendent of each other, as we have assumed in the description of the idealized scientific Experiment Prediction Observation Hypothesis Prediction Theory (model) Law Theory modified as needed FIGURE 1.5 The various parts of the scientific method.
Robert Boyle (1627–1691) was born in Ireland. He became especially interested in experiments involving air and developed an air pump with which he produced evacuated cylinders. He used these cylinders to show that a feather and a lump of lead fall at the same rate in the absence of air resistance and that sound cannot be produced in a vacuum. His most famous experiments involved careful measurements of the volume of a gas as a function of pressure. In his book The Skeptical Chymist, Boyle urged that the ancient view of elements as mystical substances should be abandoned and that an element should in-stead be defined as anything that cannot be broken down into simpler substances. This conception was an important step in the development of modern chemistry.
1.2 The Scientific Method 7 CHEMICAL IMPACT A Note-able Achievement P ost-it Notes, a product of the 3M Corporation, revolu-tionized casual written communications and personal reminders. Introduced in the United States in 1980, these sticky-but-not-too-sticky notes have now found countless uses in offices, cars, and homes throughout the world.
The invention of sticky notes occurred over a period of about 10 years and involved a great deal of serendipity. The adhesive for Post-it Notes was discovered by Dr. Spencer F.
Silver of 3M in 1968. Silver found that when an acrylate polymer material was made in a particular way, it formed cross-linked microspheres. When suspended in a solvent and sprayed on a sheet of paper, this substance formed a “sparse monolayer” of adhesive after the solvent evaporated. Scan-ning electron microscope images of the adhesive show that it has an irregular surface, a little like the surface of a gravel road. In contrast, the adhesive on cellophane tape looks smooth and uniform, like a superhighway. The bumpy sur-face of Silver’s adhesive caused it to be sticky but not so sticky to produce permanent adhesion, because the number of contact points between the binding surfaces was limited.
When he invented this adhesive, Silver had no specific ideas for its use, so he spread the word of his discovery to his fellow employees at 3M to see if anyone had an appli-cation for it. In addition, over the next several years devel-opment was carried out to improve the adhesive’s proper-ties. It was not until 1974 that the idea for Post-it Notes popped up. One Sunday Art Fry, a chemical engineer for 3M, was singing in his church choir when he became an-noyed that the bookmark in his hymnal kept falling out. He thought to himself that it would be nice if the bookmark were sticky enough to stay in place but not so sticky that it couldn’t be moved. Luckily, he remembered Silver’s glue— and the Post-it Note was born.
For the next three years Fry worked to overcome the manufacturing obstacles associated with the product. By 1977 enough Post-it Notes were being produced to supply 3M’s corporate headquarters, where the employees quickly became addicted to their many uses. Post-it Notes are now available in 62 colors and 25 shapes.
In the years since their introduction, 3M has heard some remarkable stories connected to the use of these notes. For example, a Post-it Note was applied to the nose of a corpo-rate jet, where it was intended to be read by the plane’s Las Vegas ground crew. Someone forgot to remove it, however.
The note was still on the nose of the plane when it landed in Minneapolis, having survived a take-off and landing and speeds of 500 miles per hour at temperatures as low as 56F. Stories on the 3M Web site also describe how a Post-it Note on the front door of a home survived the 140 mile per hour winds of Hurricane Hugo and how a foreign official accepted Post-it Notes in lieu of cash when a small bribe was needed to cut through bureaucratic hassles.
Post-it Notes have definitely changed the way we com-municate and remember things.
method. The coupling of observations and hypotheses occurs because once we begin to proceed down a given theoretical path, our hypotheses are unavoidably couched in the language of that theory. In other words, we tend to see what we expect to see and often fail to notice things that we do not expect. Thus the theory we are testing helps us be-cause it focuses our questions. However, at the very same time, this focusing process may limit our ability to see other possible explanations.
It is also important to keep in mind that scientists are human. They have prejudices; they misinterpret data; they become emotionally attached to their theories and thus lose objectivity; and they play politics. Science is affected by profit motives, budgets, fads, wars, and religious beliefs. Galileo, for example, was forced to recant his astronomical observations in the face of strong religious resistance. Lavoisier, the father of modern chemistry, was beheaded because of his political affiliations. Great progress in the chem-istry of nitrogen fertilizers resulted from the desire to produce explosives to fight wars.
The progress of science is often affected more by the frailties of humans and their institutions than by the limitations of scientific measuring devices. The scientific meth-ods are only as effective as the humans using them. They do not automatically lead to progress.
8 Chapter One Chemical Foundations 1.3 Units of Measurement Making observations is fundamental to all science. A quantitative observation, or mea-surement, always consists of two parts: a number and a scale (called a unit). Both parts must be present for the measurement to be meaningful.
In this textbook we will use measurements of mass, length, time, temperature, elec-tric current, and the amount of a substance, among others. Scientists recognized long ago that standard systems of units had to be adopted if measurements were to be useful. If every scientist had a different set of units, complete chaos would result. Unfortunately, different standards were adopted in different parts of the world. The two major systems are the English system used in the United States and the metric system used by most of the rest of the industrialized world. This duality causes a good deal of trouble; for exam-ple, parts as simple as bolts are not interchangeable between machines built using the two systems. As a result, the United States has begun to adopt the metric system.
Most scientists in all countries have for many years used the metric system. In 1960, an international agreement set up a system of units called the International System (le Système International in French), or the SI system. This system is based on the metric system and units derived from the metric system. The fundamental SI units are listed in Table 1.1. We will discuss how to manipulate these units later in this chapter.
Because the fundamental units are not always convenient (expressing the mass of a pin in kilograms is awkward), prefixes are used to change the size of the unit. These are listed in Table 1.2. Some common objects and their measurements in SI units are listed in Table 1.3.
Soda is commonly sold in 2-liter bottles— an example of the use of SI units in every-day life.
CHEMICAL IMPACT Critical Units!
H ow important are conversions from one unit to another?
If you ask the National Aeronautics and Space Admin-istration (NASA), very important! In 1999 NASA lost a $125 million Mars Climate Orbiter because of a failure to convert from English to metric units.
The problem arose because two teams working on the Mars mission were using different sets of units. NASA’s sci-entists at the Jet Propulsion Laboratory in Pasadena, Cali-fornia, assumed that the thrust data for the rockets on the Orbiter they received from Lockheed Martin Astronautics in Denver, which built the spacecraft, were in metric units. In reality, the units were English. As a result the Orbiter dipped 100 kilometers lower into the Mars atmosphere than planned and the friction from the atmosphere caused the craft to burn up.
NASA’s mistake refueled the controversy over whether Congress should require the United States to switch to the metric system. About 95% of the world now uses the met-ric system, and the United States is slowly switching from English to metric. For example, the automobile industry has adopted metric fasteners and we buy our soda in two-liter bottles.
Units can be very important. In fact, they can mean the difference between life and death on some occasions. In 1983, for example, a Canadian jetliner almost ran out of fuel when someone pumped 22,300 pounds of fuel into the aircraft in-stead of 22,300 kilograms. Remember to watch your units!
Artist’s conception of the lost Mars Climate Orbiter. 1.3 Units of Measurement 9 TABLE 1.1 The Fundamental SI Units Physical Quantity Name of Unit Abbreviation Mass kilogram kg Length meter m Time second s Temperature kelvin K Electric current ampere A Amount of substance mole mol Luminous intensity candela cd TABLE 1.3 Some Examples of Commonly Used Units Length A dime is 1 mm thick.
A quarter is 2.5 cm in diameter. The average height of an adult man is 1.8 m.
Mass A nickel has a mass of about 5 g. A 120-lb person has a mass of about 55 kg.
Volume A 12-oz can of soda has a volume of about 360 mL.
TABLE 1.2 The Prefixes Used in the SI System (Those most commonly encountered are shown in blue.) Exponential Prefix Symbol Meaning Notation exa E 1,000,000,000,000,000,000 1018 peta P 1,000,000,000,000,000 1015 tera T 1,000,000,000,000 1012 giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 103 hecto h 100 102 deka da 10 101 — — 1 100 deci d 0.1 101 centi c 0.01 102 milli m 0.001 103 micro 0.000001 106 nano n 0.000000001 109 pico p 0.000000000001 1012 femto f 0.000000000000001 1015 atto a 0.000000000000000001 1018 See Appendix 1.1 if you need a review of exponential notation.
1 dm3 = 1 L 1 cm 1 cm 1 cm3 = 1 mL 1 m3 FIGURE 1.6 The largest cube has sides 1 m in length and a volume of 1 m3. The middle-sized cube has sides 1 dm in length and a vol-ume of 1 dm3, or 1 L. The smallest cube has sides 1 cm in length and a volume of 1 cm3, or 1 mL.
One physical quantity that is very important in chemistry is volume, which is not a fun-damental SI unit but is derived from length. A cube that measures 1 meter (m) on each edge is represented in Fig. 1.6. This cube has a volume of (1 m)3 1 m3. Recognizing that there are 10 decimeters (dm) in a meter, the volume of this cube is (1 m)3 (10 dm)3 1000 dm3. A cubic decimeter, that is (1 dm)3, is commonly called a liter (L), which is a unit of volume slightly larger than a quart. As shown in Fig. 1.6, 1000 liters are contained in a cube with a volume of 1 cubic meter. Similarly, since 1 decimeter equals 10 centimeters (cm), the liter can be divided into 1000 cubes each with a volume of 1 cubic centimeter: Also, since 1 cm3 1 milliliter (mL), Thus 1 liter contains 1000 cubic centimeters, or 1000 milliliters.
Chemical laboratory work frequently requires measurement of the volumes of liquids.
Several devices for the accurate determination of liquid volume are shown in Fig. 1.7.
An important point concerning measurements is the relationship between mass and weight. Although these terms are sometimes used interchangeably, they are not the same.
1 liter 1000 cm3 1000 mL 1 liter 11 dm23 110 cm23 1000 cm3 10 Chapter One Chemical Foundations Mass is a measure of the resistance of an object to a change in its state of motion. Mass is measured by the force necessary to give an object a certain acceleration. On earth we use the force that gravity exerts on an object to measure its mass. We call this force the object’s weight. Since weight is the response of mass to gravity, it varies with the strength of the gravitational field. Therefore, your body mass is the same on the earth or on the moon, but your weight would be much less on the moon than on earth because of the moon’s smaller gravitational field.
Because weighing something on a chemical balance (see Fig. 1.8) involves compar-ing the mass of that object to a standard mass, the terms weight and mass are sometimes used interchangeably, although this is incorrect.
1.4 Uncertainty in Measurement The number associated with a measurement is obtained using some measuring device. For example, consider the measurement of the volume of a liquid using a buret (shown in Fig. 1.9 with the scale greatly magnified). Notice that the meniscus of the liquid occurs at about 20.15 milliliters. This means that about 20.15 mL of liquid has been delivered from the bu-ret (if the initial position of the liquid meniscus was 0.00 mL). Note that we must estimate the last number of the volume reading by interpolating between the 0.1-mL marks. Since the last number is estimated, its value may be different if another person makes the same measurement. If five different people read the same volume, the results might be as follows: 100 mL 0 1 2 3 4 50 49 48 47 46 45 44 mL 90 80 70 60 50 40 30 20 10 100-mL graduated cylinder 250-mL volumetric flask 50-mL buret 25-mL pipet Calibration mark indicates 25-mL volume Valve (stopcock) controls the liquid flow Calibration mark indicates 250-mL volume FIGURE 1.7 Common types of laboratory equipment used to measure liquid volume.
FIGURE 1.8 An electronic analytical balance.
Person Results of Measurement 1 20.15 mL 2 20.14 mL 3 20.16 mL 4 20.17 mL 5 20.16 mL FIGURE 1.9 Measurement of volume using a buret. The volume is read at the bottom of the liquid curve (called the meniscus).
20 21 22 23 24 25 mL 20 1.4 Uncertainty in Measurement 11 Bathroom Scale Balance Grapefruit 1 1.5 lb 1.476 lb Grapefruit 2 1.5 lb 1.518 lb These results show that the first three numbers (20.1) remain the same regardless of who makes the measurement; these are called certain digits. However, the digit to the right of the 1 must be estimated and therefore varies; it is called an uncertain digit. We custom-arily report a measurement by recording all the certain digits plus the first uncertain digit.
In our example it would not make any sense to try to record the volume of thousandths of a milliliter because the value for hundredths of a milliliter must be estimated when using the buret.
It is very important to realize that a measurement always has some degree of uncer-tainty. The uncertainty of a measurement depends on the precision of the measuring de-vice. For example, using a bathroom scale, you might estimate the mass of a grapefruit to be approximately 1.5 pounds. Weighing the same grapefruit on a highly precise bal-ance might produce a result of 1.476 pounds. In the first case, the uncertainty occurs in the tenths of a pound place; in the second case, the uncertainty occurs in the thousandths of a pound place. Suppose we weigh two similar grapefruits on the two devices and obtain the following results: Do the two grapefruits have the same mass? The answer depends on which set of results you consider. Thus a conclusion based on a series of measurements depends on the cer-tainty of those measurements. For this reason, it is important to indicate the uncertainty in any measurement. This is done by always recording the certain digits and the first un-certain digit (the estimated number). These numbers are called the significant figures of a measurement.
The convention of significant figures automatically indicates something about the un-certainty in a measurement. The uncertainty in the last number (the estimated number) is usually assumed to be 1 unless otherwise indicated. For example, the measurement 1.86 kilograms can be taken to mean 1.86 0.01 kilograms.
Uncertainty in Measurement In analyzing a sample of polluted water, a chemist measured out a 25.00-mL water sample with a pipet (see Fig. 1.7). At another point in the analysis, the chemist used a graduated cylinder (see Fig. 1.7) to measure 25 mL of a solution. What is the difference between the measurements 25.00 mL and 25 mL?
Solution Even though the two volume measurements appear to be equal, they really convey different information. The quantity 25 mL means that the volume is between 24 mL and 26 mL, whereas the quantity 25.00 mL means that the volume is between 24.99 mL and 25.01 mL. The pipet measures volume with much greater precision than does the grad-uated cylinder.
See Question 1.8.
When making a measurement, it is important to record the results to the appropriate number of significant figures. For example, if a certain buret can be read to 0.01 mL, A measurement always has some degree of uncertainty.
Uncertainty in measurement is discussed in more detail in Appendix 1.5.
Sample Exercise 1.1 12 Chapter One Chemical Foundations you should record a reading of twenty-five milliliters as 25.00 mL, not 25 mL. This way at some later time when you are using your results to do calculations, the uncertainty in the measurement will be known to you.
Precision and Accuracy Two terms often used to describe the reliability of measurements are precision and accu-racy. Although these words are frequently used interchangeably in everyday life, they have different meanings in the scientific context. Accuracy refers to the agreement of a par-ticular value with the true value. Precision refers to the degree of agreement among sev-eral measurements of the same quantity. Precision reflects the reproducibility of a given type of measurement. The difference between these terms is illustrated by the results of three different dart throws shown in Fig. 1.10.
Two different types of errors are illustrated in Fig. 1.10. A random error (also called an indeterminate error) means that a measurement has an equal probability of being high or low. This type of error occurs in estimating the value of the last digit of a measure-ment. The second type of error is called systematic error (or determinate error). This type of error occurs in the same direction each time; it is either always high or always low. Figure 1.10(a) indicates large random errors (poor technique). Figure 1.10(b) indi-cates small random errors but a large systematic error, and Figure 1.10(c) indicates small random errors and no systematic error.
In quantitative work, precision is often used as an indication of accuracy; we assume that the average of a series of precise measurements (which should “average out” the ran-dom errors because of their equal probability of being high or low) is accurate, or close to the “true” value. However, this assumption is valid only if systematic errors are absent.
Suppose we weigh a piece of brass five times on a very precise balance and obtain the following results: Weighing Result 1 2.486 g 2 2.487 g 3 2.485 g 4 2.484 g 5 2.488 g (a) (b) (c) FIGURE 1.10 The results of several dart throws show the difference between precise and accurate.
(a) Neither accurate nor precise (large ran-dom errors). (b) Precise but not accurate (small random errors, large systematic error).
(c) Bull’s-eye! Both precise and accurate (small random errors, no systematic error).
Sample Exercise 1.2 Normally, we would assume that the true mass of the piece of brass is very close to 2.486 grams, which is the average of the five results: However, if the balance has a defect causing it to give a result that is consistently 1.000 gram too high (a systematic error of 1.000 gram), then the measured value of 2.486 grams would be seriously in error. The point here is that high precision among several mea-surements is an indication of accuracy only if systematic errors are absent.
Precision and Accuracy To check the accuracy of a graduated cylinder, a student filled the cylinder to the 25-mL mark using water delivered from a buret (see Fig. 1.7) and then read the volume deliv-ered. Following are the results of five trials: 2.486 g 2.487 g 2.485 g 2.484 g 2.488 g 5 2.486 g 1.5 Significant Figures and Calculations 13 Volume Shown by Volume Shown Trial Graduated Cylinder by the Buret 1 25 mL 26.54 mL 2 25 mL 26.51 mL 3 25 mL 26.60 mL 4 25 mL 26.49 mL 5 25 mL 26.57 mL Average 25 mL 26.54 mL Is the graduated cylinder accurate?
Solution The results of the trials show very good precision (for a graduated cylinder). The student has good technique. However, note that the average value measured using the buret is sig-nificantly different from 25 mL. Thus this graduated cylinder is not very accurate. It pro-duces a systematic error (in this case, the indicated result is low for each measurement).
See Question 1.11.
1.5 Significant Figures and Calculations Calculating the final result for an experiment usually involves adding, subtracting, multi-plying, or dividing the results of various types of measurements. Since it is very impor-tant that the uncertainty in the final result is known correctly, we have developed rules for counting the significant figures in each number and for determining the correct number of significant figures in the final result.
Rules for Counting Significant Figures 1. Nonzero integers. Nonzero integers always count as significant figures.
2. Zeros. There are three classes of zeros: a. Leading zeros are zeros that precede all the nonzero digits. These do not count as significant figures. In the number 0.0025, the three zeros simply indicate the po-sition of the decimal point. This number has only two significant figures.
b. Captive zeros are zeros between nonzero digits. These always count as significant figures. The number 1.008 has four significant figures.
c. Trailing zeros are zeros at the right end of the number. They are significant only if the number contains a decimal point. The number 100 has only one significant figure, whereas the number 1.00 102 has three significant figures. The number one hundred written as 100. also has three significant figures.
3. Exact numbers. Many times calculations involve numbers that were not obtained us-ing measuring devices but were determined by counting: 10 experiments, 3 apples, 8 molecules. Such numbers are called exact numbers. They can be assumed to have an infinite number of significant figures. Other examples of exact numbers are the 2 in 2r (the circumference of a circle) and the 4 and the 3 in (the volume of a sphere). Exact numbers also can arise from definitions. For example, one inch is defined as exactly 2.54 centimeters. Thus, in the statement 1 in 2.54 cm, neither the 2.54 nor the 1 limits the number of significant figures when used in a calculation.
Note that the number 1.00 102 above is written in exponential notation. This type of notation has at least two advantages: the number of significant figures can be easily 4 3pr 3 Precision is an indication of accuracy only if there are no systematic errors.
Leading zeros are never significant figures.
Captive zeros are always significant figures.
Exact numbers never limit the number of significant figures in a calculation.
Trailing zeros are sometimes significant figures.
Exponential notation is reviewed in Appendix 1.1.
14 Chapter One Chemical Foundations indicated, and fewer zeros are needed to write a very large or very small number. For example, the number 0.000060 is much more conveniently represented as 6.0 1025. (The number has two significant figures.) Significant Figures Give the number of significant figures for each of the following results.
a. A student’s extraction procedure on tea yields 0.0105 g of caffeine.
b. A chemist records a mass of 0.050080 g in an analysis.
c. In an experiment a span of time is determined to be 8.050 103 s.
Solution a. The number contains three significant figures. The zeros to the left of the 1 are lead-ing zeros and are not significant, but the remaining zero (a captive zero) is significant.
b. The number contains five significant figures. The leading zeros (to the left of the 5) are not significant. The captive zeros between the 5 and the 8 are significant, and the trail-ing zero to the right of the 8 is significant because the number contains a decimal point.
c. This number has four significant figures. Both zeros are significant.
See Exercises 1.25 through 1.28.
To this point we have learned to count the significant figures in a given number. Next, we must consider how uncertainty accumulates as calculations are carried out. The detailed analysis of the accumulation of uncertainties depends on the type of calculation involved and can be complex. However, in this textbook we will employ the following simple rules that have been developed for determining the appropriate number of significant figures in the result of a calculation.
Rules for Significant Figures in Mathematical Operations 1. For multiplication or division, the number of significant figures in the result is the same as the number in the least precise measurement used in the calculation. For example, consider the calculation 8 88888n 6.4 h Corrected h Limiting term has Two significant two significant figures figures The product should have only two significant figures, since 1.4 has two significant figures.
2. For addition or subtraction, the result has the same number of decimal places as the least precise measurement used in the calculation. For example, consider the sum 12.11 18.0 m Limiting term has one decimal place 1.013 Corrected 31.123 8 88888n 31.1 h One decimal place The correct result is 31.1, since 18.0 has only one decimal place.
4.56 1.4 6.38 Sample Exercise 1.3 Although these simple rules work well for most cases, they can give misleading results in certain cases.
For more information, see L. M. Schwartz, “Propagation of Significant Figures,” J. Chem. Ed. 62 (1985): 693; and H. Bradford Thompson, “Is 8C equal to 50F?” J. Chem. Ed. 68 (1991): 400.
1.5 Significant Figures and Calculations 15 Note that for multiplication and division, significant figures are counted. For addition and subtraction, the decimal places are counted.
In most calculations you will need to round numbers to obtain the correct number of significant figures. The following rules should be applied when rounding.
Rules for Rounding 1. In a series of calculations, carry the extra digits through to the final result, then round.
2. If the digit to be removed a. is less than 5, the preceding digit stays the same. For example, 1.33 rounds to 1.3.
b. is equal to or greater than 5, the preceding digit is increased by 1. For example, 1.36 rounds to 1.4.
Although rounding is generally straightforward, one point requires special emphasis.
As an illustration, suppose that the number 4.348 needs to be rounded to two significant figures. In doing this, we look only at the first number to the right of the 3: 4.348 h Look at this number to round to two significant figures.
The number is rounded to 4.3 because 4 is less than 5. It is incorrect to round sequentially.
For example, do not round the 4 to 5 to give 4.35 and then round the 3 to 4 to give 4.4.
When rounding, use only the first number to the right of the last significant figure.
It is important to note that Rule 1 above usually will not be followed in the Sam-ple Exercises in this text because we want to show the correct number of significant figures in each step of a problem. This same practice is followed for the detailed solutions given in the Solutions Guide. However, as stated in Rule 1, the best proce-dure is to carry extra digits throughout a series of calculations and round to the correct number of significant figures only at the end. This is the practice you should follow.
The fact that your rounding procedures are different from those used in this text must be taken into account when you check your answer with the one given at the end of the book or in the Solutions Guide. Your answer (based on rounding only at the end of a calculation) may differ in the last place from that given here as the “correct” answer because we have rounded after each step. To help you understand the differ-ence between these rounding procedures, we will consider them further in Sample Exercise 1.4.
Significant Figures in Mathematical Operations Carry out the following mathematical operations, and give each result with the correct number of significant figures.
a. 1.05 103 6.135 b. 21 13.8 c. As part of a lab assignment to determine the value of the gas constant (R), a student measured the pressure (P), volume (V), and temperature (T ) for a sample of gas, where The following values were obtained: P 2.560, T 275.15, and V 8.8. (Gases will be discussed in detail in Chapter 5; we will not be concerned at this time about the units for these quantities.) Calculate R to the correct number of significant figures.
R PV T Rule 2 is consistent with the operation of electronic calculators.
Do not round sequentially. The number 6.8347 rounded to three significant figures is 6.83, not 6.84.
Sample Exercise 1.4 16 Chapter One Chemical Foundations Solution a. The result is 1.71 104, which has three significant figures because the term with the least precision (1.05 103) has three significant figures.
b. The result is 7 with no decimal point because the number with the least number of decimal places (21) has none.
c.
The correct procedure for obtaining the final result can be represented as follows: The final result must be rounded to two significant figures because 8.8 (the least precise measurement) has two significant figures. To show the effects of rounding at intermedi-ate steps, we will carry out the calculation as follows: Rounded to two significant figures g Now we proceed with the next calculation: Rounded to two significant figures, this result is Note that intermediate rounding gives a significantly different result than was obtained by rounding only at the end. Again, we must reemphasize that in your calculations you should round only at the end. However, because rounding is carried out at intermediate steps in this text (to always show the correct number of significant figures), the final answer given in the text may differ slightly from the one you obtain (rounding only at the end).
See Exercises 1.31 through 1.34.
There is a useful lesson to be learned from part c of Sample Exercise 1.4. The stu-dent measured the pressure and temperature to greater precision than the volume. A more precise value of R (one with more significant figures) could have been obtained if a more precise measurement of V had been made. As it is, the efforts expended to measure P and T very precisely were wasted. Remember that a series of measurements to obtain some final result should all be done to about the same precision.
1.6 Dimensional Analysis It is often necessary to convert a given result from one system of units to another. The best way to do this is by a method called the unit factor method, or more commonly dimensional analysis. To illustrate the use of this method, we will consider several unit conversions. Some equivalents in the English and metric systems are listed in Table 1.4.
A more complete list of conversion factors given to more significant figures appears in Appendix 6.
0.084 8.4 102 23 275.15 0.0835908 12.560218.82 275.15 22.528 275.15 23 275.15 0.082 8.2 102 R 12.560218.82 275.15 22.528 275.15 0.0818753 R PV T 12.560218.82 275.15 This number must be rounded to two significant figures.
TABLE 1.4 English–Metric Equivalents Length 1 m 1.094 yd 2.54 cm 1 in Mass 1 kg 2.205 lb 453.6 g 1 lb Volume 1 L 1.06 qt 1 ft3 28.32 L 1.6 Dimensional Analysis 17 Consider a pin measuring 2.85 centimeters in length. What is its length in inches? To accomplish this conversion, we must use the equivalence statement If we divide both sides of this equation by 2.54 centimeters, we get This expression is called a unit factor. Since 1 inch and 2.54 centimeters are exactly equiv-alent, multiplying any expression by this unit factor will not change its value.
The pin has a length of 2.85 centimeters. Multiplying this length by the appropriate unit factor gives Note that the centimeter units cancel to give inches for the result. This is exactly what we wanted to accomplish. Note also that the result has three significant figures, as required by the number 2.85. Recall that the 1 and 2.54 in the conversion factor are exact numbers by definition.
Unit Conversions I A pencil is 7.00 in long. What is its length in centimeters?
Solution In this case we want to convert from inches to centimeters. Therefore, we must use the reciprocal of the unit factor used above to do the opposite conversion: Here the inch units cancel, leaving centimeters, as requested.
See Exercises 1.37 and 1.38.
Note that two unit factors can be derived from each equivalence statement. For example, from the equivalence statement 2.54 cm 1 in, the two unit factors are How do you choose which one to use in a given situation? Simply look at the direction of the required change. To change from inches to centimeters, the inches must cancel.
Thus the factor 2.54 cm/1 in is used. To change from centimeters to inches, centimeters must cancel, and the factor 1 in/2.54 cm is appropriate.
Converting from One Unit to Another To convert from one unit to another, use the equivalence statement that relates the two units.
Derive the appropriate unit factor by looking at the direction of the required change (to cancel the unwanted units).
Multiply the quantity to be converted by the unit factor to give the quantity with the desired units.
2.54 cm 1 in and 1 in 2.54 cm 7.00 in 2.54 cm 1 in 17.00212.542 cm 17.8 cm 2.85 cm 1 in 2.54 cm 2.85 2.54 in 1.12 in 1 1 in 2.54 cm 2.54 cm 1 in Sample Exercise 1.5 Consider the direction of the required change to select the correct unit factor.
18 Chapter One Chemical Foundations Unit Conversions II You want to order a bicycle with a 25.5-in frame, but the sizes in the catalog are given only in centimeters. What size should you order?
Solution You need to go from inches to centimeters, so 2.54 cm 1 in is appropriate: See Exercises 1.37 and 1.38.
To ensure that the conversion procedure is clear, a multistep problem is considered in Sample Exercise 1.7.
Unit Conversions III A student has entered a 10.0-km run. How long is the run in miles?
Solution This conversion can be accomplished in several different ways. Since we have the equiv-alence statement 1 m 1.094 yd, we will proceed by a path that uses this fact. Before we start any calculations, let us consider our strategy. We have kilometers, which we want to change to miles. We can do this by the following route: To proceed in this way, we need the following equivalence statements: To make sure the process is clear, we will proceed step by step: Kilometers to Meters: Meters to Yards: Note that we should have only three significant figures in the result. Since this is an interme-diate result, however, we will carry the extra digit. Remember, round off only the final result.
Yards to Miles: Note in this case that 1 mi equals exactly 1760 yd by designation. Thus 1760 is an exact number.
Since the distance was originally given as 10.0 km, the result can have only three sig-nificant figures and should be rounded to 6.22 mi. Thus 10.0 km 6.22 mi 1.094 104 yd 1 mi 1760 yd 6.216 mi 1.00 104 m 1.094 yd 1 m 1.094 104 yd 10.0 km 1000 m 1 km 1.00 104 m 1760 yd 1 mi 1 m 1.094 yd 1 km 1000 m kilometers ¡ meters ¡ yards ¡ miles 25.5 in 2.54 cm 1 in 64.8 cm Sample Exercise 1.6 Sample Exercise 1.7 Normally we round to the correct number of significant figures after each step.
However, you should round only at the end.
1.7 Temperature 19 Alternatively, we can combine the steps: See Exercises 1.37 and 1.38.
In using dimensional analysis, your verification that everything has been done cor-rectly is that you end up with the correct units. In doing chemistry problems, you should always include the units for the quantities used. Always check to see that the units cancel to give the correct units for the final result. This provides a very valuable check, espe-cially for complicated problems.
Study the procedures for unit conversions in the following Sample Exercises.
Unit Conversion IV The speed limit on many highways in the United States is 55 mi/h. What number would be posted in kilometers per hour?
Solution Note that all units cancel except the desired kilometers per hour.
See Exercises 1.43 through 1.45.
Unit Conversions V A Japanese car is advertised as having a gas mileage of 15 km/L. Convert this rating to miles per gallon.
Solution See Exercise 1.46.
1.7 Temperature Three systems for measuring temperature are widely used: the Celsius scale, the Kelvin scale, and the Fahrenheit scale. The first two temperature systems are used in the physi-cal sciences, and the third is used in many of the engineering sciences. Our purpose here is to define the three temperature scales and show how conversions from one scale to another can be performed. Although these conversions can be carried out routinely on most calculators, we will consider the process in some detail here to illustrate methods of problem solving.
The three temperature scales are defined and compared in Fig. 1.11. Note that the size of the temperature unit (the degree) is the same for the Kelvin and Celsius scales.
15 km L 1000 m 1 km 1.094 yd 1 m 1 mi 1760 yd 1 L 1.06 qt 4 qt 1 gal 35 mi/gal 55 mi h 1760 yd 1 mi 1 m 1.094 yd 1 km 1000 m 88 km/h 10.0 km 1000 m 1 km 1.094 yd 1 m 1 mi 1760 yd 6.22 mi Sample Exercise 1.8 Sample Exercise 1.9 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 88888n Result obtained by rounding only at the end of the calculation Result obtained by rounding only at the end of the calculation n ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 20 Chapter One Chemical Foundations The fundamental difference between these two temperature scales is in their zero points.
Conversion between these two scales simply requires an adjustment for the different zero points.
or For example, to convert 300.00 K to the Celsius scale, we do the following calculation: Note that in expressing temperature in Celsius units, the designation C is used.
The degree symbol is not used when writing temperature in terms of the Kelvin scale. The unit of temperature on this scale is called a kelvin and is symbolized by the letter K.
Converting between the Fahrenheit and Celsius scales is somewhat more complicated because both the degree sizes and the zero points are different. Thus we need to consider two adjustments: one for degree size and one for the zero point. First, we must account for the difference in degree size. This can be done by reconsidering Fig. 1.11. Notice that since 212F 100C and 32F 0C, Thus 180 on the Fahrenheit scale is equivalent to 100 on the Celsius scale, and the unit factor is or the reciprocal, depending on the direction in which we need to go.
Next, we must consider the different zero points. Since 32F 0C, we obtain the cor-responding Celsius temperature by first subtracting 32 from the Fahrenheit temperature 180°F 100°C or 9°F 5°C 212 32 180 Fahrenheit degrees 100 0 100 Celsius degrees 300.00 273.15 26.85°C Temperature 1Celsius2 temperature 1Kelvin2 273.15 Temperature 1Kelvin2 temperature 1Celsius2 273.15 TK TC 273.15 TC TK 273.15 180 Fahrenheit degrees Boiling point of water 32°F 212°F Freezing point of water Fahrenheit 100 Celsius degrees 0°C –40°C 100°C Celsius 273.15 K 233.15 K 373.15 K Kelvin –40°F 100 kelvins FIGURE 1.11 The three major temperature scales.
1.7 Temperature 21 to account for the different zero points. Then the unit factor is applied to adjust for the difference in the degree size. This process is summarized by the equation (1.1) where TF and TC represent a given temperature on the Fahrenheit and Celsius scales, respectively. In the opposite conversion, we first correct for degree size and then correct for the different zero point. This process can be summarized in the following general equation: (1.2) Equations (1.1) and (1.2) are really the same equation in different forms. See if you can obtain Equation (1.2) by starting with Equation (1.1) and rearranging.
At this point it is worthwhile to weigh the two alternatives for learning to do tem-perature conversions: You can simply memorize the equations, or you can take the time to learn the differences between the temperature scales and to understand the processes involved in converting from one scale to another. The latter approach may take a little more effort, but the understanding you gain will stick with you much longer than the mem-orized formulas. This choice also will apply to many of the other chemical concepts. Try to think things through!
Temperature Conversions I Normal body temperature is 98.6°F. Convert this temperature to the Celsius and Kelvin scales.
Solution Rather than simply using the formulas to solve this problem, we will proceed by think-ing it through. The situation is diagramed in Fig. 1.12. First, we want to convert 98.6°F to the Celsius scale. The number of Fahrenheit degrees between 32.0°F and 98.6°F is 66.6°F. We must convert this difference to Celsius degrees: 66.6°F 5°C 9°F 37.0°C TF TC 9°F 5°C 32°F 1TF 32°F2 5°C 9°F TC Understand the process of converting from one temperature scale to another; do not simply memorize the equations.
66.6°F 32°F 98.6°F Fahrenheit 0°C ?°C Celsius 273.15 K ? K Kelvin 5°C 9°F × 66.6°F = 37.0°C + 37.0 = 310.2 K 273.15 K FIGURE 1.12 Normal body temperature on the Fahren-heit, Celsius, and Kelvin scales.
Sample Exercise 1.10 A physician taking the temperature of a patient.
22 Chapter One Chemical Foundations Thus 98.6°F corresponds to 37.0°C.
Now we can convert to the Kelvin scale: Note that the final answer has only one decimal place (37.0 is limiting).
See Exercises 1.49, 1.51, and 1.52.
Temperature Conversions II One interesting feature of the Celsius and Fahrenheit scales is that 40°C and 40°F represent the same temperature, as shown in Fig. 1.11. Verify that this is true.
Solution The difference between 32°F and 40°F is 72°F. The difference between 0°C and 40°C is 40°C. The ratio of these is as required. Thus 40°C is equivalent to 40°F.
See Challenge Problem 1.86.
Since, as shown in Sample Exercise 1.11, 40 on both the Fahrenheit and Celsius scales represents the same temperature, this point can be used as a reference point (like 0C and 32F) for a relationship between the two scales: Number of Fahrenheit degrees Number of Celsius degrees TF 1402 TC 1402 9°F 5°C 72°F 40°C 8 9°F 8 5°C 9°F 5°C TK TC 273.15 37.0 273.15 310.2 K Sample Exercise 1.11 CHEMICAL IMPACT Faux Snow S kiing is challenging and fun, but it is also big business.
Both skiers and ski operators want the season to last as long as possible. The major factor in maximizing the length of the ski season and in salvaging dry periods dur-ing the winter is the ability to “make snow.” Machine-made snow is now a required part of maintaining ideal conditions at major ski areas such as Aspen, Breckenridge, and Taos.
Snow is relatively easy to make if the air is cold enough.
To manufacture snow, water is cooled to just above 0C and then pumped at high pressure through a “gun” that produces a fine mist of water droplets that freeze before dropping to the ground. As might be expected, atmospheric conditions are critical when making snow. With an air temperature of 8C (18F) or less, untreated water can be used in the snow guns. However, the ideal type of snow for skiing is “powder”—fluffy snow made up of small, individual crys-tals. To achieve powdery snow requires sufficient nucleation sites—that is, sites where crystal growth is initiated. This condition can be achieved by “doping” the water with ions such as calcium or magnesium or with fine particles of clay.
Also, when the air temperature is between 0C and 8C, materials such as silver iodide, detergents, and organic ma-terials may be added to the water to seed the snow.
A discovery at the University of Wisconsin in the 1970s led to the additive most commonly used for snow making.
The Wisconsin scientists found that a bacterium (Pseudo-manas syringae) commonly found in nature makes a pro-tein that acts as a very effective nucleation site for ice for-mation. In fact, this discovery helped to explain why ice forms at 0C on the blossoms of fruit trees instead of the water supercooling below 0C, as pure water does when the temperature is lowered slowly below the freezing point. To help protect fruit blossoms from freeze damage, this bac-terium has been genetically modified to remove the ice nucleation protein. As a result, fruit blossoms can survive 1.7 Temperature 23 or (1.3) where TF and TC represent the same temperature (but not the same number). This equation can be used to convert Fahrenheit temperatures to Celsius, and vice versa, and may be easier to remember than Equations (1.1) and (1.2).
Temperature Conversions III Liquid nitrogen, which is often used as a coolant for low-temperature experiments, has a boiling point of 77 K. What is this temperature on the Fahrenheit scale?
Solution We will first convert 77 K to the Celsius scale: To convert to the Fahrenheit scale, we will use Equation (1.3): See Exercises 1.49, 1.51, and 1.52.
TF 281°F 40 321°F TF 40 9°F 5°C 1156°C2 281°F TF 40 196°C 40 TF 40 156°C 9°F 5°C TF 40 TC 40 9°F 5°C TC TK 273.15 77 273.15 196°C TF 40 TC 40 9°F 5°C Sample Exercise 1.12 Liquid nitrogen is so cold that water condenses out of the surrounding air, forming a cloud as the nitrogen is poured.
intact even if the temperature briefly falls below 0C. (See the Chemical Impact on Organisms and Ice Formation on page 516.) For snow-making purposes, this protein forms the basis for Snowmax (prepared and sold by York Snow of Victor, New York), which is the most popular additive for snow making.
Obviously, snow cannot be made in the summer, so what is a skiing fanatic to do during the warm months?
The answer is “dryslope” skiing. Although materials for dryslopes can be manufactured in a variety of ways, poly-mers are most commonly used for this application. One company that makes a multilayer polymer for artificial ski slopes is Briton Engineering Developments (Yorkshire, England), the producer of Snowflex. Snowflex consists of a slippery polymer fiber placed on top of a shock-absorbing base and lubricated by misting water through holes in its surface. Of course, this virtual skiing is not much like the real thing but it does provide some relief for summer ski withdrawal.
As artificial and synthetic snow amply demonstrate, chemistry makes life more fun.
A freestyle ski area at Sheffield Ski Village, in England, uses Snowflex “virtual snow” for year-round fun.
24 Chapter One Chemical Foundations Compound Density in g/cm3 at 20C Chloroform 1.492 Diethyl ether 0.714 Ethanol 0.789 Isopropyl alcohol 0.785 Toluene 0.867 Sample Exercise 1.13 There are two ways of indicating units that occur in the denominator. For example, we can write g/cm3 or g cm3.
Although we will use the former system here, the other system is widely used.
1.8 Density A property of matter that is often used by chemists as an “identification tag” for a sub-stance is density, the mass of substance per unit volume of the substance: The density of a liquid can be determined easily by weighing an accurately known volume of liquid. This procedure is illustrated in Sample Exercise 1.13.
Determining Density A chemist, trying to identify the main component of a compact disc cleaning fluid, finds that 25.00 cm3 of the substance has a mass of 19.625 g at 20°C. The following are the names and densities of the compounds that might be the main component: Density mass volume Which of these compounds is the most likely to be the main component of the compact disc cleaner?
Solution To identify the unknown substance, we must determine its density. This can be done by using the definition of density: This density corresponds exactly to that of isopropyl alcohol, which is therefore the most likely main component of the cleaner. However, note that the density of ethanol is also very close. To be sure that the compound is isopropyl alcohol, we should run several more density experiments. (In the modern laboratory, many other types of tests could be done to distinguish between these two liquids.) See Exercises 1.55 and 1.56.
Besides being a tool for the identification of substances, density has many other uses.
For example, the liquid in your car’s lead storage battery (a solution of sulfuric acid) changes density because the sulfuric acid is consumed as the battery discharges. In a fully charged battery, the density of the solution is about 1.30 g/cm3. If the density falls below 1.20 g/cm3, the battery will have to be recharged. Density measurement is also used to determine the amount of antifreeze, and thus the level of protection against freezing, in the cooling system of a car.
The densities of various common substances are given in Table 1.5.
Density mass volume 19.625g 25.00 cm3 0.7850 g/cm3 1.9 Classification of Matter 25 1.9 Classification of Matter Before we can hope to understand the changes we see going on around us—the growth of plants, the rusting of steel, the aging of people, rain becoming more acidic—we must find out how matter is organized. Matter, best defined as anything occupying space and having mass, is the material of the universe. Matter is complex and has many levels of organization. In this section we introduce basic ideas about the structure of matter and its behavior.
We will start by considering the definitions of the fundamental properties of matter.
Matter exists in three states: solid, liquid, and gas. A solid is rigid; it has a fixed volume and shape. A liquid has a definite volume but no specific shape; it assumes the shape of its container. A gas has no fixed volume or shape; it takes on the shape and volume of its container. In contrast to liquids and solids, which are only slightly compressible, gases are highly compressible; it is relatively easy to decrease the volume of a gas. Molecular-level pictures of the three states of water are given in Fig. 1.13. The different properties of ice, liquid water, and steam are determined by the different arrangements of the mol-ecules in these substances. Table 1.5 gives the states of some common substances at 20C and 1 atmosphere of pressure.
Most of the matter around us consists of mixtures of pure substances. Wood, gaso-line, wine, soil, and air are all mixtures. The main characteristic of a mixture is that it has variable composition. For example, wood is a mixture of many substances, the propor-tions of which vary depending on the type of wood and where it grows. Mixtures can be classified as homogeneous (having visibly indistinguishable parts) or heterogeneous (having visibly distinguishable parts).
A homogeneous mixture is called a solution. Air is a solution consisting of a mix-ture of gases. Wine is a complex liquid solution. Brass is a solid solution of copper and zinc. Sand in water and iced tea with ice cubes are examples of heterogeneous mixtures. Heterogeneous mixtures usually can be separated into two or more homo-geneous mixtures or pure substances (for example, the ice cubes can be separated from the tea).
Mixtures can be separated into pure substances by physical methods. A pure sub-stance is one with constant composition. Water is a good illustration of these ideas.
As we will discuss in detail later, pure water is composed solely of H2O molecules, TABLE 1.5 Densities of Various Common Substances at 20C Substance Physical State Density (g/cm3) Oxygen Gas 0.00133 Hydrogen Gas 0.000084 Ethanol Liquid 0.789 Benzene Liquid 0.880 Water Liquid 0.9982 Magnesium Solid 1.74 Salt (sodium chloride) Solid 2.16 Aluminum Solid 2.70 Iron Solid 7.87 Copper Solid 8.96 Silver Solid 10.5 Lead Solid 11.34 Mercury Liquid 13.6 Gold Solid 19.32 At 1 atmosphere pressure Visualizations: Structure of a Gas Structure of a Liquid Structure of a Solid Visualization: Comparison of a Compound and a Mixture Visualization: Homogeneous Mixtures: Air and Brass) Visualization: Comparison of a Solution and a Mixture 26 Chapter One Chemical Foundations The term volatile refers to the ease with which a substance can be changed to its vapor.
but the water found in nature (groundwater or the water in a lake or ocean) is really a mixture. Seawater, for example, contains large amounts of dissolved minerals. Boiling seawater produces steam, which can be condensed to pure water, leaving the minerals behind as solids. The dissolved minerals in seawater also can be separated out by freez-ing the mixture, since pure water freezes out. The processes of boiling and freezing are physical changes: When water freezes or boils, it changes its state but remains water; it is still composed of H2O molecules. A physical change is a change in the form of a substance, not in its chemical composition. A physical change can be used to separate a mixture into pure compounds, but it will not break compounds into elements.
One of the most important methods for separating the components of a mixture is distillation, a process that depends on differences in the volatility (how readily substances become gases) of the components. In simple distillation, a mixture is heated in a device such as that shown in Fig. 1.14. The most volatile component vaporizes at the lowest tem-perature, and the vapor passes through a cooled tube (a condenser), where it condenses back into its liquid state.
The simple, one-stage distillation apparatus shown in Fig. 1.14 works very well when only one component of the mixture is volatile. For example, a mixture of water and sand is easily separated by boiling off the water. Water containing dissolved minerals behaves in much the same way. As the water is boiled off, the minerals remain behind as nonvolatile solids. Simple distillation of seawater using the sun as the heat source is an excellent way to desalinate (remove the minerals from) seawater.
However, when a mixture contains several volatile components, the one-step distilla-tion does not give a pure substance in the receiving flask, and more elaborate methods are required.
Another method of separation is simple filtration, which is used when a mixture con-sists of a solid and a liquid. The mixture is poured onto a mesh, such as filter paper, which passes the liquid and leaves the solid behind.
Solid (Ice) Liquid (Water) Gas (Steam) (a) (b) (c) FIGURE 1.13 The three states of water (where red spheres represent oxygen atoms and blue spheres represent hydrogen atoms).
(a) Solid: the water molecules are locked into rigid positions and are close together.
(b) Liquid: the water molecules are still close together but can move around to some extent. (c) Gas: the water molecules are far apart and move randomly.
1.9 Classification of Matter 27 A third method of separation is called chromatography. Chromatography is the gen-eral name applied to a series of methods that employ a system with two phases (states) of matter: a mobile phase and a stationary phase. The stationary phase is a solid, and the mobile phase is either a liquid or a gas. The separation process occurs because the com-ponents of the mixture have different affinities for the two phases and thus move through the system at different rates. A component with a high affinity for the mobile phase moves relatively quickly through the chromatographic system, whereas one with a high affinity for the solid phase moves more slowly.
One simple type of chromatography, paper chromatography, employs a strip of porous paper, such as filter paper, for the stationary phase. A drop of the mixture to be separated is placed on the paper, which is then dipped into a liquid (the mobile phase) that travels up the paper as though it were a wick (see Fig. 1.15). This method of separat-ing a mixture is often used by biochemists, who study the chemistry of living systems.
It should be noted that when a mixture is separated, the absolute purity of the separated components is an ideal. Because water, for example, inevitably comes into contact with other materials when it is synthesized or separated from a mixture, it is never absolutely pure.
With great care, however, substances can be obtained in very nearly pure form.
Pure substances contain compounds (combinations of elements) or free elements. A compound is a substance with constant composition that can be broken down into elements by chemical processes. An example of a chemical process is the electrolysis of water, in which an electric current is passed through water to break it down into the free elements hydrogen and oxygen. This process produces a chemical change because the water molecules have been broken down. The water is gone, and in its place we have the free elements hydrogen and oxygen. A chemical change is one in which a given sub-stance becomes a new substance or substances with different properties and different Burner Distilling flask Water out Cool water in Receiving flask Condenser Distillate Vapors Thermometer FIGURE 1.14 Simple laboratory distillation apparatus. Cool water circulates through the outer portion of the condenser, causing vapors from the distilling flask to condense into a liquid. The nonvolatile component of the mixture remains in the distilling flask.
28 Chapter One Chemical Foundations composition. Elements are substances that cannot be decomposed into simpler substances by chemical or physical means.
We have seen that the matter around us has various levels of organization. The most fundamental substances we have discussed so far are elements. As we will see in later chapters, elements also have structure: They are composed of atoms, which in turn are composed of nuclei and electrons. Even the nucleus has structure: It is composed of pro-tons and neutrons. And even these can be broken down further, into elementary particles called quarks. However, we need not concern ourselves with such details at this point.
Figure 1.16 summarizes our discussion of the organization of matter.
Heterogeneous mixtures Physical methods Homogeneous mixtures (solutions) Atoms Compounds Nucleus Electrons Pure substances Physical methods Chemical methods Matter Elements Protons Neutrons Quarks Quarks FIGURE 1.16 The organization of matter.
The element mercury (top left) combines with the element iodine (top right) to form the compound mercuric iodide (bottom).
This is an example of a chemical change.
FIGURE 1.15 Paper chromatography of ink. (a) A line of the mixture to be separated is placed at one end of a sheet of porous paper. (b) The paper acts as a wick to draw up the liquid.
(c) The component with the weakest attrac-tion for the paper travels faster than the components that cling to the paper.
(a) (b) (c) For Review 29 For Review Scientific method Make observations Formulate hypotheses Perform experiments Models (theories) are explanation of why nature behaves in a particular way.
They are subject to modification over time and sometimes fail.
Quantitative observations are called measurements.
Consist of a number and a unit Involve some uncertainty Uncertainty is indicated by using significant figures • Rules to determine significant figures • Calculations using significant figures Preferred system is SI Temperature conversions TK TC 273 Density Matter can exist in three states: Solid Liquid Gas Mixtures can be separated by methods involving only physical changes: Distillation Filtration Chromatography Compounds can be decomposed to elements only through chemical changes.
REVIEW QUESTIONS 1. Define and explain the differences between the following terms.
a. law and theory b. theory and experiment c. qualitative and quantitative d. hypothesis and theory 2. Is the scientific method suitable for solving problems only in the sciences? Explain.
3. Which of the following statements (hypotheses) could be tested by quantitative measurement?
a. Ty Cobb was a better hitter than Pete Rose.
b. Ivory soap is pure.
c. Rolaids consumes 47 times its weight in excess stomach acid.
99 44 100% Density mass volume TF TC a 5°F 9°Cb 32°F TC 1TF 32°F2 a 5°C 9°F b Key Terms Section 1.2 scientific method measurement hypothesis theory model natural law law of conservation of mass Section 1.3 SI system mass weight Section 1.4 uncertainty significant figures accuracy precision random error systematic error Section 1.5 exponential notation Section 1.6 unit factor method dimensional analysis Section 1.8 density Section 1.9 matter states (of matter) homogeneous mixture heterogeneous mixture solution pure substance physical change distillation filtration chromatography paper chromatography compound chemical change element 4. For each of the following pieces of glassware, provide a sample measurement and discuss the number of significant figures and uncertainty.
5. A student performed an analysis of a sample for its calcium content and got the following results: 14.92% 14.91% 14.88% 14.91% The actual amount of calcium in the sample is 15.70%. What conclusions can you draw about the accuracy and precision of these results?
6. Compare and contrast the multiplication/division significant figure rule to the significant figure rule applied for addition/subtraction mathematical operations.
7. Explain how density can be used as a conversion factor to convert the volume of an object to the mass of the object, and vice versa.
8. On which temperature scale (F, C, or K) does 1 degree represent the smallest change in temperature?
9. Distinguish between physical changes and chemical changes.
10. Why is the separation of mixtures into pure or relatively pure substances so im-portant when performing a chemical analysis?
30 20 10 5 a.
b.
c.
11 10 4 3 2 1 Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. a. There are 365 days per year, 24 hours per day, 12 months per year, and 60 minutes per hour. Use these data to determine how many minutes are in a month.
b. Now use the following data to calculate the number of min-utes in a month: 24 hours per day, 60 minutes per hour, 7 days per week, and 4 weeks per month.
c. Why are these answers different? Which (if any) is more correct? Why?
2. You go to a convenience store to buy candy and find the owner to be rather odd. He allows you to buy pieces in multiples of four, and to buy four, you need $0.23. He only allows you to do this by using 3 pennies and 2 dimes. You have a bunch of pennies and dimes, and instead of counting them, you decide to weigh them.
You have 636.3 g of pennies, and each penny weighs 3.03 g. Each dime weighs 2.29 g. Each piece of candy weighs 10.23 g.
a. How many pennies do you have?
b. How many dimes do you need to buy as much candy as possible?
c. How much should all these dimes weigh?
d. How many pieces of candy could you buy? (number of dimes from part b) e. How much would this candy weigh?
f. How many pieces of candy could you buy with twice as many dimes?
3. When a marble is dropped into a beaker of water, it sinks to the bottom. Which of the following is the best explanation?
a. The surface area of the marble is not large enough to be held up by the surface tension of the water.
b. The mass of the marble is greater than that of the water.
c. The marble weighs more than an equivalent volume of the water.
d. The force from dropping the marble breaks the surface tension of the water.
e. The marble has greater mass and volume than the water.
30 Chapter One Chemical Foundations Questions 31 Justify your choice, and for choices you did not pick, explain what is wrong about them.
4. You have two beakers, one filled to the 100-mL mark with sugar (the sugar has a mass of 180.0 g) and the other filled to the 100-mL mark with water (the water has a mass of 100.0 g). You pour all the sugar and all the water together in a bigger beaker and stir until the sugar is completely dissolved.
a. Which of the following is true about the mass of the solution?
Explain.
i. It is much greater than 280.0 g.
ii. It is somewhat greater than 280.0 g.
iii. It is exactly 280.0 g.
iv. It is somewhat less than 280.0 g.
v. It is much less than 280.0 g.
b. Which of the following is true about the volume of the solu-tion? Explain.
i. It is much greater than 200.0 mL.
ii. It is somewhat greater than 200.0 mL.
iii. It is exactly 200.0 mL.
iv. It is somewhat less than 200.0 mL.
v. It is much less than 200.0 mL.
5. You may have noticed that when water boils, you can see bubbles that rise to the surface of the water.
a. What is inside these bubbles?
i. air ii. hydrogen and oxygen gas iii. oxygen gas iv. water vapor v. carbon dioxide gas b. Is the boiling of water a chemical or physical change? Explain.
6. If you place a glass rod over a burning candle, the glass appears to turn black. What is happening to each of the following (phys-ical change, chemical change, both, or neither) as the candle burns? Explain each answer.
a. the wax b. the wick c. the glass rod 7. Which characteristics of a solid, a liquid, and a gas are exhibited by each of the following substances? How would you classify each substance?
a. a bowl of pudding b. a bucketful of sand 8. You have water in each graduated cylinder shown: 1 mL 0.5 5 4 mL 2 1 3 You then add both samples to a beaker. How would you write the number describing the total volume? What limits the precision of this number?
9. Paracelsus, a sixteenth-century alchemist and healer, adopted as his slogan: “The patients are your textbook, the sickbed is your study.” Is this view consistent with using the scientific method?
10. What is wrong with the following statement?
“The results of the experiment do not agree with the theory. Some-thing must be wrong with the experiment.” 11. Why is it incorrect to say that the results of a measurement were accurate but not precise?
12. What data would you need to estimate the money you would spend on gasoline to drive your car from New York to Chicago? Pro-vide estimates of values and a sample calculation.
13. Sketch two pieces of glassware: one that can measure volume to the thousandths place and one that can measure volume only to the ones place.
14. You have a 1.0-cm3 sample of lead and a 1.0-cm3 sample of glass.
You drop each in separate beakers of water. How do the volumes of water displaced by each sample compare? Explain.
15. Sketch a magnified view (showing atoms/molecules) of each of the following and explain: a. a heterogeneous mixture of two different compounds b. a homogeneous mixture of an element and a compound 16. You are driving 65 mi/h and take your eyes off the road for “just a second.” What distance (in feet) do you travel in this time?
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 17. The difference between a law and a theory is the difference between what and why. Explain.
18. Explain the fundamental steps of the scientific method.
19. A measurement is a quantitative observation involving both a number and a unit. What is a qualitative observation? What are the SI units for mass, length, and volume? What is the assumed uncertainty in a number (unless stated otherwise)? The uncertainty of a measurement depends on the precision of the measuring device. Explain.
20. To determine the volume of a cube, a student measured one of the dimensions of the cube several times. If the true dimension of the cube is 10.62 cm, give an example of four sets of measure-ments that would illustrate the following.
a. imprecise and inaccurate data b. precise but inaccurate data c. precise and accurate data Give a possible explanation as to why data can be imprecise or inaccurate. What is wrong with saying a set of measurements is imprecise but accurate?
21. What are significant figures? Show how to indicate the number one thousand to 1 significant figure, 2 significant figures, 3 significant figures, and 4 significant figures. Why is the answer, to the correct number of significant figures, not 1.0 for the following calculation?
22. What is the volume per unit mass equal to? What unit conversion would the volume per unit mass be useful for?
23. When the temperature in degrees Fahrenheit (TF) is plotted ver-sus the temperature in degrees Celsius (TC), a straight line plot re-sults. A straight line plot also results when TC is plotted versus TK (the temperature in degrees Kelvin). Reference Appendix A1.3 and determine the slope and y-intercept of each of these two plots.
24. Give four examples illustrating each of the following terms.
a. homogeneous mixture d. element b. heterogeneous mixture e. physical change c. compound f. chemical change Exercises In this section similar exercises are paired.
Significant Figures and Unit Conversions 25. Which of the following are exact numbers?
a. There are 100 cm in 1 m.
b. One meter equals 1.094 yard.
c. We can use the equation to convert from Celsius to Fahrenheit temperature. Are the numbers and 32 exact or inexact?
d. 3.1415927.
26. Indicate the number of significant figures in each of the following: a. This book contains more than 1000 pages.
b. A mile is about 5300 ft.
c. A liter is equivalent to 1.059 qt.
d. The population of the United States is approaching 3.0 102 million.
e. A kilogram is 1000 g.
f. The Boeing 747 cruises at around 600 mi/h.
27. How many significant figures are there in each of the following values?
a. 6.07 1015 e. 463.8052 b. 0.003840 f. 300 c. 17.00 g. 301 d. 8 108 h. 300.
28. How many significant figures are in each of the following?
a. 100 e. 0.0048 b. 1.0 102 f. 0.00480 c. 1.00 103 g. 4.80 103 d. 100.
h. 4.800 103 29. Round off each of the following numbers to the indicated num-ber of significant digits and write the answer in standard scien-tific notation.
a. 0.00034159 to three digits b. 103.351 102 to four digits c. 17.9915 to five digits d. 3.365 105 to three digits 9 5 °F 9 5°C 32 1.5 1.0 0.50 30. Use exponential notation to express the number 480 to a. one significant figure b. two significant figures c. three significant figures d. four significant figures 31. Evaluate each of the following and write the answer to the ap-propriate number of significant figures.
a. 212.2 26.7 402.09 b. 1.0028 0.221 0.10337 c. 52.331 26.01 0.9981 d. 2.01 102 3.014 103 e. 7.255 6.8350 32. Perform the following mathematical operations, and express each result to the correct number of significant figures.
a.
b. 0.14 6.022 1023 c. 4.0 104 5.021 103 7.34993 102 d.
33. Perform the following mathematical operations and express the result to the correct number of significant figures.
a.
b. (6.404 2.91)(18.7 17.1) c. 6.071 105 8.2 106 0.521 104 d. (3.8 1012 4.0 1013)(4 1012 6.3 1013) e.
(Assume that this operation is taking the average of four num-bers. Thus 4 in the denominator is exact.) f.
(This type of calculation is done many times in calculating a percentage error. Assume that this example is such a calcula-tion; thus 100 can be considered to be an exact number.) 34. Perform the following mathematical operations, and express the result to the correct number of significant figures.
a. 6.022 1023 1.05 102 b.
c. 1.285 102 1.24 103 1.879 101 d. 1.285 102 1.24 103 e.
f.
g.
35. Perform each of the following conversions.
a. 8.43 cm to millimeters b. 2.41 102 cm to meters c. 294.5 nm to centimeters d. 1.445 104 m to kilometers 9.42 102 8.234 102 1.625 103 3 13 is exact2 9.875 102 9.795 102 9.875 102 100 1100 is exact2 11.00866 1.007282 6.02205 1023 6.6262 1034 2.998 108 2.54 109 8.925 8.905 8.925 100 9.5 4.1 2.8 3.175 4 2.526 3.1 0.470 0.623 80.705 0.4326 2.00 106 3.00 107 0.102 0.0821 273 1.01 32 Chapter One Chemical Foundations Exercises 33 e. 235.3 m to millimeters f. 903.3 nm to micrometers 36. a. How many kilograms are in one teragram?
b. How many nanometers are in 6.50 102 terameters?
c. How many kilograms are in 25 femtograms?
d. How many liters are in 8.0 cubic decimeters?
e. How many microliters are in one milliliter?
f. How many picograms are in one microgram?
37. Perform the following unit conversions.
a. Congratulations! You and your spouse are the proud parents of a new baby, born while you are studying in a country that uses the metric system. The nurse has informed you that the baby weighs 3.91 kg and measures 51.4 cm. Convert your baby’s weight to pounds and ounces and her length to inches (rounded to the nearest quarter inch).
b. The circumference of the earth is 25,000 mi at the equator.
What is the circumference in kilometers? in meters?
c. A rectangular solid measures 1.0 m by 5.6 cm by 2.1 dm. Ex-press its volume in cubic meters, liters, cubic inches, and cu-bic feet.
38. Perform the following unit conversions.
a. 908 oz to kilograms b. 12.8 L to gallons c. 125 mL to quarts d. 2.89 gal to milliliters e. 4.48 lb to grams f. 550 mL to quarts 39. Use the following exact conversion factors to perform the stated calculations: a. The Kentucky Derby race is 1.25 miles. How long is the race in rods, furlongs, meters, and kilometers?
b. A marathon race is 26 miles, 385 yards. What is this distance in rods, furlongs, meters, and kilometers?
40. Although the preferred SI unit of area is the square meter, land is often measured in the metric system in hectares (ha). One hectare is equal to 10,000 m2. In the English system, land is often mea-sured in acres (1 acre 160 rod2). Use the exact conversions and those given in Exercise 39 to calculate the following.
a. 1 ha _ km2.
b. The area of a 5.5-acre plot of land in hectares, square meters, and square kilometers.
c. A lot with dimensions 120 ft by 75 ft is to be sold for $6500.
What is the price per acre? What is the price per hectare?
41. Precious metals and gems are measured in troy weights in the English system: a. The most common English unit of mass is the pound avoirdu-pois. What is one troy pound in kilograms and in pounds? 1 carat 0.200 gram 1 grain 0.0648 gram 12 troy ounces 1 troy pound 1exact2 20 pennyweight 1 troy ounce 1exact2 24 grains 1 pennyweight 1exact2 8 furlongs 1 mile 40 rods 1 furlong 51 2 yards 1 rod b. What is the mass of a troy ounce of gold in grams and in carats?
c. The density of gold is 19.3 g/cm3. What is the volume of a troy pound of gold?
42. Apothecaries (druggists) use the following set of measures in the English system: a. Is an apothecary grain the same as a troy grain? (See Exer-cise 41.) b. 1 oz ap ________ oz troy.
c. An aspirin tablet contains 5.00 102 mg of active ingredient.
What mass in grains ap of active ingredient does it contain?
What mass in scruples?
d. What is the mass of 1 scruple in grams?
43. Science fiction often uses nautical analogies to describe space travel. If the starship U.S.S. Enterprise is traveling at warp factor 1.71, what is its speed in knots and in miles per hour? (Warp 1.71 5.00 times the speed of light; speed of light 3.00 108 m/s; 1 knot 2000 yd/h, exactly.) 44. The world record for the hundred meter dash is 9.77 s. What is the corresponding average speed in units of m/s, km/h, ft/s, and mi/h?
At this speed, how long would it take to run 1.00 102 yards?
45. Would a car traveling at a constant speed of 65 km/h violate a 40. mi/h speed limit?
46. You pass a road sign saying “New York 112 km.” If you drive at a constant speed of 65 mi/h, how long should it take you to reach New York? If your car gets 28 miles to the gallon, how many liters of gasoline are necessary to travel 112 km?
47. If you put 8.21 gallons of gas in your car and it cost you a total of $17.25, what is the cost of gas per liter in Canadian dollars?
Assume 0.82 dollar U.S. 1.00 dollar Canadian.
48. A children’s pain relief elixir contains 80. mg acetaminophen per 0.50 teaspoon. The dosage recommended for a child who weighs between 24 and 35 lb is 1.5 teaspoons. What is the range of acetaminophen dosages, expressed in mg acetaminophen/kg body weight, for children who weigh between 24 and 35 lb?
Temperature 49. Convert the following Fahrenheit temperatures to the Celsius and Kelvin scales.
a. 459F, an extremely low temperature b. 40.F, the answer to a trivia question c. 68F, room temperature d. 7 107 F, temperature required to initiate fusion reactions in the sun 50. A thermometer gives a reading of 96.1F 0.2F. What is the temperature in C? What is the uncertainty?
51. Convert the following Celsius temperatures to Kelvin and to Fahrenheit degrees.
a. the temperature of someone with a fever, 39.2C b. a cold wintery day, 25C c. the lowest possible temperature, 273C d. the melting-point temperature of sodium chloride, 801C 1 dram ap 3.888 g 8 dram ap 1 oz ap 1exact2 3 scruples 1 dram ap 1exact2 20 grains ap 1 scruple 1exact2 52. Convert the following Kelvin temperatures to Celsius and Fahren-heit degrees.
a. the temperature that registers the same value on both the Fahrenheit and Celsius scales, 233 K b. the boiling point of helium, 4 K c. the temperature at which many chemical quantities are deter-mined, 298 K d. the melting point of tungsten, 3680 K Density 53. A material will float on the surface of a liquid if the material has a density less than that of the liquid. Given that the density of wa-ter is approximately 1.0 g/mL, will a block of material having a volume of 1.2 104 in3 and weighing 350 lb float or sink when placed in a reservoir of water?
54. For a material to float on the surface of water, the material must have a density less than that of water (1.0 g/mL) and must not re-act with the water or dissolve in it. A spherical ball has a radius of 0.50 cm and weighs 2.0 g. Will this ball float or sink when placed in water? (Note: Volume of a sphere r3.) 55. A star is estimated to have a mass of 2 1036 kg. Assuming it to be a sphere of average radius 7.0 105 km, calculate the aver-age density of the star in units of grams per cubic centimeter.
56. A rectangular block has dimensions 2.9 cm 3.5 cm 10.0 cm.
The mass of the block is 615.0 g. What are the volume and den-sity of the block?
57. Diamonds are measured in carats, and 1 carat 0.200 g. The den-sity of diamond is 3.51 g/cm3. What is the volume of a 5.0-carat diamond?
58. The volume of a diamond is found to be 2.8 mL. What is the mass of the diamond in carats? (See Exercise 57.) 59. A sample containing 33.42 g of metal pellets is poured into a grad-uated cylinder initially containing 12.7 mL of water, causing the water level in the cylinder to rise to 21.6 mL. Calculate the den-sity of the metal.
60. The density of pure silver is 10.5 g/cm3 at 20C. If 5.25 g of pure silver pellets is added to a graduated cylinder containing 11.2 mL of water, to what volume level will the water in the cylinder rise?
61. In each of the following pairs, which has the greater mass? (See Table 1.5.) a. 1.0 kg of feathers or 1.0 kg of lead b. 1.0 mL of mercury or 1.0 mL of water c. 19.3 mL of water or 1.00 mL of gold d. 75 mL of copper or 1.0 L of benzene 62. Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes. If the amount of mer-cury in a polluted lake is 0.4 g Hg/mL, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100 mi2 and an average depth of 20 ft.) 63. In each of the following pairs, which has the greater volume?
a. 1.0 kg of feathers or 1.0 kg of lead b. 100 g of gold or 100 g of water c. 1.0 L of copper or 1.0 L of mercury 4 3 64. Using Table 1.5, calculate the volume of 25.0 g of each of the fol-lowing substances at 1 atm.
a. hydrogen gas b. water c. iron Chapter 5 discusses the properties of gases. One property unique to gases is that they contain mostly empty space. Explain using the results of your calculations.
65. The density of osmium (the densest metal) is 22.57 g/cm3. If a 1.00-kg rectangular block of osmium has two dimensions of 4.00 cm 4.00 cm, calculate the third dimension of the block.
66. A copper wire (density 8.96 g/cm3) has a diameter of 0.25 mm.
If a sample of this copper wire has a mass of 22 g, how long is the wire?
Classification and Separation of Matter 67. Match each description below with the following microscopic pic-tures. More than one picture may fit each description. A picture may be used more than once or not used at all.
a. a gaseous compound b. a mixture of two gaseous elements c. a solid element d. a mixture of a gaseous element and a gaseous compound 68. Define the following terms: solid, liquid, gas, pure substance, el-ement, compound, homogeneous mixture, heterogeneous mixture, solution, chemical change, physical change.
69. What is the difference between homogeneous and heteroge-neous matter? Classify each of the following as homogeneous or heterogeneous.
a. a door b. the air you breathe c. a cup of coffee (black) d. the water you drink e. salsa f. your lab partner 70. Classify each of the following as a mixture or a pure substance.
a. water f. uranium b. blood g. wine c. the oceans h. leather d. iron i. table salt (NaCl) e. brass Of the pure substances, which are elements and which are compounds?
i ii iii iv v vi 34 Chapter One Chemical Foundations Challenge Problems 35 71. Classify following as physical or chemical changes.
a. Moth balls gradually vaporize in a closet.
b. Hydrofluoric acid attacks glass, and is used to etch calibration marks on glass laboratory utensils.
c. A French chef making a sauce with brandy is able to burn off the alcohol from the brandy, leaving just the brandy flavoring.
d. Chemistry majors sometimes get holes in the cotton jeans they wear to lab because of acid spills.
72. The properties of a mixture are typically averages of the proper-ties of its components. The properties of a compound may differ dramatically from the properties of the elements that combine to produce the compound. For each process described below, state whether the material being discussed is most likely a mixture or a compound, and state whether the process is a chemical change or a physical change.
a. An orange liquid is distilled, resulting in the collection of a yellow liquid and a red solid.
b. A colorless, crystalline solid is decomposed, yielding a pale yellow-green gas and a soft, shiny metal.
c. A cup of tea becomes sweeter as sugar is added to it.
Additional Exercises 73. For a pharmacist dispensing pills or capsules, it is often easier to weigh the medication to be dispensed rather than to count the in-dividual pills. If a single antibiotic capsule weighs 0.65 g, and a pharmacist weighs out 15.6 g of capsules, how many capsules have been dispensed?
74. In Shakespeare’s Richard III, the First Murderer says: “Take that, and that! [Stabs Clarence] If that is not enough, I’ll drown you in a malmsey butt within!” Given that 1 butt 126 gal, in how many liters of malmsey (a foul brew similar to mead) was the unfortunate Clarence about to be drowned?
75. The contents of one 40. lb bag of topsoil will cover 10. square feet of ground to a depth of 1.0 inch. What number of bags are needed to cover a plot that measures 200. by 300. m to a depth of 4.0 cm?
76. In the opening scenes of the movie Raiders of the Lost Ark, Indiana Jones tries to remove a gold idol from a booby-trapped pedestal. He replaces the idol with a bag of sand of approxi-mately equal volume. (Density of gold 19.32 g/cm3; density of sand 2 g/cm3.) a. Did he have a reasonable chance of not activating the mass-sensitive booby trap?
b. In a later scene he and an unscrupulous guide play catch with the idol. Assume that the volume of the idol is about 1.0 L. If it were solid gold, what mass would the idol have? Is playing catch with it plausible?
77. A column of liquid is found to expand linearly on heating 5.25 cm for a 10.0F rise in temperature. If the initial temperature of the liquid is 98.6F, what will the final temperature be in C if the liquid has expanded by 18.5 cm?
78. A 25.00-g sample of a solid is placed in a graduated cylinder and then the cylinder is filled to the 50.0 mL mark with benzene. The mass of benzene and solid together is 58.80 g. Assuming that the solid is insoluble in benzene and that the density of benzene is 0.880 g/cm3, calculate the density of the solid.
79. For each of the following, decide which block is more dense: the orange block, the blue block, or it cannot be determined. Explain your answers.
80. According to the Official Rules of Baseball, a baseball must have a circumference not more than 9.25 in or less than 9.00 in and a mass not more than 5.25 oz or less than 5.00 oz. What range of densities can a baseball be expected to have? Express this range as a single number with an accompanying uncertainty limit.
81. The density of an irregularly shaped object was determined as follows. The mass of the object was found to be 28.90 g 0.03 g.
A graduated cylinder was partially filled with water. The reading of the level of the water was 6.4 cm3 0.1 cm3. The object was dropped in the cylinder, and the level of the water rose to 9.8 cm3 0.1 cm3.
What is the density of the object with appropriate error limits? (See Appendix 1.5.) Challenge Problems 82. Draw a picture showing the markings (graduations) on glassware that would allow you to make each of the following volume meas-urements of water and explain your answers (the numbers given are as precise as possible).
a. 128.7 mL b. 18 mL c. 23.45 mL If you made the measurements of three samples of water and then poured all of the water together in one container, what total vol-ume of water should you report? Support your answer.
83. Many times errors are expressed in terms of percentage. The per-cent error is the absolute value of the difference of the true value and the experimental value, divided by the true value, and multi-plied by 100.
Percent error 0true value experimental value0 true value 100 a.
b.
c.
d.
Calculate the percent error for the following measurements.
a. The density of an aluminum block determined in an experi-ment was 2.64 g/cm3. (True value 2.70 g/cm3.) b. The experimental determination of iron in iron ore was 16.48%.
(True value 16.12%.) c. A balance measured the mass of a 1.000-g standard as 0.9981 g.
84. A person weighed 15 pennies on a balance and recorded the fol-lowing masses: Curious about the results, he looked at the dates on each penny.
Two of the light pennies were minted in 1983 and one in 1982. The dates on the 12 heavier pennies ranged from 1970 to 1982. Two of the 12 heavier pennies were minted in 1982.
a. Do you think the Bureau of the Mint changed the way it made pennies? Explain.
b. The person calculated the average mass of the 12 heavy pen-nies. He expressed this average as 3.0828 g 0.0482 g. What is wrong with the numbers in this result, and how should the value be expressed?
85. On October 21, 1982, the Bureau of the Mint changed the com-position of pennies (see Exercise 84). Instead of an alloy of 95% Cu and 5% Zn by mass, a core of 99.2% Zn and 0.8% Cu with a thin shell of copper was adopted. The overall composition of the new penny was 97.6% Zn and 2.4% Cu by mass. Does this ac-count for the difference in mass among the pennies in Exercise 84? Assume the volume of the individual metals that make up each penny can be added together to give the overall volume of the penny, and assume each penny is the same size. (Density of Cu 8.96 g/cm3; density of Zn 7.14 g/cm3.) 86. Ethylene glycol is the main component in automobile antifreeze.
To monitor the temperature of an auto cooling system, you intend to use a meter that reads from 0 to 100. You devise a new tem-perature scale based on the approximate melting and boiling points of a typical antifreeze solution (45C and 115C). You wish these points to correspond to 0A and 100A, respectively.
a. Derive an expression for converting between A and C.
b. Derive an expression for converting between F and A.
c. At what temperature would your thermometer and a Celsius thermometer give the same numerical reading?
d. Your thermometer reads 86A. What is the temperature in C and in F?
e. What is a temperature of 45C in A?
87. Sterling silver is a solid solution of silver and copper. If a piece of a sterling silver necklace has a mass of 105.0 g and a volume of 10.12 mL, calculate the mass percent of copper in the piece of necklace. Assume that the volume of silver present plus the vol-ume of copper present equals the total volume. Refer to Table 1.5.
Mass percent of copper mass of copper total mass 100 3.112 g 3.109 g 3.059 g 2.467 g 3.079 g 2.518 g 3.129 g 2.545 g 3.050 g 3.053 g 3.054 g 3.072 g 3.081 g 3.131 g 3.064 g 88. Use molecular-level (microscopic) drawings for each of the fol-lowing.
a. Show the differences between a gaseous mixture that is a ho-mogeneous mixture of two different compounds, and a gaseous mixture that is a homogeneous mixture of a compound and an element.
b. Show the differences among a gaseous element, a liquid ele-ment, and a solid element.
89. Confronted with the box shown in the diagram, you wish to dis-cover something about its internal workings. You have no tools and cannot open the box. You pull on rope B, and it moves rather freely. When you pull on rope A, rope C appears to be pulled slightly into the box. When you pull on rope C, rope A almost disappears into the box. a. Based on these observations, construct a model for the interior mechanism of the box.
b. What further experiments could you do to refine your model?
90. An experiment was performed in which an empty 100-mL grad-uated cylinder was weighed. It was weighed once again after it had been filled to the 10.0-mL mark with dry sand. A 10-mL pipet was used to transfer 10.00 mL of methanol to the cylinder. The sand–methanol mixture was stirred until bubbles no longer emerged from the mixture and the sand looked uniformly wet.
The cylinder was then weighed again. Use the data obtained from this experiment (and displayed at the end of this problem) to find the density of the dry sand, the density of methanol, and the den-sity of sand particles. Does the bubbling that occurs when the methanol is added to the dry sand indicate that the sand and methanol are reacting?
Mass of cylinder plus wet sand 45.2613 g Mass of cylinder plus dry sand 37.3488 g Mass of empty cylinder 22.8317 g Volume of dry sand 10.0 mL Volume of sand methanol 17.6 mL Volume of methanol 10.00 mL A B C 36 Chapter One Chemical Foundations From Yoder, Suydam, and Snavely, Chemistry (New York: Harcourt Brace Jovanovich, 1975), pp. 9–11.
Marathon Problem 37 Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
91. The U.S. trade deficit at the beginning of 2005 was $475,000,000.
If the wealthiest 1.00 percent of the U.S. population (297,000,000) contributed an equal amount of money to bring the trade deficit to $0, how many dollars would each person contribute? If one of these people were to pay their share in nickels only, how many nickels are needed? Another person living abroad at the time decides to pay in pounds sterling (£). How many pounds sterling does this person contribute (assume a conversion rate of 1 £ $ 1.869)?
92. The density of osmium is reported by one source to be 22610 kg/m3. What is this density in g/cm3? What is the mass of a block of osmium measuring 10.0 cm 8.0 cm 9.0 cm?
93. At the Amundsen-Scott South Pole base station in Antarctica, when the temperature is 100.0F, researchers who live there can join the “300 Club” by stepping into a sauna heated to 200.0F then quickly running outside and around the pole that marks the South Pole. What are these temperatures in C? What are these temperatures in K? If you measured the temperatures only in C and K, can you become a member of the “300 Club” (that is, is there a 300.-degree difference between the temperature extremes when measured in C and K?) Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
94. A cylindrical bar of gold that is 1.5 in high and 0.25 in in diam-eter has a mass of 23.1984 g, as determined on an analytical bal-ance. An empty graduated cylinder is weighed on a triple-beam balance and has a mass of 73.47 g. After pouring a small amount of a liquid into the graduated cylinder, the mass is 79.16 g. When the gold cylinder is placed in the graduated cylinder (the liquid covers the top of the gold cylinder), the volume indicated on the graduated cylinder is 8.5 mL. Assume that the temperature of the gold bar and the liquid are 86F. If the density of the liquid de-creases by 1.0% for each 10.C rise in temperature (over the range 0 to 50C), determine a. the density of the gold at 86F.
b. the density of the liquid at 40.F.
Note: Parts a and b can be answered independently.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
38 2 Atoms, Molecules, and Ions Contents 2.1 The Early History of Chemistry 2.2 Fundamental Chemical Laws 2.3 Dalton’s Atomic Theory 2.4 Early Experiments to Characterize the Atom • The Electron • Radioactivity • The Nuclear Atom 2.5 The Modern View of Atomic Structure: An Introduction 2.6 Molecules and Ions 2.7 An Introduction to the Periodic Table 2.8 Naming Simple Compounds • Binary Ionic Compounds (Type I) • Formulas from Names • Binary Ionic Compounds (Type II) • Ionic Compounds with Polyatomic Ions • Binary Covalent Compounds (Type III) • Acids A worker in Thailand piles up salt crystals.
Where does one start in learning chemistry? Clearly we must consider some es-sential vocabulary and something about the origins of the science before we can proceed very far. Thus, while Chapter 1 provided background on the fundamental ideas and pro-cedures of science in general, Chapter 2 covers the specific chemical background necessary for understanding the material in the next few chapters. The coverage of these topics is necessarily brief at this point. We will develop these ideas more fully as it becomes appropriate to do so. A major goal of this chapter is to present the systems for naming chemical compounds to provide you with the vocabulary necessary to understand this book and to pursue your laboratory studies.
Because chemistry is concerned first and foremost with chemical changes, we will proceed as quickly as possible to a study of chemical reactions (Chapters 3 and 4). How-ever, before we can discuss reactions, we must consider some fundamental ideas about atoms and how they combine.
2.1 The Early History of Chemistry Chemistry has been important since ancient times. The processing of natural ores to produce metals for ornaments and weapons and the use of embalming fluids are just two applications of chemical phenomena that were utilized prior to 1000 B.C.
The Greeks were the first to try to explain why chemical changes occur. By about 400 B.C. they had proposed that all matter was composed of four fundamental substances: fire, earth, water, and air. The Greeks also considered the question of whether matter is continuous, and thus infinitely divisible into smaller pieces, or composed of small, indi-visible particles. Supporters of the latter position were Demokritos of Abdera (c. 460–c.
370 B.C.) and Leucippos, who used the term atomos (which later became atoms) to de-scribe these ultimate particles. However, because the Greeks had no experiments to test their ideas, no definitive conclusion could be reached about the divisibility of matter.
The next 2000 years of chemical history were dominated by a pseudoscience called alchemy. Some alchemists were mystics and fakes who were obsessed with the idea of turning cheap metals into gold. However, many alchemists were serious scientists, and this period saw important advances: The alchemists discovered several elements and learned to prepare the mineral acids.
The foundations of modern chemistry were laid in the sixteenth century with the devel-opment of systematic metallurgy (extraction of metals from ores) by a German, Georg Bauer (1494–1555), and the medicinal application of minerals by a Swiss alchemist/physician known as Paracelsus (full name: Philippus Theophrastus Bombastus von Hohenheim [1493–1541]).
The first “chemist” to perform truly quantitative experiments was Robert Boyle (1627–1691), who carefully measured the relationship between the pressure and volume of air. When Boyle published his book The Skeptical Chymist in 1661, the quantitative sciences of physics and chemistry were born. In addition to his results on the quantitative behavior of gases, Boyle’s other major contribution to chemistry consisted of his ideas about the chemical elements. Boyle held no preconceived notion about the number of elements. In his view, a substance was an element unless it could be broken down into two or more simpler substances. As Boyle’s experimental definition of an element became generally accepted, the list of known elements began to grow, and the Greek system of Democritus is an alternate spelling.
39 40 Chapter Two Atoms, Molecules, and Ions CHEMICAL IMPACT There’s Gold in Them There Plants!
G old has always held a strong allure. For example, the al-chemists were obsessed with finding a way to transform cheap metals into gold. Also, when gold was discovered in California in 1849, a frantic rush occurred to that area and other areas in the west. Although gold is still valuable, most of the high-grade gold ores have been exhausted. This leaves the low-grade ores—ores with low concentrations of gold that are expensive to process relative to the amount of gold finally obtained.
Now two scientists have come across a novel way to con-centrate the gold from low-grade ores. Christopher Anderson and Robert Brooks of Massey University in Palmerston North, New Zealand, have found plants that accumulate gold atoms as they grow in soil containing gold ore [Nature 395 (1998): 553]. The plants brassica (of the mustard family) and chicory seem especially effective as botanical “gold miners.” When these plants are dried and burned (after having grown in gold-rich soil), the resulting ash contains approximately 150 ppm (parts per million) of gold. (1 ppm gold represents 1 g of gold in 106 g of sample.) The New Zealand scientists were able to double the amount of gold taken from the soil by the plants by treating the soil with ammonium thiocyanate (NH4SCN). The thio-cyanate, which reacts with the gold, making it more avail-able to the plants, subsequently breaks down in the soil and therefore poses no environmental hazard.
Thus plants seem to hold great promise as gold min-ers. They are efficient and reliable and will never go on strike.
This plant from the mustard family is a newly discovered source of gold.
four elements finally died. Although Boyle was an excellent scientist, he was not always right. For example, he clung to the alchemists’ views that metals were not true elements and that a way would eventually be found to change one metal into another.
The phenomenon of combustion evoked intense interest in the seventeenth and eigh-teenth centuries. The German chemist Georg Stahl (1660–1734) suggested that a substance he called “phlogiston” flowed out of the burning material. Stahl postulated that a substance burning in a closed container eventually stopped burning because the air in the container became saturated with phlogiston. Oxygen gas, discovered by Joseph Priestley (1733–1804), an English clergyman and scientist (Fig. 2.1), was found to support vigorous combustion and was thus supposed to be low in phlogiston. In fact, oxygen was originally called “dephlogisticated air.” Oxygen gas was actually first observed by the Swedish chemist Karl W. Scheele (1742–1786), but because his results were published after Priestley’s, the latter is commonly credited with the discovery of oxygen.
FIGURE 2.1 The Priestley Medal is the highest honor given by the American Chemical Society. It is named for Joseph Priestley, who was born in England on March 13, 1733. He performed many important scientific experi-ments, among them the discovery that a gas later identified as carbon dioxide could be dissolved in water to produce seltzer. Also, as a result of meeting Benjamin Franklin in London in 1766, Priestley became inter-ested in electricity and was the first to observe that graphite was an electrical conductor. However, his great-est discovery occurred in 1774 when he isolated oxygen by heating mercuric oxide.
Because of his nonconformist political views, Priestley was forced to leave England. He died in the United States in 1804.
2.2 Fundamental Chemical Laws 41 2.2 Fundamental Chemical Laws By the late eighteenth century, combustion had been studied extensively; the gases car-bon dioxide, nitrogen, hydrogen, and oxygen had been discovered; and the list of elements continued to grow. However, it was Antoine Lavoisier (1743–1794), a French chemist (Fig. 2.2), who finally explained the true nature of combustion, thus clearing the way for the tremendous progress that was made near the end of the eighteenth century.
Lavoisier, like Boyle, regarded measurement as the essential operation of chemistry. His experiments, in which he carefully weighed the reactants and products of various reactions, suggested that mass is neither created nor destroyed. Lavoisier’s verification of this law of conservation of mass was the basis for the developments in chemistry in the nineteenth century.
Lavoisier’s quantitative experiments showed that combustion involved oxygen (which Lavoisier named), not phlogiston. He also discovered that life was supported by a process that also involved oxygen and was similar in many ways to combustion. In 1789 Lavoisier published the first modern chemistry textbook, Elementary Treatise on Chemistry, in which he presented a unified picture of the chemical knowledge assembled up to that time.
Unfortunately, in the same year the text was published, the French Revolution broke out.
Lavoisier, who had been associated with collecting taxes for the government, was executed on the guillotine as an enemy of the people in 1794.
After 1800, chemistry was dominated by scientists who, following Lavoisier’s lead, performed careful weighing experiments to study the course of chemical reactions and to determine the composition of various chemical compounds. One of these chemists, a Frenchman, Joseph Proust (1754–1826), showed that a given compound always contains exactly the same proportion of elements by mass. For example, Proust found that the sub-stance copper carbonate is always 5.3 parts copper to 4 parts oxygen to 1 part carbon (by mass). The principle of the constant composition of compounds, originally called “Proust’s law,” is now known as the law of definite proportion.
Proust’s discovery stimulated John Dalton (1766–1844), an English schoolteacher (Fig. 2.3), to think about atoms as the particles that might compose elements. Dalton reasoned that if elements were composed of tiny individual particles, a given compound should always contain the same combination of these atoms. This concept explained why the same relative masses of elements were always found in a given compound.
Oxygen is from the French oxygène, meaning “generator of acid,” because it was initially considered to be an integral part of all acids.
42 Chapter Two Atoms, Molecules, and Ions Dalton noted that compound II contains twice as much oxygen per gram of carbon as com-pound I, a fact that could easily be explained in terms of atoms. Compound I might be CO, and compound II might be CO2. This principle, which was found to apply to compounds of other elements as well, became known as the law of multiple proportions: When two elements form a series of compounds, the ratios of the masses of the second element that combine with 1 gram of the first element can always be reduced to small whole numbers.
To make sure the significance of this observation is clear, in Sample Exercise 2.1 we will consider data for a series of compounds consisting of nitrogen and oxygen.
Illustrating the Law of Multiple Proportions The following data were collected for several compounds of nitrogen and oxygen: Show how these data illustrate the law of multiple proportions.
Solution For the law of multiple proportions to hold, the ratios of the masses of nitrogen combin-ing with 1 gram of oxygen in each pair of compounds should be small whole numbers.
We therefore compute the ratios as follows: These results support the law of multiple proportions.
See Exercises 2.27 and 2.28.
A C 1.750 0.4375 4 1 B C 0.875 0.4375 2 1 A B 1.750 0.875 2 1 Mass of Nitrogen That Combines with 1 g of Oxygen Compound A 1.750 g Compound B 0.8750 g Compound C 0.4375 g Sample Exercise 2.1 Mass of Oxygen That Combines with 1 g of Carbon Compound I 1.33 g Compound II 2.66 g Subscripts are used to show the numbers of atoms present. The number 1 is understood (not written).
The symbols for the elements and the writing of chemical formulas will be illustrated further in Sections 2.6 and 2.7.
But Dalton discovered another principle that convinced him even more of the ex-istence of atoms. He noted, for example, that carbon and oxygen form two different compounds that contain different relative amounts of carbon and oxygen, as shown by the following data: 2.3 Dalton’s Atomic Theory 43 The significance of the data in Sample Exercise 2.1 is that compound A contains twice as much nitrogen (N) per gram of oxygen (O) as does compound B and that compound B contains twice as much nitrogen per gram of oxygen as does compound C.
These data can be explained readily if the substances are composed of molecules made up of nitrogen atoms and oxygen atoms. For example, one set of possibilities for compounds A, B, and C is Now we can see that compound A contains two atoms of N for every atom of O, whereas compound B contains one atom of N per atom of O. That is, compound A contains twice as much nitrogen per given amount of oxygen as does compound B. Similarly, since com-pound B contains one N per O and compound C contains one N per two O’s, the nitro-gen content of compound C per given amount of oxygen is half that of compound B.
Another set of compounds that fits the data in Sample Exercise 2.1 is Verify for yourself that these compounds satisfy the requirements.
Still another set that works is See if you can come up with still another set of compounds that satisfies the data in Sam-ple Exercise 2.1. How many more possibilities are there?
In fact, an infinite number of other possibilities exists. Dalton could not deduce ab-solute formulas from the available data on relative masses. However, the data on the com-position of compounds in terms of the relative masses of the elements supported his hypothesis that each element consisted of a certain type of atom and that compounds were formed from specific combinations of atoms.
2.3 Dalton’s Atomic Theory In 1808 Dalton published A New System of Chemical Philosophy, in which he presented his theory of atoms: 1. Each element is made up of tiny particles called atoms.
2. The atoms of a given element are identical; the atoms of different elements are different in some fundamental way or ways.
3. Chemical compounds are formed when atoms of different elements combine with each other. A given compound always has the same relative numbers and types of atoms.
4. Chemical reactions involve reorganization of the atoms—changes in the way they are bound together. The atoms themselves are not changed in a chemical reaction.
A: N O = 2 2 C: N O = 2 4 B: N O = 4 2 A: N O = 1 2 C: N O = 1 4 B: N O = 1 1 A: N O = 2 1 C: N O = 1 2 B: N O = 1 1 FIGURE 2.3 John Dalton (1766–1844), an Englishman, began teaching at a Quaker school when he was 12. His fascination with science in-cluded an intense interest in meteorology, which led to an interest in the gases of the air and their ultimate components, atoms.
Dalton is best known for his atomic the-ory, in which he postulated that the funda-mental differences among atoms are their masses. He was the first to prepare a table of relative atomic weights.
Dalton was a humble man with several apparent handicaps: He was not articulate and he was color-blind, a terrible problem for a chemist. Despite these disadvantages, he helped to revolutionize the science of chemistry.
These statements are a modern paraphrase of Dalton’s ideas.
44 Chapter Two Atoms, Molecules, and Ions It is instructive to consider Dalton’s reasoning on the relative masses of the atoms of the various elements. In Dalton’s time water was known to be composed of the elements hydrogen and oxygen, with 8 grams of oxygen present for every 1 gram of hydrogen. If the formula for water were OH, an oxygen atom would have to have 8 times the mass of a hydrogen atom. However, if the formula for water were H2O (two atoms of hydrogen for every oxygen atom), this would mean that each atom of oxygen is 16 times as mas-sive as each atom of hydrogen (since the ratio of the mass of one oxygen to that of two hydrogens is 8 to 1). Because the formula for water was not then known, Dalton could not specify the relative masses of oxygen and hydrogen unambiguously. To solve the prob-lem, Dalton made a fundamental assumption: He decided that nature would be as simple as possible. This assumption led him to conclude that the formula for water should be OH. He thus assigned hydrogen a mass of 1 and oxygen a mass of 8.
Using similar reasoning for other compounds, Dalton prepared the first table of atomic masses (sometimes called atomic weights by chemists, since mass is often determined by comparison to a standard mass—a process called weighing). Many of the masses were later proved to be wrong because of Dalton’s incorrect assumptions about the formulas of cer-tain compounds, but the construction of a table of masses was an important step forward.
Although not recognized as such for many years, the keys to determining absolute formulas for compounds were provided in the experimental work of the French chemist Joseph Gay-Lussac (1778–1850) and by the hypothesis of an Italian chemist named Amadeo Avogadro (1776–1856). In 1809 Gay-Lussac performed experiments in which he measured (under the same conditions of temperature and pressure) the volumes of gases that reacted with each other. For example, Gay-Lussac found that 2 volumes of hydrogen react with 1 volume of oxygen to form 2 volumes of gaseous water and that 1 volume of hydrogen reacts with 1 volume of chlorine to form 2 volumes of hydrogen chloride. These results are represented schematically in Fig. 2.4.
In 1811 Avogadro interpreted these results by proposing that at the same temperature and pressure, equal volumes of different gases contain the same number of particles. This assumption (called Avogadro’s hypothesis) makes sense if the distances between the par-ticles in a gas are very great compared with the sizes of the particles. Under these condi-tions, the volume of a gas is determined by the number of molecules present, not by the size of the individual particles.
If Avogadro’s hypothesis is correct, Gay-Lussac’s result, 2 volumes of hydrogen react with 1 volume of oxygen 2 volumes of water vapor can be expressed as follows: 2 molecules of hydrogen react with 1 molecule of oxygen 2 molecules of water ¡ ¡ Joseph Louis Gay-Lussac, a French physicist and chemist, was remarkably versatile.
Although he is now known primarily for his studies on the combining of volumes of gases, Gay-Lussac was instrumental in the studies of many of the other properties of gases. Some of Gay-Lussac’s motivation to learn about gases arose from his passion for ballooning. In fact, he made ascents to heights of over 4 miles to collect air sam-ples, setting altitude records that stood for about 50 years. Gay-Lussac also was the codiscoverer of boron and the developer of a process for manufacturing sulfuric acid. As chief assayer of the French mint, Gay-Lussac developed many techniques for chemical analysis and invented many types of glassware now used routinely in labs.
Gay-Lussac spent his last 20 years as a lawmaker in the French government.
FIGURE 2.4 A representation of some of Gay-Lussac’s experimental results on combining gas volumes.
2 volumes hydrogen 1 volume oxygen 2 volumes gaseous water combines with + 1 volume hydrogen combines with + 1 volume chlorine 2 volumes hydrogen chloride to form to form A molecule is a collection of atoms (see Section 2.6).
2.4 Early Experiments to Characterize the Atom 45 These observations can best be explained by assuming that gaseous hydrogen, oxygen, and chlorine are all composed of diatomic (two-atom) molecules: H2, O2, and Cl2, respectively. Gay-Lussac’s results can then be represented as shown in Fig. 2.5. (Note that this reasoning suggests that the formula for water is H2O, not OH as Dalton believed.) Unfortunately, Avogadro’s interpretations were not accepted by most chemists, and a half-century of confusion followed, in which many different assumptions were made about formulas and atomic masses.
During the nineteenth century, painstaking measurements were made of the masses of various elements that combined to form compounds. From these experiments a list of relative atomic masses could be determined. One of the chemists involved in contributing to this list was a Swede named Jöns Jakob Berzelius (1779–1848), who discovered the elements cerium, selenium, silicon, and thorium and developed the modern symbols for the elements used in writing the formulas of compounds.
2.4 Early Experiments to Characterize the Atom On the basis of the work of Dalton, Gay-Lussac, Avogadro, and others, chemistry was be-ginning to make sense. The concept of atoms was clearly a good idea. Inevitably, scien-tists began to wonder about the nature of the atom. What is an atom made of, and how do the atoms of the various elements differ?
The Electron The first important experiments that led to an understanding of the composition of the atom were done by the English physicist J. J. Thomson (Fig. 2.6), who studied electrical discharges in partially evacuated tubes called cathode-ray tubes (Fig. 2.7) during the pe-riod from 1898 to 1903. Thomson found that when high voltage was applied to the tube, a “ray” he called a cathode ray (because it emanated from the negative electrode, or cath-ode) was produced. Because this ray was produced at the negative electrode and was re-pelled by the negative pole of an applied electric field (see Fig. 2.8), Thomson postulated that the ray was a stream of negatively charged particles, now called electrons. From ex-periments in which he measured the deflection of the beam of electrons in a magnetic field, Thomson determined the charge-to-mass ratio of an electron: where e represents the charge on the electron in coulombs (C) and m represents the electron mass in grams.
e m 1.76 108 C/g + + O O H H H H H H H H O H H O Cl Cl Cl H Cl H FIGURE 2.5 A representation of combining gases at the molecular level. The spheres represent atoms in the molecules.
The Italian chemist Stanislao Cannizzaro (1826–1910) cleared up the confusion in 1860 by doing a series of molar mass determinations that convinced the scien-tific community that the correct atomic mass of carbon is 12. For more informa-tion, see From Caveman to Chemist by Hugh Salzberg (American Chemical Society, 1991), p. 223.
46 Chapter Two Atoms, Molecules, and Ions One of Thomson’s primary goals in his cathode-ray tube experiments was to gain an understanding of the structure of the atom. He reasoned that since electrons could be pro-duced from electrodes made of various types of metals, all atoms must contain electrons.
Since atoms were known to be electrically neutral, Thomson further assumed that atoms also must contain some positive charge. Thomson postulated that an atom consisted of a Source of electrical potential Stream of negative particles (electrons) Metal electrode (–) (+) Partially evacuated glass tube Metal electrode FIGURE 2.7 A cathode-ray tube. The fast-moving electrons excite the gas in the tube, causing a glow between the electrodes. The green color in the photo is due to the response of the screen (coated with zinc sulfide) to the electron beam.
CHEMICAL IMPACT Berzelius, Selenium, and Silicon J öns Jakob Berzelius was probably the best experimental chemist of his generation and, given the crudeness of his laboratory equipment, maybe the best of all time. Unlike Lavoisier, who could afford to buy the best laboratory equip-ment available, Berzelius worked with minimal equipment in very plain sur-roundings. One of Berzelius’s students described the Swed-ish chemist’s work-place: “The labora-tory consisted of two ordinary rooms with the very simplest arrangements; there were neither furnaces nor hoods, neither water system nor gas.
Against the walls stood some closets with the chemicals, in the middle the mer-cury trough and the blast lamp table. Beside this was the sink consisting of a stone water holder with a stopcock and a pot standing under it. [Next door in the kitchen] stood a small heating furnace.” In these simple facilities Berzelius performed more than 2000 experiments over a 10-year period to determine accurate atomic masses for the 50 elements then known. His success can be seen from the data in the table at left. These remarkably accurate values attest to his experimental skills and patience.
Besides his table of atomic masses, Berzelius made many other major contributions to chemistry. The most important of these was the invention of a simple set of symbols for the el-ements along with a system for writing the formulas of com-pounds to replace the awkward symbolic representations of the alchemists. Although some chemists, including Dalton, ob-jected to the new system, it was gradually adopted and forms the basis of the system we use today.
In addition to these accomplishments, Berzelius discov-ered the elements cerium, thorium, selenium, and silicon. Of these elements, selenium and silicon are particularly impor-tant in today’s world. Berzelius discovered selenium in 1817 in connection with his studies of sulfuric acid. For years se-lenium’s toxicity has been known, but only recently have we become aware that it may have a positive effect on human Comparison of Several of Berzelius’s Atomic Masses with the Modern Values Atomic Mass Berzelius’s Current Element Value Value Chlorine 35.41 35.45 Copper 63.00 63.55 Hydrogen 1.00 1.01 Lead 207.12 207.2 Nitrogen 14.05 14.01 Oxygen 16.00 16.00 Potassium 39.19 39.10 Silver 108.12 107.87 Sulfur 32.18 32.07 Visualization: Cathode-Ray Tube 2.4 Early Experiments to Characterize the Atom 47 diffuse cloud of positive charge with the negative electrons embedded randomly in it. This model, shown in Fig. 2.9, is often called the plum pudding model because the electrons are like raisins dispersed in a pudding (the positive charge cloud), as in plum pudding, a favorite English dessert.
In 1909 Robert Millikan (1868–1953), working at the University of Chicago, per-formed very clever experiments involving charged oil drops. These experiments allowed Applied electrical field Metal electrode (–) (+) (+) (–) Metal electrode Spherical cloud of positive charge Electrons FIGURE 2.8 Deflection of cathode rays by an applied electric field.
FIGURE 2.9 The plum pudding model of the atom.
health. Studies have shown that trace amounts of selenium in the diet may pro-tect people from heart disease and cancer. One study based on data from 27 countries showed an inverse relation-ship between the cancer death rate and the selenium content of soil in a particular region (low cancer death rate in areas with high selenium content). Another research paper re-ported an inverse relationship between the selenium content of the blood and the incidence of breast cancer in women. A study reported in 1998 used the toenail clippings of 33,737 men to show that selenium seems to pro-tect against prostate cancer. Selenium is also found in the heart muscle and may play an important role in proper heart function. Because of these and other studies, selenium’s rep-utation has improved, and many scientists are now studying its function in the human body.
Silicon is the second most abundant element in the earth’s crust, exceeded only by oxygen. As we will see in Chapter 10, compounds involving silicon bonded to oxygen make up most of the earth’s sand, rock, and soil. Berzelius prepared silicon in its pure form in 1824 by heating silicon tetrafluor-ide (SiF4) with potassium metal. Today, silicon forms the ba-sis for the modern microelectronics industry centered near San Francisco in a place that has come to be known as “Silicon Valley.” The technology of the silicon chip (see figure) with its printed circuits has transformed computers from room-sized mon-sters with thou-sands of unreliable vacuum tubes to desktop and note-book-sized units with trouble-free “solid-state” circuitry.
See E. J. Holmyard, Alchemy (New York: Penguin Books, 1968).
The Alchemists’ Symbols for Some Common Elements and Compounds Substance Alchemists’ Symbol Silver Lead Tin Platinum Sulfuric acid Alcohol Sea salt A silicon chip.
48 Chapter Two Atoms, Molecules, and Ions him to determine the magnitude of the electron charge (see Fig. 2.10). With this value and the charge-to-mass ratio determined by Thomson, Millikan was able to calculate the mass of the electron as 9.11 1031 kilogram.
Radioactivity In the late nineteenth century scientists discovered that certain elements produce high-energy radiation. For example, in 1896 the French scientist Henri Becquerel found ac-cidentally that a piece of a mineral containing uranium could produce its image on a photographic plate in the absence of light. He attributed this phenomenon to a sponta-neous emission of radiation by the uranium, which he called radioactivity. Studies in the early twentieth century demonstrated three types of radioactive emission: gamma () rays, beta () particles, and alpha () particles. A ray is high-energy “light”; a particle is a high-speed electron; and an particle has a 2 charge, that is, a charge twice that of the electron and with the opposite sign. The mass of an particle is 7300 times that of the electron. More modes of radioactivity are now known, and we will discuss them in Chapter 18. Here we will consider only particles because they were used in some crucial early experiments.
The Nuclear Atom In 1911 Ernest Rutherford (Fig. 2.11), who performed many of the pioneering experi-ments to explore radioactivity, carried out an experiment to test Thomson’s plum pudding model. The experiment involved directing particles at a thin sheet of metal foil, as il-lustrated in Fig. 2.12. Rutherford reasoned that if Thomson’s model were accurate, the massive particles should crash through the thin foil like cannonballs through gauze, as shown in Fig. 2.13(a). He expected the particles to travel through the foil with, at the most, very minor deflections in their paths. The results of the experiment were very dif-ferent from those Rutherford anticipated. Although most of the particles passed straight through, many of the particles were deflected at large angles, as shown in Fig. 2.13(b), and some were reflected, never hitting the detector. This outcome was a great surprise to Rutherford. (He wrote that this result was comparable with shooting a howitzer at a piece of paper and having the shell reflected back.) Atomizer to produce oil droplets Microscope Electrically charged plates X rays produce charges on the oil drops (+) (–) Oil spray FIGURE 2.10 A schematic representation of the apparatus Millikan used to determine the charge on the electron. The fall of charged oil droplets due to gravity can be halted by adjusting the voltage across the two plates. This voltage and the mass of the oil drop can then be used to calculate the charge on the oil drop. Millikan’s experiments showed that the charge on an oil drop is always a whole-number multiple of the electron charge.
A technician using a scanner to monitor the uptake of radioactive iodine in a patient’s thyroid.
FIGURE 2.11 Ernest Rutherford (1871–1937) was born on a farm in New Zealand. In 1895 he placed second in a scholarship competition to at-tend Cambridge University but was awarded the scholarship when the winner decided to stay home and get married. As a scientist in England, Rutherford did much of the early work on characterizing radioactivity. He named the and particles and the ray and coined the term half-life to describe an important attribute of radioactive elements.
His experiments on the behavior of parti-cles striking thin metal foils led him to pos-tulate the nuclear atom. He also invented the name proton for the nucleus of the hy-drogen atom. He received the Nobel Prize in chemistry in 1908.
Visualization: Millikan’s Oil Drop Experiment 2.5 The Modern View of Atomic Structure: An Introduction 49 Rutherford knew from these results that the plum pudding model for the atom could not be correct. The large deflections of the particles could be caused only by a center of concentrated positive charge that contains most of the atom’s mass, as illustrated in Fig. 2.13(b). Most of the particles pass directly through the foil because the atom is mostly open space. The deflected particles are those that had a “close encounter” with the massive positive center of the atom, and the few reflected particles are those that made a “direct hit” on the much more massive positive center.
In Rutherford’s mind these results could be explained only in terms of a nuclear atom—an atom with a dense center of positive charge (the nucleus) with electrons moving around the nucleus at a distance that is large relative to the nuclear radius.
2.5 The Modern View of Atomic Structure: An Introduction In the years since Thomson and Rutherford, a great deal has been learned about atomic structure. Because much of this material will be covered in detail in later chapters, only an introduction will be given here. The simplest view of the atom is that it consists of a tiny nucleus (with a diameter of about 1013 cm) and electrons that move about the nu-cleus at an average distance of about 108 cm from it (see Fig. 2.14).
As we will see later, the chemistry of an atom mainly results from its electrons. For this reason, chemists can be satisfied with a relatively crude nuclear model. The nucleus is assumed to contain protons, which have a positive charge equal in magnitude to the electron’s negative charge, and neutrons, which have virtually the same mass as a pro-ton but no charge. The masses and charges of the electron, proton, and neutron are shown in Table 2.1.
Source of α particles Beam of α particles Some α particles are scattered Most particles pass straight through foil Thin metal foil Screen to detect scattered α particles FIGURE 2.12 Rutherford’s experiment on -particle bombardment of metal foil.
Electrons scattered throughout Diffuse positive charge – – – – – – – – – – (a) (b) – – – – – – – – – – n+ FIGURE 2.13 (a) The expected results of the metal foil experiment if Thomson’s model were cor-rect. (b) Actual results.
The forces that bind the positively charged protons in the nucleus will be discussed in Chapter 18.
Visualization: Gold Foil Experiment 50 Chapter Two Atoms, Molecules, and Ions Two striking things about the nucleus are its small size compared with the overall size of the atom and its extremely high density. The tiny nucleus accounts for almost all the atom’s mass. Its great density is dramatically demonstrated by the fact that a piece of nuclear material about the size of a pea would have a mass of 250 million tons!
An important question to consider at this point is, “If all atoms are composed of these same components, why do different atoms have different chemical properties?” The an-swer to this question lies in the number and the arrangement of the electrons. The elec-trons constitute most of the atomic volume and thus are the parts that “intermingle” when atoms combine to form molecules. Therefore, the number of electrons possessed by a given atom greatly affects its ability to interact with other atoms. As a result, the atoms of different elements, which have different numbers of protons and electrons, show dif-ferent chemical behavior.
A sodium atom has 11 protons in its nucleus. Since atoms have no net charge, the number of electrons must equal the number of protons. Therefore, a sodium atom has 11 electrons moving around its nucleus. It is always true that a sodium atom has 11 protons and 11 electrons. However, each sodium atom also has neutrons in its nucleus, and dif-ferent types of sodium atoms exist that have different numbers of neutrons. For example, consider the sodium atoms represented in Fig. 2.15. These two atoms are isotopes, or atoms with the same number of protons but different numbers of neutrons. Note that the symbol for one particular type of sodium atom is written Mass number Element symbol Atomic number where the atomic number Z (number of protons) is written as a subscript, and the mass number A (the total number of protons and neutrons) is written as a superscript. (The particular atom represented here is called “sodium twenty-three.” It has 11 electrons, 11 protons, and 12 neutrons.) Because the chemistry of an atom is due to its electrons, iso-topes show almost identical chemical properties. In nature most elements contain mix-tures of isotopes.
d 23 11Na TABLE 2.1 The Mass and Charge of the Electron, Proton, and Neutron Particle Mass Charge Electron 9.11 1031 kg 1 Proton 1.67 1027 kg 1 Neutron 1.67 1027 kg None The magnitude of the charge of the electron and the proton is 1.60 1019 C.
~10 –13cm Nucleus ~10 –8cm FIGURE 2.14 A nuclear atom viewed in cross section.
Note that this drawing is not to scale.
¡ ¡ Mass number Z AX Element symbol Atomic number 88n 8n → Nucleus 23Na 11 11 electrons 11 electrons 24Na 11 11 protons 12 neutrons Nucleus 11 protons 13 neutrons FIGURE 2.15 Two isotopes of sodium. Both have 11 pro-tons and 11 electrons, but they differ in the number of neutrons in their nuclei.
If the atomic nucleus were the size of this ball bearing, a typical atom would be the size of this stadium.
The chemistry of an atom arises from its electrons.
2.5 The Modern View of Atomic Structure: An Introduction 51 CHEMICAL IMPACT Reading the History of Bogs S cientists often “read” the history of the earth and its in-habitants using very different “books” than traditional historians. For example, the disappearance of the dinosaurs 65 million years ago in an “instant” of geological time was a great mystery until unusually high iridium and osmium levels were discovered at a position in the earth’s crust cor-responding to that time. These high levels of iridium and osmium suggested that an extraterrestrial object had struck the earth 65 million years ago with catastrophic results for the dinosaurs. Since then, the huge buried crater caused by the object has been discovered on the Yucatan Peninsula, and virtually everyone is now convinced that this is the correct explanation for the disappear-ing dinosaurs.
History is also being “read” by scientists study-ing ice cores from glaciers in Iceland. Now Swiss sci-entists have found that ancient peat bogs can furnish a reliable historical record. Geochemist William Shotyk of the University of Bern has found a 15,000-year window on history by analyzing the lead con-tent of core samples from a Swiss mountainside peat bog [Science 281 (1998): 1635]. Various parts of the core samples were dated by 14C dating techniques (see Chapter 18, Section 18.4, for more information) and analyzed for their scandium and lead contents.
Also, the 206Pb207Pb ratio was measured for each sample. These data are represented in the accompa-nying figure. Notice that the 206Pb207Pb ratio remains very close to 1.20 (see the red band in the figure) from 14,000 years to 3200 years. The value of 1.20 is the same as the average 206Pb207Pb ratio in the earth’s soil.
The core also reveals that the total lead and scan-dium levels increased simultaneously at the 6000-year mark but that the 206Pb207Pb ratio remained close to 1.20. This coincides with the beginning of agriculture in Europe, which caused more soil dust to enter the atmosphere.
Significantly, about 3000 years ago the 206Pb207Pb ratio decreased markedly. This also corresponds in the core sample to an increase in total lead content out of proportion to the increase in scandium. This indicates the lead no longer resulted from soil dust but from other activities of humans—lead mining had begun. Since the 3000-year mark, the 206Pb207Pb ra-tio has remained well below 1.20, indicating that hu-man use of lead ores has become the dominant source of airborne lead. This is confirmed by the sharp decline in the ratio beginning 200 years ago that corresponds to the importation into England of Australian lead ores having low 206Pb207Pb ratios.
So far only lead has been used to read the history in the bog. However, Shotyk’s group is also measuring the changes in the levels of copper, zinc, cadmium, arsenic, mercury, and antimony. More interesting stories are sure to follow.
Geochemist William Shotyk’s analysis of the lead content of ice core samples reveals a 15,000-year history of lead levels. (Note: Dates are based on cali-brated radiocarbon dating. Because the core was retrieved in two segments, a break in data occurs between 2060 and 3200 years before present.) 52 Chapter Two Atoms, Molecules, and Ions Writing the Symbols for Atoms Write the symbol for the atom that has an atomic number of 9 and a mass number of 19.
How many electrons and how many neutrons does this atom have?
Solution The atomic number 9 means the atom has 9 protons. This element is called fluorine, symbolized by F. The atom is represented as and is called “fluorine nineteen.” Since the atom has 9 protons, it also must have 9 elec-trons to achieve electrical neutrality. The mass number gives the total number of protons and neutrons, which means that this atom has 10 neutrons.
See Exercises 2.43 through 2.46.
2.6 Molecules and Ions From a chemist’s viewpoint, the most interesting characteristic of an atom is its ability to combine with other atoms to form compounds. It was John Dalton who first recog-nized that chemical compounds are collections of atoms, but he could not determine the structure of atoms or their means for binding to each other. During the twentieth century we learned that atoms have electrons and that these electrons participate in bonding one atom to another. We will discuss bonding thoroughly in Chapters 8 and 9; here we will introduce some simple bonding ideas that will be useful in the next few chapters.
The forces that hold atoms together in compounds are called chemical bonds. One way that atoms can form bonds is by sharing electrons. These bonds are called covalent bonds, and the resulting collection of atoms is called a molecule. Molecules can be rep-resented in several different ways. The simplest method is the chemical formula, in which the symbols for the elements are used to indicate the types of atoms present and subscripts are used to indicate the relative numbers of atoms. For example, the formula for carbon dioxide is CO2, meaning that each molecule contains 1 atom of carbon and 2 atoms of oxygen.
Examples of molecules that contain covalent bonds are hydrogen (H2), water (H2O), oxygen (O2), ammonia (NH3), and methane (CH4). More information about a molecule is given by its structural formula, in which the individual bonds are shown (indicated by lines). Structural formulas may or may not indicate the actual shape of the molecule. For example, water might be represented as The structure on the right shows the actual shape of the water molecule. Scientists know from experimental evidence that the molecule looks like this. (We will study the shapes of molecules further in Chapter 8.) The structural formula for ammonia is shown in the margin at left.
Note that atoms connected to the central atom by dashed lines are behind the plane of the paper, and atoms connected to the central atom by wedges are in front of the plane of the paper.
In a compound composed of molecules, the individual molecules move around as independent units. For example, a molecule of methane gas can be represented in several ways. The structural formula for methane (CH4) is shown in Fig. 2.16. The space-filling H O H H H O or 19 9F Sample Exercise 2.2 H H H N Ammonia Visualization: Covalent Bonding 2.6 Molecules and Ions 53 model of methane, which shows the relative sizes of the atoms as well as their relative orientation in the molecule, is given in Fig. 2.17. Ball-and-stick models are also used to represent molecules. The ball-and-stick structure of methane is shown in Fig. 2.18.
A second type of chemical bond results from attractions among ions. An ion is an atom or group of atoms that has a net positive or negative charge. The best-known ionic compound is common table salt, or sodium chloride, which forms when neutral chlorine and sodium react.
To see how the ions are formed, consider what happens when an electron is transferred from a sodium atom to a chlorine atom (the neutrons in the nuclei will be ignored): With one electron stripped off, the sodium, with its 11 protons and only 10 electrons, now has a net 1 charge—it has become a positive ion. A positive ion is called a cation. The sodium ion is written as Na, and the process can be represented in short-hand form as Na ¡ Na e Neutral sodium atom (Na) 11 electrons Sodium ion (Na+) 11+ 11+ 10 electrons Minus 1 electron H H H H C Methane FIGURE 2.17 Space-filling model of methane. This type of model shows both the relative sizes of the atoms in the molecule and their spatial relationships.
FIGURE 2.18 Ball-and-stick model of methane.
FIGURE 2.16 The structural formula for methane.
Na is usually called the sodium ion rather than the sodium cation. Also Cl is called the chloride ion rather than the chloride anion. In general, when a spe-cific ion is referred to, the word ion rather than cation or anion is used.
54 Chapter Two Atoms, Molecules, and Ions If an electron is added to chlorine, the 18 electrons produce a net 1 charge; the chlorine has become an ion with a nega-tive charge—an anion. The chloride ion is written as Cl, and the process is represented as Because anions and cations have opposite charges, they attract each other. This force of attraction between oppositely charged ions is called ionic bonding. As illustrated in Fig. 2.19, sodium metal and chlorine gas (a green gas composed of Cl2 molecules) react Cl e ¡ Cl Chloride ion (Cl–) 18 electrons Neutral chlorine atom (Cl) 17+ 17+ 17 electrons Plus 1 electron Cl– Cl Cl Cl– Na+ Na+ Na Na FIGURE 2.19 Sodium metal (which is so soft it can be cut with a knife and which consists of individual sodium atoms) reacts with chlorine gas (which contains Cl2 molecules) to form solid sodium chloride (which contains Na and Cl ions packed together).
2.7 An Introduction to the Periodic Table 55 to form solid sodium chloride, which contains many Na and Cl ions packed together and forms the beautiful colorless cubic crystals shown in Fig. 2.19.
A solid consisting of oppositely charged ions is called an ionic solid, or a salt.
Ionic solids can consist of simple ions, as in sodium chloride, or of polyatomic (many atom) ions, as in ammonium nitrate (NH4NO3), which contains ammonium ions (NH4 ) and nitrate ions (NO3 ). The ball-and-stick models of these ions are shown in Fig. 2.20.
2.7 An Introduction to the Periodic Table In a room where chemistry is taught or practiced, a chart called the periodic table is almost certain to be found hanging on the wall. This chart shows all the known elements and gives a good deal of information about each. As our study of chemistry progresses, the usefulness of the periodic table will become more obvious. This section will simply introduce it to you.
A simplified version of the periodic table is shown in Fig. 2.21. The letters in the boxes are the symbols for the elements; these abbreviations are based on the current element names or the original names (see Table 2.2). The number shown above each symbol is the atomic number (number of protons) for that element. For example, carbon (C) has atomic number 6, and lead (Pb) has atomic number 82. Most of the elements are metals. Metals have characteristic physical properties such as efficient conduction of heat and electricity, malleability (they can be hammered into thin sheets), ductility (they can be pulled into wires), and (often) a lustrous appearance. Chemi-cally, metals tend to lose electrons to form positive ions. For example, copper is a typical metal. It is lustrous (although it tarnishes readily); it is an excellent conductor of electricity (it is widely used in electrical wires); and it is readily formed into vari-ous shapes, such as pipes for water systems. Copper is also found in many salts, such as the beautiful blue copper sulfate, in which copper is present as Cu2 ions. Copper is a member of the transition metals—the metals shown in the center of the periodic table.
The relatively few nonmetals appear in the upper-right corner of the table (to the right of the heavy line in Fig. 2.21), except hydrogen, a nonmetal that resides in the upper-left corner. The nonmetals lack the physical properties that characterize the metals.
Chemically, they tend to gain electrons in reactions with metals to form negative ions.
Nonmetals often bond to each other by forming covalent bonds. For example, chlorine is a typical nonmetal. Under normal conditions it exists as Cl2 molecules; it reacts with metals to form salts containing Cl ions (NaCl, for example); and it forms covalent bonds with nonmetals (for example, hydrogen chloride gas, HCl).
The periodic table is arranged so that elements in the same vertical columns (called groups or families) have similar chemical properties. For example, all of the alkali metals, members of Group 1A—lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr)—are very active elements that readily form ions with a 1 charge when they react with nonmetals. The members of Group 2A—beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra)—are called the alkaline earth metals. They all form ions with a 2 charge when they react with nonmetals. The halogens, the members of Group 7A—fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At)—all form diatomic molecules. Fluorine, chlorine, bromine, and iodine all react with metals to form salts containing ions with a 1 charge (F, Cl, Br, and I). The members of Group 8A—helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn)—are known as the noble gases. They all exist under normal conditions as monatomic (single-atom) gases and have little chemical reactivity.
FIGURE 2.20 Ball-and-stick models of the ammonium ion (NH4 ) and the nitrate ion (NO3 ).
Metals tend to form positive ions; non-metals tend to form negative ions.
Elements in the same vertical column in the periodic table form a group (or family) and generally have similar properties.
Visualization: Comparison of a Molecular Compound and an Ionic Compound Samples of chlorine gas, liquid bromine, and solid iodine.
TABLE 2.2 The Symbols for the Elements That Are Based on the Original Names Current Name Original Name Symbol Antimony Stibium Sb Copper Cuprum Cu Iron Ferrum Fe Lead Plumbum Pb Mercury Hydrargyrum Hg Potassium Kalium K Silver Argentum Ag Sodium Natrium Na Tin Stannum Sn Tungsten Wolfram W 56 Chapter Two Atoms, Molecules, and Ions FIGURE 2.21 The periodic table.
1 H 3 Li 11 Na 19 K 37 Rb 55 Cs 87 Fr 4 Be 12 Mg 20 Ca 38 Sr 56 Ba 88 Ra 21 Sc 39 Y 57 La 89 Ac† 22 Ti 40 Zr 72 Hf 104 Rf 23 V 41 Nb 73 Ta 105 Db 24 Cr 42 Mo 74 W 106 Sg 25 Mn 43 Tc 75 Re 107 Bh 26 Fe 44 Ru 76 Os 108 Hs 27 Co 45 Rh 77 Ir 109 Mt 110 Ds 111 Rg 28 Ni 46 Pd 78 Pt 29 Cu 47 Ag 79 Au 3 4 5 6 7 8 9 10 11 12 31 Ga 49 In 81 Tl 5 B 13 Al 32 Ge 50 Sn 82 Pb 6 C 14 Si 33 As 51 Sb 83 Bi 7 N 15 P 34 Se 52 Te 84 Po 8 O 16 S 9 F 17 Cl 35 Br 53 I 85 At 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn 2 He 58 Ce 90 Th 59 Pr 91 Pa 60 Nd 92 U 61 Pm 93 Np 62 Sm 94 Pu 63 Eu 95 Am 64 Gd 96 Cm 65 Tb 97 Bk 66 Dy 98 Cf 67 Ho 99 Es 68 Er 100 Fm 69 Tm 101 Md 70 Yb 102 No 71 Lu 103 Lr 1A 2A Transition metals 3A 4A 5A 6A 7A 8A 1 2 13 14 15 16 17 18 Alkali metals Alkaline earth metals Halogens Noble gases Lanthanides †Actinides 112 Uub 113 Uut 114 Uuq 115 Uup 30 Zn 48 Cd 80 Hg 2.8 Naming Simple Compounds 57 Note from Fig. 2.21 that alternate sets of symbols are used to denote the groups. The symbols 1A through 8A are the traditional designations, whereas the numbers 1 to 18 have been suggested recently. In this text the 1A to 8A designations will be used.
The horizontal rows of elements in the periodic table are called periods. Horizontal row 1 is called the first period (it contains H and He); row 2 is called the second period (elements Li through Ne); and so on.
We will learn much more about the periodic table as we continue with our study of chemistry. Meanwhile, when an element is introduced in this text, you should always note its position on the periodic table.
2.8 Naming Simple Compounds When chemistry was an infant science, there was no system for naming compounds.
Names such as sugar of lead, blue vitrol, quicklime, Epsom salts, milk of magnesia, gypsum, and laughing gas were coined by early chemists. Such names are called com-mon names. As chemistry grew, it became clear that using common names for compounds would lead to unacceptable chaos. Nearly 5 million chemical compounds are currently known. Memorizing common names for these compounds would be an impossible task.
The solution, of course, is to adopt a system for naming compounds in which the name tells something about the composition of the compound. After learning the system, a chemist given a formula should be able to name the compound or, given a name, should be able to construct the compound’s formula. In this section we will specify the most im-portant rules for naming compounds other than organic compounds (those based on chains of carbon atoms).
We will begin with the systems for naming inorganic binary compounds— compounds composed of two elements—which we classify into various types for easier recognition. We will consider both ionic and covalent compounds.
CHEMICAL IMPACT Hassium Fits Right in H assium, element 108, does not exist in nature but must be made in a particle accelerator. It was first created in 1984 and can be made by shooting magnesium-26 (26 12Mg) atoms at curium-248 (248 96Cm) atoms. The collisions between these atoms produce some hassium-265 (265 108Hs) atoms. The position of hassium in the periodic table (see Fig. 2.21) in the vertical column containing iron, ruthenium, and osmium suggests that hassium should have chemical properties sim-ilar to these metals. However, it is not easy to test this pre-diction—only a few atoms of hassium can be made at a given time and they last for only about 9 seconds. Imagine having to get your next lab experiment done in 9 seconds!
Amazingly, a team of chemists from the Lawrence Berkeley National Laboratory in California, the Paul Scherrer Institute and the University of Bern in Switzerland, and the Institute of Nuclear Chemistry in Germany have done experiments to characterize the chemical behavior of hassium.
For example, they have observed that hassium atoms react with oxygen to form a hassium oxide compound of the type expected from its position on the periodic table. The team has also measured other properties of hassium, including the energy released as it undergoes nuclear decay to another atom.
This work would have surely pleased Dmitri Mendeleev (see Fig. 7.23), who originally developed the periodic table and showed its power to predict chemical properties.
Another format of the periodic table will be discussed in Section 7.11.
58 Chapter Two Atoms, Molecules, and Ions Binary Ionic Compounds (Type I) Binary ionic compounds contain a positive ion (cation) always written first in the for-mula and a negative ion (anion). In naming these compounds, the following rules apply: 1. The cation is always named first and the anion second.
2. A monatomic (meaning “one-atom”) cation takes its name from the name of the element. For example, Na is called sodium in the names of compounds containing this ion.
3. A monatomic anion is named by taking the root of the element name and adding -ide.
Thus the Cl ion is called chloride.
Some common monatomic cations and anions and their names are given in Table 2.3.
The rules for naming binary ionic compounds are illustrated by the following examples: Compound Ions Present Name NaCl Na, Cl Sodium chloride KI K, I Potassium iodide CaS Ca2, S2 Calcium sulfide Li3N Li, N3 Lithium nitride CsBr Cs, Br Cesium bromide MgO Mg2, O2 Magnesium oxide Naming Type I Binary Compounds Name each binary compound.
a. CsF b. AlCl3 c. LiH Solution a. CsF is cesium fluoride.
b. AlCl3 is aluminum chloride.
c. LiH is lithium hydride.
Notice that, in each case, the cation is named first, and then the anion is named.
See Exercise 2.55.
In formulas of ionic compounds, simple ions are represented by the element symbol: Cl means Cl, Na means Na, and so on.
TABLE 2.3 Common Monatomic Cations and Anions Cation Name Anion Name H Hydrogen H Hydride Li Lithium F Fluoride Na Sodium Cl Chloride K Potassium Br Bromide Cs Cesium I Iodide Be2 Beryllium O2 Oxide Mg2 Magnesium S2 Sulfide Ca2 Calcium N3 Nitride Ba2 Barium P3 Phosphide Al3 Aluminum Ag Silver A monatomic cation has the same name as its parent element.
Sample Exercise 2.3 2.8 Naming Simple Compounds 59 Formulas from Names So far we have started with the chemical formula of a compound and decided on its systematic name. The reverse process is also important. For example, given the name calcium hydroxide, we can write the formula as Ca(OH)2 because we know that calcium forms only Ca2 ions and that, since hydroxide is OH, two of these anions will be required to give a neutral compound.
Binary Ionic Compounds (Type II) In the binary ionic compounds considered earlier (Type I), the metal present forms only a single type of cation. That is, sodium forms only Na, calcium forms only Ca2, and so on. However, as we will see in more detail later in the text, there are many metals that form more than one type of positive ion and thus form more than one type of ionic com-pound with a given anion. For example, the compound FeCl2 contains Fe2 ions, and the compound FeCl3 contains Fe3 ions. In a case such as this, the charge on the metal ion must be specified. The systematic names for these two iron compounds are iron(II) chlo-ride and iron(III) chloride, respectively, where the Roman numeral indicates the charge of the cation.
Another system for naming these ionic compounds that is seen in the older literature was used for metals that form only two ions. The ion with the higher charge has a name ending in -ic, and the one with the lower charge has a name ending in -ous. In this system, for example, Fe3 is called the ferric ion, and Fe2 is called the ferrous ion. The names for FeCl3 and FeCl2 are then ferric chloride and ferrous chloride, respectively. In this text we will use the system that employs Roman numerals. Table 2.4 lists the systematic names for many common type II cations.
Formulas from Names for Type I Binary Compounds Given the following systematic names, write the formula for each compound: a. potassium iodide b. calcium oxide c. gallium bromide Solution TABLE 2.4 Common Type II Cations Ion Systematic Name Fe3 Iron(III) Fe2 Iron(II) Cu2 Copper(II) Cu Copper(I) Co3 Cobalt(III) Co2 Cobalt(II) Sn4 Tin(IV) Sn2 Tin(II) Pb4 Lead(IV) Pb2 Lead(II) Hg2 Mercury(II) Hg2 2 Mercury(I) Ag Silver† Zn2 Zinc† Cd2 Cadmium† Note that mercury(I) ions always occur bound together to form Hg2 2 ions.
†Although these are transition metals, they form only one type of ion, and a Roman numeral is not used.
Sample Exercise 2.4 Sample Exercise 2.5 See Exercise 2.55.
Naming Type II Binary Compounds 1. Give the systematic name for each of the following compounds: a. CuCl b. HgO c. Fe2O3 2. Given the following systematic names, write the formula for each compound: a. Manganese(IV) oxide b. Lead(II) chloride Visualization: Formation of Ionic Compounds Name Formula Comments a. potassium iodide KI Contains K and I.
b. calcium oxide CaO Contains Ca2 and O2.
c. gallium bromide GaBr3 Contains Ga3 and Br.
Must have 3Br to balance charge of Ga3.
60 Chapter Two Atoms, Molecules, and Ions A compound containing a transition metal usually requires a Roman numeral in its name.
Crystals of copper(Il) sulfate.
Sample Exercise 2.6 Type II binary ionic compounds contain a metal that can form more than one type of cation.
A compound must be electrically neutral.
See Exercise 2.56.
Note that the use of a Roman numeral in a systematic name is required only in cases where more than one ionic compound forms between a given pair of elements. This case most commonly occurs for compounds containing transition metals, which often form more than one cation. Elements that form only one cation do not need to be identified by a Roman numeral. Common metals that do not require Roman numerals are the Group 1A elements, which form only 1 ions; the Group 2A elements, which form only 2 ions; and aluminum, which forms only Al3. The element silver deserves special mention at this point. In virtually all its compounds silver is found as the Ag ion. Therefore, although silver is a transition metal (and can potentially form ions other than Ag), sil-ver compounds are usually named without a Roman numeral. Thus AgCl is typically called silver chloride rather than silver(I) chloride, although the latter name is technically cor-rect. Also, a Roman numeral is not used for zinc compounds, since zinc forms only the Zn2 ion.
As shown in Sample Exercise 2.5, when a metal ion is present that forms more than one type of cation, the charge on the metal ion must be determined by balancing the pos-itive and negative charges of the compound. To do this you must be able to recognize the common cations and anions and know their charges (see Tables 2.3 and 2.5).
Naming Binary Compounds 1. Give the systematic name for each of the following compounds: a. CoBr2 b. CaCl2 c. Al2O3 2. Given the following systematic names, write the formula for each compound: a. Chromium(III) chloride b. Gallium iodide 2.
Solution All of these compounds include a metal that can form more than one type of cation. Thus we must first determine the charge on each cation. This can be done by recognizing that a compound must be electrically neutral; that is, the positive and negative charges must exactly balance.
1.
Formula Name Comments a. CuCl Copper(I) chloride Because the anion is Cl, the cation must be Cu (for charge balance), which requires a Roman numeral I.
b. HgO Mercury(II) oxide Because the anion is O2–, the cation must be Hg2 [mercury(II)].
c. Fe2O3 Iron(III) oxide The three O2– ions carry a total charge of 6, so two Fe3 ions [iron(III)] are needed to give a 6 charge.
Name Formula Comments a. Manganese(IV) oxide MnO2 Two O2– ions (total charge 4) are required by the Mn4 ion [manganese(IV)].
b. Lead(II) chloride PbCl2 Two Cl ions are required by the Pb2 ion [lead(II)] for charge balance.
2.8 Naming Simple Compounds 61 Solution 1.
2.
See Exercises 2.57 and 2.58.
The following flowchart is useful when you are naming binary ionic compounds: The common Type I and Type II ions are summarized in Fig. 2.22. Also shown in Fig. 2.22 are the common monatomic ions.
Name the cation using the element name.
Type II Type I Does the compound contain Type I or Type II cations?
Using the principle of charge balance, determine the cation charge.
Include in the cation name a Roman numeral indicating the charge.
Various chromium compounds dissolved in water. From left to right: CrCl2, K2Cr2O7, Cr(NO3)3, CrCl3, K2CrO4.
Formula Name Comments a. CoBr2 Cobalt(II) bromide Cobalt is a transition metal; the compound name must have a Roman numeral. The two Br ions must be balanced by a Co2 ion.
b. CaCl2 Calcium chloride Calcium, an alkaline earth metal, forms only the Ca2 ion. A Roman numeral is not necessary.
c. Al2O3 Aluminum oxide Aluminum forms only the Al3 ion. A Roman numeral is not necessary.
Name Formula Comments a. Chromium(III) chloride CrCl3 Chromium(III) indicates that Cr3 is present, so 3 Cl ions are needed for charge balance.
b. Gallium iodide GaI3 Gallium always forms 3 ions, so 3 I ions are required for charge balance.
FIGURE 2.22 The common cations and anions.
1A Li+ Common Type I cations Common Type II cations Common monatomic anions Mg2+ Na+ K+ Rb+ Cs+ Ca2+ N3– O2– S2– F – Cl– Br– I– Ag+ Zn2+ Cd2+ Cr3+ Cr2+ Mn3+ Mn2+ Fe3+ Fe2+ Co3+ Co2+ Cu2+ Cu+ Pb4+ Al3+ Pb2+ Sn4+ Sn2+ Hg2+ Hg2 2+ Sr2+ Ba2+ 2A 3A 4A 5A 6A 7A 8A 62 Chapter Two Atoms, Molecules, and Ions Ionic Compounds with Polyatomic Ions We have not yet considered ionic compounds that contain polyatomic ions. For example, the compound ammonium nitrate, NH4NO3, contains the polyatomic ions NH4 and NO3 .
Polyatomic ions are assigned special names that must be memorized to name the com-pounds containing them. The most important polyatomic ions and their names are listed in Table 2.5.
Note in Table 2.5 that several series of anions contain an atom of a given ele-ment and different numbers of oxygen atoms. These anions are called oxyanions.
When there are two members in such a series, the name of the one with the smaller number of oxygen atoms ends in -ite and the name of the one with the larger number ends in -ate—for example, sulfite (SO3 2) and sulfate (SO4 2). When more than two oxyanions make up a series, hypo- (less than) and per- (more than) are used as pre-fixes to name the members of the series with the fewest and the most oxygen atoms, respectively. The best example involves the oxyanions containing chlorine, as shown in Table 2.5.
Naming Compounds Containing Polyatomic Ions 1. Give the systematic name for each of the following compounds: a. Na2SO4 b. KH2PO4 c. Fe(NO3)3 d. Mn(OH)2 e. Na2SO3 f. Na2CO3 2. Given the following systematic names, write the formula for each compound: a. Sodium hydrogen carbonate b. Cesium perchlorate Polyatomic ion formulas must be memorized.
Sample Exercise 2.7 TABLE 2.5 Common Polyatomic Ions Ion Name Ion Name Hg2 2 Mercury(I) NCS Thiocyanate NH4 Ammonium CO3 2 Carbonate NO2 Nitrite HCO3 Hydrogen carbonate NO3 Nitrate (bicarbonate is a widely SO3 2 Sulfite used common name) SO4 2 Sulfate ClO Hypochlorite HSO4 Hydrogen sulfate ClO2 Chlorite (bisulfate is a widely ClO3 Chlorate used common name) ClO4 Perchlorate OH Hydroxide C2H3O2 Acetate CN Cyanide MnO4 Permanganate PO4 3 Phosphate Cr2O7 2 Dichromate HPO4 2 Hydrogen phosphate CrO4 2 Chromate H2PO4 Dihydrogen phosphate O2 2 Peroxide C2O4 2 Oxalate 2.8 Naming Simple Compounds 63 c. Sodium hypochlorite d. Sodium selenate e. Potassium bromate Solution 1.
2.
See Exercises 2.59 and 2.60.
Binary Covalent Compounds (Type III) Binary covalent compounds are formed between two nonmetals. Although these compounds do not contain ions, they are named very similarly to binary ionic compounds.
In the naming of binary covalent compounds, the following rules apply: 1. The first element in the formula is named first, using the full element name.
2. The second element is named as if it were an anion.
3. Prefixes are used to denote the numbers of atoms present. These prefixes are given in Table 2.6.
4. The prefix mono- is never used for naming the first element. For example, CO is called carbon monoxide, not monocarbon monoxide.
In binary covalent compounds, the element names follow the same rules as for binary ionic compounds.
Formula Name Comments a. Na2SO4 Sodium sulfate b. KH2PO4 Potassium dihydrogen phosphate c. Fe(NO3)3 Iron(III) nitrate Transition metal—name must contain a Roman numeral. The Fe3 ion balances three NO3 ions.
d. Mn(OH)2 Manganese(II) hydroxide Transition metal—name must contain a Roman numeral. The Mn2 ion balances three OH ions.
e. Na2SO3 Sodium sulfite f. Na2CO3 Sodium carbonate Name Formula Comments a. Sodium hydrogen NaHCO3 Often called sodium bicarbonate.
carbonate b. Cesium perchlorate CsClO4 c. Sodium hypochlorite NaOCl d. Sodium selenate Na2SeO4 Atoms in the same group, like sulfur and selenium, often form similar ions that are named similarly. Thus SeO4 2– is selenate, like SO4 2– (sulfate).
e. Potassium bromate KBrO3 As above, BrO3 is bromate, like ClO3 (chlorate).
64 Chapter Two Atoms, Molecules, and Ions To see how these rules apply, we will now consider the names of the several cova-lent compounds formed by nitrogen and oxygen: Compound Systematic Name Common Name N2O Dinitrogen monoxide Nitrous oxide NO Nitrogen monoxide Nitric oxide NO2 Nitrogen dioxide N2O3 Dinitrogen trioxide N2O4 Dinitrogen tetroxide N2O5 Dinitrogen pentoxide Notice from the preceding examples that to avoid awkward pronunciations, we often drop the final o or a of the prefix when the element begins with a vowel. For example, N2O4 is called dinitrogen tetroxide, not dinitrogen tetraoxide, and CO is called carbon monoxide, not carbon monooxide.
Some compounds are always referred to by their common names. The two best examples are water and ammonia. The systematic names for H2O and NH3 are never used.
Naming Type III Binary Compounds 1. Name each of the following compounds: a. PCl5 b. PCl3 c. SO2 2. From the following systematic names, write the formula for each compound: a. Sulfur hexafluoride b. Sulfur trioxide c. Carbon dioxide Solution 1.
2.
See Exercises 2.61 and 2.62.
The rules for naming binary compounds are summarized in Fig. 2.23. Prefixes to indicate the number of atoms are used only in Type III binary compounds (those containing two nonmetals). An overall strategy for naming compounds is given in Fig. 2.24.
TABLE 2.6 Prefixes Used to Indicate Number in Chemical Names Prefix Number Indicated mono-1 di-2 tri-3 tetra-4 penta-5 hexa-6 hepta-7 octa-8 nona-9 deca-10 Sample Exercise 2.8 Formula Name a. PCl5 Phosphorus pentachloride b. PCl3 Phosphorus trichloride c. SO2 Sulfur dioxide Name Formula a. Sulfur hexafluoride SF6 b. Sulfur trioxide SO3 c. Carbon dioxide CO2 2.8 Naming Simple Compounds 65 FIGURE 2.23 A flowchart for naming binary compounds.
FIGURE 2.24 Overall strategy for naming chemical compounds.
No Yes Type III: Use prefixes.
Does the metal form more than one cation?
No Yes Type I: Use the element name for the cation.
Type II: Determine the charge of the cation; use a Roman numeral after the element name for the cation.
Metal present?
Yes Binary compound?
No Yes No Yes Yes No Binary compound?
No Yes Yes No Polyatomic ion or ions present?
Use the strategy summarized in Figure 2.23.
This is a compound for which naming procedures have not yet been considered.
Name the compound using procedures similar to those for naming binary ionic compounds.
Naming Various Types of Compounds 1. Give the systematic name for each of the following compounds: a. P4O10 b. Nb2O5 c. Li2O2 d. Ti(NO3)4 2. Given the following systematic names, write the formula for each compound: a. Vanadium(V) fluoride b. Dioxygen difluoride c. Rubidium peroxide d. Gallium oxide Sample Exercise 2.9 66 Chapter Two Atoms, Molecules, and Ions Solution 1.
2.
See Exercises 2.63, 2.65, and 2.66.
Acids When dissolved in water, certain molecules produce a solution containing free H ions (protons). These substances, acids, will be discussed in detail in Chapters 4, 14, and 15.
Here we will simply present the rules for naming acids.
An acid can be viewed as a molecule with one or more H ions attached to an anion.
The rules for naming acids depend on whether the anion contains oxygen. If the anion does not contain oxygen, the acid is named with the prefix hydro- and the suffix -ic. For exam-ple, when gaseous HCl is dissolved in water, it forms hydrochloric acid. Similarly, HCN and H2S dissolved in water are called hydrocyanic and hydrosulfuric acids, respectively.
When the anion contains oxygen, the acidic name is formed from the root name of the anion with a suffix of -ic or -ous, depending on the name of the anion.
1. If the anion name ends in -ate, the suffix -ic is added to the root name. For example, H2SO4 contains the sulfate anion (SO4 2) and is called sulfuric acid; H3PO4 contains the phosphate anion (PO4 3) and is called phosphoric acid; and HC2H3O2 contains the acetate ion (C2H3O2 ) and is called acetic acid.
2. If the anion has an -ite ending, the -ite is replaced by -ous. For example, H2SO3, which contains sulfite (SO3 2), is named sulfurous acid; and HNO2, which contains nitrite (NO2 ), is named nitrous acid.
Acids can be recognized by the hydrogen that appears first in the formula.
Compound Name Comment a. P4O10 Tetraphosphorus Binary covalent compound (Type III), so decaoxide prefixes are used. The a in deca- is sometimes dropped.
b. Nb2O5 Niobium(V) oxide Type II binary compound containing Nb5 and O2 ions. Niobium is a transition metal and requires a Roman numeral.
c. Li2O2 Lithium peroxide Type I binary compound containing the Li and O2 2 (peroxide) ions.
d. Ti(NO3)4 Titanium(IV) nitrate Not a binary compound. Contains the Ti4 and NO3 ions. Titanium is a transition metal and requires a Roman numeral.
Name Chemical Formula Comment a. Vanadium(V) fluoride VF5 The compound contains V5 ions and requires five F ions for charge balance.
b. Dioxygen difluoride O2F2 The prefix di- indicates two of each atom.
c. Rubidium peroxide Rb2O2 Because rubidium is in Group 1A, it forms only 1 ions. Thus two Rb ions are needed to balance the 2 charge on the peroxide ion (O2 2).
d. Gallium oxide Ga2O3 Because gallium is in Group 3A, like aluminum, it forms only 3 ions.
Two Ga3 ions are required to balance the charge on three O2 ions.
For Review 67 The application of these rules can be seen in the names of the acids of the oxyanions of chlorine: Acid Anion Name HClO4 Perchlorate Perchloric acid HClO3 Chlorate Chloric acid HClO2 Chlorite Chlorous acid HClO Hypochlorite Hypochlorous acid The names of the most important acids are given in Tables 2.7 and 2.8. An overall strategy for naming acids is shown in Fig. 2.25.
TABLE 2.7 Names of Acids That Do Not Contain Oxygen Acid Name HF Hydrofluoric acid HCl Hydrochloric acid HBr Hydrobromic acid HI Hydroiodic acid HCN Hydrocyanic acid H2S Hydrosulfuric acid Note that these acids are aqueous solu-tions containing these substances.
TABLE 2.8 Names of Some Oxygen-Containing Acids Acid Name HNO3 Nitric acid HNO2 Nitrous acid H2SO4 Sulfuric acid H2SO3 Sulfurous acid H3PO4 Phosphoric acid HC2H3O2 Acetic Acid FIGURE 2.25 A flowchart for naming acids. An acid is best considered as one or more H ions attached to an anion.
-ate -ite hydro-+ anion root + -ic hydro(anion root)ic acid anion or element root + -ous (root)ous acid anion or element root + -ic (root)ic acid Does the anion contain oxygen?
Check the ending of the anion.
No Yes Yes No Key Terms Section 2.2 law of conservation of mass law of definite proportion law of multiple proportions Section 2.3 atomic masses atomic weights Avogadro’s hypothesis Section 2.4 cathode-ray tube electron radioactivity nuclear atom nucleus Section 2.5 proton neutron isotopes atomic number mass number For Review Fundamental laws Conservation of mass Definite proportion Multiple proportions Dalton’s atomic theory All elements are composed of atoms.
All atoms of a given element are identical.
Chemical compounds are formed when atoms combine.
Atoms are not changed in chemical reactions but the way they are bound together changes.
Early atomic experiments and models Thomson model Millikan experiment Rutherford experiment Nuclear model 68 Chapter Two Atoms, Molecules, and Ions Atomic structure Small dense nucleus contains protons and neutrons.
• Protons—positive charge • Neutrons—no charge Electrons reside outside the nucleus in the relatively large remaining atomic volume.
• Electrons—negative charge, small mass (11840 of proton) Isotopes have the same atomic number but different mass numbers.
Atoms combine to form molecules by sharing electrons to form covalent bonds.
Molecules are described by chemical formulas.
Chemical formulas show number and type of atoms.
• Structural formula • Ball-and-stick model • Space-filling model Formation of ions Cation—formed by loss of an electron, positive charge Anion—formed by gain of an electron, negative charge Ionic bonds—formed by interaction of cations and anions The periodic table organizes elements in order of increasing atomic number.
Elements with similar properties are in columns, or groups.
Metals are in the majority and tend to form cations.
Nonmetals tend to form anions.
Compounds are named using a system of rules depending on the type of compound.
Binary compounds • Type I—contain a metal that always forms the same cation • Type II—contain a metal that can form more than one cation • Type III—contain two nonmetals Compounds containing a polyatomic ion REVIEW QUESTIONS 1. Use Dalton’s atomic theory to account for each of the following.
a. the law of conservation of mass b. the law of definite proportion c. the law of multiple proportions 2. What evidence led to the conclusion that cathode rays had a negative charge?
3. What discoveries were made by J. J. Thomson, Henri Becquerel, and Lord Rutherford? How did Dalton’s model of the atom have to be modified to ac-count for these discoveries?
4. Consider Ernest Rutherford’s alpha-particle bombardment experiment illustrated in Figure 2.12. How did the results of this experiment lead Rutherford away from the plum pudding model of the atom to propose the nuclear model of the atom?
5. Do the proton and the neutron have exactly the same mass? How do the masses of the proton and neutron compare to the mass of the electron? Which particles make the greatest contribution to the mass of an atom? Which particles make the greatest contribution to the chemical properties of an atom?
6. What is the distinction between atomic number and mass number? Between mass number and atomic mass?
7. Distinguish between the terms family and period in connection with the peri-odic table. For which of these terms is the term group also used?
8. The compounds AlCl3, CrCl3, and ICl3 have similar formulas, yet each follows a different set of rules to name it. Name these compounds, and then compare and contrast the nomenclature rules used in each case.
Section 2.6 chemical bond covalent bond molecule chemical formula structural formula space-filling model ball-and-stick model ion cation anion ionic bond ionic solid (salt) polyatomic ion Section 2.7 periodic table metal nonmetal group (family) alkali metals alkaline earth metals halogens noble gases period Section 2.8 binary compounds binary ionic compounds oxyanions binary covalent compounds acid Questions 69 9. When metals react with nonmetals, an ionic compound generally results. What is the predicted general formula for the compound formed between an alkali metal and sulfur? Between an alkaline earth metal and nitrogen? Between alu-minum and a halogen?
10. How would you name HBrO4, KIO3, NaBrO2, and HIO? Refer to Table 2.5 and the acid nomenclature discussion in the text.
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Which of the following is true about an individual atom? Explain.
a. An individual atom should be considered to be a solid.
b. An individual atom should be considered to be a liquid.
c. An individual atom should be considered to be a gas.
d. The state of the atom depends on which element it is.
e. An individual atom cannot be considered to be a solid, liquid, or gas.
Justify your choice, and for choices you did not pick, explain what is wrong with them.
2. How would you go about finding the number of “chalk molecules” it takes to write your name on the board? Provide an explanation of all you would need to do and a sample calculation.
3. These questions concern the work of J. J. Thomson.
a. From Thomson’s work, which particles do you think he would feel are most important for the formation of compounds (chem-ical changes) and why?
b. Of the remaining two subatomic particles, which do you place second in importance for forming compounds and why?
c. Propose three models that explain Thomson’s findings and evaluate them. To be complete you should include Thomson’s findings.
4. Heat is applied to an ice cube in a closed container until only steam is present. Draw a representation of this process, assuming you can see it at an extremely high level of magnification. What happens to the size of the molecules? What happens to the total mass of the sample?
5. You have a chemical in a sealed glass container filled with air.
The setup is sitting on a balance as shown below. The chemical is ignited by means of a magnifying glass focusing sunlight on the reactant. After the chemical has completely burned, which of the following is true? Explain your answer.
a. The balance will read less than 250.0 g.
b. The balance will read 250.0 g.
c. The balance will read greater than 250.0 g.
d. Cannot be determined without knowing the identity of the chemical.
6. You take three compounds consisting of two elements and de-compose them. To determine the relative masses of X, Y, and Z, you collect and weigh the elements, obtaining the following data: a. What are the assumptions in solving this problem?
b. What are the relative masses of X, Y, and Z?
c. What are the chemical formulas of the three compounds?
d. If you decompose 21 g of compound XY, how much of each element is present?
7. The vitamin niacin (nicotinic acid, C6H5NO2) can be isolated from a variety of natural sources such as liver, yeast, milk, and whole grain. It also can be synthesized from commercially available ma-terials. Which source of nicotinic acid, from a nutritional view, is best for use in a multivitamin tablet? Why?
8. One of the best indications of a useful theory is that it raises more questions for further experimentation than it originally answered.
Does this apply to Dalton’s atomic theory? Give examples.
9. Dalton assumed that all atoms of the same element were identi-cal in all their properties. Explain why this assumption is not valid.
10. Evaluate each of the following as an acceptable name for water: a. dihydrogen oxide c. hydrogen hydroxide b. hydroxide hydride d. oxygen dihydride 11. Why do we call Ba(NO3)2 barium nitrate, but we call Fe(NO3)2 iron(II) nitrate?
12. Why is calcium dichloride not the correct systematic name for CaCl2?
13. The common name for NH3 is ammonia. What would be the sys-tematic name for NH3? Support your answer.
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 14. What refinements had to be made in Dalton’s atomic theory to ac-count for Gay-Lussac’s results on the combining volumes of gases?
Elements in Compound Masses of Elements X and Y X 0.4 g, Y 4.2 g Y and Z Y 1.4 g, Z 1.0 g X and Y X 2.0 g, Y 7.0 g 250.0g 70 Chapter Two Atoms, Molecules, and Ions 15. When hydrogen is burned in oxygen to form water, the composi-tion of water formed does not depend on the amount of oxygen reacted. Interpret this in terms of the law of definite proportion.
16. The two most reactive families of elements are the halogens and the alkali metals. How do they differ in their reactivities?
17. Explain the law of conservation of mass, the law of definite proportion, and the law of multiple proportions.
18. Section 2.3 describes the postulates of Dalton’s atomic theory.
With some modifications, these postulates hold up very well regarding how we view elements, compounds, and chemical reactions today. Answer the following questions concerning Dalton’s atomic theory and the modifications made today.
a. The atom can be broken down into smaller parts. What are the smaller parts?
b. How are atoms of hydrogen identical to each other and how can they be different from each other?
c. How are atoms of hydrogen different from atoms of helium?
How can H atoms be similar to He atoms?
d. How is water different from hydrogen peroxide (H2O2) even though both compounds are composed of only hydrogen and oxygen?
e. What happens in a chemical reaction and why is mass con-served in a chemical reaction?
19. The contributions of J. J. Thomson and Ernest Rutherford led the way to today’s understanding of the structure of the atom. What were their contributions?
20. What is the modern view of the structure of the atom?
21. The number of protons in an atom determines the identity of the atom.
What does the number and arrangement of the electrons in an atom determine? What does the number of neutrons in an atom determine?
22. Distinguish between the following terms.
a. molecule versus ion b. covalent bonding versus ionic bonding c. molecule versus compound d. anion versus cation 23. Which of the following statements are true? For the false state-ments, correct them.
a. Most of the known elements are metals.
b. Element 118 should be a nonmetal.
c. Hydrogen has mostly metallic properties.
d. A family of elements is also known as a period of elements.
e. When an alkaline earth metal, A, reacts with a halogen, X, the formula of the covalent compound formed should be A2X.
24. Each of the following compounds has three possible names listed for it. For each compound, what is the correct name and why aren’t the other names used?
a. N2O: nitrogen oxide, nitrogen(I) oxide, dinitrogen monoxide b. Cu2O: copper oxide, copper(I) oxide, dicopper monoxide c. Li2O: lithium oxide, lithium(I) oxide, dilithium monoxide Exercises In this section similar exercises are paired.
Development of the Atomic Theory 25. When mixtures of gaseous H2 and gaseous Cl2 react, a product forms that has the same properties regardless of the relative amounts of H2 and Cl2 used.
a. How is this result interpreted in terms of the law of definite proportion?
b. When a volume of H2 reacts with an equal volume of Cl2 at the same temperature and pressure, what volume of product having the formula HCl is formed?
26. A reaction of 1 liter of chlorine gas (Cl2) with 3 liters of fluorine gas (F2) yields 2 liters of a gaseous product. All gas volumes are at the same temperature and pressure. What is the formula of the gaseous product?
27. Hydrazine, ammonia, and hydrogen azide all contain only nitro-gen and hydrogen. The mass of hydrogen that combines with 1.00 g of nitrogen for each compound is 1.44 101 g, 2.16 101 g, and 2.40 102 g, respectively. Show how these data illustrate the law of multiple proportions.
28. Consider 100.0-g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon and the other has 42.9 g of carbon. How can these data support the law of multiple proportions if 42.9 is not a multiple of 27.2? Show that these data support the law of multiple pro-portions.
29. Early tables of atomic weights (masses) were generated by meas-uring the mass of a substance that reacts with 1.00 g of oxygen.
Given the following data and taking the atomic mass of hydrogen as 1.00, generate a table of relative atomic masses for oxygen, sodium, and magnesium.
How do your values compare with those in the periodic table?
How do you account for any differences?
30. Indium oxide contains 4.784 g of indium for every 1.000 g of oxy-gen. In 1869, when Mendeleev first presented his version of the periodic table, he proposed the formula In2O3 for indium oxide.
Before that time it was thought that the formula was InO. What values for the atomic mass of indium are obtained using these two formulas? Assume that oxygen has an atomic mass of 16.00.
The Nature of the Atom 31. From the information in this chapter on the mass of the proton, the mass of the electron, and the sizes of the nucleus and the atom, calculate the densities of a hydrogen nucleus and a hydrogen atom.
32. If you wanted to make an accurate scale model of the hydrogen atom and decided that the nucleus would have a diameter of 1 mm, what would be the diameter of the entire model?
33. In an experiment it was found that the total charge on an oil drop was 5.93 1018 C. How many negative charges does the drop contain?
34. A chemist in a galaxy far, far away performed the Millikan oil drop experiment and got the following results for the charges on Mass That Combines with Element 1.00 g Oxygen Assumed Formula Hydrogen 0.126 g HO Sodium 2.875 g NaO Magnesium 1.500 g MgO Exercises 71 various drops. Use these data to calculate the charge of the elec-tron in zirkombs.
2.56 1012 zirkombs 7.68 1012 zirkombs 3.84 1012 zirkombs 6.40 1013 zirkombs 35. What are the symbols of the following metals: sodium, radium, iron, gold, manganese, lead.
36. What are the symbols of the following nonmetals: fluorine, chlo-rine, bromine, sulfur, oxygen, phosphorus?
37. Give the names of the metals that correspond to the following symbols: Sn, Pt, Hg, Mg, K, Ag.
38. Give the names of the nonmetals that correspond to the follow-ing symbols: As, I, Xe, He, C, Si.
39. a. Classify the following elements as metals or nonmetals: Mg Si Rn Ti Ge Eu Au B Am Bi At Br b. The distinction between metals and nonmetals is really not a clear one. Some elements, called metalloids, are intermediate in their properties. Which of these elements would you reclassify as metalloids? What other elements in the periodic table would you expect to be metalloids?
40. a. List the noble gas elements. Which of the noble gases has only radioactive isotopes? (This situation is indicated on most peri-odic tables by parentheses around the mass of the element. See inside front cover.) b. Which lanthanide element and which transition element have only radioactive isotopes?
41. In the periodic table, how many elements are found in a. Group 2A?
c. the nickel group?
b. the oxygen family?
d. Group 8A?
42. In the periodic table, how many elements are found a. in the halogen group?
b. in the alkali family?
c. in the lanthanide series?
d. classified as transition metals?
43. How many protons and neutrons are in the nucleus of each of the following atoms? In a neutral atom of each element, how many electrons are present?
a.
79Br d. 133Cs b. 81Br e.
3H c.
239Pu f.
56Fe 44. What number of protons and neutrons are contained in the nucleus of each of the following atoms? Assuming each atom is uncharged, what number of electrons are present? a.
d.
b.
e.
c.
f.
45. Write the atomic symbol (A ZX) for each of the following isotopes.
a. Z 8, number of neutrons 9 b. the isotope of chlorine in which A 37 41 20Ca 57 26Fe 86 37Rb 13 6C 208 82Pb 235 92U c. Z 27, A 60 d. number of protons 26, number of neutrons 31 e. the isotope of I with a mass number of 131 f. Z 3, number of neutrons 4 46. Write the atomic symbol (A ZX) for each of the isotopes described below.
a. number of protons 27, number of neutrons 31 b. the isotope of boron with mass number 10 c. Z 12, A 23 d. atomic number 53, number of neutrons 79 e. Z 9, number of neutrons 10 f. number of protons 29, mass number 65 47. What is the symbol for an ion with 63 protons, 60 electrons, and 88 neutrons? If an ion contains 50 protons, 68 neutrons, and 48 electrons, what is its symbol?
48. What is the symbol of an ion with 16 protons, 18 neutrons, and 18 electrons? What is the symbol for an ion that has 16 protons, 16 neutrons, and 18 electrons?
49. Complete the following table: 50.
51. For each of the following sets of elements, label each as either noble gases, halogens, alkali metals, alkaline earth metals, or tran-sition metals.
a. Ti, Fe, Ag d. Ne, Kr, Xe b. Mg, Sr, Ba e. F, Br, I c. Li, K, Rb 52. Consider the elements of Group 4A (the “carbon family”): C, Si, Ge, Sn, and Pb. What is the trend in metallic character as one goes down this group? What is the trend in metallic character going from left to right across a period in the periodic table?
Number of Number of Protons in Neutrons in Number of Symbol Nucleus Nucleus Electrons Net Charge 53 26Fe2 26 33 3 85 125 86 13 14 10 76 54 2 Number of Number of Protons in Neutrons in Number of Net Symbol Nucleus Nucleus Electrons Charge 238 92U 20 20 2 23 28 20 89 39Y 35 44 36 15 16 3 72 Chapter Two Atoms, Molecules, and Ions 53. Would you expect each of the following atoms to gain or lose electrons when forming ions? What ion is the most likely in each case?
a. Ra c. P e. Br b. In d. Te f. Rb 54. For each of the following atomic numbers, use the periodic table to write the formula (including the charge) for the simple ion that the element is most likely to form in ionic compounds.
a. 13 c. 56 e. 87 b. 34 d. 7 f. 35 Nomenclature 55. Name the compounds in parts a–d and write the formulas for the compounds in parts e–h.
a. NaBr e. strontium fluoride b. Rb2O f. aluminum selenide c. CaS g. potassium nitride d. AlI3 h. magnesium phosphide 56. Name the compounds in parts a–d and write the formulas for the compounds in parts e–h.
a. Hg2O e. tin(II) nitride b. FeBr3 f. cobalt(III) iodide c. CoS g. mercury(II) oxide d. TiCl4 h. chromium(VI) sulfide 57. Name each of the following compounds: a. CsF c. Ag2S e. TiO2 b. Li3N d. MnO2 f. Sr3P2 58. Write the formula for each of the following compounds: a. zinc chloride d. aluminum sulfide b. tin(IV) fluoride e. mercury(I) selenide c. calcium nitride f. silver iodide 59. Name each of the following compounds: a. BaSO3 c. KMnO4 b. NaNO2 d. K2Cr2O7 60. Write the formula for each of the following compounds: a. chromium(III) hydroxide c. lead(IV) carbonate b. magnesium cyanide d. ammonium acetate 61. Name each of the following compounds: a.
b.
c. SO2 d. P2S5 62. Write the formula for each of the following compounds: a. diboron trioxide c. dinitrogen monoxide b. arsenic pentafluoride d. sulfur hexachloride 63. Name each of the following compounds: a. CuI c. CoI2 b. CuI2 d. Na2CO3 I Cl O N e. NaHCO3 h. NaOCl f. S4N4 i. BaCrO4 g. SF6 j. NH4NO3 64. Name each of the following compounds: a. HC2H3O2 g. H2SO4 b. NH4NO2 h. Sr3N2 c. Co2S3 i. Al2(SO3)3 d. ICl j. SnO2 e. Pb3(PO4)2 k. Na2CrO4 f. KIO3 l. HClO 65. Write the formula for each of the following compounds: a. sulfur difluoride b. sulfur hexafluoride c. sodium dihydrogen phosphate d. lithium nitride e. chromium(III) carbonate f. tin(II) fluoride g. ammonium acetate h. ammonium hydrogen sulfate i. cobalt(III) nitrate j. mercury(I) chloride k. potassium chlorate l. sodium hydride 66. Write the formula for each of the following compounds: a. chromium(VI) oxide b. disulfur dichloride c. nickel(II) fluoride d. potassium hydrogen phosphate e. aluminum nitride f. ammonia g. manganese(IV) sulfide h. sodium dichromate i. ammonium sulfite j. carbon tetraiodide 67. Write the formula for each of the following compounds: a. sodium oxide h. copper(I) chloride b. sodium peroxide i.
gallium arsenide c. potassium cyanide j.
cadmium selenide d. copper(II) nitrate k. zinc sulfide e. selenium tetrabromide l.
nitrous acid f. iodous acid m. diphosphorus pentoxide g. lead(IV) sulfide 68. Write the formula for each of the following compounds: a. ammonium hydrogen phosphate b. mercury(I) sulfide c. silicon dioxide d. sodium sulfite e. aluminum hydrogen sulfate f. nitrogen trichloride g. hydrobromic acid h. bromous acid i. perbromic acid j. potassium hydrogen sulfide k. calcium iodide l. cesium perchlorate 69. Name the following acids illustrated below.
70. Each of the following compounds is incorrectly named. What is wrong with each name, and what is the correct name for each compound?
a. FeCl3, iron chloride b. NO2, nitrogen(IV) oxide c. CaO, calcium(II) monoxide d. Al2S3, dialuminum trisulfide e. Mg(C2H3O2)2, manganese diacetate f. FePO4, iron(II) phosphide g. P2S5, phosphorous sulfide h. Na2O2, sodium oxide i. HNO3, nitrate acid j. H2S, sulfuric acid Additional Exercises 71. Chlorine has two natural isotopes: 37 17Cl and 35 17Cl. Hydrogen reacts with chlorine to form the compound HCl. Would a given amount of hydrogen react with different masses of the two chlorine iso-topes? Does this conflict with the law of definite proportion? Why or why not?
72. Which of the following statements is(are) true? For the false state-ments, correct them.
a. All particles in the nucleus of an atom are charged.
b. The atom is best described as a uniform sphere of matter in which electrons are embedded.
c. The mass of the nucleus is only a very small fraction of the mass of the entire atom.
d. The volume of the nucleus is only a very small fraction of the total volume of the atom.
e. The number of neutrons in a neutral atom must equal the num-ber of electrons.
73. The isotope of an unknown element, X, has a mass number of 79.
The most stable ion of the isotope has 36 electrons and forms a binary compound with sodium having a formula of Na2X. Which of the following statements is(are) true? For the false statements, correct them.
a. The binary compound formed between X and fluorine will be a covalent compound.
b. The isotope of X contains 38 protons.
c. The isotope of X contains 41 neutrons.
d. The identity of X is strontium, Sr.
74. For each of the following ions, indicate the total number of protons and electrons in the ion. For the positive ions in the list, predict a.
b.
c.
d.
e.
Cl O N H P S C the formula of the simplest compound formed between each positive ion and the oxide ion. For the negative ions in the list, predict the formula of the simplest compound formed between each negative ion and the aluminum ion.
a. Fe2 e. S2 b. Fe3 f. P3 c. Ba2 g. Br d. Cs h. N3 75. The formulas and common names for several substances are given below. Give the systematic names for these substances.
a. sugar of lead Pb(C2H3O2)2 b. blue vitrol CuSO4 c. quicklime CaO d. Epsom salts MgSO4 e. milk of magnesia Mg(OH)2 f. gypsum CaSO4 g. laughing gas N2O 76. Identify each of the following elements: a. a member of the same family as oxygen whose most stable ion contains 54 electrons b. a member of the alkali metal family whose most stable ion contains 36 electrons c. a noble gas with 18 protons in the nucleus d. a halogen with 85 protons and 85 electrons 77. An element’s most stable ion forms an ionic compound with bromine, having the formula XBr2. If the ion of element X has a mass number of 230 and has 86 electrons, what is the identity of the element, and how many neutrons does it have?
78. A certain element has only two naturally occurring isotopes: one with 18 neutrons and the other with 20 neutrons. The element forms 1 charged ions when in ionic compounds. Predict the iden-tity of the element. What number of electrons does the 1 charged ion have?
79. The designations 1A through 8A used for certain families of the periodic table are helpful for predicting the charges on ions in bi-nary ionic compounds. In these compounds, the metals generally take on a positive charge equal to the family number, while the nonmetals take on a negative charge equal to the family number minus eight. Thus the compound between sodium and chlorine contains Na ions and Cl ions and has the formula NaCl. Pre-dict the formula and the name of the binary compound formed from the following pairs of elements.
a. Ca and N e. Ba and I b. K and O f. Al and Se c. Rb and F g. Cs and P d. Mg and S h. In and Br 80. By analogy with phosphorous compounds, name the following: Na3AsO4, H3AsO4, Mg3(SbO4)2.
81. A sample of H2SO4 contains 2.02 g of hydrogen, 32.07 g of sul-fur, and 64.00 g of oxygen. How many grams of sulfur and grams of oxygen are present in a second sample of H2SO4 containing 7.27 g of hydrogen?
82. In a reaction, 34.0 g of chromium(III) oxide reacts with 12.1 g of aluminum to produce chromium and aluminum oxide. If 23.3 g of chromium is produced, what mass of aluminum oxide is produced?
Additional Exercises 73 74 Chapter Two Atoms, Molecules, and Ions Challenge Problems 83. The elements in one of the groups in the periodic table are often called the coinage metals. Identify the elements in this group based on your own experience.
84. Reaction of 2.0 L of hydrogen gas with 1.0 L of oxygen gas yields 2.0 L of water vapor. All gases are at the same temperature and pressure. Show how these data support the idea that oxygen gas is a diatomic molecule. Must we consider hydrogen to be a di-atomic molecule to explain these results?
85. A combustion reaction involves the reaction of a substance with oxygen gas. The complete combustion of any hydrocarbon (binary compound of carbon and hydrogen) produces carbon dioxide and water as the only products. Octane is a hydrocarbon that is found in gasoline. Complete combustion of octane produces 8 liters of carbon dioxide for every 9 liters of water vapor (both measured at the same temperature and pressure). What is the ratio of car-bon atoms to hydrogen atoms in a molecule of octane?
86. A chemistry instructor makes the following claim: “Consider that if the nucleus were the size of a grape, the electrons would be about 1 mile away on average.” Is this claim reasonably accurate?
Provide mathematical support.
87. Two elements, R and Q, combine to form two binary compounds.
In the first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound, 7.00 g of R combines with 4.50 g of Q.
Show that these data are in accord with the law of multiple pro-portions. If the formula of the second compound is RQ, what is the formula of the first compound?
88. The early alchemists used to do an experiment in which water was boiled for several days in a sealed glass container. Eventually, some solid residue would appear in the bottom of the flask, which was interpreted to mean that some of the water in the flask had been converted into “earth.” When Lavoisier repeated this exper-iment, he found that the water weighed the same before and af-ter heating and the mass of the flask plus the solid residue equaled the original mass of the flask. Were the alchemists correct? Ex-plain what really happened. (This experiment is described in the article by A. F. Scott in Scientific American, January 1984.) 89. Each of the following statements is true, but Dalton might have had trouble explaining some of them with his atomic theory. Give explanations for the following statements.
a. The space-filling models for ethyl alcohol and dimethyl ether are shown below.
These two compounds have the same composition by mass (52% carbon, 13% hydrogen, and 35% oxygen), yet the two have different melting points, boiling points, and solubilities in water.
b. Burning wood leaves an ash that is only a small fraction of the mass of the original wood.
c. Atoms can be broken down into smaller particles.
C O H d. One sample of lithium hydride is 87.4% lithium by mass, while another sample of lithium hydride is 74.9% lithium by mass.
However, the two samples have the same properties.
90. You have two distinct gaseous compounds made from element X and element Y. The mass percents are as follows: Compound I: 30.43% X, 69.57% Y Compound II: 63.64% X, 36.36% Y In their natural standard states, element X and element Y exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to de-termine.) When you react “gas X” with “gas Y” to make the prod-ucts, you get the following data (all at standard pressure and temperature): 1 volume “gas X” 2 volumes “gas Y” ¡ 2 volumes compound I 2 volumes “gas X” 1 volume “gas Y” ¡ 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element X and element Y.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
91. What is the systematic name of Ta2O5? If the charge on the metal remained constant and then sulfur was substituted for oxygen, how would the formula change? What is the difference in the total num-ber of protons between Ta2O5 and its sulfur analog?
92. A binary ionic compound is known to contain a cation with 51 pro-tons and 48 electrons. The anion contains one-third the number of protons as the cation. The number of electrons in the anion is equal to the number of protons plus 1. What is the formula of this compound? What is the name of this compound?
93. Using the information in Table 2.1, answer the following ques-tions. In an ion with an unknown charge, the total mass of all the electrons was determined to be 2.55 1026 g, while the total mass of its protons was 5.34 1023 g. What is the identity and charge of this ion? What is the symbol and mass number of a neu-tral atom whose total mass of its electrons is 3.92 1026 g, while its neutrons have a mass of 9.35 1023 g?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
94. You have gone back in time and are working with Dalton on a table of relative masses. Following are his data.
0.602 g gas A reacts with 0.295 g gas B 0.172 g gas B reacts with 0.401 g gas C 0.320 g gas A reacts with 0.374 g gas C a. Assuming simplest formulas (AB, BC, and AC), construct a table of relative masses for Dalton.
Marathon Problem 75 Write the simplest balanced equations, and find the actual relative masses of the elements. Explain your reasoning.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
b. Knowing some history of chemistry, you tell Dalton that if he determines the volumes of the gases reacted at constant tem-perature and pressure, he need not assume simplest formulas.
You collect the following data: 6 volumes gas A 1 volume gas B S 4 volumes product 1 volume gas B 4 volumes gas C S 4 volumes product 3 volumes gas A 2 volumes gas C S 6 volumes product 3 Stoichiometry Chemical reactions have a profound effect on our lives. There are many examples: Food is converted to energy in the human body; nitrogen and hydrogen are combined to form ammonia, which is used as a fertilizer; fuels and plastics are produced from petro-leum; the starch in plants is synthesized from carbon dioxide and water using energy from sunlight; human insulin is produced in laboratories by bacteria; cancer is induced in hu-mans by substances from our environment; and so on, in a seemingly endless list. The central activity of chemistry is to understand chemical changes such as these, and the study of reactions occupies a central place in this book. We will examine why reactions occur, how fast they occur, and the specific pathways they follow.
In this chapter we will consider the quantities of materials consumed and produced in chemical reactions. This area of study is called chemical stoichiometry (pronounced stoy ke – om etry). To understand chemical stoichiometry, you must first understand the concept of relative atomic masses.
3.1 Counting by Weighing Suppose you work in a candy store that sells gourmet jelly beans by the bean. People come in and ask for 50 beans, 100 beans, 1000 beans, and so on, and you have to count them out—a tedious process at best. As a good problem solver, you try to come up with a better system. It occurs to you that it might be far more efficient to buy a scale and count the jelly beans by weighing them. How can you count jelly beans by weighing them?
What information about the individual beans do you need to know?
Assume that all of the jelly beans are identical and that each has a mass of 5 g. If a customer asks for 1000 jelly beans, what mass of jelly beans would be required? Each bean has a mass of 5 g, so you would need 1000 beans 5 g/bean, or 5000 g (5 kg). It takes just a few seconds to weigh out 5 kg of jelly beans. It would take much longer to count out 1000 of them.
In reality, jelly beans are not identical. For example, let’s assume that you weigh 10 beans individually and get the following results: Bean Mass 1 5.1 g 2 5.2 g 3 5.0 g 4 4.8 g 5 4.9 g 6 5.0 g 7 5.0 g 8 5.1 g 9 4.9 g 10 5.0 g 77 Jelly beans can be counted by weighing.
78 Chapter Three Stoichiometry The average mass of a jelly bean is 5.0 g. Thus, to count out 1000 beans, we need to weigh out 5000 g of beans. This sample of beans, in which the beans have an average mass of 5.0 g, can be treated exactly like a sample where all of the beans ae identical.
Objects do not need to have identical masses to be counted by weighing. We simply need to know the average mass of the objects. For purposes of counting, the objects behave as though they were all identical, as though they each actually had the average mass.
We count atoms in exactly the same way. Because atoms are so small, we deal with samples of matter that contain huge numbers of atoms. Even if we could see the atoms it would not be possible to count them directly. Thus we determine the number of atoms in a given sample by finding its mass. However, just as with jelly beans, to relate the mass to a number of atoms, we must know the average mass of the atoms.
3.2 Atomic Masses As we saw in Chapter 2, the first quantitative information about atomic masses came from the work of Dalton, Gay-Lussac, Lavoisier, Avogadro, and Berzelius. By observing the proportions in which elements combine to form various compounds, nineteenth-century chemists calculated relative atomic masses. The modern system of atomic masses, insti-tuted in 1961, is based on 12C (“carbon twelve”) as the standard. In this system, 12C is as-signed a mass of exactly 12 atomic mass units (amu), and the masses of all other atoms are given relative to this standard.
The most accurate method currently available for comparing the masses of atoms in-volves the use of the mass spectrometer. In this instrument, diagramed in Fig. 3.1, atoms or molecules are passed into a beam of high-speed electrons, which knock electrons off the atoms or molecules being analyzed and change them into positive ions. An applied Can we count these nonidentical beans by weighing? Yes. The key piece of information we need is the average mass of the jelly beans. Let’s compute the average mass for our 10-bean sample.
50.0 10 5.0 g 5.1 g 5.2 g 5.0 g 4.8 g 4.9 g 5.0 g 5.0 g 5.1 g 4.9 g 5.0 g 10 Average mass total mass of beans number of beans Sample Positive ions Electron beam Heating device to vaporize sample Ion-accelerating electric field Accelerated ion beam Magnetic field Detector plate Least massive ions Most massive ions Slits FIGURE 3.1 (left) A scientist injecting a sample into a mass spectrometer. (above) Schematic diagram of a mass spectrometer.
3.2 Atomic Masses 79 electric field then accelerates these ions into a magnetic field. Because an accelerating ion produces its own magnetic field, an interaction with the applied magnetic field occurs, which tends to change the path of the ion. The amount of path deflection for each ion de-pends on its mass—the most massive ions are deflected the smallest amount—which causes the ions to separate, as shown in Fig. 3.1. A comparison of the positions where the ions hit the detector plate gives very accurate values of their relative masses. For example, when 12C and 13C are analyzed in a mass spectrometer, the ratio of their masses is found to be Since the atomic mass unit is defined such that the mass of 12C is exactly 12 atomic mass units, then on this same scale, h Exact number by definition The masses of other atoms can be determined in a similar fashion.
The mass for each element is given in the table inside the front cover of this text.
This value, even though it is actually a mass, is (for historical reasons) sometimes called the atomic weight for each element.
Look at the value of the atomic mass of carbon given in this table. You might ex-pect to see 12, since we said the system of atomic masses is based on 12C. However, the number given for carbon is not 12 but 12.01. Why? The reason for this apparent discrepancy is that the carbon found on earth (natural carbon) is a mixture of the iso-topes 12C, 13C, and 14C. All three isotopes have six protons, but they have six, seven, and eight neutrons, respectively. Because natural carbon is a mixture of isotopes, the atomic mass we use for carbon is an average value reflecting the average of the iso-topes composing it.
The average atomic mass for carbon is computed as follows: It is known that nat-ural carbon is composed of 98.89% 12C atoms and 1.11% 13C atoms. The amount of 14C is negligibly small at this level of precision. Using the masses of 12C (exactly 12 amu) and 13C (13.003355 amu), we can calculate the average atomic mass for natural carbon as follows: In this text we will call the average mass for an element the average atomic mass or, simply, the atomic mass for that element.
Even though natural carbon does not contain a single atom with mass 12.01, for stoi-chiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon.
Recall from Section 3.1 that counting by weighing works if you know the average mass of the units being counted. Counting by weighing works just the same for atoms as for jelly beans. For natural carbon with an average mass of 12.01 atomic mass units, to obtain 1000 atoms would require weighing out 12,010 atomic mass units of natural carbon (a mixture of 12C and 13C).
As in the case of carbon, the mass for each element listed in the table inside the front cover of the text is an average value based on the isotopic composition of the naturally occurring element. For instance, the mass listed for hydrogen (1.008) is the average mass for natural hydrogen, which is a mixture of 1H and 2H (deuterium). No atom of hydrogen actually has the mass 1.008.
10.98892112 amu2 10.01112113.0034 amu2 12.01 amu 98.89% of 12 amu 1.11% of 13.0034 amu Mass of 13C 11.08361292112 amu2 13.003355 amu Mass 13C Mass 12C 1.0836129 Most elements occur in nature as mixtures of isotopes; thus atomic masses are usually average values.
It is much easier to weigh out 600 hex nuts than count them one by one.
80 Chapter Three Stoichiometry CHEMICAL IMPACT Buckyballs Teach Some History A bout 250 million years ago, 90% of life on earth was destroyed in some sort of cataclysmic event. This event, which ended the Permian period and began the Triassic (the P-T boundary), is the most devastating mass extinction in the earth’s history—far surpassing the catas-trophe 65 million years ago that wiped out the dinosaurs (the K-T boundary). In 1979 geologist Walter Alvarez and his Nobel Prize–winning physicist father Luis Alvarez sug-gested that unusually high concentrations of iridium in rocks laid down at the K-T boundary meant that an aster-oid had hit the earth, causing tremendous devastation. In the last 20 years much evidence has accumulated to sup-port this hypothesis, including identification of the loca-tion of the probable crater caused by the impact in the ocean near Mexico.
Were the P-T boundary extinctions also caused by an extraterrestrial object or by some event on earth, such as a massive volcano explosion? Recent discoveries by geo-chemists Luann Becker of the University of Washington and Robert J. Poreda of the University of Rochester seem to strongly support the impact theory. Examining sediment from China and Japan, the team found fullerenes encapsu-lating argon and helium gas atoms whose isotopic compo-sition indicates that they are extraterrestrial in origin. For example, the ratio of He to He found in the fullerenes is 100 times greater than the ratio for helium found in the earth’s atmosphere. Likewise, the isotopic composition of the fullerene-trapped argon atoms is quite different from that found on earth.
Fullerenes include spherical C60 carbon molecules (“buckyballs”) whose cavities can trap other atoms such as helium and argon. (See the accompanying figure.) The sci-entists postulate that the fullerenes originated in stars or collapsing gas clouds where the noble gas atoms were trapped as the fullerenes formed. These fullerenes were then somehow incorporated into the object that eventually hit the earth. Based on the isotopic compositions, the geochemists estimate that the impacting body must have 4 2 3 2 Ion beam intensity at detector (b) Relative number of atoms 0 Mass number 20 21 22 20 40 60 80 100 91 .3 9 (c) 18 Mass number 19 20 21 22 23 24 FIGURE 3.2 (a) Neon gas glowing in a discharge tube. The relative intensities of the signals recorded when natural neon is injected into a mass spectrometer, represented in terms of (b) “peaks” and (c) a bar graph. The relative areas of the peaks are 0.9092 (20Ne), 0.00257 (21Ne), and 0.0882 (22Ne); natural neon is therefore 90.92% 20Ne, 0.257% 21Ne, and 8.82% 22Ne.
In addition to being useful for determining accurate mass values for individual atoms, the mass spectrometer is used to determine the isotopic composition of a natural element.
For example, when a sample of natural neon is injected into a mass spectrometer, the mass spectrum shown in Fig. 3.2 is obtained. The areas of the “peaks” or the heights of the bars indicate the relative abundances of 20 10Ne, 21 10Ne, and 22 10Ne atoms.
(a) 3.2 Atomic Masses 81 Isotope ratios of the noble gas atoms inside celestial buckyballs indicate that these ancient carbon cages formed in a stellar environment, not on earth.
been 10 kilometers in diameter, which is comparable in size to the asteroid that is assumed to have killed the dinosaurs.
One factor that had previously cast doubt on an as-teroid collision as the cause of the P-T catastrophe was the lack of iridium found in sediments from this period.
However, Becker and other scientists argue that this ab-sence probably means the impacting object may have been a comet rather than an asteroid. It is also possible that such a blow could have intensified the volcanism already under way on earth at that time, delivering a “one-two punch” that almost obliterated life on earth, according to Becker.
It is ironic that “buckyballs,” which made big news when they were recently synthesized for the first time in the laboratory, actually have been around for millions of years and have some very interesting history to teach us.
Figure from Chemical and Engineering News, Feb. 26, 2001, p. 9.
Reprinted by permission of Joseph Wilmhoff.
The Average Mass of an Element When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in Fig. 3.3 are obtained. Use these data to compute the average mass of natural copper. (The mass values for 63Cu and 65Cu are 62.93 amu and 64.93 amu, respectively.) Solution As shown by the graph, of every 100 atoms of natural copper, 69.09 are 63Cu and 30.91 are 65Cu. Thus the mass of 100 atoms of natural copper is The average mass of a copper atom is This mass value is used in doing calculations involving the reactions of copper and is the value given in the table inside the front cover of this book.
Reality Check: When you finish a calculation, you should always check whether your answer makes sense. In this case our answer of 63.55 amu is between the masses of the atoms that make up natural copper. This makes sense. The answer could not be smaller than 62.93 amu or larger than 64.93 amu.
See Exercises 3.27 and 3.28.
6355 amu 100 atoms 63.55 amu/atom 169.09 atoms2a62.93 amu atomb 130.91 atoms2a64.93 amu atomb 6355 amu Sample Exercise 3.1 Copper nugget.
Relative number of atoms 0 Mass number 63 65 20 40 60 80 100 69.09 30.91 FIGURE 3.3 Mass spectrum of natural copper.
82 Chapter Three Stoichiometry 3.3 The Mole Because samples of matter typically contain so many atoms, a unit of measure called the mole has been established for use in counting atoms. For our purposes, it is most conve-nient to define the mole (abbreviated mol) as the number equal to the number of carbon atoms in exactly 12 grams of pure 12C. Techniques such as mass spectrometry, which count atoms very precisely, have been used to determine this number as 6.02214 1023 (6.022 1023 will be sufficient for our purposes). This number is called Avogadro’s number to honor his contributions to chemistry. One mole of something consists of 6.022 1023 units of that substance. Just as a dozen eggs is 12 eggs, a mole of eggs is 6.022 1023 eggs.
The magnitude of the number 6.022 1023 is very difficult to imagine. To give you some idea, 1 mole of seconds represents a span of time 4 million times as long as the earth has already existed, and 1 mole of marbles is enough to cover the entire earth to a depth of 50 miles! However, since atoms are so tiny, a mole of atoms or molecules is a perfectly manageable quantity to use in a reaction (see Fig. 3.4).
How do we use the mole in chemical calculations? Recall that Avogadro’s number is defined as the number of atoms in exactly 12 grams of 12C. This means that 12 grams of 12C contains 6.022 1023 atoms. It also means that a 12.01-gram sample of natural carbon contains 6.022 1023 atoms (a mixture of 12C, 13C, and 14C atoms, with an average atomic mass of 12.01). Since the ratio of the masses of the samples (12 g12.01 g) is the same as the ratio of the masses of the individual components (12 amu12.01 amu), the two sam-ples contain the same number of atoms (6.022 1023).
To be sure this point is clear, think of oranges with an average mass of 0.5 pound each and grapefruit with an average mass of 1.0 pound each. Any two sacks for which the sack of grapefruit weighs twice as much as the sack of oranges will contain the same number of pieces of fruit. The same idea extends to atoms. Compare natural carbon (av-erage mass of 12.01) and natural helium (average mass of 4.003). A sample of 12.01 grams of natural carbon contains the same number of atoms as 4.003 grams of natural helium.
Both samples contain 1 mole of atoms (6.022 1023). Table 3.1 gives more examples that illustrate this basic idea.
Thus the mole is defined such that a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms. This definition FIGURE 3.4 Proceeding clockwise from the top, samples containing one mole each of copper, aluminum, iron, sulfur, iodine, and (in the center) mercury.
The mass of 1 mole of an element is equal to its atomic mass in grams.
The SI definition of the mole is the amount of a substance that contains as many entities as there are in exactly 12 g of carbon-12.
Avogadro’s number is 6.022 1023. One mole of anything is 6.022 1023 units of that substance.
3.3 The Mole 83 also fixes the relationship between the atomic mass unit and the gram. Since 6.022 1023 atoms of carbon (each with a mass of 12 amu) have a mass of 12 g, then and h Exact number This relationship can be used to derive the unit factor needed to convert between atomic mass units and grams.
Determining the Mass of a Sample of Atoms Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms.
Solution From the table inside the front cover of the text, we note that one americium atom has a mass of 243 amu. Thus the mass of six atoms is Using the relationship we write the conversion factor for converting atomic mass units to grams: The mass of six americium atoms in grams is Reality Check: Since this sample contains only six atoms, the mass should be very small as the amount 2.42 1021 g indicates.
See Exercise 3.33.
1.46 103 amu 1 g 6.022 1023 amu 2.42 1021 g 1 g 6.022 1023 amu 6.022 1023 amu 1 g 6 atoms 243 amu atom 1.46 103 amu 6.022 1023 amu 1 g 16.022 1023 atoms2a12 amu atom b 12 g TABLE 3.1 Comparison of 1 Mole Samples of Various Elements Element Number of Atoms Present Mass of Sample (g) Aluminum 6.022 1023 26.98 Copper 6.022 1023 63.55 Iron 6.022 1023 55.85 Sulfur 6.022 1023 32.07 Iodine 6.022 1023 126.9 Mercury 6.022 1023 200.6 Sample Exercise 3.2 84 Chapter Three Stoichiometry CHEMICAL IMPACT Elemental Analysis Catches Elephant Poachers I n an effort to combat the poaching of elephants by con-trolling illegal exports of ivory, scientists are now using the isotopic composition of ivory trinkets and elephant tusks to identify the region of Africa where the elephant lived. Using a mass spectrometer, scientists analyze the ivory for the relative amounts of 12C, 13C, 14N, 15N, 86Sr, and 87Sr to determine the diet of the elephant and thus its place of origin. For example, because grasses use a differ-ent photosynthetic pathway to produce glucose than do trees, grasses have a slightly different 13C12C ratio from that of trees. They have different ratios because each time a carbon atom is added in going from simpler to more com-plex compounds, the more massive 13C is disfavored rela-tive to 12C because it reacts more slowly. Because trees use more steps to build up glucose, they end up with a smaller 13C12C ratio in their leaves relative to grasses, and this difference is then reflected in the tissues of elephants. Thus scientists can tell whether a particular tusk came from a savanna-dwelling elephant (grass-eating) or from a tree-browsing elephant.
Similarly, because the ratios of 15N14N and 87Sr86Sr in elephant tusks also vary depending on the region of Africa the elephant inhabits, they can be used to trace the elephant’s origin. In fact, using these techniques, scientists have re-ported being able to discriminate between elephants living only about 100 miles apart.
There is now international concern about the dwindling elephant populations in Africa—their numbers have de-creased significantly in recent years. This concern has led to bans in the export of ivory from many countries in Africa.
However, a few nations still allow ivory to be exported. Thus, to enforce the trade restrictions, the origin of a given piece of ivory must be established. It is hoped that the “isotope signature” of the ivory can be used for this purpose.
(left) Pure aluminum. (right) Aluminum al-loys are used for many high-quality bicycle components, such as this chain wheel.
To do chemical calculations, you must understand what the mole means and how to determine the number of moles in a given mass of a substance. These procedures are il-lustrated in Sample Exercises 3.3 and 3.4.
Determining Moles of Atoms Aluminum (Al) is a metal with a high strength-to-mass ratio and a high resistance to cor-rosion; thus it is often used for structural purposes. Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.
Solution The mass of 1 mole (6.022 1023 atoms) of aluminum is 26.98 g. The sample we are considering has a mass of 10.0 g. Since the mass is less than 26.98 g, this sample contains less than 1 mole of aluminum atoms. We can calculate the number of moles of aluminum atoms in 10.0 g as follows: 10.0 g Al 1 mol Al 26.98 g Al 0.371 mol Al atoms Sample Exercise 3.3 3.3 The Mole 85 The number of atoms in 10.0 g (0.371 mol) of aluminum is Reality Check: One mole of Al has a mass of 26.98 g and contains 6.022 1023 atoms. Our sample is 10.0 g, which is roughly 13 of 26.98. Thus the calculated amount should be on the order of 13 of 6 1023, which it is.
See Exercise 3.34.
Calculating Numbers of Atoms A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg.
How many silicon (Si) atoms are present in the chip?
Solution The strategy for doing this problem is to convert from milligrams of silicon to grams of silicon, then to moles of silicon, and finally to atoms of silicon: Reality Check: Note that 5.68 mg of silicon is clearly much less than 1 mol of silicon (which has a mass of 28.09 g), so the final answer of 1.22 1020 atoms (compared with 6.022 1023 atoms) is in the right direction.
See Exercise 3.35.
Calculating the Number of Moles and Mass Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Cal-culate both the number of moles in a sample of cobalt containing 5.00 1020 atoms and the mass of the sample.
Solution Note that the sample of 5.00 1020 atoms of cobalt is less than 1 mole (6.022 1023 atoms) of cobalt. What fraction of a mole it represents can be determined as follows: Since the mass of 1 mole of cobalt atoms is 58.93 g, the mass of 5.00 1020 atoms can be determined as follows: Reality Check: In this case the sample contains 5 1020 atoms, which is approximately 11000 of a mole. Thus the sample should have a mass of about (11000)(58.93) 0.06.
Our answer of 0.05 makes sense.
See Exercise 3.36.
8.30 104 mol Co 58.93 g Co 1 mol Co 4.89 102 g Co 5.00 1020 atoms Co 1 mol Co 6.022 1023 atoms Co 8.30 104 mol Co 2.02 104 mol Si 6.022 1023 atoms 1 mol Si 1.22 1020 atoms 5.68 103 g Si 1 mol Si 28.09 g Si 2.02 104 mol Si 5.68 mg Si 1 g Si 1000 mg Si 5.68 103 g Si 0.371 mol Al 6.022 1023 atoms 1 mol Al 2.23 1023 atoms Sample Exercise 3.4 Always check to see if your answer is sensible.
Paying careful attention to units and making sure the answer is reasonable can help you detect an inverted conversion factor or a number that was incorrectly entered in your calculator.
Sample Exercise 3.5 Fragments of cobalt metal.
86 Chapter Three Stoichiometry 3.4 Molar Mass A chemical compound is, ultimately, a collection of atoms. For example, methane (the major component of natural gas) consists of molecules that each contain one carbon and four hydrogen atoms (CH4). How can we calculate the mass of 1 mole of methane; that is, what is the mass of 6.022 1023 CH4 molecules? Since each CH4 molecule contains one carbon atom and four hydrogen atoms, 1 mole of CH4 molecules contains 1 mole of carbon atoms and 4 moles of hydrogen atoms. The mass of 1 mole of methane can be found by summing the masses of carbon and hydrogen present: Because 16.04 g represents the mass of 1 mole of methane molecules, it makes sense to call it the molar mass for methane. Thus the molar mass of a substance is the mass in grams of one mole of the compound. Traditionally, the term molecular weight has been used for this quantity. However, we will use molar mass exclusively in this text. The mo-lar mass of a known substance is obtained by summing the masses of the component atoms as we did for methane.
Calculating Molar Mass I Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. The formula for juglone is C10H6O3.
a. Calculate the molar mass of juglone.
b. A sample of 1.56 102 g of pure juglone was extracted from black walnut husks.
How many moles of juglone does this sample represent?
Solution a. The molar mass is obtained by summing the masses of the component atoms. In 1 mole of juglone there are 10 moles of carbon atoms, 6 moles of hydrogen atoms, and 3 moles of oxygen atoms: The mass of 1 mole of juglone is 174.1 g, which is the molar mass.
b. The mass of 1 mole of this compound is 174.1 g; thus 1.56 102 g is much less than a mole. The exact fraction of a mole can be determined as follows: See Exercises 3.39 through 3.42.
Calculating Molar Mass II Calcium carbonate (CaCO3), also called calcite, is the principal mineral found in lime-stone, marble, chalk, pearls, and the shells of marine animals such as clams.
1.56 102 g juglone 1 mol juglone 174.1 g juglone 8.96 105 mol juglone Mass of 1 mol C10H6O3 174.1 g 3 O: 3 16.00 g 48.00 g 6 H: 6 1.008 g 6.048 g 10 C: 10 12.01 g 120.1 g Mass of 1 mol CH4 16.04 g Mass of 4 mol H 4 1.008 g Mass of 1 mol C 12.01 g In this case, the term 12.01 limits the number of significant figures.
A substance’s molar mass is the mass in grams of 1 mole of the substance.
Sample Exercise 3.6 Sample Exercise 3.7 Juglone 3.4 Molar Mass 87 a. Calculate the molar mass of calcium carbonate.
b. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? What is the mass of the CO3 2 ions present?
Solution a. Calcium carbonate is an ionic compound composed of Ca2 and CO3 2 ions. In 1 mole of calcium carbonate there are 1 mole of Ca2 ions and 1 mole of CO3 2 ions. The molar mass is calculated by summing the masses of the components: Thus the mass of 1 mole of CaCO3 (1 mol Ca2 plus 1 mol CO3 2) is 100.09 g. This is the molar mass.
b. The mass of 1 mole of CaCO3 is 100.09 g. The sample contains nearly 5 moles, or close to 500 g. The exact amount is determined as follows: 4.86 mol CaCO3 100.09 g CaCO3 1 mol CaCO3 486 g CaCO3 Mass of 1 mol CaCO3 100.09 g 3 O: 3 16.00 g 48.00 g 1 C: 1 12.01 g 12.01 g 1 CO3 2: 1 Ca2: 1 40.08 g 40.08 g CHEMICAL IMPACT Measuring the Masses of Large Molecules, or Making Elephants Fly W hen a chemist produces a new molecule, one crucial property for making a positive identification is the molecule’s mass. There are many ways to determine the mo-lar mass of a compound, but one of the fastest and most ac-curate methods involves mass spectrometry. This method requires that the substance be put into the gas phase and ion-ized. The deflection that the resulting ion exhibits as it is ac-celerated through a magnetic field can be used to obtain a very precise value of its mass. One drawback of this method is that it is difficult to use with large molecules because they are difficult to vaporize. That is, substances that contain large molecules typically have very high boiling points, and these molecules are often damaged when they are vaporized at such high temperatures. A case in point involves proteins, an extremely important class of large biologic molecules that are quite fragile at high temperatures. Typical methods used to obtain the masses of protein molecules are slow and tedious.
Mass spectrometry has not been used previously to ob-tain protein masses because proteins decompose at the temperatures necessary to vaporize them. However, a new technique called matrix-assisted laser desorption has been developed that allows mass spectrometric determination of protein molar masses. In this technique, the large “target” molecule is embedded in a matrix of smaller molecules. The matrix is then placed in a mass spectrometer and blasted with a laser beam, which causes its disintegration. Disinte-gration of the matrix frees the large target molecule, which is then swept into the mass spectrometer. One researcher in-volved in this project likened this method to an elephant on top of a tall building: “The elephant must fly if the building is suddenly turned into fine grains of sand.” This technique allows scientists to determine the mass of huge molecules. So far researchers have measured pro-teins with masses up to 350,000 daltons (1 dalton is the mass of a hydrogen atom). This method, which makes mass spec-trometry a routine tool for the determination of protein masses, probably will be extended to even larger molecules such as DNA and could be a revolutionary development in the characterization of biomolecules.
Calcite crystals.
88 Chapter Three Stoichiometry To find the mass of carbonate ions (CO3 2) present in this sample, we must real-ize that 4.86 moles of CaCO3 contains 4.86 moles of Ca2 ions and 4.86 moles of CO3 2 ions. The mass of 1 mole of CO3 2 ions is Thus the mass of 4.86 moles of CO3 2 ions is See Exercises 3.43 through 3.46.
Molar Mass and Numbers of Molecules Isopentyl acetate (C7H14O2) is the compound responsible for the scent of bananas. A mo-lecular model of isopentyl acetate is shown in the margin below. Interestingly, bees release about 1 g (1 106 g) of this compound when they sting. The resulting scent attracts other bees to join the attack. How many molecules of isopentyl acetate are released in a typical bee sting? How many atoms of carbon are present?
Solution Since we are given a mass of isopentyl acetate and want to find the number of molecules, we must first compute the molar mass: This means that 1 mole of isopentyl acetate (6.022 1023 molecules) has a mass of 130.18 g.
To find the number of molecules released in a sting, we must first determine the num-ber of moles of isopentyl acetate in 1 106 g: Since 1 mole is 6.022 1023 units, we can determine the number of molecules: To determine the number of carbon atoms present, we must multiply the number of mol-ecules by 7, since each molecule of isopentyl acetate contains seven carbon atoms: Note: In keeping with our practice of always showing the correct number of significant figures, we have rounded after each step. However, if extra digits are carried throughout this problem, the final answer rounds to 3 1016.
See Exercises 3.47 through 3.52.
5 1015 molecules 7 carbon atoms molecule 4 1016 carbon atoms 8 109 mol C7H14O2 6.022 1023 molecules 1 mol C7H14O2 5 1015 molecules 1 106 g C7H14O2 1 mol C7H14O2 130.18 g C7H14O2 8 109 mol C7H14O2 2 mol O 16.00 g mol 32.00 g O 130.18 g 14 mol H 1.008 g mol 14.11 g H 7 mol C 12.01 g mol 84.07 g C 4.86 mol CO3 2 60.01 g CO3 2 1 mol CO3 2 292 g CO3 2 Mass of 1 mol CO3 2 60.01 g 3 O: 3 16.00 48.00 g 1 C: 1 12.01 12.01 g Sample Exercise 3.8 Isopentyl acetate is released when a bee stings.
Carbon Oxygen Hydrogen Isopentyl acetate To show the correct number of significant figures in each calculation, we round af-ter each step. In your calculations, always carry extra significant figures through to the end; then round.
3.5 Percent Composition of Compounds 89 3.5 Percent Composition of Compounds There are two common ways of describing the composition of a compound: in terms of the numbers of its constituent atoms and in terms of the percentages (by mass) of its el-ements. We can obtain the mass percents of the elements from the formula of the com-pound by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole of the compound.
For example, for ethanol, which has the formula C2H5OH, the mass of each element present and the molar mass are obtained as follows: The mass percent (often called the weight percent) of carbon in ethanol can be computed by comparing the mass of carbon in 1 mole of ethanol to the total mass of 1 mole of ethanol and multiplying the result by 100: The mass percents of hydrogen and oxygen in ethanol are obtained in a similar manner: Reality Check: Notice that the percentages add up to 100.00%; this provides a check that the calculations are correct.
Calculating Mass Percent I Carvone is a substance that occurs in two forms having different arrangements of the atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.
Solution The masses of the elements in 1 mole of carvone are Mass of C in 1 mol 10 mol 12.01 g mol 120.1 g 16.00 g 46.07 g 100% 34.73% Mass percent of O mass of O in 1 mol C2H5OH mass of 1 mol C2H5OH 100% 6.048 g 46.07 g 100% 13.13% Mass percent of H mass of H in 1 mol C2H5OH mass of 1 mol C2H5OH 100% 24.02 g 46.07 g 100% 52.14% Mass percent of C mass of C in 1 mol C2H5OH mass of 1 mol C2H5OH 100% Mass of 1 mol C2H5OH 46.07 g Mass of O 1 mol 16.00 g mol 16.00 g Mass of H 6 mol 1.008 g mol 6.048 g Mass of C 2 mol 12.01 g mol 24.02 g Sample Exercise 3.9 90 Chapter Three Stoichiometry Next we find the fraction of the total mass contributed by each element and convert it to a percentage: Reality Check: Sum the individual mass percent values—they should total to 100% within round-off errors. In this case, the percentages add up to 100.00%.
See Exercises 3.59 and 3.60.
Calculating Mass Percent II Penicillin, the first of a now large number of antibiotics (antibacterial agents), was dis-covered accidentally by the Scottish bacteriologist Alexander Fleming in 1928, but he was never able to isolate it as a pure compound. This and similar antibiotics have saved mil-lions of lives that might have been lost to infections. Penicillin F has the formula C14H20N2SO4. Compute the mass percent of each element.
Solution The molar mass of penicillin F is computed as follows: Mass percent of S 32.07 g S 312.4 g C14H20N2SO4 100% 10.27% Mass percent of N 28.02 g N 312.4 g C14H20N2SO4 100% 8.969% Mass percent of H 20.16 g H 312.4 g C14H20N2SO4 100% 6.453% Mass percent of C 168.1 g C 312.4 g C14H20N2SO4 100% 53.81% Mass of 1 mol C14H20N2SO4 312.4 g O: 4 mol 16.00 g mol 64.00 g S: 1 mol 32.07 g mol 32.07 g N: 2 mol 14.01 g mol 28.02 g H: 20 mol 1.008 g mol 120.16 g C: 14 mol 12.01 g mol 168.1 g Mass percent of O 16.00 g O 150.2 g C10H14O 100% 10.65% Mass percent of H 14.11 g H 150.2 g C10H14O 100% 9.394% Mass percent of C 120.1 g C 150.2 g C10H14O 100% 79.96% Mass of 1 mol C10H14O 150.2 g Mass of O in 1 mol 1 mol 16.00 g mol 16.00 g Mass of H in 1 mol 14 mol 1.008 g mol 14.11 g Carvone Sample Exercise 3.10 Although Fleming is commonly given credit for the discovery of penicillin, there is good evidence that penicillium mold ex-tracts were used in the nineteenth century by Lord Joseph Lister to cure infections.
Penicillin is isolated from a mold that can be grown in large quantities in fermenta-tion tanks.
3.6 Determining the Formula of a Compound 91 CO2 H2O Furnace O2 and other gases CO2, H2O, O2, and other gases O2 H2O absorber such as Mg(ClO4)2 CO2 absorber such as NaOH Sample FIGURE 3.5 A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen. The sample is burned in the presence of excess oxygen, which converts all its carbon to carbon dioxide and all its hydrogen to water. These products are collected by absorption using appropriate materials, and their amounts are determined by measuring the increase in masses of the absorbents.
Reality Check: The percentages add up to 99.99%.
See Exercises 3.61 and 3.62.
3.6 Determining the Formula of a Compound When a new compound is prepared, one of the first items of interest is the formula of the compound. This is most often determined by taking a weighed sample of the compound and either decomposing it into its component elements or reacting it with oxygen to pro-duce substances such as CO2, H2O, and N2, which are then collected and weighed. A device for doing this type of analysis is shown in Fig. 3.5. The results of such analyses provide the mass of each type of element in the compound, which can be used to determine the mass percent of each element.
We will see how information of this type can be used to compute the formula of a compound. Suppose a substance has been prepared that is composed of carbon, hydro-gen, and nitrogen. When 0.1156 gram of this compound is reacted with oxygen, 0.1638 gram of carbon dioxide (CO2) and 0.1676 gram of water (H2O) are collected. Assuming that all the carbon in the compound is converted to CO2, we can determine the mass of carbon originally present in the 0.1156-gram sample. To do this, we must use the fraction (by mass) of carbon in CO2. The molar mass of CO2 is The fraction of carbon present by mass is This factor can now be used to determine the mass of carbon in 0.1638 gram of CO2: Remember that this carbon originally came from the 0.1156-gram sample of unknown compound. Thus the mass percent of carbon in this compound is 0.04470 g C 0.1156 g compound 100% 38.67% C 0.1638 g CO2 12.01 g C 44.01 g CO2 0.04470 g C Mass of C Total mass of CO2 12.01 g C 44.01 g CO2 Molar mass of CO2 44.01 g/mol O: 2 mol 16.00 g mol 32.00 g C: 1 mol 12.01 g mol 12.01 g Mass percent of O 64.00 g O 312.4 g C14H20N2SO4 100% 20.49% 92 Chapter Three Stoichiometry The same procedure can be used to find the mass percent of hydrogen in the unknown compound. We assume that all the hydrogen present in the original 0.1156 gram of com-pound was converted to H2O. The molar mass of H2O is 18.02 grams, and the fraction of hydrogen by mass in H2O is Therefore, the mass of hydrogen in 0.1676 gram of H2O is The mass percent of hydrogen in the compound is The unknown compound contains only carbon, hydrogen, and nitrogen. So far we have de-termined that it is 38.67% carbon and 16.22% hydrogen. The remainder must be nitrogen: h h % C % H We have determined that the compound contains 38.67% carbon, 16.22% hydrogen, and 45.11% nitrogen. Next we use these data to obtain the formula.
Since the formula of a compound indicates the numbers of atoms in the compound, we must convert the masses of the elements to numbers of atoms. The easiest way to do this is to work with 100.00 grams of the compound. In the present case, 38.67% carbon by mass means 38.67 grams of carbon per 100.00 grams of compound; 16.22% hydrogen means 16.22 grams of hydrogen per 100.00 grams of compound; and so on. To determine the formula, we must calculate the number of carbon atoms in 38.67 grams of carbon, the number of hydrogen atoms in 16.22 grams of hydrogen, and the number of nitrogen atoms in 45.11 grams of nitrogen. We can do this as follows: Thus 100.00 grams of this compound contains 3.220 moles of carbon atoms, 16.09 moles of hydrogen atoms, and 3.219 moles of nitrogen atoms.
We can find the smallest whole-number ratio of atoms in this compound by dividing each of the mole values above by the smallest of the three: Thus the formula might well be CH5N. However, it also could be C2H10N2 or C3H15N3, and so on—that is, some multiple of the smallest whole-number ratio. Each of these al-ternatives also has the correct relative numbers of atoms. That is, any molecule that can N: 3.219 3.220 1.000 1 H: 16.09 3.220 4.997 5 C: 3.220 3.220 1.000 1 45.11 g N 1 mol N 14.01 g N 3.219 mol N 16.22 g H 1 mol H 1.008 g H 16.09 mol H 38.67 g C 1 mol C 12.01 g C 3.220 mol C 100.00% 138.67% 16.22%2 45.11% N 0.01875 g H 0.1156 g compound 100% 16.22% H 0.1676 g H2O 2.016 g H 18.02 g H2O 0.01875 g H Mass of H Mass of H2O 2.016 g H 18.02 g H2O 3.6 Determining the Formula of a Compound 93 Molecular formula (empirical formula)n, where n is an integer.
C6H6 = (CH)6 S8 = (S)8 C6H12O6 = (CH2O)6 FIGURE 3.6 Examples of substances whose empirical and molecular formulas differ. Notice that molecular formula (empirical formula)n, where n is an integer.
Sample Exercise 3.11 be represented as (CH5N)n, where n is an integer, has the empirical formula CH5N. To be able to specify the exact formula of the molecule involved, the molecular formula, we must know the molar mass.
Suppose we know that this compound with empirical formula CH5N has a molar mass of 31.06 g/mol. How do we determine which of the possible choices represents the mo-lecular formula? Since the molecular formula is always a whole number multiple of the empirical formula, we must first find the empirical formula mass for CH5N: This is the same as the known molar mass of the compound. Thus in this case the empirical formula and the molecular formula are the same; this substance consists of molecules with the formula CH5N. It is quite common for the empirical and molecular formulas to be different; some examples where this is the case are shown in Fig. 3.6.
Determining Empirical and Molecular Formulas I Determine the empirical and molecular formulas for a compound that gives the following percentages upon analysis (in mass percents): 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96 g/mol.
Solution First, we convert the mass percents to masses in grams. In 100.00 g of this compound there are 71.65 g of chlorine, 24.27 g of carbon, and 4.07 g of hydrogen. We use these masses to compute the moles of atoms present: Dividing each mole value by 2.021 (the smallest number of moles present), we obtain the empirical formula ClCH2. 4.07 g H 1 mol H 1.008 g H 4.04 mol H 24.27 g C 1 mol C 12.01 g C 2.021 mol C 71.65 g Cl 1 mol Cl 35.45 g Cl 2.021 mol Cl Formula mass of CH5N 31.06 g/mol 1 N: 1 14.01 g 14.01 g 5 H: 5 1.008 g 5.040 g 1 C: 1 12.01 g 12.01 g 94 Chapter Three Stoichiometry To determine the molecular formula, we must compare the empirical formula mass with the molar mass. The empirical formula mass is 49.48 g/mol (confirm this). The mo-lar mass is known to be 98.96 g/mol.
This substance is composed of molecules with the formula Cl2C2H4.
Notice that the method we employ here allows us to determine the molecular formula of a compound but not its structural formula. The compound Cl2C2H4 is called dichloroethane. There are two forms of this compound, shown in Fig. 3.7. The form on the right was formerly used as an additive in leaded gasoline.
See Exercises 3.57 and 3.58.
Determining Empirical and Molecular Formulas II A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxy-gen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?
Solution In 100.00 g of this compound there are 43.64 g of phosphorus and 56.36 g of oxygen. In terms of moles, in 100.00 g of the compound we have Dividing both mole values by the smaller one gives This yields the formula PO2.5. Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers. To obtain the simplest set of whole numbers, we multiply both numbers by 2 to give the empirical for-mula P2O5.
To obtain the molecular formula, we must compare the empirical formula mass to the molar mass. The empirical formula mass for P2O5 is 141.94.
The molecular formula is (P2O5)2, or P4O10.
The structural formula of this interesting compound is given in Fig. 3.8.
See Exercise 3.59.
In Sample Exercises 3.11 and 3.12 we found the molecular formula by comparing the empirical formula mass with the molar mass. There is an alternate way to obtain the molecular formula. For example, in Sample Exercise 3.11 we know the molar mass of the compound is 98.96 g/mol. This means that 1 mole of the compound weighs 98.96 grams.
Molar mass Empirical formula mass 283.88 141.94 2 1.409 1.409 1 P and 3.523 1.409 2.5 O 56.36 g O 1 mol O 16.00 g O 3.523 mol O 43.64 g P 1 mol P 30.97 g P 1.409 mol P Molecular formula 1ClCH222 Cl2C2H4 Molar mass Empirical formula mass 98.96 g/mol 49.48 g/mol 2 Sample Exercise 3.12 FIGURE 3.8 The structure of P4O10. Note that some of the oxygen atoms act as “bridges” between the phosphorus atoms. This compound has a great affinity for water and is often used as a desiccant, or drying agent.
FIGURE 3.7 The two forms of dichloroethane.
3.6 Determining the Formula of a Compound 95 Since we also know the mass percentages of each element, we can compute the mass of each element present in 1 mole of compound: Now we can compute moles of atoms present per mole of compound: Thus 1 mole of the compound contains 2 mol Cl atoms, 2 mol C atoms, and 4 mol H atoms, and the molecular formula is Cl2C2H4, as obtained in Sample Exercise 3.11.
Determining a Molecular Formula Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon, 5.15% hy-drogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol.
Determine the molecular formula of caffeine.
Solution We will first determine the mass of each element in 1 mole (194.2 g) of caffeine: Now we will convert to moles: Rounding the numbers to integers gives the molecular formula for caffeine: C8H10N4O2.
See Exercise 3.76. Oxygen: 32.02 g O mol caffeine 1 mol O 16.00 g O 2.001 mol O mol caffeine Nitrogen: 56.07 g N mol caffeine 1 mol N 14.01 g N 4.002 mol N mol caffeine Hydrogen: 10.0 g H mol caffeine 1 mol H 1.008 g H 9.92 mol H mol caffeine Carbon: 96.09 g C mol caffeine 1 mol C 12.01 g C 8.001 mol C mol caffeine 16.49 g O 100.0 g caffeine 194.2 g mol 32.02 g O mol caffeine 28.87 g N 100.0 g caffeine 194.2 g mol 56.07 g N mol caffeine 5.15 g H 100.0 g caffeine 194.2 g mol 10.0 g H mol caffeine 49.48 g C 100.0 g caffeine 194.2 g mol 96.09 g C mol caffeine Hydrogen: 4.03 g H mol compound 1 mol H 1.008 g H 4.00 mol H mol compound Carbon: 24.02 g C mol compound 1 mol C 12.01 g C 2.000 mol C mol compound Chlorine: 70.90 g Cl mol compound 1 mol Cl 35.45 g Cl 2.000 mol Cl mol compound Hydrogen: 4.07 g H 100.0 g compound 98.96 g mol 4.03 g H mol compound Carbon: 24.27 g C 100.0 g compound 98.96 g mol 24.02 g C mol compound Chlorine: 71.65 g Cl 100.0 g compound 98.96 g mol 70.90 g Cl mol compound Sample Exercise 3.13 Computer-generated molecule of caffeine.
96 Chapter Three Stoichiometry 3.7 Chemical Equations Chemical Reactions A chemical change involves a reorganization of the atoms in one or more substances. For example, when the methane (CH4) in natural gas combines with oxygen (O2) in the air and burns, carbon dioxide (CO2) and water (H2O) are formed. This process is represented Molecular Formula Determination Method One Obtain the empirical formula.
Compute the mass corresponding to the empirical formula.
Calculate the ratio The integer from the previous step represents the number of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by this inte-ger, the molecular formula results. This procedure is summarized by the equation: Method Two Using the mass percentages and the molar mass, determine the mass of each element present in one mole of compound.
Determine the number of moles of each element present in one mole of compound.
The integers from the previous step represent the subscripts in the molecular formula.
Molecular formula 1empirical formula2 molar mass empirical formula mass Molar mass Empirical formula mass Note that method two assumes that the molar mass of the compound is known accurately.
Numbers very close to whole numbers, such as 9.92 and 1.08, should be rounded to whole numbers. Numbers such as 2.25, 4.33, and 2.72 should not be rounded to whole numbers.
The methods for obtaining empirical and molecular formulas are summarized as follows: Empirical Formula Determination Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element.
Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present.
Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula.
If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers.
3.7 Chemical Equations 97 by a chemical equation with the reactants (here methane and oxygen) on the left side of an arrow and the products (carbon dioxide and water) on the right side: Reactants Products Notice that the atoms have been reorganized. Bonds have been broken, and new ones have been formed. It is important to recognize that in a chemical reaction, atoms are neither created nor destroyed. All atoms present in the reactants must be accounted for among the products. In other words, there must be the same number of each type of atom on the product side and on the reactant side of the arrow. Making sure that this rule is obeyed is called balancing a chemical equation for a reaction.
The equation (shown above) for the reaction between CH4 and O2 is not balanced.
We can see this from the following representation of the reaction: Notice that the number of oxygen atoms (in O2) on the left of the arrow is two, while on the right there are three O atoms (in CO2 and H2O). Also, there are four hydrogen atoms (in CH4) on the left and only two (in H2O) on the right. Remember that a chemical reac-tion is simply a rearrangement of the atoms (a change in the way they are organized).
Atoms are not created or destroyed in a chemical reaction. Thus the reactants and prod-ucts must occur in numbers that give the same number of each type of atom among both the reactants and products. Simple trial and error will allow us to figure this out for the reaction of methane with oxygen. The needed numbers of molecules are Notice that now we have the same number of each type of atom represented among the reactants and the products.
We can represent the preceding situation in a shorthand manner by the following chemical equation: We can check that the equation is balanced by comparing the number of each type of atom on both sides: h h h 1 C 4 H 1 C 4 H 4 O 2 O 2 O To summarize, we have CH4 2O2 ¡ CO2 2H2O CH4 2O2 ¡ CO2 2H2O + + + + CH4 O2 ¡ CO2 H2O p 88n 8 8n 8 8n Reactants Products 1 C 1 C 4 H 4 H 4 O 4 O Methane reacts with oxygen to produce the flame in a Bunsen burner.
98 Chapter Three Stoichiometry TABLE 3.2 Information Conveyed by the Balanced Equation for the Combustion of Methane Reactants Products 80 g products ¡ 80 g reactants 16 g 2 132 g2 44 g 2 118 g2 6.022 1023 molecules 2 16.022 1023 molecules2 ¡ 6.022 1023 molecules 2 16.022 1023 molecules2 1 mole 2 moles ¡ 1 mole 2 moles 1 molecule 2 molecules ¡ 1 molecule 2 molecules CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 Hydrochloric acid reacts with solid sodium hydrogen carbonate to produce gaseous carbon dioxide.
State Symbol Solid (s) Liquid (l) Gas (g) Dissolved in water (in aqueous solution) (aq) The Meaning of a Chemical Equation The chemical equation for a reaction gives two important types of information: the nature of the reactants and products and the relative numbers of each.
The reactants and products in a specific reaction must be identified by experiment.
Besides specifying the compounds involved in the reaction, the equation often gives the physical states of the reactants and products: For example, when hydrochloric acid in aqueous solution is added to solid sodium hy-drogen carbonate, the products carbon dioxide gas, liquid water, and sodium chloride (which dissolves in the water) are formed: The relative numbers of reactants and products in a reaction are indicated by the co-efficients in the balanced equation. (The coefficients can be determined because we know that the same number of each type of atom must occur on both sides of the equation.) For example, the balanced equation can be interpreted in several equivalent ways, as shown in Table 3.2. Note that the total mass is 80 grams for both reactants and products. We expect the mass to remain constant, since chemical reactions involve only a rearrangement of atoms. Atoms, and therefore mass, are conserved in a chemical reaction.
From this discussion you can see that a balanced chemical equation gives you a great deal of information.
3.8 Balancing Chemical Equations An unbalanced chemical equation is of limited use. Whenever you see an equation, you should ask yourself whether it is balanced. The principle that lies at the heart of the balancing process is that atoms are conserved in a chemical reaction. The same number of each type of atom must be found among the reactants and products. It is also important to recognize that the identities of the reactants and products of a reaction are determined by experimental obser-vation. For example, when liquid ethanol is burned in the presence of sufficient oxygen gas, the products are always carbon dioxide and water. When the equation for this reaction is CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 HCl1aq2 NaHCO31s2 ¡ CO21g2 H2O1l2 NaCl1aq2 Visualization: Oxygen, Hydrogen, Soap Bubbles, and Balloons Visualization: Conservation of Mass and Balancing Equations 3.8 Balancing Chemical Equations 99 balanced, the identities of the reactants and products must not be changed. The formulas of the compounds must never be changed in balancing a chemical equation.That is, the subscripts in a formula cannot be changed, nor can atoms be added or subtracted from a formula.
Most chemical equations can be balanced by inspection, that is, by trial and error. It is always best to start with the most complicated molecules (those containing the great-est number of atoms). For example, consider the reaction of ethanol with oxygen, given by the unbalanced equation which can be represented by the following molecular models: Notice that the carbon and hydrogen atoms are not balanced. There are two carbon atoms on the left and one on the right, and there are six hydrogens on the left and two on the right.
We need to find the correct numbers of reactants and products so that we have the same number of all types of atoms among the reactants and products. We will balance the equation “by inspection” (a systematic trial-and-error procedure).
The most complicated molecule here is C2H5OH. We will begin by balancing the products that contain the atoms in C2H5OH. Since C2H5OH contains two carbon atoms, we place the coefficient 2 before the CO2 to balance the carbon atoms: Since C2H5OH contains six hydrogen atoms, the hydrogen atoms can be balanced by placing a 3 before the H2O: Last, we balance the oxygen atoms. Note that the right side of the preceding equation con-tains seven oxygen atoms, whereas the left side has only three. We can correct this by putting a 3 before the O2 to produce the balanced equation: Now we check: The equation is balanced.
The balanced equation can be represented as follows: + + + + C2H5OH1l2 O21g2 ¡ CO21g2 H2O1g2 In balancing equations, start with the most complicated molecule.
100 Chapter Three Stoichiometry You can see that all the elements balance.
Writing and Balancing the Equation for a Chemical Reaction ➥ 1 Determine what reaction is occurring. What are the reactants, the products, and the physical states involved?
➥ 2 Write the unbalanced equation that summarizes the reaction described in step 1.
➥ 3 Balance the equation by inspection, starting with the most complicated mole-cule(s). Determine what coefficients are necessary so that the same number of each type of atom appears on both reactant and product sides. Do not change the identities (formulas) of any of the reactants or products.
Balancing a Chemical Equation I Chromium compounds exhibit a variety of bright colors. When solid ammonium dichro-mate, (NH4)2Cr2O7, a vivid orange compound, is ignited, a spectacular reaction occurs, as shown in the two photographs on the next page. Although the reaction is actually somewhat more complex, let’s assume here that the products are solid chromium(III) oxide, nitrogen gas (consisting of N2 molecules), and water vapor. Balance the equation for this reaction.
Solution ➥ 1 From the description given, the reactant is solid ammonium dichromate, (NH4)2Cr2O7(s), and the products are nitrogen gas, N2(g), water vapor, H2O(g), and solid chromium(III) oxide, Cr2O3(s). The formula for chromium(III) oxide can be determined by recognizing that the Roman numeral III means that Cr3 ions are present. For a neutral compound, the formula must then be Cr2O3, since each oxide ion is O2.
➥ 2 The unbalanced equation is ➥ 3 Note that nitrogen and chromium are balanced (two nitrogen atoms and two chromium atoms on each side), but hydrogen and oxygen are not. A coefficient of 4 for H2O balances the hydrogen atoms: (4 2) H (4 2) H Note that in balancing the hydrogen we also have balanced the oxygen, since there are seven oxygen atoms in the reactants and in the products.
Reality Check: Reactant Product atoms atoms The equation is balanced.
See Exercises 3.81 and 3.82.
2 N, 8 H, 2 Cr, 7 O S 2 N, 8 H, 2 Cr, 7 O 1NH422Cr2O71s2 S Cr2O31s2 N21g2 4H2O1g2 1NH422Cr2O71s2 S Cr2O31s2 N21g2 H2O1g2 Sample Exercise 3.14 Chromate and dichromate compounds are carcinogens (cancer-inducing agents) and should be handled very carefully.
3.8 Balancing Chemical Equations 101 Balancing a Chemical Equation II At 1000C, ammonia gas, NH3(g), reacts with oxygen gas to form gaseous nitric oxide, NO(g), and water vapor. This reaction is the first step in the commercial production of nitric acid by the Ostwald process. Balance the equation for this reaction.
Solution The unbalanced equation for the reaction is Because all the molecules in this equation are of about equal complexity, where we start in balancing it is rather arbitrary. Let’s begin by balancing the hydrogen. A coefficient of 2 for NH3 and a coefficient of 3 for H2O give six atoms of hydrogen on both sides: The nitrogen can be balanced with a coefficient of 2 for NO: Finally, note that there are two atoms of oxygen on the left and five on the right. The oxygen can be balanced with a coefficient of for O2: However, the usual custom is to have whole-number coefficients. We simply multiply the entire equation by 2.
Reality Check: There are 4 N, 12 H, and 10 O on both sides, so the equation is balanced.
4NH31g2 5O21g2 S 4NO1g2 6H2O1g2 2NH31g2 5 2 O21g2 S 2NO1g2 3H2O1g2 5 2 2NH31g2 O21g2 S 2NO1g2 3H2O1g2 2NH31g2 O21g2 S NO1g2 3H2O1g2 NH31g2 O21g2 S NO1g2 H2O1g2 Decomposition of ammonium dichromate.
Sample Exercise 3.15 The Ostwald process is described in Section 20.2.
102 Chapter Three Stoichiometry We can represent this balanced equation visually as See Exercises 3.83 through 3.88.
3.9 Stoichiometric Calculations: Amounts of Reactants and Products As we have seen in previous sections of this chapter, the coefficients in chemical equa-tions represent numbers of molecules, not masses of molecules. However, when a reac-tion is to be run in a laboratory or chemical plant, the amounts of substances needed cannot be determined by counting molecules directly. Counting is always done by weighing. In this section we will see how chemical equations can be used to determine the masses of reacting chemicals.
To develop the principles for dealing with the stoichiometry of reactions, we will con-sider the reaction of propane with oxygen to produce carbon dioxide and water. We will consider the question: “What mass of oxygen will react with 96.1 grams of propane?” In doing stoichiometry, the first thing we must do is write the balanced chemical equation for the reaction. In this case the balanced equation is which can be visualized as This equation means that 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O. To use this equation to find the masses of reactants and prod-ucts, we must be able to convert between masses and moles of substances. Thus we must first ask: “How many moles of propane are present in 96.1 grams of propane?” The molar + + C3H81g2 5O21g2 ¡ 3CO21g2 4H2O1g2 + + Before doing any calculations involving a chemical reaction, be sure the equation for the reaction is balanced.
3.9 Stoichiometric Calculations: Amounts of Reactants and Products 103 CHEMICAL IMPACT High Mountains—Low Octane T he next time that you visit a gas station, take a moment to note the octane rating that accompanies the grade of gaso-line that you are purchasing. The gasoline is priced according to its octane rating—a measure of the fuel’s antiknock prop-erties. In a conventional internal combustion engine, gasoline vapors and air are drawn into the combustion cylinder on the downward stroke of the piston. This air–fuel mixture is com-pressed on the upward piston stroke (compression stroke), and a spark from the sparkplug ignites the mix. The rhythmic com-bustion of the air–fuel mix occurring sequentially in several cylinders furnishes the power to propel the vehicle down the road. Excessive heat and pressure (or poor-quality fuel) within the cylinder may cause the premature combustion of the mix-ture—commonly known as engine “knock” or “ping.” Over time, this engine knock can damage the engine, resulting in inefficient performance and costly repairs.
A consumer typically is faced with three choices of gasoline, with octane ratings of 87 (regular), 89 (midgrade), and 93 (premium). But if you happen to travel or live in the higher elevations of the Rocky Mountain states, you might be surprised to find different octane ratings at the gasoline pumps. The reason for this provides a lesson in stoichiom-etry. At higher elevations the air is less dense—the volume of oxygen per unit volume of air is smaller. Most engines are designed to achieve a 14:1 oxygen-to-fuel ratio in the cylinder prior to combustion. If less oxygen is available, then less fuel is required to achieve this optimal ratio. In turn, the lower volumes of oxygen and fuel result in a lower pressure in the cylinder. Because high pressure tends to promote knocking, the lower pressure within engine cylinders at higher elevations promotes a more controlled combustion of the air–fuel mixture, and therefore, octane requirements are lower. While consumers in the Rocky Mountain states can purchase three grades of gasoline, the octane ratings of these fuel blends are different from those in the rest of the United States. In Denver, Colorado, regular gasoline is 85 octane, midgrade is 87 octane, and premium is 91 octane—2 points lower than gasoline sold in most of the rest of the country.
mass of propane to three significant figures is 44.1 (that is, 3 12.01 8 1.008). The moles of propane can be calculated as follows: Next we must take into account the fact that each mole of propane reacts with 5 moles of oxygen. The best way to do this is to use the balanced equation to construct a mole ratio. In this case we want to convert from moles of propane to moles of oxygen. From the balanced equation we see that 5 moles of O2 is required for each mole of C3H8, so the appropriate ratio is Multiplying the number of moles of C3H8 by this factor gives the number of moles of O2 required: Notice that the mole ratio is set up so that the moles of C3H8 cancel out, and the units that result are moles of O2.
Since the original question asked for the mass of oxygen needed to react with 96.1 grams of propane, the 10.9 moles of O2 must be converted to grams. Since the molar mass of O2 is 32.0 g/mol, Therefore, 349 grams of oxygen is required to burn 96.1 grams of propane.
10.9 mol O2 32.0 g O2 1 mol O2 349 g O2 2.18 mol C3H8 5 mol O2 1 mol C3H8 10.9 mol O2 5 mol O2 1 mol C3H8 96.1 g C3H8 1 mol C3H8 44.1 g C3H8 2.18 mol C3H8 104 Chapter Three Stoichiometry This example can be extended by asking: “What mass of carbon dioxide is produced when 96.1 grams of propane is combusted with oxygen?” In this case we must convert between moles of propane and moles of carbon dioxide. This can be accomplished by looking at the balanced equation, which shows that 3 moles of CO2 is produced for each mole of C3H8 re-acted. The mole ratio needed to convert from moles of propane to moles of carbon dioxide is The conversion is Then, using the molar mass of CO2 (44.0 g/mol), we calculate the mass of CO2 produced: We will now summarize the sequence of steps needed to carry out stoichiometric calculations.
Calculating Masses of Reactants and Products in Chemical Reactions ➥ 1 Balance the equation for the reaction.
➥ 2 Convert the known mass of the reactant or product to moles of that substance.
➥ 3 Use the balanced equation to set up the appropriate mole ratios.
➥ 4 Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product.
➥ 5 Convert from moles back to grams if required by the problem.
These steps are summarized by the following diagram: 288 g CO2 1 mol CO2 44.0 g CO2 96.1 g C3H8 2.18 mol C3H8 6.54 mol CO2 1 mol C3H8 44.1 g C3H8 3 mol CO2 1 mol C3H8 6.54 mol CO2 44.0 g CO2 1 mol CO2 288 g CO2 2.18 mol C3H8 3 mol CO2 1 mol C3H8 6.54 mol CO2 3 mol CO2 1 mol C3H8 Use mole ratio to convert Convert to moles Find appropriate mole ratio Mass of known substance Moles of known substance Convert to grams Moles of desired substance Mass of desired substance Moles desired substance Moles known substance Balanced chemical equation 3.9 Stoichiometric Calculations: Amounts of Reactants and Products 105 Chemical Stoichiometry I Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
Solution ➥ 1 Using the description of the reaction, we can write the unbalanced equation: The balanced equation is ➥ 2 We convert the given mass of LiOH to moles, using the molar mass of LiOH (6.941 16.00 1.008 23.95 g/mol): ➥ 3 Since we want to determine the amount of CO2 that reacts with the given amount of LiOH, the appropriate mole ratio is ➥ 4 We calculate the moles of CO2 needed to react with the given mass of LiOH using this mole ratio: ➥ 5 Next we calculate the mass of CO2, using its molar mass (44.0 g/mol): Thus 920. g of CO2(g) will be absorbed by 1.00 kg of LiOH(s).
See Exercises 3.89 and 3.90.
Chemical Stoichiometry II Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach: Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid: Which is the more effective antacid per gram, NaHCO3 or Mg(OH)2?
Solution To answer the question, we must determine the amount of HCl neutralized per gram of NaHCO3 and per gram of Mg(OH)2. Using the molar mass of NaHCO3 (84.01 g/mol), we can determine the moles of NaHCO3 in 1.00 g of NaHCO3: 1.00 g NaHCO3 1 mol NaHCO3 84.01 g NaHCO3 1.19 102 mol NaHCO3 Mg1OH221s2 2HCl1aq2 ¡ 2H2O1l2 MgCl21aq2 NaHCO31s2 HCl1aq2 ¡ NaCl1aq2 H2O1l2 CO21aq2 20.9 mol CO2 44.0 g CO2 1 mol CO2 9.20 102 g CO2 41.8 mol LiOH 1 mol CO2 2 mol LiOH 20.9 mol CO2 1 mol CO2 2 mol LiOH 1.00 kg LiOH 1000 g LiOH 1 kg LiOH 1 mol LiOH 23.95 g LiOH 41.8 mol LiOH 2LiOH1s2 CO21g2 ¡ Li2CO31s2 H2O1l2 LiOH1s2 CO21g2 ¡ Li2CO31s2 H2O1l2 Sample Exercise 3.16 Astronaut Sidney M. Gutierrez changes the lithium hydroxide cannisters on space shuttle Columbia. The lithium hydroxide is used to purge carbon dioxide from the air in the shuttle’s cabin.
Sample Exercise 3.17 106 Chapter Three Stoichiometry Next we determine the moles of HCl using the mole ratio 1 mol HCl/1 mol NaHCO3: Thus 1.00 g of NaHCO3 will neutralize 1.19 102 mol HCl.
Using the molar mass of Mg(OH)2 (58.32 g/mol), we determine the moles of Mg(OH)2 in 1.00 g: To determine the moles of HCl that will react with this amount of Mg(OH)2, we use the mole ratio 2 mol HCl/1 mol Mg(OH)2: Thus 1.00 g of Mg(OH)2 will neutralize 3.42 102 mol HCl. It is a better antacid per gram than NaHCO3.
See Exercises 3.91 and 3.92.
3.10 Calculations Involving a Limiting Reactant When chemicals are mixed together to undergo a reaction, they are often mixed in stoi-chiometric quantities, that is, in exactly the correct amounts so that all reactants “run out” (are used up) at the same time. To clarify this concept, let’s consider the production of hydrogen for use in the manufacture of ammonia by the Haber process. Ammonia, a very important fertilizer itself and a starting material for other fertilizers, is made by com-bining nitrogen (from the air) with hydrogen according to the equation Hydrogen can be obtained from the reaction of methane with water vapor: We can illustrate what we mean by stoichiometric quantities by first visualizing the balanced equation as follows: Since this reaction involves one molecule of methane reacting with one molecule of wa-ter, to have stoichiometric amounts of methane and water we must have equal numbers of them, as shown in Fig. 3.9, where several stoichiometric mixtures are shown.
Suppose we want to calculate the mass of water required to react exactly with 2.50 103 kilograms of methane. That is, how much water will just consume all the 2.50 103 kilograms of methane, leaving no methane or water remaining?
To do this calculation, we need to recognize that we need equal numbers of methane and water molecules. Therefore, we first need to find the number of moles of methane molecules in 2.50 103 kg (2.50 106 g) of methane: h molar mass of CH4 2.50 106 g CH4 1 mol CH4 16.04 g CH4 1.56 105 mol CH4 molecules + + CH41g2 H2O1g2 ¡ 3H21g2 CO1g2 N21g2 3H21g2 ¡ 2NH31g2 1.71 102 mol Mg1OH22 2 mol HCl 1 mol Mg1OH22 3.42 102 mol HCl 1.00 g Mg1OH22 1 mol Mg1OH22 58.32 g Mg1OH22 1.71 102 mol Mg1OH22 1.19 102 mol NaHCO3 1 mol HCl 1 mol NaHCO3 1.19 102 mol HCl Milk of magnesia contains a suspension of Mg(OH)2(s).
The details of the Haber process are discussed in Section 19.2.
Visualization: Limiting Reactant 3.10 Calculations Involving a Limiting Reactant 107 This same number of water molecules has a mass determined as follows: Thus, if 2.50 103 kilograms of methane is mixed with 2.81 103 kilograms of water, both reactants will “run out” at the same time. The reactants have been mixed in stoi-chiometric quantities.
If, on the other hand, 2.50 103 kilograms of methane is mixed with 3.00 103 kilograms of water, the methane will be consumed before the water runs out. The water will be in excess; that is, there will be more water molecules than methane molecules in the reaction mixture. What is the implication of this with respect to the number of prod-uct molecules that can form?
To answer this question, consider the situation on a smaller scale. Assume we mix 10 CH4 molecules and 17 H2O molecules and let them react. How many H2 and CO mol-ecules can form?
First picture the mixture of CH4 and H2O molecules as shown in Fig. 3.10.
Then imagine that groups consisting of one CH4 molecule and one H2O molecule (Fig. 3.10) will react to form three H2 and one CO molecules (Fig. 3.11).
Notice that products can form only when both CH4 and H2O are available to react. Once the 10 CH4 molecules are used up by reacting with 10 H2O molecules, the remaining water 1.56 105 mol H2O 18.02 g mol H2O 2.81 106 g H2O 2.81 103 kg H2O FIGURE 3.9 Three different stoichiometric mixtures of methane and water, which react one-to-one.
FIGURE 3.10 A mixture of CH4 and H2O molecules.
FIGURE 3.11 Methane and water have reacted to form products according to the equation CH4 H2O ¡ 3H2 CO.
108 Chapter Three Stoichiometry molecules cannot react. They are in excess. Thus the number of products that can form is limited by the methane. Once the methane is consumed, no more products can be formed, even though some water still remains. In this situation the amount of methane limits the amount of products that can be formed. This brings us to the concept of the limiting reactant (or limiting reagent), which is the reactant that is consumed first and that there-fore limits the amounts of products that can be formed. In any stoichiometry calculation involving a chemical reaction, it is essential to determine which reactant is limiting so as to calculate correctly the amounts of products that will be formed.
To further explore the idea of a limiting reactant, consider the ammonia synthesis reaction: Assume that 5 N2 molecules and 9 H2 molecules are placed in a flask. Is this a stoichio-metric mixture of reactants, or will one of them be consumed before the other runs out?
From the balanced equation we know that each N2 molecule requires 3 H2 molecules for the reaction to occur: Thus the required H2N2 ratio is 3H21N2. In our experiment we have 9 H2 and 5 N2, or a ratio of 9H25N2 1.8H21N2.
Since the actual ratio (1.8:1) of H2N2 is less than the ratio required by the balanced equation (3:1), there is not enough hydrogen to react with all the nitrogen. That is, the hydrogen will run out first, leaving some unreacted N2 molecules. We can visualize this as shown in Fig. 3.12.
Figure 3.12 shows that 3 of the N2 molecules react with the 9 H2 molecules to pro-duce 6 NH3 molecules: This leaves 2 N2 molecules unreacted—the nitrogen is in excess.
What we have shown here is that in this experiment the hydrogen is the limiting re-actant. The amount of H2 initially present determines the amount of NH3 that can form.
The reaction was not able to use up all the N2 molecules because the H2 molecules were all consumed by the first 3 N2 molecules to react.
3N2 9H2 ¡ 6NH3 + N21g2 3H21g2 ¡ 2NH31g2 Ammonia is dissolved in irrigation water to provide fertilizer for a field of corn.
FIGURE 3.12 Hydrogen and nitrogen react to form am-monia according to the equation N2 3H2 ¡ 2NH3.
3.10 Calculations Involving a Limiting Reactant 109 Another way to look at this is to determine how much H2 would be required by 5 N2 molecules. Multiplying the balanced equation by 5 gives Thus 5 N2 molecules would require 15 H2 molecules and we have only 9. This tells us the same thing we learned earlier—the hydrogen is limiting.
The most important point here is this: The limiting reactant limits the amount of prod-uct that can form. The reaction that actually occurred was not Thus 6 NH3 were formed, not 10 NH3, because the H2, not the N2, was limiting.
In the laboratory or chemical plant we work with much larger quantities than the few molecules of the preceding example. Therefore, we must learn to deal with limiting reactants using moles. The ideas are exactly the same, except that we are using moles of molecules instead of individual molecules. For example, suppose 25.0 kilograms of nitrogen and 5.00 kilograms of hydrogen are mixed and reacted to form ammonia. How do we calculate the mass of ammonia produced when this reaction is run to completion (until one of the reactants is completely consumed)?
As in the preceding example, we must use the balanced equation to determine whether nitrogen or hydrogen is the limiting reactant and then to determine the amount of ammonia that is formed. We first calculate the moles of reactants present: Since 1 mol N2 reacts with 3 mol H2, the number of moles of H2 that will react exactly with 8.93 102 mol N2 is Thus 8.93 102 mol N2 requires 2.68 103 mol H2 to react completely. However, in this case, only 2.48 103 mol H2 is present. This means that the hydrogen will be consumed before the nitrogen. Thus hydrogen is the limiting reactant in this particular situation, and we must use the amount of hydrogen to compute the quantity of ammonia formed: Converting moles to kilograms gives 1.65 103 mol NH3 17.0 g NH3 1 mol NH3 2.80 104 g NH3 28.0 kg NH3 2.48 103 mol H2 2 mol NH3 3 mol H2 1.65 103 mol NH3 8.93 102 mol N2 3 mol H2 1 mol N2 2.68 103 mol H2 5.00 kg H2 1000 g H2 1 kg H2 1 mol H2 2.016 g H2 2.48 103 mol H2 25.0 kg N2 1000 g N2 1 kg N2 1 mol N2 28.0 g N2 8.93 102 mol N2 N21g2 3H21g2 ¡ 2NH31g2 5N21g2 15H21g2 ¡ 10NH31g2 3N21g2 9H21g2 ¡ 6NH31g2 5N21g2 15H21g2 ¡ 10NH31g2 N21g2 3H21g2 ¡ 2NH31g2 Always determine which reactant is limiting.
110 Chapter Three Stoichiometry Note that to determine the limiting reactant, we could have started instead with the given amount of hydrogen and calculated the moles of nitrogen required: Thus 2.48 103 mol H2 requires 8.27 102 mol N2. Since 8.93 102 mol N2 is actu-ally present, the nitrogen is in excess. The hydrogen will run out first, and thus again we find that hydrogen limits the amount of ammonia formed.
A related but simpler way to determine which reactant is limiting is to compare the mole ratio of the substances required by the balanced equation with the mole ratio of reactants actually present. For example, in this case the mole ratio of H2 to N2 required by the balanced equation is That is, In this experiment we have 2.48 103 mol H2 and 8.93 102 mol N2. Thus the ratio Since 2.78 is less than 3, the actual mole ratio of H2 to N2 is too small, and H2 must be limiting. If the actual H2 to N2 mole ratio had been greater than 3, then the H2 would have been in excess and the N2 would be limiting.
Stoichiometry: Limiting Reactant Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor.
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limit-ing reactant? How many grams of N2 will be formed?
Solution From the description of the problem, we can obtain the following balanced equation: Next we must compute the moles of NH3 (molar mass 17.03 g/mol) and of CuO (molar mass 79.55 g/mol): To determine the limiting reactant, we use the mole ratio for CuO and NH3: Thus 1.59 mol CuO is required to react with 1.06 mol NH3. Since only 1.14 mol CuO is actually present, the amount of CuO is limiting; CuO will run out before NH3 does. We 1.06 mol NH3 3 mol CuO 2 mol NH3 1.59 mol CuO 90.4 g CuO 1 mol CuO 79.55 g CuO 1.14 mol CuO 18.1 g NH3 1 mol NH3 17.03 g NH3 1.06 mol NH3 2NH31g2 3CuO1s2 ¡ N21g2 3Cu1s2 3H2O1g2 mol H2 mol N2 1actual2 2.48 103 8.93 102 2.78 mol H2 mol N2 1required2 3 1 3 3 mol H2 1 mol N2 2.48 103 mol H2 1 mol N2 3 mol H2 8.27 102 mol N2 Sample Exercise 3.18 3.10 Calculations Involving a Limiting Reactant 111 can verify this conclusion by comparing the mole ratio of CuO and NH3 required by the balanced equation with the mole ratio actually present Since the actual ratio is too small (smaller than 1.5), CuO is the limiting reactant.
Because CuO is the limiting reactant, we must use the amount of CuO to calcu-late the amount of N2 formed. From the balanced equation, the mole ratio between CuO and N2 is Using the molar mass of N2 (28.0 g/mol), we can calculate the mass of N2 produced: See Exercises 3.99 through 3.101.
The amount of a product formed when the limiting reactant is completely consumed is called the theoretical yield of that product. In Sample Exercise 3.18, 10.6 grams of ni-trogen represents the theoretical yield. This is the maximum amount of nitrogen that can be produced from the quantities of reactants used. Actually, the amount of product pre-dicted by the theoretical yield is seldom obtained because of side reactions (other reac-tions that involve one or more of the reactants or products) and other complications. The actual yield of product is often given as a percentage of the theoretical yield. This is called the percent yield: For example, if the reaction considered in Sample Exercise 3.18 actually gave 6.63 grams of nitrogen instead of the predicted 10.6 grams, the percent yield of nitrogen would be Calculating Percent Yield Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is re-acted with 8.60 kg H2(g). Calculate the theoretical yield of methanol. If 3.57 104 g CH3OH is actually produced, what is the percent yield of methanol?
Solution First, we must find out which reactant is limiting. The balanced equation is 2H21g2 CO1g2 ¡ CH3OH1l2 6.63 g N2 10.6 g N2 100% 62.5% Actual yield Theoretical yield 100% percent yield 0.380 mol N2 28.0 g N2 1 mol N2 10.6 g N2 1.14 mol CuO 1 mol N2 3 mol CuO 0.380 mol N2 1 mol N2 3 mol CuO mol CuO mol NH3 1actual2 1.14 1.06 1.08 mol CuO mol NH3 1required2 3 2 1.5 Percent yield is important as an indicator of the efficiency of a particular laboratory or industrial reaction.
Sample Exercise 3.19 Methanol 112 Chapter Three Stoichiometry Next we must calculate the moles of reactants: To determine which reactant is limiting, we compare the mole ratio of H2 and CO re-quired by the balanced equation with the actual mole ratio Since the actual mole ratio of H2 to CO is smaller than the required ratio, H2 is limiting.
We therefore must use the amount of H2 and the mole ratio between H2 and CH3OH to determine the maximum amount of methanol that can be produced: Using the molar mass of CH3OH (32.04 g/mol), we can calculate the theoretical yield in grams: Thus, from the amount of reactants given, the maximum amount of CH3OH that can be formed is 6.86 104 g. This is the theoretical yield.
The percent yield is See Exercises 3.103 and 3.104.
Actual yield 1grams2 Theoretical yield 1grams2 100 3.57 104 g CH3OH 6.86 104 g CH3OH 100% 52.0% 2.14 103 mol CH3OH 32.04 g CH3OH 1 mol CH3OH 6.86 104 g CH3OH 4.27 103 mol H2 1 mol CH3OH 2 mol H2 2.14 103 mol CH3OH mol H2 mol CO 1actual2 4.27 103 2.44 103 1.75 mol H2 mol CO 1required2 2 1 2 8.60 kg H2 1000 g H2 1 kg H2 1 mol H2 2.016 g H2 4.27 103 mol H2 68.5 kg CO 1000 g CO 1 kg CO 1 mol CO 28.02 g CO 2.44 103 mol CO Methanol is used as a fuel in Indianapolis-type racing cars.
For Review 113 Solving a Stoichiometry Problem Involving Masses of Reactants and Products ➥ 1 Write and balance the equation for the reaction.
➥ 2 Convert the known masses of substances to moles.
➥ 3 Determine which reactant is limiting.
➥ 4 Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.
➥ 5 Convert from moles to grams, using the molar mass.
This process is summarized in the diagram below: Use mole ratio to convert Convert to moles Find appropriate mole ratio Find limiting reactant Masses of known substances Moles of known substances Convert to grams Moles of desired product Mass of desired product Moles limiting reactant Moles desired substance Moles limiting reactant Balanced chemical equation For Review Stoichiometry Deals with the amounts of substances consumed and/or produced in a chemical reaction.
We count atoms by measuring the mass of the sample.
To relate mass and the number of atoms, the average atomic mass is required.
Mole The amount of carbon atoms in exactly 12 g of pure 12C 6.022 1023 units of a substance The mass of one mole of an element the atomic mass in grams Molar mass Mass (g) of one mole of a compound or element Obtained for a compound by finding the sum of the average masses of its constituent atoms Key Terms chemical stoichiometry Section 3.2 mass spectrometer average atomic mass Section 3.3 mole Avogadro’s number Section 3.4 molar mass Section 3.5 mass percent Section 3.6 empirical formula molecular formula 114 Chapter Three Stoichiometry Percent composition The mass percent of each element in a compound Empirical formula The simplest whole-number ratio of the various types of atoms in a compound Can be obtained from the mass percent of elements in a compound Molecular formula For molecular substances: • The formula of the constituent molecules • Always an integer multiple of the empirical formula For ionic substances: • The same as the empirical formula Chemical reactions Reactants are turned into products.
Atoms are neither created nor destroyed.
All of the atoms present in the reactants must also be present in the products.
Characteristics of a chemical equation Represents a chemical reaction Reactants on the left side of the arrow, products on the right side When balanced, gives the relative numbers of reactant and product molecules or ions Stoichiometry calculations Amounts of reactants consumed and products formed can be determined from the balanced chemical equation.
The limiting reactant is the one consumed first, thus limiting the amount of prod-uct that can form.
Yield The theoretical yield is the maximum amount that can be produced from a given amount of the limiting reactant.
The actual yield, the amount of product actually obtained, is always less than the theoretical yield.
REVIEW QUESTIONS 1. Explain the concept of “counting by weighing” using marbles as your example.
2. Atomic masses are relative masses. What does this mean?
3. The atomic mass of boron (B) is given in the periodic table as 10.81, yet no single atom of boron has a mass of 10.81 amu. Explain.
4. What three conversion factors and in what order would you use them to convert the mass of a compound into atoms of a particular element in that compound— for example, from 1.00 g aspirin (C9H8O4) to number of hydrogen atoms in the 1.00-g sample?
5. Figure 3.5 illustrates a schematic diagram of a combustion device used to ana-lyze organic compounds. Given that a certain amount of a compound containing carbon, hydrogen, and oxygen is combusted in this device, explain how the data relating to the mass of CO2 produced and the mass of H2O produced can be manipulated to determine the empirical formula.
Percent yield actual yield 1g2 theoretical yield 1g2 100% Mass percent mass of element in 1 mole of substance mass of 1 mole of substance 100% Section 3.7 chemical equation reactants products balancing a chemical equation Section 3.9 mole ratio Section 3.10 stoichiometric quantities Haber process limiting reactant (reagent) theoretical yield percent yield Active Learning Questions 115 6. What is the difference between the empirical and molecular formulas of a compound? Can they ever be the same? Explain.
7. Consider the hypothetical reaction between A2 and AB pictured below.
What is the balanced equation? If 2.50 mol A2 is reacted with excess AB, what amount (moles) of product will form? If the mass of AB is 30.0 amu and the mass of A2 is 40.0 amu, what is the mass of the product? If 15.0 g of AB is reacted, what mass of A2 is required to react with all of the AB, and what mass of product is formed?
8. What is a limiting reactant problem? Explain two different strategies that can be used to solve limiting reactant problems.
9. Consider the following mixture of SO2(g) and O2(g).
If SO2(g) and O2(g) react to form SO3(g), draw a representation of the product mixture assuming the reaction goes to completion. What is the limiting reactant in the reaction? If 96.0 g of SO2 reacts with 32.0 g O2, what mass of product will form?
10. Why is the actual yield of a reaction often less than the theoretical yield?
O2 SO2 ?
A2B AB A2 Active Learning Questions These questions are designed to be used by groups of students in class.
The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. The following are actual student responses to the question: Why is it necessary to balance chemical equations?
a. The chemicals will not react until you have added the correct mole ratios.
b. The correct products will not be formed unless the right amount of reactants have been added.
c. A certain number of products cannot be formed without a cer-tain number of reactants.
d. The balanced equation tells you how much reactant you need and allows you to predict how much product you’ll make.
e. A mole-to-mole ratio must be established for the reaction to occur as written.
Justify the best choice, and for choices you did not pick, explain what is wrong with them.
2. What information do we get from a formula? From an equation?
116 Chapter Three Stoichiometry 3. You are making cookies and are missing a key ingredient— eggs. You have most of the other ingredients needed to make the cookies, except you have only 1.33 cups of butter and no eggs.
You note that the recipe calls for 2 cups of butter and 3 eggs (plus the other ingredients) to make 6 dozen cookies. You call a friend and have him bring you some eggs.
a. What number of eggs do you need?
b. If you use all the butter (and get enough eggs), what number of cookies will you make?
Unfortunately, your friend hangs up before you tell him how many eggs you need. When he arrives, he has a surprise for you— to save time, he has broken them all in a bowl for you. You ask him how many he brought, and he replies, “I can’t remember.” You weigh the eggs and find that they weigh 62.1 g. Assuming that an average egg weighs 34.21 g, a. What quantity of butter is needed to react with all the eggs?
b. What number of cookies can you make?
c. Which will you have left over, eggs or butter?
d. What quantity is left over?
4. Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3).
Consider the mixture of N2 ( ) and H2 ( ) in a closed container as illustrated below: Assuming the reaction goes to completion, draw a representa-tion of the product mixture. Explain how you arrived at this representation.
5. For the preceding question, which of the following equations best represents the reaction?
a.
b.
c.
d.
e.
Justify your choice, and for choices you did not pick, explain what is wrong with them.
6. You know that chemical A reacts with chemical B. You react 10.0 g A with 10.0 g B. What information do you need to determine the amount of product that will be produced?
Explain.
7. A new grill has a mass of 30.0 kg. You put 3.0 kg of charcoal in the grill. You burn all the charcoal and the grill has a mass of 30.0 kg. What is the mass of the gases given off?
Explain.
2N2 6H2 ¡ 4NH3 N2 3H2 ¡ 2NH3 N 3H ¡ NH3 N2 H2 ¡ NH3 6N2 6H2 ¡ 4NH3 4N2 8. Consider an iron bar on a balance as shown.
As the iron bar rusts, which of the following is true? Explain your answer.
a. The balance will read less than 75.0 g.
b. The balance will read 75.0 g.
c. The balance will read greater than 75.0 g.
d. The balance will read greater than 75.0 g, but if the bar is removed, the rust is scraped off, and the bar replaced, the balance will read 75.0 g.
9. You may have noticed that water sometimes drips from the exhaust of a car as it is running. Is this evidence that there is at least a small amount of water originally present in the gasoline? Explain.
Questions 10 and 11 deal with the following situation: You react chemical A with chemical B to make one product. It takes 100 g of A to react completely with 20 g B.
10. What is the mass of the product?
a. less than 10 g b. between 20 and 100 g c. between 100 and 120 g d. exactly 120 g e. more than 120 g 11. What is true about the chemical properties of the product?
a. The properties are more like chemical A.
b. The properties are more like chemical B.
c. The properties are an average of those of chemical A and chemical B.
d. The properties are not necessarily like either chemical A or chemical B.
e. The properties are more like chemical A or more like chem-ical B, but more information is needed.
Justify your choice, and for choices you did not pick, explain what is wrong with them.
12. Is there a difference between a homogeneous mixture of hydrogen and oxygen in a 2:1 mole ratio and a sample of water vapor?
Explain.
13. Chlorine exists mainly as two isotopes, 37Cl and 35Cl. Which is more abundant? How do you know?
14. The average mass of a carbon atom is 12.011. Assuming you could pick up one carbon atom, estimate the chance that you would ran-domly get one with a mass of 12.011. Support your answer.
15. Can the subscripts in a chemical formula be fractions? Explain.
Can the coefficients in a balanced chemical equation be frac-tions? Explain. Changing the subscripts of chemicals can bal-ance the equations mathematically. Why is this unacceptable?
16. Consider the equation If you mix 1.0 mol of A with 1.0 mol of B, what amount (moles) of A2B can be produced?
2A B ¡ A2B.
75.0g Exercises 117 17. According to the law of conservation of mass, mass cannot be gained or destroyed in a chemical reaction. Why can’t you sim-ply add the masses of two reactants to determine the total mass of product?
18. Which of the following pairs of compounds have the same empirical formula?
a. acetylene, C2H2, and benzene, C6H6 b. ethane, C2H6, and butane, C4H10 c. nitrogen dioxide, NO2, and dinitrogen tetroxide, N2O4 d. diphenyl ether, C12H10O, and phenol, C6H5OH A blue question or exercise number indicates that the answer to that question or exercise appears at the back of the book and a solution appears in the Solutions Guide.
Questions 19. Reference section 3.2 to find the atomic masses of 12C and 13C, the relative abundance of 12C and 13C in natural carbon, and the average mass (in amu) of a carbon atom. If you had a sample of natural carbon containing exactly 10,000 atoms, determine the number of 12C and 13C atoms present. What would be the aver-age mass (in amu) and the total mass (in amu) of the carbon atoms in this 10,000-atom sample? If you had a sample of nat-ural carbon containing 6.0221 1023 atoms, determine the num-ber of 12C and 13C atoms present. What would be the average mass (in amu) and the total mass (in amu) of this 6.0221 1023 atom sample? Given that 1 g 6.0221 1023 amu, what is the total mass of 1 mol of natural carbon in units of grams?
20. Avogadro’s number, molar mass, and the chemical formula of a compound are three useful conversion factors. What unit con-versions can be accomplished using these conversion factors?
21. If you had a mol of U.S. dollar bills and equally distributed the money to all of the people of the world, how rich would every person be? Assume a world population of 6 billion.
22. What is the difference between the molar mass and the empiri-cal formula mass of a compound? When are these masses the same and when are they different? When different, how is the molar mass related to the empirical formula mass?
23. How is the mass percent of elements in a compound different for a 1.0-g sample versus a 100.-g sample versus a 1-mol sam-ple of the compound?
24. A balanced chemical equation contains a large amount of infor-mation. What information is given in a balanced equation?
25. Consider the following generic reaction: What steps and information are necessary to perform the fol-lowing determinations assuming that 1.00 104 molecules of A2B2 are reacted with excess C?
a. mass of CB produced b. atoms of A produced c. mol of C reacted d. percent yield of CB 26. Consider the following generic reaction: Y2 2XY ¡ 2XY2 A2B2 2C ¡ 2CB and 2A In a limiting reactant problem, a certain quantity of each reactant is given and you are usually asked to calculate the mass of prod-uct formed. If 10.0 g of Y2 is reacted with 10.0 g of XY, outline two methods you could use the determine which reactant is lim-iting (runs out first) and thus determines the mass of product formed. A method sometimes used to solve limiting reactant prob-lems is to assume each reactant is limiting and then calculate the mass of product formed from each given quantity of reactant. How does this method work in determining which reactant is limiting?
Exercises In this section similar exercises are paired.
Atomic Masses and the Mass Spectrometer 27. An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.9759 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass and identify the element.
28. An element “X” has five major isotopes, which are listed below along with their abundances. What is the element?
29. The element rhenium (Re) has two naturally occurring isotopes, 185Re and 187Re, with an average atomic mass of 186.207 amu.
Rhenium is 62.60% 187Re, and the atomic mass of 187Re is 186.956 amu. Calculate the mass of 185Re.
30. Assume silicon has three major isotopes in nature as shown in the table below. Fill in the missing information.
31. The mass spectrum of bromine (Br2) consists of three peaks with the following characteristics: How do you interpret these data?
Mass (amu) Relative Size 157.84 0.2534 159.84 0.5000 161.84 0.2466 Isotope Mass (amu) Abundance 28Si 27.98 ——––— 29Si ——––— 4.70% 32Si 29.97 3.09% Isotope Percent Natural Abundance Atomic Mass 46X 8.00% 45.95269 47X 7.30% 46.951764 48X 73.80% 47.947947 49X 5.50% 48.947841 50X 5.40% 49.944792 118 Chapter Three Stoichiometry 32. Gallium arsenide, GaAs, has gained widespread use in semi-conductor devices that convert light and electrical signals in fiber-optic communications systems. Gallium consists of 60.% 69Ga and 40.% 71Ga. Arsenic has only one naturally occurring isotope, 75As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and Ga2As2.
What would the distribution of peaks look like for these two fragments?
Moles and Molar Masses 33. Calculate the mass of 500. atoms of iron (Fe).
34. What number of Fe atoms and what amount (moles) of Fe atoms are in 500.0 g of iron?
35. Diamond is a natural form of pure carbon. What number of atoms of carbon are in a 1.00-carat diamond (1.00 carat 0.200 g)?
36. A diamond contains 5.0 1021 atoms of carbon. What amount (moles) of carbon and what mass (grams) of carbon are in this diamond?
37. Aluminum metal is produced by passing an electric current through a solution of aluminum oxide (Al2O3) dissolved in molten cryolite (Na3AlF6). Calculate the molar masses of Al2O3 and Na3AlF6.
38. The Freons are a class of compounds containing carbon, chlo-rine, and fluorine. While they have many valuable uses, they have been shown to be responsible for depletion of the ozone in the upper atmosphere. In 1991, two replacement compounds for Freons went into production: HFC-134a (CH2FCF3) and HCFC-124 (CHClFCF3). Calculate the molar masses of these two compounds.
39. Calculate the molar mass of the following substances.
a.
b.
c. (NH4)2Cr2O7 40. Calculate the molar mass of the following substances.
a.
b. Ca3(PO4)2 c. Na2HPO4 41. What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 39?
42. What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 40?
43. What mass of compound is present in 5.00 mol of each of the compounds in Exercise 39?
44. What mass of compound is present in 5.00 mol of each of the compounds in Exercise 40?
P O N H H N 45. What mass of nitrogen is present in 5.00 mol of each of the com-pounds in Exercise 39?
46. What mass of phosphorus is present in 5.00 mol of each of the compounds in Exercise 40?
47. What number of molecules (or formula units) are present in 1.00 g of each of the compounds in Exercise 39?
48. What number of molecules (or formula units) are present in 1.00 g of each of the compounds in Exercise 40?
49. What number of atoms of nitrogen are present in 1.00 g of each of the compounds in Exercise 39?
50. What number of atoms of phosphorus are present in 1.00 g of each of the compounds in Exercise 40?
51. Ascorbic acid, or vitamin C (C6H8O6), is an essential vitamin. It cannot be stored by the body and must be present in the diet.
What is the molar mass of ascorbic acid? Vitamin C tablets are taken as a dietary supplement. If a typical tablet contains 500.0 mg of vitamin C, what amount (moles) and what number of molecules of vitamin C does it contain?
52. The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is C9H8O4.
a. Calculate the molar mass of aspirin.
b. A typical aspirin tablet contains 500. mg of C9H8O4. What amount (moles) of C9H8O4 molecules and what number of molecules of acetylsalicylic acid are in a 500.-mg tablet?
53. What amount (moles) are represented by each of these samples?
a. 150.0 g Fe2O3 c. 1.5 1016 molecules of BF3 b. 10.0 mg NO2 54. What amount (moles) is represented by each of these samples?
a. 20.0 mg caffeine, C8H10N4O2 b. 2.72 1021 molecules of ethanol, C2H5OH c. 1.50 g of dry ice, CO2 55. What number of atoms of nitrogen are present in 5.00 g of each of the following?
a. glycine, C2H5O2N c. calcium nitrate b. magnesium nitride d. dinitrogen tetroxide 56. Complete the following table.
57. Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed Total Mass of Moles of Molecules Atoms in Sample Sample in Sample Sample 4.24 g C6H6 ——––— ——––— ——––— ——––— 0.224 mol H2O ——––— ——––— ——––— ——––— 2.71 1022 ——––— molecules CO2 ——––— ——––— ——––— 3.35 1022 total atoms in CH3OH sample Exercises 119 as Nutra-Sweet. The molecular formula of aspartame is C14H18N2O5.
a. Calculate the molar mass of aspartame.
b. What amount (moles) of molecules are present in 10.0 g as-partame?
c. Calculate the mass in grams of 1.56 mol aspartame.
d. What number of molecules are in 5.0 mg aspartame?
e. What number of atoms of nitrogen are in 1.2 g aspartame?
f. What is the mass in grams of 1.0 109 molecules of aspartame?
g. What is the mass in grams of one molecule of aspartame?
58. Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a seda-tive and hypnotic. It is the compound used to make “Mickey Finns” in detective stories.
a. Calculate the molar mass of chloral hydrate.
b. What amount (moles) of C2H3Cl3O2 molecules are in 500.0 g chloral hydrate?
c. What is the mass in grams of 2.0 102 mol chloral hydrate?
d. What number of chlorine atoms are in 5.0 g chloral hydrate?
e. What mass of chloral hydrate would contain 1.0 g Cl?
f. What is the mass of exactly 500 molecules of chloral hydrate?
Percent Composition 59. Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers: a. C3H4O2 (acrylic acid, from which acrylic plastics are made) b. C4H6O2 (methyl acrylate, from which Plexiglas is made) c. C3H3N (acrylonitrile, from which Orlon is made) 60. Anabolic steroids are performance enhancement drugs whose use has been banned from most major sporting activities. One ana-bolic steroid is fluoxymesterone (C20H29FO3). Calculate the per-cent composition by mass of fluoxymesterone.
61. Several important compounds contain only nitrogen and oxygen.
Place the following compounds in order of increasing mass per-cent of nitrogen.
a. NO, a gas formed by the reaction of N2 with O2 in internal combustion engines b. NO2, a brown gas mainly responsible for the brownish color of photochemical smog c. N2O4, a colorless liquid used as fuel in space shuttles d. N2O, a colorless gas sometimes used as an anesthetic by den-tists (known as laughing gas) 62. Arrange the following substances in order of increasing mass percent of carbon.
a. caffeine, C8H10N4O2 b. sucrose, C12H22O11 c. ethanol, C2H5OH 63. Vitamin B12, cyanocobalamin, is essential for human nutrition.
It is concentrated in animal tissue but not in higher plants.
Although nutritional requirements for the vitamin are quite low, people who abstain completely from animal products may develop a deficiency anemia. Cyanocobalamin is the form used in vitamin supplements. It contains 4.34% cobalt by mass.
Calculate the molar mass of cyanocobalamin, assuming that there is one atom of cobalt in every molecule of cyanocobalamin.
64. Fungal laccase, a blue protein found in wood-rotting fungi, is 0.390% Cu by mass. If a fungal laccase molecule contains 4 cop-per atoms, what is the molar mass of fungal laccase?
Empirical and Molecular Formulas 65. Express the composition of each of the following compounds as the mass percents of its elements.
a. formaldehyde, CH2O b. glucose, C6H12O6 c. acetic acid, HC2H3O2 66. Considering your answer to Exercise 65, which type of formula, empirical or molecular, can be obtained from elemental analy-sis that gives percent composition?
67. Give the empirical formula for each of the compounds repre-sented below.
68. Determine the molecular formulas to which the following empirical formulas and molar masses pertain.
a. SNH (188.35 g/mol) b. NPCl2 (347.64 g/mol) c. CoC4O4 (341.94 g/mol) d. SN (184.32 g/mol) 69. The compound adrenaline contains 56.79% C, 6.56% H, 28.37% O, and 8.28% N by mass. What is the empirical formula for adrenaline?
70. The most common form of nylon (nylon-6) is 63.68% carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.14% oxygen. Calcu-late the empirical formula for nylon-6.
71. There are two binary compounds of mercury and oxygen. Heat-ing either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating 0.6498 g of one of the com-pounds leaves a residue of 0.6018 g. Heating 0.4172 g of the other compound results in a mass loss of 0.016 g. Determine the empirical formula of each compound.
a.
c.
d.
b.
O N C H P 120 Chapter Three Stoichiometry 72. A sample of urea contains 1.121 g N, 0.161 g H, 0.480 g C, and 0.640 g O. What is the empirical formula of urea?
73. A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?
74. Determine the molecular formula of a compound that contains 26.7% P, 12.1% N, and 61.2% Cl, and has a molar mass of 580 g/mol.
75. Adipic acid is an organic compound composed of 49.31% C, 43.79% O, and the rest hydrogen. If the molar mass of adipic acid is 146.1 g/mol, what are the empirical and molecular for-mulas for adipic acid?
76. Maleic acid is an organic compound composed of 41.39% C, 3.47% H, and the rest oxygen. If 0.129 mol of maleic acid has a mass of 15.0 g, what are the empirical and molecular formulas of maleic acid?
77. Many homes in rural America are heated by propane gas, a compound that contains only carbon and hydrogen. Complete combustion of a sample of propane produced 2.641 g of carbon dioxide and 1.442 g of water as the only products. Find the em-pirical formula of propane.
78. A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?
79. Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chem-ical industry. Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water. The molar mass of cumene is between 115 and 125 g/mol. Determine the empirical and molecular formulas.
80. A compound contains only carbon, hydrogen, and oxygen.
Combustion of 10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?
Balancing Chemical Equations 81. Give the balanced equation for each of the following chemical reactions: a. Glucose (C6H12O6) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor.
b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.
c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate (NH4SCN).
82. Give the balanced equation for each of the following.
a. The combustion of ethanol (C2H5OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas.
b. Aqueous solutions of lead(II) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of lead(II) phosphate with aqueous sodium nitrate as the other product.
c. Solid zinc reacts with aqueous HCl to form aqueous zinc chloride and hydrogen gas.
d. Aqueous strontium hydroxide reacts with aqueous hydro-bromic acid to produce water and aqueous strontium bromide.
83. Balance the following equations: a.
b.
c.
84. Balance each of the following chemical equations.
a.
b.
c.
d.
e.
f.
g.
85. Balance the following equations representing combustion reactions: a.
b.
c.
d.
e.
86. Balance the following equations: a.
b.
c.
d.
87. Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction.
a.
b. Silicon tetrachloride is reacted with very pure magnesium, producing silicon and magnesium chloride.
c.
88. Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, CaSiO3. Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium sil-icate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often gradu-ated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate.
CaSiO31s2 HF1aq2 S CaF21aq2 SiF41g2 H2O1l2 Na2SiF61s2 Na1s2 S Si1s2 NaF1s2 Si1s2 CO1g2 — —¡ Electric arc furnace SiO21s2 C1s2 Eu1s2 HF1g2 S EuF31s2 H21g2 KClO31s2 ¡ Heat KCl1s2 O21g2 NaHCO31s2 ¡ Heat Na2CO31s2 CO21g2 H2O1g2 Cr1s2 S81s2 S Cr2S31s2 FeO1s2 O21g2 S Fe2O31s2 Fe1s2 O21g2 S Fe2O31s2 C12H22O111s2 O21g2 S CO21g2 H2O1g2 + + (g) (g) (g) (g) + + (g) (l) (g) (g) H O C FeCO31s2 H2CO31aq2 SFe1HCO3221aq2 MoS21s2 O21g2 SMoO31s2 SO21g2 CaO1s2 C1s2 SCaC21s2 CO21g2 PCl51l2 H2O1l2 SH3PO41aq2 HCl1g2 NH31g2 O21g2 SNO1g2 H2O1g2 Fe2O31s2 HNO31aq2 SFe1NO3231aq2 H2O1l2 KO21s2 H2O1l2 SKOH1aq2 O21g2 H2O21aq2 AgNO31aq2 H2SO41aq2 SAg2SO41s2 HNO31aq2 Al1OH231s2 HCl1aq2 S AlCl31aq2 H2O1l2 Ca1OH221aq2 H3PO41aq2 S H2O1l2 Ca31PO4221s2 Exercises 121 Reaction Stoichiometry 89. Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is What masses of iron(III) oxide and aluminum must be used to produce 15.0 g iron? What is the maximum mass of aluminum oxide that could be produced?
90. The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate (KClO3) with excess red phosphorus, what mass of tetraphosphorus decaoxide (P4O10) would be produced?
91. The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is What mass of NH4ClO4 should be used in the fuel mixture for every kilogram of Al?
92. One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(OH)2 indicates the presence of eight water molecules. This compound is called bar-ium hydroxide octahydrate.
a. Balance the equation.
b. What mass of ammonium thiocyanate (NH4SCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?
93. Bacterial digestion is an economical method of sewage treat-ment. The reaction bacterial tissue is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. What mass of bacterial tissue is pro-duced in a treatment plant for every 1.0 104 kg of wastewater containing 3.0% NH4 ions by mass? Assume that 95% of the ammonium ions are consumed by the bacteria.
94. Phosphorus can be prepared from calcium phosphate by the fol-lowing reaction: Phosphorite is a mineral that contains Ca3(PO4)2 plus other non-phosphorus-containing compounds. What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phorphorite sample is 75% Ca3(PO4)2 by mass? Assume an excess of the other reactants.
6CaSiO31s2 P41s2 10CO1g2 2Ca31PO4221s2 6SiO21s2 10C1s2 ¡ C5H7O2N1s2 54NO2 1aq2 52H2O1l2 109H1aq2 5CO21g2 55NH4 1aq2 76O21g2 — ¡ bacteria 8H2O 8H2O Ba1SCN221s2 H2O1l2 NH31g2 Ba1OH22 8H2O1s2 NH4SCN1s2 ¡ Al2O31s2 AlCl31s2 3NO1g2 6H2O1g2 3Al1s2 3NH4ClO41s2 ¡ KClO31s2 P41s2 ¡ P4O101s2 KCl1s2 1unbalanced2 Fe2O31s2 2Al1s2 ¡ 2Fe1l2 Al2O31s2 95. Aspirin (C9H8O4) is synthesized by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3). The balanced equation is a. What mass of acetic anhydride is needed to completely con-sume 1.00 102 g salicylic acid?
b. What is the maximum mass of aspirin (the theoretical yield) that could be produced in this reaction?
96. The space shuttle environmental control system handles excess CO2 (which the astronauts breathe out; it is 4.0% by mass of ex-haled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, Li2CO3, and water. If there are 7 as-tronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 25,000 g of LiOH pellets available for each shuttle mission?
Assume the density of air is 0.0010 g mL.
Limiting Reactants and Percent Yield 97. Consider the reaction between NO(g) and O2(g) represented below.
What is the balanced equation for this reaction and what is the limiting reactant?
98. Consider the following reaction: If a container were to have 10 molecules of O2 and 10 molecules of NH3 initially, how many total molecules (reactants plus prod-ucts) would be present in the container after this reaction goes to completion?
99. Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes upon contact with blood, releasing elemental oxygen gas (which in-hibits the growth of anaerobic microorganisms); and it foams upon contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be pre-pared by the action of an acid on an alkaline earth metal perox-ide, such as barium peroxide: What mass of hydrogen peroxide should result when 1.50 g of barium peroxide is treated with 25.0 mL of hydrochloric acid solution containing 0.0272 g of HCl per mL? What mass of which reagent is left unreacted?
100. Consider the following unbalanced equation: Ca31PO4221s2 H2SO41aq2 ¡ CaSO41s2 H3PO41aq2 BaO21s2 2HCl1aq2 ¡ H2O21aq2 BaCl21aq2 4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 C7H6O3 C4H6O3 ¡ C9H8O4 HC2H3O2 NO NO2 O2 122 Chapter Three Stoichiometry of 1000. kg/h. What mass of water must be evaporated per hour if the final product contains only 20.% water?
109. Consider the reaction Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of H2 and 25 molecules of O2 b. 100 molecules of H2 and 40 molecules of O2 c. 100 molecules of H2 and 100 molecules of O2 d. 0.50 mol H2 and 0.75 mol O2 e. 0.80 mol H2 and 0.75 mol O2 f. 1.0 g H2 and 0.25 mol O2 g. 5.00 g H2 and 56.00 g O2 110. Some bismuth tablets, a medication used to treat upset stomachs, contain 262 mg of bismuth subsalicylate, C7H5BiO4, per tablet.
Assuming two tablets are digested, calculate the mass of bismuth consumed.
111. The empirical formula of styrene is CH; the molar mass of styrene is 104.14 g/mol. What number of H atoms are present in a 2.00-g sample of styrene?
112. Terephthalic acid is an important chemical used in the manufac-ture of polyesters and plasticizers. It contains only C, H, and O.
Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO2 and 6.45 mg H2O. If 0.250 mol of terephthalic acid has a mass of 41.5 g, determine the molecular formula for terephthalic acid.
113. A sample of a hydrocarbon (a compound consisting of only car-bon and hydrogen) contains 2.59 1023 atoms of hydrogen and is 17.3% hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound are present, and what is the mass of the sample?
114. A binary compound between an unknown element E and hy-drogen contains 91.27% E and 8.73% H by mass. If the formula of the compound is E3H8, calculate the atomic mass of E.
115. A 0.755-g sample of hydrated copper(II) sulfate was heated carefully until it had changed completely to anhy-drous copper(II) sulfate (CuSO4) with a mass of 0.483 g. De-termine the value of x. [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the num-ber of water molecules per formula unit of CuSO4 in the hydrated crystal.] 116. ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile (C3H3N), butadiene (C4H6), and styrene (C8H8).
a. A sample of ABS plastic contains 8.80% N by mass. It took 0.605 g of Br2 to react completely with a 1.20-g sample of ABS plastic. Bromine reacts 1:1 (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer?
b. What are the relative numbers of each of the monomer units in this polymer?
117. A sample of LSD (D-lysergic acid diethylamide, C24H30N3O) is added to some table salt (sodium chloride) to form a mixture.
Given that a 1.00-g sample of the mixture undergoes combustion CuSO4 xH2O 2H21g2 O21g2 ¡ 2H2O1g2 What masses of calcium sulfate and phosphoric acid can be pro-duced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?
101. Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: If 5.00 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
102. Acrylonitrile (C3H3N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction.
If 15.0 g C3H6, 10.0 g O2, and 5.00 g NH3 are reacted, what mass of acrylonitrile can be produced, assuming 100% yield?
103. A student prepared aspirin in a laboratory experiment using the reaction in Exercise 95. The student reacted 1.50 g salicylic acid with 2.00 g acetic anhydride. The yield was 1.50 g aspirin.
Calculate the theoretical yield and the percent yield for this experiment.
104. DDT, an insecticide harmful to fish, birds, and humans, is pro-duced by the following reaction: chlorobenzene chloral DDT In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral.
a. What mass of DDT is formed?
b. Which reactant is limiting? Which is in excess?
c. What mass of the excess reactant is left over?
d. If the actual yield of DDT is 200.0 g, what is the percent yield?
105. Bornite (Cu3FeS3) is a copper ore used in the production of cop-per. When heated, the following reaction occurs: If 2.50 metric tons of bornite is reacted with excess O2 and the process has an 86.3% yield of copper, what mass of copper is produced?
106. Consider the following unbalanced reaction: What mass of F2 is needed to produce 120. g of PF3 if the re-action has a 78.1% yield?
Additional Exercises 107. A given sample of a xenon fluoride compound contains mole-cules of the type XeFn, where n is some whole number. Given that 9.03 1020 molecules of XeFn weighs 0.368 g, determine the value for n in the formula.
108. Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. A ce-real product containing 58% H2O by mass is produced at the rate P41s2 F21g2 ¡ PF31g2 2Cu3FeS31s2 7O21g2 ¡ 6Cu1s2 2FeO1s2 6SO21g2 2C6H5Cl C2HOCl3 ¡ C14H9Cl5 H2O 2C3H61g2 2NH31g2 3O21g2 ¡ 2C3H3N1g2 6H2O1g2 2NH31g2 3O21g2 2CH41g2 ¡ 2HCN1g2 6H2O1g2 Challenge Problems 123 empirical formula of the compound? Hint: Combustion involves reacting with excess O2. Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O. Also assume that all the nitrogen ends up in the NH3 in the second experiment.
126. Nitric acid is produced commercially by the Ostwald process, represented by the following equations: What mass of NH3 must be used to produce 1.0 106 kg HNO3 by the Ostwald process? Assume 100% yield in each reaction and assume that the NO produced in the third step is not recycled.
127. Consider a 5.430-g mixture of FeO and Fe3O4. You react this mixture with an excess of oxygen to form 5.779 g Fe2O3. Cal-culate the percent by mass of FeO in the original mixture.
128. A 9.780-g gaseous mixture contains ethane (C2H6) and propane (C3H8). Complete combustion to form carbon dioxide and water requires 1.120 mol of oxygen. Calculate the mass percent of ethane in the original mixture.
129. Zinc and magnesium metal each react with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas.
A 10.00-g mixture of zinc and magnesium produces 0.5171 g of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.
130. A 2.077-g sample of an element, which has an atomic mass between 40 and 55, reacts with oxygen to form 3.708 g of an ox-ide. Determine the formula of the oxide (and identify the element).
131. Consider a gaseous binary compound with a molar mass of 62.09 g mol. When 1.39 g of this compound is completely burned in excess oxygen, 1.21 g of water is formed. Determine the formula of the compound. Assume water is the only product that con-tains hydrogen.
132. A 2.25-g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?
133. In the production of printed circuit boards for the electronics in-dustry, a 0.60-mm layer of copper is laminated onto an insulat-ing plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is A plant needs to manufacture 10,000 printed circuit boards, each 8.0 16.0 cm in area. An average of 80.% of the copper is re-moved from each board (density of copper 8.96 g/cm3). What masses of Cu(NH3)4Cl2 and NH3 are needed to do this? Assume 100% yield.
134. The aspirin substitute, acetaminophen (C8H9O2N), is produced by the following three-step synthesis: I.
C6H8ONCl1s2 2H2O1l2 C6H5O3N1s2 3H21g2 HCl1aq2 ¡ Cu1NH324Cl21aq2 4NH31aq2 Cu1s2 ¡ 2Cu1NH324Cl1aq2 3NO21g2 H2O1l2 ¡ 2HNO31aq2 NO1g2 2NO1g2 O21g2 ¡ 2NO21g2 4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 to produce 1.20 g of CO2, what is the mass percentage of LSD in the mixture?
118. Methane (CH4) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products.
a. Write the balanced chemical equation for the reaction of methane and sulfur.
b. Calculate the theoretical yield of carbon disulfide when 120. g of methane is reacted with an equal mass of sulfur.
119. A potential fuel for rockets is a combination of B5H9 and O2.
The two react according to the following balanced equation: If one tank in a rocket holds 126 g of B5H9 and another tank holds 192 g of O2, what mass of water can be produced when the entire contents of each tank react together?
120. Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If 25.0 g of Ag2O is reacted with 50.0 g of C10H10N4SO2, what mass of silver sulfadiazine, AgC10H9N4SO2, can be produced, assuming 100% yield?
121. An iron ore sample contains Fe2O3 plus other impurities. A 752-g sample of impure iron ore is heated with excess carbon, producing 453 g of pure iron by the following reaction: What is the mass percent of Fe2O3 in the impure iron ore sam-ple? Assume that Fe2O3 is the only source of iron and that the reaction is 100% efficient.
122. Commercial brass, an alloy of Zn and Cu, reacts with hydro-chloric acid as follows: (Cu does not react with HCl.) When 0.5065 g of a certain brass alloy is reacted with excess HCl, 0.0985 g ZnCl2 is eventually isolated.
a. What is the composition of the brass by mass?
b. How could this result be checked without changing the above procedure?
123. Vitamin A has a molar mass of 286.4 g/mol and a general molecular formula of CxHyE, where E is an unknown element.
If vitamin A is 83.86% C and 10.56% H by mass, what is the molecular formula of vitamin A?
Challenge Problems 124. Natural rubidium has the average mass of 85.4678 and is com-posed of isotopes 85Rb (mass 84.9117) and 87Rb. The ratio of atoms 85Rb/87Rb in natural rubidium is 2.591. Calculate the mass of 87Rb.
125. A compound contains only carbon, hydrogen, nitrogen, and oxy-gen. Combustion of 0.157 g of the compound produced 0.213 g CO2 and 0.0310 g H2O. In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3. What is the Zn1s2 2HCl1aq2 ¡ ZnCl21aq2 H21g2 Fe2O31s2 3C1s2 ¡ 2Fe1s2 3CO1g2 Ag2O1s2 2C10H10N4SO21s2 ¡ 2AgC10H9N4SO21s2 H2O1l2 2B5H91l2 12O21g2 ¡ 5B2O31s2 9H2O1g2 124 Chapter Three Stoichiometry b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface?
c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part b is used, what number of ruthenium atoms is needed to construct the surface?
142. Tetrodotoxin is a toxic chemical found in fugu pufferfish, a pop-ular but rare delicacy in Japan. This compound has a LD50 (the amount of substance that is lethal to 50.% of a population sample) of 10. g per kg of body mass. Tetrodotoxin is 41.38% carbon by mass, 13.16% nitrogen by mass, and 5.37% hydrogen by mass, with the remaining amount consisting of oxygen. What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin has a mass of 1.59 1021 g, what is the molecu-lar formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the LD50 dosage for a person weighing 165 lb?
143. An ionic compound MX3 is prepared according to the following unbalanced chemical equation.
A 0.105-g sample of X2 contains 8.92 1020 molecules. The compound MX3 consists of 54.47% X by mass. What are the identities of M and X, and what is the correct name for MX3?
Starting with 1.00 g each of M and X2, what mass of MX3 can be prepared?
144. The compound As2I4 is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic (d 5.72 g/cm3) that is 3.00 cm on edge is allowed to react with 1.01 1024 molecules of arsenic triiodide, how much As2I4 can be prepared? If the percent yield of As2I4 was 75.6%, what mass of As2I4 was actually isolated?
Marathon Problems These problems are designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
145. From the information below, determine the mass of substance C that will be formed if 45.0 grams of substance A reacts with 23.0 grams of substance B. (Assume that the reaction between A and B goes to completion.) a. Substance A is a gray solid that consists of an alkaline earth metal and carbon (37.5% by mass). It reacts with substance B to produce substances C and D. Forty million trillion formula units of A have a mass of 4.26 milligrams.
b. 47.9 grams of substance B contains 5.36 grams of hydrogen and 42.5 grams of oxygen.
c. When 10.0 grams of C is burned in excess oxygen, 33.8 grams of carbon dioxide and 6.92 grams of water are produced. A mass spectrum of substance C shows a parent molecular ion with a mass-to-charge ratio of 26.
d. Substance D is the hydroxide of the metal in substance A.
M X2 ¡ MX3 II.
III.
The first two reactions have percent yields of 87% and 98% by mass, respectively. The overall reaction yields 3 mol of aceta-minophen product for every 4 mol of C6H5O3N reacted.
a. What is the percent yield by mass for the overall process?
b. What is the percent yield by mass of step III?
135. An element X forms both a dichloride (XCl2) and a tetrachloride (XCl4). Treatment of 10.00 g XCl2 with excess chlorine forms 12.55 g XCl4. Calculate the atomic mass of X, and identify X.
136. When M2S3(s) is heated in air, it is converted to MO2(s).
A 4.000-g sample of M2S3(s) shows a decrease in mass of 0.277 g when it is heated in air. What is the average atomic mass of M?
137. When aluminum metal is heated with an element from Group 6A of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group 6A element, the product is 18.56% Al by mass. What is the formula of the compound?
138. A sample of a mixture containing only sodium chloride and potassium chloride has a mass of 4.000 g. When this sample is dissolved in water and excess silver nitrate is added, a white solid (silver chloride) forms. After filtration and drying, the solid sil-ver chloride has the mass 8.5904 g. Calculate the mass percent of each mixture component.
139. Ammonia reacts with O2 to form either NO(g) or NO2(g) ac-cording to these unbalanced equations: In a certain experiment 2.00 mol of NH3(g) and 10.00 mol of O2(g) are contained in a closed flask. After the reaction is com-plete, 6.75 mol of O2(g) remains. Calculate the number of moles of NO(g) in the product mixture: (Hint:You cannot do this prob-lem by adding the balanced equations, because you cannot as-sume that the two reactions will occur with equal probability.) 140. You take 1.00 g of an aspirin tablet (a compound consisting solely of carbon, hydrogen, and oxygen), burn it in air, and collect 2.20 g CO2 and 0.400 g H2O. You know that the molar mass of aspirin is between 170 and 190 g/mol. Reacting 1 mole of sali-cylic acid with 1 mole of acetic anhydride (C4H6O3) gives you 1 mole of aspirin and 1 mole of acetic acid (C2H4O2). Use this information to determine the molecular formula of salicylic acid.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
141. With the advent of techniques such as scanning tunneling mi-croscopy, it is now possible to “write” with individual atoms by manipulating and arranging atoms on an atomic surface.
a. If an image is prepared by manipulating iron atoms and their total mass is 1.05 1020 g, what number of iron atoms were used?
NH31g2 O21g2 ¡ NO21g2 H2O1g2 NH31g2 O21g2 ¡ NO1g2 H2O1g2 C8H9O2N1s2 HC2H3O21l2 C6H7ON1s2 C4H6O31l2 ¡ C6H7ON1s2 H2O1l2 NaCl1aq2 C6H8ONCl1s2 NaOH1aq2 ¡ Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
Marathon Problems 125 146. Consider the following balanced chemical equation: a. Equal masses of A and B are reacted. Complete each of the following with either “A is the limiting reactant because ”; “B is the limiting reactant because ”; or “we cannot determine the limiting reactant because ”.
i. If the molar mass of A is greater than the molar mass of B, then ii. If the molar mass of B is greater than the molar mass of A, then A 5B ¡ 3C 4D b. The products of the reaction are carbon dioxide (C) and wa-ter (D). Compound A has the same molar mass as carbon diox-ide. Compound B is a diatomic molecule. Identify compound B and support your answer.
c. Compound A is a hydrocarbon that is 81.71% carbon by mass.
Determine its empirical and molecular formulas.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
126 4 Types of Chemical Reactions and Solution Stoichiometry Contents 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes • Strong Electrolytes • Weak Electrolytes • Nonelectrolytes 4.3 The Composition of Solutions • Dilution 4.4 Types of Chemical Reactions 4.5 Precipitation Reactions 4.6 Describing Reactions in Solution 4.7 Stoichiometry of Precipitation Reactions 4.8 Acid–Base Reactions • Acid–Base Titrations 4.9 Oxidation–Reduction Reactions • Oxidation States • The Characteristics of Oxidation–Reduction Reactions 4.10 Balancing Oxidation– Reduction Equations • The Half-Reaction Method for Balancing Oxidation–Reduction Reactions in Aqueous Solutions Yellow lead(II) iodide is produced when lead(II) nitrate is mixed with potassium iodide.
Much of the chemistry that affects each of us occurs among substances dissolved in water. For example, virtually all the chemistry that makes life possible occurs in an aqueous environment. Also, various medical tests involve aqueous reactions, depending heavily on analyses of blood and other body fluids. In addition to the common tests for sugar, cholesterol, and iron, analyses for specific chemical markers allow detection of many diseases before obvious symptoms occur.
Aqueous chemistry is also important in our environment. In recent years, contami-nation of the groundwater by substances such as chloroform and nitrates has been widely publicized. Water is essential for life, and the maintenance of an ample supply of clean water is crucial to all civilization.
To understand the chemistry that occurs in such diverse places as the human body, the atmosphere, the groundwater, the oceans, the local water treatment plant, your hair as you shampoo it, and so on, we must understand how substances dissolved in water react with each other.
However, before we can understand solution reactions, we need to discuss the nature of solutions in which water is the dissolving medium, or solvent. These solutions are called aqueous solutions. In this chapter we will study the nature of materials after they are dis-solved in water and various types of reactions that occur among these substances. You will see that the procedures developed in Chapter 3 to deal with chemical reactions work very well for reactions that take place in aqueous solutions. To understand the types of reactions that occur in aqueous solutions, we must first explore the types of species present.
This requires an understanding of the nature of water.
4.1 Water, the Common Solvent Water is one of the most important substances on earth. It is essential for sustaining the reactions that keep us alive, but it also affects our lives in many indirect ways. Water helps moderate the earth’s temperature; it cools automobile engines, nuclear power plants, and many industrial processes; it provides a means of transportation on the earth’s surface and a medium for the growth of a myriad of creatures we use as food; and much more.
One of the most valuable properties of water is its ability to dissolve many different substances. For example, salt “disappears” when you sprinkle it into the water used to cook vegetables, as does sugar when you add it to your iced tea. In each case the “dis-appearing” substance is obviously still present—you can taste it. What happens when a solid dissolves? To understand this process, we need to consider the nature of water. Liquid water consists of a collection of H2O molecules. An individual H2O molecule is “bent” or V-shaped, with an HOOOH angle of approximately 105 degrees: The OOH bonds in the water molecule are covalent bonds formed by electron shar-ing between the oxygen and hydrogen atoms. However, the electrons of the bond are not shared equally between these atoms. For reasons we will discuss in later chapters, oxy-gen has a greater attraction for electrons than does hydrogen. If the electrons were shared equally between the two atoms, both would be electrically neutral because, on average, the number of electrons around each would equal the number of protons in that nucleus.
H H O 105˚ 127 128 Chapter Four Types of Chemical Reactions and Solution Stoichiometry However, because the oxygen atom has a greater attraction for electrons, the shared electrons tend to spend more time close to the oxygen than to either of the hydrogens.
Thus the oxygen atom gains a slight excess of negative charge, and the hydrogen atoms become slightly positive. This is shown in Fig. 4.1, where (delta) indicates a partial charge (less than one unit of charge). Because of this unequal charge distribution, water is said to be a polar molecule. It is this polarity that gives water its great ability to dissolve compounds.
A schematic of an ionic solid dissolving in water is shown in Fig. 4.2. Note that the “positive ends” of the water molecules are attracted to the negatively charged anions and that the “negative ends” are attracted to the positively charged cations. This process is called hydration. The hydration of its ions tends to cause a salt to “fall apart” in the wa-ter, or to dissolve. The strong forces present among the positive and negative ions of the solid are replaced by strong water–ion interactions.
It is very important to recognize that when ionic substances (salts) dissolve in water, they break up into the individual cations and anions. For instance, when ammonium nitrate (NH4NO3) dissolves in water, the resulting solution contains NH4 and NO3 ions moving around independently. This process can be represented as where (aq) designates that the ions are hydrated by unspecified numbers of water molecules.
The solubility of ionic substances in water varies greatly. For example, sodium chlo-ride is quite soluble in water, whereas silver chloride (contains Ag and Cl ions) is only very slightly soluble. The differences in the solubilities of ionic compounds in water typically depend on the relative attractions of the ions for each other (these forces hold the solid together) and the attractions of the ions for water molecules (which cause the solid to disperse [dissolve] in water). Solubility is a complex topic that we will explore in much more detail in Chapter 11. However, the most important thing to remember at NH4NO31s2 –¡ H2O1l 2 NH4 1aq2 NO3 1aq2 FIGURE 4.1 (top) The water molecule is polar. (bottom) A space-filling model of the water molecule.
Cation Anion 2δ– δ+ δ+ 2δ– δ+ δ+ – – – – – – + + + + + + – – – – – – + + + + + + + FIGURE 4.2 Polar water molecules interact with the positive and negative ions of a salt, assisting in the dissolving process.
Visualization: The Dissolution of a Solid in a Liquid 2δ– 105˚ δ+ δ+ O H H 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 129 this point is that when an ionic solid does dissolve in water, the ions become hydrated and are dispersed (move around independently).
Water also dissolves many nonionic substances. Ethanol (C2H5OH), for example, is very soluble in water. Wine, beer, and mixed drinks are aqueous solutions of ethanol and other substances. Why is ethanol so soluble in water? The answer lies in the structure of the alcohol molecules, which is shown in Fig. 4.3(a). The molecule contains a polar OOH bond like those in water, which makes it very compatible with water. The interaction of water with ethanol is represented in Fig. 4.3(b).
Many substances do not dissolve in water. Pure water will not, for example, dissolve animal fat, because fat molecules are nonpolar and do not interact effectively with polar water molecules. In general, polar and ionic substances are expected to be more soluble in water than nonpolar substances. “Like dissolves like” is a useful rule for predicting sol-ubility. We will explore the basis for this generalization when we discuss the details of solution formation in Chapter 11.
4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes As we discussed in Chapter 2, a solution is a homogeneous mixture. It is the same through-out (the first sip of a cup of coffee is the same as the last), but its composition can be varied by changing the amount of dissolved substances (one can make weak or strong cof-fee). In this section we will consider what happens when a substance, the solute, is dissolved in liquid water, the solvent.
One useful property for characterizing a solution is its electrical conductivity, its abil-ity to conduct an electric current. This characteristic can be checked conveniently by us-ing an apparatus like the ones shown in Figure 4.4. If the solution in the container conducts electricity, the bulb lights. Pure water is not an electrical conductor. However, some aque-ous solutions conduct current very efficiently, and the bulb shines very brightly; these so-lutions contain strong electrolytes. Other solutions conduct only a small current, and the bulb glows dimly; these solutions contain weak electrolytes. Some solutions permit no current to flow, and the bulb remains unlit; these solutions contain nonelectrolytes.
The basis for the conductivity properties of solutions was first correctly identified by Svante Arrhenius (1859–1927), then a Swedish graduate student in physics, who carried out research on the nature of solutions at the University of Uppsala in the early 1880s.
Arrhenius came to believe that the conductivity of solutions arose from the presence of ions, an idea that was at first scorned by the majority of the scientific establishment. How-ever, in the late 1890s when atoms were found to contain charged particles, the ionic theory suddenly made sense and became widely accepted.
As Arrhenius postulated, the extent to which a solution can conduct an electric current depends directly on the number of ions present. Some materials, such as sodium chloride, readily produce ions in aqueous solution and thus are strong electrolytes. Other substances, Polar bond (a) (b) δ– H C C H H O δ+ H H H H C C H H O H H H H O H FIGURE 4.3 (a) The ethanol molecule contains a polar O—H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O—H bond in ethanol. This is a case of “like dissolving like.” An electrolyte is a substance that when dissolved in water produces a solution that can conduct electricity.
Visualization: Electrolytes Visualization: Electrolyte Behavior 130 Chapter Four Types of Chemical Reactions and Solution Stoichiometry NaCl(s) dissolves Na+ Cl– FIGURE 4.5 When solid NaCl dissolves, the Na and Cl ions are randomly dispersed in the water.
such as acetic acid, produce relatively few ions when dissolved in water and are weak electrolytes. A third class of materials, such as sugar, form virtually no ions when dissolved in water and are nonelectrolytes.
Strong Electrolytes Strong electrolytes are substances that are completely ionized when they are dissolved in water, as represented in Fig. 4.4(a). We will consider several classes of strong electrolytes: (1) soluble salts, (2) strong acids, and (3) strong bases.
As shown in Fig. 4.2, a salt consists of an array of cations and anions that separate and become hydrated when the salt dissolves. For example, when NaCl dissolves in water, it produces hydrated Na and Cl ions in the solution (see Fig. 4.5). Virtually no NaCl (a) (c) (b) – – – – – – + + + + + + FIGURE 4.4 Electrical conductivity of aqueous solutions.
The circuit will be completed and will allow current to flow only when there are charge carriers (ions) in the solution. Note: Water molecules are present but not shown in these pictures. (a) A hydrochloric acid solu-tion, which is a strong electrolyte, contains ions that readily conduct the current and give a brightly lit bulb. (b) An acetic acid solution, which is a weak electrolyte, con-tains only a few ions and does not conduct as much current as a strong electrolyte.
The bulb is only dimly lit. (c) A sucrose solution, which is a nonelectrolyte, contains no ions and does not conduct a current.
The bulb remains unlit.
4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 131 units are present. Thus NaCl is a strong electrolyte. It is important to recognize that these aqueous solutions contain millions of water molecules that we will not include in our molecular-level drawings.
One of Arrhenius’s most important discoveries concerned the nature of acids. Acid-ity was first associated with the sour taste of citrus fruits. In fact, the word acid comes directly from the Latin word acidus, meaning “sour.” The mineral acids sulfuric acid (H2SO4) and nitric acid (HNO3), so named because they were originally obtained by the treatment of minerals, were discovered around 1300.
Although acids were known for hundreds of years before the time of Arrhenius, no one had recognized their essential nature. In his studies of solutions, Arrhenius found that when the substances HCl, HNO3, and H2SO4 were dissolved in water, they behaved as strong electrolytes. He postulated that this was the result of ionization reactions in water, for example: Thus Arrhenius proposed that an acid is a substance that produces H ions (protons) when it is dissolved in water.
Studies of conductivity show that when HCl, HNO3, and H2SO4 are placed in water, virtually every molecule ionizes. These substances are strong electrolytes and are thus called strong acids. All three are very important chemicals, and much more will be said about them as we proceed. However, at this point the following facts are important: Sulfuric acid, nitric acid, and hydrochloric acid are aqueous solutions and should be written in chemical equations as H2SO4(aq), HNO3(aq), and HCl(aq), respectively, although they often appear without the (aq) symbol.
A strong acid is one that completely dissociates into its ions. Thus, if 100 molecules of HCl are dissolved in water, 100 H ions and 100 Cl ions are produced. Virtually no HCl molecules exist in aqueous solutions (see Fig. 4.6).
Sulfuric acid is a special case. The formula H2SO4 indicates that this acid can pro-duce two H ions per molecule when dissolved in water. However, only the first H ion is completely dissociated. The second H ion can be pulled off under certain con-ditions, which we will discuss later. Thus an aqueous solution of H2SO4 contains mostly H ions and HSO4 ions.
Another important class of strong electrolytes consists of the strong bases, soluble ionic compounds containing the hydroxide ion (OH). When these compounds are dis-solved in water, the cations and OH ions separate and move independently. Solutions containing bases have a bitter taste and a slippery feel. The most common basic solutions are those produced when solid sodium hydroxide (NaOH) or potassium hydroxide (KOH) is dissolved in water to produce ions, as follows (see Fig. 4.7): Weak Electrolytes Weak electrolytes are substances that exhibit a small degree of ionization in water. That is, they produce relatively few ions when dissolved in water, as shown in Fig. 4.4(b). The most common weak electrolytes are weak acids and weak bases.
KOH1s2 ¡ H2O K1aq2 OH1aq2 NaOH1s2 ¡ H2O Na1aq2 OH1aq2 H2SO4 ¡ H2O H1aq2 HSO4 1aq2 HNO3 ¡ H2O H1aq2 NO3 1aq2 HCl ¡ H2O H1aq2 Cl1aq2 The Arrhenius definition of an acid is a substance that produces H ions in solution.
Perchloric acid, HClO4(aq), is another strong acid.
Strong electrolytes dissociate (ionize) completely in aqueous solution.
H+ Cl– + – + + + + + + + + + + – – – – – – – – – – – – – + + OH– Na+ + + + + + + + + + – – – – – – – – – – FIGURE 4.6 HCl(aq) is completely ionized.
FIGURE 4.7 An aqueous solution of sodium hydroxide.
Weak electrolytes dissociate (ionize) only to a small extent in aqueous solution.
132 Chapter Four Types of Chemical Reactions and Solution Stoichiometry The main acidic component of vinegar is acetic acid (HC2H3O2). The formula is written to indicate that acetic acid has two chemically distinct types of hydrogen atoms.
Formulas for acids are often written with the acidic hydrogen atom or atoms (any that will produce H ions in solution) listed first. If any nonacidic hydrogens are present, they are written later in the formula. Thus the formula HC2H3O2 indicates one acidic and three nonacidic hydrogen atoms. The dissociation reaction for acetic acid in water can be writ-ten as follows: Acetic acid is very different from the strong acids because only about 1% of its molecules dissociate in aqueous solutions at typical concentrations. For example, in a solution con-taining 0.1 mole of HC2H3O2 per liter, for every 100 molecules of HC2H3O2 originally dissolved in water, approximately 99 molecules of HC2H3O2 remain intact (see Fig. 4.8).
That is, only one molecule out of every 100 dissociates (to produce one H ion and one C2H3O2 ion).
Because acetic acid is a weak electrolyte, it is called a weak acid. Any acid, such as acetic acid, that dissociates (ionizes) only to a slight extent in aqueous solutions is called a weak acid. In Chapter 14 we will explore the subject of weak acids in detail.
The most common weak base is ammonia (NH3). When ammonia is dissolved in water, it reacts as follows: The solution is basic because OH ions are produced. Ammonia is called a weak base because the resulting solution is a weak electrolyte; that is, very few ions are formed. In fact, in a solution containing 0.1 mole of NH3 per liter, for every 100 molecules of NH3 NH31aq2 H2O1l2 ¡ NH4 1aq2 OH1aq2 HC2H3O21aq2 ∆ H2O H1aq2 C2H3O2 1aq2 + Carbon Oxygen Hydrogen – FIGURE 4.8 Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Only a small percentage of the molecules are ionized.
CHEMICAL IMPACT Arrhenius: A Man with Solutions S cience is a human endeavor, subject to human frailties and governed by personalities, politics, and prejudices.
One of the best illustrations of the often bumpy path of the advancement of scientific knowledge is the story of Swedish chemist Svante Arrhenius.
When Arrhenius began studies toward his doctorate at the University of Uppsala around 1880, he chose to in-vestigate the passage of electricity through solutions, a mystery that had baffled scientists for a century. The first experiments had been done in the 1770s by Cavendish, who compared the conductivity of salt solution with that of rain water using his own physiologic reaction to the electric shocks he received! Arrhenius had an array of instruments to measure electric current, but the process of carefully weighing, measuring, and recording data from a multitude of experiments was a tedious one.
After his long series of experiments was performed, Arrhenius quit his laboratory bench and returned to his country Svante August Arrhenius.
4.3 The Composition of Solutions 133 originally dissolved, only one NH4 ion and one OH ion are produced; 99 molecules of NH3 remain unreacted (see Fig. 4.9).
Nonelectrolytes Nonelectrolytes are substances that dissolve in water but do not produce any ions, as shown in Fig. 4.4(c). An example of a nonelectrolyte is ethanol (see Fig. 4.3 for the structural formula).
When ethanol dissolves, entire C2H5OH molecules are dispersed in the water. Since the molecules do not break up into ions, the resulting solution does not conduct an electric current.
Another common nonelectrolyte is table sugar (sucrose, C12H22O11), which is very soluble in water but which produces no ions when it dissolves. The sucrose molecules remain intact.
4.3 The Composition of Solutions Chemical reactions often take place when two solutions are mixed. To perform stoi-chiometric calculations in such cases, we must know two things: (1) the nature of the re-action, which depends on the exact forms the chemicals take when dissolved, and (2) the amounts of chemicals present in the solutions, usually expressed as concentrations.
The concentration of a solution can be described in many different ways, as we will see in Chapter 11. At this point we will consider only the most commonly used expression of concentration, molarity (M), which is defined as moles of solute per volume of solution in liters: A solution that is 1.0 molar (written as 1.0 M) contains 1.0 mole of solute per liter of solution.
M molarity moles of solute liters of solution Nitrogen Oxygen Hydrogen + – FIGURE 4.9 The reaction of NH3 in water.
his new theory as he defended his dissertation. His diplo-macy paid off: He was awarded his degree, albeit reluctantly, because the professors still did not believe his model and considered him to be a marginal scientist, at best.
Such a setback could have ended his scientific career, but Arrhenius was a crusader; he was determined to see his theory triumph. He promptly embarked on a political cam-paign, enlisting the aid of several prominent scientists, to get his theory accepted.
Ultimately, the ionic theory triumphed. Arrhenius’s fame spread, and honors were heaped on him, culminating in the Nobel Prize in chemistry in 1903. Not one to rest on his lau-rels, Arrhenius turned to new fields, including astronomy; he formulated a new theory that the solar system may have come into being through the collision of stars. His exceptional ver-satility led him to study the use of serums to fight disease, energy resources and conservation, and the origin of life.
Additional insight on Arrhenius and his scientific career can be obtained from his address on receiving the Willard Gibbs Award.
See Journal of the American Chemical Society 36 (1912): 353.
home to try to formulate a model that could account for his data. He wrote, “I got the idea in the night of the 17th of May in the year 1883, and I could not sleep that night until I had worked through the whole problem.” His idea was that ions were responsible for conducting electricity through a solution.
Back at Uppsala, Arrhenius took his doctoral disserta-tion containing the new theory to his advisor, Professor Cleve, an eminent chemist and the discoverer of the ele-ments holmium and thulium. Cleve’s uninterested response was what Arrhenius had expected. It was in keeping with Cleve’s resistance to new ideas—he had not even accepted Mendeleev’s periodic table, introduced 10 years earlier.
It is a long-standing custom that before a doctoral de-gree is granted, the dissertation must be defended before a panel of professors. Although this procedure is still followed at most universities today, the problems are usually worked out in private with the evaluating professors before the actual defense. However, when Arrhenius did it, the dissertation defense was an open debate, which could be rancorous and humiliating. Knowing that it would be unwise to antagonize his professors, Arrhenius downplayed his convictions about 134 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Calculation of Molarity I Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.
Solution To find the molarity of the solution, we first compute the number of moles of solute using the molar mass of NaOH (40.00 g/mol): Then we divide by the volume of the solution in liters: See Exercises 4.21 and 4.22.
Calculation of Molarity II Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
Solution First we calculate the number of moles of HCl (molar mass 36.46 g/mol): Next we must change the volume of the solution to liters: Finally, we divide the moles of solution by the liters of solution: See Exercises 4.21 and 4.22.
It is important to realize that the conventional description of a solution’s concentration may not accurately reflect the true composition of the solution. Solution concentration is always given in terms of the form of the solute before it dissolves. For example, when a solution is described as being 1.0 M NaCl, this means that the solution was prepared by dissolving 1.0 mole of solid NaCl in enough water to make 1.0 liter of solution; it does not mean that the solution contains 1.0 mole of NaCl units. Actually, the solution contains 1.0 mole of Na ions and 1.0 mole of Cl ions. This situation is further illustrated in Sample Exercise 4.3.
Concentrations of Ions I Give the concentration of each type of ion in the following solutions: a. 0.50 M Co(NO3)2 b. 1 M Fe(ClO4)3 Molarity 4.28 102 mol HCl 2.68 102 L solution 1.60 M HCl 26.8 mL 1 L 1000 mL 2.68 102 L 1.56 g HCl 1 mol HCl 36.46 g HCl 4.28 102 mol HCl Molarity mol solute L solution 0.288 mol NaOH 1.50 L solution 0.192 M NaOH 11.5 g NaOH 1 mol NaOH 40.00 g NaOH 0.288 mol NaOH Sample Exercise 4.1 Sample Exercise 4.2 Sample Exercise 4.3 4.3 The Composition of Solutions 135 Solution a. When solid Co(NO3)2 dissolves, the cobalt(II) cation and the nitrate anions separate: For each mole of Co(NO3)2 that is dissolved, the solution contains 1 mol Co2 ions and 2 mol NO3 ions. Thus a solution that is 0.50 M Co(NO3)2 contains 0.50 M Co2 and (2 0.50) M NO3 or 1.0 M NO3 .
b. When solid Fe(ClO4)3 dissolves, the iron(III) cation and the perchlorate anions separate: Thus a solution that is described as 1 M Fe(ClO4)3 actually contains 1 M Fe3 ions and 3 M ClO4 ions.
See Exercises 14.23 and 14.24.
Often chemists need to determine the number of moles of solute present in a given volume of a solution of known molarity. The procedure for doing this is easily derived from the definition of molarity. If we multiply the molarity of a solution by the volume (in liters) of a particular sample of the solution, we get the moles of solute present in that sample: This procedure is demonstrated in Sample Exercises 4.4 and 4.5.
Concentrations of Ions II Calculate the number of moles of Cl ions in 1.75 L of 1.0 103 M ZnCl2.
Solution When solid ZnCl2 dissolves, it produces ions as follows: Thus a 1.0 103 M ZnCl2 solution contains 1.0 103 M Zn2 ions and 2.0 103 M Cl ions. To calculate the moles of Cl ions in 1.75 L of the 1.0 103 M ZnCl2 so-lution, we must multiply the volume times the molarity: See Exercise 4.25.
Concentration and Volume Typical blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg NaCl?
Solution We must first determine the number of moles represented by 1.0 mg NaCl (molar mass 58.45 g/mol): 1.0 mg NaCl 1 g NaCl 1000 mg NaCl 1 mol NaCl 58.45 g NaCl 1.7 105 mol NaCl 3.5 103 mol Cl 1.75 L solution 2.0 103 M Cl 1.75 L solution 2.0 103 mol Cl L solution ZnCl21s2 ¡ H2O Zn21aq2 2Cl1aq2 Liters of solution molarity liters of solution moles of solute liters of solution moles of solute Fe1ClO4231s2 ¡ H2O Fe31aq2 3ClO4 1aq2 Co1NO3221s2 ¡ H2O Co21aq2 2NO3 1aq2 An aqueous solution of Co(NO3)2.
M moles of solute liters of solution Sample Exercise 4.4 Sample Exercise 4.5 136 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Next, we must determine what volume of 0.14 M NaCl solution contains 1.7 105 mol NaCl. There is some volume, call it V, that when multiplied by the molarity of this solu-tion will yield 1.7 105 mol NaCl. That is: We want to solve for the volume: Thus 0.12 mL of blood contains 1.7 105 mol NaCl or 1.0 mg NaCl.
See Exercises 4.27 and 4.28.
A standard solution is a solution whose concentration is accurately known. Stan-dard solutions, often used in chemical analysis, can be prepared as shown in Fig. 4.10 and in Sample Exercise 4.6.
Solutions of Known Concentration To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be weighed out to make this solution?
Solution We must first determine the moles of K2Cr2O7 required: 1.00 L solution 0.200 mol K2Cr2O7 L solution 0.200 mol K2Cr2O7 V 1.7 105 mol NaCl 0.14 mol NaCl L solution 1.2 104 L solution V 0.14 mol NaCl L solution 1.7 105 mol NaCl Sample Exercise 4.6 Volume marker (calibration mark) Wash bottle (a) Weighed amount of solute (b) (c) FIGURE 4.10 Steps involved in the preparation of a standard aqueous solution. (a) Put a weighed amount of a substance (the solute) into the volumetric flask, and add a small quantity of water. (b) Dissolve the solid in the water by gently swirling the flask (with the stopper in place). (c) Add more water (with gentle swirling) until the level of the solution just reaches the mark etched on the neck of the flask. Then mix the solution thoroughly by inverting the flask several times.
4.3 The Composition of Solutions 137 This amount can be converted to grams using the molar mass of K2Cr2O7 (294.18 g/mol).
Thus, to make 1.00 L of 0.200 M K2Cr2O7, the chemist must weigh out 58.8 g K2Cr2O7, transfer it to a 1.00-L volumetric flask, and add distilled water to the mark on the flask.
See Exercises 4.29a and c and 4.30c and e.
Dilution To save time and space in the laboratory, routinely used solutions are often purchased or prepared in concentrated form (called stock solutions). Water is then added to achieve the molarity desired for a particular solution. This process is called dilution. For example, the common acids are purchased as concentrated solutions and diluted as needed. A typ-ical dilution calculation involves determining how much water must be added to an amount of stock solution to achieve a solution of the desired concentration. The key to doing these calculations is to remember that because only water (no solute) is added to accomplish the dilution.
For example, suppose we need to prepare 500. mL of 1.00 M acetic acid (HC2H3O2) from a 17.4 M stock solution of acetic acid. What volume of the stock solution is required?
The first step is to determine the number of moles of acetic acid in the final solution by multiplying the volume by the molarity (remembering that the volume must be changed to liters): Thus we need to use a volume of 17.4 M acetic acid that contains 0.500 mol HC2H3O2.
That is, Solving for V gives Thus, to make 500 mL of a 1.00 M acetic acid solution, we can take 28.7 mL of 17.4 M acetic acid and dilute it to a total volume of 500 mL with distilled water.
A dilution procedure typically involves two types of glassware: a pipet and a volumetric flask. A pipet is a device for accurately measuring and transferring a given volume of solution. There are two common types of pipets: volumetric (or transfer) pipets and measuring pipets, as shown in Fig. 4.11. Volumetric pipets come in specific sizes, such as 5 mL, 10 mL, 25 mL, and so on. Measuring pipets are used to measure volumes for which a volumetric pipet is not available. For example, we would use a measuring pipet as shown in Fig. 4.12 on page 139 to deliver 28.7 mL of 17.4 M acetic acid into a 500-mL volumetric flask and then add water to the mark to perform the dilution described above.
V 0.500 mol HC2H3O2 17.4 mol HC2H3O2 L solution 0.0287 L or 28.7 mL solution V 17.4 mol HC2H3O2 L solution 0.500 mol HC2H3O2 500. mL solution 1 L solution 1000 mL solution 1.00 mol HC2H3O2 L solution 0.500 mol HC2H3O2 Moles of solute after dilution moles of solute before dilution 0.200 mol K2Cr2O7 294.20 g K2Cr2O7 mol K2Cr2O7 58.8 g K2Cr2O7 Dilution with water does not alter the numbers of moles of solute present.
(a) (b) Calibration mark FIGURE 4.11 (a) A measuring pipet is graduated and can be used to measure various volumes of liquid accurately. (b) A volumetric (transfer) pipet is designed to measure one volume accurately. When filled to the mark, it de-livers the volume indicated on the pipet.
Visualization: Dilution 138 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Concentration and Volume What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?
Solution We must first determine the moles of H2SO4 in 1.5 L of 0.10 M H2SO4: 1.5 L solution 0.10 mol H2SO4 L solution 0.15 mol H2SO4 CHEMICAL IMPACT Tiny Laboratories O ne of the major impacts of modern technology is to make things smaller. The best example is the computer.
Calculations that 30 years ago required a machine the size of a large room now can be carried out on a hand-held cal-culator. This tendency toward miniaturization is also having a major impact on the science of chemical analysis. Using the techniques of computer chip makers, researchers are now constructing minuscule laboratories on the surface of a tiny chip made of silicon, glass, or plastic (see photo). Instead of electrons, 106 to 109 L of liquids moves between re-action chambers on the chip through tiny capillaries. The chips typically contain no moving parts. Instead of conven-tional pumps, the chip-based laboratories use voltage dif-ferences to move liquids that contain ions from one reaction chamber to another.
Microchip laboratories have many advantages. They require only tiny amounts of sample. This is especially ad-vantageous for expensive, difficult-to-prepare materials or in cases such as criminal investigations, where only small amounts of evidence may exist. The chip laboratories also minimize contamination because they represent a “closed system” once the material has been introduced to the chip.
In addition, the chips can be made to be disposable to prevent cross-contamination of different samples.
The chip laboratories present some difficulties not found in macroscopic laboratories. The main problem concerns the large surface area of the capillaries and reaction chambers relative to the sample volume. Molecules or biological cells in the sample solution encounter so much “wall” that they may undergo unwanted reactions with the wall materials.
Glass seems to present the least of these problems, and the walls of silicon chip laboratories can be protected by formation of relatively inert silicon dioxide. Because plas-tic is inexpensive, it seems a good choice for disposable chips, but plastic also is the most reactive with the samples and the least durable of the available materials.
Caliper Technologies Corporation, of Palo Alto, Cali-fornia, is working toward creating a miniature chemistry laboratory about the size of a toaster that can be used with “plug-in” chip-based laboratories. Various chips would be furnished with the unit that would be appropriate for differ-ent types of analyses. The entire unit would be connected to a computer to collect and analyze the data. There is even the possibility that these “laboratories” could be used in the home to perform analyses such as blood sugar and blood cholesterol and to check for the presence of bacteria such as E. coli and many others. This would revolutionize the health care industry.
Plastic chips such as this one made by Caliper Technologies are being used to per-form laboratory procedures traditionally done with test tubes.
Adapted from “The Incredible Shrinking Laboratory,” by Corinna Wu, as appeared in Science News, Vol. 154,August 15, 1998, p. 104.
Sample Exercise 4.7 4.3 The Composition of Solutions 139 Next we must find the volume of 16 M H2SO4 that contains 0.15 mol H2SO4: Solving for V gives Thus, to make 1.5 L of 0.10 M H2SO4 using 16 M H2SO4, we must take 9.4 mL of the con-centrated acid and dilute it with water to 1.5 L. The correct way to do this is to add the 9.4 mL of acid to about 1 L of distilled water and then dilute to 1.5 L by adding more water.
See Exercises 4.29b and d and 4.30a, b, and d.
As noted earlier, the central idea in performing the calculations associated with dilu-tions is to recognize that the moles of solute are not changed by the dilution. Another way to express this condition is by the following equation: where M1 and V1 represent the molarity and volume of the original solution (before dilu-tion) and M2 and V2 represent the molarity and volume of the diluted solution. This equa-tion makes sense because mol solute after dilution M2 V2 M1 V1 mol solute before dilution M1V1 M2V2 V 0.15 mol H2SO4 16 mol H2SO4 1 L solution 9.4 103 L or 9.4 mL solution V 16 mol H2SO4 L solution 0.15 mol H2SO4 In diluting an acid, “Do what you oughta, always add acid to water.” (a) (b) (c) 500 mL Rubber bulb FIGURE 4.12 (a) A measuring pipet is used to transfer 28.7 mL of 17.4 M acetic acid solution to a volumetric flask. (b) Water is added to the flask to the calibration mark. (c) The result-ing solution is 1.00 M acetic acid.
140 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Repeat Sample Exercise 4.7 using the equation M1V1 M2V2. Note that in doing so and V1 is the unknown quantity sought. The equation M1V1 M2V2 always holds for a dilu-tion. This equation will be easy for you to remember if you understand where it comes from.
4.4 Types of Chemical Reactions Although we have considered many reactions so far in this text, we have examined only a tiny fraction of the millions of possible chemical reactions. To make sense of all these reactions, we need some system for grouping reactions into classes. Although there are many different ways to do this, we will use the system most commonly used by practic-ing chemists: M1 16 M M2 0.10 M V2 1.5 L Types of Solution Reactions Precipitation reactions Acid–base reactions Oxidation–reduction reactions Virtually all reactions can be put into one of these classes. We will define and illustrate each type in the following sections.
4.5 Precipitation Reactions When two solutions are mixed, an insoluble substance sometimes forms; that is, a solid forms and separates from the solution. Such a reaction is called a precipitation reaction, and the solid that forms is called a precipitate. For example, a precipitation reaction occurs when an aqueous solution of potassium chromate, K2CrO4(aq), which is yellow, is added to a colorless aqueous solution containing barium nitrate, Ba(NO3)2(aq). As shown in Fig. 4.13, when these solutions are mixed, a yellow solid forms. What is the equation that describes this chemical change? To write the equation, we must know the identities of the reactants and products. The reactants have already been described: K2CrO4(aq) and Ba(NO3)2(aq). Is there some way we can predict the identities of the products? In partic-ular, what is the yellow solid?
The best way to predict the identity of this solid is to think carefully about what prod-ucts are possible. To do this, we need to know what species are present in the solution after the two reactant solutions are mixed. First, let’s think about the nature of each reactant solution. The designation Ba(NO3)2(aq) means that barium nitrate (a white solid) has been dissolved in water. Notice that barium nitrate contains the Ba2 and NO3 ions. Remem-ber: In virtually every case, when a solid containing ions dissolves in water, the ions separate and move around independently. That is, Ba(NO3)2(aq) does not contain Ba(NO3)2 units; it contains separated Ba2 and NO3 ions. See Fig. 4.14(a).
Similarly, since solid potassium chromate contains the K and CrO4 2 ions, an aque-ous solution of potassium chromate (which is prepared by dissolving solid K2CrO4 in water) contains these separated ions, as shown in Fig. 4.14(b).
We can represent the mixing of K2CrO4(aq) and Ba(NO3)2(aq) in two ways. First, we can write K2CrO41aq2 Ba1NO3221aq2 ¡ products A precipitation reaction also can be called a double displacement reaction.
The quantitative aspects of precipitation reactions are covered in Chapter 15.
When ionic compounds dissolve in water, the resulting solution contains the sepa-rated ions.
FIGURE 4.13 When yellow aqueous potassium chromate is added to a colorless barium nitrate solu-tion, yellow barium chromate precipitates.
Visualization: Precipitation Reactions 4.5 Precipitation Reactions 141 However, a much more accurate representation is The ions in The ions in K2CrO4(aq) Ba(NO3)2(aq) Thus the mixed solution contains the ions: as illustrated in Fig. 4.15(a).
How can some or all of these ions combine to form a yellow solid? This is not an easy question to answer. In fact, predicting the products of a chemical reaction is one of the hardest things a beginning chemistry student is asked to do. Even an experienced chemist, when confronted with a new reaction, is often not sure what will happen.
The chemist tries to think of the various possibilities, considers the likelihood of each K CrO4 2 Ba2 NO3 2K1aq2 CrO4 21aq2 Ba21aq2 2NO3 1aq2 ¡ products ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (a) (b) CrO4 2– K+ Ba2+ NO3 – FIGURE 4.14 Reactant solutions: (a) Ba(NO3)2(aq) and (b) K2CrO4(aq).
CrO4 2– K+ (a) (b) Ba2+ NO3 – (c) FIGURE 4.15 The reaction of K2CrO4(aq) and Ba(NO3)2(aq). (a) The molecular-level “picture” of the mixed solution before any reaction has occurred.
(b) The molecular-level “picture” of the solution after the reaction has occurred to form BaCrO4(s). Note: BaCrO4(s) is not molecular. It actually contains Ba2 and CrO4 2 ions packed together in a lattice. (c) A photo of the solution after the reaction has occurred, show-ing the solid BaCrO4 on the bottom.
142 Chapter Four Types of Chemical Reactions and Solution Stoichiometry possibility, and then makes a prediction (an educated guess). Only after identifying each product experimentally is the chemist sure what reaction has taken place. However, an educated guess is very useful because it provides a place to start. It tells us what kinds of products we are most likely to find. We already know some things that will help us predict the products of the above reaction.
1. When ions form a solid compound, the compound must have a zero net charge. Thus the products of this reaction must contain both anions and cations. For example, K and Ba2 could not combine to form the solid, nor could CrO4 2 and NO3 .
2. Most ionic materials contain only two types of ions: one type of cation and one type of anion (for example, NaCl, KOH, Na2SO4, K2CrO4, Co(NO3)2, NH4Cl, Na2CO3).
The possible combinations of a given cation and a given anion from the list of ions K, CrO4 2, Ba2, and NO3 are Which of these possibilities is most likely to represent the yellow solid? We know it’s not K2CrO4 or Ba(NO3)2. They are the reactants. They were present (dissolved) in the sepa-rate solutions that were mixed. The only real possibilities for the solid that formed are To decide which of these most likely represents the yellow solid, we need more facts. An experienced chemist knows that the K ion and the NO3 ion are both colorless. Thus, if the solid is KNO3, it should be white, not yellow. On the other hand, the CrO4 2 ion is yellow (note in Fig. 4.14 that K2CrO4(aq) is yellow). Thus the yellow solid is almost cer-tainly BaCrO4. Further tests show that this is the case.
So far we have determined that one product of the reaction between K2CrO4(aq) and Ba(NO3)2(aq) is BaCrO4(s), but what happened to the K and NO3 ions? The answer is that these ions are left dissolved in the solution; KNO3 does not form a solid when the K and NO3 ions are present in this much water. In other words, if we took solid KNO3 and put it in the same quantity of water as is present in the mixed solution, it would dis-solve. Thus, when we mix K2CrO4(aq) and Ba(NO3)2(aq), BaCrO4(s) forms, but KNO3 is left behind in solution (we write it as KNO3(aq)). Thus the overall equation for this precipitation reaction using the formulas of the reactants and products is As long as water is present, the KNO3 remains dissolved as separated ions. (See Fig. 4.15 to help visualize what is happening in this reaction. Note the solid BaCrO4 on the bottom of the container, while the K and NO3 ions remain dispersed in the solution.) If we removed the solid BaCrO4 and then evaporated the water, white solid KNO3 would be obtained; the K and NO3 ions would assemble themselves into solid KNO3 when the water is removed.
Now let’s consider another example. When an aqueous solution of silver nitrate is added to an aqueous solution of potassium chloride, a white precipitate forms, as shown in Fig. 4.16. We can represent what we know so far as Remembering that when ionic substances dissolve in water, the ions separate, we can write In silver In potassium Combined solution, nitrate chloride before reaction solution solution Since we know the white solid must contain both positive and negative ions, the possible compounds that can be assembled from this collection of ions are AgNO3 KCl AgCl KNO3 Ag, NO3 K, Cl ¡ Ag, NO3 , K, Cl ¡ white solid AgNO31aq2 KCl1aq2 ¡ unknown white solid K2CrO41aq2 Ba1NO3221aq2 ¡ BaCrO41s2 2KNO31aq2 KNO3 and BaCrO4 K2CrO4 KNO3 BaCrO4 Ba1NO322 FIGURE 4.16 Precipitation of silver chloride by mixing solutions of silver nitrate and potassium chloride. The K and NO3 ions remain in solution.
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 4.5 Precipitation Reactions 143 Since AgNO3 and KCl are the substances dissolved in the two reactant solutions, we know that they do not represent the white solid product. Therefore, the only real possibilities are From the first example considered, we know that KNO3 is quite soluble in water. Thus solid KNO3 will not form when the reactant solids are mixed. The product must be AgCl(s) (which can be proved by experiment to be true). The overall equation for the reaction now can be written Figure 4.17 shows the result of mixing aqueous solutions of AgNO3 and KCl, including a microscopic visualization of the reaction.
Notice that in these two examples we had to apply both concepts (solids must have a zero net charge) and facts (KNO3 is very soluble in water, CrO4 2 is yellow, and so on).
Doing chemistry requires both understanding ideas and remembering key information. Pre-dicting the identity of the solid product in a precipitation reaction requires knowledge of the solubilities of common ionic substances. As an aid in predicting the products of pre-cipitation reactions, some simple solubility rules are given in Table 4.1. You should mem-orize these rules.
The phrase slightly soluble used in the solubility rules in Table 4.1 means that the tiny amount of solid that dissolves is not noticeable. The solid appears to be insoluble to the naked eye. Thus the terms insoluble and slightly soluble are often used interchangeably.
Note that the information in Table 4.1 allows us to predict that AgCl is the white solid formed when solutions of AgNO3 and KCl are mixed. Rules 1 and 2 indicate that KNO3 is soluble, and Rule 3 states that AgCl is insoluble.
AgNO31aq2 KCl1aq2 ¡ AgCl1s2 KNO31aq2 AgCl and KNO3 FIGURE 4.17 Photos and accompanying molecular-level representations illustrating the reaction of KCl(aq) with AgNO3(aq) to form AgCl(s). Note that it is not possible to have a photo of the mixed solution before the reaction occurs, because it is an imaginary step that we use to help visualize the reaction. Actually, the reaction occurs immediately when the two solutions are mixed.
Visualization: Reactions of Silver I K+ Cl– Ag+ Ag+ NO3– Solutions are mixed 144 Chapter Four Types of Chemical Reactions and Solution Stoichiometry When solutions containing ionic substances are mixed, it will be helpful in determining the products if you think in terms of ion interchange. For example, in the preceding discussion we considered the results of mixing AgNO3(aq) and KCl(aq). In determining the products, we took the cation from one reactant and combined it with the anion of the other reactant: r p Possible solid products The solubility rules in Table 4.1 allow us to predict whether either product forms as a solid.
The key to dealing with the chemistry of an aqueous solution is first to focus on the actual components of the solution before any reaction occurs and then to figure out how these components will react with each other. Sample Exercise 4.8 illustrates this process for three different reactions.
Predicting Reaction Products Using the solubility rules in Table 4.1, predict what will happen when the following pairs of solutions are mixed.
a. KNO3(aq) and BaCl2(aq) b. Na2SO4(aq) and Pb(NO3)2(aq) c. KOH(aq) and Fe(NO3)3(aq) Solution a. The formula KNO3(aq) represents an aqueous solution obtained by dissolving solid KNO3 in water to form a solution containing the hydrated ions K(aq) and NO3 (aq).
Likewise, BaCl2(aq) represents a solution formed by dissolving solid BaCl2 in water to produce Ba2(aq) and Cl(aq). When these two solutions are mixed, the resulting solution contains the ions K, NO3 , Ba2, and Cl. All ions are hydrated, but the (aq) is omitted for simplicity. To look for possible solid products, combine the cation from one reactant with the anion from the other: r p Possible solid products K NO3 Ba2 Cl ¡ Ag NO3 K Cl ¡ To begin, focus on the ions in solution before any reaction occurs.
Sample Exercise 4.8 Lead sulfate is a white solid.
TABLE 4.1 Simple Rules for the Solubility of Salts in Water 1. Most nitrate (NO3 ) salts are soluble.
2. Most salts containing the alkali metal ions (Li, Na, K, Cs, Rb) and the ammonium ion (NH4 ) are soluble.
3. Most chloride, bromide, and iodide salts are soluble. Notable exceptions are salts contain-ing the ions Ag, Pb2, and Hg2 2.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, Hg2SO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH and KOH. The compounds Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble.
6. Most sulfide (S2), carbonate (CO3 2), chromate (CrO4 2), and phosphate (PO4 3) salts are only slightly soluble.
Visualization: Solubility Rules 4.6 Describing Reactions in Solution 145 Note from Table 4.1 that the rules predict that both KCl and Ba(NO3)2 are soluble in water. Thus no precipitate forms when KNO3(aq) and BaCl2(aq) are mixed. All the ions remain dissolved in solution. No chemical reaction occurs.
b. Using the same procedures as in part a, we find that the ions present in the combined solution before any reaction occurs are Na, SO4 2, Pb2, and NO3 . The possible salts that could form precipitates are The compound NaNO3 is soluble, but PbSO4 is insoluble (see Rule 4 in Table 4.1).
When these solutions are mixed, PbSO4 will precipitate from the solution. The balanced equation is c. The combined solution (before any reaction occurs) contains the ions K, OH, Fe3, and NO3 . The salts that might precipitate are KNO3 and Fe(OH)3. The solubility rules in Table 4.1 indicate that both K and NO3 salts are soluble. However, Fe(OH)3 is only slightly soluble (Rule 5) and hence will precipitate. The balanced equation is See Exercises 4.37 and 4.38.
4.6 Describing Reactions in Solution In this section we will consider the types of equations used to represent reactions in solution. For example, when we mix aqueous potassium chromate with aqueous barium nitrate, a reaction occurs to form a precipitate (BaCrO4) and dissolved potassium nitrate.
So far we have written the overall or formula equation for this reaction: Although the formula equation shows the reactants and products of the reaction, it does not give a correct picture of what actually occurs in solution. As we have seen, aqueous solutions of potassium chromate, barium nitrate, and potassium nitrate contain individual ions, not collections of ions, as implied by the formula equation. Thus the complete ionic equation better represents the actual forms of the reactants and products in solution. In a complete ionic equation, all substances that are strong electrolytes are represented as ions.
The complete ionic equation reveals that only some of the ions participate in the reaction. The K and NO3 ions are present in solution both before and after the reac-tion. The ions that do not participate directly in the reaction are called spectator ions.
The ions that participate in this reaction are the Ba2 and CrO4 2 ions, which combine to form solid BaCrO4: This equation, called the net ionic equation, includes only those solution components directly involved in the reaction. Chemists usually write the net ionic equation for a reaction in solution because it gives the actual forms of the reactants and products and includes only the species that undergo a change.
Ba21aq2 CrO4 21aq2 ¡ BaCrO41s2 BaCrO41s2 2K1aq2 2NO3 1aq2 2K1aq2 CrO4 21aq2 Ba21aq2 2NO3 1aq2 ¡ K2CrO41aq2 Ba1NO3221aq2 ¡ BaCrO41s2 2KNO31aq2 3KOH1aq2 Fe1NO3231aq2 ¡ Fe1OH231s2 3KNO31aq2 Na2SO41aq2 Pb1NO3221aq2 ¡ PbSO41s2 2NaNO31aq2 Na SO4 2 Pb2 NO3 ¡ Solid Fe(OH)3 forms when aqueous KOH and Fe(NO3)3 are mixed.
A strong electrolyte is a substance that completely breaks apart into ions when dissolved in water.
Net ionic equations include only those components that undergo changes in the reaction.
146 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Sample Exercise 4.9 Three Types of Equations Are Used to Describe Reactions in Solution The formula equation gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution.
The complete ionic equation represents as ions all reactants and products that are strong electrolytes.
The net ionic equation includes only those solution components undergoing a change.
Spectator ions are not included.
Writing Equations for Reactions For each of the following reactions, write the formula equation, the complete ionic equa-tion, and the net ionic equation.
a. Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate.
b. Aqueous potassium hydroxide is mixed with aqueous iron(III) nitrate to form a precipitate of iron(III) hydroxide and aqueous potassium nitrate.
Solution a. Formula Equation Complete Ionic Equation (Remember: Any ionic compound dissolved in water will be present as the separated ions.) h h h h h Spectator Spectator Solid, Spectator Spectator ion ion not written ion ion as separate ions Canceling the spectator ions gives the following net ionic equation.
Net Ionic Equation b. Formula Equation Complete Ionic Equation Net Ionic Equation See Exercises 4.39 through 4.44.
3OH1aq2 Fe31aq2 ¡ Fe1OH231s2 Fe1OH231s2 3K1aq2 3NO3 1aq2 3K1aq2 3OH1aq2 Fe31aq2 3NO3 1aq2 ¡ 3KOH1aq2 Fe1NO3231aq2 ¡ Fe1OH231s2 3KNO31aq2 Cl1aq2 Ag1aq2 ¡ AgCl1s2 K1aq2 Cl1aq2 Ag1aq2 NO3 1aq2 ¡ AgCl1s2 K1aq2 NO3 1aq2 K1aq2 Cl1aq2 Ag1aq2 NO3 1aq2 ¡ AgCl1s2 K1aq2 NO3 1aq2 KCl1aq2 AgNO31aq2 ¡ AgCl1s2 KNO31aq2 4.7 Stoichiometry of Precipitation Reactions 147 4.7 Stoichiometry of Precipitation Reactions In Chapter 3 we covered the principles of chemical stoichiometry: the procedures for cal-culating quantities of reactants and products involved in a chemical reaction. Recall that in performing these calculations we first convert all quantities to moles and then use the coefficients of the balanced equation to assemble the appropriate mole ratios. In cases where reactants are mixed we must determine which reactant is limiting, since the reactant that is consumed first will limit the amounts of products formed. These same principles apply to reactions that take place in solutions. However, two points about solution reactions need special emphasis. The first is that it is sometimes difficult to tell immediately what reaction will occur when two solutions are mixed. Usually we must do some thinking about the various possibilities and then decide what probably will happen. The first step in this process always should be to write down the species that are actually present in the solution, as we did in Section 4.5. The second special point about solution reactions is that to obtain the moles of reactants we must use the volume of the solution and its molarity.
This procedure was covered in Section 4.3.
We will introduce stoichiometric calculations for reactions in solution in Sample Exercise 4.10.
Determining the Mass of Product Formed Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 so-lution to precipitate all the Ag ions in the form of AgCl.
Solution When added to the AgNO3 solution (which contains Ag and NO3 ions), the solid NaCl dissolves to yield Na and Cl ions. Thus the mixed solution contains the ions Note from Table 4.1 that NaNO3 is soluble and AgCl is insoluble. Therefore, solid AgCl forms according to the following net ionic equation: In this case we must add enough Cl ions to react with all the Ag ions present. Thus we must calculate the moles of Ag ions present in 1.50 L of a 0.100 M AgNO3 solution (re-member that a 0.100 M AgNO3 solution contains 0.100 M Ag ions and 0.100 M NO3 ions): Because Ag and Cl react in a 1:1 ratio, 0.150 mol Cl ions and thus 0.150 mol NaCl are required. We calculate the mass of NaCl required as follows: See Exercise 4.47.
Notice from Sample Exercise 4.10 that the procedures for doing stoichiometric calcula-tions for solution reactions are very similar to those for other types of reactions. It is useful to think in terms of the following steps for reactions in solution.
0.150 mol NaCl 58.45 g NaCl mol NaCl 8.77 g NaCl 1.50 L 0.100 mol Ag L 0.150 mol Ag Ag1aq2 Cl1aq2 ¡ AgCl1s2 Ag NO3 Na Cl Sample Exercise 4.10 Determine moles of products Check units of products Determine moles of reactants Identify limiting reactant Write the reaction Balanced net ionic equation Species present 148 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Solving Stoichiometry Problems for Reactions in Solution ➥ 1 Identify the species present in the combined solution, and determine what reac-tion occurs.
➥ 2 Write the balanced net ionic equation for the reaction.
➥ 3 Calculate the moles of reactants.
➥ 4 Determine which reactant is limiting.
➥ 5 Calculate the moles of product or products, as required.
➥ 6 Convert to grams or other units, as required.
Determining the Mass of Product Formed When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.
Solution ➥1 Identify the species present in the combined solution, and determine what reaction occurs. When the aqueous solutions of Na2SO4 (containing Na and SO4 2 ions) and Pb(NO3)2 (containing Pb2 and NO3 ions) are mixed, the resulting solution contains the ions Na, SO4 2, Pb2, and NO3 . Since NaNO3 is soluble and PbSO4 is insoluble (see Rule 4 in Table 4.1), solid PbSO4 will form.
➥2 Write the balanced net ionic equation for the reaction. The net ionic equation is ➥3 Calculate the moles of reactants. Since 0.0500 M Pb(NO3)2 contains 0.0500 M Pb2 ions, we can calculate the moles of Pb2 ions in 1.25 L of this solution as follows: The 0.0250 M Na2SO4 solution contains 0.0250 M SO4 2 ions, and the number of moles of SO4 2 ions in 2.00 L of this solution is ➥4 Determine which reactant is limiting. Because Pb2 and SO4 2 react in a 1:1 ratio, the amount of SO4 2 will be limiting (0.0500 mol SO4 2 is less than 0.0625 mol Pb2).
➥5 Calculate the moles of product. Since the Pb2 ions are present in excess, only 0.0500 mol of solid PbSO4 will be formed.
➥6 Convert to grams of product. The mass of PbSO4 formed can be calculated using the molar mass of PbSO4 (303.3 g/mol): See Exercises 4.49 and 4.50.
0.0500 mol PbSO4 303.3 g PbSO4 1 mol PbSO4 15.2 g PbSO4 2.00 L 0.0250 mol SO4 2 L 0.0500 mol SO4 2 1.25 L 0.0500 mol Pb2 L 0.0625 mol Pb2 Pb21aq2 SO4 21aq2 ¡ PbSO41s2 Sample Exercise 4.11 Determine moles of products Convert to grams Determine moles of reactants Write the reaction Na+ Pb2+ Pb2+(aq) + SO4 2–(aq) PbSO4(s) SO4 2– SO4 2– is limiting NO3 – Grams needed 15.2 g PbSO4 4.8 Acid–Base Reactions 149 4.8 Acid–Base Reactions Earlier in this chapter we considered Arrhenius’s concept of acids and bases: An acid is a substance that produces H ions when dissolved in water, and a base is a substance that produces OH ions. Although these ideas are fundamentally correct, it is convenient to have a more general definition of a base, which includes substances that do not contain OH ions. Such a definition was provided by Johannes N. Brønsted (1879–1947) and Thomas M. Lowry (1874–1936), who defined acids and bases as follows: An acid is a proton donor.
A base is a proton acceptor.
How do we know when to expect an acid–base reaction? One of the most difficult tasks for someone inexperienced in chemistry is to predict what reaction might occur when two solutions are mixed. With precipitation reactions, we found that the best way to deal with this problem is to focus on the species actually present in the mixed solution. This idea also applies to acid–base reactions. For example, when an aqueous solution of hydrogen chloride (HCl) is mixed with an aqueous solution of sodium hydroxide (NaOH), the combined solution contains the ions H, Cl, Na, and OH. The separated ions are present because HCl is a strong acid and NaOH is a strong base. How can we predict what reaction occurs, if any? First, will NaCl precipitate? From Table 4.1 we can see that NaCl is soluble in water and thus will not precipitate. Therefore, the Na and Cl ions are spectator ions. On the other hand, because water is a nonelectrolyte, large quantities of H and OH ions cannot coexist in solution. They react to form H2O molecules: This is the net ionic equation for the reaction that occurs when aqueous solutions of HCl and NaOH are mixed.
Next, consider mixing an aqueous solution of acetic acid (HC2H3O2) with an aque-ous solution of potassium hydroxide (KOH). In our earlier discussion of conductivity we said that an aqueous solution of acetic acid is a weak electrolyte. This tells us that acetic acid does not dissociate into ions to any great extent. In fact, in 0.1 M HC2H3O2 approx-imately 99% of the HC2H3O2 molecules remain undissociated. However, when solid KOH is dissolved in water, it dissociates completely to produce K and OH ions. Therefore, in the solution formed by mixing aqueous solutions of HC2H3O2 and KOH, before any reaction occurs, the principal species are HC2H3O2, K, and OH. What reaction will occur? A possible precipitation reaction could occur between K and OH. However, we know that KOH is soluble, so precipitation does not occur. Another possibility is a reaction involving the hydroxide ion (a proton acceptor) and some proton donor. Is there a source of protons in the solution? The answer is yes—the HC2H3O2 molecules. The OH ion has such a strong affinity for protons that it can strip them from the HC2H3O2 molecules. The net ionic equation for this reaction is This reaction illustrates a very important general principle: The hydroxide ion is such a strong base that for purposes of stoichiometric calculations it can be assumed to react completely with any weak acid that we will encounter. Of course, OH ions also react completely with the H ions in solutions of strong acids.
We will now deal with the stoichiometry of acid–base reactions in aqueous solutions.
The procedure is fundamentally the same as that used previously for precipitation reactions.
Performing Calculations for Acid–Base Reactions ➥ 1 List the species present in the combined solution before any reaction occurs, and decide what reaction will occur.
OH1aq2 HC2H3O21aq2 ¡ H2O1l2 C2H3O2 1aq2 H1aq2 OH1aq2 ¡ H2O1l2 The Brønsted–Lowry concept of acids and bases will be discussed in detail in Chapter 14.
Determine moles of products Check units of products Determine moles of reactants Identify limiting reactant Write the reaction Balanced net ionic equation Species present Visualization: Proton Transfer 150 Chapter Four Types of Chemical Reactions and Solution Stoichiometry ➥ 2 Write the balanced net ionic equation for this reaction.
➥ 3 Calculate the moles of reactants. For reactions in solution, use the volumes of the original solutions and their molarities.
➥ 4 Determine the limiting reactant where appropriate.
➥ 5 Calculate the moles of the required reactant or product.
➥ 6 Convert to grams or volume (of solution), as required.
An acid–base reaction is often called a neutralization reaction. When just enough base is added to react exactly with the acid in a solution, we say the acid has been neutralized.
Neutralization Reactions I What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?
Solution ➥1 List the species present in the combined solution before any reaction occurs, and decide what reaction will occur. The species present in the mixed solutions before any reaction occurs are From HCl(aq) From NaOH(aq) What reaction will occur? The two possibilities are Since we know that NaCl is soluble, the first reaction does not take place (Na and Cl are spectator ions). However, as we have seen before, the reaction of the H and OH ions to form H2O does occur.
➥2 Write the balanced net ionic equation. The balanced net ionic equation for this reaction is ➥3 Calculate the moles of reactants. The number of moles of OH ions in the 25.0-mL sample of 0.350 M NaOH is ➥4 Determine the limiting reactant. This problem requires the addition of just enough H ions to react exactly with the OH ions present. Thus we need not be concerned with determining a limiting reactant.
➥5 Calculate the moles of reactant needed. Since H and OH ions react in a 1:1 ratio, 8.75 103 mol H ions is required to neutralize the OH ions present.
➥6 Convert to volume required. The volume V of 0.100 M HCl required to furnish 8.75 103 mol H ions can be calculated as follows: Solving for V gives V 8.75 103 mol H 0.100 mol H L 8.75 102 L V 0.100 mol H L 8.75 103 mol H 25.0 mL NaOH 1 L 1000 mL 0.350 mol OH L NaOH 8.75 103 mol OH H1aq2 OH1aq2 ¡ H2O1l2 H1aq2 OH1aq2 ¡ H2O1l2 Na1aq2 Cl1aq2 ¡ NaCl1s2 Na OH H Cl Sample Exercise 4.12 ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Convert to volume Moles OH– 8.75 × 10–3 Write the reaction H+ Na+ H+(aq) + OH–(aq) H2O(l) Cl– No limiting reactant OH– Volume needed 87.5 mL of 0.100 M HCl needed Moles H+ 8.75 × 10–3 4.8 Acid–Base Reactions 151 Thus 8.75 102 L (87.5 mL) of 0.100 M HCl is required to neutralize 25.0 mL of 0.350 M NaOH.
See Exercises 4.59 and 4.60.
Neutralization Reactions II In a certain experiment, 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H or OH ions in excess after the reaction goes to completion?
Solution The species available for reaction are From HNO3 From KOH solution solution Since KNO3 is soluble, K and NO3 are spectator ions, so the net ionic equation is We next compute the amounts of H and OH ions present: Since H and OH react in a 1:1 ratio, the limiting reactant is H. This means that 7.00 103 mol H ions will react with 7.00 103 mol OH ions to form 7.00 103 mol H2O.
The amount of OH ions in excess is obtained from the following difference: The volume of the combined solution is the sum of the individual volumes: Thus the molarity of OH ions in excess is See Exercises 4.61 and 4.62.
Acid–Base Titrations Volumetric analysis is a technique for determining the amount of a certain substance by doing a titration. A titration involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance be-ing analyzed (the analyte). The titrant contains a substance that reacts in a known manner with the analyte. The point in the titration where enough titrant has been added to react ex-actly with the analyte is called the equivalence point or the stoichiometric point. This point is often marked by an indicator, a substance added at the beginning of the titration that changes color at (or very near) the equivalence point. The point where the indicator mol OH L solution 1.00 102 mol OH 8.10 102 L 0.123 M OH 28.0 mL 53.0 mL 81.0 mL 8.10 102 L Original volume of HNO3 original volume of KOH total volume 1.70 102 mol OH 7.00 103 mol OH 1.00 102 mol OH Original amount amount consumed amount in excess 53.0 mL KOH 1 L 1000 mL 0.320 mol OH L KOH 1.70 102 mol OH 28.0 mL HNO3 1 L 1000 mL 0.250 mol H L HNO3 7.00 103 mol H H1aq2 OH1aq2 ¡ H2O1l2 K OH H NO3 Sample Exercise 4.13 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Find excess OH– concentration Find moles H+, OH– Find moles OH– that react Write the reaction H+ K+ H+(aq) + OH–(aq) H2O(l) NO3 – Limiting reactant is H+ OH– Concentration of OH– needed 0.123 M OH– Ideally, the endpoint and stoichiometric point should coincide.
Visualization: Neutralization of a Strong Acid by a Strong Base 152 Chapter Four Types of Chemical Reactions and Solution Stoichiometry actually changes color is called the endpoint of the titration. The goal is to choose an in-dicator such that the endpoint (where the indicator changes color) occurs exactly at the equivalence point (where just enough titrant has been added to react with all the analyte).
The following three requirements must be met for a titration to be successful: 1. The exact reaction between titrant and analyte must be known (and rapid).
2. The stoichiometric (equivalence) point must be marked accurately.
3. The volume of titrant required to reach the stoichiometric point must be known accurately.
When the analyte is a base or an acid, the required titrant is a strong acid or strong base, respectively. This procedure is called an acid–base titration. An indicator very com-monly used for acid–base titrations is phenolphthalein, which is colorless in an acidic so-lution and pink in a basic solution. Thus, when an acid is titrated with a base, the phenolphthalein remains colorless until after the acid is consumed and the first drop of excess base is added. In this case, the endpoint (the solution changes from colorless to pink) occurs approximately one drop of base beyond the stoichiometric point. This type of titration is illustrated in the three photos in Fig. 4.18.
We will deal with the acid–base titrations only briefly here but will return to the topic of titrations and indicators in more detail in Chapter 15. The titration of an acid with a standard solution containing hydroxide ions is described in Sample Exercise 4.15. In Sample Exercise 4.14 we show how to determine accurately the concentration of a sodium hydroxide solution. This procedure is called standardizing the solution.
FIGURE 4.18 The titration of an acid with a base. (a) The titrant (the base) is in the buret, and the flask contains the acid solution along with a small amount of indicator. (b) As base is added drop by drop to the acid solution in the flask during the titration, the indicator changes color, but the color disappears on mixing. (c) The stoichiometric (equivalence) point is marked by a permanent indicator color change. The volume of base added is the difference between the final and initial buret readings.
Visualization: Titrations 4.8 Acid–Base Reactions 153 Neutralization Titration A student carries out an experiment to standardize (determine the exact concentration of) a sodium hydroxide solution. To do this, the student weighs out a 1.3009-g sample of potassium hydrogen phthalate (KHC8H4O4, often abbreviated KHP). KHP (molar mass 204.22 g/mol) has one acidic hydrogen. The student dissolves the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution with the sodium hydroxide solution to the phenolphthalein endpoint. The difference between the final and initial buret readings indicates that 41.20 mL of the sodium hydroxide solution is required to react exactly with the 1.3009 g KHP. Calculate the concentration of the sodium hydroxide solution.
Solution Aqueous sodium hydroxide contains the Na and OH ions, and KHC8H4O4 dissolves in water to give the K and HC8H4O4 ions. As the titration proceeds, the mixed solution contains the following ions: K, HC8H4O4 , Na, and OH. The OH will remove an H from the HC8H4O4 to give the following net ionic reaction: Since the reaction exhibits 1:1 stoichiometry, we know that 41.20 mL of the sodium hy-droxide solution must contain exactly the same number of moles of OH as there are moles of HC8H4O4 in 1.3009 g KHC8H4O4.
We calculate the moles of KHC8H4O4 in the usual way: This means that 6.3701 103 mol OH must be added to react with the 6.3701 103 mol HC8H4O4 . Thus 41.20 mL (4.120 102 L) of the sodium hydroxide solution must contain 6.3701 103 mol OH (and Na), and the concentration of the sodium hy-droxide solution is This standard sodium hydroxide solution can now be used in other experiments (see Sample Exercise 4.15).
See Exercises 4.63 and 4.66.
Neutralization Analysis An environmental chemist analyzed the effluent (the released waste material) from an in-dustrial process known to produce the compounds carbon tetrachloride (CCl4) and ben-zoic acid (HC7H5O2), a weak acid that has one acidic hydrogen atom per molecule. A sample of this effluent weighing 0.3518 g was shaken with water, and the resulting aque-ous solution required 10.59 mL of 0.1546 M NaOH for neutralization. Calculate the mass percent of HC7H5O2 in the original sample.
Solution In this case, the sample was a mixture containing CCl4 and HC7H5O2, and it was titrated with OH ions. Clearly, CCl4 is not an acid (it contains no hydrogen atoms), so we can assume it does not react with OH ions. However, HC7H5O2 is an acid that donates one H ion per molecule to react with an OH ion as follows: HC7H5O21aq2 OH1aq2 ¡ H2O1l2 C7H5O2 1aq2 0.1546 M Molarity of NaOH mol NaOH L solution 6.3701 103 mol NaOH 4.120 102 L 1.3009 g KHC8H4O4 1 mol KHC8H4O4 204.22 g KHC8H4O4 6.3701 103 mol KHC8H4O4 HC8H4O4 1aq2 OH1aq2 ¡ H2O1l2 C8H4O4 21aq2 Sample Exercise 4.14 HC8H4O4 – K+ Sample Exercise 4.15 154 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Although HC7H5O2 is a weak acid, the OH ion is such a strong base that we can assume that each OH ion added will react with a HC7H5O2 molecule until all the benzoic acid is consumed.
We must first determine the number of moles of OH ions required to react with all the HC7H5O2: This number is also the number of moles of HC7H5O2 present. The number of grams of the acid is calculated using its molar mass (122.12 g/mol): The mass percent of HC7H5O2 in the original sample is See Exercise 4.65.
In doing problems involving titrations, you must first decide what reaction is occur-ring. Sometimes this seems difficult because the titration solution contains several components. The key to success is to first write down all the components in the solution and focus on the chemistry of each one. We have been emphasizing this approach in deal-ing with the reactions between ions in solution. Make it a habit to write down the components of solutions before trying to decide what reaction(s) might take place as you attempt the end-of-chapter problems involving titrations.
4.9 Oxidation–Reduction Reactions We have seen that many important substances are ionic. Sodium chloride, for example, can be formed by the reaction of elemental sodium and chlorine: In this reaction, solid sodium, which contains neutral sodium atoms, reacts with chlorine gas, which contains diatomic Cl2 molecules, to form the ionic solid NaCl, which contains Na and Cl ions. This process is represented in Fig. 4.19. Reactions like this one, in which one or more electrons are transferred, are called oxidation–reduction reactions or redox reactions.
Many important chemical reactions involve oxidation and reduction. Photosynthesis, which stores energy from the sun in plants by converting carbon dioxide and water to sugar, is a very important oxidation–reduction reaction. In fact, most reactions used for energy production are redox reactions. In humans, the oxidation of sugars, fats, and proteins provides the energy necessary for life. Combustion reactions, which provide most of the energy to power our civilization, also involve oxidation and reduction. An example is the reaction of methane with oxygen: Even though none of the reactants or products in this reaction is ionic, the reaction is still assumed to involve a transfer of electrons from carbon to oxygen. To explain this, we must introduce the concept of oxidation states.
CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 energy 2Na1s2 Cl21g2 ¡ 2NaCl1s2 0.1999 g 0.3518 g 100 56.82% 1.637 103 mol HC7H5O2 122.12 g HC7H5O2 1 mol HC7H5O2 0.1999 g HC7H5O2 10.59 mL NaOH 1 L 1000 mL 0.1546 mol OH L NaOH 1.637 103 mol OH The first step in the analysis of a complex solution is to write down the components and focus on the chemistry of each one. When a strong electrolyte is present, write it as separated ions.
Visualization: Zinc and Iodine Visualization: Dry Ice and Magnesium Visualization: Barking Dogs: Reaction of Phosphorus Visualization: Sugar and Potassium Chlorate 4.9 Oxidation–Reduction Reactions 155 Oxidation States The concept of oxidation states (also called oxidation numbers) provides a way to keep track of electrons in oxidation–reduction reactions, particularly redox reactions involving covalent substances. Recall that electrons are shared by atoms in covalent bonds. The ox-idation states of atoms in covalent compounds are obtained by arbitrarily assigning the electrons (which are actually shared) to particular atoms. We do this as follows: For a covalent bond between two identical atoms, the electrons are split equally between the two. In cases where two different atoms are involved (and the electrons are thus shared unequally), the shared electrons are assigned completely to the atom that has the stronger attraction for electrons. For example, recall from the discussion of the water molecule in Section 4.1 that oxygen has a greater attraction for electrons than does hydrogen. There-fore, in assigning the oxidation state of oxygen and hydrogen in H2O, we assume that the oxygen atom actually possesses all the electrons. Recall that a hydrogen atom has one electron. Thus, in water, oxygen has formally “taken” the electrons from two hydrogen atoms. This gives the oxygen an excess of two electrons (its oxidation state is 2) and leaves each hydrogen with no electrons (the oxidation state of each hydrogen is thus 1).
We define the oxidation states (or oxidation numbers) of the atoms in a covalent compound as the imaginary charges the atoms would have if the shared electrons were divided equally between identical atoms bonded to each other or, for different atoms, were all assigned to the atom in each bond that has the greater attraction for electrons. Of course, Cl– Cl Cl Cl– Na+ Na+ 2Na(s) Sodium 2NaCl(s) Sodium chloride Cl2(g) Chlorine + Na Na FIGURE 4.19 The reaction of solid sodium and gaseous chlorine to form solid sodium chloride.
156 Chapter Four Types of Chemical Reactions and Solution Stoichiometry for ionic compounds containing monatomic ions, the oxidation states of the ions are equal to the ion charges.
These considerations lead to a series of rules for assigning oxidation states that are summarized in Table 4.2. Application of these simple rules allows the assignment of oxidation states in most compounds. To apply these rules recognize that the sum of the oxidation states must be zero for an electrically neutral compound. For an ion, the sum of the oxidation states must equal the charge of the ion. The principles are illustrated by Sample Exercise 4.16.
TABLE 4.2 Rules for Assigning Oxidation States The Oxidation State of . . .
Summary Examples • An atom in an element Element: 0 is zero • A monatomic ion is the Monatomic ion: same as its charge charge of ion • Fluorine is 1 in its Fluorine: 1 compounds • Oxygen is usually 2 in Oxygen: 2 its compounds Exception: peroxides (containing O2 2) in which oxygen is 1 • Hydrogen is 1 in its Hydrogen: 1 covalent compounds H2O, HCl, NH3 H2O, CO2 HF, PF3 Na, Cl Na1s2, O21g2, O31g2, Hg1l2 Oxidation of copper metal by nitric acid.
The copper atoms lose two electrons to form Cu2 ions, which give a deep green color that becomes turquoise when diluted with water.
CHEMICAL IMPACT Iron Zeroes in on Pollution T reating groundwater contaminated with pollutants is typ-ically very complicated and very expensive. However, chemists have discovered a low-tech, economical method for treating the contaminated groundwater near a former semi-conductor manufacturing plant in Sunnyvale, California.
They have replaced the elaborate decontamination machin-ery used at the site for more than a decade with 220 tons of iron filings buried in a giant trough. Because there are no pumps to maintain and no electricity to purchase, this simple system will save approximately $300,000 per year. The prop-erty, which was thought to be unusable for the 30-year lifetime of the old clean-up process because of the need for constant monitoring and access, can now be used immediately.
A schematic of the iron treatment method is shown in the accompanying figure. At Sunnyvale, the iron barrier is 40 feet long, 4 feet wide, and 20 feet deep. In the 4 days it takes for contaminated water to seep through the wall of iron, the chlorinated organic contaminants are degraded into prod-ucts that are then themselves decomposed to simpler sub-stances. According to engineers on the site, the polluted water that seeps through the wall meets Environmental Protection Agency (EPA) standards when it emerges on the other side.
How does iron metal clean up contaminated ground-water? It’s a result of the ability of iron metal (oxidation state 0) to act as a reducing agent toward the chlorine-containing organic pollutant molecules. The reaction can be represented as follows: where RCl represents a chlorinated organic molecule. The reaction appears to involve a direct reaction between the metal and an RCl molecule adsorbed on the metal surface.
Fe21aq2 RH1aq2 Cl1aq2 Fe1s2 RCl1aq2 H1aq2 ¡ 4.9 Oxidation–Reduction Reactions 157 It is worthwhile to note at this point that the convention is to write actual charges on ions as n or n, the number being written before the plus or minus sign. On the other hand, oxidation states (not actual charges) are written n or n, the number being written after the plus or minus sign.
Assigning Oxidation States Assign oxidation states to all atoms in the following.
a. CO2 b. SF6 c. NO3 Solution a. Since we have a specific rule for the oxidation state of oxygen, we will assign its value first. The oxidation state of oxygen is 2. The oxidation state of the carbon atom can be determined by recognizing that since CO2 has no charge, the sum of the oxidation states for oxygen and carbon must be zero. Since each oxygen is 2 and there are two oxygen atoms, the carbon atom must be assigned an oxidation state of 4: p r 4 2 for each oxygen We can check the assigned oxidation states by noting that when the number of atoms is taken into account, the sum is zero as required: p h No. of C No. of O atoms atoms 1142 2122 0 CO2 Sample Exercise 4.16 In addition to decomposing chlori-nated organic contaminants, iron appears to be useful against other pollutants as well. Iron can degrade dye wastes from textile mills and can reduce soluble Cr(VI) compounds to insoluble Cr(III) products, which are much less harmful.
Iron’s reducing abilities also appear use-ful in removing radioactive technetium, a common pollutant at nuclear processing facilities. Iron also appears to be effective for removing nitrates from the soil.
Other metals, such as zinc, tin, and palladium, have shown promise for use in groundwater clean-up, too. These metals generally react more quickly than iron but are more expensive and pose their own environmental hazards.
Inexpensive and environmentally benign, iron seems to be the metal of choice for most groundwater clean-up. It’s cheap, it’s effective, it’s almost a miracle!
158 Chapter Four Types of Chemical Reactions and Solution Stoichiometry b. Since we have no rule for sulfur, we first assign the oxidation state of each fluorine as 1. The sulfur must then be assigned an oxidation state of 6 to balance the total of 6 from the fluorine atoms: p r 6 1 for each fluorine Reality Check: c. Oxygen has an oxidation state of 2. Because the sum of the oxidation states of the three oxygens is 6 and the net charge on the NO3 ion is 1, the nitrogen must have an oxidation state of 5: p r 5 2 for each oxygen Reality Check: Note that in this case the sum must be 1 (the overall charge on the ion).
See Exercises 4.67 through 4.70.
We need to make one more point about oxidation states, and this can be illustrated by the compound Fe3O4, which is the main component in magnetite, an iron ore that accounts for the reddish color of many types of rocks and soils. To determine the oxidation states in Fe3O4, we first assign each oxygen atom its usual oxidation state of 2. The three iron atoms must yield a total of 8 to balance the total of 8 from the four oxygens. This means that each iron atom has an oxidation state of A noninteger value for the oxidation state may seem strange because charge is expressed in whole numbers. However, although they are rare, noninteger oxidation states do occur because of the rather arbitrary way that electrons are divided up by the rules in Table 4.2. For Fe3O4, for example, the rules as-sume that all the iron atoms are equal, when in fact this compound can best be viewed as containing four O2 ions, two Fe3 ions, and one Fe2 ion per formula unit. (Note that the “average” charge on iron works out to be which is equal to the oxidation state we determined above.) Noninteger oxidation states should not intimidate you. They are used in the same way as integer oxidation states—for keeping track of electrons.
The Characteristics of Oxidation–Reduction Reactions Oxidation–reduction reactions are characterized by a transfer of electrons. In some cases, the transfer occurs in a literal sense to form ions, such as in the reaction However, sometimes the transfer is less obvious. For example, consider the combustion of methane (the oxidation state for each atom is given): Oxidation h h h state 4 1 0 4 2 1 2 (each H) (each O)(each H) Note that the oxidation state for oxygen in O2 is 0 because it is in elemental form. In this reaction there are no ionic compounds, but we can still describe the process in terms of a transfer of electrons. Note that carbon undergoes a change in oxidation state from 4 in CH4 to 4 in CO2. Such a change can be accounted for by a loss of eight electrons (the symbol e stands for an electron); h h 4 4 CH4 ¡ CO2 8e CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 2Na1s2 Cl21g2 ¡ 2NaCl1s2 8 3, 8 3.
5 3122 1 NO3 6 6112 0 SF6 Magnetite is a magnetic ore containing Fe3O4. Note that the compass needle points toward the ore.
8n 88 n 88n 8n 4.9 Oxidation–Reduction Reactions 159 On the other hand, each oxygen changes from an oxidation state of 0 in O2 to 2 in H2O and CO2, signifying a gain of two electrons per atom. Since four oxygen atoms are in-volved, this is a gain of eight electrons: h 0 4(2) 8 No change occurs in the oxidation state of hydrogen, and it is not formally involved in the electron-transfer process.
With this background, we can now define some important terms. Oxidation is an in-crease in oxidation state (a loss of electrons). Reduction is a decrease in oxidation state (a gain of electrons). Thus in the reaction h h h 0 0 1 1 sodium is oxidized and chlorine is reduced. In addition, Cl2 is called the oxidizing agent (electron acceptor), and Na is called the reducing agent (electron donor). These terms are summarized in Fig. 4.20.
Concerning the reaction h h h h 4 1 0 4 2 1 2 CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 2Na1s2 Cl21g2 ¡ 2NaCl1s2 2O2 8e ¡ CO2 2H2O loses electrons oxidation state increases reducing agent gains electrons oxidation state decreases oxidizing agent M X M+ X – electron transfer Reduced Oxidized FIGURE 4.20 A summary of an oxidation–reduction process, in which M is oxidized and X is reduced.
8 n 88n 8 n 8 n 8n 8n CHEMICAL IMPACT Pearly Whites P eople have long been concerned about the “whiteness” of their teeth. In the Middle Ages the local barber-surgeon would whiten teeth using nitric acid—a procedure fraught with dangers, including the fact that nitric acid dis-solves tooth enamel, which in turn leads to massive tooth decay. Today many safer procedures are available for keep-ing teeth sparkling white.
The outer layer of teeth, the enamel, consists of the min-eral hydroxyapatite, which contains calcium phosphate.
Underneath the enamel is dentin, an off-white mixture of calcium phosphate and collagen that protects the nerves and blood vessels at the center of the tooth.
The discoloration of teeth is usually due to colored mol-ecules in our diet from sources such as blueberries, red wine, and coffee. The tar from cigarettes also stains teeth. Aging is another factor. As we get older, chemical changes occur that cause the dentin to become more yellow.
The stains produced when colored molecules are ad-sorbed to the surfaces of teeth can be removed by brushing.
Toothpastes contain abrasives such as tiny particles of silica, aluminum oxide, calcium carbonate, or calcium phosphate to help scrub off adsorbed stains.
Stains due to molecules lying below the surface are usu-ally attacked with an oxidizing agent, hydrogen peroxide (H2O2). As H2O2 breaks down into water and oxygen, in-termediates are produced that react with and decompose the molecules that produce teeth discoloration.
Off-the-shelf teeth whiteners typically contain carbamide peroxide (a 1:1 mixture of urea and hydrogen peroxide), glyc-erin, stannate and pyrophosphate salts (preservatives), and flavoring agents. These whiteners come in a form that can be brushed directly onto the teeth or are embedded in a plastic strip that can be stuck to the teeth. Because these products have a low strength for safety reasons, it may take several weeks of applying them for full whitening to occur.
Whitening treatments by dentists often involve the ap-plication of substances containing more than 30% hydrogen peroxide. These substances must be used with the appropri-ate protection of the tissues surrounding the teeth. Keeping your teeth white is another example of chemistry in action.
160 Chapter Four Types of Chemical Reactions and Solution Stoichiometry we can say the following: Carbon is oxidized because there has been an increase in its oxidation state (carbon has formally lost electrons).
Oxygen is reduced because there has been a decrease in its oxidation state (oxygen has formally gained electrons).
CH4 is the reducing agent.
O2 is the oxidizing agent.
Note that when the oxidizing or reducing agent is named, the whole compound is specified, not just the element that undergoes the change in oxidation state.
Oxidation–Reduction Reactions I When powdered aluminum metal is mixed with pulverized iodine crystals and a drop of water is added to help the reaction get started, the resulting reaction produces a great deal of energy. The mixture bursts into flames, and a purple smoke of I2 vapor is produced from the excess iodine. The equation for the reaction is For this reaction, identify the atoms that are oxidized and reduced, and specify the oxi-dizing and reducing agents.
Solution The first step is to assign oxidation states: h h h 0 0 3 1 (each I) Free elements AlI3(s) is a salt that contains Al3 and I ions 2Al1s2 3I21s2 ¡ 2AlI31s2 2Al1s2 3I21s2 ¡ 2AlI31s2 8n Finely ground aluminum and iodine are mixed and react vigorously to form alu-minum iodide after a drop of water is added. The purple cloud is excess iodine vaporized by the heat of the reaction.
Sample Exercise 4.17 Oxidation is an increase in oxidation state. Reduction is a decrease in oxidation state.
A helpful mnemonic device is OIL RIG (Oxidation Involves Loss; Reduction Involves Gain). Another common mnemonic is LEO says GER. (Loss of Electrons, Oxidation; Gain of Electrons, Reduction).
An oxidizing agent is reduced and a reducing agent is oxidized in a redox reaction.
CHEMICAL IMPACT Aging: Does It Involve Oxidation?
A lthough aging is supposed to bring wisdom, almost no one wants to get old. Along with wisdom, aging brings wrinkles, loss of physical strength, and greater susceptibil-ity to disease.
Why do we age? No one knows for certain, but many scientists think that oxidation plays a major role. The oxy-gen molecule and other oxidizing agents in the body appar-ently can extract single electrons from the large molecules that make up cell membranes, thus making them very reac-tive. Subsequently, these activated molecules can link up, changing the properties of the cell membrane. At some point, enough of these reactions have occurred that the body’s im-mune system comes to view the changed cell as an “enemy” and destroys it. This is particularly detrimental to the or-ganism when the cells involved are irreplaceable. Nerve cells, for example, fall into this category. They rarely re-generate in an adult.
The body has defenses against oxidation, such as vita-min E, a well-known antioxidant. Studies have shown that red blood cells age much faster than normal when they are deficient in vitamin E. Based on studies such as these, some have suggested large doses of vitamin E as a preventive meas-ure against aging, but there is no solid evidence that this practice has any impact on aging.
Another protective antioxidant found in our bodies is su-peroxide dismutase (SOD), which protects us from the superoxide ion O2 , a powerful oxidizing agent that is par-ticularly damaging to vital enzymes. The importance of SOD in opposing the aging process is indicated from the results of a study by Dr. Richard Cutler at the Gerontology Research 4.9 Oxidation–Reduction Reactions 161 Since each aluminum atom changes its oxidation state from 0 to 3 (an increase in oxi-dation state), aluminum is oxidized. On the other hand, the oxidation state of each iodine atom decreases from 0 to 1, and iodine is reduced. Since Al furnishes electrons for the reduction of iodine, it is the reducing agent; I2 is the oxidizing agent.
See Exercises 4.71 and 4.72.
Oxidation–Reduction Reactions II Metallurgy, the process of producing a metal from its ore, always involves oxidation–reduction reactions. In the metallurgy of galena (PbS), the principal lead-containing ore, the first step is the conversion of lead sulfide to its oxide (a process called roasting): The oxide is then treated with carbon monoxide to produce the free metal: For each reaction, identify the atoms that are oxidized and reduced, and specify the oxidizing and reducing agents.
Solution For the first reaction, we can assign the following oxidation states: h h h h 2 2 0 2 2 4 2 (each O) The oxidation state for the sulfur atom increases from 2 to 4. Thus sulfur is oxidized.
The oxidation state for each oxygen atom decreases from 0 to 2. Oxygen is reduced. The 2PbS1s2 3O21g2 ¡ 2PbO1s2 2SO21g2 PbO1s2 CO1g2 ¡ Pb1s2 CO21g2 2PbS1s2 3O21g2 ¡ 2PbO1s2 2SO21g2 Sample Exercise 4.18 8 n 8 n 8 n Center of the National Institutes of Health in Baltimore that showed a strong correlation between the life spans of a dozen mammalian species and their levels of SOD. Human SOD is now being produced by the techniques of biotechnology in amounts that will enable scientists to carefully study its effects on aging and on various diseases such as rheumatoid arthritis and muscular dystrophy. Although SOD is available in health food stores in forms to be taken orally, this practice is useless because the SOD is digested (broken down into simpler substances) before it can reach the bloodstream.
Research does indicate that consuming certain foods may retard the aging process. For example, a recent study of 8000 male Harvard graduates found that chocolate and candy eaters live almost a year longer than those who abstain. Al-though the researchers from Harvard School of Public Health are not certain of the mechanism for this effect, they suggest that the antioxidants present in chocolate may provide the health benefits. For example, chocolate contains phenols, antioxidants that are also present in wine, another substance that seems to promote good health if used in moderation.
Oxidation is only one possible cause for aging. Research continues on many fronts to try to discover why we get “older” as time passes.
Can eating chocolate slow down the aging process?
162 Chapter Four Types of Chemical Reactions and Solution Stoichiometry oxidizing agent (that accepts the electrons) is O2, and the reducing agent (that donates electrons) is PbS.
For the second reaction we have h h h h 2 2 2 2 0 4 2 (each O) Lead is reduced (its oxidation state decreases from 2 to 0), and carbon is oxidized (its oxidation state increases from 2 to 4). PbO is the oxidizing agent, and CO is the re-ducing agent.
See Exercises 4.71 and 4.72.
4.10 Balancing Oxidation–Reduction Equations Oxidation–reduction reactions in aqueous solutions are often complicated, which means that it can be difficult to balance their equations by simple inspection. In this section we will discuss a special technique for balancing the equations of redox reactions that occur in aqueous solutions. It is called the half-reaction method.
The Half-Reaction Method for Balancing Oxidation–Reduction Reactions in Aqueous Solutions For oxidation–reduction reactions that occur in aqueous solution, it is useful to separate the reaction into two half-reactions: one involving oxidation and the other involving reduction. For example, consider the unbalanced equation for the oxidation–reduction reaction between cerium(IV) ion and tin(II) ion: This reaction can be separated into a half-reaction involving the substance being reduced, and one involving the substance being oxidized, The general procedure is to balance the equations for the half-reactions separately and then to add them to obtain the overall balanced equation. The half-reaction method for balancing oxidation–reduction equations differs slightly depending on whether the reaction takes place in acidic or basic solution.
The Half-Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution ➥ 1 Write separate equations for the oxidation and reduction half-reactions.
➥ 2 For each half-reaction, a. Balance all the elements except hydrogen and oxygen.
b. Balance oxygen using H2O.
Sn21aq2 ¡ Sn41aq2 Ce41aq2 ¡ Ce31aq2 Ce41aq2 Sn21aq2 ¡ Ce31aq2 Sn41aq2 PbO1s2 CO1g2 ¡ Pb1s2 CO21g2 8 n 8 n 8 n 4.10 Balancing Oxidation–Reduction Equations 163 We will illustrate this method by balancing the equation for the reaction between permanganate and iron(II) ions in acidic solution: This reaction can be used to analyze iron ore for its iron content.
➥ 1 Identify and write equations for the half-reactions. The oxidation states for the half-reaction involving the permanganate ion show that manganese is reduced: h h 7 2 (each O) 2 This is the reduction half-reaction. The other half-reaction involves the oxidation of iron(II) to iron(III) ion and is the oxidation half-reaction: h h 2 3 Fe2 ¡ Fe3 MnO4 ¡ Mn2 MnO4 1aq2 Fe21aq2 ¡ Fe31aq2 Mn21aq2 c. Balance hydrogen using H.
d. Balance the charge using electrons.
➥ 3 If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions.
➥ 4 Add the half-reactions, and cancel identical species.
➥ 5 Check that the elements and charges are balanced.
These steps are summarized by the following flowchart: Balance Balance Equalize electrons transferred Equalize electrons transferred Cancel identical species Balancing order a. elements (except H,O) b. oxygen (use H2O) c. hydrogen (use H+) d. charge (use electrons) Check that elements and charges are balanced Write separate half-reactions Oxidation half-reaction Balanced oxidation half-reaction Reduction half-reaction Balanced reduction half-reaction Add half-reactions Acid 8 n 164 Chapter Four Types of Chemical Reactions and Solution Stoichiometry ➥2 Balance each half-reaction. For the reduction reaction, we have a. The manganese is balanced.
b. We balance oxygen by adding 4H2O to the right side of the equation: c. Next, we balance hydrogen by adding 8H to the left side: d. All the elements have been balanced, but we need to balance the charge using electrons. At this point we have the following overall charges for reactants and products in the reduction half-reaction: 8 1 2 0 7 2 We can equalize the charges by adding five electrons to the left side: 2 2 Both the elements and the charges are now balanced, so this represents the balanced reduction half-reaction. The fact that five electrons appear on the reactant side of the equation makes sense, since five electrons are required to reduce MnO4 (Mn has an oxidation state of 7) to Mn2 (Mn has an oxidation state of 2).
For the oxidation reaction the elements are balanced, and we must simply balance the charge: 2 3 One electron is needed on the right side to give a net 2 charge on both sides: 2 2 ➥ 3 Equalize the electron transfer in the two half-reactions. Since the reduction half-reaction involves a transfer of five electrons and the oxidation half-reaction involves a transfer of only one electron, the oxidation half-reaction must be multiplied by 5: ➥ 4 Add the half-reactions. The half-reactions are added to give Note that the electrons cancel (as they must) to give the final balanced equation: 5Fe21aq2 MnO4 1aq2 8H1aq2 ¡ 5Fe31aq2 Mn21aq2 4H2O1l2 5Fe31aq2 Mn21aq2 4H2O1l2 5e 5e 5Fe21aq2 MnO4 1aq2 8H1aq2 ¡ 5Fe21aq2 ¡ 5Fe31aq2 5e Fe21aq2 ¡ Fe31aq2 e Fe21aq2 ¡ Fe31aq2 Fe21aq2 ¡ Fe31aq2 5e 8H1aq2 MnO4 1aq2 ¡ Mn21aq2 4H2O1l2 8H1aq2 MnO4 1aq2 ¡ Mn21aq2 4H2O1l2 8H1aq2 MnO4 1aq2 ¡ Mn21aq2 4H2O1l2 MnO4 1aq2 ¡ Mn21aq2 4H2O1l2 MnO4 1aq2 ¡ Mn21aq2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ The number of electrons gained in the reduction half-reaction must equal the number of electrons lost in the oxidation half-reaction.
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 4.10 Balancing Oxidation–Reduction Equations 165 ➥ 5 Check that elements and charges are balanced.
Elements balance: Charges balance: The equation is balanced.
Balancing Oxidation–Reduction Reactions (Acidic) Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blue-violet solution of Cr3 ions. Under certain conditions, K2Cr2O7 reacts with ethyl alcohol (C2H5OH) as follows: Balance this equation using the half-reaction method.
Solution ➥ 1 The reduction half-reaction is Chromium is reduced from an oxidation state of 6 in Cr2O7 2 to one of 3 in Cr3.
The oxidation half-reaction is Carbon is oxidized from an oxidation state of 2 in C2H5OH to 4 in CO2.
➥ 2 Balancing all elements except hydrogen and oxygen in the first half-reaction, we have Balancing oxygen using H2O, we have Balancing hydrogen using H, we have Balancing the charge using electrons, we have 6e 14H1aq2 Cr2O7 21aq2 ¡ 2Cr31aq2 7H2O1l2 14H1aq2 Cr2O7 21aq2 ¡ 2Cr31aq2 7H2O1l2 Cr2O7 21aq2 ¡ 2Cr31aq2 7H2O1l2 Cr2O7 21aq2 ¡ 2Cr31aq2 C2H5OH1l2 ¡ CO21g2 Cr2O7 21aq2 ¡ Cr31aq2 H1aq2 Cr2O7 21aq2 C2H5OH1l2 ¡ Cr31aq2 CO21g2 H2O1l2 5132 122 0 17 5122 112 8112 17 ¡ 5Fe, 1Mn, 4O, 8H ¡ 5Fe, 1Mn, 4O, 8H Sample Exercise 4.19 When potassium dichromate reacts with ethanol, a blue-violet solution containing Cr3 is formed. 166 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Next, we turn to the oxidation half-reaction Balancing carbon, we have Balancing oxygen using H2O, we have Balancing hydrogen using H, we have We then balance the charge by adding 12e to the right side: ➥ 3 In the reduction half-reaction there are 6 electrons on the left-hand side, and there are 12 electrons on the right-hand side of the oxidation half-reaction. Thus we multiply the reduction half-reaction by 2 to give ➥ 4 Adding the half-reactions and canceling identical species, we have ➥ 5 Check that elements and charges are balanced.
See Exercises 4.73 and 4.74.
Oxidation–reduction reactions can occur in basic solutions (the reactions involve OH ions) as well as in acidic solution (the reactions involve H ions). The half-reaction method for balancing equations is slightly different for the two cases.
The Half-Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution ➥ 1 Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H ions were present.
➥ 2 To both sides of the equation obtained above, add a number of OH ions that is equal to the number of H ions. (We want to eliminate H by form-ing H2O.) ➥ 3 Form H2O on the side containing both H and OH ions, and eliminate the number of H2O molecules that appear on both sides of the equation.
➥ 4 Check that elements and charges are balanced.
Charges balance: 16 2122 0 12 ¡ 4132 0 0 12 Elements balance: 22H, 4Cr, 15O, 2C ¡ 22H, 4Cr, 15O, 2C 16H1aq2 2Cr2O7 21aq2 C2H5OH1l2 ¡ 4Cr3 11H2O1l2 2CO21g2 C2H5OH1l2 3H2O1l2 ¡ 2CO21g2 12H1aq2 12e 12e 28H1aq2 2Cr2O7 21aq2 ¡ 4Cr31aq2 14H2O1l2 12e 28H1aq2 2Cr2O7 21aq2 ¡ 4Cr31aq2 14H2O1l2 C2H5OH1l2 3H2O1l2 ¡ 2CO21g2 12H1aq2 12e C2H5OH1l2 3H2O1l2 ¡ 2CO21g2 12H1aq2 C2H5OH1l2 3H2O1l2 ¡ 2CO21g2 C2H5OH1l2 ¡ 2CO21g2 C2H5OH1l2 ¡ CO21g2 Reduction Half-Reaction: Oxidation Half-Reaction: Complete Reaction: 4.10 Balancing Oxidation–Reduction Equations 167 We will illustrate how this method is applied in Sample Exercise 4.20.
Balancing Oxidation–Reduction Reactions (Basic) Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. An aqueous solution containing cyanide ion is often used to extract the silver using the following reaction that occurs in basic solution: Balance this equation using the half-reaction method.
Solution ➥1 Balance the equation as if H ions were present. Balance the oxidation half-reaction: Balance carbon and nitrogen: Balance the charge: 2CN1aq2 Ag1s2 ¡ Ag1CN22 1aq2 e 2CN1aq2 Ag1s2 ¡ Ag1CN22 1aq2 CN1aq2 Ag1s2 ¡ Ag1CN22 1aq2 Ag1s2 CN1aq2 O21g2 — ¡ Basic Ag1CN22 1aq2 Balance Balance Equalize electrons transferred Equalize electrons transferred Cancel identical species Balancing order a. elements (except H,O) b. oxygen (use H2O) c. hydrogen (use H+) d. charge (use electrons) Add OH– to both sides of equation (equal to H+) Eliminate number of H2O appearing on both sides Write separate half-reactions Oxidation half-reaction Balanced oxidation half-reaction Reduction half-reaction Balanced reduction half-reaction Add half-reactions Check that elements and charges are balanced Form H2O on the side containing H+ and OH– ions Check that elements and charges are balanced Sample Exercise 4.20 This method is summarized by the following flowchart: Key Terms aqueous solution Section 4.1 polar molecule hydration solubility Section 4.2 solute solvent electrical conductivity strong electrolyte weak electrolyte nonelectrolyte acid strong acid strong base weak acid weak base For Review Chemical reactions in solution are very important in everyday life.
Water is a polar solvent that dissolves many ionic and polar substances.
Electrolytes Strong electolyte: 100% dissociated to produce separate ions; strongly conducts an electric current Weak electrolyte: Only a small percentage of dissolved molecules produce ions; weakly conducts an electric current Nonelectrolyte: Dissolved substance produces no ions; does not conduct an electric current Acids and bases Arrhenius model • Acid: produces H • Base: produces OH 168 Chapter Four Types of Chemical Reactions and Solution Stoichiometry Balance the reduction half-reaction: Balance oxygen: Balance hydrogen: Balance the charge: Multiply the balanced oxidation half-reaction by 4: Add the half-reactions, and cancel identical species: ➥ 2 Add OH ions to both sides of the balanced equation to eliminate the H ions.
We need to add 4OH to each side: 4H2O(l) ➥ 3 Eliminate as many H2O molecules as possible: ➥ 4 Check that elements and charges are balanced.
See Exercises 4.75 and 4.76.
Charges balance: 8112 0 0 0 8 ¡ 4112 4112 8 Elements balance: 8C, 8N, 4Ag, 4O, 4H ¡ 8C, 8N, 4Ag, 4O, 4H 8CN1aq2 4Ag1s2 O21g2 2H2O1l2 ¡ 4Ag1CN22 1aq2 4OH1aq2 4Ag1CN22 1aq2 2H2O1l2 4OH1aq2 8CN1aq2 4Ag1s2 O21g2 4H1aq2 4OH1aq2 ¡ 8CN1aq2 4Ag1s2 O21g2 4H1aq2 ¡ 4Ag1CN22 1aq2 2H2O1l2 4e O21g2 4H1aq2 ¡ 2H2O1l2 8CN1aq2 4Ag1s2 ¡ 4Ag1CN22 1aq2 4e 8CN1aq2 4Ag1s2 ¡ 4Ag1CN22 1aq2 4e 4e O21g2 4H1aq2 ¡ 2H2O1l2 O21g2 4H1aq2 ¡ 2H2O1l2 O21g2 ¡ 2H2O1l2 O21g2 ¡ Oxidation Half-Reaction: Reduction Half-Reaction: Complete Reaction: ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ For Review 169 Section 4.3 molarity standard solution dilution Section 4.5 precipitation reaction precipitate Section 4.6 formula equation complete ionic equation spectator ions net ionic equation Section 4.8 acid base neutralization reaction volumetric analysis titration stoichiometric (equivalence) point indicator endpoint Section 4.9 oxidation–reduction (redox) reaction oxidation state oxidation reduction oxidizing agent (electron acceptor) reducing agent (electron donor) Section 4.10 half-reactions Brønsted–Lowry model • Acid: proton donor • Base: proton acceptor Strong acid: completely dissociates into separated H and anions Weak acid: dissociates to a slight extent Molarity One way to describe solution composition Moles solute volume of solution (L) molarity Standard solution: molarity is accurately known Dilution Solvent is added to reduce the molarity Moles of solute after dilution moles of solute before dilution Types of equations that describe solution reactions Formula equation: all reactants and products are written as complete formulas Complete ionic equation: all reactants and products that are strong electrolytes are written as separated ions Net ionic equation: only those compounds that undergo a change are written; spectator ions are not included Solubility rules Based on experiment observation Help predict the outcomes of precipitation reactions Important types of solution reactions Acid–base reactions: involve a transfer of H ions Precipitation reactions: formation of a solid occurs Oxidation–reduction reactions: involve electron transfer Titrations Measures the volume of a standard solution (titrant) needed to react with a substance in solution Stoichiometric (equivalence) point: the point at which the required amount of titrant has been added to exactly react with the substance being analyzed Endpoint: the point at which a chemical indicator changes color Oxidation–reduction reactions Oxidation states are assigned using a set of rules to keep track of electron flow Oxidation: increase in oxidation state (a loss of electrons) Reduction: decrease in oxidation state (a gain of electrons) Oxidizing agent: gains electrons (is reduced) Reducing agent: loses electrons (is oxidized) Equations for oxidation–reduction reactions are usually balanced by the half-reaction method REVIEW QUESTIONS 1. The (aq) designation listed after a solute indicates the process of hydration. Us-ing KBr(aq) and C2H5OH(aq) as your examples, explain the process of hydra-tion for soluble ionic compounds and for soluble covalent compounds.
2. Characterize strong electrolytes versus weak electrolytes versus nonelectrolytes.
Give examples of each. How do you experimentally determine whether a solu-ble substance is a strong electrolyte, weak electrolyte, or nonelectrolyte?
M1V1 M2V2 Molarity 1M2 moles of solute volume of solution 1L2 3. You have a sugar solution (solution A) with concentration x. You pour one-fourth of this solution into a beaker, and add an equiv-alent volume of water (solution B).
a. What is the ratio of sugar in solutions A and B?
b. Compare the volumes of solutions A and B.
c. What is the ratio of the concentrations of sugar in solutions A and B?
4. You add an aqueous solution of lead nitrate to an aqueous solu-tion of potassium iodide. Draw highly magnified views of each solution individually, and the mixed solution including any prod-uct that forms. Write the balanced equation for the reaction.
5. Order the following molecules from lowest to highest oxidation state of the nitrogen atom: HNO3, NH4Cl, N2O, NO2, NaNO2.
6. Why is it that when something gains electrons, it is said to be reduced? What is being reduced?
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Assume you have a highly magnified view of a solution of HCl that allows you to “see” the HCl. Draw this magnified view. If you dropped in a piece of magnesium, the magnesium would disappear and hydrogen gas would be released. Represent this change using symbols for the elements, and write out the balanced equation.
2. You have a solution of table salt in water. What happens to the salt concentration (increases, decreases, or stays the same) as the solution boils? Draw pictures to explain your answer.
3. Distinguish between the terms slightly soluble and weak electrolyte.
4. Molarity is a conversion factor relating moles of solute in solution to the vol-ume of the solution. How does one use molarity as a conversion factor to con-vert from moles of solute to volume of solution, and from volume of solution to moles of solute present?
5. What is a dilution? What stays constant in a dilution? Explain why the equation M1V1 M2V2 works for dilution problems.
6. When the following beakers are mixed, draw a molecular-level representation of the product mixture (see Fig. 4.17).
7. Differentiate between the formula equation, the complete ionic equation, and the net ionic equation. For each reaction in Question 6, write all three balanced equations.
8. What is an acid–base reaction? Strong bases are soluble ionic compounds that contain the hydroxide ion. List the strong bases. When a strong base reacts with an acid, what is always produced? Explain the terms titration, stoichiometric point, neutralization, and standardization.
9. Define the terms oxidation, reduction, oxidizing agent, and reducing agent.
Given a chemical reaction, how can you tell if it is a redox reaction?
10. What is a half-reaction? Why must the number of electrons lost in the oxidation equal the number of electrons gained in a reduction? Summarize briefly the steps in the half-reaction method for balancing redox reactions. What two items must be balanced in a redox reaction (or any reaction)?
Na+ + + Pb2+ NO3 – K+ Cl– Al3+ Br– OH– Exercises 171 7. Consider separate aqueous solutions of HCl and H2SO4 with the same molar concentrations. You wish to neutralize an aqueous solution of NaOH. For which acid solution would you need to add more volume (in milliliters) to neutralize the base?
a. the HCl solution b. the H2SO4 solution c. You need to know the acid concentrations to answer this question.
d. You need to know the volume and concentration of the NaOH solution to answer this question.
e. c and d Explain.
8. Draw molecular-level pictures to differentiate between concen-trated and dilute solutions.
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 9. Differentiate between what happens when the following are dis-solved in water.
a. polar solute versus nonpolar solute b. KF versus C6H12O6 c. RbCl versus AgCl d. HNO3 versus CO 10. A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 M NaOH is avail-able, how would the student prepare the solution? To help insure three significant figures in the NaOH molarity, to how many sig-nificant figures should the volumes and mass be determined?
11. List the formulas of three soluble bromide salts and three insol-uble bromide salts. Do the same exercise for sulfate salts, hy-droxide salts, and phosphate salts (list three soluble salts and three insoluble salts). List the formulas for six insoluble Pb2 salts and one soluble Pb2 salt.
12. When 1.0 mol of solid lead nitrate is added to 2.0 mol of aqueous potassium iodide, a yellow precipitate forms. After the precipi-tate settles to the bottom, does the solution above the precipitate conduct electricity? Explain. Write the complete ionic equation to help you answer this question.
13. What is an acid and what is a base? An acid–base reaction is sometimes called a proton-transfer reaction. Explain.
14. A student had 1.00 L of a 1.00 M acid solution. Much to the sur-prise of the student, it took 2.00 L of 1.00 M NaOH solution to react completely with the acid. Explain why it took twice as much NaOH to react with all of the acid.
In a different experiment, a student had 10.0 mL of 0.020 M HCl. Again, much to the surprise of the student, it took only 5.00 mL of 0.020 M strong base to react completely with the HCl. Explain why it took only half as much strong base to react with all of the HCl.
15. Differentiate between the following terms.
a. species reduced versus the reducing agent b. species oxidized versus the oxidizing agent c. oxidation state versus actual charge 16. When balancing reactions in Chapter 3, we did not mention that reactions must be charge balanced as well as mass balanced.
What do charge balanced and mass balanced mean? How are redox reactions charge balanced?
Exercises In this section similar exercises are paired.
Aqueous Solutions: Strong and Weak Electrolytes 17. Show how each of the following strong electrolytes “breaks up” into its component ions upon dissolving in water by drawing molecular-level pictures.
a. NaBr f. FeSO4 b. MgCl2 g. KMnO4 c. Al(NO3)3 h. HClO4 d. (NH4)2SO4 i. NH4C2H3O2 (ammonium acetate) e. NaOH 18. Match each name below with the following microscopic pictures of that compound in aqueous solution.
a. barium nitrate c. potassium carbonate b. sodium chloride d. magnesium sulfate Which picture best represents HNO3(aq)? Why aren’t any of the pictures a good representation of HC2H3O2(aq)?
19. Calcium chloride is a strong electrolyte and is used to “salt” streets in the winter to melt ice and snow. Write a reaction to show how this substance breaks apart when it dissolves in water.
20. Commercial cold packs and hot packs are available for treating athletic injuries. Both types contain a pouch of water and a dry chemical. When the pack is struck, the pouch of water breaks, dissolving the chemical, and the solution becomes either hot or cold. Many hot packs use magnesium sulfate, and many cold packs use ammonium nitrate. Write reactions to show how these strong electrolytes break apart when they dissolve in water.
Solution Concentration: Molarity 21. Calculate the molarity of each of these solutions.
a. A 5.623-g sample of NaHCO3 is dissolved in enough water to make 250.0 mL of solution.
b. A 184.6-mg sample of K2Cr2O7 is dissolved in enough water to make 500.0 mL of solution.
c. A 0.1025-g sample of copper metal is dissolved in 35 mL of concentrated HNO3 to form Cu2 ions and then water is added to make a total volume of 200.0 mL. (Calculate the molarity of Cu2.) 22. A solution of ethanol (C2H5OH) in water is prepared by dis-solving 75.0 mL of ethanol (density 0.79 g/cm3) in enough water to make 250.0 mL of solution. What is the molarity of the ethanol in this solution?
2+ i ii iii iv 2+ 2+ 2+ 2– 2– 2– 2– + + + + + + + – – – – – – – 172 Chapter Four Types of Chemical Reactions and Solution Stoichiometry For solution B, 10.00 mL of solution A was diluted to 250.0 mL.
For solution C, 10.00 mL of solution B was diluted to 500.0 mL.
Calculate the concentrations of the stock solution and solutions A, B, and C.
Precipitation Reactions 35. On the basis of the general solubility rules given in Table 4.1, predict which of the following substances are likely to be solu-ble in water.
a. aluminum nitrate b. magnesium chloride c. rubidium sulfate d. nickel(II) hydroxide e. lead(II) sulfide f. magnesium hydroxide g. iron(III) phosphate 36. On the basis of the general solubility rules given in Table 4.1, predict which of the following substances are likely to be soluble in water.
a. zinc chloride b. lead(II) nitrate c. lead(II) sulfate d. sodium iodide e. cobalt(III) sulfide f. chromium(III) hydroxide g. magnesium carbonate h. ammonium carbonate 37. When the following solutions are mixed together, what precipi-tate (if any) will form?
a.
b.
c.
d.
38. When the following solutions are mixed together, what precipi-tate (if any) will form?
a.
b.
c.
d.
39. For the reactions in Exercise 37, write the balanced formula equation, complete ionic equation, and net ionic equation. If no precipitate forms, write “No reaction.” 40. For the reactions in Exercise 38, write the balanced formula equation, complete ionic equation, and net ionic equation. If no precipitate forms, write “No reaction.” 41. Write the balanced formula and net ionic equation for the reaction that occurs when the contents of the two beakers are added to-gether. What colors represent the spectator ions in each reaction?
Cu2+ SO4 2– Na+ S2– a.
+ Na2CrO41aq2 AlBr31aq2 K2CO31aq2 MgI21aq2 Ni1NO3221aq2 CaCl21aq2 Hg21NO3221aq2 CuSO41aq2 K2S1aq2 Ni1NO3221aq2 CaCl21aq2 Na2SO41aq2 Al1NO3231aq2 Ba1OH221aq2 FeSO41aq2 KCl1aq2 23. Calculate the concentration of all ions present in each of the fol-lowing solutions of strong electrolytes.
a. 0.100 mol of Ca(NO3)2 in 100.0 mL of solution b. 2.5 mol of Na2SO4 in 1.25 L of solution c. 5.00 g of NH4Cl in 500.0 mL of solution d. 1.00 g K3PO4 in 250.0 mL of solution 24. Calculate the concentration of all ions present in each of the fol-lowing solutions of strong electrolytes.
a. 0.0200 mol of sodium phosphate in 10.0 mL of solution b. 0.300 mol of barium nitrate in 600.0 mL of solution c. 1.00 g of potassium chloride in 0.500 L of solution d. 132 g of ammonium sulfate in 1.50 L of solution 25. Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: 100.0 mL of 0.30 M AlCl3, 50.0 mL of 0.60 M MgCl2, or 200.0 mL of 0.40 M NaCl?
26. Which of the following solutions of strong electrolytes contains the largest number of ions: 100.0 mL of 0.100 M NaOH, 50.0 mL of 0.200 M BaCl2, or 75.0 mL of 0.150 M Na3PO4?
27. What mass of NaOH is contained in 250.0 mL of a 0.400 M sodium hydroxide solution?
28. If 10. g of AgNO3 is available, what volume of 0.25 M AgNO3 solution can be prepared?
29. Describe how you would prepare 2.00 L of each of the follow-ing solutions.
a. 0.250 M NaOH from solid NaOH b. 0.250 M NaOH from 1.00 M NaOH stock solution c. 0.100 M K2CrO4 from solid K2CrO4 d. 0.100 M K2CrO4 from 1.75 M K2CrO4 stock solution 30. How would you prepare 1.00 L of a 0.50 M solution of each of the following?
a. H2SO4 from “concentrated” (18 M) sulfuric acid b. HCl from “concentrated” (12 M) reagent c. NiCl2 from the salt NiCl2 6H2O d. HNO3 from “concentrated” (16 M) reagent e. Sodium carbonate from the pure solid 31. A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water.
Calculate the concentration of ammonium ions and sulfate ions in the final solution.
32. Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate.
33. A standard solution is prepared for the analysis of fluoxy-mesterone (C20H29FO3), an anabolic steroid. A stock solution is first prepared by dissolving 10.0 mg of fluoxymesterone in enough water to give a total volume of 500.0 mL. A 100.0-L aliquot (portion) of this solution is diluted to a final volume of 100.0 mL. Calculate the concentration of the final solution in terms of molarity.
34. A stock solution containing Mn2 ions was prepared by dis-solving 1.584 g pure manganese metal in nitric acid and dilut-ing to a final volume of 1.000 L. The following solutions were then prepared by dilution: For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL.
Exercises 173 51. How many grams of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.
52. The drawings below represent aqueous solutions. Solution A is 2.00 L of a 2.00 M aqueous solution of copper(II) nitrate. Solution B is 2.00 L of a 3.00 M aqueous solution of potassium hydroxide.
A B a. Draw a picture of the solution made by mixing solutions A and B together after the precipitation reaction takes place.
Make sure this picture shows the correct relative volume com-pared to solutions A and B, and the correct relative number of ions, along with the correct relative amount of solid formed.
b. Determine the concentrations (in M) of all ions left in solu-tion (from part a) and the mass of solid formed.
53. A 1.42-g sample of a pure compound, with formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.
54. You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You pro-duce a white solid, which you then collect, dry, and measure.
The white solid has a mass of 0.641 g.
a. If you had an extremely magnified view of the solution (to the atomic-molecular level), list the species you would see (include charges, if any).
b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges.
c. Calculate the percent sodium chloride in the original unknown mixture.
Acid–Base Reactions 55. Write the balanced formula, complete ionic, and net ionic equa-tions for each of the following acid–base reactions.
a.
b.
c.
56. Write the balanced formula, complete ionic, and net ionic equa-tions for each of the following acid–base reactions.
a.
b.
c.
57. Write the balanced formula, complete ionic, and net ionic equa-tions for the reactions that occur when the following are mixed.
a. potassium hydroxide (aqueous) and nitric acid b. barium hydroxide (aqueous) and hydrochloric acid c. perchloric acid [HClO4(aq)] and solid iron(III) hydroxide Ca1OH221aq2 HCl1aq2 S HC2H3O21aq2 KOH1aq2 S HNO31aq2 Al1OH231s2 S HCl1aq2 NaOH1aq2 S HCN1aq2 NaOH1aq2 S HClO41aq2 Mg1OH221s2 S Cu2+ NO3 – K+ OH– 42. Give an example how each of the following insoluble ionic com-pounds could be produced using a precipitation reaction. Write the balanced formula equation for each reaction.
a. Fe(OH)3(s) c. PbSO4(s) b. Hg2Cl2(s) d. BaCrO4(s) 43. Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed.
a. ammonium sulfate and barium nitrate b. lead(II) nitrate and sodium chloride c. sodium phosphate and potassium nitrate d. sodium bromide and rubidium chloride e. copper(II) chloride and sodium hydroxide 44. Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed.
a. chromium(III) chloride and sodium hydroxide b. silver nitrate and ammonium carbonate c. copper(II) sulfate and mercury(I) nitrate d. strontium nitrate and potassium iodide 45. Separate samples of a solution of an unknown soluble ionic compound are treated with KCl, Na2SO4, and NaOH. A precip-itate forms only when Na2SO4 is added. Which cations could be present in the unknown soluble ionic compound?
46. A sample may contain any or all of the following ions: Hg2 2, Ba2, and Mn2.
a. No precipitate formed when an aqueous solution of NaCl was added to the sample solution.
b. No precipitate formed when an aqueous solution of Na2SO4 was added to the sample solution.
c. A precipitate formed when the sample solution was made basic with NaOH.
Which ion or ions are present in the sample solution?
47. What mass of Na2CrO4 is required to precipitate all of the sil-ver ions from 75.0 mL of a 0.100 M solution of AgNO3?
48. What volume of 0.100 M Na3PO4 is required to precipitate all the lead(II) ions from 150.0 mL of 0.250 M Pb(NO3)2?
49. What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100 M KOH?
50. What mass of barium sulfate can be produced when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron(III) sulfate?
Ag+ NO3 – K+ I– c.
+ Co2+ Cl– Na+ OH– b.
+ 174 Chapter Four Types of Chemical Reactions and Solution Stoichiometry 68. Assign oxidation states for all atoms in each of the following compounds.
a. UO2 2 f. Mg2P2O7 b. As2O3 g. Na2S2O3 c. NaBiO3 h. Hg2Cl2 d. As4 i. Ca(NO3)2 e. HAsO2 69. Assign the oxidation state for nitrogen in each of the following.
a. Li3N f. NO2 b. NH3 g. NO2 c. N2H4 h. NO3 d. NO i. N2 e. N2O 70. Assign oxidation numbers to all the atoms in each of the following.
a. SrCr2O7 g. PbSO3 b. CuCl2 h. PbO2 c. O2 i. Na2C2O4 d. H2O2 j. CO2 e. MgCO3 k. (NH4)2Ce(SO4)3 f. Ag l. Cr2O3 71. Specify which of the following are oxidation–reduction reactions, and identify the oxidizing agent, the reducing agent, the sub-stance being oxidized, and the substance being reduced.
a.
b.
c.
d.
e.
72. Specify which of the following equations represent oxidation– reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced.
a.
b.
c.
d.
73. Balance the following oxidation–reduction reactions that occur in acidic solution.
a.
b.
c.
d.
e.
74. Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method.
a.
b.
c.
d.
e.
75. Balance the following oxidation–reduction reactions that occur in basic solution.
a.
b.
c. NO2 1aq2 Al1s2 S NH31g2 AlO2 1aq2 Cl21g2 S Cl1aq2 OCl1aq2 Al1s2 MnO4 1aq2 S MnO21s2 Al1OH24 1aq2 H3AsO41aq2 Zn1s2 S AsH31g2 Zn21aq2 Mn21aq2 NaBiO31s2 S Bi31aq2 MnO4 1aq2 Pb1s2 PbO21s2 H2SO41aq2 S PbSO41s2 Cr2O7 21aq2 Cl1aq2 S Cr31aq2 Cl21g2 Cu1s2 NO3 1aq2 S Cu21aq2 NO1g2 CH3OH1aq2 Cr2O7 21aq2 S CH2O1aq2 Cr31aq2 Br1aq2 MnO4 1aq2 S Br21l2 Mn21aq2 As2O31s2 NO3 1aq2 S H3AsO41aq2 NO1g2 I1aq2 ClO1aq2 S I3 1aq2 Cl1aq2 Zn1s2 HCl1aq2 S Zn21aq2 H21g2 Cl1aq2 2H1aq2 2CrO4 21aq2 S Cr2O7 21aq2 H2O1l2 Zn1s2 2HCl1aq2 S ZnCl21aq2 H21g2 2AgNO31aq2 Cu1s2 S Cu1NO3221aq2 2Ag1s2 CH41g2 H2O1g2 S CO1g2 3H21g2 Al1OH24 1aq2 S AlO2 1aq2 2H2O1l2 SiCl41l2 2Mg1s2 S 2MgCl21s2 Si1s2 SiCl41l2 2H2O1l2 S 4HCl1aq2 SiO21s2 HCl1g2 NH31g2 S NH4Cl1s2 Cu1s2 2Ag1aq2 S 2Ag1s2 Cu21aq2 58. Write the balanced formula, complete ionic, and net ionic equations for the reactions that occur when the following are mixed.
a. solid silver hydroxide and hydrobromic acid b. aqueous strontium hydroxide and hydroiodic acid c. solid chromium(III) hydroxide and nitric acid 59. What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH?
a. 0.100 M HCl b. 0.150 M HNO3 c. 0.200 M HC2H3O2 (1 acidic hydrogen) 60. What volume of each of the following bases will react com-pletely with 25.00 mL of 0.200 M HCl?
a. 0.100 M NaOH b. 0.0500 M Ba(OH)2 c. 0.250 M KOH 61. Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the excess H or OH ions left in this solution?
62. A student mixes four reagents together, thinking that the solu-tions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Is the resulting solution neutral? If not, calculate the concentration of excess H or OH ions left in solution.
63. A 25.00-mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutral-ization. What is the concentration of the original hydrochloric acid solution?
64. What volume of 0.0200 M calcium hydroxide is required to neu-tralize 35.00 mL of 0.0500 M nitric acid?
65. A student titrates an unknown amount of potassium hydrogen phthalate (KHC8H4O4, often abbreviated KHP) with 20.46 mL of a 0.1000 M NaOH solution. KHP (molar mass 204.22 g/mol) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?
66. The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of 204.22 g/mol. In the titration, 34.67 mL of the sodium hydroxide solution was re-quired to react with 0.1082 g KHP. Calculate the molarity of the sodium hydroxide.
Oxidation–Reduction Reactions 67. Assign oxidation states for all atoms in each of the following compounds.
a. KMnO4 f. Fe3O4 b. NiO2 g. XeOF4 c. Na4Fe(OH)6 h. SF4 d. (NH4)2HPO4 i. CO e. P4O6 j. C6H12O6 Additional Exercises 175 which was precipitated by adding an excess of barium chloride solution. The mass of BaSO4 obtained was 0.5032 g. What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?
86. A mixture contains only NaCl and Fe(NO3)3. A 0.456-g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Fe(OH)3. The precipitate is filtered, dried, and weighed. Its mass is 0.107 g. Calculate the following.
a. the mass of iron in the sample b. the mass of Fe(NO3)3 in the sample c. the mass percent of Fe(NO3)3 in the sample 87. A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution.
88. What acid and what strong base would react in aqueous solution to produce the following salts in the formula equation? Write the balanced formula equation for each reaction.
a. potassium perchlorate b. cesium nitrate c. calcium iodide 89. A 10.00-mL sample of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with 0.5062 M NaOH, and 16.58 mL is required to reach the equivalence point.
a. What is the molarity of the acetic acid?
b. If the density of the vinegar is 1.006 g/cm3, what is the mass percent of acetic acid in the vinegar?
90. When hydrochloric acid reacts with magnesium metal, hydro-gen gas and aqueous magnesium chloride are produced. What volume of 5.0 M HCl is required to react completely with 3.00 g of magnesium?
91. A 2.20-g sample of an unknown acid (empirical formula C3H4O3) is dissolved in 1.0 L of water. A titration required 25.0 mL of 0.500 M NaOH to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?
92. Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen.
It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and 4.09% H by mass. A titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 g carminic acid.
Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?
93. A 30.0-mL sample of an unknown strong base is neutralized after the addition of 12.0 mL of a 0.150 M HNO3 solution. If the unknown base concentration is 0.0300 M, give some possible identities for the unknown base.
94. Many oxidation–reduction reactions can be balanced by inspec-tion. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized.
a.
b.
c.
d. Cu1s2 Ag1aq2 S Ag1s2 Cu21aq2 C3H81g2 O21g2 S CO21g2 H2O1l2 CH41g2 S1s2 S CS21l2 H2S1g2 Al1s2 HCl1aq2 S AlCl31aq2 H21g2 76. Balance the following oxidation–reduction reactions that occur in basic solution.
a.
b.
c.
77. Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The re-action is Balance this equation.
78. Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the AuCl4 ion and gaseous NO. Write a bal-anced equation for the dissolution of gold in aqua regia.
Additional Exercises 79. Which of the following statements is (are) true? For the false statements, correct them.
a. A concentrated solution in water will always contain a strong or weak electrolyte.
b. A strong electrolyte will break up into ions when dissolved in water.
c. An acid is a strong electrolyte.
d. All ionic compounds are strong electrolytes in water.
80. A 230.-mL sample of a 0.275 M CaCl2 solution is left on a hot plate overnight; the following morning, the solution is 1.10 M. What volume of water evaporated from the 0.275 M CaCl2 solution?
81. Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions.
a. chloride ion d. sulfate ion b. calcium ion e. mercury(I) ion, Hg2 2 c. iron(III) ion f. silver ion 82. Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500 M silver nitrate is added dropwise until precipitate formation is complete.
The mass of the white precipitate formed is 0.641 g.
a. Calculate the mass percent of magnesium chloride in the mixture.
b. Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.
83. A 1.00-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 1.38 g of silver chloride. Identify the metal.
84. A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of Al(OH)3. The precipitate is fil-tered, dried, and weighed. The mass of the precipitate is 0.107 g. What is the mass percent of Al2(SO4)3 in the sample?
85. Saccharin (C7H5NO3S) is sometimes dispensed in tablet form.
Ten tablets with a total mass of 0.5894 g were dissolved in wa-ter. They were oxidized to convert all the sulfur to sulfate ion, Na2SO41aq2 MnCl21aq2 H2O1l2 Cl21g2 NaCl1aq2 H2SO41aq2 MnO21s2 ¡ CN1aq2 MnO4 1aq2 S CNO1aq2 MnO21s2 MnO4 1aq2 S21aq2 S MnS1s2 S1s2 Cr1s2 CrO4 21aq2 S Cr1OH231s2 176 Chapter Four Types of Chemical Reactions and Solution Stoichiometry the same. When the solutions are added together, a blood-red precipitate forms. After the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g.
a. Calculate the concentration of the potassium ions in the orig-inal potassium chromate solution.
b. Calculate the concentration of the chromate ions in the final solution.
100. A sample is a mixture of KCl and KBr. When 0.1024 g of the sample is dissolved in water and reacted with excess silver nitrate, 0.1889 g solid is obtained. What is the composition by mass percent of the original mixture?
101. You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of BaCl2 is added. The BaSO4 that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of SO4 2 ion is in the sample? What is the mass percent of SO4 2 ion in the sample? What are the percent compositions by mass of Na2SO4 and K2SO4 in the sample?
102. Zinc and magnesium metal each react with hydrochloric acid ac-cording to the following equations: A 10.00-g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mix-ture is then reacted with 156 mL of 3.00 M silver nitrate to pro-duce the maximum possible amount of silver chloride.
a. Determine the percent magnesium by mass in the original mixture.
b. If 78.0 mL of HCl was added, what was the concentration of the HCl?
103. You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chlo-ride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II) nitrate solution?
104. Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations.
You place 87.7 mL of a 0.500 M solution of copper(II) sulfate in a beaker. You then add 2.00 g of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you iso-late 2.27 g of copper. Which equation above describes the reac-tion that occurred? Support your answer.
105. Consider an experiment in which two burets, Y and Z, are simul-taneously draining into a beaker that initially contained 275.0 mL of 0.300 M HCl. Buret Y contains 0.150 M NaOH and buret Z contains 0.250 M KOH. The stoichiometric point in the titration is reached 60.65 minutes after Y and Z were started simultane-ously. The total volume in the beaker at the stoichiometric point is 655 mL. Calculate the flow rates of burets Y and Z. Assume the flow rates remain constant during the experiment.
copper1s2 iron1III2 sulfate1aq2 copper1II2 sulfate1aq2 iron1s2 ¡ copper1s2 iron1II2 sulfate1aq2 copper1II2 sulfate1aq2 iron1s2 ¡ Mg1s2 2HCl1aq2 ¡ MgCl21aq2 H21g2 Zn1s2 2HCl1aq2 ¡ ZnCl21aq2 H21g2 95. One of the classical methods for the determination of the man-ganese content in steel is to convert all the manganese to the deeply colored permanganate ion and then to measure the ab-sorption of light. The steel is dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both these steps.
Challenge Problems 96. The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every 106 parts of solution.
Mathematically, by mass: In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to 1.0 g of solute per 1.0 mL, which equals 1.0 g solution. Parts per billion is defined in a similar fashion.
Calculate the molarity of each of the following aqueous solutions.
a. 5.0 ppb Hg in H2O b. 1.0 ppb CHCl3 in H2O c. 10.0 ppm As in H2O d. 0.10 ppm DDT (C14H9Cl5) in H2O 97. In most of its ionic compounds, cobalt is either Co(II) or Co(III).
One such compound, containing chloride ion and waters of hy-dration, was analyzed, and the following results were obtained.
A 0.256-g sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sam-ple of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g.
a. What is the percent composition, by mass, of the compound?
b. Assuming the compound contains one cobalt atom per for-mula unit, what is the molecular formula?
c. Write balanced equations for the three reactions described.
98. Polychlorinated biphenyls (PCBs) have been used extensively as dielectric materials in electrical transformers. Because PCBs have been shown to be potentially harmful, analysis for their presence in the environment has become very important. PCBs are manufactured according to the following generic reaction: This reaction results in a mixture of PCB products. The mixture is analyzed by decomposing the PCBs and then precipitating the resulting Cl as AgCl.
a. Develop a general equation that relates the average value of n to the mass of a given mixture of PCBs and the mass of AgCl produced.
b. A 0.1947-g sample of a commercial PCB yielded 0.4791 g of AgCl. What is the average value of n for this sample?
99. You have two 500.0 mL aqueous solutions. Solution A is a so-lution of silver nitrate, and solution B is a solution of potassium chromate. The masses of the solutes in each of the solutions are C12H10 nCl2 S C12H10nCln nHCl ppm mg solute g solution mg solute kg solution Marathon Problems 177 115. In a 1-L beaker, 203 mL of 0.307 M ammonium chromate was mixed with 137 mL of 0.269 M chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write the balanced chemical reaction occurring here. If the percent yield of the reac-tion was 88.0%, how much chromium(III) chromate was isolated?
116. The vanadium in a sample of ore is converted to VO2. The VO2 ion is subsequently titrated with MnO4 in acidic solution to form V(OH)4 and manganese(II) ion. To titrate the solution, 26.45 mL of 0.02250 M MnO4 was required. If the mass per-cent of vanadium in the ore was 58.1%, what was the mass of the ore sample? Which of the four transition metal ions in this titration has the highest oxidation state?
117. The unknown acid H2X can be neutralized completely by OH according to the following (unbalanced) equation: The ion formed as a product, X2, was shown to have 36 total electrons. What is element X? Propose a name for H2X? To completely neutralize a sample of H2X, 35.6 mL of 0.175 M OH solution was required. What was the mass of the H2X sample used?
Marathon Problems These problems are designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
118. Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472-g sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precip-itate had been filtered and dried, it weighed 0.2327 g.
Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium.
Chris thought it was sodium. Randy reported that it was gallium.
What formula did each student assign to the sulfate salt?
Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the CRC Hand-book of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?
119. You have two 500.0-mL aqueous solutions. Solution A is a so-lution of a metal nitrate that is 8.246% nitrogen by mass. The ionic compound in solution B consists of potassium, chromium, and oxygen; chromium has an oxidation state of 6 and there are 2 potassiums and 1 chromium in the formula. The masses of the solutes in each of the solutions are the same. When the so-lutions are added together, a blood-red precipitate forms. After the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g.
a. Identify the ionic compounds in solution A and solution B.
b. Identify the blood-red precipitate.
c. Calculate the concentration (molarity) of all ions in the orig-inal solutions.
d. Calculate the concentration (molarity) of all ions in the final solution.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
H2X1aq2 OH ¡ X2 H2O 106. Complete and balance each acid–base reaction.
a.
Contains three acidic hydrogens b.
Contains two acidic hydrogens c.
Contains two acidic hydrogens d.
Contains two acidic hydrogens 107. What volume of 0.0521 M Ba(OH)2 is required to neutralize ex-actly 14.20 mL of 0.141 M H3PO4? Phosphoric acid contains three acidic hydrogens.
108. A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid?
Sulfuric acid contains two acidic hydrogens.
109. Some of the substances commonly used in stomach antacids are MgO, Mg(OH)2, and Al(OH)3.
a. Write a balanced equation for the neutralization of hy-drochloric acid by each of these substances.
b. Which of these substances will neutralize the greatest amount of 0.10 M HCl per gram?
110. A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete reaction. Determine the molar mass of the acid.
111. Citric acid, which can be obtained from lemon juice, has the molecular formula C6H8O7. A 0.250-g sample of citric acid dis-solved in 25.0 mL of water requires 37.2 mL of 0.105 M NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?
112. Balance the following equations by the half-reaction method.
a.
b.
c.
d.
e.
f.
113. It took 25.06 0.05 mL of a sodium hydroxide solution to titrate a 0.4016-g sample of KHP (see Exercise 65). Calculate the concen-tration and uncertainty in the concentration of the sodium hydroxide solution. (See Appendix 1.5.) Neglect any uncertainty in the mass.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
114. Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal com-pound. It is composed solely of C, F, and B; it is 42.23% C by mass and 55.66% F by mass.
a. What is the empirical formula of BARF?
b. A 2.251-g sample of BARF dissolved in 347.0 mL of solution produces a 0.01267 M solution. What is the molecular for-mula of BARF?
Fe1OH221s2 H2O21aq2 ¡ Base Fe1OH231s2 Ce1OH231s2 Fe1OH231s2 CO3 21aq2 NO3 1aq2 Fe1CN26 41aq2 Ce41aq2 ¡ Base CrO4 21aq2 IO4 1aq2 Cl1aq2 CrI31s2 Cl21g2 ¡ Base Cr31aq2 Ce31aq2 NO3 1aq2 CO21g2 SO4 21aq2 Cr1NCS26 41aq2 Ce41aq2 ¡ Acid IO3 1aq2 I1aq2 ¡ Acid I3 1aq2 Fe1s2 HCl1aq2 S HFeCl41aq2 H21g2 H2C2O41aq2 NaOH1aq2 S H2Se1aq2 Ba1OH221aq2 S H2SO41aq2 Al1OH231s2 S H3PO41aq2 NaOH1aq2 S 178 5 Gases Contents 5.1 Pressure • Units of Pressure 5.2 The Gas Laws of Boyle, Charles, and Avogadro • Boyle’s Law • Charles’s Law • Avogadro’s Law 5.3 The Ideal Gas Law 5.4 Gas Stoichiometry • Molar Mass of a Gas 5.5 Dalton’s Law of Partial Pressures • Collecting a Gas over Water 5.6 The Kinetic Molecular Theory of Gases • Pressure and Volume (Boyle’s Law) • Pressure and Temperature • Volume and Temperature (Charles’s Law) • Volume and Number of Moles (Avogadro’s Law) • Mixture of Gases (Dalton’s Law) • Deriving the Ideal Gas Law • The Meaning of Temperature • Root Mean Square Velocity 5.7 Effusion and Diffusion • Effusion • Diffusion 5.8 Real Gases 5.9 Characteristics of Several Real Gases 5.10 Chemistry in the Atmosphere The steaming fumaroles located in Bjarnarflag, Iceland release a variety of gases.
Matter exists in three distinct physical states: gas, liquid, and solid. Although rel-atively few substances exist in the gaseous state under typical conditions, gases are very important. For example, we live immersed in a gaseous solution. The earth’s atmosphere is a mixture of gases that consists mainly of elemental nitrogen (N2) and oxygen (O2).
The atmosphere both supports life and acts as a waste receptacle for the exhaust gases that accompany many industrial processes. The chemical reactions of these waste gases in the atmosphere lead to various types of pollution, including smog and acid rain. The gases in the atmosphere also shield us from harmful radiation from the sun and keep the earth warm by reflecting heat radiation back toward the earth. In fact, there is now great concern that an increase in atmospheric carbon dioxide, a product of the combustion of fossil fuels, is causing a dangerous warming of the earth.
In this chapter we will look carefully at the properties of gases. First we will see how measurements of gas properties lead to various types of laws—statements that show how the properties are related to each other. Then we will construct a model to explain why gases behave as they do. This model will show how the behavior of the individual particles of a gas leads to the observed properties of the gas itself (a collection of many, many particles).
The study of gases provides an excellent example of the scientific method in action.
It illustrates how observations lead to natural laws, which in turn can be accounted for by models.
5.1 Pressure A gas uniformly fills any container, is easily compressed, and mixes completely with any other gas. One of the most obvious properties of a gas is that it exerts pressure on its sur-roundings. For example, when you blow up a balloon, the air inside pushes against the elastic sides of the balloon and keeps it firm.
As mentioned earlier, the gases most familiar to us form the earth’s atmosphere. The pressure exerted by this gaseous mixture that we call air can be dramatically demonstrated by the experiment shown in Fig. 5.1. A small volume of water is placed in a metal can, 179 As a gas, water occupies 1200 times as much space as it does as a liquid at 25C and atmospheric pressure.
FIGURE 5.1 The pressure exerted by the gases in the at-mosphere can be demonstrated by boiling water in a large metal can (a) and then turning off the heat and sealing the can. As the can cools, the water vapor condenses, lowering the gas pressure inside the can.
This causes the can to crumple (b).
Visualization: Collapsing Can 180 Chapter Five Gases and the water is boiled, which fills the can with steam. The can is then sealed and allowed to cool. Why does the can collapse as it cools? It is the atmospheric pressure that crum-ples the can. When the can is cooled after being sealed so that no air can flow in, the wa-ter vapor (steam) condenses to a very small volume of liquid water. As a gas, the water filled the can, but when it is condensed to a liquid, the liquid does not come close to fill-ing the can. The H2O molecules formerly present as a gas are now collected in a very small volume of liquid, and there are very few molecules of gas left to exert pressure out-ward and counteract the air pressure. As a result, the pressure exerted by the gas mole-cules in the atmosphere smashes the can.
A device to measure atmospheric pressure, the barometer, was invented in 1643 by an Italian scientist named Evangelista Torricelli (1608–1647), who had been a student of Galileo. Torricelli’s barometer is constructed by filling a glass tube with liquid mercury and inverting it in a dish of mercury, as shown in Fig. 5.2. Notice that a large quantity of mercury stays in the tube. In fact, at sea level the height of this column of mercury aver-ages 760 mm. Why does this mercury stay in the tube, seemingly in defiance of gravity?
Figure 5.2 illustrates how the pressure exerted by the atmospheric gases on the surface of mercury in the dish keeps the mercury in the tube.
Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity—in other words, it results from the weight of the air. Changing weather conditions cause the atmospheric pressure to vary, so the height of the column of Hg supported by the atmosphere at sea level varies; it is not always 760 mm. The mete-orologist who says a “low” is approaching means that the atmospheric pressure is going to decrease. This condition often occurs in conjunction with a storm.
Atmospheric pressure also varies with altitude. For example, when Torricelli’s ex-periment is done in Breckenridge, Colorado (elevation 9600 feet), the atmosphere sup-ports a column of mercury only about 520 mm high because the air is “thinner.” That is, there is less air pushing down on the earth’s surface at Breckenridge than at sea level.
Units of Pressure Because instruments used for measuring pressure, such as the manometer (Fig. 5.3), of-ten contain mercury, the most commonly used units for pressure are based on the height h = 760 mm Hg for standard atmosphere Vacuum FIGURE 5.2 A torricellian barometer. The tube, com-pletely filled with mercury, is inverted in a dish of mercury. Mercury flows out of the tube until the pressure of the column of mercury (shown by the black arrow) “stand-ing on the surface” of the mercury in the dish is equal to the pressure of the air (shown by the purple arrows) on the rest of the surface of the mercury in the dish.
h (b) Gas pressure (P gas) greater than atmospheric pressure Atmospheric pressure (P atm) h (a) Gas pressure (P gas) less than atmospheric pressure P gas = P atm – h P gas = P atm + h Atmospheric pressure (P atm) FIGURE 5.3 A simple manometer, a device for measur-ing the pressure of a gas in a container. The pressure of the gas is given by h (the differ-ence in mercury levels) in units of torr (equivalent to mm Hg). (a) Gas pressure atmospheric pressure h. (b) Gas pressure atmospheric pressure h.
Soon after Torricelli died, a German physicist named Otto von Guericke in-vented an air pump. In a famous demon-stration for the King of Prussia in 1663, Guericke placed two hemispheres to-gether, pumped the air out of the result-ing sphere through a valve, and showed that teams of horses could not pull the hemispheres apart. Then, after secretly opening the air valve, Guericke easily separated the hemispheres by hand. The King of Prussia was so impressed that he awarded Guericke a lifetime pension!
5.2 The Gas Laws of Boyle, Charles, and Avogadro 181 of the mercury column (in millimeters) that the gas pressure can support. The unit mm Hg (millimeter of mercury) is often called the torr in honor of Torricelli. The terms torr and mm Hg are used interchangeably by chemists. A related unit for pressure is the standard atmosphere (abbreviated atm): However, since pressure is defined as force per unit area, the fundamental units of pressure involve units of force divided by units of area. In the SI system, the unit of force is the newton (N) and the unit of area is meters squared (m2).
(For a review of the SI system, see Chapter 1.) Thus the unit of pressure in the SI system is newtons per meter squared (N/m2) and is called the pascal (Pa). In terms of pascals, the standard atmosphere is Thus 1 atmosphere is about 105 pascals. Since the pascal is so small, and since it is not commonly used in the United States, we will use it sparingly in this book. However, converting from torrs or atmospheres to pascals is straightforward, as shown in Sample Exercise 5.1.
Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this pressure in both atmospheres and pascals.
Solution See Exercises 5.27 and 5.28.
5.2 The Gas Laws of Boyle, Charles, and Avogadro In this section we will consider several mathematical laws that relate the properties of gases. These laws derive from experiments involving careful measurements of the rele-vant gas properties. From these experimental results, the mathematical relationships among the properties can be discovered. These relationships are often represented pictorially by means of graphs (plots).
We will take a historical approach to these laws to give you some perspective on the scientific method in action.
Boyle’s Law The first quantitative experiments on gases were performed by an Irish chemist, Robert Boyle (1627–1691). Using a J-shaped tube closed at one end (Fig. 5.4), which he report-edly set up in the multistory entryway of his house, Boyle studied the relationship between the pressure of the trapped gas and its volume. Representative values from Boyle’s experiments are given in Table 5.1. These data show that the product of the pressure and 6.4 102 atm 101,325 Pa 1 atm 6.5 103 Pa 49 torr 1 atm 760 torr 6.4 102 atm 1 standard atmosphere 101,325 Pa Pressure force area 1 standard atmosphere 1 atm 760 mm Hg 760 torr Checking tire pressure.
Sample Exercise 5.1 1 atm 760 mm Hg 760 torr 101,325 Pa 29.92 in Hg 14.7 lb/in2 Visualization: Boyle’s Law: A Graphical View 182 Chapter Five Gases volume for the trapped air sample is constant within the accuracies of Boyle’s measurements (note the third column in Table 5.1). This behavior can be represented by the equation which is called Boyle’s law and where k is a constant for a given sample of air at a spe-cific temperature.
It is convenient to represent the data in Table 5.1 by using two different plots. The first type of plot, P versus V, forms a curve called a hyperbola shown in Fig. 5.5(a). Looking at this plot, note that as the volume drops by about half (from 58.8 to 29.1), the pressure dou-bles (from 24.0 to 48.0). In other words, there is an inverse relationship between pressure and volume. The second type of plot can be obtained by rearranging Boyle’s law to give which is the equation for a straight line of the type where m represents the slope and b the intercept of the straight line. In this case, y V, x 1P, m k, and b 0. Thus a plot of V versus 1P using Boyle’s data gives a straight line with an intercept of zero, as shown in Fig. 5.5(b).
In the three centuries since Boyle carried out his studies, the sophistication of measur-ing techniques has increased tremendously. The results of highly accurate measurements show that Boyle’s law holds precisely only at very low pressures. Measurements at higher pres-sures reveal that PV is not constant but varies as the pressure is varied. Results for several gases at pressures below 1 atm are shown in Fig. 5.6. Note the very small changes that occur in the product PV as the pressure is changed at these low pressures. Such changes become y mx b V k P k1 P PV k h h Mercury Gas Gas Mercury added FIGURE 5.4 A J-tube similar to the one used by Boyle.
TABLE 5.1 Actual Data from Boyle’s Experiment Volume Pressure Pressure Volume (in3) (in Hg) (in Hg in3) 117.5 12.0 14.1 102 87.2 16.0 14.0 102 70.7 20.0 14.1 102 58.8 24.0 14.1 102 44.2 32.0 14.1 102 35.3 40.0 14.1 102 29.1 48.0 14.0 102 FIGURE 5.5 Plotting Boyle’s data from Table 5.1. (a) A plot of P versus V shows that the volume doubles as the pressure is halved. (b) A plot of V versus 1P gives a straight line. The slope of this line equals the value of the constant k.
Boyle’s law: at constant temperature.
Graphing is reviewed in Appendix 1.3.
V r 1 P P (in Hg) 0 V (in3) 20 40 60 50 100 P P 2 V 2V (a) V (in3) 0 0 1/P (in Hg) 0.01 0.02 0.03 20 40 slope = k (b) 5.2 The Gas Laws of Boyle, Charles, and Avogadro 183 more significant at much higher pressures, where the complex nature of the dependence of PV on pressure becomes more obvious. We will discuss these deviations and the reasons for them in detail in Section 5.8. A gas that strictly obeys Boyle’s law is called an ideal gas. We will describe the characteristics of an ideal gas more completely in Section 5.3.
One common use of Boyle’s law is to predict the new volume of a gas when the pres-sure is changed (at constant temperature), or vice versa. Because deviations from Boyle’s law are so slight at pressures close to 1 atm, in our calculations we will assume that gases obey Boyle’s law (unless stated otherwise).
Boyle’s Law I Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is found in the exhaust of automobiles and power plants. Consider a 1.53-L sample of gaseous SO2 at a pressure of 5.6 103 Pa. If the pressure is changed to 1.5 104 Pa at a constant temperature, what will be the new volume of the gas?
Solution We can solve this problem using Boyle’s law, which also can be written as where the subscripts 1 and 2 represent two states (conditions) of the gas (both at the same temperature). In this case, We can solve the preceding equation for V2: The new volume will be 0.57 L.
See Exercise 5.33.
The fact that the volume decreases in Sample Exercise 5.2 makes sense because the pressure was increased. To help eliminate errors, make it a habit to check whether an an-swer to a problem makes physical sense.
We mentioned before that Boyle’s law is only approximately true for real gases. To determine the significance of the deviations, studies of the effect of changing pressure on the volume of a gas are often done, as shown in Sample Exercise 5.3.
Boyle’s Law II In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume measure-ments were made at various pressures, using 1.0 mol NH3 gas at a temperature of 0C. Using the results listed on the following page, calculate the Boyle’s law constant for NH3 at the various pressures.
V2 P1V1 P2 5.6 103 Pa 1.53 L 1.5 104 Pa 0.57 L V1 1.53 L V2 ?
P1 5.6 103 Pa P2 1.5 104 Pa P1V1 k P2V2 or P1V1 P2V2 PV k PV (L.atm) 0 22.25 P (atm) 1.00 0.75 0.50 0.25 22.30 22.35 22.40 22.45 Ideal Ne O2 CO2 FIGURE 5.6 A plot of PV versus P for several gases at pressures below 1 atm. An ideal gas is expected to have a con-stant value of PV, as shown by the dotted line. Carbon dioxide shows the largest change in PV, and this change is actually quite small: PV changes from about 22.39 L atm at 0.25 atm to 22.26 L atm at 1.00 atm. Thus Boyle’s law is a good approximation at these relatively low pressures.
As pressure increases, the volume of SO2 decreases.
Sample Exercise 5.2 V = 1.53 L V = ?
5.6 × 103 Pa 1.5 × 104 Pa Boyle’s law also can be written as P1V1 P2V2 Always check that your answer makes physical (common!) sense.
Sample Exercise 5.3 184 Chapter Five Gases Solution To determine how closely NH3 gas follows Boyle’s law under these conditions, we cal-culate the value of k (in L atm) for each set of values: Experiment Pressure (atm) Volume (L) 1 0.1300 172.1 2 0.2500 89.28 3 0.3000 74.35 4 0.5000 44.49 5 0.7500 29.55 6 1.000 22.08 Experiment 1 2 3 4 5 6 k PV 22.37 22.32 22.31 22.25 22.16 22.08 Although the deviations from true Boyle’s law behavior are quite small at these low pressures, note that the value of k changes regularly in one direction as the pressure is increased. Thus, to calculate the “ideal” value of k for NH3, we can plot PV versus P, as shown in Fig. 5.7, and extrapolate (extend the line beyond the experimental points) back to zero pressure, where, for reasons we will discuss later, a gas behaves most ideally. The value of k obtained by this extrapolation is 22.41 L atm. Notice that this is the same value obtained from similar plots for the gases CO2, O2, and Ne at 0C, as shown in Fig. 5.6.
See Exercise 5.97.
Charles’s Law In the century following Boyle’s findings, scientists continued to study the properties of gases. One of these scientists was a French physicist, Jacques Charles (1746–1823), who was the first person to fill a balloon with hydrogen gas and who made the first solo bal-loon flight. Charles found in 1787 that the volume of a gas at constant pressure increases linearly with the temperature of the gas. That is, a plot of the volume of a gas (at con-stant pressure) versus its temperature (C) gives a straight line. This behavior is shown for samples of several gases in Fig. 5.8. The slopes of the lines in this graph are different PV (L • atm) 0 P (atm) 0.20 1.00 22.1 22.6 22.2 22.3 22.4 22.5 0.40 0.60 0.80 FIGURE 5.7 A plot of PV versus P for 1 mol of ammo-nia. The dashed line shows the extrapola-tion of the data to zero pressure to give the “ideal” value of PV of 22.41 L atm.
V (L) –300 T (°C) –200 –100 0 100 200 1 2 3 6 4 5 300 –273.2 °C N2O H2 H2O CH4 He FIGURE 5.8 Plots of V versus T (C) for several gases.
The solid lines represent experimental mea-surements on gases. The dashed lines repre-sent extrapolation of the data into regions where these gases would become liquids or solids. Note that the samples of the various gases contain different numbers of moles.
A snowmaking machine, in which water is blown through nozzles by compressed air. The mixture is cooled by expansion to form ice crystals of snow.
Visualization: Liquid Nitrogen and Balloons 5.2 The Gas Laws of Boyle, Charles, and Avogadro 185 because the samples contain different numbers of moles of gas. A very interesting feature of these plots is that the volumes of all the gases extrapolate to zero at the same temper-ature, 273.2C. On the Kelvin temperature scale this point is defined as 0 K, which leads to the following relationship between the Kelvin and Celsius scales: When the volumes of the gases shown in Fig. 5.8 are plotted versus temperature on the Kelvin scale, the plots in Fig. 5.9 result. In this case, the volume of each gas is directly proportional to temperature and extrapolates to zero when the temperature is 0 K. This behavior is represented by the equation known as Charles’s law, where T is in kelvins and b is a proportionality constant.
Before we illustrate the uses of Charles’s law, let us consider the importance of 0 K.
At temperatures below this point, the extrapolated volumes would become negative. The fact that a gas cannot have a negative volume suggests that 0 K has a special significance.
In fact, 0 K is called absolute zero, and there is much evidence to suggest that this temperature cannot be attained. Temperatures of approximately 0.000001 K have been produced in laboratories, but 0 K has never been reached.
Charles’s Law A sample of gas at 15C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38C and 1 atm?
Solution Charles’s law, which describes the dependence of the volume of a gas on temperature at constant pressure, can be used to solve this problem. Charles’s law in the form V bT can be rearranged to An equivalent statement is where the subscripts 1 and 2 represent two states for a given sample of gas at constant pressure. In this case, we are given the following (note that the temperature values must be changed to the Kelvin scale): Solving for V2 gives Reality Check: The new volume is greater than the initial volume, which makes physical sense because the gas will expand as it is heated.
See Exercise 5.35.
Avogadro’s Law In Chapter 2 we noted that in 1811 the Italian chemist Avogadro postulated that equal volumes of gases at the same temperature and pressure contain the same number of V2 aT2 T1 b V1 a311 K 288 Kb 2.58 L 2.79 L V1 2.58 L V2 ?
T1 15°C 273 288 K T2 38°C 273 311 K V1 T1 b V2 T2 V T b V bT K °C 273 V (L) 1 2 3 6 4 5 T (K) 73 173 273 373 473 573 0 N2O H2 H2O CH4 He FIGURE 5.9 Plots of V versus T as in Fig. 5.8, except here the Kelvin scale is used for temperature.
Charles’s law: (expressed in K) of constant pressure.
V r T Sample Exercise 5.4 Charles’s law also can be written as V1 T1 V2 T2 Visualization: Charles’s Law: A Graphical View 186 Chapter Five Gases “particles.” This observation is called Avogadro’s law, which is illustrated by Fig. 5.10.
Stated mathematically, Avogadro’s law is where V is the volume of the gas, n is the number of moles of gas particles, and a is a proportionality constant. This equation states that for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas. This rela-tionship is obeyed closely by gases at low pressures.
Avogadro’s Law Suppose we have a 12.2-L sample containing 0.50 mol oxygen gas (O2) at a pressure of 1 atm and a temperature of 25C. If all this O2 were converted to ozone (O3) at the same temperature and pressure, what would be the volume of the ozone?
Solution The balanced equation for the reaction is To calculate the moles of O3 produced, we must use the appropriate mole ratio: Avogadro’s law states that V an, which can be rearranged to give Since a is a constant, an alternative representation is where V1 is the volume of n1 moles of O2 gas and V2 is the volume of n2 moles of O3 gas.
In this case we have Solving for V2 gives Reality Check: Note that the volume decreases, as it should, since fewer moles of gas molecules will be present after O2 is converted to O3.
See Exercises 5.35 and 5.36.
5.3 The Ideal Gas Law We have considered three laws that describe the behavior of gases as revealed by exper-imental observations: Avogadro’s law: V an 1at constant T and P2 Charles’s law: V bT 1at constant P and n2 Boyle’s law: V k P 1at constant T and n2 V2 an2 n1 b V1 a0.33 mol 0.50 molb 12.2 L 8.1 L V1 12.2 L V2 ?
n1 0.50 mol n2 0.33 mol V1 n1 a V2 n2 V n a 0.50 mol O2 2 mol O3 3 mol O2 0.33 mol O3 3O21g2 ¡ 2O31g2 V an N2 H2 Ar CH4 FIGURE 5.10 These balloons each hold 1.0 L of gas at 25C and 1 atm. Each balloon contains 0.041 mol of gas, or 2.5 1022 molecules.
Sample Exercise 5.5 Avogadro’s law also can be written as V1 n1 V2 n2 5.3 The Ideal Gas Law 187 These relationships, which show how the volume of a gas depends on pressure, temper-ature, and number of moles of gas present, can be combined as follows: where R is the combined proportionality constant called the universal gas constant. When the pressure is expressed in atmospheres and the volume in liters, R has the value 0.08206 L atmK mol. The preceding equation can be rearranged to the more familiar form of the ideal gas law: The ideal gas law is an equation of state for a gas, where the state of the gas is its condition at a given time. A particular state of a gas is described by its pressure, volume, temperature, and number of moles. Knowledge of any three of these properties is enough to completely define the state of a gas, since the fourth property can then be determined from the equation for the ideal gas law.
It is important to recognize that the ideal gas law is an empirical equation—it is based on experimental measurements of the properties of gases. A gas that obeys this equation is said to behave ideally. The ideal gas equation is best regarded as a limit-ing law—it expresses behavior that real gases approach at low pressures and high tem-peratures. Therefore, an ideal gas is a hypothetical substance. However, most gases obey the ideal gas equation closely enough at pressures below 1 atm that only mini-mal errors result from assuming ideal behavior. Unless you are given information to the contrary, you should assume ideal gas behavior when solving problems involving gases in this text.
The ideal gas law can be used to solve a variety of problems. Sample Exercise 5.6 demonstrates one type, where you are asked to find one property characterizing the state of a gas, given the other three.
Ideal Gas Law I A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0C and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.
Solution Solving the ideal gas law for n gives In this case P 1.5 atm, V 8.56 L, T 0C 273 273 K, and R 0.08206 L atm/K mol. Thus See Exercises 5.37 through 5.42.
The ideal gas law is also used to calculate the changes that will occur when the con-ditions of the gas are changed.
n 11.5 atm218.56 L2 a0.08206 L atm K molb1273 K2 0.57 mol n PV RT PV nRT V RaTn P b R 0.08206 L atm K mol The ideal gas law applies best at pressures smaller than 1 atm.
Sample Exercise 5.6 The reaction of zinc with hydrochloric acid to produce bubbles of hydrogen gas.
Visualization: The Ideal Gas Law, PV nRT 188 Chapter Five Gases Ideal Gas Law II Suppose we have a sample of ammonia gas with a volume of 7.0 mL at a pressure of 1.68 atm. The gas is compressed to a volume of 2.7 mL at a constant temperature. Use the ideal gas law to calculate the final pressure.
Solution The basic assumption we make when using the ideal gas law to describe a change in state for a gas is that the equation applies equally well to both the initial and the final states.
In dealing with a change in state, we always place the variables that change on one side of the equals sign and the constants on the other. In this case the pressure and volume change, and the temperature and the number of moles remain constant (as does R, by definition). Thus we write the ideal gas law as p r Change Remain constant Since n and T remain the same in this case, we can write P1V1 nRT and P2V2 nRT.
Combining these gives We are given P1 1.68 atm, V1 7.0 mL, and V2 2.7 mL. Solving for P2 thus gives Reality Check: Does this answer make sense? The volume decreased (at constant temperature), so the pressure should increase, as the result of the calculation indicates.
Note that the calculated final pressure is 4.4 atm. Most gases do not behave ideally above 1 atm. Therefore, we might find that if we measured the pressure of this gas sample, the observed pressure would differ slightly from 4.4 atm.
See Exercises 5.43 and 5.44.
Ideal Gas Law III A sample of methane gas that has a volume of 3.8 L at 5C is heated to 86C at constant pressure. Calculate its new volume.
Solution To solve this problem, we take the ideal gas law and segregate the changing variables and the constants by placing them on opposite sides of the equation. In this case, volume and temperature change, and the number of moles and pressure (and, of course, R) remain constant. Thus PV nRT becomes which leads to Combining these gives V1 T1 nR P V2 T2 or V1 T1 V2 T2 V1 T1 nR P and V2 T2 nR P V T nR P P2 aV1 V2 bP1 a7.0 mL 2.7 mLb 1.68 atm 4.4 atm P1V1 nRT P2V2 or P1V1 P2V2 PV nRT Sample Exercise 5.7 10 9 8 7 6 5 4 3 2 1 mL 0 5 atm 1 2 3 4 2.7 ml 7.0 ml 1.68 4.4 atm 10 9 8 7 6 5 4 3 2 1 mL 0 5 1 2 3 4 As pressure increases, the volume decreases.
Sample Exercise 5.8 5.3 The Ideal Gas Law 189 We are given Thus Reality Check: Is the answer sensible? In this case the temperature increased (at constant pressure), so the volume should increase. Thus the answer makes sense.
See Exercises 5.45 and 5.46.
The problem in Sample Exercise 5.8 could be described as a “Charles’s law problem,” whereas the problem in Sample Exercise 5.7 might be called a “Boyle’s law problem.” In both cases, however, we started with the ideal gas law. The real advantage of using the ideal gas law is that it applies to virtually any problem dealing with gases and is easy to remember.
Ideal Gas Law IV A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of 15C and a volume of 3.48 L. If condi-tions are changed so that the temperature is 36C and the pressure is 468 torr, what will be the volume of the sample?
Solution Since, for this sample, pressure, temperature, and volume all change while the number of moles remains constant, we use the ideal gas law in the form which leads to Then We have Thus See Exercises 5.47 and 5.48.
Since the equation used in Sample Exercise 5.9 involves a ratio of pressures, it was unnecessary to convert pressures to units of atmospheres. The units of torrs cancel. (You V2 1309 K21345 torr213.48 L2 1258 K21468 torr2 3.07 L V1 3.48 L V2 ?
T1 15°C 273 258 K T2 36°C 273 309 K P1 345 torr P2 468 torr V2 T2P1V1 T1P2 P1V1 T1 nR P2V2 T2 or P1V1 T1 P2V2 T2 PV T nR V2 T2V1 T1 1359 K213.8 L2 278 K 4.9 L V1 3.8 L V2 ?
T1 5°C 273 278 K T2 86°C 273 359 K Sample Exercise 5.9 Visualization: Changes in Gas Volume, Pressure, and Concentration Always convert the temperature to the Kelvin scale when applying the ideal gas law.
190 Chapter Five Gases will obtain the same answer by inserting and into the equation.) How-ever, temperature must always be converted to the Kelvin scale; since this conversion in-volves addition of 273, the conversion factor does not cancel. Be careful.
One of the many other types of problems dealing with gases that can be solved us-ing the ideal gas law is illustrated in Sample Exercise 5.10.
Ideal Gas Law V A sample containing 0.35 mol argon gas at a temperature of 13C and a pressure of 568 torr is heated to 56C and a pressure of 897 torr. Calculate the change in volume that occurs.
Solution We use the ideal gas law to find the volume for each set of conditions: P2 468 760 P1 345 760 Sample Exercise 5.10 Argon glowing in a discharge tube.
State 1 State 2 T2 56°C 273 329 K T1 13°C 273 286 K P2 897 torr 1 atm 760 torr 1.18 atm P1 568 torr 1 atm 760 torr 0.747 atm n2 0.35 mol n1 0.35 mol Solving the ideal gas law for volume gives and Thus, in going from state 1 to state 2, the volume changes from 11 L to 8.0 L. The change in volume, V ( is the Greek capital letter delta), is then The change in volume is negative because the volume decreases. Note that for this prob-lem (unlike Sample Exercise 5.9) the pressures must be converted from torrs to atmos-pheres, as required by the atmosphere part of the units for R, since each volume was found separately and the conversion factor does not cancel.
See Exercise 5.49.
5.4 Gas Stoichiometry Suppose we have 1 mole of an ideal gas at 0C (273.2 K) and 1 atm. From the ideal gas law, the volume of the gas is given by V nRT P 11.000 mol210.08206 L atm/K mol21273.2 K2 1.000 atm 22.42 L ¢V V2 V1 8.0 L 11 L 3 L V2 n2RT2 P2 10.35 mol210.08206 L atm/K mol21329 K2 11.18 atm2 8.0 L V1 n1RT1 P1 10.35 mol210.08206 L atm/K mol21286 K2 10.747 atm2 11 L When 273.15 K is used in this calcula-tion, the molar volume obtained in Sample Exercise 5.3 is the same value as 22.41 L.
5.4 Gas Stoichiometry 191 This volume of 22.42 liters is the molar volume of an ideal gas (at 0C and 1 atm). The measured molar volumes of several gases are listed in Table 5.2. Note that the molar volumes of some of the gases are very close to the ideal value, while others deviate sig-nificantly. Later in this chapter we will discuss some of the reasons for the deviations.
The conditions 0C and 1 atm, called standard temperature and pressure (abbre-viated STP), are common reference conditions for the properties of gases. For example, the molar volume of an ideal gas is 22.42 liters at STP (see Fig. 5.11).
Gas Stoichiometry I A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?
Solution We could solve this problem by using the ideal gas equation, but we can take a shortcut by using the molar volume of an ideal gas at STP. Since 1 mole of an ideal gas at STP has a volume of 22.42 L, 1.75 L of N2 at STP will contain less than 1 mole. We can find how many moles using the ratio of 1.75 L to 22.42 L: See Exercises 5.51 and 5.52.
Many chemical reactions involve gases. By assuming ideal behavior for these gases, we can carry out stoichiometric calculations if the pressure, volume, and temperature of the gases are known.
Gas Stoichiometry II Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3).
Calculate the volume of CO2 at STP produced from the decomposition of 152 g CaCO3 by the reaction CaCO31s2 ¡ CaO1s2 CO21g2 1.75 L N2 1 mol N2 22.42 L N2 7.81 102 mol N2 STP: 0C and 1 atm TABLE 5.2 Molar Volumes for Various Gases at 0C and 1 atm Molar Gas Volume (L) Oxygen (O2) 22.397 Nitrogen (N2) 22.402 Hydrogen (H2) 22.433 Helium (He) 22.434 Argon (Ar) 22.397 Carbon dioxide (CO2) 22.260 Ammonia (NH3) 22.079 FIGURE 5.11 22.4 L of a gas would just fit into this box.
Sample Exercise 5.11 Sample Exercise 5.12 192 Chapter Five Gases Solution We employ the same strategy we used in the stoichiometry problems earlier in this book.
That is, we compute the number of moles of CaCO3 consumed and the number of moles of CO2 produced. The moles of CO2 can then be converted to volume using the molar volume of an ideal gas.
Using the molar mass of CaCO3 (100.09 g/mol), we can calculate the number of moles of CaCO3: Since each mole of CaCO3 produces a mole of CO2, 1.52 mol CO2 will be formed. We can compute the volume of CO2 at STP by using the molar volume: Thus the decomposition of 152 g CaCO3 produces 34.1 L CO2 at STP.
See Exercises 5.53 through 5.56.
Note that in Sample Exercise 5.12 the final step involved calculation of the volume of gas from the number of moles. Since the conditions were specified as STP, we were able to use the molar volume of a gas at STP. If the conditions of a problem are different from STP, the ideal gas law must be used to compute the volume.
Gas Stoichiometry III A sample of methane gas having a volume of 2.80 L at 25C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31C and 1.25 atm. The mix-ture was then ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125C.
Solution From the description of the reaction, the unbalanced equation is which can be balanced to give Next, we must find the limiting reactant, which requires calculating the numbers of moles of each reactant. We convert the given volumes of methane and oxygen to moles using the ideal gas law as follows: In the balanced equation for the combustion reaction, 1 mol CH4 requires 2 mol O2. Thus the moles of O2 required by 0.189 mol CH4 can be calculated as follows: 0.189 mol CH4 2 mol O2 1 mol CH4 0.378 mol O2 nO2 PV RT 11.25 atm2135.0 L2 10.08206 L atm/K mol21304 K2 1.75 mol nCH4 PV RT 11.65 atm212.80 L2 10.08206 L atm/K mol21298 K2 0.189 mol CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 CH41g2 O21g2 ¡ CO21g2 H2O1g2 1.52 mol CO2 22.42 L CO2 1 mol CO2 34.1 L CO2 152 g CaCO3 1 mol CaCO3 100.09 g CaCO3 1.52 mol CaCO3 Remember that the molar volume of an ideal gas is 22.42 L when measured at STP.
Sample Exercise 5.13 5.4 Gas Stoichiometry 193 Since 1.75 mol O2 is available, O2 is in excess. The limiting reactant is CH4. The number of moles of CH4 available must be used to calculate the number of moles of CO2 produced: Since the conditions stated are not STP, we must use the ideal gas law to calculate the volume: In this case n 0.189 mol, T 125C 273 398 K, P 2.50 atm, and R 0.08206 L atm/K mol. Thus This represents the volume of CO2 produced under these conditions.
See Exercises 5.57 and 5.58.
Molar Mass of a Gas One very important use of the ideal gas law is in the calculation of the molar mass (mo-lecular weight) of a gas from its measured density. To see the relationship between gas density and molar mass, consider that the number of moles of gas n can be expressed as Substitution into the ideal gas equation gives However, mV is the gas density d in units of grams per liter. Thus or (5.1) Thus, if the density of a gas at a given temperature and pressure is known, its molar mass can be calculated.
Gas Density/Molar Mass The density of a gas was measured at 1.50 atm and 27C and found to be 1.95 g/L.
Calculate the molar mass of the gas.
Solution Using Equation (5.1), we calculate the molar mass as follows: Molar mass dRT P a1.95 g Lba0.08206 L atm K molb1300. K2 1.50 atm 32.0 g/mol Molar mass dRT P P dRT molar mass P nRT V 1m molar mass2RT V m1RT2 V1molar mass2 n grams of gas molar mass mass molar mass m molar mass V 10.189 mol210.08206 L atm/K mol21398 K2 2.50 atm 2.47 L V nRT P 0.189 mol CH4 1 mol CO2 1 mol CH4 0.189 mol CO2 Density mass volume Sample Exercise 5.14 194 Chapter Five Gases Reality Check: These are the units expected for molar mass.
See Exercises 5.61 through 5.64.
You could memorize the equation involving gas density and molar mass, but it is better simply to remember the total gas equation, the definition of density, and the rela-tionship between number of moles and molar mass. You can then derive the appropriate equation when you need it. This approach ensures that you understand the concepts and means one less equation to memorize.
5.5 Dalton’s Law of Partial Pressures Among the experiments that led John Dalton to propose the atomic theory were his stud-ies of mixtures of gases. In 1803 Dalton summarized his observations as follows: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. This statement, known as Dalton’s law of partial pressures, can be expressed as follows: where the subscripts refer to the individual gases (gas 1, gas 2, and so on). The symbols P1, P2, P3, and so on represent each partial pressure, the pressure that a particular gas would exert if it were alone in the container.
Assuming that each gas behaves ideally, the partial pressure of each gas can be cal-culated from the ideal gas law: The total pressure of the mixture PTOTAL can be represented as where nTOTAL is the sum of the numbers of moles of the various gases. Thus, for a mixture of ideal gases, it is the total number of moles of particles that is important, not the identity or composition of the involved gas particles. This idea is illustrated in Fig. 5.12.
nTOTALaRT V b 1n1 n2 n3 p 2aRT V b PTOTAL P1 P2 P3 p n1RT V n2RT V n3RT V p P1 n1RT V , P2 n2RT V , P3 n3RT V , p PTOTAL P1 P2 P3 p FIGURE 5.12 The partial pressure of each gas in a mix-ture of gases in a container depends on the number of moles of that gas. The total pressure is the sum of the partial pressures and depends on the total moles of gas par-ticles present, no matter what they are.
5.5 Dalton’s Law of Partial Pressures 195 This important observation indicates some fundamental characteristics of an ideal gas. The fact that the pressure exerted by an ideal gas is not affected by the identity (composition) of the gas particles reveals two things about ideal gases: (1) the volume of the individual gas particle must not be important, and (2) the forces among the particles must not be important. If these factors were important, the pressure exerted by the gas would depend on the nature of the individual particles. These observations will strongly influence the model that we will eventually construct to explain ideal gas behavior.
Dalton’s Law I Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25C and 1.0 atm and 12 L O2 at 25C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25C.
Solution The first step is to calculate the number of moles of each gas using the ideal gas law in the form: The tank containing the mixture has a volume of 5.0 L, and the temperature is 25C. We can use these data and the ideal gas law to calculate the partial pressure of each gas: The total pressure is the sum of the partial pressures: See Exercises 5.65 and 5.66.
At this point we need to define the mole fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. The Greek lowercase letter chi () is used to symbolize the mole fraction. For example, for a given component in a mixture, the mole fraction 1 is x1 n1 nTOTAL n1 n1 n2 n3 p PTOTAL PHe PO2 9.3 atm 2.4 atm 11.7 atm PO2 10.49 mol210.08206 L atm/K mol21298 K2 5.0 L 2.4 atm PHe 11.9 mol210.08206 L atm/K mol21298 K2 5.0 L 9.3 atm P nRT V nO2 11.0 atm2112 L2 10.08206 L atm/K mol21298 K2 0.49 mol nHe 11.0 atm2146 L2 10.08206 L atm/K mol21298 K2 1.9 mol n PV RT Sample Exercise 5.15 196 Chapter Five Gases From the ideal gas equation we know that the number of moles of a gas is directly pro-portional to the pressure of the gas, since That is, for each component in the mixture, Therefore, we can represent the mole fraction in terms of pressures: n1 n1 n2 n3 In fact, the mole fraction of each component in a mixture of ideal gases is directly related to its partial pressure: x2 n2 nTOTAL P2 PTOTAL P1 P1 P2 P3 p P1 PTOTAL 1V RT 2P1 1V RT 21P1 P2 P3 p2 x1 n1 nTOTAL P11V RT 2 P11V RT 2 P21V RT 2 P31V RT 2 p n1 P1a V RTb, n2 P2a V RTb, p n P a V RTb CHEMICAL IMPACT Separating Gases A ssume you work for an oil company that owns a huge natural gas reservoir containing a mixture of methane and nitrogen gases. In fact, the gas mixture contains so much nitrogen that it is unusable as a fuel. Your job is to separate the nitrogen (N2) from the methane (CH4). How might you accomplish this task? You clearly need some sort of “mo-lecular filter” that will stop the slightly larger methane mole-cules (size 430 pm) and allow the nitrogen molecules (size 410 pm) to pass through. To accomplish the sepa-ration of molecules so similar in size will require a very precise “filter.” The good news is that such a filter exists. Recent work by Steven Kuznick and Valerie Bell at Engelhard Corpora-tion in New Jersey and Michael Tsapatsis at the University of Massachusetts has produced a “molecular sieve” in which the pore (passage) sizes can be adjusted precisely enough to separate N2 molecules from CH4 molecules. The material involved is a special hydrated titanosilicate (contains H2O, Ti, Si, O, and Sr) compound patented by Engelhard known as ETS-4 (Engelhard TitanoSilicate-4). When sodium ions are substituted for the strontium ions in ETS-4 and the new material is carefully dehydrated, a uniform and controllable pore-size reduction occurs (see figure). The researchers have shown that the material can be used to separate N2 ( 410 pm) from O2 ( 390 pm). They have also shown that it is possible to reduce the nitrogen content of natural gas from 18% to less than 5% with a 90% recovery of methane.
Molecular sieve framework of titanium (blue), silicon (green), and oxygen (red) atoms contracts on heating—at room temperature (left), d 4.27 Å; at 250C (right), d 3.94 Å.
⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ Dehydration d d 5.5 Dalton’s Law of Partial Pressures 197 Dalton’s Law II The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.
Solution The mole fraction of O2 can be calculated from the equation Note that the mole fraction has no units.
See Exercise 5.69.
The expression for the mole fraction, can be rearranged to give That is, the partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure.
P1 x1 PTOTAL x1 P1 PTOTAL xO2 PO2 PTOTAL 156 torr 743 torr 0.210 Sample Exercise 5.16 CHEMICAL IMPACT The Chemistry of Air Bags M ost experts agree that air bags represent a very impor-tant advance in automobile safety. These bags, which are stored in the auto’s steering wheel or dash, are designed to inflate rapidly (within about 40 ms) in the event of a crash, cushioning the front-seat occupants against impact. The bags then deflate immediately to allow vision and movement af-ter the crash. Air bags are activated when a severe deceler-ation (an impact) causes a steel ball to compress a spring and electrically ignite a detonator cap, which, in turn, causes sodium azide (NaN3) to decompose explosively, forming sodium and nitrogen gas: This system works very well and requires a relatively small amount of sodium azide (100 g yields 56 L N2(g) at 25C and 1.0 atm).
When a vehicle containing air bags reaches the end of its useful life, the sodium azide present in the activators must be given proper disposal. Sodium azide, besides being ex-plosive, has a toxicity roughly equal to that of sodium 2NaN31s2 ¡ 2Na1s2 3N21g2 cyanide. It also forms hydrazoic acid (HN3), a toxic and explosive liquid, when treated with acid.
The air bag represents an application of chemistry that has already saved thousands of lives.
Inflated air bags.
198 Chapter Five Gases Dalton’s Law III The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr.
Solution The partial pressure of N2 can be calculated as follows: See Exercise 5.70.
Collecting a Gas over Water A mixture of gases results whenever a gas is collected by displacement of water. For ex-ample, Fig. 5.13 shows the collection of oxygen gas produced by the decomposition of solid potassium chlorate. In this situation, the gas in the bottle is a mixture of water va-por and the oxygen being collected. Water vapor is present because molecules of water escape from the surface of the liquid and collect in the space above the liquid. Molecules of water also return to the liquid. When the rate of escape equals the rate of return, the number of water molecules in the vapor state remains constant, and thus the pressure of water vapor remains constant. This pressure, which depends on temperature, is called the vapor pressure of water.
Gas Collection over Water A sample of solid potassium chlorate (KClO3) was heated in a test tube (see Fig. 5.13) and decomposed by the following reaction: The oxygen produced was collected by displacement of water at 22C at a total pressure of 754 torr. The volume of the gas collected was 0.650 L, and the vapor pressure of wa-ter at 22C is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed.
Solution First we find the partial pressure of O2 from Dalton’s law of partial pressures: PTOTAL PO2 PH2O PO2 21 torr 754 torr 2KClO31s2 ¡ 2KCl1s2 3O21g2 PN2 xN2 PTOTAL 0.7808 760. torr 593 torr FIGURE 5.13 The production of oxygen by thermal de-composition of KClO3. The MnO2 is mixed with the KClO3 to make the reaction faster.
KClO3(MnO2) O2(g), H2O(g) H2O Vapor pressure will be discussed in detail in Chapter 10. A table of water vapor pressure values is given in Section 10.8.
Sample Exercise 5.18 Sample Exercise 5.17 5.6 The Kinetic Molecular Theory of Gases 199 Thus Now we use the ideal gas law to find the number of moles of O2: In this case, Thus Next we will calculate the moles of KClO3 needed to produce this quantity of O2. From the balanced equation for the decomposition of KClO3, we have a mole ratio of 2 mol KClO33 mol O2. The moles of KClO3 can be calculated as follows: Using the molar mass of KClO3 (122.6 g/mol), we calculate the grams of KClO3: Thus the original sample contained 2.12 g KClO3.
See Exercises 5.71 through 5.73.
5.6 The Kinetic Molecular Theory of Gases We have so far considered the behavior of gases from an experimental point of view. Based on observations from different types of experiments, we know that at pressures of less than 1 atm most gases closely approach the behavior described by the ideal gas law. Now we want to construct a model to explain this behavior.
Before we do this, let’s briefly review the scientific method. Recall that a law is a way of generalizing behavior that has been observed in many experiments. Laws are very useful, since they allow us to predict the behavior of similar systems. For example, if a chemist prepares a new gaseous compound, a measurement of the gas density at known pressure and temperature can provide a reliable value for the compound’s molar mass.
However, although laws summarize observed behavior, they do not tell us why nature behaves in the observed fashion. This is the central question for scientists. To try to an-swer this question, we construct theories (build models). The models in chemistry consist of speculations about what the individual atoms or molecules (microscopic particles) might be doing to cause the observed behavior of the macroscopic systems (collections of very large numbers of atoms and molecules).
A model is considered successful if it explains the observed behavior in question and predicts correctly the results of future experiments. It is important to understand that a model can never be proved absolutely true. In fact, any model is an approximation by its 1.73 102 mol KClO3 122.6 g KClO3 1 mol KClO3 2.12 g KClO3 2.59 102 mol O2 2 mol KClO3 3 mol O2 1.73 102 mol KClO3 nO2 10.964 atm210.650 L2 10.08206 L atm/K mol21295 K2 2.59 102 mol R 0.08206 L atm/K mol T 22°C 273 295 K V 0.650 L PO2 733 torr 733 torr 760 torr/atm 0.964 atm nO2 PO2V RT PO2 754 torr 21 torr 733 torr 200 Chapter Five Gases very nature and is bound to fail at some point. Models range from the simple to the ex-traordinarily complex. We use simple models to predict approximate behavior and more complicated models to account very precisely for observed quantitative behavior. In this text we will stress simple models that provide an approximate picture of what might be happening and that fit the most important experimental results.
An example of this type of model is the kinetic molecular theory (KMT), a simple model that attempts to explain the properties of an ideal gas. This model is based on spec-ulations about the behavior of the individual gas particles (atoms or molecules). The pos-tulates of the kinetic molecular theory as they relate to the particles of an ideal gas can be stated as follows: 1. The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). See Fig. 5.14.
2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.
3. The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other.
4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.
Of course, the molecules in a real gas have finite volumes and do exert forces on each other. Thus real gases do not conform to these assumptions. However, we will see that these postulates do indeed explain ideal gas behavior.
The true test of a model is how well its predictions fit the experimental observations.
The postulates of the kinetic molecular model picture an ideal gas as consisting of parti-cles having no volume and no attractions for each other, and the model assumes that the gas produces pressure on its container by collisions with the walls.
Let’s consider how this model accounts for the properties of gases as summarized by the ideal gas law: PV nRT.
Pressure and Volume (Boyle’s Law) We have seen that for a given sample of gas at a given temperature (n and T are constant) that if the volume of a gas is decreased, the pressure increases: h Constant P 1nRT2 1 V FIGURE 5.14 (a) One mole of N2(l) has a volume of approximately 35 mL and a density of 0.81 g/mL. (b) One mole of N2(g) has a volume of 22.4 L (STP) and a density of 1.2 103 g/mL. Thus the ratio of the volumes of gaseous N2 and liquid N2 is 22.40.035 640 and the spacing of the molecules is 9 times farther apart in N2(g).
(a) (b) Visualization: Visualizing Molecular Motion: Many Molecules Visualization: Boyle’s Law: A Molecular-Level View Visualization: Visualizing Molecular Motion: Single Molecule 5.6 The Kinetic Molecular Theory of Gases 201 This makes sense based on the kinetic molecular theory, since a decrease in volume means that the gas particles will hit the wall more often, thus increasing pressure, as illustrated in Fig. 5.15.
Pressure and Temperature From the ideal gas law we can predict that for a given sample of an ideal gas at a con-stant volume, the pressure will be directly proportional to the temperature: h Constant The KMT accounts for this behavior because when the temperature of a gas increases, the speeds of its particles increase, the particles hitting the wall with greater force and greater frequency. Since the volume remains the same, this would result in increased gas pressure, as illustrated in Fig. 5.16.
Volume and Temperature (Charles’s Law) The ideal gas law indicates that for a given sample of gas at a constant pressure, the vol-ume of the gas is directly proportional to the temperature in kelvins: h Constant This can be visualized from the KMT, as shown in Fig. 5.17. When the gas is heated to a higher temperature, the speeds of its molecules increase and thus they hit the walls more often and with more force. The only way to keep the pressure constant in this situation is to increase the volume of the container. This compensates for the increased particle speeds.
V anR P bT P anR V bT FIGURE 5.15 The effects of decreasing the volume of a sample of gas at constant temperature.
FIGURE 5.16 The effects of increasing the temperature of a sample of gas at constant volume.
Volume is decreased Temperature is increased Visualization: Charles’s Law: A Molecular-Level View Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
202 Chapter Five Gases Volume and Number of Moles (Avogadro’s Law) The ideal gas law predicts that the volume of a gas at a constant temperature and pressure depends directly on the number of gas particles present: h Constant This makes sense in terms of the KMT, because an increase in the number of gas particles at the same temperature would cause the pressure to increase if the volume were held constant (see Fig. 5.18). The only way to return the pressure to its original value is to increase the volume.
It is important to recognize that the volume of a gas (at constant P and T) depends only on the number of gas particles present. The individual volumes of the particles are not a factor because the particle volumes are so small compared with the distances between the particles (for a gas behaving ideally).
Mixture of Gases (Dalton’s Law) The observation that the total pressure exerted by a mixture of gases is the sum of the pressures of the individual gases is expected because the KMT assumes that all gas par-ticles are independent of each other and that the volumes of the individual particles are unimportant. Thus the identities of the gas particles do not matter.
Deriving the Ideal Gas Law We have shown qualitatively that the assumptions of the KMT successfully account for the observed behavior of an ideal gas. We can go further. By applying the principles of physics to the assumptions of the KMT, we can in effect derive the ideal gas law.
V aRT P bn FIGURE 5.17 The effects of increasing the temperature of a sample of gas at constant pressure.
FIGURE 5.18 The effects of increasing the number of moles of gas particles at constant tempera-ture and pressure.
Temperature is increased Gas cylinder Moles of gas increases Increase volume to return to original pressure 5.6 The Kinetic Molecular Theory of Gases 203 As shown in detail in Appendix 2, we can apply the definitions of velocity, momen-tum, force, and pressure to the collection of particles in an ideal gas and derive the fol-lowing expression for pressure: where P is the pressure of the gas, n is the number of moles of gas, NA is Avogadro’s number, m is the mass of each particle, is the average of the square of the velocities of the particles, and V is the volume of the container.
The quantity represents the average kinetic energy of a gas particle. If the average kinetic energy of an individual particle is multiplied by NA, the number of particles in a mole, we get the average kinetic energy for a mole of gas particles: Using this definition, we can rewrite the expression for pressure as The fourth postulate of the kinetic molecular theory is that the average kinetic energy of the particles in the gas sample is directly proportional to the temperature in Kelvins.
Thus, since (KE)avg T, we can write Note that this expression has been derived from the assumptions of the kinetic molecular theory. How does it compare to the ideal gas law—the equation obtained from experi-ment? Compare the ideal gas law, From experiment with the result from the kinetic molecular theory, From theory PV n r T PV n RT PV n 2 3 1KE2avg r T or PV n r T r P 2 3 c n1KE2avg V d or PV n 2 3 1KE2avg 1KE2avg NA11 2mu22 1 2mu2 u2 P 2 3 c nNA11 2mu22 V d Kinetic energy (KE) given by the equation is the energy due to the motion of a particle. We will discuss this further in Section 6.1.
KE 1 2mu2 (a) A balloon filled with air at room temperature. (b) The balloon is dipped into liquid nitrogen at 77 K.
(c) The balloon collapses as the molecules inside slow down due to the decreased temperature. Slower molecules produce a lower pressure.
Visualization: Liquid Nitrogen and Balloons 204 Chapter Five Gases These expressions have exactly the same form if R, the universal gas constant, is considered the proportionality constant in the second case.
The agreement between the ideal gas law and the predictions of the kinetic molecular theory gives us confidence in the validity of the model. The characteristics we have assumed for ideal gas particles must agree, at least under certain conditions, with their actual behavior.
The Meaning of Temperature We have seen from the kinetic molecular theory that the Kelvin temperature indicates the average kinetic energy of the gas particles. The exact relationship between temperature and average kinetic energy can be obtained by combining the equations: which yields the expression This is a very important relationship. It summarizes the meaning of the Kelvin tempera-ture of a gas: The Kelvin temperature is an index of the random motions of the particles of a gas, with higher temperature meaning greater motion. (As we will see in Chapter 10, temperature is an index of the random motions in solids and liquids as well as in gases.) Root Mean Square Velocity In the equation from the kinetic molecular theory, the average velocity of the gas parti-cles is a special kind of average. The symbol means the average of the squares of the particle velocities. The square root of is called the root mean square velocity and is symbolized by urms: We can obtain an expression for urms from the equations Combination of these equations gives Taking the square root of both sides of the last equation produces In this expression m represents the mass in kilograms of a single gas particle. When NA, the number of particles in a mole, is multiplied by m, the product is the mass of a mole of gas particles in kilograms. We will call this quantity M. Substituting M for NAm in the equation for urms, we obtain Before we can use this equation, we need to consider the units for R. So far we have used 0.08206 L atm/K mol as the value of R. But to obtain the desired units (meters urms B 3RT M 2u2 urms B 3RT NAm NA11 2mu22 3 2RT or u2 3RT NAm 1KE2avg NA11 2mu22 and 1KE2avg 3 2RT urms 2u2 u2 u2 1KE2avg 3 2RT PV n RT 2 3 1KE2avg 5.6 The Kinetic Molecular Theory of Gases 205 per second) for urms, R must be expressed in different units. As we will see in more de-tail in Chapter 6, the energy unit most often used in the SI system is the joule (J). A joule is defined as a kilogram meter squared per second squared (kg m2/s2). When R is con-verted to include the unit of joules, it has the value 8.3145 J/K mol. When R in these units is used in the expression , urms is obtained in the units of meters per second as desired.
Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of helium gas at 25C.
Solution The formula for root mean square velocity is In this case T 25C 273 298 K, R 8.3145 J/K mol, and M is the mass of a mole of helium in kilograms: Thus Since the units of J are kg m2/s2, this expression becomes Note that the resulting units (m/s) are appropriate for velocity.
See Exercises 5.79 and 5.80.
So far we have said nothing about the range of velocities actually found in a gas sam-ple. In a real gas there are large numbers of collisions between particles. For example, as we will see in the next section, when an odorous gas such as ammonia is released in a room, it takes some time for the odor to permeate the air. This delay results from colli-sions between the NH3 molecules and the O2 and N2 molecules in the air, which greatly slow the mixing process.
If the path of a particular gas particle could be monitored, it would look very erratic, something like that shown in Fig. 5.19. The average distance a particle travels between collisions in a particular gas sample is called the mean free path. It is typically a very small distance (1 107 m for O2 at STP). One effect of the many collisions among gas particles is to produce a large range of velocities as the particles collide and exchange ki-netic energy. Although urms for oxygen gas at STP is approximately 500 meters per sec-ond, the majority of O2 molecules do not have this velocity. The actual distribution of molecular velocities for oxygen gas at STP is shown in Fig. 5.20. This figure shows the relative number of gas molecules having each particular velocity.
We are also interested in the effect of temperature on the velocity distribution in a gas.
Figure 5.21 shows the velocity distribution for nitrogen gas at three temperatures. Note that as the temperature is increased, the curve peak moves toward higher values and the range B1.86 106 kg m2 kg s2 1.36 103 m/s urms b 3a8.3145 J K molb1298 K2 4.00 103 kg mol B1.86 106 J kg M 4.00 g mol 1 kg 1000 g 4.00 103 kg/mol urms B 3RT M 13RT M R 8.3145 J K mol R 0.08206 L atm K mol Sample Exercise 5.19 FIGURE 5.19 Path of one particle in a gas. Any given par-ticle will continuously change its course as a result of collisions with other particles, as well as with the walls of the container.
FIGURE 5.20 A plot of the relative number of O2 mole-cules that have a given velocity at STP.
Relative number of O2 molecules with given velocity 0 Molecular velocity (m/s) 4 × 102 8 × 102 Visualization: Kinetic-Molecular Theory/Heat Transfer 206 Chapter Five Gases of velocities becomes much larger. The peak of the curve reflects the most probable velocity (the velocity found most often as we sample the movement of the various particles in the gas). Because the kinetic energy increases with temperature, it makes sense that the peak of the curve should move to higher values as the temperature of the gas is increased.
5.7 Effusion and Diffusion We have seen that the postulates of the kinetic molecular theory, when combined with the appropriate physical principles, produce an equation that successfully fits the experimen-tally observed behavior of gases as they approach ideal behavior. Two phenomena involving gases provide further tests of this model.
Diffusion is the term used to describe the mixing of gases. When a small amount of pungent-smelling ammonia is released at the front of a classroom, it takes some time be-fore everyone in the room can smell it, because time is required for the ammonia to mix with the air. The rate of diffusion is the rate of the mixing of gases. Effusion is the term used to describe the passage of a gas through a tiny orifice into an evacuated chamber, as shown in Fig. 5.22. The rate of effusion measures the speed at which the gas is transferred into the chamber.
Effusion Thomas Graham (1805–1869), a Scottish chemist, found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.
Stated in another way, the relative rates of effusion of two gases at the same temperature and pressure are given by the inverse ratio of the square roots of the masses of the gas particles: where M1 and M2 represent the molar masses of the gases. This equation is called Graham’s law of effusion.
Effusion Rates Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors (see Fig. 5.23).
Rate of effusion for gas 1 Rate of effusion for gas 2 1M2 1M1 FIGURE 5.22 The effusion of a gas into an evacuated chamber. The rate of effusion (the rate at which the gas is transferred across the barrier through the pin hole) is inversely proportional to the square root of the mass of the gas molecules.
FIGURE 5.21 A plot of the relative number of N2 molecules that have a given velocity at three tempera-tures. Note that as the temperature increases, both the average velocity and the spread of velocities increase.
Relative number of N2 molecules with given velocity 0 Velocity (m/s) 1000 2000 3000 273 K 1273 K 2273 K Gas Vacuum Pinhole In Graham’s law the units for molar mass can be g/mol or kg/mol, since the units cancel in the ratio .
1M2 1M1 Sample Exercise 5.20 Visualization: Effusion of a Gas 5.7 Effusion and Diffusion 207 Solution First we need to compute the molar masses: Molar mass of H2 2.016 g/mol, and molar mass of UF6 352.02 g/mol. Using Graham’s law, The effusion rate of the very light H2 molecules is about 13 times that of the massive UF6 molecules.
See Exercises 5.85 through 5.88.
Does the kinetic molecular model for gases correctly predict the relative effusion rates of gases summarized by Graham’s law? To answer this question, we must recognize that the effusion rate for a gas depends directly on the average velocity of its particles.
The faster the gas particles are moving, the more likely they are to pass through the ef-fusion orifice. This reasoning leads to the following prediction for two gases at the same pressure and temperature (T): This equation is identical to Graham’s law. Thus the kinetic molecular model does fit the experimental results for the effusion of gases.
Diffusion Diffusion is frequently illustrated by the lecture demonstration represented in Fig. 5.24, in which two cotton plugs soaked in ammonia and hydrochloric acid are simultaneously placed at the ends of a long tube. A white ring of ammonium chloride (NH4Cl) forms where the NH3 and HCl molecules meet several minutes later: White solid NH31g2 HCl1g2 ¡ NH4Cl1s2 Effusion rate for gas 1 Effusion rate for gas 2 urms for gas 1 urms for gas 2 B 3RT M1 B 3RT M2 1M2 1M1 Rate of effusion for H2 Rate of effusion for UF6 1MUF6 1MH2 B 352.02 2.016 13.2 FIGURE 5.23 Relative molecular speed distribution of H2 and UF6.
Percentage of molecules 0.04 0.03 0.02 0.01 0 1000 0 2000 Speed 3000 UF6 at 273 K H2 at 273 K Cotton wet with NH3(aq) Cotton wet with HCl(aq) Air Air Glass tube HCl dNH3 NH3 dHCl White ring of NH4Cl(s) forms where the NH3 and HCl meet FIGURE 5.24 (above right) When HCl(g) and NH3(g) meet in the tube, a white ring of NH4Cl(s) forms. (above left) A demonstration of the relative diffusion rates of NH3 and HCl molecules through air. Two cotton plugs, one dipped in HCl(aq) and one dipped in NH3(aq), are simultaneously inserted into the ends of the tube.
Gaseous NH3 and HCl vaporizing from the cotton plugs diffuse toward each other and, where they meet, react to form NH4Cl(s).
Visualization: Diffusion of Gases Visualization: Gaseous Ammo-nia and Hydrochloric Acid 208 Chapter Five Gases As a first approximation we might expect that the distances traveled by the two gases are related to the relative velocities of the gas molecules: However, careful experiments produce an observed ratio of less than 1.5, indicating that a quantitative analysis of diffusion requires a more complex analysis.
The diffusion of the gases through the tube is surprisingly slow in light of the fact that the velocities of HCl and NH3 molecules at 25C are about 450 and 660 meters per sec-ond, respectively. Why does it take several minutes for the NH3 and HCl molecules to meet?
The answer is that the tube contains air and thus the NH3 and HCl molecules undergo many collisions with O2 and N2 molecules as they travel through the tube. Because so many collisions occur when gases mix, diffusion is quite complicated to describe theoretically.
5.8 Real Gases An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law, although many gases come very close at low pressures and/or high temperatures. Thus ideal gas behavior can best be thought of as the behavior approached by real gases under certain conditions.
We have seen that a very simple model, the kinetic molecular theory, by making some rather drastic assumptions (no interparticle interactions and zero volume for the gas par-ticles), successfully explains ideal behavior. However, it is important that we examine real gas behavior to see how it differs from that predicted by the ideal gas law and to deter-mine what modifications are needed in the kinetic molecular theory to explain the ob-served behavior. Since a model is an approximation and will inevitably fail, we must be ready to learn from such failures. In fact, we often learn more about nature from the failures of our models than from their successes.
We will examine the experimentally observed behavior of real gases by measuring the pressure, volume, temperature, and number of moles for a gas and noting how the quantity PVnRT depends on pressure. Plots of PVnRT versus P are shown for several gases in Fig. 5.25. For an ideal gas, PVnRT equals 1 under all conditions, but notice that for real gases, PVnRT approaches 1 only at very low pressures (typically below 1 atm).
To illustrate the effect of temperature, PVnRT is plotted versus P for nitrogen gas at several temperatures in Fig. 5.26. Note that the behavior of the gas appears to become more nearly ideal as the temperature is increased. The most important conclusion to be drawn from these figures is that a real gas typically exhibits behavior that is closest to ideal behavior at low pressures and high temperatures.
One of the most important procedures in science is correcting our models as we col-lect more data. We will understand more clearly how gases actually behave if we can fig-ure out how to correct the simple model that explains the ideal gas law so that the new model fits the behavior we actually observe for gases. So the question is: How can we modify the assumptions of the kinetic molecular theory to fit the behavior of real gases?
The first person to do important work in this area was Johannes van der Waals (1837–1923), a physics professor at the University of Amsterdam who in 1910 received a Nobel Prize for his work. To follow his analysis, we start with the ideal gas law, Remember that this equation describes the behavior of a hypothetical gas consisting of volumeless entities that do not interact with each other. In contrast, a real gas consists of atoms or molecules that have finite volumes. Therefore, the volume available to a given particle in a real gas is less than the volume of the container because the gas par-ticles themselves take up some of the space. To account for this discrepancy, van der Waals P nRT V Distance traveled by NH3 Distance traveled by HCl urms for NH3 urms for HCl B MHCl MNH3 B 36.5 17 1.5 FIGURE 5.25 Plots of PVnRT versus P for several gases (200 K). Note the significant deviations from ideal behavior (PVnRT 1). The behavior is close to ideal only at low pressures (less than 1 atm).
FIGURE 5.26 Plots of PVnRT versus P for nitrogen gas at three temperatures. Note that although nonideal behavior is evident in each case, the deviations are smaller at the higher temperatures.
0 0 P (atm) 200 400 600 800 1.0 2.0 PV nRT H2 CH4 Ideal gas CO2 N2 1000 0 0.6 200 400 600 800 1.0 1.4 1.8 P (atm) PV nRT 293 K 673 K Ideal gas 203 K 5.8 Real Gases 209 represented the actual volume as the volume of the container V minus a correction factor for the volume of the molecules nb, where n is the number of moles of gas and b is an empirical constant (one determined by fitting the equation to the experimental results). Thus the volume actually available to a given gas molecule is given by the difference V nb.
This modification of the ideal gas equation leads to the equation The volume of the gas particles has now been taken into account.
The next step is to allow for the attractions that occur among the particles in a real gas. The effect of these attractions is to make the observed pressure Pobs smaller than it would be if the gas particles did not interact: This effect can be understood using the following model. When gas particles come close together, attractive forces occur, which cause the particles to hit the wall very slightly less often than they would in the absence of these interactions (see Fig. 5.27).
The size of the correction factor depends on the concentration of gas molecules defined in terms of moles of gas particles per liter (nV). The higher the concentration, the more likely a pair of gas particles will be close enough to attract each other. For large numbers of particles, the number of interacting pairs of particles depends on the square of the num-ber of particles and thus on the square of the concentration, or (nV)2. This can be justi-fied as follows: In a gas sample containing N particles, there are N 1 partners available for each particle, as shown in Fig. 5.28. Since the 1 2 pair is the same as the 2 1 pair, this analysis counts each pair twice. Thus, for N particles, there are N(N 1)2 pairs.
If N is a very large number, N 1 approximately equals N, giving N 22 possible pairs.
Thus the pressure, corrected for the attractions of the particles, has the form where a is a proportionality constant (which includes the factor of from N 22). The value of a for a given real gas can be determined from observing the actual behavior of that gas. Inserting the corrections for both the volume of the particles and the attractions of the particles gives the equation Observed Volume Volume Pressure pressure of the correction correction container Pobs nRT V nb a a n Vb 2 1 2 Pobs P¿ a a n Vb 2 p p Pobs 1P¿ correction factor2 a nRT V nb correction factorb P¿ nRT V nb FIGURE 5.27 (a) Gas at low concentration—relatively few interactions between particles. The indi-cated gas particle exerts a pressure on the wall close to that predicted for an ideal gas.
(b) Gas at high concentration—many more interactions between particles. The indi-cated gas particle exerts a much lower pres-sure on the wall than would be expected in the absence of interactions.
(a) Wall (b) Wall P is corrected for the finite volume of the particles. The attractive forces have not yet been taken into account.
The attractive forces among molecules will be discussed in Chapter 10.
We have now corrected for both the finite volume and the attractive forces of the particles.
FIGURE 5.28 Illustration of pairwise interactions among gas particles. In a sample with 10 particles, each particle has 9 possible partners, to give 10(9)2 45 distinct pairs. The factor of arises because when particle 1 is the particle of interest we count the ➀ ➁pair, and when particle ➁is the particle of interest we count the ➁ ➀ pair. However, ➀ ➁and ➁ ➀are the same pair that we have counted twice. Therefore, we must divide by 2 to get the actual number of pairs.
p p p p 1 2 1 3 2 4 5 6 10 7 8 9 Gas sample with ten particles Given particle { 8888n 8 n 8 n 88 n ⎧ ⎨ ⎩ 210 Chapter Five Gases This equation can be rearranged to give the van der Waals equation: Corrected pressure Corrected volume Pideal Videal The values of the weighting factors a and b are determined for a given gas by fitting experimental behavior. That is, a and b are varied until the best fit of the observed pres-sure is obtained under all conditions. The values of a and b for various gases are given in Table 5.3.
Experimental studies indicate that the changes van der Waals made in the basic as-sumptions of the kinetic molecular theory correct the major flaws in the model. First, con-sider the effects of volume. For a gas at low pressure (large volume), the volume of the container is very large compared with the volumes of the gas particles. That is, in this case the volume available to the gas is essentially equal to the volume of the container, and the gas behaves ideally. On the other hand, for a gas at high pressure (small container volume), the volume of the particles becomes significant so that the volume available to the gas is significantly less than the container volume. These cases are illustrated in Fig. 5.29. Note from Table 5.3 that the volume correction constant b generally increases with the size of the gas molecule, which gives further support to these arguments.
The fact that a real gas tends to behave more ideally at high temperatures also can be explained in terms of the van der Waals model. At high temperatures the particles are moving so rapidly that the effects of interparticle interactions are not very important.
The corrections to the kinetic molecular theory that van der Waals found necessary to explain real gas behavior make physical sense, which makes us confident that we un-derstand the fundamentals of gas behavior at the particle level. This is significant because so much important chemistry takes place in the gas phase. In fact, the mixture of gases called the atmosphere is vital to our existence. In Section 5.10 we consider some of the important reactions that occur in the atmosphere.
5.9 Characteristics of Several Real Gases We can understand gas behavior more completely if we examine the characteristics of sev-eral common gases. Note from Figure 5.25 that the gases H2, N2, CH4, and CO2 show dif-ferent behavior when the compressibility ( ) is plotted versus P. For example, notice that the plot for H2(g) never drops below the ideal value (1.0) in contrast to all the other gases. What is special about H2 compared to these other gases? Recall from Section 5.8 that the reason that the compressibility of a real gas falls below 1.0 is that the actual (ob-served) pressure is lower than the pressure expected for an ideal gas due to the intermo-lecular attractions that occur in real gases. This must mean that H2 molecules have very low attractive forces for each other. This idea is borne out by looking at the van der Waals PV nRT c Pobs a a n Vb 2 d 1V nb2 nRT Pobs is usually called just P.
FIGURE 5.29 The volume taken up by the gas particles themselves is less important at (a) large container volume (low pressure) than at (b) small container volume (high pressure).
⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ TABLE 5.3 Values of the van der Waals Constants for Some Common Gases Gas He 0.0341 0.0237 Ne 0.211 0.0171 Ar 1.35 0.0322 Kr 2.32 0.0398 Xe 4.19 0.0511 H2 0.244 0.0266 N2 1.39 0.0391 O2 1.36 0.0318 Cl2 6.49 0.0562 CO2 3.59 0.0427 CH4 2.25 0.0428 NH3 4.17 0.0371 H2O 5.46 0.0305 ba L molb aaatm L2 mol2 b (a) (b) 5.10 Chemistry in the Atmosphere 211 a value for H2 in Table 5.3. Note that H2 has the lowest value among the gases H2, N2, CH4, and CO2. Remember that the value of a reflects how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure: A low value for a reflects weak intermolecular forces among the gas molecules.
Also notice that although the compressibility for N2 dips below 1.0, it does not show as much deviation as that for CH4, which in turn does not show as much deviation as the compressibility for CO2. Based on this behavior we can surmise that the importance of intermolecular interactions increases in this order: This order is reflected by the relative a values for these gases in Table 5.3. In Section 10.1, we will see how these variations in intermolecular interactions can be explained. The main point to be made here is that real gas behavior can tell us about the relative importance of intermolecular attractions among gas molecules.
5.10 Chemistry in the Atmosphere The most important gases to us are those in the atmosphere that surrounds the earth’s surface. The principal components are N2 and O2, but many other important gases, such as H2O and CO2, are also present. The average composition of the earth’s atmosphere near sea level, with the water vapor removed, is shown in Table 5.4. Because of gravitational effects, the composition of the earth’s atmosphere is not constant; heavier molecules tend to be near the earth’s surface, and light molecules tend to migrate to higher altitudes, with some eventually escaping into space. The atmosphere is a highly complex and dynamic system, but for convenience we divide it into several layers based on the way the tem-perature changes with altitude. (The lowest layer, called the troposphere, is shown in Fig. 5.30.) Note that in contrast to the complex temperature profile of the atmosphere, the pressure decreases in a regular way with increasing altitude.
The chemistry occurring in the higher levels of the atmosphere is mostly determined by the effects of high-energy radiation and particles from the sun and other sources in space. In fact, the upper atmosphere serves as an important shield to prevent this high-energy radiation from reaching the earth, where it would damage the relatively fragile molecules sustaining life. In particular, the ozone in the upper atmosphere helps prevent high-energy ultraviolet radiation from penetrating to the earth. Intensive research is in progress to determine the natural factors that control the ozone concentration and how it is affected by chemicals released into the atmosphere.
The chemistry occurring in the troposphere, the layer of atmosphere closest to the earth’s surface, is strongly influenced by human activities. Millions of tons of gases and particulates are released into the troposphere by our highly industrial civilization. Actu-ally, it is amazing that the atmosphere can absorb so much material with relatively small permanent changes (so far).
Significant changes, however, are occurring. Severe air pollution is found around many large cities, and it is probable that long-range changes in our planet’s weather are taking place. We will discuss some of the long-range effects of pollution in Chapter 6. In this section we will deal with short-term, localized effects of pollution.
The two main sources of pollution are transportation and the production of electric-ity. The combustion of petroleum in vehicles produces CO, CO2, NO, and NO2, along with unburned molecules from petroleum. When this mixture is trapped close to the ground in stagnant air, reactions occur producing chemicals that are potentially irritating and harmful to living systems.
H2 6 N2 6 CH4 6 CO2 Pideal Pobserved a a n Vb 2 FIGURE 5.30 The variation of temperature (blue) and pressure (dashed lines) with altitude. Note that the pressure steadily decreases with al-titude, but the temperature increases and decreases.
TABLE 5.4 Atmospheric Composition Near Sea Level (Dry Air) Component Mole Fraction N2 0.78084 O2 0.20948 Ar 0.00934 CO2 0.000345 Ne 0.00001818 He 0.00000524 CH4 0.00000168 Kr 0.00000114 H2 0.0000005 NO 0.0000005 Xe 0.000000087 The atmosphere contains various amounts of water vapor depending on conditions.
Pressure (atm) –100 Temperature (°C) –50 0 50 100 10 –1 Troposphere 10 –3 10 –8 10 –13 0 1 2 5 10 20 50 100 200 500 1000 Altitude (km) 1 212 Chapter Five Gases The complex chemistry of polluted air appears to center around the nitrogen oxides (NOx). At the high temperatures found in the gasoline and diesel engines of cars and trucks, N2 and O2 react to form a small quantity of NO that is emitted into the air with the exhaust gases (see Fig. 5.31). This NO is immediately oxidized in air to NO2, which, in turn, absorbs energy from sunlight and breaks up into nitric oxide and free oxygen atoms: Radiant Oxygen atoms are very reactive and can combine with O2 to form ozone: Ozone is also very reactive and can react directly with other pollutants, or the ozone can absorb light and break up to form an energetically excited O2 molecule (O2) and an en-ergetically excited oxygen atom (O). The latter species readily reacts with a water mol-ecule to form two hydroxyl radicals (OH): The hydroxyl radical is a very reactive oxidizing agent. For example, OH can react with NO2 to form nitric acid: The OH radical also can react with the unburned hydrocarbons in the polluted air to produce chemicals that cause the eyes to water and burn and are harmful to the respiratory system.
OH NO2 ¡ HNO3 O H2O ¡ 2OH O1g2 O21g2 ¡ O31g2 NO21g2 — ¡ energy NO1g2 O1g2 FIGURE 5.31 Concentration (in molecules per million molecules of “air”) for some smog compo-nents versus time of day. (From “Photochemistry of Air Pollution,” by P. A.
Leighton, in Physical Chemistry: A Series of Mono-graphs, edited by Eric Hutchinson and P. Van Rysselberghe, copyright 1961 and renewed 1989, Elsevier Science (USA), reproduced by permission of the publisher.) CHEMICAL IMPACT Acid Rain: A Growing Problem the forests are under stress in Germany, and buildings and statues are deteriorating all over the world.
The Field Museum in Chicago contains more white Georgia marble than any other structure in the world. But nearly 70 years of exposure to the elements has taken such a toll on it that the building has recently undergone a multimillion-dollar renovation to replace the damaged marble with freshly quarried material.
What is the chemistry of the deterioration of marble by sulfuric acid? Marble is produced by geologic processes at high temperatures and pressures from lime-stone, a sedimentary rock formed by slow deposition of calcium carbonate from the shells of marine organisms.
Limestone and marble are chemically identical (CaCO3) but differ in physical properties; limestone is composed of smaller particles of calcium carbonate and is thus more porous and more workable. Although both limestone and marble are used for buildings, marble can be polished to a higher sheen and is often preferred for decorative purposes.
Both marble and limestone react with sulfuric acid to form calcium sulfate. The process can be represented most R ainwater, even in pristine wilderness areas, is slightly acidic because some of the carbon dioxide present in the atmosphere dissolves in the raindrops to produce H ions by the following reaction: This process produces only very small concentrations of H ions in the rainwater. However, gases such as NO2 and SO2, which are by-products of energy use, can produce sig-nificantly higher H concentrations. Nitrogen dioxide re-acts with water to give a mixture of nitrous acid and nitric acid: Sulfur dioxide is oxidized to sulfur trioxide, which then re-acts with water to form sulfuric acid: The damage caused by the acid formed in polluted air is a growing worldwide problem. Lakes are dying in Norway, SO31g2 H2O1l2 ¡ H2SO41aq2 2SO21g2 O21g2 ¡ 2SO31g2 2NO21g2 H2O1l2 ¡ HNO21aq2 HNO31aq2 H2O1l2 CO21g2 ¡ H1aq2 HCO3 1aq2 Concentration (ppm) 4:00 0 Time of day 0.1 0.2 0.3 0.4 0.5 6:00 8:00 10:00 Noon 2:00 4:00 6:00 NO NO2 O3 Molecules of unburned fuel (petroleum) Other pollutants The OH radical has no charge [it has one fewer electron than the hydroxide ion (OH)].
5.10 Chemistry in the Atmosphere 213 The end product of this whole process is often referred to as photochemical smog, so called because light is required to initiate some of the reactions. The production of pho-tochemical smog can be understood more clearly by examining as a group the reactions discussed above: Net reaction: Note that the NO2 molecules assist in the formation of ozone without being themselves used up. The ozone formed then leads to the formation of OH and other pollutants.
We can observe this process by analyzing polluted air at various times during a day (see Fig. 5.31). As people drive to work between 6 and 8 a.m., the amounts of NO, NO2, and unburned molecules from petroleum increase. Later, as the decomposition of NO2 oc-curs, the concentration of ozone and other pollutants builds up. Current efforts to combat the formation of photochemical smog are focused on cutting down the amounts of mole-cules from unburned fuel in automobile exhaust and designing engines that produce less nitric oxide.
The other major source of pollution results from burning coal to produce electricity.
Much of the coal found in the Midwest contains significant quantities of sulfur, which, when burned, produces sulfur dioxide: S 1in coal2 O21g2 ¡ SO21g2 3 2O21g2 ¡ O31g2 NO1g2 1 2O21g2 ¡ NO21g2 O1g2 O21g2 ¡ O31g2 NO21g2 ¡ NO1g2 O1g2 Although represented here as O2, the ac-tual oxidant for NO is OH or an organic peroxide such as CH3COO, formed by oxidation of organic pollutants.
simply as In this equation the calcium sulfate is represented by sepa-rate hydrated ions because calcium sulfate is quite water sol-uble and dissolves in rainwater. Thus, in areas bathed by rainwater, the marble slowly dissolves away.
In areas of the building protected from the rain, the cal-cium sulfate can form the mineral gypsum (CaSO4 2H2O).
The 2H2O in the formula of gypsum indicates the presence of two water molecules (called waters of hydration) for each CaSO4 formula unit in the solid. The smooth surface of the marble is thus replaced by a thin layer of gypsum, a more porous material that binds soot and dust.
What can be done to protect limestone and marble struc-tures from this kind of damage? Of course, one approach is to lower sulfur dioxide emissions from power plants (see Fig. 5.33). In addition, scientists are experimenting with coatings to protect marble from the acidic atmosphere. How-ever, a coating can do more harm than good unless it “breathes.” If moisture trapped beneath the coating freezes, the expanding ice can fracture the marble. Needless to say, it is difficult to find a coating that will allow water, but not acid, to pass—but the search continues.
Ca21aq2 SO4 21aq2 H2O1l2 CO21g2 CaCO31s2 H2SO41aq2 ¡ The damaging effects of acid rain can be seen by comparing these photos of a decorative statue on the Field Museum in Chicago. The first photo was taken about 1920, the second in 1990.
214 Chapter Five Gases A further oxidation reaction occurs when sulfur dioxide is changed to sulfur trioxide in the air: The production of sulfur trioxide is significant because it can combine with droplets of water in the air to form sulfuric acid: Sulfuric acid is very corrosive to both living things and building materials. Another result of this type of pollution is acid rain. In many parts of the northeastern United States and southeastern Canada, acid rain has caused some freshwater lakes to become too acidic to support any life (Fig. 5.32).
The problem of sulfur dioxide pollution is made more complicated by the energy cri-sis. As petroleum supplies dwindle and the price increases, our dependence on coal will probably grow. As supplies of low-sulfur coal are used up, high-sulfur coal will be uti-lized. One way to use high-sulfur coal without further harming the air quality is to re-move the sulfur dioxide from the exhaust gas by means of a system called a scrubber before it is emitted from the power plant stack. A common method of scrubbing is to blow powdered limestone (CaCO3) into the combustion chamber, where it is decomposed to lime and carbon dioxide: The lime then combines with the sulfur dioxide to form calcium sulfite: To remove the calcium sulfite and any remaining unreacted sulfur dioxide, an aqueous suspension of lime is injected into the exhaust gases to produce a slurry (a thick suspen-sion), as shown in Fig. 5.33.
Unfortunately, there are many problems associated with scrubbing. The systems are complicated and expensive and consume a great deal of energy. The large quantities of calcium sulfite produced in the process present a disposal problem. With a typical scrub-ber, approximately 1 ton of calcium sulfite per year is produced per person served by the power plant. Since no use has yet been found for this calcium sulfite, it is usually buried in a landfill. As a result of these difficulties, air pollution by sulfur dioxide continues to be a major problem, one that is expensive in terms of damage to the environment and hu-man health as well as in monetary terms.
CaO1s2 SO21g2 ¡ CaSO31s2 CaCO31s2 ¡ CaO1s2 CO21g2 SO31g2 H2O1l2 ¡ H2SO41aq2 2SO21g2 O21g2 ¡ 2SO31g2 FIGURE 5.32 An environmental officer in Wales tests the pH of water.
FIGURE 5.33 A schematic diagram of the process for scrubbing sulfur dioxide from stack gases in power plants.
This reaction is very slow unless solid particles are present. See Chapter 12 for a discussion.
Coal CaCO3 Air To smokestack S + O2 SO2 CaSO3 unreacted SO2 + CO2 + CaO Combustion chamber CaSO3 slurry Water + CaO Scrubber For Review 215 Key Terms Section 5.1 barometer manometer mm Hg torr standard atmosphere pascal Section 5.2 Boyle’s law ideal gas Charles’s law absolute zero Avogadro’s law Section 5.3 universal gas constant ideal gas law Section 5.4 molar volume standard temperature and pressure (STP) Section 5.5 Dalton’s law of partial pressures partial pressure mole fraction Section 5.6 kinetic molecular theory (KMT) root mean square velocity joule Section 5.7 diffusion effusion Graham’s law of effusion Section 5.8 real gas van der Waals equation Section 5.10 atmosphere air pollution photochemical smog acid rain For Review State of a gas The state of a gas can be described completely by specifying its pressure (P), volume (V), temperature (T) and the amount (moles) of gas present (n) Pressure • Common units • SI unit: pascal Gas laws Discovered by observing the properties of gases Boyle’s law: PV k Charles’s law: V bT Avogadro’s law: V an Ideal gas law: PV nRT Dalton’s law of partial pressures: Ptotal P1 P2 P3 , where Pn represents the partial pressure of component n in a mixture of gases Kinetic molecular theory (KMT) Model that accounts for ideal gas behavior Postulates of the KMT: • Volume of gas particles is zero • No particle interactions • Particles are in constant motion, colliding with the container walls to produce pressure • The average kinetic energy of the gas particles is directly proportional to the temperature of the gas in kelvins Gas properties The particles in any gas sample have a range of velocities The root mean square (rms) velocity for a gas represents the average of the squares of the particle velocities Diffusion: the mixing of two or more gases Effusion: the process in which a gas passes through a small hole into an empty chamber Real gas behavior Real gases behave ideally only at high temperatures and low pressures Understanding how the ideal gas equation must be modified to account for real gas behavior helps us understand how gases behave on a molecular level Van der Waals found that to describe real gas behavior we must consider particle interactions and particle volumes REVIEW QUESTIONS 1. Explain how a barometer and a manometer work to measure the pressure of the atmosphere or the pressure of a gas in a container.
urms B 3RT M p 1 atm 101,325 Pa 1 atm 760 torr 1 torr 1 mm Hg 216 Chapter Five Gases 2. What are Boyle’s law, Charles’s law, and Avogadro’s law? What plots do you make to show a linear relationship for each law?
3. Show how Boyle’s law, Charles’s law, and Avogadro’s law are special cases of the ideal gas law. Using the ideal gas law, determine the relationship between P and n (at constant V and T ) and between P and T (at constant V and n).
4. Rationalize the following observations.
a. Aerosol cans will explode if heated.
b. You can drink through a soda straw.
c. A thin-walled can will collapse when the air inside is removed by a vacuum pump.
d. Manufacturers produce different types of tennis balls for high and low elevations.
5. Consider the following balanced equation in which gas X forms gas X2: Equal moles of X are placed in two separate containers. One container is rigid so the volume cannot change; the other container is flexible so the volume changes to keep the internal pressure equal to the external pressure. The above reaction is run in each container. What happens to the pressure and density of the gas inside each container as reactants are converted to products?
6. Use the postulates of the kinetic molecular theory (KMT) to explain why Boyle’s law, Charles’s law, Avogadro’s law, and Dalton’s law of partial pressures hold true for ideal gases. Use the KMT to explain the P versus n (at constant V and T) relationship and the P versus T (at constant V and n) relationship.
7. Consider the following velocity distribution curves A and B.
a. If the plots represent the velocity distribution of 1.0 L of He(g) at STP ver-sus 1.0 L of Cl2(g) at STP, which plot corresponds to each gas? Explain your reasoning.
b. If the plots represent the velocity distribution of 1.0 L of O2(g) at tempera-tures of 273 K versus 1273 K, which plot corresponds to each temperature?
Explain your reasoning. Under which temperature condition would the O2(g) sample behave most ideally? Explain.
8. Briefly describe two methods one might use to find the molar mass of a newly synthesized gas for which a molecular formula was not known.
9. In the van der Waals equation, why is a term added to the observed pressure and why is a term subtracted from the container volume to correct for nonideal gas behavior?
10. Why do real gases not always behave ideally? Under what conditions does a real gas behave most ideally? Why?
Velocity (m/s) A B Relative number of molecules 2X1g2 S X21g2 4. As you increase the temperature of a gas in a sealed, rigid con-tainer, what happens to the density of the gas? Would the results be the same if you did the same experiment in a container with a piston at constant pressure? (See Figure 5.17.) 5. A diagram in a chemistry book shows a magnified view of a flask of air as follows: What do you suppose is between the dots (the dots represent air molecules)?
a. air b. dust c. pollutants d. oxygen e. nothing 6. If you put a drinking straw in water, place your finger over the opening, and lift the straw out of the water, some water stays in the straw. Explain.
7. A chemistry student relates the following story: I noticed my tires were a bit low and went to the gas station. As I was filling the tires, I thought about the kinetic molecular theory (KMT). I noticed the tires because the volume was low, and I realized that I was increasing both the pressure and volume of the tires.
“Hmmm,” I thought, “that goes against what I learned in chem-istry, where I was told pressure and volume are inversely pro-portional.” What is the fault in the logic of the chemistry student in this situation? Explain why we think pressure and volume to be inversely related (draw pictures and use the KMT).
8. Chemicals X and Y (both gases) react to form the gas XY, but it takes a bit of time for the reaction to occur. Both X and Y are placed in a container with a piston (free to move), and you note the volume. As the reaction occurs, what happens to the volume of the container? (See Fig. 5.18.) 9. Which statement best explains why a hot-air balloon rises when the air in the balloon is heated?
a. According to Charles’s law, the temperature of a gas is directly related to its volume. Thus the volume of the balloon increases, making the density smaller. This lifts the balloon.
b. Hot air rises inside the balloon, and this lifts the balloon.
c. The temperature of a gas is directly related to its pressure.
The pressure therefore increases, and this lifts the balloon.
d. Some of the gas escapes from the bottom of the balloon, thus decreasing the mass of gas in the balloon. This decreases the density of the gas in the balloon, which lifts the balloon.
e. Temperature is related to the root mean square velocity of the gas molecules. Thus the molecules are moving faster, hitting the balloon more, and thus lifting the balloon.
Justify your choice, and for the choices you did not pick, explain what is wrong with them.
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Consider the following apparatus: a test tube covered with a non-permeable elastic membrane inside a container that is closed with a cork. A syringe goes through the cork.
a. As you push down on the syringe, how does the membrane covering the test tube change?
b. You stop pushing the syringe but continue to hold it down.
In a few seconds, what happens to the membrane?
2. Figure 5.2 shows a picture of a barometer. Which of the following statements is the best explanation of how this barometer works?
a. Air pressure outside the tube causes the mercury to move in the tube until the air pressure inside and outside the tube is equal.
b. Air pressure inside the tube causes the mercury to move in the tube until the air pressure inside and outside the tube is equal.
c. Air pressure outside the tube counterbalances the weight of the mercury in the tube.
d. Capillary action of the mercury causes the mercury to go up the tube.
e. The vacuum that is formed at the top of the tube holds up the mercury.
Justify your choice, and for the choices you did not pick, explain what is wrong with them. Pictures help!
3. The barometer below shows the level of mercury at a given at-mospheric pressure. Fill all the other barometers with mercury for that same atmospheric pressure. Explain your answer.
Hg(l) Syringe Membrane Cork 217 218 Chapter Five Gases 19. Boyle’s law can be represented graphically in several ways.
Which of the following plots does not correctly represent Boyle’s law (assuming constant T and n)? Explain.
20. As weather balloons rise from the earth’s surface, the pressure of the atmosphere becomes less, tending to cause the volume of the balloons to expand. However, the temperature is much lower in the upper atmosphere than at sea level. Would this temperature effect tend to make such a balloon expand or contract? Weather balloons do, in fact, expand as they rise. What does this tell you?
21. Which noble gas has the smallest density at STP? Explain.
22. Consider two different containers, each filled with 2 moles of Ne(g). One of the containers is rigid and has constant volume.
The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.
23. Do all the molecules in a 1-mol sample of CH4(g) have the same kinetic energy at 273 K? Do all molecules in a 1-mol sample of N2(g) have the same velocity at 546 K? Explain.
24. Consider the following samples of gases at the same temperature.
Arrange each of these samples in order from lowest to highest: a. pressure b. average kinetic energy c. density d. root mean square velocity Note: Some samples of gases may have equal values for these attributes. Assume the larger containers have a volume twice the volume of the smaller containers and assume the mass of an ar-gon atom is twice the mass of a neon atom.
25. As NH3(g) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of NH3 reacted. Explain. As NH3(g) i ii iii iv Ne Ar v vi vii viii P PV V P 1/P V 1/V P 10. Draw a highly magnified view of a sealed, rigid container filled with a gas. Then draw what it would look like if you cooled the gas significantly but kept the temperature above the boiling point of the substance in the container. Also draw what it would look like if you heated the gas significantly. Finally, draw what each situation would look like if you evacuated enough of the gas to decrease the pressure by a factor of 2.
11. If you release a helium balloon, it soars upward and eventually pops. Explain this behavior.
12. If you have any two gases in different containers that are the same size at the same pressure and same temperature, what is true about the moles of each gas? Why is this true?
13. Explain the following seeming contradiction: You have two gases, A and B, in two separate containers of equal volume and at equal pressure and temperature. Therefore, you must have the same number of moles of each gas. Because the two tempera-tures are equal, the average kinetic energies of the two samples are equal. Therefore, since the energy given such a system will be converted to translational motion (that is, move the mole-cules), the root mean square velocities of the two are equal, and thus the particles in each sample move, on average, with the same relative speed. Since A and B are different gases, they each must have a different molar mass. If A has higher molar mass than B, the particles of A must be hitting the sides of the container with more force. Thus the pressure in the container of gas A must be higher than that in the container with gas B. However, one of our initial assumptions was that the pressures were equal.
14. You have a balloon covering the mouth of a flask filled with air at 1 atm. You apply heat to the bottom of the flask until the vol-ume of the balloon is equal to that of the flask.
a. Which has more air in it, the balloon or the flask? Or do both have the same amount? Explain.
b. In which is the pressure greater, the balloon or the flask? Or is the pressure the same? Explain.
15. How does Dalton’s law of partial pressures help us with our model of ideal gases? That is, what postulates of the kinetic mo-lecular theory does it support?
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of the book and a solution appears in the Solutions Guide.
Questions 16. At room temperature, water is a liquid with a molar volume of 18 mL. At 105C and 1 atm pressure, water is a gas and has a molar volume of over 30 L. Explain the large difference in molar volumes.
17. If a barometer were built using water (d 1.0 g/cm3) instead of mercury (d 13.6 g/cm3), would the column of water be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain.
18. A bag of potato chips is packed and sealed in Los Angeles, California, and then shipped to Lake Tahoe, Nevada, during ski season. It is noticed that the volume of the bag of potato chips has increased upon its arrival in Lake Tahoe. What external conditions would most likely cause the volume increase?
Exercises 219 is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of NH3?
26. Which of the following statements is (are) true? For the false statements, correct them.
a. At constant temperature, the lighter the gas molecules, the faster the average velocity of the gas molecules.
b. At constant temperature, the heavier the gas molecules, the larger the average kinetic energy of the gas molecules.
c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving relatively quickly.
d. As temperature increases, the effect of interparticle interac-tions on gas behavior is increased.
e. At constant V and T, as gas molecules are added into a con-tainer, the number of collisions per unit area increases re-sulting in a higher pressure.
f. The kinetic molecular theory predicts that pressure is inversely proportional to temperature at constant volume and mol of gas.
Exercises In this section similar exercises are paired.
Pressure 27. Freon-12 (CF2Cl2) is commonly used as the refrigerant in cen-tral home air conditioners. The system is initially charged to a pressure of 4.8 atm. Express this pressure in each of the fol-lowing units (1 atm 14.7 psi).
a. mm Hg b. torr c. Pa d. psi 28. A gauge on a compressed gas cylinder reads 2200 psi (pounds per square inch; 1 atm 14.7 psi). Express this pressure in each of the following units.
a. standard atmospheres b. megapascals (MPa) c. torr 29. A sealed-tube manometer (as shown below) can be used to meas-ure pressures below atmospheric pressure. The tube above the mercury is evacuated. When there is a vacuum in the flask, the mercury levels in both arms of the U-tube are equal. If a gaseous sample is introduced into the flask, the mercury levels are dif-ferent. The difference h is a measure of the pressure of the gas inside the flask. If h is equal to 6.5 cm, calculate the pressure in the flask in torr, pascals, and atmospheres.
Gas h 30. If the sealed-tube manometer in Exercise 29 had a height dif-ference of 20.0 inches between the mercury levels, what is the pressure in the flask in torr and atmospheres?
31. A diagram for an open-tube manometer is shown below.
If the flask is open to the atmosphere, the mercury levels are equal. For each of the following situations where a gas is con-tained in the flask, calculate the pressure in the flask in torr, at-mospheres, and pascals.
c. Calculate the pressures in the flask in parts a and b (in torr) if the atmospheric pressure is 635 torr.
32. a. If the open-tube manometer in Exercise 31 contains a non-volatile silicone oil (density 1.30 g/cm3) instead of mercury (density 13.6 g/cm3), what are the pressures in the flask as shown in parts a and b in torr, atmospheres, and pascals?
b. What advantage would there be in using a less dense fluid than mercury in a manometer used to measure relatively small differences in pressure?
Gas Laws 33. A particular balloon is designed by its manufacturer to be in-flated to a volume of no more than 2.5 L. If the balloon is filled with 2.0 L of helium at sea level, is released, and rises to an al-titude at which the atmospheric pressure is only 500. mm Hg, will the balloon burst? (Assume temperature is constant.) 34. A balloon is filled to a volume of 7.00 102 mL at a tempera-ture of 20.0C. The balloon is then cooled at constant pressure to a temperature of 1.00 102 K. What is the final volume of the balloon?
Atmosphere (760. torr) Atmosphere (760. torr) Flask a.
b.
118 mm Flask 215 mm Atmosphere 220 Chapter Five Gases increases the temperature of the tire to 58C. The volume of the tire increases by 4.0%. What is the new pressure in the bicycle tire?
45. Consider two separate gas containers at the following conditions: How is the pressure in container B related to the pressure in con-tainer A?
46. A container is filled with an ideal gas to a pressure of 40.0 atm at 0C.
a. What will be the pressure in the container if it is heated to 45C?
b. At what temperature would the pressure be 1.50 102 atm?
c. At what temperature would the pressure be 25.0 atm?
47. An ideal gas is contained in a cylinder with a volume of 5.0 102 mL at a temperature of 30.C and a pressure of 710. torr.
The gas is then compressed to a volume of 25 mL, and the temperature is raised to 820.C. What is the new pressure of the gas?
48. A compressed gas cylinder contains 1.00 103 g of argon gas.
The pressure inside the cylinder is 2050. psi (pounds per square inch) at a temperature of 18C. How much gas remains in the cylinder if the pressure is decreased to 650. psi at a temperature of 26C?
49. A sealed balloon is filled with 1.00 L of helium at 23C and 1.00 atm. The balloon rises to a point in the atmosphere where the pressure is 220. torr and the temperature is 31C. What is the change in volume of the balloon as it ascends from 1.00 atm to a pressure of 220. torr?
50. A hot-air balloon is filled with air to a volume of 4.00 103 m3 at 745 torr and 21C. The air in the balloon is then heated to 62C, causing the balloon to expand to a volume of 4.20 103 m3.
What is the ratio of the number of moles of air in the heated bal-loon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.) Gas Density, Molar Mass, and Reaction Stoichiometry 51. Consider the following reaction: It takes 2.00 L of pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of alu-minum reacted?
4Al1s2 3O21g2 S 2Al2O31s2 Container A Container B Contents: SO2(g) Contents: unknown gas Pressure PA Pressure PB Moles of gas 1.0 mol Moles of gas 2.0 mol Volume 1.0 L Volume 2.0 L Temperature 7C Temperature 287C 35. An 11.2-L sample of gas is determined to contain 0.50 mol of N2. At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?
36. Consider the following chemical equation.
If 25.0 mL of NO2 gas is completely converted to N2O4 gas un-der the same conditions, what volume will the N2O4 occupy?
37. Complete the following table for an ideal gas.
2NO21g2 ¡ N2O41g2 P(atm) V(L) n(mol) T a.
5.00 2.00 155C b.
0.300 2.00 155 K c.
4.47 25.0 2.01 d.
2.25 10.5 75C 38. Complete the following table for an ideal gas.
P V n T a.
7.74 103 Pa 12.2 mL 25C b.
43.0 mL 0.421 mol 223 K c.
455 torr 4.4 102 mol 331C d.
745 mm Hg 11.2 L 0.401 mol 39. Suppose two 200.0-L tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 135 atm in its respective tank at 24C?
40. A flask that can withstand an internal pressure of 2500 torr, but no more, is filled with a gas at 21.0C and 758 torr and heated.
At what temperature will it burst?
41. A 2.50-L container is filled with 175 g argon.
a. If the pressure is 10.0 atm, what is the temperature?
b. If the temperature is 225 K, what is the pressure?
42. A person accidentally swallows a drop of liquid oxygen, O2(l), which has a density of 1.149 g/mL. Assuming the drop has a volume of 0.050 mL, what volume of gas will be produced in the person’s stomach at body temperature (37C) and a pressure of 1.0 atm?
43. A gas sample containing 1.50 mol at 25C exerts a pressure of 400. torr. Some gas is added to the same container and the tem-perature is increased to 50.C. If the pressure increases to 800.
torr, how many moles of gas were added to the container? As-sume a constant-volume container.
44. A bicycle tire is filled with air to a pressure of 100. psi at a temperature of 19C. Riding the bike on asphalt on a hot day Exercises 221 59. Hydrogen cyanide is prepared commercially by the reaction of methane, CH4(g), ammonia, NH3(g), and oxygen, O2(g), at high temperature. The other product is gaseous water.
a. Write a chemical equation for the reaction.
b. What volume of HCN(g) can be obtained from 20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g)? The volumes of all gases are measured at the same temperature and pressure.
60. Methanol, CH3OH, can be produced by the following reaction: Hydrogen at STP flows into a reactor at a rate of 16.0 L/min.
Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30 g of methanol is produced per minute, what is the percent yield of the reaction?
61. An unknown diatomic gas has a density of 3.164 g/L at STP.
What is the identity of the gas?
62. A compound has the empirical formula CHCl. A 256-mL flask, at 373 K and 750. torr, contains 0.800 g of the gaseous com-pound. Give the molecular formula.
63. Uranium hexafluoride is a solid at room temperature, but it boils at 56C. Determine the density of uranium hexafluoride at 60.C and 745 torr.
64. Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions 78% N2, 21% O2, and 1.0% Ar, what is the density of air at standard temperature and pressure?
Partial Pressure 65. A piece of solid carbon dioxide, with a mass of 7.8 g, is placed in a 4.0-L otherwise empty container at 27C. What is the pres-sure in the container after all the carbon dioxide vaporizes? If 7.8 g solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?
66. A mixture of 1.00 g H2 and 1.00 g He is placed in a 1.00-L con-tainer at 27C. Calculate the partial pressure of each gas and the total pressure.
67. Consider the flasks in the following diagram. What are the final partial pressures of H2 and N2 after the stopcock between the two flasks is opened? (Assume the final volume is 3.00 L.) What is the total pressure (in torr)?
2.00 L H2 475 torr 1.00 L N2 0.200 atm CO1g2 2H21g2 ¡ CH3OH1g2 52. A student adds 4.00 g of dry ice (solid CO2) to an empty bal-loon. What will be the volume of the balloon at STP after all the dry ice sublimes (converts to gaseous CO2)?
53. Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively accord-ing to the following reaction: What mass of NaN3(s) must be reacted to inflate an air bag to 70.0 L at STP?
54. Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): What volume of pure O2(g), collected at 27C and 746 torr, would be generated by decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution? Ignore any water vapor that may be present.
55. In 1897 the Swedish explorer Andreé tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hydrogen gas was prepared from iron splints and diluted sulfuric acid. The reaction is The volume of the balloon was 4800 m3 and the loss of hydrogen gas during filling was estimated at 20.%. What mass of iron splints and 98% (by mass) H2SO4 were needed to ensure the complete filling of the balloon? Assume a temper-ature of 0C, a pressure of 1.0 atm during filling, and 100% yield.
56. Sulfur trioxide, SO3, is produced in enormous quantities each year for use in the synthesis of sulfuric acid.
What volume of O2(g) at 350.C and a pressure of 5.25 atm is needed to completely convert 5.00 g of sulfur to sulfur trioxide?
57. Consider the reaction between 50.0 mL of liquid methyl alco-hol, CH3OH (density 0.850 g/mL), and 22.8 L of O2 at 27C and a pressure of 2.00 atm. The products of the reaction are CO2(g) and H2O(g). Calculate the number of moles of H2O formed if the reaction goes to completion.
58. Urea (H2NCONH2) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of am-monia and carbon dioxide: Ammonia gas at 223C and 90. atm flows into a reactor at a rate of 500. L/min. Carbon dioxide at 223C and 45 atm flows into the reactor at a rate of 600. L/min. What mass of urea is pro-duced per minute by this reaction assuming 100% yield?
2NH31g2 CO21g2 —¡ Heat Pressure H2NCONH21s2 H2O1g2 2SO21g2 O21g2 S 2SO31g2 S1s2 O21g2 S SO21g2 Fe1s2 H2SO41aq2 ¡ FeSO41aq2 H21g2 2H2O21aq2 S 2H2O1l2 O21g2 2NaN31s2 ¡ 2Na1s2 3N21g2 222 Chapter Five Gases 76. Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine [(CH3)2N2H2] mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: If 150 g of dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 27C in an evac-uated 250-L tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has 100% yield?
Kinetic Molecular Theory and Real Gases 77. Calculate the average kinetic energies of CH4 and N2 molecules at 273 K and 546 K.
78. A 100.-L flask contains a mixture of methane, CH4, and argon at 25C. The mass of argon present is 228 g and the mole frac-tion of methane in the mixture is 0.650. Calculate the total kinetic energy of the gaseous mixture.
79. Calculate the root mean square velocities of CH4 and N2 mole-cules at 273 K and 546 K.
80. Consider separate 1.0-L samples of He(g) and UF6(g), both at 1.00 atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?
81. Consider a 1.0-L container of neon gas at STP. Will the aver-age kinetic energy, average velocity, and frequency of colli-sions of gas molecules with the walls of the container increase, decrease, or remain the same under each of the following conditions?
a. The temperature is increased to 100C.
b. The temperature is decreased to 50C.
c. The volume is decreased to 0.5 L.
d. The number of moles of neon is doubled.
82. Consider two gases, A and B, each in a 1.0-L container with both gases at the same temperature and pressure. The mass of gas A in the container is 0.34 g and the mass of gas B in the container is 0.48 g.
a. Which gas sample has the most molecules present? Explain.
b. Which gas sample has the largest average kinetic energy?
Explain.
A 0.34 g B 0.48 g 1CH322N2H21l2 2N2O41l2 ¡ 3N21g2 4H2O1g2 2CO21g2 68. Consider the flask apparatus in Exercise 67, which now contains 2.00 L of H2 at a pressure of 360. torr and 1.00 L of N2 at an unknown pressure. If the total pressure in the flasks is 320. torr after the stopcock is opened, determine the initial pressure of N2 in the 1.00-L flask.
69. The partial pressure of CH4(g) is 0.175 atm and that of O2(g) is 0.250 atm in a mixture of the two gases.
a. What is the mole fraction of each gas in the mixture?
b. If the mixture occupies a volume of 10.5 L at 65C, calcu-late the total number of moles of gas in the mixture.
c. Calculate the number of grams of each gas in the mixture.
70. A 1.00-L gas sample at 100.C and 600. torr contains 50.0% he-lium and 50.0% xenon by mass. What are the partial pressures of the individual gases?
71. Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.
Typically, the hydrogen gas is bubbled through water for col-lection and becomes saturated with water vapor. Suppose 240.
mL of hydrogen gas is collected at 30.C and has a total pres-sure of 1.032 atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at 30C.) 72. Helium is collected over water at 25C and 1.00 atm total pres-sure. What total volume of gas must be collected to obtain 0.586 g of helium? (At 25C the vapor pressure of water is 23.8 torr.) 73. At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8765-g sample of impure sodium chlorate was heated until the production of oxy-gen gas ceased. The oxygen gas collected over water occupied 57.2 mL at a temperature of 22C and a pressure of 734 torr.
Calculate the mass percent of NaClO3 in the original sample.
(At 22C the vapor pressure of water is 19.8 torr.) 74. Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to 400C in a nickel reaction vessel. A 100.0-mL nickel container is filled with xenon and fluorine, giving partial pressures of 1.24 atm and 10.10 atm, respectively, at a temperature of 25C. The reaction vessel is heated to 400C to cause a reaction to occur and then cooled to a temperature at which F2 is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining F2 gas is transferred to another 100.0-mL nickel container, where the pressure of F2 at 25C is 7.62 atm. Assuming all of the xenon has reacted, what is the formula of the product?
75. Hydrogen azide, HN3, decomposes on heating by the following unbalanced reaction: If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the par-tial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.
HN31g2 ¡ N21g2 H21g2 Zn1s2 2HCl1aq2 S ZnCl21aq2 H21g2 Additional Exercises 223 92. A 1.0-L sample of air is collected at 25C at sea level (1.00 atm).
Estimate the volume this sample of air would have at an altitude of 15 km (see Fig. 5.30).
93. Write reactions to show how nitric and sulfuric acids are pro-duced in the atmosphere.
94. Write reactions to show how the nitric and sulfuric acids in acid rain react with marble and limestone. (Both marble and lime-stone are primarily calcium carbonate.) Additional Exercises 95. Draw a qualitative graph to show how the first property varies with the second in each of the following (assume 1 mol of an ideal gas and T in kelvins).
a. PV versus V with constant T b. P versus T with constant V c. T versus V with constant P d. P versus V with constant T e. P versus 1V with constant T f. PVT versus P 96. At STP, 1.0 L Br2 reacts completely with 3.0 L F2, producing 2.0 L of a product. What is the formula of the product? (All sub-stances are gases.) 97. A form of Boyle’s law is PV k (at constant T and n). Table 5.1 contains actual data from pressure–volume experiments conducted by Robert Boyle. The value of k in most experiments is 14.1 102 in Hg in3. Express k in units of atm L. In Sample Exercise 5.3, k was determined for NH3 at various pressures and volumes. Give some reasons why the k values differ so dramatically between Sample Exercise 5.3 and Table 5.1.
98. An ideal gas at 7C is in a spherical flexible container having a radius of 1.00 cm. The gas is heated at constant pressure to 88C.
Determine the radius of the spherical container after the gas is heated. (Volume of a sphere 43
r 3.) 99. A 2.747-g sample of manganese metal is reacted with excess HCl gas to produce 3.22 L of H2(g) at 373 K and 0.951 atm and a manganese chloride compound (MnClx). What is the formula of the manganese chloride compound produced in the reaction?
100. Equal moles of hydrogen gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the forma-tion of gaseous water. Assuming that the reaction goes to com-pletion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are meas-ured at the same temperature and pressure?
101. A 15.0-L tank is filled with H2 to a pressure of 2.00 102 atm.
How many balloons (each 2.00 L) can be inflated to a pressure of 1.00 atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below 1.00 atm pressure.
102. A spherical glass container of unknown volume contains helium gas at 25C and 1.960 atm. When a portion of the helium is with-drawn and adjusted to 1.00 atm at 25C, it is found to have a c. Which gas sample has the fastest average velocity? Explain.
d. How can the pressure in the two containers be equal to each other since the larger gas B molecules collide with the con-tainer walls more forcefully?
83. Consider three identical flasks filled with different gases.
Flask A: CO at 760 torr and 0C Flask B: N2 at 250 torr and 0C Flask C: H2 at 100 torr and 0C a. In which flask will the molecules have the greatest average kinetic energy?
b. In which flask will the molecules have the greatest average velocity?
84. Consider separate 1.0-L gaseous samples of H2, Xe, Cl2, and O2 all at STP.
a. Rank the gases in order of increasing average kinetic energy.
b. Rank the gases in order of increasing average velocity.
c. How can separate 1.0-L samples of O2 and H2 each have the same average velocity?
85. Freon-12 is used as a refrigerant in central home air condition-ers. The rate of effusion of Freon-12 to Freon-11 (molar mass 137.4 g/mol) is 1.07:1. The formula of Freon-12 is one of the following: CF4, CF3Cl, CF2Cl2, CFCl3, or CCl4. Which formula is correct for Freon-12?
86. The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effu-sion of pure methane (CH4) gas is 47.8 mL/min. What is the mo-lar mass of the unknown gas?
87. One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of 12C16O, 12C17O, and 12C18O. Name some advantages and disad-vantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.
88. It took 4.5 minutes for 1.0 L helium to effuse through a porous barrier. How long will it take for 1.0 L Cl2 gas to effuse under identical conditions?
89. Calculate the pressure exerted by 0.5000 mol N2 in a 1.0000-L container at 25.0C a. using the ideal gas law.
b. using the van der Waals equation.
c. Compare the results.
90. Calculate the pressure exerted by 0.5000 mol N2 in a 10.000-L container at 25.0C a. using the ideal gas law.
b. using the van der Waals equation.
c. Compare the results.
d. Compare the results with those in Exercise 89.
Atmosphere Chemistry 91. Use the data in Table 5.4 to calculate the partial pressure of He in dry air assuming that the total pressure is 1.0 atm. Assuming a temperature of 25C, calculate the number of He atoms per cu-bic centimeter.
224 Chapter Five Gases would be the pressure of CO2 inside the wine bottle at 25C?
(The density of ethanol is 0.79 g/cm3.) 108. One of the chemical controversies of the nineteenth century con-cerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming Be3 ions) and that it gave an oxide with the formula Be2O3. This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming Be2 ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of 9.0. In 1894, A. Combes (Comptes Rendus 1894, p. 1221) reacted beryllium with the anion C5H7O2 and measured the density of the gaseous product. Combes’s data for two different exper-iments are as follows: If beryllium is a divalent metal, the molecular formula of the product will be Be(C5H7O2)2; if it is trivalent, the formula will be Be(C5H7O2)3. Show how Combes’s data help to confirm that beryllium is a divalent metal.
109. The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot CuO(s): The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH so-lution, the gas contains N2 and is saturated with water vapor. In a given experiment a 0.253-g sample of a compound produced 31.8 mL N2 saturated with water vapor at 25C and 726 torr.
What is the mass percent of nitrogen in the compound? (The vapor pressure of water at 25C is 23.8 torr.) 110. A compound containing only C, H, and N yields the following data.
i. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O.
ii. A 65.2-mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 109), giving 35.6 mL of N2 at 740. torr and 25C.
iii. The effusion rate of the compound as a gas was measured and found to be 24.6 mL/min. The effusion rate of argon gas, under identical conditions, is 26.4 mL/min.
What is the molecular formula of the compound?
111. An organic compound contains C, H, N, and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g of CO2 and 0.0991 g of H2O. A sample of 0.4831 g of the com-pound was analyzed for nitrogen by the Dumas method (see Ex-ercise 109). At STP, 27.6 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to Compound —¡ Hot CuO1s2 N21g2 CO21g2 H2O1g2 I II Mass 0.2022 g 0.2224 g Volume 22.6 cm3 26.0 cm3 Temperature 13C 17C Pressure 765.2 mm Hg 764.6 mm volume of 1.75 cm3. The gas remaining in the first container shows a pressure of 1.710 atm. Calculate the volume of the spher-ical container.
103. A 2.00-L sample of O2(g) was collected over water at a total pressure of 785 torr and 25C. When the O2(g) was dried (wa-ter vapor removed), the gas had a volume of 1.94 L at 25C and 785 torr. Calculate the vapor pressure of water at 25C.
104. A 20.0-L stainless steel container was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at 25C? at 125C?
105. Metallic molybdenum can be produced from the mineral molyb-denite, MoS2. The mineral is first oxidized in air to molybde-num trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The bal-anced equations are Calculate the volumes of air and hydrogen gas at 17C and 1.00 atm that are necessary to produce 1.00 103 kg of pure molyb-denum from MoS2. Assume air contains 21% oxygen by volume and assume 100% yield for each reaction.
106. Nitric acid is produced commercially by the Ostwald process. In the first step ammonia is oxidized to nitric oxide: Assume this reaction is carried out in the apparatus diagramed below.
The stopcock between the two reaction containers is opened, and the reaction proceeds using proper catalysts. Calculate the par-tial pressure of NO after the reaction is complete. Assume 100% yield for the reaction, assume the final container volume is 3.00 L, and assume the temperature is constant.
107. In the “Méthode Champenoise,” grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is Fermentation of 750. mL grape juice (density 1.0 g/cm3) is allowed to take place in a bottle with a total volume of 825 mL until 12% by volume is ethanol (C2H5OH). Assuming that the CO2 is insoluble in H2O (actually, a wrong assumption), what C6H12O61aq2 ¡ 2C2H5OH1aq2 2CO21g2 2.00 L NH3 0.500 atm 1.00 L O2 1.50 atm 4NH31g2 5O21g2 S 4NO1g2 6H2O1g2 MoO31s2 3H21g2 S Mo1s2 3H2O1l2 MoS21s2 7 2O21g2 S MoO31s2 2SO21g2 Challenge Problems 225 117. Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon com-busting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 K.
This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.
118. You have an equimolar mixture of the gases SO2 and O2, along with some He, in a container fitted with a piston. The density of this mixture at STP is 1.924 g/L. Assume ideal behavior and constant temperature and pressure.
a. What is the mole fraction of He in the original mixture?
b. The SO2 and O2 react to completion to form SO3. What is the density of the gas mixture after the reaction is complete?
119. Methane (CH4) gas flows into a combustion chamber at a rate of 200. L/min at 1.50 atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited.
a. To ensure complete combustion of CH4 to CO2(g) and H2O(g), three times as much oxygen as is necessary is reacted. As-suming air is 21 mole percent O2 and 79 mole percent N2, calculate the flow rate of air necessary to deliver the required amount of oxygen.
b. Under the conditions in part a, combustion of methane was not complete as a mixture of CO2(g) and CO(g) was produced. It was determined that 95.0% of the carbon in the exhaust gas was present in CO2. The remainder was pre-sent as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of CO, CO2, O2, N2, and H2O. Assume CH4 is completely reacted and N2 is unreacted.
120. A steel cylinder contains 5.00 mol of graphite (pure carbon) and 5.00 mol of O2. The mixture is ignited and all the graphite re-acts. Combustion produces a mixture of CO gas and CO2 gas.
After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0%.
Calculate the mole fractions of CO, CO2, and O2 in the final gaseous mixture.
121. The total mass that can be lifted by a balloon is given by the dif-ference between the mass of air displaced by the balloon and the mass of the gas inside the balloon. Consider a hot-air balloon that approximates a sphere 5.00 m in diameter and contains air heated to 65C. The surrounding air temperature is 21C. The pressure in the balloon is equal to the atmospheric pressure, which is 745 torr.
a. What total mass can the balloon lift? Assume that the average molar mass of air is 29.0 g/mol. (Hint: Heated air is less dense than cool air.) b. If the balloon is filled with enough helium at 21C and 745 torr to achieve the same volume as in part a, what total mass can the balloon lift?
c. What mass could the hot-air balloon in part a lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is 630. torr?
122. You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is 1.00 atm and the temperature is be 4.02 g/L at 127C and 256 torr. What are the empirical and molecular formulas of the compound?
112. Consider the following diagram: Container A (with porous walls) is filled with air at STP. It is then inserted into a large enclosed container (B), which is then flushed with H2(g). What will happen to the pressure inside con-tainer A? Explain your answer.
113. Without looking at tables of values, which of the following gases would you expect to have the largest value of the van der Waals constant b: H2, N2, CH4, C2H6, or C3H8? From the values in Table 5.3 for the van der Waals constant a for the gases H2, CO2, N2, and CH4, predict which of these gas molecules show the strongest intermolecular attractions.
Challenge Problems 114. An important process for the production of acrylonitrile (C3H3N) is given by the following reaction: A 150.-L reactor is charged to the following partial pressures at 25C: What mass of acrylonitrile can be produced from this mixture (Mpa 106 Pa)?
115. A chemist weighed out 5.14 g of a mixture containing unknown amounts of BaO(s) and CaO(s) and placed the sample in a 1.50-L flask containing CO2(g) at 30.0C and 750. torr. After the reaction to form BaCO3(s) and CaCO3(s) was completed, the pressure of CO2(g) remaining was 230. torr. Calculate the mass percentages of CaO(s) and BaO(s) in the mixture.
116. A mixture of chromium and zinc weighing 0.362 g was re-acted with an excess of hydrochloric acid. After all the met-als in the mixture reacted, 225 mL of dry hydrogen gas was collected at 27C and 750. torr. Determine the mass percent Zn in the metal sample. [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium(III) chloride and hydrogen gas.] PO2 1.500 MPa PNH3 0.800 MPa PC3H6 0.500 MPa 2C3H61g2 2NH31g2 3O21g2 ¡ 2C3H3N1g2 6H2O1g2 B A H2 226 Chapter Five Gases If 2.55 102 mL of NO(g) is isolated at 29C and 1.5 atm, what amount (moles) of UO2 was used in the reaction?
128. Silane, SiH4, is the silicon analogue of methane, CH4. It is prepared industrially according to the following equations: a. If 156 mL of HSiCl3 (d 1.34 g/mL) is isolated when 15.0 L of HCl at 10.0 atm and 35C is used, what is the percent yield of HSiCl3?
b. When 156 mL of HSiCl3 is heated, what volume of SiH4 at 10.0 atm and 35C will be obtained if the percent yield of the reaction is 93.1%?
129. Solid thorium(IV) fluoride has a boiling point of 1680C. What is the density of a sample of gaseous thorium(IV) fluoride at its boiling point under a pressure of 2.5 atm in a 1.7-L container?
Which gas will effuse faster at 1680C, thorium(IV) fluoride or uranium(III) fluoride? How much faster?
130. Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6). A typical mixture might have methane 0.915 and ethane 0.085. What are the partial pres-sures of the two gases in a 15.00-L container of natural gas at 20.C and 1.44 atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
131. Use the following information to identify element A and com-pound B, then answer questions a and b.
An empty glass container has a mass of 658.572 g. It has a mass of 659.452 g after it has been filled with nitrogen gas at a pressure of 790. torr and a temperature of 15C. When the con-tainer is evacuated and refilled with a certain element (A) at a pressure of 745 torr and a temperature of 26C, it has a mass of 660.59 g.
Compound B, a gaseous organic compound that consists of 85.6% carbon and 14.4% hydrogen by mass, is placed in a stain-less steel vessel (10.68 L) with excess oxygen gas. The vessel is placed in a constant-temperature bath at 22C. The pressure in the vessel is 11.98 atm. In the bottom of the vessel is a container that is packed with Ascarite and a desiccant. Ascarite is asbestos impregnated with sodium hydroxide; it quantitatively absorbs carbon dioxide: 2NaOH1s2 CO21g2 ¡ Na2CO31s2 H2O1l2 4HSiCl31l2 ¡ SiH41g2 3SiCl41l2 Si1s2 3HCl1g2 ¡ HSiCl31l2 H21g2 25C. The air has a mole fraction of nitrogen of 0.790, the rest being oxygen.
a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. Be complete.
b. Above what temperature would you heat the balloon so that it would float?
123. You have a helium balloon at 1.00 atm and 25C. You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0% nitrogen, 21.0% oxygen by volume. The “lift” of a balloon is given by the difference be-tween the mass of air displaced by the balloon and the mass of gas inside the balloon.
a. Will the temperature in the hot-air balloon have to be higher or lower than 25C? Explain.
b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and 25C. Assume atmospheric conditions are 1.00 atm and 25C.
124. We state that the ideal gas law tends to hold best at low pres-sures and high temperatures. Show how the van der Waals equa-tion simplifies to the ideal gas law under these conditions.
125. Atmospheric scientists often use mixing ratios to express the con-centrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): On a recent autumn day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached 3.0 102 ppmv. The atmospheric pressure at that time was 628 torr, and the temperature was 0C.
a. What was the partial pressure of CO?
b. What was the concentration of CO in molecules per cubic centimeter?
126. Nitrogen gas (N2) reacts with hydrogen gas (H2) to form am-monia gas (NH3). You have nitrogen and hydrogen gases in a 15.0-L container fitted with a movable piston (the piston allows the container volume to change so as to keep the pressure con-stant inside the container). Initially the partial pressure of each reactant gas is 1.00 atm. Assume the temperature is constant and that the reaction goes to completion.
a. Calculate the partial pressure of ammonia in the container af-ter the reaction has reached completion.
b. Calculate the volume of the container after the reaction has reached completion.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
127. In the presence of nitric acid, UO2 undergoes a redox process.
It is converted to UO2 2 and nitric oxide (NO) gas is produced according to the following unbalanced equation: NO3 1aq2 UO21aq2 ¡ NO1g2 UO2 21aq2 ppmv of X vol. of X at STP total vol. of air at STP 106 Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
Marathon Problem 227 The desiccant is anhydrous magnesium perchlorate, which quantitatively absorbs the water produced by the combustion re-action as well as the water produced by the above reaction. Nei-ther the Ascarite nor the desiccant reacts with compound B or oxygen. The total mass of the container with the Ascarite and desiccant is 765.3 g.
The combustion reaction of compound B is initiated by a spark. The pressure immediately rises, then begins to decrease, and finally reaches a steady value of 6.02 atm. The stainless steel vessel is carefully opened, and the mass of the container inside the vessel is found to be 846.7 g.
A and B react quantitatively in a 1:1 mole ratio to form one mole of the single product, gas C.
a. How many grams of C will be produced if 10.0 L of A and 8.60 L of B (each at STP) are reacted by opening a stopcock connecting the two samples?
b. What will be the total pressure in the system?
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
228 6 Thermochemistry Contents 6.1 The Nature of Energy • Chemical Energy 6.2 Enthalpy and Calorimetry • Enthalpy • Calorimetry 6.3 Hess’s Law • Characteristics of Enthalpy Changes 6.4 Standard Enthalpies of Formation 6.5 Present Sources of Energy • Petroleum and Natural Gas • Coal • Effects of Carbon Dioxide on Climate 6.6 New Energy Sources • Coal Conversion • Hydrogen as a Fuel • Other Energy Alternatives Hot lava flowing into the ocean in Hawaii Volcanoes National Park creates clouds of steam.
Energy is the essence of our very existence as individuals and as a society. The food that we eat furnishes the energy to live, work, and play, just as the coal and oil consumed by manufacturing and transportation systems power our modern industrialized civilization.
In the past, huge quantities of carbon-based fossil fuels have been available for the taking. This abundance of fuels has led to a world society with a voracious appetite for energy, consuming millions of barrels of petroleum every day. We are now dangerously dependent on the dwindling supplies of oil, and this dependence is an important source of tension among nations in today’s world. In an incredibly short time we have moved from a period of ample and cheap supplies of petroleum to one of high prices and uncertain supplies. If our present standard of living is to be maintained, we must find alternatives to petroleum. To do this, we need to know the relationship between chemistry and energy, which we explore in this chapter.
There are additional problems with fossil fuels. The waste products from burning fossil fuels significantly affect our environment. For example, when a carbon-based fuel is burned, the carbon reacts with oxygen to form carbon dioxide, which is released into the atmosphere. Although much of this carbon dioxide is consumed in various natural processes such as photosynthesis and the formation of carbonate materials, the amount of carbon dioxide in the atmosphere is steadily increasing. This increase is significant because atmospheric carbon dioxide absorbs heat radiated from the earth’s surface and radiates it back toward the earth. Since this is an important mechanism for controlling the earth’s temperature, many scientists fear that an increase in the concentration of carbon dioxide will warm the earth, causing significant changes in climate. In addition, impurities in the fossil fuels react with components of the air to produce air pollution. We discussed some aspects of this problem in Chapter 5.
Just as energy is important to our society on a macroscopic scale, it is critically important to each living organism on a microscopic scale. The living cell is a miniature chemical factory powered by energy from chemical reactions. The process of cellular respiration extracts the energy stored in sugars and other nutrients to drive the various tasks of the cell. Although the extraction process is more complex and more subtle, the energy obtained from “fuel” molecules by the cell is the same as would be obtained from burning the fuel to power an internal combustion engine.
Whether it is an engine or a cell that is converting energy from one form to another, the processes are all governed by the same principles, which we will begin to explore in this chapter. Additional aspects of energy transformation will be covered in Chapter 16.
6.1 The Nature of Energy Although the concept of energy is quite familiar, energy itself is rather difficult to define precisely. We will define energy as the capacity to do work or to produce heat. In this chapter we will concentrate specifically on the heat transfer that accompanies chemical processes.
One of the most important characteristics of energy is that it is conserved. The law of conservation of energy states that energy can be converted from one form to another but can be neither created nor destroyed. That is, the energy of the universe is constant.
Energy can be classified as either potential or kinetic energy. Potential energy is energy due to position or composition. For example, water behind a dam has potential energy that can be converted to work when the water flows down through turbines, thereby creating 229 One interesting definition of energy is that which is needed to oppose natural attractions (for example, gravity and electrostatic attractions).
The total energy content of the universe is constant.
230 Chapter Six Thermochemistry electricity. Attractive and repulsive forces also lead to potential energy. The energy released when gasoline is burned results from differences in attractive forces between the nuclei and electrons in the reactants and products. The kinetic energy of an object is energy due to the motion of the object and depends on the mass of the object m and its velocity v: .
Energy can be converted from one form to another. For example, consider the two balls in Fig. 6.1(a). Ball A, because of its higher position initially, has more potential energy than ball B. When A is released, it moves down the hill and strikes B. Eventually, the arrangement shown in Fig. 6.1(b) is achieved. What has happened in going from the initial to the final arrangement? The potential energy of A has decreased, but since energy is conserved, all the energy lost by A must be accounted for. How is this energy distributed?
Initially, the potential energy of A is changed to kinetic energy as the ball rolls down the hill. Part of this kinetic energy is then transferred to B, causing it to be raised to a higher final position. Thus the potential energy of B has been increased. However, since the final position of B is lower than the original position of A, some of the energy is still unaccounted for. Both balls in their final positions are at rest, so the missing energy can-not be due to their motions. What has happened to the remaining energy?
The answer lies in the interaction between the hill’s surface and the ball. As ball A rolls down the hill, some of its kinetic energy is transferred to the surface of the hill as heat. This transfer of energy is called frictional heating. The temperature of the hill in-creases very slightly as the ball rolls down.
Before we proceed further, it is important to recognize that heat and temperature are decidedly different. As we saw in Chapter 5, temperature is a property that reflects the random motions of the particles in a particular substance. Heat, on the other hand, involves the transfer of energy between two objects due to a temperature difference. Heat is not a substance contained by an object, although we often talk of heat as if this were true.
Note that in going from the initial to the final arrangements in Fig. 6.1, ball B gains potential energy because work was done by ball A on B. Work is defined as force acting over a distance. Work is required to raise B from its original position to its final one. Part of the original energy stored as potential energy in A has been transferred through work to B, thereby increasing B’s potential energy. Thus there are two ways to transfer energy: through work and through heat.
In rolling to the bottom of the hill shown in Fig. 6.1, ball A will always lose the same amount of potential energy. However, the way that this energy transfer is divided between work and heat depends on the specific conditions—the pathway. For example, the surface of the hill might be so rough that the energy of A is expended completely through frictional heating; A is moving so slowly when it hits B that it cannot move B to the next level. In this case, no work is done. Regardless of the condition of the hill’s surface, the total energy trans-ferred will be constant. However, the amounts of heat and work will differ. Energy change is independent of the pathway; however, work and heat are both dependent on the pathway.
This brings us to a very important concept: the state function or state property. A state function refers to a property of the system that depends only on its present state. A state func-tion (property) does not depend in any way on the system’s past (or future). In other words, the value of a state function does not depend on how the system arrived at the present state; it depends only on the characteristics of the present state. This leads to a very important char-acteristic of a state function: A change in this function (property) in going from one state to another state is independent of the particular pathway taken between the two states.
A nonscientific analogy that illustrates the difference between a state function and a nonstate function is elevation on the earth’s surface and distance between two points.
In traveling from Chicago (elevation 674 ft) to Denver (elevation 5280 ft), the change in elevation is always 5280 674 4606 ft regardless of the route taken between the two cities. The distance traveled, however, depends on how you make the trip. Thus ele-vation is a function that does not depend on the route (pathway) but distance is pathway dependent. Elevation is a state function and distance is not.
KE 1 2mv2 A B A B (a) Initial (b) Final Held in place FIGURE 6.1 (a) In the initial positions, ball A has a higher potential energy than ball B. (b) After A has rolled down the hill, the poten-tial energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.
Heat involves a transfer of energy.
This infrared photo of a house shows where energy leaks occur. The more red the color, the more energy (heat) is leaving the house.
Visualization: Coffee Creamer Flammability 6.1 The Nature of Energy 231 Of the functions considered in our present example, energy is a state function, but work and heat are not state functions.
Chemical Energy The ideas we have just illustrated using mechanical examples also apply to chemical sys-tems. The combustion of methane, for example, is used to heat many homes in the United States: To discuss this reaction, we divide the universe into two parts: the system and the sur-roundings. The system is the part of the universe on which we wish to focus attention; the surroundings include everything else in the universe. In this case we define the system as the reactants and products of the reaction. The surroundings consist of the reaction container (a furnace, for example), the room, and anything else other than the reactants and products.
When a reaction results in the evolution of heat, it is said to be exothermic (exo- is a prefix meaning “out of”); that is, energy flows out of the system. For example, in the combustion of methane, energy flows out of the system as heat. Reactions that absorb en-ergy from the surroundings are said to be endothermic. When the heat flow is into a sys-tem, the process is endothermic. For example, the formation of nitric oxide from nitrogen and oxygen is endothermic: Where does the energy, released as heat, come from in an exothermic reaction? The answer lies in the difference in potential energies between the products and the reactants.
Which has lower potential energy, the reactants or the products? We know that total en-ergy is conserved and that energy flows from the system into the surroundings in an exothermic reaction. This means that the energy gained by the surroundings must be equal to the energy lost by the system. In the combustion of methane, the energy content of the system decreases, which means that 1 mole of CO2 and 2 moles of H2O molecules (the products) possess less potential energy than do 1 mole of CH4 and 2 moles of O2 mole-cules (the reactants). The heat flow into the surroundings results from a lowering of the potential energy of the reaction system. This always holds true. In any exothermic reac-tion, some of the potential energy stored in the chemical bonds is being converted to ther-mal energy (random kinetic energy) via heat.
The energy diagram for the combustion of methane is shown in Fig. 6.2, where (PE) represents the change in potential energy stored in the bonds of the products as compared with the bonds of the reactants. In other words, this quantity represents the difference between N21g2 O21g2 energy 1heat2 ¡ 2NO1g2 CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 energy 1heat2 Energy is a state function; work and heat are not.
2 mol O2 1 mol CH4 (Reactants) 2 mol H2O 1 mol CO2 (Products) Potential energy System Surroundings ∆(PE) Energy released to the surroundings as heat FIGURE 6.2 The combustion of methane releases the quantity of energy (PE) to the surround-ings via heat flow. This is an exothermic process.
Visualization: Sugar and Potassium Chlorate 232 Chapter Six Thermochemistry the energy required to break the bonds in the reactants and the energy released when the bonds in the products are formed. In an exothermic process, the bonds in the products are stronger (on average) than those of the reactants. That is, more energy is released by form-ing the new bonds in the products than is consumed to break the bonds in the reactants. The net result is that the quantity of energy (PE) is transferred to the surroundings through heat.
For an endothermic reaction, the situation is reversed, as shown in Fig. 6.3. Energy that flows into the system as heat is used to increase the potential energy of the system. In this case the products have higher potential energy (weaker bonds on average) than the reactants.
The study of energy and its interconversions is called thermodynamics. The law of conservation of energy is often called the first law of thermodynamics and is stated as follows: The energy of the universe is constant.
The internal energy E of a system can be defined most precisely as the sum of the kinetic and potential energies of all the “particles” in the system. The internal energy of a system can be changed by a flow of work, heat, or both. That is, where E represents the change in the system’s internal energy, q represents heat, and w represents work.
Thermodynamic quantities always consist of two parts: a number, giving the magni-tude of the change, and a sign, indicating the direction of the flow. The sign reflects the system’s point of view. For example, if a quantity of energy flows into the system via heat (an endothermic process), q is equal to x, where the positive sign indicates that the sys-tem’s energy is increasing. On the other hand, when energy flows out of the system via heat (an exothermic process), q is equal to x, where the negative sign indicates that the system’s energy is decreasing.
¢E q w 1 mol N2 1 mol O2 (Reactants) 2 mol NO (Products) Potential energy System Surroundings ∆(PE) Heat absorbed from the surroundings FIGURE 6.3 The energy diagram for the reaction of nitrogen and oxygen to form nitric oxide.
This is an endothermic process: Heat [equal in magnitude to (PE)] flows into the system from the surroundings.
Surroundings ∆E < 0 System Surroundings ∆E > 0 System Energy Energy 6.1 The Nature of Energy 233 In this text the same conventions also apply to the flow of work. If the system does work on the surroundings (energy flows out of the system), w is negative. If the sur-roundings do work on the system (energy flows into the system), w is positive. We define work from the system’s point of view to be consistent for all thermodynamic quantities.
That is, in this convention the signs of both q and w reflect what happens to the system; thus we use .
In this text we always take the system’s point of view. This convention is not fol-lowed in every area of science. For example, engineers are in the business of designing machines to do work, that is, to make the system (the machine) transfer energy to its sur-roundings through work. Consequently, engineers define work from the surroundings’ point of view. In their convention, work that flows out of the system is treated as positive because the energy of the surroundings has increased. The first law of thermodynamics is then written E q w , where w signifies work from the surroundings’ point of view.
Internal Energy Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.
Solution We use the equation where q 15.6 kJ, since the process is endothermic, and w 1.4 kJ, since work is done on the system. Thus The system has gained 17.0 kJ of energy.
See Exercises 6.21 and 6.22.
A common type of work associated with chemical processes is work done by a gas (through expansion) or work done to a gas (through compression). For example, in an automobile engine, the heat from the combustion of the gasoline expands the gases in the cylinder to push back the piston, and this motion is then translated into the motion of the car.
Suppose we have a gas confined to a cylindrical container with a movable piston as shown in Fig. 6.4, where F is the force acting on a piston of area A. Since pressure is defined as force per unit area, the pressure of the gas is Work is defined as force applied over a distance, so if the piston moves a distance h, as shown in Fig. 6.4, then the work done is Since P FA or F P A, then Since the volume of a cylinder equals the area of the piston times the height of the cylinder (Fig. 6.4), the change in volume V resulting from the piston moving a distance h is ¢V final volume initial volume A ¢h Work F ¢h P A ¢h Work force distance F ¢h P F A ¢E 15.6 kJ 1.4 kJ 17.0 kJ ¢E q w ¢E q w The convention in this text is to take the system’s point of view; q x denotes an exothermic process, and q x denotes an endothermic one.
The joule (J) is the fundamental SI unit for energy: One kilojoule (kJ) 103 J.
J kg m2 s2 Sample Exercise 6.1 ∆h P = F A Initial state P = F A Final state ∆h Area = A ∆V (a) (b) ∆V FIGURE 6.4 (a) The piston, moving a distance h against a pressure P, does work on the sur-roundings. (b) Since the volume of a cylin-der is the area of the base times its height, the change in volume of the gas is given by h A V.
Visualization: Work versus Energy Flow 234 Chapter Six Thermochemistry Substituting V A h into the expression for work gives This gives us the magnitude (size) of the work required to expand a gas V against a pres-sure P.
What about the sign of the work? The gas (the system) is expanding, moving the pis-ton against the pressure. Thus the system is doing work on the surroundings, so from the system’s point of view the sign of the work should be negative.
For an expanding gas, V is a positive quantity because the volume is increasing.
Thus V and w must have opposite signs, which leads to the equation Note that for a gas expanding against an external pressure P, w is a negative quantity as required, since work flows out of the system. When a gas is compressed, V is a nega-tive quantity (the volume decreases), which makes w a positive quantity (work flows into the system).
PV Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a con-stant external pressure of 15 atm.
Solution For a gas at constant pressure, In this case P 15 atm and V 64 46 18 L. Hence Note that since the gas expands, it does work on its surroundings.
Reality Check: Energy flows out of the gas, so w is a negative quantity.
See Exercises 6.25 through 6.27.
In dealing with “PV work,” keep in mind that the P in P V always refers to the ex-ternal pressure—the pressure that causes a compression or that resists an expansion.
Internal Energy, Heat, and Work A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 106 L to 4.50 106 L by the addition of 1.3 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. (To convert between L atm and J, use 1 L atm 101.3 J.) Solution To calculate E, we use the equation Since the problem states that 1.3 108 J of energy is added as heat, q 1.3 108 J ¢E q w w 15 atm 18 L 270 L atm w P¢V w P¢V Work P A ¢h P¢V Sample Exercise 6.2 Sample Exercise 6.3 w and P V have opposite signs because when the gas expands ( V is positive), work flows into the surroundings (w is negative).
For an ideal gas, work can occur only when its volume changes. Thus, if a gas is heated at constant volume, the pressure increases but no work occurs.
6.2 Enthalpy and Calorimetry 235 The work done can be calculated from the expression In this case P 1.0 atm and Thus Note that the negative sign for w makes sense, since the gas is expanding and thus doing work on the surroundings.
To calculate E, we must sum q and w. However, since q is given in units of J and w is given in units of L atm, we must change the work to units of joules: Then Reality Check: Since more energy is added through heating than the gas expends doing work, there is a net increase in the internal energy of the gas in the balloon. Hence E is positive.
See Exercises 6.28 through 6.30.
6.2 Enthalpy and Calorimetry Enthalpy So far we have discussed the internal energy of a system. A less familiar property of a system is its enthalpy, H, which is defined as where E is the internal energy of the system, P is the pressure of the system, and V is the volume of the system.
Since internal energy, pressure, and volume are all state functions, enthalpy is also a state function. But what exactly is enthalpy? To help answer this question, consider a process carried out at constant pressure and where the only work allowed is pressure– volume work (w P V ). Under these conditions, the expression becomes or where qP is the heat at constant pressure.
We will now relate qP to a change in enthalpy. The definition of enthalpy is H E PV. Therefore, we can say Change in H 1change in E2 1change in PV2 qP ¢E P¢V ¢E qP P¢V ¢E qP w H E PV ¢E q w 11.3 108 J2 15.1 107 J2 8 107 J w 5.0 105 L atm 101.3 J L atm 5.1 107 J w 1.0 atm 5.0 105 L 5.0 105 L atm 4.50 106 L 4.00 106 L 0.50 106 L 5.0 105 L ¢V Vfinal Vinitial w P¢V A propane burner is used to heat the air in a hot-air balloon.
Enthalpy is a state function. A change in enthalpy does not depend on the pathway between two states.
Recall from the previous section that w and P V have opposite signs: w P V 236 Chapter Six Thermochemistry or Since P is constant, the change in PV is due only to a change in volume. Thus and This expression is identical to the one we obtained for qP: Thus, for a process carried out at constant pressure and where the only work allowed is that from a volume change, we have At constant pressure (where only PV work is allowed), the change in enthalpy H of the system is equal to the energy flow as heat. This means that for a reaction studied at con-stant pressure, the flow of heat is a measure of the change in enthalpy for the system. For this reason, the terms heat of reaction and change in enthalpy are used interchangeably for reactions studied at constant pressure.
For a chemical reaction, the enthalpy change is given by the equation In a case in which the products of a reaction have a greater enthalpy than the reactants, H will be positive. Thus heat will be absorbed by the system, and the reaction is endothermic. On the other hand, if the enthalpy of the products is less than that of the reactants, H will be negative. In this case the overall decrease in enthalpy is achieved by the generation of heat, and the reaction is exothermic.
Enthalpy When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is re-leased as heat. Calculate H for a process in which a 5.8-g sample of methane is burned at constant pressure.
Solution At constant pressure, 890 kJ of energy per mole of CH4 is produced as heat: Note that the minus sign indicates an exothermic process. In this case, a 5.8-g sample of CH4 (molar mass 16.0 g/mol) is burned. Since this amount is smaller than 1 mole, less than 890 kJ will be released as heat. The actual value can be calculated as follows: and Thus, when a 5.8-g sample of CH4 is burned at constant pressure, See Exercises 6.35 through 6.38.
¢H heat flow 320 kJ 0.36 mol CH4 890 kJ mol CH4 320 kJ 5.8 g CH4 1 mol CH4 16.0 g CH4 0.36 mol CH4 qP ¢H 890 kJ/mol CH4 ¢H Hproducts Hreactants ¢H qP qP ¢E P¢V ¢H ¢E P¢V ¢1PV2 P¢V ¢H ¢E ¢1PV 2 H q only at constant pressure.
The change in enthalpy of a system has no easily interpreted meaning except at constant pressure, where H heat.
At constant pressure, exothermic means H is negative; endothermic means H is positive.
Sample Exercise 6.4 6.2 Enthalpy and Calorimetry 237 Calorimetry The device used experimentally to determine the heat associated with a chemical reaction is called a calorimeter. Calorimetry, the science of measuring heat, is based on observ-ing the temperature change when a body absorbs or discharges energy as heat. Substances respond differently to being heated. One substance might require a great deal of heat en-ergy to raise its temperature by one degree, whereas another will exhibit the same tem-perature change after absorbing relatively little heat. The heat capacity C of a substance, which is a measure of this property, is defined as When an element or a compound is heated, the energy required will depend on the amount of the substance present (for example, it takes twice as much energy to raise the temperature of two grams of water by one degree than it takes to raise the temperature of one gram of water by one degree). Thus, in defining the heat capacity of a substance, the amount of substance must be specified. If the heat capacity is given per gram of substance, it is called the specific heat capacity, and its units are J/C g or J/K g. If the heat capacity is given per mole of the substance, it is called the molar heat capacity, and it has the units J/C mol or J/K mol. The specific heat capacities of some common substances are given in Table 6.1. Note from this table that the heat capacities of metals are very different from that of water. It takes much less energy to change the temperature of a gram of a metal by 1C than for a gram of water.
Although the calorimeters used for highly accurate work are precision instruments, a very simple calorimeter can be used to examine the fundamentals of calorimetry. All we need are two nested Styrofoam cups with a cover through which a stirrer and thermome-ter can be inserted, as shown in Fig. 6.5. This device is called a “coffee-cup calorimeter.” The outer cup is used to provide extra insulation. The inner cup holds the solution in which the reaction occurs.
The measurement of heat using a simple calorimeter such as that shown in Fig. 6.5 is an example of constant-pressure calorimetry, since the pressure (atmo-spheric pressure) remains constant during the process. Constant-pressure calorimetry is used in determining the changes in enthalpy (heats of reactions) for reactions occurring in solution. Recall that under these conditions, the change in enthalpy equals the heat.
For example, suppose we mix 50.0 mL of 1.0 M HCl at 25.0C with 50.0 mL of 1.0 M NaOH also at 25C in a calorimeter. After the reactants are mixed by stirring, the temperature is observed to increase to 31.9C. As we saw in Section 4.8, the net ionic equation for this reaction is When these reactants (each originally at the same temperature) are mixed, the tem-perature of the mixed solution is observed to increase. Therefore, the chemical reaction must be releasing energy as heat. This released energy increases the random motions of the solution components, which in turn increases the temperature. The quantity of energy released can be determined from the temperature increase, the mass of solution, and the specific heat capacity of the solution. For an approximate result, we will assume that the calorimeter does not absorb or leak any heat and that the solution can be treated as if it were pure water with a density of 1.0 g/mL.
We also need to know the heat required to raise the temperature of a given amount of water by 1C. Table 6.1 lists the specific heat capacity of water as 4.18 J/C g.
This means that 4.18 J of energy is required to raise the temperature of 1 gram of water by 1C.
H1aq2 OH1aq2 ¡ H2O1l2 C heat absorbed increase in temperature Specific heat capacity: the energy required to raise the temperature of one gram of a substance by one degree Celsius.
Molar heat capacity: the energy required to raise the temperature of one mole of a substance by one degree Celsius.
TABLE 6.1 The Specific Heat Capacities of Some Common Substances Specific Heat Capacity Substance (J/C g) H2O(l) 4.18 H2O(s) 2.03 Al(s) 0.89 Fe(s) 0.45 Hg(l) 0.14 C(s) 0.71 Thermometer Styrofoam cover Styrofoam cups Stirrer FIGURE 6.5 A coffee-cup calorimeter made of two Styrofoam cups.
238 Chapter Six Thermochemistry From these assumptions and definitions, we can calculate the heat (change in enthalpy) for the neutralization reaction: Energy released by the reaction In this case the increase in temperature ( T ) 31.9C 25.0C 6.9C, and the mass of solution (m) 100.0 mL 1.0 g/mL 1.0 102 g. Thus How much energy would have been released if twice these amounts of solutions had been mixed? The answer is that twice as much energy would have been produced. The heat of a reaction is an extensive property; it depends directly on the amount of substance, in this case on the amounts of reactants. In contrast, an intensive property is not related to the amount of a substance. For example, temperature is an intensive property.
Enthalpies of reaction are often expressed in terms of moles of reacting substances.
The number of moles of H ions consumed in the preceding experiment is Thus 2.9 103 J heat was released when 5.0 102 mol H ions reacted, or 2.9 103 J 5.0 102 mol H 5.8 104 J/mol 50.0 mL 1 L 1000 mL 1.0 mol L H 5.0 102 mol H 2.9 103 J a4.18 J °C gb11.0 102 g216.9°C2 Energy released s m ¢T s m ¢T specific heat capacity mass of solution increase in temperature energy absorbed by the solution If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution.
CHEMICAL IMPACT Nature Has Hot Plants T he voodoo lily is a beautiful, seductive—and foul-smelling—plant. The exotic-looking lily features an elab-orate reproductive mechanism—a purple spike that can reach nearly 3 feet in length and is cloaked by a hoodlike leaf. But approach to the plant reveals bad news—it smells terrible!
Despite its antisocial odor, this putrid plant has fas-cinated biologists for many years because of its ability to generate heat. At the peak of its metabolic activity, the plant’s blossom can be as much as 15C above its am-bient temperature. To generate this much heat, the meta-bolic rate of the plant must be close to that of a flying hummingbird!
What’s the purpose of this intense heat production? For a plant faced with limited food supplies in the very com-petitive tropical climate where it grows, heat production seems like a great waste of energy. The answer to this mystery is that the voodoo lily is pollinated mainly by carrion-loving insects. Thus the lily prepares a malodorous mixture of chemicals characteristic of rotting meat, which it then “cooks” off into the surrounding air to attract flesh-feeding beetles and flies. Then, once the insects enter the pollination chamber, the high temperatures there (as high as 110F) cause the insects to remain very active to better carry out their pollination duties.
The voodoo lily is only one of many such thermogenic (heat-producing) plants. Another interesting example is the eastern skunk cabbage, which produces enough heat to bloom inside of a snow bank by creating its own ice caves.
These plants are of special interest to biologists because they provide opportunities to study metabolic reactions that are quite subtle in “normal” plants. For example, recent studies have shown that salicylic acid, the active form of aspirin, is probably very important in producing the metabolic bursts in thermogenic plants.
Besides studying the dramatic heat effects in thermo-genic plants, biologists are also interested in calorimetric 6.2 Enthalpy and Calorimetry 239 of heat released per 1.0 mol H ions neutralized. Thus the magnitude of the enthalpy change per mole for the reaction is 58 kJ/mol. Since heat is evolved, H 58 kJ/mol.
Constant-Pressure Calorimetry When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25C in a calorimeter, the white solid BaSO4 forms and the tempera-ture of the mixture increases to 28.1C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/C g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.
Solution The ions present before any reaction occurs are Ba2, NO3 , Na , and SO4 2. The Na and NO3 ions are spectator ions, since NaNO3 is very soluble in water and will not precipitate under these conditions. The net ionic equation for the reaction is therefore Since the temperature increases, formation of the solid BaSO4 must be exothermic; H will be negative.
Heat evolved by reaction specific heat capacity mass of solution increase in temperature heat absorbed by solution Ba21aq2 SO4 21aq2 ¡ BaSO41s2 H1aq2 OH1aq2 ¡ H2O1l2 Notice that in this example we mentally keep track of the direction of the energy flow and assign the correct sign at the end of the calculation.
Sample Exercise 6.5 studies of regular plants. For example, very precise calorimeters have been designed that can be used to study the heat produced, and thus the metabolic activities, of clumps of cells no larger than a bread crumb. Several sci-entists have suggested that a single calorimetric measure-ment taking just a few minutes on a tiny plant might be use-ful in predicting the growth rate of the mature plant throughout its lifetime. If true, this would provide a very efficient method for selecting the plants most likely to thrive as adults.
Because the study of the heat production by plants is an excellent way to learn about plant metabolism, this continues to be a “hot” area of research.
The voodoo lily attracts pollinating insects with its foul odor.
240 Chapter Six Thermochemistry Since 1.00 L of each solution is used, the total solution volume is 2.00 L, and Thus Since 1.0 L of 1.0 M Ba(NO3)2 contains 1 mol Ba2 ions and 1.0 L of 1.0 M Na2SO4 contains 1.0 mol SO4 2 ions, 1.0 mol solid BaSO4 is formed in this experiment. Thus the enthalpy change per mole of BaSO4 formed is See Exercises 6.51 through 6.54.
Calorimetry experiments also can be performed at constant volume. For example, when a photographic flashbulb flashes, the bulb becomes very hot, because the reaction of the zirconium or magnesium wire with the oxygen inside the bulb is exothermic. The reaction occurs inside the flashbulb, which is rigid and does not change volume. Under these conditions, no work is done (because the volume must change for pressure–volume work to be performed). To study the energy changes in reactions under conditions of con-stant volume, a “bomb calorimeter” (Fig. 6.6) is used. Weighed reactants are placed in-side a rigid steel container (the “bomb”) and ignited. The energy change is determined by measuring the increase in the temperature of the water and other calorimeter parts. For a constant-volume process, the change in volume V is equal to zero, so work (which is P V ) is also equal to zero. Therefore, (constant volume) ¢E q w q qV ¢H 2.6 104 J/mol 26 kJ/mol q qP ¢H 2.6 104 J Heat evolved 14.18 J/°C g212.0 103 g213.1°C2 2.6 104 J Temperature increase 28.1°C 25.0°C 3.1°C Mass of solution 2.00 L 1000 mL 1 L 1.0 g mL 2.0 103 g Water Insulating container Stirrer Thermometer Ignition wires Reactants in sample cup Steel bomb FIGURE 6.6 A bomb calorimeter. The reaction is carried out inside a rigid steel “bomb” (photo of actual disassembled “bomb’’ shown on right), and the heat evolved is absorbed by the surrounding water and other calorimeter parts. The quantity of energy pro-duced by the reaction can be calculated from the temperature increase.
6.2 Enthalpy and Calorimetry 241 Suppose we wish to measure the energy of combustion of octane (C8H18), a compo-nent of gasoline. A 0.5269-g sample of octane is placed in a bomb calorimeter known to have a heat capacity of 11.3 kJ/C. This means that 11.3 kJ of energy is required to raise the temperature of the water and other parts of the calorimeter by 1C. The octane is ig-nited in the presence of excess oxygen, and the temperature increase of the calorimeter is 2.25C. The amount of energy released is calculated as follows: Energy released by the reaction temperature increase energy required to change the temperature by 1C T heat capacity of calorimeter This means that 25.4 kJ of energy was released by the combustion of 0.5269 g octane.
2.25°C 11.3 kJ/°C 25.4 kJ CHEMICAL IMPACT Firewalking: Magic or Science?
F or millennia people have been amazed at the ability of Eastern mystics to walk across beds of glowing coals without any apparent discomfort. Even in the United States, thousands of people have per-formed feats of firewalking as part of mo-tivational seminars. How is this possible?
Do firewalkers have supernatural powers?
Actually, there are good scientific ex-planations, based on the concepts covered in this chapter, of why firewalking is pos-sible. The first important factor concerns the heat capacity of feet. Because human tissue is mainly composed of water, it has a relatively large specific heat capacity.
This means that a large amount of energy must be transferred from the coals to sig-nificantly change the temperature of the feet. During the brief contact between feet and coals, there is relatively little time for energy flow so the feet do not reach a high enough temperature to cause damage.
Second, although the surface of the coals has a very high temperature, the red hot layer is very thin. Therefore, the quantity of energy avail-able to heat the feet is smaller than might be expected. This factor points to the difference between temperature and heat.
Temperature reflects the intensity of the random kinetic en-ergy in a given sample of matter. The amount of energy avail-able for heat flow, on the other hand, depends on the quan-tity of matter at a given temperature—10 grams of matter at a given temperature contains 10 times as much thermal energy as 1 gram of the same matter. This is why the tiny spark from a sparkler does not hurt when it hits your hand.
The spark has a very high temperature but has so little mass that no significant energy transfer occurs to your hand. This same argument applies to the very thin hot layer on the coals.
Thus, although firewalking is an impressive feat, there are several sound scientific reasons why it is possible (with the proper training and a properly prepared bed of coals).
A group of firewalkers in Japan.
242 Chapter Six The number of moles of octane is Since 25.4 kJ of energy was released for 4.614 103 mol octane, the energy released per mole is Since the reaction is exothermic, E is negative: Note that since no work is done in this case, E is equal to the heat.
Thus q 5.50 103 kJ/mol.
Constant-Volume Calorimetry It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principally methane). To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/C. When a 1.50-g sample of methane gas was burned with ex-cess oxygen in the calorimeter, the temperature increased by 7.3C. When a 1.15-g sam-ple of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3C.
Calculate the energy of combustion (per gram) for hydrogen and methane.
Solution We calculate the energy of combustion for methane using the heat capacity of the calorime-ter (11.3 kJ/C) and the observed temperature increase of 7.3C: Similarly, for hydrogen The energy released in the combustion of 1 g hydrogen is approximately 2.5 times that for 1 g methane, indicating that hydrogen gas is a potentially useful fuel.
See Exercises 6.55 and 6.56.
6.3 Hess’s Law Since enthalpy is a state function, the change in enthalpy in going from some initial state to some final state is independent of the pathway. This means that in going from a par-ticular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. This principle is known as Hess’s law and can be illustrated by examining the oxidation of nitrogen to produce nitrogen dioxide. The overall reaction can be written in one step, where the enthalpy change is represented by H1.
N21g2 2O21g2 ¡ 2NO21g2 ¢H1 68 kJ Energy released in the combustion of 1 g H2 162 kJ 1.15 g 141 kJ/g 162 kJ Energy released in the combustion of 1.15 g H2 111.3 kJ/°C2114.3°C2 Energy released in the combustion of 1 g CH4 83 kJ 1.5 g 55 kJ/g 83 kJ Energy released in the combustion of 1.5 g CH4 111.3 kJ/°C217.3°C2 ¢E q w q since w 0 ¢Ecombustion 5.50 103 kJ/mol 25.4 kJ 4.614 103 mol 5.50 103 kJ/mol 0.5269 g octane 1 mol octane 114.2 g octane 4.614 103 mol octane Sample Exercise 6.6 Hydrogen’s potential as a fuel is discussed in Section 6.6.
The direction of energy flow is indicated by words in this example. Using signs to designate the direction of energy flow: Ecombustion 55 kJ/g for methane and Ecombustion 141 kJ/g for hydrogen. H is not dependent on the reaction pathway.
6.3 Hess’s Law 243 This reaction also can be carried out in two distinct steps, with enthalpy changes desig-nated by H2 and H3: Note that the sum of the two steps gives the net, or overall, reaction and that The principle of Hess’s law is shown schematically in Fig. 6.7.
Characteristics of Enthalpy Changes To use Hess’s law to compute enthalpy changes for reactions, it is important to under-stand two characteristics of H for a reaction: 1. If a reaction is reversed, the sign of H is also reversed.
2. The magnitude of H is directly proportional to the quantities of reactants and prod-ucts in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of H is multiplied by the same integer.
Both these rules follow in a straightforward way from the properties of enthalpy changes.
The first rule can be explained by recalling that the sign of H indicates the direction of the heat flow at constant pressure. If the direction of the reaction is reversed, the direc-tion of the heat flow also will be reversed. To see this, consider the preparation of xenon tetrafluoride, which was the first binary compound made from a noble gas: This reaction is exothermic, and 251 kJ of energy flows into the surroundings as heat. On the other hand, if the colorless XeF4 crystals are decomposed into the elements, accord-ing to the equation the opposite energy flow occurs because 251 kJ of energy must be added to the system to produce this endothermic reaction. Thus, for this reaction, H 251 kJ.
The second rule comes from the fact that H is an extensive property, depending on the amount of substances reacting. For example, since 251 kJ of energy is evolved for the reaction Xe1g2 2F21g2 ¡ XeF41s2 XeF41s2 ¡ Xe1g2 2F21g2 Xe1g2 2F21g2 ¡ XeF41s2 ¢H 251 kJ ¢H1 ¢H2 ¢H3 68 kJ Net reaction: N21g2 2O21g2 ¡ 2NO21g2 ¢H2 ¢H3 68 kJ 2NO1g2 O21g2 ¡ 2NO21g2 ¢H3 112 kJ N21g2 O21g2 ¡ 2NO1g2 ¢H2 180 kJ O2(g), 2NO(g) ∆H2 = 180 kJ N2(g), 2O2(g) 2NO2(g) 68 kJ O2(g), 2NO(g) ∆H3 = –112 kJ 2NO2(g) N2(g), 2O2(g) ∆H1 = 68 kJ = ∆H2 + ∆H3 = 180 kJ – 112 kJ Two-step reaction One-step reaction H (kJ) FIGURE 6.7 The principle of Hess’s law. The same change in enthalpy occurs when nitrogen and oxygen react to form nitrogen dioxide, regardless of whether the reaction occurs in one (red) or two (blue) steps.
Crystals of xenon tetrafluoride, the first reported binary compound containing a noble gas element.
Reversing the direction of a reaction changes the sign of H.
Visualization: Hess’s Law 244 Chapter Six then for a preparation involving twice the quantities of reactants and products, or twice as much heat would be evolved: Hess’s Law I Two forms of carbon are graphite, the soft, black, slippery material used in “lead” pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (394 kJ/mol) and diamond ( 396 kJ/mol), calculate H for the conversion of graphite to diamond: Solution The combustion reactions are Note that if we reverse the second reaction (which means we must change the sign of H) and sum the two reactions, we obtain the desired reaction: Thus 2 kJ of energy is required to change 1 mol graphite to diamond. This process is endothermic.
See Exercises 6.57 and 6.58. Cgraphite1s2 ¡ Cdiamond1s2 ¢H 2 kJ CO21g2 ¡ Cdiamond1s2 O21g2 ¢H 1396 kJ2 Cgraphite1s2 O21g2 ¡ CO21g2 ¢H 394 kJ Cdiamond1s2 O21g2 ¡ CO21g2 ¢H 396 kJ Cgraphite1s2 O21g2 ¡ CO21g2 ¢H 394 kJ Cgraphite1s2 ¡ Cdiamond1s2 ¢H 21251 kJ2 502 kJ 2Xe1g2 4F21g2 ¡ 2XeF41s2 Sample Exercise 6.7 (left) graphite; (right) diamond.
6.3 Hess’s Law 245 Hess’s Law II Diborane (B2H6) is a highly reactive boron hydride that was once considered as a possi-ble rocket fuel for the U.S. space program. Calculate H for the synthesis of diborane from its elements, according to the equation using the following data: Reaction H (a) 1273 kJ (b) 2035 kJ (c) 286 kJ (d) 44 kJ Solution To obtain H for the required reaction, we must somehow combine equations (a), (b), (c), and (d) to produce that reaction and add the corresponding H values. This can best be done by focusing on the reactants and products of the required reaction. The reactants are B(s) and H2(g), and the product is B2H6(g). How can we obtain the correct equation? Re-action (a) has B(s) as a reactant, as needed in the required equation. Thus reaction (a) will be used as it is. Reaction (b) has B2H6(g) as a reactant, but this substance is needed as a product. Thus reaction (b) must be reversed, and the sign of H must be changed ac-cordingly. Up to this point we have H2O1l2 ¡ H2O1g2 H21g2 1 2O21g2 ¡ H2O1l2 B2H61g2 3O21g2 ¡ B2O31s2 3H2O1g2 2B1s2 3 2O21g2 ¡ B2O31s2 2B1s2 3H21g2 ¡ B2H61g2 Sample Exercise 6.8 Sum: B2O31s2 2B1s2 3 2O21g2 3H2O1g2 ¡ B2O31s2 B2H61g2 3O21g2 ¢H 762 kJ 1b2 B2O31s2 3H2O1g2 ¡ B2H61g2 3O21g2 ¢H 12035 kJ2 1a2 2B1s2 3 2O21g2 ¡ B2O31s2 ¢H 1273 kJ Deleting the species that occur on both sides gives We are closer to the required reaction, but we still need to remove H2O(g) and O2(g) and introduce H2(g) as a reactant. We can do this using reactions (c) and (d). If we multiply reaction (c) and its H value by 3 and add the result to the preceding equation, we have 2B1s2 3H2O1g2 ¡ B2H61g2 3 2O21g2 ¢H 762 kJ Sum: 2B1s2 3H21g2 3 2O21g2 3H2O1g2 ¡ B2H61g2 3 2O21g2 3H2O1l2 ¢H 96 kJ 3 1c2 33H21g2 1 2O21g2 ¡ H2O1l2 4 ¢H 31286 kJ2 2B1s2 3H2O1g2 ¡ B2H61g2 3 2O21g2 ¢H 762 kJ We can cancel the on both sides, but we cannot cancel the H2O because it is gaseous on one side and liquid on the other. This can be solved by adding reaction (d), multiplied by 3: 3 2O21g2 2B1s2 3H21g2 3H2O1g2 3H2O1l2 ¡ B2H61g2 3H2O1l2 3H2O1g2 ¢H 36 kJ 3 1d2 33H2O1l2 ¡ H2O1g2 4 ¢H 3144 kJ2 2B1s2 3H21g2 3H2O1g2 ¡ B2H61g2 3H2O1l2 ¢H 96 kJ This gives the reaction required by the problem: Thus H for the synthesis of 1 mol diborane from the elements is 36 kJ.
See Exercises 6.59 through 6.64.
2B1s2 3H21g2 ¡ B2H61g2 ¢H 36 kJ 246 Chapter Six Thermochemistry Hints for Using Hess’s Law Calculations involving Hess’s law typically require that several reactions be manipulated and combined to finally give the reaction of interest. In doing this procedure you should • Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal • Reverse any reactions as needed to give the required reactants and products • Multiply reactions to give the correct numbers of reactants and products This process involves some trial and error, but it can be very systematic if you always allow the final reaction to guide you.
6.4 Standard Enthalpies of Formation For a reaction studied under conditions of constant pressure, we can obtain the enthalpy change using a calorimeter. However, this process can be very difficult. In fact, in some cases it is impossible, since certain reactions do not lend themselves to such study. An ex-ample is the conversion of solid carbon from its graphite form to its diamond form: The value of H for this process cannot be obtained by direct measurement in a calorime-ter because the process is much too slow under normal conditions. However, as we saw in Sample Exercise 6.7, H for this process can be calculated from heats of combustion.
This is only one example of how useful it is to be able to calculate H values for chem-ical reactions. We will next show how to do this using standard enthalpies of formation.
The standard enthalpy of formation ( Hf ) of a compound is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.
A degree symbol on a thermodynamic function, for example, H , indicates that the corresponding process has been carried out under standard conditions. The standard state for a substance is a precisely defined reference state. Because thermodynamic functions often depend on the concentrations (or pressures) of the substances involved, we must use a common reference state to properly compare the thermodynamic properties of two sub-stances. This is especially important because, for most thermodynamic properties, we can measure only changes in the property. For example, we have no method for determining absolute values of enthalpy. We can measure enthalpy changes ( H values) only by per-forming heat-flow experiments.
Conventional Definitions of Standard States For a Compound The standard state of a gaseous substance is a pressure of exactly 1 atmosphere.
For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid.
For a substance present in a solution, the standard state is a concentration of exactly 1 M.
For an Element The standard state of an element is the form in which the element exists under con-ditions of 1 atmosphere and 25C. (The standard state for oxygen is O2(g) at a pres-sure of 1 atmosphere; the standard state for sodium is Na(s); the standard state for mercury is Hg(l); and so on.) ° Cgraphite1s2 ¡ Cdiamond1s2 Recently, the International Union of Pure and Applied Chemists (IUPAC) has adopted 1 bar (100,000 Pa) as the stan-dard pressure instead of 1 atm (101,305 Pa). Both standards are now in wide use.
Standard state is not the same as the standard temperature and pressure (STP) for a gas (discussed in Section 5.4).
6.4 Standard Enthalpies of Formation 247 Several important characteristics of the definition of the enthalpy of formation will become clearer if we again consider the formation of nitrogen dioxide from the elements in their standard states: Note that the reaction is written so that both elements are in their standard states, and 1 mole of product is formed. Enthalpies of formation are always given per mole of product with the product in its standard state.
The formation reaction for methanol is written as The standard state of carbon is graphite, the standard states for oxygen and hydrogen are the diatomic gases, and the standard state for methanol is the liquid.
The Hf values for some common substances are shown in Table 6.2. More values are found in Appendix 4. The importance of the tabulated H f values is that enthalpies for many reactions can be calculated using these numbers. To see how this is done, we will calculate the standard enthalpy change for the combustion of methane: Enthalpy is a state function, so we can invoke Hess’s law and choose any convenient path-way from reactants to products and then sum the enthalpy changes along the chosen path-way. A convenient pathway, shown in Fig. 6.8, involves taking the reactants apart to the respective elements in their standard states in reactions (a) and (b) and then forming the products from these elements in reactions (c) and (d). This general pathway will work for any reaction, since atoms are conserved in a chemical reaction.
Note from Fig. 6.8 that reaction (a), where methane is taken apart into its elements, is just the reverse of the formation reaction for methane: Since reversing a reaction means changing the sign of H but keeping the magnitude the same, H for reaction (a) is Hf , or 75 kJ. Thus H(a) 75 kJ.
Next we consider reaction (b). Here oxygen is already an element in its standard state, so no change is needed. Thus H(b) 0.
C1s2 2H21g2 ¡ CH41g2 ¢H° f 75 kJ/mol CH41g2 ¡ C1s2 2H21g2 CH41g2 2O21g2 ¡ CO21g2 2H2O1l2 C1s2 2H21g2 1 2O21g2 ¡ CH3OH1l2 ¢H° f 239 kJ/mol 1 2N21g2 O21g2 ¡ NO21g2 ¢H° f 34 kJ/mol Brown nitrogen dioxide gas.
TABLE 6.2 Standard Enthalpies of Formation for Several Compounds at 25C Compound H f (kJ/mol) NH3(g) 46 NO2(g) 34 H2O(l) 286 Al2O3(s) 1676 Fe2O3(s) 826 CO2(g) 394 CH3OH(l) 239 C8H18(l) 269 CH4(g) C(s) CO2(g) 2H2(g) 2H2O(l) 2O2(g) 2O2(g) Reactants Elements Products (a) (b) (d) (c) FIGURE 6.8 In this pathway for the combustion of methane, the reactants are first taken apart in reactions (a) and (b) to form the con-stituent elements in their standard states, which are then used to assemble the prod-ucts in reactions (c) and (d).
248 Chapter Six Thermochemistry The next steps, reactions (c) and (d), use the elements formed in reactions (a) and (b) to form the products. Note that reaction (c) is simply the formation reaction for carbon dioxide: and Reaction (d) is the formation reaction for water: However, since 2 moles of water are required in the balanced equation, we must form 2 moles of water from the elements: Thus We have now completed the pathway from the reactants to the products. The change in enthalpy for the reaction is the sum of the H values (including their signs) for the steps: This process is diagramed in Fig. 6.9. Notice that the reactants are taken apart and converted to elements [not necessary for O2(g)] that are then used to form products. You can see that this is a very exothermic reaction because very little energy is required to convert the reactants to the respective elements but a great deal of energy is released when these elements form the products. This is why this reaction is so useful for producing heat to warm homes and offices.
Let’s examine carefully the pathway we used in this example. First, the reactants were broken down into the elements in their standard states. This process involved reversing 891 kJ 175 kJ2 0 1394 kJ2 1572 kJ2 3¢H° f for CH41g2 4 0 3 ¢H° f for CO21g2 4 32 ¢H° f for H2O1l24 ¢H°1a2 ¢H°1b2 ¢H°1c2 ¢H°1d2 ¢H°reaction ¢H°1d2 2 ¢H° f for H2O1l2 21286 kJ2 572 kJ 2H21g2 O21g2 ¡ 2H2O1l2 H21g2 1 2O21g2 ¡ H2O1l2 ¢H° f 286 kJ/mol ¢H°1c2 ¢H° f for CO21g2 394 kJ C1s2 O21g2 ¡ CO21g2 ¢H° f 394 kJ/mol Reactants Step 1 (a) Step 2 (c) Elements Products ∆Ha = 75 kJ (b) ∆Hb = 0 kJ (d) ∆Hd = –572 kJ ∆Hc = –394 kJ FIGURE 6.9 A schematic diagram of the energy changes for the reaction CH4(g) 2O2(g) → CO2(g) 2H2O(l).
6.4 Standard Enthalpies of Formation 249 the formation reactions and thus switching the signs of the enthalpies of formation. The products were then constructed from these elements. This involved formation reactions and thus enthalpies of formation. We can summarize this entire process as follows: The enthalpy change for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products. Remember to multiply the enthalpies of formation by integers as required by the balanced equation. This statement can be represented symbolically as follows: (6.1) where the symbol (sigma) means “to take the sum of the terms,” and np and nr repre-sent the moles of each product or reactant, respectively.
Elements are not included in the calculation because elements require no change in form. We have in effect defined the enthalpy of formation of an element in its standard state as zero, since we have chosen this as our reference point for calculating enthalpy changes in reactions.
Keep in Mind the Following Key Concepts When Doing Enthalpy Calculations: When a reaction is reversed, the magnitude of H remains the same, but its sign changes.
When the balanced equation for a reaction is multiplied by an integer, the value of H for that reaction must be multiplied by the same integer.
The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: Elements in their standard states are not included in the Hreaction calculations. That is, Hf for an element in its standard state is zero.
Enthalpies from Standard Enthalpies of Formation I Using the standard enthalpies of formation listed in Table 6.2, calculate the standard en-thalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid.
Solution We will use the pathway in which the reactants are broken down into elements in their standard states, which are then used to form the products (see Fig. 6.10).
➥ 1 Decomposition of NH3(g) into elements (reaction (a) in Fig. 6.10). The first step is to decompose 4 moles of NH3 into N2 and H2: The preceding reaction is 4 times the reverse of the formation reaction for NH3: Thus ¢H°1a2 4 mol3146 kJ/mol24 184 kJ 1 2N21g2 3 2H21g2 ¡ NH31g2 ¢H° f 46 kJ/mol 4NH31g2 ¡ 2N21g2 6H21g2 4NH31g2 7O21g2 ¡ 4NO21g2 6H2O1l2 ¢H°reaction ©np¢H° f 1products2 ©nr¢H° f 1reactants2 ¢H°reaction ©np¢H° f1products2 ©nr¢H° f1reactants2 Subtraction means to reverse the sign and add.
Elements in their standard states are not included in enthalpy calculations using H f values.
Sample Exercise 6.9 250 Chapter Six Thermochemistry ➥ 2 Elemental oxygen (reaction (b) in Fig. 6.10). Since O2(g) is an element in its stan-dard state, H(b) 0.
We now have the elements N2(g), H2(g), and O2(g), which can be combined to form the products of the overall reaction.
➥ 3 Synthesis of NO2(g) from elements (reaction (c) in Fig. 6.10). The overall reaction equation has 4 moles of NO2. Thus the required reaction is 4 times the formation reaction for NO2: and From Table 6.2, H f for NO2(g) 34 kJ/mol and ➥ 4 Synthesis of H2O(l) from elements (reaction (d) in Fig. 6.10). Since the overall equation for the reaction has 6 moles of H2O(l), the required reaction is 6 times the forma-tion reaction for H2O(l): and From Table 6.2, Hf for H2O(l) 286 kJ/mol and To summarize, we have done the following: Elements in their standard states 6H2O1l2 7O21g2 7O21g2 4NO21g2 2N21g2 6H21g2 4NH31g2 ¢H°1d2 6 mol 1286 kJ/mol2 1716 kJ ¢H°1d2 6 ¢H° f for H2O1l2 6 3H21g2 1 2O21g2 ¡ H2O1l2 4 ¢H°1c2 4 mol 34 kJ/mol 136 kJ ¢H°1c2 4 ¢H° f for NO21g2 4 31 2N21g2 O21g2 ¡ NO21g2 4 4NH3(g) 2N2(g) 4NO2(g) 6H2(g) 6H2O(l) 7O2(g) 7O2(g) Reactants Elements Products (a) (b) (d) (c) FIGURE 6.10 A pathway for the combustion of ammonia. H(a) 88888n H(b) 88888n H(c) 88888n H(d) 88888n ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ 6.4 Standard Enthalpies of Formation 251 We add the H values for the steps to get H for the overall reaction: Remember that elemental reactants and products do not need to be included, since Hf for an element in its standard state is zero. Note that we have again obtained Equa-tion (6.1). The final solution is See Exercises 6.67 and 6.68.
Now that we have shown the basis for Equation (6.1), we will make direct use of it to calculate H for reactions in succeeding exercises.
Enthalpies from Standard Enthalpies of Formation II Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction: This reaction occurs when a mixture of powdered aluminum and iron(III) oxide is ignited with a magnesium fuse.
Solution We use Equation (6.1): where Thus This reaction is so highly exothermic that the iron produced is initially molten. This process is often used as a lecture demonstration and also has been used in welding massive steel objects such as ships’ propellers.
See Exercises 6.71 and 6.72.
1676 kJ 1826 kJ2 850. kJ ¢H°reaction ¢H° f for Al2O31s2 ¢H° f for Fe2O31s2 ¢H° f for Al1s2 ¢H° f for Fe1s2 0 ¢H° f for Al2O31s2 1676 kJ/mol ¢H° f for Fe2O31s2 826 kJ/mol ¢H° ©np¢H° f 1products2 ©nr¢H° f1reactants2 2Al1s2 Fe2O31s2 ¡ Al2O31s2 2Fe1s2 1396 kJ ¢H°reaction 34 134 kJ24 36 1286 kJ2 4 34 146 kJ24 ©np¢H° f1products2 ©nr¢H° f1reactants2 34 ¢H° f for NH31g2 4 34 ¢H° f for NO21g2 4 36 ¢H° f for H2O1l24 36 ¢H° f for H2O1l24 34 ¢H° f for NH31g24 0 34 ¢H° f for NO21g24 ¢H°reaction ¢H°1a2 ¢H°1b2 ¢H°1c2 ¢H°1d2 Sample Exercise 6.10 The thermite reaction is one of the most energetic chemical reactions known.
Visualization: Thermite Reaction 252 Chapter Six Thermochemistry Enthalpies from Standard Enthalpies of Formation III Methanol (CH3OH) is often used as a fuel in high-performance engines in race cars. Us-ing the data in Table 6.2, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of compounds, but assume for this problem that gasoline is pure liquid octane (C8H18).
Solution The combustion reaction for methanol is Using the standard enthalpies of formation from Table 6.2 and Equation (6.1), we have Thus 1454 kJ of heat is evolved when 2 moles of methanol burn. The molar mass of methanol is 32.0 g/mol. This means that 1454 kJ of energy is produced when 64.0 g methanol burns. The enthalpy of combustion per gram of methanol is The combustion reaction for octane is Using the standard enthalpies of information from Table 6.2 and Equation (6.1), we have This is the amount of heat evolved when 2 moles of octane burn. Since the molar mass of octane is 114.2 g/mol, the enthalpy of combustion per gram of octane is The enthalpy of combustion per gram of octane is approximately twice that per gram of methanol. On this basis, gasoline appears to be superior to methanol for use in a rac-ing car, where weight considerations are usually very important. Why, then, is methanol used in racing cars? The answer is that methanol burns much more smoothly than gaso-line in high-performance engines, and this advantage more than compensates for its weight disadvantage.
See Exercise 6.77.
6.5 Present Sources of Energy Woody plants, coal, petroleum, and natural gas hold a vast amount of energy that originally came from the sun. By the process of photosynthesis, plants store energy that can be claimed by burning the plants themselves or the decay products that have been converted 1.09 104 kJ 21114.2 g2 47.8 kJ/g 1.09 104 kJ 16 1394 kJ2 18 1286 kJ2 2 1269 kJ2 2 ¢H° f for C8H181l2 ¢H°reaction 16 ¢H° f for CO21g2 18 ¢H° f for H2O1l2 2C8H181l2 25O21g2 ¡ 16CO21g2 18H2O1l2 1454 kJ 64.0 g 22.7 kJ/g 1454 kJ 2 1394 kJ2 4 1286 kJ2 2 1239 kJ2 2 ¢H° f for CH3OH1l2 ¢H°reaction 2 ¢H° f for CO21g2 4 ¢H° f for H2O1l2 2CH3OH1l2 3O21g2 ¡ 2CO21g2 4H2O1l2 Sample Exercise 6.11 6.5 Present Sources of Energy 253 over millions of years to fossil fuels. Although the United States currently depends heav-ily on petroleum for energy, this dependency is a relatively recent phenomenon, as shown in Fig. 6.11. In this section we discuss some sources of energy and their effects on the environment.
Petroleum and Natural Gas Although how they were produced is not completely understood, petroleum and natural gas were most likely formed from the remains of marine organisms that lived approxi-mately 500 million years ago. Petroleum is a thick, dark liquid composed mostly of com-pounds called hydrocarbons that contain carbon and hydrogen. (Carbon is unique among elements in the extent to which it can bond to itself to form chains of various lengths.) Table 6.3 gives the formulas and names for several common hydrocarbons. Natural gas, usually associated with petroleum deposits, consists mostly of methane, but it also con-tains significant amounts of ethane, propane, and butane.
The composition of petroleum varies somewhat, but it consists mostly of hydrocar-bons having chains that contain from 5 to more than 25 carbons. To be used efficiently, the petroleum must be separated into fractions by boiling. The lighter molecules (having the lowest boiling points) can be boiled off, leaving the heavier ones behind. The com-mercial uses of various petroleum fractions are shown in Table 6.4.
The petroleum era began when the demand for lamp oil during the Industrial Revolution outstripped the traditional sources: animal fats and whale oil. In response to this increased demand, Edwin Drake drilled the first oil well in 1859 at Titusville, Pennsylvania. The petroleum from this well was refined to produce kerosene (fraction C10–C18), which served as an excellent lamp oil. Gasoline (fraction C5–C10) had limited use and was often discarded. However, this situation soon changed. The development of the electric light decreased the need for kerosene, and the advent of the “horseless carriage” with its gasoline-powered engine signaled the birth of the gasoline age.
As gasoline became more important, new ways were sought to increase the yield of gasoline obtained from each barrel of petroleum. William Burton invented a process at Standard Oil of Indiana called pyrolytic (high-temperature) cracking. In this process, the heavier molecules of the kerosene fraction are heated to about 700C, causing them to break (crack) into the smaller molecules of hydrocarbons in the gasoline fraction. As cars became larger, more efficient internal combustion engines were designed. Because of the uneven burning of the gasoline then available, these engines “knocked,” producing unwanted noise and even engine damage. Intensive research to find additives that would promote smoother burning produced tetraethyl lead, (C2H5)4Pb, a very effective “anti-knock” agent.
1850 1900 1950 1975 2000 91% 9% 21% 71% 5% 3% 36% 52% 6% 6% 18% 73% 23% 62% 11% 4% 6% 3% Coal Wood Petroleum/natural gas Hydro and nuclear FIGURE 6.11 Energy sources used in the United States.
This oil rig in Norway is the largest in the world.
TABLE 6.3 Names and Formulas for Some Common Hydrocarbons Formula Name CH4 Methane C2H6 Ethane C3H8 Propane C4H10 Butane C5H12 Pentane C6H14 Hexane C7H16 Heptane C8H18 Octane 254 Chapter Six Thermochemistry The addition of tetraethyl lead to gasoline became a common practice, and by 1960, gasoline contained as much as 3 grams of lead per gallon. As we have discovered so of-ten in recent years, technological advances can produce environmental problems. To pre-vent air pollution from automobile exhaust, catalytic converters have been added to car exhaust systems. The effectiveness of these converters, however, is destroyed by lead. The use of leaded gasoline also greatly increased the amount of lead in the environment, where it can be ingested by animals and humans. For these reasons, the use of lead in gasoline has been phased out, requiring extensive (and expensive) modifications of engines and of the gasoline refining process.
Coal Coal was formed from the remains of plants that were buried and subjected to high pres-sure and heat over long periods of time. Plant materials have a high content of cellulose, a complex molecule whose empirical formula is CH2O but whose molar mass is around 500,000 g/mol. After the plants and trees that flourished on the earth at various times and places died and were buried, chemical changes gradually lowered the oxygen and hydro-gen content of the cellulose molecules. Coal “matures” through four stages: lignite, sub-bituminous, bituminous, and anthracite. Each stage has a higher carbon-to-oxygen and carbon-to-hydrogen ratio; that is, the relative carbon content gradually increases. Typical elemental compositions of the various coals are given in Table 6.5. The energy available from the combustion of a given mass of coal increases as the carbon content increases.
Therefore, anthracite is the most valuable coal, and lignite the least valuable.
Coal is an important and plentiful fuel in the United States, currently furnishing ap-proximately 23% of our energy. As the supply of petroleum dwindles, the share of the en-ergy supply from coal is expected to increase. However, coal is expensive and dangerous to mine underground, and the strip mining of fertile farmland in the Midwest or of scenic land in the West causes obvious problems. In addition, the burning of coal, especially high-sulfur coal, yields air pollutants such as sulfur dioxide, which, in turn, can lead to acid rain, as we learned in Chapter 5. However, even if coal were pure carbon, the car-bon dioxide produced when it was burned would still have significant effects on the earth’s climate.
Effects of Carbon Dioxide on Climate The earth receives a tremendous quantity of radiant energy from the sun, about 30% of which is reflected back into space by the earth’s atmosphere. The remaining energy passes through the atmosphere to the earth’s surface. Some of this energy is absorbed by plants for photosynthesis and some by the oceans to evaporate water, but most of it is absorbed by soil, rocks, and water, increasing the temperature of the earth’s surface. This energy is in turn radiated from the heated surface mainly as infrared radiation, often called heat radiation.
TABLE 6.4 Uses of the Various Petroleum Fractions Petroleum Fraction in Terms of Numbers of Carbon Atoms Major Uses C5–C10 Gasoline C10–C18 Kerosene Jet fuel C15–C25 Diesel fuel Heating oil Lubricating oil C25 Asphalt Coal has variable composition depending on both its age and location.
The electromagnetic spectrum, including visible and infrared radiation, is dis-cussed in Chapter 7.
TABLE 6.5 Elemental Composition of Various Types of Coal Mass Percent of Each Element Type of Coal C H O N S Lignite 71 4 23 1 1 Subbituminous 77 5 16 1 1 Bituminous 80 6 8 1 5 Anthracite 92 3 3 1 1 6.5 Present Sources of Energy 255 The atmosphere, like window glass, is transparent to visible light but does not allow all the infrared radiation to pass back into space. Molecules in the atmosphere, principally H2O and CO2, strongly absorb infrared radiation and radiate it back toward the earth, as shown in Fig. 6.12, so a net amount of thermal energy is retained by the earth’s atmos-phere, causing the earth to be much warmer than it would be without its atmosphere. In a way, the atmosphere acts like the glass of a greenhouse, which is transparent to visible light but absorbs infrared radiation, thus raising the temperature inside the building. This greenhouse effect is seen even more spectacularly on Venus, where the dense atmosphere is thought to be responsible for the high surface temperature of that planet.
Thus the temperature of the earth’s surface is controlled to a significant extent by the carbon dioxide and water content of the atmosphere. The effect of atmospheric moisture (humidity) is apparent in the Midwest. In summer, when the humidity is high, the heat of the sun is retained well into the night, giving very high nighttime temperatures. On the other hand, in winter, the coldest temperatures always occur on clear nights, when the low humidity allows efficient radiation of energy back into space.
The atmosphere’s water content is controlled by the water cycle (evaporation and pre-cipitation), and the average remains constant over the years. However, as fossil fuels have been used more extensively, the carbon dioxide concentration has increased by about 16% from 1880 to 1980. Comparisons of satellite data have now produced evidence that the greenhouse effect has significantly warmed the earth’s atmosphere. The data compare the same areas in both 1979 and 1997. The analysis shows that more infrared radiation was blocked by CO2, methane, and other greenhouse gases. This could increase the earth’s av-erage temperature by as much as 3C, causing dramatic changes in climate and greatly affecting the growth of food crops.
How well can we predict long-term effects? Because weather has been studied for a period of time that is minuscule compared with the age of the earth, the factors that con-trol the earth’s climate in the long range are not clearly understood. For example, we do not understand what causes the earth’s periodic ice ages. So it is difficult to estimate the impact of the increasing carbon dioxide levels.
In fact, the variation in the earth’s average temperature over the past century is some-what confusing. In the northern latitudes during the past century, the average temperature rose by 0.8C over a period of 60 years, then cooled by 0.5C during the next 25 years, and finally warmed by 0.2C in the succeeding 15 years. Such fluctuations do not match the steady increase in carbon dioxide. However, in southern latitudes and near the equa-tor during the past century, the average temperature showed a steady rise totaling 0.4C.
Infrared radiated by the earth CO2 and H2O molecules Earth’s atmosphere Visible light from the sun Earth FIGURE 6.12 The earth’s atmosphere is transparent to visible light from the sun. This visible light strikes the earth, and part of it is changed to infrared radiation. The infrared radiation from the earth’s surface is strongly ab-sorbed by CO2, H2O, and other molecules present in smaller amounts (for example, CH4 and N2O) in the atmosphere. In effect, the atmosphere traps some of the energy, acting like the glass in a greenhouse and keeping the earth warmer than it would otherwise be.
The average temperature of the earth’s surface is 298 K. It would be 255 K without the “greenhouse gases.” Sheep grazing on a ranch in Australia.
256 Chapter Six Thermochemistry This figure is in reasonable agreement with the predicted effect of the increasing carbon dioxide concentration over that period. Another significant fact is that the past 10 years constitute the warmest decade on record.
Although the exact relationship between the carbon dioxide concentration in the at-mosphere and the earth’s temperature is not known at present, one thing is clear: The in-crease in the atmospheric concentration of carbon dioxide is quite dramatic (see Fig. 6.13).
We must consider the implications of this increase as we consider our future energy needs.
Methane is another greenhouse gas that is 21 times more potent than carbon dioxide.
This fact is particularly significant for countries with lots of animals, because methane is produced by methanogenic archae that live in the animals’ rumen. For example, sheep and cattle produce about 14% of Australia’s total greenhouse emissions. To reduce this level, Australia has initiated a program to vaccinate sheep and cattle to lower the number of ar-chae present in their digestive systems. It is hoped that this effort will reduce by 20% the amount of methane emitted by these animals.
6.6 New Energy Sources As we search for the energy sources of the future, we need to consider economic, climatic, and supply factors. There are several potential energy sources: the sun (solar), nuclear processes (fission and fusion), biomass (plants), and synthetic fuels. Direct use of the sun’s radiant energy to heat our homes and run our factories and transportation systems seems a sensible long-term goal. But what do we do now? Conservation of fossil fuels is one obvious step, but substitutes for fossil fuels also must be found. We will discuss some alternative sources of energy here. Nuclear power will be considered in Chapter 21.
Coal Conversion One alternative energy source involves using a traditional fuel—coal—in new ways. Since transportation costs for solid coal are high, more energy-efficient fuels are being devel-oped from coal. One possibility is to produce a gaseous fuel. Substances like coal that contain large molecules have high boiling points and tend to be solids or thick liquids. To convert coal from a solid to a gas therefore requires reducing the size of the molecules; the coal structure must be broken down in a process called coal gasification. This is done by treating the coal with oxygen and steam at high temperatures to break many of the carbon–carbon bonds. These bonds are replaced by carbon–hydrogen and carbon–oxygen bonds as the coal fragments react with the water and oxygen. The process is represented in Fig. 6.14. The desired product is a mixture of carbon monoxide and hydrogen called synthetic gas, or syngas, and methane (CH4) gas. Since all the components of this prod-uct can react with oxygen to release heat in a combustion reaction, this gas is a useful fuel.
One of the most important considerations in designing an industrial process is efficient use of energy. In coal gasification, some of the reactions are exothermic: Other gasification reactions are endothermic, for example: If such conditions as the rate of feed of coal, air, and steam are carefully controlled, the correct temperature can be maintained in the process without using any external energy source. That is, an energy balance is maintained.
C1s2 H2O1g2 ¡ H21g2 CO1g2 ¢H° 131 kJ C1s2 O21g2 ¡ CO21g2 ¢H° 394 kJ C1s2 1 2O21g2 ¡ CO1g2 ¢H° 111 kJ C1s2 2H21g2 ¡ CH41g2 ¢H° 75 kJ CO2 concentration (ppmv) 200 300 400 500 1750 1800 1850 Year (A.D.) 1900 1950 2000 Global CO2 Global temperature FIGURE 6.13 The atmospheric CO2 concentration and the average global temperature over the last 250 years. Note the significant increase in CO2 concentration in the last 50 years.
(Source: National Assessment Synthesis Team, Climate Change Impacts on the United States: The Potential Consequences of Climate, Variability and Change, Overview, Report for the U.S. Global Change Research Program, Cambridge University Press, Cambridge, UK, p. 13, 2000.) An industrial process must be energy efficient.
6.6 New Energy Sources 257 Presently only a few plants in the United States use syngas produced on site to produce electricity. These plants are being used to evaluate the economic feasibility of producing electrical power by coal gasification.
Although syngas can be used directly as a fuel, it is also important as a raw mate-rial to produce other fuels. For example, syngas can be converted directly to methanol: Methanol is used in the production of synthetic fibers and plastics and also can be used as a fuel. In addition, it can be converted directly to gasoline. Approximately half of South Africa’s gasoline supply comes from methanol produced from syngas.
In addition to coal gasification, the formation of coal slurries is another new use of coal. A slurry is a suspension of fine particles in a liquid, and coal must be pulverized and mixed with water to form a slurry. The slurry can be handled, stored, and burned in ways similar to those used for residual oil, a heavy fuel oil from petroleum accounting for al-most 15% of U.S. petroleum imports. One hope is that coal slurries might replace solid coal and residual oil as fuels for electricity-generating power plants. However, the water needed for slurries might place an unacceptable burden on water resources, especially in the western states.
Hydrogen as a Fuel If you have ever seen a lecture demonstration where hydrogen–oxygen mixtures were ig-nited, you have witnessed a demonstration of hydrogen’s potential as a fuel. The com-bustion reaction is As we saw in Sample Exercise 6.6, the heat of combustion of H2(g) per gram is approx-imately 2.5 times that of natural gas. In addition, hydrogen has a real advantage over fossil fuels in that the only product of hydrogen combustion is water; fossil fuels also produce carbon dioxide. However, even though it appears that hydrogen is a very logical choice as a major fuel for the future, there are three main problems: the cost of production, storage, and transport.
H21g2 1 2O21g2 ¡ H2O1l2 ¢H° 286 kJ CO1g2 2H21g2 ¡ CH3OH1l2 Heat Separate CO(g) + H2O(g) CO2(g) + H2(g) CO(g) + 3H2(g) CH4(g) + H2O(g) CH4(g) Remove CO2, H2O, impurities CH4(g) Syngas [CO(g), H2(g)] Coal (C) + steam [H2O(g)] + air [O2(g)] CH4(g), CO(g), CO2(g), H2(g), H2O(g) + sulfur-containing impurities (sulfur compounds) FIGURE 6.14 Coal gasification. Reaction of coal with a mixture of steam and air breaks down the large hydrocarbon molecules in the coal to smaller gaseous molecules, which can be used as fuels.
The main engines in the space shuttle Endeavour use hydrogen and oxygen as fuel.
258 Chapter Six Thermochemistry First let’s look at the production problem. Although hydrogen is very abundant on earth, virtually none of it exists as the free gas. Currently, the main source of hydrogen gas is from the treatment of natural gas with steam: We can calculate H for this reaction using Equation (6.1): Note that this reaction is highly endothermic; treating methane with steam is not an efficient way to obtain hydrogen for fuel. It would be much more economical to burn the methane directly.
A virtually inexhaustible supply of hydrogen exists in the waters of the world’s oceans.
However, the reaction requires 286 kJ of energy per mole of liquid water, and under current circumstances, large-scale production of hydrogen from water is not economically feasible. However, several methods for such production are currently being studied: electrolysis of water, thermal de-composition of water, thermochemical decomposition of water, and biological decompo-sition of water.
Electrolysis of water involves passing an electric current through it, as shown in Fig. 1.16 in Chapter 1. The present cost of electricity makes the hydrogen produced by electrolysis too expensive to be competitive as a fuel. However, if in the future we develop more efficient sources of electricity, this situation could change.
Recent research at the University of Minnesota by Lanny Schmidt and his cowork-ers suggests that corn could be a feasible source of hydrogen. In this process the starch from the corn is fermented to produce alcohol, which is then decomposed in a special H2O1l2 ¡ H21g2 1 2O21g2 111 kJ 175 kJ2 1242 kJ2 206 kJ ¢H° f for CO1g2 ¢H° f for CH41g2 ¢H° f for H2O1g2 ¢H° ©np¢H° f 1products2 ©nr¢H° f1reactants2 CH41g2 H2O1g2 ¡ 3H21g2 CO1g2 Electrolysis will be discussed in Chapter 17.
CHEMICAL IMPACT Farming the Wind I n the Midwest the wind blows across fields of corn, soy-beans, wheat, and wind turbines—wind turbines? It turns out that the wind that seems to blow almost continuously across the plains is now becoming the latest cash crop. One of these new-breed wind farmers is Daniel Juhl, who re-cently erected 17 wind turbines on six acres of land near Woodstock, Minnesota. These turbines can generate as much as 10 megawatts (MW) of electricity, which Juhl sells to the local electrical utility.
There is plenty of untapped wind-power in the United States. Wind mappers rate regions on a scale of 1 to 6 (with 6 being the best) to indicate the quality of the wind resource.
Wind farms are now being developed in areas rated from 4 to 6. The farmers who own the land welcome the increased income derived from the wind blowing across their land.
Economists estimate that each acre devoted to wind turbines can pay royalties to the farmers of as much as $8000 per year, or many times the revenue from growing corn on that same land. Daniel Juhl claims that farmers who construct the turbines themselves can realize as much as $20,000 per year per turbine. Globally, wind generation of electricity has nearly quadrupled in the last five years and is expected to increase by about 60% per year in the United States. The economic feasibility of wind-generated electricity has greatly improved in the last 30 years as the wind turbines have become more efficient. Today’s turbines can produce elec-tricity that costs about the same as that from other sources.
The most impressive thing about wind power is the magni-tude of the supply. According to the American Wind Energy Association in Washington, D.C., the wind-power potential in the United States is comparable or larger than the energy resources under the sands of Saudi Arabia.
6.6 New Energy Sources 259 reactor at 140C with a rhodium and cerium oxide catalyst to give hydrogen. These sci-entists indicate that enough hydrogen gas can be obtained from a few ounces of ethanol to generate electricity to run six 60-watt bulbs for an hour.
Thermal decomposition is another method for producing hydrogen from water. This involves heating the water to several thousand degrees, where it spontaneously decom-poses into hydrogen and oxygen. However, attaining temperatures in this range would be very expensive even if a practical heat source and a suitable reaction container were available.
In the thermochemical decomposition of water, chemical reactions, as well as heat, are used to “split” water into its components. One such system involves the following re-actions (the temperature required for each is given in parentheses): Note that the HI is not consumed in the net reaction. Note also that the maximum tem-perature required is 825C, a temperature that is feasible if a nuclear reactor is used as a heat source. A current research goal is to find a system for which the required tempera-tures are low enough that sunlight can be used as the energy source.
But what about the organisms that decompose water without the aid of electricity or high temperatures? In the process of photosynthesis, green plants absorb carbon dioxide and water and use them along with energy from the sun to produce the substances needed for growth. Scientists have studied photosynthesis for years, hoping to get answers to hu-manity’s food and energy shortages. At present, much of this research involves attempts to modify the photosynthetic process so that plants will release hydrogen gas from water instead of using the hydrogen to produce complex compounds. Small-scale experiments have shown that under certain conditions plants do produce hydrogen gas, but the yields Net reaction: H2O ¡ H2 1 2O2 H2SO4 ¡ SO2 H2O 1 2O2 1825°C2 2H2O SO2 I2 ¡ H2SO4 2HI 190°C2 2HI ¡ I2 H2 1425°C2 The biggest hurdle that must be overcome before wind power can become a significant electricity producer in the United States is construction of the transmission infrastruc-ture—the power lines needed to move the electricity from the rural areas to the cities where most of the power is used.
For example, the hundreds of turbines planned in southwest Minnesota in a development called Buffalo Ridge could sup-ply enough electricity to power 1 million homes if trans-mission problems can be solved.
Another possible scenario for wind farms is to use the electrical power generated to decompose water to produce hydrogen gas that could be carried to cities by pipelines and used as a fuel. One real benefit of hydrogen is that it pro-duces water as its only combustion product. Thus, it is essentially pollution-free.
Within a few years wind power could be a major source of electricity. There could be a fresh wind blowing across the energy landscape of the United States in the near future.
This State Line Wind Project along the Oregon-Washington border uses approximately 399 wind turbines to create enough electricity to power some 70,000 households.
260 Chapter Six Thermochemistry are far from being commercially useful. At this point the economical production of hydrogen gas remains unrealized.
The storage and transportation of hydrogen also present problems. First, on metal surfaces the H2 molecule decomposes to atoms. Since the atoms are so small, they can migrate into the metal, causing structural changes that make it brittle. This might lead to a pipeline failure if hydrogen were pumped under high pressure.
An additional problem is the relatively small amount of energy that is available per unit volume of hydrogen gas. Although the energy available per gram of hydrogen is significantly greater than that per gram of methane, the energy available per given volume of hydrogen is about one-third that available from the same volume of methane. This is demonstrated in Sample Exercise 6.12.
Although the use of hydrogen as a fuel solves some of the problems associated with fossil fuels, it does present some potential environmental problems of its own. Studies by John M. Eiler and his colleagues at California Institute of Technology indicate that, if hydrogen becomes a major source of energy, accidental leakage of the gas into the atmosphere could pose a threat. The Cal Tech scientists calculate that leakage could raise the concentration of H2 in the atmosphere from its natural level of 0.5 part per million to more than 2 parts per million. As some of the H2 eventually finds its way into the upper atmosphere, it would react with O2 to form water, which would increase the number of ice crystals. This could lead to the destruction of some of the protective ozone because many of the chemical reactions that destroy ozone occur on the surfaces of ice crystals.
However, as is the usual case with environmental issues, the situation is complicated. The scenario suggested by Eiler’s team may not happen because the leaked H2 could be con-sumed by soil microbes that use hydrogen as a nutrient. In fact, Eiler’s studies show that 90% of the H2 emitted into the atmosphere today from sources such as motor vehicles and forest fires is eventually absorbed by soil organisms.
The evaluation of hydrogen as a fuel illustrates how complex and interconnected the economic and environmental issues are.
Enthalpies of Combustion Compare the energy available from the combustion of a given volume of methane and the same volume of hydrogen at the same temperature and pressure.
Solution In Sample Exercise 6.6 we calculated the heat released for the combustion of methane and hydrogen: 55 kJ/g CH4 and 141 kJ/g H2. We also know from our study of gases that 1 mol H2(g) has the same volume as 1 mol CH4(g) at the same temperature and pressure (assuming ideal behavior). Thus, for molar volumes of both gases under the same condi-tions of temperature and pressure, Thus about three times the volume of hydrogen gas is needed to furnish the same energy as a given volume of methane.
See Exercise 6.78.
285 882 1 3 1141 kJ/g212.02 g H2/mol H22 155 kJ/g2116.04 g CH4/mol CH42 enthalpy of combustion per mole of H2 enthalpy of combustion per mole of CH4 Enthalpy of combustion of 1 molar volume of H21g2 Enthalpy of combustion of 1 molar volume of CH41g2 Sample Exercise 6.12 6.6 New Energy Sources 261 Could hydrogen be considered as a potential fuel for automobiles? This is an in-triguing question. The internal combustion engines in automobiles can be easily adapted to burn hydrogen. In fact, BMW is now experimenting with a fleet of cars powered by hydrogen-burning internal combustion engines. However, the primary difficulty is the stor-age of enough hydrogen to give an automobile a reasonable range. This is illustrated by Sample Exercise 6.13.
Comparing Enthalpies of Combustion Assuming that the combustion of hydrogen gas provides three times as much energy per gram as gasoline, calculate the volume of liquid H2 (density 0.0710 g/mL) required to furnish the energy contained in 80.0 L (about 20 gal) of gasoline (density 0.740 g/mL).
Calculate also the volume that this hydrogen would occupy as a gas at 1.00 atm and 25C.
Solution The mass of 80.0 L gasoline is Since H2 furnishes three times as much energy per gram as gasoline, only a third as much liquid hydrogen is needed to furnish the same energy: Since density massvolume, then volume massdensity, and the volume of H2(l) needed is Thus 277 L of liquid H2 is needed to furnish the same energy of combustion as 80.0 L of gasoline.
To calculate the volume that this hydrogen would occupy as a gas at 1.00 atm and 25C, we use the ideal gas law: In this case P 1.00 atm, T 273 25C 298 K, and .
Also, Thus At 1 atm and 25C, the hydrogen gas needed to replace 20 gal of gasoline occupies a vol-ume of 238,000 L.
See Exercises 6.79 and 6.80.
2.38 105 L 238,000 L V nRT P 19.75 103 mol210.08206 L atm/K mol21298 K2 1.00 atm n 19,700 g H2 1 mol H2 2.02 g H2 9.75 103 mol H2 R 0.08206 L atm/K mol PV nRT 2.77 105 mL 277 L V 19,700 g 0.0710 g/mL Mass of H21l2 needed 59,200 g 3 19,700 g 80.0 L 1000 mL 1 L 0.740 g mL 59,200 g Sample Exercise 6.13 262 Chapter Six Thermochemistry You can see from Sample Exercise 6.13 that an automobile would need a huge tank to hold enough hydrogen gas (at 1 atm) to have a typical mileage range. Clearly, hydro-gen must be stored as a liquid or in some other way. Is this feasible? Because of its very low boiling point (20 K), storage of liquid hydrogen requires a superinsulated container that can withstand high pressures. Storage in this manner would be both expensive and hazardous because of the potential for explosion. Thus storage of hydrogen in the indi-vidual automobile as a liquid does not seem practical.
A much better alternative seems to be the use of metals that absorb hydrogen to form solid metal hydrides: To use this method of storage, hydrogen gas would be pumped into a tank containing the solid metal in powdered form, where it would be absorbed to form the hydride, whose volume would be little more than that of the metal alone. This hydrogen would then be available for combustion in the engine by release of H2(g) from the hydride as needed: Several types of solids that absorb hydrogen to form hydrides are being studied for use in hydrogen-powered vehicles. The most likely use of hydrogen in automobiles will be to power fuel cells (see Section 17.5). Ford, Honda, and Toyota are all experimenting with cars powered by hydrogen fuel cells.
Other Energy Alternatives Many other energy sources are being considered for future use. The western states, espe-cially Colorado, contain huge deposits of oil shale, which consists of a complex carbon-based material called kerogen contained in porous rock formations. These deposits have the potential of being a larger energy source than the vast petroleum deposits of the Mid-dle East. The main problem with oil shale is that the trapped fuel is not fluid and cannot MH21s2 ¡ M1s2 H21g2 H21g2 M1s2 ¡ MH21s2 Metal hydrides are discussed in Chapter 18.
CHEMICAL IMPACT Veggie Gasoline?
G asoline usage is as high as ever, and world petroleum supplies will even-tually dwindle. One possible alternative to petroleum as a source of fuels and lubricants is vegetable oil—the same vegetable oil we now use to cook french fries. Researchers believe that the oils from soybeans, corn, canola, and sun-flowers all have the potential to be used in cars as well as on salads.
The use of vegetable oil for fuel is not a new idea. Rudolf Diesel reportedly used peanut oil to run one of his engines at the Paris Exposition in 1900. In addi-tion, ethyl alcohol has been used widely as a fuel in South America and as a fuel additive in the United States.
This promotion bus both advertises biodiesel and demonstrates its usefulness 6.6 New Energy Sources 263 be pumped. To recover the fuel, the rock must be heated to a temperature of 250C or higher to decompose the kerogen to smaller molecules that produce gaseous and liquid products. This process is expensive and yields large quantities of waste rock, which have a negative environmental impact.
Ethanol (C2H5OH) is another fuel with the potential to supplement, if not replace, gasoline. The most common method of producing ethanol is fermentation, a process in which sugar is changed to alcohol by the action of yeast. The sugar can come from virtually any source, including fruits and grains, although fuel-grade ethanol would probably come mostly from corn. Car engines can burn pure alcohol or gasohol, an alcohol–gasoline mix-ture (10% ethanol in gasoline), with little modification. Gasohol is now widely available in the United States. The use of pure alcohol as a motor fuel is not feasible in most of the United States because it does not vaporize easily when temperatures are low. However, pure ethanol could be a very practical fuel in warm climates. For example, in Brazil, large quantities of ethanol fuel are being produced for cars.
Methanol (CH3OH), an alcohol similar to ethanol, which has been used successfully for many years in race cars, is now being evaluated as a motor fuel in California. A major gasoline retailer has agreed to install pumps at 25 locations to dispense a fuel that is 85% methanol and 15% gasoline for use in specially prepared automobiles. The California Energy Commission feels that methanol has great potential for providing a secure, long-term energy supply that would alleviate air quality problems. Arizona and Colorado are also considering methanol as a major source of portable energy.
Another potential source of liquid fuels is oil squeezed from seeds (seed oil). For ex-ample, some farmers in North Dakota, South Africa, and Australia are now using sun-flower oil to replace diesel fuel. Oil seeds, found in a wide variety of plants, can be processed to produce an oil composed mainly of carbon and hydrogen, which of course reacts with oxygen to produce carbon dioxide, water, and heat. It is hoped that oil-seed plants can be developed that will thrive under soil and climatic conditions unsuitable for corn and wheat. The main advantage of seed oil as a fuel is that it is renewable. Ideally, fuel would be grown just like food crops.
The sugars in corn are fermented and used to produce ethanol, an additive for gasoline.
Biodiesel, a fuel made by esterifying the fatty acids found in vegetable oil, has some real advantages over reg-ular diesel fuel. Biodiesel produces fewer pollutants such as particulates, carbon monoxide, and complex organic molecules, and since vegetable oils have no sulfur, there is no noxious sulfur dioxide in the exhaust gases. Also, biodiesel can run in existing engines with little modifica-tion. In addition, biodiesel is much more biodegradable than petroleum-based fuels, so spills cause less environ-mental damage.
Of course, biodiesel also has some serious drawbacks.
The main one is that it costs about three times as much as regular diesel fuel. Biodiesel also produces more nitrogen oxides in the exhaust than conventional diesel fuel and is less stable in storage. Biodiesel also can leave more gummy deposits in engines and must be “winterized” by removing components that tend to solidify at low temperatures.
The best solution may be to use biodiesel as an addi-tive to regular diesel fuel. One such fuel is known as B20 because it is 20% biodiesel and 80% conventional diesel fuel. B20 is especially attractive because of the higher lu-bricating ability of vegetable oils, thus reducing diesel en-gine wear.
Vegetable oils are also being looked at as replacements for motor oils and hydraulic fluids. Tests of a sunflower seed–based engine lubricant manufactured by Renewable Lubricants of Hartville, Ohio, have shown satisfactory lu-bricating ability while lowering particle emissions. In addi-tion, Lou Honary and his colleagues at the University of Northern Iowa have developed BioSOY, a vegetable oil–based hydraulic fluid for use in heavy machinery.
Veggie oil fuels and lubricants seem to have a growing market as petroleum supplies wane and as environmental laws become more stringent. In Germany’s Black Forest re-gion, for example, environmental protection laws require that farm equipment use only vegetable oil fuels and lubri-cants. In the near future there may be veggie oil in your garage as well as in your kitchen.
Adapted from “Fill ’Er Up . . . with Veggie Oil,” by Corinna Wu, as appeared in Science News, Vol. 154, December 5, 1998, p. 364.
264 Chapter Six Thermochemistry Key Terms Section 6.1 energy law of conservation of energy potential energy kinetic energy heat work pathway state function (property) system surroundings exothermic endothermic thermodynamics first law of thermodynamics internal energy Section 6.2 enthalpy calorimeter calorimetry heat capacity specific heat capacity molar heat capacity constant-pressure calorimetry constant-volume calorimetry Section 6.3 Hess’s law Section 6.4 standard enthalpy of formation standard state Section 6.5 fossil fuels petroleum natural gas coal greenhouse effect Section 6.6 syngas For Review Energy The capacity to do work or produce heat Is conserved (first law of thermodynamics) Can be converted from one form to another Is a state function Potential energy: stored energy Kinetic energy: energy due to motion The internal energy for a system is the sum of its potential and kinetic energies The internal energy of a system can be changed by work and heat: E q w Work Force applied over a distance For an expanding/contracting gas Not a state function w P V Heat Energy flow due to a temperature difference Exothermic: energy as heat flows out of a system Endothermic: energy as heat flows into a system Not a state function Measured for chemical reactions by calorimetry Enthalpy H E PV Is a state function Hess’s law: the change in enthalpy in going from a given set of reactants to a given set of products is the same whether the process takes place in one step or a series of steps Standard enthalpies of formation ( H f) can be used to calculate H for a chemical reaction Energy use Energy sources from fossil fuels are associated with difficult supply and environ-mental impact issues The greenhouse effect results from release into the atmosphere of gases, including carbon dioxide, that strongly absorb infrared radiation, thus warming the earth Alternative fuels are being sought to replace fossil fuels: • Hydrogen • Syngas from coal • Biofuels from plants such as corn and certain seed-producing plants REVIEW QUESTIONS 1. Define the following terms: potential energy, kinetic energy, path-dependent function, state function, system, surroundings.
¢H° reaction a np¢H° f 1products2 a nr¢H° 1reactants2 Active Learning Questions 265 2. Consider the following potential energy diagrams for two different reactions.
Which plot represents an exothermic reaction? In plot a, do the reactants on average have stronger or weaker bonds than the products? In plot b, reactants must gain potential energy to convert to products. How does this occur?
3. What is the first law of thermodynamics? How can a system change its internal energy, E? What are the sign conventions for thermodynamic quantities used in this text?
4. When a gas expands, what is the sign of w? Why? When a gas contracts, what is the sign of w? Why? What are the signs of q and w for the process of boiling water?
5. What is the heat gained/released at constant pressure equal to (qP ?)? What is the heat gained/released at constant volume equal to (qV ?)? Explain why H is obtained directly from a coffee-cup calorimeter, whereas E is obtained di-rectly from a bomb calorimeter.
6. High-quality audio amplifiers generate large amounts of heat. To dissipate the heat and prevent damage to the electronic components, heat-radiating metal fins are used. Would it be better to make these fins out of iron or aluminum? Why?
(See Table 6.1 for specific heat capacities.) 7. Explain how calorimetry works to calculate H or E for a reaction. Does the temperature of the calorimeter increase or decrease for an endothermic reaction?
For an exothermic reaction? Explain.
8. What is Hess’s law? When a reaction is reversed, what happens to the sign and magnitude of H for that reversed reaction? When the coefficients in a balanced reaction are multiplied by a factor n, what happens to the sign and magnitude of H for that multiplied reaction?
9. Define the standard enthalpy of formation. What are standard states for elements and for compounds? Using Hess’s law, illustrate why the formula (products) (reactants) works to calculate H for a reaction.
10. What are some of the problems associated with the world’s dependence on fossil fuels? What are some alternative fuels for petroleum products?
nr¢H° f ©np¢H° f ¢H° reaction Reactants Products Potential energy a.
Products Reactants Potential energy b.
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Objects placed together eventually reach the same temperature.
When you go into a room and touch a piece of metal in that room, it feels colder than a piece of plastic. Explain.
2. What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid wa-ter? How do you know? Which of the two is more stable? How do you know?
3. A fire is started in a fireplace by striking a match and lighting crumpled paper under some logs. Explain all the energy trans-fers in this scenario using the terms exothermic, endothermic, system, surroundings, potential energy, and kinetic energy in the discussion.
266 Chapter Six Thermochemistry 4. Liquid water turns to ice. Is this process endothermic or exother-mic? Explain what is occurring using the terms system, surroundings, heat, potential energy, and kinetic energy in the discussion.
5. Consider the following statements: “Heat is a form of energy, and energy is conserved. The heat lost by a system must be equal to the amount of heat gained by the surroundings. Therefore, heat is conserved.” Indicate everything you think is correct in these statements. Indicate everything you think is incorrect.
Correct the incorrect statements and explain.
6. Consider 5.5 L of a gas at a pressure of 3.0 atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 L.
a. Calculate the work done, and indicate the correct sign.
b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 L.
The second step results in a final volume of 10.5 L. Calcu-late the work done, and indicate the correct sign.
c. Calculate the work done if after the first step the volume is 8.0 L and the second step leads to a volume of 10.5 L. Does the work differ from that in part b? Explain.
7. In Question 6 the work calculated for the different conditions in the various parts of the question was different even though the system had the same initial and final conditions. Based on this information, is work a state function?
a. Explain how you know that work is not a state function.
b. Why does the work increase with an increase in the number of steps?
c. Which two-step process resulted in more work, when the first step had the bigger change in volume or when the second step had the bigger change in volume? Explain.
8. Photosynthetic plants use the following reaction to produce glucose, cellulose, and so forth: How might extensive destruction of forests exacerbate the green-house effect?
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 9. Consider an airplane trip from Chicago, Illinois to Denver, Colorado. List some path-dependent functions and some state functions for the plane trip 10. How is average bond strength related to relative potential ener-gies of the reactants and the products?
11. Assuming gasoline is pure C8H18(l), predict the signs of q and w for the process of combusting gasoline into CO2(g) and H2O(g).
12. What is the difference between H and E?
13. The enthalpy of combustion of CH4(g) when H2O(l) is formed is 891 kJ/mol and the enthalpy of combustion of CH4(g) when H2O(g) is formed is 803 kJ/mol. Use these data and Hess’s law to determine the enthalpy of vaporization for water.
6CO21g2 6H2O1l2 ¬¡ Sunlight C6H12O61s2 6O21g2 14. Standard enthalpies of formation are relative values. What are H° f values relative to?
15. What is incomplete combustion of fossil fuels? Why can this be a problem?
16. Explain the advantages and disadvantages of hydrogen as an alternative fuel.
Exercises In this section similar exercises are paired.
Potential and Kinetic Energy 17. Calculate the kinetic energy of a baseball (mass 5.25 oz) with a velocity of 1.0 102 mi/h.
18. Calculate the kinetic energy of a 1.0 105-g object with a velocity of 2.0 105 cm/s.
19. Which has the greater kinetic energy, an object with a mass of 2.0 kg and a velocity of 1.0 m/s or an object with a mass of 1.0 kg and a velocity of 2.0 m/s?
20. Consider the accompanying diagram. Ball A is allowed to fall and strike ball B. Assume that all of ball A’s energy is transferred to ball B, at point I, and that there is no loss of energy to other sources. What is the kinetic energy and the potential energy of ball B at point II? The potential energy is given by PE mgz, where m is the mass in kilograms, g is the gravitational constant (9.81 m/s2), and z is the distance in meters.
Heat and Work 21. Calculate E for each of the following.
a. q 47 kJ, w 88 kJ b. q 82 kJ, w 47 kJ c. q 47 kJ, w 0 d. In which of these cases do the surroundings do work on the system?
22. A system undergoes a process consisting of the following two steps: Step 1: The system absorbs 72 J of heat while 35 J of work is done on it.
B 4.00 kg I II 10.0 m 3.00 m A 2.00 kg Exercises 267 Step 2: The system absorbs 35 J of heat while performing 72 J of work.
Calculate E for the overall process.
23. If the internal energy of a thermodynamic system is increased by 300. J while 75 J of expansion work is done, how much heat was transferred and in which direction, to or from the system?
24. Calculate the internal energy change for each of the following.
a. One hundred (100.) joules of work are required to compress a gas. At the same time, the gas releases 23 J of heat.
b. A piston is compressed from a volume of 8.30 L to 2.80 L against a constant pressure of 1.90 atm. In the process, there is a heat gain by the system of 350. J.
c. A piston expands against 1.00 atm of pressure from 11.2 L to 29.1 L. In the process, 1037 J of heat is absorbed.
25. A sample of an ideal gas at 15.0 atm and 10.0 L is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature. Calculate the work in units of kJ for the gas expansion. (Hint: Boyle’s law applies.) 26. A piston performs work of 210. L atm on the surroundings, while the cylinder in which it is placed expands from 10. L to 25 L.
At the same time, 45 J of heat is transferred from the surround-ings to the system. Against what pressure was the piston working?
27. Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650. torr if all the energy of combustion is converted into work to push back the piston?
28. As a system increases in volume, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.500 atm. The final volume of the system is 58.0 L. What was the initial volume of the system if the internal energy of the system decreased by 102.5 J?
29. A balloon filled with 39.1 mol helium has a volume of 876 L at 0.0C and 1.00 atm pressure. The temperature of the balloon is increased to 38.0C as it expands to a volume of 998 L, the pres-sure remaining constant. Calculate q, w, and E for the helium in the balloon. (The molar heat capacity for helium gas is .) 30. One mole of H2O(g) at 1.00 atm and 100.C occupies a volume of 30.6 L. When one mole of H2O(g) is condensed to one mole of H2O(l) at 1.00 atm and 100.C, 40.66 kJ of heat is released.
If the density of H2O(l) at this temperature and pressure is 0.996 g/cm3, calculate E for the condensation of one mole of water at 1.00 atm and 100.C.
Properties of Enthalpy 31. One of the components of polluted air is NO. It is formed in the high-temperature environment of internal combustion engines by the following reaction: Why are high temperatures needed to convert N2 and O2 to NO?
N21g2 O21g2 ¡ 2NO1g2 ¢H 180 kJ 20.8 J/°C mol 32. The reaction is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is 227 kJ. In designing a sul-furic acid plant, is it necessary to provide for heating or cooling of the reaction mixture? Explain.
33. Are the following processes exothermic or endothermic?
a. When solid KBr is dissolved in water, the solution gets colder.
b. Natural gas (CH4) is burned in a furnace.
c. When concentrated H2SO4 is added to water, the solution gets very hot.
d. Water is boiled in a teakettle.
34. Are the following processes exothermic or endothermic?
a. the combustion of gasoline in a car engine b. water condensing on a cold pipe c.
d.
35. The overall reaction in a commercial heat pack can be repre-sented as a. How much heat is released when 4.00 mol iron is reacted with excess O2?
b. How much heat is released when 1.00 mol Fe2O3 is produced?
c. How much heat is released when 1.00 g iron is reacted with excess O2?
d. How much heat is released when 10.0 g Fe and 2.00 g O2 are reacted?
36. Consider the following reaction: a. How much heat is evolved for the production of 1.00 mol of H2O(l)?
b. How much heat is evolved when 4.03 g of hydrogen is re-acted with excess oxygen?
c. How much heat is evolved when 186 g of oxygen is reacted wih excess hydrogen?
d. The total volume of hydrogen gas needed to fill the Hinden-burg was 2.0 108 L at 1.0 atm and 25C. How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?
37. Consider the combustion of propane: Assume that all the heat in Sample Exercise 6.3 comes from the combustion of propane. What mass of propane must be burned to furnish this amount of energy assuming the heat transfer process is 60.% efficient?
38. Consider the following reaction: Calculate the enthalpy change for each of the following cases: a. 1.00 g methane is burned in excess oxygen.
b. 1.00 103 L methane gas at 740. torr and 25C is burned in excess oxygen.
CH41g2 2O21g2 ¡ CO21g2 2H2O1l2 ¢H 891 kJ C3H81g2 5O21g2 ¡ 3CO21g2 4H2O1l2 ¢H 2221 kJ 2H21g2 O21g2 ¡ 2H2O1l2 ¢H 572 kJ 4Fe1s2 3O21g2 ¡ 2Fe2O31s2 ¢H 1652 kJ F21g2 ¡ 2F1g2 CO21s2 ¡ CO21g2 SO31g2 H2O1l2 ¡ H2SO41aq2 268 Chapter Six Thermochemistry 39. For the process at 298 K and 1.0 atm, H is more positive than E by 2.5 kJ/mol. What does the 2.5 kJ/mol quantity represent?
40. For the following reactions at constant pressure, predict if H E, H E, or H E.
a.
b.
c.
Calorimetry and Heat Capacity 41. Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of 25.0 g of the substance from 15.0C to 37.0C? Calculate the energy.
Which substance in Table 6.1 has the largest temperature change when 550. g of the substance absorbs 10.7 kJ of energy?
Calculate the temperature change.
42. The specific heat capacity of silver is a. Calculate the energy required to raise the temperature of 150.0 g Ag from 273 K to 298 K.
b. Calculate the energy required to raise the temperature of 1.0 mol Ag by 1.0C (called the molar heat capacity of silver).
c. It takes 1.25 kJ of energy to heat a sample of pure silver from 12.0C to 15.2C. Calculate the mass of the sample of silver.
43. A 5.00-g sample of one of the substances listed in Table 6.1 was heated from 25.2C to 55.1C, requiring 133 J to do so. What substance was it?
44. It takes 585 J of energy to raise the temperature of 125.6 g mer-cury from 20.0C to 53.5C. Calculate the specific heat capacity and the molar heat capacity of mercury.
45. A 30.0-g sample of water at 280. K is mixed with 50.0 g of water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.
46. A biology experiment requires the preparation of a water bath at 37.0C (body temperature). The temperature of the cold tap water is 22.0C, and the temperature of the hot tap water is 55.0C. If a student starts with 90.0 g of cold water, what mass of hot water must be added to reach 37.0C?
47. A 5.00-g sample of aluminum pellets (specific heat capacity 0.89 J/C g) and a 10.00-g sample of iron pellets (specific heat capacity 0.45 J/C g) are heated to 100.0C. The mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.0C. Calculate the final temperature of the metal and wa-ter mixture, assuming no heat loss to the surroundings.
48. Hydrogen gives off 120. J/g of energy when burned in oxy-gen, and methane gives off 50. J/g under the same circum-stances. If a mixture of 5.0 g of hydrogen and 10. g of methane is burned, and the heat released is transferred to 50.0 g of water at 25.0C, what final temperature will be reached by the water?
49. A 150.0-g sample of a metal at 75.0C is added to 150.0 g of H2O at 15.0C. The temperature of the water rises to 18.3C.
Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.
0.24 J/°C g.
4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 N21g2 3H21g2 ¡ 2NH31g2 2HF1g2 ¡ H21g2 F21g2 H2O1l2 ¡ H2O1g2 50. A 110.-g sample of copper (specific heat capacity 0.20 J/C g) is heated to 82.4C and then placed in a container of water at 22.3C. The final temperature of the water and copper is 24.9C.
What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?
51. In a coffee-cup calorimeter, 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed to yield the following reaction: The two solutions were initially at 22.60C, and the final tem-perature is 23.40C. Calculate the heat that accompanies this reaction in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 g and a specific heat capacity of 4.18 J/C g.
52. In a coffee-cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00C. After dis-solution of the salt, the final temperature of the calorimeter con-tents is 23.34C. Assuming the solution has a heat capacity of 4.18 J/C g and assuming no heat loss to the calorimeter, cal-culate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.
53. Consider the dissolution of CaCl2: An 11.0-g sample of CaCl2 is dissolved in 125 g of water, with both substances at 25.0C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assum-ing the solution has a specific heat capacity of 4.18 J/C g.
54. Consider the reaction Calculate the heat when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.100 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0C and that the final mixture has a mass of 400.0 g and a specific heat capacity of 4.18 J/C g, calculate the final temperature of the mixture.
55. The heat capacity of a bomb calorimeter was determined by burn-ing 6.79 g of methane (energy of combustion 802 kJ/mol CH4) in the bomb. The temperature changed by 10.8C.
a. What is the heat capacity of the bomb?
b. A 12.6-g sample of acetylene, C2H2, produced a temperature increase of 16.9C in the same calorimeter. What is the en-ergy of combustion of acetylene (in kJ/mol)?
56. A 0.1964-g sample of quinone (C6H4O2) is burned in a bomb calorimeter that has a heat capacity of 1.56 kJ/C. The temper-ature of the calorimeter increases by 3.2C. Calculate the energy of combustion of quinone per gram and per mole.
Hess’s Law 57. The enthalpy of combustion of solid carbon to form carbon diox-ide is 393.7 kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is 283.3 kJ/mol CO.
Use these data to calculate H for the reaction 2C1s2 O21g2 ¡ 2CO1g2 ¢H 118 kJ 2HCl1aq2 Ba1OH221aq2 ¡ BaCl21aq2 2H2O1l2 CaCl21s2 ¡ Ca21aq2 2Cl1aq2 ¢H 81.5 kJ Ag1aq2 Cl1aq2 ¡ AgCl1s2 Exercises 269 58. Combustion reactions involve reacting a substance with oxygen.
When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (2341 kJ/mol), C4H8 (2755 kJ/mol), and H2 (286 kJ/mol), calculate H for the reaction 59. Given the following data calculate H for the reaction On the basis of the enthalpy change, is this a useful reaction for the synthesis of ammonia?
60. Given the following data calculate H for the reaction 61. Given the following data calculate H for the reaction 62. The bombardier beetle uses an explosive discharge as a de-fensive measure. The chemical reaction involved is the oxida-tion of hydroquinone by hydrogen peroxide to produce quinone and water: Calculate H for this reaction from the following data: 63. Given the following data calculate H for the reaction CaC21s2 2H2O1l2 ¡ Ca1OH221aq2 C2H21g2 C1graphite2 O21g2 ¡ CO21g2 ¢H 393.5 kJ C2H21g2 5 2O21g2 ¡ 2CO21g2 H2O1l2 ¢H 1300. kJ CaO1s2 H2O1l2 ¡ Ca1OH221aq2 ¢H 653.1 kJ Ca1s2 1 2O21g2 ¡ CaO1s2 ¢H 635.5 kJ Ca1s2 2C1graphite2 ¡ CaC21s2 ¢H 62.8 kJ H2O1g2 ¡ H2O1l2 ¢H 43.8 kJ H21g2 1 2O21g2 ¡ H2O1g2 ¢H 241.8 kJ H21g2 O21g2 ¡ H2O21aq2 ¢H 191.2 kJ ¢H 177.4 kJ C6H41OH221aq2 ¡ C6H4O21aq2 H21g2 C6H41OH221aq2 H2O21aq2 ¡ C6H4O21aq2 2H2O1l2 NO1g2 O1g2 ¡ NO21g2 NO1g2 O31g2 ¡ NO21g2 O21g2 ¢H 199 kJ O21g2 ¡ 2O1g2 ¢H 495 kJ 2O31g2 ¡ 3O21g2 ¢H 427 kJ ClF1g2 F21g2 ¡ ClF31g2 2F21g2 O21g2 ¡ 2F2O1g2 ¢H 43.4 kJ 2ClF31g2 2O21g2 ¡ Cl2O1g2 3F2O1g2 ¢H 341.4 kJ 2ClF1g2 O21g2 ¡ Cl2O1g2 F2O1g2 ¢H 167.4 kJ 2N21g2 6H2O1g2 ¡ 3O21g2 4NH31g2 2H21g2 O21g2 ¡ 2H2O1g2 ¢H 484 kJ NH31g2 ¡ 1 2N21g2 3 2 H21g2 ¢H 46 kJ C4H41g2 2H21g2 ¡ C4H81g2 64. Given the following data calculate H for the reaction Standard Enthalpies of Formation 65. Give the definition of the standard enthalpy of formation for a substance. Write separate reactions for the formation of NaCl, H2O, C6H12O6, and PbSO4 that have H values equal to Hf for each compound.
66. Write reactions for which the enthalpy change will be a. H f for solid aluminum oxide.
b. The standard enthalpy of combustion of liquid ethanol, C2H5OH(l).
c. The standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid.
d. Hf for gaseous vinyl chloride, C2H3Cl(g).
e. The enthalpy of combustion of liquid benzene, C6H6(l).
f. The enthalpy of solution of solid ammonium bromide.
67. Use the values of H f in Appendix 4 to calculate H for the following reactions.
a.
b.
c.
68. Use the values of H f in Appendix 4 to calculate H for the following reactions. (See Exercise 67.) a.
b.
c.
69. The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: a. Use the values of H f in Appendix 4 to calculate the value of H for each of the preceding reactions.
b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations.
(Water is also a product.) Is the overall reaction exothermic or endothermic? 3NO21g2 H2O1l2 ¡ 2HNO31aq2 NO1g2 2NO1g2 O21g2 ¡ 2NO21g2 4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 MgO1s2 H2O1l2 ¡ Mg1OH221s2 SiCl41l2 2H2O1l2 ¡ SiO21s2 4HCl1aq2 + + (l) (g) (g) (g) NH31g2 HCl1g2 ¡ NH4Cl1s2 Ca31PO4221s2 3H2SO41l2 ¡ 3CaSO41s2 2H3PO41l2 + + + (g) (g) (g) (g) (g) N H O C P4O101s2 6PCl51g2 ¡ 10Cl3PO1g2 PCl31g2 1 2O21g2 ¡ Cl3PO1g2 ¢H 285.7 kJ PCl31g2 Cl21g2 ¡ PCl51g2 ¢H 84.2 kJ P41s2 5O21g2 ¡ P4O101s2 ¢H 2967.3 kJ P41s2 6Cl21g2 ¡ 4PCl31g2 ¢H 1225.6 kJ 270 Chapter Six Thermochemistry 70. Calculate H for each of the following reactions using the data in Appendix 4: Explain why a water or carbon dioxide fire extinguisher might not be effective in putting out a sodium fire.
71. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction is Calculate H for this reaction.
72. The space shuttle orbiter utilizes the oxidation of methylhy-drazine by dinitrogen tetroxide for propulsion: Calculate H for this reaction.
73. Consider the reaction Calculate Hf for ClF3(g).
74. The standard enthalpy of combustion of ethene gas, C2H4(g), is 1411.1 kJ/mol at 298 K. Given the following enthalpies of formation, calculate Hf for C2H4(g).
Energy Consumption and Sources 75. Ethanol (C2H5OH) has been proposed as an alternative fuel.
Calculate the standard of enthalpy of combustion per gram of liquid ethanol.
76. Methanol (CH3OH) has also been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid methanol and compare this answer to that for ethanol in Exercise 75.
77. Some automobiles and buses have been equipped to burn propane (C3H8). Compare the amounts of energy that can be obtained per gram of C3H8(g) and per gram of gasoline, assuming that gaso-line is pure octane, C8H18(l). (See Sample Exercise 6.11.) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?
78. Acetylene (C2H2) and butane (C4H10) are gaseous fuels with en-thalpies of combustion of 49.9 kJ/g and 49.5 kJ/g, respectively.
Compare the energy available from the combustion of a given vol-ume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.
79. Assume that 4.19 106 kJ of energy is needed to heat a home.
If this energy is derived from the combustion of methane (CH4), what volume of methane, measured at STP, must be burned? ( Hcombustion for CH4 891 kJ/mol) H2O1l2 285.8 kJ/mol CO21g2 393.5 kJ/mol ¢H° 1196 kJ 2ClF31g2 2NH31g2 ¡ N21g2 6HF1g2 Cl21g2 4N2H3CH31l2 5N2O41l2 ¡ 12H2O1g2 9N21g2 4CO21g2 ¡ Al2O31s2 AlCl31s2 3NO1g2 6H2O1g2 3Al1s2 3NH4ClO41s2 2Na1s2 CO21g2 ¡ Na2O1s2 CO1g2 2Na1s2 2H2O1l2 ¡ 2NaOH1aq2 H21g2 4Na1s2 O21g2 ¡ 2Na2O1s2 80. The complete combustion of acetylene, C2H2(g), produces 1300. kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal of water by 10.0C if the process is 80.0% efficient? Assume the density of water is 1.00 g/cm3.
Additional Exercises 81. Three gas-phase reactions were run in a constant-pressure piston apparatus as illustrated below. For each reaction, give the bal-anced reaction and predict the sign of w (the work done) for the reaction.
If just the balanced reactions were given, how could you predict the sign of w for a reaction?
82. Consider the following changes: a.
b.
c.
d.
e.
At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?
I21s2 ¡ I21g2 2CH3OH1l2 3O21g2 ¡ 2CO21g2 4H2O1l2 Ca3P21s2 6H2O1l2 ¡ 3Ca1OH221s2 2PH31g2 CO1g2 H2O1g2 ¡ H21g2 CO21g2 N21g2 ¡ N21l2 1 atm O N 1 atm c.
1 atm Cl C O b.
1 atm 1 atm S O a.
1 atm Challenge Problems 271 83. Consider the following cyclic process carried out in two steps on a gas: Step 1: 45 J of heat is added to the gas, and 10. J of expansion work is performed.
Step 2: 60. J of heat is removed from the gas as the gas is com-pressed back to the initial state.
Calculate the work for the gas compression in Step 2.
84. Calculate H for the reaction A 5.00-g chunk of potassium is dropped into 1.00 kg water at 24.0C. What is the final temperature of the water after the pre-ceding reaction occurs? Assume that all the heat is used to raise the temperature of the water. (Never run this reaction. It is very dangerous; it bursts into flame!) 85. The enthalpy of neutralization for the reaction of a strong acid with a strong base is 56 kJ/mol of water produced. How much energy will be released when 200.0 mL of 0.400 M HCl is mixed with 150.0 mL of 0.500 M NaOH?
86. When 1.00 L of 2.00 M Na2SO4 solution at 30.0C is added to 2.00 L of 0.750 M Ba(NO3)2 solution at 30.0C in a calorime-ter, a white solid (BaSO4) forms. The temperature of the mixture increases to 42.0C. Assuming that the specific heat capacity of the solution is 6.37 J/C g and that the density of the final solution is 2.00 g/mL, calculate the enthalpy change per mole of BaSO4 formed.
87. If a student performs an endothermic reaction in a calorimeter, how does the calculated value of H differ from the actual value if the heat exchanged with the calorimeter is not taken into account?
88. In a bomb calorimeter, the reaction vessel is surrounded by wa-ter that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains 1.00 kg water and has a total heat capacity of 10.84 kJ/C, what is the heat capacity of the calorimeter components?
89. The bomb calorimeter in Exercise 88 is filled with 987 g of wa-ter. The initial temperature of the calorimeter contents is 23.32C.
A 1.056-g sample of benzoic acid ( Ecomb 26.42 kJ/g) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?
90. Given the following data calculate H for the reaction 91. At 298 K, the standard enthalpies of formation for C2H2(g) and C6H6(l) are 227 kJ/mol and 49 kJ/mol, respectively.
a. Calculate H for C6H61l2 ¡ 3C2H21g2 FeO1s2 CO1g2 ¡ Fe1s2 CO21g2 Fe3O41s2 CO1g2 ¡ 3FeO1s2 CO21g2 ¢H° 18 kJ 3Fe2O31s2 CO1g2 ¡ 2Fe3O41s2 CO21g2 ¢H° 39 kJ Fe2O31s2 3CO1g2 ¡ 2Fe1s2 3CO21g2 ¢H° 23 kJ 2K1s2 2H2O1l2 ¡ 2KOH1aq2 H21g2 b. Both acetylene (C2H2) and benzene (C6H6) can be used as fu-els. Which compound would liberate more energy per gram when combusted in air?
92. Using the following data, calculate the standard heat of forma-tion of ICl(g) in kJ/mol: 93. Calculate H for each of the following reactions, which occur in the atmosphere.
a.
b.
c.
d.
Challenge Problems 94. Consider 2.00 mol of an ideal gas that is taken from state A (PA 2.00 atm, VA 10.0 L) to state B (PB 1.00 atm, VB 30.0 L) by two different pathways: These pathways are summarized on the following graph of P versus V: Calculate the work (in units of J) associated with the two path-ways. Is work a state function? Explain.
95. Combustion of table sugar produces CO2(g) and H2O(l). When 1.46 g of table sugar is combusted in a constant-volume (bomb) calorimeter, 24.00 kJ of heat is liberated.
a. Assuming that table sugar is pure sucrose, C12H22O11(s), write the balanced equation for the combustion reaction.
P (atm) 0 V (L) A B C 1 4 D 2 3 1 2 10 20 30 a VD 10.0 L PD 1.00 atmb a VB 30.0 L PB 1.00 atmb a VA 10.0 L PA 2.00 atmb a VC 30.0 L PC 2.00 atmb 2NO1g2 O21g2 ¡ 2NO21g2 SO31g2 H2O1l2 ¡ H2SO41aq2 O31g2 NO1g2 ¡ NO21g2 O21g2 C2H41g2 O31g2 ¡ CH3CHO1g2 O21g2 I21s2 ¡ I21g2 ¢H° 62.8 kJ ICl1g2 ¡ I1g2 Cl1g2 ¢H° 211.3 kJ I21g2 ¡ 2I1g2 ¢H° 151.0 kJ Cl21g2 ¡ 2Cl1g2 ¢H° 242.3 kJ 88n 1 88n 2 88n 3 88n 4 State A State B 272 Chapter Six Thermochemistry b. Calculate E in kJ/mol C12H22O11 for the combustion reac-tion of sucrose.
c. Calculate H in kJ/mol C12H22O11 for the combustion reac-tion of sucrose at 25C.
96. The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt 1 J/s). The plants in an agricul-tural field produce the equivalent of 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha 10,000 m2). Assuming that sucrose is produced by the reaction calculate the percentage of sunlight used to produce the su-crose—that is, determine the efficiency of photosynthesis.
97. The best solar panels currently available are about 13% efficient in converting sunlight to electricity. A typical home will use about 40. kWh of electricity per day (1 kWh 1 kilowatt hour; 1 kW 1000 J/s). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home’s electricity. (See Exercise 96 for the energy rate supplied by the sun.) 98. On Easter Sunday, April 3, 1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: a. Calculate H for this reaction. Approximately 2.0 104 gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0% HNO3 by mass with a density of 1.42 g/cm3. How much sodium carbonate was required for complete neutralization of the spill, and how much heat was evolved? ( Hf for NaNO3(aq) 467 kJ/mol) b. According to The Denver Post for April 4, 1983, authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of H, what was their major concern?
99. A piece of chocolate cake contains about 400 Calories. A nutri-tional Calorie is equal to 1000 calories (thermochemical calo-ries), which is equal to 4.184 kJ. How many 8-in-high steps must a 180-lb man climb to expend the 400 Cal from the piece of cake? See Exercise 20 for the formula for potential energy.
100. The standard enthalpy of formation of H2O(l) at 298 K is 285.8 kJ/mol. Calculate the change in internal energy for the following process at 298 K and 1 atm: (Hint: Using the ideal gas equation, derive an expression for work in terms of n, R, and T.) 101. You have a 1.00-mol sample of water at 30.C and you heat it until you have gaseous water at 140.C. Calculate q for the entire process. Use the following data. H2O1l2 ¡ H2O1g2 ¢Hvaporization 40.7 kJ mol 1at 100.°C2 H2O1s2 ¡ H2O1l2 ¢Hfusion 6.02 kJ mol 1at 0°C2 Specific heat capacity of steam 2.02 J °C g Specific heat capacity of water 4.18 J °C g Specific heat capacity of ice 2.03 J °C g H2O1l2 ¡ H21g2 1 2O21g2 ¢E° ?
2NaNO31aq2 H2O1l2 CO21g2 ¡ 2HNO31aq2 Na2CO31s2 ¢H 5640 kJ 12CO21g2 11H2O1l2 ¡ C12H22O111s2 12O21g2 102. A 500.0-g sample of an element at 195C is dropped into an ice–water mixture; 109.5 g of ice melts and an ice–water mix-ture remains. Calculate the specific heat of the element. See Exercise 101 for pertinent information.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
103. The preparation of NO2(g) from N2(g) and O2(g) is an en-dothermic reaction: The enthalpy change of reaction for the balanced equation (with lowest whole-number coefficients) is H 67.7 kJ. If 2.50 102 mL of N2(g) at 100.C and 3.50 atm and 4.50 102 mL of O2(g) at 100.C and 3.50 atm are mixed, what amount of heat is necessary to synthesize NO2(g)?
104. Nitromethane, CH3NO2, can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly a. The standard enthalpy change of reaction ( Hrxn) for the bal-anced reaction (with lowest whole-number coefficients) is 1288.5 kJ. Calculate the H f for nitromethane.
b. A 15.0-L flask containing a sample of nitromethane is filled with O2 and the flask is heated to 100.ºC. At this tempera-ture, and after the reaction is complete, the total pressure of all the gases inside the flask is 950. torr. If the mole frac-tion of nitrogen ( ) is 0.134 after the reaction is com-plete, what mass of nitrogen was produced?
105. A cubic piece of uranium metal (specific heat capacity ) at 200.0C is dropped into 1.00 L of deuterium oxide (“heavy water,” specific heat capacity at 25.5C. The final temperature of the uranium and deuterium oxide mixture is 28.5C. Given the densities of uranium (19.05 g/cm3) and deuterium oxide (1.11 g/mL), what is the edge length of the cube of uranium?
Marathon Problems These problems are designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
106. A sample consisting of 22.7 g of a nongaseous, unstable com-pound X is placed inside a metal cylinder with a radius of 8.00 cm, and a piston is carefully placed on the surface of the compound so that, for all practical purposes, the distance be-tween the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is plugged so that nothing in- 4.211 J/°C g2 0.117 J/°C g xnitrogen CH3NO21l2 O21g2 ¡ CO21g2 N21g2 H2O1g2 N21g2 O21g2 ¡ NO21g2 1unbalanced2 Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
Marathon Problems 273 side the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in 10.00 kg of water at 25.00C. The baro-metric pressure is 778 torr.
When the compound spontaneously decomposes, the pis-ton moves up, the temperature of the water reaches a maxi-mum of 29.52C, and then it gradually decreases as the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is 59.8 cm. Chemical analysis shows that the cylinder contains 0.300 mol carbon dioxide, 0.250 mol liquid water, 0.025 mol oxygen gas, and an undetermined amount of a gaseous element A.
It is known that the enthalpy change for the decomposition of X, according to the reaction described above, is 1893 kJ/mol X. The standard enthalpies of formation for gaseous carbon diox-ide and liquid water are 393.5 kJ/mol and 286 kJ/mol, respectively. The heat capacity for water is 4.184 J/C g. The conversion factor between L atm and J can be determined from the two values for the gas constant R, namely, 0.08206 L atm/mol K and 8.3145 J/mol K. The vapor pressure of water at 29.5C is 31 torr. Assume that the heat capacity of the piston-and-cylinder apparatus is negligible and that the piston has negligible mass.
Given the preceding information, determine a. The formula for X.
b. The pressure–volume work (in kJ) for the decomposition of the 22.7-g sample of X.
c. The molar change in internal energy for the decomposition of X and the approximate standard enthalpy of formation for X.
107. A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine H f for the hydrocarbon: Density of CO2 and H2O product mixture at 1.00 atm, 200.C 0.751g/L The density of the hydrocarbon is less than the density of Kr at the same conditions.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at College.hmco.com/ PIC/Zumdahl7e.
¢H° f 1H2O2 242 kJ mol ¢H° f 1CO22 393.5 kJ mol ¢H° rxn 2044.5 kJ mol hydrocarbon 274 7 Atomic Structure and Periodicity Contents 7.1 Electromagnetic Radiation 7.2 The Nature of Matter • The Photoelectric Effect 7.3 The Atomic Spectrum of Hydrogen 7.4 The Bohr Model 7.5 The Quantum Mechanical Model of the Atom • The Physical Meaning of a Wave Function 7.6 Quantum Numbers 7.7 Orbital Shapes and Energies 7.8 Electron Spin and the Pauli Principle 7.9 Polyelectronic Atoms 7.10 The History of the Periodic Table 7.11 The Aufbau Principle and the Periodic Table 7.12 Periodic Trends in Atomic Properties • Ionization Energy • Electron Affinity • Atomic Radius 7.13 The Properties of a Group: The Alkali Metals • Information Contained in the Periodic Table • The Alkali Metals Light refracted through a prism.
In the past 200 years, a great deal of experimental evidence has accumulated to sup-port the atomic model. This theory has proved to be both extremely useful and physically reasonable. When atoms were first suggested by the Greek philosophers Democritus and Leucippus about 400 B.C., the concept was based mostly on intuition. In fact, for the fol-lowing 20 centuries, no convincing experimental evidence was available to support the existence of atoms. The first real scientific data were gathered by Lavoisier and others from quantitative measurements of chemical reactions. The results of these stoichiomet-ric experiments led John Dalton to propose the first systematic atomic theory. Dalton’s theory, although crude, has stood the test of time extremely well.
Once we came to “believe in” atoms, it was logical to ask: What is the nature of an atom? Does an atom have parts, and if so, what are they? In Chapter 2 we considered some of the experiments most important for shedding light on the nature of the atom. Now we will see how the atomic theory has evolved to its present state.
One of the most striking things about the chemistry of the elements is the periodic repetition of properties. There are several groups of elements that show great similarities in chemical behavior. As we saw in Chapter 2, these similarities led to the development of the periodic table of the elements. In this chapter we will see that the modern theory of atomic structure accounts for periodicity in terms of the electron arrangements in atoms.
However, before we examine atomic structure, we must consider the revolution that took place in physics in the first 30 years of the twentieth century. During that time, ex-periments were carried out, the results of which could not be explained by the theories of classical physics developed by Isaac Newton and many others who followed him. A rad-ical new theory called quantum mechanics was developed to account for the behavior of light and atoms. This “new physics” provides many surprises for humans who are used to the macroscopic world, but it seems to account flawlessly (within the bounds of nec-essary approximations) for the behavior of matter.
As the first step in our exploration of this revolution in science we will consider the properties of light, more properly called electromagnetic radiation.
7.1 Electromagnetic Radiation One of the ways that energy travels through space is by electromagnetic radiation. The light from the sun, the energy used to cook food in a microwave oven, the X rays used by dentists, and the radiant heat from a fireplace are all examples of electromagnetic radiation. Although these forms of radiant energy seem quite different, they all exhibit the same type of wavelike behavior and travel at the speed of light in a vacuum.
Waves have three primary characteristics: wavelength, frequency, and speed. Wave-length (symbolized by the lowercase Greek letter lambda, ) is the distance between two consecutive peaks or troughs in a wave, as shown in Fig. 7.1. The frequency (symbol-ized by the lowercase Greek letter nu, ) is defined as the number of waves (cycles) per second that pass a given point in space. Since all types of electromagnetic radiation travel at the speed of light, short-wavelength radiation must have a high frequency. You can see this in Fig. 7.1, where three waves are shown traveling between two points at constant speed. Note that the wave with the shortest wavelength ( 3) has the highest frequency and the wave with the longest wavelength ( 1) has the lowest frequency. This implies an inverse relationship between wavelength and frequency, that is, 1 , or ln c r 275 Wavelength and frequency are inversely related.
c speed of light 2.9979 108 m/s 276 Chapter Seven Atomic Structure and Periodicity where is the wavelength in meters, is the frequency in cycles per second, and c is the speed of light (2.9979108 m/s). In the SI system, cycles is understood, and the unit per second becomes 1/s, or s1, which is called the hertz (abbreviated Hz).
Electromagnetic radiation is classified as shown in Fig. 7.2. Radiation provides an important means of energy transfer. For example, the energy from the sun reaches the earth mainly in the form of visible and ultraviolet radiation, whereas the glowing coals of a fireplace transmit heat energy by infrared radiation. In a microwave oven the wa-ter molecules in food absorb microwave radiation, which increases their motions. This energy is then transferred to other types of molecules via collisions, causing an increase in the food’s temperature. As we proceed in the study of chemistry, we will consider many of the classes of electromagnetic radiation and the ways in which they affect matter.
1 second λ1 ν1 = 4 cycles/second = 4 hertz ν2 = 8 cycles/second = 8 hertz λ2 λ3 ν3 = 16 cycles/second = 16 hertz Although the waves associated with light are not obvious to the naked eye, ocean waves provide a familiar source of recreation.
FIGURE 7.1 The nature of waves. Note that the radia-tion with the shortest wavelength has the highest frequency.
10 –12 10 –10 10 –8 7 × 10 –7 10 –4 10 –2 1 102 104 4 × 10 –7 Gamma rays X rays Ultraviolet Visible Infrared Microwaves Radio waves FM Shortwave AM 4 × 10 –7 5 × 10 –7 6 × 10 –7 7 × 10 –7 Wavelength in meters FIGURE 7.2 Classification of electromagnetic radiation.
Spectrum adapted by permission from C. W. Keenan, D. C. Kleinfelter, and J. H. Wood, General College Chemistry, 6th ed. (New York: Harper & Row, 1980).
Visualization: Electromagnetic Wave 7.2 The Nature of Matter 277 Frequency of Electromagnetic Radiation The brilliant red colors seen in fireworks are due to the emission of light with wavelengths around 650 nm when strontium salts such as Sr(NO3)2 and SrCO3 are heated. (This can be easily demonstrated in the lab by dissolving one of these salts in methanol that contains a little water and igniting the mixture in an evaporating dish.) Calculate the frequency of red light of wavelength 6.50 102 nm.
Solution We can convert wavelength to frequency using the equation where c 2.9979 108 m/s. In this case 6.50 102 nm. Changing the wavelength to meters, we have and See Exercises 7.31 and 7.32.
7.2 The Nature of Matter It is probably fair to say that at the end of the nineteenth century, physicists were feeling rather smug. Theories could explain phenomena as diverse as the motions of the planets and the dispersion of visible light by a prism. Rumor has it that students were being dis-couraged from pursuing physics as a career because it was felt that all the major prob-lems had been solved, or at least described in terms of the current physical theories.
At the end of the nineteenth century, the idea prevailed that matter and energy were distinct. Matter was thought to consist of particles, whereas energy in the form of light (electromagnetic radiation) was described as a wave. Particles were things that had mass and whose position in space could be specified. Waves were described as massless and delocalized; that is, their position in space could not be specified. It also was assumed that there was no intermingling of matter and light. Everything known before 1900 seemed to fit neatly into this view.
n c l 2.9979 108 m/s 6.50 107 m 4.61 1014 s1 4.61 1014 Hz 6.50 102 nm 1 m 109 nm 6.50 107 m ln c or n c l CHEMICAL IMPACT Flies That Dye M editerranean and Mexican fruit flies are formidable pests that have the potential to seriously damage several im-portant fruit crops. Because of this, there have been several widely publicized sprayings of residential areas in southern California with the pesticide malathion to try to control fruit flies. Now there may be a better way to kill fruit flies—with a blend of two common dyes (red dye no. 28 and yellow dye no. 8) long used to color drugs and cosmetics. One of the most interesting things about this new pesticide is that it is activated by light. After an insect eats the blend of dyes, the molecules absorb light (through the insect’s transparent body), which causes them to generate oxidizing agents that attack the proteins and cell membranes in the bug’s body.
Death occurs within 12 hours.
The sunlight that turns on the dye’s toxicity after the fly ingests it also degrades the dye in the environment, making it relatively safe. It appears likely that in the near future the fruit fly will “dye” with little harm to the environment.
Sample Exercise 7.1 When a strontium salt is dissolved in methanol (with a little water) and ignited, it gives a brilliant red flame. The red color is produced by emission of light when electrons, excited by the energy of the burning methanol, fall back to their ground states.
Visualization: Electrified Pickle 278 Chapter Seven Atomic Structure and Periodicity At the beginning of the twentieth century, however, certain experimental results sug-gested that this picture was incorrect. The first important advance came in 1900 from the German physicist Max Planck (1858–1947). Studying the radiation profiles emitted by solid bodies heated to incandescence, Planck found that the results could not be explained in terms of the physics of his day, which held that matter could absorb or emit any quantity of en-ergy. Planck could account for these observations only by postulating that energy can be gained or lost only in whole-number multiples of the quantity h , where h is a constant called Planck’s constant, determined by experiment to have the value 6.626 1034 J s.
That is, the change in energy for a system E can be represented by the equation where n is an integer (1, 2, 3, . . .), h is Planck’s constant, and is the frequency of the electromagnetic radiation absorbed or emitted.
Planck’s result was a real surprise. It had always been assumed that the energy of mat-ter was continuous, which meant that the transfer of any quantity of energy was possible.
Now it seemed clear that energy is in fact quantized and can occur only in discrete units of size h . Each of these small “packets” of energy is called a quantum. A system can transfer energy only in whole quanta. Thus energy seems to have particulate properties.
The Energy of a Photon The blue color in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200C. Then the compound emits blue light having a wavelength of 450 nm. What is the increment of energy (the quantum) that is emitted at 4.50 102 nm by CuCl?
Solution The quantum of energy can be calculated from the equation The frequency for this case can be calculated as follows: So A sample of CuCl emitting light at 450 nm can lose energy only in increments of 4.41 1019 J, the size of the quantum in this case.
See Exercises 7.33 and 7.34.
The next important development in the knowledge of atomic structure came when Al-bert Einstein (see Fig. 7.3) proposed that electromagnetic radiation is itself quantized. Ein-stein suggested that electromagnetic radiation can be viewed as a stream of “particles” called photons. The energy of each photon is given by the expression where h is Planck’s constant, is the frequency of the radiation, and is the wavelength of the radiation.
Ephoton hn hc l ¢E hn 16.626 1034 J s216.66 1014 s12 4.41 1019 J n c l 2.9979 108 m/s 4.50 107 m 6.66 1014 s1 ¢E hn ¢E nhn Energy can be gained or lost only in integer multiples of h.
Planck’s constant 6.626 1034 J s.
When alternating current at 110 volts is applied to a dill pickle, a glowing discharge occurs. The current flowing between the electrodes (forks), which is supported by the Na and Cl ions present, apparently causes some sodium atoms to form in an excited state. When these atoms relax to the ground state, they emit visible light at 589 nm, producing the yellow glow reminiscent of sodium vapor lamps.
Sample Exercise 7.2 Visualization: Photoelectric Effect The Photoelectric Effect Einstein arrived at this conclusion through his analysis of the photoelectric effect (for which he later was awarded the Nobel Prize). The photoelectric effect refers to the phenomenon in which electrons are emitted from the surface of a metal when light strikes it. The following observations characterize the photoelectric effect.
1. Studies in which the frequency of the light is varied show that no electrons are emitted by a given metal below a specific threshold frequency 0.
2. For light with frequency lower than the threshold frequency, no electrons are emitted regardless of the intensity of the light.
3. For light with frequency greater than the threshold frequency, the number of electrons emitted increases with the intensity of the light.
4. For light with frequency greater than the threshold frequency, the kinetic energy, of the emitted electrons increases linearly with the frequency of the light.
These observations can be explained by assuming that electromagnetic radiation is quantized (consists of photons), and that the threshold frequency represents the minimum energy required to remove the electron from the metal’s surface.
Minimum energy required to remove an electron E0 h 0 Because a photon with energy less than E0 ( 0) cannot remove an electron, light with a frequency less than the threshold frequency produces no electrons. On the other hand, for light where 0, the energy in excess of that required to remove the electron is given to the electron as kinetic energy (KE): h h Mass of Velocity Energy of Energy required electron of incident to remove electron electron photon from metal’s surface Because in this picture the intensity of light is a measure of the number of photons pres-ent in a given part of the beam, a greater intensity means that more photons are available to release electrons (as long as 0 for the radiation).
In a related development, Einstein derived the famous equation in his special theory of relativity published in 1905. The main significance of this equation is that energy has mass. This is more apparent if we rearrange the equation in the following form: m Energy Mass Speed of light m E c2 E mc2 KEelectron 1 2my2 hn hn0 FIGURE 7.3 Albert Einstein (1879–1955) was born in Germany. Nothing in his early development suggested genius; even at the age of 9 he did not speak clearly, and his parents feared that he might be handicapped. When asked what profession Einstein should follow, his school principal replied, “It doesn’t matter; he’ll never make a success of anything.” When he was 10, Einstein entered the Luitpold Gymnasium (high school), which was typical of German schools of that time in being harshly disciplinarian. There he developed a deep suspicion of authority and a skepticism that encouraged him to question and doubt—valuable qualities in a scientist. In 1905, while a patent clerk in Switzerland, Einstein published a paper explaining the photoelectric effect via the quantum theory. For this revolutionary thinking he received a Nobel Prize in 1921. Highly regarded by this time, he worked in Germany until 1933, when Hitler’s persecution of the Jews forced him to come to the United States. He worked at the Institute for Advanced Studies in Princeton, New Jersey, until his death in 1955.
Einstein was undoubtedly the greatest physicist of our age. Even if someone else had derived the theory of relativity, his other work would have ensured his ranking as the second greatest physicist of his time. Our concepts of space and time were radically changed by ideas he first proposed when he was 26 years old. From then until the end of his life, he attempted unsuccessfully to find a single unifying theory that would explain all physical events.
h h h A h A 279 280 Chapter Seven Atomic Structure and Periodicity Using this form of the equation, we can calculate the mass associated with a given quantity of energy. For example, we can calculate the apparent mass of a photon. For electromagnetic radiation of wavelength , the energy of each photon is given by the expression Then the apparent mass of a photon of light with wavelength is given by Does a photon really have mass? The answer appears to be yes. In 1922 American physicist Arthur Compton (1892–1962) performed experiments involving collisions of X rays and electrons that showed that photons do exhibit the apparent mass calculated from the preceding equation. However, it is clear that photons do not have mass in the classi-cal sense. A photon has mass only in a relativistic sense—it has no rest mass.
m E c2 hc l c2 h lc Ephoton hc l Note that the apparent mass of a photon depends on its wavelength. The mass of a photon at rest is thought to be zero, although we never observe it at rest.
Light as a wave phenomenon Light as a stream of photons FIGURE 7.4 Electromagnetic radiation exhibits wave properties and particulate properties. The energy of each photon of the radiation is related to the wavelength and frequency by the equation Ephoton h hc.
I n the animal world, the ability to see at night provides predators with a distinct advantage over their prey. The same advantage can be gained by military forces and law enforcement agencies around the world through the use of recent advances in night vision technology.
All types of night vision equipment are electro-optical de-vices that amplify existing light. A lens collects light and fo-cuses it on an image intensifier. The image intensifier is based on the photoelectric effect—materials that give off electrons when light is shined on them. Night vision intensifiers use semiconductor-based materials to produce large numbers of electrons for a given input of photons. The emitted electrons are then directed onto a screen covered with compounds that phosphoresce (glow when struck by electrons). While televi-sion tubes use various phosphors to produce color pictures, night vision devices use phosphors that appear green, because the human eye can distinguish more shades of green than any other color. The viewing screen shows an image that otherwise would be invisible to the naked eye during nighttime viewing.
Current night vision devices use gallium arsenide (GaAs)–based intensifiers that can amplify input light as much as 50,000 times. These devices are so sensitive they can use starlight to produce an image. It is also now possi-ble to use light (infrared) that cannot be sensed with the human eye to create an image.
This technology, while developed originally for military and law enforcement applications, is now becoming avail-able to the consumer. For example, Cadillac included night vision as an option on its cars for the year 2000. As night-imaging technology improves and costs become less prohib-itive, a whole new world is opening up for the technophile— after the sun goes down.
CHEMICAL IMPACT Chemistry That Doesn’t Leave You in the Dark A night vision photo of the midair refueling of a U.S. Air Force plane.
7.2 The Nature of Matter 281 We can summarize the important conclusions from the work of Planck and Einstein as follows: Energy is quantized. It can occur only in discrete units called quanta.
Electromagnetic radiation, which was previously thought to exhibit only wave proper-ties, seems to show certain characteristics of particulate matter as well. This phenom-enon is sometimes referred to as the dual nature of light and is illustrated in Fig. 7.4.
Thus light, which previously was thought to be purely wavelike, was found to have certain characteristics of particulate matter. But is the opposite also true? That is, does matter that is normally assumed to be particulate exhibit wave properties? This question was raised in 1923 by a young French physicist named Louis de Broglie (1892–1987). To see how de Broglie supplied the answer to this question, recall that the relationship between mass and wavelength for electromagnetic radiation is m h c. For a particle with velocity , the corresponding expression is Rearranging to solve for , we have This equation, called de Broglie’s equation, allows us to calculate the wavelength for a particle, as shown in Sample Exercise 7.3.
Calculations of Wavelength Compare the wavelength for an electron (mass 9.11 1031 kg) traveling at a speed of 1.0 107 m/s with that for a ball (mass 0.10 kg) traveling at 35 m/s.
Solution We use the equation hm , where since For the electron, For the ball, See Exercises 7.41 through 7.44.
Notice from Sample Exercise 7.3 that the wavelength associated with the ball is in-credibly short. On the other hand, the wavelength of the electron, although still quite small, happens to be on the same order as the spacing between the atoms in a typical crystal.
This is important because, as we will see presently, it provides a means for testing de Broglie’s equation.
lb 6.626 1034 kg m m s 10.10 kg2135 m/s2 1.9 1034 m le 6.626 1034 kg m m s 19.11 1031 kg211.0 107 m/s2 7.27 1011 m 1 J 1 kg m2/s2 h 6.626 1034 J s or 6.626 1034 kg m2/s l h my m h ly Do not confuse (frequency) with (velocity).
Sample Exercise 7.3 282 Chapter Seven Atomic Structure and Periodicity CHEMICAL IMPACT Thin Is In S ince the beginning of television about 75 years ago, TV sets have been built around cathode ray tubes (CRTs) in which a “gun” fires electrons at a screen containing phosphors (compounds that emit colored light when excited by some en-ergy source). Although CRT televisions produce excellent pic-tures, big-screen TVs are very thick and very heavy. Several new technologies are now being used that reduce the bulk of color monitors. One such approach involves a plasma flat-panel display. As the name suggests, the major advantage of these screens is that they are very thin and relatively light.
All color monitors work by manipulating millions of pixels, each of which contains red, blue, and green color-producing phosphors. By combining these three fundamen-tal colors with various weightings, all colors of the rainbow can be generated, thereby producing color images on the monitor. The various types of monitors differ in the energy source used to excite the phosphors. Whereas a CRT monitor uses an electron gun as the energy source, a plasma monitor uses an applied voltage to produce gas-phase ions and elec-trons, which, when they recombine, emit ultraviolet light.
This light, in turn, excites the phosphors.
Plasma monitors have pixel compartments that contain xenon and neon gas. Each pixel consists of three subpixels: one containing a red phosphor, one with a green phosphor, and one with a blue phosphor. Two perpendicular sets of electrodes define a matrix around the subpixels: Electrodes Electrodes Diffraction results when light is scattered from a regular array of points or lines.
You may have noticed the diffraction of light from the ridges and grooves of a com-pact disc. The colors result because the various wavelengths of visible light are not all scattered in the same way. The colors are “separated,” giving the same effect as light passing through a prism. Just as a regular arrangement of ridges and grooves produces diffraction, so does a regular array of atoms or ions in a crystal, as shown in the photographs below. For example, when X rays are directed onto a crystal of sodium chloride, with its regular array of Na and Cl ions, the scattered radiation produces a diffraction pattern of bright spots and dark areas on a photographic plate, as shown in Fig. 7.5(a). This occurs because the scattered light can interfere constructively (the peaks and troughs of the beams are in phase) to produce a bright spot [Fig. 7.5(b)] or destructively (the peaks and troughs are out of phase) to produce a dark area [Fig. 7.5(c)].
A diffraction pattern can only be explained in terms of waves. Thus this phenom-enon provides a test for the postulate that particles such as electrons have wavelengths.
As we saw in Sample Exercise 7.3, an electron with a velocity of 107 m/s (easily achieved by acceleration of the electron in an electric field) has a wavelength of about 1010 m, which is roughly the distance between the ions in a crystal such as sodium chloride.
This is important because diffraction occurs most efficiently when the spacing between the scattering points is about the same as the wavelength of the wave being diffracted.
Thus, if electrons really do have an associated wavelength, a crystal should diffract electrons. An experiment to test this idea was carried out in 1927 by C. J. Davisson and (top) The pattern produced by electron diffraction of a titanium/nickel alloy. (bottom) Pattern produced by X-ray diffraction of a beryl crystal.
7.2 The Nature of Matter 283 L. H. Germer at Bell Laboratories. When they directed a beam of electrons at a nickel crystal, they observed a diffraction pattern similar to that seen from the diffraction of X rays. This result verified de Broglie’s relationship, at least for electrons. Larger chunks of matter, such as balls, have such small wavelengths (see Sample Exercise 7.3) that they are impossible to verify experimentally. However, we believe that all matter obeys de Broglie’s equation.
Now we have come full circle. Electromagnetic radiation, which at the turn of the twen-tieth century was thought to be a pure waveform, was found to possess particulate properties.
Conversely, electrons, which were thought to be particles, were found to have a wavelength associated with them. The significance of these results is that matter and energy are not distinct.
One set of the electrodes is above the pixels, and the perpendicular set is below the pixels. When the computer managing the image places a voltage difference across a given subpixel, electrons are removed from the xenon and neon atoms present to form a plasma (cations and elec-trons). When the cations recombine with the electrons, pho-tons of light are emitted that are absorbed by the phosphor compound, which then emits red, green, or blue light. By controlling the size of the voltage on a given subpixel, a given pixel can produce a variety of colors. When all of the pixels are excited appropriately, a color image is produced.
The plasma display makes it possible to have a large, yet relatively thin screen. Since each pixel is energized in-dividually, this display looks bright and clear from almost any angle. The main disadvantage of this technology is its relatively high cost. However, as advances are being made, the price is falling significantly. CRT monitors may soon be of interest only to antique collectors.
A plasma display from Sony.
FIGURE 7.5 (a) Diffraction occurs when electromagnetic radiation is scattered from a regular array of objects, such as the ions in a crystal of sodium chloride. The large spot in the center is from the main incident beam of X rays. (b) Bright spots in the diffraction pattern result from constructive interference of waves. The waves are in phase; that is, their peaks match. (c) Dark areas result from destructive interference of waves. The waves are out of phase; the peaks of one wave coincide with the troughs of another wave.
X rays NaCl crystal (a) Detector screen Diffraction pattern on detector screen (front view) Waves in phase (peaks on one wave match peaks on the other wave) Increased intensity (bright spot) Waves out of phase (troughs and peaks coincide) Decreased intensity (dark spot) Constructive interference Destructive interference Trough Peak (b) (c) 284 Chapter Seven Atomic Structure and Periodicity Energy is really a form of matter, and all matter shows the same types of properties. That is, all matter exhibits both particulate and wave properties. Large pieces of matter, such as base-balls, exhibit predominantly particulate properties. The associated wavelength is so small that it is not observed. Very small “bits of matter,” such as photons, while showing some particu-late properties, exhibit predominantly wave properties. Pieces of matter with intermediate mass, such as electrons, show clearly both the particulate and wave properties of matter.
7.3 The Atomic Spectrum of Hydrogen As we saw in Chapter 2, key information about the atom came from several experiments carried out in the early twentieth century, in particular Thomson’s discovery of the elec-tron and Rutherford’s discovery of the nucleus. Another important experiment was the study of the emission of light by excited hydrogen atoms. When a sample of hydrogen gas receives a high-energy spark, the H2 molecules absorb energy, and some of the HOH bonds are broken. The resulting hydrogen atoms are excited; that is, they contain excess energy, which they release by emitting light of various wavelengths to produce what is called the emission spectrum of the hydrogen atom.
To understand the significance of the hydrogen emission spectrum, we must first de-scribe the continuous spectrum that results when white light is passed through a prism, as shown in Fig. 7.6(a). This spectrum, like the rainbow produced when sunlight is – + Prism Slit Continuous spectrum Electric arc (white light source) (a) – + Prism Detector (photographic plate) Hydrogen gas discharge tube (b) High voltage 410 nm 434 nm 486 nm 656 nm Detector (photographic plate) I B G Y O R Arc Slit V A beautiful rainbow.
FIGURE 7.6 (a) A continuous spectrum containing all wavelengths of visible light (indicated by the initial letters of the colors of the rain-bow). (b) The hydrogen line spectrum con-tains only a few discrete wavelengths.
Spectrum adapted by permission from C. W. Keenan, D. C. Kleinfelter, and J. H. Wood, General College Chemistry, 6th ed. (New York: Harper & Row, 1980).
Visualization: Refraction of White Light Visualization: The Line Spectrum of Hydrogen Visualization: Flame Tests 7.4 The Bohr Model 285 dispersed by raindrops, contains all the wavelengths of visible light. In contrast, when the hydrogen emission spectrum in the visible region is passed through a prism, as shown in Fig. 7.6(b), we see only a few lines, each of which corresponds to a discrete wavelength.
The hydrogen emission spectrum is called a line spectrum.
What is the significance of the line spectrum of hydrogen? It indicates that only cer-tain energies are allowed for the electron in the hydrogen atom. In other words, the en-ergy of the electron in the hydrogen atom is quantized. This observation ties in perfectly with the postulates of Max Planck discussed in Section 7.2. Changes in energy between discrete energy levels in hydrogen will produce only certain wavelengths of emitted light, as shown in Fig. 7.7. For example, a given change in energy from a high to a lower level would give a wavelength of light that can be calculated from Planck’s equation: Wavelength of light Change in Frequency of emitted energy light emitted The discrete line spectrum of hydrogen shows that only certain energies are possible; that is, the electron energy levels are quantized. In contrast, if any energy level were allowed, the emission spectrum would be continuous.
7.4 The Bohr Model In 1913, a Danish physicist named Niels Bohr (1885–1962), aware of the experimental results we have just discussed, developed a quantum model for the hydrogen atom. Bohr proposed that the electron in a hydrogen atom moves around the nucleus only in certain allowed circular orbits. He calculated the radii for these allowed orbits by using the the-ories of classical physics and by making some new assumptions.
From classical physics Bohr knew that a particle in motion tends to move in a straight line and can be made to travel in a circle only by application of a force toward the cen-ter of the circle. Thus Bohr reasoned that the tendency of the revolving electron to fly off the atom must be just balanced by its attraction for the positively charged nucleus. But classical physics also decreed that a charged particle under acceleration should radiate en-ergy. Since an electron revolving around the nucleus constantly changes its direction, it is constantly accelerating. Therefore, the electron should emit light and lose energy—and thus be drawn into the nucleus. This, of course, does not correlate with the existence of stable atoms.
Clearly, an atomic model based solely on the theories of classical physics was un-tenable. Bohr also knew that the correct model had to account for the experimental spec-trum of hydrogen, which showed that only certain electron energies were allowed. The experimental data were absolutely clear on this point. Bohr found that his model would fit the experimental results if he assumed that the angular momentum of the electron (an-gular momentum equals the product of mass, velocity, and orbital radius) could occur only in certain increments. It was not clear why this should be true, but with this assumption, Bohr’s model gave hydrogen atom energy levels consistent with the hydrogen emission spectrum. The model is represented pictorially in Fig. 7.8.
¢E hn hc l E Various energy levels in the hydrogen atom ∆E3 = hc λ3 ∆E2 = hc λ2 ∆E1 = hc λ1 FIGURE 7.7 A change between two discrete energy levels emits a photon of light.
m8 m8 m 88 E 4 5 3 2 1 n (a) n = 4 n = 5 n = 3 n = 2 n = 1 (b) Wavelength Line spectrum (c) FIGURE 7.8 Electronic transitions in the Bohr model for the hydrogen atom. (a) An energy-level diagram for elec-tronic transitions. (b) An orbit-transition diagram, which accounts for the experimental spectrum. (Note that the orbits shown are schematic. They are not drawn to scale.) (c) The resulting line spectrum on a photographic plate. Note that the lines in the visible region of the spectrum correspond to transitions from higher levels to the n 2 level.
286 Chapter Seven Atomic Structure and Periodicity Although we will not show the derivation here, the most important equation to come from Bohr’s model is the expression for the energy levels available to the electron in the hydrogen atom: (7.1) in which n is an integer (the larger the value of n, the larger is the orbit radius) and Z is the nuclear charge. Using Equation (7.1), Bohr was able to calculate hydrogen atom en-ergy levels that exactly matched the values obtained by experiment.
The negative sign in Equation (7.1) simply means that the energy of the electron bound to the nucleus is lower than it would be if the electron were at an infinite distance ( ) from the nucleus, where there is no interaction and the energy is zero: The energy of the electron in any orbit is negative relative to this reference state.
Equation (7.1) can be used to calculate the change in energy of an electron when the electron changes orbits. For example, suppose an electron in level n 6 of an excited hy-drogen atom falls back to level n 1 as the hydrogen atom returns to its lowest possible energy state, its ground state. We use Equation (7.1) with Z 1, since the hydrogen nucleus contains a single proton. The energies corresponding to the two states are as follows: Note that for n 1 the electron has a more negative energy than it does for n 6, which means that the electron is more tightly bound in the smallest allowed orbit.
The change in energy E when the electron falls from n 6 to n 1 is The negative sign for the change in energy indicates that the atom has lost energy and is now in a more stable state. The energy is carried away from the atom by the production (emission) of a photon.
The wavelength of the emitted photon can be calculated from the equation where E represents the change in energy of the atom, which equals the energy of the emitted photon. We have Note that for this calculation the absolute value of E is used (we have not included the negative sign). In this case we indicate the direction of energy flow by saying that a photon l hc ¢E 16.626 1034 J s212.9979 108 m/s2 2.117 1018 J 9.383 108 m ¢E h a c lb or l hc ¢E 2.117 1018 J E1 E6 12.178 1018 J2 16.050 1020 J2 ¢E energy of final state energy of initial state For n 1: E1 2.178 1018 Ja12 12b 2.178 1018 J For n 6: E6 2.178 1018 Ja12 62b 6.050 1020 J E 2.178 1018 JaZ 2 q b 0 n q E 2.178 1018 JaZ 2 n2 b The J in Equation (7.1) stands for joules.
Niels Hendrik David Bohr (1885–1962) as a boy lived in the shadow of his younger brother Harald, who played on the 1908 Danish Olympic Soccer Team and later became a distinguished mathematician. In school, Bohr received his poorest marks in composition and struggled with writing during his entire life. In fact, he wrote so poorly that he was forced to dictate his Ph.D. thesis to his mother. Nevertheless, Bohr was a brilliant physicist. After receiv-ing his Ph.D. in Denmark, he constructed a quantum model for the hydrogen atom by the time he was 27. Even though his model later proved to be incorrect, Bohr remained a central figure in the drive to understand the atom. He was awarded the Nobel Prize in physics in 1922.
7.4 The Bohr Model 287 of wavelength 9.383 108 m has been emitted from the hydrogen atom. Simply plug-ging the negative value of E into the equation would produce a negative value for , which is physically meaningless.
Energy Quantization in Hydrogen Calculate the energy required to excite the hydrogen electron from level n 1 to level n 2. Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground state to reach this excited state. Solution Using Equation (7.1) with Z 1, we have The positive value for E indicates that the system has gained energy. The wavelength of light that must be absorbed to produce this change is See Exercises 7.45 and 7.46.
At this time we must emphasize two important points about the Bohr model: 1. The model correctly fits the quantized energy levels of the hydrogen atom and pos-tulates only certain allowed circular orbits for the electron.
2. As the electron becomes more tightly bound, its energy becomes more negative rel-ative to the zero-energy reference state (corresponding to the electron being at infinite distance from the nucleus). As the electron is brought closer to the nucleus, energy is released from the system.
Using Equation (7.1), we can derive a general equation for the electron moving from one level (ninitial) to another level (nfinal): (7.2) Equation (7.2) can be used to calculate the energy change between any two energy levels in a hydrogen atom, as shown in Sample Exercise 7.5.
2.178 1018 Ja 1 nfinal 2 1 ninitial 2b 12.178 1018 J2 a 12 nfinal 2b 12.178 1018 J2a 12 ninitial 2b Efinal Einitial ¢E energy of level nfinal energy of level ninitial 1.216 107 m l hc ¢E 16.626 1034 J s212.9979 108 m/s2 1.633 1018 J ¢E E2 E1 15.445 1019 J2 12.178 1018 J2 1.633 1018 J E2 2.178 1018 Ja12 22b 5.445 1019 J E1 2.178 1018 Ja12 12b 2.178 1018 J Sample Exercise 7.4 After this exercise we will no longer show cancellation marks. However, the same process for canceling units applies throughout this text.
Note from Fig. 7.2 that the light required to produce the transition from the n 1 to n 2 level in hydrogen lies in the ultraviolet region.
288 Chapter Seven Atomic Structure and Periodicity T he art of using mixtures of chemicals to produce explo-sives is an ancient one. Black powder—a mixture of potassium nitrate, charcoal, and sulfur—was being used in China well before 1000 A.D. and has been used subsequently through the centuries in military explosives, in construction blasting, and for fireworks. The DuPont Company, now a major chemical manufacturer, started out as a manufacturer of black powder. In fact, the founder, Eleuthère duPont, learned the manufacturing technique from none other than Lavoisier.
Before the nineteenth century, fireworks were confined mainly to rockets and loud bangs. Orange and yellow colors came from the presence of charcoal and iron filings.
However, with the great advances in chemistry in the nineteenth century, new compounds found their way into fireworks. Salts of copper, strontium, and barium added brilliant colors. Magnesium and aluminum metals gave a dazzling white light. Fireworks, in fact, have changed very little since then.
How do fireworks produce their brilliant colors and loud bangs? Actually, only a handful of different chemicals are responsible for most of the spectacular effects. To produce the noise and flashes, an oxidizer (an oxidizing agent) and a fuel (a reducing agent) are used. A common mixture in-volves potassium perchlorate (KClO4) as the oxidizer and aluminum and sulfur as the fuel. The perchlorate oxidizes the fuel in a very exothermic reaction, which produces a bril-liant flash, due to the aluminum, and a loud report from the rapidly expanding gases produced. For a color effect, an el-ement with a colored emission spectrum is included. Recall that the electrons in atoms can be raised to higher-energy orbitals when the atoms absorb energy. The excited atoms can then release this excess energy by emitting light of spe-cific wavelengths, often in the visible region. In fireworks, the energy to excite the electrons comes from the reaction between the oxidizer and fuel.
Yellow colors in fireworks are due to the 589-nm emis-sion of sodium ions. Red colors come from strontium salts emitting at 606 nm and from 636 to 688 nm. This red color is familiar from highway safety flares. Barium salts give a green color in fireworks, due to a series of emission lines between 505 and 535 nm. A really good blue color, however, is hard to obtain. Copper salts give a blue color, emitting in the 420- to 460-nm region. But difficulties occur because the oxidizing agent, potassium chlorate (KClO3), reacts with copper salts to form copper chlorate, a highly explosive com-pound that is dangerous to store. (The use of KClO3 in fire-works has been largely abandoned because of its explosive hazards.) Paris green, a copper salt containing arsenic, was once used extensively but is now considered to be too toxic.
In recent years the colors produced by fireworks have become more intense because of the formation of metal chlo-rides during the burning process. These gaseous metal chlo-ride molecules produce colors much more brilliant than do the metal atoms by themselves. For example, strontium chlo-ride produces a much brighter red than do strontium atoms.
CHEMICAL IMPACT Fireworks A typical aerial shell used in fireworks displays. Time-delayed fuses cause a shell to explode in stages. In this case a red starburst occurs first, followed by a blue starburst, and finally a flash and loud report.
(Reprinted with permission from Chemical & Engineering News, June 29, 1981, p. 24.
Copyright © 1981, American Chemical Society.) Twine 7.4 The Bohr Model 289 Thus, chlorine-donating compounds are now included in many fireworks shells.
A typical aerial shell is shown in the diagram. The shell is launched from a mortar (a steel cylinder) using black powder as the propellant. Time-delayed fuses are used to fire the shell in stages. A list of chemicals com-monly used in fireworks is given in the table.
Although you might think that the chemistry of fireworks is simple, the achievement of the vivid white flashes and the brilliant colors requires complex combi-nations of chemicals. For example, because the white flashes produce high flame temperatures, the colors tend to wash out. Thus oxidizers such as KClO4 are commonly used with fuels that produce relatively low flame tem-peratures. An added difficulty, however, is that perchlo-rates are very sensitive to accidental ignition and are therefore quite hazardous. Another problem arises from the use of sodium salts. Because sodium produces an ex-tremely bright yellow emission, sodium salts cannot be used when other colors are desired. Carbon-based fuels also give a yellow flame that masks other colors, and this limits the use of organic compounds as fuels. You can see that the manufacture of fireworks that produce the de-sired effects and are also safe to handle requires careful selection of chemicals. And, of course, there is still the dream of a deep blue flame.
Fireworks in Washington, D.C.
Chemicals Commonly Used in the Manufacture of Fireworks Oxidizers Fuels Special Effects Potassium nitrate Aluminum Red flame: strontium nitrate, strontium carbonate Potassium chlorate Magnesium Green flame: barium nitrate, barium chlorate Potassium perchlorate Titanium Blue flame: copper carbonate, copper sulfate, copper oxide Ammonium perchlorate Charcoal Yellow flame: sodium oxalate, cryolite (Na3AlF6) Barium nitrate Sulfur White flame: magnesium, aluminum Barium chlorate Antimony sulfide Gold sparks: iron filings, charcoal Strontium nitrate Dextrin White sparks: aluminum, magnesium, aluminum–magnesium alloy, titanium Red gum Whistle effect: potassium benzoate or sodium salicylate Polyvinyl chloride White smoke: mixture of potassium nitrate and sulfur Colored smoke: mixture of potassium chlorate, sulfur, and organic dye 290 Chapter Seven Atomic Structure and Periodicity Electron Energies Calculate the energy required to remove the electron from a hydrogen atom in its ground state.
Solution Removing the electron from a hydrogen atom in its ground state corresponds to taking the electron from ninitial 1 to Thus The energy required to remove the electron from a hydrogen atom in its ground state is 2.178 1018 J.
See Exercises 7.51 and 7.52.
At first Bohr’s model appeared to be very promising. The energy levels calculated by Bohr closely agreed with the values obtained from the hydrogen emission spectrum. How-ever, when Bohr’s model was applied to atoms other than hydrogen, it did not work at all.
Although some attempts were made to adapt the model using elliptical orbits, it was con-cluded that Bohr’s model is fundamentally incorrect. The model is, however, very im-portant historically, because it showed that the observed quantization of energy in atoms could be explained by making rather simple assumptions. Bohr’s model paved the way for later theories. It is important to realize, however, that the current theory of atomic structure is in no way derived from the Bohr model. Electrons do not move around the nucleus in circular orbits, as we shall see later in this chapter.
7.5 The Quantum Mechanical Model of the Atom By the mid-1920s it had become apparent that the Bohr model could not be made to work.
A totally new approach was needed. Three physicists were at the forefront of this effort: Werner Heisenberg (1901–1976), Louis de Broglie (1892–1987), and Erwin Schrödinger (1887–1961). The approach they developed became known as wave mechanics or, more commonly, quantum mechanics. As we have already seen, de Broglie originated the idea that the electron, previously considered to be a particle, also shows wave properties.
Pursuing this line of reasoning, Schrödinger, an Austrian physicist, decided to attack the problem of atomic structure by giving emphasis to the wave properties of the electron. To Schrödinger and de Broglie, the electron bound to the nucleus seemed similar to a stand-ing wave, and they began research on a wave mechanical description of the atom.
The most familiar example of standing waves occurs in association with musical instruments such as guitars or violins, where a string attached at both ends vibrates to pro-duce a musical tone. The waves are described as “standing” because they are stationary; 2.178 1018 J10 12 2.178 1018 J 2.178 1018 Ja 1 q 1 12b ¢E 2.178 1018 Ja 1 nfinal 2 1 ninitial 2b nfinal q.
Although Bohr’s model fits the energy levels for hydrogen, it is a fundamentally incorrect model for the hydrogen atom.
Unplucked string 1 half-wavelength 2 half-wavelengths 3 half-wavelengths FIGURE 7.9 The standing waves caused by the vibration of a guitar string fastened at both ends.
Each dot represents a node (a point of zero displacement).
Wave-generating apparatus.
Sample Exercise 7.5 Visualization: Flame Tests 7.5 The Quantum Mechanical Model of the Atom 291 the waves do not travel along the length of the string. The motions of the string can be explained as a combination of simple waves of the type shown in Fig. 7.9. The dots in this figure indicate the nodes, or points of zero lateral (sideways) displacement, for a given wave. Note that there are limitations on the allowed wavelengths of the standing wave.
Each end of the string is fixed, so there is always a node at each end. This means that there must be a whole number of half wavelengths in any of the allowed motions of the string (see Fig. 7.9). Standing waves can be illustrated using the wave generator shown in the photo below.
A similar situation results when the electron in the hydrogen atom is imagined to be a standing wave. As shown in Fig. 7.10, only certain circular orbits have a circumference into which a whole number of wavelengths of the standing electron wave will “fit.” All other orbits would produce destructive interference of the standing electron wave and are not allowed. This seemed like a possible explanation for the observed quantization of the hydrogen atom, so Schrödinger worked out a model for the hydrogen atom in which the electron was assumed to behave as a standing wave.
It is important to recognize that Schrödinger could not be sure that this idea would work. The test had to be whether or not the model would correctly fit the experimental data on hydrogen and other atoms. The physical principles for describing standing waves were well known in 1925 when Schrödinger decided to treat the electron in this way. His mathematical treatment is too complicated to be detailed here. However, the form of Schrödinger’s equation is where , called the wave function, is a function of the coordinates (x, y, and z) of the electron’s position in three-dimensional space and ˆ H represents a set of mathematical in-structions called an operator. In this case, the operator contains mathematical terms that produce the total energy of the atom when they are applied to the wave function. E rep-resents the total energy of the atom (the sum of the potential energy due to the attraction between the proton and electron and the kinetic energy of the moving electron). When this equation is analyzed, many solutions are found. Each solution consists of a wave func-tion that is characterized by a particular value of E. A specific wave function is often called an orbital.
To illustrate the most important ideas of the quantum (wave) mechanical model of the atom, we will first concentrate on the wave function corresponding to the lowest en-ergy for the hydrogen atom. This wave function is called the 1s orbital. The first point of interest is to explore the meaning of the word orbital. As we will see, this is not a trivial matter. One thing is clear: An orbital is not a Bohr orbit. The electron in the hydrogen 1s orbital is not moving around the nucleus in a circular orbit. How, then, is the electron moving? The answer is quite surprising: We do not know. The wave function gives us no information about the detailed pathway of the electron. This is somewhat disturbing. When we solve problems involving the motions of particles in the macroscopic world, we are able to predict their pathways. For example, when two billiard balls with known veloci-ties collide, we can predict their motions after the collision. However, we cannot predict the electron’s motion from the 1s orbital function. Does this mean that the theory is wrong?
Not necessarily: We have already learned that an electron does not behave much like a billiard ball, so we must examine the situation closely before we discard the theory.
To help us understand the nature of an orbital, we need to consider a principle dis-covered by Werner Heisenberg, one of the primary developers of quantum mechanics.
Heisenberg’s mathematical analysis led him to a surprising conclusion: There is a funda-mental limitation to just how precisely we can know both the position and momentum of a particle at a given time. This is a statement of the Heisenberg uncertainty principle.
Stated mathematically, the uncertainty principle is ¢x ¢1my2 h 4p H ˆc Ec n = 4 n = 5 n = 4 1 3 (a) (b) (c) Mismatch FIGURE 7.10 The hydrogen electron visualized as a standing wave around the nucleus. The circumference of a particular circular orbit would have to correspond to a whole number of wavelengths, as shown in (a) and (b), or else destructive interference occurs, as shown in (c). This is consistent with the fact that only certain electron energies are allowed; the atom is quantized.
(Although this idea encouraged scientists to use a wave theory, it does not mean that the electron really travels in circular orbits.) 292 Chapter Seven Atomic Structure and Periodicity where x is the uncertainty in a particle’s position, (m) is the uncertainty in a particle’s momentum, and h is Planck’s constant. Thus the minimum uncertainty in the product is h4. What this equation really says is that the more accurately we know a particle’s position, the less accurately we can know its momentum, and vice versa. This limitation is so small for large particles such as baseballs or billiard balls that it is unno-ticed. However, for a small particle such as the electron, the limitation becomes quite im-portant. Applied to the electron, the uncertainty principle implies that we cannot know the exact motion of the electron as it moves around the nucleus. It is therefore not appropri-ate to assume that the electron is moving around the nucleus in a well-defined orbit, as in the Bohr model.
The Physical Meaning of a Wave Function Given the limitations indicated by the uncertainty principle, what then is the physical meaning of a wave function for an electron? That is, what is an atomic orbital? Although the wave function itself has no easily visualized meaning, the square of the function does have a definite physical significance. The square of the function indicates the probability of finding an electron near a particular point in space. For example, suppose we have two positions in space, one defined by the coordinates x1, y1, and z1 and the other by the coordinates x2, y2, and z2. The relative probability of finding the electron at positions 1 and 2 is given by substituting the values of x, y, and z for the two positions into the wave function, squaring the function value, and computing the following ratio: The quotient N1N2 is the ratio of the probabilities of finding the electron at positions 1 and 2. For example, if the value of the ratio N1N2 is 100, the electron is 100 times more likely to be found at position 1 than at position 2. The model gives no information concerning when the electron will be at either position or how it moves between the positions. This vagueness is consistent with the concept of the Heisenberg uncertainty principle.
The square of the wave function is most conveniently represented as a probability distribution, in which the intensity of color is used to indicate the probability value near a given point in space. The probability distribution for the hydrogen 1s wave function (or-bital) is shown in Fig. 7.11(a). The best way to think about this diagram is as a three-dimensional time exposure with the electron as a tiny moving light. The more times the electron visits a particular point, the darker the negative becomes. Thus the darkness of a point indicates the probability of finding an electron at that position. This diagram is also known as an electron density map; electron density and electron probability mean the same thing. When a chemist uses the term atomic orbital, he or she is probably picturing an electron density map of this type.
Another way of representing the electron probability distribution for the 1s wave func-tion is to calculate the probability at points along a line drawn outward in any direction from the nucleus. The result is shown in Fig. 7.11(b). Note that the probability of finding the electron at a particular position is greatest close to the nucleus and drops off rapidly as the distance from the nucleus increases. We are also interested in knowing the total probability of finding the electron in the hydrogen atom at a particular distance from the nucleus. Imagine that the space around the hydrogen nucleus is made up of a series of thin spherical shells (rather like layers in an onion), as shown in Fig. 7.12(a). When the total probability of finding the electron in each spherical shell is plotted versus the distance from the nucleus, the plot in Fig. 7.12(b) is obtained. This graph is called the radial probability distribution.
The maximum in the curve occurs because of two opposing effects. The probability of finding an electron at a particular position is greatest near the nucleus, but the volume 3c1x1, y1, z12 4 2 3c1x2, y2, z22 4 2 N1 N2 ¢ ¢1my2 Probability (R2) Distance from nucleus (r) (b) (a) FIGURE 7.11 (a) The probability distribution for the hydrogen 1s orbital in three-dimensional space. (b) The probability of finding the electron at points along a line drawn from the nucleus outward in any direction for the hydrogen 1s orbital.
Probability is the likelihood, or odds, that something will occur.
7.6 Quantum Numbers 293 of the spherical shell increases with distance from the nucleus. Therefore, as we move away from the nucleus, the probability of finding the electron at a given position decreases, but we are summing more positions. Thus the total probability increases to a certain radius and then decreases as the electron probability at each position becomes very small. For the hydrogen 1s orbital, the maximum radial probability (the distance at which the elec-tron is most likely to be found) occurs at a distance of 5.29 102 nm or 0.529 Å from the nucleus. Interestingly, this is exactly the radius of the innermost orbit in the Bohr model. Note that in Bohr’s model the electron is assumed to have a circular path and so is always found at this distance. In the quantum mechanical model, the specific electron motions are unknown, and this is the most probable distance at which the electron is found.
One more characteristic of the hydrogen 1s orbital that we must consider is its size.
As we can see from Fig. 7.11, the size of this orbital cannot be defined precisely, since the probability never becomes zero (although it drops to an extremely small value at large values of r). So, in fact, the hydrogen 1s orbital has no distinct size. However, it is use-ful to have a definition of relative orbital size. The definition most often used by chemists to describe the size of the hydrogen 1s orbital is the radius of the sphere that encloses 90% of the total electron probability. That is, 90% of the time the electron is inside this sphere.
So far we have described only the lowest-energy wave function in the hydrogen atom, the 1s orbital. Hydrogen has many other orbitals, which we will describe in the next section. However, before we proceed, we should summarize what we have said about the meaning of an atomic orbital. An orbital is difficult to define precisely at an introductory level. Technically, an orbital is a wave function. However, it is usually most helpful to picture an orbital as a three-dimensional electron density map. That is, an electron “in” a particular atomic orbital is assumed to exhibit the electron probability indicated by the orbital map.
7.6 Quantum Numbers When we solve the Schrödinger equation for the hydrogen atom, we find many wave func-tions (orbitals) that satisfy it. Each of these orbitals is characterized by a series of num-bers called quantum numbers, which describe various properties of the orbital: The principal quantum number (n) has integral values: 1, 2, 3, . . . . The principal quantum number is related to the size and energy of the orbital. As n increases, the orbital becomes larger and the electron spends more time farther from the nucleus.
An increase in n also means higher energy, because the electron is less tightly bound to the nucleus, and the energy is less negative.
The angular momentum quantum number () has integral values from 0 to n 1 for each value of n. This quantum number is related to the shape of atomic orbitals.
The value of for a particular orbital is commonly assigned a letter: 0 is called s; 1 Å 1010 m; the angstrom is most often used as the unit for atomic radius because of its convenient size. Another convenient unit is the picometer: 1 pm 1012 m Radial probability (4πr2R2) Distance from nucleus (r) (b) (a) FIGURE 7.12 (a) Cross section of the hydrogen 1s orbital probability distribution divided into successive thin spherical shells. (b) The radial probability distribution. A plot of the total probability of finding the electron in each thin spherical shell as a function of distance from the nucleus.
Visualization: 1s Orbital 294 Chapter Seven Atomic Structure and Periodicity 1 is called p; 2 is called d; 3 is called f. This system arises from early spectral studies and is summarized in Table 7.1.
The magnetic quantum number (m) has integral values between and , including zero. The value of m is related to the orientation of the orbital in space relative to the other orbitals in the atom.
The first four levels of orbitals in the hydrogen atom are listed with their quantum numbers in Table 7.2. Note that each set of orbitals with a given value of (some-times called a subshell) is designated by giving the value of n and the letter for . Thus an orbital where n 2 and 1 is symbolized as 2p. There are three 2p orbitals, which have different orientations in space. We will describe these orbitals in the next section.
Electron Subshells For principal quantum level n 5, determine the number of allowed subshells (different values of ), and give the designation of each.
Solution For n 5, the allowed values of run from 0 to 4 (n 1 5 1). Thus the subshells and their designations are See Exercises 7.57 through 7.59.
5s 5p 5d 5f 5g / 0 / 1 / 2 / 3 / 4 n 1, 2, 3, . .
0, 1, . . . (n 1) m , . . . 0, . . . / / / / Number of Orbitals per Subshell s 1 p 3 d 5 f 7 g 9 TABLE 7.2 Quantum Numbers for the First Four Levels of Orbitals in the Hydrogen Atom Orbital n Designation m Number of Orbitals 1 0 1s 0 1 2 0 2s 0 1 1 2p 1, 0, 1 3 3 0 3s 0 1 1 3p 1, 0, 1 3 2 3d 2, 1, 0, 1, 2 5 4 0 4s 0 1 1 4p 1, 0, 1 3 2 4d 2, 1, 0, 1, 2 5 3 4f 3, 2, 1, 0, 1, 2, 3 7 / / Sample Exercise 7.6 TABLE 7.1 The Angular Momentum Quantum Numbers and Corresponding Letters Used to Designate Atomic Orbitals Value of 0 1 2 3 4 Letter Used s p d f g / 7.7 Orbital Shapes and Energies 295 7.7 Orbital Shapes and Energies We have seen that the meaning of an orbital is represented most clearly by a probability distribution. Each orbital in the hydrogen atom has a unique probability distribution. We also saw that another means of representing an orbital is by the surface that surrounds 90% of the total electron probability. These two types of representations for the hydrogen 1s, 2s, and 3s orbitals are shown in Fig. 7.13. Note the characteristic spherical shape of each of the s orbitals. Note also that the 2s and 3s orbitals contain areas of high proba-bility separated by areas of zero probability. These latter areas are called nodal surfaces, or simply nodes. The number of nodes increases as n increases. For s orbitals, the num-ber of nodes is given by n 1. For our purposes, however, we will think of s orbitals only in terms of their overall spherical shape, which becomes larger as the value of n increases.
The two types of representations for the 2p orbitals (there are no 1p orbitals) are shown in Fig. 7.14. Note that the p orbitals are not spherical like s orbitals but have two lobes separated by a node at the nucleus. The p orbitals are labeled according to the axis of the xyz coordinate system along which the lobes lie. For example, the 2p orbital with lobes centered along the x axis is called the 2px orbital.
At this point it is useful to remember that mathematical functions have signs. For ex-ample, a simple sine wave (see Fig. 7.1) oscillates from positive to negative and repeats this pattern. Atomic orbital functions also have signs. The functions for s orbitals are positive everywhere in three-dimensional space. That is, when the s orbital function is evaluated at any point in space, it results in a positive number. In contrast, the p orbital functions have different signs in different regions of space. For example, the pz orbirtal has a positive sign in all the regions of space in which z is positive and has a negative sign when z is negative. This behavior is indicated in Fig. 7.14(b) by the positive and neg-ative signs inside their boundary surfaces. It is important to understand that these are math-ematical signs, not charges. Just as a sine wave has alternating positive and negative phases, so too p orbitals have positive and negative phases. The phases of the px, py, and pz orbitals are indicated in Fig. 7.14(b).
As you might expect from our discussion of the s orbitals, the 3p orbitals have a more complex probability distribution than that of the 2p orbitals (see Fig. 7.15), but they can still be represented by the same boundary surface shapes. The surfaces just grow larger as the value of n increases.
There are no d orbitals that correspond to principal quantum levels n 1 and n 2. The d orbitals ( 2) first occur in level n 3. The five 3d orbitals have the shapes shown in Fig. 7.16. The d orbitals have two different fundamental shapes. Four of the orbitals (dxz, dyz, dxy, and ) have four lobes centered in the plane indicated in the orbital label. Note that dxy and are both centered in the xy plane; however, the lobes of lie along the x and y axes, while the lobes of dxy lie between the axes.
The fifth orbital, has a unique shape with two lobes along the z axis and a belt cen-tered in the xy plane. The d orbitals for levels n 3 look like the 3d orbitals but have larger lobes.
The f orbitals first occur in level n 4, and as might be expected, they have shapes even more complex than those of the d orbitals. Figure 7.17 shows representations of the 4f orbitals ( 3) along with their designations. These orbitals are not involved in the bonding in any of the compounds we will consider in this text. Their shapes and labels are simply included for completeness.
So far we have talked about the shapes of the hydrogen atomic orbitals but not about their energies. For the hydrogen atom, the energy of a particular orbital is determined by its value of n. Thus all orbitals with the same value of n have the same energy—they are said to be degenerate. This is shown in Fig. 7.18, where the energies for the orbitals in the first three quantum levels for hydrogen are shown.
dz2, dx2y2 dx2y2 dx2y2 Nodes Node 1s 2s 3s (a) 1s 2s 3s (b) FIGURE 7.13 Two representations of the hydrogen 1s, 2s, and 3s orbitals. (a) The electron probability distribution. (b) The surface that contains 90% of the total electron probability (the size of the orbital, by definition).
n value g 2px m orientation in space h value / Visualization: Orbital Energies Visualization: 2px, 2py, 2pz Orbitals Visualization: Orbitals 3dyz, 3dz2 3dx2y2, 3dxy, 3dxz, 296 Chapter Seven Atomic Structure and Periodicity Hydrogen’s single electron can occupy any of its atomic orbitals. However, in the lowest energy state, the ground state, the electron resides in the 1s orbital. If energy is put into the atom, the electron can be transferred to a higher-energy orbital, producing an excited state.
A Summary of the Hydrogen Atom In the quantum (wave) mechanical model, the electron is viewed as a standing wave.
This representation leads to a series of wave functions (orbitals) that describe the pos-sible energies and spatial distributions available to the electron.
In agreement with the Heisenberg uncertainty principle, the model cannot specify the detailed electron motions. Instead, the square of the wave function represents the probability distribution of the electron in that orbital. This allows us to picture or-bitals in terms of probability distributions, or electron density maps.
The size of an orbital is arbitrarily defined as the surface that contains 90% of the total electron probability.
The hydrogen atom has many types of orbitals. In the ground state, the single elec-tron resides in the 1s orbital. The electron can be excited to higher-energy orbitals if energy is put into the atom.
7.8 Electron Spin and the Pauli Principle The concept of electron spin was developed by Samuel Goudsmit and George Uhlenbeck while they were graduate students at the University of Leyden in the Netherlands. They found that a fourth quantum number (in addition to n, , and m) was necessary to account for the details of the emission spectra of atoms. The spectral data indicate that the elec-tron has a magnetic moment with two possible orientations when the atom is placed in an external magnetic field. Since they knew from classical physics that a spinning charge produces a magnetic moment, it seemed reasonable to assume that the electron could have two spin states, thus producing the two oppositely directed magnetic moments FIGURE 7.15 A cross section of the electron probability distribution for a 3p orbital.
FIGURE 7.14 Representation of the 2p orbitals. (a) The electron probability distribution for a 2p orbital. (Generated from a program by Robert Allendoerfer on Project SERAPHIM disk PC 2402; reprinted with permission.) (b) The boundary surface representations of all three 2p orbitals. Note that the signs inside the surface indicate the phases (signs) of the orbital in that region of space.
(a) x x y z x y z 2px 2py 2pz (b) z y + + + – – – 7.8 Electron Spin and the Pauli Principle 297 FIGURE 7.16 Representation of the 3d orbitals. (a) Electron density plots of selected 3d orbitals. (Generated from a program by Robert Allendoerfer on Project SERAPHIM disk PC 2402; reprinted with permission.) (b) The boundary surfaces of all five 3d orbitals, with the signs (phases) indicated.
z z z y x z z z z fz3 – — 3 5 zr2 fxyz fy(x2 – z2) fx(z2 – y2) fz(x2 – y2) y x y x y x y x y x fx3 – — 3 5 xr2 fy3 – — 3 5 yr2 y x FIGURE 7.17 Representation of the 4f orbitals in terms of their boundary surfaces.
x y z dyz x y z x y z y z x y z dxz dxy dx2 – y2 dz2 x (b) + + + + + – – + + – – – – – – + – + + (a) 298 Chapter Seven Atomic Structure and Periodicity (see Fig. 7.19). The new quantum number adopted to describe this phenomenon, called the electron spin quantum number (ms), can have only one of two values, and We can interpret this to mean that the electron can spin in one of two opposite directions, although other interpretations also have been suggested.
For our purposes, the main significance of electron spin is connected with the postu-late of Austrian physicist Wolfgang Pauli (1900–1958): In a given atom no two electrons can have the same set of four quantum numbers (n, , m, and ms). This is called the Pauli exclusion principle. Since electrons in the same orbital have the same values of n, , and m, this postulate says that they must have different values of ms. Then, since only two values of ms are allowed, an orbital can hold only two electrons, and they must have opposite spins. This principle will have important consequences as we use the atomic model to account for the electron arrangements of the atoms in the periodic table.
7.9 Polyelectronic Atoms The quantum mechanical model gives a description of the hydrogen atom that agrees very well with experimental data. However, the model would not be very useful if it did not account for the properties of all the other atoms as well.
To see how the model applies to polyelectronic atoms, that is, atoms with more than one electron, let’s consider helium, which has two protons in its nucleus and two electrons: e 2 e Three energy contributions must be considered in the description of the helium atom: (1) the kinetic energy of the electrons as they move around the nucleus, (2) the potential energy of attraction between the nucleus and the electrons, and (3) the potential energy of repulsion between the two electrons.
Although the helium atom can be readily described in terms of the quantum mechanical model, the Schrödinger equation that results cannot be solved exactly. The difficulty arises in dealing with the repulsions between the electrons. Since the electron pathways are unknown, the electron repulsions cannot be calculated exactly. This is called the electron correlation problem.
The electron correlation problem occurs with all polyelectronic atoms. To treat these systems using the quantum mechanical model, we must make approximations. Most com-monly, the approximation used is to treat each electron as if it were moving in a field of charge that is the net result of the nuclear attraction and the average repulsions of all the other electrons.
For example, consider the sodium atom, which has 11 electrons: Now let’s single out the outermost electron and consider the forces this electron feels. The electron clearly is attracted to the highly charged nucleus. However, the electron also feels the repulsions caused by the other 10 electrons. The net effect is that the electron is not l l+ l le– 1 2.
1 2 ms or Each orbital can hold a maximum of two electrons.
1 2 1 2 E 3s 2s 1s 3p 2p 3d e– N S (b) e– S N (a) FIGURE 7.19 A picture of the spinning electron. Spinning in one direction, the electron produces the magnetic field oriented as shown in (a).
Spinning in the opposite direction, it gives a magnetic field of the opposite orientation, as shown in (b).
FIGURE 7.18 Orbital energy levels for the hydrogen atom.
7.10 The History of the Periodic Table 299 bound nearly as tightly to the nucleus as it would be if the other electrons were not present. We say that the electron is screened or shielded from the nuclear charge by the repulsions of the other electrons.
This picture of polyelectronic atoms leads to hydrogenlike orbitals for these atoms.
They have the same general shapes as the orbitals for hydrogen, but their sizes and ener-gies are different. The differences occur because of the interplay between nuclear attrac-tion and the electron repulsions.
One especially important difference between polyelectronic atoms and the hydrogen atom is that for hydrogen all the orbitals in a given principal quantum level have the same energy (they are said to be degenerate). This is not the case for polyelectronic atoms, where we find that for a given principal quantum level the orbitals vary in energy as follows: In other words, when electrons are placed in a particular quantum level, they “pre-fer” the orbitals in the order s, p, d, and then f. Why does this happen? Although the con-cept of orbital energies is a complicated matter, we can qualitatively understand why the 2s orbital has a lower energy than the 2p orbital in a polyelectronic atom by looking at the probability profiles of these orbitals (see Fig. 7.20). Notice that the 2p orbital has its maximum probability closer to the nucleus than for the 2s. This might lead us to predict that the 2p would be preferable (lower energy) to the 2s orbital. However, notice the small hump of electron density that occurs in the 2s profile very near the nucleus. This means that although an electron in the 2s orbital spends most of its time a little farther from the nucleus than does an electron in the 2p orbital, it spends a small but very significant amount of time very near the nucleus. We say that the 2s electron penetrates to the nucleus more than one in the 2p orbital. This penetration effect causes an electron in a 2s orbital to be attracted to the nucleus more strongly than an electron in a 2p orbital. That is, the 2s or-bital is lower in energy than the 2p orbitals in a polyelectronic atom.
The same thing happens in the other principal quantum levels as well. Figure 7.21 shows the radial probability profiles for the 3s, 3p, and 3d orbitals. Note again the hump in the 3s profile very near the nucleus. The innermost hump for the 3p is farther out, which causes the energy of the 3s orbital to be lower than that of the 3p. Notice that the 3d or-bital has its maximum probability closer to the nucleus than either the 3s or 3p does, but its absence of probability near the nucleus causes it to be highest in energy of the three orbitals. The relative energies of the orbitals for n 3 are In general, the more effectively an orbital allows its electron to penetrate the shielding electrons to be close to the nuclear charge, the lower is the energy of that orbital.
A summary diagram of the orders of the orbital energies for polyelectronic atoms is represented in Fig. 7.22. We will use these orbitals in Section 7.11 to show how the elec-trons are arranged in polyelectronic atoms.
7.10 The History of the Periodic Table The modern periodic table contains a tremendous amount of useful information. In this section we will discuss the origin of this valuable tool; later we will see how the quan-tum mechanical model for the atom explains the periodicity of chemical properties. Cer-tainly the greatest triumph of the quantum mechanical model is its ability to account for the arrangement of the elements in the periodic table.
The periodic table was originally constructed to represent the patterns observed in the chemical properties of the elements. As chemistry progressed during the eighteenth and nineteenth centuries, it became evident that the earth is composed of a great many E3s 6 E3p 6 E3d Ens 6 Enp 6 End 6 Enf Radial probability Distance from nucleus 2s 2p FIGURE 7.20 A comparison of the radial probability distributions of the 2s and 2p orbitals.
Radial probability Distance from the nucleus Penetration Most probable distance from the nucleus Radial probability Distance from the nucleus Penetration 3s 3p 3d (a) (b) FIGURE 7.21 (a) The radial probability distribution for an electron in a 3s orbital. Although a 3s electron is mostly found far from the nucleus, there is a small but significant probability (shown by the arrows) of its being found close to the nucleus. The 3s electron penetrates the shield of inner electrons. (b) The radial probability distribution for the 3s, 3p, and 3d orbitals.
The arrows indicate that the s orbital (red arrow) allows greater electron penetration than the p orbital (yellow arrow) does; the d orbital allows minimal electron penetration.
300 Chapter Seven Atomic Structure and Periodicity elements with very different properties. Things are much more complicated than the sim-ple model of earth, air, fire, and water suggested by the ancients. At first, the array of elements and properties was bewildering. Gradually, however, patterns were noticed.
The first chemist to recognize patterns was Johann Dobereiner (1780–1849), who found several groups of three elements that have similar properties, for example, chlorine, bromine, and iodine. However, as Dobereiner attempted to expand this model of triads (as he called them) to the rest of the known elements, it became clear that it was severely limited.
The next notable attempt was made by the English chemist John Newlands, who in 1864 suggested that elements should be arranged in octaves, based on the idea that cer-tain properties seemed to repeat for every eighth element in a way similar to the musical scale, which repeats for every eighth tone. Even though this model managed to group sev-eral elements with similar properties, it was not generally successful.
The present form of the periodic table was conceived independently by two chemists: the German Julius Lothar Meyer (1830–1895) and Dmitri Ivanovich Mendeleev (1834–1907), a Russian (Fig. 7.23). Usually Mendeleev is given most of the credit, be-cause it was he who emphasized how useful the table could be in predicting the existence and properties of still unknown elements. For example, in 1872 when Mendeleev first pub-lished his table (see Fig. 7.24), the elements gallium, scandium, and germanium were un-known. Mendeleev correctly predicted the existence and properties of these elements from gaps in his periodic table. The data for germanium (which Mendeleev called “ekasilicon”) are shown in Table 7.3. Note the excellent agreement between the actual values and Mendeleev’s predictions, which were based on the properties of other members in the group of elements similar to germanium.
Using his table, Mendeleev also was able to correct several values for atomic masses.
For example, the original atomic mass of 76 for indium was based on the assumption that indium oxide had the formula InO. This atomic mass placed indium, which has metallic properties, among the nonmetals. Mendeleev assumed the atomic mass was probably in-correct and proposed that the formula of indium oxide was really In2O3. Based on this correct formula, indium has an atomic mass of approximately 113, placing the element among the metals. Mendeleev also corrected the atomic masses of beryllium and uranium.
Because of its obvious usefulness, Mendeleev’s periodic table was almost universally adopted, and it remains one of the most valuable tools at the chemist’s disposal. For ex-ample, it is still used to predict the properties of elements recently discovered, as shown in Table 7.4.
A current version of the periodic table is shown inside the front cover of this book.
The only fundamental difference between this table and that of Mendeleev is that it lists the elements in order by atomic number rather than by atomic mass. The reason for this will become clear later in this chapter as we explore the electron arrangements of the atom. Another recent format of the table is discussed in the following section.
FIGURE 7.23 Dmitri Ivanovich Mendeleev (1834–1907), born in Siberia as the youngest of 17 children, taught chem-istry at the University of St. Petersburg. In 1860 Mendeleev heard the Italian chemist Cannizzaro lecture on a reliable method for determining the correct atomic masses of the elements. This important develop-ment paved the way for Mendeleev’s own brilliant contribution to chemistry—the periodic table. In 1861 Mendeleev returned to St. Petersburg, where he wrote a book on organic chemistry. Later Mendeleev also wrote a book on inorganic chemistry, and he was struck by the fact that the systematic approach charac-terizing organic chemistry was lacking in inorganic chemistry. In attempting to systematize inorganic chemistry, he eventually arranged the elements in the form of the periodic table.
Mendeleev was a versatile genius who was interested in many fields of science. He worked on many problems associated with Russia’s natural resources, such as coal, salt, and various metals. Being particu-larly interested in the petroleum industry, he visited the United States in 1876 to study the Pennsylvania oil fields. His interests also included meteorology and hot-air balloons. In 1887 he made an ascent in a balloon to study a total eclipse of the sun.
E 3s 2s 1s 3p 2p 3d FIGURE 7.22 The orders of the energies of the orbitals in the first three levels of polyelectronic atoms.
7.10 The History of the Periodic Table 301 FIGURE 7.24 Mendeleev’s early periodic table, published in 1872. Note the spaces left for missing elements with atomic masses 44, 68, 72, and 100.
(From Annalen der Chemie und Pharmacie, VIII, Supplementary Volume for 1872, page 511.) TABLE 7.3 Comparison of the Properties of Germanium as Predicted by Mendeleev and as Actually Observed Properties of Predicted Observed in Germanium in 1871 1886 Atomic weight 72 72.3 Density 5.5 g/cm3 5.47 g/cm3 Specific heat 0.31 J/(C g) 0.32 J/(C g) Melting point Very high 960C Oxide formula RO2 GeO2 Oxide density 4.7 g/cm3 4.70 g/cm3 Chloride formula RCl4 GeCl4 bp of chloride 100C 86C TABLE 7.4 Predicted Properties of Elements 113 and 114 Property Element 113 Element 114 Chemically like Thallium Lead Atomic mass 297 298 Density 16 g/mL 14 g/mL Melting point 430C 70C Boiling point 1100C 150C 302 Chapter Seven Atomic Structure and Periodicity 7.11 The Aufbau Principle and the Periodic Table We can use the quantum mechanical model of the atom to show how the electron arrange-ments in the hydrogenlike atomic orbitals of the various atoms account for the organiza-tion of the periodic table. Our main assumption here is that all atoms have the same type of orbitals as have been described for the hydrogen atom. As protons are added one by one to the nucleus to build up the elements, electrons are similarly added to these hy-drogenlike orbitals. This is called the aufbau principle.
Hydrogen has one electron, which occupies the 1s orbital in its ground state. The con-figuration for hydrogen is written as 1s1, which can be represented by the following orbital diagram: The arrow represents an electron spinning in a particular direction.
The next element, helium, has two electrons. Since two electrons with opposite spins can occupy an orbital, according to the Pauli exclusion principle, the electrons for helium are in the 1s orbital with opposite spins, producing a 1s2 configuration: Lithium has three electrons, two of which can go into the 1s orbital before the orbital is filled. Since the 1s orbital is the only orbital for n 1, the third electron will occupy the lowest-energy orbital with n 2, or the 2s orbital, giving a 1s22s1 configuration: Li: 1s22s1 1s 2s 2p He: 1s2 1s 2s 2p H: 1s1 1s 2s 2p Aufbau is German for “building up.” H (Z 1) He (Z 2) Li (Z 3) Be (Z 4) B (Z 5) etc.
(Z atomic number) CHEMICAL IMPACT The Growing Periodic Table T he periodic table of the elements has undergone signif-icant changes since Mendeleev published his first ver-sion in 1869. In particular, in the past 60 years we have added 20 new elements beyond uranium. These so-called transuranium elements all have been synthesized using par-ticle accelerators.
Edwin M. McMillan and Phillip H. Abelson succeeded in synthesizing the first transuranium element, neptunium (element 93), at the University of California, Berkeley, in 1940. In 1941, Glenn T. Seaborg synthesized and identified element 94 (plutonium), and over the next several years, researchers under his direction at UC Berkeley discovered nine other transuranium elements. In 1945 Seaborg sug-gested that the elements heavier than element 89 (actinium) were misplaced as transition metals and should be relocated on the periodic table in a series below the transition metals (the actinide series). Seaborg was awarded a Nobel Prize in chemistry in 1951 for his contributions.
Dr. Glenn Seaborg.
7.11 The Aufbau Principle and the Periodic Table 303 The next element, beryllium, has four electrons, which occupy the 1s and 2s orbitals: Boron has five electrons, four of which occupy the 1s and 2s orbitals. The fifth elec-tron goes into the second type of orbital with n 2, the 2p orbitals: Since all the 2p orbitals have the same energy (are degenerate), it does not matter which 2p orbital the electron occupies.
Carbon is the next element and has six electrons. Two electrons occupy the 1s orbital, two occupy the 2s orbital, and two occupy 2p orbitals. Since there are three 2p orbitals with the same energy, the mutually repulsive electrons will occupy separate 2p orbitals.
This behavior is summarized by Hund’s rule (named for the German physicist F. H.
Hund), which states that the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particu-lar set of degenerate orbitals. By convention, the unpaired electrons are represented as having parallel spins (with spin “up”).
The configuration for carbon could be written 1s22s22p12p1 to indicate that the electrons occupy separate 2p orbitals. However, the configuration is usually given as 1s22s22p2, and it is understood that the electrons are in different 2p orbitals. The orbital diagram for carbon is Note that the unpaired electrons in the 2p orbitals are shown with parallel spins.
1s 2s 2p C: 1s22s22p2 1s 2s 2p B: 1s22s22p1 Be: 1s22s2 1s 2s 2p For an atom with unfilled subshells, the lowest energy is achieved by electrons occupying separate orbitals with parallel spins, as far as allowed by the Pauli exclusion principle.
In recent years, three major research facilities have taken the lead in synthesizing new elements. Along with UC Berkeley, Nuclear Research in Dubna, Russia, and GSI in Darmstadt, Germany, were responsible for synthesizing el-ements 104–112 by the end of 1996.
As it turned out, naming the new elements has caused more controversy than anything else connected with their discovery. Traditionally, the discoverer of an element is allowed to name it. However, because there is some dis-pute among the researchers at Berkeley, Darmstadt, and Dubna about who really discovered the various elements, competing names were submitted. After years of contro-versy, the International Union of Pure and Applied Chem-istry (IUPAC) finally settled on the names listed in the accompanying table.
The name for element 106 in honor of Glenn Seaborg caused special controversy because an element had never before been named for a living person (Dr. Seaborg died in 1999). However, because of Seaborg’s commanding stature in the scientific community, the name seaborgium was adopted.
Names for the elements beyond 111 have not been de-cided, and these elements are represented on many periodic tables with three letters that symbolize their atomic num-bers. More traditional names will no doubt be assigned in due time (hopefully with a minimum of controversy).
Atomic Number Name Symbol 104 Rutherfordium Rf 105 Dubium Db 106 Seaborgium Sg 107 Bohrium Bh 108 Hassium Hs 109 Meitnerium Mt 110 Darmstadtium Ds 111 Roentgenium Rg 304 Chapter Seven Atomic Structure and Periodicity The configuration for nitrogen, which has seven electrons, is 1s22s22p3. The three electrons in the 2p orbitals occupy separate orbitals with parallel spins: The configuration for oxygen, which has eight electrons, is 1s22s22p4. One of the 2p orbitals is now occupied by a pair of electrons with opposite spins, as required by the Pauli exclusion principle: The orbital diagrams and electron configurations for fluorine (nine electrons) and neon (ten electrons) are as follows: With neon, the orbitals with n 1 and n 2 are now completely filled.
For sodium, the first ten electrons occupy the 1s, 2s, and 2p orbitals, and the eleventh electron must occupy the first orbital with n 3, the 3s orbital. The electron configura-tion for sodium is 1s22s22p63s1. To avoid writing the inner-level electrons, this configura-tion is often abbreviated as [Ne]3s1, where [Ne] represents the electron configuration of neon, 1s22s22p6.
The next element, magnesium, has the configuration 1s22s22p63s2, or [Ne]3s2. Then the next six elements, aluminum through argon, have configurations obtained by filling the 3p orbitals one electron at a time. Figure 7.25 summarizes the electron configurations of the first 18 elements by giving the number of electrons in the type of orbital occupied last.
At this point it is useful to introduce the concept of valence electrons, the electrons in the outermost principal quantum level of an atom. The valence electrons of the nitro-gen atom, for example, are the 2s and 2p electrons. For the sodium atom, the valence elec-tron is the electron in the 3s orbital, and so on. Valence electrons are the most important electrons to chemists because they are involved in bonding, as we will see in the next two chapters. The inner electrons are known as core electrons.
Note in Fig. 7.25 that a very important pattern is developing: The elements in the same group (vertical column of the periodic table) have the same valence electron configuration. Remember that Mendeleev originally placed the elements in groups based on similarities in chemical properties. Now we understand the reason behind these 1s 2s 2p Ne: 1s22s22p6 1s22s22p5 F: 1s 2s 2p O: 1s22s22p4 1s 2s 2p N: 1s22s22p3 A vial containing potassium metal. The sealed vial contains an inert gas to protect the potassium from reacting with oxygen.
H 1s1 Li 2s1 Be 2s2 B 2p1 C 2p2 N 2p3 O 2p4 F 2p5 Ne 2p6 He 1s2 Na 3s1 Mg 3s2 Al 3p1 Si 3p2 P 3p3 S 3p4 Cl 3p5 Ar 3p6 FIGURE 7.25 The electron configurations in the type of orbital occupied last for the first 18 elements.
Sodium metal is so reactive that it is stored under kerosene to protect it from the oxygen in the air.
[Ne] is shorthand for 1s22s22p6.
7.11 The Aufbau Principle and the Periodic Table 305 groupings. Elements with the same valence electron configuration show similar chemical behavior.
The element after argon is potassium. Since the 3p orbitals are fully occupied in ar-gon, we might expect the next electron to go into a 3d orbital (recall that for n 3 the orbitals are 3s, 3p, and 3d). However, the chemistry of potassium is clearly very similar to that of lithium and sodium, indicating that the last electron in potassium occupies the 4s orbital instead of one of the 3d orbitals, a conclusion confirmed by many types of experiments. The electron configuration of potassium is The next element is calcium: The next element, scandium, begins a series of 10 elements (scandium through zinc) called the transition metals, whose configurations are obtained by adding electrons to the five 3d orbitals. The configuration of scandium is That of titanium is And that of vanadium is Chromium is the next element. The expected configuration is [Ar]4s23d 4. However, the observed configuration is The explanation for this configuration of chromium is beyond the scope of this book. In fact, chemists are still disagreeing over the exact cause of this anomaly. Note, however, that the observed configuration has both the 4s and 3d orbitals half-filled. This is a good way to remember the correct configuration.
The next four elements, manganese through nickel, have the expected configurations: The configuration for copper is expected to be [Ar]4s23d 9. However, the observed configuration is In this case, a half-filled 4s orbital and a filled set of 3d orbitals characterize the actual configuration.
Zinc has the expected configuration: The configurations of the transition metals are shown in Fig. 7.26. After that, the next six elements, gallium through krypton, have configurations that correspond to filling the 4p orbitals (see Fig. 7.26).
The entire periodic table is represented in Fig. 7.27 in terms of which orbitals are be-ing filled. The valence electron configurations are given in Fig. 7.28. From these two fig-ures, note the following additional points: 1. The (n 1)s orbitals always fill before the nd orbitals. For example, the 5s orbitals fill in rubidium and strontium before the 4d orbitals fill in the second row of transition Zn: 3Ar44s23d10 Cu: 3Ar44s13d10 Fe: 3Ar44s23d 6 Ni : 3Ar44s23d 8 Mn: 3Ar44s23d 5 Co : 3Ar44s23d7 Cr: 3Ar44s13d5 V: 3Ar44s23d3 Ti: 3Ar44s23d2 Sc: 3Ar44s23d1 Ca: 3Ar44s2 K: 1s22s22p63s23p64s1 or 3Ar44s1 Calcium metal.
Chromium is often used to plate bumpers and hood ornaments, such as this statue of Mercury found on a 1929 Buick.
The (n 1)s orbital fills before the nd orbitals.
306 Chapter Seven Atomic Structure and Periodicity metals (yttrium through cadmium). This early filling of the s orbitals can be explained by the penetration effect. For example, the 4s orbital allows for so much more pene-tration to the vicinity of the nucleus that it becomes lower in energy than the 3d or-bital. Thus the 4s fills before the 3d. The same things can be said about the 5s and 4d, the 6s and 5d, and the 7s and 6d orbitals.
2. After lanthanum, which has the configuration [Xe]6s25d1, a group of 14 elements called the lanthanide series, or the lanthanides, occurs. This series of elements cor-responds to the filling of the seven 4f orbitals. Note that sometimes an electron occupies a 5d orbital instead of a 4f orbital. This occurs because the energies of the 4f and 5d orbitals are very similar.
3. After actinium, which has the configuration [Rn]7s26d1, a group of 14 elements called the actinide series, or the actinides, occurs. This series corresponds to the filling of the seven 5f orbitals. Note that sometimes one or two electrons occupy the 6d orbitals instead of the 5f orbitals, because these orbitals have very similar energies.
FIGURE 7.27 The orbitals being filled for elements in various parts of the periodic table. Note that in going along a horizontal row (a period), the (n 1)s orbital fills before the nd orbital. The group labels indicate the number of valence electrons (ns plus np electrons) for the elements in each group.
Lanthanides are elements in which the 4f orbitals are being filled.
Actinides are elements in which the 5f orbitals are being filled.
K 4s1 Ca 4s2 Sc 3d1 Ti 3d2 V 3d3 Mn 3d5 Fe 3d6 Co 3d7 Ni 3d8 Zn 3d10 Ga 4p1 Ge 4p2 As 4p3 Se 4p4 Br 4p5 Kr 4p6 Cr 4s1 3d5 Cu 4s13d10 FIGURE 7.26 Electron configurations for potassium through krypton. The transition metals (scandium through zinc) have the general configuration [Ar]4s23dn, except for chromium and copper.
1A 2A 3A 4A 5A 6A 7A 8A 6p 5d 4s 5s 6s 7s 2s 3s 1s La Ac Group 4f 5f Period 4 5 6 7 2 3 1 6d 4d 3d 5p 4p 3p 2p 1s 7.11 The Aufbau Principle and the Periodic Table 307 4. The group labels for Groups 1A, 2A, 3A, 4A, 5A, 6A, 7A, and 8A indicate the total number of valence electrons for the atoms in these groups. For example, all the elements in Group 5A have the configuration ns2np3. (The d electrons fill one period late and are usually not counted as valence electrons.) The meaning of the group labels for the transition metals is not as clear as for the Group A elements, and these will not be used in this text.
5. The groups labeled 1A, 2A, 3A, 4A, 5A, 6A, 7A, and 8A are often called the main-group, or representative, elements. Every member of these groups has the same va-lence electron configuration.
The International Union of Pure and Applied Chemistry (IUPAC), a body of scientists organized to standardize scientific conventions, has recommended a new form for the pe-riodic table, which the American Chemical Society has adopted (see the blue numbers in Fig. 7.28). In this new version the group number indicates the number of s, p, and d elec-trons added since the last noble gas. We will not use the new format in this book, but you The group label tells the total number of valence electrons for that group.
FIGURE 7.28 The periodic table with atomic symbols, atomic numbers, and partial electron configurations.
1 H 1s1 3 Li 2s1 11 Na 3s1 19 K 4s1 37 Rb 5s1 55 Cs 6s1 87 Fr 7s1 4 Be 2s2 12 Mg 3s2 20 Ca 4s2 38 Sr 5s2 56 Ba 6s2 88 Ra 7s2 21 Sc 4s23d1 39 Y 5s24d1 57 La 6s25d1 89 Ac 7s26d1 22 Ti 4s23d2 40 Zr 5s24d2 72 Hf 4f146s25d2 104 Rf 7s26d2 23 V 4s23d3 41 Nb 5s14d4 73 Ta 6s25d3 105 Db 7s26d3 24 Cr 4s13d5 42 Mo 5s14d5 74 W 6s25d4 106 Sg 7s26d4 25 Mn 4s23d5 43 Tc 5s14d6 75 Re 6s25d5 107 Bh 7s26d5 26 Fe 4s23d6 44 Ru 5s14d7 76 Os 6s25d6 108 Hs 7s26d6 110 Ds 7s26d8 111 Rg 7s16d10 112 Uub 7s26d10 113 Uut 7s26d107p1 114 Uuq 7s26d107p2 115 Uup 7s26d107p3 27 Co 4s23d7 45 Rh 5s14d8 77 Ir 6s25d7 109 Mt 7s26d7 28 Ni 4s23d8 46 Pd 4d10 78 Pt 6s15d9 29 Cu 4s13d10 47 Ag 5s14d10 79 Au 6s15d10 31 Ga 4s24p1 49 In 5s25p1 81 Tl 6s26p1 5 B 2s22p1 13 Al 3s23p1 32 Ge 4s24p2 50 Sn 5s25p2 82 Pb 6s26p2 6 C 2s22p2 14 Si 3s23p2 33 As 4s24p3 51 Sb 5s25p3 83 Bi 6s26p3 7 N 2s22p3 15 P 3s23p3 34 Se 4s24p4 52 Te 5s25p4 84 Po 6s26p4 8 O 2s22p4 16 S 3s23p4 9 F 2s22p5 17 Cl 3s23p5 35 Br 4s24p5 53 I 5s25p5 85 At 6s26p5 10 Ne 2s22p6 18 Ar 3s23p6 36 Kr 4s24p6 54 Xe 5s25p6 86 Rn 6s26p6 2 He 1s2 58 Ce 6s24f15d1 90 Th 7s25f 06d2 59 Pr 6s24f 35d0 91 Pa 7s25f 26d1 60 Nd 6s24f45d0 92 U 7s25f 36d1 61 Pm 6s24f 55d0 93 Np 7s25f46d1 62 Sm 6s24f65d0 94 Pu 7s25f66d0 63 Eu 6s24f 75d0 95 Am 7s25f 76d0 64 Gd 6s24f 75d1 96 Cm 7s25f 76d1 65 Tb 6s24f 95d0 97 Bk 7s25f 96d0 66 Dy 6s24f105d0 98 Cf 7s25f106d0 67 Ho 6s24f115d0 99 Es 7s25f116d0 68 Er 6s24f125d0 100 Fm 7s25f126d0 69 Tm 6s24f135d0 101 Md 7s25f136d0 70 Yb 6s24f145d0 102 No 7s25f146d0 71 Lu 6s24f145d1 103 Lr 7s25f146d1 d-Transition Elements Period number, highest occupied electron level Noble Gases Lanthanides Actinides Representative Elements 1A ns1 Group numbers 2A ns2 3A ns2np1 4A ns2np2 5A ns2np3 6A ns2np4 7A ns2np5 8A ns2np6 Representative Elements f-Transition Elements 1 2 3 4 5 6 7 30 Zn 4s23d10 48 Cd 5s24d10 80 Hg 6s25d10 1 13 14 15 16 17 8 9 10 11 12 3 4 5 6 7 18 2 308 Chapter Seven Atomic Structure and Periodicity should be aware that the familiar periodic table may be soon replaced by this or a similar format.
The results considered in this section are very important. We have seen that the quan-tum mechanical model can be used to explain the arrangement of the elements in the periodic table. This model allows us to understand that the similar chemistry exhibited by the members of a given group arises from the fact that they all have the same valence electron configuration. Only the principal quantum number of the valence orbitals changes in going down a particular group.
It is important to be able to give the electron configuration for each of the main-group elements. This is most easily done by using the periodic table. If you understand how the table is organized, it is not necessary to memorize the order in which the orbitals fill. Re-view Figs. 7.27 and 7.28 to make sure that you understand the correspondence between the orbitals and the periods and groups.
Predicting the configurations of the transition metals (3d, 4d, and 5d elements), the lanthanides (4f elements), and the actinides (5f elements) is somewhat more difficult because there are many exceptions of the type encountered in the first-row transition met-als (the 3d elements). You should memorize the configurations of chromium and copper, the two exceptions in the first-row transition metals, since these elements are often encountered.
Electron Configurations Give the electron configurations for sulfur (S), cadmium (Cd), hafnium (Hf), and radium (Ra) using the periodic table inside the front cover of this book.
Solution Sulfur is element 16 and resides in Period 3, where the 3p orbitals are being filled (see Fig. 7.29). Since sulfur is the fourth among the “3p elements,” it must have four 3p elec-trons. Its configuration is Cadmium is element 48 and is located in Period 5 at the end of the 4d transition met-als, as shown in Fig. 7.29. It is the tenth element in the series and thus has 10 electrons in the 4d orbitals, in addition to the 2 electrons in the 5s orbital. The configuration is Hafnium is element 72 and is found in Period 6, as shown in Fig. 7.29. Note that it occurs just after the lanthanide series. Thus the 4f orbitals are already filled. Hafnium is Cd: 1s22s22p63s23p64s23d104p65s24d10 or 3Kr45s24d10 S: 1s22s22p63s23p4 or 3Ne43s23p4 Sample Exercise 7.7 FIGURE 7.29 The positions of the elements considered in Sample Exercise 7.7.
When an electron configuration is given in this text, the orbitals are listed in the order in which they fill.
Cr: [Ar]4s13d 5 Cu: [Ar]4s13d10 Period 1 2 3 4 5 6 7 Group 1A 2A 3A 4A 5A 6A 7A 8A 1s 2s 3s 4s 5s 6s 7s La Ac Ra 3d 4d 5d Cd 2p 3p 4p 5p 6p S 1s 4f 5f Hf 6d 7.12 Periodic Trends in Atomic Properties 309 the second member of the 5d transition series and has two 5d electrons. The configura-tion is Radium is element 88 and is in Period 7 (and Group 2A), as shown in Fig. 7.29. Thus radium has two electrons in the 7s orbital, and the configuration is See Exercises 7.69 through 7.72.
7.12 Periodic Trends in Atomic Properties We have developed a fairly complete picture of polyelectronic atoms. Although the model is rather crude because the nuclear attractions and electron repulsions are simply lumped together, it is very successful in accounting for the periodic table of elements. We will next use the model to account for the observed trends in several important atomic prop-erties: ionization energy, electron affinity, and atomic size.
Ionization Energy Ionization energy is the energy required to remove an electron from a gaseous atom or ion: where the atom or ion is assumed to be in its ground state.
To introduce some of the characteristics of ionization energy, we will consider the energy required to remove several electrons in succession from aluminum in the gaseous state. The ionization energies are Several important points can be illustrated from these results. In a stepwise ionization process, it is always the highest-energy electron (the one bound least tightly) that is removed first. The first ionization energy I1 is the energy required to remove the highest-energy electron of an atom. The first electron removed from the aluminum atom comes from the 3p orbital (Al has the electron configuration [Ne]3s23p1). The second electron comes from the 3s orbital (since Al has the configuration [Ne]3s2). Note that the value of I1 is considerably smaller than the value of I2, the second ionization energy.
This makes sense for several reasons. The primary factor is simply charge. Note that the first electron is removed from a neutral atom (Al), whereas the second electron is re-moved from a 1 ion (Al). The increase in positive charge binds the electrons more firmly, and the ionization energy increases. The same trend shows up in the third (I3) and fourth (I4) ionization energies, where the electron is removed from the Al2 and Al3 ions, respectively.
The increase in successive ionization energies for an atom also can be interpreted us-ing our simple model for polyelectronic atoms. The increase in ionization energy from I1 to I2 makes sense because the first electron is removed from a 3p orbital that is higher in energy than the 3s orbital from which the second electron is removed. The largest jump in ionization energy by far occurs in going from the third ionization energy (I3) to the fourth (I4). This is so because I4 corresponds to removing a core electron (Al3 has the configu-ration 1s22s22p6), and core electrons are bound much more tightly than valence electrons. Al31g2 ¡ Al41g2 e I4 11,600 kJ/mol Al21g2 ¡ Al31g2 e I3 2740 kJ/mol Al1g2 ¡ Al21g2 e I2 1815 kJ/mol Al1g2 ¡ Al1g2 e I1 580 kJ/mol X1g2 ¡ X1g2 e Ra: 1s22s22p63s23p64s23d104p65s24d105p66s24f 145d106p67s2 or 3Rn47s2 Hf: 1s22s22p63s23p64s23d104p65s24d105p66s24f 145d2 or 3Xe46s24f 145d2 Ionization energy results in the formation of a positive ion.
Setting the aluminum cap on the Washington Monument in 1884. At that time, aluminum was regarded as a precious metal.
Visualization: Periodic Table Trends 310 Chapter Seven Atomic Structure and Periodicity Table 7.5 gives the values of ionization energies for all the Period 3 elements. Note the large jump in energy in each case in going from removal of valence electrons to removal of core electrons.
The values of the first ionization energies for the elements in the first six periods of the periodic table are graphed in Fig. 7.30. Note that in general as we go across a period from left to right, the first ionization energy increases. This is consistent with the idea that electrons added in the same principal quantum level do not completely shield the in-creasing nuclear charge caused by the added protons. Thus electrons in the same princi-pal quantum level are generally more strongly bound as we move to the right on the pe-riodic table, and there is a general increase in ionization energy values as electrons are added to a given principal quantum level.
On the other hand, first ionization energy decreases in going down a group. This can be seen most clearly by focusing on the Group 1A elements (the alkali metals) and the Group 8A elements (the noble gases), as shown in Table 7.6. The main reason for the de-crease in ionization energy in going down a group is that the electrons being removed are, on average, farther from the nucleus. As n increases, the size of the orbital increases, and the electron is easier to remove.
First ionization energy increases across a period and decreases down a group.
FIGURE 7.30 The values of first ionization energy for the elements in the first six periods. In general, ionization energy decreases in going down a group. For example, note the decrease in values for Group 1A and Group 8A. In general, ionization energy increases in going left to right across a period. For example, note the sharp increase going across Period 2 from lithium through neon.
TABLE 7.6 First Ionization Energies for the Alkali Metals and Noble Gases Atom I1(kJ/mol) Group 1A Li 520 Na 495 K 419 Rb 409 Cs 382 Group 8A He 2377 Ne 2088 Ar 1527 Kr 1356 Xe 1176 Rn 1042 General decrease General increase TABLE 7.5 Successive Ionization Energies in Kilojoules per Mole for the Elements in Period 3 Element I1 I2 I3 I4 I5 I6 I7 Na 495 4560 Mg 735 1445 7730 Core electrons Al 580 1815 2740 11,600 Si 780 1575 3220 4350 16,100 P 1060 1890 2905 4950 6270 21,200 S 1005 2260 3375 4565 6950 8490 27,000 Cl 1255 2295 3850 5160 6560 9360 11,000 Ar 1527 2665 3945 5770 7230 8780 12,000 Note the large jump in ionization energy in going from removal of valence electrons to removal of core electrons.
Ionization energy (kJ/mol) 10 Atomic number 0 Li Na K Rb Cs H B Be C O N F Mg Al Cl S P Zn As Br Cd Tl He Ne Ar Kr Xe Rn 18 36 54 86 500 1000 1500 2000 2500 Period 2 Period 3 Period 4 Period 5 Period 6 7.12 Periodic Trends in Atomic Properties 311 In Fig. 7.30 we see that there are some discontinuities in ionization energy in going across a period. For example, for Period 2, discontinuities occur in going from beryllium to boron and from nitrogen to oxygen. These exceptions to the normal trend can be ex-plained in terms of electron repulsions. The decrease in ionization energy in going from beryllium to boron reflects the fact that the electrons in the filled 2s orbital provide some shielding for electrons in the 2p orbital from the nuclear charge. The decrease in ioniza-tion energy in going from nitrogen to oxygen reflects the extra electron repulsions in the doubly occupied oxygen 2p orbital.
The ionization energies for the representative elements are summarized in Fig. 7.31.
Trends in Ionization Energies The first ionization energy for phosphorus is 1060 kJ/mol, and that for sulfur is 1005 kJ/mol. Why?
Solution Phosphorus and sulfur are neighboring elements in Period 3 of the periodic table and have the following valence electron configurations: Phosphorus is 3s23p3, and sulfur is 3s23p4.
Ordinarily, the first ionization energy increases as we go across a period, so we might expect sulfur to have a greater ionization energy than phosphorus. However, in this case the fourth p electron in sulfur must be placed in an already occupied orbital. The electron–electron repulsions that result cause this electron to be more easily removed than might be expected.
See Exercises 7.93 and 7.94.
Ionization Energies Consider atoms with the following electron configurations: Which atom has the largest first ionization energy, and which one has the smallest second ionization energy? Explain your choices.
1s22s22p63s2 1s22s22p63s1 1s22s22p6 FIGURE 7.31 Trends in ionization energies (kJ/mol) for the representative elements.
Sample Exercise 7.8 Sample Exercise 7.9 H 1311 1 2 3 4 5 6 1A 2A 3A 4A 5A 6A 7A 8A Be 899 B 800 C 1086 N 1402 S 1005 Te 869 Se 941 O 1314 He 2377 Si 780 Pb 715 Bi 703 Xe 1176 Kr 1356 Ar 1527 Ne 2088 F 1681 Cl 1255 P 1060 As 947 Rn 1042 Sb 834 Po 813 At (926) I 1009 Br 1143 Sn 708 Ge 761 Al 580 Ga 579 In 558 Tl 589 Mg 735 Ca 590 Sr 549 Ba 503 Li 520 Na 495 K 419 Rb 409 Cs 382 312 Chapter Seven Atomic Structure and Periodicity Solution The atom with the largest value of I1 is the one with the configuration 1s22s22p6 (this is the neon atom), because this element is found at the right end of Period 2. Since the 2p electrons do not shield each other very effectively, I1 will be relatively large. The other configurations given include 3s electrons. These electrons are effectively shielded by the core electrons and are farther from the nucleus than the 2p electrons in neon. Thus I1 for these atoms will be smaller than for neon.
The atom with the smallest value of I2 is the one with the configuration 1s22s22p63s2 (the magnesium atom). For magnesium, both I1 and I2 involve valence electrons. For the atom with the configuration 1s22s22p63s1 (sodium), the second electron lost (correspond-ing to I2) is a core electron (from a 2p orbital).
See Exercises 7.121 and 7.123.
Electron Affinity Electron affinity is the energy change associated with the addition of an electron to a gaseous atom: Because two different conventions have been used, there is a good deal of confusion in the chemical literature about the signs for electron affinity values. Electron affinity has been defined in many textbooks as the energy released when an electron is added to a gaseous atom. This convention requires that a positive sign be attached to an exothermic addition of an electron to an atom, which opposes normal thermodynamic conventions.
Therefore, in this book we define electron affinity as a change in energy, which means that if the addition of the electron is exothermic, the corresponding value for electron affin-ity will carry a negative sign.
Figure 7.32 shows the electron affinity values for the atoms among the first 20 ele-ments that form stable, isolated negative ions—that is, the atoms that undergo the addition of an electron as shown above. As expected, all these elements have negative (exothermic) electron affinities. Note that the more negative the energy, the greater the quantity of en-ergy released. Although electron affinities generally become more negative from left to right across a period, there are several exceptions to this rule in each period. The depen-dence of electron affinity on atomic number can be explained by considering the changes in electron repulsions as a function of electron configurations. For example, the fact that the nitrogen atom does not form a stable, isolated N(g) ion, whereas carbon forms C(g), reflects the difference in the electron configurations of these atoms. An electron added to nitrogen (1s22s22p3) to form the N(g) ion (1s22s22p4) would have to occupy a 2p orbital that already contains one electron. The extra repulsion between the electrons in this dou-bly occupied orbital causes N(g) to be unstable. When an electron is added to carbon (1s22s22p2) to form the C(g) ion (1s22s22p3), no such extra repulsions occur.
In contrast to the nitrogen atom, the oxygen atom can add one electron to form the sta-ble O(g) ion. Presumably oxygen’s greater nuclear charge compared with that of nitrogen X1g2 e ¡ X1g2 Electron affinity is associated with the production of a negative ion.
The sign convention for electron affinity values follows the convention for energy changes used in Chapter 6.
Electron affinity (kJ/mol) Atomic number 2 4 6 8 10 –300 –200 –100 0 12 14 16 18 20 H Li B C O F Na Al Si P Ca S Cl K FIGURE 7.32 The electron affinity values for atoms among the first 20 elements that form stable, isolated X ions. The lines shown connect adjacent elements. The absence of a line indicates missing elements (He, Be, N, Ne, Mg, and Ar) whose atoms do not add an electron exothermically and thus do not form stable, isolated X ions.
7.12 Periodic Trends in Atomic Properties 313 is sufficient to overcome the repulsion associated with putting a second electron into an al-ready occupied 2p orbital. However, it should be noted that a second electron cannot be added to an oxygen atom [O(g) e n O2(g)] to form an isolated oxide ion. This out-come seems strange in view of the many stable oxide compounds (MgO, Fe2O3, and so on) that are known. As we will discuss in detail in Chapter 8, the O2 ion is stabilized in ionic compounds by the large attractions that occur among the positive ions and the oxide ions.
When we go down a group, electron affinity should become more positive (less energy released), since the electron is added at increasing distances from the nucleus. Although this is generally the case, the changes in electron affinity in going down most groups are rela-tively small, and numerous exceptions occur. This behavior is demonstrated by the electron affinities of the Group 7A elements (the halogens) shown in Table 7.7. Note that the range of values is quite small compared with the changes that typically occur across a period. Also note that although chlorine, bromine, and iodine show the expected trend, the energy re-leased when an electron is added to fluorine is smaller than might be expected. This smaller energy release has been attributed to the small size of the 2p orbitals. Because the electrons must be very close together in these orbitals, there are unusually large electron–electron re-pulsions. In the other halogens with their larger orbitals, the repulsions are not as severe.
Atomic Radius Just as the size of an orbital cannot be specified exactly, neither can the size of an atom.
We must make some arbitrary choices to obtain values for atomic radii. These values can be obtained by measuring the distances between atoms in chemical compounds. For ex-ample, in the bromine molecule, the distance between the two nuclei is known to be 228 pm. The bromine atomic radius is assumed to be half this distance, or 114 pm, as shown in Fig. 7.33. These radii are often called covalent atomic radii because of the way they are determined (from the distances between atoms in covalent bonds).
For nonmetallic atoms that do not form diatomic molecules, the atomic radii are es-timated from their various covalent compounds. The radii for metal atoms (called metal-lic radii) are obtained from half the distance between metal atoms in solid metal crystals.
The values of the atomic radii for the representative elements are shown in Fig. 7.34.
Note that these values are significantly smaller than might be expected from the 90% electron density volumes of isolated atoms, because when atoms form bonds, their electron “clouds” interpenetrate. However, these values form a self-consistent data set that can be used to discuss the trends in atomic radii.
Note from Fig. 7.34 that the atomic radii decrease in going from left to right across a period. This decrease can be explained in terms of the increasing effective nuclear charge (decreasing shielding) in going from left to right. This means that the valence electrons are drawn closer to the nucleus, decreasing the size of the atom.
Atomic radius increases down a group, because of the increases in the orbital sizes in successive principal quantum levels.
Trends in Radii Predict the trend in radius for the following ions: Be2, Mg2, Ca2, and Sr2.
Solution All these ions are formed by removing two electrons from an atom of a Group 2A element.
In going from beryllium to strontium, we are going down the group, so the sizes increase: h h Smallest radius Largest radius See Exercises 7.85, 7.86, and 7.89.
Be2 6 Mg2 6 Ca2 6 Sr2 TABLE 7.7 Electron Affinities of the Halogens Electron Affinity Atom (kJ/mol) F 327.8 Cl 348.7 Br 324.5 I 295.2 Sample Exercise 7.10 FIGURE 7.33 The radius of an atom (r) is defined as half the distance between the nuclei in a mole-cule consisting of identical atoms.
Br Br 2r Visualization: Determining the Atomic Radius of a Nonmetal (Chlorine) Visualization: Determining the Atomic Radius of a Nonmetal (Carbon) 314 Chapter Seven Atomic Structure and Periodicity 7.13 The Properties of a Group: The Alkali Metals We have seen that the periodic table originated as a way to portray the systematic properties of the elements. Mendeleev was primarily responsible for first showing its usefulness in correlating and predicting the elemental properties. In this section we will summarize much of the information available from the table. We also will illustrate the usefulness of the table by discussing the properties of a representative group, the alkali metals.
Information Contained in the Periodic Table 1. The essence of the periodic table is that the groups of representative elements exhibit similar chemical properties that change in a regular way. The quantum mechanical model of the atom has allowed us to understand the basis for the similarity of properties in a group—that each group member has the same valence electron configuration. It is the number and type of valence electrons that primarily determine an atom’s chemistry.
2. One of the most valuable types of information available from the periodic table is the electron configuration of any representative element. If you understand the organization FIGURE 7.34 Atomic radii (in picometers) for selected atoms. Note that atomic radius decreases going across a period and increases going down a group. The values for the noble gases are estimated, because data from bonded atoms are lacking.
H 37 He 32 Li 152 Be 113 B 88 C 77 N 70 O 66 F 64 Ne 69 Na 186 Mg 160 Al 143 Si 117 P 110 S 104 Cl 99 Ar 97 K 227 Ca 197 Ga 122 Ge 122 As 121 Se 117 Br 114 Kr 110 Rb 247 Sr 215 In 163 Sn 140 Sb 141 Te 143 I 133 Xe 130 Cs 265 Ba 217 Tl 170 Pb 175 Bi 155 Po 167 At 140 Rn 145 8A 7A 6A 5A 4A 3A 2A 1A Atomic radius increases Atomic radius decreases 7.13 The Properties of a Group: The Alkali Metals 315 of the table, you will not need to memorize electron configurations for these elements.
Although the predicted electron configurations for transition metals are sometimes in-correct, this is not a serious problem. You should, however, memorize the configura-tions of two exceptions, chromium and copper, since these 3d transition elements are found in many important compounds.
3. As we mentioned in Chapter 2, certain groups in the periodic table have special names.
These are summarized in Fig. 7.35. Groups are often referred to by these names, so you should learn them.
4. The most basic division of the elements in the periodic table is into metals and non-metals. The most important chemical property of a metal atom is the tendency to give up one or more electrons to form a positive ion; metals tend to have low ion-ization energies. The metallic elements are found on the left side of the table, as shown in Fig. 7.35. The most chemically reactive metals are found on the lower left-hand portion of the table, where the ionization energies are smallest. The most distinctive chemical property of a nonmetal atom is the ability to gain one or more electrons to form an anion when reacting with a metal. Thus nonmetals are elements FIGURE 7.35 Special names for groups in the periodic table.
Metals and nonmetals were first discussed in Chapter 2.
Alkali metals 1A H 2A Alkaline earth metals Transition elements Halogens 3A 4A 5A 6A 7A Noble gases 8A Lanthanides Actinides 1A 2A 3A 4A 5A 6A 7A 8A Metals Nonmetals Metalloids Lanthanides Actinides 316 Chapter Seven Atomic Structure and Periodicity with large ionization energies and the most negative electron affinities. The nonmetals are found on the right side of the table, with the most reactive ones in the upper right-hand corner, except for the noble gas elements, which are quite unreactive. The division into metals and nonmetals shown in Fig. 7.35 is only approximate. Many elements along the division line exhibit both metallic and nonmetallic properties un-der certain circumstances. These elements are often called metalloids, or sometimes semimetals.
The Alkali Metals The metals of Group 1A, the alkali metals, illustrate very well the relationships among the properties of the elements in a group. Lithium, sodium, potassium, rubidium, cesium, and francium are the most chemically reactive of the metals. We will not discuss fran-cium here because it occurs in nature in only very small quantities. Although hydrogen is found in Group 1A of the periodic table, it behaves as a nonmetal, in contrast to the other members of that group. The fundamental reason for hydrogen’s nonmetallic character is its very small size (see Fig. 7.34). The electron in the small 1s orbital is bound tightly to the nucleus.
Some important properties of the first five alkali metals are shown in Table 7.8. The data in Table 7.8 show that in going down the group, the first ionization energy decreases and the atomic radius increases. This agrees with the general trends discussed in Sec-tion 7.12.
The overall increase in density in going down Group 1A is typical of all groups. This occurs because atomic mass generally increases more rapidly than atomic size. Thus there is more mass per unit volume for each succeeding element.
The smooth decrease in melting point and boiling point in going down Group 1A is not typical; in most other groups more complicated behavior occurs. Note that the melt-ing point of cesium is only 29C. Cesium can be melted readily using only the heat from your hand. This is very unusual—metals typically have rather high melting points. For example, tungsten melts at 3410C. The only other metals with low melting points are mercury (mp 38C) and gallium (mp 30C).
The chemical property most characteristic of a metal is the ability to lose its valence electrons. The Group 1A elements are very reactive. They have low ionization energies and react with nonmetals to form ionic solids. A typical example involves the reaction of sodium with chlorine to form sodium chloride: where sodium chloride contains Na and Cl ions. This is an oxidation–reduction reac-tion in which chlorine oxidizes sodium. In the reactions between metals and nonmetals, 2Na1s2 Cl21g2 ¡ 2NaCl1s2 Hydrogen will be discussed further in Chapter 19.
TABLE 7.8 Properties of Five Alkali Metals First Atomic Valence Density Ionization (covalent) Ionic (M) Electron at 25C mp bp Energy Radius Radius Element Configuration (g/cm3) (C) (C) (kJ/mol) (pm) (pm) Li 2s1 0.53 180 1330 520 152 60 Na 3s1 0.97 98 892 495 186 95 K 4s1 0.86 64 760 419 227 133 Rb 5s1 1.53 39 668 409 247 148 Cs 6s1 1.87 29 690 382 265 169 Other groups will be discussed in Chapters 19 and 20.
Oxidation–reduction reactions were discussed in Chapter 4.
7.13 The Properties of a Group: The Alkali Metals 317 it is typical for the nonmetal to behave as the oxidizing agent and the metal to behave as the reducing agent, as shown by the following reactions: Contains Na and S2 ions Contains Li and N3 ions Contains Na and O2 2 ions For reactions of the types just shown, the relative reducing powers of the alkali met-als can be predicted from the first ionization energies listed in Table 7.8. Since it is much easier to remove an electron from a cesium atom than from a lithium atom, cesium should be the better reducing agent. The expected trend in reducing ability is This order is observed experimentally for direct reactions between the solid alkali metals and nonmetals. However, this is not the order for reducing ability found when the alkali Cs 7 Rb 7 K 7 Na 7 Li 2Na1s2 O21g2 ¡ Na2O21s2 6Li1s2 N21g2 ¡ 2Li3N1s2 2Na1s2 S1s2 ¡ Na2S1s2 Potassium reacts violently with water.
CHEMICAL IMPACT Potassium—Too Much of a Good Thing Can Kill You P otassium is widely recognized as an essential element. In fact, our daily requirement for potassium is more than twice that for sodium. Because most foods contain potassium, serious defi-ciency of this element in humans is rare.
However, potassium deficiency can be caused by kidney malfunction or by the use of certain diuretics. Potassium deficiency leads to muscle weakness, ir-regular heartbeat, and depression.
Potassium is found in the fluids of the body as the K ion, and its presence is essential to the operation of our ner-vous system. The passage of impulses along the nerves requires the flow of K (and Na) through channels in the mem-branes of the nerve cells. Failure of this ion flow prevents nerve transmissions and results in death. For example, the black mamba snake kills its victims by injecting a venom that blocks the potas-sium channels in the nerve cells.
Although a steady intake of potassium is essential to pre-serve life, ironically, too much potassium can be lethal. In fact, the deadly ingredient in the drug mixture used for executing criminals is potassium chloride. Injection of a large amount of a potassium chloride solution produces an excess of K ion in the fluids surrounding the cells and prevents the essential flow of K out the cells to allow nerve impulses to occur. This causes the heart to stop beating. Unlike other forms of execu-tion, death by lethal injection of potassium chloride does not harm the organs of the body. Thus condemned criminals who are executed in this manner could potentially donate their or-gans for transplants. However, this idea is very controversial.
The black mamba snake’s venom kills by blocking the potassium channels in the nerve cells of victims.
318 Chapter Seven Atomic Structure and Periodicity metals react in aqueous solution. For example, the reduction of water by an alkali metal is very vigorous and exothermic: The order of reducing abilities observed for this reaction for the first three group members is In the gas phase potassium loses an electron more easily than sodium, and sodium more easily than lithium. Thus it is surprising that lithium is the best reducing agent toward water.
This reversal occurs because the formation of the M ions in aqueous solution is strongly influenced by the hydration of these ions by the polar water molecules. The hy-dration energy of an ion represents the change in energy that occurs when water mole-cules attach to the M ion. The hydration energies for the Li, Na, and K ions (shown in Table 7.9) indicate that the process is exothermic in each case. However, nearly twice as much energy is released by the hydration of the Li ion as for the K ion. This difference is caused by size effects; the Li ion is much smaller than the K ion, and thus its charge density (charge per unit volume) is also much greater. This means that the polar water molecules are more strongly attracted to the small Li ion. Because the Li ion is so strongly hydrated, its formation from the lithium atom occurs more readily than the formation of the K ion from the potassium atom. Although a potassium atom in the gas phase loses its valence electron more readily than a lithium atom in the gas phase, the opposite is true in aqueous solution. This anomaly is an example of the importance of the polarity of the water molecule in aqueous reactions.
There is one more surprise involving the highly exothermic reactions of the alkali metals with water. Experiments show that in water lithium is the best reducing agent, so we might expect that lithium should react the most violently with water. However, this is not true. Sodium and potassium react much more vigorously. Why is this so? The answer lies in the relatively high melting point of lithium. When sodium and potassium react with water, the heat evolved causes them to melt, giving a larger area of contact with water.
Lithium, on the other hand, does not melt under these conditions and reacts more slowly.
This illustrates the important principle (which we will discuss in detail in Chapter 12) that the energy change for a reaction and the rate at which it occurs are not necessarily related.
In this section we have seen that the trends in atomic properties summarized by the periodic table can be a great help in understanding the chemical behavior of the elements.
This fact will be emphasized over and over as we proceed in our study of chemistry.
Li 7 K 7 Na 2M1s2 2H2O1l2 ¡ H21g2 2M1aq2 2OH1aq2 energy Key Terms Section 7.1 electromagnetic radiation wavelength frequency Section 7.2 Planck’s constant quantization photon photoelectric effect E mc2 dual nature of light diffraction diffraction pattern Section 7.3 continuous spectrum line spectrum For Review Electromagnetic radiation Characterized by its wavelength ( ), frequency ( ), and speed (c 2.9979 108 m/s) c Can be viewed as a stream of “particles” called photons, each with energy h, where h is Planck’s constant (6.626 1034 J s) Photoelectric effect When light strikes a metal surface, electrons are emitted Analysis of the kinetic energy and numbers of the emitted electrons led Einstein to suggest that electromagnetic radiation can be viewed as a stream of photons Hydrogen spectrum The emission spectrum of hydrogen shows discrete wavelengths Indicates that hydrogen has discrete energy levels TABLE 7.9 Hydration Energies for Li, Na, and K Ions Hydration Energy Ion (kJ/mol) Li 510 Na 402 K 314 For Review 319 Section 7.4 quantum model ground state Section 7.5 standing wave wave function orbital quantum (wave) mechanical model Heisenberg uncertainty principle probability distribution radial probability distribution Section 7.6 quantum numbers principal quantum number (n) angular momentum quantum number () magnetic quantum number (m) subshell Section 7.7 nodal surface node degenerate orbital Section 7.8 electron spin electron spin quantum number Pauli exclusion principle Section 7.9 polyelectronic atoms Section 7.11 aufbau principle Hund’s rule valence electrons core electrons transition metals lanthanide series actinide series main-group elements (representative elements) Section 7.12 first ionization energy second ionization energy electron affinity atomic radii Section 7.13 metalloids (semimetals) Bohr model of the hydrogen atom Using the data from the hydrogen spectrum and assuming angular momentum to be quantized, Bohr devised a model in which the electron traveled in circular orbits Although an important pioneering effort, this model proved to be entirely incorrect Wave (quantum) mechanical model An electron is described as a standing wave The square of the wave function (often called an orbital) gives a probability distri-bution for the electron position The exact position of the electron is never known, which is consistent with the Heisen-berg uncertainty principle: it is impossible to know accurately both the position and the momentum of a particle simultaneously Probability maps are used to define orbital shapes Orbitals are characterized by the quantum numbers n, , and m Electron spin Described by the spin quantum number ms which can have values of Pauli exclusion principle: no two electrons in a given atom can have the same set of quantum numbers n, , m, and ms Only two electrons with opposite spins can occupy a given orbital Periodic table By populating the orbitals from the wave mechanical model (the aufbau principle), the form of the periodic table can be explained According to the wave mechanical model, atoms in a given group have the same va-lence (outer) electron configuration The trends in properties such as ionization energies and atomic radii can be explained in terms of the concepts of nuclear attraction, electron repulsions, shielding, and penetration REVIEW QUESTIONS 1. Four types of electromagnetic radiation (EMR) are ultraviolet, microwaves, gamma rays, and visible. All of these types of EMR can be characterized by wavelength, frequency, photon energy, and speed of travel. Define these terms and rank the four types of electromagnetic radiation in order of increasing wave-length, frequency, photon energy, and speed.
2. Characterize the Bohr model of the atom. In the Bohr model, what do we mean when we say something is quantized? How does the Bohr model of the hydrogen atom explain the hydrogen emission spectrum? Why is the Bohr model fundamentally incorrect?
3. What experimental evidence supports the quantum theory of light? Explain the wave-particle duality of all matter. For what size particles must one consider both the wave and the particle properties?
4. List the most important ideas of the quantum mechanical model of the atom. In-clude in your discussion the terms or names wave function, orbital, Heisenberg uncertainty principle, de Broglie, Schrödinger, and probability distribution.
5. What are quantum numbers? What information do we get from the quantum numbers n, , and m? We define a spin quantum number (ms), but do we know that an electron literally spins?
6. How do 2p orbitals differ from each other? How do 2p and 3p orbitals differ from each other? What is a nodal surface in an atomic orbital? What is wrong with 1p, 1d, 2d, 1f, 2f, and 3f orbitals? Explain what we mean when we say that a 4s electron is more penetrating than a 3d electron.
1 2 320 Chapter Seven Atomic Structure and Periodicity 7. Four blocks of elements in a periodic table refer to various atomic orbitals being filled. What are the four blocks and the corresponding orbitals? How do you get the energy ordering of the atomic orbitals from the periodic table? What is the aufbau principle? Hund’s rule? The Pauli exclusion principle? There are two common exceptions to the ground-state electron configuration for elements 1–36 as predicted by the periodic table. What are they?
8. What is the difference between core electrons and valence electrons? Why do we emphasize the valence electrons in an atom when discussing atomic proper-ties? What is the relationship between valence electrons and elements in the same group of the periodic table?
9. Using the element phosphorus as an example, write the equation for a process in which the energy change will correspond to the ionization energy and to the electron affinity.
Explain why the first ionization energy tends to increase as one proceeds from left to right across a period. Why is the first ionization energy of aluminum lower than that of magnesium, and the first ionization energy of sulfur lower than that of phosphorus?
Why do the successive ionization energies of an atom always increase? Note the successive ionization energies for silicon given in Table 7.5. Would you ex-pect to see any large jumps between successive ionization energies of silicon as you removed all the electrons, one by one, beyond those shown in the table?
10. The radius trend and the ionization energy trend are exact opposites. Does this make sense? Define electron affinity. Electron affinity values are both exother-mic (negative) and endothermic (positive). However, ionization energy values are always endothermic (positive). Explain.
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. What does it mean for something to have wavelike properties?
Particulate properties? Electromagnetic radiation can be discussed in terms of both particles and waves. Explain the experimental verification for each of these views.
2. Defend and criticize Bohr’s model. Why was it reasonable that such a model was proposed, and what evidence was there that it “works”? Why do we no longer “believe” in it?
3. The first four ionization energies for the elements X and Y are shown below. The units are not kJ/mol.
Identify the elements X and Y. There may be more than one cor-rect answer, so explain completely.
X Y First 170 200 Second 350 400 Third 1800 3500 Fourth 2500 5000 4. Compare the first ionization energy of helium to its second ion-ization energy, remembering that both electrons come from the 1s orbital. Explain the difference without using actual numbers from the text.
5. Which has the larger second ionization energy, lithium or beryl-lium? Why?
6. Explain why a graph of ionization energy versus atomic number (across a row) is not linear. Where are the exceptions? Why are there exceptions?
7. Without referring to your text, predict the trend of second ion-ization energies for the elements sodium through argon. Com-pare your answer with Table 7.5. Explain any differences.
8. Account for the fact that the line that separates the metals from the nonmetals on the periodic table is diagonal downward to the right instead of horizontal or vertical.
9. Explain electron from a quantum mechanical perspective, in-cluding a discussion of atomic radii, probabilities, and orbitals.
10. Choose the best response for the following. The ionization en-ergy for the chlorine atom is equal in magnitude to the electron affinity for a. the Cl atom.
b. the Cl ion.
c. the Cl ion.
d. the F atom.
e. none of these.
Exercises 321 Explain each choice. Justify your choice, and for the choices you did not select, explain what is incorrect about them.
11. Consider the following statement: “The ionization energy for the potassium atom is negative, because when K loses an electron to become K, it achieves a noble gas electron configuration.” Indicate everything that is correct in this statement. Indicate everything that is incorrect. Correct the incorrect information and explain.
12. In going across a row of the periodic table, electrons are added and ionization energy generally increases. In going down a col-umn of the periodic table, electrons are also being added but ion-ization energy decreases. Explain.
13. How does probability fit into the description of the atom?
14. What is meant by an orbital?
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of the book and a solution appears in the Solutions Guide.
Questions 15. What type of relationship (direct or inverse) exists between wave-length, frequency, and photon energy? What does a photon en-ergy unit of a Joule equal?
16. Explain the photoelectric effect.
17. How does the wavelength of a fast-pitched baseball compare to the wavelength of an electron traveling at 1/10 the speed of light? What is the significance of this comparison? See Sample Exercise 7.3.
18. The Bohr model only works for one electron species. Why do we discuss it in this text (what’s good about it)?
19. Describe the significance of the radial probability distribution shown in Fig. 7.12(b).
20. The periodic table consists of four blocks of elements which cor-respond to s, p, d, and f orbitals being filled. After f orbitals come g and h orbitals. In theory, if a g block and an h block of elements existed, how long would the rows of g and h elements be in this theoretical periodic table?
21. Many times the claim is made that subshells half-filled with elec-trons are particularly stable. Can you suggest a possible physi-cal basis for this claim?
22. Diagonal relationships in the periodic table exist as well as the vertical relationships. For example, Be and Al are similar in some of their properties, as are B and Si. Rationalize why these diag-onal relationships hold for properties such as size, ionization en-ergy, and electron affinity.
23. Elements with very large ionization energies also tend to have highly exothermic electron affinities. Explain. Which group of elements would you expect to be an exception to this statement?
24. The changes in electron affinity as one goes down a group in the periodic table are not nearly as large as the variations in ioniza-tion energies. Why?
25. Why is it much harder to explain the line spectra of polyelec-tronic atoms and ions than it is to explain the line spectra of hy-drogen and hydrogenlike ions?
26. Scientists use emission spectra to confirm the presence of an element in materials of unknown composition. Why is this possible?
27. Does the minimization of electron–electron repulsions correlate with Hund’s rule?
28. In the hydrogen atom, what is the physical significance of the state for which n and E 0?
29. The work function is the energy required to remove an electron from an atom on the surface of a metal. How does this defini-tion differ from that for ionization energy?
30. Many more anhydrous lithium salts are hygroscopic (readily ab-sorb water) than are those of the other alkali metals. Explain.
Exercises In this section similar exercises are paired.
Light and Matter 31. Photosynthesis uses 660-nm light to convert CO2 and H2O into glucose and O2. Calculate the frequency of this light.
32. An FM radio station broadcasts at 99.5 MHz. Calculate the wave-length of the corresponding radio waves.
33. Microwave radiation has a wavelength on the order of 1.0 cm.
Calculate the frequency and the energy of a single photon of this radiation. Calculate the energy of an Avogadro’s number of pho-tons (called an einstein) of this radiation.
34. A photon of ultraviolet (UV) light possesses enough energy to mutate a strand of human DNA. What is the energy of a single UV photon and a mole of UV photons having a wavelength of 25 nm?
35. Consider the following waves representing electromagnetic radiation: Which wave has the longer wavelength? Calculate the wave-length. Which wave has the higher frequency and larger photon energy? Calculate these values. Which wave has the faster ve-locity? What type of electromagnetic radiation are illustrated?
36. One type of electromagnetic radiation has a frequency of 107.1 MHz, another type has a wavelength of 2.12 1010 m, and another type of electromagnetic radiation has photons with energy equal to 3.97 1019 J/photon. Identify each type of electromagnetic radiation and place them in order of increasing photon energy and increasing frequency.
37. Carbon absorbs energy at a wavelength of 150. nm. The total amount of energy emitted by a carbon sample is 1.98 105 J.
Calculate the number of carbon atoms present in the sample, as-suming that each atom emits one photon.
1.6 x 10–3 m Wave a Wave b q 322 Chapter Seven Atomic Structure and Periodicity 49. Does a photon of visible light ( 400 to 700 nm) have suffi-cient energy to excite an electron in a hydrogen atom from the n 1 to the n 5 energy state? from the n 2 to the n 6 energy state?
50. An electron is excited from the n 1 ground state to the n 3 state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true.
a. It takes more energy to ionize (completely remove) the elec-tron from n 3 than from the ground state.
b. The electron is farther from the nucleus on average in the n 3 state than in the n 1 state.
c. The wavelength of light emitted if the electron drops from n 3 to n 2 will be shorter than the wavelength of light emitted if the electron falls from n 3 to n 1.
d. The wavelength of light emitted when the electron returns to the ground state from n 3 will be the same as the wave-length of light absorbed to go from n 1 to n 3.
e. For n 3, the electron is in the first excited state.
51. Calculate the maximum wavelength of light capable of remov-ing an electron for a hydrogen atom from the energy state char-acterized by n 1. by n 2.
52. Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can com-pletely remove (ionize) the electron from the H atom is 1460 nm.
What is the initial excited state for the electron (n ?)?
53. An excited hydrogen atom with an electron in the n 5 state emits light having a frequency of 6.90 1014 s1. Determine the principal quantum level for the final state in this electronic transition.
54. An excited hydrogen atom emits light with a wavelength of 397.2 nm to reach the energy level for which n 2. In which principal quantum level did the electron begin?
Quantum Mechanics, Quantum Numbers, and Orbitals 55. Using the Heisenberg uncertainty principle, calculate x for each of the following.
a. an electron with 0.100 m/s b. a baseball (mass 145 g) with 0.100 m/s c. How does the answer in part a compare with the size of a hydrogen atom?
d. How does the answer in part b correspond to the size of a baseball?
56. The Heisenberg uncertainty principle can be expressed in the form where E represents energy and t represents time. Show that the units for this form are the same as the units for the form used in this chapter: 57. What are the possible values for the quantum numbers n, , and m?
¢x ¢1my2 h 4p ¢E ¢t h 4p 38. A carbon–oxygen double bond in a certain organic molecule ab-sorbs radiation that has a frequency of 6.0 1013 s1.
a. What is the wavelength of this radiation?
b. To what region of the spectrum does this radiation belong?
c. What is the energy of this radiation per photon? Per mole of photons?
d. A carbon–oxygen bond in a different molecule absorbs radi-ation with frequency equal to 5.4 1013 s1. Is this radiation more or less energetic?
39. The work function of an element is the energy required to re-move an electron from the surface of the solid element. The work function for lithium is 279.7 kJ/mol (that is, it takes 279.7 kJ of energy to remove one mole of electrons from one mole of Li atoms on the surface of Li metal). What is the maximum wave-length of light that can remove an electron from an atom on the surface of lithium metal?
40. It takes 208.4 kJ of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the sur-face of solid rubidium? What is the maximum wavelength of light capable of doing this?
41. Calculate the de Broglie wavelength for each of the following.
a. an electron with a velocity 10.% of the speed of light b. a tennis ball (55 g) served at 35 m/s (80 mi/h) 42. Neutron diffraction is used in determining the structures of molecules.
a. Calculate the de Broglie wavelength of a neutron moving at 1.00% of the speed of light.
b. Calculate the velocity of a neutron with a wavelength of 75 pm (1 pm 1012 m).
43. A particle has a velocity that is 90.% of the speed of light. If the wavelength of the particle is 1.5 1015 m, calculate the mass of the particle.
44. Calculate the velocities of electrons with de Broglie wavelengths of 1.0 102 nm and 1.0 nm, respectively.
Hydrogen Atom: The Bohr Model 45. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition?
a.
b.
c.
46. Calculate the wavelength of light emitted when each of the fol-lowing transitions occur in the hydrogen atom. What type of elec-tromagnetic radiation is emitted in each transition?
a.
b.
c.
47. Using vertical lines, indicate the transitions from Exercise 45 on an energy-level diagram for the hydrogen atom (see Fig. 7.8).
48. Using vertical lines, indicate the transitions from Exercise 46 on an energy-level diagram for the hydrogen atom (see Fig. 7.8).
n 5 S n 3 n 5 S n 4 n 4 S n 3 n 2 S n 1 n 4 S n 2 n 3 S n 2 Exercises 323 58. Which of the following orbital designations are incorrect: 1s, 1p, 7d, 9s, 3f, 4f, 2d?
59. Which of the following sets of quantum numbers are not allowed in the hydrogen atom? For the sets of quantum numbers that are incorrect, state what is wrong in each set.
a.
b.
c.
d.
60. Which of the following sets of quantum numbers are not al-lowed? For each incorrect set, state why it is incorrect.
a.
b.
c.
d.
e.
f.
61. What is the physical significance of the value of 2 at a partic-ular point in an atomic orbital?
62. In defining the sizes of orbitals, why must we use an arbitrary value, such as 90% of the probability of finding an electron in that region?
Polyelectronic Atoms 63. How many orbitals in an atom can have the designation 5p, 4d, n 5, n 4?
64. How many electrons in an atom can have the designation 1p, 4f, 7py, 2s, n 3?
65. Give the maximum number of electrons in an atom that can have these quantum numbers: a.
b.
c.
d.
e.
66. Give the maximum number of electrons in an atom that can have these quantum numbers: a.
b.
c.
d.
e.
67. Draw atomic orbital diagrams representing the ground-state elec-tron configuration for each of the following elements.
a. Na b. Co c. Kr How many unpaired electrons are present in each element?
68. For elements 1–36, there are two exceptions to the filling order as predicted from the periodic table. Draw the atomic orbital di-agrams for the two exceptions and indicate how many unpaired electrons are present.
n 1, / 0, m/ 0 n 2, / 2 n 3, ms 1 2 n 2, / 1, m/ 1, ms 1 2 n 0, / 0, m/ 0 n 2, / 1 n 3, / 2 n 5, ms 1 2 n 5, m/ 1 n 4 6dx2y2, 3dz2, n 3, / 1, m/ 2, ms 1 2 n 5, / 4, m/ 2, ms 1 2 n 2, / 1, m/ 1, ms 1 n 4, / 1, m/ 1, ms 1 2 n 4, / 3, m/ 2, ms 1 2 n 3, / 3, m/ 0, ms 1 2 n 2, / 1, m/ 1 n 0, / 0, m/ 0 n 4, / 3, m/ 4 n 3, / 2, m/ 2 69. The elements Si, Ga, As, Ge, Al, Cd, S, and Se are all used in the manufacture of various semiconductor devices. Write the ex-pected electron configuration for these atoms.
70. The elements Cu, O, La, Y, Ba, Tl, and Bi are all found in high-temperature ceramic superconductors. Write the expected elec-tron configuration for these atoms.
71. Write the expected electron configurations for each of the fol-lowing atoms: Sc, Fe, P, Cs, Eu, Pt, Xe, Br.
72. Write the expected electron configurations for each of the fol-lowing atoms: Cl, Sb, Sr, W, Pb, Cf.
73. Write the expected ground-state electron configuration for the following.
a. the element with one unpaired 5p electron that forms a co-valent with compound fluorine b. the (as yet undiscovered) alkaline earth metal after radium c. the noble gas with electrons occupying 4f orbitals d. the first-row transition metal with the most unpaired electrons 74. Using only the periodic table inside the front cover of the text, write the expected ground-state electron configurations for a. the third element in Group 5A.
b. element number 116.
c. an element with three unpaired 5d electrons.
d. the halogen with electrons in the 6p atomic orbitals.
75. In the ground state of mercury, Hg, a. how many electrons occupy atomic orbitals with n 3?
b. how many electrons occupy d atomic orbitals?
c. how many electrons occupy pz atomic orbitals?
d. how many electrons have spin “up” ( )?
76. In the ground state of element 115, Uup, a. how many electrons have n 5 as one of their quantum numbers?
b. how many electrons have 3 as one of their quantum numbers?
c. how many electrons have m 1 as one of their quantum numbers?
d. how many electrons have as one of their quantum numbers?
77. Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.
78. Give a possible set of values of the four quantum numbers for the 4s and 3d electrons in titanium.
79. A certain oxygen atom has the electron configuration 1s22s22px 22py 2. How many unpaired electrons are present? Is this an excited state of oxygen? In going from this state to the ground state would energy be released or absorbed?
80. Which of the following electron configurations correspond to an excited state? Identify the atoms and write the ground-state elec-tron configuration where appropriate.
a.
b.
c.
d. [Ar] How many unpaired electrons are present in each of these species?
4s23d54p1 1s22s22p43s1 1s22s22p6 1s22s23p1 ms 1 2 ms 1 2 324 Chapter Seven Atomic Structure and Periodicity a. What will be its electron configuration?
b. What element will it most resemble chemically?
c. What will be the formula of the neutral binary compounds it forms with sodium, magnesium, carbon, and oxygen?
d. What oxyanions would you expect Uus to form?
93. The first ionization energies of As and Se are 0.947 and 0.941 MJ/mol, respectively. Rationalize these values in terms of elec-tron configurations.
94. Rank the elements Be, B, C, N, and O in order of increasing first ionization energy. Explain your reasoning.
95. For each of the following pairs of elements pick the atom with a. more favorable (exothermic) electron affinity.
b. higher ionization energy.
c. larger size.
96. For each of the following pairs of elements pick the atom with a. more favorable (exothermic) electron affinity.
b. higher ionization energy.
c. larger size.
97. The electron affinities of the elements from aluminum to chlo-rine are 44, 120, 74, 200.4, and 384.7 kJ/mol, respec-tively. Rationalize the trend in these values.
98. The electron affinity for sulfur is more exothermic than that for oxygen. How do you account for this?
99. Order each of the following sets from the least exothermic elec-tron affinity to the most exothermic electron affinity.
a. F, Cl, Br, I b. N, O, F 100. Which has the more negative electron affinity, the oxygen atom or the O ion? Explain your answer.
101. Write equations corresponding to the following.
a. The fourth ionization energy of Se b. The electron affinity of S c. The electron affinity of Fe3 d. The ionization energy of Mg 102. Using data from the text, determine the following values (justify your answer): a. the electron affinity of Mg2 b. the ionization energy of Cl c. the electron affinity of Cl d. the ionization energy of Mg (Electron affinity of Mg 230 kJ/mol) Alkali Metals 103. An ionic compound of potassium and oxygen has the empirical formula KO. Would you expect this compound to be potas-sium(II) oxide or potassium peroxide? Explain.
104. Give the name and formula of each of the binary compounds formed from the following elements.
a. Li and N b. Na and Br c. K and S 1Mg and K2 1F and Cl2 1C and N2 1Ar and Br2 81. Which of elements 1–36 have two unpaired electrons in the ground state?
82. Which of elements 1–36 have one unpaired electron in the ground state?
83. One bit of evidence that the quantum mechanical model is “correct” lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired elec-trons present in the atom. Consider the ground-state electron configurations for Li, N, Ni, Te, Ba, and Hg. Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?
84. How many unpaired electrons are present in each of the follow-ing in the ground state: O, O, O, Os, Zr, S, F, Ar?
The Periodic Table and Periodic Properties 85. Arrange the following groups of atoms in order of increasing size.
a. Te, S, Se b. K, Br, Ni c. Ba, Si, F 86. Arrange the following groups of atoms in order of increasing size.
a. Rb, Na, Be b. Sr, Se, Ne c. Fe, P, O 87. Arrange the atoms in Exercise 85 in order of increasing first ion-ization energy.
88. Arrange the atoms in Exercise 86 in order of increasing first ion-ization energy.
89. In each of the following sets, which atom or ion has the small-est radius?
a. H, He b. Cl, In, Se c. element 120, element 119, element 117 d. Nb, Zn, Si e. Na, Na, Na 90. In each of the following sets, which atom or ion has the small-est ionization energy?
a. Ca, Sr, Ba b. K, Mn, Ga c. N, O, F d. S2, S, S2 e. Cs, Ge, Ar 91. Element 106 has been named seaborgium, Sg, in honor of Glenn Seaborg, discoverer of the first transuranium element.
a. Write the expected electron configuration for element 106.
b. What other element would be most like element 106 in its properties?
c. Write the formula for a possible oxide and a possible oxyan-ion of element 106.
92. Predict some of the properties of element 117 (the symbol is Uus, following conventions proposed by the International Union of Pure and Applied Chemistry, or IUPAC).
Additional Exercises 325 105. Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in 1859. The name came from the Latin caesius (“sky blue”) because of the prominent blue line observed for this element at 455.5 nm. Calculate the frequency and energy of a photon of this light.
106. The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the fre-quency and the energy of a photon of light at each of these wave-lengths? What are the energies in kJ/mol?
107. Does the information on alkali metals in Table 7.8 of the text confirm the general periodic trends in ionization energy and atomic radius? Explain.
108. Predict the atomic number of the next alkali metal after fran-cium and give its ground-state electron configuration.
109. Complete and balance the equations for the following reactions.
a.
b.
110. Complete and balance the equations for the following reactions.
a.
b.
Additional Exercises 111. Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the AgCl particles, the following reaction occurs: The silver metal that is formed causes the lenses to darken. The enthalpy change for this reaction is 3.10 102 kJ/mol. Assum-ing all this energy must be supplied by light, what is the maxi-mum wavelength of light that can cause this reaction?
112. A certain microwave oven delivers 750. watts (joule/s) of power to a coffee cup containing 50.0 g of water at 25.0C. If the wave-length of microwaves in the oven is 9.75 cm, how long does it take, and how many photons must be absorbed, to make the wa-ter boil? The specific heat capacity of water is 4.18 J/C g and assume only the water absorbs the energy of the microwaves.
113. Mars is roughly 60 million km from earth. How long does it take for a radio signal originating from earth to reach Mars?
114. Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of 3.59 1019 J, what color of visible light is emitted?
115. One of the visible lines in the hydrogen emission spectrum cor-responds to the n 6 to n 2 electronic transition. What color light is this transition? See Exercise 114.
AgCl ¡ Ag Cl Na1s2 Cl21g2 S Cs1s2 H2O1l2 S Rb1s2 S1s2 S Li1s2 N21g2 S 116. Using Fig. 7.28, list the elements (ignore the lanthanides and ac-tinides) that have ground-state electron configurations that dif-fer from those we would expect from their positions in the periodic table.
117. Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms?
a. The principal quantum number completely determines the energy of a given electron.
b. The angular momentum quantum number, , determines the shapes of the atomic orbitals.
c. The magnetic quantum number, m, determines the direction that the atomic orbitals point in space.
118. Although no currently known elements contain electrons in g or-bitals in the ground state, it is possible that these elements will be found or that electrons in excited states of known elements could be in g orbitals. For g orbitals, the value of is 4. What is the lowest value of n for which g orbitals could exist? What are the possible values of m? How many electrons could a set of g orbitals hold?
119. Consider the representations of the p and d atomic orbitals in Figs. 7.14 and 7.16. What do the and signs indicate?
120. Total radial probability distributions for the helium, neon, and argon atoms are shown in the following graph. How can one in-terpret the shapes of these curves in terms of electron configu-rations, quantum numbers, and nuclear charges?
121. The following graph plots the first, second, and third ionization energies for Mg, Al, and Si.
Without referencing the text, which plot corresponds to which element? In one of the plots, there is a huge jump in energy Number of electrons removed 1 2 3 Ionization energy (kJ/mol) Radial electron density 0 Distance from nucleus (Å) 0.5 1.0 Ar Ne He hv Infrared Red Orange Yellow Green Blue Violet Ultraviolet 7 x 10–5 6 x 10–5 5 x 10–5 4 x 10–5 cm Wavelength 326 Chapter Seven Atomic Structure and Periodicity between I2 and I3, unlike in the other two plots. Explain this phenomenon.
122. An ion having a 4 charge and a mass of 49.9 amu has 2 elec-trons with principal quantum number n 1, 8 electrons with n 2, and 10 electrons with n 3. Supply as many of the prop-erties for the ion as possible from the information given. Hint: In forming ions for this species, the 4s electrons are lost before the 3d electrons.
a. the atomic number b. total number of s electrons c. total number of p electrons d. total number of d electrons e. the number of neutrons in the nucleus f. the ground-state electron configuration of the neutral atom 123. The successive ionization energies for an unknown element are To which family in the periodic table does the unknown element most likely belong?
124. An unknown element is a nonmetal and has a valence electron configuration of ns2np4.
a. How many valence electrons does this element have?
b. What are some possible identities for this element?
c. What is the formula of the compound this element would form with potassium?
d. Would this element have a larger or smaller radius than barium?
e. Would this element have a greater or smaller ionization energy than fluorine?
125. Using data from this chapter, calculate the change in energy expected for each of the following processes.
a.
b.
c.
d.
Challenge Problems 126. One of the emission spectral lines for Be3 has a wavelength of 253.4 nm for an electronic transition that begins in the state with n 5. What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one-electron ions. Don’t forget the Z factor: Z nuclear charge atomic number.) 127. The figure below represents part of the emission spectrum for a one-electron ion in the gas phase. All the lines result from electronic transitions from excited states to the n 3 state. (See Exercise 126.) Wavelength A B Mg1g2 2F1g2 S Mg21g2 2F1g2 Mg1g2 F1g2 S Mg21g2 F1g2 Mg1g2 F1g2 S Mg1g2 F1g2 Na1g2 Cl1g2 S Na1g2 Cl1g2 I4 17,948 kJ/mol I3 14,807 kJ/mol I2 1752 kJ/mol I1 896 kJ/mol a. What electronic transitions correspond to lines A and B?
b. If the wavelength of line B is 142.5 nm, calculate the wave-length of line A.
128. When the excited electron in a hydrogen atom falls from n 5 to n 2, a photon of blue light is emitted. if an excited elec-tron in He falls from n 4, to which energy level must it fall so that a similar blue light (as with the hydrogen) is emitted?
Prove it. (See Exercise 126.) 129. The wave function for the 2pz orbital in the hydrogen atom is where a0 is the value for the radius of the first Bohr orbit in me-ters (5.29 1011), is Z(ra0), r is the value for the distance from the nucleus in meters, and is an angle. Calculate the value of 2pz 2 at r a0 for 0 (z axis) and for 90 (xy plane).
130. Answer the following questions assuming that ms could have three values rather than two and that the rules for n, , and m are the normal ones.
a. How many electrons would an orbital be able to hold?
b. How many elements would the first and second periods in the periodic table contain?
c. How many elements would be contained in the first transi-tion metal series?
d. How many electrons would the set of 4f orbitals be able to hold?
131. Assume that we are in another universe with different physical laws. Electrons in this universe are described by four quantum numbers with meanings similar to those we use. We will call these quantum numbers p, q, r, and s. The rules for these quan-tum numbers are as follows: p 1, 2, 3, 4, 5, . . . .
q takes on positive odd integer and q p.
r takes on all even integer values from q to q. (Zero is considered an even number.) s or a. Sketch what the first four periods of the periodic table will look like in this universe.
b. What are the atomic numbers of the first four elements you would expect to be least reactive?
c. Give an example, using elements in the first four rows, of ionic compounds with the formulas XY, XY2, X2Y, XY3, and X2Y3.
d. How many electrons can have p 4, q 3?
e. How many electrons can have p 3, q 0, r 0?
f. How many electrons can have p 6?
132. Without looking at data in the text, sketch a qualitative graph of the third ionization energy versus atomic number for the ele-ments Na through Ar, and explain your graph.
133. The following numbers are the ratios of second ionization en-ergy to first ionization energy: Na: 9.2 Mg: 2.0 Al: 3.1 Si: 2.0 1 2 1 2 c2pz 1 422p a Z a0 b 3 2 ses 2 cosu Marathon Problem 327 139. Francium, Fr, is a radioactive element found in some uranium minerals and is formed as a result of the decay of actinium.
a. What are the electron configurations of francium and its pre-dicted most common ion?
b. It has been estimated that at any one time, there is only one (1.0) ounce of francium on earth. Assuming this is true, what number of francium atoms exist on earth?
c. The longest-lived isotope of francium is 223Fr. What is the total mass in grams of the neutrons in one atom of this isotope?
140. Answer the following questions based on the given electron con-figurations and identify the elements.
a. Arrange these atoms in order of increasing size: [Kr]5s24d105p6; [Kr]5s24d105p1; [Kr]5s24d105p3.
b. Arrange these atoms in order of decreasing first ionization energy: [Ne]3s23p5; [Ar]4s23d104p3; [Ar]4s23d104p5.
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
141. From the information below, identify element X.
a. The wavelength of the radio waves sent by an FM station broadcasting at 97.1 MHz is 30.0 million (3.00 107) times greater than the wavelength corresponding to the energy dif-ference between a particular excited state of the hydrogen atom and the ground state.
b. Let V represent the principal quantum number for the valence shell of element X. If an electron in the hydrogen atom falls from shell V to the inner shell corresponding to the ex-cited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of 570. m/s.
c. The number of unpaired electrons for element X in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations n 2, m 1, and d. Let A equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of A also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
ms 1 2.
P: 1.8 S: 2.3 Cl: 1.8 Ar: 1.8 Explain these relative numbers.
134. We expect the atomic radius to increase going down a group in the periodic table. Can you suggest why the atomic radius of hafnium breaks this rule? (See data below.) 135. Consider the following ionization energies for aluminum: a. Account for the trend in the values of the ionization energies.
b. Explain the large increase between I3 and I4.
c. Which one of the four ions has the greatest electron affinity?
Explain.
d. List the four aluminum ions given in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.) 136. While Mendeleev predicted the existence of several undiscov-ered elements, he did not predict the existence of the noble gases, the lanthanides, or the actinides. Propose reasons why Mendeleev was not able to predict the existence of the noble gases.
137. An atom of a particular element is traveling at 1.00% of the speed of light. The de Broglie wavelength is found to be 3.31 103 pm.
Which element is this? Prove it.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
138. As the weapons officer aboard the Starship Chemistry, it is your duty to configure a photon torpedo to remove an electron from the outer hull of an enemy vessel. You know that the work func-tion (the binding energy of the electron) of the hull of the enemy ship is 7.52 1019 J.
a. What wavelength does your photon torpedo need to be to eject an electron?
b. You find an extra photon torpedo with a wavelength of 259 nm and fire it at the enemy vessel. Does this photon torpedo do any damage to the ship (does it eject an electron)?
c. If the hull of the enemy vessel is made of the element with an electron configuration of [Ar]4s13d10, what metal is this? Al31g2 ¡ Al41g2 e I4 11,600 kJ/mol Al21g2 ¡ Al31g2 e I3 2740 kJ/mol Al1g2 ¡ Al21g2 e I2 1815 kJ/mol Al1g2 ¡ Al1g2 e I1 580 kJ/mol Atomic Radii, in pm Sc 157 Ti 147.7 Y 169.3 Zr 159.3 La 191.5 Hf 147.6 Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
328 8 Bonding: General Concepts Contents 8.1 Types of Chemical Bonds 8.2 Electronegativity 8.3 Bond Polarity and Dipole Moments 8.4 Ions: Electron Configurations and Sizes • Predicting Formulas of Ionic Compounds • Sizes of Ions 8.5 Energy Effects in Binary Ionic Compounds • Lattice Energy Calculations 8.6 Partial Ionic Character of Covalent Bonds 8.7 The Covalent Chemical Bond: A Model • Models: An Overview 8.8 Covalent Bond Energies and Chemical Reactions • Bond Energy and Enthalpy 8.9 The Localized Electron Bonding Model 8.10 Lewis Structures 8.11 Exceptions to the Octet Rule 8.12 Resonance • Odd-Electron Molecules • Formal Charge 8.13 Molecular Structure: The VSEPR Model • The VSEPR Model and Multiple Bonds • Molecules Containing No Single Central Atom • The VSEPR Model—-How Well Does It Work?
Carbon forms very stable spherical C60 molecules.
As we examine the world around us, we find it to be composed almost entirely of compounds and mixtures of compounds: Rocks, coal, soil, petroleum, trees, and human bodies are all complex mixtures of chemical compounds in which different kinds of atoms are bound together. Substances composed of unbound atoms do exist in nature, but they are very rare. Examples are the argon in the atmosphere and the helium mixed with nat-ural gas reserves.
The manner in which atoms are bound together has a profound effect on chemical and physical properties. For example, graphite is a soft, slippery material used as a lu-bricant in locks, and diamond is one of the hardest materials known, valuable both as a gemstone and in industrial cutting tools. Why do these materials, both composed solely of carbon atoms, have such different properties? The answer, as we will see, lies in the bonding in these substances.
Silicon and carbon are next to each other in Group 4A of the periodic table. From our knowledge of periodic trends, we might expect SiO2 and CO2 to be very similar. But SiO2 is the empirical formula of silica, which is found in sand and quartz, and carbon dioxide is a gas, a product of respiration. Why are they so different? We will be able to answer this question after we have developed models for bonding.
Molecular bonding and structure play the central role in determining the course of all chemical reactions, many of which are vital to our survival. Later in this book we will demonstrate their importance by showing how enzymes facilitate complex chemical re-actions, how genetic characteristics are transferred, and how hemoglobin in the blood car-ries oxygen throughout the body. All of these fundamental biological reactions hinge on the geometric structures of molecules, sometimes depending on very subtle differences in molecular shape to channel the chemical reaction one way rather than another.
Many of the world’s current problems require fundamentally chemical answers: disease and pollution control, the search for new energy sources, the development of new fertilizers to increase crop yields, the improvement of the protein content in various staple grains, and many more. To understand the behavior of natural materials, we must under-stand the nature of chemical bonding and the factors that control the structures of 329 Quartz grows in beautiful, regular crystals.
330 Chapter Eight Bonding: General Concepts compounds. In this chapter we will present various classes of compounds that illustrate the different types of bonds and then develop models to describe the structure and bond-ing that characterize materials found in nature. Later these models will be useful in un-derstanding chemical reactions.
8.1 Types of Chemical Bonds What is a chemical bond? There is no simple and yet complete answer to this question.
In Chapter 2 we defined bonds as forces that hold groups of atoms together and make them function as a unit.
There are many types of experiments we can perform to determine the fundamental nature of materials. For example, we can study physical properties such as melting point, hardness, and electrical and thermal conductivity. We can also study solubility character-istics and the properties of the resulting solutions. To determine the charge distribution in a molecule, we can study its behavior in an electric field. We can obtain information about the strength of a bonding interaction by measuring the bond energy, which is the energy required to break the bond.
There are several ways in which atoms can interact with one another to form aggre-gates. We will consider several specific examples to illustrate the various types of chem-ical bonds.
Earlier, we saw that when solid sodium chloride is dissolved in water, the resulting solution conducts electricity, a fact that helps to convince us that sodium chloride is com-posed of Na and Cl ions. Therefore, when sodium and chlorine react to form sodium chloride, electrons are transferred from the sodium atoms to the chlorine atoms to form Na and Cl ions, which then aggregate to form solid sodium chloride. Why does this happen? The best simple answer is that the system can achieve the lowest possible energy by behaving in this way. The attraction of a chlorine atom for the extra electron and the very strong mutual attractions of the oppositely charged ions provide the driving forces for the process. The resulting solid sodium chloride is a very sturdy material; it has a melt-ing point of approximately 800C. The bonding forces that produce this great thermal sta-bility result from the electrostatic attractions of the closely packed, oppositely charged ions. This is an example of ionic bonding. Ionic substances are formed when an atom that loses electrons relatively easily reacts with an atom that has a high affinity for elec-trons. That is, an ionic compound results when a metal reacts with a nonmetal.
The energy of interaction between a pair of ions can be calculated using Coulomb’s law in the form where E has units of joules, r is the distance between the ion centers in nanometers, and Q1 and Q2 are the numerical ion charges.
For example, in solid sodium chloride the distance between the centers of the Na and Cl ions is 2.76 Å (0.276 nm), and the ionic energy per pair of ions is where the negative sign indicates an attractive force. That is, the ion pair has lower en-ergy than the separated ions.
Coulomb’s law also can be used to calculate the repulsive energy when two like-charged ions are brought together. In this case the calculated value of the energy will have a positive sign.
We have seen that a bonding force develops when two different types of atoms re-act to form oppositely charged ions. But how does a bonding force develop between two E 12.31 1019 J nm2 c 112112 0.276 nm d 8.37 1019 J E 12.31 1019 J nm2 aQ1Q2 r b Na+ Cl– 0.276 nm 8.1 Types of Chemical Bonds 331 identical atoms? Let’s explore this situation from a very simple point of view by con-sidering the energy terms that result when two hydrogen atoms are brought close together, as shown in Fig. 8.1(a). When hydrogen atoms are brought close together, there are two unfavorable potential energy terms, proton–proton repulsion and electron–electron repulsion, and one favorable term, proton–electron attraction. Under what conditions will the H2 molecule be favored over the separated hydrogen atoms? That is, what conditions will favor bond formation? The answer lies in the strong tendency in nature for any system to achieve the lowest possible energy. A bond will form (that is, the two hydro-gen atoms will exist as a molecular unit) if the system can lower its total energy in the process.
In this case, then, the hydrogen atoms will position themselves so that the system will achieve the lowest possible energy; the system will act to minimize the sum of the posi-tive (repulsive) energy terms and the negative (attractive) energy term. The distance where the energy is minimal is called the bond length. The total energy of this system as a function of distance between the hydrogen nuclei is shown in Fig. 8.1(b). Note several important features of this diagram: The energy terms involved are the net potential energy that results from the attrac-tions and repulsions among the charged particles and the kinetic energy due to the motions of the electrons.
The zero point of energy is defined with the atoms at infinite separation.
At very short distances the energy rises steeply because of the importance of the re-pulsive forces when the atoms are very close together.
The bond length is the distance at which the system has minimum energy.
In the H2 molecule, the electrons reside primarily in the space between the two nu-clei, where they are attracted simultaneously by both protons. This positioning is precisely what leads to the stability of the H2 molecule compared with two separated hydrogen A bond will form if the energy of the aggregate is lower than that of the separated atoms.
Potential energy was discussed in Chapter 6.
FIGURE 8.1 (a) The interaction of two hydrogen atoms. (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms.
As the atoms approach each other (right side of graph), the energy decreases until the distance reaches 0.074 nm (0.74 Å) and then begins to increase again due to repulsions.
Sufficiently far apart to have no interaction Energy (kJ/mol) 0 Internuclear distance (nm) 0.074 –458 0 (H—H bond length) H H H H H H H H (a) (b) + + H atom H atom The atoms begin to interact as they move closer together.
+ H atom + H atom H2 molecule + + Optimum distance to achieve lowest overall energy of system 332 Chapter Eight Bonding: General Concepts atoms. The potential energy of each electron is lowered because of the increased attrac-tive forces in this area. When we say that a bond is formed between the hydrogen atoms, we mean that the H2 molecule is more stable than two separated hydrogen atoms by a cer-tain quantity of energy (the bond energy).
We can also think of a bond in terms of forces. The simultaneous attraction of each electron by the protons generates a force that pulls the protons toward each other and that just balances the proton–proton and electron–electron repulsive forces at the distance cor-responding to the bond length.
The type of bonding we encounter in the hydrogen molecule and in many other mol-ecules in which electrons are shared by nuclei is called covalent bonding.
So far we have considered two extreme types of bonding. In ionic bonding the par-ticipating atoms are so different that one or more electrons are transferred to form oppo-sitely charged ions, which then attract each other. In covalent bonding two identical atoms share electrons equally. The bonding results from the mutual attraction of the two nuclei for the shared electrons. Between these extremes are intermediate cases in which the atoms are not so different that electrons are completely transferred but are different enough that unequal sharing results, forming what is called a polar covalent bond. An example of this type of bond occurs in the hydrogen fluoride (HF) molecule. When a sample of hy-drogen fluoride gas is placed in an electric field, the molecules tend to orient themselves as shown in Fig. 8.2, with the fluoride end closest to the positive pole and the hydrogen end closest to the negative pole. This result implies that the HF molecule has the follow-ing charge distribution: where (lowercase delta) is used to indicate a fractional charge. This same effect was noted in Chapter 4, where many of water’s unusual properties were attributed to the po-lar bonds in the H2O molecule.
The most logical explanation for the development of the partial positive and nega-tive charges on the atoms (bond polarity) in such molecules as HF and H2O is that the O¬H d d H¬F Ionic and covalent bonds are the extreme bond types.
CHEMICAL IMPACT No Lead Pencils D id you ever wonder why the part of a pencil that makes the mark is called the “lead”? Pencils have no lead in them now—and they never have. Apparently the association between writing and the element lead arose during the Ro-man Empire, when lead rods were used as writing utensils because they leave a gray mark on paper. Many centuries later, in 1564, a deposit of a black substance found to be very useful for writing was discovered in Borrowdale, England.
This substance, originally called “black lead,” was shown in 1879 by Swedish chemist Carl Scheele to be a form of carbon and was subsequently named graphite (after the Greek graphein, meaning “to write”).
Originally, chunks of graphite from Borrowdale, called marking stones, were used as writing instruments. Later, sticks of graphite were used. Because graphite is brittle, the sticks needed reinforcement. At first they were wrapped in string, which was unwound as the core wore down. Even-tually, graphite rods were tied between two wooden slats or inserted into hollowed-out wooden sticks to form the first crude pencils.
Although Borrowdale graphite was pure enough to use directly, most graphite must be mixed with other materials to be useful for writing instruments. In 1795, the French chemist Nicolas-Jaques Conté invented a process in which graphite is mixed with clay and water to produce pencil “lead,” a recipe that is still used today. In modern pencil manufacture, graphite and clay are mixed and crushed into a fine powder to which water is added. After the gray sludge Visualization: Polar Molecules 8.2 Electronegativity 333 FIGURE 8.2 The effect of an electric field on hydrogen fluoride molecules. (a) When no electric field is present, the molecules are randomly oriented. (b) When the field is turned on, the molecules tend to line up with their negative ends toward the positive pole and their positive ends toward the negative pole.
electrons in the bonds are not shared equally. For example, we can account for the po-larity of the HF molecule by assuming that the fluorine atom has a stronger attraction for the shared electrons than the hydrogen atom. Likewise, in the H2O molecule the oxy-gen atom appears to attract the shared electrons more strongly than the hydrogen atoms do. Because bond polarity has important chemical implications, we find it useful to quan-tify the ability of an atom to attract shared electrons. In the next section we show how this is done.
8.2 Electronegativity The different affinities of atoms for the electrons in a bond are described by a property called electronegativity: the ability of an atom in a molecule to attract shared electrons to itself.
is blended for several days, it is dried, ground up again, and mixed with more water to give a gray paste. The paste is extruded through a metal tube to form thin rods, which are then cut into pencil-length pieces called “leads.” These leads are heated in an oven to 1000C until they are smooth and hard. The ratio of clay to graphite is adjusted to vary the hardness of the lead—-the more clay in the mix, the harder the lead and the lighter the line it makes.
Pencils are made from a slat of wood with several grooves cut in it to hold the leads. A similar grooved slat is then placed on top and glued to form a “sandwich” from which individual pencils are cut, sanded smooth, and painted. Although many types of wood have been used over the years to make pencils, the current favorite is incense cedar from the Sierra Nevada Mountains of California.
Modern pencils are simple but amazing instruments.
The average pencil can write approximately 45,000 words, which is equivalent to a line 35 miles long. The graphite in a pencil is easily transferred to paper because graphite con-tains layers of carbon atoms bound together in a “chicken-wire” structure. Although the bonding within each layer is very strong, the bonding between layers is weak, giving graphite its slippery, soft nature. In this way, graphite is much different from diamond, the other common elemental form of carbon. In diamond the carbon atoms are bound tightly in all three dimensions, making it extremely hard— the hardest natural substance.
Pencils are very useful—-especially for doing chemistry problems—-because we can erase our mistakes. Most pencils used in the United States have erasers (first attached to pen-cils in 1858), although most European pencils do not. Laid end-to end, the number of pencils made in the United States each year would circle the earth about 15 times. Pencils il-lustrate how useful a simple substance like graphite can be.
δ+ (a) δ– – + (b) H F δ+ δ– 334 Chapter Eight Bonding: General Concepts The most widely accepted method for determining values of electronegativity is that of Linus Pauling (1901–1995), an American scientist who won the Nobel Prizes for both chemistry and peace. To understand Pauling’s model, consider a hypothetical molecule HX. The relative electronegativities of the H and X atoms are determined by comparing the measured HOX bond energy with the “expected” HOX bond energy, which is an av-erage of the HOH and XOX bond energies: The difference ( ) between the actual (measured) and expected bond energies is If H and X have identical electronegativities, (HOX)act and (HOX)exp are the same, and is 0. On the other hand, if X has a greater electronegativity than H, the shared elec-tron(s) will tend to be closer to the X atom. The molecule will be polar, with the follow-ing charge distribution: Note that this bond can be viewed as having an ionic as well as a covalent compo-nent. The attraction between the partially (and oppositely) charged H and X atoms will lead to a greater bond strength. Thus (HOX)act will be larger than (HOX)exp. The greater is the difference in the electronegativities of the atoms, the greater is the ionic component of the bond and the greater is the value of . Thus the relative electronegativities of H and X can be assigned from the values.
Electronegativity values have been determined by this process for virtually all the el-ements; the results are given in Fig. 8.3. Note that electronegativity generally increases going from left to right across a period and decreases going down a group for the repre-sentative elements. The range of electronegativity values is from 4.0 for fluorine to 0.7 for cesium.
The relationship between electronegativity and bond type is shown in Table 8.1. For identical atoms (an electronegativity difference of zero), the electrons in the bond are shared equally, and no polarity develops. When two atoms with very different electronegativities d d H¬X ¢ 1H¬X2act 1H¬X2exp Expected H¬X bond energy H¬H bond energy X¬X bond energy 2 FIGURE 8.3 The Pauling electronegativity values. Electronegativity generally increases across a period and decreases down a group.
H 2.1 Li 1.0 Be 1.5 Na 0.9 Mg 1.2 K 0.8 Ca 1.0 Rb 0.8 Sr 1.0 Cs 0.7 Ba 0.9 Fr 0.7 Ra 0.9 Sc 1.3 Y 1.2 La–Lu 1.0–1.2 Ac 1.1 Ti 1.5 Zr 1.4 Hf 1.3 Th 1.3 V 1.6 Nb 1.6 Ta 1.5 Pa 1.4 Cr 1.6 Mo 1.8 W 1.7 U 1.4 Mn 1.5 Tc 1.9 Re 1.9 Np–No 1.4–1.3 Fe 1.8 Ru 2.2 Os 2.2 Co 1.9 Rh 2.2 Ir 2.2 Ni 1.9 Pd 2.2 Pt 2.2 Cu 1.9 Ag 1.9 Au 2.4 Zn 1.6 Cd 1.7 Hg 1.9 Ga 1.6 In 1.7 Tl 1.8 Al 1.5 B 2.0 Ge 1.8 Sn 1.8 Pb 1.9 Si 1.8 C 2.5 As 2.0 Sb 1.9 Bi 1.9 P 2.1 N 3.0 Se 2.4 Te 2.1 Po 2.0 S 2.5 O 3.5 Br 2.8 I 2.5 At 2.2 Cl 3.0 F 4.0 Increasing electronegativity Decreasing electronegativity 8.3 Bond Polarity and Dipole Moments 335 interact, electron transfer can occur to form the ions that make up an ionic substance.
Intermediate cases give polar covalent bonds with unequal electron sharing.
Relative Bond Polarities Order the following bonds according to polarity: HOH, OOH, ClOH, SOH, and FOH.
Solution The polarity of the bond increases as the difference in electronegativity increases. From the electronegativity values in Fig. 8.3, the following variation in bond polarity is expected (the electronegativity value appears in parentheses below each element): Electronegativity difference See Exercises 8.31 and 8.32.
8.3 Bond Polarity and Dipole Moments We have seen that when hydrogen fluoride is placed in an electric field, the molecules have a preferential orientation (see Fig. 8.2). This follows from the charge distribution in the HF molecule, which has a positive end and a negative end. A molecule such as HF that has a center of positive charge and a center of negative charge is said to be dipolar, or to have a dipole moment. The dipolar character of a molecule is often represented by an arrow pointing to the negative charge center with the tail of the arrow indicating the positive center of charge: δ+ δ− H F 14.0212.12 13.5212.12 13.0212.12 12.5212.12 12.1212.12 H¬H 6 S¬H 6 Cl¬H 6 O¬H 6 F¬H TABLE 8.1 The Relationship Between Electronegativity and Bond Type Electronegativity Difference Bond in the Bonding Atoms Type Zero Covalent Intermediate Polar covalent Large Ionic ⎯ ⎯ ⎯ → ⎯ ⎯ ⎯ → ⎯ ⎯ ⎯ → ⎯ ⎯ ⎯ → Sample Exercise 8.1 Covalent character Ionic character 336 Chapter Eight Bonding: General Concepts Another way to represent the charge distribution in HF is by an electrostatic poten-tial diagram (see Fig. 8.4). For this representation the colors of visible light are used to show the variation in charge distribution. Red indicates the most electron-rich region of the molecule and blue indicates the most electron-poor region.
Of course, any diatomic (two-atom) molecule that has a polar bond also will show a molecular dipole moment. Polyatomic molecules also can exhibit dipolar behavior. For example, because the oxygen atom in the water molecule has a greater electronegativity than the hydrogen atoms, the molecular charge distribution is that shown in Fig. 8.5(a).
Because of this charge distribution, the water molecule behaves in an electric field as if it had two centers of charge—one positive and one negative—as shown in Fig. 8.5(b). The water molecule has a dipole moment. The same type of behavior is observed for the NH3 molecule (Fig. 8.6). Some molecules have polar bonds but do not have a dipole moment.
This occurs when the individual bond polarities are arranged in such a way that they can-cel each other out. An example is the CO2 molecule, which is a linear molecule that has the charge distribution shown in Fig. 8.7. In this case the opposing bond polarities can-cel out, and the carbon dioxide molecule does not have a dipole moment. There is no pref-erential way for this molecule to line up in an electric field. (Try to find a preferred ori-entation to make sure you understand this concept.) There are many cases besides that of carbon dioxide where the bond polarities op-pose and exactly cancel each other. Some common types of molecules with polar bonds but no dipole moment are shown in Table 8.2.
FIGURE 8.5 (a) The charge distribution in the water molecule. (b) The water molecule in an electric field. (c) The electrostatic potential diagram of the water molecule.
FIGURE 8.6 (a) The structure and charge distribution of the ammonia molecule. The polarity of the NOH bonds occurs because nitrogen has a greater electronegativity than hydrogen.
(b) The dipole moment of the ammonia molecule oriented in an electric field.
(c) The electrostatic potential diagram for ammonia.
FIGURE 8.7 (a) The carbon dioxide molecule. (b) The opposed bond polarities cancel out, and the carbon dioxide molecule has no dipole moment. (c) The electrostatic potential diagram for carbon dioxide.
– + H H N H 3δ– δ+ δ+ δ+ (a) (b) (c) ∆+ ∆– 2δ– δ+ δ+ – + ∆– ∆+ (a) (b) (c) H O H H O H 2δ+ C δ– δ– (a) O O (b) (c) FIGURE 8.4 An electrostatic potential map of HF. Red indicates the most electron-rich area (the fluorine atom) and blue indicates the most electron-poor region (the hydrogen atom).
H F 8.3 Bond Polarity and Dipole Moments 337 Bond Polarity and Dipole Moment For each of the following molecules, show the direction of the bond polarities and indi-cate which ones have a dipole moment: HCl, Cl2, SO3 (a planar molecule with the oxy-gen atoms spaced evenly around the central sulfur atom), CH4 [tetrahedral (see Table 8.2) with the carbon atom at the center], and H2S (V-shaped with the sulfur atom at the point).
Solution The HCl molecule: In Fig. 8.3, we see that the electronegativity of chlorine (3.0) is greater than that of hydrogen (2.1). Thus the chlorine will be partially negative, and the hydro-gen will be partially positive. The HCl molecule has a dipole moment: The Cl2 molecule: The two chlorine atoms share the electrons equally. No bond po-larity occurs, and the Cl2 molecule has no dipole moment.
The SO3 molecule: The electronegativity of oxygen (3.5) is greater than that of sul-fur (2.5). This means that each oxygen will have a partial negative charge, and the sulfur will have a partial positive charge: δ– S O O 3δ+ δ– δ– O H δ– δ+ Cl TABLE 8.2 Types of Molecules with Polar Bonds but No Resulting Dipole Moment Cancellation Type of Polar Bonds Example Ball-and-Stick Model Linear molecules with BOAOB CO2 two identical bonds Planar molecules with three identical SO3 bonds 120 degrees apart Tetrahedral molecules with four identical CCl4 bonds 109.5 degrees apart A B B B 120 Sample Exercise 8.2 A B B B B S O O O H Cl Atoms in stable compounds usually have a noble gas electron configuration.
338 Chapter Eight Bonding: General Concepts The bond polarities arranged symmetrically as shown cancel, and the molecule has no di-pole moment. This molecule is the second type shown in Table 8.2.
The CH4 molecule: Carbon has a slightly higher electronegativity (2.5) than does hy-drogen (2.1). This leads to small partial positive charges on the hydrogen atoms and a small partial negative charge on the carbon: The presence of polar bonds does not always yield a polar molecule.
This case is similar to the third type in Table 8.2, and the bond polarities cancel. The mol-ecule has no dipole moment.
The H2S molecule: Since the electronegativity of sulfur (2.5) is slightly greater than that of hydrogen (2.1), the sulfur will have a partial negative charge, and the hydrogen atoms will have a partial positive charge, which can be represented as This case is analogous to the water molecule, and the polar bonds result in a dipole mo-ment oriented as shown: See Exercise 8.114.
8.4 Ions: Electron Configurations and Sizes The description of the electron arrangements in atoms that emerged from the quantum mechanical model has helped a great deal in our understanding of what constitutes a sta-ble compound. In virtually every case the atoms in a stable compound have a noble gas arrangement of electrons. Nonmetallic elements achieve a noble gas electron configura-tion either by sharing electrons with other nonmetals to form covalent bonds or by tak-ing electrons from metals to form ions. In the second case, the nonmetals form anions, and the metals form cations. The generalizations that apply to electron configurations in stable compounds are as follows: • When two nonmetals react to form a covalent bond, they share electrons in a way that completes the valence electron configurations of both atoms. That is, both nonmetals attain noble gas electron configurations.
δ+ δ+H H 4δ– δ+ H δ+ H C S 2δ– δ+ δ+ H H S H H H H H H C S H H 8.4 Ions: Electron Configurations and Sizes 339 • When a nonmetal and a representative-group metal react to form a binary ionic com-pound, the ions form so that the valence electron configuration of the nonmetal achieves the electron configuration of the next noble gas atom and the valence orbitals of the metal are emptied. In this way both ions achieve noble gas electron configurations.
These generalizations apply to the vast majority of compounds and are important to remember. We will deal with covalent bonds more thoroughly later, but now we will con-sider what implications these rules hold for ionic compounds.
Predicting Formulas of Ionic Compounds At the beginning of this discussion it should be emphasized that when chemists use the term ionic compound, they are usually referring to the solid state of that compound. In the solid state the ions are close together. That is, solid ionic compounds contain a large collection of positive and negative ions packed together in a way that minimizes theEQE and BQB repulsions and maximizes the BQE attractions. This situation stands in contrast to the gas phase of an ionic substance, where the ions are quite far apart on av-erage. In the gas phase, a pair of ions may get close enough to interact, but large collec-tions of ions do not exist. Thus, when we speak in this text of the stability of an ionic compound, we are referring to the solid state, where the large attractive forces present among oppositely charged ions tend to stabilize (favor the formation of) the ions. For ex-ample, as we mentioned in the preceding chapter, the O2 ion is not stable as an isolated, gas-phase species but, of course, is very stable in many solid ionic compounds. That is, MgO(s), which contains Mg2 and O2 ions, is very stable, but the isolated, gas-phase ion pair Mg2 . . O2 is not energetically favorable in comparison with the separate neutral gaseous atoms. Thus you should keep in mind that in this section, and in most other cases where we are describing the nature of ionic compounds, the discussion usually refers to the solid state, where many ions are simultaneously interacting.
To illustrate the principles of electron configurations in stable, solid ionic compounds, we will consider the formation of an ionic compound from calcium and oxygen. We can predict what compound will form by considering the valence electron configurations of the two atoms: From Fig. 8.3 we see that the electronegativity of oxygen (3.5) is much greater than that of calcium (1.0). Because of this large difference, electrons will be transferred from cal-cium to oxygen to form oxygen anions and calcium cations in the compound. How many electrons are transferred? We can base our prediction on the observation that noble gas configurations are generally the most stable. Note that oxygen needs two electrons to fill its 2s and 2p valence orbitals and to achieve the configuration of neon (1s22s22p6). And by losing two electrons, calcium can achieve the configuration of argon. Two electrons are therefore transferred: To predict the formula of the ionic compound, we simply recognize that chemical compounds are always electrically neutral—they have the same quantities of positive and negative charges. In this case we have equal numbers of Ca2 and O2 ions, and the em-pirical formula of the compound is CaO.
The same principles can be applied to many other cases. For example, consider the compound formed between aluminum and oxygen. Because aluminum has the configura-tion [Ne]3s23p1, it loses three electrons to form the Al3 ion and thus achieves the neon 2e Ca O ¡ Ca2 O2 O: 3He42s22p4 Ca: 3Ar44s2 In the solid state of an ionic compound the ions are relatively close together, and many ions are simultaneously interacting: In the gas phase of an ionic substance the ions would be relatively far apart and would not contain large groups of ions: + + + --+ ---repulsion attraction + + + ---Visualization: Formation of Ionic Compounds 340 Chapter Eight Bonding: General Concepts configuration. Therefore, the Al3 and O2 ions form in this case. Since the compound must be electrically neutral, there must be three O2 ions for every two Al3 ions, and the compound has the empirical formula Al2O3.
Table 8.3 shows common elements that form ions with noble gas electron configura-tions in ionic compounds. In losing electrons to form cations, metals in Group 1A lose one electron, those in Group 2A lose two electrons, and those in Group 3A lose three electrons. In gaining electrons to form anions, nonmetals in Group 7A (the halogens) gain one electron, and those in Group 6A gain two electrons. Hydrogen typically behaves as a nonmetal and can gain one electron to form the hydride ion (H), which has the elec-tron configuration of helium.
There are some important exceptions to the rules discussed here. For example, tin forms both Sn2 and Sn4 ions, and lead forms both Pb2 and Pb4 ions. Also, bismuth forms Bi3 and Bi5 ions, and thallium forms Tl and Tl3 ions. There are no simple explanations for the behavior of these ions. For now, just note them as exceptions to the very useful rule that ions generally adopt noble gas electron configurations in ionic compounds. Our discussion here refers to representative metals. The transition metals exhibit more complicated behavior, forming a variety of ions that will be considered in Chapter 21.
Sizes of Ions Ion size plays an important role in determining the structure and stability of ionic solids, the properties of ions in aqueous solution, and the biologic effects of ions. As with atoms, it is impossible to define precisely the sizes of ions. Most often, ionic radii are determined A bauxite mine. Bauxite contains Al2O3, the main source of aluminum.
TABLE 8.3 Common Ions with Noble Gas Configurations in Ionic Compounds Electron Group 1A Group 2A Group 3A Group 6A Group 7A Configuration H, Li Be2 [He] Na Mg2 Al3 O2 F [Ne] K Ca2 S2 Cl [Ar] Rb Sr2 Se2 Br [Kr] Cs Ba2 Te2 I [Xe] 8.4 Ions: Electron Configurations and Sizes 341 from the measured distances between ion centers in ionic compounds. This method, of course, involves an assumption about how the distance should be divided up between the two ions. Thus you will note considerable disagreement among ionic sizes given in vari-ous sources. Here we are mainly interested in trends and will be less concerned with ab-solute ion sizes.
Various factors influence ionic size. We will first consider the relative sizes of an ion and its parent atom. Since a positive ion is formed by removing one or more electrons from a neutral atom, the resulting cation is smaller than its parent atom. The opposite is true for negative ions; the addition of electrons to a neutral atom produces an anion sig-nificantly larger than its parent atom.
It is also important to know how the sizes of ions vary depending on the positions of the parent elements in the periodic table. Figure 8.8 shows the sizes of the most impor-tant ions (each with a noble gas configuration) and their position in the periodic table.
Note that ion size increases down a group. The changes that occur horizontally are com-plicated because of the change from predominantly metals on the left-hand side of the pe-riodic table to nonmetals on the right-hand side. A given period thus contains both elements that give up electrons to form cations and ones that accept electrons to form anions.
One trend worth noting involves the relative sizes of a set of isoelectronic ions— ions containing the same number of electrons. Consider the ions O2, F, Na, Mg2, and Al3. Each of these ions has the neon electron configuration. How do the sizes of these ions vary? In general, there are two important facts to consider in predicting the relative sizes of ions: the number of electrons and the number of protons. Since these ions are isoelectronic, the number of electrons is 10 in each case. Electron repulsions therefore should be about the same in all cases. However, the number of protons increases from 8 to 13 as we go from the O2 ion to the Al3 ion. Thus, in going from O2 to Al3, the 10 electrons experience a greater attraction as the positive charge on the nucleus in-creases. This causes the ions to become smaller. You can confirm this by looking at Fig.
8.8. In general, for a series of isoelectronic ions, the size decreases as the nuclear charge Z increases.
FIGURE 8.8 Sizes of ions related to positions of the elements on the periodic table. Note that size generally increases down a group. Also note that in a series of isoelectronic ions, size decreases with increasing atomic number. The ionic radii are given in units of picometers.
Li+ 60 Be2+ 31 O2– 140 F– 136 Na+ 95 Mg2+ 65 Al3+ 50 S2– 184 Cl– 181 K+ 133 Ca2+ 99 Ga3+ 62 Se2– 198 Br– 195 Rb+ 148 Sr2+ 113 In3+ 81 71 Sb5+ 62 Te2– 221 I– 216 Sn4+ Visualization: Ionic Radii For isoelectronic ions, size decreases as Z increases.
342 Chapter Eight Bonding: General Concepts Relative Ion Size I Arrange the ions Se2, Br, Rb, and Sr2 in order of decreasing size.
Solution This is an isoelectronic series of ions with the krypton electron configuration. Since these ions all have the same number of electrons, their sizes will depend on the nuclear charge.
The Z values are 34 for Se2, 35 for Br, 37 for Rb, and 38 for Sr2. Since the nuclear charge is greatest for Sr2, it is the smallest of these ions. The Se2 ion is largest: Largest Smallest See Exercises 8.37 and 8.38.
Relative Ion Size II Choose the largest ion in each of the following groups.
a. Li, Na, K, Rb, Cs b. Ba2, Cs, I, Te2 Solution a. The ions are all from Group 1A elements. Since size increases down a group (the ion with the greatest number of electrons is largest), Cs is the largest ion.
b. This is an isoelectronic series of ions, all of which have the xenon electron configura-tion. The ion with the smallest nuclear charge is largest: See Exercises 8.39 and 8.40.
8.5 Energy Effects in Binary Ionic Compounds In this section we will introduce the factors that influence the stability and the structures of solid binary ionic compounds. We know that metals and nonmetals react by transfer-ring electrons to form cations and anions that are mutually attractive. The resulting ionic solid forms because the aggregated oppositely charged ions have a lower energy than the original elements. Just how strongly the ions attract each other in the solid state is indi-cated by the lattice energy—the change in energy that takes place when separated gaseous ions are packed together to form an ionic solid: The lattice energy is often defined as the energy released when an ionic solid forms from its ions. However, in this book the sign of an energy term is always determined from the system’s point of view: negative if the process is exothermic, positive if endothermic.
Thus, using this convention, the lattice energy has a negative sign.
We can illustrate the energy changes involved in the formation of an ionic solid by considering the formation of solid lithium fluoride from its elements: To see the energy terms associated with this process, we take advantage of the fact that energy is a state function and break this reaction into steps, the sum of which gives the overall reaction.
Li1s2 1 2F21g2 ¡ LiF1s2 M1g2 X1g2 ¡ MX1s2 Z 52 Z 53 Z 55 Z 56 Ba2 7 Cs 7 I Te2 7 c c Se2 7 Br 7 Rb 7 Sr2 The structures of ionic solids will be discussed in detail in Chapter 10.
Sample Exercise 8.3 Sample Exercise 8.4 8.5 Energy Effects in Binary Ionic Compounds 343 ➥ 1 Sublimation of solid lithium. Sublimation involves taking a substance from the solid state to the gaseous state: The enthalpy of sublimation for Li(s) is 161 kJ/mol.
➥ 2 Ionization of lithium atoms to form Li ions in the gas phase: This process corresponds to the first ionization energy for lithium, which is 520 kJ/mol.
➥ 3 Dissociation of fluorine molecules. We need to form a mole of fluorine atoms by breaking the F—F bonds in a half mole of F2 molecules: The energy required to break this bond is 154 kJ/mol. In this case we are breaking the bonds in a half mole of fluorine, so the energy required for this step is (154 kJ)2, or 77 kJ.
➥ 4 Formation of F ions from fluorine atoms in the gas phase: The energy change for this process corresponds to the electron affinity of fluorine, which is 328 kJ/mol.
➥ 5 Formation of solid lithium fluoride from the gaseous Li and F ions: This corresponds to the lattice energy for LiF, which is 1047 kJ/mol.
Since the sum of these five processes yields the desired overall reaction, the sum of the individual energy changes gives the overall energy change: Li1g2 F1g2 ¡ LiF1s2 F1g2 e ¡ F1g2 1 2F21g2 ¡ F1g2 Li1g2 ¡ Li1g2 e Li1s2 ¡ Li1g2 Lithium fluoride.
Process Energy Change (kJ) 161 520 77 328 1047 617 kJ (per mole of LiF) Overall: Li1s2 1 2F21g2 S LiF1s2 Li1g2 F1g2 S LiF1s2 F1g2 e S F1g2 1 2F21g2 S F1g2 Li1g2 S Li1g2 e Li1s2 S Li1g2 In doing this calculation, we have ignored the small difference between Hsub and Esub.
This process is summarized by the energy diagram in Fig. 8.9. Note that the forma-tion of solid lithium fluoride from its elements is highly exothermic, mainly because of the very large negative lattice energy. A great deal of energy is released when the ions combine to form the solid. In fact, note that the energy released when an electron is added to a fluorine atom to form the F ion (328 kJ/mol) is not enough to remove an electron from lithium (520 kJ/mol). That is, when a metallic lithium atom reacts with a nonmetallic fluorine atom to form separated ions, the process is endothermic and thus unfavorable. Clearly, then, the main impetus for the formation of an ionic compound rather than a covalent compound results from the strong mutual attractions among the Li and F ions in the solid. The lattice energy is the dom-inant energy term.
Li1g2 F1g2 ¡ Li1g2 F1g2 344 Chapter Eight Bonding: General Concepts The structure of solid lithium fluoride is represented in Fig. 8.10. Note the alternat-ing arrangement of the Li and F ions. Also note that each Li is surrounded by six F ions, and each F ion is surrounded by six Li ions. This structure can be rationalized by assuming that the ions behave as hard spheres that pack together in a way that both maximizes the attractions among the oppositely charged ions and minimizes the repul-sions among the identically charged ions.
All the binary ionic compounds formed by an alkali metal and a halogen have the structure shown in Fig. 8.10, except for the cesium salts. The arrangement of ions shown in Fig. 8.10 is often called the sodium chloride structure, after the most common sub-stance that possesses it.
Lattice Energy Calculations In discussing the energetics of the formation of solid lithium fluoride, we emphasized the importance of lattice energy in contributing to the stability of the ionic solid. Lattice en-ergy can be represented by a modified form of Coulomb’s law: where k is a proportionality constant that depends on the structure of the solid and the electron configurations of the ions, Q1 and Q2 are the charges on the ions, and r is the shortest distance between the centers of the cations and anions. Note that the lattice en-ergy has a negative sign when Q1 and Q2 have opposite signs. This result is expected, since bringing cations and anions together is an exothermic process. Also note that the process becomes more exothermic as the ionic charges increase and as the distances be-tween the ions in the solid decrease.
The importance of the charges in ionic solids can be illustrated by comparing the en-ergies involved in the formation of NaF(s) and MgO(s). These solids contain the isoelec-tronic ions Na, F, Mg2, and O2. The energy diagram for the formation of the two solids is given in Fig. 8.11. Note several important features: The energy released when the gaseous Mg2 and O2 ions combine to form solid MgO is much greater (more than four times greater) than that released when the gaseous Na and F ions combine to form solid NaF.
The energy required to remove two electrons from the magnesium atom (735 kJ/mol for the first and 1445 kJ/mol for the second, yielding a total of 2180 kJ/mol) is much greater than the energy required to remove one electron from a sodium atom (495 kJ/mol).
Lattice energy kaQ1Q2 r b Li+(g) + F(g) 77 kJ(3) Li+(g) + F2(g) 1 2 520 kJ(2) Li(g) + F2(g) 1 2 161 kJ(1) Li(s) + F2(g) 1 2 Li+(g) + F–(g) –328 kJ(4) –1047 kJ(5) LiF(s) E –617 kJ Overall change (a) (b) FIGURE 8.9 The energy changes involved in the forma-tion of solid lithium fluoride from its ele-ments. The numbers in parentheses refer to the reaction steps discussed in the text.
FIGURE 8.10 The structure of lithium fluoride. (a) Repre-sented by ball-and-stick model. Note that each Li ion is surrounded by six F ions, and each F ion is surrounded by six Li ions. (b) Represented with the ions shown as spheres. The structure is determined by packing the spherical ions in a way that both maximizes the ionic attractions and minimizes the ionic repulsions.
Visualization: Born–Haber Cycle for NaCl(s) 8.5 Energy Effects in Binary Ionic Compounds 345 Energy (737 kJ/mol) is required to add two electrons to the oxygen atom in the gas phase. Addition of the first electron is exothermic (141 kJ/mol), but addition of the second electron is quite endothermic (878 kJ/mol). This latter energy must be obtained indirectly, since the O2(g) ion is not stable.
In view of the facts that twice as much energy is required to remove the second elec-tron from magnesium as to remove the first and that addition of an electron to the gaseous O ion is quite endothermic, it seems puzzling that magnesium oxide contains Mg2 and O2 ions rather than Mg and O ions. The answer lies in the lattice energy. Note that the lattice energy for combining gaseous Mg2 and O2 ions to form MgO(s) is 3000 kJ/mol FIGURE 8.11 Comparison of the energy changes involved in the formation of solid sodium fluoride and solid magnesium oxide. Note the large lattice energy for magnesium oxide (where doubly charged ions are combining) compared with that for sodium fluoride (where singly charged ions are combining).
E Mg2+(g) + O2–(g) 737 Electron affinity Mg2+(g) + O(g) Mg2+(g) + O2(g) 1 2 247 2180 Ionization energy 150 Mg(g) + O2(g) 1 2 Mg(s) + O2(g) 1 2 –602 –570 Overall energy change MgO(s) NaF(s) –923 Lattice energy –328 Electron affinity –3916 Lattice energy Na(s) + F2(g) 1 2 Na(g) + F2(g) 1 2 109 495 Ionization energy Na+(g) + F2(g) 1 2 Na+(g) + F–(g) Na+(g) + F(g) 77 346 Chapter Eight Bonding: General Concepts more negative than that for combining gaseous Na and F ions to form NaF(s). Thus the energy released in forming a solid containing Mg2 and O2 ions rather than Mg and O ions more than compensates for the energies required for the processes that produce the Mg2 and O2 ions.
If there is so much lattice energy to be gained in going from singly charged to doubly charged ions in the case of magnesium oxide, why then does solid sodium fluoride contain Na and F ions rather than Na2 and F2 ions? We can answer this question by recognizing that both Na and F ions have the neon electron configura-tion. Removal of an electron from Na requires an extremely large quantity of energy (4560 kJ/mol) because a 2p electron must be removed. Conversely, the addition of an electron to F would require use of the relatively high-energy 3s orbital, which is also an unfavorable process. Thus we can say that for sodium fluoride the extra energy required to form the doubly charged ions is greater than the gain in lattice energy that would result.
This discussion of the energies involved in the formation of solid ionic compounds illustrates that a variety of factors operate to determine the composition and structure of these compounds. The most important of these factors involve the balancing of the ener-gies required to form highly charged ions and the energy released when highly charged ions combine to form the solid.
8.6 Partial Ionic Character of Covalent Bonds Recall that when atoms with different electronegativities react to form molecules, the elec-trons are not shared equally. The possible result is a polar covalent bond or, in the case of a large electronegativity difference, a complete transfer of one or more electrons to form ions. The cases are summarized in Fig. 8.12.
How well can we tell the difference between an ionic bond and a polar covalent bond?
The only honest answer to this question is that there are probably no totally ionic bonds between discrete pairs of atoms. The evidence for this statement comes from calculations of the percent ionic character for the bonds of various binary compounds in the gas phase.
These calculations are based on comparisons of the measured dipole moments for mole-cules of the type X—Y with the calculated dipole moments for the completely ionic case, XY. The percent ionic character of a bond can be defined as Application of this definition to various compounds (in the gas phase) gives the results shown in Fig. 8.13, where percent ionic character is plotted versus the difference in the electronegativity values of X and Y. Note from this plot that ionic character increases with electronegativity difference, as expected. However, none of the bonds reaches 100% ionic character, even though compounds with the maximum possible electronegativity differ-ences are considered. Thus, according to this definition, no individual bonds are com-pletely ionic. This conclusion is in contrast to the usual classification of many of these compounds (as ionic solids). All the compounds shown in Fig. 8.13 with more than 50% ionic character are normally considered to be ionic solids. Recall, however, the results in Fig. 8.13 are for the gas phase, where individual XY molecules exist. These results can-not necessarily be assumed to apply to the solid state, where the existence of ions is fa-vored by the multiple ion interactions.
Another complication in identifying ionic compounds is that many substances contain polyatomic ions. For example, NH4Cl contains NH4 and Cl ions, and Na2SO4 contains Na and SO4 2 ions. The ammonium and sulfate ions are held together by covalent bonds. Thus, calling NH4Cl and Na2SO4 ionic compounds is somewhat ambiguous.
Percent ionic character of a bond a measured dipole moment of X¬Y calculated dipole moment of XYb 100% FIGURE 8.12 The three possible types of bonds: (a) a covalent bond formed between identical F atoms; (b) the polar covalent bond of HF, with both ionic and covalent components; and (c) an ionic bond with no electron sharing.
(a) (b) (c) – + H F F F Since the equation for lattice energy con-tains the product Q1Q2, the lattice energy for a solid with 2 and 2 ions should be four times that for a solid with 1 and 1 ions. That is, For MgO and NaF, the observed ratio of lattice energies (see Fig. 8.11) is 3916 kJ 923 kJ 4.24 122122 112112 4 8.7 The Covalent Chemical Bond: A Model 347 We will avoid these problems by adopting an operational definition of ionic com-pounds: Any compound that conducts an electric current when melted will be classified as ionic.
8.7 The Covalent Chemical Bond: A Model Before we develop specific models for covalent chemical bonding, it will be helpful to summarize some of the concepts introduced in this chapter.
What is a chemical bond? Chemical bonds can be viewed as forces that cause a group of atoms to behave as a unit.
Why do chemical bonds occur? There is no principle of nature that states that bonds are favored or disfavored. Bonds are neither inherently “good” nor inherently “bad” as far as nature is concerned; bonds result from the tendency of a system to seek its lowest pos-sible energy. From a simplistic point of view, bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms. For example, approximately 1652 kJ of energy is required to break a mole of methane (CH4) molecules into separate C and H atoms. Or, taking the opposite view, 1652 kJ of energy is released when 1 mole of methane is formed from 1 mole of gaseous C atoms and 4 moles of gaseous H atoms. Thus we can say that 1 mole of CH4 molecules in the gas phase is 1652 kJ lower in energy than 1 mole of carbon atoms plus 4 moles of hydrogen atoms. Methane is therefore a stable molecule relative to its separated atoms.
We find it useful to interpret molecular stability in terms of a model called a chemi-cal bond. To understand why this model was invented, let’s continue with methane, which consists of four hydrogen atoms arranged at the corners of a tetrahedron around a carbon atom: FIGURE 8.13 The relationship between the ionic charac-ter of a covalent bond and the electronega-tivity difference of the bonded atoms. Note that the compounds with ionic character greater than 50% are normally considered to be ionic compounds.
A tetrahedron has four equal triangular faces.
Percent ionic character 0 0 Electronegativity difference 1 2 3 25 50 75 100 LiF KF CsF CsCl KCl NaCl KBr LiCl LiBr Kl CsI HF LiI HCl HBr ICl HI IBr H H C H H Given this structure, it is natural to envision four individual COH interactions (we call them bonds). The energy of stabilization of CH4 is divided equally among the four bonds to give an average COH bond energy per mole of COH bonds: 1652 kJ/mol 4 413 kJ/mol Bonding is a model proposed to explain molecular stability.
348 Chapter Eight Bonding: General Concepts Molten NaCl conducts an electric current, indicating the presence of mobile Na and Cl ions.
Next, consider methyl chloride, which consists of CH3Cl molecules having the structure Experiments have shown that approximately 1578 kJ of energy is required to break down 1 mole of gaseous CH3Cl molecules into gaseous carbon, chlorine, and hydrogen atoms.
The reverse process can be represented as A mole of gaseous methyl chloride is lower in energy by 1578 kJ than its separate gaseous atoms. Thus a mole of methyl chloride is held together by 1578 kJ of energy. Again, it is very useful to divide this energy into individual bonds. Methyl chloride can be visual-ized as containing one C—Cl bond and three C—H bonds. If we assume arbitrarily that a C—H interaction represents the same quantity of energy in any situation (that is, that the strength of a C—H bond is independent of its molecular environment), we can do the following bookkeeping: These assumptions allow us to associate given quantities of energy with C—H and C—Cl bonds.
It is important to note that the bond concept is a human invention. Bonds provide a method for dividing up the energy evolved when a stable molecule is formed from its component atoms. Thus in this context a bond represents a quantity of energy obtained from the overall molecular energy of stabilization in a rather arbitrary way. This is not to say that the concept of individual bonds is a bad idea. In fact, the modern concept of the chemical bond, conceived by the American chemists G. N. Lewis and Linus Pauling, is one of the most useful ideas chemists have ever developed.
Models: An Overview The framework of chemistry, like that of any science, consists of models—attempts to explain how nature operates on the microscopic level based on experiences in the macro-scopic world. To understand chemistry, one must understand its models and how they are used. We will use the concept of bonding to reemphasize the important characteristics of models, including their origin, structure, and uses.
Models originate from our observations of the properties of nature. For example, the concept of bonds arose from the observations that most chemical processes involve col-lections of atoms and that chemical reactions involve rearrangements of the ways the atoms are grouped. Therefore, to understand reactions, we must understand the forces that bind atoms together.
In natural processes there is a tendency toward lower energy. Collections of atoms therefore occur because the aggregated state has lower energy than the separated atoms.
Why? As we saw earlier in this chapter, the best explanations for the energy change involve C¬Cl bond energy 1578 1239 339 kJ/mol C¬Cl bond energy 31413 kJ/mol2 1578 kJ C¬Cl bond energy 31average C¬H bond energy2 1578 kJ 1 mol C¬Cl bonds plus 3 mol C¬H bonds 1578 kJ C1g2 Cl1g2 3H1g2 ¡ CH3Cl1g2 1578 kJ/mol H H C H Cl 8.7 The Covalent Chemical Bond: A Model 349 atoms sharing electrons or atoms transferring electrons to become ions. In the case of elec-tron sharing, we find it convenient to assume that individual bonds occur between pairs of atoms. Let’s explore the validity of this assumption and see how it is useful.
In a diatomic molecule such as H2, it is natural to assume that a bond exists between the atoms, holding them together. It is also useful to assume that individual bonds are present in polyatomic molecules such as CH4. Therefore, instead of thinking of CH4 as a unit with a sta-bilization energy of 1652 kJ per mole, we choose to think of CH4 as containing four C—H bonds, each worth 413 kJ of energy per mole of bonds. Without this concept of individual bonds in molecules, chemistry would be hopelessly complicated. There are millions of different chemical compounds, and if each of these compounds had to be considered as an entirely new entity, the task of understanding chemical behavior would be overwhelming.
The bonding model provides a framework to systematize chemical behavior by en-abling us to think of molecules as collections of common fundamental components. For example, a typical biomolecule, such as a protein, contains hundreds of atoms and might seem discouragingly complex. However, if we think of a protein as constructed of indi-vidual bonds, COC, COH, CON, COO, NOH, and so on, it helps tremendously in predicting and understanding the protein’s behavior. The essential idea is that we expect a given bond to behave about the same in any molecular environment. Used in this way, the model of the chemical bond has helped chemists to systematize the reactions of the millions of existing compounds.
In addition to being useful, the bonding model is physically sensible. It makes sense that atoms can form stable groups by sharing electrons; shared electrons give a lower en-ergy state because they are simultaneously attracted by two nuclei.
Also, as we will see in the next section, bond energy data support the existence of discrete bonds that are relatively independent of the molecular environment. It is very important to remember, however, that the chemical bond is only a model. Although our concept of discrete bonds in molecules agrees with many of our observations, some mo-lecular properties require that we think of a molecule as a whole, with the electrons free to move through the entire molecule. This is called delocalization of the electrons, a concept that will be discussed more completely in the next chapter.
The concept of individual bonds makes it much easier to deal with complex molecules such as DNA. A small segment of a DNA molecule is shown here.
350 Chapter Eight Bonding: General Concepts Fundamental Properties of Models Models are human inventions, always based on an incomplete understanding of how nature works. A model does not equal reality.
Models are often wrong. This property derives from the first property. Models are based on speculation and are always oversimplifications.
Models tend to become more complicated as they age. As flaws are discovered in our models, we “patch” them and thus add more detail.
It is very important to understand the assumptions inherent in a particular model be-fore you use it to interpret observations or to make predictions. Simple models usu-ally involve very restrictive assumptions and can be expected to yield only qualitative information. Asking for a sophisticated explanation from a simple model is like expecting to get an accurate mass for a diamond using a bathroom scale.
For a model to be used effectively, we must understand its strengths and weak-nesses and ask only appropriate questions. An illustration of this point is the simple aufbau principle used to account for the electron configurations of the elements. Al-though this model correctly predicts the configuration for most atoms, chromium and copper, for example, do not agree with the predictions. Detailed studies show that the configurations of chromium and copper result from complex electron interactions that are not taken into account in the simple model. However, this does not mean that we should discard the simple model that is so useful for most atoms. Instead, we must apply it with caution and not expect it to be correct in every case.
When a model is wrong, we often learn much more than when it is right. If a model makes a wrong prediction, it usually means we do not understand some fundamen-tal characteristics of nature. We often learn by making mistakes. (Try to remember this when you get back your next chemistry test.) 8.8 Covalent Bond Energies and Chemical Reactions In this section we will consider the energies associated with various types of bonds and see how the bonding concept is useful in dealing with the energies of chemical reactions.
One important consideration is to establish the sensitivity of a particular type of bond to its molecular environment. For example, consider the stepwise decomposition of methane: Process Energy Required (kJ/mol) CH4(g) S CH3(g) H(g) 435 CH3(g) S CH2(g) H(g) 453 CH2(g) S CH(g) H(g) 425 CH(g) S C(g) H(g) 339 Total 1652 Average 413 Although a COH bond is broken in each case, the energy required varies in a non-systematic way. This example shows that the COH bond is somewhat sensitive to its environment. We use the average of these individual bond dissociation energies even though this quantity only approximates the energy associated with a COH bond in a par-ticular molecule. The degree of sensitivity of a bond to its environment also can be seen 1652 4 8.8 Covalent Bond Energies and Chemical Reactions 351 from experimental measurements of the energy required to break the COH bond in the following molecules: Measured COH Bond Molecule Energy (kJ/mol) HCBr3 380 HCCl3 380 HCF3 430 C2H6 410 These data show that the COH bond strength varies significantly with its environment, but the concept of an average COH bond strength remains useful to chemists. The aver-age values of bond energies for various types of bonds are listed in Table 8.4.
So far we have discussed bonds in which one pair of electrons is shared. This type of bond is called a single bond. As we will see in more detail later, atoms sometimes share two pairs of electrons, forming a double bond, or share three pairs of electrons, forming a triple bond. The bond energies for these multiple bonds are also given in Table 8.4.
A relationship also exists between the number of shared electron pairs and the bond length. As the number of shared electrons increases, the bond length shortens. This rela-tionship is shown for selected bonds in Table 8.5.
Bond Energy and Enthalpy Bond energy values can be used to calculate approximate energies for reactions. To illustrate how this is done, we will calculate the change in energy that accompanies the following reaction: This reaction involves breaking one HOH and one FOF bond and forming two HOF bonds. For bonds to be broken, energy must be added to the system—-an endothermic H21g2 F21g2 ¡ 2HF1g2 TABLE 8.4 Average Bond Energies (kJ/mol) Single Bonds Multiple Bonds HOH 432 NOH 391 IOI 149 CPC 614 HOF 565 NON 160 IOCl 208 CqC 839 HOCl 427 NOF 272 IOBr 175 OPO 495 HOBr 363 NOCl 200 CPO 745 HOI 295 NOBr 243 SOH 347 CqO 1072 NOO 201 SOF 327 NPO 607 COH 413 OOH 467 SOCl 253 NPN 418 COC 347 OOO 146 SOBr 218 NqN 941 CON 305 OOF 190 SOS 266 CqN 891 COO 358 OOCl 203 CPN 615 COF 485 OOI 234 SiOSi 340 COCl 339 SiOH 393 COBr 276 FOF 154 SiOC 360 COI 240 FOCl 253 SiOO 452 COS 259 FOBr 237 ClOCl 239 ClOBr 218 BrOBr 193 CPO(CO2) 799 352 Chapter Eight Bonding: General Concepts process. Consequently, the energy terms associated with bond breaking have positive signs.
The formation of a bond releases energy, an exothermic process, so the energy terms as-sociated with bond making carry a negative sign. We can write the enthalpy change for a reaction as follows: sum of the energies required to break old bonds (positive signs) plus the sum of the energies released in the formation of new bonds (negative signs) This leads to the expression Energy required Energy released where represents the sum of terms, and D represents the bond energy per mole of bonds.
(D always has a positive sign.) In the case of the formation of HF, Thus, when 1 mol H2(g) and 1 mol F2(g) react to form 2 mol HF(g), 544 kJ of energy should be released.
This result can be compared with the calculation of H for this reaction from the standard enthalpy of formation for HF (271 kJ/mol): Thus the use of bond energies to calculate H works quite well in this case.
H from Bond Energies Using the bond energies listed in Table 8.4, calculate H for the reaction of methane with chlorine and fluorine to give Freon-12 (CF2Cl2).
Solution The idea here is to break the bonds in the gaseous reactants to give individual atoms and then assemble these atoms into the gaseous products by forming new bonds: Reactants atoms products ¬¡ Energy released ¬¡ Energy required CH41g2 2Cl21g2 2F21g2 ¡ CF2Cl21g2 2HF1g2 2HCl1g2 ¢H° 2 mol 1271 kJ/mol2 542 kJ 544 kJ 1 mol 432 kJ mol 1 mol 154 kJ mol 2 mol 565 kJ mol ¢H DH¬H DF¬F 2DH¬F ¢H ©D 1bonds broken2 ©D 1bonds formed2 ¢H TABLE 8.5 Bond Lengths for Selected Bonds Bond Bond Type Bond Length (pm) Bond Energy (kJ/mol) COC Single 154 347 CPC Double 134 614 CqC Triple 120 839 COO Single 143 358 CPO Double 123 745 CON Single 143 305 CPN Double 138 615 CqN Triple 116 891 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ Sample Exercise 8.5 8.9 The Localized Electron Bonding Model 353 We then combine the energy changes to calculate H: where the minus sign gives the correct sign to the energy terms for the exothermic processes.
Reactant Bonds Broken: Product Bonds Formed: and We now can calculate H: Since the sign of the value for the enthalpy change is negative, this means that 1194 kJ of energy is released per mole of CF2Cl2 formed.
See Exercises 8.53 through 8.60.
8.9 The Localized Electron Bonding Model So far we have discussed the general characteristics of the chemical bonding model and have seen that properties such as bond strength and polarity can be assigned to individ-ual bonds. In this section we introduce a specific model used to describe covalent bonds.
We need a simple model that can be applied easily even to very complicated molecules and that can be used routinely by chemists to interpret and organize the wide variety of chemical phenomena. The model that serves this purpose is the localized electron (LE) model, which assumes that a molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Electron pairs in the molecule are assumed to be localized on a particular atom or in the space between two atoms. Those pairs of electrons localized on an atom are called lone pairs, and those found in the space between the atoms are called bonding pairs.
1194 kJ 2438 kJ 3632 kJ ¢H energy required to break bonds energy released when bonds form Total energy released 3632 kJ HCl: 2 mol H¬Cl 2 mol 427 kJ mol 3854 kJ HF: 2 mol H¬F 2 mol 565 kJ mol 1130 kJ 2 mol C¬Cl 2 mol 339 kJ mol 3678 kJ CF2Cl2: 2 mol C¬F 2 mol 485 kJ mol 3970 kJ Total energy required 2438 kJ 2F2: 2 mol F¬FF 2 mol 154 kJ mol 2308 kJ 2Cl2: 2 mol Cl¬Cl 2 mol 239 kJ mol 1478 kJ CH4: 4 mol C¬H 4 mol 413 kJ mol 1652 kJ ¢H energy required to break bonds energy released when bonds form 354 Chapter Eight Bonding: General Concepts As we will apply it, the LE model has three parts: 1. Description of the valence electron arrangement in the molecule using Lewis struc-tures (will be discussed in the next section).
2. Prediction of the geometry of the molecule using the valence shell electron-pair repulsion (VSEPR) model (will be discussed in Section 8.13).
3. Description of the type of atomic orbitals used by the atoms to share electrons or hold lone pairs (will be discussed in Chapter 9).
8.10 Lewis Structures The Lewis structure of a molecule shows how the valence electrons are arranged among the atoms in the molecule. These representations are named after G. N. Lewis (Fig. 8.14).
The rules for writing Lewis structures are based on observations of thousands of molecules. From experiment, chemists have learned that the most important requirement for the formation of a stable compound is that the atoms achieve noble gas electron configurations.
We have already seen that when metals and nonmetals react to form binary ionic com-pounds, electrons are transferred and the resulting ions typically have noble gas electron configurations. An example is the formation of KBr, where the K ion has the [Ar] elec-tron configuration and the Br ion has the [Kr] electron configuration. In writing Lewis structures, the rule is that only the valence electrons are included. Using dots to represent electrons, the Lewis structure for KBr is No dots are shown on the K ion because it has no valence electrons. The Br ion is shown with eight electrons because it has a filled valence shell.
Next we will consider Lewis structures for molecules with covalent bonds, involving elements in the first and second periods. The principle of achieving a noble gas electron configuration applies to these elements as follows: • Hydrogen forms stable molecules where it shares two electrons. That is, it follows a duet rule. For example, when two hydrogen atoms, each with one electron, combine to form the H2 molecule, we have By sharing electrons, each hydrogen in H2, in effect, has two electrons; that is, each hydrogen has a filled valence shell.
• Helium does not form bonds because its valence orbital is already filled; it is a noble gas.
Helium has the electron configuration 1s2 and can be represented by the Lewis structure He : Lewis structures show only valence electrons.
FIGURE 8.14 G. N. Lewis (1875–1946).
8.10 Lewis Structures 355 • The second-row nonmetals carbon through fluorine form stable molecules when they are surrounded by enough electrons to fill the valence orbitals, that is, the 2s and the three 2p orbitals. Since eight electrons are required to fill these orbitals, these elements typically obey the octet rule; they are surrounded by eight electrons. An example is the F2 molecule, which has the following Lewis structure: Note that each fluorine atom in F2 is, in effect, surrounded by eight electrons, two of which are shared with the other atom. This is a bonding pair of electrons, as discussed earlier. Each fluorine atom also has three pairs of electrons not involved in bonding.
These are the lone pairs.
• Neon does not form bonds because it already has an octet of valence electrons (it is a noble gas). The Lewis structure is Note that only the valence electrons of the neon atom (2s22p6) are represented by the Lewis structure. The 1s2 electrons are core electrons and are not shown.
From the preceding discussion we can formulate the following rules for writing the Lewis structures of molecules containing atoms from the first two periods.
Steps for Writing Lewis Structures ➥ 1 Sum the valence electrons from all the atoms. Do not worry about keeping track of which electrons come from which atoms. It is the total number of electrons that is important.
➥ 2 Use a pair of electrons to form a bond between each pair of bound atoms.
➥ 3 Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the second-row elements.
To see how these steps are applied, we will draw the Lewis structures of a few mol-ecules. We will first consider the water molecule and follow the previous steps.
➥ 1 We sum the valence electrons for H2O as shown: p p p H H O ➥ 2 Using a pair of electrons per bond, we draw in the two OOH single bonds: Note that a line instead of a pair of dots is used to indicate each pair of bonding electrons.
This is the standard notation.
➥ 3 We distribute the remaining electrons to achieve a noble gas electron configuration for each atom. Since four electrons have been used in forming the two bonds, four elec-trons (8 4) remain to be distributed. Hydrogen is satisfied with two electrons (duet rule), but oxygen needs eight electrons to have a noble gas configuration. Thus the remaining H¬O¬H 1 1 6 8 valence electrons Carbon, nitrogen, oxygen, and fluorine always obey the octet rule in stable molecules.
356 Chapter Eight Bonding: General Concepts four electrons are added to oxygen as two lone pairs. Dots are used to represent the lone pairs: This is the correct Lewis structure for the water molecule. Each hydrogen has two elec-trons and the oxygen has eight, as shown below: As a second example, let’s write the Lewis structure for carbon dioxide. Summing the valence electrons gives p p p C O O After forming a bond between the carbon and each oxygen, the remaining electrons are distributed to achieve noble gas configurations on each atom.
In this case we have 12 electrons (16 4) remaining after the bonds are drawn. The dis-tribution of these electrons is determined by a trial-and-error process. We have 6 pairs of electrons to distribute. Suppose we try 3 pairs on each oxygen to give Is this correct? To answer this question, we need to check two things: 1. The total number of electrons. There are 16 valence electrons in this structure, which is the correct number.
2. The octet rule for each atom. Each oxygen has 8 electrons, but the carbon has only 4. This cannot be the correct Lewis structure.
How can we arrange the 16 available electrons to achieve an octet for each atom?
Suppose there are 2 shared pairs between the carbon and each oxygen: Now each atom is surrounded by 8 electrons, and the total number of electrons is 16, as required. This is the correct Lewis structure for carbon dioxide, which has two double bonds and four lone pairs.
Finally, let’s consider the Lewis structure of the CN (cyanide) ion. Summing the va-lence electrons, we have 4 5 1 10 888n CN O¬C¬O 4 6 6 16 888n 888n represents HOOOH represents H:O:H O O C O O C 8.10 Lewis Structures 357 Note that the negative charge means an extra electron is present. After drawing a single bond (C—N), we distribute the remaining electrons to achieve a noble gas configuration for each atom. Eight electrons remain to be distributed. We can try various possibilities, for example: This structure is incorrect because C and N have only six electrons each instead of eight.
The correct arrangement is (Satisfy yourself that both carbon and nitrogen have eight electrons.) Writing Lewis Structures Give the Lewis structure for each of the following.
a. HF d. CH4 b. N2 e. CF4 c. NH3 f. NO Solution In each case we apply the three steps for writing Lewis structures. Recall that lines are used to indicate shared electron pairs and that dots are used to indicate nonbonding pairs (lone pairs). We have the following tabulated results: Sample Exercise 8.6 See Exercises 8.67 and 8.68.
Calculate Use Remaining Total Number of Electrons to Valence Draw Single Electrons Achieve Noble Check Number Electrons Bonds Remaining Gas Configurations of Electrons a. HF 1 7 8 HOF 6 H, 2 F, 8 b. N2 5 5 10 8 N, 8 c. NH3 5 3(1) 8 2 H, 2 N, 8 d. CH4 4 4(1) 8 0 H, 2 C, 8 e. CF4 4 4(7) 32 24 F, 8 C, 8 f. NO 5 6 1 10 8 N, 8 O, 8 N O [ ] N O F F F C F F F F C F H H H C H H H H C H H H H N H H H N N N N N H F 358 Chapter Eight Bonding: General Concepts When writing Lewis structures, do not worry about which electrons come from which atoms in a molecule. The best way to look at a molecule is to regard it as a new entity that uses all the available valence electrons of the atoms to achieve the lowest possible energy. The valence electrons belong to the molecule, rather than to the individual atoms.
Simply distribute all valence electrons so that the various rules are satisfied, without re-gard for the origin of each particular electron.
8.11 Exceptions to the Octet Rule The localized electron model is a simple but very successful model, and the rules we have used for Lewis structures apply to most molecules. However, with such a simple model, some exceptions are inevitable. Boron, for example, tends to form compounds in which the boron atom has fewer than eight electrons around it—it does not have a complete octet. Boron trifluoride (BF3), a gas at normal temperatures and pressures, reacts very T he element nitrogen exists at normal temperatures and pressures as a gas containing N2, a molecule with a very strong triple bond. In the gas phase the diatomic molecules move around independently with almost no tendency to associate with each other. Under intense pressure, however, nitrogen changes to a dramatically different form. This conclusion was reached at the Carnegie Institution in Washington, D.C., by Mikhail Erements and his colleagues, who subjected nitrogen to a pressure of 2.4 million atmos-pheres in a special diamond anvil press. Under this tremen-dous pressure the bonds of the N2 molecules break and a substance containing an aggregate of nitrogen atoms forms.
In other words, under great pressure elemental nitrogen changes from a substance containing diatomic molecules to one containing many nitrogen atoms bonded to each other.
Interestingly, this substance remains intact even after the pressure is released—-as long as the temperature remains at 100 K. This new form of nitrogen has a very high potential energy relative to N2. Thus this substance would be an extraordinarily powerful propellant or explosive if enough of it could be made. This new form of nitrogen is also a semiconductor for electricity; normal nitrogen gas is an insulator.
The newly discovered form of nitrogen is significant for several reasons. For one thing, it may help us understand the nature of the interiors of the giant gas planets such as Jupiter. Also, their success in changing nitrogen to an atomic solid encourages high-pressure scientists who are trying to accomplish the same goal with hydrogen. It is surprising that nitrogen, which has diatomic molecules containing bonds more than twice as strong as those in hydrogen, will form an atomic solid at these pressures but hydrogen does not.
Hydrogen remains a molecular solid at far greater pressures than nitrogen can endure.
CHEMICAL IMPACT Nitrogen Under Pressure A diamond anvil cell used to study materials at very high pressures.
In a sense this approach corrects for the fact that the localized electron model overemphasizes that a mol-ecule is simply a sum of its parts—that is, that the atoms retain their individual identities in the molecule.
8.11 Exceptions to the Octet Rule 359 energetically with molecules such as water and ammonia that have available electron pairs (lone pairs). The violent reactivity of BF3 with electron-rich molecules arises because the boron atom is electron-deficient. Boron trifluoride has 24 valence electrons. The Lewis structure often drawn for BF3 is Note that in this structure boron has only 6 electrons around it. The octet rule for boron can be satisfied by drawing a structure with a double bond, such as Recent studies indicate that double bonding may be important in BF3. However, the boron atom in BF3 certainly behaves as if it is electron-deficient, as indicated by the re-activity of BF3 toward electron-rich molecules, for example, toward NH3 to form H3NBF3: In this stable compound, boron has an octet of electrons.
It is characteristic of boron to form molecules in which the boron atom is electron-deficient. On the other hand, carbon, nitrogen, oxygen, and fluorine can be counted on to obey the octet rule.
Some atoms exceed the octet rule. This behavior is observed only for those elements in Period 3 of the periodic table and beyond. To see how this arises, we will consider the Lewis structure for sulfur hexafluoride (SF6), a well-known and very stable molecule. The sum of the valence electrons is Indicating the single bonds gives the structure on the left below: We have used 12 electrons to form the SOF bonds, which leaves 36 electrons. Since fluor-ine always follows the octet rule, we complete the six fluorine octets to give the structure on the right above. This structure uses all 48 valence electrons for SF6, but sulfur has 12 electrons around it; that is, sulfur exceeds the octet rule. How can this happen?
To answer this question, we need to consider the different types of valence orbitals characteristic of second- and third-period elements. The second-row elements have 2s and 2p valence orbitals, and the third-row elements have 3s, 3p, and 3d orbitals. The 3s and 3p orbitals fill with electrons in going from sodium to argon, but the 3d orbitals remain empty. For example, the valence orbital diagram for a sulfur atom is 3s 3p 3d 6 6172 48 electrons 360 Chapter Eight Bonding: General Concepts The localized electron model assumes that the empty 3d orbitals can be used to accom-modate extra electrons. Thus the sulfur atom in SF6 can have 12 electrons around it by using the 3s and 3p orbitals to hold 8 electrons, with the extra 4 electrons placed in the formerly empty 3d orbitals.
Lewis Structures: Comments About the Octet Rule The second-row elements C, N, O, and F should always be assumed to obey the octet rule.
The second-row elements B and Be often have fewer than eight electrons around them in their compounds. These electron-deficient compounds are very reactive.
The second-row elements never exceed the octet rule, since their valence orbitals (2s and 2p) can accommodate only eight electrons.
Third-row and heavier elements often satisfy the octet rule but can exceed the octet rule by using their empty valence d orbitals.
When writing the Lewis structure for a molecule, satisfy the octet rule for the atoms first. If electrons remain after the octet rule has been satisfied, then place them on the elements having available d orbitals (elements in Period 3 or beyond).
Lewis Structures for Molecules That Violate the Octet Rule I Write the Lewis structure for PCl5.
Solution We can follow the same stepwise procedure we used above for sulfur hexafluoride.
➥1 Sum the valence electrons.
h h P Cl ➥2 Indicate single bonds between bound atoms.
➥3 Distribute the remaining electrons. In this case, 30 electrons (40 10) remain.
These are used to satisfy the octet rule for each chlorine atom. The final Lewis structure is Note that phosphorus, which is a third-row element, has exceeded the octet rule by two electrons.
See Exercises 8.71 and 8.72.
In the PCl5 and SF6 molecules, the central atoms (P and S, respectively) must have the extra electrons. However, in molecules having more than one atom that can exceed the octet rule, it is not always clear which atom should have the extra electrons. Consider 5 5172 40 electrons Third-row elements can exceed the octet rule.
Whether the atoms that exceed the octet rule actually place the extra electrons in their d orbitals is a matter of controversy among theoretical chemists. We will not consider this issue in this text.
Sample Exercise 8.7 8.11 Exceptions to the Octet Rule 361 the Lewis structure for the triiodide ion (I3 ), which has h h 1 1 charge Indicating the single bonds gives IOIOI. At this point, 18 electrons (22 4) remain.
Trial and error will convince you that one of the iodine atoms must exceed the octet rule, but which one?
The rule we will follow is that when it is necessary to exceed the octet rule for one of several third-row (or higher) elements, assume that the extra electrons should be placed on the central atom.
Thus for I3 the Lewis structure is where the central iodine exceeds the octet rule. This structure agrees with known properties of I3 .
Lewis Structures for Molecules That Violate the Octet Rule II Write the Lewis structure for each molecule or ion.
a. ClF3 b. XeO3 c. RnCl2 d. BeCl2 e. ICl4 Solution a. The chlorine atom (third row) accepts the extra electrons.
b. All atoms obey the octet rule.
c. Radon, a noble gas in Period 6, accepts the extra electrons.
d. Beryllium is electron-deficient.
e. Iodine exceeds the octet rule.
See Exercises 8.71 and 8.72.
3172 1 22 valence electrons Sample Exercise 8.8 362 Chapter Eight Bonding: General Concepts 8.12 Resonance Sometimes more than one valid Lewis structure (one that obeys the rules we have out-lined) is possible for a given molecule. Consider the Lewis structure for the nitrate ion (NO3 ), which has 24 valence electrons. To achieve an octet of electrons around each atom, a structure like this is required: If this structure accurately represents the bonding in NO3 , there should be two types of N777O bonds observed in the molecule: one shorter bond (the double bond) and two iden-tical longer ones (the two single bonds). However, experiments clearly show that NO3 exhibits only one type of N777O bond with a length and strength between those expected for a single bond and a double bond. Thus, although the structure we have shown above is a valid Lewis structure, it does not correctly represent the bonding in NO3 . This is a serious problem, and it means that the model must be modified.
Look again at the proposed Lewis structure for NO3 . There is no reason for choos-ing a particular oxygen atom to have the double bond. There are really three valid Lewis structures: Is any of these structures a correct description of the bonding in NO3 ? No, because NO3 does not have one double and two single bonds—it has three equivalent bonds. We can solve this problem by making the following assumption: The correct description of NO3 is not given by any one of the three Lewis structures but is given only by the superposi-tion of all three.
The nitrate ion does not exist as any of the three extreme structures but exists as an average of all three. Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule. The resulting electron structure of the molecule is given by the average of these resonance structures. This situation is usually represented by double-headed arrows as follows: Note that in all these resonance structures the arrangement of the nuclei is the same. Only the placement of the electrons differs. The arrows do not mean that the molecule “flips” from one resonance to another. They simply show that the actual structure is an average of the three resonance structures.
The concept of resonance is necessary because the localized electron model postu-lates that electrons are localized between a given pair of atoms. However, nature does not really operate this way. Electrons are really delocalized—they can move around the entire molecule. The valence electrons in the NO3 molecule distribute themselves to provide equivalent N777O bonds. Resonance is necessary to compensate for the defective assump-tion of the localized electron model. However, this model is so useful that we retain the 8.12 Resonance 363 concept of localized electrons and add resonance to allow the model to treat species such as NO3 .
Resonance Structures Describe the electron arrangement in the nitrite anion (NO2 ) using the localized electron model.
Solution We will follow the usual procedure for obtaining the Lewis structure for the NO2 ion.
In NO2 there are 5 2(6) 1 18 valence electrons. Indicating the single bonds gives the structure The remaining 14 electrons (18 4) can be distributed to produce these structures: This is a resonance situation. Two equivalent Lewis structures can be drawn. The elec-tronic structure of the molecule is correctly represented not by either resonance structure but by the average of the two. There are two equivalent N777O bonds, each one intermediate between a single and a double bond.
See Exercises 8.73 through 8.78.
Odd-Electron Molecules Relatively few molecules formed from nonmetals contain odd numbers of electrons. One common example is nitric oxide (NO), which is formed when nitrogen and oxygen gases react at the high temperatures in automobile engines. Nitric oxide is emitted into the air, where it immediately reacts with oxygen to form gaseous nitrogen dioxide (NO2), another odd-electron molecule.
Since the localized electron model is based on pairs of electrons, it does not handle odd-electron cases in a natural way. To treat odd-electron molecules, a more sophisticated model is needed.
Formal Charge Molecules or polyatomic ions containing atoms that can exceed the octet rule often have many nonequivalent Lewis structures, all of which obey the rules for writing Lewis struc-tures. For example, as we will see in detail below, the sulfate ion has a Lewis structure with all single bonds and several Lewis structures that contain double bonds. How do we decide which of the many possible Lewis structures best describes the actual bonding in sulfate? One method is to estimate the charge on each atom in the various possible Lewis structures and use these charges to select the most appropriate structure(s). We will see below how this is done, but first we must decide on a method to assign atomic charges in molecules.
In Chapter 4 we discussed one system for obtaining charges, called oxidation states.
However, in assigning oxidation states, we always count both the shared electrons as belonging to the more electronegative atom in a bond. This practice leads to highly exaggerated estimates of charge. In other words, although oxidation states are useful for bookkeeping electrons in redox reactions, they are not realistic estimates of the actual O¬N¬O Sample Exercise 8.9 Equivalent Lewis structures contain the same numbers of single and multiple bonds. For example, the resonance structures for O3 are equivalent Lewis structures. These are equally important in describing the bonding in O3. Nonequivalent Lewis structures contain different numbers of single and multiple bonds.
O O O O O O and 364 Chapter Eight Bonding: General Concepts charges on individual atoms in a molecule, so they are not suitable for judging the appropriateness of Lewis structures. Another definition of the charge on an atom in a mol-ecule, the formal charge, can, however, be used to evaluate Lewis structures. As we will see below, the formal charge of an atom in a molecule is the difference between the num-ber of valence electrons on the free atom and the number of valence electrons assigned to the atom in the molecule.
Therefore, to determine the formal charge of a given atom in a molecule, we need to know two things: 1. The number of valence electrons on the free neutral atom (which has zero net charge because the number of electrons equals the number of protons) 2. The number of valence electrons “belonging” to the atom in a molecule We then compare these numbers. If in the molecule the atom has the same number of va-lence electrons as it does in the free state, the positive and negative charges just balance, and it has a formal charge of zero. If the atom has one more valence electron in a mole-cule than it has as a free atom, it has a formal charge of 1, and so on. Thus the formal charge on an atom in a molecule is defined as To compute the formal charge of an atom in a molecule, we assign the valence electrons in the molecule to the various atoms, making the following assumptions: 1. Lone pair electrons belong entirely to the atom in question.
2. Shared electrons are divided equally between the two sharing atoms.
Thus the number of valence electrons assigned to a given atom is calculated as follows: We will illustrate the procedure for calculating formal charges by considering two of the possible Lewis structures for the sulfate ion, which has 32 valence electrons. For the Lewis structure each oxygen atom has 6 lone pair electrons and shares 2 electrons with the sulfur atom.
Thus, using the preceding assumptions, each oxygen is assigned 7 valence electrons.
h h Lone Shared pair electrons electrons h Valence electrons on a free O atom Valence electrons assigned to each O in SO4 2 Formal charge on oxygen 6 minus 7 1 Valence electrons assigned to each oxygen 6 plus 1 2 122 7 1 2 1number of shared electrons2 1Valence electrons2assigned 1number of lone pair electrons2 1number of valence electrons assigned to the atom in the molecule2 Formal charge 1number of valence electrons on free atom2 h A A A 8.12 Resonance 365 The formal charge on each oxygen is 1.
For the sulfur atom there are no lone pair electrons, and eight electrons are shared with the oxygen atoms. Thus, for sulfur, h h Lone Shared pair electrons electrons h Valence electrons on a free S atom Valence electrons assigned to S in A second possible Lewis structure is In this case the formal charges are as follows: For oxygen atoms with single bonds: For oxygen atoms with double bonds: h Each double bond has 4 electrons For the sulfur atom: We will use two fundamental assumptions about formal charges to evaluate Lewis structures: 1. Atoms in molecules try to achieve formal charges as close to zero as possible.
2. Any negative formal charges are expected to reside on the most electronegative atoms.
We can use these principles to evaluate the two Lewis structures for sulfate given previ-ously. Notice that in the structure with only single bonds, each oxygen has a formal charge of 1, while the sulfur has a formal charge of 2. In contrast, in the structure with two double bonds and two single bonds, the sulfur and two oxygen atoms have a formal charge of 0, while two oxygens have a formal charge of 1. Based on the assumptions given above, the structure with two double bonds is preferred—it has lower formal charges and Formal charge 6 6 0 Valence electrons assigned 0 1 21122 6 Formal charge 6 6 0 Valence electrons assigned 4 1 2142 6 Formal charge 6 7 1 Valence electrons assigned 6 1 2122 7 SO4 2 Formal charge on sulfur 6 minus 4 2 Valence electrons assigned to sulfur 0 plus 1 2182 4 h A A A 366 Chapter Eight Bonding: General Concepts the 1 formal charges are on electronegative oxygen atoms. Thus, for the sulfate ion, we might expect resonance structures such as to more closely describe the bonding than the Lewis structure with only single bonds.
Rules Governing Formal Charge To calculate the formal charge on an atom: 1. Take the sum of the lone pair electrons and one-half the shared electrons. This is the number of valence electrons assigned to the atom in the molecule.
2. Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom to obtain the formal charge.
The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species.
If nonequivalent Lewis structures exist for a species, those with formal charges clos-est to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion.
Formal Charges Give possible Lewis structures for XeO3, an explosive compound of xenon. Which Lewis structure or structures are most appropriate according to the formal charges?
Solution For XeO3 (26 valence electrons) we can draw the following possible Lewis structures (for-mal charges are indicated in parentheses): Based on the ideas of formal charge, we would predict that the Lewis structures with the lower values of formal charge would be most appropriate for describing the bonding in XeO3.
See Exercises 8.81 through 8.86.
As a final note, there are a couple of cautions about formal charge to keep in mind.
First, although formal charges are closer to actual atomic charges in molecules than are oxidation states, formal charges still provide only estimates of charge—they should not Sample Exercise 8.10 8.13 Molecular Structure: The VSEPR Model 367 be taken as actual atomic charges. Second, the evaluation of Lewis structures using for-mal charge ideas can lead to erroneous predictions. Tests based on experiments must be used to make the final decisions on the correct description of the bonding in a molecule or polyatomic ion.
8.13 Molecular Structure: The VSEPR Model The structures of molecules play a very important role in determining their chemical prop-erties. As we will see later, this is particularly important for biological molecules; a slight change in the structure of a large biomolecule can completely destroy its usefulness to a cell or may even change the cell from a normal one to a cancerous one.
Many accurate methods now exist for determining molecular structure, the three-dimensional arrangement of the atoms in a molecule. These methods must be used if precise information about structure is required. However, it is often useful to be able to predict the approximate molecular structure of a molecule. In this section we consider a simple model that allows us to do this. This model, called the valence shell electron-pair repulsion (VSEPR) model, is useful in predicting the geometries of molecules formed from nonmetals. The main postulate of this model is that the structure around a given atom is determined principally by minimizing electron-pair repulsions. The idea here is that the bonding and nonbonding pairs around a given atom will be positioned as far apart as possible. To see how this model works, we will first consider the molecule BeCl2, which has the Lewis structure Note that there are two pairs of electrons around the beryllium atom. What arrangement of these electron pairs allows them to be as far apart as possible to minimize the repul-sions? Clearly, the best arrangement places the pairs on opposite sides of the beryllium atom at 180 degrees from each other: This is the maximum possible separation for two electron pairs. Once we have determined the optimal arrangement of the electron pairs around the central atom, we can specify the molecular structure of BeCl2, that is, the positions of the atoms. Since each electron pair on beryllium is shared with a chlorine atom, the molecule has a linear structure with a 180-degree bond angle: Next, let’s consider BF3, which has the Lewis structure Here the boron atom is surrounded by three pairs of electrons. What arrangement will minimize the repulsions? The electron pairs are farthest apart at angles of 120 degrees: Cl Cl Be 180 BeCl2 has only four electrons around Be and is expected to be very reactive with electron-pair donors.
Visualization:VSEPR Visualization: VSEPR: Two Electron Pairs Visualization: VSEPR: Three Electron Pairs 368 Chapter Eight Bonding: General Concepts Since each of the electron pairs is shared with a fluorine atom, the molecular structure will be This is a planar (flat) and triangular molecule, which is commonly described as a trigo-nal planar structure.
Next, let’s consider the methane molecule, which has the Lewis structure There are four pairs of electrons around the central carbon atom. What arrangement of these electron pairs best minimizes the repulsions? First, let’s try a square planar arrangement: The carbon atom and the electron pairs are centered in the plane of the paper, and the an-gles between the pairs are all 90 degrees.
Is there another arrangement with angles greater than 90 degrees that would put the electron pairs even farther away from each other? The answer is yes. The tetrahedral structure has angles of 109.5 degrees: It can be shown that this is the maximum possible separation of four pairs around a given atom. This means that whenever four pairs of electrons are present around an atom, they should always be arranged tetrahedrally.
Now that we have the electron-pair arrangement that gives the least repulsion, we can determine the positions of the atoms and thus the molecular structure of CH4. In methane, each of the four electron pairs is shared between the carbon atom and a hydrogen atom.
Thus the hydrogen atoms are placed as in Fig. 8.15, and the molecule has a tetrahedral structure with the carbon atom at the center.
Recall that the main idea of the VSEPR model is to find the arrangement of electron pairs around the central atom that minimizes the repulsions. Then we can determine the molecular structure from knowing how the electron pairs are shared with the peripheral atoms. Use the following steps to predict the structure of a molecule using the VSEPR model.
C C C 109.5˚ H H C H H FIGURE 8.15 The molecular structure of methane. The tetrahedral arrangement of electron pairs produces a tetrahedral arrangement of hydrogen atoms.
C Visualization: VSEPR: Four Electron Pairs 8.13 Molecular Structure: The VSEPR Model 369 Steps to Apply the VSEPR Model ➥ 1 Draw the Lewis structure for the molecule.
➥ 2 Count the electron pairs and arrange them in the way that minimizes repulsion (that is, put the pairs as far apart as possible).
➥ 3 Determine the positions of the atoms from the way the electron pairs are shared.
➥ 4 Determine the name of the molecular structure from the positions of the atoms.
We will predict the structure of ammonia (NH3) using this stepwise approach.
➥ 1 Draw the Lewis structure: ➥ 2 Count the pairs of electrons and arrange them to minimize repulsions. The NH3 molecule has four pairs of electrons: three bonding pairs and one nonbonding pair. From the discussion of the methane molecule, we know that the best arrangement of four elec-tron pairs is a tetrahedral array, as shown in Fig. 8.16(a).
➥ 3 Determine the positions of the atoms. The three H atoms share electron pairs, as shown in Fig. 8.16(b).
➥ 4 Name the molecular structure. It is very important to recognize that the name of the molecular structure is always based on the positions of the atoms. The placement of the electron pairs determines the structure, but the name is based on the positions of the atoms. Thus it is incorrect to say that the NH3 molecule is tetrahedral. It has a tetrahedral arrangement of electron pairs but not a tetrahedral arrangement of atoms. The molecular structure of ammonia is a trigonal pyramid (one side is different from the other three) rather than a tetrahedron, as shown in Fig. 8.16(c).
Prediction of Molecular Structure I Describe the molecular structure of the water molecule.
Solution The Lewis structure for water is There are four pairs of electrons: two bonding pairs and two nonbonding pairs. To mini-mize repulsions, these are best arranged in a tetrahedral array, as shown in Fig. 8.17(a).
H H O H H H N H H N (a) (b) H Lone pair (c) N When four uniform balloons are tied together, they naturally form a tetrahedral shape.
FIGURE 8.16 (a) The tetrahedral arrangement of electron pairs around the nitrogen atom in the am-monia molecule. (b) Three of the electron pairs around nitrogen are shared with hy-drogen atoms as shown and one is a lone pair. Although the arrangement of electron pairs is tetrahedral, as in the methane mole-cule, the hydrogen atoms in the ammonia molecule occupy only three corners of the tetrahedron. A lone pair occupies the fourth corner. (c) Note that molecular geometry is trigonal pyramidal, not tetrahedral.
Sample Exercise 8.11 370 Chapter Eight Bonding: General Concepts Although H2O has a tetrahedral arrangement of electron pairs, it is not a tetrahedral mol-ecule. The atoms in the H2O molecule form a V shape, as shown in Fig. 8.17(b) and (c).
See Exercises 8.91 and 8.92.
From Sample Exercise 8.11 we see that the H2O molecule is V-shaped, or bent, be-cause of the presence of the lone pairs. If no lone pairs were present, the molecule would be linear, the polar bonds would cancel, and the molecule would have no dipole moment.
This would make water very different from the polar substance so familiar to us.
From the previous discussion we would predict that the HOXOH bond angle (where X is the central atom) in CH4, NH3, and H2O should be the tetrahedral angle of 109.5 degrees. Experimental studies, however, show that the actual bond angles are those given in Fig. 8.18. What significance do these results have for the VSEPR model? One possi-ble point of view is that we should be pleased to have the observed angles so close to the tetrahedral angle. The opposite view is that the deviations are significant enough to require modification of the simple model so that it can more accurately handle similar cases. We will take the latter view.
Let us examine the following data: One interpretation of the trend observed here is that lone pairs require more space than bonding pairs; in other words, as the number of lone pairs increases, the bonding pairs are increasingly squeezed together.
O (a) (b) Lone pair (c) O Lone pair H H Bonding pair Bonding pair FIGURE 8.17 (a) The tetrahedral arrangement of the four electron pairs around oxygen in the water molecule. (b) Two of the electron pairs are shared between oxygen and the hydrogen atoms and two are lone pairs. (c) The V-shaped molecular structure of the water molecule.
FIGURE 8.18 The bond angles in the CH4, NH3, and H2O molecules. Note that the bond angle between bonding pairs decreases as the number of lone pairs increases. Note that all of the angles in CH4 are 109.5 degrees and all of the angles in NH3 are 107 degrees.
CH4 NH3 H2O Number of Lone Pairs 0 1 2 Bond Angle 109.5 107 104.5 H H O H 109.5˚ 104.5˚ 107˚ H C H Methane Ammonia Water H H H H N 8.13 Molecular Structure: The VSEPR Model 371 H X X (a) (b) FIGURE 8.19 (a) In a bonding pair of electrons, the electrons are shared by two nuclei. (b) In a lone pair, both electrons must be close to a single nucleus and tend to take up more of the space around that atom.
This interpretation seems to make physical sense if we think in the following terms.
A bonding pair is shared between two nuclei, and the electrons can be close to either nucleus. They are relatively confined between the two nuclei. A lone pair is localized on only one nucleus, and both electrons will be close only to that nucleus, as shown schemat-ically in Fig. 8.19. These pictures help us understand why a lone pair may require more space near an atom than a bonding pair.
As a result of these observations, we make the following addition to the original postulate of the VSEPR model: Lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs.
So far we have considered cases with two, three, and four electron pairs around the central atom. These are summarized in Table 8.6.
Table 8.7 summarizes the structures possible for molecules in which there are four electron pairs around the central atom with various numbers of atoms bonded to it. Note that molecules with four pairs of electrons around the central atom can be tetrahedral (AB4), trigonal pyramidal (AB3), and V-shaped (AB2).
For five pairs of electrons, there are several possible choices. The one that produces minimum repulsion is a trigonal bipyramid. Note from Table 8.6 that this arrangement TABLE 8.6 Arrangements of Electron Pairs Around an Atom Yielding Minimum Repulsion Number of Electron Pairs Arrangement of Electron Pairs Example 2 Linear 3 Trigonal planar 4 Tetrahedral 5 Trigonal bipyramidal 6 Octahedral A A A 90˚ A 120˚ 90˚ A 90˚ 372 Chapter Eight Bonding: General Concepts has two different angles, 90 degrees and 120 degrees. As the name suggests, the struc-ture formed by this arrangement of pairs consists of two trigonal-based pyramids that share a common base. Table 8.8 summarizes the structures possible for molecules in which there are five electron pairs around the central atom with various numbers of atoms bonded to it. Note that molecules with five pairs of electrons around the central atom can be trigonal bipyramidal (AB5), see-saw (AB4), T-shaped (AB3), and linear (AB2).
Six pairs of electrons can best be arranged around a given atom with 90-degree angles to form an octahedral structure, as shown in Table 8.6.
To use the VSEPR model to determine the geometric structures of molecules, you should memorize the relationships between the number of electron pairs and their best arrangement.
TABLE 8.7 Structures of Molecules That Have Four Electron Pairs Around the Central Atom Electron-Pair Arrangement Molecular Structure TABLE 8.8 Structures of Molecules with Five Electron Pairs Around the Central Atom Electron-Pair Arrangement Molecular Structure B B A B B B B A B B A B B B A B B Tetrahedral B B A B Trigonal pyramid B A B V-shaped (bent) B B B A B B B B B A B B Trigonal bipyramidal B B B B A B B A B B A B B B A B B “See-saw” B B A B T-structure B A B Linear 8.13 Molecular Structure: The VSEPR Model 373 Prediction of Molecular Structure II When phosphorus reacts with excess chlorine gas, the compound phosphorus pentachlo-ride (PCl5) is formed. In the gaseous and liquid states, this substance consists of PCl5 molecules, but in the solid state it consists of a 1 : 1 mixture of PCl4 and PCl6 ions.
Predict the geometric structures of PCl5, PCl4 , and PCl6 .
Solution The Lewis structure for PCl5 is shown. Five pairs of electrons around the phosphorus atom require a trigonal bipyramidal arrangement (see Table 8.6). When the chlorine atoms are included, a trigonal bipyramidal molecule results: Cl Cl Cl P P Cl Cl Cl Cl Cl Cl Cl P P Cl Cl Cl + Cl Cl Cl + P Cl Cl P The Lewis structure for the PCl4 ion (5 4(7) 1 32 valence electrons) is shown below. There are four pairs of electrons surrounding the phosphorus atom in the PCl4 ion, which requires a tetrahedral arrangement of the pairs. Since each pair is shared with a chlorine atom, a tetrahedral PCl4 cation results.
Cl P Cl Cl Cl Cl P Cl Cl – – Cl Cl Cl Cl Cl P The Lewis structure for PCl6 (5 6(7) 1 48 valence electrons) is shown be-low. Since phosphorus is surrounded by six pairs of electrons, an octahedral arrangement is required to minimize repulsions, as shown below in the center. Since each electron pair is shared with a chlorine atom, an octahedral PCl6 anion is predicted.
Sample Exercise 8.13 Sample Exercise 8.12 See Exercises 8.89, 8.90, 8.93, and 8.94.
Prediction of Molecular Structure III Because the noble gases have filled s and p valence orbitals, they were not expected to be chemically reactive. In fact, for many years these elements were called inert gases because of this supposed inability to form any compounds. However, in the early 1960s several compounds of krypton, xenon, and radon were synthesized. For example, a team at the Visualization: VSEPR: Iodine Pentafluoride 374 Chapter Eight Bonding: General Concepts F leads to the structure (a) Xe F F F F 90° F Xe F F leads to the structure (b) F F F F Xe F 180° Xe F F F FIGURE 8.20 Possible electron-pair arrangements for XeF4. Since arrangement (a) has lone pairs at 90 degrees from each other, it is less favorable than arrangement (b), where the lone pairs are at 180 degrees.
Argonne National Laboratory produced the stable colorless compound xenon tetrafluoride (XeF4). Predict its structure and whether it has a dipole moment.
Solution The Lewis structure for XeF4 is The xenon atom in this molecule is surrounded by six pairs of electrons, which means an octahedral arrangement.
The structure predicted for this molecule will depend on how the lone pairs and bond-ing pairs are arranged. Consider the two possibilities shown in Fig. 8.20. The bonding pairs are indicated by the presence of the fluorine atoms. Since the structure predicted dif-fers in the two cases, we must decide which of these arrangements is preferable. The key is to look at the lone pairs. In the structure in part (a), the lone pair–lone pair angle is 90 degrees; in the structure in part (b), the lone pairs are separated by 180 degrees. Since lone pairs require more room than bonding pairs, a structure with two lone pairs at 90 degrees is unfavorable. Thus the arrangement in Fig. 8.20(b) is preferred, and the molecular structure is predicted to be square planar. Note that this molecule is not described as being octahedral. There is an octahedral arrangement of electron pairs, but the atoms form a square planar structure.
Although each Xe—F bond is polar (fluorine has a greater electronegativity than xenon), the square planar arrangement of these bonds causes the polarities to cancel.
Thus XeF4 has no dipole moment, as shown in the margin.
See Exercises 8.95 through 8.98.
Xe 8.13 Molecular Structure: The VSEPR Model 375 We can further illustrate the use of the VSEPR model for molecules or ions with lone pairs by considering the triiodide ion (I3 ).
The central iodine atom has five pairs around it, which requires a trigonal bipyramidal arrangement. Several possible arrangements of lone pairs are shown in Fig. 8.21. Note that structures (a) and (b) have lone pairs at 90 degrees, whereas in (c) all lone pairs are at 120 degrees. Thus structure (c) is preferred. The resulting molecular structure for I3 is linear: The VSEPR Model and Multiple Bonds So far in our treatment of the VSEPR model we have not considered any molecules with multiple bonds. To see how these molecules are handled by this model, let’s con-sider the NO3 ion, which requires three resonance structures to describe its electronic structure: The NO3 ion is known to be planar with 120-degree bond angles: O O 120° O N 3I¬I¬I4 I I I I – I I I (a) (b) (c) 90° 90° I I I I 90° I I 120° FIGURE 8.21 Three possible arrangements of the electron pairs in the I3 ion. Arrangement (c) is preferred because there are no 90-degree lone pair–lone pair interactions.
This planar structure is the one expected for three pairs of electrons around a central atom, which means that a double bond should be counted as one effective pair in using the VSEPR model. This makes sense because the two pairs of electrons involved in the dou-ble bond are not independent pairs. Both the electron pairs must be in the space between the nuclei of the two atoms to form the double bond. In other words, the double bond acts as one center of electron density to repel the other pairs of electrons. The same holds true for triple bonds. This leads us to another general rule: For the VSEPR model, multiple bonds count as one effective electron pair.
The molecular structure of nitrate also shows us one more important point: When a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular structure using the VSEPR model. These rules are illustrated in Sample Exercise 8.14.
Structures of Molecules with Multiple Bonds Predict the molecular structure of the sulfur dioxide molecule. Is this molecule expected to have a dipole moment?
Solution First, we must determine the Lewis structure for the SO2 molecule, which has 18 valence electrons. The expected resonance structures are To determine the molecular structure, we must count the electron pairs around the sulfur atom. In each resonance structure the sulfur has one lone pair, one pair in a single bond, and one double bond. Counting the double bond as one pair yields three effective pairs around the sulfur. According to Table 8.6, a trigonal planar arrangement is required, which yields a V-shaped molecule: Thus the structure of the SO2 molecule is expected to be V-shaped, with a 120-degree bond angle. The molecule has a dipole moment directed as shown: Since the molecule is V-shaped, the polar bonds do not cancel.
See Exercises 8.99 and 8.100.
It should be noted at this point that lone pairs that are oriented at least 120 degrees from other pairs do not produce significant distortions of bond angles. For example, the angle in the SO2 molecule is actually quite close to 120 degrees. We will follow the 376 Chapter Eight Bonding: General Concepts Sample Exercise 8.14 8.13 Molecular Structure: The VSEPR Model 377 FIGURE 8.22 The molecular structure of methanol.
(a) The arrangement of electron pairs and atoms around the carbon atom. (b) The arrangement of bonding and lone pairs around the oxygen atom. (c) The molecular structure.
O H C H H (a) (b) (c) H O H C H H H C O general principle that a 120-degree angle provides lone pairs with enough space so that distortions do not occur. Angles less than 120 degrees are distorted when lone pairs are present.
Molecules Containing No Single Central Atom So far we have considered molecules consisting of one central atom surrounded by other atoms. The VSEPR model can be readily extended to more complicated mole-cules, such as methanol (CH3OH). This molecule is represented by the following Lewis structure: The molecular structure can be predicted from the arrangement of pairs around the carbon and oxygen atoms. Note that there are four pairs of electrons around the carbon, which requires a tetrahedral arrangement, as shown in Fig. 8.22(a). The oxy-gen also has four pairs, which requires a tetrahedral arrangement. However, in this case the tetrahedron will be slightly distorted by the space requirements of the lone pairs [Fig. 8.22(b)]. The overall geometric arrangement for the molecule is shown in Fig. 8.22(c).
Summary of the VSEPR Model The rules for using the VSEPR model to predict molecular structure follow: Determine the Lewis structure(s) for the molecule.
For molecules with resonance structures, use any of the structures to predict the mo-lecular structure.
Sum the electron pairs around the central atom.
In counting pairs, count each multiple bond as a single effective pair.
The arrangement of the pairs is determined by minimizing electron-pair repulsions.
These arrangements are shown in Table 8.6.
Lone pairs require more space than bonding pairs do. Choose an arrange-ment that gives the lone pairs as much room as possible. Recognize that the lone pairs may produce a slight distortion of the structure at angles less than 120 degrees.
The VSEPR Model—How Well Does It Work?
The VSEPR model is very simple. There are only a few rules to remember, yet the model correctly predicts the molecular structures of most molecules formed from nonmetallic el-ements. Molecules of any size can be treated by applying the VSEPR model to each ap-propriate atom (those bonded to at least two other atoms) in the molecule. Thus we can use this model to predict the structures of molecules with hundreds of atoms. It does, 378 Chapter Eight Bonding: General Concepts I n this chapter we have stressed the importance of being able to predict the three-dimensional structure of a molecule. Molecular structure is impor-tant because of its effect on chemical reactivity. This is especially true in bio-logical systems, where reactions must be efficient and highly specific. Among the hundreds of types of molecules in the fluids of a typical biological sys-tem, the appropriate reactants must find and react only with each other—-they must be very discriminating. This speci-ficity depends largely on structure. The molecules are constructed so that only the appropriate partners can approach each other in a way that allows reaction.
Another area where molecular structure is central is in the use of molecules as a means of communication. Exam-ples of a chemical communication occur in humans in the conduction of nerve impulses across synapses, the control of the manufacture and storage of key chemicals in cells, and the senses of smell and taste. Plants and animals also use chemical communication. For example, ants lay down a chemical trail so that other ants can find a particular food supply. Ants also warn their fellow workers of approaching danger by emitting certain chemicals.
Molecules convey messages by fitting into appropriate receptor sites in a very specific way, which is determined by their structure. When a molecule occupies a receptor site, chemical processes are stimulated that produce the appropri-ate response. Sometimes receptors can be fooled, as in the use of artificial sweeteners—-molecules fit the sites on the taste buds that stimulate a “sweet” response in the brain, but they are not metabolized in the same way as natural sugars. Sim-ilar deception is useful in insect control. If an area is sprayed CHEMICAL IMPACT Chemical Structure and Communication: Semiochemicals The queen bee secretes a chemical that prevents the worker bees from raising a competitive sovereign.
however, fail in a few instances. For example, phosphine (PH3), which has a Lewis struc-ture analogous to that of ammonia, would be predicted to have a molecular structure similar to that for NH3, with bond an-gles of approximately 107 degrees. However, the bond angles of phosphine are actually 94 degrees. There are ways of explaining this structure, but more rules have to be added to the model.
This again illustrates the point that simple models are bound to have exceptions. In introductory chemistry we want to use simple models that fit the majority of cases; we are willing to accept a few failures rather than complicate the model. The amazing thing about the VSEPR model is that such a simple model predicts correctly the structures of so many molecules.
8.13 Molecular Structure: The VSEPR Model 379 with synthetic female sex attractant molecules, the males of that species become so confused that mating does not occur.
A semiochemical is a molecule that delivers a message between members of the same or different species of plant or animal. There are three groups of these chemical mes-sengers: allomones, kairomones, and pheromones. Each is of great ecological importance.
An allomone is defined as a chemical that somehow gives adaptive advantage to the producer. For example, leaves of the black walnut tree contain a herbicide, juglone, that appears after the leaves fall to the ground. Juglone is not toxic to grass or certain grains, but it is effective against plants such as apple trees that would compete for the available water and food supplies.
Antibiotics are also allomones, since the microorganisms produce them to inhibit other species from growing near them.
Many plants produce bad-tasting chemicals to protect themselves from plant-eating insects and animals. The fa-miliar compound nicotine deters animals from eating the to-bacco plant. The millipede sends an unmistakable “back off” message by squirting a predator with benzaldehyde and hydrogen cyanide.
Defense is not the only use of allomones, however.
Flowers use scent as a way to attract pollinating insects.
Honeybees, for instance, are guided to alfalfa and flowers by a series of sweet-scented compounds.
Kairomones are chemical messengers that bring advan-tageous news to the receiver, and the floral scents are kairomones from the honeybees’ viewpoint. Many predators are guided by kairomones emitted by their food. For exam-ple, apple skins exude a chemical that attracts the codling moth larva. In some cases kairomones help the underdog.
Certain marine mollusks can pick up the “scent” of their predators, the sea stars, and make their escape.
Pheromones are chemicals that affect receptors of the same species as the donor. That is, they are specific within a species. Releaser pheromones cause an immediate reac-tion in the receptor, and primer pheromones cause long-term effects. Examples of releaser pheromones are sex attractants of insects, generated in some species by the males and in others by the females. Sex pheromones also have been found in plants and mammals.
Alarm pheromones are highly volatile compounds (ones easily changed to a gas) released to warn of danger. Honey-bees produce isoamyl acetate (C7H14O2) in their sting glands.
Because of its high volatility, this compound does not linger after the state of alert is over. Social behavior in insects is char-acterized by the use of trail pheromones, which are used to in-dicate a food source. Social insects such as bees, ants, wasps, and termites use these substances. Since trail pheromones are less volatile compounds, the indicators persist for some time.
Primer pheromones, which cause long-term behavioral changes, are harder to isolate and identify. One example, however, is the “queen substance” produced by queen hon-eybees. All the eggs in a colony are laid by one queen bee.
If she is removed from the hive or dies, the worker bees are activated by the absence of the queen substance and begin to feed royal jelly to bee larvae so as to raise a new queen. The queen substance also prevents the development of the work-ers’ ovaries so that only the queen herself can produce eggs.
Many studies of insect pheromones are now under way in the hope that they will provide a method of controlling insects that is more efficient and safer than the current chemical pesticides.
107˚ NH3 H H H N PH3 94˚ H H H P 380 Chapter Eight Bonding: General Concepts Key Terms Section 8.1 bond energy ionic bonding ionic compound Coulomb’s law bond length covalent bonding polar covalent bond Section 8.2 electronegativity Section 8.3 dipolar dipole moment Section 8.4 isoelectronic ions Section 8.5 lattice energy Section 8.8 single bond double bond triple bond Section 8.9 localized electron (LE) model lone pair bonding pair Section 8.10 Lewis structure duet rule octet rule Section 8.12 resonance resonance structure formal charge Section 8.13 molecular structure valence shell electron-pair repulsion (VSEPR) model linear structure trigonal planar structure tetrahedral structure trigonal pyramid trigonal bipyramid octahedral structure square planar structure For Review Chemical bonds Hold groups of atoms together Occur when a group of atoms can lower its total energy by aggregating Types of chemical bonds • Ionic: electrons are transferred to form ions • Covalent: equal sharing of electrons • Polar covalent: unequal electron sharing Percent ionic character of a bond XOY Electronegativity: the relative ability of an atom to attract shared electrons • The polarity of a bond depends on the electronegativity difference of the bonded atoms The spacial arrangement of polar bonds in a molecule determines whether the molecule has a dipole moment Ionic bonding An ion has a different size than its parent atom • An anion is larger than its parent ion • A cation is smaller than its parent atom Lattice energy: the change in energy when ions are packed together to form an ionic solid Bond energy The energy necessary to break a covalent bond Increases as the number of shared pairs increases Can be used to estimate the enthalpy change for a chemical reaction Lewis structures Show how the valence electron pairs are arranged among the atoms in a molecule or polyatomic ion Stable molecules usually contain atoms that have their valence orbitals filled • Leads to a duet rule for hydrogen • Leads to an octet rule for second-row elements • The atoms of elements in the third row and beyond can exceed the octet rule Several equivalent Lewis structures can be drawn for some molecules, a concept called resonance When several nonequivalent Lewis structures can be drawn for a molecule, formal charge is often used to choose the most appropriate structure(s) VSEPR model Based on the idea that electron pairs will be arranged around a central atom in a way that minimizes the electron repulsions Can be used to predict the geometric structure of most molecules REVIEW QUESTIONS 1. Distinguish between the terms electronegativity versus electron affinity, covalent bond versus ionic bond, and pure covalent bond versus polar covalent bond.
Measured dipole moment of X¬Y Calculated dipole moment for XY 100% For Review 381 Characterize the types of bonds in terms of electronegativity difference. Energetically, why do ionic and covalent bonds form?
2. When an element forms an anion, what happens to the radius? When an element forms a cation, what happens to the radius? Why? Define the term isoelectronic.
When comparing sizes of ions, which ion has the largest radius and which ion has the smallest radius in an isoelectronic series? Why?
3. Define the term lattice energy. Why, energetically, do ionic compounds form? Figure 8.11 illustrates the energy changes involved in the formation of MgO(s) and NaF(s). Why is the lattice energy of MgO(s) so different from that of NaF(s)? The magnesium oxide is composed of Mg2 and O2 ions.
Energetically, why does Mg2O2 form and not MgO? Why doesn’t Mg3O3 form?
4. Explain how bond energies can be used to estimate H for a reaction. Why is this an estimate of H? How do the product bond strengths compare to the reac-tant bond strengths for an exothermic reaction? For an endothermic reaction?
What is the relationship between the number of bonds between two atoms and bond strength? Bond length?
5. Give a rationale for the octet rule and the duet rule for H in terms of orbitals. Give the steps for drawing a Lewis structure for a molecule or ion.
In general, molecules and ions always follow the octet rule unless it is impossible. The three types of exceptions are molecules/ions with too few electrons, molecules/ions with an odd number of electrons, and molecules/ ions with too many electrons. Which atoms sometimes have fewer than 8 electrons around them? Give an example. Which atoms sometimes have more than 8 electrons around them? Give some examples. Why are odd-electron species generally very reactive and uncommon? Give an example of an odd-electron molecule.
6. Explain the terms resonance and delocalized electrons. When a substance exhibits resonance, we say that none of the individual Lewis structures accurately portrays the bonding in the substance. Why do we draw resonance structures?
7. Define formal charge and explain how to calculate it. What is the purpose of the formal charge? Organic compounds are composed mostly of carbon and hydrogen, but also may have oxygen, nitrogen, and/or halogens in the formula.
Formal charge arguments work very well for organic compounds when drawing the best Lewis structure. How do C, H, N, O, and Cl satisfy the octet rule in or-ganic compounds so as to have a formula charge of zero?
8. Explain the main postulate of the VSEPR model. List the five base geometries (along with bond angles) that most molecules or ions adopt to minimize elec-tron-pair repulsions. Why are bond angles sometimes slightly less than predicted in actual molecules as compared to what is predicted by the VSEPR model?
9. Give two requirements that should be satisfied for a molecule to be polar.
Explain why CF4 and XeF4 are nonpolar compounds (have no dipole moments) while SF4 is polar (has a dipole moment). Is CO2 polar? What about COS?
Explain.
10. Consider the following compounds: CO2, SO2, KrF2, SO3, NF3, IF3, CF4, SF4, XeF4, PF5, IF5, and SCl6. These 12 compounds are all examples of different molecular structures. Draw the Lewis structures for each and predict the molecu-lar structure. Predict the bond angles and the polarity of each. (A polar molecule has a dipole moment, while a nonpolar molecule does not.) See Exercises 89 and 90 for the molecular structures based on the trigonal bipyramid and the oc-tahedral geometries.
382 Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Explain the electronegativity trends across a row and down a col-umn of the periodic table. Compare these trends with those of ionization energies and atomic radii. How are they related?
2. The ionic compound AB is formed. The charges on the ions may be 1, 1; 2, 2; 3, 3; or even larger. What are the fac-tors that determine the charge for an ion in an ionic compound?
3. Using only the periodic table, predict the most stable ion for Na, Mg, Al, S, Cl, K, Ca, and Ga. Arrange these from largest to smallest radius, and explain why the radius varies as it does.
Compare your predictions with Fig. 8.8.
4. The bond energy for a COH bond is about 413 kJ/mol in CH4 but 380 kJ/mol in CHBr3. Although these values are relatively close in magnitude, they are different. Explain why they are dif-ferent. Does the fact that the bond energy is lower in CHBr3 make any sense? Why?
5. Consider the following statement: “Because oxygen wants to have a negative two charge, the second electron affinity is more negative than the first.” Indicate everything that is correct in this statement. Indicate everything that is incorrect. Correct the in-correct statements and explain.
6. Which has the greater bond lengths: NO2 or NO3 ? Explain.
7. The following ions are best described with resonance structures.
Draw the resonance structures, and using formal charge argu-ments, predict the best Lewis structure for each ion.
a. NCO b. CNO 8. Would you expect the electronegativity of titanium to be the same in the species Ti, Ti2, Ti3, and Ti4? Explain.
9. The second electron affinity values for both oxygen and sulfur are unfavorable (endothermic). Explain.
10. What is meant by a chemical bond? Why do atoms form bonds with each other? Why do some elements exist as molecules in nature instead of as free atoms?
11. Why are some bonds ionic and some covalent?
12. Does a Lewis structure tell which electrons come from which atoms? Explain.
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 13. Some plant fertilizer compounds are (NH4)2SO4, Ca3(PO4)2, K2O, P2O5, and KCl. Which of these compounds contain both ionic and covalent bonds?
14. Some of the important properties of ionic compounds are as follows: i. low electrical conductivity as solids and high conductivity in solution or when molten ii. relatively high melting and boiling points iii. brittleness iv. solubility in polar solvents How does the concept of ionic bonding discussed in this chap-ter account for these properties?
15. What is the electronegativity trend? Where does hydrogen fit into the electronegativity trend for the other elements in the periodic table?
16. Give one example of a compound having a linear molecular structure that has an overall dipole moment (is polar) and one example that does not have an overall dipole moment (is non-polar). Do the same for molecules that have trigonal planar and tetrahedral molecular structures.
17. When comparing the size of different ions, the general radii trend discussed in Chapter 7 is generally not very useful. What do you concentrate on when comparing sizes of ions to each other or when comparing the size of an ion to its neutral atom?
18. In general, the higher the charge on the ions in an ionic com-pound, the more favorable the lattice energy. Why do some stable ionic compounds have 1 charged ions even though 4, 5, 6, charged ions would have a more favorable lattice energy?
19. Combustion reactions of fossil fuels provide most of the energy needs of the world. Why are combustion reactions of fossil fuels so exothermic?
20. Which of the following statements is(are) true? Correct the false statements.
a. It is impossible to satisfy the octet rule for all atoms in XeF2.
b. Because SF4 exists, then OF4 should also exist because oxy-gen is in the same family as sulfur.
c. The bond in NO should be stronger than the bond in NO.
d. As predicted from the two Lewis structures for ozone, one oxygen-oxygen bond is stronger than the other oxygen-oxygen bond.
21. Three resonance structures can be drawn for CO2. Which reso-nance structure is best from a formal charge standpoint?
22. Which of the following statements is(are) true? Correct the false statements.
a. The molecules SeS3, SeS2, PCl5, TeCl4, ICl3, and XeCl2 all exhibit at least one bond angle which is approximately .
b. The bond angle in SO2 should be similar to the bond angle in CS2 or SCl2.
c. Of the compounds CF4, KrF4, and SeF4, only SeF4 exhibits an overall dipole moment (is polar).
d. Central atoms in a molecule adopt a geometry of the bonded atoms and lone pairs about the central atom in order to max-imize electron repulsions.
Exercises In this section similar exercises are paired.
Chemical Bonds and Electronegativity 23. Without using Fig. 8.3, predict the order of increasing elec-tronegativity in each of the following groups of elements.
a. C, N, O c. Si, Ge, Sn b. S, Se, Cl d. Tl, S, Ge 120° Exercises 383 24. Without using Fig. 8.3, predict the order of increasing elec-tronegativity in each of the following groups of elements.
a. Na, K, Rb c. F, Cl, Br b. B, O, Ga d. S, O, F 25. Without using Fig. 8.3, predict which bond in each of the fol-lowing groups will be the most polar.
a. COF, SiOF, GeOF b. POCl or SOCl c. SOF, SOCl, SOBr d. TiOCl, SiOCl, GeOCl 26. Without using Fig. 8.3, predict which bond in each of the fol-lowing groups will be the most polar.
a. COH, SiOH, SnOH b. AlOBr, GaOBr, InOBr, TlOBr c. COO or SiOO d. OOF or OOCl 27. Repeat Exercises 23 and 25, this time using the values for the electronegativities of the elements given in Fig. 8.3. Are there differences in your answers?
28. Repeat Exercises 24 and 26, this time using the values for the electronegativities of the elements given in Fig. 8.3. Are there differences in your answers?
29. Which of the following incorrectly shows the bond polarity?
Show the correct bond polarity for those that are incorrect.
a. HOF d. BrOBr b. ClOI e. OOP c.
SiOS 30. Indicate the bond polarity (show the partial positive and partial negative ends) in the following bonds.
a. COO d. BrOTe b. POH e. SeOS c. HOCl 31. Hydrogen has an electronegativity value between boron and car-bon and identical to phosphorus. With this in mind, rank the fol-lowing bonds in order of decreasing polarity: POH, OOH, NOH, FOH, COH.
32. Rank the following bonds in order of increasing ionic character: NOO, CaOO, COF, BrOBr, KOF.
Ions and Ionic Compounds 33. Write electron configurations for the most stable ion formed by each of the elements Fr, Be, P, Cl, and Se (when in stable ionic compounds).
34. Write electron configurations for a. the cations Mg2, K, and Al3.
b. the anions N3, O2, F, and Te2.
35. Which of the following ions have noble gas electron configurations?
a. Fe2, Fe3, Sc3, Co3 b. Tl, Te2, Cr3 c. Pu4, Ce4, Ti4 d. Ba2, Pt2, Mn2 36. What noble gas has the same election configuration as each of the ions in the following compounds?
a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide 37. Give three ions that are isoelectronic with xenon. Place these ions in order of increasing size.
38. Consider the ions Sc3, Cl, K, Ca2, and S2. Match these ions to the following pictures that represent the relative sizes of the ions.
39. For each of the following groups, place the atoms and/or ions in order of decreasing size.
a. Cu, Cu, Cu2 b. Ni2, Pd2, Pt2 c. O, O, O2 d. La3, Eu3, Gd3, Yb3 e. Te2, I, Cs, Ba2, La3 40. For each of the following groups, place the atoms and/or ions in order of decreasing size.
a. V, V2, V3, V5 b. Na, K, Rb, Cs c. Te2, I, Cs, Ba2 d. P, P, P2, P3 e. O2, S2, Se2, Te2 41. Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound.
a. Al and S c. Mg and Cl b. K and N d. Cs and Br 42. Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound.
a. Ga and I c. Sr and F b. Na and O d. Ca and P 43. Which compound in each of the following pairs of ionic sub-stances has the most exothermic lattice energy? Justify your answers.
a. NaCl, KCl b. LiF, LiCl c. Mg(OH)2, MgO d. Fe(OH)2, Fe(OH)3 e. NaCl, Na2O f. MgO, BaS 44. Which compound in each of the following pairs of ionic sub-stances has the most exothermic lattice energy? Justify your answers.
a. LiF, CsF b. NaBr, NaI c. BaCl2, BaO d. Na2SO4, CaSO4 e. KF, K2O f. Li2O, Na2S 45. Use the following data to estimate Hf for potassium chloride.
K1s2 1 2Cl21g2 ¡ KCl1s2 384 Chapter Eight Bonding: General Concepts 51. Rationalize the following lattice energy values: 52. The lattice energies of FeCl3, FeCl2, and Fe2O3 are (in no par-ticular order) 2631, 5359, and 14,774 kJ/mol. Match the appropriate formula to each lattice energy. Explain.
Bond Energies 53. Use bond energy values (Table 8.4) to estimate H for each of the following reactions in the gas phase.
a.
b.
54. Use bond energy values (Table 8.4) to estimate H for each of the following reactions.
a.
b.
55. Use bond energies (Table 8.4) to predict H for the isomeriza-tion of methyl isocyanide to acetonitrile: 56. Acetic acid is responsible for the sour taste of vinegar. It can be manufactured using the following reaction: Use tabulated values of bond energies (Table 8.4) to estimate H for this reaction.
57. Use bond energies to predict H for the combustion of ethanol: C2H5OH(l) 3O2(g) n 2CO2(g) 3H2O(g) 58. Use bond energies to estimate H for the combustion for one mole of acetylene: 59. Use bond energies to estimate H for the following reaction: 60. The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide for propulsion: 5N2O41l2 4N2H3CH31l2 ¡ 12H2O1g2 9N21g2 4CO21g2 H2O21aq2 CH3OH1aq2 ¡ H2CO1aq2 2H2O1l2 C2H21g2 5 2O21g2 ¡ 2CO21g2 H2O1g2 q OOH(l) O(g) CH3OH(g) C CH3C B O CH3N ‚ C1g2 ¡ CH3C ‚ N1g2 4HF(g) N N(g) N N 2F2(g) (l) H H H H 2H2(g) H C N(g) H H H H H C N(g) N ‚ N 3H2 S 2NH3 H2 Cl2 S 2HCl Lattice Hsub Hf Energy I.E. of M of M Na2S 365 2203 495 109 K2S 381 2052 419 90 Rb2S 361 1949 409 82 Cs2S 360 1850 382 78 47. Consider the following energy changes: Magnesium oxide exists as Mg2O2 and not as MgO. Explain.
48. Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being “pulled” from the sodium atom to the fluorine atom exothermic or endothermic?
Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?
49. LiI(s) has a heat of formation of 272 kJ/mol and a lattice en-ergy of 753 kJ/mol. The ionization energy of Li(g) is 520.
kJ/mol, the bond energy of I2(g) is 151 kJ/mol, and the electron affinity of I(g) is 295 kJ/mol. Use these data to determine the heat of sublimation of Li(s).
50. Use the following data to estimate H for the reaction S(g) e →S2(g). Include an estimate of uncertainty.
Assume that all values are known to 1 kJ/mol.
S1g2 e ¡ S1g2 ¢H 200 kJ/mol S1s2 ¡ S1g2 ¢H 277 kJ/mol Lattice Energy Compound (kJ/mol) CaSe 2862 Na2Se 2130 CaTe 2721 Na2Te 2095 H (kJ/mol) Mg(g) Mg(g) e 735 Mg(g) Mg2(g) e 1445 O(g) e O(g) 141 O(g) e O2(g) 878 S S S S Lattice energy 690. kJ/mol Ionization energy for K 419 kJ/mol Electron affinity of Cl 349 kJ/mol Bond energy of Cl2 239 kJ/mol Enthalpy of sublimation for K 64 kJ/mol Lattice energy 3916 kJ/mol First ionization energy of Mg 735 kJ/mol Second ionization energy of Mg 1445 kJ/mol Electron affinity of F 328 kJ/mol Bond energy of F2 154 kJ/mol Enthalpy of sublimation of Mg 150. kJ/mol 46. Use the following data to estimate Hf for magnesium fluoride.
Mg1s2 F21g2 ¡ MgF21s2 Exercises 385 Use bond energies to estimate H for this reaction. The struc-tures for the reactants are: 61. Consider the following reaction: Estimate the carbon–fluorine bond energy given that the COC bond energy is 347 kJ/mol, the CPC bond energy is 614 kJ/mol, and the FOF bond energy is 154 kJ/mol.
62. Consider the following reaction: The bond energy for A2 is one-half the amount of the AB bond energy. The bond energy of B2 432 kJ/mol. What is the bond energy of A2?
63. Compare your answers from parts a and b of Exercise 53 with H values calculated for each reaction using standard enthalpies of formation in Appendix 4. Do enthalpy changes calculated from bond energies give a reasonable estimate of the actual values?
64. Compare your answer from Exercise 56 to the H value calcu-lated from standard enthalpies of formation in Appendix 4.
Explain any discrepancies.
65. The standard enthalpies of formation for S(g), F(g), SF4(g), and SF6(g) are 278.8, 79.0, 775, and 1209 kJ/mol, respectively.
a. Use these data to estimate the energy of an SOF bond.
b. Compare your calculated values to the value given in Table 8.4. What conclusions can you draw?
c. Why are the H f values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?
66. Use the following standard enthalpies of formation to estimate the NOH bond energy in ammonia: N(g), 472.7 kJ/mol; H(g), 216.0 kJ/mol; NH3(g), 46.1 kJ/mol. Compare your value to the one in Table 8.4.
Lewis Structures and Resonance 67. Write Lewis structures that obey the octet rule for each of the following.
a. HCN d. NH4 g. CO2 b. PH3 e. H2CO h. O2 c. CHCl3 f. SeF2 i. HBr Except for HCN and H2CO, the first atom listed is the central atom. For HCN and H2CO, carbon is the central atom.
68. Write Lewis structures that obey the octet rule for each of the following molecules and ions. (In each case the first atom listed is the central atom.) a.
b. NF3, SO3 2, PO3 3, ClO3 POCl3, SO4 2, XeO4, PO4 3, ClO4 A2 B2 ¡ 2AB ¢H 285 kJ 549 kJ H H H(g) F F H H C C H H H H C C (g) F2(g) N N O O O O J M O D E G N N H H H3C H O D G H c.
d. Considering your answers to parts a, b, and c, what conclu-sions can you draw concerning the structures of species con-taining the same number of atoms and the same number of valence electrons?
69. One type of exception to the octet rule are compounds with cen-tral atoms having fewer than eight electrons around them. BeH2 and BH3 are examples of this type of exception. Draw the Lewis structures for BeH2 and BH3.
70. Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reac-tants and products in the reactions described below.
a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide.
b. Boron trihydride accepts a pair of electrons from ammonia, forming BH3NH3.
Give a possible explanation for why these two reactions occur.
71. The most common type of exception to the octet rule are com-pounds or ions with central atoms having more than eight elec-trons around them. PF5, SF4, ClF3 and Br3 are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?
72. SF6, ClF5, and XeF4 are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.
73. Write Lewis structures for the following. Show all resonance structures where applicable.
a. NO2 , NO3 , N2O4 (N2O4 exists as O2NONO2.) b. OCN, SCN, N3 (Carbon is the central atom in OCN and SCN.) 74. Some of the important pollutants in the atmosphere are ozone (O3), sulfur dioxide, and sulfur trioxide. Write Lewis structures for these three molecules. Show all resonance structures where applicable.
75. Benzene (C6H6) consists of a six-membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis struc-tures for benzene, including resonance structures.
76. Borazine (B3N3H6) has often been called “inorganic” benzene.
Write Lewis structures for borazine. Borazine contains a six-membered ring of alternating boron and nitrogen atoms with one hydrogen bonded to each boron and nitrogen.
77. An important observation supporting the concept of resonance in the localized electron model was that there are only three different structures of dichlorobenzene (C6H4Cl2). How does this fact support the concept of resonance (see Exercise 75)?
78. Consider the following bond lengths: In the CO3 2 ion, all three COO bonds have identical bond lengths of 136 pm. Why?
79. Place the species below in order of shortest to longest nitrogen– nitrogen bond.
(N2F4 exists as F2NONF2, and N2F2 exists as FNONF.) N2 N2F4 N2F2 C¬O 143 pm C“O 123 pm C‚O 109 pm ClO2 , SCl2, PCl2 386 Chapter Eight Bonding: General Concepts 80. Order the following species with respect to carbon–oxygen bond length (longest to shortest).
What is the order from the weakest to the strongest carbon– oxygen bond? (CH3OH exists as H3COOH.) Formal Charge 81. Write Lewis structures that obey the octet rule for the following species. Assign the formal charge for each central atom.
a.
e.
b.
f.
c.
g.
d.
h.
82 Write Lewis structures for the species in Exercise 81 that involve minimum formal charges.
83. Write the Lewis structure for O2F2(O2F2 exists as FOOOOOF).
Assign oxidation states and formal charges to the atoms in O2F2.
This compound is a vigorous and potent oxidizing and fluori-nating agent. Are oxidation states or formal charges more use-ful in accounting for these properties of O2F2?
84. Oxidation of the cyanide ion produces the stable cyanate ion, OCN. The fulminate ion, CNO, on the other hand, is very un-stable. Fulminate salts explode when struck; Hg(CNO)2 is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable? (C is the central atom in OCN and N is the central atom in CNO.) 85. When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The structure of this compound has a formal charge of zero on all elements in the compound. Draw the Lewis structure for the vile-smelling orange liquid.
86. Nitrous oxide (N2O) has three possible Lewis structures: Given the following bond lengths, rationalize the observations that the NON bond length in N2O is 112 pm and that the NOO bond length is 119 pm. Assign for-mal charges to the resonance structures for N2O. Can you elim-inate any of the resonance structures on the basis of formal charges? Is this consistent with observation?
Molecular Structure and Polarity 87. Predict the molecular structure and bond angles for each mole-cule or ion in Exercises 67 and 73.
88. Predict the molecular structure and bond angles for each mole-cule or ion in Exercises 68 and 74.
NO4 3 PO4 3 ClO3 ClO4 XeO4 SO4 2 SO2Cl2 POCl3 CO, CO2, CO3 2, CH3OH 89. There are several molecular structures based on the trigonal bipyramid geometry (see Table 8.6). Three such structures are Which of the compounds in Exercises 71 and 72 have these mo-lecular structures?
90. Two variations of the octahedral geometry (see Table 8.6) are illustrated below.
Which of the compounds in Exercises 71 and 72 have these mo-lecular structures?
91. Predict the molecular structure (including bond angles) for each of the following.
a. SeO3 b. SeO2 92. Predict the molecular structure (including bond angles) for each of the following.
a. PCl3 b. SCl2 c. SiF4 93. Predict the molecular structure (including bond angles) for each of the following. (See Exercises 89 and 90.) a. XeCl2 b. ICl3 c. TeF4 d. PCl5 94. Predict the molecular structure (including bond angles) for each of the following. (See Exercises 89 and 90.) a. ICl5 b. XeCl4 c. SeCl6 95. Which of the molecules in Exercise 91 have dipole moments (are polar)?
96. Which of the molecules in Exercise 92 have dipole moments (are polar)?
97. Which of the molecules in Exercise 93 have dipole moments (are polar)?
98. Which of the molecules in Exercise 94 have dipole moments (are polar)?
99. Write Lewis structures and predict the molecular structures of the following. (See Exercises 89 and 90.) a. OCl2, KrF2, BeH2, SO2 b. SO3, NF3, IF3 c. CF4, SeF4, KrF4 d. IF5, AsF5 Which of these compounds are polar?
B A A A A A 90 90 90 Square pyramid B A A A A 90 90 Square planar A B A 180 A B A A A 120 A B A A 90 90 90 90 Linear T-shaped See-saw NON 167 pm NPO 115 pm NPN 120 pm NOO 147 pm NqN 110 pm Challenge Problems 387 100. Write Lewis structures and predict whether each of the follow-ing is polar or nonpolar.
a. HOCN (exists as HOOCN) b. COS c. XeF2 d. CF2Cl2 e. SeF6 f. H2CO (C is the central atom.) 101. Consider the following Lewis structure where E is an unknown element: What are some possible identities for element E? Predict the molecular structure (including bond angles) for this ion.
102. Consider the following Lewis structure where E is an unknown element: What are some possible identities for element E? Predict the molecular structure (including bond angles) for this ion. (See Exercises 89 and 90.) 103. The molecules BF3, CF4, CO2, PF5, and SF6 are all nonpolar, even though they all contain polar bonds. Why?
104. Two different compounds have the formula XeF2Cl2. Write Lewis structures for these two compounds, and describe how mea-surement of dipole moments might be used to distinguish be-tween them.
Additional Exercises 105. Arrange the following in order of increasing radius and increas-ing ionization energy.
a. N, N, N b. Se, Se, Cl, Cl c. Br, Rb, Sr2 106. For each of the following, write an equation that corresponds to the energy given.
a. lattice energy of NaCl b. lattice energy of NH4Br c. lattice energy of MgS d. OPO double bond energy beginning with O2(g) as a reactant 107. Use bond energies (Table 8.4), values of electron affinities (Table 7.7), and the ionization energy of hydrogen (1312 kJ/mol) to es-timate H for each of the following reactions.
a.
b.
c.
d.
(Electron affinity of OH(g) 180. kJ/mol.) H2O1g2 S H1g2 OH1g2 HI1g2 S H1g2 I1g2 HCl1g2 S H1g2 Cl1g2 HF1g2 S H1g2 F1g2 F D G ð š O 2– ½ k F E O ð – O E O O O O A ð ð ð ð 108. Write Lewis structures for CO3 2, HCO3 , and H2CO3. When acid is added to an aqueous solution containing carbonate or bicarbonate ions, carbon dioxide gas is formed. We generally say that carbonic acid (H2CO3) is unstable. Use bond energies to estimate H for the reaction (in the gas phase) Specify a possible cause for the instability of carbonic acid.
109. Which member of the following pairs would you expect to be more energetically stable? Justify each choice.
a. NaBr or NaBr2 b. ClO4 or ClO4 c. SO4 or XeO4 d. OF4 or SeF4 110. What do each of the following sets of compounds/ions have in common with each other?
a. SO3, NO3 , CO3 2 b. O3, SO2, NO2 111. What do each of the following sets of compounds/ions have in common with each other? See your Lewis structures for Exer-cises 91 through 94.
a. XeCl4, XeCl2 b. ICl5, TeF4, ICl3, PCl3, SCl2, SeO2 112. Although both Br3 and I3 ions are known, the F3 ion has not been observed. Explain.
113. Refer back to Exercises 81 and 82. Would you make the same prediction for the molecular structure for each case using the Lewis structure obtained in Exercise 81 as compared with the one obtained in Exercise 82?
114. Which of the following molecules have dipole moments? For the molecules that are polar, indicate the polarity of each bond and the direction of the net dipole moment of the molecule.
a. CH2Cl2, CHCl3, CCl4 b. CO2, N2O c. PH3, NH3 115. The structure of TeF5 is Draw a complete Lewis structure for TeF5 , and explain the dis-tortion from the ideal square pyramidal structure. (See Exercise 90.) Challenge Problems 116. An alternative definition of electronegativity is where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Electronegativity constant 1I.E. E.A.2 F F F F F Te 79° H2CO3 ¡ CO2 H2O 388 Chapter Eight Bonding: General Concepts Chapter 7 to calculate the (I.E. E.A.) term for F, Cl, Br, and I. Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kJ/mol, respectively.
(Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronega-tivities for the other halogens and compare to values given in the text.) 117. Calculate the standard heat of formation of the compound ICl(g) at 25C, and show your work. (Hint: Use Table 8.4 and Appendix 4.) 118. Given the following information: Heat of sublimation of Li(s) 166 kJ/mol Bond energy of HCl 427 kJ/mol Ionization energy of Li(g) 520. kJ/mol Electron affinity of Cl(g) 349 kJ/mol Lattice energy of LiCl(s) 829 kJ/mol Bond energy of H2 432 kJ/mol Calculate the net change in energy for the following reaction: 119. Use data in this chapter (and Chapter 7) to discuss why MgO is an ionic compound but CO is not an ionic compound.
120. Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) al-lowing the metal cation and nonmetal anion to come together.
a. What is the sign of the energy change for each of these three processes?
b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer.
c. What must be the sign of the sum of the three processes?
d. Given your answer to part c, why do ionic bonds occur?
e. Given your above explanations, why is NaCl stable but not Na2Cl? NaCl2? What about MgO compared to MgO2?
Mg2O?
121. The compound NF3 is quite stable, but NCl3 is very unstable (NCl3 was first synthesized in 1811 by P. L. Dulong, who lost three fingers and an eye studying its properties). The compounds NBr3 and Nl3 are unknown, although the explosive compound Nl3 NH3 is known. Account for the instability of these halides of nitrogen.
122. Three processes that have been used for the industrial manufac-ture of acrylonitrile (CH2CHCN), an important chemical used in the manufacture of plastics, synthetic rubber, and fibers, are shown below. Use bond energy values (Table 8.4) to estimate H for each of the reactions.
a.
2Li1s2 2HCl1g2 ¡ 2LiCl1s2 H21g2 b.
The nitrogen–oxygen bond energy in nitric oxide (NO) is 630.
kJ/mol.
c.
d.
Is the elevated temperature noted in parts b and c needed to provide energy to endothermic reactions?
123. The compound hexaazaisowurtzitane is the highest-energy explo-sive known (C & E News, Jan. 17, 1994, p. 26). The compound, also known as CL-20, was first synthesized in 1987. The method of synthesis and detailed performance data are still classified be-cause of CL-20’s potential military application in rocket boosters and in warheads of “smart” weapons. The structure of CL-20 is In such shorthand structures, each point where lines meet rep-resents a carbon atom. In addition, the hydrogens attached to the carbon atoms are omitted; each of the six carbon atoms has one hydrogen atom attached. Finally, assume that the two O atoms in the NO2 groups are attached to N with one single bond and one double bond.
Three possible reactions for the explosive decomposition of CL-20 are i.
ii.
iii.
a.
Use bond energies to estimate H for these three reactions.
b.
Which of the above reactions releases the largest amount of energy per kilogram of CL-20?
124. Many times extra stability is characteristic of a molecule or ion in which resonance is possible. How could this be used to explain the acidities of the following compounds? (The acidic hydrogen is marked by an asterisk.) Part c shows resonance in the C6H5 ring.
125. Peroxacetyl nitrate, or PAN, is present in photochemical smog.
Draw Lewis structures (including resonance forms) for PAN. The skeletal arrangement is H C O O C O O O O O O H H A A O A O D G N C6H6N12O121s2 S 6CO21g2 6N21g2 3H21g2 C6H6N12O121s2 S 3CO1g2 3CO21g2 6N21g2 3H2O1g2 6CO1g2 6N21g2 3H2O1g2 3 2O21g2 C6H6N12O121s2 S O2N O2N N O2N NO2 NO2 NO2 N N CL-20 N N N 2CH2“CHCN 6H2O ¡ Catalyst 425510°C 2CH2“CHCH3 2NH3 3O2 4CH2“CHCN 6H2O N2 ¡ 700°C Ag 4CH2“CHCH3 6NO Marathon Problem 389 126. Draw a Lewis structure for the N,N-dimethylformamide mole-cule. The skeletal structure is Various types of evidence lead to the conclusion that there is some double bond character to the CON bond. Draw one or more resonance structures that support this observation.
127. Predict the molecular structure for each of the following. (See Exercises 89 and 90.) a. BrFI2 b. XeO2F2 c. TeF2Cl3 For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.
128. The study of carbon-containing compounds and their properties is called organic chemistry. Besides carbon atoms, organic com-pounds also can contain hydrogen, oxygen, and nitrogen atoms (as well as other types of atoms). A common trait of simple or-ganic compounds is to have Lewis structures where all atoms have a formal charge of zero. Consider the following incomplete Lewis structure for an organic compound called histidine (an amino acid, which is the building block of all proteins found in our bodies): Draw a complete Lewis structure for histidine in which all atoms have a formal charge of zero. What should be the approximate bond angles about the carbon atom labeled 1 and the nitrogen atom labeled 2?
129. Using bond energies, estimate H for the following reaction: 130. Consider the following computer-generated model of caffeine.
Draw a Lewis structure for caffeine in which all atoms have a formal charge of zero.
H O N C H H H N H H N C H H C H C N C C O C O H 1 2 Integrative Problems These problems require the integration of multiple concepts to find the solutions.
131. A compound, XF5, is 42.81% fluorine by mass. Identify the el-ement X. What is the molecular structure of XF5?
132. A polyatomic ion is composed of C, N, and an unknown ele-ment X. The skeletal Lewis structure of this polyatomic ion is [XOCON]. The ion X2 has an electron configuration of [Ar]4s23d104p6. What is element X? Knowing the identity of X, complete the Lewis structure of the polyatomic ion, including all important resonance structures.
133. Identify the following elements based on their electron configu-rations and rank them in order of increasing electronegativity: [Ar]4s13d5; [Ne]3s23p3; [Ar]4s23d104p3; [Ne]3s23p5.
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
134. Identify the five compounds of H, N, and O described below.
For each compound, write a Lewis structure that is consistent with the information given.
a. All the compounds are electrolytes, although not all of them are strong electrolytes. Compounds C and D are ionic and compound B is covalent.
b. Nitrogen occurs in its highest possible oxidation state in com-pounds A and C; nitrogen occurs in its lowest possible oxi-dation state in compounds C, D, and E. The formal charge on both nitrogens in compound C is 1; the formal charge on the only nitrogen in compound B is 0.
c. Compounds A and E exist in solution. Both solutions give off gases. Commercially available concentrated solutions of com-pound A are normally 16 M. The commercial, concentrated solution of compound E is 15 M.
d. Commercial solutions of compound E are labeled with a misnomer that implies that a binary, gaseous compound of nitrogen and hydrogen has reacted with water to produce ammonium ions and hydroxide ions. Actually, this reaction occurs to only a slight extent.
e. Compound D is 43.7% N and 50.0% O by mass. If compound D were a gas at STP, it would have a density of 2.86 g/L.
f. A formula unit of compound C has one more oxygen than a formula unit of compound D. Compounds C and A have one ion in common when compound A is acting as a strong electrolyte.
g. Solutions of compound C are weakly acidic; solutions of com-pound A are strongly acidic; solutions of compounds B and E are basic. The titration of 0.726 g of compound B requires 21.98 mL of 1.000 M HCl for complete neutralization.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl 7e.
CH3CH2OH(aq) CH3CH2OCCH3(aq) HOCCH3(aq) H2O(l) B O B O 390 9 Covalent Bonding: Orbitals Contents 9.1 Hybridization and the Localized Electron Model • sp3 Hybridization • sp2 Hybridization • sp Hybridization • dsp3 Hybridization • d2sp3 Hybridization • The Localized Electron Model: A Summary 9.2 The Molecular Orbital Model • Bond Order 9.3 Bonding in Homonuclear Diatomic Molecules • Paramagnetism 9.4 Bonding in Heteronuclear Diatomic Molecules 9.5 Combining the Localized Electron and Molecular Orbital Models A close-up of soap bubbles reveals their geometric shapes.
In Chapter 8 we discussed the fundamental concepts of bonding and introduced the most widely used simple model for covalent bonding: the localized electron model.
We saw the usefulness of a bonding model as a means for systematizing chemistry by allowing us to look at molecules in terms of individual bonds. We also saw that mole-cular structure can be predicted by minimizing electron-pair repulsions. In this chapter we will examine bonding models in more detail, particularly focusing on the role of orbitals.
9.1 Hybridization and the Localized Electron Model As we saw in Chapter 8, the localized electron model views a molecule as a collection of atoms bound together by sharing electrons between their atomic orbitals. The arrangement of valence electrons is represented by the Lewis structure (or structures, where resonance occurs), and the molecular geometry can be predicted from the VSEPR model. In this section we will describe the atomic orbitals used to share electrons and hence to form the bonds.
sp3 Hybridization Let us reconsider the bonding in methane, which has the Lewis structure and molecular geometry shown in Fig. 9.1. In general, we assume that bonding involves only the valence orbitals. This means that the hydrogen atoms in methane use 1s orbitals. The valence or-bitals of a carbon atom are the 2s and 2p orbitals shown in Fig. 9.2. In thinking about how carbon can use these orbitals to bond to the hydrogen atoms, we can see two related problems: 1. Using the 2p and 2s atomic orbitals will lead to two different types of COH bonds: (a) those from the overlap of a 2p orbital of carbon and a 1s orbital of hydrogen (there will be three of these) and (b) those from the overlap of a 2s orbital of carbon and a 1s orbital of hydrogen (there will be one of these). This is a problem because methane is known to have four identical COH bonds.
2. Since the carbon 2p orbitals are mutually perpendicular, we might expect the three COH bonds formed with these orbitals to be oriented at 90-degree angles: However, the methane molecule is known by experiment to be tetrahedral with bond angles of 109.5 degrees.
This analysis leads to one of two conclusions: Either the simple localized electron model is wrong or carbon adopts a set of atomic orbitals other than its “native” 2s and 2p orbitals to bond to the hydrogen atoms in forming the methane molecule. The sec-ond conclusion seems more reasonable. The 2s and 2p orbitals present on an isolated carbon atom may not be the best set of orbitals for bonding; a new set of atomic 391 The valence orbitals are the orbitals associated with the highest principal quantum level that contains electrons on a given atom.
Hybridization is a modification of the localized electron model to account for the observation that atoms often seem to use special atomic orbitals in forming molecules.
sp3 hybridization gives a tetrahedral set of orbitals.
Visualization: Hybridization: sp3 392 Chapter Nine Covalent Bonding: Orbitals orbitals might better serve the carbon atom in forming molecules. To account for the known structure of methane, it makes sense to assume that the carbon atom has four equivalent atomic orbitals, arranged tetrahedrally. In fact, such a set of orbitals can be obtained quite readily by combining the carbon 2s and 2p orbitals, as shown schemat-ically in Fig. 9.3. This mixing of the native atomic orbitals to form special orbitals for bonding is called hybridization. The four new orbitals are called sp3 orbitals because they are formed from one 2s and three 2p orbitals (s1p3). We say that the carbon atom undergoes sp3 hybridization or is sp3 hybridized. The four sp3 orbitals are identical in shape, each one having a large lobe and a small lobe (see Fig. 9.4). The four or-bitals are oriented in space so that the large lobes form a tetrahedral arrangement, as shown in Fig. 9.3.
The hybridization of the carbon 2s and 2p orbitals also can be represented by an or-bital energy-level diagram, as shown in Fig. 9.5. Note that electrons have been omitted because we do not need to be concerned with the electron arrangements on the individ-ual atoms—it is the total number of electrons and the arrangement of these electrons in the molecule that are important. We are assuming that carbon’s atomic orbitals are rearranged to accommodate the best electron arrangement for the molecule as a whole.
The new sp3 atomic orbitals on carbon are used to share electron pairs with the 1s orbitals from the four hydrogen atoms, as shown in Fig. 9.6 on page 393.
At this point let’s summarize the bonding in the methane molecule. The experimen-tally known structure of this molecule can be explained if we assume that the carbon atom adopts a special set of atomic orbitals. These new orbitals are obtained by combining the 2s and the three 2p orbitals of the carbon atom to produce four identically shaped orbitals that are oriented toward the corners of a tetrahedron and are used to bond to the hydro-gen atoms. Thus the four sp3 orbitals on carbon in methane are postulated to account for its known structure.
Remember this principle: Whenever a set of equivalent tetrahedral atomic orbitals is required by an atom, this model assumes that the atom adopts a set of sp3 orbitals; the atom becomes sp3 hybridized.
It is really not surprising that an atom in a molecule might adopt a different set of atomic orbitals (called hybrid orbitals) from those it has in the free state. It does not seem unreasonable that to achieve minimum energy, an atom uses one set of atomic or-bitals in the free state and a different set in a molecule. This is consistent with the idea that a molecule is more than simply a sum of its parts. What the atoms in a molecule were FIGURE 9.2 The valence orbitals on a free carbon atom: 2s, 2px, 2py, and 2pz.
FIGURE 9.3 The “native” 2s and three 2p atomic orbitals characteristic of a free carbon atom are combined to form a new set of four sp3 orbitals. The small lobes of the orbitals are usually omitted from diagrams for clarity.
s py px pz x y z z y x y z x y z x z y z x y x x y z y x sp3 sp3 sp3 sp3 Hybridization gives a tetrahedral arrangement s py px pz x y z z y x y z x y z z x FIGURE 9.1 (a) The Lewis structure of the methane molecule. (b) The tetrahedral molecular geometry of the methane molecule.
H H C H H H C H H H (a) (b) 9.1 Hybridization and the Localized Electron Model 393 FIGURE 9.4 Cross section of an sp3 orbital. This shows a “slice” of the electron density of the sp3 orbitals illustrated in the center diagram of Fig. 9.3.
(Generated from a program by Robert Allendoerfer on Project SERAPHIM disk PC2402; printed with permission.) FIGURE 9.5 An energy-level diagram showing the formation of four sp3 orbitals.
Sample Exercise 9.1 FIGURE 9.6 The tetrahedral set of four sp3 orbitals of the carbon atom are used to share electron pairs with the four 1s orbitals of the hydrogen atoms to form the four equivalent COH bonds. This accounts for the known tetrahedral structure of the CH4 molecule.
E 2p 2s Orbitals in a free C atom sp3 Orbitals in C in the CH4 molecule Hybridization sp3 sp3 H1s H1s H1s H1s C sp3 sp3 FIGURE 9.7 The nitrogen atom in ammonia is sp3 hybridized.
sp3 sp3 H1s H1s H1s lone pair N sp3 sp3 like before the molecule was formed is not as important as how the electrons are best arranged in the molecule. Therefore, this model assumes that the individual atoms respond as needed to achieve the minimum energy for the molecule.
The Localized Electron Model I Describe the bonding in the ammonia molecule using the localized electron model.
Solution A complete description of the bonding involves three steps: 1. Writing the Lewis structure 2. Determining the arrangement of electron pairs using the VSEPR model 3. Determining the hybrid atomic orbitals needed to describe the bonding in the molecule The Lewis structure for NH3 is The four electron pairs around the nitrogen atom require a tetrahedral arrangement to min-imize repulsions. We have seen that a tetrahedral set of sp3 hybrid orbitals is obtained by combining the 2s and three 2p orbitals. In the NH3 molecule three of the sp3 orbitals are used to form bonds to the three hydrogen atoms, and the fourth sp3 orbital holds the lone pair, as shown in Fig. 9.7.
See Exercise 9.15.
sp2 Hybridization Ethylene (C2H4) is an important starting material in the manufacture of plastics. The C2H4 molecule has 12 valence electrons and the following Lewis structure: We saw in Chapter 8 that a double bond acts as one effective pair, so in the ethylene mol-ecule each carbon is surrounded by three effective pairs. This requires a trigonal planar arrangement with bond angles of 120 degrees. What orbitals do the carbon atoms in this molecule employ? The molecular geometry requires a set of orbitals in one plane at angles 394 Chapter Nine Covalent Bonding: Orbitals of 120 degrees. Since the 2s and 2p valence orbitals of carbon do not have the required arrangement, we need a set of hybrid orbitals.
The sp3 orbitals we have just considered will not work because they are at angles of 109.5 degrees rather than the required 120 degrees. In ethylene the carbon atom must hy-bridize in a different manner. A set of three orbitals arranged at 120-degree angles in the same plane can be obtained by combining one s orbital and two p orbitals, as shown in Fig. 9.8. The orbital energy-level diagram for this arrangement is shown in Fig. 9.9. Since one 2s and two 2p orbitals are used to form these hybrid orbitals, this is called sp2 hybridization. Note from Fig. 9.8 that the plane of the sp2 hybridized orbitals is deter-mined by which p orbitals are used. Since in this case we have arbitrarily decided to use the px and py orbitals, the hybrid orbitals are centered in the xy plane.
In forming the sp2 orbitals, one 2p orbital on carbon has not been used. This remaining p orbital (pz) is oriented perpendicular to the plane of the sp2 orbitals, as shown in Fig. 9.10.
Now we will see how these orbitals can be used to account for the bonds in ethylene.
The three sp2 orbitals on each carbon can be used to share electrons, as shown in Fig. 9.11.
In each of these bonds, the electron pair is shared in an area centered on a line running between the atoms. This type of covalent bond is called a sigma () bond. In the ethyl-ene molecule, the bonds are formed using sp2 orbitals on each carbon atom and the 1s orbital on each hydrogen atom.
How can we explain the double bond between the carbon atoms? In the bond the electron pair occupies the space between the carbon atoms. The second bond must there-fore result from sharing an electron pair in the space above and below the bond. This type of bond can be formed using the 2p orbital perpendicular to the sp2 hybrid orbitals on each carbon atom (refer to Fig. 9.10). These parallel p orbitals can share an electron pair, which occupies the space above and below a line joining the atoms, to form a pi () bond, as shown in Fig. 9.12.
sp2 hybridization gives a trigonal planar arrangement of atomic orbitals.
Note in Fig. 9.10 and the figures that follow that the orbital lobes are artificially narrowed to more clearly show their relative orientations.
FIGURE 9.8 The hybridization of the s, px, and py atomic orbitals results in the formation of three sp2 orbitals centered in the xy plane. The large lobes of the orbitals lie in the plane at angles of 120 degrees and point toward the corners of a triangle.
FIGURE 9.9 An orbital energy-level diagram for sp2 hy-bridization. Note that one p orbital remains unchanged.
y Hybridization gives a trigonal planar arrangement z y 120° z z x s py px x y z z y x y z x x y x y x z E Hybridization 2p 2s Orbitals in an isolated carbon atom sp2 Carbon orbitals in ethylene 2p A double bond acts as one effective electron pair.
9.1 Hybridization and the Localized Electron Model 395 Note that bonds are formed from orbitals whose lobes point toward each other, but bonds result from parallel orbitals. A double bond always consists of one bond, where the electron pair is located directly between the atoms, and one bond, where the shared pair occupies the space above and below the bond.
We can now completely specify the orbitals that this model assumes are used to form the bonds in the ethylene molecule. As shown in Fig. 9.13, the carbon atoms use sp2 hybrid orbitals to form the bonds to the hydrogen atoms and to each other, and they use p to form the bond with each other. Note that we have accounted fully for the Lewis struc-ture of ethylene with its carbon–carbon double bond and carbon–hydrogen single bonds.
This example illustrates an important general principle of this model: Whenever an atom is surrounded by three effective pairs, a set of sp2 hybrid orbitals is required.
sp Hybridization Another type of hybridization occurs in carbon dioxide, which has the following Lewis structure: In the CO2 molecule the carbon atom has two effective pairs that will be arranged at an angle of 180 degrees. We therefore need a pair of atomic orbitals oriented in opposite directions. This requires a new type of hybridization, since neither sp3 nor sp2 hybrid FIGURE 9.11 The bonds in ethylene. Note that for each bond the shared electron pair occupies the re-gion directly between the atoms.
FIGURE 9.10 When an s and two p orbitals are mixed to form a set of three sp2 orbitals, one p or-bital remains unchanged and is perpendicu-lar to the plane of the hybrid orbitals. Note that in this figure and those that follow, the orbitals are drawn with narrowed lobes to show their orientations more clearly.
p orbital sp2 orbital sp2 orbital sp2 orbital C C sp2 sp2 sp2 sp2 H1s H1s H1s H1s sp2 sp2 FIGURE 9.12 A carbon–carbon double bond consists of a bond and a bond. In the bond the shared electrons occupy the space directly between the atoms. The bond is formed from the unhybridized p orbitals on the two carbon atoms. In a bond the shared electron pair occupies the space above and below a line joining the atoms.
FIGURE 9.13 (a) The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.
sp2 sp2 sp2 sp2 H1s H1s C C 2p sp2 sp2 H C C (a) (b) H H H sigma bond pi bond C C p orbital p orbital Visualization: Hybridization: sp2 Visualization: Formation of CPC Double Bond in Ethylene Visualization: Hybridization: sp 396 Chapter Nine Covalent Bonding: Orbitals orbitals will fit this case. To obtain two hybrid orbitals arranged at 180 degrees requires sp hybridization, involving one s orbital and one p orbital, as shown in Fig. 9.14.
In terms of this model, two effective pairs around an atom will always require sp hybridization of that atom. The sp orbitals of carbon in carbon dioxide can be seen in Fig. 9.15, and the corresponding orbital energy-level diagram for their formation is given in Fig. 9.16. These sp hybrid orbitals are used to form the bonds between the carbon and the oxygen atoms. Note that two 2p orbitals remain unchanged on the sp hybridized carbon. These are used to form the bonds with the oxygen atoms.
In the CO2 molecule each oxygen atom has three effective pairs around it, requir-ing a trigonal planar arrangement of the pairs. Since a trigonal set of hybrid orbitals requires sp2 hybridization, each oxygen atom is sp2 hybridized. One p orbital on each oxygen is unchanged and is used for the bond with the carbon atom.
Now we are ready to use our model to describe the bonding in carbon dioxide. The sp orbitals on carbon form bonds with the sp2 orbitals on the two oxygen atoms (Fig. 9.15). The remaining sp2 orbitals on the oxygen atoms hold lone pairs. The bonds between the carbon atom and each oxygen atom are formed by the overlap of parallel 2p orbitals. The sp hybridized carbon atom has two unhybridized p orbitals, pictured in Fig. 9.17. Each of these p orbitals is used to form a bond with an oxygen atom (see Fig. 9.18). The total bonding picture for the CO2 molecule is shown in Fig. 9.19. Note that this picture of the bonding neatly explains the arrangement of electrons predicted by the Lewis structure.
Another molecule whose bonding can be described by sp hybridization is acetylene (C2H2), which has the systematic name ethyne. The Lewis structure for acetylene is H¬C ‚ C¬H More rigorous theoretical models of CO2 indicate that each of the oxygen atoms uses two p orbitals simultaneously to form the pi bonds to the carbon atom, thus leading to unusually strong CPO bonds.
FIGURE 9.14 When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at 180 degrees results.
FIGURE 9.15 The hybrid orbitals in the CO2 molecule.
FIGURE 9.16 The orbital energy-level diagram for the formation of sp hybrid orbitals on carbon.
We will assume that minimizing electron repulsions also is important for the peripheral atoms in a molecule and apply the VSEPR model to these atoms as well.
FIGURE 9.17 The orbitals of an sp hybridized carbon atom.
FIGURE 9.18 The orbital arrangement for an sp2 hybridized oxygen atom.
z s px y y x z z y gives a linear arrangement x y z Hybridization 180° x x x y z sp2 O C sp2 O sp2 sp sp sp2 sp2 sp2 Orbitals in the sp hybridized C in CO2 2p 2p sp sp E 2p 2s Orbitals in a free C atom Hybridization p sp sp p C p sp2 sp2 sp2 O 9.1 Hybridization and the Localized Electron Model 397 Because the triple bond counts as one effective repulsive unit, each carbon has two effective pairs, which requires a linear arrangement. Thus each carbon atom requires sp hybridiza-tion, leaving two unchanged p orbitals (see Fig. 9.16). One of the oppositely oriented (see Fig. 9.14) sp orbitals is used to form a bond to the hydrogen atom; the other sp orbital overlaps with the similar sp orbital on the other carbon to form the sigma bond. The two pi bonds are formed from the overlap of the two p orbitals on each carbon. This accounts for the triple bond (one sigma and two pi bonds) in acetylene.
The Localized Electron Model II Describe the bonding in the N2 molecule.
Solution The Lewis structure for the nitrogen molecule is where each nitrogen atom is surrounded by two effective pairs. (Remember that a multi-ple bond counts as one effective pair.) This gives a linear arrangement (180 degrees) re-quiring a pair of oppositely directed orbitals. This situation requires sp hybridization. Each nitrogen atom in the nitrogen molecule has two sp hybrid orbitals and two unchanged p orbitals, as shown in Fig. 9.20(a). The sp orbitals are used to form the bond between the nitrogen atoms and to hold lone pairs, as shown in Fig. 9.20(b). The p orbitals are used to form the two bonds [see Fig. 9.20(c)]; each pair of overlapping parallel p or-bitals holds one electron pair. Such bonding accounts for the electron arrangement given by the Lewis structure. The triple bond consists of a bond (overlap of two sp orbitals) and two bonds (each one from an overlap of two p orbitals). In addition, a lone pair occupies an sp orbital on each nitrogen atom.
See Exercises 9.17 and 9.18.
dsp3 Hybridization To illustrate the treatment of a molecule in which the central atom exceeds the octet rule, consider the bonding in the phosphorus pentachloride molecule (PCl5). The Lewis structure : N ‚ N : FIGURE 9.19 (a) The orbitals used to form the bonds in carbon dioxide. Note that the carbon– oxygen double bonds each consist of one bond and one bond. (b) The Lewis structure for carbon dioxide.
Sample Exercise 9.2 O C O sigma bond (1 pair of electrons) pi bond (1 pair of electrons) pi bond (1 pair of electrons) (a) (b) O C O 398 Chapter Nine Covalent Bonding: Orbitals shows that the phosphorus atom is surrounded by five electron pairs. Since five pairs require a trigonal bipyramidal arrangement, we need a trigonal bipyramidal set of atomic orbitals on phosphorus. Such a set of orbitals is formed by dsp3 hybridization of one d orbital, one s orbital, and three p orbitals, as shown in Fig. 9.21.
The dsp3 hybridized phosphorus atom in the PCl5 molecule uses its five dsp3 orbitals to share electrons with the five chlorine atoms. Note that a set of five effective pairs around a given atom always requires a trigonal bipyramidal arrangement, which in turn requires dsp3 hybridization of that atom.
The Lewis structure for PCl5 shows that each chlorine atom is surrounded by four electron pairs. This requires a tetrahedral arrangement, which in turn requires a set of four sp3 orbitals on each chlorine atom.
Now we can describe the bonding in the PCl5 molecule. The five POCl bonds are formed by sharing electrons between a dsp3 orbital on the phosphorus atom and an sp3 orbital on each chlorine.† The other sp3 orbitals on each chlorine hold lone pairs. This is shown in Fig. 9.22.
The Localized Electron Model III Describe the bonding in the triiodide ion (I3 ).
Solution The Lewis structure for I3 FIGURE 9.20 (a) An sp hybridized nitrogen atom. There are two sp hybrid orbitals and two unhy-bridized p orbitals. (b) The bond in the N2 molecule. (c) The two bonds in N2 are formed when electron pairs are shared between two sets of parallel p orbitals.
(d) The total bonding picture for N2.
FIGURE 9.21 A set of dsp3 hybrid orbitals on a phospho-rus atom. Note that the set of five dsp3 orbitals has a trigonal bipyramidal arrange-ment. (Each dsp3 orbital also has a small lobe that is not shown in this diagram.) There is considerable controversy about whether the d orbitals are as heavily involved in the bonding in these molecules as this model predicts. However, this matter is beyond the scope of this text.
†Although we have no way of proving conclusively that each chlorine is sp3 hybridized, we assume that minimizing electron-pair repulsions is as important for peripheral atoms as for the central atom. Thus we will apply the VSEPR model and hybridization to both central and peripheral atoms.
Sample Exercise 9.3 p sp sp p (a) lone pair sigma bond lone pair N N (b) (c) (d) N N N sp sp sp sp N N dsp3 dsp3 dsp3 dsp3 dsp3 P 9.1 Hybridization and the Localized Electron Model 399 shows that the central iodine atom has five pairs of electrons (see Section 8.11). A set of five pairs requires a trigonal bipyramidal arrangement, which in turn requires a set of dsp3 orbitals. The outer iodine atoms have four pairs of electrons, which calls for a tetrahedral arrangement and sp3 hybridization.
Thus the central iodine is dsp3 hybridized. Three of these hybrid orbitals hold lone pairs, and two of them overlap with sp3 orbitals of the other two iodine atoms to form bonds.
See Exercise 9.23.
d2sp3 Hybridization Some molecules have six pairs of electrons around a central atom; an example is sulfur hexafluoride (SF6), which has the Lewis structure This requires an octahedral arrangement of pairs and in turn an octahedral set of six hybrid orbitals, or d2sp3 hybridization, in which two d orbitals, one s orbital, and three p orbitals are combined (see Fig. 9.23). Note that six electron pairs around an atom are always arranged octahedrally and require d2sp3 hybridization of the atom. Each of the d2sp3 or-bitals on the sulfur atom is used to bond to a fluorine atom. Since there are four pairs on each fluorine atom, the fluorine atoms are assumed to be sp3 hybridized.
F F F F F F S (a) FIGURE 9.22 (a) The structure of the PCI5 molecule.
(b) The orbitals used to form the bonds in PCl5. The phosphorus uses a set of five dsp3 orbitals to share electron pairs with sp3 orbitals on the five chlorine atoms. The other sp3 orbitals on each chlorine atom hold lone pairs.
FIGURE 9.23 An octahedral set of d2sp3 orbitals on a sulfur atom. The small lobe of each hybrid orbital has been omitted for clarity.
P Cl Cl Cl Cl Cl (b) d2sp3 d2sp3 d2sp3 d2sp3 d2sp3 d2sp3 S d 2sp3 hybridization gives six orbitals arranged octahedrally.
400 Chapter Nine Covalent Bonding: Orbitals The Localized Electron Model IV How is the xenon atom in XeF4 hybridized?
Solution As seen in Sample Exercise 8.13, XeF4 has six pairs of electrons around xenon that are arranged octahedrally to minimize repulsions. An octahedral set of six atomic orbitals is required to hold these electrons, and the xenon atom is d2sp3 hybridized.
Sample Exercise 9.4 Xenon uses six d2sp3 hybrid atomic orbitals to bond to the four fluorine atoms and to hold the two lone pairs.
See Exercise 9.24.
The Localized Electron Model: A Summary The description of a molecule using the localized electron model involves three distinct steps.
Localized Electron Model ➥ 1 Draw the Lewis structure(s).
➥ 2 Determine the arrangement of electron pairs using the VSEPR model.
➥ 3 Specify the hybrid orbitals needed to accommodate the electron pairs.
It is important to do the steps in this order. For a model to be successful, it must fol-low nature’s priorities. In the case of bonding, it seems clear that the tendency for a molecule to minimize its energy is more important than the maintenance of the char-acteristics of atoms as they exist in the free state. The atoms adjust to meet the “needs” of the molecule. When considering the bonding in a particular molecule, therefore, we always start with the molecule rather than the component atoms. In the molecule the electrons will be arranged to give each atom a noble gas configuration, where possible, and to minimize electron-pair repulsions. We then assume that the atoms adjust their orbitals by hybridization to allow the molecule to adopt the structure that gives the minimum energy.
In applying the localized electron model, we must remember not to overempha-size the characteristics of the separate atoms. It is not where the valence electrons originate that is important; it is where they are needed in the molecule to achieve stability.
In the same vein, it is not the orbitals in the isolated atom that matter, but which orbitals the molecule requires for minimum energy.
The requirements for the various types of hybridization are summarized in Fig. 9.24 on the following page.
Xe F F F F 9.1 Hybridization and the Localized Electron Model 401 The Localized Electron Model V For each of the following molecules or ions, predict the hybridization of each atom, and describe the molecular structure.
a. CO b. BF4 c. XeF2 Solution a. The CO molecule has 10 valence electrons, and its Lewis structure is Each atom has two effective pairs, which means that both are sp hybridized. The triple bond consists of a bond produced by overlap of an sp orbital from each atom and FIGURE 9.24 The relationship of the number of effective pairs, their spatial arrangement, and the hybrid orbital set required.
Sample Exercise 9.5 Number of Effective Pairs Arrangement of Pairs Hybridization Required 2 Linear sp 180° 3 Trigonal planar sp2 120° 4 Tetrahedral sp3 109.5° 5 Trigonal bipyramidal dsp3 90° 120° 90° 90° 6 Octahedral d2sp3 402 Chapter Nine Covalent Bonding: Orbitals two bonds produced by overlap of 2p orbitals from each atom. The lone pairs are in sp orbitals. Since the CO molecule has only two atoms, it must be linear.
b. The BF4 ion has 32 valence electrons. The Lewis structure shows four pairs of elec-trons around the boron atom, which means a tetrahedral arrangement: This requires sp3 hybridization of the boron atom. Each fluorine atom also has four electron pairs and can be assumed to be sp3 hybridized (only one sp3 orbital is shown for each fluorine atom). The BF4 ion’s molecular structure is tetrahedral.
c. The XeF2 molecule has 22 valence electrons. The Lewis structure shows five electron pairs on the xenon atom, which requires a trigonal bipyramidal arrangement: F F Xe Xe F 5 electron pairs around Xe F F F F F sp3 sp3 B Lone pairs sp sp C O C–O sigma bond sp sp O C pi bonds 2p 2p 2p 2p 9.2 The Molecular Orbital Model 403 Note that the lone pairs are placed in the plane where they are 120 degrees apart. To accommodate five pairs at the vertices of a trigonal bipyramid requires that the xenon atom adopt a set of five dsp3 orbitals. Each fluorine atom has four electron pairs and can be assumed to be sp3 hybridized. The XeF2 molecule has a linear arrangement of atoms.
See Exercises 9.27 and 9.28.
9.2 The Molecular Orbital Model We have seen that the localized electron model is of great value in interpreting the structure and bonding of molecules. However, there are some problems with this model.
For example, it incorrectly assumes that electrons are localized, and so the concept of resonance must be added. Also, the model does not deal effectively with molecules con-taining unpaired electrons. And finally, the model gives no direct information about bond energies.
Another model often used to describe bonding is the molecular orbital model.
To introduce the assumptions, methods, and results of this model, we will consider the simplest of all molecules, H2, which consists of two protons and two electrons.
A very stable molecule, H2 is lower in energy than the separated hydrogen atoms by 432 kJ/mol.
Since the hydrogen molecule consists of protons and electrons, the same components found in separated hydrogen atoms, it seems reasonable to use a theory similar to the atomic theory discussed in Chapter 7, which assumes that the electrons in an atom exist in orbitals of a given energy. Can we apply this same type of model to the hydrogen molecule? Yes. In fact, describing the H2 molecule in terms of quantum mechanics is quite straightforward.
However, even though it is formulated rather easily, this problem cannot be solved exactly. The difficulty is the same as that in dealing with polyelectronic atoms—the electron correlation problem. Since we do not know the details of the electron movements, we can-not deal with the electron–electron interactions in a specific way. We need to make ap-proximations that allow a solution of the problem but do not destroy the model’s physi-cal integrity. The success of these approximations can only be measured by comparing predictions based on theory with experimental observations. In this case we will see that the simplified model works very well.
Just as atomic orbitals are solutions to the quantum mechanical treatment of atoms, molecular orbitals (MOs) are solutions to the molecular problem. Molecular orbitals F F sp3 dsp3 Xe Molecular orbital theory parallels the atomic theory discussed in Chapter 7.
Visualization: Bonding in H2 404 Chapter Nine Covalent Bonding: Orbitals have many of the same characteristics as atomic orbitals. Two of the most important are that they can hold two electrons with opposite spins and that the square of the molecular orbital wave function indicates electron probability.
We will now describe the bonding in the hydrogen molecule using this model. The first step is to obtain the hydrogen molecule’s orbitals, a process that is greatly simplified if we assume that the molecular orbitals can be constructed from the hydrogen 1s atomic orbitals.
When the quantum mechanical equations for the hydrogen molecule are solved, two molecular orbitals result, which can be represented as where 1sA and 1sB represent the 1s orbitals from the two separated hydrogen atoms. This process is shown schematically in Fig. 9.25.
The orbital properties of most interest are size, shape (described by the electron probability distribution), and energy. These properties for the hydrogen molecular orbitals are represented in Fig. 9.26. From Fig. 9.26 we can note several important points: 1. The electron probability of both molecular orbitals is centered along the line passing through the two nuclei. For MO1 the greatest electron probability is between the nu-clei, and for MO2 it is on either side of the nuclei. This type of electron distribution is described as sigma (
), as in the localized electron model. Accordingly, we refer to MO1 and MO2 as sigma () molecular orbitals.
2. In the molecule only the molecular orbitals are available for occupation by elec-trons. The 1s atomic orbitals of the hydrogen atoms no longer exist, because the H2 molecule—a new entity—has its own set of new orbitals.
3. MO1 is lower in energy than the 1s orbitals of free hydrogen atoms, while MO2 is higher in energy than the 1s orbitals. This fact has very important implications for the stability of the H2 molecule, since if the two electrons (one from each hydrogen atom) occupy the lower-energy MO1, they will have lower energy than they do in the two separate hydrogen atoms. This situation favors molecule formation, because nature tends to seek the lowest energy state. That is, the driving force for molecule formation is that the molecular orbital available to the two electrons has lower energy than the atomic orbitals these electrons occupy in the separated atoms. This situation is favorable to bonding, or probonding.
On the other hand, if the two electrons were forced to occupy the higher-energy MO2, they would be definitely antibonding. In this case, these electrons would have lower energy in the separated atoms than in the molecule, and the separated state would be favored. Of course, since the lower-energy MO1 is available, the two electrons occupy that MO and the molecule is stable.
We have seen that the molecular orbitals of the hydrogen molecule fall into two classes: bonding and antibonding. A bonding molecular orbital is lower in energy than the atomic orbitals of which it is composed. Electrons in this type of orbital will favor the molecule; that is, they will favor bonding. An antibonding molecular orbital is higher in energy than the atomic orbitals of which it is composed. Electrons in this type of orbital will favor the separated atoms (they are antibonding). Figure 9.27 illustrates these ideas.
4. Figure 9.26 shows that for the bonding molecular orbital in the H2 molecule the elec-trons have the greatest probability of being between the nuclei. This is exactly what we would expect, since the electrons can lower their energies by being simultane-ously attracted by both nuclei. On the other hand, the electron distribution for the MO2 1sA 1sB MO1 1sA 1sB FIGURE 9.25 The combination of hydrogen 1s atomic or-bitals to form MOs. The phases of the or-bitals are shown by signs inside the bound-ary surfaces. When the orbitals are added, the matching phases produce constructive interference, which give enhanced electron probability between the nuclei. This results in a bonding molecular orbital. When one orbital is subtracted from the other, destructive interference occurs between the opposite phases, leading to a node between the nuclei. This is an antibonding MO.
FIGURE 9.26 (a) The MO energy-level diagram for the H2 molecule. (b) The shapes of the MOs are obtained by squaring the wave functions for MO1 and MO2. The positions of the nuclei are indicated by .
1sA + – 1sA 1sB antibonding (MO2) 1sB bonding (MO1) + + + + + + – E MO2 1sA H2 HA HB 1sB MO1 Energy diagram (a) Electron probability distribution (b) Visualization: Sigma Bonding and Antibonding Orbitals 9.2 The Molecular Orbital Model 405 antibonding molecular orbital is such that the electrons are mainly outside the space between the nuclei. This type of distribution is not expected to provide any bonding force. In fact, it causes the electrons to be higher in energy than in the separated atoms.
Thus the molecular orbital model produces electron distributions and energies that agree with our basic ideas of bonding. This fact reassures us that the model is physically reasonable.
5. The labels on molecular orbitals indicate their symmetry (shape), the parent atomic orbitals, and whether they are bonding or antibonding. Antibonding character is indicated by an asterisk. For the H2 molecule, both MOs have symmetry, and both are constructed from hydrogen 1s atomic orbitals. The molecular orbitals for H2 are therefore labeled as follows: 6. Molecular electron configurations can be written in much the same way as atomic (electron) configurations. Since the H2 molecule has two electrons in the 1s molec-ular orbital, the electron configuration is 1s 2.
7. Each molecular orbital can hold two electrons, but the spins must be opposite.
8. Orbitals are conserved. The number of molecular orbitals will always be the same as the number of atomic orbitals used to construct them.
Many of the above points are summarized in Fig. 9.28.
Now suppose we could form the H2 ion from a hydride ion (H) and a hydrogen atom. Would this species be stable? Since the H ion has the configuration 1s2 and the H atom has a 1s1 configuration, we will use 1s atomic orbitals to construct the MO diagram for the H2 ion, as shown in Fig. 9.29. The electron configuration for H2 is (
1s)2(
1s)1.
The key idea is that the H2 ion will be stable if it has a lower energy than its sepa-rated parts. From Fig. 9.29 we see that in going from the separated H ion and H atom to the H2 ion, the model predicts that two electrons are lowered in energy and one elec-tron is raised in energy. In other words, two electrons are bonding and one electron is antibonding. Since more electrons favor bonding, H2 is predicted to be a stable entity— a bond has formed. But how would we expect the bond strengths in the molecules of H2 and H2 to compare?
In the formation of the H2 molecule, two electrons are lowered in energy and no elec-trons are raised in energy compared with the parent atoms. When H2 is formed, two elec-trons are lowered in energy and one is raised, producing a net lowering of the energy of only one electron. Thus the model predicts that H2 is twice as stable as H2 with respect to their separated components. In other words, the bond in the H2 molecule is predicted to be about twice as strong as the bond in the H2 ion.
Bond Order To indicate bond strength, we use the concept of bond order. Bond order is the differ-ence between the number of bonding electrons and the number of antibonding electrons divided by 2.
We divide by 2 because, from the localized electron model, we are used to thinking of bonds in terms of pairs of electrons.
Bond order number of bonding electrons number of antibonding electrons 2 MO2 s1s MO1 s1s Bonding will result if the molecule has lower energy than the separated atoms.
FIGURE 9.27 Bonding and antibonding molecular orbitals (MOs).
FIGURE 9.28 A molecular orbital energy-level diagram for the H2 molecule.
E Antibonding MO Energy of an atomic orbital in a free atom A Energy of an atomic orbital in a free atom B Bonding MO E Atomic orbitals in a free H atom σ1s σ1s H1s H1s σ1s and σ1s are molecular orbitals in the H2 molecule.
Although the model predicts that H2 should be stable, this ion has never been observed, again emphasizing the perils of simple models.
E σ1s σ1s H1s H1s FIGURE 9.29 The molecular orbital energy-level diagram for the H2 ion.
406 Chapter Nine Covalent Bonding: Orbitals Since the H2 molecule has two bonding electrons and no antibonding electrons, the bond order is The H2 ion has two bonding electrons and one antibonding electron; the bond order is Bond order is an indication of bond strength because it reflects the difference between the number of bonding electrons and the number of antibonding electrons. Larger bond order means greater bond strength.
We will now apply the molecular orbital model to the helium molecule (He2). Does this model predict that this molecule will be stable? Since the He atom has a 1s2 configuration, 1s orbitals are used to construct the molecular orbitals, and the mole-cule will have four electrons. From the diagram shown in Fig. 9.30 it is apparent that two electrons are raised in energy and two are lowered in energy. Thus the bond order is zero: This implies that the He2 molecule is not stable with respect to the two free He atoms, which agrees with the observation that helium gas consists of individual He atoms.
9.3 Bonding in Homonuclear Diatomic Molecules In this section we consider homonuclear diatomic molecules (those composed of two identical atoms) of elements in Period 2 of the periodic table. Since the lithium atom has a 1s22s1 electron configuration, it would seem that we should use the Li 1s and 2s orbitals to form the molecular orbitals of the Li2 molecule. However, the 1s orbitals on the lithium atoms are much smaller than the 2s orbitals and therefore do not overlap in space to any appreciable extent (see Fig. 9.31). Thus the two electrons in each 1s or-bital can be assumed to be localized and not to participate in the bonding. To partici-pate in molecular orbitals, atomic orbitals must overlap in space. This means that only the valence orbitals of the atoms contribute significantly to the molecular orbitals of a particular molecule.
2 2 2 0 Bond order 2 1 2 1 2 Bond order 2 0 2 1 E σ1s σ1s He1s He1s FIGURE 9.30 The molecular orbital energy-level diagram for the He2 molecule.
Li 1s 2s Li 1s 2s FIGURE 9.31 The relative sizes of the lithium 1s and 2s atomic orbitals.
E σ2s σ2s 2s 2s Li2 Li Li MO shapes FIGURE 9.32 The molecular orbital energy-level diagram for the Li2 molecule.
9.3 Bonding in Homonuclear Diatomic Molecules 407 The molecular orbital diagram of the Li2 molecule and the shapes of its bonding and antibonding MOs are shown in Fig. 9.32. The electron configuration for Li2 (valence elec-trons only) is 2s 2, and the bond order is The Li2 is a stable molecule (has lower energy than two separated lithium atoms). How-ever, this does not mean that Li2 is the most stable form of elemental lithium. In fact, at normal temperature and pressure, lithium exists as a solid containing many lithium atoms bound to each other.
For the beryllium molecule (Be2) the bonding and antibonding orbitals both contain two electrons. In this case the bond order is (2 2)2 0, and since Be2 is not more stable than two separated Be atoms, no molecule forms. However, beryllium metal contains many beryllium atoms bonded to each other and is stable for reasons we will discuss in Chapter 10.
Since the boron atom has a 1s22s22p1 configuration, we describe the B2 molecule by considering how p atomic orbitals combine to form molecular orbitals. Recall that p or-bitals have two lobes and that they occur in sets of three mutually perpendicular orbitals [see Fig. 9.33(a)]. When two B atoms approach each other, two pairs of p orbitals can overlap in a parallel fashion [Fig. 9.33(b) and (c)] and one pair can overlap head-on [Fig. 9.33(d)].
First, let’s consider the molecular orbitals from the head-on overlap, as shown in Fig. 9.34(a). Note that the electrons in the bonding MO are, as expected, concentrated between the nuclei, and the electrons in the antibonding MO are concentrated outside the area between the two nuclei. Also, both these MOs are molecular orbitals. The 2 0 2 1 Beryllium metal.
(a) (b) (c) (d) B B + + – + – + – + – + + – + – – + + – + – – – – + FIGURE 9.33 (a) The three mutually perpendicular 2p orbitals on two adjacent boron atoms. The signs indicate the orbital phases. Two pairs of parallel p orbitals can overlap, as shown in (b) and (c), and the third pair can overlap head-on, as shown in (d).
Visualization: Pi Bonding and Antibonding Orbitals 408 Chapter Nine Covalent Bonding: Orbitals p orbitals that overlap in a parallel fashion also produce bonding and antibonding or-bitals [Fig. 9.34(b)]. Since the electron probability lies above and below the line between the nuclei, both the orbitals are pi () molecular orbitals. They are designated as 2p for the bonding MO and 2p for the antibonding MO.
Let’s try to make an educated guess about the relative energies of the and mo-lecular orbitals formed from the 2p atomic orbitals. Would we expect the electrons to pre-fer the bonding orbital (where the electron probability is concentrated in the area between the nuclei) or the bonding orbital? The orbital would seem to have the lower energy, since the electrons are closest to the two nuclei. This agrees with the observation that interactions are stronger than interactions.
Figure 9.35 gives the molecular orbital energy-level diagram expected when the two sets of 2p orbitals on the boron atoms combine to form molecular orbitals. Note that there are two bonding orbitals at the same energy (degenerate orbitals) formed from the two pairs of parallel p orbitals, and there are two degenerate antibonding orbitals. The energy of the 2p orbitals is expected to be higher than that of the 2p orbital because interactions are generally stronger than interactions.
To construct the total molecular orbital diagram for the B2 molecule, we make the as-sumption that the 2s and 2p orbitals combine separately (in other words, there is no 2s–2p mixing). The resulting diagram is shown in Fig. 9.36. Note that B2 has six valence electrons. (Remember the 1s orbitals and electrons are assumed not to participate in the bonding.) This diagram predicts the bond order: Therefore, B2 should be a stable molecule.
4 2 2 1 2py 2py Bonding π2p π2p 2py 2py Antibonding 2px 2px Bonding σ2p σ2p Antibonding + – + – – – – + 2px 2px – + – + – + – – + + – + + – + + + – – + – – + – + (b) (a) FIGURE 9.34 (a) The two p orbitals on the boron atoms that overlap head-on combine to form bonding and antibonding orbitals. The bonding orbital is formed by reversing the sign of the right orbital so the positive phases of both orbitals match between the nuclei to produce constructive interference.
This leads to enhanced electron probability between the nuclei. The antibonding orbital is formed by the direct combination of the orbitals, which gives destructive interfer-ence of the positive phase of one orbital with the negative phase of the second or-bital. This produces a node between the nu-clei, which gives decreased electron proba-bility. (b) When the parallel p orbitals are combined with the positive and negative phases matched, constructive interference occurs, giving a bonding orbital. When the orbitals have opposite phases (the signs of one orbital are reversed), destructive interference occurs, resulting in an antibonding orbital.
E σ2p σ2p B2p B2p π2p π2p π2p π2p FIGURE 9.35 The expected molecular orbital energy-level diagram resulting from the combination of the 2p orbitals on two boron atoms.
9.3 Bonding in Homonuclear Diatomic Molecules 409 Paramagnetism At this point we need to discuss an additional molecular property—magnetism. Most materials have no magnetism until they are placed in a magnetic field. However, in the presence of such a field, magnetism of two types can be induced. Paramagnetism causes the substance to be attracted into the inducing magnetic field. Diamagnetism causes the substance to be repelled from the inducing magnetic field. Figure 9.37 illustrates how para-magnetism is measured. The sample is weighed with the electromagnet turned off and then weighed again with the electromagnet turned on. An increase in weight when the field is turned on indicates the sample is paramagnetic. Studies have shown that paramagnetism is associated with unpaired electrons and diamagnetism is associated with paired electrons.
Any substance that has both paired and unpaired electrons will exhibit a net paramagnet-ism, since the effect of paramagnetism is much stronger than that of diamagnetism.
The molecular orbital energy-level diagram represented in Fig. 9.36 predicts that the B2 molecule will be diamagnetic, since the MOs contain only paired electrons. However, experiments show that B2 is actually paramagnetic with two unpaired electrons. Why does the model yield the wrong prediction? This is yet another illustration of how models are developed and used. In general, we try to use the simplest possible model that accounts for all the important observations. In this case, although the simplest model successfully describes the diatomic molecules up to B2, it certainly is suspect if it cannot describe the B2 molecule correctly. This means we must either discard the model or find a way to modify it.
Let’s consider one assumption that we made. In our treatment of B2, we have as-sumed that the s and p orbitals combine separately to form molecular orbitals. Calcu-lations show that when the s and p orbitals are allowed to mix in the same molecular orbital, a different energy-level diagram results for B2 (see Fig. 9.38). Note that even though the s and p contributions to the MOs are no longer separate, we retain the simple orbital designations. The energies of 2p and 2p orbitals are reversed by p–s mixing, and the 2s and the 2s orbitals are no longer equally spaced relative to the energy of the free 2s orbital.
When the six valence electrons for the B2 molecule are placed in the modified energy-level diagram, each of the last two electrons goes into one of the degenerate 2p orbitals.
This produces a paramagnetic molecule in agreement with experimental results. Thus, when the model is extended to allow p–s mixing in molecular orbitals, it predicts the correct magnetism. Note that the bond order is (4 2)2 1, as before.
The remaining diatomic molecules of the elements in Period 2 can be described using similar ideas. For example, the C2 and N2 molecules use the same set of orbitals as for B2 (see Fig. 9.38). Because the importance of 2s–2p mixing decreases across the period, the 2p and 2p orbitals revert to the order expected in the absence of 2s–2p mixing for the molecules O2 and F2, as shown in Fig. 9.39.
Several significant points arise from the orbital diagrams, bond strengths, and bond lengths summarized in Fig. 9.39 for the Period 2 diatomics: 1. There are definite correlations between bond order, bond energy, and bond length. As the bond order predicted by the molecular orbital model increases, the bond energy increases and the bond length decreases. This is a clear indication that the bond or-der predicted by the model accurately reflects bond strength, and it strongly supports the reasonableness of the MO model.
2. Comparison of the bond energies of the B2 and F2 molecules indicates that bond or-der cannot automatically be associated with a particular bond energy. Although both molecules have a bond order of 1, the bond in B2 appears to be about twice as strong as the bond in F2. As we will see in our later discussion of the halogens, F2 has an unusually weak single bond due to larger than usual electron–electron repulsions (there are 14 valence electrons on the small F2 molecule).
E σ2p σ2p 2p π2p π2p π2p π2p 2p σ2s σ2s 2s 2s B atom B2 molecule B atom FIGURE 9.36 The expected molecular orbital energy-level diagram for the B2 molecule.
Glass tubing Sample tube Electromagnet Balance FIGURE 9.37 Diagram of the kind of apparatus used to measure the paramagnetism of a sample. A paramagnetic sample will appear heavier when the electromagnet is turned on be-cause the sample is attracted into the in-ducing magnetic field.
410 Chapter Nine Covalent Bonding: Orbitals 3. Note the very large bond energy associated with the N2 molecule, which the molecular orbital model predicts will have a bond order of 3, a triple bond. The very strong bond in N2 is the principal reason that many nitrogen-containing compounds are used as high explosives. The reactions involving these explosives give the very stable N2 molecule as a product, thus releasing large quantities of energy.
4. The O2 molecule is known to be paramagnetic. This can be very convincingly demon-strated by pouring liquid oxygen between the poles of a strong magnet, as shown in Fig. 9.40. The oxygen remains there until it evaporates. Significantly, the molecular orbital model correctly predicts oxygen’s paramagnetism, while the localized electron model predicts a diamagnetic molecule.
The Molecular Orbital Model I For the species O2, O2 , and O2 , give the electron configuration and the bond order for each. Which has the strongest bond?
Solution The O2 molecule has 12 valence electrons (6 6); O2 has 11 valence electrons (6 6 1); and O2 has 13 valence electrons (6 6 1). We will assume that the ions E σ2p π2p σ2p π2p σ2s σ2s B2 C2 N2 O2 F2 Magnetism Para– magnetic Dia– magnetic Dia– magnetic Para– magnetic Dia– magnetic Bond order 1 2 3 σ2s σ2s σ2p π2p π2p σ2p 2 1 Observed bond dissociation energy (kJ/mol) 290 620 942 495 154 Observed bond length (pm) 159 131 110 121 143 FIGURE 9.39 The molecular orbital energy-level diagrams, bond orders, bond energies, and bond lengths for the diatomic molecules B2 through F2.
Note that for O2 and F2 the 2p orbital is lower in energy than the 2p orbitals.
Sample Exercise 9.6 E σ2p σ2p 2p π2p π2p π2p π2p 2p σ2s σ2s 2s 2s B atom B2 molecule B atom FIGURE 9.38 The correct molecular orbital energy-level diagram for the B2 molecule. When p–s mixing is allowed, the energies of the 2p and 2p orbitals are reversed. The two electrons from the B 2p orbitals now occupy separate, degenerate 2p molecular orbitals and thus have parallel spins.
Therefore, this diagram explains the observed paramagnetism of B2.
Visualization: Molecular Orbital Diagram (N2) Visualization: Magnetic Properties of Liquid Nitrogen and Oxygen 9.3 Bonding in Homonuclear Diatomic Molecules 411 can be treated using the same molecular orbital diagram as for the neutral diatomic molecule: O2 O2 O2 2p –––– –––– –––– 2p — h — h — h — — hg — — h 2p — hg — hg — hg — hg — hg — hg 2p —— hg —— hg —— hg 2s —— hg —— hg —— hg 2s —— hg —— hg —— hg The electron configuration for each species can then be taken from the diagram: The bond orders are: Thus O2 is expected to have the strongest bond of the three species.
See Exercises 9.39 and 9.40.
The Molecular Orbital Model II Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules.
a. Ne2 b. P2 Solution a. The valence orbitals for Ne are 2s and 2p. Thus we can use the molecular orbitals we have already constructed for the diatomic molecules of the Period 2 elements. The Ne2 molecule has 16 valence electrons (8 from each atom). Placing these electrons in the appropriate molecular orbitals produces the following diagram: —— hg — hg — hg — hg — hg E —— hg —— hg —— hg The bond order is (8 8)2 0, and Ne2 does not exist.
b. The P2 molecule contains phosphorus atoms from the third row of the periodic table. We will assume that the diatomic molecules of the Period 3 elements can be s2s s2s s2p p2p p2p s2p For O2 : 8 5 2 1.5 For O2 : 8 3 2 2.5 For O2: 8 4 2 2 O2 : 1s2s221s2s221s2p221p2p241p2p23 O2 : 1s2s221s2s221s2p221p2p241p2p21 O2: 1s2s221s2s221s2p221p2p241p2p22 FIGURE 9.40 When liquid oxygen is poured into the space between the poles of a strong magnet, it remains there until it boils away. This attraction of liquid oxygen for the magnetic field demonstrates the paramagnetism of the O2 molecule.
Sample Exercise 9.7 h A A A A A A A A A A Visualization: Magnetic Properties of Liquid Nitrogen and Oxygen 412 Chapter Nine Covalent Bonding: Orbitals treated in a way very similar to that which we have used so far. Thus we will draw the MO diagram for P2 analogous to that for N2. The only change will be that the molecular orbitals will be formed from 3s and 3p atomic orbitals. The P2 model has 10 valence electrons (5 from each phosphorus atom). The resulting molecular orbital diagram is —— — — —— hg E — hg — hg —— hg —— hg The molecule has a bond order of 3 and is expected to be diamagnetic.
See Exercises 9.37 through 9.42.
9.4 Bonding in Heteronuclear Diatomic Molecules In this section we will deal with selected examples of heteronuclear (different atoms) diatomic molecules. A special case involves molecules containing atoms adjacent to each other in the periodic table. Since the atoms involved in such a molecule are so sim-ilar, we can use the molecular orbital diagram for homonuclear molecules. For exam-ple, we can predict the bond order and magnetism of nitric oxide (NO) by placing its 11 valence electrons (5 from nitrogen and 6 from oxygen) in the molecular orbital energy-level diagram shown in Fig. 9.41. The molecule should be paramagnetic and has a bond order of Experimentally, nitric oxide is indeed found to be paramagnetic. Notice that this odd-electron molecule is described very naturally by the MO model. In contrast, the localized electron model, in the simple form used in this text, cannot be used readily to treat such molecules.
The Molecular Orbital Model III Use the molecular orbital model to predict the magnetism and bond order of the NO and CN ions.
Solution The NO ion has 10 valence electrons (5 6 1). The CN ion also has 10 valence elec-trons (4 5 1). Both ions are therefore diamagnetic and have a bond order derived from the equation The molecular orbital diagram for these two ions is the same (see Fig. 9.42).
See Exercises 9.43 and 9.44.
8 2 2 3 8 3 2 2.5 s3s s3s p3p s3p p3p s3p h A A A A A A A A A A A E σ2p π2p σ2p π2p σ2s σ2s FIGURE 9.41 The molecular orbital energy-level diagram for the NO molecule. We assume that orbital order is the same as that for N2.
The bond order is 2.5.
Sample Exercise 9.8 9.5 Combining the Localized Electron and Molecular Orbital Models 413 When the two atoms of a diatomic molecule are very different, the energy-level dia-gram for homonuclear molecules can no longer be used. A new diagram must be devised for each molecule. We will illustrate this case by considering the hydrogen fluoride (HF) molecule. The electron configurations of the hydrogen and fluorine atoms are 1s1 and 1s22s22p5, respectively. To keep things as simple as possible, we will assume that fluorine uses only one of its 2p orbitals to bond to hydrogen. Thus the molecular orbitals for HF will be composed of fluorine 2p and hydrogen 1s orbitals. Figure 9.43 gives the partial molecular orbital energy-level diagram for HF, focusing only on the orbitals involved in the bonding. We are assuming that fluorine’s other valence electrons remain localized on the fluorine. The 2p orbital of fluorine is shown at a lower energy than the 1s orbital of hydrogen on the diagram because fluorine binds its valence electrons more tightly. Thus the 2p electron on a free fluorine atom is at lower energy than the 1s electron on a free hydrogen atom. The diagram predicts that the HF molecule should be stable because both electrons are lowered in energy relative to their energy in the free hydrogen and fluorine atoms, which is the driving force for bond formation.
Because the fluorine 2p orbital is lower in energy than the hydrogen 1s orbital, the electrons prefer to be closer to the fluorine atom. That is, the molecular orbital con-taining the bonding electron pair shows greater electron probability close to the fluorine (see Fig. 9.44). The electron pair is not shared equally. This causes the fluorine atom to have a slight excess of negative charge and leaves the hydrogen atom partially positive.
This is exactly the bond polarity observed for HF. Thus the molecular orbital model accounts in a straightforward way for the different electronegativities of hydrogen and fluorine and the resulting unequal charge distribution.
9.5 Combining the Localized Electron and Molecular Orbital Models One of the main difficulties with the localized electron model is its assumption that electrons are localized. This problem is most apparent with molecules for which several valid Lewis structures can be drawn. It is clear that none of these structures taken alone adequately describes the electronic structure of the molecule. The concept of resonance was invented to solve this problem. However, even with resonance included, the localized electron model does not describe molecules and ions such as O3 and NO3 in a very satisfying way.
It would seem that the ideal bonding model would be one with the simplicity of the localized electron model but with the delocalization characteristic of the molecular orbital model. We can achieve this by combining the two models to describe molecules that require resonance. Note that for species such as O3 and NO3 the double bond changes position in the resonance structures (see Fig. 9.45). Since a double bond involves one and one bond, there is a bond between all bound atoms in each resonance structure. It is really the bond that has different locations in the various resonance structures.
Therefore we conclude that the bonds in a molecule can be described as being localized with no apparent problems. It is the bonding that must be treated as being E σ2p π2p σ2p π2p σ2s σ2s E 1s HF molecule H atom F atom 2p σ σ H nucleus F nucleus FIGURE 9.42 The molecular orbital energy-level diagram for both the NO and CN ions.
FIGURE 9.43 A partial molecular orbital energy-level diagram for the HF molecule.
FIGURE 9.44 The electron probability distribution in the bonding molecular orbital of the HF molecule. Note the greater electron density close to the fluorine atom.
O O O O O O N O O O N O O O N O O O FIGURE 9.45 The resonance structures for O3 and NO3 .
Note that it is the double bond that occupies various positions in the resonance structures.
414 Chapter Nine Covalent Bonding: Orbitals delocalized. Thus, for molecules that require resonance, we will use the localized elec-tron model to describe the bonding and the molecular orbital model to describe the bonding. This allows us to keep the bonding model as simple as possible and yet give a more physically accurate description of such molecules.
We will illustrate the general method by considering the bonding in benzene, an im-portant industrial chemical that must be handled carefully because it is a known carcino-gen. The benzene molecule (C6H6) consists of a planar hexagon of carbon atoms with one hydrogen atom bound to each carbon atom, as shown in Fig. 9.46(a). In the molecule all six COC bonds are known to be equivalent. To explain this fact, the localized electron model must invoke resonance [see Fig. 9.46(b)].
A better description of the bonding in benzene results when we use a combination of the models, as described above. In this description it is assumed that the bonds of car-bon involve sp2 orbitals, as shown in Fig. 9.47. These bonds are all centered in the plane of the molecule.
In molecules that require resonance, it is the bonding that is most clearly delocalized.
H H H H C C C C C C H H H H H H H H H H H H H H (a) (b) FIGURE 9.46 (a) The benzene molecule consists of a ring of six carbon atoms with one hydrogen atom bound to each carbon; all atoms are in the same plane. All the COC bonds are known to be equivalent. (b) Two of the resonance structures for the benzene molecule. The localized electron model must invoke resonance to account for the six equal COC bonds.
O ne of the best things about New Mexico is the food.
Authentic New Mexican cuisine employs liberal amounts of green and red chilies—often called chili peppers. Chilies apparently originated in parts of South America and were spread north by birds. When Columbus came to North America, which he originally thought was India, he observed the natives using chilies for spicing foods. When he took chilies back to Europe, Columbus mistakenly called them peppers and the name stuck.
The spicy payload of chilies is delivered mainly by the chemical capsaicin, which has the following structure: Capsaicin was isolated as a pure substance by L. T. Thresh in 1846. Since then substituted capsaicins have also been found in chilies. The spicy power of chilies derives mostly from capsaicin and dihydrocapsaicin.
The man best known for explaining the “heat” of chilies is Wilbur Scoville, who defined the Scoville unit for meas-uring chili power. He arbitrarily established the hotness of pure capsaicin as 16 million. On this scale a typical green or red chili has a rating of about 2500 Scoville units. You may have had an encounter with habanero chilies that left you looking for a firehose to put out the blaze in your mouth—habaneros have a Scoville rating of about 500,000!
Capsaicin has found many uses outside of cooking. It is used in pepper sprays and repellant sprays for many garden pests, although birds are unaffected by capsaicin. Capsaicin also stimulates the body’s circulation and causes pain recep-tors to release endorphins, similar to the effect produced by intense exercise. Instead of jogging you may want to sit on the couch eating chilies. Either way you are going to sweat.
CHEMICAL IMPACT What’s Hot?
H H H HO C N H H H CH3O C C C H H C H H H H H H C C H C C CH3 CH3 H O 9.5 Combining the Localized Electron and Molecular Orbital Models 415 FIGURE 9.49 (a) The p orbitals used to form the bond-ing system in the NO3 ion. (b) A represen-tation of the delocalization of the electrons in the molecular orbital system of the NO3 ion.
H H H sp2 C C C C C C sp2 H1s H H H H H H H H H H H H H (a) (b) H H FIGURE 9.47 The bonding system in the benzene molecule.
FIGURE 9.48 (a) The molecular orbital system in ben-zene is formed by combining the six p or-bitals from the six sp2 hybridized carbon atoms. (b) The electrons in the resulting molecular orbitals are delocalized over the entire ring of carbon atoms, giving six equivalent bonds. A composite of these or-bitals is represented here.
O O N O (a) (b) O O N O Since each carbon atom is sp2 hybridized, a p orbital perpendicular to the plane of the ring remains on each carbon atom. These six p orbitals can be used to form molecular orbitals, as shown in Fig. 9.48(a). The electrons in the resulting molecular orbitals are delocalized above and below the plane of the ring, as shown in Fig. 9.48(b). This gives six equivalent COC bonds, as required by the known structure of the benzene molecule.
The benzene structure is often written as to indicate the delocalized bonding in the molecule.
Very similar treatments can be applied to other planar molecules for which resonance is required by the localized electron model. For example, the NO3 ion can be described using the molecular orbital system shown in Fig. 9.49. In this molecule each atom is assumed to be sp2 hybridized, which leaves one p orbital on each atom perpendicular to the plane of the ion. These p orbitals can combine to form the molecular orbital system.
Visualization: Pi Bonding in the Nitrate Ion 416 Chapter Nine Covalent Bonding: Orbitals For Review Two widely used bonding models Localized electron model Molecular orbital model Localized electron model Molecule is pictured as a group of atoms sharing electron pairs between atomic orbitals Hybrid orbitals, which are combinations of the “native” atomic orbitals, are often required to account for the molecular structure • Four electron pairs (tetrahedral arrangement) require sp3 orbitals • Three electron pairs (trigonal planar arrangement) require sp2 orbitals • Two electron pairs (linear arrangement) requires sp orbitals Two types of bonds Sigma: electrons are shared in the area centered on a line joining the atoms Pi: a shared electron pair occupies the space above and below the line joining the atoms Molecular orbital model A molecule is assumed to be a new entity consisting of positively charged nuclei and electrons The electrons in the molecule are contained in molecular orbitals, which in the simplest form of the model are constructed from the atomic orbitals of the constituent atoms The model correctly predicts relative bond strength, magnetism, and bond polarity It correctly portrays electrons as being delocalized in polyatomic molecules The main disadvantage of the model is that it is difficult to apply qualitatively to polyatomic molecules Molecular orbitals are classified in two ways: energy and shape Energy • A bonding MO is lower in energy than the atomic orbitals from which it is con-structed. Electrons in this type of MO are lower in energy in the molecule than in the separated atoms and thus favor molecule formation.
• An antibonding MO is higher in energy than the atomic orbitals from which it is constructed. Electrons in this type of MO are higher in energy in the mole-cule than in the separated atoms and thus do not favor molecule formation.
Shape (symmetry) • Sigma (
) MOs have their electron probability centered on a line passing through the nuclei • Pi () MOs have their electron probability concentrated above and below the line connecting the nuclei Bond order is an index of bond strength Molecules that require the concept of resonance in the localized electron model can be more accurately described by combining the localized electron and molecular orbital models The bonds are localized The bonds are delocalized Bond order number of bonding electrons number of antibonding electrons 2 Key Terms Section 9.1 hybridization sp3 hybridization hybrid orbitals sp2 hybridization sigma (
) bond pi () bond sp hybridization dsp3 hybridization d2sp3 hybridization Section 9.2 molecular orbital model molecular orbital (MO) sigma (
) molecular orbital bonding molecular orbital antibonding molecular orbital bond order Section 9.3 pi () molecular orbital paramagnetism diamagnetism Section 9.4 heteronuclear diatomic molecule Section 9.5 delocalized bonding Active Learning Questions 417 REVIEW QUESTIONS 1. Why do we hybridize atomic orbitals to explain the bonding in covalent com-pounds? What type of bonds form from hybrid orbitals, sigma or pi? Explain.
2. What hybridization is required for central atoms that have a tetrahedral arrange-ment of electron pairs? A trigonal planar arrangement of electron pairs? A linear arrangement of electron pairs? How many unhybridized p atomic orbitals are pres-ent when a central atom exhibits tetrahedral geometry? Trigonal planar geometry?
Linear geometry? What are the unhybridized p atomic orbitals used for?
3. Describe the bonding in H2S, CH4, H2CO, and HCN using the localized electron model.
4. What hybridization is required for central atoms exhibiting trigonal bipyramidal geometry? Octahedral geometry? Describe the bonding of PF5, SF4, SF6, and IF5 using the localized electron model.
5. Electrons in bonding molecular orbitals are most likely to be found in the region between the two bonded atoms. Why does this arrangement favor bond-ing? In a antibonding orbital, where are the electrons most likely to be found in relation to the nuclei in a bond?
6. Show how 2s orbitals combine to form bonding and antibonding molecular orbitals. Show how 2p orbitals overlap to form bonding, bonding, anti-bonding, and antibonding molecular orbitals.
7. What are the relationships among bond order, bond energy, and bond length?
Which of these can be measured? Distinguish between the terms paramagnetic and diamagnetic. What type of experiment can be done to determine if a material is paramagnetic?
8. How does molecular orbital theory explain the following observations?
a. H2 is stable, while He2 is unstable.
b. B2 and O2 are paramagnetic, while C2, N2, and F2 are diamagnetic.
c. N2 has a very large bond energy associated with it.
d. NO is more stable than NO.
9. Consider the heteronuclear diatomic molecule HF. Explain in detail how mo-lecular orbital theory is applied to describe the bonding in HF.
10. What is delocalized bonding and what does it explain? Explain the delocalized bonding system in C6H6 (benzene) and O3 (ozone).
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of the bonding and antibond-ing orbitals which is lower in energy? Explain.
2. Explain the difference between the and MOs for homonu-clear diatomic molecules. How are bonding and antibonding or-bitals different? Why are there two MOs and one MO? Why are the MOs degenerate?
3. Compare Figs. 9.36 and 9.38. Why are they different? Because B2 is known to be paramagnetic, the 2p and 2p molecular orbitals must be switched from the first prediction. What is the rationale for this? Why might one expect the 2p to be lower in energy than the 2p? Why can’t we use diatomic oxygen to help us decide whether the 2p or 2p is lower in energy?
4. Which of the following would you expect to be more favorable energetically? Explain.
a. An H2 molecule in which enough energy is added to excite one electron from the bonding to the antibonding MO b. Two separate H atoms 5. Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as or .
6. Which is the more correct statement: “The methane molecule (CH4) is a tetrahedral molecule because it is sp3 hybridized” or “The methane molecule (CH4) is sp3 hybridized because it is a tetrahedral molecule”? What, if anything, is the difference between these two statements?
418 Chapter Nine Covalent Bonding: Orbitals A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 7. In the hybrid orbital model, compare and contrast bonds versus bonds. What orbitals form the bonds and what orbitals form the bonds? Assume the z-axis is the internuclear axis.
8. In the molecular orbital model, compare and contrast versus bonds. What orbitals form the bonds and what orbitals form the bonds? Assume the z-axis is the internuclear axis.
9. Why are d orbitals sometimes used to form hybrid orbitals? Which period of elements does not use d orbitals for hybridization? If necessary, which d orbitals (3d, 4d, 5d, or 6d) would sulfur use to form hybrid orbitals requiring d atomic orbitals? Answer the same question for arsenic and for iodine.
10. The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is bro-ken. Why?
11. Compare and contrast bonding versus antibonding molecular orbitals.
12. What modification to the molecular orbital model was made from the experimental evidence that B2 is paramagnetic?
13. Why does the molecular orbital model do a better job in explaining the bonding in NO and NO than the hybrid orbital model?
14. The three NO bonds in NO3 are all equivalent in length and strength. How is this explained even though any valid Lewis structure for NO3 has one double bond and two single bonds to nitrogen?
Exercises In this section similar exercises are paired.
The Localized Electron Model and Hybrid Orbitals 15. Use the localized electron model to describe the bonding in H2O.
16. Use the localized electron model to describe the bonding in CCl4.
17. Use the localized electron model to describe the bonding in H2CO (carbon is the central atom).
18. Use the localized electron model to describe the bonding in C2H2 (exists as HCCH).
19. The space-filling models of ethane and ethanol are shown below.
Use the localized electron model to describe the bonding in ethane and ethanol.
H O C Ethane (C2H6) Ethanol (C2H5OH) 20. The space-filling models of hydrogen cyanide and phosgene are shown below.
Use the localized electron model to describe the bonding in hy-drogen cyanide and phosgene.
21. Give the expected hybridization of the central atom for the mol-ecules or ions in Exercises 67 and 73 from Chapter 8.
22. Give the expected hybridization of the central atom for the mol-ecules or ions in Exercises 68 and 74 from Chapter 8.
23. Give the expected hybridization of the central atom for the mol-ecules or ions in Exercise 71 from Chapter 8.
24. Give the expected hybridization of the central atom for the mol-ecules in Exercise 72 from Chapter 8.
25. Give the expected hybridization of the central atom for the mol-ecules in Exercises 91 and 92 from Chapter 8.
26. Give the expected hybridization of the central atom for the mol-ecules in Exercises 93 and 94 from Chapter 8.
27. For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the over-all polarity.
a. CF4 e. BeH2 i. KrF4 b. NF3 f. TeF4 j. SeF6 c. OF2 g. AsF5 k. IF5 d. BF3 h. KrF2 l. IF3 28. For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (in-cluding bond angles), and give the expected hybrid orbitals for sulfur.
a. SO2 b. SO3 c.
d.
e. SO3 2 f. SO4 2 g. SF2 h. SF4 i. SF6 j. F3SOSF k. SF5 Cl N H O C Hydrogen cyanide (HCN) Phosgene (COCl2) Exercises 419 29. Why must all six atoms in C2H4 be in the same plane?
30. The allene molecule has the following Lewis structure: Are all four hydrogen atoms in the same plane? If not, what is their spatial relationship? Explain.
31. Biacetyl and acetoin are added to margarine to make it taste more like butter.
Complete the Lewis structures, predict values for all COCOO bond angles, and give the hybridization of the carbon atoms in these two compounds. Are the four carbons and two oxygens in biacetyl in the same plane? How many bonds and how many bonds are there in biacetyl and acetoin?
32. Many important compounds in the chemical industry are deriva-tives of ethylene (C2H4). Two of them are acrylonitrile and methyl methacrylate.
Complete the Lewis structures, showing all lone pairs. Give ap-proximate values for bond angles a through f. Give the hybridiza-tion of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule lie in the same plane? How many bonds and how many bonds are there in methyl methacrylate and acrylonitrile?
33. One of the first drugs to be approved for use in treatment of ac-quired immune deficiency syndrome (AIDS) was azidothymidine (AZT). Complete the Lewis structure for AZT.
a H H C C C H H H N b c Acrylonitrile C C C O O CH3 CH3 f e d Methyl methacrylate a. How many carbon atoms are sp3 hybridized?
b. How many carbon atoms are sp2 hybridized?
c. Which atom is sp hybridized?
d. How many bonds are in the molecule?
e. How many bonds are in the molecule?
f. What is the NONON bond angle in the azide (ON3) group?
g. What is the HOOOC bond angle in the side group attached to the five-membered ring?
h. What is the hybridization of the oxygen atom in the OCH2OH group?
34. Hot and spicy foods contain molecules that stimulate pain-detecting nerve endings. Two such molecules are piperine and capsaicin: Piperine is the active compound in white and black pepper, and capsaicin is the active compound in chili peppers. The ring struc-tures in piperine and capsaicin are shorthand notation. Each point where lines meet represents a carbon atom.
a. Complete the Lewis structure for piperine and capsaicin show-ing all lone pairs of electrons.
b. How many carbon atoms are sp, sp2, and sp3 hybridized in each molecule?
c. Which hybrid orbitals are used by the nitrogen atoms in each molecule?
d. Give approximate values for the bond angles marked a through l in the above structures.
The Molecular Orbital Model 35. Which of the following are predicted by the molecular orbital model to be stable diatomic species?
a. H2 , H2, H2 , H2 2 b. He2 2, He2 , He2 36. Which of the following are predicted by the molecular orbital model to be stable diatomic species?
a. N2 2, O2 2, F2 2 b. Be2, B2, Ne2 37. Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders.
Which ones are paramagnetic?
a. Li2 b. C2 c. S2 G D B H HO (CH2)3 H3CO H H G A G D N H C O CH CH2 CH3 CH2G CH D J G CH CH3 Capsaicin g h i j k l O G D P B H C H H O H f H CH P CH O CH CH O C O N H H H H H H H H H H Piperine a b c e d O 420 Chapter Nine Covalent Bonding: Orbitals 38. Consider the following electron configuration: Give four species that, in theory, would have this electron con-figuration 39. Using molecular orbital theory, explain why the removal of one electron in O2 strengthens bonding, while the removal of one elec-tron in N2 weakens bonding.
40. Using the molecular orbital model to describe the bonding in F2 , F2, and F2 , predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?
41. Which charge(s) for the N2 molecule would give a bond order of 2.5?
42. A Lewis structure obeying the octet rule can be drawn for O2 as follows: Use the molecular orbital energy-level diagram for O2 to show that the above Lewis structure corresponds to an excited state.
43. Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders.
Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy.
a. CO b. CO c. CO2 44. Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders.
Which ones are paramagnetic? Place the species in order of in-creasing bond length and bond energy.
a. NO b. NO c. NO 45. In which of the following diatomic molecules would the bond strength be expected to weaken as an electron is removed to form the positive charged ion?
a. H2 c. C2 2 b. B2 d. OF 46. In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an elec-tron? Explain.
CN or NO O2 2 or N2 2 47. Show how two 2p atomic orbitals can combine to form a or a molecular orbital.
48. Show how a hydrogen 1s atomic orbital and a fluorine 2p atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals or molecular orbitals?
49. Use Figs. 9.43 and 9.44 to answer the following questions.
a. Would the bonding molecular orbital in HF place greater elec-tron density near the H or the F atom? Why?
b. Would the bonding molecular orbital have greater fluorine 2p character, greater hydrogen 1s character, or an equal contribu-tion from both? Why?
1s3s221s3s 221s3p221p3p241p3p 24 c. Answer the previous two questions for the antibonding mo-lecular orbital in HF.
50. The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be 97.06 pm and 424.7 kJ/mol, respectively. Assume that the OH molecule is anal-ogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lower-energy pz or-bital from oxygen with the higher-energy 1s orbital of hydrogen (the OOH bond lies along the z-axis).
a. Which of the two molecular orbitals will have the greater hy-drogen 1s character?
b. Can the 2px orbital of oxygen form molecular orbitals with the 1s orbital of hydrogen? Explain.
c. Knowing that only the 2p orbitals of oxygen will interact sig-nificantly with the 1s orbital of hydrogen, complete the mo-lecular orbital energy-level diagram for OH. Place the correct number of electrons in the energy levels.
d. Estimate the bond order for OH.
e. Predict whether the bond order of OH will be greater than, less than, or the same as that of OH. Explain.
51. Describe the bonding in the O3 molecule and the NO2 ion using the localized electron model. How would the molecular orbital model describe the bonding in these two species?
52. Describe the bonding in the CO3 2 ion using the localized elec-tron model. How would the molecular orbital model describe the bonding in this species?
Additional Exercises 53. Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the cen-tral atom) for the following.
a. XeO3 d. XeOF2 b. XeO4 e. XeO3F2 c. XeOF4 54. FClO2 and F3ClO can both gain a fluoride ion to form stable an-ions. F3ClO and F3ClO2 will both lose a fluoride ion to form sta-ble cations. Draw the Lewis structures and describe the hybrid orbitals used by chlorine in these ions.
55. Vitamin B6 is an organic compound whose deficiency in the human body can cause apathy, irritability, and an increased susceptibility to infections. Below is an incomplete Lewis structure for vitamin B6. Complete the Lewis structure and answer the following ques-tions. Hint: Vitamin B6 can be classified as an organic compound (a compound based on carbon atoms). The majority of Lewis struc-tures for simple organic compounds have all atoms with a formal charge of zero. Therefore, add lone pairs and multiple bonds to the structure below to give each atom a formal charge of zero.
H H H C O H O C C O C C C N C H C H H H H d f c b a e g Additional Exercises 421 a. How many bonds and bonds exist in vitamin B6?
b. Give approximate values for the bond angles marked a through g in the structure.
c. How many carbon atoms are sp2 hybridized?
d. How many carbon, oxygen, and nitrogen atoms are sp3 hy-bridized?
e. Does vitamin B6 exhibit delocalized bonding? Explain.
56. Aspartame is an artificial sweetener marketed under the name Nutra-Sweet. A partial Lewis structure for aspartame is shown below.
Note that the six-sided ring is shorthand notation for a benzene ring (OC6H5). Benzene is discussed in Section 9.5. Complete the Lewis structure for aspartame. How many C and N atoms exhibit sp2 hybridization? How many C and O atoms exhibit sp3 hybridization? How many and bonds are in aspartame?
Aspartame is an organic compound and the Lewis structure fol-lows the guidelines outlined in Exercise 55.
57. Using bond energies from Table 8.4, estimate the barrier to rota-tion about a carbon–carbon double bond. To do this, consider what must happen to go from to in terms of making and breaking chemical bonds; that is, what must happen in terms of the bond?
58. The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide (CO2), and carbon suboxide (C3O2). The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).
59. Complete the Lewis structures of the following molecules. Pre-dict the molecular structure, polarity, bond angles, and hybrid or-bitals used by the atoms marked by asterisks for each molecule.
a. BH3 H H H B C C OH O O C O H2N CH NH OCH3 CHCH2 CH2 b. N2F2 c. C4H6 d. ICl3 60. Complete the following resonance structures for POCl3.
a. Would you predict the same molecular structure from each res-onance structure?
b. What is the hybridization of P in each structure?
c. What orbitals can the P atom use to form the bond in struc-ture B?
d. Which resonance structure would be favored on the basis of formal charges?
61. The N2O molecule is linear and polar.
a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer.
b. On the basis of your answer to part a, write the Lewis structure of N2O (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom.
c. How would the multiple bonding in be described in terms of orbitals?
62. Describe the bonding in NO, NO, and NO using both the lo-calized electron and molecular orbital models. Account for any discrepancies between the two models.
63. Describe the bonding in the first excited state of N2 (the one closest in energy to the ground state) using the molecular or-bital model. What differences do you expect in the properties of the molecule in the ground state as compared to the first ex-cited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.) 64. Acetylene (C2H2) can be produced from the reaction of calcium carbide (CaC2) with water. Use both the localized electron and molecular orbital models to describe the bonding in the acetylide anion (C2 2).
65. Using an MO energy-level diagram, would you expect F2 to have a lower or higher first ionization energy than atomic fluorine?
Why?
66. Show how a dxz atomic orbital and a pz atomic orbital combine to form a bonding molecular orbital. Assume the x-axis is the NO q O Q S S O N Cl Cl P O Cl Cl Cl P O Cl (A) (B) H C C H H C H C H H 422 Chapter Nine Covalent Bonding: Orbitals internuclear axis. Is a or a molecular orbital formed?
Explain.
67. What type of molecular orbital would result from the in phase combination of two dxz atomic orbitals shown below? Assume the x-axis is the internuclear axis.
68. Consider three molecules: A, B, and C. Molecule A has a hybridization of sp3. Molecule B has two more effective pairs (electron pairs around the central atom) than molecule A. Mole-cule C consists of two bonds and two bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.
Challenge Problems 69. Consider your Lewis structure for the computer-generated model of caffeine shown in Exercise 130 of Chapter 8. How many C and N atoms are sp2 hybridized in your Lewis structure for caffeine?
How many C and N atoms are sp3 hybridized? sp hybridized?
How many and bonds are in your Lewis structure?
70. The space-filling model for benzoic acid is shown below.
Describe the bonding in benzoic acid using the localized electron model combined with the molecular orbital model.
71. Two structures can be drawn by cyanuric acid: a. Are these two structures the same molecule? Explain.
b. Give the hybridization of the carbon and nitrogen atoms in each structure.
c. Use bond energies (Table 8.4) to predict which form is more stable; that is, which contains the strongest bonds?
C O N C N N C H H O O H C N C N N C O O H H O H Benzoic acid (C6H5CO2H) H O C z x z x – + – + + – + – 72. Cholesterol (C27H46O) has the following structure: In such shorthand structures, each point where lines meet repre-sents a carbon atom and most H atoms are not shown. Draw the complete structure showing all carbon and hydrogen atoms. (There will be four bonds to each carbon atom.) Indicate which carbon atoms use sp2 or sp3 hybrid orbitals. Are all carbon atoms in the same plane, as implied by the structure?
73. Cyanamide (H2NCN), an important industrial chemical, is pro-duced by the following steps: Calcium cyanamide (CaNCN) is used as a direct-application fer-tilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: a. Write Lewis structures for NCN2, H2NCN, dicyandiamide, and melamine, including resonance structures where appro-priate.
b. Give the hybridization of the C and N atoms in each species.
c. How many bonds and how many bonds are in each species?
d. Is the ring in melamine planar?
e. There are three different CON bond distances in dicyandi-amide, NCNC(NH2)2, and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.
74. In Exercise 75 in Chapter 8, the Lewis structures for benzene (C6H6) were drawn. Using one of the Lewis structures, estimate Hf for C6H6(g) using bond energies and given that the standard enthalpy of formation of C(g) is 717 kJ/mol. The experimental Hf value of C6H6(g) is 83 kJ/mol. Explain the discrepancy be-tween the experimental value and the calculated Hf value for C6H6(g).
H2NCN NCNC(NH2)2 Dicyandiamide Acid H2N NH2 NH2 N C NCNC(NH2)2 Heat NH3 C C N N Melamine ( bonds not shown) CaNCN ¡ Acid H2NCN Cyanamide CaC2 N2 ¡ CaNCN C Integrative Problems 423 75. A flask containing gaseous N2 is irradiated with 25-nm light.
a. Using the following information, indicate what species can form in the flask during irradiation.
b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions?
c. Explain why the first ionization energy of N2 (1501 kJ/mol) is greater than the first ionization energy of atomic nitrogen (1402 kJ/mol).
76. As compared with CO and O2, CS and S2 are very unstable mol-ecules. Give an explanation based on the relative abilities of the sulfur and oxygen atoms to form bonds.
77. Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: Rationalize the difference in the values of H for these reactions, even though each reaction appears to involve only the breaking of one NOCl bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.) 78. Use the MO model to explain the bonding in BeH2. When con-structing the MO energy-level diagram, assume that the Be’s 1s electrons are not involved in bond formation.
79. Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that CO is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: a. On the basis of electronegativities, would you expect the car-bon atom or the oxygen atom to form bonds to metals?
b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis?
c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule, Figs. 9.43 and 9.44.) Antibonding MOs place more electron density near the less electronegative atom in the diatomic mol-ecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.
80. Arrange the following from lowest to highest ionization energy: O, O2, O2 , O2 . Explain your answer.
81. Use the MO model to determine which of the following has the smallest ionization energy: N2, O2, N2 2, N2 , O2 , Explain your answer.
M¬C‚O ONCl1g2 ¡ NO1g2 Cl1g2 ¢H 158 kJ/mol NCl31g2 ¡ NCl21g2 Cl1g2 ¢H 375 kJ/mol N1g2 ¡ N1g2 e ¢H 1402 kJ/mol N21g2 ¡ N2 1g2 e ¢H 1501 kJ/mol N21g2 ¡ 2N1g2 ¢H 941 kJ/mol 82. Given that the ionization energy of F2 is 290 kJ, do the following: a. Calculate the bond energy of F2 . You will need to look up the bond energy of F2 and ionization energy of F.
b. Explain the difference in bond energy between F2 and F2 us-ing MO theory.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
83. As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, B2.
a. What is the bond order in Li2 and B2?
b. How many electrons must be removed from B2 to make it iso-electronic with Li2 so that it might be used in the warp drive?
c. The reaction to make B2 isoelectronic with Li2 is generalized (where n number of electrons determined in part b) as follows: How much energy is needed to ionize 1.5 kg of B2 to the desired isoelectronic species?
84. An unusual category of acids known as superacids, which are de-fined as any acid stronger than 100% sulfuric acid, can be pre-pared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with SbF5 produces the superacid [H2F][SbF6]: a. What are the molecular structures of all species in this reac-tion? What are the hybridizations of the central atoms in each species?
b. What mass of [H2F][SbF6] can be prepared when 2.93 mL of anhydrous HF (density 0.975 g/mL) and 10.0 mL of SbF5 (density 3.10 g/mL) are allowed to react?
85. Determine the molecular structure and hybridization of the central atom X in the polyatomic ion XY3 given the following infor-mation:A neutral atom of X contains 36 electrons, and the element Y makes an anion with a 1– charge, which has the electron configuration 1s22s22p6.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
2HF1l2 SbF51l2 S 3H2F4 3SbF64 1l2 B2 S B2 n ne ¢H 6455 kJ/mol 424 10 Liquids and Solids Contents 10.1 Intermolecular Forces • Dipole–Dipole Forces • London Dispersion Forces 10.2 The Liquid State • Structural Model for Liquids 10.3 An Introduction to Structures and Types of Solids • X-Ray Analysis of Solids • Types of Crystalline Solids 10.4 Structure and Bonding in Metals • Bonding Models for Metals • Metal Alloys 10.5 Carbon and Silicon: Network Atomic Solids • Ceramics • Semiconductors 10.6 Molecular Solids 10.7 Ionic Solids 10.8 Vapor Pressure and Changes of State • Vapor Pressure • Changes of State 10.9 Phase Diagrams • Applications of the Phase Diagram for Water • The Phase Diagram for Carbon Dioxide Karst formation in Phang Nga Bay, Thailand at sunset. The Karst is limestone that has eroded into these formations rising out of the sea.
You have only to think about water to appreciate how different the three states of matter are. Flying, swimming, and ice skating are all done in contact with water in its various forms. Clearly, the arrangements of the water molecules must be significantly different in its gas, liquid, and solid forms.
In Chapter 5 we saw that a gas can be pictured as a substance whose component par-ticles are far apart, are in rapid random motion, and exert relatively small forces on each other. The kinetic molecular model was constructed to account for the ideal behavior that most gases approach at high temperatures and low pressures.
Solids are obviously very different from gases. A gas has low density and high com-pressibility and completely fills its container. Solids have much greater densities, are compressible only to a very slight extent, and are rigid—a solid maintains its shape irre-spective of its container. These properties indicate that the components of a solid are close together and exert large attractive forces on each other.
The properties of liquids lie somewhere between those of solids and gases but not midway between, as can be seen from some of the properties of the three states of water. For example, compare the enthalpy change for the melting of ice at (the heat of fusion) with that for vaporizing liquid water at (the heat of vaporization): These values show a much greater change in structure in going from the liquid to the gaseous state than in going from the solid to the liquid state. This suggests that there are extensive attractive forces among the molecules in liquid water, similar to but not as strong as those in the solid state.
The relative similarity of the liquid and solid states also can be seen in the densi-ties of the three states of water. As shown in Table 10.1, the densities for liquid and solid water are quite close. Compressibilities also can be used to explore the relation-ship among water’s states. At the density of liquid water changes from 0.99707 g/cm3 at a pressure of 1 atm to 1.046 g/cm3 at 1065 atm. Given the large change in pres-sure, this is a very small variation in the density. Ice also shows little variation in den-sity with increased pressure. On the other hand, at the density of gaseous water changes from g/cm3 at 1 atm pressure to 0.157 g/cm3 at 242 atm—a huge variation.
The conclusion is clear. The liquid and solid states show many similarities and are strikingly different from the gaseous state, as shown schematically in Fig. 10.1.
We must bear this in mind as we develop models for the structures of solids and liquids.
We will proceed in our study of liquids and solids by first considering the properties and structures of liquids and solids. Then we will consider the changes in state that oc-cur between solid and liquid, liquid and gas, and solid and gas.
3.26 104 400°C, 25°C, H2O1l2 ¡ H2O1g2 ¢H°vap 40.7 kJ/mol H2O1s2 ¡ H2O1l2 ¢H°fus 6.02 kJ/mol 100°C 0°C 425 Although the densities of solid and liquid water are quite similar, as is typical for most substances, water is quite unusual in that the density of its solid state is slightly less than that of its liquid state. For most substances, the density of the solid state is slightly greater than that of the liquid state.
Visualization: Intermolecular Forces: London Dispersion Forces 426 Chapter Ten Liquids and Solids 10.1 Intermolecular Forces In Chapters 8 and 9 we saw that atoms can form stable units called molecules by sharing elec-trons. This is called intramolecular (within the molecule) bonding. In this chapter we con-sider the properties of the condensed states of matter (liquids and solids) and the forces that cause the aggregation of the components of a substance to form a liquid or a solid. These forces may involve covalent or ionic bonding, or they may involve weaker interactions usu-ally called intermolecular forces (because they occur between, rather than within, molecules).
It is important to recognize that when a substance such as water changes from solid to liquid to gas, the molecules remain intact. The changes in states are due to changes in the forces among the molecules rather than in those within the molecules. In ice, as we will see later in this chapter, the molecules are virtually locked in place, although they can vibrate about their positions. If energy is added, the motions of the molecules increase, and they eventually achieve the greater movement and disorder characteristic of liquid wa-ter. The ice has melted. As more energy is added, the gaseous state is eventually reached, with the individual molecules far apart and interacting relatively little. However, the gas still consists of water molecules. It would take much energy to overcome the covalent bonds and decompose the water molecules into their component atoms. This can be seen by comparing the energy needed to vaporize 1 mole of liquid water (40.7 kJ) with that needed to break the OOH bonds in 1 mole of water molecules (934 kJ).
Dipole–Dipole Forces As we saw in Section 8.3, molecules with polar bonds often behave in an electric field as if they had a center of positive charge and a center of negative charge. That is, they exhibit a dipole moment. Molecules with dipole moments can attract each other electrostatically by lining up so that the positive and negative ends are close to each other, as shown in Fig.
10.2(a). This is called a dipole–dipole attraction. In a condensed state such as a liquid, where many molecules are in close proximity, the dipoles find the best compromise between attraction and repulsion. That is, the molecules orient themselves to maximize the B,E interactions and to minimize B,B andE,E interactions, as represented in Fig. 10.2(b).
Dipole–dipole forces are typically only about 1% as strong as covalent or ionic bonds, and they rapidly become weaker as the distance between the dipoles increases. At low pressures in the gas phase, where the molecules are far apart, these forces are relatively unimportant.
Particularly strong dipole–dipole forces, however, are seen among molecules in which hydrogen is bound to a highly electronegative atom, such as nitrogen, oxygen, or fluorine.
Two factors account for the strengths of these interactions: the great polarity of the bond and the close approach of the dipoles, allowed by the very small size of the hydrogen atom. Because dipole–dipole attractions of this type are so unusually strong, they are given a special name—hydrogen bonding. Figure 10.3 shows hydrogen bonding among water Intermolecular forces were introduced in Chapter 5 to explain nonideal gas behavior.
Remember that temperature is a mea-sure of the random motions of the parti-cles in a substance.
Dipole–dipole forces are forces that act between polar molecules.
FIGURE 10.1 Schematic representations of the three states of matter.
TABLE 10.1 Densities of the Three States of Water Density State (g/cm3) Solid (0C, 1 atm) 0.9168 Liquid (25C, 1 atm) 0.9971 Gas (400C, 1 atm) 3.26 104 Solid Liquid Gas Visualization: Intermolecular Forces: Dipole–Dipole Forces Visualization: Intermolecular Forces: Hydrogen Bonding Forces 10.1 Intermolecular Forces 427 molecules, which occurs between the partially positive H atoms and the lone pairs on ad-jacent water molecules.
Hydrogen bonding has a very important effect on physical properties. For example, the boiling points for the covalent hydrides of the elements in Groups 4A, 5A, 6A, and 7A are given in Fig. 10.4. Note that the nonpolar tetrahedral hydrides of Group 4A show a steady increase in boiling point with molar mass (that is, in going down the group), whereas, for the other groups, the lightest member has an unexpectedly high boiling point.
Why? The answer lies in the especially large hydrogen bonding interactions that exist among the smallest molecules with the most polar XOH bonds. These unusually strong hydrogen bonding forces are due primarily to two factors. One factor is the relatively large electronegativity values of the lightest elements in each group, which leads to especially polar XOH bonds. The second factor is the small size of the first element of each group, which allows for the close approach of the dipoles, further strengthening the intermolec-ular forces. Because the interactions among the molecules containing the lightest elements in Groups 5A and 6A are so strong, an unusually large quantity of energy must be supplied FIGURE 10.2 (a) The electrostatic interaction of two polar molecules. (b) The interaction of many dipoles in a condensed state.
FIGURE 10.3 (a) The polar water molecule. (b) Hydrogen bonding among water molecules. Note that the small size of the hydrogen atom allows for close interactions.
FIGURE 10.4 The boiling points of the covalent hydrides of the elements in Groups 4A, 5A, 6A, and 7A.
+ – + – + – + – – + + – + – – + – + – + Attraction Repulsion (a) (b) H (a) (b) 2δ – δ + δ + O H H H O H H O H H O H O H O H H O H O H H H O H H H H O H H O –100 0 100 –200 Boiling point (°C) Period 2 3 4 5 H2O Group 6A Group 7A Group 5A Group 4A HF NH3 CH4 SiH4 GeH4 SnH4 HI SbH3 H2Te H2S H2Se HCl AsH3 HBr PH3 428 Chapter Ten Liquids and Solids to overcome these interactions and separate the molecules to produce the gaseous state.
These molecules will remain together in the liquid state even at high temperatures—hence the very high boiling points.
Hydrogen bonding is also important in organic molecules (molecules with a carbon chain backbone). For example, the alcohols methanol (CH3OH) and ethanol (CH3CH2OH) have much higher boiling points than would be expected from their molar masses because of the polar OOH bonds in these molecules, which produce hydrogen bonding.
London Dispersion Forces Even molecules without dipole moments must exert forces on each other. We know this be-cause all substances—even the noble gases—exist in the liquid and solid states under cer-tain conditions. The forces that exist among noble gas atoms and nonpolar molecules are called London dispersion forces. To understand the origin of these forces, let’s consider a pair of noble gas atoms. Although we usually assume that the electrons of an atom are uni-formly distributed about the nucleus, this is apparently not true at every instant. As the elec-trons move about the nucleus, a momentary nonsymmetrical electron distribution can de-velop that produces a temporary dipolar arrangement of charge. The formation of this temporary dipole can, in turn, affect the electron distribution of a neighboring atom. That is, this instantaneous dipole that occurs accidentally in a given atom can then induce a sim-ilar dipole in a neighboring atom, as represented in Fig. 10.5(a). This phenomenon leads to an interatomic attraction that is relatively weak and short-lived but that can be very signif-icant especially for large atoms (see below). For these interactions to become strong enough to produce a solid, the motions of the atoms must be greatly slowed down. This explains, for instance, why the noble gas elements have such low freezing points (see Table 10.2).
Note from Table 10.2 that the freezing point rises going down the group. The prin-cipal cause for this trend is that as the atomic number increases, the number of electrons increases, and there is an increased chance of the occurrence of momentary dipole inter-actions. We describe this phenomenon using the term polarizability, which indicates the Boiling point will be defined precisely in Section 10.8.
FIGURE 10.5 (a) An instantaneous polarization can occur on atom A, creating an instantaneous di-pole. This dipole creates an induced dipole on neighboring atom B. (b) Nonpolar mole-cules such as H2 also can develop instanta-neous and induced dipoles.
TABLE 10.2 The Freezing Points of the Group 8A Elements Element Freezing Point (C) Helium 269.7 Neon 248.6 Argon 189.4 Krypton 157.3 Xenon 111.9 Helium is the only element that will not freeze by lowering its temperature at 1 atm.
Pressure must be applied to freeze helium.
δ+ δ– δ+ δ– δ+ δ– No polarization Instantaneous dipole on molecule A induces a dipole on molecule B (b) Molecule A Molecule B Molecule A Molecule B Molecule A Molecule B H H H H H H Atom A Atom B No polarization Atom A Atom B Instantaneous dipole on atom A induces a dipole on atom B Atom A Atom B (a) + + + δ+ δ– + δ+ δ– + + H H H H H H δ+ δ– 10.2 The Liquid State 429 ease with which the electron “cloud” of an atom can be distorted to give a dipolar charge distribution. Thus we say that large atoms with many electrons exhibit a higher polariz-ability than small atoms. This means that the importance of London dispersion forces in-creases greatly as the size of the atom increases.
These same ideas also apply to nonpolar molecules such as H2, CH4, CCl4, and CO2 [see Fig. 10.5(b)]. Since none of these molecules has a permanent dipole moment, their principal means of attracting each other is through London dispersion forces.
10.2 The Liquid State Liquids and liquid solutions are vital to our lives. Of course, water is the most important liquid. Besides being essential to life, water provides a medium for food preparation, for transportation, for cooling in many types of machines and industrial processes, for recre-ation, for cleaning, and for a myriad of other uses.
Liquids exhibit many characteristics that help us understand their nature. We have already mentioned their low compressibility, lack of rigidity, and high density compared with gases.
Many of the properties of liquids give us direct information about the forces that exist among the particles. For example, when a liquid is poured onto a solid surface, it tends to bead as droplets, a phenomenon that depends on the intermolecular forces. Although molecules in the interior of the liquid are completely surrounded by other molecules, those at the liquid sur-face are subject to attractions only from the side and from below (Fig. 10.6). The effect of this uneven pull on the surface molecules tends to draw them into the body of the liquid and causes a droplet of liquid to assume the shape that has the minimum surface area—a sphere.
To increase a liquid’s surface area, molecules must move from the interior of the liq-uid to the surface. This requires energy, since some intermolecular forces must be over-come. The resistance of a liquid to an increase in its surface area is called the surface tension of the liquid. As we would expect, liquids with relatively large intermolecular forces, such as those with polar molecules, tend to have relatively high surface tensions.
Polar liquids typically exhibit capillary action, the spontaneous rising of a liquid in a narrow tube. Two different types of forces are responsible for this property: cohesive forces, the intermolecular forces among the molecules of the liquid, and adhesive forces, the forces between the liquid molecules and their container. We have already seen how cohesive forces operate among polar molecules. Adhesive forces occur when a container is made of a sub-stance that has polar bonds. For example, a glass surface contains many oxygen atoms with partial negative charges that are attractive to the positive end of a polar molecule such as water. This ability of water to “wet” glass makes it creep up the walls of the tube where the water surface touches the glass. This, however, tends to increase the surface area of the wa-ter, which is opposed by the cohesive forces that try to minimize the surface. Thus, because water has both strong cohesive (intermolecular) forces and strong adhesive forces to glass, it “pulls itself” up a glass capillary tube (a tube with a small diameter) to a height where the weight of the column of water just balances the water’s tendency to be attracted to the glass surface. The concave shape of the meniscus (see Fig. 10.7) shows that water’s adhe-sive forces toward the glass are stronger than its cohesive forces. A nonpolar liquid such as mercury (see Fig. 10.7) shows a convex meniscus. This behavior is characteristic of a liq-uid in which the cohesive forces are stronger than the adhesive forces toward glass.
Another property of liquids strongly dependent on intermolecular forces is viscosity, a measure of a liquid’s resistance to flow. As might be expected, liquids with large inter-molecular forces tend to be highly viscous. For example, glycerol, whose structure is The dispersion forces in molecules with large atoms are quite significant and are often actually more important than dipole–dipole forces.
For a given volume, a sphere has a smaller surface area than any other shape.
Surface tension: The resistance of a liquid to an increase in its surface area.
The composition of glass is discussed in Section 10.5.
FIGURE 10.6 A molecule in the interior of a liquid is attracted by the molecules surrounding it, whereas a molecule at the surface of a liquid is attracted only by molecules below it and on each side.
Surface Viscosity: A measure of a liquid’s resis-tance to flow.
430 Chapter Ten Liquids and Solids has an unusually high viscosity due mainly to its high capacity to form hydrogen bonds using its OOH groups (see margin).
Molecular complexity also leads to higher viscosity because very large molecules can become entangled with each other. For example, gasoline, a nonviscous liquid, contains hydrocarbon molecules of the type CH3O(CH2)nOCH3, where n varies from about 3 to 8. However, grease, which is very viscous, contains much larger hydrocarbon molecules in which n varies from 20 to 25.
Structural Model for Liquids In many respects, the development of a structural model for liquids presents greater chal-lenges than the development of such a model for the other two states of matter. In the gaseous state the particles are so far apart and are moving so rapidly that intermolecular forces are negligible under most circumstances. This means that we can use a relatively simple model for gases. In the solid state, although the intermolecular forces are large, the molecular motions are minimal, and fairly simple models are again possible. The liq-uid state, however, has both strong intermolecular forces and significant molecular motions. Such a situation precludes the use of really simple models for liquids. Recent advances in spectroscopy, the study of the manner in which substances interact with electromagnetic radiation, make it possible to follow the very rapid changes that occur in liquids. As a result, our models of liquids are becoming more accurate. As a starting point, a typical liquid might best be viewed as containing a large number of regions where the arrangements of the components are similar to those found in the solid, but with more disorder, and a smaller number of regions where holes are present. The situation is highly dynamic, with rapid fluctuations occurring in both types of regions.
10.3 An Introduction to Structures and Types of Solids There are many ways to classify solids, but the broadest categories are crystalline solids, those with a highly regular arrangement of their components, and amorphous solids, those with considerable disorder in their structures.
Beads of water on a waxed car finish. The nonpolar component of the wax causes the water to form approximately spherical droplets.
FIGURE 10.7 Nonpolar liquid mercury forms a convex meniscus in a glass tube, whereas polar water forms a concave meniscus.
Glycerol 10.3 An Introduction to Structures and Types of Solids 431 The regular arrangement of the components of a crystalline solid at the microscopic level produces the beautiful, characteristic shapes of crystals, such as those shown in Fig. 10.8. The positions of the components in a crystalline solid are usually represented by a lattice, a three-dimensional system of points designating the positions of the com-ponents (atoms, ions, or molecules) that make up the substance. The smallest repeating unit of the lattice is called the unit cell. Thus a particular lattice can be generated by repeating the unit cell in all three dimensions to form the extended structure. Three com-mon unit cells and their lattices are shown in Fig. 10.9. Note from Fig. 10.9 that the extended structure in each case can be viewed as a series of repeating unit cells that share common faces in the interior of the solid.
Although we will concentrate on crystalline solids in this book, there are many im-portant noncrystalline (amorphous) materials. An example is common glass, which is best pictured as a solution in which the components are “frozen in place” before they can achieve an ordered arrangement. Although glass is a solid (it has a rigid shape), a great deal of disorder exists in its structure.
X-Ray Analysis of Solids The structures of crystalline solids are most commonly determined by X-ray diffraction.
Diffraction occurs when beams of light are scattered from a regular array of points in which the spacings between the components are comparable with the wavelength of the light. Diffraction is due to constructive interference when the waves of parallel beams are in phase and to destructive interference when the waves are out of phase.
When X rays of a single wavelength are directed at a crystal, a diffraction pattern is obtained, as we saw in Fig. 7.5. The light and dark areas on the photographic plate occur because the waves scattered from various atoms may reinforce or cancel each other (see Fig. 10.10). The key to whether the waves reinforce or cancel is the difference in distance traveled by the waves after they strike the atoms. The waves are in phase before they are reflected, so if the difference in distance traveled is an integral number of wavelengths, the waves will still be in phase.
Since the distance traveled depends on the distance between the atoms, the diffraction pattern can be used to determine the interatomic spacings. The exact relationship can be worked out using the diagram in Fig. 10.11, which shows two in-phase waves being reflected by atoms in two different layers in a crystal. The extra distance traveled by the lower wave is the sum of the distances xy and yz, and the waves will be in phase after reflection if (10.1) xy yz nl FIGURE 10.8 Two crystalline solids: pyrite (left), amethyst (right).
where n is an integer and is the wavelength of the X rays. Using trigonometry (see Fig. 10.11), we can show that (10.2) where d is the distance between the atoms and is the angle of incidence and reflection.
Combining Equation (10.1) and Equation (10.2) gives (10.3) Equation (10.3) is called the Bragg equation after William Henry Bragg (1862–1942) and his son William Lawrence Bragg (1890–1972), who shared the Nobel Prize in physics in 1915 for their pioneering work in X-ray crystallography.
nl 2d sin u xy yz 2d sin u 432 Chapter Ten Liquids and Solids FIGURE 10.9 Three cubic unit cells and the corresponding lattices. Note that only parts of spheres on the corners and faces of the unit cells reside inside the unit cell, as shown by the “cutoff” versions.
Simple cubic Body-centered cubic Face-centered cubic (c) (b) (a) Unit cell Lattice Space-filling unit cell Example Polonium metal Uranium metal Gold metal 10.3 An Introduction to Structures and Types of Solids 433 A diffractometer is a computer-controlled instrument used for carrying out the X-ray analysis of crystals. It rotates the crystal with respect to the X-ray beam and collects the data produced by the scattering of the X rays from the various planes of atoms in the crys-tal. The results are then analyzed by computer.
The techniques for crystal structure analysis have reached a level of sophistication that allows the determination of very complex structures, such as those important in biological systems. For example, the structures of several enzymes have been determined, thus en-abling biochemists to understand how they perform their functions. We will explore this topic further in Chapter 12. Using X-ray diffraction, we can gather data on bond lengths and angles and in so doing can test the predictions of our models of molecular geometry.
Using the Bragg Equation X rays of wavelength 1.54 Å were used to analyze an aluminum crystal. A reflection was produced at degrees. Assuming n 1, calculate the distance d between the planes of atoms producing this reflection.
Solution To determine the distance between the planes, we use Equation (10.3) with n 1, 1.54 Å, and 19.3 degrees. Since 2d sin n , See Exercises 10.41 through 10.44.
d nl 2 sin u 11211.54 Å2 12210.33052 2.33 Å 233 pm u 19.3 FIGURE 10.10 X rays scattered from two different atoms may reinforce (constructive interference) or cancel (destructive interference) one another. (a) Both the incident rays and the reflected rays are also in phase. In this case, d1 is such that the difference in the distances traveled by the two rays is a whole number of wavelengths. (b) The incident rays are in phase but the reflected rays are exactly out of phase. In this case d2 is such that the difference in distances traveled by the two rays is an odd number of half wavelengths.
Graduate student Maria Zhuravlera operat-ing an X-ray diffractometer at Michigan State University.
FIGURE 10.11 Reflection of X rays of wavelength from a pair of atoms in two different layers of a crystal. The lower wave travels an extra dis-tance equal to the sum of xy and yz. If this distance is an integral number of wave-lengths (n 1, 2, 3, . . .), the waves will reinforce each other when they exit the crystal.
Sample Exercise 10.1 d1 In-phase In-phase (a) d2 In-phase Out of phase (b) d x z y θ θ θ θ Incident rays Reflected rays 434 Chapter Ten Liquids and Solids CHEMICAL IMPACT Smart Fluids M atter seems to be getting smarter these days. Increas-ingly, we have discovered materials that can remember their initial shape after being deformed or can sense and re-spond to their environment. In particular, valuable new ma-terials have been formulated whose properties can be changed instantly by applying a magnetic or electric field.
One example of such a substance is a fluid whose flow characteristics (rheology) can be changed from free flowing to almost solid in about 0.01 second by the application of an electromagnetic field. This “magnetorheological” (MR) fluid was developed by Lord Corporation. Working in collaboration with Delphi Corporation, the company is applying the fluid in suspension control of General Motors automobiles such as Cadillacs and Corvettes. The so-called Magneride system has sensors that monitor the road surface and provide information about what suspension damping is needed. In response, a mes-sage is instantly sent to an electromagnetic coil in the shock absorbers, which adjusts the viscosity of the MR fluid to provide continuously variable damping. The result: an amaz-ingly smooth ride and unerring road-holding ability.
The MR fluid is composed of a synthetic oil in which particles of an iron-containing compound are suspended. When the magnetic field is turned off, these particles flow freely in all directions (see the figure above). When the field is turned on, the particles aggregate into chains that line up perpendicular to the flow of the fluid, thereby increasing its viscosity in proportion to the strength of the applied field.
Many other applications of MR fluids besides auto sus-pensions are under development. For example, this technology is being used in a prosthesis (see below) for above-the-knee amputees, which gives them a more natural gait and improves stair climbing. One very large-scale application is in Japan’s National Museum of Emerging Science and Innovation, where an MR fluid is being used in dampers to protect the building against earthquake damage. Large MR-fluid dampers are also being used for stabilizing bridges such as the Dong Ting Lake Bridge in China’s Hunan province to steady it in high winds.
Magnetic field off Magnetic particles flow randomly Magnetic field on Applied field (H) creates structure that increases viscosity H Flow Flow This High Intelligence Prosthesis for the knee uses an MR fluid damper to provide motion that closely duplicates the natural movement of the knee joint.
10.3 An Introduction to Structures and Types of Solids 435 Types of Crystalline Solids There are many different types of crystalline solids. For example, although both sugar and salt dissolve readily in water, the properties of the resulting solutions are quite different.
The salt solution readily conducts an electric current, whereas the sugar solution does not.
This behavior arises from the nature of the components in these two solids. Common salt (NaCl) is an ionic solid; it contains and ions. When solid sodium chloride dis-solves in the polar water, sodium and chloride ions are distributed throughout the result-ing solution and are free to conduct electric current. Table sugar (sucrose), on the other hand, is composed of neutral molecules that are dispersed throughout the water when the solid dissolves. No ions are present, and the resulting solution does not conduct electric-ity. These examples illustrate two important types of solids: ionic solids, represented by sodium chloride, and molecular solids, represented by sucrose. Ionic solids have ions at the points of the lattice that describes the structure of the solid. A molecular solid, on the other hand, has discrete covalently bonded molecules at each of its lattice points. Ice is a molecular solid that has an H2O molecule at each point (see Fig. 10.12).
A third type of solid is represented by elements such as carbon (which exists in the forms graphite, diamond, and the fullerenes), boron, silicon, and all metals. These substances all have atoms at the lattice points that describe the structure of the solid. Therefore, we call solids of this type atomic solids. Examples of these three types of solids are shown in Fig. 10.12.
Cl Na Buckminsterfullerene, C60, is a particular member of the fullerene family.
FIGURE 10.12 Examples of three types of crystalline solids. Only part of the structure is shown in each case. (a) An atomic solid. (b) An ionic solid. (c) A molecular solid. The dotted lines show the hydrogen bonding interactions among the polar water molecules. Cl– Na+ Sodium chloride (b) H2O C Diamond Ice (a) (c) 436 Chapter Ten Liquids and Solids To summarize, we find it convenient to classify solids according to what type of com-ponent occupies the lattice points. This leads to the classifications atomic solids (atoms at the lattice points), molecular solids (discrete, relatively small molecules at the lattice points), and ionic solids (ions at the lattice points). In addition, atomic solids are placed into the following subgroups based on the bonding that exists among the atoms in the solid: metal-lic solids, network solids, and Group 8A solids. In metallic solids, a special type of delo-calized nondirectional covalent bonding occurs. In network solids, the atoms bond to each other with strong directional covalent bonds that lead to giant molecules, or networks, of atoms. In the Group 8A solids, the noble gas elements are attracted to each other with London dispersion forces. The classification of solids is summarized in Table 10.3.
The markedly different bonding present in the various atomic solids leads to dra-matically different properties for the resulting solids. For example, although argon, cop-per, and diamond all are atomic solids, they have strikingly different properties. Argon (a Group 8A solid) has a very low melting point whereas diamond (a network solid) and copper (a metallic solid) melt at high temperatures (about 3500 and respectively). Copper is an excellent conductor of electricity, whereas argon and diamond are both insulators. Copper can be easily changed in shape; it is both malleable (can be formed into thin sheets) and ductile (can be pulled into a wire). Diamond, on the other hand, is the hardest natural substance known. We will explore the structure and bonding of atomic solids in the next two sections.
10.4 Structure and Bonding in Metals Metals are characterized by high thermal and electrical conductivity, malleability, and duc-tility. As we will see, these properties can be traced to the nondirectional covalent bond-ing found in metallic crystals.
A metallic crystal can be pictured as containing spherical atoms packed together and bonded to each other equally in all directions. We can model such a structure by packing uniform, hard spheres in a manner that most efficiently uses the available space. Such an arrangement is called closest packing. The spheres are packed in layers, as shown in Fig. 10.13, in which each sphere is surrounded by six others. In the second layer the spheres do not lie directly over those in the first layer. Instead, each one occupies an indentation (or dimple) formed by three spheres in the first layer. In the third layer the spheres can occupy the dimples of the second layer in two possible ways: They can occupy positions so that each sphere in the third layer lies directly over a sphere in the first layer (the aba arrangement; Fig. 10.13a), or they can occupy positions so that no sphere in the third layer lies over one in the first layer (the abc arrangement; Fig. 10.13b).
The aba arrangement has the hexagonal unit cell shown in Fig. 10.14, and the resulting structure is called the hexagonal closest packed (hcp) structure. The abc arrangement has a face-centered cubic unit cell, as shown in Fig. 10.15, and the resulting structure is 1083°C, 1189°C2, The internal forces in a solid determine the properties of the solid.
The closest packing model for metallic crystals assumes that metal atoms are uniform, hard spheres.
TABLE 10.3 Classification of Solids Atomic Solids Metallic Network Group 8A Molecular Solids Ionic Solids Components That Occupy Metal Nonmetal Group 8A Discrete molecules Ions the Lattice Points: atoms atoms atoms Bonding: Delocalized Directional covalent London Dipole–dipole and/or Ionic covalent (leading to dispersion London dispersion giant molecules) forces forces Visualization: Electron Sea Model 10.4 Structure and Bonding in Metals 437 FIGURE 10.13 The closest packing arrangement of uni-form spheres. In each layer a given sphere is surrounded by six others, creating six dimples, only three of which can be occu-pied in the next layer. (a) aba packing: The second layer is like the first, but it is dis-placed so that each sphere in the second layer occupies a dimple in the first layer.
The spheres in the third layer occupy dim-ples in the second layer so that the spheres in the third layer lie directly over those in the first layer (aba). (b) abc packing: The spheres in the third layer occupy dimples in the second layer so that no spheres in the third layer lie above any in the first layer (abc). The fourth layer is like the first.
(a) abab — Closest packing (b) abca — Closest packing Top view Top view Side view Top view Top view Top view Side view a b a Atom in third layer lies over atom in first layer.
Top view Side view Unit cell a c b a a c b a An atom in every fourth layer lies over an atom in the first layer.
Unit cell FIGURE 10.14 When spheres are closest packed so that the spheres in the third layer are directly over those in the first layer (aba), the unit cell is the hexagonal prism illustrated here in red.
FIGURE 10.15 When spheres are packed in the abc arrangement, the unit cell is face-centered cubic. To make the cubic arrangement eas-ier to see, the vertical axis has been tilted as shown.
438 Chapter Ten Liquids and Solids CHEMICAL IMPACT Seething Surfaces W hen we picture a solid, we think of the particles as be-ing packed closely together with relatively little mo-tion. The particles are thought to vibrate randomly about their positions but stay in nearly the same place. Recent re-search, however, indicates surface particles are a great deal more mobile than was previously thought. Independent teams of scientists from the University of Leiden in the Netherlands and Sandia National Laboratory in New Mexico have found a surprising amount of atom-swapping occurring on the surface of a copper crystal.
The Dutch scientist Raoul van Gastel and his colleagues used a scanning tunneling microscope (STM) to study the surface of a copper crystal containing indium atom impuri-ties. They noted that a given patch of surface would stay the same for several scans and then, suddenly, the indium atoms would appear at different places. Surprisingly, the indium atoms seemed to make “long jumps,” moving as many as five atom positions between scans. The most likely expla-nation for these movements is a “hole” created by a copper atom escaping the surface. This hole moves around as other atoms shift to fill it in succession (see accompanying fig-ure). The best analogy to the movement of the hole is the toy slide puzzle with 15 numbered pieces and one missing piece in a array. The object of the game is to slide a piece into the hole and then repeat the process until the num-bers appear in order.
The hole on the copper surface moves very fast—up to 100 million times per second—shuffling copper atoms and allowing the indium atoms to change positions. Van Gastel believes that all of the observed motion results from just a few fast-moving holes. In fact, he suggests that just one in 6 billion copper atoms is missing at a given time, analo-gous to one person in the entire earth’s population. Its absence causes a given atom on the surface to move every 30 or 40 seconds. Brian Swartzentruber of Sandia National Laboratories came to similar conclusions using an STM to track the movement of palladium atoms on a copper surface.
These results have important implications. For exam-ple, metal surfaces are often used to speed up particular re-actions. The motions on the metal surface could significantly influence the way that reactants interact with the surface.
Also, a lot of effort is now being expended to construct tiny “machines” (called nanoscale devices) by assembling indi-vidual atoms on a solid surface. These devices could be lit-erally torn apart by excess surface motions.
4 4 A toy slide puzzle.
A section of a surface containing copper atoms (red) and an indium atom (yellow). A hole due to a missing copper atom is shown on the left.
The blue line on the right shows the movement of this hole. As an atom moves to fill the hole, the hole moves as well. In the process, the indium atom jumps to a new position.
10.4 Structure and Bonding in Metals 439 (a) (b) (c) atom atom 1 2 1 8 FIGURE 10.17 The net number of spheres in a face-centered cubic unit cell. (a) Note that the sphere on a corner of the col-ored cell is shared with 7 other unit cells (a total of 8). Thus of such a sphere lies within a given unit cell.
Since there are 8 corners in a cube, there are 8 of these pieces, or 1 net sphere. (b) The sphere on the cen-ter of each face is shared by 2 unit cells, and thus each unit cell has of each of these types of spheres.
There are 6 of these spheres to give 3 net spheres. (c) Thus the face-centered cubic unit cell contains 4 net spheres (all of the pieces can be assembled to give 4 spheres).
1 2 1 2 1 8 1 8 called the cubic closest packed (ccp) structure. Note that in the hcp structure the spheres in every other layer occupy the same vertical position (ababab . . .), whereas in the ccp structure the spheres in every fourth layer occupy the same vertical position (abcabca . . .).
A characteristic of both structures is that each sphere has 12 equivalent nearest neighbors: 6 in the same layer, 3 in the layer above, and 3 in the layer below (that form the dimples).
This is illustrated for the hcp structure in Fig. 10.16.
Knowing the net number of spheres (atoms) in a particular unit cell is important for many applications involving solids. To illustrate how to find the net number of spheres in a unit cell, we will consider a face-centered cubic unit cell (Fig. 10.17).
Note that this unit cell is defined by the centers of the spheres on the cube’s corners.
Thus 8 cubes share a given sphere, so of this sphere lies inside each unit cell. Since a cube has 8 corners, there are pieces, or enough to put together 1 whole sphere.
The spheres at the center of each face are shared by 2 unit cells, so of each lies in-side a particular unit cell. Since the cube has 6 faces, we have pieces, or enough to construct 3 whole spheres. Thus the net number of spheres in a face-centered cubic unit cell is Calculating the Density of a Closest Packed Solid Silver crystallizes in a cubic closest packed structure. The radius of a silver atom is 144 pm.
Calculate the density of solid silver.
Solution Density is mass per unit volume. Thus we need to know how many silver atoms occupy a given volume in the crystal. The structure is cubic closest packed, which means the unit cell is face-centered cubic, as shown in the accompanying figure.
We must find the volume of this unit cell for silver and the net number of atoms it contains. Note that in this structure the atoms touch along the diagonals for each a8 1 8b a6 1 2b 4 6 1 2 1 2 8 1 8 1 8 b a b hcp 1 2 3 4 6 5 7 8 9 11 12 10 FIGURE 10.16 The indicated sphere has 12 nearest neighbors.
Sample Exercise 10.2 Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
face and not along the edges of the cube. Thus the length of the diagonal is r + 2r + r, or 4r. We use this fact to find the length of the edge of the cube by the Pythagorean theorem: Since r 144 pm for a silver atom, The volume of the unit cell is d3, which is (407 pm)3, or pm3. We convert this to cubic centimeters as follows: Since we know that the net number of atoms in the face-centered cubic unit cell is 4, we have 4 silver atoms contained in a volume of . The density is therefore See Exercises 10.45 through 10.48.
Examples of metals that form cubic closest packed solids are aluminum, iron, cop-per, cobalt, and nickel. Magnesium and zinc are hexagonal closest packed. Calcium and certain other metals can crystallize in either of these structures. Some metals, however, assume structures that are not closest packed. For example, the alkali metals have struc-tures characterized by a body-centered cubic (bcc) unit cell (see Fig. 10.9), where the spheres touch along the body diagonal of the cube. In this structure, each sphere has 8 near-est neighbors (count the number of atoms around the atom at the center of the unit cell), as compared with 12 in the closest packed structures. Why a particular metal adopts the structure it does is not well understood.
Bonding Models for Metals Any successful bonding model for metals must account for the typical physical proper-ties of metals: malleability, ductility, and the efficient and uniform conduction of heat and electricity in all directions. Although the shapes of most pure metals can be changed rel-atively easily, most metals are durable and have high melting points. These facts indicate that the bonding in most metals is both strong and nondirectional. That is, although it is difficult to separate metal atoms, it is relatively easy to move them, provided the atoms stay in contact with each other.
The simplest picture that explains these observations is the electron sea model, which envisions a regular array of metal cations in a “sea” of valence electrons (see Fig. 10.18). The mobile electrons can conduct heat and electricity, and the metal ions can be easily moved around as the metal is hammered into a sheet or pulled into a wire.
10.6 g/cm3 Density mass volume 14 atoms21107.9 g/mol211 mol 6.022 1023 atoms2 6.74 1023 cm3 6.74 1023 cm3 6.74 107 pm3 a1.00 1010 cm pm b 3 6.74 1023 cm3 6.74 107 d 1144 pm21182 407 pm d 28r2 r28 d2 8r2 2d2 16r2 d2 d2 14r22 Malleable: Can be pounded into thin sheets.
Ductile: Can be drawn to form a wire.
440 Chapter Ten Liquids and Solids Crystalline silver contains cubic closest packed silver atoms.
r 2r r d d d 4r 4r d d 10.4 Structure and Bonding in Metals 441 CHEMICAL IMPACT Closest Packing of M & Ms A lthough we usually think of scientists as dealing with es-oteric and often toxic materials, sometimes they surprise us. For example, scientists at several prestigious universities have lately shown a lot of interest in M & M candies.
To appreciate the scientists’interest in M & Ms, we must consider the importance of packing atoms, molecules, or mi-crocrystals in understanding the structures of solids. The most efficient use of space is the closest packing of uniform spheres, where 74% of the space is occupied by the spheres and 26% of space is left unoccupied. Although the structures of most pure metals can be explained in terms of closest packing, most other substances—such as many alloys and ceramics—consist of random arrays of microscopic particles.
For this reason, it is of interest to study how such objects pack in a random way.
When uniform spheres, such as marbles, are poured into a large container, the resulting random packing of the spheres results in only 64% of the space being occupied by the spheres. Thus it was very surprising when Princeton Uni-versity chemist Salvatore Torquato and his colleagues at Cornell and North Carolina Central Universities discovered that, when the ellipsoidal-shaped M & Ms are poured into a large container, the candies occupy 73.5% of the available space. In other words, the randomly packed M & Ms occupy space with almost the same efficiency as closest packed spheres do.
Why do randomly packed ellipsoids occupy space so much more efficiently than randomly packed spheres? The scientists speculate that because the ellipsoids can tip and ro-tate in ways that spheres cannot, they can pack more closely to their neighbors.
According to Torquato, these results are important be-cause they will help us better understand the properties of disordered materials ranging from powders to glassy solids.
He also says that M & Ms make ideal test objects because they are inexpensive and uniform and “you can eat the experiment afterward.” + e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– e– 2+ (a) (b) + + + + + + + 2+ 2+ 2+ 2+ 2+ 2+ 2+ FIGURE 10.18 The electron sea model for metals postu-lates a regular array of cations in a “sea” of valence electrons. (a) Representation of an alkali metal (Group 1A) with one valence electron. (b) Representation of an alkaline earth metal (Group 2A) with two valence electrons.
A related model that gives a more detailed view of the electron energies and motions is the band model, or molecular orbital (MO) model, for metals. In this model, the elec-trons are assumed to travel around the metal crystal in molecular orbitals formed from the valence atomic orbitals of the metal atoms (Fig. 10.19).
Recall that in the MO model for the gaseous Li2 molecule (Section 9.3), two widely spaced molecular orbital energy levels (bonding and antibonding) result when two identical atomic orbitals interact. However, when many metal atoms interact, 442 Chapter Ten Liquids and Solids as in a metal crystal, the large number of resulting molecular orbitals become more closely spaced and finally form a virtual continuum of levels, called bands, as shown in Fig. 10.19.
As an illustration, picture a magnesium metal crystal, which has an hcp structure.
Since each magnesium atom has one 3s and three 3p valence atomic orbitals, a crystal with n magnesium atoms has available n(3s) and 3n(3p) orbitals to form the molecular orbitals, as illustrated in Fig. 10.20. Note that the core electrons are localized, as shown by their presence in the energy “well” around each magnesium atom. However, the valence electrons occupy closely spaced molecular orbitals, which are only partially filled.
The existence of empty molecular orbitals close in energy to filled molecular orbitals explains the thermal and electrical conductivity of metal crystals. Metals conduct electricity and heat very efficiently because of the availability of highly mobile electrons. For ex-ample, when an electric potential is placed across a strip of metal, for current to flow, electrons must be free to move. In the band model for metals, the electrons in partially filled bonds are mobile. These conduction electrons are free to travel throughout the metal crystal as dictated by the potential imposed on the metal. The molecular orbitals occupied by these conducting electrons are called conduction bands. These mobile electrons also account for the efficiency of the conduction of heat through metals. When one end of a metal rod is heated, the mobile electrons can rapidly transmit the thermal energy to the other end.
Metal Alloys Because of the nature of the structure and bonding of metals, other elements can be in-troduced into a metallic crystal relatively easily to produce substances called alloys. An alloy is best defined as a substance that contains a mixture of elements and has metallic properties. Alloys can be conveniently classified into two types.
1s 2s 2p Magnesium atoms 3s 3p Energy Filled MOs 12+ Empty MOs 12+ 12+ 12+ 12+ FIGURE 10.20 (left) A representation of the energy levels (bands) in a magnesium crystal. The electrons in the 1s, 2s, and 2p orbitals are close to the nuclei and thus are localized on each magnesium atom as shown. However, the 3s and 3p valence orbitals overlap and mix to form molecular orbitals.
Electrons in these energy levels can travel throughout the crystal. (right) Crystals of magnesium grown from a vapor.
FIGURE 10.19 The molecular orbital energy levels pro-duced when various numbers of atomic orbitals interact. Note that for two atomic orbitals two rather widely spaced energy levels result. (Recall the description of H2 in Section 9.2.) As more atomic orbitals are available to form molecular orbitals, the resulting energy levels are more closely spaced, finally producing a band of very closely spaced orbitals.
Energy 2 4 16 6.02 × 1023 Number of interacting atomic orbitals 10.4 Structure and Bonding in Metals 443 CHEMICAL IMPACT What Sank the Titanic?
O n April 12, 1912, the steamship Titanic struck an iceberg in the North Atlantic approximately 100 miles south of the Grand Banks of Newfoundland and within 3 hours was resting on the bottom of the ocean. Of her more than 2300 passengers and crew, over 1500 lost their lives. While the tragic story of the Titanic has never faded from the minds and imaginations of the generations that followed, the 1985 discovery of the wreck by a joint Franco-American expedi-tion at a depth of 12,612 feet rekindled the world’s interest in the “greatest oceangoing vessel” ever built. The discovery also would reveal important scientific clues as to why and how the Titanic sank so quickly in the frigid waters of the North Atlantic.
The Titanic was designed to be virtually “unsinkable,” and even in the worst-case scenario, a head-on collision with another ocean liner, the ship was engineered to take from one to three days to sink. Thus its quick trip to the bottom has puzzled scientists for years. In 1991, Steve Blasco, an ocean-floor geologist for the Canadian Department of Nat-ural Resources, led a scientific expedi-tion to the wreck. On one of 17 dives to the site, Blasco’s team recovered a piece of steel that appeared to be a part of the Titanic’s hull. Unlike modern steel, which would have shown evidence of bending in a collision, the steel recov-ered from the Titanic appeared to have shattered on impact with the iceberg.
This suggested that the metal might not have been as ductile (ductility is the abil-ity to stretch without breaking) as it should have been. In 1994, tests were conducted on small pieces of metal, called coupons, cut from the recovered piece of hull. These samples shattered without bending. Further analysis showed that the steel used to construct the hull of the Titanic was high in sul-fur content, and it is known that sulfur occlusions tend to make steel more brittle. This evidence suggests that the quality of the steel used to make the hull of the Titanic may very well have been an important factor that led to the rapid sinking of the ship.
But—not so fast. The Titanic continues to provoke con-troversy. A team of naval engineers and scientists recently have concluded that it was not brittle steel but faulty rivets that doomed the Titanic. During expeditions in 1996 and 1998 conducted by RMS Titanic, Inc., more samples of Titanic’s steel and rivets were obtained for further study.
Analysis of these samples by a team headed by Tim Foecke of the National Institute of Standards and Technology (NIST) shows that the rivets contain three times the expected amount of silicate slag. Foecke and his colleagues argue that the high slag content resulted in weak rivets that snapped in large numbers when the collision occurred, mortally wound-ing the ship.
What sank the Titanic? It hit an iceberg. The details re-main to be figured out.
Bow of the Titanic under 2 miles of water.
1 2 444 Chapter Ten Liquids and Solids In a substitutional alloy some of the host metal atoms are replaced by other metal atoms of similar size. For example, in brass, approximately one-third of the atoms in the host copper metal have been replaced by zinc atoms, as shown in Fig. 10.21(a). Sterling silver (93% silver and 7% copper), pewter (85% tin, 7% copper, 6% bismuth, and 2% antimony), and plumber’s solder (95% tin and 5% antimony) are other examples of substitutional alloys.
An interstitial alloy is formed when some of the interstices (holes) in the closest packed metal structure are occupied by small atoms, as shown in Fig. 10.21(b). Steel, the best-known interstitial alloy, contains carbon atoms in the holes of an iron crystal.
The presence of the interstitial atoms changes the properties of the host metal. Pure iron is relatively soft, ductile, and malleable due to the absence of directional bonding. The spherical metal atoms can be rather easily moved with respect to each other. However, when carbon, which forms strong directional bonds, is introduced into an iron crystal, the presence of the directional carbon–iron bonds makes the resulting alloy harder, stronger, and less ductile than pure iron. The amount of carbon directly affects the prop-erties of steel. Mild steels, containing less than 0.2% carbon, are ductile and malleable and are used for nails, cables, and chains. Medium steels, containing 0.2 to 0.6% carbon, are harder than mild steels and are used in rails and structural steel beams. High-carbon steels, containing 0.6 to 1.5% carbon, are tough and hard and are used for springs, tools, and cutlery.
Many types of steel also contain elements in addition to iron and carbon. Such steels are often called alloy steels, and they can be viewed as being mixed interstitial (carbon) and substitutional (other metals) alloys. Bicycle frames, for example, are constructed from a wide variety of alloy steels. The compositions of the two brands of steel tubing most commonly used in expensive racing bicycles are given in Table 10.4.
10.5 Carbon and Silicon: Network Atomic Solids Many atomic solids contain strong directional covalent bonds to form a solid that might best be viewed as a “giant molecule.” We call these substances network solids. In con-trast to metals, these materials are typically brittle and do not efficiently conduct heat or electricity. To illustrate network solids, in this section we will discuss two very important elements, carbon and silicon, and some of their compounds.
The two most common forms of carbon, diamond and graphite, are typical network solids. In diamond, the hardest naturally occurring substance, each carbon atom is sur-rounded by a tetrahedral arrangement of other carbon atoms to form a huge molecule [see Fig. 10.22(a)]. This structure is stabilized by covalent bonds, which, in terms of the localized electron model, are formed by the overlap of sp3 hybridized carbon atomic orbitals.
It is also useful to consider the bonding among the carbon atoms in diamond in terms of the molecular orbital model. Energy-level diagrams for diamond and a typical metal are given in Fig. 10.23. Recall that the conductivity of metals can be explained by FIGURE 10.21 Two types of alloys.
copper zinc Brass (a) iron carbon Steel (b) TABLE 10.4 The Composition of the Two Brands of Steel Tubing Commonly Used to Make Lightweight Racing Bicycles Brand of Tubing % C % Si % Mn % Mo % Cr Reynolds 0.25 0.25 1.3 0.20 — Columbus 0.25 0.30 0.65 0.20 1.0 Diamond (a) Graphite (b) Weak bonding between layers FIGURE 10.22 The structures of diamond and graphite. In each case only a small part of the entire structure is shown.
Visualization: Homogeneous Mixtures: Air and Brass Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
10.5 Carbon and Silicon: Network Atomic Solids 445 postulating that electrons are excited from filled levels into the very near empty levels, or conduction bands. However, note that in the energy-level diagram for diamond there is a large gap between the filled and the empty levels. This means that electrons cannot be transferred easily to the empty conduction bands. As a result, diamond is not expected to be a good electrical conductor. In fact, this prediction of the model agrees exactly with the observed behavior of diamond, which is known to be an electrical insulator—it does not conduct an electric current.
Graphite is very different from diamond. While diamond is hard, basically colorless, and an insulator, graphite is slippery, black, and a conductor. These differences, of course, arise from the differences in bonding in the two types of solids. In contrast to the tetrahedral arrangement of carbon atoms in diamond, the structure of graphite is based on layers of carbon atoms arranged in fused six-membered rings, as shown in Fig. 10.22(b). Each car-bon atom in a particular layer of graphite is surrounded by the three other carbon atoms in a trigonal planar arrangement with 120-degree bond angles. The localized electron model predicts sp2 hybridization in this case. The three sp2 orbitals on each carbon are used to form bonds with three other carbon atoms. One 2p orbital remains unhybridized on each carbon and is perpendicular to the plane of carbon atoms, as shown in Fig. 10.24.
C C C C C C C C (a) (b) C C C C C C C C C C C C C C C C C C FIGURE 10.24 The p orbitals (a) perpendicular to the plane of the carbon ring system in graphite can combine to form (b) an extensive -bonding network.
Visualization: Network Solids E (a) Empty MOs Filled MOs E Empty MOs Filled MOs (b) FIGURE 10.23 Partial representation of the molecular orbital energies in (a) diamond and (b) a typical metal.
Graphite and diamond, two forms of carbon.
446 Chapter Ten Liquids and Solids These orbitals combine to form a group of closely spaced molecular orbitals that are important in two ways. First, they contribute significantly to the stability of the graphite layers because of the bond formation. Second, the molecular orbitals with their delocalized electrons account for the electrical conductivity of graphite. These closely spaced orbitals are exactly analogous to the conduction bands found in metal crystals.
Graphite is often used as a lubricant in locks (where oil is undesirable because it col-lects dirt). The slipperiness that is characteristic of graphite can be explained by noting that graphite has very strong bonding within the layers of carbon atoms but little bonding between the layers (the valence electrons are all used to form and bonds among car-bons within the layers). This arrangement allows the layers to slide past one another quite readily. Graphite’s layered structure is shown in Fig. 10.25. This is in contrast to diamond, which has uniform bonding in all directions in the crystal.
Because of their extreme hardness, diamonds are used extensively in industrial cut-ting implements. Thus it is desirable to convert cheaper graphite to diamond. As we might expect from the higher density of diamond (3.5 g/cm3) compared with that of graphite (2.2 g/cm3), this transformation can be accomplished by applying very high pressures to graphite. The application of 150,000 atm of pressure at converts graphite virtu-ally completely to diamond. The high temperature is required to break the strong bonds in graphite so the rearrangement can occur.
Silicon is an important constituent of the compounds that make up the earth’s crust.
In fact, silicon is to geology as carbon is to biology. Just as carbon compounds are the basis for most biologically significant systems, silicon compounds are fundamental to most of the rocks, sands, and soils found in the earth’s crust. However, although carbon and silicon are next to each other in Group 4A of the periodic table, the carbon-based compounds of biology and the silicon-based compounds of geology have markedly dif-ferent structures. Carbon compounds typically contain long strings of carbon–carbon bonds, whereas the most stable silicon compounds involve chains with silicon–oxygen bonds.
The fundamental silicon–oxygen compound is silica, which has the empirical formula SiO2. Knowing the properties of the similar compound carbon dioxide, one might expect silica to be a gas that contains discrete SiO2 molecules. In fact, nothing could be further from the truth—quartz and some types of sand are typical of the materials composed of silica. What accounts for this difference? The answer lies in the bonding.
2800°C Computer-generated model of silica.
FIGURE 10.25 Graphite consists of layers of carbon atoms.
10.5 Carbon and Silicon: Network Atomic Solids 447 Recall that the Lewis structure of CO2 is and that each C O bond can be viewed as a combination of a bond involving a carbon sp hybrid orbital and a bond involving a carbon 2p orbital. On the contrary, silicon can-not use its valence 3p orbitals to form strong bonds with oxygen, mainly because of the larger size of the silicon atom and its orbitals, which results in less effective overlap with the smaller oxygen orbitals. Therefore, instead of forming bonds, the silicon atom satisfies the octet rule by forming single bonds with four oxygen atoms, as shown in the representation of the structure of quartz in Fig. 10.26. Note that each silicon atom is at the center of a tetrahedral arrangement of oxygen atoms, which are shared with other sil-icon atoms. Although the empirical formula for quartz is SiO2, the structure is based on a network of SiO4 tetrahedra with shared oxygen atoms rather than discrete SiO2 mole-cules. It is obvious that the differing abilities of carbon and silicon to form bonds with oxygen have profound effects on the structures and properties of CO2 and SiO2.
Compounds closely related to silica and found in most rocks, soils, and clays are the silicates. Like silica, the silicates are based on interconnected SiO4 tetrahedra. However, in contrast to silica, where the O/Si ratio is 2:1, silicates have O/Si ratios greater than 2:1 and contain silicon–oxygen anions. This means that to form the neutral solid silicates, cations are needed to balance the excess negative charge. In other words, silicates are salts containing metal cations and polyatomic silicon–oxygen anions. Examples of important silicate anions are shown in Fig. 10.27.
When silica is heated above its melting point (about C) and cooled rapidly, an amorphous solid called a glass results (see Fig. 10.28). Note that a glass contains 1600° p p p p s “ FIGURE 10.26 (top) The structure of quartz (empirical for-mula SiO2). Quartz contains chains of SiO4 tetrahedra (bottom) that share oxygen atoms.
O O O O Si The bonding in the CO2 molecule was de-scribed in Section 9.1.
(a) (b) SiO44– Si2O76– Si3O96– (Si4O11)n6n– Silicon Oxygen FIGURE 10.27 Examples of silicate anions, all of which are based on tetrahedra.
SiO4 4 FIGURE 10.28 Two-dimensional representations of (a) a quartz crystal and (b) a quartz glass.
448 Chapter Ten Liquids and Solids a good deal of disorder, in contrast to the crystalline nature of quartz. Glass more closely resembles a very viscous solution than it does a crystalline solid. Common glass results when substances such as Na2CO3 are added to the silica melt, which is then cooled. The properties of glass can be varied greatly by varying the additives. For example, addition of B2O3 produces a glass (called borosilicate glass) that expands and contracts little under large temperature changes. Thus it is useful for labware and cook-ing utensils. The most common brand name for this glass is Pyrex. The addition of K2O produces an especially hard glass that can be ground to the precise shapes needed for eyeglass and contact lenses. The compositions of several types of glass are shown in Table 10.5.
Ceramics Ceramics are typically made from clays (which contain silicates) and hardened by firing at high temperatures. Ceramics are nonmetallic materials that are strong, brittle, and resistant to heat and attack by chemicals.
Like glass, ceramics are based on silicates, but with that the resemblance ends. Glass can be melted and remelted as often as desired, but once a ceramic has been hardened, it is resistant to extremely high temperatures. This behavior results from the very different structures of glasses and ceramics. A glass is a homogeneous, noncrystalline “frozen so-lution,” and a ceramic is heterogeneous. A ceramic contains two phases: minute crystals of silicates that are suspended in a glassy cement.
To understand how ceramics harden, it is necessary to know something about the structure of clays. Clays are formed by the weathering action of water and carbon diox-ide on the mineral feldspar, which is a mixture of silicates with empirical formulas such as K2O Al2O3 6SiO2 and Na2O Al2O3 6SiO2. Feldspar is really an alumi-nosilicate in which aluminum as well as silicon atoms are part of the oxygen-bridged polyanion. The weathering of feldspar produces kaolinite, consisting of tiny thin platelets with the empirical formula Al2Si2O5(OH)4. When dry, the platelets cling together; when water is present, they can slide over one another, giving clay its plas-ticity. As clay dries, the platelets begin to interlock again. When the remaining water is driven off during firing, the silicates and cations form a glass that binds the tiny crystals of kaolinite.
Ceramics have a very long history. Rocks, which are natural ceramic materials, served as the earliest tools. Later, clay vessels dried in the sun or baked in fires served as containers for food and water. These early vessels were no doubt crude and quite porous. With the discovery of glazing, which probably occurred about 3000 B.C.
in Egypt, pottery became more serviceable as well as more beautiful. Prized porce-lain is essentially the same material as crude earthenware, but specially selected clays and glazings are used for porcelain and the clay object is fired at a very high temperature.
An artist paints a ceramic vase before glazing.
A glass pitcher being manufactured.
TABLE 10.5 Compositions of Some Common Types of Glass Percentages of Various Components Type of Glass SiO2 CaO Na2O B2O3 Al2O3 K2O MgO Window (soda-lime glass) 72 11 13 — 0.3 3.8 — Cookware (aluminosilicate glass) 55 15 — — 20 — 10 Heat-resistant (borosilicate glass) 76 3 5 13 2 0.5 — Optical 69 12 6 0.3 — 12 — 10.5 Carbon and Silicon: Network Atomic Solids 449 Although ceramics have been known since antiquity, they are not obsolete materi-als. On the contrary, ceramics constitute one of the most important classes of “high-tech” materials. Because of their stability at high temperatures and resistance to corrosion, ceramics seem an obvious choice for constructing jet and automobile engines in which the greatest fuel efficiencies are possible at very high temperatures. But ceramics are brittle—they break rather than bend—which limits their usefulness. However, more flex-ible ceramics can be obtained by adding small amounts of organic polymers. Taking their cue from natural “organoceramics” such as teeth and shells of sea creatures that contain small amounts of organic polymers, materials scientists have found that incorporating tiny amounts of long organic molecules into ceramics as they form produces materials that are much less subject to fracture. These materials should be useful for lighter, more durable engine parts, as well as for flexible superconducting wire and microelectronic devices. In addition, these organoceramics hold great promise for prosthetic devices such as artificial bones.
CHEMICAL IMPACT Golfing with Glass Y ou probably can guess what material traditionally was used to construct the “woods” used in golf. Modern tech-nology has changed things. Like the bats used in college baseball, most “woods” are now made of metal. While bats are made of aluminum, golf club heads are often made of stainless steel or titanium.
Metals and their alloys usually form crystals that con-tain highly ordered arrangements of atoms. However, a com-pany called Liquidmetal Golf of Laguna Niguel, California, has begun producing golf clubs containing glass—metallic glass. The company has found that when molten mixtures of titanium, zirconium, nickel, beryllium, and copper are cooled, they solidify, forming a glass. Unlike crystalline ma-terials that contain a regular array of atoms, glasses are amor-phous—the atoms are randomly scattered throughout the solid.
These golf clubs with metallic glass inserts have some unusual characteristics. Golfers who have tried the clubs say they combine hardness with a “soft feel.” Studies show that the glass transfers more of the energy of the golf swing to the ball with less impact to the golfer’s hands than with reg-ular metal woods.
One of the fortunate properties of this five-component metallic glass (invented in 1992 by William L. Johnson and Atakan Peker at the California Institute of Technology) is that it can be cooled relatively slowly to form the glass. This allows manufacture of relatively large glass objects such as inserts for golf club heads. Most mixtures of metals that form glasses must be cooled very rapidly to obtain the glass, which results in tiny particles of glass leading to powders.
David S. Lee, head of manufacturing at Liquidmetal Golf, says that golf clubs were an obvious first applica-tion for this five-component glass because golfers are used to paying high prices for clubs that employ new technology. Liquidmetal Golf is now looking for other applications of this new glass. How about glass bicycle frames?
Golf clubs with a titanium shell and metallic glass inserts.
450 Chapter Ten Liquids and Solids Semiconductors Elemental silicon has the same structure as diamond, as might be expected from its position in the periodic table (in Group 4A directly under carbon). Recall that in diamond there is a large energy gap between the filled and empty molecular orbitals (see Fig. 10.23). This gap prevents excitation of electrons to the empty molecular orbitals (conduction bands) and makes diamond an insulator. In silicon the situation is similar, but the energy gap is smaller. A few electrons can cross the gap at mak-ing silicon a semiconducting element, or semiconductor. In addition, at higher tem-peratures, where more energy is available to excite electrons into the conduction bands, the conductivity of silicon increases. This is typical behavior for a semiconducting element and is in contrast to that of metals, whose conductivity decreases with increasing temperature.
The small conductivity of silicon can be enhanced at normal temperatures if the silicon crystal is doped with certain other elements. For example, when a small fraction of silicon atoms is replaced by arsenic atoms, each having one more valence electron than silicon, extra electrons become available for conduction, as shown in Fig.
10.29(a). This produces an n-type semiconductor, a substance whose conductivity is increased by doping it with atoms having more valence electrons than the atoms in the host crystal. These extra electrons lie close in energy to the conduction bands and can be easily excited into these levels, where they can conduct an electric current [see Fig. 10.30(a)].
We also can enhance the conductivity of silicon by doping the crystal with an ele-ment such as boron, which has only three valence electrons, one less than silicon. Because boron has one less electron than is required to form the bonds with the surrounding silicon atoms, an electron vacancy, or hole, is created, as shown in Fig. 10.29(b). As an electron fills this hole, it leaves a new hole, and this process can be repeated. Thus the hole ad-vances through the crystal in a direction opposite to movement of the electrons jumping to fill the hole. Another way of thinking about this phenomenon is that in pure silicon each atom has four valence electrons and the low-energy molecular orbitals are exactly filled. Replacing silicon atoms with boron atoms leaves vacancies in these molecular orbitals, as shown in Fig. 10.30(b). This means that there is only one electron in some of the molecular orbitals, and these unpaired electrons can function as conducting electrons.
Thus the substance becomes a better conductor. When semiconductors are doped with atoms having fewer valence electrons than the atoms of the host crystal, they are called p-type semiconductors, so named because the positive holes can be viewed as the charge carriers.
Most important applications of semiconductors involve connection of a p-type and an n-type to form a p–n junction. Figure 10.31(a) shows a typical junction; the red dots 25°C, Electrons must be in singly occupied molecular orbitals to conduct a current.
E (a) Empty MOs (Conduction bands) Filled MOs “Excess” valence electrons (•) from donor impurity E (b) Empty MOs (Conduction bands) Electron vacancies ( ) due to the doping atoms FIGURE 10.30 Energy-level diagrams for (a) an n-type semiconductor and (b) a p-type semiconductor.
Si Si Si Si Si Si Si Si Si B Si Si Si Si Si Si Si Si B Si Si Si Si Si Si As Si Si Si Si Si Si Si As Si n-type semiconductor p-type semiconductor (a) (b) FIGURE 10.29 (a) A silicon crystal doped with arsenic, which has one more valence electron than silicon. (b) A silicon crystal doped with boron, which has one less electron than silicon.
Visualization: Magnetic Levi-tation by a Superconductor 10.5 Carbon and Silicon: Network Atomic Solids 451 represent excess electrons in the n-type semiconductor, and the white circles represent holes (electron vacancies) in the p-type semiconductor. At the junction, a small number of electrons migrate from the n-type region into the p-type region, where there are va-cancies in the low-energy molecular orbitals. The effect of these migrations is to place a negative charge on the p-type region (since it now has a surplus of electrons) and a pos-itive charge on the n-type region (since it has lost electrons, leaving holes in its low-energy molecular orbitals). This charge buildup, called the contact potential, or junction poten-tial, prevents further migration of electrons.
Now suppose an external electric potential is applied by connecting the negative ter-minal of a battery to the p-type region and the positive terminal to the n-type region. The situation represented in Fig. 10.31(b) results. Electrons are drawn toward the positive terminal, and the resulting holes move toward the negative terminal—exactly opposite to the natural flow of electrons at the p–n junction. The junction resists the imposed current flow in this direction and is said to be under reverse bias. No current flows through the system.
On the other hand, if the battery is connected so that the negative terminal is con-nected to the n-type region and the positive terminal is connected to the p-type region [Fig. 10.31(c)], the movement of electrons (and holes) is in the favored direction. The junction has low resistance, and a current flows easily. The junction is said to be under forward bias.
A p–n junction makes an excellent rectifier, a device that produces a pulsating direct current (flows in one direction) from alternating current (flows in both directions alternately). When placed in a circuit where the potential is constantly reversing, a p–n junction transmits current only under forward bias, thus converting the alternat-ing current to direct current. Radios, computers, and other electronic devices formerly (+) (–) p n (a) (+) (–) p n (b) (–) (+) p n (c) To negative terminal of battery To positive terminal of battery To positive terminal of battery To negative terminal of battery FIGURE 10.31 The p–n junction involves the contact of a p-type and an n-type semiconductor. (a) The charge carriers of the p-type region are holes ( ). In the n-type region the charge carriers are electrons ( ). (b) No current flows (reverse bias). (c) Current readily flows (forward bias). Note that each elec-tron that crosses the boundary leaves a hole behind. Thus the electrons and the holes move in opposite directions.
Printed circuits are discussed in the Chemical Impact feature on page 452.
452 Chapter Ten Liquids and Solids CHEMICAL IMPACT Transistors and Printed Circuits T ransistors have had an immense impact on the technol-ogy of electronic devices for which signal amplification is needed, such as communications equipment and comput-ers. Before the invention of the transistor at Bell Laborato-ries in 1947, amplification was provided exclusively by vac-uum tubes, which were both bulky and unreliable. The first electronic digital computer, ENIAC, built at the University of Pennsylvania, had 19,000 vacuum tubes and consumed 150,000 watts of electricity. Because of the discovery and development of the transistor and the printed circuit, a hand-held calculator run by a small battery now has the same com-puting power as ENIAC.
A junction transistor is made by joining n-type and p-type semiconductors so as to form an n–p–n or a p–n–p junction. The former type is shown in Fig. 10.32. In this di-agram the input signal (to be amplified) occurs in circuit 1, which has a small resistance and a forward-biased n–p junc-tion (junction 1). As the voltage of the input signal to this circuit varies, the current in the circuit varies, which means there is a change in the number of electrons crossing the n–p junction. Circuit 2 has a relatively large resistance and is un-der reverse bias. The key to operation of the transistor is that current only flows in circuit 2 when electrons crossing junc-tion 1 also cross junction 2 and travel to the positive termi-nal. Since the current in circuit 1 determines the number of electrons crossing junction 1, the number of electrons avail-able to cross junction 2 is also directly proportional to the current in circuit 1. The current in circuit 2 therefore varies depending on the current in circuit 1.
The voltage V, current I, and resistance R in a circuit are related by the equation V IR Since circuit 2 has a large resistance, a given current in cir-cuit 2 produces a larger voltage than the same current in circuit 1, which has a small resistance. Thus a signal or vari-able voltage in circuit 1, such as might be produced by a human voice on a telephone, is reproduced in circuit 2, but with much greater voltage changes. That is, the input signal has been amplified by the junction transistor. This device, which has replaced the large vacuum tube, is a tiny compo-nent of a printed circuit on a silicon chip.
Silicon chips are really “planar” transistors constructed from thin layers of n-type and p-type regions connected by conductors. A chip less than 1 cm wide can contain hun-dreds of printed circuits and be used in computers, radios, and televisions.
A printed circuit has many n–p–n junction transistors.
Fig. 10.33 illustrates the formation of one transistor area.
The chip begins as a thin wafer of silicon that has been doped with an n-type impurity. A protective layer of silicon dioxide is then produced on the wafer by exposing it in a furnace to an oxidizing atmosphere. The next step is to produce a p-type semiconductor. To do this, the surface of the oxide is covered by a polymeric photoresist, as shown in Fig. 10.33(a). A template that only allows light to shine through in selected areas is then placed on top [Fig.
10.33(b)], and light is shown on the chip. The photoresist Amplified output signal Large resistance Variable input signal Small resistance + − − e– e– n n p Junction 1 Junction 2 Circuit 1 Circuit 2 Forward bias Reverse bias Transistor + FIGURE 10.32 A schematic of two circuits connected by a transistor. The signal in circuit 1 is amplified in circuit 2.
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10.5 Carbon and Silicon: Network Atomic Solids 453 that has been exposed to light undergoes a chemical change that causes its solubility to be different from the unexposed photoresist. The exposed photoresist is dissolved using selective solvents [Fig. 10.33(c)], and the exposed area is treated with an etching solution to dissolve the oxide coating [Fig. 10.33(d)]. When the remaining photoresist is dissolved, the silicon wafer has its oxide coating intact except at the one spot (of diameter x), as shown in Fig.
10.33(d).
Exposing the wafer to a p-type impurity such as boron at about causes a p-type semiconductor area to be formed in the exposed spot as the boron atoms diffuse into the silicon crystal [Fig. 10.33(e)]. Next, to form a small n-type area in the center of the p-type region, the wafer is again placed in the oxidizing furnace to be recoated over its entire surface with ox-ide. Then a new photoresist covering is applied, which is illuminated through a template with a transparent area indicated by y [Fig. 10.33(f)].
The photoresist and oxide are then removed from the illuminated area, and the wafer is ex-posed to an n-type impurity to form a small n-type region as shown in Fig. 10.33(g). Next, conductors are layered onto the chip giving the finished transistor [Fig. 10.33(h)], which has two circuits connected through an n–p–n junc-tion (see Fig. 10.32). This transistor then be-comes a part of a large circuit layered onto the chip and interconnected by conductors.
The method given here for producing a printed circuit does not represent the latest tech-nology in this field. The manufacture of printed circuits is a highly competitive business, and changes in methodology occur almost daily.
1000°C p p p p (a) (b) (c) (d) (e) (f) (g) (h) Photoresist Silicon dioxide n Template x x B atoms Template y n n n Electrical connections n x y Electrical connection n FIGURE 10.33 The steps for forming a transistor in a crystal of initially pure silicon.
454 Chapter Ten Liquids and Solids used bulky, unreliable vacuum tubes as rectifiers. The p–n junction has revolution-ized electronics; modern solid-state components contain p–n junctions in printed circuits.
10.6 Molecular Solids So far we have considered solids in which atoms occupy the lattice positions. In some of these substances (network solids), the solid can be considered to be one giant molecule.
In addition, there are many types of solids that contain discrete molecular units at each lattice position. A common example is ice, where the lattice positions are occupied by water molecules [see Fig. 10.12(c)]. Other examples are dry ice (solid carbon dioxide), some forms of sulfur that contain S8 molecules [Fig. 10.34(a)], and certain forms of phos-phorus that contain P4 molecules [Fig. 10.34(b)]. These substances are characterized by strong covalent bonding within the molecules but relatively weak forces between the mol-ecules. For example, it takes only 6 kJ of energy to melt 1 mole of solid water (ice) be-cause only intermolecular (H2OOH2O) interactions must be overcome. However, 470 kJ of energy is required to break 1 mole of covalent OOH bonds. The differences between the covalent bonds within the molecules and the forces between the molecules are ap-parent from the comparison of the interatomic and intermolecular distances in solids shown in Table 10.6.
The forces that exist among the molecules in a molecular solid depend on the nature of the molecules. Many molecules such as CO2, I2, P4, and S8 have no dipole moment, and the intermolecular forces are London dispersion forces. Because these forces are of-ten relatively small, we might expect all these substances to be gaseous at as is the case for carbon dioxide. However, as the size of the molecules increases, the London forces become quite large, causing many of these substances to be solids at When molecules do have dipole moments, their intermolecular forces are signifi-cantly greater, especially when hydrogen bonding is possible. Water molecules are par-ticularly well suited to interact with each other because each molecule has two polar OOH bonds and two lone pairs on the oxygen atom. This can lead to the association 25°C.
25°C, FIGURE 10.34 (a) Sulfur crystals (yellow) contain S8 mole-cules. (b) White phosphorus (containing P4 molecules) is so reactive with the oxygen in air that it must be stored under water.
(a) (b) Visualization: Molecular Solids A “steaming” piece of dry ice.
10.6 Molecular Solids 455 CHEMICAL IMPACT Explosive Sniffer T hese days security is at the top of everyone’s list of important concerns, especially for those people who are responsible for the safety of our transportation systems. In particular, airports need speedy and sensitive detectors for explosives. Plastic explosives are especially tricky to detect because they do not respond to metal detectors, and they can be shaped into innocent-looking objects to avoid X-ray de-tection. However, a team of scientists at Oak Ridge National Laboratory led by Thomas Thundat has just published a description of an inexpensive device that is extremely sensitive to two N-containing compounds found in plastic explosives. The key part of this detection device is a tiny (180-micrometer), V-shaped cantilever made of silicon. The cantilever is shown in the accompanying photo next to a human hair for size comparison.
The upper surface of the cantilever was first coated with a layer of gold and then a one-molecule-thick layer of an acid that binds to each of the two N-containing molecules to be detected: pentaerythritol tetranitrate (PETN) and hexahydro-1,3,5-triazine (RDX). When a stream of air con-taining tiny amounts of PETN or RDX passes over the can-tilever, these molecules bind to the cantilever, causing it to bend “like a diving board.” This bending is not due to the added mass of the attached PETN and RDX. Rather, the de-formation occurs because the area of the cantilever surface where binding takes place stretches relative to the unbound areas. A laser pointed at the cantilever detects the bending motion when PETN or RDX (or both) is present. The device’s sensitivity is quite remarkable: 14 parts per trillion of PETN and 30 parts per trillion of RDX.
All in all, this device appears very promising for de-tecting plastic explosives in luggage. The cantilevers are in-expensive to construct (approximately $1), and the entire device is about the size of a shoe box. Also, the Oak Ridge team can fabricate thousands of cantilevers in one device.
By putting different coatings on the cantilever arms, it should be possible to detect many other types of chemicals and possible biological agents.
This detector looks like a very promising addition to our arsenal of security devices.
When explosive compounds bind to these V-shaped cantilevers, the mi-croscopic structures, which are about the width of a hair, bend and pro-duce a signal.
TABLE 10.6 Comparison of Atomic Separations Within Molecules (Covalent Bonds) and Between Molecules (Intermolecular Interactions) Distance Between Closest Distance Between Solid Atoms in Molecule Molecules in the Solid P4 220 pm 380 pm S8 206 pm 370 pm Cl2 199 pm 360 pm The shorter distances within the molecules indicate stronger bonding.
456 Chapter Ten Liquids and Solids of four hydrogen atoms with each oxygen: two by covalent bonds and two by dipole forces: Note the two relatively short covalent oxygen–hydrogen bonds and the two longer oxygen–hydrogen dipole interactions that can be seen in the ice structure in Fig. 10.12(c).
10.7 Ionic Solids Ionic solids are stable, high-melting substances held together by the strong electrostatic forces that exist between oppositely charged ions. The principles governing the structures of ionic solids were introduced in Section 8.5. In this section we will review and extend these principles.
The structures of most binary ionic solids, such as sodium chloride, can be explained by the closest packing of spheres. Typically, the larger ions, usually the anions, are packed in one of the closest packing arrangements (hcp or ccp), and the smaller cations fit into holes among the closest packed anions. The packing is done in a way that maximizes the electrostatic attractions among oppositely charged ions and minimizes the repulsions among ions with like charges.
There are three types of holes in closest packed structures: 1. Trigonal holes are formed by three spheres in the same layer [Fig. 10.35(a)].
2. Tetrahedral holes are formed when a sphere sits in the dimple of three spheres in an adjacent layer [Fig. 10.35(b)].
3. Octahedral holes are formed between two sets of three spheres in adjoining layers of the closest packed structures [Fig. 10.35(c)].
For spheres of a given diameter, the holes increase in size in the order trigonal tetrahedral octahedral In fact, trigonal holes are so small that they are never occupied in binary ionic compounds.
Whether the tetrahedral or octahedral holes in a given binary ionic solid are occupied de-pends mainly on the relative sizes of the anion and cation. For example, in zinc sulfide the ions (ionic radius 180 pm) are arranged in a cubic closest packed structure with the smaller ions (ionic radius 70 pm) in the tetrahedral holes. The locations of the tetrahedral holes in the face-centered cubic unit cell of the ccp structure are shown in Fig. 10.36(a). Note from this figure that there are eight tetrahedral holes in the unit cell. Also recall from the discussion in Section 10.4 that there are four net spheres in the Zn2 S2 H H O O O H H H H n n n n Trigonal hole Tetrahedral hole Octahedral hole (a) (b) (c) FIGURE 10.35 The holes that exist among closest packed uniform spheres. (a) The trigonal hole formed by three spheres in a given plane.
(b) The tetrahedral hole formed when a sphere occupies a dimple formed by three spheres in an adjacent layer. (c) The octahe-dral hole formed by six spheres in two ad-jacent layers.
10.7 Ionic Solids 457 face-centered cubic unit cell. Thus there are twice as many tetrahedral holes as packed anions in the closest packed structure. Zinc sulfide must have the same number of ions and ions to achieve electrical neutrality. Thus in the zinc sulfide structure only half the tetrahedral holes contain ions, as shown in Fig. 10.36(c).
The structure of sodium chloride can be described in terms of a cubic closest packed array of ions with ions in all the octahedral holes. The locations of the octahe-dral holes in the face-centered cubic unit cell are shown in Fig. 10.37(a). The easiest octahedral hole to find in this structure is the one at the center of the cube. Note that this hole is surrounded by six spheres, as is required to form an octahedron. The remaining octahedral holes are shared with other unit cells and are more difficult to visualize.
However, it can be shown that the number of octahedral holes in the ccp structure is the same as the number of packed anions. Figure 10.37(b) shows the structure for sodium chloride that results from ions filling all the octahedral holes in a ccp array of ions.
A great variety of ionic solids exists. Our purpose in this section is not to give an ex-haustive treatment of ionic solids, but to emphasize the fundamental principles governing their structures. As we have seen, the most useful model for explaining the structures of these solids regards the ions as hard spheres that are packed to maximize attractions and minimize repulsions.
Determining the Number of Ions in a Unit Cell Determine the net number of and ions in the sodium chloride unit cell.
Solution Note from Fig. 10.37(b) that the ions are cubic closest packed and thus form a face-centered cubic unit cell. There is a ion on each corner and one at the center of each face of the cube. Thus the net number of ions present in a unit cell is 811 82 611 22 4 Cl Cl Cl Cl Na Cl Na Na Cl Zn2 Zn2 S2 Closest packed structures contain twice as many tetrahedral holes as packed spheres. Closest packed structures con-tain the same number of octahedral holes as packed spheres.
+ + (a) (b) (c) ZnS FIGURE 10.36 (a) The location (X) of a tetrahedral hole in the face-centered cubic unit cell. (b) One of the tetrahedral holes. (c) The unit cell for ZnS where the S2 ions (yellow) are closest packed with the Zn2 ions (red) in alternat-ing tetrahedral holes.
Sample Exercise 10.3 Visualization: Structure of an Ionic Solid (NaCl).
+ + + + + + + + + + + + + (a) (b) FIGURE 10.37 (a) The locations (gray X) of the octahedral holes in the face-centered cubic unit cell. (b) Representation of the unit cell for solid NaCl.
The Cl ions (green spheres) have a ccp arrangement with Na ions (gray spheres) in all the octahedral holes. Note that this repre-sentation shows the idealized closest packed structure of NaCl. In the actual structure, the Cl ions do not quite touch.
458 Chapter Ten Liquids and Solids The ions occupy the octahedral holes located in the center of the cube and mid-way along each edge. The ion in the center of the cube is contained entirely in the unit cell, whereas those on the edges are shared by four unit cells (four cubes share a common edge). Since the number of edges in a cube is 12, the net number of ions present is We have shown that the net number of ions in a unit cell is 4 ions and 4 ions, which agrees with the 1:1 stoichiometry of sodium chloride.
See Exercises 10.61 through 10.68.
In this chapter we have considered various types of solids. Table 10.7 summarizes these types of solids and some of their properties.
Cl Na 1112 1211 42 4 Na Na Na TABLE 10.7 Types and Properties of Solids Type of Solid: Atomic Molecular Ionic Network Metallic Group 8A Structural Unit: Atom Atom Atom Molecule Ion Type of Bonding: Directional Nondirectional covalent London Polar molecules: Ionic covalent bonds bonds involving dispersion dipole–dipole electrons that are forces interactions delocalized Nonpolar molecules: throughout the London dispersion crystal forces Typical Properties: Hard Wide range of Soft Hard hardness High melting Wide range of Very low Low melting point High melting point melting points melting point point Insulator Conductor Insulator Insulator Examples: Diamond Silver Argon(s) Ice (solid H2O) Sodium chloride Iron Dry ice (solid CO2) Calcium fluoride Brass ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 10.8 Vapor Pressure and Changes of State 459 Types of Solids Using Table 10.7, classify each of the following substances according to the type of solid it forms.
a. Gold b. Carbon dioxide c. Lithium fluoride d. Krypton Solution a. Solid gold is an atomic solid with metallic properties.
b. Solid carbon dioxide contains nonpolar carbon dioxide molecules and is a molecular solid.
c. Solid lithium fluoride contains Li and F ions and is a binary ionic solid.
d. Solid krypton contains krypton atoms that can interact only through London disper-sion forces. It is an atomic solid but has properties characteristic of a molecular solid with nonpolar molecules.
See Exercises 10.71 and 10.72.
10.8 Vapor Pressure and Changes of State Now that we have considered the general properties of the three states of matter, we can explore the processes by which matter changes state. One very familiar example of a change in state occurs when a liquid evaporates from an open container. This is clear ev-idence that the molecules of a liquid can escape the liquid’s surface and form a gas, a process called vaporization, or evaporation. Vaporization is endothermic because energy is required to overcome the relatively strong intermolecular forces in the liquid. The en-ergy required to vaporize 1 mole of a liquid at a pressure of 1 atm is called the heat of vaporization, or the enthalpy of vaporization, and is usually symbolized as Hvap.
The endothermic nature of vaporization has great practical significance; in fact, one of the most important roles that water plays in our world is to act as a coolant. Because of the strong hydrogen bonding among its molecules in the liquid state, water has an un-usually large heat of vaporization (40.7 kJ/mol). A significant portion of the sun’s energy that reaches earth is spent evaporating water from the oceans, lakes, and rivers rather than warming the earth. The vaporization of water is also crucial to the body’s temperature-control system through evaporation of perspiration.
Vapor Pressure When a liquid is placed in a closed container, the amount of liquid at first decreases but eventually becomes constant. The decrease occurs because there is an initial net transfer of molecules from the liquid to the vapor phase (Fig. 10.38). This evaporation process oc-curs at a constant rate at a given temperature (see Fig. 10.39). However, the reverse process is different. Initially, as the number of vapor molecules increases, so does the rate of re-turn of these molecules to the liquid. The process by which vapor molecules re-form a liquid is called condensation. Eventually, enough vapor molecules are present above the liquid so that the rate of condensation equals the rate of evaporation (see Fig. 10.39). At this point no further net change occurs in the amount of liquid or vapor because the two opposite processes exactly balance each other; the system is at equilibrium. Note that this system is highly dynamic on the molecular level—molecules are constantly escaping from and entering the liquid at a high rate. However, there is no net change because the two opposite processes just balance each other.
Vapor is the usual term for the gas phase of a substance that exists as a solid or liquid at 25C and 1 atm. Hvap for water at 100C is 40.7 kJ/mol.
FIGURE 10.38 Behavior of a liquid in a closed container.
(a) Initially, net evaporation occurs as molecules are transferred from the liquid to the vapor phase, so the amount of liquid decreases. (b) As the number of vapor molecules increases, the rate of return to the liquid (condensation) increases, until finally the rate of condensation equals the rate of evaporation. The system is at equi-librium, and no further changes occur in the amounts of vapor or liquid.
(a) (b) Sample Exercise 10.4 460 Chapter Ten Liquids and Solids The pressure of the vapor present at equilibrium is called the equilibrium vapor pressure, or more commonly, the vapor pressure of the liquid. A simple barometer can measure the vapor pressure of a liquid, as shown in Fig. 10.40(a). The liquid is injected at the bottom of the tube of mercury and floats to the surface because the mercury is so dense. A portion of the liquid evaporates at the top of the column, producing a vapor whose pressure pushes some mercury out of the tube. When the system reaches equilib-rium, the vapor pressure can be determined from the change in the height of the mercury column since Thus The vapor pressures of liquids vary widely [see Fig. 10.40(b)]. Liquids with high vapor pressures are said to be volatile—they evaporate rapidly from an open dish. The vapor pressure of a liquid is principally determined by the size of the intermolecular forces in the liquid. Liquids in which the intermolecular forces are large have relatively low vapor Pvapor Patmosphere PHg column Patmosphere Pvapor PHg column Patm = 760 torr (a) Vapor pressure Vacuum 736 H2O vapor 760 – 736 = 24 torr 695 215 C2H5OH vapor 760 – 695 = 65 torr (C2H5)2O vapor 760 – 215 = 545 torr (b) FIGURE 10.40 (a) The vapor pressure of a liquid can be measured easily using a simple barometer of the type shown here. (b) The three liq-uids, water, ethanol (C2H5OH), and diethyl ether [(C2H5)2O], have quite different vapor pressures. Ether is by far the most volatile of the three. Note that in each case a little liquid remains (floating on the mercury).
Rate Rate of evaporation Rate of condensation Rates are equal beyond this time; equilibrium vapor pressure is attained.
Time FIGURE 10.39 The rates of condensation and evaporation over time for a liquid sealed in a closed container. The rate of evaporation remains constant and the rate of condensation in-creases as the number of molecules in the vapor phase increases, until the two rates become equal. At this point, the equilib-rium vapor pressure is attained.
A system at equilibrium is dynamic on the molecular level but shows no macro-scopic changes.
10.8 Vapor Pressure and Changes of State 461 pressures because the molecules need high energies to escape to the vapor phase. For example, although water has a much lower molar mass than diethyl ether, the strong hydrogen-bonding forces that exist among water molecules in the liquid cause water’s vapor pressure to be much lower than that of diethyl ether [see Fig. 10.40(b)]. In general, substances with large molar masses have relatively low vapor pressures, mainly because of the large dispersion forces. The more electrons a substance has, the more polarizable it is, and the greater the dispersion forces are.
Measurements of the vapor pressure for a given liquid at several temperatures show that vapor pressure increases significantly with temperature. Figure 10.41 illustrates the distribution of molecular kinetic energy present in a liquid at two different temperatures.
To overcome the intermolecular forces in a liquid, a molecule must have sufficient kinetic energy. As the temperature of the liquid is increased, the fraction of molecules having the minimum energy needed to overcome these forces and escape to the vapor phase increases markedly. Thus the vapor pressure of a liquid increases dramatically with temperature.
Values for water at several temperatures are given in Table 10.8.
The quantitative nature of the temperature dependence of vapor pressure can be rep-resented graphically. Plots of vapor pressure versus temperature for water, ethanol, and diethyl ether are shown in Fig. 10.42(a). Note the nonlinear increase in vapor pressure for all the liquids as the temperature is increased. We find that a straight line can be ob-tained by plotting ln(Pvap) versus 1T, where T is the Kelvin temperature, as shown in Fig. 10.42(b). We can represent this behavior by the equation (10.4) where Hvap is the enthalpy of vaporization, R is the universal gas constant, and C is a constant characteristic of a given liquid. The symbol ln means that the natural logarithm of the vapor pressure is taken.
Equation (10.4) is the equation for a straight line of the form y mx b, where Determining Enthalpies of Vaporization Using the plots in Fig. 10.42(b), determine whether water or diethyl ether has the larger enthalpy of vaporization.
b intercept C m slope ¢Hvap R x 1 T y ln1Pvap2 ln1Pvap2 ¢Hvap R a1 Tb C TABLE 10.8 The Vapor Pressure of Water as a Function of Temperature T (C) P (torr) 0.0 4.579 10.0 9.209 20.0 17.535 25.0 23.756 30.0 31.824 40.0 55.324 60.0 149.4 70.0 233.7 90.0 525.8 T1 Kinetic energy (a) Number of molecules with a given energy Energy needed to overcome intermolecular forces in liquid T2 Kinetic energy (b) Number of molecules with a given energy Energy needed to overcome intermolecular forces in liquid FIGURE 10.41 The number of molecules in a liquid with a given energy versus kinetic energy at two temperatures. Part (a) shows a lower tem-perature than that in part (b). Note that the proportion of molecules with enough en-ergy to escape the liquid to the vapor phase (indicated by shaded areas) increases dra-matically with temperature. This causes vapor pressure to increase markedly with temperature.
Natural logarithms are reviewed in Appendix 1.2.
Sample Exercise 10.5 462 Chapter Ten Liquids and Solids Solution When ln(Pvap) is plotted versus 1T, the slope of the resulting straight line is Note from Fig. 10.42(b) that the slopes of the lines for water and diethyl ether are both negative, as expected, and that the line for ether has the smaller slope. Thus ether has the smaller value of Hvap. This makes sense because the hydrogen bonding in water causes it to have a relatively large enthalpy of vaporization.
See Exercise 10.79.
Equation (10.4) is important for several reasons. For example, we can determine the heat of vaporization for a liquid by measuring Pvap at several temperatures and then evaluating the slope of a plot of ln(Pvap) versus 1T. On the other hand, if we know the values of Hvap and Pvap at one temperature, we can use Equation (10.4) to calculate Pvap at another temperature. This can be done by recognizing that the constant C does not depend on temperature. Thus at two temperatures T1 and T2 we can solve Equation (10.4) for C and then write the equality ln1Pvap,T12 ¢Hvap RT1 C ln1Pvap,T22 ¢Hvap RT2 ¢Hvap R 0 Temperature (°C) 100 200 300 400 500 600 700 800 900 0 10 20 30 40 50 60 70 80 90 100 760 Pvap (torr) 34.6° 78.4° 100.0° 1/T (K) ln (Pvap ) Diethyl ether Ethanol Water Diethyl ether Ethanol Water (a) (b) FIGURE 10.42 (a) The vapor pressure of water, ethanol, and diethyl ether as a function of temperature. (b) Plots of In(Pvap) versus 1T (Kelvin temperature) for water, ethanol, and diethyl ether.
10.8 Vapor Pressure and Changes of State 463 This can be rearranged to or (10.5) Calculating Vapor Pressure The vapor pressure of water at is 23.8 torr, and the heat of vaporization of water at is 43.9 kJ/mol. Calculate the vapor pressure of water at Solution We will use Equation (10.5): For water we have Thus Taking the antilog (see Appendix 1.2) of both sides gives See Exercises 10.81 through 10.84.
Like liquids, solids have vapor pressures. Figure 10.43 shows iodine vapor in equi-librium with solid iodine in a closed flask. Under normal conditions iodine sublimes; that is, it goes directly from the solid to the gaseous state without passing through the liquid state. Sublimation also occurs with dry ice (solid carbon dioxide).
Changes of State What happens when a solid is heated? Typically, it will melt to form a liquid. If the heat-ing continues, the liquid will at some point boil and form the vapor phase. This process can be represented by a heating curve: a plot of temperature versus time for a process where energy is added at a constant rate.
The heating curve for water is given in Fig. 10.44. As energy flows into the ice, the random vibrations of the water molecules increase as the temperature rises. Eventually, the molecules become so energetic that they break loose from their lattice positions, and the change from solid to liquid occurs. This is indicated by a plateau at on the 0°C Pvap,T2 93.7 torr 23.8 Pvap,T2 0.254 lna 23.8 Pvap,T2 b 1.37 lna 23.8 torr Pvap,T2 1torr2 b 43,900 J/mol 8.3145 J/K mola 1 323 K 1 298 Kb R 8.3145 J/K mol ¢Hvap 43.9 KJ/mol 43,900 J/mol T2 50. 273 323 K T1 25 273 298 K Pvap,T1 23.8 torr lna Pvap,T1 Pvap,T2 b ¢Hvap R a 1 T2 1 T1 b 50.°C.
25°C 25°C lna Pvap,T1 Pvap,T2 b ¢Hvap R a 1 T2 1 T1 b ln1Pvap,T12 ln1Pvap,T22 ¢Hvap R a 1 T2 1 T1 b Sample Exercise 10.6 Equation (10.5) is called the Clausius–Clapeyron equation.
In solving this problem, we ignore the fact that Hvap is slightly temperature dependent.
Phase changes of carbon dioxide are discussed in Section 10.9.
Sublimation: A process in which a sub-stance goes directly from the solid to the gaseous state.
Visualization: Changes of State 464 Chapter Ten Liquids and Solids heating curve. At this temperature, called the melting point, all the added energy is used to disrupt the ice structure by breaking the hydrogen bonds, thus increasing the potential energy of the water molecules. The enthalpy change that occurs at the melting point when a solid melts is called the heat of fusion, or more accurately, the enthalpy of fusion, Hfus. The melting points and enthalpies of fusion for several representative solids are listed in Table 10.9.
The temperature remains constant until the solid has completely changed to liquid; then it begins to increase again. At 100C the liquid water reaches its boiling point, and the temperature then remains constant as the added energy is used to vaporize the liquid.
When the liquid is completely changed to vapor, the temperature again begins to rise. Note that changes of state are physical changes; although intermolecular forces have been over-come, no chemical bonds have been broken. If the water vapor were heated to much higher temperatures, the water molecules would break down into the individual atoms. This would FIGURE 10.43 Iodine being heated, causing it to sublime, forming crystals of I2(s) on the bottom of an evaporating dish cooled by ice.
Temperature (°C) Time –20 0 20 40 60 80 100 120 140 Steam Water and steam Water Ice and water Ice FIGURE 10.44 The heating curve (not drawn to scale) for a given quantity of wa-ter where energy is added at a constant rate. The plateau at the boiling point is longer than the plateau at the melting point be-cause it takes almost seven times more energy (and thus seven times the heating time) to vaporize liquid water than to melt ice.
The slopes of the other lines are different because the different states of water have different molar heat capacities (the energy re-quired to raise the temperature of 1 mole of a substance by 1C).
TABLE 10.9 Melting Points and Enthalpies of Fusion for Several Representative Solids Melting Point Enthalpy of Compound (°C) Fusion (kJ/mol) O2 218 0.45 HCl 114 1.99 HI 51 2.87 CCl4 23 2.51 CHCl3 64 9.20 H2O 0 6.02 NaF 992 29.3 NaCl 801 30.2 The melting and boiling points will be de-fined more precisely later in this section.
Ionic solids such as NaCl and NaF have very high melting points and enthalpies of fusion because of the strong ionic forces in these solids. At the other ex-treme is O2(s), a molecular solid con-taining nonpolar molecules with weak intermolecular forces. (See Table 10.9.) 10.8 Vapor Pressure and Changes of State 465 be a chemical change, since covalent bonds are broken. We no longer have water after this occurs.
The melting and boiling points for a substance are determined by the vapor pressures of the solid and liquid states. Figure 10.45 shows the vapor pressures of solid and liquid water as functions of temperature near 0C Note that below 0C the vapor pressure of ice is less than the vapor pressure of liquid water. Also note that the vapor pressure of ice has a larger temperature dependence than that of the liquid. That is, the vapor pressure of ice increases more rapidly for a given rise in temperature than does the vapor pressure of wa-ter. Thus, as the temperature of the solid is increased, a point is eventually reached where the liquid and solid have identical vapor pressures. This is the melting point.
These concepts can be demonstrated experimentally using the apparatus illustrated in Fig. 10.46, where ice occupies one compartment and liquid water the other. Consider the following cases.
Case 1 A temperature at which the vapor pressure of the solid is greater than that of the liquid.
At this temperature the solid requires a higher pressure than the liquid does to be in equilibrium with the vapor. Thus, as vapor is released from the solid to try to achieve equi-librium, the liquid will absorb vapor in an attempt to reduce the vapor pressure to its equilibrium value. The net effect is a conversion from solid to liquid through the vapor phase. In fact, no solid can exist under these conditions. The amount of solid will steadily decrease and the volume of liquid will increase. Finally, there will be only liquid in the right compartment, which will come to equilibrium with the water vapor, and no further changes will occur in the system. This temperature must be above the melting point of ice, since only the liquid state can exist.
Case 2 A temperature at which the vapor pressure of the solid is less than that of the liquid. This is the opposite of the situation in case 1. In this case, the liquid requires a higher pres-sure than the solid does to be in equilibrium with the vapor, so the liquid will gradually disappear, and the amount of ice will increase. Finally, only the solid will remain, which will achieve equilibrium with the vapor. This temperature must be below the melting point of ice, since only the solid state can exist.
FIGURE 10.45 The vapor pressures of solid and liquid wa-ter as a function of temperature. The data for liquid water below 0C are obtained from supercooled water. The data for solid water above 0C are estimated by extrapola-tion of vapor pressure from below 0C.
Pvap (torr) 0 Temperature (°C) 5 10 Vapor pressure of solid Vapor pressure of liquid –5 0 +5 Solid water Liquid water Water vapor FIGURE 10.46 An apparatus that allows solid and liquid water to interact only through the vapor state.
466 Chapter Ten Liquids and Solids Case 3 A temperature at which the vapor pressures of the solid and liquid are identical. In this case, the solid and liquid states have the same vapor pressure, so they can coexist in the apparatus at equilibrium simultaneously with the vapor. This temperature represents the freezing point where both the solid and liquid states can exist.
We can now describe the melting point of a substance more precisely. The normal melting point is defined as the temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure is 1 atmosphere.
Boiling occurs when the vapor pressure of a liquid becomes equal to the pressure of its environment. The normal boiling point of a liquid is the temperature at which the va-por pressure of the liquid is exactly 1 atmosphere. This concept is illustrated in Fig. 10.47.
At temperatures where the vapor pressure of the liquid is less than 1 atmosphere, no bubbles of vapor can form because the pressure on the surface of the liquid is greater than the pressure in any spaces in the liquid where the bubbles are trying to form. Only when the liquid reaches a temperature at which the pressure of vapor in the spaces in the liquid is 1 atmosphere can bubbles form and boiling occur.
However, changes of state do not always occur exactly at the boiling point or melting point. For example, water can be readily supercooled; that is, it can be cooled below 0°C at 1 atm pressure and remain in the liquid state. Supercooling occurs because, as it is cooled, the water may not achieve the degree of organization necessary to form ice at 0°C, and thus it continues to exist as the liquid. At some point the correct ordering occurs and ice rapidly forms, releasing energy in the exothermic process and bringing the temperature back up to the melting point, where the remainder of the water freezes (see Fig. 10.48).
A liquid also can be superheated, or raised to temperatures above its boiling point, especially if it is heated rapidly. Superheating can occur because bubble formation in the interior of the liquid requires that many high-energy molecules gather in the same vicin-ity, and this may not happen at the boiling point, especially if the liquid is heated rapidly.
If the liquid becomes superheated, the vapor pressure in the liquid is greater than the at-mospheric pressure. Once a bubble does form, since its internal pressure is greater than that of the atmosphere, it can burst before rising to the surface, blowing the surrounding liquid out of the container. This is called bumping and has ruined many experiments. It can be avoided by adding boiling chips to the flask containing the liquid. Boiling chips are bits of porous ceramic material containing trapped air that escapes on heating, form-ing tiny bubbles that act as “starters” for vapor bubble formation. This allows a smooth onset of boiling as the boiling point is reached.
FIGURE 10.47 Water in a closed system with a pressure of 1 atm exerted on the piston. No bubbles can form within the liquid as long as the vapor pressure is less than 1 atm.
Liquid water Movable piston Constant pressure of 1 atmosphere Boiling chip releasing air bubbles acts as a nucleating agent for the bubbles that form when water boils.
Temperature (°C) 0° Time Crystallization begins Liquid Expected behavior Liquid and solid S FIGURE 10.48 The supercooling of water. The extent of su-percooling is given by S.
10.9 Phase Diagrams 467 10.9 Phase Diagrams A phase diagram is a convenient way of representing the phases of a substance as a func-tion of temperature and pressure. For example, the phase diagram for water (Fig. 10.49) shows which state exists at a given temperature and pressure. It is important to recognize that a phase diagram describes conditions and events in a closed system of the type rep-resented in Fig. 10.47, where no material can escape into the surroundings and no air is present. Notice that the diagram is not drawn to scale (neither axis is linear). This is done to emphasize certain features of the diagram that will be discussed below.
To show how to interpret the phase diagram for water, we will consider heating ex-periments at several pressures, shown by the dashed lines in Fig. 10.50.
Experiment 1 Pressure is 1 atm. This experiment begins with the cylinder shown in Fig. 10.47 completely filled with ice at a temperature of C and the piston exerting a pressure of 1 atm di-rectly on the ice (there is no air space). Since at temperatures below the vapor pressure of ice is less than 1 atm—which is the constant external pressure on the piston—no vapor is present in the cylinder. As the cylinder is heated, ice is the only component until the temperature reaches where the ice changes to liquid water as energy is added. This is the normal melting point of water. Note that under these conditions no vapor exists in the system. The vapor pressures of the solid and liquid are equal, but this vapor pressure is less than 1 atm, so no water vapor can exist. This is true on the solid/liquid line everywhere except at the triple point (see Experiment 3 below). When the solid has completely changed to liquid, the temperature again rises. At this point, the cylinder contains only liquid water.
No vapor is present because the vapor pressure of liquid water under these conditions is less than 1 atm, the constant external pressure on the piston. Heating continues until the temperature of the liquid water reaches At this point, the vapor pressure of liquid water is 1 atm, and boiling occurs, with the liquid changing to vapor. This is the normal boiling point of water. After the liquid has been completely converted to steam, the tem-perature again rises as the heating continues. The cylinder now contains only water vapor.
Experiment 2 Pressure is 2.0 torr. Again, we start with ice as the only component in the cylinder at The pressure exerted by the piston in this case is only 2.0 torr. As heating pro-ceeds, the temperature rises to where the ice changes directly to vapor, a process known as sublimation. Sublimation occurs when the vapor pressure of ice is equal to the 10°C, 20°C.
100°C.
0°C, 0°C 20° FIGURE 10.49 The phase diagram for water. Tm represents the normal melting point; T3 and P3 denote the triple point; Tb represents the normal boiling point; Tc represents the critical temperature; Pc represents the criti-cal pressure. The negative slope of the solid/liquid line reflects the fact that the density of ice is less than that of liquid water. (Note that this line extends indefi-nitely, as indicated by the arrow.) Pc = 218 1.00 P3 = 0.0060 Tm T3 Tb 0 0.0098 100 374 Solid Liquid Gas Temperature (°C) Pressure (atm) Triple point Critical point Tc 468 Chapter Ten Liquids and Solids external pressure, which in this case is only 2.0 torr. No liquid water appears under these conditions because the vapor pressure of liquid water is always greater than 2.0 torr, and thus it cannot exist at this pressure. If liquid water were placed in a cylinder under such a low pressure, it would vaporize immediately at temperatures above or freeze at temperatures below Experiment 3 Pressure is 4.58 torr. Again, we start with ice as the only component in the cylinder at In this case the pressure exerted on the ice by the piston is 4.58 torr. As the cylinder is heated, no new phase appears until the temperature reaches (273.16 K). At this point, called the triple point, solid and liquid water have identical vapor pressures of 4.58 torr. Thus at 0.01°C (273.16 K) and 4.58 torr all three states of water are present. In fact, only under these conditions can all three states of water coexist in a closed system.
Experiment 4 Pressure is 225 atm. In this experiment we start with liquid water in the cylinder at the pressure exerted by the piston on the water is 225 atm. Liquid water can be present at this temperature because of the high external pressure. As the temperature increases, something happens that we did not see in the first three experiments: The liquid gradu-ally changes into a vapor but goes through an intermediate “fluid” region, which is nei-ther true liquid nor vapor. This is quite unlike the behavior at lower temperatures and pressures, say at and 1 atm, where the temperature remains constant while a def-inite phase change from liquid to vapor occurs. This unusual behavior occurs because the conditions are beyond the critical point for water. The critical temperature can be de-fined as the temperature above which the vapor cannot be liquefied no matter what pressure is applied. The critical pressure is the pressure required to produce liquefaction at the critical temperature. Together, the critical temperature and the critical pressure define the critical point. For water the critical point is and 218 atm. Note that the liquid/vapor line on the phase diagram for water ends at the critical point. Beyond this point the tran-sition from one state to another involves the intermediate “fluid” region just described.
Applications of the Phase Diagram for Water There are several additional interesting features of the phase diagram for water. Note that the solid/liquid boundary line has a negative slope. This means that the melting point of ice decreases as the external pressure increases. This behavior, which is opposite to that 374°C 100°C.
300°C; 0.01°C 20°C.
10°C.
10°C FIGURE 10.50 Diagrams of various heating experiments on samples of water in a closed system.
Temperature (K) Pressure (atm) 1.0 Solid Liquid Gas Expt 1 Expt 3 Expt 2 Expt 4 10.9 Phase Diagrams 469 observed for most substances, occurs because the density of ice is less than that of liquid water at the melting point. The maximum density of water occurs at when liquid water freezes, its volume increases.
We can account for the effect of pressure on the melting point of water using the fol-lowing reasoning. At the melting point, liquid and solid water coexist—they are in dy-namic equilibrium, since the rate at which ice is melting is just balanced by the rate at which the water is freezing. What happens if we apply pressure to this system? When sub-jected to increased pressure, matter reduces its volume. This behavior is most dramatic for gases but also occurs for condensed states. Since a given mass of ice at has a larger volume than the same mass of liquid water, the system can reduce its volume in response to the increased pressure by changing to liquid. Thus at and an external pres-sure greater than 1 atm, water is liquid. In other words, the freezing point of water is less than when the pressure is greater than 1 atm.
Figure 10.51 illustrates the effect of pressure on ice. At the point X on the phase diagram, ice is subjected to increased pressure at constant temperature. Note that as the pressure is increased, the solid/liquid line is crossed, indicating that the ice melts. This phenomenon may be important in ice skating. The narrow blade of the skate exerts a large pressure, since the skater’s weight is supported by the small area of the blade. Also, the frictional heating due to the moving skate contributes to the melting of the ice. After the blade passes, the liquid refreezes as normal pressure and temperature return. Without this lubrication effect due to the thawing ice, ice skating would not be the smooth, graceful activity that many people enjoy.
Ice’s lower density has other implications. When water freezes in a pipe or an engine block, it will expand and break the container. This is why water pipes are insulated in cold climates and antifreeze is used in water-cooled engines. The lower density of ice also means that ice formed on rivers and lakes will float, providing a layer of insulation that helps prevent bodies of water from freezing solid in the winter. Aquatic life can therefore continue to live through periods of freezing temperatures.
A liquid boils at the temperature where the vapor pressure of the liquid equals the external pressure. Thus the boiling point of a substance, like the melting point, depends on the external pressure. This is why water boils at different temperatures at different el-evations (see Table 10.10), and any cooking carried out in boiling water will be affected by this variation. For example, it takes longer to hard-boil an egg in Leadville, Colorado (elevation: 10,150 ft), than in San Diego, California (sea level), since water boils at a lower temperature in Leadville.
As we mentioned earlier, the phase diagram for water describes a closed system.
Therefore, we must be very cautious in using the phase diagram to explain the behavior of water in a natural setting, such as on the earth’s surface. For example, in dry climates (low humidity), snow and ice seem to sublime—a minimum amount of slush is produced.
Wet clothes put on an outside line at temperatures below freeze and then dry while frozen. However, the phase diagram (Fig. 10.47) shows that ice should not be able to sublime at normal atmospheric pressures. What is happening in these cases? Ice in the natural environment is not in a closed system. The pressure is provided by the atmosphere rather than by a solid piston. This means that the vapor produced over the ice can escape from the immediate region as soon as it is formed. The vapor does not come to equilibrium with the solid, and the ice slowly disappears. Sublimation, which seems forbidden by the phase diagram, does in fact occur under these conditions, although it is not the sublimation under equilibrium conditions described by the phase diagram.
0°C 0°C 0°C 0°C 4°C; The physics of ice skating is quite complex, and there is disagreement about whether the pressure or the frictional heating of the ice skate is most important. See “Letter to the Editor,” by R. Silberman, J. Chem.
Ed. 65 (1988): 186.
The effect of pressure on ice allows this skater to glide smoothly.
Pressure (atm) Solid/liquid line Pressure at which ice changes to water (at this temperature) Liquid Solid Temperature (K) Gas FIGURE 10.51 The phase diagram for water. At point X on the phase diagram, water is a solid. How-ever, as the external pressure is increased while the temperature remains constant (indicated by the vertical dotted line), the solid/liquid line is crossed and the ice melts.
Water boils at 89°C in Leadville, Colorado.
Visualization: Boiling Water with Ice Water 470 Chapter Ten Liquids and Solids CHEMICAL IMPACT Making Diamonds at Low Pressures: Fooling Mother Nature I n 1955 Robert H. Wentorf, Jr., accomplished something that borders on alchemy—he turned peanut butter into diamonds. He and his coworkers at the General Electric Research and Development Center also changed roofing pitch, wood, coal, and many other carbon-containing mate-rials into diamonds, using a process involving temperatures of and pressures of atm. Although the first diamonds made by this process looked like black sand be-cause of the impurities present, the process has now been developed to a point such that beautiful, clear, gem-quality diamonds can be produced. General Electric now has the ca-pacity to produce 150 million carats (30,000 kg) of diamonds annually (virtually all of which is “diamond grit” used for industrial purposes such as abrasive coatings on cutting tools). The production of large, gem-quality diamonds by this process is still too expensive to compete with the natural sources of these stones. However, this may change as meth-ods are developed for making diamonds at low pressures.
The high temperatures and pressures used in the GE process for making diamonds make sense if one looks at the accompanying phase diagram for carbon. Note that graphite—not diamond—is the most stable form of carbon 105 2000°C under ordinary conditions of temperature and pressure.
However, diamond becomes more stable than graphite at very high pressures (as one would expect from the greater 107 109 1011 0 2000 4000 6000 Temperature (K) Pressure (Pa) Diamond Graphite Liquid Vapor The phase diagram for carbon.
The Phase Diagram for Carbon Dioxide The phase diagram for carbon dioxide (Fig. 10.52) differs from that for water. The solid/ liquid line has a positive slope, since solid carbon dioxide is more dense than liquid carbon dioxide. The triple point for carbon dioxide occurs at 5.1 atm and and the critical point occurs at 72.8 atm and At a pressure of 1 atm, solid carbon dioxide sublimes 31°C.
56.6°C, TABLE 10.10 Boiling Point of Water at Various Locations Feet Above Patm Boiling Point Location Sea Level (torr) (°C) Top of Mt. Everest, Tibet 29,028 240 70 Top of Mt. McKinley, Alaska 20,320 340 79 Top of Mt. Whitney, Calif.
14,494 430 85 Leadville, Colo.
10,150 510 89 Top of Mt. Washington, N.H.
6,293 590 93 Boulder, Colo.
5,430 610 94 Madison, Wis.
900 730 99 New York City, N.Y.
10 760 100 Death Valley, Calif.
282 770 100.3 A carbon dioxide fire extinguisher.
10.9 Phase Diagrams 471 density of diamond). The high temperature used in the GE process is necessary to disrupt the bonds in graphite so that diamond (the most stable form of carbon at the high pres-sures used in the process) can form. Once the diamond is produced, the elemental carbon is “trapped” in this form at normal conditions ( 1 atm) because the reaction back to the graphite form is so slow. That is, even though graphite is more stable than diamond at and 1 atm, diamond can exist almost indefinitely because the conversion to graphite is a very slow reaction. As a result, diamonds formed at the high pressures found deep in the earth’s crust can be brought to the earth’s surface by natural geologic processes and continue to exist for millions of years. We have seen that diamond formed in the laboratory at high pressures is “trapped” in this form, but this process is very expensive. Can diamond be formed at low pressures?
The phase diagram for carbon says no. However, researchers have found that under the right conditions diamonds can be 25°C 25°C, In Morocco, a 50-km-long slab called Beni Bousera contains chunks of graphite that were probably once diamonds formed in the deposit when it was buried 150 km underground. As this slab slowly rose to the surface over millions of years, the very slow reaction changing diamond to graphite had time to occur. On the other hand, in the diamond-rich kimberlite deposits in South Africa, which rise to the sur-face much faster, the diamonds have not had sufficient time to revert to graphite.
“grown” at low pressures. The process used is called chemical vapor deposition (CVD). CVD uses an energy source to re-lease carbon atoms from a compound such as methane into a steady flow of hydrogen gas (some of which is dissociated to produce hydrogen atoms). The carbon atoms then deposit as a diamond film on a surface maintained at a temperature between 600 and Why does diamond form on this surface rather than the favored graphite? Nobody is sure, but it has been suggested that at these relatively high tempera-tures the diamond structure grows faster than the graphite structure and so diamond is favored under these conditions.
It also has been suggested that the hydrogen atoms present react much faster with graphite fragments than with diamond fragments, effectively removing any graphite from the grow-ing film. Once it forms, of course, diamond is trapped. The major advantage of CVD is that there is no need for the ex-traordinarily high pressures used in the traditional process for synthesizing diamonds.
The first products with diamond films are already on the market. Audiophiles can buy tweeters that have diaphragms coated with a thin diamond film that limits sound distortion.
Watches with diamond-coated crystals are planned, as are diamond-coated windows in infrared scanning devices used in analytical instruments and missile guidance systems.
These applications represent only the beginning for diamond-coated products.
900°C.
at a property that leads to its common name, dry ice. No liquid phase occurs un-der normal atmospheric conditions, making dry ice a convenient refrigerant.
Carbon dioxide is often used in fire extinguishers, where it exists as a liquid at under high pressures. Liquid carbon dioxide released from the extinguisher into the envi-ronment at 1 atm immediately changes to a vapor. Being heavier than air, this vapor smoth-ers the fire by keeping oxygen away from the flame. The liquid/vapor transition is highly endothermic, so cooling also results, which helps to put out the fire.
25°C 78°C, Pc = 72.8 1.00 P3 = 5.1 Tm T3 Tc Solid Liquid Gas Triple point Temperature (°C) –78 –56.6 31 Pressure (atm) Critical point FIGURE 10.52 The phase diagram for carbon dioxide. The liquid state does not exist at a pressure of 1 atm. The solid/liquid line has a positive slope, since the density of solid carbon dioxide is greater than that of liquid carbon dioxide.
472 Chapter Ten Liquids and Solids For Review Condensed states of matter: liquids and solids Held together by forces among the component molecules, atoms, or ions Liquids exhibit properties such as surface tension, capillary action, and viscosity that depend on the forces among the components Dipole–dipole forces Attractions among molecules with dipole moments Hydrogen bonding is a particularly strong form of dipole–dipole attraction • Occurs in molecules containing hydrogen bonded to a highly electronegative ele-ment such as nitrogen, oxygen, or fluorine • Produces unusually high boiling points London dispersion forces Caused by instantaneous dipoles that form in atoms or nonpolar molecules Crystalline solids Have a regular arrangement of components often represented as a lattice; the smallest repeating unit of the lattice is called the unit cell Classified by the types of components: • Atomic solids (atoms) • Ionic solids (ions) • Molecular solids (molecules) Arrangement of the components can be determined by X-ray analysis Metals Structure is modeled by assuming atoms to be uniform spheres • Closest packing • Hexagonal • Cubic Metallic bonding can be described in terms of two models • Electron sea model: valence electrons circulate freely among the metal cations • Band model: electrons are assumed to occupy molecular orbitals • Conduction bands: closely spaced molecular orbitals with empty electron spaces Alloys: mixtures with metallic properties • Substitutional • Interstitial Network solids Contain giant networks of atoms covalently bound together Examples are diamond and graphite Silicates are network solids containing bridges that form the basis for many rocks, clays, and ceramics Semiconductors Very pure silicon is “doped” with other elements • n-type: doping atoms typically contain five valence electrons (one more than silicon) • p-type: doping elements typically contain three valence electrons Modern electronics are based on devices with p–n junctions Molecular solids Components are discrete molecules Intermolecular forces are typically weak, leading to relatively low boiling and melting points Si¬O¬Si Key Terms Section 10.1 condensed states intermolecular forces dipole–dipole attraction hydrogen bonding London dispersion forces Section 10.2 surface tension capillary action viscosity Section 10.3 crystalline solid amorphous solid lattice unit cell X-ray diffraction ionic solid molecular solid atomic solid Section 10.4 closest packing hexagonal closest packed (hcp) structure cubic closest packed (ccp) structure band model molecular orbital (MO) model alloy substitutional alloy interstitial alloy Section 10.5 network solid silica silicate glass ceramic semiconductor n-type semiconductor p-type semiconductor p–n junction Section 10.8 vaporization (evaporation) heat of vaporization enthalpy of vaporization ( ) condensation equilibrium equilibrium vapor pressure sublimation heating curve enthalpy (heat) of fusion ( ) normal melting point normal boiling point supercooled superheated ¢Hfus ¢Hvap For Review 473 Section 10.9 phase diagram triple point critical temperature critical pressure critical point Ionic solids Components are ions Interionic forces are relatively strong, leading to solids with high melting and boiling points Many structures consist of closest packing of the larger ions with the smaller ions in tetrahedral or octahedral holes Phase changes The change from liquid to gas (vapor) is called vaporization or evaporation Condensation is the reverse of vaporization Equilibrium vapor pressure: the pressure that occurs over a liquid or solid in a closed system when the rate of evaporation equals the rate of condensation • Liquids whose components have high intermolecular forces have relatively low vapor pressures • Normal boiling point: the temperature at which the vapor pressure of a liquid equals one atmosphere • Normal melting point: the temperature at which a solid and its liquid have the same vapor pressure (at 1 atm external pressure) Phase diagram • Shows what state exists at a given temperature and pressure in a closed system • Triple point: temperature at which all three phases exist simultaneously • Critical point: defined by the critical temperature and pressure • Critical temperature: the temperature above which the vapor cannot be lique-fied no matter the applied pressure • Critical pressure: the pressure required to produce liquefaction at the critical temperature REVIEW QUESTIONS 1. What are intermolecular forces? How do they differ from intramolecular forces?
What are dipole–dipole forces? How do typical dipole–dipole forces differ from hydrogen-bonding interactions? In what ways are they similar? What are London dispersion forces? How do typical London dispersion forces differ from dipole– dipole forces? In what ways are they similar? Describe the relationship between molecular size and strength of London dispersion forces. Place the major types of intermolecular forces in order of increasing strength. Is there some overlap?
That is, can the strongest London dispersion forces be greater than some dipole–dipole forces? Give an example of such an instance.
2. Define the following terms and describe how each depends on the strength of the intermolecular forces.
a. surface tension b. viscosity c. melting point d. boiling point e. vapor pressure 3. Compare and contrast solids versus liquids versus gases.
4. Distinguish between the items in the following pairs.
a. crystalline solid; amorphous solid b. ionic solid; molecular solid c. molecular solid; network solid d. metallic solid; network solid 5. What is a lattice? What is a unit cell? Describe a simple cubic unit cell. How many net atoms are contained in a simple cubic unit cell? How is the radius of the atom 474 Chapter Ten Liquids and Solids Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. It is possible to balance a paper clip on the surface of water in a beaker. If you add a bit of soap to the water, however, the pa-per clip sinks. Explain how the paper clip can float and why it sinks when soap is added.
2. Consider a sealed container half-filled with water. Which state-ment best describes what occurs in the container?
a. Water evaporates until the air is saturated with water vapor; at this point, no more water evaporates.
b. Water evaporates until the air is overly saturated (supersatu-rated) with water, and most of this water recondenses; this cycle continues until a certain amount of water vapor is pres-ent, and then the cycle ceases.
c. Water does not evaporate because the container is sealed.
d. Water evaporates, and then water evaporates and recondenses simultaneously and continuously.
related to the cube edge length for a simple cubic unit cell? Answer the same ques-tions for the body-centered cubic unit cell and for the face-centered unit cell.
6. What is closest packing? What is the difference between hexagonal closest pack-ing and cubic closest packing? What is the unit cell for each closest packing?
7. Use the band model to describe differences among insulators, conductors, and semiconductors. Also use the band model to explain why each of the following increases the conductivity of a semiconductor.
a. increasing the temperature b. irradiating with light c. adding an impurity How do conductors and semiconductors differ as to the effect of temperature on electrical conductivity? How can an n-type semiconductor be produced from pure germanium? How can a p-type semiconductor be produced from pure germanium?
8. Describe, in general, the structures of ionic solids. Compare and contrast the structure of sodium chloride and zinc sulfide. How many tetrahedral holes and octahedral holes are there per closest packed anion? In zinc sulfide, why are only one-half of the tetrahedral holes filled with cations?
9. Define each of the following.
a. evaporation b. condensation c. sublimation d. boiling e. melting f. enthalpy of vaporization g. enthalpy of fusion h. heating curve Why is the enthalpy of vaporization for water much greater than its enthalpy of fusion? What does this say about the changes in intermolecular forces in going from solid to liquid to vapor? What do we mean when we say that a liquid is volatile? Do volatile liquids have large or small vapor pressures at room temper-ature? What strengths of intermolecular forces occur in highly volatile liquids?
10. Compare and contrast the phase diagrams of water versus carbon dioxide. Why doesn’t CO2 have a normal melting point and a normal boiling point, whereas water does? The slopes of the solid–liquid lines in the phase diagrams of H2O and CO2 are different. What do the slopes of the solid–liquid lines indicate in terms of the relative densities of the solid and liquid states for each substance?
How do the melting points of H2O and CO2 depend on pressure? How do the boiling points of H2O and CO2 depend on pressure? Rationalize why the critical temperature for H2O is greater than that for CO2.
Exercises 475 e. Water evaporates until it is eventually all in vapor form.
Explain each choice. Justify your choice, and for choices you did not pick, explain what is wrong with them.
3. Explain the following: You add 100 mL of water to a 500-mL round-bottom flask and heat the water until it is boiling. You remove the heat and stopper the flask, and the boiling stops.
You then run cool water over the neck of the flask, and the boiling begins again. It seems as though you are boiling water by cooling it.
4. Is it possible for the dispersion forces in a particular substance to be stronger than the hydrogen bonding forces in another sub-stance? Explain your answer.
5. Does the nature of intermolecular forces change when a sub-stance goes from a solid to a liquid, or from a liquid to a gas?
What causes a substance to undergo a phase change?
6. Why do liquids have a vapor pressure? Do all liquids have vapor pressures? Explain. Do solids exhibit vapor pressure? Explain.
How does vapor pressure change with changing temperature?
Explain.
7. Water in an open beaker evaporates over time. As the water is evaporating, is the vapor pressure increasing, decreasing, or stay-ing the same? Why?
8. What is the vapor pressure of water at How do you know?
9. Refer to Fig. 10.44. Why doesn’t temperature increase continu-ously over time? That is, why does the temperature stay constant for periods of time?
10. Which are stronger, intermolecular or intramolecular forces for a given molecule? What observation(s) have you made that sup-port this? Explain.
11. Why does water evaporate?
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 12. The nonpolar hydrocarbon C25H52 is a solid at room temperature. Its boiling point is greater than Which has the stronger intermolecular forces, C25H52 or H2O? Explain your answer.
13. Atoms are assumed to touch in closest packed structures, yet every closest packed unit cell contains a significant amount of empty space. Why?
14. Define critical temperature and critical pressure. In terms of the kinetic molecular theory, why is it impossible for a substance to exist as a liquid above its critical temperature?
15. Use the kinetic molecular theory to explain why a liquid gets cooler as it evaporates from an insulated container.
16. Will a crystalline solid or an amorphous solid give a simpler X-ray diffraction pattern? Why?
17. What is an alloy? Explain the differences in structure between substitutional and interstitial alloys. Give an example of each type.
18. Describe what is meant by a dynamic equilibrium in terms of the vapor pressure of a liquid.
400°C.
100°C?
19. How does each of the following affect the rate of evaporation of a liquid in an open dish?
a. intermolecular forces b. temperature c. surface area 20. When a person has a severe fever, one therapy used to reduce the fever is an “alcohol rub.” Explain how the evaporation of al-cohol from a person’s skin removes heat energy from the body.
21. When wet laundry is hung on a clothesline on a cold winter day, it will freeze but eventually dry. Explain.
22. Why is a burn from steam typically much more severe than a burn from boiling water?
23. You have three covalent compounds with three very different boiling points. All of the compounds have similar molar mass and relative shape. Explain how these three compounds could have very different boiling points.
24. Compare and contrast the structures of the following solids.
a. diamond versus graphite b. silica versus silicates versus glass 25. Compare and contrast the structures of the following solids.
a. CO2(s) versus H2O(s) b. NaCl(s) versus CsCl(s); See Exercise 61 for the structures.
26. Silicon carbide (SiC) is an extremely hard substance that acts as an electrical insulator. Propose a structure for SiC.
27. A plot of ln Pvap versus 1 T (K) is linear with a negative slope.
Why is this the case?
28. Iodine, like most substances, exhibits only three phases; solid, liquid, and vapor. The triple point of iodine is at 90 torr and . Which of the following statements concerning liquid I2 must be true? Explain your answer.
a. I2(l) is more dense than I2(g).
b. I2(l) cannot exist above .
c. I2(l) cannot exist at 1 atmosphere pressure.
d. I2(l) cannot have a vapor pressure greater than 90 torr.
e. I2(l) cannot exist at a pressure of 10 torr.
Exercises In this section similar exercises are paired.
Intermolecular Forces and Physical Properties 29. Identify the most important types of interparticle forces present in the solids of each of the following substances.
a. Ar e. CH4 b. HCl f. CO c. HF g. NaNO3 d. CaCl2 30. Identify the most important types of interparticle forces present in the solids of each of the following substances.
a. NH4Cl b. Teflon, CF3(CF2CF2)nCF3 c. Polyethylene, CH3(CH2CH2)nCH3 d. CHCl3 e. NH3 f. NO g. BF3 115°C 115°C 476 Chapter Ten Liquids and Solids 31. Predict which substance in each of the following pairs would have the greater intermolecular forces.
a. CO2 or OCS b. SeO2 or SO2 c. CH3CH2CH2NH2 or H2NCH2CH2NH2 d. CH3CH3 or H2CO e. CH3OH or H2CO 32. Consider the compounds Cl2, HCl, F2, NaF, and HF. Which com-pound has a boiling point closest to that of argon? Explain.
33. Rationalize the difference in boiling points for each of the fol-lowing pairs of substances: a. n-pentane CH3CH2CH2CH2CH3 C neopentane b. HF C HCl C c. HCl C LiCl C d. n-pentane CH3CH2CH2CH2CH3 C n-hexane CH3CH2CH2CH2CH2CH3 34. Consider the following compounds and formulas. (Note: The for-mulas are written in such a way as to give you an idea of the structure.) ethanol: CH3CH2OH dimethyl ether: CH3OCH3 propane: CH3CH2CH3 The boiling points of these compounds are (in no particular order) , and . Match the boiling points to the correct compounds.
35. In each of the following groups of substances, pick the one that has the given property. Justify your answer.
a. highest boiling point: HBr, Kr, or Cl2 b. highest freezing point: H2O, NaCl, or HF c. lowest vapor pressure at C: Cl2, Br2, or I2 d. lowest freezing point: N2, CO, or CO2 e. lowest boiling point: CH4, CH3CH3, or CH3CH2CH3 f. highest boiling point: HF, HCl, or HBr O B g. lowest vapor pressure at 25ºC: CH3CH2CH3, CH3CCH3, or CH3CH2CH2OH 36. In each of the following groups of substances, pick the one that has the given property. Justify each answer.
a. highest boiling point: CCl4, CF4, CBr4 b. lowest freezing point: LiF, F2, HCl c. smallest vapor pressure at C: CH3OCH3, CH3CH2OH, CH3CH2CH3 d. greatest viscosity: H2S, HF, H2O2 25° 25° 78.5°C 23°C 42.1°C, 69°C 36.2° 1360° 85° 85° 20° H3C CH3 9.5°C CH3 CH3 C 36.2° e. greatest heat of vaporization: H2CO, CH3CH3, CH4 f. smallest enthalpy of fusion: I2, CsBr, CaO Properties of Liquids 37. The shape of the meniscus of water in a glass tube is different from that of mercury in a glass tube. Why?
38. Explain why water forms into beads on a waxed car finish.
39. Hydrogen peroxide (H2O2) is a syrupy liquid with a relatively low vapor pressure and a normal boiling point of Rationalize the differences of these physical properties from those of water.
40. Carbon diselenide (CSe2) is a liquid at room temperature. The normal boiling point is and the melting point is Carbon disulfide (CS2) is also a liquid at room temperature with normal boiling and melting points of and re-spectively. How do the strengths of the intermolecular forces vary from CO2 to CS2 to CSe2? Explain.
Structures and Properties of Solids 41. X rays from a copper X-ray tube ( 154 pm) were diffracted at an angle of 14.22 degrees by a crystal of silicon. Assuming first-order diffraction (n 1 in the Bragg equation), what is the interplanar spacing in silicon?
42. The second-order diffraction (n 2) for a gold crystal is at an angle of for X rays of 154 pm. What is the spacing between these crystal planes?
43. A topaz crystal has an interplanar spacing (d) of 1.36 Å (1 Å m). Calculate the wavelength of the X ray that should be used if (assume n 1).
44. X rays of wavelength 2.63 Å were used to analyze a crystal.
The angle of first-order diffraction (n 1 in the Bragg equa-tion) was 15.55 degrees. What is the spacing between crystal planes, and what would be the angle for second-order diffrac-tion (n 2)?
45. Calcium has a cubic closest packed structure as a solid. Assum-ing that calcium has an atomic radius of 197 pm, calculate the density of solid calcium.
46. Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 g/cm3. Calculate a value for the atomic radius of nickel.
u 15.0° 1 1010 22.20° l 111.6°C, 46.5°C 45.5°C.
125°C, 152.2°C.
H2O in glass Hg in glass Exercises 477 47. A certain form of lead has a cubic closest packed structure with an edge length of 492 pm. Calculate the value of the atomic radius and the density of lead.
48. You are given a small bar of an unknown metal X. You find the density of the metal to be 10.5 g/cm3. An X-ray diffraction ex-periment measures the edge of the face-centered cubic unit cell as 4.09 Å (1 Å 1010 m). Identify X.
49. Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm3. Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.) 50. Barium has a body-centered cubic structure. If the atomic radius of barium is 222 pm, calculate the density of solid barium.
51. The radius of gold is 144 pm, and the density is 19.32 g/cm3.
Does elemental gold have a face-centered cubic structure or a body-centered cubic structure?
52. The radius of tungsten is 137 pm and the density is 19.3 g/cm3.
Does elemental tungsten have a face-centered cubic structure or a body-centered cubic structure?
53. What fraction of the total volume of a cubic closest packed struc-ture is occupied by atoms? ( .) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers.
54. Iron has a density of 7.86 g/cm3 and crystallizes in a body-centered cubic lattice. Show that only 68% of a body-centered lattice is actually occupied by atoms, and determine the atomic radius of iron.
55. Explain how doping silicon with either phosphorus or gal-lium increases the electrical conductivity over that of pure silicon.
56. Explain how a p–n junction makes an excellent rectifier.
57. Selenium is a semiconductor used in photocopying machines.
What type of semiconductor would be formed if a small amount of indium impurity is added to pure selenium?
58. The Group 3A/Group 5A semiconductors are composed of equal amounts of atoms from Group 3A and Group 5A—for example, InP and GaAs. These types of semiconductors are used in light-emitting diodes and solid-state lasers. What would you add to make a p-type semiconductor from pure GaAs? How would you dope pure GaAs to make an n-type semiconductor?
59. The band gap in aluminum phosphide (AlP) is 2.5 electron-volts ( ). What wavelength of light is emitted by an AlP diode?
60. An aluminum antimonide solid-state laser emits light with a wavelength of 730. nm. Calculate the band gap in joules.
61. The structures of some common crystalline substances are shown below. Show that the net composition of each unit cell corre-sponds to the correct formula of each substance.
1eV 1.6 1019 J Hint: Vsphere 4 3pr3 62. The unit cell for nickel arsenide is shown below. What is the for-mula of this compound?
63. Cobalt fluoride crystallizes in a closest packed array of fluoride ions with the cobalt ions filling one-half of the octahedral holes.
What is the formula of this compound?
64. The compounds Na2O, CdS, and ZrI4 all can be described as cu-bic closest packed anions with the cations in tetrahedral holes.
What fraction of the tetrahedral holes is occupied for each case?
65. What is the formula for the compound that crystallizes with a cubic closest packed array of sulfur ions, and that contains zinc ions in of the tetrahedral holes and aluminum ions in of the octahedral holes?
66. Assume the two-dimensional structure of an ionic compound, MxAy, is What is the empirical formula of this ionic compound?
1 2 1 8 Ni As Cl Na Cl Cs S Zn O Ti 478 Chapter Ten Liquids and Solids 67. A certain metal fluoride crystallizes in such a way that the fluor-ide ions occupy simple cubic lattice sites, while the metal ions occupy the body centers of half the cubes. What is the formula of the metal fluoride?
68. The structure of manganese fluoride can be described as a sim-ple cubic array of manganese ions with fluoride ions at the cen-ter of each edge of the cubic unit cell. What is the charge of the manganese ions in this compound?
69. The unit cell of MgO is shown below.
Does MgO have a structure like that of NaCl or ZnS? If the den-sity of MgO is 3.58 g/cm3, estimate the radius (in centimeters) of the anions and the cations.
70. The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 61).
Given that the density of cesium chloride is 3.97 g/cm3, and as-suming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent and ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of is 169 pm, and the ionic radius of is 181 pm.
71. What type of solid will each of the following substances form?
a. CO2 e. Ru i. NaOH b. SiO2 f. I2 j. U c. Si g. KBr k. CaCO3 d. CH4 h. H2O l. PH3 72. What type of solid will each of the following substances form?
a. diamond e. KCl i. Ar b. PH3 f. quartz j. Cu c. H2 g. NH4NO3 k. C6H12O6 d. Mg h. SF2 73. The memory metal, nitinol, is an alloy of nickel and titanium. It is called a memory metal because after being deformed, a piece of nitinol wire will return to its original shape. The structure of nitinol consists of a simple cubic array of Ni atoms and an in-ner penetrating simple cubic array of Ti atoms. In the extended lattice, a Ti atom is found at the center of a cube of Ni atoms; the reverse is also true.
a. Describe the unit cell for nitinol.
b. What is the empirical formula of nitinol?
c. What are the coordination numbers (number of nearest neigh-bors) of Ni and Ti in nitinol?
Cl Cs Cl Cs Mg2 O2 74. The unit cell for a pure xenon fluoride compound is shown below.
What is the formula of the compound?
75. Perovskite is a mineral containing calcium, titanium, and oxygen.
Two different representations of the unit cell are shown below.
Show that both these representations give the same formula and the same number of oxygen atoms around each titanium atom.
76. A mineral crystallizes in a cubic closest packed array of oxygen ions with aluminum ions in some of the octahedral holes and magnesium ions in some of the tetrahedral holes. Deduce the formula of this mineral and predict the fraction of octahedral holes and tetrahedral holes that are filled by the various cations.
77. Materials containing the elements Y, Ba, Cu, and O that are su-perconductors (electrical resistance equals zero) at temperatures Barium Oxygen Copper Yttrium (a) (b) Actual structure of superconductor Ideal perovskite structure Titanium Calcium Oxygen Xenon Fluorine Exercises 479 above that of liquid nitrogen were recently discovered. The struc-tures of these materials are based on the perovskite structure. Were they to have the ideal perovskite structure, the superconductor would have the structure shown in part (a) of the figure above.
a. What is the formula of this ideal perovskite material?
b. How is this structure related to the perovskite structure shown in Exercise 75?
These materials, however, do not act as superconductors unless they are deficient in oxygen. The structure of the actual super-conducting phase appears to be that shown in part (b) of the figure.
c. What is the formula of this material?
78. The structures of another class of ceramic, high-temperature su-perconductors are shown in the figure below.
a. Determine the formula of each of these four superconductors.
b. One of the structural features that appears to be essential for high-temperature superconductivity is the presence of planar sheets of copper and oxygen atoms. As the number of sheets in each unit cell increases, the temperature for the onset of superconductivity increases. Order the four structures from lowest to the highest superconducting temperature.
c. Assign oxidation states to Cu in each structure assuming Tl exists as Tl3. The oxidation states of Ca, Ba, and O are assumed to be 2, 2, and 2, respectively.
d. It also appears that copper must display a mixture of oxida-tion states for a material to exhibit superconductivity. Explain how this occurs in these materials as well as in the super-conductor in Exercise 77.
(a) Ca (b) (c) (d) Ba Cu O Ti 81. In Breckenridge, Colorado, the typical atmospheric pressure is 520. torr. What is the boiling point of water ( kJ/mol) in Breckenridge?
82. What pressure would have to be applied to steam at to condense the steam to liquid water ( kJ/mol)?
83. Carbon tetrachloride, CCl4, has a vapor pressure of 213 torr at and 836 torr at What is the normal boiling point of CCl4?
84. The normal boiling point for acetone is At an elevation of 5300 ft the atmospheric pressure is 630. torr. What would be the boiling point of acetone ( kJ/mol) at this ele-vation? What would be the vapor pressure of acetone at at this elevation?
85. A substance, X, has the following properties: 25.0°C ¢Hvap 32.0 56.5°C.
80.°C.
40.°C ¢Hvap 40.7 350.°C ¢Hvap 40.7 Sketch a heating curve for substance X starting at 86. Given the data in Exercise 85 on substance X, calculate the en-ergy that must be removed to convert 250. g of substance X from a gas at to a solid at Assume X has a molar mass of 75.0 g/mol.
50.°C.
100.°C 50.°C.
Phase Changes and Phase Diagrams 79. Plot the following data and determine Hvap for magnesium and lithium. In which metal is the bonding stronger?
¢ Vapor Pressure Temperature (°C) (mm Hg) Li Mg 1.
750.
620.
10.
890.
740.
100.
1080.
900.
400.
1240.
1040.
760.
1310.
1110.
Temperature (°C) Vapor Pressure (mm Hg) 0.
14.4 10.
26.6 20.
47.9 30.
81.3 40.
133 50.
208 80.
670.
Specific Heat Capacities 20. kJ/mol C(s) 3.0 5.0 kJ/mol C(l) 2.5 bp C C(g) 1.0 mp C 15° J/g °C 75° J/g °C ¢Hfus J/g °C ¢Hvap 80. From the following data for liquid nitric acid, determine its heat of vaporization and normal boiling point.
480 Chapter Ten Liquids and Solids 87. How much energy does it take to convert 0.500 kg ice at to steam at C? Specific heat capacities: ice, 2.03 ; liquid, 4.2 ; steam, 2.0 , kJ/mol, kJ/mol.
88. Consider a 75.0-g sample of H2O(g) at What phase or phases are present when 215 kJ of energy is removed from this sample? (See Exercise 87.) 89. An ice cube tray contains enough water at to make 18 ice cubes that each have a mass of 30.0 g. The tray is placed in a freezer that uses CF2Cl2 as a refrigerant. The heat of vaporiza-tion of CF2Cl2 is 158 J/g. What mass of CF2Cl2 must be vapor-ized in the refrigeration cycle to convert all the water at to ice at ? The heat capacities for H2O(s) and H2O(l) are 2.03 and 4.18 respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
90. A 0.250-g chunk of sodium metal is cautiously dropped into a mixture of 50.0 g of water and 50.0 g of ice, both at The reaction is Will the ice melt? Assuming the final mixture has a specific heat capacity of 4.18 calculate the final temperature. The en-thalpy of fusion for ice is 6.02 kJ/mol.
91. Consider the phase diagram given below. What phases are present at points A through H? Identify the triple point, normal boiling point, normal freezing point, and critical point. Which phase is denser, solid or liquid?
92. Sulfur exhibits two solid phases, rhombic and monoclinic. Use the accompanying phase diagram for sulfur to answer the fol-lowing questions. (The phase diagram is not to scale.) Temperature Pressure 1 atm 153˚C, 1420 atm 95.39˚C, 1 atm 115.21˚C, 1 atm 444.6˚C, 1 atm 115.18˚C, 3.2 10–5 atm 95.31˚C, 5.1 10–6 atm Rhombic Monoclinic Liquid Gas 1.0 atm D E A B F G H C J/g °C, 2Na1s2 2H2O1l2 ¡ 2NaOH1aq2 H21g2 ¢H 368 kJ 0°C.
J/g °C, J/g °C 5.0°C 22.0°C 22.0°C 125°C.
¢Hfus 6.02 ¢Hvap 40.7 J/g °C J/g °C J/g °C 250.° 20.°C a. How many triple points are in the phase diagram?
b. What phases are in equilibrium at each of the triple points?
c. What is the stable phase at 1 atm and ?
d. What are the normal melting point and the normal boiling point of sulfur?
e. Which is the densest phase?
f. At a pressure of atm, can rhombic sulfur sublime?
g. What phase changes occur when the pressure on a sample of sulfur at is increased from atm to 1500 atm?
93. Use the accompanying phase diagram for carbon to answer the following questions.
a. How many triple points are in the phase diagram?
b. What phases can coexist at each triple point?
c. What happens if graphite is subjected to very high pressures at room temperature?
d. If we assume that the density increases with an increase in pressure, which is more dense, graphite or diamond?
94. Like most substances, bromine exists in one of the three typical phases. Br2 has a normal melting point of and a normal boiling point of The triple point for Br2 is and 40 torr, and the critical point is and 100 atm. Using this information, sketch a phase diagram for bromine indicating the points described above. Based on your phase diagram, order the three phases from least dense to most dense. What is the stable phase of Br2 at room temperature and 1 atm? Under what tem-perature conditions can liquid bromine never exist? What phase changes occur as the temperature of a sample of bromine at 0.10 atm is increased from to ?
95. The melting point of a fictional substance X is at 10.0 atm.
If the density of the solid phase of X is 2.67 g/cm3 and the den-sity of the liquid phase is 2.78 g/cm3 at 10.0 atm, predict whether the normal melting point of X will be less than, equal to, or greater than Explain.
96. Consider the following data for xenon: Triple point: 280 torr Normal melting point: C Normal boiling point: C Which is more dense, Xe(s), or Xe(l)? How do the melting point and boiling point of xenon depend on pressure?
107° 112° 121°C, 225°C.
225°C 200°C 50°C 320°C 7.3°C 59°C.
7.2°C 107 109 1011 0 2000 4000 6000 Temperature (K) Pressure (Pa) Diamond Graphite Liquid Vapor 1.0 108 100.°C 1.0 105 100.°C Challenge Problems 481 Additional Exercises 97. Rationalize why chalk (calcium carbonate) has a higher melting point than motor oil (large compounds made from carbon and hydrogen), which has a higher melting point than water, which engages in relatively strong hydrogen-bonding interactions.
98. Rationalize the differences in physical properties in terms of in-termolecular forces for the following organic compounds. Com-pare the first three substances with each other, compare the last three with each other, and then compare all six. Can you account for any anomalies?
99. Consider the following vapor pressure versus temperature plot for three different substances A, B, and C.
If the three substances are CH4, SiH4, and NH3, match each curve to the correct substance.
100. Consider the following enthalpy changes: How do the strengths of hydrogen bonds vary with the elec-tronegativity of the element to which hydrogen is bonded? Where in the preceding series would you expect hydrogen bonds of the following type to fall?
¢H 21 kJ/mol H2O1g2 HOH1g2 ¡ H2O ---HOH 1in ice2 ¢H 46 kJ/mol 1CH322C“O HF ¡ 1CH322C“O ---HF F HF ¡ FHF ¢H 155 kJ/mol Pvap (torr) Temperature (°C) A B C Label the four substances as either ionic, network, metallic, or molecular solids.
104. A 20.0-g sample of ice at is mixed with 100.0 g of wa-ter at . Calculate the final temperature of the mixture as-suming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.03 and 4.18 respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
105. In regions with dry climates, evaporative coolers are used to cool air. A typical electric air conditioner is rated at Btu/h (1 Btu, or British thermal unit amount of energy to raise the temperature of 1 lb of water by ). How much water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner?
106. The critical point of NH3 is and 111 atm, and the critical point of N2 is and 34 atm. Which of these substances cannot be liquefied at room temperature no matter how much pressure is applied? Explain.
Challenge Problems 107. When 1 mol of benzene is vaporized at a constant pressure of 1.00 atm and its boiling point of 353.0 K, 30.79 kJ of energy (heat) is absorbed and the volume change is 28.90 L. What are and for this process?
108. You and a friend each synthesize a compound with the formula XeCl2F2. Your compound is a liquid and your friend’s compound is a gas (at the same conditions of temperature and pressure).
Explain how the two compounds with the same formulas can exist in different phases at the same conditions of pressure and temperature.
109. Using the heats of fusion and vaporization for water given in Ex-ercise 87, calculate the change in enthalpy for the sublimation of water: H2O1s2 ¡ H2O1g2 ¢H ¢E 147°C 132°C 1°F 1.00 104 J/g °C, 80.0°C 10.0°C 101. How could you tell experimentally if TiO2 is an ionic solid or a network solid?
102. Boron nitride (BN) exists in two forms. The first is a slippery solid formed from the reaction of BCl3 with NH3, followed by heating in an ammonia atmosphere at Subjecting the first form of BN to a pressure of 85,000 atm at produces a second form that is the second hardest substance known. Both forms of BN remain solids to Suggest structures for the two forms of BN.
103. Consider the following data concerning four different substances.
3000°C.
1800°C 750°C.
bp (°C) mp (°C) (kJ/mol) Benzene, C6H6 80 6 33.9 Naphthalene, C10H8 218 80 51.5 Carbon tetra-chloride 76 23 31.8 Acetone, CH3COCH3 56 95 31.8 Acetic acid, CH3CO2H 118 17 39.7 Benzoic acid, C6H5CO2H 249 122 68.2 ¢Hvap Conducts Electricity Compound as a Solid Other Properties B2H6 no gas at SiO2 no high mp CsI no aqueous solution conducts electricity W yes high mp 25°C 482 Chapter Ten Liquids and Solids Using the value given in Exercise 100 and the number of hydrogen bonds formed with each water molecule, estimate what portion of the intermolecular forces in ice can be accounted for by hydrogen bonding.
110. Oil of wintergreen, or methyl salicylate, has the following structure: Methyl-4-hydroxybenzoate is another molecule with exactly the same molecular formula; it has the following structure: Account for the large difference in the melting points of the two substances.
111. Consider the following melting point data: O OH mp 8°C C OCH3 ¢H 115. Mn crystallizes in the same type of cubic unit cell as Cu.
Assuming that the radius of Mn is 5.6% larger than the radius of Cu and the density of copper is 8.96 g/cm3, calculate the density of Mn.
116. You are asked to help set up a historical display in the park by stacking some cannonballs next to a Revolutionary War cannon.
You are told to stack them by starting with a triangle in which each side is composed of four touching cannonballs. You are to continue stacking them until you have a single ball on the top centered over the middle of the triangular base.
a. How many cannonballs do you need?
b. What type of closest packing is displayed by the cannonballs?
c. The four corners of the pyramid of cannonballs form the corners of what type of regular geometric solid?
117. Some water is placed in a sealed glass container connected to a vacuum pump (a device used to pump gases from a container), and the pump is turned on. The water appears to boil and then freezes. Explain these changes using the phase diagram for wa-ter. What would happen to the ice if the vacuum pump was left on indefinitely?
118. The molar enthalpy of vaporization of water at 373 K and 1.00 atm is 40.7 kJ/mol. What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.) 119. For a simple cubic array, solve for the volume of an interior sphere (cubic hole) in terms of the radius of a sphere in the array.
120. Consider two different compounds, each with the formula C2H6O.
One of these compounds is a liquid at room conditions and the other is a gas. Write Lewis structures consistent with this observa-tion and explain your answer. Hint: the oxygen atom in both struc-tures satisfies the octet rule with two bonds and two lone pairs.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
121. A 0.132-mol sample of an unknown semiconducting material with the formula XY has a mass of 19.0 g. The element X has an electron configuration of [Kr]5s24d10. What is this semicon-ducting material? A small amount of the Y atoms in the semi-conductor is replaced with an equivalent amount of atoms with and electron configuration of [Ar]4s23d104p5. Does this corre-spond to n-type or p-type doping?
122. A metal burns in air at under high pressure to form an oxide with formula MO2. This compound is 23.72% oxy-gen by mass. The distance between touching atoms in a cubic closest packed crystal of this metal is 269.0 pm. What is this metal? What is its density?
123. One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at followed by condensation of the mercury vapor. Given the heat of vaporiza-tion of mercury (296 J/g) and the vapor pressure of mercury at ( torr), what is the vapor pressure of the condensed mercury at C? How many atoms of mercury are present in the mercury vapor at if the reaction is con-ducted in a closed 15.0-L container?
300.°C 300.° 2.56 103 25.0°C 600.°C 600°C Compound: NaCl MgCl2 AlCl3 SiCl4 PCl3 SCl2 Cl2 mp : 801 708 190 Compound: NaF MgF2 AlF3 SiF4 PF5 SF6 F2 mp : 997 1396 1040 220 56 94 90 1°C2 101 78 91 70 1°C2 Account for the trends in melting points in terms of interparti-cle forces.
112. MnO has either the NaCl type structure or the CsCl type struc-ture (see Exercise 70). The edge length of the MnO unit cell is cm and the density of MnO is 5.28 g/cm3.
a. Does MnO crystallize in the NaCl or the CsCl type structure?
b. Assuming that the ionic radius of oxygen is 140. pm, esti-mate the ionic radius of manganese.
113. Some ionic compounds contain a mixture of different charged cations. For example, some titanium oxides contain a mixture of Ti2 and Ti3 ions. Consider a certain oxide of titanium that is 28.31% oxygen by mass and contains a mixture of Ti2 and Ti3 ions. Determine the formula of the compound and the relative numbers of Ti2 and Ti3 ions.
114. Spinel is a mineral that contains 37.9% aluminum, 17.1% mag-nesium, and 45.0% oxygen, by mass, and has a density of 3.57 g/cm3. The edge of the cubic unit cell measures 809 pm.
How many of each type of ion are present in the unit cell?
4.47 108 Marathon Problem 483 Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
124. General Zod has sold Lex Luthor what Zod claims to be a new copper-colored form of kryptonite, the only substance that can harm Superman. Lex, not believing in honor among thieves, de-cided to carry out some tests on the supposed kryptonite. From previous tests, Lex knew that kryptonite is a metal having a spe-cific heat capacity of 0.082 and a density of 9.2 g/cm3.
Lex Luthor’s first experiment was an attempt to find the specific heat capacity of kryptonite. He dropped a sample of the metal into a boiling water bath at a temperature of He waited until the metal had reached the bath temperature and then quickly transferred it to of water that was contained in a calorimeter at an initial tem-perature of The final temperature of the metal and water was Based on these results, is it possible to distinguish between copper and kryptonite? Explain.
25.2°C.
25.0°C 0.2°C.
100 g 3 g 100.0°C 0.2°C.
10 g 3 g J/g °C, When Lex found that his results from the first experiment were inconclusive, he decided to determine the density of the sample. He managed to steal a better balance and determined the mass of another portion of the purported kryptonite to be 4 g 1 g. He dropped this sample into water contained in a 25-mL graduated cylinder and found that it displaced a volume of 0.42 mL 0.02 mL. Is the metal copper or kryptonite? Explain.
Lex was finally forced to determine the crystal structure of the metal General Zod had given him. He found that the cubic unit cell contained 4 atoms and had an edge length of 600. pm.
Explain how this information enabled Lex to identify the metal as copper or kryptonite.
Will Lex be going after Superman with the kryptonite or seeking revenge on General Zod? What improvements could he have made in his experimental techniques to avoid performing the crystal structure determination?
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
484 11Properties of Solutions Contents 11.1 Solution Composition 11.2 The Energies of Solution Formation 11.3 Factors Affecting Solubility • Structure Effects • Pressure Effects • Temperature Effects (for Aqueous Solutions) 11.4 The Vapor Pressures of Solutions • Nonideal Solutions 11.5 Boiling-Point Elevation and Freezing-Point Depression • Boiling-Point Elevation • Freezing-Point Depression 11.6 Osmotic Pressure • Reverse Osmosis 11.7 Colligative Properties of Electrolyte Solutions 11.8 Colloids Opals are formed from colloidal suspensions of silica when the liquid evaporates.
Most of the substances we encounter in daily life are mixtures: Wood, milk, gasoline, champagne, seawater, shampoo, steel, and air are common examples. When the components of a mixture are uniformly intermingled—that is, when a mixture is homogeneous—it is called a solution. Solutions can be gases, liquids, or solids, as shown in Table 11.1. However, we will be concerned in this chapter with the properties of liquid solutions, particularly those containing water. As we saw in Chapter 4, many essential chemical reactions occur in aqueous solutions because water is capable of dissolving so many substances.
11.1 Solution Composition Because a mixture, unlike a chemical compound, has a variable composition, the relative amounts of substances in a solution must be specified. The qualitative terms dilute (rela-tively little solute present) and concentrated (relatively large amount of solute) are often used to describe solution content, but we need to define solution composition more pre-cisely to perform calculations. For example, in dealing with the stoichiometry of solution reactions in Chapter 4, we found it useful to describe solution composition in terms of molarity, or the number of moles of solute per liter of solution (symbolized by M).
Other ways of describing solution composition are also useful. Mass percent (some-times called weight percent) is the percent by mass of the solute in the solution: Another way of describing solution composition is the mole fraction (symbolized by the Greek lowercase letter chi, ), the ratio of the number of moles of a given component to the total number of moles of solution. For a two-component solution, where nA and nB represent the number of moles of the two components, Mole fraction of component A xA nA nA nB x Mass percent a mass of solute mass of solutionb 100% 485 A solute is the substance being dissolved.
The solvent is the dissolving medium.
When liquids are mixed, the liquid present in the largest amount is called the solvent.
Molarity moles of solute liters of solution TABLE 11.1 Various Types of Solutions State of State of State of Example Solution Solute Solvent Air, natural gas Gas Gas Gas Vodka in water, Liquid Liquid Liquid antifreeze Brass Solid Solid Solid Carbonated water Liquid Gas Liquid (soda) Seawater, sugar Liquid Solid Liquid solution Hydrogen in Solid Gas Solid platinum 486 Chapter Eleven Properties of Solutions Still another way of describing solution composition is molality (symbolized by m), the number of moles of solute per kilogram of solvent: Various Methods for Describing Solution Composition A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molal-ity of ethanol in this solution.
Solution Molarity: The moles of ethanol can be obtained from its molar mass (46.07 g/mol): Mass percent: Mole fraction: Molality: See Exercises 11.25 through 11.27.
0.217 m 2.17 102 mol 0.1000 kg Molality of C2H5OH moles of C2H5OH kilogram of H2O 2.17 102 mol 100.0 g 1 kg 1000 g 2.17 102 5.58 0.00389 xC2H5OH 2.17 102 mol 2.17 102 mol 5.56 mol nH2O 100.0 g H2O 1 mol H2O 18.0 g H2O 5.56 mol Mole fraction of C2H5OH nC2H5OH nC2H5OH nH2O 0.990% C2H5OH a 1.00 g C2H5OH 100.0 g H2O 1.00 g C2H5OHb 100% Mass percent C2H5OH amass of C2H5OH mass of solution b 100% 0.215 M Molarity of C2H5OH moles of C2H5OH liters of solution 2.17 102 mol 0.101 L Volume 101 mL 1 L 1000 mL 0.101 L 1.00 g C2H5OH 1 mol C2H5OH 46.07 g C2H5OH 2.17 102 mol C2H5OH Molality moles of solute kilogram of solvent Since molarity depends on the volume of the solution, it changes slightly with temperature. Molality is independent of temperature because it depends only on mass.
Sample Exercise 11.1 In very dilute aqueous solutions, the magnitude of the molality and the molar-ity are almost the same.
11.1 Solution Composition 487 Another concentration measure sometimes encountered is normality (symbolized by N). Normality is defined as the number of equivalents per liter of solution, where the definition of an equivalent depends on the reaction taking place in the solution. For an acid–base reaction, the equivalent is the mass of acid or base that can furnish or accept exactly 1 mole of protons (H ions). In Table 11.2 note, for example, that the equivalent mass of sulfuric acid is the molar mass divided by 2, since each mole of H2SO4 can fur-nish 2 moles of protons. The equivalent mass of calcium hydroxide is also half the mo-lar mass, since each mole of Ca(OH)2 contains 2 moles of OH ions that can react with 2 moles of protons. The equivalent is defined so that 1 equivalent of acid will react with exactly 1 equivalent of base.
For oxidation–reduction reactions, the equivalent is defined as the quantity of oxidizing or reducing agent that can accept or furnish 1 mole of electrons. Thus 1 equivalent of reducing agent will react with exactly 1 equivalent of oxidizing agent. The equivalent mass of an ox-idizing or reducing agent can be calculated from the number of electrons in its half-reaction.
For example, MnO4 reacting in acidic solution absorbs five electrons to produce Mn2: Since the MnO4 ion present in 1 mole of KMnO4 consumes 5 moles of electrons, the equivalent mass is the molar mass divided by 5: Calculating Various Methods of Solution Composition from the Molarity The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality, and normality of the sulfuric acid.
Solution The density of the solution in grams per liter is Thus 1 liter of this solution contains 1230. g of the mixture of sulfuric acid and water.
Since the solution is 3.75 M, we know that 3.75 mol H2SO4 is present per liter of solu-tion. The number of grams of H2SO4 present is 3.75 mol 98.1 g H2SO4 1 mol 368 g H2SO4 1.230 g mL 1000 mL 1 L 1.230 103 g/L Equivalent mass of KMnO4 molar mass 5 158 g 5 31.6 g MnO4 5e 8H ¡ Mn2 4H2O The definition of an equivalent depends on the reaction taking place in the solution.
The quantity we call equivalent mass here traditionally has been called equiva-lent weight.
Oxidation–reduction half-reactions were discussed in Section 4.10.
TABLE 11.2 The Molar Mass, Equivalent Mass, and Relationship of Molarity and Normality for Several Acids and Bases Molar Equivalent Relationship of Molarity Acid or Base Mass Mass and Normality HCl 36.5 36.5 1 M 1 N H2SO4 98 49 1 M 2 N NaOH 40 40 1 M 1 N Ca(OH)2 74 37 1 M 2 N 74 2 98 2 Sample Exercise 11.2 A modern 12-volt lead storage battery of the type used in automobiles.
488 Chapter Eleven Properties of Solutions The amount of water present in 1 liter of solution is obtained from the difference Since we now know the masses of the solute and solvent, we can calculate the mass percent.
From the moles of solute and the mass of solvent we can calculate the molality.
Since each sulfuric acid molecule can furnish two protons, 1 mol H2SO4 represents 2 equivalents. Thus a solution with 3.75 mol H2SO4 per liter contains equivalents per liter, and the normality is 7.50 N.
See Exercise 11.31.
11.2 The Energies of Solution Formation Dissolving solutes in liquids is very common. We dissolve salt in the water used to cook vegetables, sugar in iced tea, stains in cleaning fluid, gaseous carbon dioxide in water to make soda water, ethanol in gasoline to make gasohol, and so on.
2 3.75 7.50 3.75 mol H2SO4 862 g H2O 1 kg H2O 1000 g H2O 4.35 m Molality of H2SO4 moles H2SO4 kilogram of H2O 29.9% H2SO4 Mass percent H2SO4 mass of H2SO4 mass of solution 100% 368 g 1230. g 100% 1230. g solution 368 g H2SO4 862 g H2O CHEMICAL IMPACT Electronic Ink T he printed page has been a primary means of commu-nication for over 3000 years, and researchers at the Massachusetts Institute of Technology (MIT) believe they have discovered why. It seems that the brain responds pos-itively to fixed images on a sheet of paper, particularly those areas of the brain that store and process “spatial maps.” In comparison, information displayed on computer screens or TV screens seems to lack some of the visual signals that stimulate the learning centers of the brain to retain knowledge. While modern technology provides us with many other media by which we can communicate, the appeal of written words on a piece of paper remains.
Surprisingly, the technology of printing has changed very little since the invention of the printing press—that is, until now.
In the past several years Joseph M. Jacobson and his students at MIT have developed a prototype of a self-printing page. The key to this self-printing “paper” is microencap-sulation technology—the same technology that is used in “carbonless” carbon paper and “scratch-and-sniff” cologne and perfume advertisements in magazines. Jacobson’s sys-tem involves the use of millions of transparent fluid-filled capsules containing microscopic particles. These particles are colored and positively charged on one side and white and negatively charged on the other. When an electric field is selectively applied to the capsules, the white side of the microparticles can be oriented upward or the colored side can be caused to flip up. Appropriate application of an electric field can orient the particles in such a way as to produce words, and once the words have been created, virtually no more energy is needed to keep the particles in place. An image can be maintained on a page with con-sumption of only 50 millionths of an amp of power! The entire display is about 200 m thick (2.5 times that of paper) m 11.2 The Energies of Solution Formation 489 Solubility is important in other ways. For example, because the pesticide DDT is fat-soluble, it is retained and concentrated in animal tissues, where it causes detrimental effects. This is why DDT, even though it is effective for killing mosquitos, has been banned in the United States. Also, the solubility of various vitamins is important in determining correct dosages. The insolubility of barium sulfate means it can be used safely to improve X rays of the gastrointestinal tract, even though Ba2 ions are quite toxic.
What factors affect solubility? The cardinal rule of solubility is like dissolves like. We find that we must use a polar solvent to dissolve a polar or ionic solute and a nonpolar solvent to dissolve a nonpolar solute. Now we will try to understand why this behavior occurs. To simplify the discussion, we will assume that the formation of a liquid solution takes place in three distinct steps.
➥ 1 Separating the solute into its individual components (expanding the solute).
➥ 2 Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).
➥ 3 Allowing the solute and solvent to interact to form the solution.
These steps are illustrated in Fig. 11.1. Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. In other words, steps 1 and 2 are endothermic, and step 3 is often exothermic. The enthalpy change associated with the formation of the solution, called the enthalpy (heat) of solu-tion ( ), is the sum of the values for the steps: where may have a positive sign (energy absorbed) or a negative sign (energy re-leased), as shown in Fig. 11.2.
¢Hsoln ¢Hsoln ¢H1 ¢H2 ¢H3 ¢H ¢Hsoln and is so flexible and durable that it can be curled around a pencil and can operate at temperatures from 4 to Presently, print resolution is not as good as a modern laser printer, but reduction of the microencapsulated particles from 50 to 40 m should produce print that rivals the quality of the laser printer.
The first commercial applications of this technology are expected to appear in retail stores across the country in the form of electronic signs that can be updated instantly from a central location. The present technology is a long way from being able to create electronic books, but this is the eventual goal of Jacobson’s research team. It seems very likely that this electronic ink technology will con-tribute greatly to the evolution of the printed page over the next century.
m 158°F.
Signs like this one created by E Ink are the first to use electronic ink, which can be updated from a computer inside the store or from a remote location.
DDT The enthalpy of solution is the sum of the energies used in expanding both solvent and solute and the energy of solvent– solute interaction.
Polar solvents dissolve polar solutes; nonpolar solvents dissolve nonpolar solutes.
FIGURE 11.2 The heat of solution (a) Hsoln has a nega-tive sign (the process is exothermic) if step 3 releases more energy than that required by steps 1 and 2. (b) Hsoln has a positive sign (the process is endothermic) if steps 1 and 2 require more energy than is released in step 3. (If the energy changes for steps 1 and 2 equal that for step 3, then Hsoln is zero.) 490 Chapter Eleven Properties of Solutions To illustrate the importance of the various energy terms in the equation for , we will consider two specific cases. First, we know that oil is not soluble in water. When oil tankers leak, the petroleum forms an oil slick that floats on the water and is eventually carried onto the beaches. We can explain the immiscibility of oil and water by consider-ing the energy terms involved. Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. We expect to be small for a typical nonpolar solute, but it will be relatively large for the large oil molecules. The term H3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, H2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus will be large and positive because of the and terms. Since a large amount of energy would have to be expended to form an oil–water solution, this process does not occur to any appreciable extent. These same arguments hold true for any nonpolar solute and polar solvent—the combination of a non-polar solute and a highly polar solvent is not expected to produce a solution.
¢H2 ¢H1 ¢Hsoln ¢H1 ¢Hsoln H1 is expected to be small for nonpolar solutes but can be large for large molecules.
Expanded solute Solute Solvent Expanded solvent Solution Step 1 Step 2 ∆H1 + ∆H2 + ∆H3 ∆H1 ∆Hsoln ∆H2 Step 3 ∆H3 FIGURE 11.1 The formation of a liquid solution can be divided into three steps: (1) expanding the solute, (2) expanding the solvent, and (3) combining the expanded solute and solvent to form the solution.
Energy (H) ∆H1 + ∆H2 ∆Hsoln Energy of separated solute and solvent Energy of solution ∆H3 (a) Energy (H) ∆H1 + ∆H2 ∆Hsoln Energy of separated solute and solvent Energy of solution ∆H3 (b) 11.2 The Energies of Solution Formation 491 As a second case, let’s consider the solubility of an ionic solute, such as sodium chlo-ride, in water. Here the term is large and positive because the strong ionic forces in the crystal must be overcome, and is large and positive because hydrogen bonds must be broken in the water. Finally, is large and negative because of the strong interac-tions between the ions and the water molecules. In fact, the exothermic and endothermic terms essentially cancel, as shown from the known values: Here the enthalpy (heat) of hydration ( Hhyd) combines the terms H2 (for expand-ing the solvent) and H3 (for solvent–solute interactions). The heat of hydration represents the enthalpy change associated with the dispersal of a gaseous solute in water. Thus the heat of solution for dissolving sodium chloride is the sum of H1 and Hhyd: Note that Hsoln is small but positive; the dissolving process requires a small amount of energy. Then why is NaCl so soluble in water? The answer lies in nature’s tendency toward higher probability of the mixed state. That is, processes naturally run in the di-rection that leads to the most probable state. For example, imagine equal numbers of or-ange and yellow spheres separated by a partition, as shown in Fig. 11.3(a). If we remove the partition and shake the container, the spheres will mix [Fig. 11.3(b)], and no amount of shaking will cause them to return to the state of separated orange and yellow. Why?
The mixed state is simply much more likely to occur (more probable) than the original separate state because there are many more ways of placing the spheres to give a mixed state than a separated state. This is a general principle. One factor that favors a process is an increase in probability.
But energy considerations are also important. Processes that require large amounts of energy tend not to occur. Since dissolving 1 mole of solid NaCl requires only a small amount of energy, the solution forms, presumably because of the large increase in the probability of the state when the solute and solvent are mixed.
The various possible cases for solution formation are summarized in Table 11.3. Note that in two cases, polar–polar and nonpolar–nonpolar, the heat of solution is expected to be small. In these cases, the solution forms because of the increase in the probability of the mixed state. In the other cases (polar–nonpolar and nonpolar–polar), the heat of so-lution is expected to be large and positive, and the large quantity of energy required acts to prevent the solution from forming. Although this discussion has greatly oversimplified the complex driving forces for solubility, these ideas are a useful starting point for un-derstanding the observation that like dissolves like.
¢ ¢Hsoln 786 kJ/mol 783 kJ/mol 3 kJ/mol ¢ ¢ ¢ ¢ ¢ 783 kJ/mol H2O1l2 Na1g2 Cl1g2 ¡ Na1aq2 Cl1aq2 ¢Hhyd ¢H2 ¢H3 NaCl1s2 ¡ Na1g2 Cl1g2 ¢H1 786 kJ/mol ¢H3 ¢H2 ¢H1 The factors that act as driving forces for a process are discussed more fully in Chapter 16.
(a) (b) FIGURE 11.3 (a) Orange and yellow spheres separated by a partition in a closed container. (b) The spheres after the partition is removed and the container has been shaken for some time.
Gasoline floating on water. Since gasoline is nonpolar, it is immiscible with water, be-cause water contains polar molecules.
492 Chapter Eleven Properties of Solutions Differentiating Solvent Properties Decide whether liquid hexane (C6H14) or liquid methanol (CH3OH) is the more appro-priate solvent for the substances grease (C20H42) and potassium iodide (KI).
Solution Hexane is a nonpolar solvent because it contains COH bonds. Thus hexane will work best for the nonpolar solute grease. Methanol has an OOH group that makes it signifi-cantly polar. Thus it will serve as the better solvent for the ionic solid KI.
See Exercises 11.37 through 11.39.
11.3 Factors Affecting Solubility Structure Effects In the last section we saw that solubility is favored if the solute and solvent have similar polarities. Since it is the molecular structure that determines polarity, there should be a definite connection between structure and solubility. Vitamins provide an excellent ex-ample of the relationship among molecular structure, polarity, and solubility.
Recently, there has been considerable publicity about the pros and cons of consum-ing large quantities of vitamins. For example, large doses of vitamin C have been advo-cated to combat various illnesses, including the common cold. Vitamin E has been ex-tolled as a youth-preserving elixir and a protector against the carcinogenic (cancer-causing) effects of certain chemicals. However, there are possible detrimental effects from taking large amounts of some vitamins, depending on their solubilities.
Vitamins can be divided into two classes: fat-soluble (vitamins A, D, E, and K) and water-soluble (vitamins B and C). The reason for the differing solubility characteristics can be seen by comparing the structures of vitamins A and C (Fig. 11.4). Vitamin A, com-posed mostly of carbon and hydrogen atoms that have similar electronegativities, is vir-tually nonpolar. This causes it to be soluble in nonpolar materials such as body fat, which is also largely composed of carbon and hydrogen, but not soluble in polar solvents such as water. On the other hand, vitamin C has many polar O—H and C—O bonds, making the molecule polar and thus water-soluble. We often describe nonpolar materials such as vitamin A as hydrophobic (water-fearing) and polar substances such as vitamin C as hy-drophilic (water-loving).
Because of their solubility characteristics, the fat-soluble vitamins can build up in the fatty tissues of the body. This has both positive and negative effects. Since these vitamins Sample Exercise 11.3 TABLE 11.3 The Energy Terms for Various Types of Solutes and Solvents H1 H2 H3 Hsoln Outcome Polar solute, Large Large Large, Small Solution polar solvent negative forms Nonpolar solute, Small Large Small Large, No solution polar solvent positive forms Nonpolar solute, Small Small Small Small Solution nonpolar solvent forms Polar solute, Large Small Small Large, No solution nonpolar solvent positive forms Hexane Liquid methanol Grease Visualization: Ammonia Fountain Visualization: Micelle Formation: The Cleansing Action of Soap 11.3 Factors Affecting Solubility 493 can be stored, the body can tolerate for a time a diet deficient in vitamins A, D, E, or K.
Conversely, if excessive amounts of these vitamins are consumed, their buildup can lead to the illness hypervitaminosis.
In contrast, the water-soluble vitamins are excreted by the body and must be con-sumed regularly. This fact was first recognized when the British navy discovered that scurvy, a disease often suffered by sailors, could be prevented if the sailors regularly ate fresh limes (which are a good source of vitamin C) when aboard ship (hence the name “limey” for the British sailor).
Pressure Effects While pressure has little effect on the solubilities of solids or liquids, it does signifi-cantly increase the solubility of a gas. Carbonated beverages, for example, are always bottled at high pressures of carbon dioxide to ensure a high concentration of car-bon dioxide in the liquid. The fizzing that occurs when you open a can of soda re-sults from the escape of gaseous carbon dioxide because under these conditions the pressure of CO2 above the solution is now much lower than that used in the bottling process.
The increase in gas solubility with pressure can be understood from Fig. 11.5.
Figure 11.5(a) shows a gas in equilibrium with a solution; that is, the gas molecules are entering and leaving the solution at the same rate. If the pressure is suddenly increased [Fig. 11.5(b)], the number of gas molecules per unit volume increases, and the gas enters the solution at a higher rate than it leaves. As the concentration of dis-solved gas increases, the rate of the escape of the gas also increases until a new equi-librium is reached [Fig. 11.5(c)], where the solution contains more dissolved gas than before.
CH3 CH3 CH3 CH3 H CH3 C C C C C C O H C C C H H H H H H H H H H H H C C C C C C H O O C C C C C C H O O H O H H H H O H Vitamin A Vitamin C H FIGURE 11.4 The molecular structures of (a) vitamin A (nonpolar, fat-soluble) and (b) vitamin C (polar, water-soluble). The circles in the structural formulas indicate polar bonds. Note that vitamin C contains far more polar bonds than vitamin A.
Carbonation in a bottle of soda.
494 Chapter Eleven Properties of Solutions The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law: where C represents the concentration of the dissolved gas, k is a constant characteristic of a particular solution, and P represents the partial pressure of the gaseous solute above the solution. In words, Henry’s law states that the amount of a gas dissolved in a solu-tion is directly proportional to the pressure of the gas above the solution.
Henry’s law is obeyed most accurately for dilute solutions of gases that do not dis-sociate in or react with the solvent. For example, Henry’s law is obeyed by oxygen gas in water, but it does not correctly represent the behavior of gaseous hydrogen chloride in water because of the dissociation reaction HCl1g2 ¡ H2O H1aq2 Cl1aq2 C kP CHEMICAL IMPACT Ionic Liquids?
S o far in this text, you have seen that ionic substances are stable solids with high melting points. For example, sodium chloride has a melting point near One of the “hottest” areas of current chemical research is ionic liquids—substances composed of ions that are liquids at nor-mal temperatures and pressures. This unusual behavior re-sults from the differences in the sizes of the anions and cations in the ionic liquids. Dozens of small anions, such as (tetrafluoroborate) or (hexafluorophosphate), can be paired with thousands of large cations, such as 1-hexyl-3-methylimidazolium or 1-butyl-3-methylimidazolium (parts a and b respectively, in the accompanying figure). These sub-stances remain liquids because the bulky, asymmetrical cations do not pack together efficiently with the smaller, PF6 BF4 800°C.
symmetrical anions. In contrast, in sodium chloride the ions can pack very efficiently to form a compact, orderly arrange-ment, leading to maximum cation–anion attractions and thus a high melting point.
The excitement being generated by these ionic liquids arises from many factors. For one thing, almost an infinite variety of ionic liquids are possible due to the large variety of bulky cations and small anions available. According to Kenneth R. Seddon, Director of QUILL (Queen’s University Ionic Liquid Laboratory) in Northern Ireland, a trillion ionic liquids are possible. Another great advantage of these liquids is their long liquid range, typically from to .
In addition, the cations in the liquids can be designed to perform specific functions. For example, chemist James 200°C 100°C William Henry (1774–1836), a close friend of John Dalton, formulated his law in 1801.
Henry’s law holds only when there is no chemical reaction between the solute and solvent.
(a) (b) (c) Solution FIGURE 11.5 (a) A gaseous solute in equilibrium with a solution. (b) The piston is pushed in, which increases the pressure of the gas and the number of gas molecules per unit volume.
This causes an increase in the rate at which the gas enters the solution, so the concen-tration of dissolved gas increases. (c) The greater gas concentration in the solution causes an increase in the rate of escape. A new equilibrium is reached.
11.3 Factors Affecting Solubility 495 Calculations Using Henry’s Law A certain soft drink is bottled so that a bottle at contains CO2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO2 in the atmosphere is atm, calculate the equilibrium concentrations of CO2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO2 in aqueous solution is mol/L atm at Solution We can write Henry’s law for CO2 as where mol/L atm. In the unopened bottle, 5.0 atm and In the opened bottle, the CO2 in the soda eventually reaches equilibrium with the atmos-pheric CO2, so atm and Note the large change in concentration of CO2. This is why soda goes “flat” after being open for a while.
See Exercises 11.43 and 11.44.
Temperature Effects (for Aqueous Solutions) Everyday experiences of dissolving substances such as sugar may lead you to think that solubility always increases with temperature. This is not the case. The dissolving of a solid occurs more rapidly at higher temperatures, but the amount of solid that can be dis-solved may increase or decrease with increasing temperature. The effect of temperature CCO2 kCO2PCO2 a3.1 102 mol L atmb14.0 104 atm2 1.2 105 mol/L PCO2 4.0 104 CCO2 kCO2PCO2 13.1 102 mol/L atm215.0 atm2 0.16 mol/L PCO2 3.1 102 kCO2 CCO2 kCO2PCO2 25°C.
3.1 102 4.0 104 25°C H. Davis, of the University of South Alabama in Mobile, has designed various cations that will attract potentially harm-ful ions such as mercury, cadmium, uranium, and americium (the latter two are commonly found in nuclear waste mate-rials) and leach them out of contaminated solutions. Davis has also developed cations that will remove H2S (which pro-duces SO2 when the gas is burned) and CO2 (which does not burn) from natural gas. Potentially, these ionic solutions might also be used to remove CO2 from the exhaust gases of fossil-fuel–burning power plants to lessen the “green-house effect.” The biggest obstacle to the widespread use of ionic liq-uids is their cost. Normal organic solvents used in industry typically cost a few cents per liter, but ionic liquids can cost hundreds of times that amount. However, the environmen-tally friendly nature of ionic liquids (they produce no va-pors because the ions are not volatile) and the flexibility of (a) (b) these substances as reaction media make them very attrac-tive. As a consequence, efforts are under way to make their use economically feasible.
The term ionic liquid may have seemed like an oxy-moron in the past, but these substances have a very promis-ing future.
Sample Exercise 11.4 H soln refers to the formation of a 1.0 M ideal solution and is not necessarily rele-vant to the process of dissolving a solid in a saturated solution. Thus H soln is of limited use in predicting the variation of solubility with temperature.
496 Chapter Eleven Properties of Solutions on the solubility in water of several solids is shown in Fig. 11.6. Note that although the solubility of most solids in water increases with temperature, the solubilities of some substances (such as sodium sulfate and cerium sulfate) decrease with increasing temperature.
Predicting the temperature dependence of solubility is very difficult. For example, al-though there is some correlation between the sign of H soln and the variation of solubil-ity with temperature, important exceptions exist. The only sure way to determine the temperature dependence of a solid’s solubility is by experiment.
The behavior of gases dissolving in water appears less complex. The solubility of a gas in water typically decreases with increasing temperature,† as is shown for several cases in Fig. 11.7. This temperature effect has important environmental implications be-cause of the widespread use of water from lakes and rivers for industrial cooling. After being used, the water is returned to its natural source at a higher than ambient tempera-ture (thermal pollution has occurred). Because it is warmer, this water contains less than the normal concentration of oxygen and is also less dense; it tends to “float” on the colder water below, thus blocking normal oxygen absorption. This effect can be especially im-portant in deep lakes. The warm upper layer can seriously decrease the amount of oxy-gen available to aquatic life in the deeper layers of the lake.
The decreasing solubility of gases with increasing temperature is also responsible for the formation of boiler scale. As we will see in more detail in Chapter 14, the bicarbonate ion is formed when carbon dioxide is dissolved in water containing the carbonate ion: When the water also contains Ca2 ions, this reaction is especially important—calcium bicarbonate is soluble in water, but calcium carbonate is insoluble. When the water is heated, the carbon dioxide is driven off. For the system to replace the lost carbon diox-ide, the reverse reaction must occur: This reaction, however, also increases the concentration of carbonate ions, causing solid calcium carbonate to form. This solid is the boiler scale that coats the walls of contain-ers such as industrial boilers and tea kettles. Boiler scale reduces the efficiency of heat transfer and can lead to blockage of pipes (see Fig. 11.8).
2HCO3 1aq2 ¡ H2O1l2 CO21aq2 CO3 21aq2 CO3 21aq2 CO21aq2 H2O1l2 ¡ 2HCO3 1aq2 For more information see R. S. Treptow, “Le Châtelier’s Principle Applied to the Temperature Depen-dence of Solubility,” J. Chem. Ed. 61 (1984): 499.
†The opposite behavior is observed for most nonaqueous solvents.
0 20 60 100 140 180 220 260 300 Sugar (C12H22O11) KNO3 NaNO3 NaBr KBr KCl Na2SO4 Ce2(SO4)3 0 20 40 60 80 100 Temperature (°C) Solubility (g solute/100 g H2O) FIGURE 11.6 The solubilities of several solids as a func-tion of temperature. Note that while most substances become more soluble in water with increasing temperature, sodium sulfate and cerium sulfate become less soluble.
Methane Oxygen Carbon monoxide Nitrogen Helium 0 10 20 30 Temperature (°C) 2.0 1.0 Solubility (10 –3 mol/L) FIGURE 11.7 The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.
FIGURE 11.8 A pipe with accumulated mineral de-posits. The cross section clearly indicates the reduction in pipe capacity.
11.4 The Vapor Pressures of Solutions 497 CHEMICAL IMPACT The Lake Nyos Tragedy O n August 21, 1986, a cloud of gas suddenly boiled from Lake Nyos in Cameroon, killing nearly 2000 people.
Although at first it was speculated that the gas was hydro-gen sulfide, it now seems clear it was carbon dioxide. What would cause Lake Nyos to emit this huge, suffocating cloud of CO2? Although the answer may never be known for cer-tain, many scientists believe that the lake suddenly “turned over,” bringing to the surface water that contained huge quantities of dissolved carbon dioxide. Lake Nyos is a deep lake that is thermally stratified: Layers of warm, less dense water near the surface float on the colder, denser water lay-ers near the lake’s bottom. Under normal conditions the lake stays this way; there is little mixing among the different lay-ers. Scientists believe that over hundreds or thousands of years, carbon dioxide gas had seeped into the cold water at the lake’s bottom and dissolved in great amounts because of the large pressure of CO2 present (in accordance with Henry’s law). For some reason on August 21, 1986, the lake apparently suffered an overturn, possibly due to wind or to unusual cooling of the lake’s surface by monsoon clouds.
This caused water that was greatly supersaturated with CO2 to reach the surface and release tremendous quantities of gaseous CO2 that suffocated thousands of humans and ani-mals before they knew what hit them—a tragic, monumen-tal illustration of Henry’s law.
Since 1986 the scientists studying Lake Nyos and nearby Lake Monoun have observed a rapid recharging of the CO2 levels in the deep waters of these lakes, causing concern that another deadly gas release could occur at any time. Apparently the only way to prevent such a disaster is to pump away the CO2-charged deep water in the two lakes.
Scientists at a conference to study this problem in 1994 rec-ommended such a solution, but it has not yet been funded by Cameroon.
Lake Nyos in Cameroon.
11.4 The Vapor Pressures of Solutions Liquid solutions have physical properties significantly different from those of the pure solvent, a fact that has great practical importance. For example, we add antifreeze to the water in a car’s cooling system to prevent freezing in winter and boiling in summer. We also melt ice on sidewalks and streets by spreading salt. These preventive measures work because of the solute’s effect on the solvent’s properties.
To explore how a nonvolatile solute affects a solvent, we will consider the experi-ment represented in Fig. 11.9, in which a sealed container encloses a beaker containing an aqueous sulfuric acid solution and a beaker containing pure water. Gradually, the volume of the sulfuric acid solution increases and the volume of the pure water decreases. Why?
We can explain this observation if the vapor pressure of the pure solvent is greater than that of the solution. Under these conditions, the pressure of vapor necessary to achieve equilibrium with the pure solvent is greater than that required to reach equilibrium with the aqueous acid solution. Thus, as the pure solvent emits vapor to attempt to reach equi-librium, the aqueous sulfuric acid solution absorbs vapor to try to lower the vapor pres-sure toward its equilibrium value. This process results in a net transfer of water from the pure water through the vapor phase to the sulfuric acid solution. The system can reach an equilibrium vapor pressure only when all the water is transferred to the solution. This A nonvolatile solute has no tendency to escape from solution into the vapor phase.
498 Chapter Eleven Properties of Solutions experiment is just one of many observations indicating that the presence of a nonvolatile solute lowers the vapor pressure of a solvent.
We can account for this behavior in terms of the simple model shown in Fig. 11.10.
The dissolved nonvolatile solute decreases the number of solvent molecules per unit vol-ume and it should proportionately lower the escaping tendency of the solvent molecules.
For example, in a solution consisting of half nonvolatile solute molecules and half sol-vent molecules, we might expect the observed vapor pressure to be half that of the pure solvent, since only half as many molecules can escape. In fact, this is what is observed.
Detailed studies of the vapor pressures of solutions containing nonvolatile solutes were carried out by François M. Raoult (1830–1901). His results are described by the equation known as Raoult’s law: where Psoln is the observed vapor pressure of the solution, is the mole fraction of solvent, and is the vapor pressure of the pure solvent. Note that for a solution of half solute and half solvent molecules, is 0.5, so the vapor pressure of the solution is half that of the pure solvent. On the other hand, for a solution in which three-fourths of the solution molecules are solvent, and The idea is that the nonvolatile solute simply dilutes the solvent.
Psoln 0.75P0 solvent.
xsolvent 3 4 0.75, xsolvent P0 solvent xsolvent Psoln xsolventP0 solvent Visualization: Vapor Pressure Lowering: Liquid/Vapor Equilibrium Visualization: Vapor Pressure Lowering: Addition of a Solute Visualization: Vapor Pressure Lowering: Solution/Vapor Equilibrium Water vapor Water Aqueous solution Aqueous solution (a) (b) FIGURE 11.9 An aqueous solution and pure water in a closed environment. (a) Initial stage.
(b) After a period of time, the water is transferred to the solution.
Pure solvent Solution with a nonvolatile solute 1 2 3 4 5 6 1 2 3 4 5 6 FIGURE 11.10 The presence of a nonvolatile solute in-hibits the escape of solvent molecules from the liquid and so lowers the vapor pressure of the solvent.
11.4 The Vapor Pressures of Solutions 499 Raoult’s law is a linear equation of the form where , , and Thus a plot of versus gives a straight line with a slope equal to , as shown in Fig. 11.11.
Calculating the Vapor Pressure of a Solution Calculate the expected vapor pressure at for a solution prepared by dissolving 158.0 g of common table sugar (sucrose, molar mass 342.3 g/mol) in 643.5 cm3 of water. At the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.
Solution We will use Raoult’s law in the form To calculate the mole fraction of water in the solution, we must first determine the num-ber of moles of sucrose: To determine the moles of water present, we first convert volume to mass using the density: The number of moles of water is therefore The mole fraction of water in the solution is Then Thus the vapor pressure of water has been lowered from 23.76 torr in the pure state to 23.46 torr in the solution. The vapor pressure has been lowered by 0.30 torr.
See Exercises 11.45 and 11.46.
The phenomenon of the lowering of the vapor pressure gives us a convenient way to “count” molecules and thus provides a means for experimentally determining molar masses. Suppose a certain mass of a compound is dissolved in a solvent and the vapor pressure of the resulting solution is measured. Using Raoult’s law, we can determine the number of moles of solute present. Since the mass of this number of moles is known, we can calculate the molar mass.
We also can use vapor pressure measurements to characterize solutions. For exam-ple, 1 mole of sodium chloride dissolved in water lowers the vapor pressure approximately twice as much as expected because the solid has two ions per formula unit, which sepa-rate when it dissolves. Thus vapor pressure measurements can give valuable information about the nature of the solute after it dissolves.
Psoln xH2OP0 H2O 10.98732123.76 torr2 23.46 torr 35.63 mol 36.09 mol 0.9873 xH2O mol H2O mol H2O mol sucrose 35.63 mol 35.63 mol 0.4616 mol 641.6 g H2O 1 mol H2O 18.01 g H2O 35.63 mol H2O 643.5 cm3 H2O 0.9971 g H2O cm3 H2O 641.6 g H2O 0.4616 mol sucrose Moles of sucrose 158.0 g sucrose 1 mol sucrose 342.3 g sucrose Psoln xH2OP0 H2O 25°C, 25°C P0 solvent xsolvent Psoln b 0.
m P0 solvent xsolvent x y Psoln, y mx b, Sample Exercise 11.5 0 1 Psoln Vapor pressure of pure solvent Mole fraction of solvent χsolvent Solution vapor pressure FIGURE 11.11 For a solution that obeys Raoult’s law, a plot of Psoln versus solvent gives a straight line.
Raoult’s law states that the vapor pres-sure of a solution is directly proportional to the mole fraction of solvent present.
The lowering of vapor pressure depends on the number of solute particles present in the solution.
500 Chapter Eleven Properties of Solutions Calculating the Vapor Pressure of a Solution Containing Ionic Solute Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4 (molar mass 142 g/mol) with 175 g water at . The vapor pressure of pure water at is 23.76 torr.
Solution First, we need to know the mole fraction of H2O.
It is essential to recognize that when 1 mole of solid Na2SO4 dissolves, it produces 2 mol Na+ ions and 1 mol ions. Thus the number of solute particles present in this solu-tion is three times the number of moles of solute dissolved: xH2O nH2O nsolute nH2O 9.72 mol 0.738 mol 9.72 mol 9.72 10.458 0.929 nsolute 310.2462 0.738 mol SO4 2 nNa2SO4 35.0 g Na2SO4 1 mol Na2SO4 142 g Na2SO4 0.246 mol Na2SO4 nH2O 175 g H2O 1 mol H2O 18.0 g H2O 9.72 mol H2O 25°C 25°C CHEMICAL IMPACT Spray Power P roducts in aerosol cans are widely used in our society.
We use hairsprays, mouth sprays, shaving cream, whipped cream, spray paint, spray cleaners, and many oth-ers. As in the case of most consumer products, chemistry plays an important role in making aerosol products work.
An aerosol is a mixture of small particles (solids or liq-uids) dispersed in some sort of medium (a gas or a liquid). An inspection of the ingredients in an aerosol can reveals a long list of chemical substances, all of which fall into one of three categories: (1) an active ingredient, (2) an inactive ingredient, or (3) a propellant. The active ingredients perform the func-tions for which the product was purchased (for example, the resins in hairspray). It is very important that the contents of an aerosol can be chemically compatible. If an undesired chem-ical reaction were to occur inside the can, it is likely that the product would be unable to perform its function. The inactive ingredients serve to keep the product properly mixed and pre-vent chemical reactions within the can prior to application. The propellant delivers the product out of the can.
Most aerosol products contain liquefied hydrocarbon pro-pellants such as propane (C3H8) and butane (C4H10). While these molecules are extremely flammable, they are excellent propellants, and they also help to disperse and mix the com-ponents of the aerosol can as they are delivered. These pro-pellants have critical temperatures above room temperature, For foods delivered by aerosol cans, propane and butane are obvi-ously not appropriate propellants. For substances such as whipped cream, the propellant N2O is often used.
Insecticide is sprayed from an aerosol can.
Sample Exercise 11.6 11.4 The Vapor Pressures of Solutions 501 Now we can use Raoult’s law to predict the vapor pressure: See Exercise 11.48.
Nonideal Solutions So far we have assumed that the solute is nonvolatile and so does not contribute to the vapor pressure over the solution. However, for liquid–liquid solutions where both com-ponents are volatile, a modified form of Raoult’s law applies: where PTOTAL represents the total vapor pressure of a solution containing A and B, and are the mole fractions of A and B, P0 A and P 0 B are the vapor pressures of pure A and pure B, and PA and PB are the partial pressures resulting from molecules of A and of B in the vapor above the solution (see Fig. 11.12).
xB xA PTOTAL PA PB xAP0 A xBP0 B Psoln xH2OP0 H2O 10.9292123.76 torr2 22.1 torr which means that the intermolecular forces among their mol-ecules are strong enough to form a liquid when pressure is applied. In the highly pressurized aerosol can, the liquid phase of the propellant is in equilibrium with the gaseous phase of the propellant in the head space of the can. The ability of the propellant to maintain this equilibrium is the key to how the aerosol can works. All aerosol cans are constructed in a sim-ilar way (see accompanying diagram). At the top of the can is a valve (acts to open and seal the can) and an actuator (to open the valve). Pushing the actuator opens the valve, and the propellant gas escapes through a long tube (the dip tube) that extends from the bottom of the can. With the valve open, the propellant, at a greater pressure than the atmosphere, escapes through the dip tube, carrying the active ingredient(s) with it.
The rapidly expanding gas propels the contents from the can and in some instances (for example, shaving cream, carpet shampoo) produces a foam. After each use, the remaining pro-pellant in the can reestablishes equilibrium between the liq-uid and gaseous phases, keeping the pressure constant within the can as long as sufficient propellant remains. The trick is to have the active and inactive ingredients and the propellant run out at the same time. Given the nature of the most com-mon propellants, you can understand the warning about not putting the “empty” cans in a fire.
Aqueous solution containing ingredients to be delivered Liquid propellant Propellant vapor An aerosol can for delivery of an active ingredient dissolved in an aqueous solution.
FIGURE 11.12 When a solution contains two volatile components, both contribute to the total vapor pressure. Note that in this case the solution contains equal numbers of the components and but the vapor contains more than . This means that component is more volatile (has a higher vapor pressure as a pure liquid) than component .
502 Chapter Eleven Properties of Solutions A liquid–liquid solution that obeys Raoult’s law is called an ideal solution. Raoult’s law is to solutions what the ideal gas law is to gases. As with gases, ideal behavior for solutions is never perfectly achieved but is sometimes closely approached. Nearly ideal behavior is often observed when the solute–solute, solvent–solvent, and solute–solvent interactions are very similar. That is, in solutions where the solute and solvent are very much alike, the solute simply acts to dilute the solvent. However, if the solvent has a spe-cial affinity for the solute, such as if hydrogen bonding occurs, the tendency of the solvent molecules to escape will be lowered more than expected. The observed vapor pressure will be lower than the value predicted by Raoult’s law; there will be a negative deviation from Raoult’s law.
When a solute and solvent release large quantities of energy in the formation of a solution, that is, when Hsoln is large and negative, we can assume that strong interactions exist between the solute and solvent. In this case we expect a negative deviation from Raoult’s law, because both components will have a lower escaping tendency in the solu-tion than in the pure liquids. This behavior is illustrated by an acetone–water solution, where the molecules can hydrogen-bond effectively: In contrast, if two liquids mix endothermically, it indicates that the solute–solvent interactions are weaker than the interactions among the molecules in the pure liquids.
More energy is required to expand the liquids than is released when the liquids are mixed.
In this case the molecules in the solution have a higher tendency to escape than expected, and positive deviations from Raoult’s law are observed (see Fig. 11.13). An example of this case is provided by a solution of ethanol and hexane, whose Lewis structures are as follows: A O O O O A A A O Q H C O H H H H H C A O O O A A A C H H H H H C O A A H H O A A H H O A A H H O A A H H H C C C C Ethanol Hexane G D P , D O H O C CH3 H CH3 O Vapor pressure of pure B Vapor pressure of pure A Vapor pressure of solution Partial pressure B Partial pressure A Vapor pressure χB Vapor pressure of solution Vapor pressure of solution (a) (b) (c) χA χB χA χB χA FIGURE 11.13 Vapor pressure for a solution of two volatile liquids. (a) The behavior predicted for an ideal liquid–liquid solution by Raoult’s law. (b) A solution for which PTOTAL is larger than the value calculated from Raoult’s law. This solution shows a positive deviation from Raoult’s law. (c) A solution for which PTOTAL is smaller than the value calculated from Raoult’s law. This solution shows a negative deviation from Raoult’s law.
Strong solute–solvent interaction gives a vapor pressure lower than that predicted by Raoult’s law.
11.4 The Vapor Pressures of Solutions 503 The polar ethanol and the nonpolar hexane molecules are not able to interact effectively. Thus the enthalpy of solution is positive, as is the deviation from Raoult’s law.
Finally, for a solution of very similar liquids, such as benzene and toluene (shown in margin), the enthalpy of solution is very close to zero, and thus the solution closely obeys Raoult’s law (ideal behavior).
A summary of the behavior of various types of solutions is given in Table 11.4.
Calculating the Vapor Pressure of a Solution Containing Two Liquids A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass 119.4 g/mol). At this solution has a total vapor pressure of 260. torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at are 345 and 293 torr, respectively.
Solution To decide whether this solution behaves ideally, we first calculate the expected vapor pres-sure using Raoult’s law: where A stands for acetone and C stands for chloroform. The calculated value can then be compared with the observed vapor pressure.
First, we must calculate the number of moles of acetone and chloroform: Since the solution contains equal numbers of moles of acetone and chloroform, that is, the expected vapor pressure is PTOTAL 10.50021345 torr2 10.50021293 torr2 319 torr x A 0.500 and x C 0.500 11.9 g chloroform 1 mol chloroform 119 g chloroform 0.100 mol chloroform 5.81 g acetone 1 mol acetone 58.1 g acetone 0.100 mol acetone PTOTAL xAP0 A xCP0 C 35°C 35°C, CH3 Benzene Toluene TABLE 11.4 Summary of the Behavior of Various Types of Solutions Interactive Forces Between Solute (A) T for and Solvent (B) Solution Deviation from Particles Hsoln Formation Raoult’s Law Example A ↔A, B ↔B q A ↔B Zero Zero None (ideal Benzene– solution) toluene A ↔A, B ↔B A ↔B Negative Positive Negative Acetone– (exothermic) water A ↔A, B ↔B A ↔B Positive Negative Positive Ethanol– (endothermic) hexane Acetone Chloroform Sample Exercise 11.7 FIGURE 11.14 Phase diagrams for pure water (red lines) and for an aqueous solution containing a nonvolatile solute (blue lines). Note that the boiling point of the solution is higher than that of pure water. Conversely, the freezing point of the solution is lower than that of pure water. The effect of a non-volatile solute is to extend the liquid range of a solvent.
504 Chapter Eleven Properties of Solutions Comparing this value with the observed pressure of 260. torr shows that the solution does not behave ideally. The observed value is lower than that expected. This negative devia-tion from Raoult’s law can be explained in terms of the hydrogen bonding interaction which lowers the tendency of these molecules to escape from the solution.
See Exercises 11.55 and 11.56.
11.5 Boiling-Point Elevation and Freezing-Point Depression In the preceding section we saw how a solute affects the vapor pressure of a liquid sol-vent. Because changes of state depend on vapor pressure, the presence of a solute also af-fects the freezing point and boiling point of a solvent. Freezing-point depression, boiling-point elevation, and osmotic pressure (discussed in Section 11.6) are called colligative properties. As we will see, they are grouped together because they depend only on the number, and not on the identity, of the solute particles in an ideal solution. Because of their direct relationship to the number of solute particles, the colligative properties are very useful for characterizing the nature of a solute after it is dissolved in a solvent and for determining molar masses of substances.
Boiling-Point Elevation The normal boiling point of a liquid occurs at the temperature where the vapor pres-sure is equal to 1 atmosphere. We have seen that a nonvolatile solute lowers the vapor pressure of the solvent. Therefore, such a solution must be heated to a higher temper-ature than the boiling point of the pure solvent to reach a vapor pressure of 1 atmos-phere. This means that a nonvolatile solute elevates the boiling point of the solvent.
Figure 11.14 shows the phase diagram for an aqueous solution containing a nonvolatile solute. Note that the liquid/vapor line is shifted to higher temperatures than those for pure water.
O G D P , G D O Cl H O C CH3 Cl Cl CH3 C Acetone Chloroform Visualization: Boiling-Point Elevation: Liquid/Vapor Equilibrium Visualization: Boiling-Point Elevation: Addition of a Solute Visualization: Boiling-Point Elevation: Solution/Vapor Equilibrium 1 atm Pressure (atm) ∆T f ∆Tb Freezing point of solution Freezing point of water Boiling point of water Boiling point of solution Temperature (°C) Vapor pressure of pure water Vapor pressure of solution Normal boiling point was defined in Section 10.8.
In this case the usually nonpolar COH bond is strongly polarized by the three attached, highly electronegative chlorine atoms, thus producing hydrogen bonding.
11.5 Boiling-Point Elevation and Freezing-Point Depression 505 As you might expect, the magnitude of the boiling-point elevation depends on the con-centration of the solute. The change in boiling point can be represented by the equation where is the boiling-point elevation, or the difference between the boiling point of the solution and that of the pure solvent, Kb is a constant that is characteristic of the solvent and is called the molal boiling-point elevation constant, and msolute is the molality of the solute in the solution.
Values of Kb for some common solvents are given in Table 11.5. The molar mass of a solute can be determined from the observed boiling-point elevation, as shown in Sam-ple Exercise 11.8.
Calculating the Molar Mass by Boiling-Point Elevation A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of . Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution.
Solution We make use of the equation where From Table 11.5, for water Kb 0.51. The molality of this solution then can be calcu-lated by rearranging the boiling-point elevation equation to give The solution was prepared using 0.1500 kg water. Using the definition of molality, we can find the number of moles of glucose in the solution.
Thus 0.10 mol glucose has a mass of 18.00 g, and 1.0 mol glucose has a mass of 180 g (10 18.00 g). The molar mass of glucose is 180 g/mol.
See Exercise 11.58.
nglucose 10.67 mol/kg210.1500 kg2 0.10 mol msolute 0.67 mol/kg mol solute kg solvent nglucose 0.1500 kg msolute ¢T Kb 0.34°C 0.51°C kg/mol 0.67 mol/kg ¢T 100.34°C 100.00°C 0.34°C ¢T Kbmsolute 100.34°C ¢T ¢T Kbmsolute TABLE 11.5 Molal Boiling-Point Elevation Constants (Kb) and Freezing-Point Depression Constants (Kf) for Several Solvents Boiling Freezing Point Kb Point Kf Solvent (C) (C kg/mol) (C) (C kg/mol) Water (H2O) 100.0 0.51 0 1.86 Carbon tetrachloride (CCl4) 76.5 5.03 22.99 30.
Chloroform (CHCl3) 61.2 3.63 63.5 4.70 Benzene (C6H6) 80.1 2.53 5.5 5.12 Carbon disulfide (CS2) 46.2 2.34 111.5 3.83 Ethyl ether (C4H10O) 34.5 2.02 116.2 1.79 Camphor (C10H16O) 208.0 5.95 179.8 40.
Sample Exercise 11.8 Sugar dissolved in water to make candy causes the boiling point to be elevated above 100ºC.
Spreading salt on a highway.
506 Chapter Eleven Properties of Solutions Freezing-Point Depression When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. Why? Recall that the vapor pressures of ice and liquid water are the same at Suppose a solute is dissolved in water. The resulting solution will not freeze at because the water in the solution has a lower vapor pressure than that of pure ice. No ice will form under these conditions. However, the vapor pressure of ice de-creases more rapidly than that of liquid water as the temperature decreases. Therefore, as the solution is cooled, the vapor pressure of the ice and that of the liquid water in the so-lution will eventually become equal. The temperature at which this occurs is the new freez-ing point of the solution and is below The freezing point has been depressed.
We can account for this behavior in terms of the simple model shown in Fig. 11.15.
The presence of the solute lowers the rate at which molecules in the liquid return to the solid state. Thus, for an aqueous solution, only the liquid state is found at As the so-lution is cooled, the rate at which water molecules leave the solid ice decreases until this rate and the rate of formation of ice become equal and equilibrium is reached. This is the freezing point of the water in the solution.
Because a solute lowers the freezing point of water, compounds such as sodium chlo-ride and calcium chloride are often spread on streets and sidewalks to prevent ice from forming in freezing weather. Of course, if the outside temperature is lower than the freez-ing point of the resulting salt solution, ice forms anyway. So this procedure is not effec-tive at extremely cold temperatures.
The solid/liquid line for an aqueous solution is shown on the phase diagram for wa-ter in Fig. 11.14. Since the presence of a solute elevates the boiling point and depresses the freezing point of the solvent, adding a solute has the effect of extending the liquid range.
The equation for freezing-point depression is analogous to that for boiling-point el-evation: where is the freezing-point depression, or the difference between the freezing point of the pure solvent and that of the solution, and Kf is a constant that is characteristic of a particular solvent and is called the molal freezing-point depression constant. Values of Kf for common solvents are listed in Table 11.5.
Like the boiling-point elevation, the observed freezing-point depression can be used to determine molar masses and to characterize solutions.
Freezing-Point Depression What mass of ethylene glycol (C2H6O2, molar mass 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at ( )? Assume the density of water is exactly 1 g/mL.
23.3°C 10.0°F ¢T ¢T Kfmsolute 0°C.
0°C.
0°C 0°C.
(a) (b) FIGURE 11.15 (a) Ice in equilibrium with liquid water.
(b) Ice in equilibrium with liquid water con-taining a dissolved solute (shown in pink).
Visualization: Freezing-Point Depression: Solid/Liquid Equilibrium Visualization: Freezing-Point Depression: Addition of a Solute Visualization: Freezing-Point Depression: Solid/Solution Equilibrium Sample Exercise 11.9 Ethylene glycol Melting point and freezing point both re-fer to the temperature where the solid and liquid coexist.
11.5 Boiling-Point Elevation and Freezing-Point Depression 507 Solution The freezing point must be lowered from to To determine the molality of ethylene glycol needed to accomplish this, we can use the equation where and Kf 1.86 (from Table 11.5). Solving for the molality gives This means that 12.5 mol ethylene glycol must be added per kilogram of water. We have 10.0 L, or 10.0 kg, of water. Therefore, the total number of moles of ethylene glycol needed is The mass of ethylene glycol needed is See Exercises 11.61 and 11.62.
Determining Molar Mass by Freezing-Point Depression A chemist is trying to identify a human hormone that controls metabolism by determin-ing its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing-point depression was determined to be Calculate the molar mass of the hormone.
Solution From Table 11.5, Kf for benzene is so the molality of the hormone is The moles of hormone can be obtained from the definition of molality: or Since 0.546 g hormone was dissolved, mol hormone has a mass of 0.546 g, and Thus the molar mass of the hormone is 776 g/mol.
See Exercises 11.63 and 11.64.
x 776 g/mol 0.546 g 7.04 104 mol x g 1.00 mol 7.04 104 mol hormone a4.69 102 mol kg b 10.0150 kg2 7.04 104 mol 4.69 102 mol/kg msolute mol hormone 0.0150 kg benzene mhormone ¢T Kf 0.240°C 5.12°C kg/mol 4.69 102 mol/kg 5.12°C kg/mol, 0.240°C.
1.25 102 mol 62.1 g mol 7.76 103 g 1or 7.76 kg2 12.5 mol kg 10.0 kg 1.25 102 mol msolute ¢T Kf 23.3°C 1.86°C kg/mol 12.5 mol/kg ¢T 23.3°C ¢T Kfmsolute 23.3°C.
0°C The addition of antifreeze lowers the freezing point of water in a car’s radiator.
Sample Exercise 11.10 11.6 Osmotic Pressure Osmotic pressure, another of the colligative properties, can be understood from Fig. 11.16.
A solution and pure solvent are separated by a semipermeable membrane, which allows solvent but not solute molecules to pass through. As time passes, the volume of the solu-tion increases and that of the solvent decreases. This flow of solvent into the solution through the semipermeable membrane is called osmosis. Eventually the liquid levels stop changing, indicating that the system has reached equilibrium. Because the liquid levels are different at this point, there is a greater hydrostatic pressure on the solution than on the pure solvent. This excess pressure is called the osmotic pressure.
We can take another view of this phenomenon, as illustrated in Fig. 11.17. Osmosis can be prevented by applying a pressure to the solution. The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution. A simple model to explain osmotic pressure can be constructed as shown in Fig. 11.18. The membrane allows only solvent mol-ecules to pass through. However, the initial rates of solvent transfer to and from the solu-tion are not the same. The solute particles interfere with the passage of solvent, so the rate of transfer is slower from the solution to the solvent than in the reverse direction. Thus there is a net transfer of solvent molecules into the solution, which causes the solution volume to increase. As the solution level rises in the tube, the resulting pressure exerts an extra “push” on the solvent molecules in the solution, forcing them back through the membrane. Even-tually, enough pressure develops so that the solvent transfer becomes equal in both direc-tions. At this point, equilibrium is achieved and the levels stop changing.
Osmotic pressure can be used to characterize solutions and determine molar masses, as can the other colligative properties, but osmotic pressure is particularly useful because a small concentration of solute produces a relatively large osmotic pressure.
Experiments show that the dependence of the osmotic pressure on solution concen-tration is represented by the equation where is the osmotic pressure in atmospheres, M is the molarity of the solution, R is the gas law constant, and T is the Kelvin temperature.
A molar mass determination using osmotic pressure is illustrated in Sample Exercise 11.11.
ß ß MRT FIGURE 11.16 A tube with a bulb on the end that is cov-ered by a semipermeable membrane. The solution is inside the tube and is bathed in the pure solvent. There is a net transfer of solvent molecules into the solution until the hydrostatic pressure equalizes the sol-vent flow in both directions.
508 Chapter Eleven Properties of Solutions Time Time Net movement of solvent (at equilibrium) Osmotic pressure Pure solvent Membrane Solution Applied pressure, needed to stop osmosis Semipermeable membrane Solution Pure solvent FIGURE 11.17 The normal flow of solvent into the solution (osmosis) can be prevented by applying an external pressure to the solution. The minimum pressure required to stop the osmosis is equal to the osmotic pressure of the solution.
Visualization: Osmosis 11.6 Osmotic Pressure 509 Semipermeable membrane Solution Pure solvent Solution Pure solvent Semipermeable membrane Osmotic pressure (a) (b) FIGURE 11.18 (a) A pure solvent and its solution (contain-ing a nonvolatile solute) are separated by a semipermeable membrane through which solvent molecules (blue) can pass but solute molecules (green) cannot. The rate of sol-vent transfer is greater from solvent to solu-tion than from solution to solvent. (b) The system at equilibrium, where the rate of solvent transfer is the same in both directions.
Determining Molar Mass from Osmotic Pressure To determine the molar mass of a certain protein, g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at Calculate the molar mass of the protein.
Solution We use the equation In this case we have Note that the osmotic pressure must be converted to atmospheres because of the units of R.
Solving for M gives Since g protein was dissolved in 1 mL solution, the mass of protein per liter of solution is 1.00 g. The solution’s concentration is mol/L. This concentration is produced from g protein per milliliter, or 1.00 g/L. Thus mol protein has a mass of 1.00 g and The molar mass of the protein is . This molar mass may seem very large, but it is relatively small for a protein.
See Exercise 11.66.
In osmosis, a semipermeable membrane prevents transfer of all solute particles. A similar phenomenon, called dialysis, occurs at the walls of most plant and animal cells.
1.66 104 g/mol x 1.66 104 g 1.00 g 6.01 105 mol x g 1.00 mol 6.01 105 1.00 103 6.01 105 1.00 103 M 1.47 103 atm 10.08206 L atm/K mol21298 K2 6.01 105 mol/L T 25.0 273 298 K R 0.08206 L atm/K mol ß 1.12 torr 1 atm 760 torr 1.47 103 atm ß MRT 25.0°C.
1.00 103 Sample Exercise 11.11 Measurements of osmotic pressure gen-erally give much more accurate molar mass values than those from freezing-point or boiling-point changes.
510 Chapter Eleven Properties of Solutions However, in this case the membrane allows transfer of both solvent molecules and small solute molecules and ions. One of the most important applications of dialysis is the use of artificial kidney machines to purify the blood. The blood is passed through a cellophane tube, which acts as the semipermeable membrane. The tube is immersed in a dialyzing solution (see Fig. 11.19). This “washing” solution contains the same concentrations of ions and small molecules as blood but has none of the waste products normally removed by the kidneys. The resulting dialysis (movement of waste molecules into the washing so-lution) cleanses the blood.
Solutions that have identical osmotic pressures are said to be isotonic solutions. Flu-ids administered intravenously must be isotonic with body fluids. For example, if red blood cells are bathed in a hypertonic solution, which is a solution having an osmotic pressure higher than that of the cell fluids, the cells will shrivel because of a net transfer of water out of the cells. This phenomenon is called crenation. The opposite phenomenon, called hemolysis, occurs when cells are bathed in a hypotonic solution, a solution with an os-motic pressure lower than that of the cell fluids. In this case, the cells rupture because of the flow of water into the cells.
We can use the phenomenon of crenation to our advantage. Food can be preserved by treating its surface with a solute that gives a solution that is hypertonic to bacteria cells.
Bacteria on the food then tend to shrivel and die. This is why salt can be used to protect meat and sugar can be used to protect fruit.
Isotonic Solutions What concentration of sodium chloride in water is needed to produce an aqueous solu-tion isotonic with blood ( at ?
Solution We can calculate the molarity of the solute from the equation M 7.70 atm 10.08206 L atm/K mol21298 K2 0.315 mol/L ß MRT or M ß RT 25°C2 ß 7.70 atm Patient undergoing dialysis.
The brine used in pickling causes the cucumbers to shrivel.
Sample Exercise 11.12 FIGURE 11.19 Representation of the functioning of an artificial kidney.
Waste products dialyze out into the washing solution Dialyzing solution Impure blood in Purified blood out Essential ions and molecules remain in blood 11.6 Osmotic Pressure 511 This represents the total molarity of solute particles. But NaCl gives two ions per formula unit. Therefore, the concentration of NaCl needed is That is, See Exercise 11.68.
Reverse Osmosis If a solution in contact with pure solvent across a semipermeable membrane is subjected to an external pressure larger than its osmotic pressure, reverse osmosis occurs. The pressure will cause a net flow of solvent from the solution to the solvent, as shown in Fig. 11.20.
In reverse osmosis, the semipermeable membrane acts as a “molecular filter” to remove solute particles. This fact is applicable to the desalination (removal of dissolved salts) of seawater, which is highly hypertonic to body fluids and thus is not drinkable.
As the population of the Sun Belt areas of the United States increases, more demand will be placed on the limited supplies of fresh water there. One obvious source of fresh water is from the desalination of seawater. Various schemes have been suggested, includ-ing solar evaporation, reverse osmosis, and even a plan for towing icebergs from Antarc-tica. The problem, of course, is that all the available processes are expensive. However, as water shortages increase, desalination is becoming necessary. For example, the first full-time public desalination plant in the United States started operations on Catalina Island, just off the coast of California (see Fig. 11.21). This plant, which can produce 132,000 gallons of drinkable water from the Pacific Ocean every day, operates by reverse osmosis. Powerful pumps, developing over 800 lb/in2 of pressure, are employed to force seawater through synthetic semipermeable membranes.
Catalina Island’s plant may be just the beginning. The city of Santa Barbara opened a $40 million desalination plant in 1992 that can produce 8 million gallons of drinking water per day, and other plants are in the planning stages.
A small-scale, manually operated reverse osmosis desalinator has been developed by the U.S. Navy to provide fresh water on life rafts. Potable water can be supplied by this desalinator at the rate of 1.25 gallons of water per hour—enough to keep 25 people alive.
This compact desalinator, which weighs only 10 pounds, can now replace the bulky cases of fresh water formerly stored in Navy life rafts.
0.315 M 0.1575 M 0.1575 M 0.1575 M NaCl ¡ Na Cl 0.315 M 2 0.1575 M 0.158 M.
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ Red blood cells in three stages of osmosis. (a) The normal shape of a red blood cell. (b) This cell has shrunk because water moved out of it by osmosis.
(c) This cell is swollen with water that has moved into it by osmosis.
Semipermeable membrane Solution Pure solvent Pressure greater than πsoln π FIGURE 11.20 Reverse osmosis. A pressure greater than the osmotic pressure of the solution is ap-plied, which causes a net flow of solvent molecules (blue) from the solution to the pure solvent. The solute molecules (green) remain behind.
512 Chapter Eleven Properties of Solutions 11.7 Colligative Properties of Electrolyte Solutions As we have seen previously, the colligative properties of solutions depend on the total concentration of solute particles. For example, a 0.10 m glucose solution shows a freezing point depression of On the other hand, a 0.10 m sodium chloride solution should show a freezing-point depression of since the solution is 0.10 m Na ions and 0.10 m Cl ions. There-fore, the solution contains a total of 0.20 m solute particles, and T (1.86C kg/mol) (0.20 mol/kg) 0.37C.
The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed using the van’t Hoff factor, i: The expected value for i can be calculated for a salt by noting the number of ions per formula unit. For example, for NaCl, i is 2; for K2SO4, i is 3; and for Fe3(PO4)2, i is 5.
These calculated values assume that when a salt dissolves, it completely dissociates into its component ions, which then move around independently. This assumption is not al-ways true. For example, the freezing-point depression observed for 0.10 m NaCl is 1.87 times that for 0.10 m glucose rather than twice as great. That is, for a 0.10 m NaCl solu-tion the observed value for i is 1.87 rather than 2. Why? The best explanation is that ion pairing occurs in solution (see Fig. 11.22). At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle. In general, ion pairing is most important in concentrated solutions. As the solution becomes more dilute, i moles of particles in solution moles of solute dissolved 0.37°C, ¢T Kfm 11.86°C kg/mol210.100 mol/kg2 0.186°C 0.186°C: Salt water pumped from underground wells.
1.
Salt water is forced through 20-micron and 5-micron filters at a pressure of 800 pounds per square inch.
2.
Fresh water is forced through additional filters.
3.
Fresh water is pumped into water supply.
4.
Brine is pumped into ocean.
5.
FIGURE 11.21 (a) Residents of Catalina Island off the coast of southern California are benefiting from a new desalination plant that can supply 132,000 gallons a day, or one-third of the island’s daily needs. (b) Machinery in the desalination plant for Catalina Island.
Dutch chemist J. H. van’t Hoff (1852–1911) received the first Nobel Prize in chemistry in 1901.
(a) (b) Ion pair Ion pair + – – – – – – + + + + + FIGURE 11.22 In an aqueous solution a few ions aggre-gate, forming ion pairs that behave as a unit.
11.7 Colligative Properties of Electrolyte Solutions 513 the ions are farther apart and less ion pairing occurs. For example, in a 0.0010 m NaCl solution, the observed value of i is 1.97, which is very close to the expected value.
Ion pairing occurs to some extent in all electrolyte solutions. Table 11.6 shows ex-pected and observed values of i for a given concentration of various electrolytes. Note that the deviation of i from the expected value tends to be greatest where the ions have multiple charges. This is expected because ion pairing ought to be most important for highly charged ions.
The colligative properties of electrolyte solutions are described by including the van’t Hoff factor in the appropriate equation. For example, for changes in freezing and boiling points, the modified equation is where K represents the freezing-point depression or boiling-point elevation constant for the solvent.
For the osmotic pressure of electrolyte solutions, the equation is Osmotic Pressure The observed osmotic pressure for a 0.10 M solution of Fe(NH4)2(SO4)2 at is 10.8 atm. Compare the expected and experimental values for i.
Solution The ionic solid Fe(NH4)2(SO4)2 dissociates in water to produce 5 ions: Thus the expected value for i is 5. We can obtain the experimental value for i by using the equation for osmotic pressure: where 10.8 atm, M 0.10 mol/L, R 0.08206 L atm/K mol, and T 25 273 298 K. Substituting these values into the equation gives The experimental value for i is less than the expected value, presumably because of ion pairing.
See Exercises 11.73 and 11.74.
i ß MRT 10.8 atm 10.10 mol/L210.08206 L atm/K mol21298 K2 4.4 ß iMRT or i ß MRT Fe1NH4221SO422 ¡ H2O Fe2 24 2SO4 2 25°C ß iMRT ¢T imK TABLE 11.6 Expected and Observed Values of the van’t Hoff Factor for 0.05 m Solutions of Several Electrolytes Electrolyte i (expected) i (observed) aCl 2.0 1.9 MgCl2 3.0 2.7 MgSO4 2.0 1.3 FeCl3 4.0 3.4 HCl 2.0 1.9 Glucose 1.0 1.0 A nonelectrolyte shown for comparison.
Sample Exercise 11.13 514 Chapter Eleven Properties of Solutions 11.8 Colloids Mud can be suspended in water by vigorous stirring. When the stirring stops, most of the particles rapidly settle out, but even after several days some of the smallest parti-cles remain suspended. Although undetected in normal lighting, their presence can be demonstrated by shining a beam of intense light through the suspension. The beam is visible from the side because the light is scattered by the suspended particles (Fig.
11.23). In a true solution, on the other hand, the beam is invisible from the side because the individual ions and molecules dispersed in the solution are too small to scatter visible light.
The scattering of light by particles is called the Tyndall effect and is often used to distinguish between a suspension and a true solution.
A suspension of tiny particles in some medium is called a colloidal dispersion, or a colloid. The suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Colloids are classified according to the states of the dispersed phase and the dispersing medium. Table 11.7 summarizes various types of colloids.
What stabilizes a colloid? Why do the particles remain suspended rather than form-ing larger aggregates and precipitating out? The answer is complicated, but the main fac-tor seems to be electrostatic repulsion. A colloid, like all other macroscopic substances, is electrically neutral. However, when a colloid is placed in an electric field, the dispersed particles all migrate to the same electrode and thus must all have the same charge. How is this possible? The center of a colloidal particle (a tiny ionic crystal, a group of mole-cules, or a single large molecule) attracts from the medium a layer of ions, all of the same charge. This group of ions, in turn, attracts another layer of oppositely charged ions, as shown in Fig. 11.24. Because the colloidal particles all have an outer layer of ions with the same charge, they repel each other and do not easily aggregate to form particles that are large enough to precipitate.
FIGURE 11.23 The Tyndall effect.
+ – + + + + + + + – – – – – – – + – + + + + + + + – – – – – – – FIGURE 11.24 A representation of two colloidal particles.
In each the center particle is surrounded by a layer of positive ions, with negative ions in the outer layer. Thus, although the parti-cles are electrically neutral, they still repel each other because of their outer negative layer of ions.
CHEMICAL IMPACT The Drink of Champions—Water I n1965, the University of Florida football team, the Gators, participated in a research program to test a sports drink formula containing a mixture of carbohydrates and elec-trolytes. The drink was used to help prevent dehydration caused by extreme workouts in the hot Florida climate. The Gators’ success that season was in part attributed to their use of the sports drink formula. In 1967, a modified form of this formula was marketed with the name Gatorade. Today, Gatorade leads sales in sports drinks, but many other brands have entered a market where annual sales exceed $700 million!
During moderate- to high-intensity exercise, glycogen (a fuel reserve that helps maintain normal body processes) can be depleted within 60 to 90 minutes. Blood sugar lev-els drop as the glycogen reserves are used up, and lactic acid (a by-product of glucose metabolism) builds up in muscle tissue causing fatigue and muscle cramps. Muscles also generate a large amount of heat that must be dissipated. Wa-ter, which has a large specific heat capacity, is used to take heat away from these muscles. Sweating and evaporative cooling help the body maintain a constant temperature, but at a huge cost. During a high-intensity workout in hot weather, anywhere from 1 to 3 quarts of water can be lost from sweating per hour. Sweating away more than 2% of your body weight—a quart for every 100 pounds—can put a large stress on the heart, increasing body temperature and decreasing performance. Excessive sweating also results in the loss of sodium and potassium ions—two very important electrolytes that are present in the fluids inside and outside cells.
All the major sports drinks contain three main ingre-dients—carbohydrates in the form of simple sugars such as 11.8 Colloids 515 TABLE 11.7 Types of Colloids Dispersing Dispersed Examples Medium Substance Colloid Type Fog, aerosol sprays Gas Liquid Aerosol Smoke, airborne bacteria Gas Solid Aerosol Whipped cream, soap suds Liquid Gas Foam Milk, mayonnaise Liquid Liquid Emulsion Paint, clays, gelatin Liquid Solid Sol Marshmallow, polystyrene foam Solid Gas Solid foam Butter, cheese Solid Liquid Solid emulsion Ruby glass Solid Solid Solid sol sucrose, glucose, and fructose; electrolytes, including sodium and potassium ions; and water. Because these are the three major substances lost through sweating, good scientific reasoning suggests that drinking sports drinks should improve performance. But just how effectively do sports drinks deliver on their promises?
Recent studies have confirmed that athletes who eat a balanced diet and drink plenty of water are just as well off as those who consume sports drinks. A sports drink may have only one advantage over drinking water—it tastes bet-ter than water to most athletes. And if a drink tastes better, it will encourage more consumption, thus keeping cells hydrated.
Since most of the leading sports drinks contain the same ingredients in similar concentrations, taste may be the sin-gle most important factor in choosing your drink. If you are not interested in any particular sports drink, drink plenty of water. The key to quality performance is to keep your cells hydrated.
Adapted with permission from “Sports Drinks: Don’t Sweat the Small Stuff,” by Tim Graham, ChemMatters, February 1999, p. 11.
For healthy athletes, drinking water during exercise may be as effective as drinking sports drinks.
The destruction of a colloid, called coagulation, usually can be accomplished either by heating or by adding an electrolyte. Heating increases the velocities of the colloidal particles, causing them to collide with enough energy that the ion barriers are pene-trated and the particles can aggregate. Because this process is repeated many times, the particle grows to a point where it settles out. Adding an electrolyte neutralizes the adsorbed ion layers. This is why clay suspended in rivers is deposited where the river reaches the ocean, forming the deltas characteristic of large rivers like the Mississippi. The high salt content of the seawater causes the colloidal clay particles to coagulate.
The removal of soot from smoke is another example of the coagulation of a colloid.
When smoke is passed through an electrostatic precipitator (Fig. 11.25), the suspended solids are removed. The use of precipitators has produced an immense improvement in the air quality of heavily industrialized cities.
Point electrodes Soot-free gases escape High DC voltage Plate electrodes Soot-laden smoke Soot particles removed here Ground FIGURE 11.25 The Cottrell precipitator installed in a smokestack. The charged plates attract the colloidal particles because of their ion lay-ers and thus remove them from the smoke.
516 Chapter Eleven Properties of Solutions CHEMICAL IMPACT Organisms and Ice Formation T he ice-cold waters of the polar oceans are teeming with fish that seem immune to freezing. One might think that these fish have some kind of antifreeze in their blood. How-ever, studies show that they are protected from freezing in a very different way from the way antifreeze protects our cars. As we have seen in this chapter, solutes such as sugar, salt, and ethylene glycol lower the temperature at which the solid and liquid phases of water can coexist. However, the fish could not tolerate high concentrations of solutes in their blood because of the osmotic pressure effects. Instead, they are protected by proteins in their blood. These proteins al-low the water in the bloodstream to be supercooled—exist below —without forming ice. They apparently coat the surface of each tiny ice crystal, as soon as it begins to form, preventing it from growing to a size that would cause bio-logic damage.
Although it might at first seem surprising, this research on polar fish has attracted the attention of ice cream manu-facturers. Premium quality ice cream is smooth; it does not have large ice crystals in it. The makers of ice cream would like to incorporate these polar fish proteins, or molecules that behave similarly, into ice cream to prevent the growth of ice crystals during storage.
Fruit and vegetable growers have a similar interest: They also want to prevent ice formation that damages their crops during an unusual cold wave. However, this is a very different kind of problem than keeping polar fish from freez-ing. Many types of fruits and vegetables are colonized by 0°C bacteria that manufacture a protein that encourages freezing by acting as a nucleating agent to start an ice crystal.
Chemists have identified the offending protein in the bacte-ria and the gene that is responsible for making it. They have learned to modify the genetic material of these bacteria in a way that removes their ability to make the protein that en-courages ice crystal formation. If testing shows that these modified bacteria have no harmful effects on the crop or the environment, the original bacteria strain will be replaced with the new form so that ice crystals will not form so read-ily when a cold snap occurs.
An Antarctic fish, Chaerophalus aceratus.
For Review Solution composition Molarity (M): moles solute per liter of solution Mass percent: ratio of mass of solute to mass of solution times 100% Mole fraction ( ): ratio of moles of a given component to total moles of all components Molality (m): moles solute per mass of solvent (in kg) Normality (N): number of equivalents per liter of solution Enthalpy of solution ( Hsoln) The enthalpy change accompanying solution formation Can be partitioned into • The energy required to overcome the solute–solute interactions • The energy required to “make holes” in the solvent • The energy associated with solute–solvent interactions x Key Terms Section 11.1 molarity mass percent mole fraction molality normality Section 11.2 enthalpy (heat) of solution enthalpy (heat) of hydration Section 11.3 Henry’s law thermal pollution Section 11.4 Raoult’s law ideal solution For Review 517 Section 11.5 colligative properties molal boiling-point elevation constant molal freezing-point depression constant Section 11.6 semipermeable membrane osmosis osmotic pressure dialysis isotonic solution reverse osmosis desalination Section 11.7 van’t Hoff factor ion pairing Section 11.8 Tyndall effect colloid (colloidal dispersion) coagulation Factors That affect solubility Polarity of solute and solvent • “Like dissolves like” is a useful generalization Pressure increases the solubility of gases in a solvent • Henry’s law: Temperature effects • Increased temperature decreases the solubility of a gas in water • Most solids are more soluble at higher temperatures but important exceptions exist Vapor pressure of solutions A solution containing a nonvolatile solute has a lower vapor pressure than a solu-tion of the pure solvent Raoult’s law defines an ideal solution • Solutions in which the solute–solvent attractions differ from the solute–solute and solvent–solvent attractions violate Raoult’s law Colligative properties Depend on the number of solute particles present Boiling-point elevation: Freezing-point lowering: Osmotic pressure: • Osmosis occurs when a solution and pure solvent are separated by a semiperme-able membrane that allows solvent molecules to pass but not solute particles • Reverse osmosis occurs when the applied pressure is greater than the osmotic pressure of the solution Because colligative properties depend on the number of particles, solutes that break into several ions when they dissolve have an effect proportional to the num-ber of ions produced • The van’t Hoff factor i represents the number of ions produced by each formula unit of solute Colloids A suspension of tiny particles stabilized by electrostatic repulsion among the ion layers surrounding the individual particles Can be coagulated (destroyed) by heating or adding an electrolyte REVIEW QUESTIONS 1. The four most common ways to describe solution composition are mass percent, mole fraction, molarity, and molality. Define each of these solution composition terms. Why is molarity temperature-dependent, whereas the other three solution composition terms are temperature-independent?
2. Using KF as an example, write equations that refer to and Lattice energy was defined in Chapter 8 as for the reaction K(g) F(g) KF(s). Show how you would utilize Hess’s law to calculate from and for KF, where for KF, as for other solu-ble ionic compounds, is a relatively small number. How can this be since and are relatively large negative numbers?
3. What does the axiom “like dissolves like” mean? There are four types of solute/ solvent combinations: polar solutes in polar solvents, nonpolar solutes in polar solvents, and so on. For each type of solution, discuss the magnitude of 4. Structure, pressure, and temperature all have an effect on solubility.
Discuss each of their effects. What is Henry’s law? Why does Henry’s law not work for HCl(g)? What do the terms hydrophobic and hydrophilic mean?
¢Hsoln.
¢HLE ¢Hhyd ¢Hsoln ¢HLE lattice energy ¢HLE ¢Hhyd ¢Hsoln ¡ ¢H ¢Hhyd.
¢Hsoln ß MRT ¢T K fm solute ¢T Kbm solute P soln vapor xsolventP solvent vapor C kP 5. Define the terms in Raoult’s law. Figure 11.9 illustrates the net transfer of water molecules from pure water to an aqueous solution of a nonvolatile solute. Ex-plain why eventually all of the water from the beaker of pure water will transfer to the aqueous solution. If the experiment illustrated in Fig. 11.9 was performed using a volatile solute, what would happen? How do you calculate the total vapor pressure when both the solute and solvent are volatile?
6. In terms of Raoult’s law, distinguish between an ideal liquid–liquid solution and a nonideal liquid–liquid solution. If a solution is ideal, what is true about for the solution formation, and the interactive forces within the pure solute and pure solvent as compared to the interactive forces within the solution. Give an example of an ideal solution. Answer the previous two questions for solutions that exhibit either negative or positive deviations from Raoult’s law.
7. Vapor-pressure lowering is a colligative property, as are freezing-point depres-sion and boiling-point elevation. What is a colligative property? Why is the freezing point depressed for a solution as compared to the pure solvent? Why is the boiling point elevated for a solution as compared to the pure solvent?
Explain how to calculate for a freezing-point depression problem or a boiling-point elevation problem. Of the solvents listed in Table 11.5, which would have the largest freezing-point depression for a 0.50 molal solution? Which would have the smallest boiling-point elevation for a 0.50 molal solution?
A common application of freezing-point depression and boiling-point elevation experiments is to provide a means to calculate the molar mass of a nonvolatile solute. What data are needed to calculate the molar mass of a non-volatile solute? Explain how you would manipulate these data to calculate the molar mass of the nonvolatile solute.
8. What is osmotic pressure? How is osmotic pressure calculated? Molarity units are used in the osmotic pressure equation. When does the molarity of a solution approximately equal the molality of the solution? Before refrigeration was com-mon, many foods were preserved by salting them heavily, and many fruits were preserved by mixing them with a large amount of sugar (fruit preserves). How do salt and sugar act as preservatives? Two applications of osmotic pressure are dialysis and desalination. Explain these two processes.
9. Distinguish between a strong electrolyte, a weak electrolyte, and a nonelec-trolyte. How can colligative properties be used to distinguish between them?
What is the van’t Hoff factor? Why is the observed freezing-point depression for electrolyte solutions sometimes less than the calculated value? Is the discrepancy greater for concentrated or dilute solutions?
10. What is a colloidal dispersion? Give some examples of colloids. The Tyndall effect is often used to distinguish between a colloidal suspension and a true solution. Explain. The destruction of a colloid is done through a process called coagulation. What is coagulation?
¢T ¢T ¢Hsoln, 518 Chapter Eleven Properties of Solutions Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Consider Fig. 11.9. According to the caption and picture, water seems to go from one beaker to another.
a. Explain why this occurs.
b. The explanation in the text uses terms such as vapor pressure and equilibrium. Explain what these have to do with the phe-nomenon. For example, what is coming to equilibrium?
c. Does all the water end up in the second beaker?
d. Is water evaporating from the beaker containing the solution?
If so, is the rate of evaporation increasing, decreasing, or stay-ing constant?
Draw pictures to illustrate your explanations.
Questions 519 2. Once again, consider Fig. 11.9. Suppose instead of having a nonvolatile solute in the solvent in one beaker, the two beakers con-tain different volatile liquids. That is, suppose one beaker contains liquid A ( ) and the other beaker contains liquid B ( ). Explain what happens as time passes. How is this similar to the first case (shown in the figure)? How is it different?
3. Assume that you place a freshwater plant into a saltwater solu-tion and examine it under a microscope. What happens to the plant cells? What if you placed a saltwater plant in pure water?
Explain. Draw pictures to illustrate your explanations.
4. How does relate to deviations from Raoult’s law? Explain.
5. You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: “The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil.” What do you say to your friend?
6. You drop an ice cube (made from pure water) into a saltwater solution at Explain what happens and why.
7. Using the phase diagram for water and Raoult’s law, explain why salt is spread on the roads in winter (even when it is below freezing).
8. You and your friend are each drinking cola from separate 2-L bottles. Both colas are equally carbonated. You are able to drink 1 L of cola, but your friend can drink only about half a liter. You each close the bottles and place them in the refrigerator. The next day when you each go to get the colas, whose will be more car-bonated and why?
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Solution Review If you have trouble with these exercises, review Sections 4.1 to 4.3 in Chapter 4.
9. Rubbing alcohol contains 585 g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molarity.
10. What volume of a 0.580 M solution of CaCl2 contains 1.28 g of solute?
11. Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicar-bonate.
12. Write equations showing the ions present after the following strong electrolytes are dissolved in water.
a. HNO3 d. SrBr2 g. NH4NO3 b. Na2SO4 e. KClO4 h. CuSO4 c. Al(NO3)3 f. NH4Br i.
NaOH Questions 13. Rationalize the temperature dependence of the solubility of a gas in water in terms of the kinetic molecular theory.
14. The weak electrolyte NH3(g) does not obey Henry’s law. Why?
O2(g) obeys Henry’s law in water but not in blood (an aqueous solution). Why?
0°C.
¢Hsoln Pvap 100 torr Pvap 50 torr 15. The two beakers in the sealed container illustrated below con-tain pure water and an aqueous solution of a volatile solute.
If the solute is less volatile than water, explain what will hap-pen to the volumes in the two containers as time passes.
16. The following plot shows the vapor pressure of various solutions of components A and B at some temperature.
Which of the following statements is false concerning solutions of A and B?
a. The solutions exhibit negative deviations from Raoult’s law.
b.
for the solutions should be exothermic.
c. The intermolecular forces are stronger in solution than in ei-ther pure A or pure B.
d. Pure liquid B is more volatile than pure liquid A.
e. The solution with will have a lower boiling point than either pure A or pure B.
17. When pure methanol is mixed with water, the resulting solution feels warm. Would you expect this solution to be ideal? Explain.
18. Detergent molecules can stabilize the emulsion of oil in water as well as remove dirt from soiled clothes. A typical detergent is sodium dodecylsulfate, or SDS, and it has a formula of . In aqueous solution, SDS suspends oil or dirt by forming small aggregates of detergent anions called micelles. Propose a structure for micelles.
19. For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentra-tion units different?
20. In order for sodium chloride to dissolve in water, a small amount of energy must be added during solution formation. This is not energetically favorable. Why is NaCl so soluble in water?
CH31CH2210CH2SO4 Na xB 0.6 ¢Hmix 0 PA 0 PB 0 1 Mole fraction χB Vapor pressure (torr) Water Aqueous solution 21. Which of the following statements is(are) true? Correct the false statements.
a. The vapor pressure of a solution is directly related to the mole fraction of solute.
b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at .
c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present.
d. When sugar is added to water, the boiling point of the solu-tion increases above because sugar has a higher boil-ing point than water.
22. Is the following statement true of false? Explain your answer.
When determining the molar mass of a solute using boiling point of freezing point data, camphor would be the best solvent choice of all of the solvents listed in Table 11.5.
23. Explain the terms isotonic solution, crenation, and hemolysis.
24. What is ion pairing?
Exercises In this section similar exercises are paired.
Concentration of Solutions 25. A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 104 mL.
Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/cm3.
26. An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3.Calculate the molality, molarity, and mole fraction of the ethylene glycol.
27. Common commercial acids and bases are aqueous solutions with the following properties: 100°C 0°C 30. A bottle of wine contains 12.5% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality.
31. A 1.37 M solution of citric acid (H3C6H5O7) in water has a den-sity of 1.10 g/cm3. Calculate the mass percent, molality, mole frac-tion, and normality of the citric acid. Citric acid has three acidic protons.
32. Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Den-sity of acetone 0.788 g/cm3; density of ethanol 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Energetics of Solutions and Solubility 33. The lattice energy of NaI is 686 kJ/mol, and the enthalpy of hydration is 694 kJ/mol. Calculate the enthalpy of solution per mole of solid NaI. Describe the process to which this enthalpy change applies.
34. a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide.
Density Mass Percent (g/cm3) of Solute Hydrochloric acid 1.19 38 Nitric acid 1.42 70.
Sulfuric acid 1.84 95 Acetic acid 1.05 99 Ammonia 0.90 28 Lattice energy was defined in Chapter 8 as the energy change for the process M(g) X(g) n MX(s).
520 Chapter Eleven Properties of Solutions Calculate the molarity, molality, and mole fraction of each of the preceding reagents.
28. In lab you need to prepare at least 100 mL of each of the fol-lowing solutions. Explain how you would proceed using the given information.
a. 2.0 m KCl in water (density of ) b. 15% NaOH by mass in water ( ) c. 25% NaOH by mass in d. 0.10 mole fraction of C6H12O6 in water ( ) 29. A solution is prepared by mixing 25 mL pentane ( ) with 45 mL hexane ( ). As-suming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane.
C6H14, d 0.66 g/cm3 0.63 g/cm3 C5H12, d d 1.00 g/cm3 CH3OH 1d 0.79 g/cm32 d 1.00 g/cm3 H2O 1.00 g/cm3 Lattice Energy Hsoln CaCl2(s) 2247 kJ/mol 46 kJ/mol CaI2(s) 2059 kJ/mol 104 kJ/mol b. Based on your answers to part a, which ion, or , is more strongly attracted to water?
35. Although Al(OH)3 is insoluble in water, NaOH is very soluble.
Explain in terms of lattice energies.
36. The high melting points of ionic solids indicate that a lot of en-ergy must be supplied to separate the ions from one another. How is it possible that the ions can separate from one another when soluble ionic compounds are dissolved in water, often with es-sentially no temperature change?
37. Which solvent, water or carbon tetrachloride, would you choose to dissolve each of the following?
a. KrF2 e. MgF2 b. SF2 f. CH2O c. SO2 g.
d. CO2 38. Which solvent, water or hexane ( ), would you choose to dissolve each of the following?
a. NaCl c. octane (C8H18) b. HF d. (NH4)2SO4 39. What factors cause one solute to be more strongly attracted to water than another? For each of the following pairs, predict which substance would be more soluble in water.
a. CH3CH2OH or CH3CH2CH3 b. CHCl3 or CCl4 c. CH3CH2OH or CH3(CH2)14CH2OH C6H14 CH2“CH2 I Cl Exercises 521 40. Which ion in each of the following pairs would you expect to be more strongly hydrated? Why?
a.
d.
b.
e.
c.
f.
41. Rationalize the trend in water solubility for the following sim-ple alcohols: ClO4 or SO4 2 Fe2 or Fe3 Cl or ClO4 Mg2 or Be2 F or Br Na or Mg2 46. The vapor pressure of a solution containing 53.6 g glycerin (C3H8O3) in 133.7 g ethanol (C2H5OH) is 113 torr at . Cal-culate the vapor pressure of pure ethanol at assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.
47. At a certain temperature, the vapor pressure of pure benzene (C6H6) is 0.930 atm. A solution was prepared by dissolving 10.0 g of a nondissociating, nonvolatile solute in 78.11 g of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 atm. Assuming the solution behaves ideally, deter-mine the molar mass of the solute.
48. A solution of sodium chloride in water has a vapor pressure of 19.6 torr at . What is the mole fraction of NaCl solute parti-cles in this solution? What would be the vapor pressure of this so-lution at ? The vapor pressure of pure water is 23.8 torr at and 71.9 torr at and assume sodium chloride exists as and ions in solution.
49. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At the vapor pressures of pentane and hexane are 511 and 150. torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 45 mL hexane (density, 0.66 g/mL).
a. What is the vapor pressure of the resulting solution?
b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?
50. A solution is prepared by mixing 0.0300 mol CH2Cl2 and 0.0500 mol CH2Br2 at . Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at .
At , the vapor pressures of pure CH2Cl2 and pure CH2Br2 are 133 and 11.4 torr, respectively.
51. What is the composition of a methanol (CH3OH)–propanol (CH3CH2CH2OH) solution that has a vapor pressure of 174 torr at ? At , the vapor pressures of pure methanol and pure propanol are 303 and 44.6 torr, respectively. Assume the solu-tion is ideal.
52. Benzene and toluene form an ideal solution. Consider a solution of benzene and toluene prepared at . Assuming the mole fractions of benzene and toluene in the vapor phase are equal, calculate the composition of the solution. At the vapor pres-sures of benzene and toluene are 95 and 28 torr, respectively.
53. Which of the following will have the lowest total vapor pressure at ?
a. pure water (vapor pressure 23.8 torr at ) b. a solution of glucose in water with c. a solution of sodium chloride in water with d. a solution of methanol in water with (Consider the vapor pressure of both methanol [143 torr at ] and water.) 54. Which of the choices in Exercise 53 has the highest vapor pressure?
55. A solution is made by mixing 50.0 g acetone (CH3COCH3) and 50.0 g methanol (CH3OH). What is the vapor pressure of this solution at ? What is the composition of the vapor expressed as a mole fraction? Assume ideal solution and gas behavior. (At the vapor pressures of pure acetone and pure methanol are 271 and 143 torr, respectively.) The actual vapor pressure of this solution is 161 torr. Explain any discrepancies.
25°C 25°C 25°C xCH3OH 0.2 xNaCl 0.01 xC6H12O6 0.01 25°C 25°C 25°C 25°C 40°C 40°C 25°C 25°C 25°C 25°C Cl Na 45°C 25°C 45°C 25°C 40°C 40°C Solubility (g/100 g H2O Alcohol at 20ºC) Methanol, CH3OH Soluble in all proportions Ethanol, CH3CH2OH Soluble in all proportions Propanol, CH3CH2CH2OH Soluble in all proportions Butanol, CH3(CH2)2CH2OH 8.14 Pentanol, CH3(CH2)3CH2OH 2.64 Hexanol, CH3(CH2)4CH2OH 0.59 Heptanol, CH3(CH2)5CH2OH 0.09 42. The solubility of benzoic acid (HC7H5O2), is 0.34 g/100 mL in water at and is 10.0 g/100 mL in ben-zene (C6H6) at . Rationalize this solubility behavior. (Hint: Benzoic acid forms a dimer in benzene.) Would benzoic acid be more or less soluble in a 0.1 M NaOH solution than it is in wa-ter? Explain.
43. The solubility of nitrogen in water is at when the N2 pressure above water is 0.790 atm. Calculate the Henry’s law constant for N2 in units of for Henry’s law in the form where C is the gas concentration in mol/L. Calculate the solubility of N2 in water when the partial pressure of nitrogen above water is 1.10 atm at .
44. In Exercise 107 in Chapter 5, the pressure of CO2 in a bottle of sparkling wine was calculated assuming that the CO2 was in-soluble in water. This was a bad assumption. Redo this problem by assuming that CO2 obeys Henry’s law. Use the data given in that problem to calculate the partial pressure of CO2 in the gas phase and the solubility of CO2 in the wine at . The Henry’s law constant for CO2 is at with Henry’s law in the form C kP, where C is the concentration of the gas in mol/L.
Vapor Pressures of Solutions 45. Glycerin, C3H8O3, is a nonvolatile liquid. What is the vapor pres-sure of a solution made by adding 164 g of glycerin to 338 mL of H2O at ? The vapor pressure of pure water at is 54.74 torr and its density is 0.992 g/cm3.
39.8°C 39.8°C 25°C 3.1 102 mol/L atm 25°C 0°C C kP, mol/L atm 0°C 8.21 104 mol/L 25°C 25°C 522 Chapter Eleven Properties of Solutions 65. a. Calculate the freezing-point depression and osmotic pressure at of an aqueous solution containing 1.0 g/L of a protein ( ) if the density of the solution is 1.0 g/cm3.
b. Considering your answer to part a, which colligative property, freezing-point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.
66. An aqueous solution of 10.00 g of catalase, an enzyme found in the liver, has a volume of 1.00 L at . The solution’s osmotic pressure at is found to be 0.74 torr. Calculate the molar mass of catalase.
67. If the human eye has an osmotic pressure of 8.00 atm at , what concentration of solute particles in water will provide an iso-tonic eyedrop solution (a solution with equal osmotic pressure)?
68. How would you prepare 1.0 L of an aqueous solution of sodium chloride having an osmotic pressure of 15 atm at ? Assume sodium chloride exists as and ions in solution.
Properties of Electrolyte Solutions 69. Consider the following solutions: 0.010 m Na3PO4 in water 0.020 m CaBr2 in water 0.020 m KCl in water 0.020 m HF in water (HF is a weak acid.) a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as 0.040 m C6H12O6 in water? C6H12O6 is a nonelectrolyte.
b. Which solution would have the highest vapor pressure at ?
c. Which solution would have the largest freezing-point depression?
70. From the following: pure water solution of C12H22O11 (m 0.01) in water solution of NaCl (m 0.01) in water solution of CaCl2 (m 0.01) in water choose the one with the a. highest freezing point.
d. lowest boiling point.
b. lowest freezing point.
e. highest osmotic pressure.
c. highest boiling point.
71. Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation.) a. 0.050 m MgCl2 b. 0.050 m FeCl3 72. A water desalination plant is set up near a salt marsh containing water that is 0.10 M NaCl. Calculate the minimum pressure that must be applied at to purify the water by reverse osmo-sis. Assume NaCl is completely dissociated.
73. Use the following data for three aqueous solutions of CaCl2 to calculate the apparent value of the van’t Hoff factor.
20.°C 28°C Cl Na 22°C 25°C 27°C 27°C molar mass 9.0 104 g/mol 25°C 56. The vapor pressures of several solutions of water–propanol (CH3CH2CH2OH) were determined at various compositions, with the following data collected at : 45°C Vapor pressure H2O (torr) 0 74.0 0.15 77.3 0.37 80.2 0.54 81.6 0.69 80.6 0.83 78.2 1.00 71.9 x Molality Freezing-Point Depression (°C) 0.0225 0.110 0.0910 0.440 0.278 1.330 a. Are solutions of water and propanol ideal? Explain.
b. Predict the sign of for water–propanol solutions.
c. Are the interactive forces between propanol and water mole-cules weaker than, stronger than, or equal to the interactive forces between the pure substances? Explain.
d. Which of the solutions in the data would have the lowest nor-mal boiling point?
Colligative Properties 57. A solution is prepared by dissolving 27.0 g of urea, ( ) in 150.0 g of water. Calculate the boiling point of the solution.
Urea is a nonelectrolyte.
58. A 2.00-g sample of a large biomolecule was dissolved in 15.0 g of carbon tetrachloride. The boiling point of this solution was determined to be Calculate the molar mass of the bio-molecule. For carbon tetrachloride, the boiling-point constant is and the boiling point of pure carbon tetrachlo-ride is .
59. What mass of glycerin (C3H8O3), a nonelectrolyte, must be dis-solved in 200.0 g water to give a solution with a freezing point of ?
60. The freezing point of t-butanol is and Kf is Usually t-butanol absorbs water on exposure to air. If the freez-ing point of a 10.0-g sample of t-butanol is , how many grams of water are present in the sample?
61. Calculate the freezing point and boiling point of an antifreeze solution that is 50.0% by mass of ethylene glycol (HOCH2CH2OH) in water. Ethylene glycol is a nonelectrolyte.
62. What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 15.0 L of water to produce an antifreeze solution with a freezing point of ? What is the boiling point of this solution? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.) 63. Thyroxine, an important hormone that controls the rate of me-tabolism in the body, can be isolated from the thyroid gland.
When 0.455 g of thyroxine is dissolved in 10.0 g of benzene, the freezing point of the solution is depressed by What is the molar mass of thyroxine? See Table 11.5.
64. Anthraquinone contains only carbon, hydrogen, and oxygen and has an empirical formula of C7H4O. The freezing point of camphor is lowered by when 1.32 g anthraquinone is dissolved in 11.4 g camphor. Determine the molecular formula of anthraquinone.
22.3°C 0.300°C.
25.0°C 24.59°C 9.1°C kg/mol.
25.50°C 1.50°C 76.50°C 5.03°C kg/mol, 77.85°C.
2CO, NH2 ¢Hsoln Challenge Problems 523 74. Calculate the freezing point and the boiling point of each of the following solutions using the observed van’t Hoff factors in Table 11.6.
a. 0.050 m MgCl2 b. 0.050 m FeCl3 75. In the winter of 1994, record low temperatures were registered throughout the United States. For example, in Champaign, Illi-nois, a record low of was registered. At this temperature can salting icy roads with CaCl2 be effective in melting the ice?
a. Assume i 3.00 for CaCl2.
b. Assume the average value of i from Exercise 73.
(The solubility of CaCl2 in cold water is 74.5 g per 100.0 g of water.) 76. A 0.500-g sample of a compound is dissolved in enough water to form 100.0 mL of solution. This solution has an osmotic pres-sure of 2.50 atm at . If each molecule of the solute disso-ciates into two particles (in this solvent), what is the molar mass of this solute?
Additional Exercises 77. In a coffee-cup calorimeter, 1.60 g of NH4NO3 was mixed with 75.0 g of water at an initial temperature of . After dis-solution of the salt, the final temperature of the calorimeter con-tents was .
a. Assuming the solution has a heat capacity of , and assuming no heat loss to the calorimeter, calculate the enthalpy of solution ( ) for the dissolution of NH4NO3 in units of kJ/mol.
b. If the enthalpy of hydration for NH4NO3 is calculate the lattice energy of NH4NO3.
78. In flushing and cleaning columns used in liquid chromatography to remove adsorbed contaminants, a series of solvents is used.
Hexane (C6H14), chloroform (CHCl3), methanol (CH3OH), and water are passed through the column in that order. Rationalize the order in terms of intermolecular forces and the mutual sol-ubility (miscibility) of the solvents.
79. Explain the following on the basis of the behavior of atoms and/or ions.
a. Cooking with water is faster in a pressure cooker than in an open pan.
b. Salt is used on icy roads.
c. Melted sea ice from the Artic Ocean produces fresh water.
d. CO2(s) (dry ice) does not have a normal boiling point under normal atmospheric conditions, even though CO2 is a liquid in fire extinguishers.
e. Adding a solute to a solvent extends the liquid phase over a larger temperature range.
80. The term “proof ” is defined as twice the percent by volume of pure ethanol in solution. Thus, a solution that is 95% (by volume) ethanol is 190 proof. What is the molarity of ethanol in a 92 proof ethanol/water solution? Assume the density of ethanol, C2H5OH, is 0.79 g/cm3 and the density of water is 1.0 g/cm3.
81. At , the vapor in equilibrium with a solution containing car-bon disulfide and acetonitrile has a total pressure of 263 torr and is 85.5 mole percent carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At , the vapor pressure of carbon disulfide is 375 torr. Assume the solution and vapor exhibit ideal behavior.
25°C 25°C 630. kJ/mol, ¢Hsoln 4.18 J/g °C 23.34°C 25.00°C 25°C 29°F 82. If the fluid inside a tree is about 0.1 M more concentrated in solute than the groundwater that bathes the roots, how high will a column of fluid rise in the tree at ? Assume that the density of the fluid is 1.0 g/cm3. (The density of mercury is 13.6 g/cm3.) 83. An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass per-cents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous so-lution. A freezing point of is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water.
Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.
84. Consider the following: What would happen to the level of liquid in the two arms if the semipermeable membrane separating the two liquids were per-meable to a. H2O only?
b.
85. Consider an aqueous solution containing sodium chloride that has a density of 1.01 g/mL. Assume the solution behaves ide-ally. The freezing point of this solution at 1.0 atm is .
Calculate the percent composition of this solution (by mass).
86. What stabilizes a colloidal suspension? Explain why adding heat or adding an electrolyte can cause the suspended particles to set-tle out.
87. The freezing point of an aqueous solution is a. Determine the boiling point of this solution.
b. Determine the vapor pressure (in mm Hg) of this solution at (the vapor pressure of pure water at is 23.76 mm Hg).
c. Explain any assumptions you make in solving parts a and b.
Challenge Problems 88. The vapor pressure of pure benzene is 750.0 torr and the vapor pressure of toluene is 300.0 torr at a certain temperature. You make a solution by pouring “some” benzene with “some” toluene. You then place this solution in a closed container and wait for the vapor to come into equilibrium with the solution.
Next, you condense the vapor. You put this liquid (the condensed vapor) in a closed container and wait for the vapor to come into equilibrium with the solution. You then condense this vapor and find the mole fraction of benzene in this vapor to be 0.714.
Determine the mole fraction of benzene in the original solution assuming the solution behaves ideally.
89. Liquid A has vapor pressure x, and liquid B has vapor pres-sure y. What is the mole fraction of the liquid mixture if the vapor above the solution is 30.% A by moles? 50.% A? 80.% A?
(Calculate in terms of x and y.) Liquid A has vapor pressure x, liquid B has vapor pressure y.
What is the mole fraction of the vapor above the solution if the liquid mixture is 30.% A by moles? 50.% A? 80.% A? (Calculate in terms of x and y.) 25°C 25°C 2.79°C.
1.28°C H2O, Na, and Cl?
5.20°C 25°C 524 Chapter Eleven Properties of Solutions 90. Erythrocytes are red blood cells containing hemoglobin. In a saline solution they shrivel when the salt concentration is high and swell when the salt concentration is low. In a aqueous solution of NaCl, whose freezing point is erythro-cytes neither swell nor shrink. If we want to calculate the osmotic pressure of the solution inside the erythrocytes under these con-ditions, what do we need to assume? Why? Estimate how good (or poor) of an assumption this is. Make this assumption and calculate the osmotic pressure of the solution inside the erythrocytes.
91. You make 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.00 kg of water. The freezing point of this solution is found to be Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.
92. An aqueous solution is 1.00% NaCl by mass and has a density of 1.071 g/cm3 at . The observed osmotic pressure of this solution is 7.83 atm at .
a. What fraction of the moles of NaCl in this solution exist as ion pairs?
b. Calculate the freezing point that would be observed for this solution.
93. The vapor in equilibrium with a pentane–hexane solution at has a mole fraction of pentane equal to 0.15 at . What is the mole fraction of pentane in the solution? (See Exercise 49 for the vapor pressures of the pure liquids.) 94. A forensic chemist is given a white solid that is suspected of be-ing pure cocaine ( ). She dissolves g of the solid in g benzene.
The freezing point is lowered by a. What is the molar mass of the substance? Assuming that the percent uncertainty in the calculated molar mass is the same as the percent uncertainty in the temperature change, calcu-late the uncertainty in the molar mass.
b. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine ( )?
c. Assuming that the absolute uncertainties in the measure-ments of temperature and mass remain unchanged, how could the chemist improve the precision of her results?
95. A 1.60-g sample of a mixture of naphthalene (C10H8) and an-thracene (C14H10) is dissolved in 20.0 g benzene (C6H6). The freezing point of the solution is . What is the composi-tion as mass percent of the sample mixture? The freezing point of benzene is , and Kf is 96. A solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at is observed to be 0.3950 atm.
What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.) 97. Formic acid (HCO2H) is a monoprotic acid that ionizes only par-tially in aqueous solutions. A 0.10 M formic acid solution is 4.2% ionized. Assuming that the molarity and molality of the solution 25.0°C 5.12°C kg/mol.
5.51°C 2.81°C 299.36 g/mol C18H21NO3, molar mass 1.32 0.04°C.
15.60 0.01 1.22 0.01 C17H21NO4, molar mass 303.35 g/mol 25°C 25°C 25°C 25°C 0.426°C.
0.406°C, 25°C are the same, calculate the freezing point and the boiling point of 0.10 M formic acid.
98. Specifications for lactated Ringer’s solution, which is used for intravenous (IV) injections, are as follows to reach 100. mL of solution: 285–315 mg Na 14.1–17.3 mg K 4.9–6.0 mg Ca2 368–408 mg Cl 231–261 mg lactate, C3H5O3 a. Specify the amounts of NaCl, KCl, CaCl2 2H2O, and NaC3H5O3 needed to prepare 100. mL of lactated Ringer’s solution.
b. What is the range of the osmotic pressure of the solution at given the above specifications?
99. In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water contains about 560 g of dissolved solids per milliliter.
Reverse osmosis units are marketed in this area to soften water.
A typical unit exerts a pressure of 8.0 atm and can produce 45 L of water per day.
a. Assuming all of the dissolved solids are MgCO3 and assum-ing a temperature of , what total volume of water must be processed to produce 45 L of pure water?
b. Would the same system work for purifying seawater? (As-sume seawater is 0.60 M NaCl.) Integrative Problems These problems require the integration of multiple concepts to find the solutions.
100. Creatinine, C4H7N3O, is a by-product of muscle metabolism, and creatinine levels in the body are known to be a fairly re-liable indicator of kidney function. The normal level of crea-tinine in the blood for adults is approximately 1.0 mg per deciliter (dL) of blood. If the density of blood is 1.025 g/mL, calculate the molality of a normal creatinine level in a 10.0-mL blood sample. What is the osmotic pressure of this solution at ?
101. An aqueous solution containing 0.250 mol of Q, a strong elec-trolyte, in g of water freezes at . What is the van’t Hoff factor for Q? The molal freezing-point depression constant for water is 1.86 What is the formula of Q if it is 38.68% chlorine by mass and there are twice as many an-ions as cations in one formula unit of Q?
102. Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound.
A 0.378 m solution is prepared by dissolving 38.4 g of sodium diatrizoate (NaDTZ) in mL of water at (the density of water at is 0.995 g/mL). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at is 34.1 torr?
31.2°C 31.2°C 31.2°C 1.60 102 °C kg/mol.
2.79°C 5.00 102 25.0°C 27°C m 37°C, Marathon Problem 525 Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
103. Using the following information, identify the strong electrolyte whose general formula is Ignore the effect of interionic attractions in the solution.
a.
is a common oxyanion. When 30.0 mg of the anhydrous sodium salt containing this oxyanion (NanA, where n 1, 2, or 3) is reduced, 15.26 mL of 0.02313 M reducing agent is required to react completely with the NanA present. Assume a 1:1 mole ratio in the reaction.
An Mx1A2y zH2O b. The cation is derived from a silvery white metal that is relatively expensive. The metal itself crystallizes in a body-centered cubic unit cell and has an atomic radius of 198.4 pm. The solid, pure metal has a density of 5.243 g/cm3.
The oxidation number of M in the strong electrolyte in ques-tion is 3.
c. When 33.45 mg of the compound is present (dissolved) in 10.0 mL of aqueous solution at , the solution has an osmotic pressure of 558 torr.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
25°C This Marathon Problem was developed by James H. Burness, Penn State University, York Campus. Reprinted with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
526 12 Chemical Kinetics Contents 12.1 Reaction Rates 12.2 Rate Laws: An Introduction • Types of Rate Laws 12.3 Determining the Form of the Rate Law • Method of Initial Rates 12.4 The Integrated Rate Law • First-Order Rate Laws • Half-Life of a First-Order Reaction • Second-Order Rate Laws • Zero-Order Rate Laws • Integrated Rate Laws for Reactions with More Than One Reactant 12.5 Rate Laws: A Summary 12.6 Reaction Mechanisms 12.7 A Model for Chemical Kinetics 12.8 Catalysis • Heterogeneous Catalysis • Homogeneous Catalysis The kinetic energy of these world championship runners is evident in the 800-meter race at Saint-Denis, France.
The applications of chemistry focus largely on chemical reactions, and the commer-cial use of a reaction requires knowledge of several of its characteristics, including its stoichiometry, energetics, and rate. A reaction is defined by its reactants and products, whose identity must be learned by experiment. Once the reactants and products are known, the equation for the reaction can be written and balanced, and stoichiometric calculations can be carried out. Another very important characteristic of a reaction is its spontaneity.
Spontaneity refers to the inherent tendency for the process to occur; however, it implies nothing about speed. Spontaneous does not mean fast. There are many spontaneous reac-tions that are so slow that no apparent reaction occurs over a period of weeks or years at normal temperatures. For example, there is a strong inherent tendency for gaseous hydrogen and oxygen to combine, that is, but in fact the two gases can coexist indefinitely at Similarly, the gaseous reactions are both highly likely to occur from a thermodynamic standpoint, but we observe no re-actions under normal conditions. In addition, the process of changing diamond to graphite is spontaneous but is so slow that it is not detectable.
To be useful, reactions must occur at a reasonable rate. To produce the 20 million tons of ammonia needed each year for fertilizer, we cannot simply mix nitrogen and hy-drogen gases at and wait for them to react. It is not enough to understand the stoi-chiometry and thermodynamics of a reaction; we also must understand the factors that govern the rate of the reaction. The area of chemistry that concerns reaction rates is called chemical kinetics.
One of the main goals of chemical kinetics is to understand the steps by which a reaction takes place. This series of steps is called the reaction mechanism. Under-standing the mechanism allows us to find ways to facilitate the reaction. For example, the Haber process for the production of ammonia requires high temperatures to achieve commercially feasible reaction rates. However, even higher temperatures (and more cost) would be required without the use of iron oxide, which speeds up the reaction.
In this chapter we will consider the main ideas of chemical kinetics. We will explore rate laws, reaction mechanisms, and simple models for chemical reactions.
12.1 Reaction Rates To introduce the concept of the rate of a reaction, we will consider the decomposition of nitrogen dioxide, a gas that causes air pollution. Nitrogen dioxide decomposes to nitric oxide and oxygen as follows: Suppose in a particular experiment we start with a flask of nitrogen dioxide at and measure the concentrations of nitrogen dioxide, nitric oxide, and oxygen as the nitrogen dioxide decomposes. The results of this experiment are summarized in Table 12.1, and the data are plotted in Fig. 12.1.
300°C 2NO21g2 ¡ 2NO1g2 O21g2 25°C N21g2 3H21g2 ¡ 2NH31g2 H21g2 Cl21g2 ¡ 2HCl1g2 25°C.
2H21g2 O21g2 ¡ 2H2O1l2 527 The kinetics of air pollution is discussed in Section 12.8.
Visualization: Coffee Creamer Flammability 528 Chapter Twelve Chemical Kinetics Note from these results that the concentration of the reactant (NO2) decreases with time and the concentrations of the products (NO and O2) increase with time (see Fig. 12.2).
Chemical kinetics deals with the speed at which these changes occur. The speed, or rate, of a process is defined as the change in a given quantity over a specific period of time.
For chemical reactions, the quantity that changes is the amount or concentration of a reactant or product. So the reaction rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time: ¢ 3A4 ¢t Rate concentration of A at time t2 concentration of A at time t1 t2 t1 The energy required for athletic exertion, the breaching of an Orca whale, and the combustion of fuel in a race car all result from chemical reactions.
[A] means concentration of A in mol/L.
12.1 Reaction Rates 529 TABLE 12.1 Concentrations of Reactant and Products as a Function of Time for the Reaction 2NO2(g) 2NO(g) O2(g) (at 300ºC) Concentration (mol/L) Time ( 1 s) NO2 NO O2 0 0.0100 0 0 50 0.0079 0.0021 0.0011 100 0.0065 0.0035 0.0018 150 0.0055 0.0045 0.0023 200 0.0048 0.0052 0.0026 250 0.0043 0.0057 0.0029 300 0.0038 0.0062 0.0031 350 0.0034 0.0066 0.0033 400 0.0031 0.0069 0.0035 S 0.0003 70 s O2 0.0025 0.005 0.0075 0.0100 0.0006 70 s 0.0026 110 s NO2 NO 50 100 150 200 250 300 350 400 Concentrations (mol/L) Time (s) ∆[NO2] ∆t FIGURE 12.1 Starting with a flask of nitrogen dioxide at 300°C, the concentrations of nitrogen dioxide, nitric oxide, and oxygen are plotted versus time.
530 Chapter Twelve Chemical Kinetics where A is the reactant or product being considered, and the square brackets indicate con-centration in mol/L. As usual, the symbol indicates a change in a given quantity. Note that a change can be positive (increase) or negative (decrease), thus leading to a positive or negative reaction rate by this definition. However, for convenience, we will always define the rate as a positive quantity, as we will see.
Now let us calculate the average rate at which the concentration of NO2 changes over the first 50 seconds of the reaction using the data given in Table 12.1.
Note that since the concentration of NO2 decreases with time, is a negative quan-tity. Because it is customary to work with positive reaction rates, we define the rate of this particular reaction as Since the concentrations of reactants always decrease with time, any rate expression in-volving a reactant will include a negative sign. The average rate of this reaction from 0 to 50 seconds is then The average rates for this reaction during several other time intervals are given in Table 12.2. Note that the rate is not constant but decreases with time. The rates given in Table 12.2 are average rates over 50-second time intervals. The value of the rate at a par-ticular time (the instantaneous rate) can be obtained by computing the slope of a line tangent to the curve at that point. Figure 12.1 shows a tangent drawn at t 100 seconds.
The slope of this line gives the rate at t 100 seconds as follows: ¢ 3NO24 ¢t Slope of the tangent line change in y change in x 4.2 105 mol/L s 14.2 105 mol/L s2 Rate ¢ 3NO24 ¢t Rate ¢ 3NO24 ¢t ¢[NO2] 4.2 105 mol/L s 0.0079 mol/L 0.0100 mol/L 50. s 3NO24 t50 3NO24 t0 50. s 0 s Change in 3NO24 Time elapsed ¢ 3NO24 ¢t FIGURE 12.2 Representation of the reaction 2NO2(g) 2NO(g) O2(g). (a) The reaction at the very beginning ( ). (b) and (c) As time passes, NO2 is converted to NO and O2.
t 0 S TABLE 12.2 Average Rate (in ) of Decomposition of Nitrogen Dioxide as a Function of Time Time Period (s) Note that the rate decreases with time.
200 S 250 1.0 105 150 S 200 1.4 105 100 S 150 2.0 105 50 S 100 2.8 105 0 S 50 4.2 105 ¢[NO2] ¢t mol/L s Appendix 1.3 reviews slopes of straight lines.
Time (a) (b) (c) 12.1 Reaction Rates 531 But Therefore, So far we have discussed the rate of this reaction only in terms of the reactant. The rate also can be defined in terms of the products. However, in doing so we must take into account the coefficients in the balanced equation for the reaction, because the stoichiom-etry determines the relative rates of consumption of reactants and generation of products.
For example, in the reaction we are considering, both the reactant NO2 and the product NO have a coefficient of 2, so NO is produced at the same rate as NO2 is consumed. We can verify this from Fig. 12.1. Note that the curve for NO is the same shape as the curve for NO2, except that it is inverted, or flipped over.
This means that, at any point in time, the slope of the tangent to the curve for NO will be the negative of the slope to the curve for NO2. (Verify this at the point t 100 sec-onds on both curves.) In the balanced equation, the product O2 has a coefficient of 1, which means it is produced half as fast as NO, since NO has a coefficient of 2. That is, the rate of NO production is twice the rate of O2 production.
We also can verify this fact from Fig. 12.1. For example, at seconds, The slope at t 250 seconds on the NO curve is twice the slope of that point on the O2 curve, showing that the rate of production of NO is twice that of O2.
4.3 106 mol/L s Slope of the tangent to the O2 curve 3.0 104 mol/L 70. s 8.6 106 mol/L s Slope of the tangent to the NO curve 6.0 104 mol/L 70. s t 250 2NO21g2 ¡ 2NO1g2 O21g2 2.4 105 mol/L s a0.0026 mol/L 110 s b Rate 1slope of the tangent line2 Rate ¢ 3NO24 ¢t Los Angeles on a clear day, and on a day when air pollution is significant.
532 Chapter Twelve Chemical Kinetics The rate information can be summarized as follows: We have seen that the rate of a reaction is not constant, but that it changes with time.
This is so because the concentrations change with time (Fig. 12.1).
Because the reaction rate changes with time, and because the rate is different (by fac-tors that depend on the coefficients in the balanced equation) depending on which reac-tant or product is being studied, we must be very specific when we describe a rate for a chemical reaction.
12.2 Rate Laws: An Introduction Chemical reactions are reversible. In our discussion of the decomposition of nitrogen diox-ide, we have so far considered only the forward reaction, as shown here: However, the reverse reaction also can occur. As NO and O2 accumulate, they can react to re-form NO2: When gaseous NO2 is placed in an otherwise empty container, initially the dominant reaction is and the change in the concentration of NO2 depends only on the forward re-action. However, after a period of time, enough products accumulate so that the reverse reaction becomes important. Now depends on the difference in the rates of the forward and reverse reactions. This complication can be avoided if we study the rate of a reaction under conditions where the reverse reaction makes only a negligible contribution.
Typically, this means that we must study a reaction at a point soon after the reactants are mixed, before the products have had time to build up to significant levels.
If we choose conditions where the reverse reaction can be neglected, the reaction rate will depend only on the concentrations of the reactants. For the decomposition of nitrogen dioxide, we can write (12.1) Such an expression, which shows how the rate depends on the concentrations of reactants, is called a rate law. The proportionality constant k, called the rate constant, and n, called the order of the reactant, must both be determined by experiment. The order of a reac-tant can be an integer (including zero) or a fraction. For the relatively simple reactions we will consider in this book, the orders will often be positive integers.
Note two important points about Equation (12.1): 1. The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.
2. The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.
Rate k3NO24 n ¢[NO2] (¢[NO2]) 2NO21g2 ¡ 2NO1g2 O21g2 O21g2 2NO1g2 ¡ 2NO21g2 2NO21g2 ¡ 2NO1g2 O21g2 2a ¢ 3O24 ¢t b ¢ 3NO4 ¢t ¢ 3NO24 ¢t 2(rate of production of O2) rate of production of NO Rate of consumption of NO2 When forward and reverse reaction rates are equal, there will be no changes in the concentrations of reactants or products.
This is called chemical equilibrium and is discussed fully in Chapter 13.
12.2 Rate Laws: An Introduction 533 Before we go further we must define exactly what we mean by the term rate in Equa-tion (12.1). In Section 12.1 we saw that reaction rate means a change in concentration per unit time. However, which reactant or product concentration do we choose in defining the rate? For example, for the decomposition of NO2 to produce O2 and NO considered in Section 12.1, we could define the rate in terms of any of these three species. However, since O2 is produced only half as fast as NO, we must be careful to specify which species we are talking about in a given case. For instance, we might choose to define the reaction rate in terms of the consumption of NO2: On the other hand, we could define the rate in terms of the production of O2: Note that because 2NO2 molecules are consumed for every O2 molecule produced, or and Thus the value of the rate constant depends on how the rate is defined.
In this text we will always be careful to define exactly what is meant by the rate for a given reaction so that there will be no confusion about which specific rate constant is being used.
Types of Rate Laws Notice that the rate law we have used to this point expresses rate as a function of con-centration. For example, for the decomposition of NO2 we have defined which tells us (once we have determined the value of n) exactly how the rate depends on the concentration of the reactant, NO2. A rate law that expresses how the rate depends on concentration is technically called the differential rate law, but it is often simply called the rate law. Thus when we use the term the rate law in this text, we mean the expres-sion that gives the rate as a function of concentration.
A second kind of rate law, the integrated rate law, also will be important in our study of kinetics. The integrated rate law expresses how the concentrations depend on time. Although we will not consider the details here, a given differential rate law is al-ways related to a certain type of integrated rate law, and vice versa. That is, if we deter-mine the differential rate law for a given reaction, we automatically know the form of the integrated rate law for the reaction. This means that once we determine experimentally either type of rate law for a reaction, we also know the other one.
Which rate law we choose to determine by experiment often depends on what types of data are easiest to collect. If we can conveniently measure how the rate changes as the concentrations are changed, we can readily determine the differential (rate/concen-tration) rate law. On the other hand, if it is more convenient to measure the concentra-tion as a function of time, we can determine the form of the integrated (concentration/ time) rate law. We will discuss how rate laws are actually determined in the next several sections.
Why are we interested in determining the rate law for a reaction? How does it help us? It helps us because we can work backward from the rate law to infer the steps by Rate ¢ 3NO24 ¢t k3NO24 n k 2 k¿ k3NO24 n 2k¿ 3NO24 n Rate 2 rate¿ Rate¿ ¢ 3O24 ¢t k¿ 3NO24 n Rate ¢ 3NO24 ¢t k3NO24 n The name differential rate law comes from a mathematical term. We will regard it simply as a label. The terms differential rate law and rate law will be used inter-changeably in this text.
534 Chapter Twelve Chemical Kinetics which the reaction occurs. Most chemical reactions do not take place in a single step but result from a series of sequential steps. To understand a chemical reaction, we must learn what these steps are. For example, a chemist who is designing an insecticide may study the reactions involved in the process of insect growth to see what type of molecule might interrupt this series of reactions. Or an industrial chemist may be trying to make a given reaction occur faster. To accomplish this, he or she must know which step is slowest, be-cause it is that step that must be speeded up. Thus a chemist is usually not interested in a rate law for its own sake but because of what it reveals about the steps by which a re-action occurs. We will develop a process for finding the reaction steps in this chapter.
Rate Laws: A Summary There are two types of rate laws.
1. The differential rate law (often called simply the rate law) shows how the rate of a reaction depends on concentrations.
2. The integrated rate law shows how the concentrations of species in the reaction depend on time.
Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants.
Because the differential and integrated rate laws for a given reaction are related in a well-defined way, the experimental determination of either of the rate laws is sufficient.
Experimental convenience usually dictates which type of rate law is determined experimentally.
Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.
12.3 Determining the Form of the Rate Law The first step in understanding how a given chemical reaction occurs is to determine the form of the rate law. That is, we need to determine experimentally the power to which each reactant concentration must be raised in the rate law. In this section we will explore ways to obtain the differential rate law for a reaction. First, we will consider the decom-position of dinitrogen pentoxide in carbon tetrachloride solution: Data for this reaction at C are listed in Table 12.3 and plotted in Fig. 12.3. In this re-action the oxygen gas escapes from the solution and thus does not react with the nitro-gen dioxide, so we do not have to be concerned about the effects of the reverse reaction at any time over the life of the reaction. That is, the reverse reaction is negligible at all times over the course of this reaction.
Evaluation of the reaction rates at concentrations of N2O5 of 0.90 M and 0.45 M, by taking the slopes of the tangents to the curve at these points (see Fig. 12.3), yields the following data: 45° 2N2O51soln2 ¡ 4NO21soln2 O21g2 TABLE 12.3 Concentration/ Time Data for the Reaction (at 45ºC) [N2O5] (mol/L) Time (s) 1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000 O2( g) (soln) 2N2O5(soln) S 4NO2 [N2O5] Rate (mol/L s) 0.90 M 0.45 M 2.7 104 5.4 104 12.3 Determining the Form of the Rate Law 535 Note that when [N2O5] is halved, the rate is also halved. This means that the rate of this reaction depends on the concentration of N2O5 to the first power. In other words, the (differential) rate law for this reaction is Thus the reaction is first order in N2O5. Note that for this reaction the order is not the same as the coefficient of N2O5 in the balanced equation for the reaction. This reempha-sizes the fact that the order of a particular reactant must be obtained by observing how the reaction rate depends on the concentration of that reactant.
We have seen that by determining the instantaneous rate at two different reactant con-centrations, the rate law for the decomposition of N2O5 is shown to have the form where A represents N2O5.
Method of Initial Rates One common method for experimentally determining the form of the rate law for a reac-tion is the method of initial rates. The initial rate of a reaction is the instantaneous rate determined just after the reaction begins (just after ). The idea is to determine the instantaneous rate before the initial concentrations of reactants have changed significantly.
Several experiments are carried out using different initial concentrations, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations. This allows the form of the rate law to be deter-mined. We will illustrate the method of initial rates using the following equation: Table 12.4 gives initial rates obtained from three experiments involving different initial concentrations of reactants. The general form of the rate law for this reaction is We can determine the values of n and m by observing how the initial rate depends on the initial concentrations of and In Experiments 1 and 2, where the initial NO2 .
NH4 Rate ¢ 3NH4 4 ¢t k3NH4 4 n3NO2 4 m NH4 1aq2 NO2 1aq2 ¡ N21g2 2H2O1l2 t 0 Rate ¢ 3A4 ¢t k3A4 Rate ¢ 3N2O54 ¢t k3N2O54 1 k3N2O54 .20 .40 .60 .80 1.00 400 800 1200 1600 2000 Rate = 5.4 × 10 –4 mol/L.s Rate = 2.7 × 10 –4 mol/L. s [N2O5] (mol/L) Time (s) FIGURE 12.3 A plot of the concentration of N2O5 as a function of time for the reaction (at 45°C). Note that the reaction rate at is twice that at .
[N2O5] 0.45 M [N2O5] 0.90 M 2N2O5(soln) S 4NO2(soln) O2(g) First order: [A]. Doubling the concentration of A doubles the reaction rate.
rate k The value of the initial rate is determined for each experiment at the same value of t as close to as possible.
t 0 Visualization: Decomposition of N2O5 Visualization: Reaction Rate and Concentration 536 Chapter Twelve Chemical Kinetics concentration of remains the same but the initial concentration of doubles, the observed initial rate also doubles. Since we have for Experiment 1 and for Experiment 2 The ratio of these rates is 2.00 Thus which means the value of m is 1. The rate law for this reaction is first order in the reac-tant A similar analysis of the results for Experiments 2 and 3 yields the ratio The value of n is also 1.
We have shown that the values of n and m are both 1 and the rate law is This rate law is first order in both and . Note that it is merely a coincidence that n and m have the same values as the coefficients of and in the balanced equation for the reaction.
The overall reaction order is the sum of n and m. For this reaction, The reaction is second order overall.
n m 2.
NO2 NH4 NH4 NO2 Rate k3NH4 4 3NO2 4 2.00 a0.200 0.100b n 12.002n Rate 3 Rate 2 5.40 107 mol/L s 2.70 107 mol/L s 10.200 mol/L2n 10.100 mol/L2n NO2 .
Rate 2 Rate 1 2.00 12.02m 10.010 mol/L2m 10.0050 mol/L2m 12.02m Rate 2 Rate 1 2.70 107 mol/L s 1.35 107 mol/L s k10.100 mol/L2n10.010 mol/L2m k10.100 mol/L2n10.0050 mol/L2m Rate 2.70 107 mol/L s k10.100 mol/L2n10.010 mol/L2m Rate 1.35 107 mol/L s k10.100 mol/L2n10.0050 mol/L2m Rate k3NH4 4n3NO2 4m NO2 NH4 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ TABLE 12.4 Initial Rates from Three Experiments for the Reaction Initial Initial Concentration Concentration Initial Experiment of NH4 of NH2 Rate (mol/L s) 1 0.100 M 0.0050 M 2 0.100 M 0.010 M 3 0.200 M 0.010 M 5.40 107 2.70 107 1.35 107 NH4 (aq) NO2 (aq) S N2(g) 2H2O(l) Rates 1, 2, and 3 were determined at the same value of t (very close to ).
t 0 Overall reaction order is the sum of the orders for the various reactants.
12.3 Determining the Form of the Rate Law 537 The value of the rate constant k can now be calculated using the results of any of the three experiments shown in Table 12.4. From the data for Experiment 1, we know that Then Determining a Rate Law The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation Table 12.5 gives the results from four experiments. Using these data, determine the or-ders for all three reactants, the overall reaction order, and the value of the rate constant.
Solution The general form of the rate law for this reaction is We can determine the values of n, m, and p by comparing the rates from the various ex-periments. To determine the value of n, we use the results from Experiments 1 and 2, in which only changes: Thus n is equal to 1.
To determine the value of m, we use the results from Experiments 2 and 3, in which only changes: Thus m is equal to 1.
2.0 a0.20 mol/L 0.10 mol/Lb m 12.02m Rate 3 Rate 2 3.2 103 mol/L s 1.6 103 mol/L s k10.20 mol/L2n10.20 mol/L2m10.10 mol/L2 p k10.20 mol/L2n10.10 mol/L2m10.10 mol/L2 p [Br] 2.0 a0.20 mol/L 0.10 mol/Lb n 12.02n Rate 2 Rate 1 1.6 103 mol/L s 8.0 104 mol/L s k10.20 mol/L2n10.10 mol/L2m10.10 mol/L2 p k10.10 mol/L2n10.10 mol/L2m10.10 mol/L2 p [BrO3 ] Rate k3BrO3 4 n3Br4 m3H4 p BrO3 1aq2 5Br1aq2 6H1aq2 ¡ 3Br21l2 3H2O1l2 k 1.35 107 mol/L s 10.100 mol/L210.0050 mol/L2 2.7 104 L/mol s 1.35 107 mol/L s k10.100 mol/L210.0050 mol/L2 Rate k3NH4 4 3NO2 4 Sample Exercise 12.1 TABLE 12.5 The Results from Four Experiments to Study the Reaction Initial Initial Initial Measured Concentration Concentration Concentration Initial of of of Rate Experiment (mol/L) (mol/L) (mol/L) (mol/L s) 1 0.10 0.10 0.10 2 0.20 0.10 0.10 3 0.20 0.20 0.10 4 0.10 0.10 0.20 3.2 103 3.2 103 1.6 103 8.0 104 H Br BrO3 BrO3 (aq) 5Br(aq) 6H(aq) S 3Br2(l) 3H2O(l) 538 Chapter Twelve Chemical Kinetics To determine the value of p, we use the results from Experiments 1 and 4, in which and are constant but differs: Thus p is equal to 2.
The rate of this reaction is first order in and and second order in The overall reaction order is The rate law can now be written The value of the rate constant k can be calculated from the results of any of the four ex-periments. For Experiment 1, the initial rate is mol/L s and and Using these values in the rate law gives Reality Check: Verify that the same value of k can be obtained from the results of the other experiments.
See Exercises 12.25 through 12.28.
12.4 The Integrated Rate Law The rate laws we have considered so far express the rate as a function of the reactant con-centrations. It is also useful to be able to express the reactant concentrations as a func-tion of time, given the (differential) rate law for the reaction. In this section we show how this is done.
We will proceed by first looking at reactions involving a single reactant: all of which have a rate law of the form We will develop the integrated rate laws individually for the cases (first order), (second order), and (zero order).
First-Order Rate Laws For the reaction 2N2O51soln2 ¡ 4NO21soln2 O21g2 n 0 n 2 n 1 Rate ¢3A4 ¢t k3A4n aA ¡ products k 8.0 104 mol/L s 1.0 104 mol4/L4 8.0 L3/mol3 s 8.0 104 mol/L s k11.0 104 mol4/L42 8.0 104 mol/L s k10.10 mol/L210.10 mol/L210.10 mol/L22 [H] 0.10 M.
[Br] 0.10 M, 0.100 M, [BrO3 ] 8.0 104 Rate k3BrO3 4 3Br4 3H4 2 n m p 4.
H.
Br BrO3 4.0 12.02 p 12.022 4.0 a0.20 mol/L 0.10 mol/Lb p Rate 4 Rate 1 3.2 103 mol/L s 8.0 104 mol/L s k10.10 mol/L2n10.10 mol/L2m10.20 mol/L2 p k10.10 mol/L2n10.10 mol/L2m10.10 mol/L2 p [H] [Br] [BrO3 ] 12.4 The Integrated Rate Law 539 we have found that the rate law is Since the rate of this reaction depends on the concentration of to the first power, it is a first-order reaction. This means that if the concentration of in a flask were suddenly doubled, the rate of production of and also would double. This rate law can be put into a different form using a calculus operation known as integration, which yields the expression where ln indicates the natural logarithm, t is the time, is the concentration of at time t, and is the initial concentration of (at the start of the experiment). Note that such an equation, called the integrated rate law, expresses the concentration of the reactant as a function of time.
For a chemical reaction of the form where the kinetics are first order in [A], the rate law is and the integrated first-order rate law is (12.2) There are several important things to note about Equation (12.2): 1. The equation shows how the concentration of A depends on time. If the initial con-centration of A and the rate constant k are known, the concentration of A at any time can be calculated.
2. Equation (12.2) is of the form where a plot of y versus x is a straight line with slope m and intercept b. In Equation (12.2), Thus, for a first-order reaction, plotting the natural logarithm of concentration versus time always gives a straight line. This fact is often used to test whether a reaction is first order. For the reaction the reaction is first order in A if a plot of ln[A] versus t is a straight line. Conversely, if this plot is not a straight line, the reaction is not first order in A.
3. This integrated rate law for a first-order reaction also can be expressed in terms of a ratio of [A] and [A]0 as follows: First-Order Rate Laws I The decomposition of in the gas phase was studied at constant temperature.
2N2O51g2 ¡ 4NO21g2 O21g2 N2O5 lna 3A4 0 3A4 b kt aA ¡ products y ln3A4 x t m k b ln3A40 y mx b, ln3A4 kt ln3A4 0 Rate ¢3A4 ¢t k3A4 aA ¡ products t 0, N2O5 [N2O5]0 N2O5 [N2O5] ln3N2O54 kt ln3N2O540 O2 NO2 N2O5 N2O5 Rate ¢3N2O54 ¢t k3N2O54 Sample Exercise 12.2 Appendix 1.2 contains a review of logarithms.
An integrated rate law relates concentration to reaction time.
For a first-order reaction, a plot of ln[A] versus t is always a straight line.
540 Chapter Twelve Chemical Kinetics The following results were collected: [N2O5] (mol/L) Time (s) 0.1000 0 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 ln[N2O5] Time (s) 0 50 100 200 300 400 5.075 4.382 3.689 2.996 2.649 2.303 Using these data, verify that the rate law is first order in , and calculate the value of the rate constant, where the .
Solution We can verify that the rate law is first order in by constructing a plot of versus time. The values of at various times are given in the table above and the plot of versus time is shown in Fig. 12.4. The fact that the plot is a straight line confirms that the reaction is first order in , since it follows the equation Since the reaction is first order, the slope of the line equals , where Since the first and last points are exactly on the line, we will use these points to cal-culate the slope: See Exercise 12.31.
First-Order Rate Laws II Using the data given in Sample Exercise 12.2, calculate at 150 s after the start of the reaction.
Solution We know from Sample Exercise 12.2 that mol/L at 100 s and mol/L at 200 s. Since 150 s is halfway between 100 and 200 s, it is tempting to 0.0250 [N2O5] [N2O5] 0.0500 [N2O5] k 1slope2 6.93 103 s1 Slope 5.075 12.3032 400. s 0 s 2.772 400. s 6.93 103 s1 Slope change in y change in x ¢y ¢x ¢1ln3N2O54 2 ¢t k kt ln[N2O5]0.
ln[N2O5] N2O5 ln[N2O5] ln[N2O5] ln[N2O5] [N2O5] rate ¢[N2O5] ¢t [N2O5] –6.0 0 ln [N2O5] Time (s) 100 200 300 400 –4.0 –2.0 FIGURE 12.4 A plot of ln[N2O5] versus time.
Sample Exercise 12.3 12.4 The Integrated Rate Law 541 assume that we can simply use an arithmetic average to obtain at that time. This is incorrect because it is , not , that is directly proportional to t. To calculate after 150 s, we use Equation (12.2): where s, (as determined in Sample Exercise 12.2), and mol/L.
Note that this value of is not halfway between 0.0500 and 0.0250 mol/L.
See Exercise 12.31.
Half-Life of a First-Order Reaction The time required for a reactant to reach half its original concentration is called the half-life of a reactant and is designated by the symbol . For example, we can calculate the half-life of the decomposition reaction discussed in Sample Exercise 12.2. The data plot-ted in Fig. 12.5 show that the half-life for this reaction is 100 seconds. We can see this by considering the following numbers: t1 2 [N2O5] 3N2O54 t150 antilog13.3432 0.0353 mol/L 1.040 2.303 3.343 ln13N2O542t150 16.93 103 s121150. s2 ln10.1002 [N2O5]0 0.1000 k 6.93 103 s1 t 150.
ln3N2O54 kt ln3N2O54 0 [N2O5] [N2O5] ln[N2O5] [N2O5] The antilog operation means to exponentiate (see Appendix 1.2).
[N2O5]0 0.1000 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 [N2O5]0 2 [N2O5]0 4 [N2O5]0 8 50 150 250 350 100 200 300 400 t1/2 t1/2 t1/2 Time (s) [N2O5] (mol/L) FIGURE 12.5 A plot of [N2O5] versus time for the decom-position reaction of N2O5.
Visualization: Half-Life of Reactions N2O5 t (s) 0.100 0 t 100 s; 0.0500 100 t 100 s; 0.0250 200 t 100 s; 0.0125 300 3N2O54 t300 3N2O54 t200 0.0125 0.0250 1 2 3N2O54 t200 3N2O54 t100 0.025 0.050 1 2 3N2O54 t100 3N2O54 t0 0.050 0.100 1 2 ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ 542 Chapter Twelve Chemical Kinetics Note that it always takes 100 seconds for to be halved in this reaction.
A general formula for the half-life of a first-order reaction can be derived from the integrated rate law for the general reaction If the reaction is first order in [A], By definition, when Then, for the integrated rate law becomes or .
Substituting the value of ln(2) and solving for gives (12.3) This is the general equation for the half-life of a first-order reaction. Equation (12.3) can be used to calculate if k is known or k if is known. Note that for a first-order re-action, the half-life does not depend on concentration.
Half-Life for First-Order Reaction A certain first-order reaction has a half-life of 20.0 minutes.
a. Calculate the rate constant for this reaction.
b. How much time is required for this reaction to be 75% complete?
Solution a. Solving Equation (12.3) for k gives b. We use the integrated rate law in the form If the reaction is 75% complete, 75% of the reactant has been consumed, leaving 25% in the original form: This means that Then lna 3A4 0 3A4 b ln14.02 kt a3.47 102 min bt 3A4 3A4 0 0.25 or 3A4 0 3A4 1 0.25 4.0 3A4 3A4 0 100% 25% lna 3A40 3A4 b kt k 0.693 t1 2 0.693 20.0 min 3.47 102 min1 t1 2 t1 2 t1 2 0.693 k t1 2 ln122 kt1 2 lna 3A4 0 3A4 0 2b kt1 2 t t1 2, 3A4 3A4 0 2 t t1 2, lna 3A40 3A4 b kt aA ¡ products [N2O5] Sample Exercise 12.4 For a first-order reaction, is independent of the initial concentration.
t1/2 12.4 The Integrated Rate Law 543 and Thus it takes 40. minutes for this particular reaction to reach 75% completion.
Let’s consider another way of solving this problem using the definition of half-life. After one half-life the reaction has gone 50% to completion. If the initial con-centration were 1.0 mol/L, after one half-life the concentration would be 0.50 mol/L.
One more half-life would produce a concentration of 0.25 mol/L. Comparing 0.25 mol/L with the original 1.0 mol/L shows that 25% of the reactant is left after two half-lives. This is a general result. (What percentage of reactant remains after three half-lives?) Two half-lives for this reaction is 2(20.0 min), or 40.0 min, which agrees with the preceding answer.
See Exercises 12.32 and 12.42 through 12.44.
Second-Order Rate Laws For a general reaction involving a single reactant, that is, that is second order in A, the rate law is (12.4) The integrated second-order rate law has the form (12.5) Note the following characteristics of Equation (12.5): 1. A plot of 1[A] versus t will produce a straight line with a slope equal to k.
2. Equation (12.5) shows how [A] depends on time and can be used to calculate [A] at any time t, provided k and [A]0 are known.
When one half-life of the second-order reaction has elapsed by definition, Equation (12.5) then becomes Solving for gives the expression for the half-life of a second-order reaction: (12.6) t1 2 1 k3A40 t1 2 1 3A4 0 kt1 2 2 3A4 0 1 3A4 0 kt1 2 1 3A4 0 2 kt1 2 1 3A4 0 3A4 3A40 2 (t t1 2), 1 3A4 kt 1 3A40 Rate ¢ 3A4 ¢t k3A4 2 aA ¡ products t ln14.02 3.47 102 min 40. min For second-order reactions, a plot of 1[A] versus t will be linear.
Second order: [A]2. Doubling the concentration of A quadruples the reaction rate; tripling the concentration of A increases the rate by nine times.
rate k 544 Chapter Twelve Chemical Kinetics Determining Rate Laws Butadiene reacts to form its dimer according to the equation The following data were collected for this reaction at a given temperature: 2C4H61g2 ¡ C8H121g2 Sample Exercise 12.5 When two identical molecules combine, the resulting molecule is called a dimer.
[C4H6] (molL) Time (1 s) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200 t (s) 0 100 1000 160 1800 210 2800 270 3600 320 4400 370 5200 415 6200 481 6.175 6.028 5.915 5.767 5.599 5.348 5.075 4.605 In[C4H4] 1 [C4H6] a. Is this reaction first order or second order?
b. What is the value of the rate constant for the reaction?
c. What is the half-life for the reaction under the conditions of this experiment?
Solution a. To decide whether the rate law for this reaction is first order or second order, we must see whether the plot of ln[C4H6] versus time is a straight line (first order) or the plot of 1[C4H6] versus time is a straight line (second order). The data necessary to make these plots are as follows: The resulting plots are shown in Fig. 12.6. Since the ln[C4H6] versus t plot [Fig. 12.6(a)] is not a straight line, the reaction is not first order. The reaction is, how-ever, second order, as shown by the linearity of the 1[C4H6] versus t plot [Fig. 12.6(b)].
Thus we can now write the rate law for this second-order reaction: b. For a second-order reaction, a plot of 1[C4H6] versus t produces a straight line of slope k. In terms of the standard equation for a straight line, we have and Thus the slope of the line can be expressed as follows: Slope ¢y ¢x ¢a 1 3C4H64 b ¢t x t.
y 1 [C4H6] y mx b, Rate ¢ 3C4H64 ¢t k3C4H64 2 Butadiene (C4H6) 12.4 The Integrated Rate Law 545 Using the points at and we can find the rate constant for the reaction: c. The expression for the half-life of a second-order reaction is In this case (from part b) and [A]0 [C4H6]0 0.01000 M (the concentration at ). Thus The initial concentration of C4H6 is halved in 1630 s.
See Exercises 12.33, 12.34, 12.45, and 12.46.
It is important to recognize the difference between the half-life for a first-order reac-tion and the half-life for a second-order reaction. For a second-order reaction, t12 depends on both k and [A]0; for a first-order reaction, t12 depends only on k. For a first-order re-action, a constant time is required to reduce the concentration of the reactant by half, and then by half again, and so on, as the reaction proceeds. From Sample Exercise 12.5 we can see that this is not true for a second-order reaction. For that second-order reaction, we found that the first half-life (the time required to go from to is 1630 seconds. We can estimate the second half-life from the con-centration data as a function of time. Note that to reach 0.0024 M C4H6 (approximately 0.00502) requires 5200 seconds of reaction time. Thus to get from 0.0050 M C4H6 to 0.0024 M C4H6 takes 3570 seconds (5200 1630). The second half-life is much longer than the first. This pattern is characteristic of second-order reactions. In fact, for a second-order reaction, each successive half-life is double the preceding one (provided the effects [C4H6] 0.0050 M) [C4H6] 0.010 M t1 2 1 16.14 102 L/mol s211.000 102 mol/L2 1.63 103 s t 0 k 6.14 102 L/mol s t1 2 1 k3A40 k slope 1481 1002 L/mol 16200. 02 s 381 6200. L/mol s 6.14 102 L/mol s t 6200, t 0 0 2000 4000 6000 –6.000 –5.000 Time (s) ln [C4H6] 0 2000 4000 6000 100 400 Time (s) 1 [C4H6] 200 300 (a) (b) FIGURE 12.6 (a) A plot of ln[C4H6] versus t. (b) A plot of 1[C4H6] versus t.
For a second-order reaction, t12 is de-pendent on [A]0. For a first-order reaction, t12 is independent of [A]0.
546 Chapter Twelve Chemical Kinetics of the reverse reaction can be ignored, as we are assuming here). Prove this to yourself by examining the equation Zero-Order Rate Laws Most reactions involving a single reactant show either first-order or second-order kinet-ics. However, sometimes such a reaction can be a zero-order reaction. The rate law for a zero-order reaction is For a zero-order reaction, the rate is constant. It does not change with concentration as it does for first-order or second-order reactions.
The integrated rate law for a zero-order reaction is (12.7) In this case a plot of [A] versus t gives a straight line of slope as shown in Fig. 12.7.
The expression for the half-life of a zero-order reaction can be obtained from the in-tegrated rate law. By definition, [A] [A]02 when so or Solving for gives (12.8) Zero-order reactions are most often encountered when a substance such as a metal surface or an enzyme is required for the reaction to occur. For example, the decomposi-tion reaction occurs on a hot platinum surface. When the platinum surface is completely covered with N2O molecules, an increase in the concentration of N2O has no effect on the rate, since only those N2O molecules on the surface can react. Under these conditions, the rate is a constant because it is controlled by what happens on the platinum surface rather than by the total concentration of N2O, as illustrated in Fig. 12.8. This reaction also can occur at high temperatures with no platinum surface present, but under these conditions, it is not zero order.
Integrated Rate Laws for Reactions with More Than One Reactant So far we have considered the integrated rate laws for simple reactions with only one re-actant. Special techniques are required to deal with more complicated reactions. Let’s con-sider the reaction From experimental evidence we know that the rate law is Rate ¢3BrO3 4 ¢t k3BrO3 4 3Br4 3H42 BrO3 1aq2 5Br1aq2 6H1aq2 ¡ 3Br21l2 3H2O1l2 2N2O1g2 ¡ 2N21g2 O21g2 t1 2 3A4 0 2k t1 2 kt1 2 3A4 0 2k 3A4 0 2 kt1 2 3A4 0 t t1 2, k, 3A4 kt 3A4 0 Rate k3A4 0 k112 k t1 2 1 (k[A]0).
For each successive half-life, [A]0 is halved. Since t12 1k[A]0, t12 doubles.
A zero-order reaction has a constant rate.
[A]0 [A] 0 t ∆t ∆[A] Slope = ∆[A] = –k ∆t FIGURE 12.7 A plot of [A] versus t for a zero- order reaction.
12.4 The Integrated Rate Law 547 Suppose we run this reaction under conditions where [BrO3 ]0 1.0 103 M, [Br]0 1.0 M, and [H]0 1.0 M. As the reaction proceeds, [BrO3 ] decreases signif-icantly, but because the Br ion and H ion concentrations are so large initially, relatively little of these two reactants is consumed. Thus [Br] and [H] remain approximately constant. In other words, under the conditions where the Br ion and H ion concentra-tions are much larger than the BrO3 ion concentration, we can assume that throughout the reaction This means that the rate law can be written where, since [Br]0 and [H]0 are constant, The rate law is first order. However, since this law was obtained by simplifying a more complicated one, it is called a pseudo-first-order rate law. Under the conditions of this experi-ment, a plot of ln[BrO3 ] versus t will give a straight line where the slope is equal to k . Since [Br]0 and [H]0 are known, the value of k can be calculated from the equation which can be rearranged to give Note that the kinetics of complicated reactions can be studied by observing the be-havior of one reactant at a time. If the concentration of one reactant is much smaller than the concentrations of the others, then the amounts of those reactants present in large con-centrations will not change significantly and can be regarded as constant. The change in concentration with time of the reactant present in a relatively small amount can then be used to determine the order of the reaction in that component. This technique allows us to determine rate laws for complex reactions.
k k¿ 3Br4 03H4 0 2 k¿ k3Br4 03H4 0 2 Rate k¿ 3BrO3 4 k¿ k3Br4 03H4 0 2 Rate k3Br4 03H4 0 23BrO3 4 k¿ 3BrO3 4 3Br4 3Br4 0 and 3H4 3H4 0 Pt N2O (a) (b) Pt FIGURE 12.8 The decomposition reaction 2N2O(g) n 2N2(g) O2(g) takes place on a platinum surface. Although [N2O] is twice as great in (b) as in (a), the rate of decomposition of N2O is the same in both cases because the platinum surface can accommodate only a certain number of molecules. As a result, this reaction is zero order.
548 Chapter Twelve Chemical Kinetics 12.5 Rate Laws: A Summary In the last several sections we have developed the following important points: 1. To simplify the rate laws for reactions, we have always assumed that the rate is be-ing studied under conditions where only the forward reaction is important. This pro-duces rate laws that contain only reactant concentrations.
2. There are two types of rate laws.
a. The differential rate law (often called the rate law) shows how the rate depends on the concentrations. The forms of the rate laws for zero-order, first-order, and second-order kinetics of reactions with single reactants are shown in Table 12.6.
b. The integrated rate law shows how concentration depends on time. The integrated rate laws corresponding to zero-order, first-order, and second- order kinetics of one-reactant reactions are given in Table 12.6.
3. Whether we determine the differential rate law or the integrated rate law depends on the type of data that can be collected conveniently and accurately. Once we have experimentally determined either type of rate law, we can write the other for a given reaction.
4. The most common method for experimentally determining the differential rate law is the method of initial rates. In this method several experiments are run at different ini-tial concentrations and the instantaneous rates are determined for each at the same value of t (as close to as possible). The point is to evaluate the rate before the concentrations change significantly from the initial values. From a comparison of the initial rates and the initial concentrations the dependence of the rate on the concen-trations of various reactants can be obtained—that is, the order in each reactant can be determined.
5. To experimentally determine the integrated rate law for a reaction, concentrations are measured at various values of t as the reaction proceeds. Then the job is to see which integrated rate law correctly fits the data. Typically this is done visually by ascer-taining which type of plot gives a straight line. A summary for one-reactant reactions is given in Table 12.6. Once the correct straight-line plot is found, the correct inte-grated rate law can be chosen and the value of k obtained from the slope. Also, the (differential) rate law for the reaction can then be written.
t 0 TABLE 12.6 Summary of the Kinetics for Reactions of the Type aA Products That Are Zero, First, or Second Order in [A] Order Zero First Second Rate Law: Integrated Rate Law: Plot Needed to Give a Straight Line: Relationship of Rate Constant to the Slope of Straight Line: Half-Life: t1 2 1 k3A4 0 t1 2 0.693 k t1 2 3A40 2k Slope k Slope k Slope k 1 3A4 versus t ln3A4 versus t 3A4 versus t 1 3A4 kt 1 3A4 0 ln3A4 kt ln3A4 0 3A4 kt 3A4 0 Rate k3A42 Rate k3A4 Rate k S Visualization: Rate Laws 12.6 Reaction Mechanisms 549 6. The integrated rate law for a reaction that involves several reactants can be treated by choosing conditions such that the concentration of only one reactant varies in a given experiment. This is done by having the concentration of one reactant remain small compared with the concentrations of all the others, causing a rate law such as to reduce to where and and . The value of n is obtained by determining whether a plot of [A] versus t is linear ( ), a plot of ln[A] versus t is linear ( ), or a plot of 1[A] versus t is linear ( ). The value of is de-termined from the slope of the appropriate plot. The values of m, p, and k can be found by determining the value of at several different concentrations of B and C.
12.6 Reaction Mechanisms Most chemical reactions occur by a series of steps called the reaction mechanism. To understand a reaction, we must know its mechanism, and one of the main purposes for studying kinetics is to learn as much as possible about the steps involved in a reaction. In this section we explore some of the fundamental characteristics of reaction mechanisms.
Consider the reaction between nitrogen dioxide and carbon monoxide: The rate law for this reaction is known from experiment to be As we will see below, this reaction is more complicated than it appears from the balanced equation. This is quite typical; the balanced equation for a reaction tells us the reactants, the products, and the stoichiometry but gives no direct information about the reaction mechanism.
For the reaction between nitrogen dioxide and carbon monoxide, the mechanism is thought to involve the following steps: where k1 and k2 are the rate constants of the individual reactions. In this mechanism, gaseous NO3 is an intermediate, a species that is neither a reactant nor a product but that is formed and consumed during the reaction sequence. This reaction is illustrated in Fig. 12.9.
NO31g2 CO1g2 ¡ k2 NO21g2 CO21g2 NO21g2 NO21g2 ¡ k1 NO31g2 NO1g2 Rate k3NO24 2 NO21g2 CO1g2 ¡ NO1g2 CO21g2 k¿ k¿ n 2 n 1 n 0 [C]0 [A]0 [B]0 [A]0 k¿ k[B]0 m[C]0 p Rate k¿ 3A4 n Rate k3A4 n3B4 m3C4p A balanced equation does not tell us how the reactants become products.
Visualization: Oscillating Reaction + Step 1 Step 2 + + + FIGURE 12.9 A molecular representation of the elemen-tary steps in the reaction of NO2 and CO.
An intermediate is formed in one step and used up in a subsequent step and so is never seen as a product.
A reaction is only as fast as its slowest step.
550 Chapter Twelve Chemical Kinetics Each of these two reactions is called an elementary step, a reaction whose rate law can be written from its molecularity. Molecularity is defined as the number of species that must collide to produce the reaction indicated by that step. A reaction in-volving one molecule is called a unimolecular step. Reactions involving the collision of two and three species are termed bimolecular and termolecular, respectively. Ter-molecular steps are quite rare, because the probability of three molecules colliding si-multaneously is very small. Examples of these three types of elementary steps and the corresponding rate laws are shown in Table 12.7. Note from Table 12.7 that the rate law for an elementary step follows directly from the molecularity of that step. For ex-ample, for a bimolecular step the rate law is always second order, either of the form for a step with a single reactant or of the form for a step involving two reactants.
We can now define a reaction mechanism more precisely. It is a series of elementary steps that must satisfy two requirements: 1. The sum of the elementary steps must give the overall balanced equation for the reaction.
2. The mechanism must agree with the experimentally determined rate law.
To see how these requirements are applied, we will consider the mechanism given above for the reaction of nitrogen dioxide and carbon monoxide. First, note that the sum of the two steps gives the overall balanced equation: The first requirement for a correct mechanism is met. To see whether the mechanism meets the second requirement, we need to introduce a new idea: the rate-determining step.
Multistep reactions often have one step that is much slower than all the others. Reactants can become products only as fast as they can get through this slowest step. That is, the overall reaction can be no faster than the slowest, or rate-determining, step in the sequence.
An analogy for this situation is the pouring of water rapidly into a container through a funnel. The water collects in the container at a rate that is essentially determined by the size of the funnel opening and not by the rate of pouring.
Which is the rate-determining step in the reaction of nitrogen dioxide and carbon monoxide? Let’s assume that the first step is rate-determining and the second step is relatively fast: Slow (rate-determining) Fast NO31g2 CO1g2 ¡ NO21g2 CO21g2 NO21g2 NO21g2 ¡ NO31g2 ONO1g2 Overall reaction: NO21g2 CO1g2 ¡ NO1g2 CO21g2 NO21g2 NO21g2 NO31g2 CO1g2 ¡ NO31g2 NO1g2 NO21g2 CO21g2 NO31g2 CO1g2 ¡ NO21g2 CO21g2 NO21g2 NO21g2 ¡ NO31g2 NO1g2 k[A][B] k[A]2 The prefix uni- means one, bi- means two, and ter- means three.
A unimolecular elementary step is always first order, a bimolecular step is always second order, and so on.
TABLE 12.7 Examples of Elementary Steps Elementary Step Molecularity Rate Law Unimolecular Bimolecular Bimolecular Termolecular Termolecular Rate k[A][B][C] A B C S products 12A B S products2 Rate k[A]2[B] A A B S products Rate k[A][B] A B S products 12A S products2 Rate k[A]2 A A S products Rate k[A] A S products 12.6 Reaction Mechanisms 551 What we have really assumed here is that the formation of NO3 occurs much more slowly than its reaction with CO. The rate of CO2 production is then controlled by the rate of formation of NO3 in the first step. Since this is an elementary step, we can write the rate law from the molecularity. The bimolecular first step has the rate law Since the overall reaction rate can be no faster than the slowest step, Note that this rate law agrees with the experimentally determined rate law given earlier.
The mechanism we assumed above satisfies the two requirements stated earlier and may be the correct mechanism for the reaction.
How does a chemist deduce the mechanism for a given reaction? The rate law is al-ways determined first. Then, using chemical intuition and following the two rules given on the previous page, the chemist constructs possible mechanisms and tries, with further experiments, to eliminate those that are least likely. A mechanism can never be proved ab-solutely. We can only say that a mechanism that satisfies the two requirements is possi-bly correct. Deducing mechanisms for chemical reactions can be difficult and requires skill and experience. We will only touch on this process in this text.
Reaction Mechanisms The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is The experimentally determined rate law is A suggested mechanism for this reaction is Slow Fast Is this an acceptable mechanism? That is, does it satisfy the two requirements?
Solution The first requirement for an acceptable mechanism is that the sum of the steps should give the balanced equation: The first requirement is met.
The second requirement is that the mechanism must agree with the experimentally determined rate law. Since the proposed mechanism states that the first step is rate-determining, the overall reaction rate must be that of the first step. The first step is bimolecular, so the rate law is Rate k13NO24 3F24 Overall reaction: 2NO2 F2 ¡ 2NO2F 2NO2 F2 F ¡ 2NO2F F F NO2 ¡ NO2F NO2 F2 ¡ NO2F F F NO2 ¡ k2 NO2F NO2 F2 ¡ k1 NO2F F Rate k3NO24 3F24 2NO21g2 F21g2 ¡ 2NO2F1g2 Overall rate k13NO24 2 Rate of formation of NO3 ¢ 3NO34 ¢t k13NO24 2 Sample Exercise 12.6 + + + 552 Chapter Twelve Chemical Kinetics This has the same form as the experimentally determined rate law. The proposed mecha-nism is acceptable because it satisfies both requirements. (Note that we have not proved that it is the correct mechanism.) See Exercises 12.51 and 12.52.
Although the mechanism given in Sample Exercise 12.6 has the correct stoichiome-try and fits the observed rate law, other mechanisms may also satisfy these requirements.
For example, the mechanism might be Slow Fast Fast Fast To decide on the most probable mechanism for the reaction, the chemist doing the study would have to perform additional experiments.
12.7 A Model for Chemical Kinetics How do chemical reactions occur? We already have given some indications. For example, we have seen that the rates of chemical reactions depend on the concentrations of the re-acting species. The initial rate for the reaction can be described by the rate law where the order of each reactant depends on the detailed reaction mechanism. This explains why reaction rates depend on concentration. But what about some of the other factors affecting reaction rates? For example, how does temperature affect the speed of a reaction?
We can answer this question qualitatively from our experience. We have refrigerators because food spoilage is retarded at low temperatures. The combustion of wood occurs at a measurable rate only at high temperatures. An egg cooks in boiling water much faster at sea level than in Leadville, Colorado (elevation 10,000 ft), where the boiling point of water is approximately . These observations and others lead us to conclude that chem-ical reactions speed up when the temperature is increased. Experiments have shown that virtually all rate constants show an exponential increase with absolute temperature, as rep-resented in Fig. 12.10.
In this section we discuss a model used to account for the observed characteris-tics of reaction rates. This model, called the collision model, is built around the central idea that molecules must collide to react. We have already seen how this assumption explains the concentration dependence of reaction rates. Now we need to consider whether this model can account for the observed temperature dependence of reaction rates.
The kinetic molecular theory of gases predicts that an increase in temperature raises molecular velocities and so increases the frequency of collisions between molecules. This idea agrees with the observation that reaction rates are greater at higher temperatures.
Thus there is qualitative agreement between the collision model and experimental obser-vations. However, it is found that the rate of reaction is much smaller than the calculated collision frequency in a collection of gas particles. This must mean that only a small frac-tion of the collisions produces a reaction. Why?
90°C Rate k3A4n3B4 m aA bB ¡ products NO3 NOF ¡ NO2F NO2 NOF2 NO2 ¡ NO2F NOF NO2 O ¡ NO3 NO2 F2 ¡ NOF2 O k T (K) FIGURE 12.10 A plot showing the exponential dependence of the rate constant on absolute tempera-ture. The exact temperature dependence of k is different for each reaction. This plot represents the behavior of a rate constant that doubles for every increase in tempera-ture of 10 K.
12.7 A Model for Chemical Kinetics 553 This question was first addressed in the 1880s by Svante Arrhenius. He proposed the existence of a threshold energy, called the activation energy, that must be overcome to produce a chemical reaction. Such a proposal makes sense, as we can see by considering the decomposition of BrNO in the gas phase: In this reaction two bonds must be broken and one bond must be formed.
Breaking a bond requires considerable energy (243 kJ/mol), which must come from somewhere. The collision model postulates that the energy comes from the kinetic energies possessed by the reacting molecules before the collision. This kinetic energy is changed into potential energy as the molecules are distorted during a collision to break bonds and rearrange the atoms into the product molecules.
We can envision the reaction progress as shown in Fig. 12.11. The arrangement of atoms found at the top of the potential energy “hill,” or barrier, is called the activated complex, or transition state. The conversion of BrNO to NO and is exothermic, as indicated by the fact that the products have lower potential energy than the reactant. How-ever, has no effect on the rate of the reaction. Rather, the rate depends on the size of the activation energy The main point here is that a certain minimum energy is required for two BrNO mol-ecules to “get over the hill” so that products can form. This energy is furnished by the energy of the collision. A collision between two BrNO molecules with small kinetic energies will not have enough energy to get over the barrier. At a given temperature only a certain fraction of the collisions possesses enough energy to be effective (to result in product formation).
We can be more precise by recalling from Chapter 5 that a distribution of velocities exists in a sample of gas molecules. Therefore, a distribution of collision energies also exists, as shown in Fig. 12.12 for two different temperatures. Figure 12.12 also shows the activation energy for the reaction in question. Only collisions with energy greater than Ea.
¢E Br2 Br¬N Br¬Br Br¬N 2BrNO1g2 ¡ 2NO1g2 Br21g2 The higher the activation energy, the slower the reaction at a given temperature.
(products) 2NO + Br2 Ea Potential energy ON Br ON Br .
.
2BrNO (reactant) (transition state) Reaction progress (a) (b) ∆E ....
....
....
....
....
FIGURE 12.11 (a) The change in potential energy as a function of reaction progress for the reaction 2BrNO n 2NO Br2. The activation energy Ea represents the energy needed to disrupt the BrNO molecules so that they can form products. The quantity E represents the net change in energy in going from reactant to products. (b) A molecular representation of the reaction.
Visualization: Transition States and Activation Energy 554 Chapter Twelve Chemical Kinetics the activation energy are able to react (get over the barrier). At the lower temperature, T1, the fraction of effective collisions is quite small. However, as the temperature is increased to T2, the fraction of collisions with the required activation energy increases dramatically.
When the temperature is doubled, the fraction of effective collisions much more than dou-bles. In fact, the fraction of effective collisions increases exponentially with temperature.
This is encouraging for our theory; remember that rates of reactions are observed to in-crease exponentially with temperature. Arrhenius postulated that the number of collisions having an energy greater than or equal to the activation energy is given by the expression: where Ea is the activation energy, R is the universal gas constant, and T is the Kelvin tem-perature. The factor represents the fraction of collisions with energy Ea or greater at temperature T.
We have seen that not all molecular collisions are effective in producing chemical re-actions because a minimum energy is required for the reaction to occur. There is, how-ever, another complication. Experiments show that the observed reaction rate is consid-erably smaller than the rate of collisions with enough energy to surmount the barrier. This means that many collisions, even though they have the required energy, still do not pro-duce a reaction. Why not?
The answer lies in the molecular orientations during collisions. We can illustrate this using the reaction between two BrNO molecules, as shown in Fig. 12.13. Some col-lision orientations can lead to reaction, and others cannot. Therefore, we must include a correction factor to allow for collisions with nonproductive molecular orientations.
To summarize, two requirements must be satisfied for reactants to collide success-fully (to rearrange to form products): 1. The collision must involve enough energy to produce the reaction; that is, the colli-sion energy must equal or exceed the activation energy.
2. The relative orientation of the reactants must allow formation of any new bonds nec-essary to produce products.
Taking these factors into account, we can represent the rate constant as k zpeEa RT eEa RT 1total number of collisions2eEa RT Number of collisions with the activation energy O N Br Br O N Br Br O N O N O N Br Br O N Br Br N O O N O N Br Br O N O N Br Br O N No reaction (a) (b) (c) FIGURE 12.13 Several possible orientations for a collision between two BrNO molecules. Orientations (a) and (b) can lead to a reaction, but ori-entation (c) cannot.
FIGURE 12.12 Plot showing the number of collisions with a particular energy at T1 and T2, where T2 T1.
T1 T2 0 0 Ea Energy Number of collisions T2 > T1 Visualization: The Gas Phase Reaction of NO and Cl2 12.7 A Model for Chemical Kinetics 555 where z is the collision frequency, p is called the steric factor (always less than 1) and reflects the fraction of collisions with effective orientations, and represents the frac-tion of collisions with sufficient energy to produce a reaction. This expression is most of-ten written in form (12.9) which is called the Arrhenius equation. In this equation, A replaces zp and is called the frequency factor for the reaction.
Taking the natural logarithm of each side of the Arrhenius equation gives (12.10) Equation (12.10) is a linear equation of the type where Thus, for a reaction where the rate constant obeys the Arrhenius equation, a plot of ln(k) versus 1T gives a straight line.
The slope and intercept can be used to determine, respectively, the values of Ea and A characteristic of that reaction. The fact that most rate constants obey the Arrhenius equa-tion to a good approximation indicates that the collision model for chemical reactions is physically reasonable.
Determining Activation Energy I The reaction was studied at several temperatures, and the following values of k were obtained: 2N2O51g2 ¡ 4NO21g2 O21g2 m Ea R slope, x 1 T, and b ln(A) intercept.
y ln(k), y mx b, ln1k2 Ea R a1 Tb ln1A2 k AeEa RT eEa RT A snowy tree cricket. The frequency of a cricket’s chirps depends on the temperature of the cricket.
Sample Exercise 12.7 k (s1) T (C) 20 30 40 50 60 2.9 103 9.1 104 2.7 104 7.3 105 2.0 105 Calculate the value of Ea for this reaction.
Solution To obtain the value of Ea, we need to construct a plot of ln(k) versus 1T. First, we must calculate values of ln(k) and 1T, as shown below: T (C) T (K) 1/T (K) k (s1) ln(k) 20 293 30 303 40 313 50 323 60 333 5.84 2.9 103 3.00 103 7.00 9.1 104 3.10 103 8.22 2.7 104 3.19 103 9.53 7.3 105 3.30 103 10.82 2.0 105 3.41 103 556 Chapter Twelve Chemical Kinetics The plot of ln(k) versus 1T is shown in Fig. 12.14, where the slope is found to be The value of Ea can be determined by solving the follow-ing equation: Thus the value of the activation energy for this reaction is See Exercises 12.57 and 12.58.
The most common procedure for finding Ea for a reaction involves measuring the rate constant k at several temperatures and then plotting ln(k) versus 1T, as shown in Sam-ple Exercise 12.7. However, Ea also can be calculated from the values of k at only two temperatures by using a formula that can be derived as follows from Equation (12.10).
At temperature T1, where the rate constant is k1, At temperature T2, where the rate constant is k2, ln1k22 Ea RT2 ln1A2 ln1k12 Ea RT1 ln1A2 1.0 105J/mol.
1.0 10 5 J/mol Ea R1slope2 18.3145 J/K mol211.2 104 K2 Slope Ea R 1.2 104 K.
¢ln1k2 ¢a1 Tb ln(k) 3.00 3.25 3.50 –11.00 –10.00 –9.00 –8.00 –7.00 –6.00 Slope = ∆ ln(k) ∆(1/T) = –1.2 104 K 10–3 1/T (K) 10–3 10–3 ∆ ln(k) ∆(1/T) FIGURE 12.14 Plot of ln(k) versus for the reaction 2N2O5(g) 4NO2(g) O2(g). The value of the activation energy for this reaction can be obtained from the slope of the line, which equals EaR.
S 1 T 12.8 Catalysis 557 Subtracting the first equation from the second gives And (12.11) Therefore, the values of k1 and k2 measured at temperatures T1 and T2 can be used to calculate Ea, as shown in Sample Exercise 12.8.
Determining Activation Energy II The gas-phase reaction between methane and diatomic sulfur is given by the equation At the rate constant for this reaction is 1.1 L/mol s, and at the rate con-stant is 6.4 L/mol s. Using these values, calculate Ea for this reaction.
Solution The relevant data are shown in the following table: 625°C 550°C CH41g2 2S21g2 ¡ CS21g2 2H2S1g2 lnak2 k1 b Ea R a 1 T1 1 T2 b Ea RT2 Ea RT1 ln1k22 ln1k12 c Ea RT2 ln1A2 d c Ea RT1 ln1A2 d Sample Exercise 12.8 k (Lmol s) T (C) T (K) 550 625 898 T2 6.4 k2 823 T1 1.1 k1 Substituting these values into Equation (12.11) gives Solving for Ea gives See Exercises 12.59 through 12.62.
12.8 Catalysis We have seen that the rate of a reaction increases dramatically with temperature. If a par-ticular reaction does not occur fast enough at normal temperatures, we can speed it up by raising the temperature. However, sometimes this is not feasible. For example, living cells can survive only in a rather narrow temperature range, and the human body is designed to operate at an almost constant temperature of . But many of the complicated bio-chemical reactions keeping us alive would be much too slow at this temperature without intervention. We exist only because the body contains many substances called enzymes, which increase the rates of these reactions. In fact, almost every biologically important reaction is assisted by a specific enzyme.
98.6°F 1.4 105 J/mol Ea 18.3145 J/K mol2lna6.4 1.1b a 1 823 K 1 898 Kb lna6.4 1.1b Ea 8.3145 J/K mol a 1 823 K 1 898 Kb 558 Chapter Twelve Chemical Kinetics Although it is possible to use higher temperatures to speed up commercially impor-tant reactions, such as the Haber process for synthesizing ammonia, this is very expen-sive. In a chemical plant an increase in temperature means significantly increased costs for energy. The use of an appropriate catalyst allows a reaction to proceed rapidly at a relatively low temperature and can therefore hold down production costs.
A catalyst is a substance that speeds up a reaction without being consumed itself. Just as virtually all vital biologic reactions are assisted by enzymes (biologic catalysts), almost all industrial processes also involve the use of catalysts. For example, the production of sulfuric acid uses vanadium(V) oxide, and the Haber process uses a mixture of iron and iron oxide.
How does a catalyst work? Remember that for each reaction a certain energy barrier must be surmounted. How can we make a reaction occur faster without raising the tem-perature to increase the molecular energies? The solution is to provide a new pathway for the reaction, one with a lower activation energy. This is what a catalyst does, as is shown in Fig. 12.15. Because the catalyst allows the reaction to occur with a lower activation energy, a much larger fraction of collisions is effective at a given temperature, and the re-action rate is increased. This effect is illustrated in Fig. 12.16. Note from this diagram that although a catalyst lowers the activation energy Ea for a reaction, it does not affect the energy difference between products and reactants.
Catalysts are classified as homogeneous or heterogeneous. A homogeneous catalyst is one that is present in the same phase as the reacting molecules. A heterogeneous cat-alyst exists in a different phase, usually as a solid.
Heterogeneous Catalysis Heterogeneous catalysis most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption refers to the collection of one substance on the surface of another substance; absorption refers to the penetration of one substance into another. Water is absorbed by a sponge.
An important example of heterogeneous catalysis occurs in the hydrogenation of un-saturated hydrocarbons, compounds composed mainly of carbon and hydrogen with some carbon–carbon double bonds. Hydrogenation is an important industrial process used to change unsaturated fats, occurring as oils, to saturated fats (solid shortenings such as Crisco) in which the bonds have been converted to bonds through addition of hydrogen.
A simple example of hydrogenation involves ethylene: This reaction is quite slow at normal temperatures, mainly because the strong bond in the hydrogen molecule results in a large activation energy for the reaction. However, the C¬C C“C ¢E FIGURE 12.15 Energy plots for a catalyzed and an uncat-alyzed pathway for a given reaction.
∆E Reactants Products Catalyzed pathway Uncatalyzed pathway Reaction progress Energy Ea (uncatalyzed) Energy Effective collisions (uncatalyzed) Effective collisions (catalyzed) Ea (catalyzed) Energy (a) (b) Number of collisions with a given energy Number of collisions with a given energy FIGURE 12.16 Effect of a catalyst on the number of reac-tion-producing collisions. Because a catalyst provides a reaction pathway with a lower activation energy, a much greater fraction of the collisions is effective for the cat-alyzed pathway (b) than for the uncatalyzed pathway (a) (at a given temperature). This allows reactants to become products at a much higher rate, even though there is no temperature increase.
These cookies contain partially hydrogenated vegetable oil.
Visualization: Heterogeneous Catalysis 12.8 Catalysis 559 reaction rate can be greatly increased by using a solid catalyst of platinum, palladium, or nickel. The hydrogen and ethylene adsorb on the catalyst surface, where the reaction occurs. The main function of the catalyst apparently is to allow formation of metal– hydrogen interactions that weaken the HOH bonds and facilitate the reaction. The mechanism is illustrated in Fig. 12.17.
Typically, heterogeneous catalysis involves four steps: 1. Adsorption and activation of the reactants 2. Migration of the adsorbed reactants on the surface 3. Reaction of the adsorbed substances 4. Escape, or desorption, of the products Heterogeneous catalysis also occurs in the oxidation of gaseous sulfur dioxide to gaseous sulfur trioxide. This process is especially interesting because it illustrates both positive and negative consequences of chemical catalysis.
The negative side is the formation of damaging air pollutants. Recall that sulfur diox-ide, a toxic gas with a choking odor, is formed whenever sulfur-containing fuels are burned.
However, it is sulfur trioxide that causes most of the environmental damage, mainly through the production of acid rain. When sulfur trioxide combines with a droplet of water, sulfuric acid is formed: This sulfuric acid can cause considerable damage to vegetation, buildings and statues, and fish populations.
Sulfur dioxide is not rapidly oxidized to sulfur trioxide in clean, dry air. Why, then, is there a problem? The answer is catalysis. Dust particles and water droplets catalyze the reaction between SO2 and O2 in the air.
On the positive side, the heterogeneous catalysis of the oxidation of SO2 is used to advantage in the manufacture of sulfuric acid, where the reaction of O2 and SO2 to form SO3 is catalyzed by a solid mixture of platinum and vanadium(V) oxide.
Heterogeneous catalysis is also utilized in the catalytic converters in automobile ex-haust systems. The exhaust gases, containing compounds such as nitric oxide, carbon monoxide, and unburned hydrocarbons, are passed through a converter containing beads of solid catalyst (see Fig. 12.18). The catalyst promotes the conversion of carbon monoxide to carbon dioxide, hydrocarbons to carbon dioxide and water, and nitric oxide to nitrogen gas to lessen the environmental impact of the exhaust gases. However, this beneficial catal-ysis can, unfortunately, be accompanied by the unwanted catalysis of the oxidation of SO2 to SO3, which reacts with the moisture present to form sulfuric acid.
Because of the complex nature of the reactions that take place in the converter, a mix-ture of catalysts is used. The most effective catalytic materials are transition metal oxides and noble metals such as palladium and platinum.
Homogeneous Catalysis A homogeneous catalyst exists in the same phase as the reacting molecules. There are many examples in both the gas and liquid phases. One such example is the unusual catalytic behavior of nitric oxide toward ozone. In the troposphere, that part of the atmosphere clos-est to earth, nitric oxide catalyzes ozone production. However, in the upper atmosphere it catalyzes the decomposition of ozone. Both these effects are unfortunate environmentally.
In the lower atmosphere, NO is produced in any high-temperature combustion process where N2 is present. The reaction N21g2 O21g2 ¡ 2NO1g2 H2O1l2 SO31g2 ¡ H2SO41aq2 Metal surface Carbon Hydrogen (a) (b) (c) (d) FIGURE 12.17 Heterogeneous catalysis of the hydrogena-tion of ethylene. (a) The reactants above the metal surface. (b) Hydrogen is adsorbed onto the metal surface, forming metal– hydrogen bonds and breaking the HOH bonds. The bond in ethylene is broken and metal–carbon bonds are formed during adsorption. (c) The adsorbed molecules and atoms migrate toward each other on the metal surface, forming new COH bonds.
(d) The C atoms in ethane (C2H6) have com-pletely saturated bonding capacities and so cannot bind strongly to the metal surfaces.
The C2H6 molecule thus escapes.
p 560 Chapter Twelve Chemical Kinetics is very slow at normal temperatures because of the very strong and bonds.
However, at elevated temperatures, such as those found in the internal combustion engines of automobiles, significant quantities of NO form. Some of this NO is converted back to N2 in the catalytic converter, but significant amounts escape into the atmosphere to react with oxygen: In the atmosphere, NO2 can absorb light and decompose as follows: The oxygen atom is very reactive and can combine with oxygen molecules to form ozone: Ozone is a powerful oxidizing agent that can react with other air pollutants to form sub-stances irritating to the eyes and lungs, and is itself very toxic.
In this series of reactions, nitric oxide is acting as a true catalyst because it assists the production of ozone without being consumed itself. This can be seen by summing the reactions: 3 2O21g2 ¡ O31g2 O21g2 O1g2 ¡ O31g2 NO21g2 ¬ ¡ Light NO1g2 O1g2 NO1g2 1 2O21g2 ¡ NO21g2 O21g2 O1g2 ¡ O31g2 NO21g2 ¬ ¡ Light NO1g2 O1g2 2NO1g2 O21g2 ¡ 2NO21g2 O“O N‚N CHEMICAL IMPACT Automobiles: Air Purifiers?
O utlandish as it may seem, a new scheme has been pro-posed to turn automobiles into air purifiers, devouring the pollutants ozone and carbon monoxide. Engelhard Cor-poration, an Iselin, New Jersey, company that specializes in the manufacture of catalytic converters for automotive ex-haust systems, has developed a catalyst that decomposes ozone to oxygen and converts carbon monoxide to carbon dioxide. Engelhard proposes to paint the catalyst on auto-mobile radiators and air-conditioner compressors where fans draw large volumes of air for cooling purposes. The cata-lyst works well at the warm temperatures present on the surfaces of these devices. The idea is to let cars destroy pol-lutants using nothing but the catalyst and waste radiator heat.
It’s an intriguing idea. The residents of Los Angeles drive nearly 300 million miles every day. At that rate, they could process a lot of air.
Catalytic converter CO2 N2 CO NO Engine Exhaust gases Exhaust gases FIGURE 12.18 The exhaust gases from an automobile engine are passed through a catalytic con-verter to minimize environmental damage.
Although O2 is represented here as the oxidizing agent for NO, the actual oxidiz-ing agent is probably some type of per-oxide compound produced by reaction of oxygen with pollutants. The direct reac-tion of NO and O2 is very slow.
Visualization: Homogeneous Catalysis In the upper atmosphere, the presence of nitric oxide has the opposite effect—the de-pletion of ozone. The series of reactions involved is Nitric oxide is again catalytic, but here its effect is to change O3 to O2. This is a poten-tial problem because O3, which absorbs ultraviolet light, is necessary to protect us from the harmful effects of this high-energy radiation. That is, we want O3 in the upper at-mosphere to block ultraviolet radiation from the sun but not in the lower atmosphere, where we would have to breathe it and its oxidation products.
The ozone layer is also threatened by Freons, a group of stable, noncorrosive com-pounds, until recently, used as refrigerants and as propellants in aerosol cans. The most commonly used substance of this type was Freon-12 (CCl2F2). The chemical inertness of Freons makes them valuable but also creates a problem, since they remain in the envi-ronment a long time. Eventually, they migrate into the upper atmosphere to be decom-posed by high-energy light. Among the decomposition products are chlorine atoms: These chlorine atoms can catalyze the decomposition of ozone: The problem of Freons has been brought into strong focus by the discovery of a mys-terious “hole” in the ozone layer in the stratosphere over Antarctica. Studies performed there to find the reason for the hole have found unusually high levels of chlorine monox-ide (ClO). This strongly implicates the Freons in the atmosphere as being responsible for the ozone destruction.
Because they pose environmental problems, Freons have been banned by international agreement. Substitute compounds are now being used. O1g2 O31g2 ¡ 2O21g2 O1g2 ClO1g2 ¡ Cl1g2 O21g2 Cl1g2 O31g2 ¡ ClO1g2 O21g2 CCl2F21g2 ¬ ¡ Light CClF21g2 Cl1g2 O1g2 O31g2 ¡ 2O21g2 O1g2 NO21g2 ¡ NO1g2 O21g2 NO1g2 O31g2 ¡ NO21g2 O21g2 This graphic shows data from the Total Ozone Mapping Spectrometer (TOMS) Earth Probe.
Freon-12 Ozone 12.8 Catalysis 561 562 Chapter Twelve Chemical Kinetics CHEMICAL IMPACT Enzymes: Nature’s Catalysts T he most impressive examples of homogeneous catalysis occur in nature, where the complex reactions necessary for plant and animal life are made possible by enzymes. En-zymes are large molecules specifically tailored to facilitate a given type of reaction. Usually enzymes are proteins, an important class of biomolecules constructed from -amino acids that have the general structure where R represents any one of 20 different substituents.
These amino acid molecules can be “hooked together” to form a polymer (a word meaning “many parts”) called a pro-tein. The general structure of a protein can be represented as follows: N H H R H OH O C C boring CON bond to be broken much more easily. When the reaction is completed, the remaining portion of the sub-strate protein and the newly formed amino acid are released by the enzyme.
The process just described for carboxypeptidase-A is characteristic of the behavior of other enzymes. Enzyme catalysis can be represented by the series of reactions shown below: where E represents the enzyme, S represents the substrate, represents the enzyme–substrate complex, and P rep-resents the products. The enzyme and substrate form a com-plex, where the reaction occurs. The enzyme then releases the product and is ready to repeat the process. The most amazing thing about enzymes is their efficiency. Because an enzyme plays its catalytic role over and over and very rapidly, only a tiny amount of enzyme is required. This makes the isolation of enzymes for study quite difficult.
E S E S ¡ E P E S ¡ E S H N H C H C O N H C H C O N H C H C O O H Many amino acid fragments Fragment from an amino acid with sub-stituent R Fragment from an amino acid with sub-stituent R' Fragment from an amino acid with sub-stituent R'' N H R R' R'' Since specific proteins are needed by the human body, the proteins in food must be broken into their constituent amino acids, which are then used to construct new proteins in the body’s cells. The reaction in which a protein is bro-ken down one amino acid at a time is shown in Fig. 12.19.
Note that in this reaction a water molecule reacts with a pro-tein molecule to produce an amino acid and a new protein containing one less amino acid. Without the enzymes found in human cells, this reaction would be much too slow to be useful. One of these enzymes is carboxypeptidase-A, a zinc-containing protein (Fig. 12.20).
Carboxypeptidase-A captures the protein to be acted on (called the substrate) in a special groove and positions the substrate so that the end is in the active site, where the catal-ysis occurs (Fig. 12.21). Note that the ion bonds to the oxygen of the (carbonyl) group. This polarizes the electron density in the carbonyl group, allowing the neigh-C“O Zn2 R'' R' R'' C O O H C H N H C O O H H C H N H N H C O O H C H N C C H N H N H H H H O O H R' Water molecule Protein New protein Amino acid H FIGURE 12.19 The removal of the end amino acid from a protein by reaction with a molecule of water. The products are an amino acid and a new, smaller protein.
12.8 Catalysis 563 (a) (b) FIGURE 12.20 (a) The structure of the enzyme carboxypeptidase-A, which contains 307 amino acids. The zinc ion is shown above as a black sphere in the center. (b) Carboxypeptidase-A with a substrate (pink) in place.
FIGURE 12.21 Protein–substrate interaction. The substrate is shown in black and red, with the red representing the terminal amino acid. Blue indicates side chains from the enzyme that help bind the substrate.
OH CH2 CH CO– 2 +NH2 C NH2 HN HO C O Zn2+ O H H –O C O CHR NH C O 564 Chapter Twelve Chemical Kinetics For Review Chemical kinetics The study of the factors that control the rate (speed) of a chemical reaction • Rate is defined in terms of the change in concentration of a given reaction component per unit time • Kinetic measurements are often made under conditions where the reverse reaction is insignificant The kinetic and thermodynamic properties of a reaction are not fundamentally related Rate laws Differential rate law: describes the rate as a function of concentration • k is the rate constant • n is the order; not related to the coefficients in the balanced equation Integrated rate law: describes the concentration as a function of time • For a reaction of the type for which n 0: n 1: n 2: • The value of k can be determined from the plot of the appropriate function of [A] versus t Reaction mechanism Series of elementary steps by which an overall reaction occurs • Elementary step: rate law for the step can be written from the molecularity of the reaction Two requirements for an acceptable mechanism: • The elementary steps sum to give the correct overall balanced equation • The mechanism agrees with the experimentally determined rate law Simple reactions can have an elementary step that is slower than all of the other steps; which is called the rate-determining step.
Kinetic models The simplest model to account for reaction kinetics is the collision model • Molecules must collide to react • The collision kinetic energy furnishes the potential energy needed to enable the reactants to rearrange to form products t1 2 1 k3A4 0 n 2: 1 3A4 kt 1 3A4 0 t1 2 0.693 k n 1: ln3A4 kt ln3A4 0 t1 2 3A4 0 2k n 0: 3A4 kt 3A40 Rate k3A4 n aA ¡ products Rate ¢ 3A4 ¢t k3A4 n Key Terms chemical kinetics Section 12.1 reaction rate instantaneous rate Section 12.2 rate law rate constant order (differential) rate law integrated rate law Section 12.3 method of initial rates initial rate overall reaction order Section 12.4 first-order reaction integrated first-order rate law half-life of a reactant integrated second-order rate law zero-order reaction integrated zero-order rate law pseudo-first-order rate law Section 12.6 reaction mechanism intermediate elementary step molecularity unimolecular step bimolecular step termolecular step rate-determining step Section 12.7 collision model activation energy activated complex (transition state) molecular orientations steric factor Arrhenius equation frequency factor Section 12.8 enzyme catalyst homogeneous catalyst heterogeneous catalyst adsorption For Review 565 • A certain threshold energy called the activation energy (Ea) is necessary for a re-action to occur • The relative orientations of the colliding reactants are also a determining factor in the reaction rate • This model leads to the Arrhenius equation: • A depends on the collision frequency and relative orientation of the molecules • The value of Ea can be found by obtaining the values of k at several temperatures Catalyst Speeds up a reaction without being consumed Works by providing a lower-energy pathway for the reaction Enzymes are biological catalysts Catalysts can be classified as homogeneous or heterogeneous • Homogeneous: exist in the same phase as the reactants • Heterogeneous: exist in a different phase than the reactants REVIEW QUESTIONS 1. Define reaction rate. Distinguish between the initial rate, average rate, and in-stantaneous rate of a chemical reaction. Which of these rates is usually fastest?
The initial rate is the rate used by convention. Give a possible explanation as to why.
2. Distinguish between the differential rate law and the integrated rate law. Which of these is often called just the “rate law”? What is k in a rate law, and what are orders in a rate law? Explain.
3. One experimental procedure that can be used to determine the rate law of a reaction is the method of initial rates. What data are gathered in the method of initial rates, and how are these data manipulated to determine k and the orders of the species in the rate law? Are the units for k, the rate constant, the same for all rate laws? Explain. If a reaction is first order in A, what happens to the rate if [A] is tripled? If the initial rate for a reaction increases by a factor of 16 when [A] is quadrupled, what is the order of n? If a reaction is third order in A and [A] is doubled, what happens to the initial rate? If a reaction is zero order, what effect does [A] have on the initial rate of a reaction?
4. The initial rate for a reaction is equal to the slope of the tangent line at in a plot of [A] versus time. From calculus, Therefore, the differential rate law for a reaction is Assuming you have some calculus in your background, derive the zero-, first-, and second-order integrated rate laws using the differential rate law.
5. Consider the zero-, first-, and second-order integrated rate laws. If you have con-centration versus time data for some species in a reaction, what plots would you make to “prove” a reaction is either zero, first, or second order? How would the rate constant, k, be determined from such a plot? What does the y-intercept equal in each plot? When a rate law contains the concentration of two or more species, how can plots be used to determine k and the orders of the species in the rate law?
6. Derive expressions for the half-life of zero-, first-, and second-order reactions using the integrated rate law for each order. How does each half-life depend on Rate d3A4 dt k3A4 n.
initial rate d3A4 dt .
t 0 k AeEa RT 566 Chapter Twelve Chemical Kinetics Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Define stability from both a kinetic and thermodynamic perspec-tive. Give examples to show the differences in these concepts.
2. Describe at least two experiments you could perform to determine a rate law.
3. Make a graph of [A] versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.
4. How does temperature affect k, the rate constant? Explain.
5. Consider the following statements: “In general, the rate of a chemical reaction increases a bit at first because it takes a while for the reaction to get ‘warmed up.’ After that, however, the rate of the reaction decreases because its rate is dependent on the concentrations of the reactants, and these are decreasing.” Indi-cate everything that is correct in these statements, and indicate everything that is incorrect. Correct the incorrect statements and explain.
6. For the reaction explain at least two ways in which the rate law could be zero order in chemical A.
7. A friend of yours states, “A balanced equation tells us how chem-icals interact. Therefore, we can determine the rate law directly from the balanced equation.” What do you tell your friend?
8. Provide a conceptual rationale for the differences in the half-lives of zero-, first-, and second-order reactions.
A B S C, concentration? If the half-life for a reaction is 20. seconds, what would be the second half-life assuming the reaction is either zero, first, or second order?
7. Define each of the following.
a. elementary step b. molecularity c. reaction mechanism d. intermediate e. rate-determining step What two requirements must be met to call a mechanism plausible? Why say a “plausible” mechanism instead of the “correct” mechanism? Is it true that most reactions occur by a one-step mechanism? Explain.
8. What is the premise underlying the collision model? How is the rate affected by each of the following?
a. activation energy b. temperature c. frequency of collisions d. orientation of collisions Sketch a potential energy versus reaction progress plot for an endothermic reac-tion and for an exothermic reaction. Show and in both plots. When con-centrations and temperatures are equal, would you expect the rate of the forward reaction to be equal to, greater than, or less than the rate of the reverse reaction if the reaction is exothermic? Endothermic?
9. Give the Arrhenius equation. Take the natural log of both sides and place this equation in the form of a straight-line equation . What data would you need and how would you graph those data to get a linear relationship using the Arrhenius equation? What does the slope of the straight line equal? What does the y-intercept equal? What are the units of R in the Arrhenius equation?
Explain how if you know the rate constant value at two different temperatures, you can determine the activation energy for the reaction.
10. Why does a catalyst increase the rate of a reaction? What is the difference between a homogeneous catalyst and a heterogeneous catalyst? Would a given reaction necessarily have the same rate law for both a catalyzed and an uncatalyzed pathway? Explain.
(y mx b) Ea ¢E In the Questions and the Exercises, the term rate law always refers to the differential rate law.
Exercises 567 A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 9. Define what is meant by unimolecular and bimolecular steps. Why are termolecular steps infrequently seen in chemical reactions?
10. Hydrogen reacts explosively with oxygen. However, a mixture of H2 and O2 can exist indefinitely at room temperature. Explain why H2 and O2 do not react under these conditions.
11. For the reaction the observed rate law is Which of the changes listed below would affect the value of the rate constant k?
a. increasing the partial pressure of hydrogen gas b. changing the temperature c. using an appropriate catalyst 12. The rate law for a reaction can be determined only from experiment and not from the balanced equation. Two experimental procedures were outlined in Chapter 12. What are these two procedures? Ex-plain how each method is used to determine rate laws.
13. Table 12.2 illustrates how the average rate of a reaction decreases with time. Why does the average rate decrease with time? How does the instantaneous rate of a reaction depend on time? Why are initial rates used by convention?
14. The type of rate law for a reaction, either the differential rate law or the integrated rate, is usually determined by which data is easiest to collect. Explain.
15. The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant?
If a reactant has a 1 order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?
16. Reactions that require a metal catalyst are often zero order after a certain amount of reactant(s) are present. Explain.
17. The central idea of the collision model is that molecules must col-lide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.
18. Would the slope of a ln k versus 1/T (K) plot for a catalyzed re-action be more of less negative than the slope of the ln k versus 1/T (K) plot for the uncatalyzed reaction? Explain.
Exercises In this section similar exercises are paired.
Reaction Rates 19. Consider the reaction If, in a certain experiment, over a specific time period, 0.0048 mol PH3 is consumed in a 2.0-L container each second of reaction, what are the rates of production of P4 and H2 in this experiment?
4PH31g2 ¡ P41g2 6H21g2 Rate k3NO4 23H24 2H21g2 2NO1g2 ¡ N21g2 2H2O1g2 20. In the Haber process for the production of ammonia, what is the relationship between the rate of production of ammo-nia and the rate of consumption of hydrogen?
21. At will decompose according to the following reaction: The following data were collected for the concentration of at various times.
a. Calculate the average rate of decomposition of between 0 and s. Use this rate to calculate the average rate of production of over the same time period.
b. What are these rates for the time period s to s?
22. Consider the general reaction and the following average rate data over some time period : Determine a set of possible coefficients to balance this general reaction.
23. What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds?
a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law 24. The rate law for the reaction is What are the units for k, assuming time in seconds and concen-tration in mol/L?
Rate Laws from Experimental Data: Initial Rates Method 25. The reaction 2NO1g2 Cl21g2 ¡ 2NOCl1g2 Rate k3Cl24 1 23CHCl34 Cl21g2 CHCl31g2 ¡ HCl1g2 CCl41g2 ¢C ¢t 0.0160 mol/L s ¢B ¢t 0.0120 mol/L s ¢A ¢t 0.0080 mol/L s ¢t aA bB ¡ cC 4.32 104 2.16 104 O2(g) 2.16 104 H2O2 Time (s) [H2O2] (mol/L) 0 1.000 2.16 104 0.500 4.32 104 0.250 H2O2 2H2O21aq2 ¡ 2H2O1l2 O21g2 H2O2(aq) 40°C, N21g2 3H21g2 ¡ 2NH31g2 568 Chapter Twelve Chemical Kinetics was studied at . The following results were obtained where Rate ¢3Cl24 ¢t 10°C c. Calculate the rate constant when concentrations are given in moles per liter.
28. The following data were obtained for the gas-phase decomposi-tion of dinitrogen pentoxide, 2N2O51g2 ¡ 4NO21g2 O21g2 [I]o [S2O8 2]o Initial Rate (mol/L) (mol/L) (mol/L s) 0.080 0.040 12.5 106 0.040 0.040 6.25 106 0.080 0.020 6.25 106 0.032 0.040 5.00 106 0.060 0.030 7.00 106 [Hb]0 [CO]0 Initial Rate (mol/L) (mol/L) (mol/L s) 2.21 1.00 0.619 4.42 1.00 1.24 4.42 3.00 3.71 [ClO2]0 [OH]0 Initial Rate (mol/L) (mol/L) (mol/L s) 0.0500 0.100 5.75 102 0.100 0.100 2.30 101 0.100 0.0500 1.15 101 [NOCl]0 Initial Rate (molecules/cm3) (molecules/cm3 s) 3.0 1016 5.98 104 2.0 1016 2.66 104 1.0 1016 6.64 103 4.0 1016 1.06 105 [N2O5]0 Initial Rate (mol/L) (mol/L s) 0.0750 8.90 104 0.190 2.26 103 0.275 3.26 103 0.410 4.85 103 a. What is the rate law?
b. What is the value of the rate constant?
26. The reaction was studied at . The following results were obtained where Rate ¢3S2O8 24 ¢t 25°C 2I1aq2 S2O8 21aq2 ¡ I21aq2 2SO4 21aq2 a. Determine the rate law.
b. Calculate a value for the rate constant for each experiment and an average value for the rate constant.
27. The decomposition of nitrosyl chloride was studied: The following data were obtained where Rate ¢3NOCl4 ¢t 2NOCl1g2 ∆2NO1g2 Cl21g2 Defining the rate as write the rate law and calculate the value of the rate constant.
29. The rate of the reaction between hemoglobin (Hb) and carbon monoxide (CO) was studied at . The following data were collected with all concentration units in mol/L. (A hemoglobin concentration of 2.21 mol/L is equal to ) 2.21 106 mol/L.
m 20°C ¢ [N2O5] ¢t, a. Determine the orders of this reaction with respect to Hb and CO.
b. Determine the rate law.
c. Calculate the value of the rate constant.
d. What would be the initial rate for an experiment with and ?
30. The following data were obtained for the reaction where Rate ¢ 3ClO24 ¢t 2ClO21aq2 2OH1aq2 ¡ ClO3 1aq2 ClO2 1aq2 H2O1l2 [CO] 0 2.40 mmol/L [Hb] 0 3.36 mmol/L Initial Rate [NO]0 (mol/L) [Cl2]0 (mol/L) (mol/L min) 0.10 0.10 0.18 0.10 0.20 0.36 0.20 0.20 1.45 a. What is the rate law?
b. Calculate the rate constant.
a.
Determine the rate law and the value of the rate constant.
b.
What would be the initial rate for an experiment with ?
0.175 mol/L and [OH]0 0.0844 mol/L [ClO2]0 Exercises 569 Integrated Rate Laws 31. The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Since the pressure of a gas is directly proportional to the con-centration of gas, we can express the rate law for a gaseous reaction in terms of partial pressures. Using the above data, de-duce the rate law, the integrated rate law, and the value of the rate constant, all in terms of pressure units in atm and time in seconds.
Predict the pressure of C2H5OH after 900. s from the start of the reaction. (Hint: To determine the order of the reaction with re-spect to C2H5OH, compare how the pressure of C2H5OH decreases with each time listing.) Determine the rate law, the integrated law, and the value of the rate constant. Calculate after the start of the reaction.
34. A certain reaction has the following general form: At a particular temperature and concen-tration versus time data were collected for this reaction, and a plot of 1[A] versus time resulted in a straight line with a slope value of a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.
b. Calculate the half-life for this reaction.
c. How much time is required for the concentration of A to de-crease to 35. The decomposition of ethanol (C2H5OH) on an alumina (Al2O3) surface was studied at 600 K. Concentration versus time data were col-lected for this reaction, and a plot of [A] versus time resulted in a straight line with a slope of a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.
b. If the initial concentration of C2H5OH was calculate the half-life for this reaction.
c. How much time is required for all the C2H5OH to decompose?
36. At 500 K in the presence of a copper surface, ethanol decomposes according to the equation The pressure of C2H5OH was measured as a function of time and the following data were obtained: C2H5OH1g2 ¡ CH3CHO1g2 H21g2 1.25 102 M 1.25 102 M, 4.00 105 mol/L s.
C2H5OH1g2 ¡ C2H41g2 H2O1g2 7.00 104 M?
3.60 102 L/mol s.
[A]0 2.80 103 M, aA ¡ bB [NO2] at 2.70 104 s Time (s) [NO2] (mol/L) 0 0.500 1.20 103 0.444 3.00 103 0.381 4.50 103 0.340 9.00 103 0.250 1.80 104 0.174 Time (s) (torr) 0 250.
100.
237 200.
224 300.
211 400.
198 500.
185 PC2H5OH Time (s) [H2O2] (mol/L) 0 1.00 120 1 0.91 300 1 0.78 600 1 0.59 1200 1 0.37 1800 1 0.22 2400 1 0.13 3000 1 0.082 3600 1 0.050 Assuming that determine the rate law, the integrated rate law, and the value of the rate constant. Calculate [H2O2] at 4000. s after the start of the reaction.
32. A certain reaction has the following general form: At a particular temperature and concen-tration versus time data were collected for this reaction, and a plot of ln[A] versus time resulted in a straight line with a slope value of a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.
b. Calculate the half-life for this reaction.
c. How much time is required for the concentration of A to de-crease to ?
33. The rate of the reaction depends only on the concentration of nitrogen dioxide below . At a temperature below the following data were collected: 225°C, 225°C NO21g2 CO1g2 ¡ NO1g2 CO21g2 2.50 103 M 2.97 102 min1.
[A]0 2.00 102 M, aA ¡ bB Rate ¢3H2O24 ¢t 570 Chapter Twelve Chemical Kinetics 37. The dimerization of butadiene was studied at 500. K, and the following data were obtained: 2C4H61g2 ¡ C8H121g2 What is the order of the reaction with respect to A and what is the initial concentration of A?
40. Consider the data plotted in Exercise 39 when answering the fol-lowing questions.
a. What is the concentration of A after 9 s?
b. What are the first three half-lives for this experiment?
41. The reaction is known to be zero order in A and to have a rate constant of at An experiment was run at where a. Write the integrated rate law for this reaction.
b. Calculate the half-life for the reaction.
c. Calculate the concentration of B after s has elapsed.
42. The radioactive isotope 32P decays by first-order kinetics and has a half-life of 14.3 days. How long does it take for 95.0% of a sample of 32P to decay?
43. A first-order reaction is 75.0% complete in 320. s.
a. What are the first and second half-lives for this reaction?
b. How long does it take for 90.0% completion?
44. The rate law for the decomposition of phosphine (PH3) is It takes 120. s for 1.00 M PH3 to decrease to 0.250 M. How much time is required for 2.00 M PH3 to decrease to a concentration of 0.350 M?
45. Consider the following initial rate data for the decomposition of compound AB to give A and B: Rate ¢ 3PH34 ¢t k3PH34 5.0 103 [A]0 1.0 103 M.
25°C 25°C.
5.0 102 mol/L s A ¡ B C Determine the half-life for the decomposition reaction initially having 1.00 M AB present.
46. The rate law for the reaction 2NOBr1g2 ¡ 2NO1g2 Br21g2 Time (s) [O] (atoms/cm3) 0 5.0 109 1.0 102 1.9 109 2.0 102 6.8 108 3.0 102 2.5 108 [AB]0 Initial Rate (mol/L) (mol/L s) 0.200 0.400 0.600 2.88 102 1.28 102 3.20 103 0 100 80 60 40 20 0 2 Time (s) 1/[A] 4 6 0 0.05 0.04 0.03 0.02 0.01 0 2 Time (s) [A] 4 6 0 2 4 6 –3.0 –3.5 –4.0 –4.5 –5.0 Time (s) ln[A] Assuming that determine the form of the rate law, the integrated rate law, and the rate constant for this reaction. (These are actual experimental data, so they may not give a perfectly straight line.) 38. The rate of the reaction was studied at a certain temperature.
a. In the first set of experiments, NO2 was in large excess, at a concentration of molecules/cm3 with the following data collected: 1.0 1013 O1g2 NO21g2 ¡ NO1g2 O21g2 Rate ¢3C4H64 ¢t Time (s) [C4H6] (mol/L) 195 1.6 102 604 1.5 102 1246 1.3 102 2180 1.1 102 6210 0.68 102 What is the order of the reaction with respect to oxygen atoms?
b. The reaction is known to be first order with respect to NO2.
Determine the overall rate law and the value of the rate constant.
39. Experimental data for the reaction have been plotted in the following three different ways (with con-centration units in mol/L): A ¡ 2B C Exercises 571 at some temperature is a. If the half-life for this reaction is 2.00 s when [NOBr]0 0.900 M, calculate the value of k for this reaction.
b. How much time is required for the concentration of NOBr to decrease to 0.100 M?
47. For the reaction products, successive half-lives are observed to be 10.0, 20.0, and 40.0 min for an experiment in which Calculate the concentration of A at the following times.
a. 80.0 min b. 30.0 min 48. Consider the hypothetical reaction where the rate law is An experiment is carried out where The reaction is started, and after 8.0 seconds, the concentration of A is a. Calculate k for this reaction.
b. Calculate the half-life for this experiment.
c. Calculate the concentration of A after 13.0 seconds.
d. Calculate the concentration of C after 13.0 seconds.
Reaction Mechanisms 49. Write the rate laws for the following elementary reactions.
a.
b.
c.
c.
50. The mechanisms shown below have been proposed to explain the kinetics of the reaction considered in Question 11. Which of the following are acceptable mechanisms? Explain.
Mechanism I: Mechanism II: Slow Fast Fast Mechanism III: Slow Fast 51. A proposed mechanism for a reaction is Slow Fast Fast C4H9OH2 H2O ¡ C4H9OH H3O C4H9 H2O ¡ C4H9OH2 C4H9Br ¡ C4H9 Br N2O1g2 H21g2 ¡ N21g2 H2O1g2 H21g2 2NO1g2 ¡ N2O1g2 H2O1g2 H21g2 O1g2 ¡ H2O1g2 N1g2 NO1g2 ¡ N21g2 O1g2 H21g2 NO1g2 ¡ H2O1g2 N1g2 2H21g2 2NO1g2 ¡ N21g2 2H2O1g2 O31g2 O1g2 S 2O21g2 O31g2 S O21g2 O1g2 O31g2 NO1g2 S O21g2 NO21g2 CH3NC1g2 S CH3CN1g2 3.8 103 M.
[B]0 3.0 M, and [C]0 2.0 M.
[A]0 1.0 102 M, Rate ¢3A4 ¢t k3A4 3B4 2 A B 2C ¡ 2D 3E [A]0 0.10 M.
A S Rate ¢3NOBr4 ¢t k3NOBr4 2 Write the rate law expected for this mechanism. What is the over-all balanced equation for the reaction? What are the intermediates in the proposed mechanism?
52. The mechanism for the reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be Slow Fast Write the rate law expected for this mechanism. What is the over-all balanced equation for the reaction?
Temperature Dependence of Rate Constants and the Collision Model 53. For the following reaction profile, indicate a. the positions of reactants and products.
b. the activation energy.
c.
for the reaction.
54. Draw a rough sketch of the energy profile for each of the fol-lowing cases: a.
b.
c.
55. The activation energy for the reaction is 125 kJ/mol, and for the reaction is kJ/mol. What is the activation energy for the reverse reaction [ ]?
56. For a certain process, the activation energy is greater for the for-ward reaction than for the reverse reaction. Does this reaction have a positive or negative value for ?
57. The rate constant for the gas-phase decomposition of N2O5, has the following temperature dependence: N2O5 ¡ 2NO2 1 2O2 ¢E NO2(g) CO(g) NO(g) CO2(g) ¡ 216 ¢E NO21g2 CO1g2 ¡ NO1g2 CO21g2 ¢E 50 kJ/mol, Ea 50 kJ/mol ¢E 10 kJ/mol, Ea 50 kJ/mol ¢E 10 kJ/mol, Ea 25 kJ/mol Reaction coordinate E ¢E NO3 CO ¡ NO2 CO2 NO2 NO2 ¡ NO3 NO T (K) k (s1) 338 4.9 103 318 5.0 104 298 3.5 105 572 Chapter Twelve Chemical Kinetics Make the appropriate graph using these data, and determine the activation energy for this reaction.
58. The reaction in a certain solvent is first order with respect to and zero order with respect to In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) versus 1T was constructed resulting in a straight line with a slope value of and y-intercept of 33.5. Assume k has units of a. Determine the activation energy for this reaction.
b. Determine the value of the frequency factor A.
c. Calculate the value of k at 59. The activation energy for the decomposition of HI(g) to H2(g) and I2(g) is 186 kJ/mol. The rate constant at 555 K is L/mol s. What is the rate constant at 645 K?
60. A first-order reaction has rate constants of and at and , respectively. What is the value of the activation energy?
61. A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased from to a higher temperature, the rate constant increases by a factor of 7.00. Calculate the higher temperature.
62. Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the ac-tivation energy be for this statement to be true for a temperature increase from 25 to ?
63. Which of the following reactions would you expect to proceed at a faster rate at room temperature? Why? (Hint: Think about which reaction would have the lower activation energy.) 64. One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is 210 kJ/mol, which is less than either the SiOSi or the SiOH bond energy. Why would a similar mechanism not be expected to play a very important role in the decomposition of long chains of carbon atoms as seen in organic compounds? H3O1aq2 OH1aq2 ¡ 2H2O1l2 2Ce41aq2 Hg2 21aq2 ¡ 2Ce31aq2 2Hg21aq2 35°C 22°C 20.°C 0°C 8.1 102 s1 4.6 102 s1 3.52 107 25°C.
s1.
1.10 104 K OH.
(CH3)3CBr 1CH323CBr OH ¡ 1CH323COH Br Catalysts 65. One mechanism for the destruction of ozone in the upper atmos-phere is Slow Fast a. Which species is a catalyst?
b. Which species is an intermediate?
c. Ea for the uncatalyzed reaction is 14.0 kJ. Ea for the same reaction when catalyzed is 11.9 kJ.
What is the ratio of the rate constant for the catalyzed reac-tion to that for the uncatalyzed reaction at ? Assume that the frequency factor A is the same for each reaction.
66. One of the concerns about the use of Freons is that they will mi-grate to the upper atmosphere, where chlorine atoms can be gen-erated by the following reaction: Freon-12 Chlorine atoms can act as a catalyst for the destruction of ozone.
The activation energy for the reaction is 2.1 kJ/mol. Which is the more effective catalyst for the de-struction of ozone, Cl or NO? (See Exercise 65.) 67. Assuming that the mechanism for the hydrogenation of C2H4 given in Section 12.8 is correct, would you predict that the product of the reaction of C2H4 with D2 would be CH2DOCH2D or CHD2OCH3? How could the reaction of C2H4 with D2 be used to confirm the mechanism for the hydrogenation of C2H4 given in Section 12.8?
68. The decomposition of NH3 to N2 and H2 was studied on two surfaces: Cl O3 ¡ ClO O2 CCl2F2 ¡ hv CF2Cl Cl 25°C O31g2 O1g2 ¡ 2O2 Overall reaction O31g2 O1g2 ¡ 2O21g2 NO21g2 O1g2 ¡ NO1g2 O21g2 O31g2 NO1g2 ¡ NO21g2 O21g2 Without a catalyst, the activation energy is 335 kJ/mol.
a. Which surface is the better heterogeneous catalyst for the decomposition of NH3? Why?
b. How many times faster is the reaction at 298 K on the W sur-face compared with the reaction with no catalyst present? As-sume that the frequency factor A is the same for each reaction.
c. The decomposition reaction on the two surfaces obeys a rate law of the form Rate k 3NH34 3H24 Surface Ea (kJ/mol) W 163 Os 197 Additional Exercises 573 How can you explain the inverse dependence of the rate on the H2 concentration?
69. A famous chemical demonstration is the “magic genie” procedure, in which hydrogen peroxide decomposes to water and oxygen gas with the aid of a catalyst. The activation energy of this (uncat-alyzed) reaction is 70.0 kJ/mol. When the catalyst is added, the activation energy (at ) is 42.0 kJ/mol. Theoretically, to what temperature ( ) would one have to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at ? Assume the frequency fac-tor A is constant and assume the initial concentrations are the same.
70. The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency fac-tor A is constant and assume the initial concentrations are the same.
Additional Exercises 71. The reaction was studied, and the following data were obtained where Rate ¢3O24 ¢t 2NO1g2 O21g2 ¡ 2NO21g2 20.°C °C 20.°C What would be the initial rate for an experiment where [NO]0 molecules/cm3 and [O2]0 molecules/cm3?
72. Sulfuryl chloride (SO2Cl2) decomposes to sulfur dioxide (SO2) and chlorine (Cl2) by reaction in the gas phase. The following pressure data were obtained when a sample containing mol sulfuryl chloride was heated to 600. K in a -L container.
5.00 101 5.00 102 7.36 1018 6.21 1018 Defining the rate as a. determine the value of the rate constant for the decomposition of sulfuryl chloride at 600. K.
b. what is the half-life of the reaction?
c. what fraction of the sulfuryl chloride remains after 20.0 h?
¢3SO2Cl24 ¢t , Calculate Ea for this reaction.
74. Experimental values for the temperature dependence of the rate constant for the gas-phase reaction are as follows: NO O3 ¡ NO2 O2 73. For the reaction the following data were collected, where Rate ¢ 3N2O54 ¢t 2N2O51g2 ¡ 4NO21g2 O21g2 [NO]0 [O2]0 Initial Rate (molecules/cm3) (molecules/cm3) (molecules/cm3 s) 1.00 1018 1.00 1018 2.00 1016 3.00 1018 1.00 1018 1.80 1017 2.50 1018 2.50 1018 3.13 1017 T 338 K T 318 K Time (s) [N2O5] [N2O5] 0 1.00 101 M 1.00 101 M 100.
6.14 102 M 9.54 102 M 300.
2.33 102 M 8.63 102 M 600.
5.41 103 M 7.43 102 M 900.
1.26 103 M 6.39 102 M T (K) k (L/mol s) 195 1.08 109 230.
2.95 109 260.
5.42 109 298 12.0 109 369 35.5 109 Time (hours): 0.00 1.00 2.00 4.00 8.00 16.00 (atm): 4.93 4.26 3.52 2.53 1.30 0.34 PSO2Cl2 Make the appropriate graph using these data, and determine the activation energy for this reaction.
75. For enzyme-catalyzed reactions that follow the mechanism a graph of the rate as a function of [S], the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.
Rate [S] E S ∆E P E S ∆E S 574 Chapter Twelve Chemical Kinetics 76. The activation energy of a certain uncatalyzed biochemical re-action is 50.0 kJ/mol. In the presence of a catalyst at the rate constant for the reaction increases by a factor of as compared with the uncatalyzed reaction. Assuming the fre-quency factor A is the same for both the catalyzed and uncat-alyzed reactions, calculate the activation energy for the catalyzed reaction.
77. Consider the reaction where the rate law is defined as An experiment is carried out where and a. If after 3.00 min, calculate the value of k.
b. Calculate the half-life for this experiment.
c. Calculate the concentration of B and the concentration of A after 10.0 min.
Challenge Problems 78. Consider a reaction of the type in which the rate law is found to be (termolecular reactions are im-probable but possible). If the first half-life of the reaction is found to be 40. s, what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: 79. A study was made of the effect of the hydroxide concentration on the rate of the reaction The following data were obtained: I1aq2 OCl1aq2 ¡ IO1aq2 Cl1aq2 Rate d3A4 dt k3A4 3 rate k[A]3 aA S products, [A] 3.26 105 M, [A]0 1.00 104 M.
[B]0 [C]0 1.00 M ¢3A4 ¢t k3A4 23B4 3C4 3A B C S D E 2.50 103 37°C, Both processes are known to be second order in reactant, and k1 is known to be at . In a particular experi-ment A and B were placed in separate containers at , where and . It was found that after each reaction had progressed for 3.00 min, . In this case the rate laws are defined as a. Calculate the concentration of A2 after 3.00 min.
b. Calculate the value of k2.
c. Calculate the half-life for the experiment involving A.
81. The reaction was studied by performing two experiments. In the first experiment the rate of disappearance of NO was followed in the presence of a large excess of O3. The results were as follows ([O3] remains effectively constant at molecules/cm3): 1.0 1014 NO1g2 O31g2 ¡ NO21g2 O21g2 Rate ¢ 3B4 ¢t k23B4 2 Rate ¢ 3A4 ¢t k13A4 2 3.00[B] [A] [B]0 2.50 102 M [A]0 1.00 102 M 25°C 25°C 0.250 L/mol s [I]0 [OCl]0 [OH]0 Initial Rate (mol/L) (mol/L) (mol/L) (mol/L s) 0.0013 0.012 0.10 9.4 103 0.0026 0.012 0.10 18.7 103 0.0013 0.0060 0.10 4.7 103 0.0013 0.018 0.10 14.0 103 0.0013 0.012 0.050 18.7 103 0.0013 0.012 0.20 4.7 103 0.0013 0.018 0.20 7.0 103 Time [NO] (ms) (molecules/cm3) 0 6.0 108 100 1 5.0 108 500 1 2.4 108 700 1 1.7 108 1000 1 9.9 107 Time [O3] (ms) (molecules/cm3) 0 1.0 1010 50 1 8.4 109 100 1 7.0 109 200 1 4.9 109 300 1 3.4 109 Determine the rate law and the value of the rate constant for this reaction.
80. Two isomers (A and B) of a given compound dimerize as follows: 2B ¡ k2 B2 2A ¡ k1 A2 In the second experiment [NO] was held constant at molecules/cm3. The data for the disappearance of O3 are as follows: 2.0 1014 a. What is the order with respect to each reactant?
b. What is the overall rate law?
c. What is the value of the rate constant from each set of ex-periments?
d. What is the value of the rate constant for the overall rate law?
Rate k3NO4 x3O34 y Rate k¿ 3NO4 x Rate k– 3O34 y Challenge Problems 575 On the energy profile, indicate a. The positions of reactants and products.
b. The activation energy for the overall reaction.
c.
for the reaction.
d. Which point on the plot represents the energy of the interme-diate in the two-step reaction?
e. Which step in the mechanism for this reaction is rate deter-mining, the first or the second step? Explain.
83. Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average in-terval between flashes of individual insects was 16.3 s at and 13.0 s at .
a. What is the apparent activation energy of the reaction that con-trols the flashing?
b. What would be the average interval between flashes of an in-dividual firefly at ?
c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.
84. The decomposition of NO2(g) occurs by the following bimolecular elementary reaction: The rate constant at 273 K is and the ac-tivation energy is 111 kJ/mol. How long will it take for the con-centration of NO2(g) to decrease from an initial partial pressure of 2.5 atm to 1.5 atm at 500. K? Assume ideal gas behavior.
85. The following data were collected in two studies of the reaction 2A B ¡ C D 2.3 1012 L/mol s, 2NO21g2 ¡ 2NO1g2 O21g2 30.0°C 27.8°C 21.0°C ¢E In experiment 1, In experiment 2, a. Why is [B] much greater than [A]?
b. Give the rate law and value for k for this reaction.
86. The following data were collected in two studies of the reaction 2H21g2 2NO1g2 ¡ N21g2 2H2O1g2 Rate ¢ 3A4 ¢t [B]0 10.0 M.
[B]0 5.0 M.
Experiment 1 Experiment 2 Time (s) [A] (mol/L) 102 [A] (mol/L) 102 0 10.0 10.0 20.
6.67 5.00 40.
5.00 3.33 60.
4.00 2.50 80.
3.33 2.00 100.
2.86 1.67 120.
2.50 1.43 Experiment 1 Experiment 2 Time (s) [H2] (mol/L) [H2] (mol/L) 0.
10.
20.
30.
?
40.
6.3 104 5.0 103 1.3 103 2.5 103 7.1 103 5.0 103 8.4 103 1.0 102 1.0 102 [B] (mol/L) Time (s) 6.0 105 5.8 105 5.0 105 6.9 105 4.0 105 8.5 105 3.0 105 1.1 104 2.0 105 1.6 104 1.0 105 2.7 104 Reaction coordinate E In experiment 1, In experiment 2, a. Use the concentration versus time data to determine the rate law for the reaction.
b. Solve for the rate constant (k) for the reaction. Include units.
c. Calculate the concentration of H2 in experiment 1 at 87. Consider the hypothetical reaction In a study of this reaction three experiments were run at the same temperature. The rate is defined as Experiment 1: 3C4 0 1.0 M 3B4 0 1.0 103 M 3A4 0 2.0 M ¢ 3B4 ¢t.
A B 2C ¡ 2D 3E t 30. s.
Rate ¢ 3H24 ¢t [NO]0 20.0 M.
[NO]0 10.0 M.
Experiment 2: 3C4 0 1.0 M 3B4 0 3.0 M 3A4 0 1.0 102 M 82. Most reactions occur by a series of steps. The energy profile for a certain reaction that proceeds by a two-step mechanism is 576 Chapter Twelve Chemical Kinetics Experiment 3: 3C4 0 5.0 101 M 3A40 10.0 M 3B4 0 5.0 M [C] (mol/L) Time (s) 0.43 0.36 0.29 0.22 0.15 0.08 6.0 102 5.0 102 4.0 102 3.0 102 2.0 102 1.0 102 [I]0 [H]0 Slope (mol/L) (mol/L) (min1) 0.1000 0.0400 0.3000 0.0400 0.4000 0.0400 0.0750 0.0200 0.0750 0.0800 0.0750 0.1600 0.174 0.118 0.0760 0.480 0.360 0.120 [CH3X] (mol/L) Time (h) 7.08 103 1.0 4.52 103 1.5 2.23 103 2.3 4.76 104 4.0 8.44 105 5.7 2.75 105 7.0 Write the rate law for this reaction, and calculate the rate constant.
88. Hydrogen peroxide and the iodide ion react in acidic solution as follows: The kinetics of this reaction were studied by following the decay of the concentration of and constructing plots of versus time. All the plots were linear and all solutions had mol/L. The slopes of these straight lines depended on the initial concentrations of and The results follow: H.
I [H2O2]0 8.0 104 ln[H2O2] H2O2 H2O21aq2 3I1aq2 2H1aq2 ¡ I3 1aq2 2H2O1l2 The rate law for this reaction has the form a. Specify the order of this reaction with respect to and[I].
b. Calculate the values of the rate constants, and k2.
k1 [H2O2] Rate ¢3H2O24 ¢t 1k1 k23H 42 3I4 m3H2O24 n c. What reason could there be for the two-term dependence of the rate on ?
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
89. Sulfuryl chloride undergoes first-order decomposition at 320.ºC with a half-life of 8.75 h.
What is the value of the rate constant, k, in s1? If the initial pressure of SO2Cl2 is 791 torr and the decomposition occurs in a 1.25-L container, how many molecules of SO2Cl2 remain after 12.5 h?
90. Upon dissolving InCl(s) in HCl, In(aq) undergoes a dispro-portionation reaction according to the following unbalanced equation: This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of In(aq) after 1.25 h if the initial solution of In(aq) was prepared by dissolving 2.38 g of InCl(s) in 5.00 102 mL of dilute HCl? What mass of ln(s) is formed after 1.25 h?
91. The decomposition of iodoethane in the gas phase proceeds ac-cording to the following equation: At 660. K, k 7.2 104 s1; at 720. K, k 1.7 102 s1.
What is the rate constant for this first-order decomposition at 325ºC? If the initial pressure of iodoethane is 894 torr at 245ºC, what is the pressure of iodoethane after three half-lives?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
92. Consider the following reaction: At the following two experiments were run, yielding the following data: Experiment 1: M [Y]0 3.0 25°C, CH3X Y ¡ CH3Y X C2H5I1g2 ¡ C2H41g2 HI1g2 In1aq2 ¡ In1s2 In31aq2 SO2Cl21g2 ¡ SO21g2 Cl21g2 [H] [A] (mol/L) Time (s) 1.0 3.0 5.0 8.0 10.0 13.0 2.0 103 2.9 103 3.8 103 5.5 103 7.1 103 8.9 103 Marathon Problem 577 Experiment 2: M [Y]0 4.5 a. Determine the rate law and the value of k for this reaction at .
b. Determine the half-life at .
c. Determine Ea for the reaction.
d. Given that the COX bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
85°C 25°C [CH3X] (mol/L) Time (h) 4.50 103 0 1.70 103 1.0 4.19 104 2.5 1.11 104 4.0 2.81 105 5.5 Experiments also were run at . The value of the rate constant at was found to be (with the time in units of hours), where .
[CH3X]0 1.0 102 M and [Y]0 3.0 M 7.88 108 85°C 85°C 578 13 Chemical Equilibrium Contents 13.1 The Equilibrium Condition • The Characteristics of Chemical Equilibrium 13.2 The Equilibrium Constant 13.3 Equilibrium Expressions Involving Pressures 13.4 Heterogeneous Equilibria 13.5 Applications of the Equilibrium Constant • The Extent of a Reaction • Reaction Quotient • Calculating Equilibrium Pressures and Concentrations 13.6 Solving Equilibrium Problems • Treating Systems That Have Small Equilibrium Constants 13.7 Le Châtelier’s Principle • The Effect of a Change in Concentration • The Effect of a Change in Pressure • The Effect of a Change in Temperature In doing stoichiometry calculations we assumed that reactions proceed to completion, that is, until one of the reactants runs out. Many reactions do proceed essentially to com-pletion. For such reactions it can be assumed that the reactants are quantitatively con-verted to products and that the amount of limiting reactant that remains is negligible. On the other hand, there are many chemical reactions that stop far short of completion. An example is the dimerization of nitrogen dioxide: The reactant, NO2, is a dark brown gas, and the product, N2O4, is a colorless gas. When NO2 is placed in an evacuated, sealed glass vessel at the initial dark brown color decreases in intensity as it is converted to colorless N2O4. However, even over a long pe-riod of time, the contents of the reaction vessel do not become colorless. Instead, the in-tensity of the brown color eventually becomes constant, which means that the concentration of NO2 is no longer changing. This is illustrated on the molecular level in Fig. 13.1. This observation is a clear indication that the reaction has stopped short of completion. In fact, the system has reached chemical equilibrium, the state where the concentrations of all reactants and products remain constant with time.
Any chemical reactions carried out in a closed vessel will reach equilibrium. For some reactions the equilibrium position so favors the products that the reaction appears to have gone to completion. We say that the equilibrium position for such reactions lies far to the right (in the direction of the products). For example, when gaseous hydrogen and oxygen are mixed in stoichiometric quantities and react to form water vapor, the reaction proceeds essentially to completion. The amounts of the reactants that remain when the system reaches equilibrium are so tiny as to be negligible. By contrast, some reactions occur only to a slight extent. For example, when solid CaO is placed in a closed vessel at the decomposition to solid Ca and gaseous O2 is virtually undetectable. In cases like this, the equilibrium position is said to lie far to the left (in the direction of the reactants).
In this chapter we will discuss how and why a chemical system comes to equilibrium and the characteristics of equilibrium. In particular, we will discuss how to calculate the concentrations of the reactants and products present for a given system at equilibrium.
13.1 The Equilibrium Condition Since no changes occur in the concentrations of reactants or products in a reaction sys-tem at equilibrium, it may appear that everything has stopped. However, this is not the case. On the molecular level, there is frantic activity. Equilibrium is not static but is a highly dynamic situation. The concept of chemical equilibrium is analogous to the flow of cars across a bridge connecting two island cities. Suppose the traffic flow on the bridge 25°C, 25°C, NO21g2 NO21g2 ¡ N2O41g2 579 The effect of temperature on the endothermic, aqueous equilibrium: Pink Blue The violet solution in the center is at 25°C and contains significant quantities of both pink Co(H2O)6 2 and blue CoCl4 2. When the solution is cooled, it turns pink because the equilibrium is shifted to the left. Heating the solution favors the blue CoCl4 2 ions.
Co(H2O)6 2 4Cl ∆CoCl4 2 6H2O Equilibrium is a dynamic situation.
580 Chapter Thirteen Chemical Equilibrium is the same in both directions. It is obvious that there is motion, since one can see the cars traveling back and forth across the bridge, but the number of cars in each city is not chang-ing because equal numbers of cars are entering and leaving. The result is no net change in the car population.
To see how this concept applies to chemical reactions, consider the reaction between steam and carbon monoxide in a closed vessel at a high temperature where the reaction takes place rapidly: Assume that the same number of moles of gaseous CO and gaseous H2O are placed in a closed vessel and allowed to react. The plots of the concentrations of reactants and prod-ucts versus time are shown in Fig. 13.2. Note that since CO and H2O were originally pres-ent in equal molar quantities, and since they react in a 1:1 ratio, the concentrations of the two gases are always equal. Also, since H2 and CO2 are formed in equal amounts, they are always present in the same concentrations.
Figure 13.2 is a profile of the progress of the reaction. When CO and H2O are mixed, they immediately begin to react to form H2 and CO2. This leads to a decrease in the con-centrations of the reactants, but the concentrations of the products, which were initially at zero, are increasing. Beyond a certain time, indicated by the dashed line in Fig. 13.2, the concentrations of reactants and products no longer change—equilibrium has been reached. Unless the system is somehow disturbed, no further changes in concentrations will occur. Note that although the equilibrium position lies far to the right, the concen-trations of reactants never go to zero; the reactants will always be present in small but constant concentrations. This is shown on the microscopic level in Fig. 13.3.
What would happen to the gaseous equilibrium mixture of reactants and products represented in Fig. 13.3, parts (c) and (d), if we injected some H2O(g) into the box? To answer this question, we need to be sure we understand the equilibrium condition: The H2O1g2 CO1g2 ∆H21g2 CO21g2 FIGURE 13.2 The changes in concentrations with time for the reaction H2O(g) CO(g) H2(g) CO2(g) when equimolar quantities of H2O(g) and CO(g) are mixed.
∆ Concentration Time [CO] or [H2O] [CO2] or [H2] Equilibrium FIGURE 13.1 A molecular representation of the reaction 2NO2(g) n N2O4(g) over time in a closed vessel. Note that the numbers of NO2 and N2O4 in the container become constant (c and d) after sufficient time has passed.
Time (a) (b) (c) (d) 13.1 The Equilibrium Condition 581 concentrations of reactants and products remain constant at equilibrium because the forward and reverse reaction rates are equal. If we inject some H2O molecules, what will happen to the forward reaction: It will speed up because more H2O mole-cules means more collisions between H2O and CO molecules. This in turn will form more products and will cause the reverse reaction to speed up. Thus the system will change until the forward and reverse reaction rates again become equal. Will this new equilibrium position contain more or fewer product molecules than are shown in Fig. 13.3(c) and (d)? Think about this carefully. If you are not sure of the answer now, keep reading. We will consider this type of situation in more detail later in this chapter.
Why does equilibrium occur? We saw in Chapter 12 that molecules react by collid-ing with one another, and the more collisions, the faster the reaction. This is why reac-tion rates depend on concentrations. In this case the concentrations of H2O and CO are lowered by the forward reaction: As the concentrations of the reactants decrease, the forward reaction slows down (Fig.
13.4). As in the bridge traffic analogy, there is also a reverse direction: Initially in this experiment no H2 and CO2 were present, and this reverse reaction could not occur. However, as the forward reaction proceeds, the concentrations of H2 and CO2 build up, and the rate of the reverse reaction increases (Fig. 13.4) as the forward reaction slows down. Eventually, the concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction. The system has reached equilibrium.
H2O CO — H2 CO2 H2O CO ¡ H2 CO2 H2O CO d H2 CO2 H2O CO S H2 CO2?
Time (a) (b) (c) (d) FIGURE 13.3 (a) H2O and CO are mixed in equal numbers and begin to react (b) to form CO2 and H2. After time has passed, equilibrium is reached (c) and the numbers of reactant and product molecules then remain constant over time (d).
FIGURE 13.4 The changes with time in the rates of forward and reverse reactions for when equimolar quantities of and are mixed. The rates do not change in the same way with time because the forward reaction has a much larger rate constant than the reverse reaction.
CO( g) H2O( g) H2O(g) CO(g) ∆H2(g) CO2(g) Reaction rates Time Equilibrium Forward rate Reverse rate Forward rate = Reverse rate A double arrow (w x) is used to show that a reaction can occur in either direction.
582 Chapter Thirteen Chemical Equilibrium The equilibrium position of a reaction—left, right, or somewhere in between—is determined by many factors: the initial concentrations, the relative energies of the reac-tants and products, and the relative degree of “organization” of the reactants and prod-ucts. Energy and organization come into play because nature tries to achieve minimum energy and maximum disorder, as we will show in detail in Chapter 16. For now, we will simply view the equilibrium phenomenon in terms of the rates of opposing reactions.
The Characteristics of Chemical Equilibrium To explore the important characteristics of chemical equilibrium, we will consider the syn-thesis of ammonia from elemental nitrogen and hydrogen: This process is of great commercial value because ammonia is an important fertilizer for the growth of corn and other crops. Ironically, this beneficial process was discovered in Germany just before World War I in a search for ways to produce nitrogen-based explo-sives. In the course of this work, German chemist Fritz Haber (1868–1934) pioneered the large-scale production of ammonia.
When gaseous nitrogen, hydrogen, and ammonia are mixed in a closed vessel at no apparent change in the concentrations occurs over time, regardless of the original amounts of the gases. Why? There are two possible reasons why the concentrations of the reactants and products of a given chemical reaction remain unchanged when mixed.
1. The system is at chemical equilibrium.
2. The forward and reverse reactions are so slow that the system moves toward equilib-rium at a rate that cannot be detected.
The second reason applies to the nitrogen, hydrogen, and ammonia mixture at As we saw in Chapters 8 and 9, the N2 molecule has a very strong triple bond (941 kJ/mol) and thus is very unreactive. Also, the H2 molecule has an unusually strong single bond (432 kJ/mol). Therefore, mixtures of N2, H2, and NH3 at can exist with no apparent change over long periods of time, unless a catalyst is introduced to speed up the forward and reverse reactions. Under appropriate conditions, the system does reach equilibrium, as shown in Fig. 13.5. Note that because of the reaction stoichiometry, H2 disappears three times as fast as N2 does and NH3 forms twice as fast as N2 disappears.
13.2 The Equilibrium Constant Science is fundamentally empirical—it is based on experiment. The development of the equilibrium concept is typical. From their observations of many chemical reactions, two Norwegian chemists, Cato Maximilian Guldberg (1836–1902) and Peter Waage (1833–1900), proposed in 1864 the law of mass action as a general description of the equilibrium con-dition. Guldberg and Waage postulated that for a reaction of the type where A, B, C, and D represent chemical species and j, k, l, and m are their coefficients in the balanced equation, the law of mass action is represented by the following equilib-rium expression: The square brackets indicate the concentrations of the chemical species at equilibrium, and K is a constant called the equilibrium constant.
K 3C4 l3D4 m 3A4 j3B4k jA kB ∆lC mD 25°C 25°C.
25°C, N21g2 3H21g2 ∆2NH31g2 The United States produces about 20 million tons of ammonia annually.
Molecules with strong bonds produce large activation energies and tend to react slowly at 25C.
The law of mass action is based on experimental observation.
H2 NH3 N2 Time Concentration Equilibrium FIGURE 13.5 A concentration profile for the reaction when only and are mixed initially.
H2(g) N2(g) N2(g) 3H2( g) ∆2NH3( g) The relationship between equilibrium and thermodynamics is explored in Section 16.8.
13.2 The Equilibrium Constant 583 Writing Equilibrium Expressions Write the equilibrium expression for the following reaction: Solution Applying the law of mass action gives Coefficient Coefficient of NO2 of H2O Coefficient of O2 Coefficient of NH3 See Exercise 13.17.
The value of the equilibrium constant at a given temperature can be calculated if we know the equilibrium concentrations of the reaction components, as illustrated in Sample Exercise 13.2.
It is very important to note at this point that the equilibrium constants are customar-ily given without units. The reason for this is beyond the scope of this text, but it involves corrections for the nonideal behavior of the substances taking part in the reaction. When these corrections are made, the units cancel out and the corrected K has no units. Thus we will not use units for K in this text.
Calculating the Values of K The following equilibrium concentrations were observed for the Haber process at a. Calculate the value of K at for this reaction.
b. Calculate the value of the equilibrium constant at for the reaction c. Calculate the value of the equilibrium constant at for the reaction given by the equation Solution a. The balanced equation for the Haber process is Thus Note that K is written without units.
3.8 104 K 3NH34 2 3N24 3H24 3 13.1 10222 18.5 101213.1 10323 N21g2 3H21g2 ∆2NH31g2 1 2N21g2 3 2H21g2 ∆NH31g2 127°C 2NH31g2 ∆N21g2 3H21g2 127°C 127°C 3H24 3.1 103 mol/L 3N24 8.5 101 mol/L 3NH34 3.1 102 mol/L 127°C: K 3NO24 43H2O4 6 3NH34 43O24 7 4NH31g2 7O21g2 ∆4NO21g2 6H2O1g2 Sample Exercise 13.1 Sample Exercise 13.2 The square brackets indicate concentra-tion in units of mol/L.
8888n 8888n 888n 888n 584 Chapter Thirteen Chemical Equilibrium b.
This reaction is written in the reverse order from the equation given in part a. This leads to the equilibrium expression which is the reciprocal of the expression used in part a. Therefore, c.
We use the law of mass action: If we compare this expression to that obtained in part a, we see that since Thus See Exercises 13.19 and 13.21 through 13.24.
We can draw some important conclusions from the results of Sample Exercise 13.2.
For a reaction of the form the equilibrium expression is If this reaction is reversed, then the new equilibrium expression is If the original reaction is multiplied by some factor n to give the equilibrium expression becomes We Can Summarize These Conclusions About the Equilibrium Expression as Follows: The equilibrium expression for a reaction is the reciprocal of that for the reaction writ-ten in reverse.
When the balanced equation for a reaction is multiplied by a factor n, the equilib-rium expression for the new reaction is the original expression raised to the nth power.
Thus Knew (Koriginal)n.
K values are customarily written without units.
K– 3C4 nl3D4 nm 3A4 nj3B4nk Kn njA nkB ∆nlC nmD K¿ 3A4j3B4 k 3C4 l3D4 m 1 K K 3C4 l3D4 m 3A4 j3B4k jA kB ∆lC mD K– K1 2 13.8 10421 2 1.9 102 K– K1 2 3NH34 3N24 1 23H24 3 2 a 3NH34 2 3N24 3H24 3b 1 2 K– 3NH34 3N24 1 23H24 3 2 K¿ 3N24 3H24 3 3NH34 2 1 K 1 3.8 104 2.6 105 K¿ 3N24 3H24 3 3NH34 2 13.2 The Equilibrium Constant 585 The law of mass action is widely applicable. It correctly describes the equilib-rium behavior of an amazing variety of chemical systems in solution and in the gas phase. Although, as we will see later, corrections must be applied in certain cases, such as for concentrated aqueous solutions and for gases at high pressures, the law of mass action provides a remarkably accurate description of all types of chemical equilibria.
Consider again the ammonia synthesis reaction. The equilibrium constant K always has the same value at a given temperature. At the value of K is When-ever N2, H2, and NH3 are mixed together at this temperature, the system will always come to an equilibrium position such that This expression has the same value at regardless of the amounts of the gases that are mixed together initially.
Although the special ratio of products to reactants defined by the equilibrium ex-pression is constant for a given reaction system at a given temperature, the equilibrium concentrations will not always be the same. Table 13.1 gives three sets of data for the syn-thesis of ammonia, showing that even though the individual sets of equilibrium concen-trations are quite different for the different situations, the equilibrium constant, which depends on the ratio of the concentrations, remains the same (within experimental error).
Note that subscript zeros indicate initial concentrations.
Each set of equilibrium concentrations is called an equilibrium position. It is es-sential to distinguish between the equilibrium constant and the equilibrium positions for a given reaction system. There is only one equilibrium constant for a particular system at a particular temperature, but there are an infinite number of equilibrium positions. The specific equilibrium position adopted by a system depends on the initial concentrations, but the equilibrium constant does not.
Equilibrium Positions The following results were collected for two experiments involving the reaction at between gaseous sulfur dioxide and oxygen to form gaseous sulfur trioxide: 600°C 500°C, 3NH34 2 3N24 3H24 3 6.0 102 6.0 102.
500°C The law of mass action applies to solution and gaseous equilibria.
A cross section showing how anhydrous ammonia is injected into the soil to act as a fertilizer.
For a reaction at a given temperature, there are many equilibrium positions but only one value for K.
Experiment 1 Experiment 2 Initial Equilibrium Initial Equilibrium 3SO34 0.260 M 3SO34 0 0.350 M 3SO34 3.50 M 3SO34 0 3.00 M 3O24 0.0450 M 3O24 0 0 3O24 1.25 M 3O24 0 1.50 M 3SO24 0.590 M 3SO24 0 0.500 M 3SO24 1.50 M 3SO24 0 2.00 M Sample Exercise 13.3 Show that the equilibrium constant is the same in both cases.
Solution The balanced equation for the reaction is From the law of mass action, K 3SO34 2 3SO24 23O24 2SO21g2 O21g2 ∆2SO31g2 586 Chapter Thirteen Chemical Equilibrium For Experiment 1, For Experiment 2, The value of K is constant, within experimental error.
See Exercise 13.24.
13.3 Equilibrium Expressions Involving Pressures So far we have been describing equilibria involving gases in terms of concentrations. Equi-libria involving gases also can be described in terms of pressures. The relationship be-tween the pressure and the concentration of a gas can be seen from the ideal gas equation: or where C equals nV, or the number of moles n of gas per unit volume V. Thus C represents the molar concentration of the gas.
For the ammonia synthesis reaction, the equilibrium expression can be written in terms of concentrations, that is, or in terms of the equilibrium partial pressures of the gases, that is, Both the symbols K and Kc are used commonly for an equilibrium constant in terms of concentrations. We will always use K in this book. The symbol Kp represents an equilib-rium constant in terms of partial pressures.
Kp PNH3 2 1PN221PH2 32 K 3NH342 3N24 3H243 CNH3 2 1CN221CH2 32 Kc P a n VbRT CRT PV nRT K2 10.26022 10.5902210.04502 4.32 K1 13.5022 11.502211.252 4.36 The ideal gas equation was discussed in Section 5.3.
K involves concentrations; Kp involves pressures. In some books, the symbol Kc is used instead of K.
TABLE 13.1 Results of Three Experiments for the Reaction N2(g) 3H2(g) 2NH3(g) Initial Equilibrium Experiment Concentrations Concentrations I II III 3NH34 1.82 M 3NH34 0 3.00 M K 6.02 102 3H24 2.77 M 3H24 0 1.00 M 3N24 2.59 M 3N24 0 2.00 M 3NH34 0.203 M 3NH34 0 1.000 M K 6.02 102 3H24 1.197 M 3H24 0 0 3N24 0.399 M 3N24 0 0 3NH34 0.157 M 3NH34 0 0 K 6.02 102 3H24 0.763 M 3H24 0 1.000 M 3N24 0.921 M 3N24 0 1.000 M K [NH3]2 [N2][H2]3 ∆ 13.3 Equilibrium Expressions Involving Pressures 587 Calculating Values of Kp The reaction for the formation of nitrosyl chloride was studied at The pressures at equilibrium were found to be Calculate the value of Kp for this reaction at Solution For this reaction, See Exercises 13.25 and 13.26.
The relationship between K and Kp for a particular reaction follows from the fact that for an ideal gas, For example, for the ammonia synthesis reaction, However, for the synthesis of hydrogen fluoride from its elements, the relationship between K and Kp is given by Thus, for this reaction, K is equal to Kp. This equality occurs because the sum of the coefficients on either side of the balanced equation is identical, so the terms in RT can-cel out. In the equilibrium expression for the ammonia synthesis reaction, the sum of the powers in the numerator is different from that in the denominator, and K does not equal Kp.
Kp aPHF RT b 2 a PH2 RT ba PF2 RTb PHF 2 1PH221PF22 K 3HF42 3H24 3F24 CHF 2 1CH221CF22 H21g2 F21g2 ∆2HF1g2 Kp1RT22 PNH3 2 1PN221PH2 32 1RT22 a PNH3 RT b 2 a PN2 RT ba PH2 RT b 3 PNH3 2 1PN221PH2 32 a 1 RTb 2 a 1 RTb 4 K 3NH34 2 3N24 3H24 3 CNH3 2 1CN221CH2 32 C P RT.
1.9 103 Kp PNOCl 2 1PNO2221PCl22 11.222 15.0 1022213.0 1012 25°C.
PCl2 3.0 101 atm PNO 5.0 102 atm PNOCl 1.2 atm 25°C.
2NO1g2 Cl21g2 ∆2NOCl1g2 Sample Exercise 13.4 P CRT or C P RT 588 Chapter Thirteen Chemical Equilibrium For the general reaction the relationship between K and Kp is where is the sum of the coefficients of the gaseous products minus the sum of the co-efficients of the gaseous reactants. This equation is quite easy to derive from the definitions of K and Kp and the relationship between pressure and concentration. For the preceding general reaction, where the difference in the sums of the coefficients for the gaseous products and reactants.
Calculating K from Kp Using the value of Kp obtained in Sample Exercise 13.4, calculate the value of K at for the reaction Solution From the value of Kp, we can calculate K using where and Sum of Sum of product reactant coefficients coefficients n g Thus and See Exercises 13.27 and 13.28.
13.4 Heterogeneous Equilibria So far we have discussed equilibria only for systems in the gas phase, where all reactants and products are gases. These are homogeneous equilibria. However, many equilibria in-volve more than one phase and are called heterogeneous equilibria. For example, the 4.6 104 11.9 103210.08206212982 K Kp1RT2 Kp K1RT21 K RT ¢n 2 12 12 1 T 25 273 298 K Kp K1RT2¢n 2NO1g2 Cl21g2 ∆2NOCl1g2 25°C ¢n (l m) ( j k), K1RT2¢n 1CC l21CD m2 1CA j21CB k2 1RT2lm 1RT2jk K1RT21lm21jk2 Kp 1PC l21PD m2 1PA j21PB k2 1CC RT2l1CD RT2m 1CA RT2 j1CB RT2k ¢n Kp K1RT2¢n jA kB ∆lC mD Sample Exercise 13.5 n n n always involves products minus reactants.
13.4 Heterogeneous Equilibria 589 thermal decomposition of calcium carbonate in the commercial preparation of lime occurs by a reaction involving both solid and gas phases: h Lime Straightforward application of the law of mass action leads to the equilibrium expression However, experimental results show that the position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present (see Fig. 13.6). The fun-damental reason for this behavior is that the concentrations of pure solids and liquids can-not change. Thus the equilibrium expression for the decomposition of solid calcium carbonate might be represented as where C1 and C2 are constants representing the concentrations of the solids CaO and CaCO3, respectively. This expression can be rearranged to give We can generalize from this result as follows: If pure solids or pure liquids are involved in a chemical reaction, their concentrations are not included in the equilib-rium expression for the reaction. This simplification occurs only with pure solids or liquids, not with solutions or gases, since in these last two cases the concentrations can vary.
For example, in the decomposition of liquid water to gaseous hydrogen and oxygen, where water is not included in either equilibrium expression because it is a pure liquid. How-ever, if the reaction were carried out under conditions where the water is a gas rather than a liquid, that is, 2H2O1g2 ∆2H21g2 O21g2 K 3H2423O24 and Kp 1PH2 221PO22 2H2O1l2 ∆2H21g2 O21g2 C2K¿ C1 K 3CO24 K¿ 3CO24C1 C2 K¿ 3CO24 3CaO4 3CaCO34 CaCO31s2 ∆CaO1s2 CO21g2 Lime is among the top five chemicals manufactured in the United States in terms of the amount produced.
The concentrations of pure liquids and solids are constant.
The Seven Sisters chalk cliffs in East Sussex, England. The chalk is made up of compressed calcium carbonate skeletons of microscopic algae from the late Cretaceous Period.
FIGURE 13.6 The position of the equilibrium CaCO3(s) CaO(s) CO2(g) does not depend on the amounts of CaCO3(s) and CaO(s) present.
∆ CaO CaCO3 CO2 (a) (b) 590 Chapter Thirteen Chemical Equilibrium then because the concentration or pressure of water vapor can change.
Equilibrium Expressions for Heterogeneous Equilibria Write the expressions for K and Kp for the following processes: a. Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chlo-rine gas.
b. Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate.
Solution a. The reaction is The equilibrium expressions are In this case neither the pure solid PCl5 nor the pure liquid PCl3 is included in the equi-librium expressions.
b. The reaction is The equilibrium expressions are The solids are not included.
See Exercise 13.29.
K 3H2O4 5 and Kp 1PH2O25 CuSO4 5H2O1s2 ∆CuSO41s2 5H2O1g2 K 3Cl24 and Kp PCl2 PCl51s2 ∆PCl31l2 Cl21g2 K 3H2423O24 3H2O42 and Kp 1PH2 221PO22 PH2O 2 Sample Exercise 13.6 Hydrated copper(II) sulfate on the left.
Water applied to anhydrous copper(II) sulfate, on the right, forms the hydrated compound.
13.5 Applications of the Equilibrium Constant 591 13.5 Applications of the Equilibrium Constant Knowing the equilibrium constant for a reaction allows us to predict several impor-tant features of the reaction: the tendency of the reaction to occur (but not the speed of the reaction), whether a given set of concentrations represents an equilibrium con-dition, and the equilibrium position that will be achieved from a given set of initial concentrations.
To introduce some of these ideas, we will first consider the reaction where and represent two different types of atoms. Assume that this reaction has an equilibrium constant equal to 16.
In a given experiment, the two types of molecules are mixed together in the follow-ing amounts: After the system reacts and comes to equilibrium, what will the system look like? We know that at equilibrium the ratio must be satisfied, where each N represents the number of molecules of each type. We orig-inally have 9 molecules and 12 molecules. As a place to start, let’s just assume that 5 molecules disappear for the system to reach equilibrium. Since equal numbers of the and molecules react, this means that 5 molecules also will disappear.
This also means that 5 molecules and 5 molecules will be formed. We can sum-marize as follows: Do the new conditions represent equilibrium for this reaction system? We can find out by taking the ratio of the numbers of molecules: Thus this is not an equilibrium position because the ratio is not 16, as required for equilibrium. In which direction must the system move to achieve equilibrium? Since the (N (5)(5) )(N ) (N (4)(7) ) (N ) = = 0.9 Initial Conditions 9 molecules 12 molecules 0 molecules 0 molecules New Conditions 9 − 5 = 4 molecules 12 − 5 = 7 molecules 0 + 5 = 5 molecules 0 + 5 = 5 molecules (N )(N ) (N ) (N ) = 16 + + 592 Chapter Thirteen Chemical Equilibrium observed ratio is smaller than 16, we must increase the numerator and decrease the denominator: The system needs to move to the right (toward more products) to achieve equilibrium. That is, more than 5 of the original reactant molecules must disappear to reach equilibrium for this system. How can we find the correct number? Since we do not know the number of molecules that need to disappear to reach equilibrium, let’s call this number x. Now we can set up a table similar to the one we used earlier: For the system to be at equilibrium, we know that the following ratio must be satisfied: The easiest way to solve for x here is by trial and error. From our previous discussion we know that x is greater than 5. Also, we know that it must be less than 9 because we have only 9 molecules to start. We can’t use all of them or we will have a zero in the de-nominator, which causes the ratio to be infinitely large. By trial and error, we find that because The equilibrium mixture can be pictured as follows: Note that it constains 8 molecules, 8 molecules, 1 molecule, and 4 molecules as required.
This pictorial example should help you understand the fundamental ideas of equilib-rium. Now we will proceed to a more systematic quantitative treatment of chemical equilibrium.
The Extent of a Reaction The inherent tendency for a reaction to occur is indicated by the magnitude of the equi-librium constant. A value of K much larger than 1 means that at equilibrium the reaction 1x21x2 19 x2112 x2 182182 19 82112 82 64 4 16 x 8 (N )(N ) (N ) (N ) (x)(x) (9 −x)(12 −x) = 16 = x disappear x disappear x form x form Initial Conditions 9 molecules 12 molecules 0 molecules 0 molecules Equilibrium Conditions 9 −x molecules 12 −x molecules x molecules x molecules 13.5 Applications of the Equilibrium Constant 593 system will consist of mostly products—the equilibrium lies to the right. Another way of saying this is that reactions with very large equilibrium constants go essentially to com-pletion. On the other hand, a very small value of K means that the system at equilibrium will consist of mostly reactants—the equilibrium position is far to the left. The given re-action does not occur to any significant extent.
It is important to understand that the size of K and the time required to reach equi-librium are not directly related. The time required to achieve equilibrium depends on the reaction rate, which is determined by the size of the activation energy. The size of K is determined by thermodynamic factors such as the difference in energy between products and reactants. This difference is represented in Fig. 13.7 and will be discussed in detail in Chapter 16.
Reaction Quotient When the reactants and products of a given chemical reaction are mixed, it is useful to know whether the mixture is at equilibrium or, if not, the direction in which the system must shift to reach equilibrium. If the concentration of one of the reactants or products is zero, the system will shift in the direction that produces the missing component. How-ever, if all the initial concentrations are nonzero, it is more difficult to determine the di-rection of the move toward equilibrium. To determine the shift in such cases, we use the reaction quotient, Q. The reaction quotient is obtained by applying the law of mass ac-tion using initial concentrations instead of equilibrium concentrations. For example, for the synthesis of ammonia the expression for the reaction quotient is where the subscript zeros indicate initial concentrations.
To determine in which direction a system will shift to reach equilibrium, we compare the values of Q and K. There are three possible cases: 1. Q is equal to K. The system is at equilibrium; no shift will occur.
2. Q is greater than K. In this case, the ratio of initial concentrations of products to ini-tial concentrations of reactants is too large. To reach equilibrium, a net change of products to reactants must occur. The system shifts to the left, consuming products and forming reactants, until equilibrium is achieved.
3. Q is less than K. In this case, the ratio of initial concentrations of products to initial concentrations of reactants is too small. The system must shift to the right, consum-ing reactants and forming products, to attain equilibrium.
Using the Reaction Quotient For the synthesis of ammonia at the equilibrium constant is Predict the direction in which the system will shift to reach equilibrium in each of the following cases: a.
b.
c. [NH3]0 1.0 104 M; [N2]0 5.0 M; [H2]0 1.0 102 M [NH3]0 2.00 104 M; [N2]0 1.50 105 M; [H2]0 3.54 101 M [NH3]0 1.0 103 M; [N2]0 1.0 105 M; [H2]0 2.0 103 M 6.0 102.
500°C, Q 3NH340 2 3N2403H240 3 N21g2 3H21g2 ∆2NH31g2 Ea ∆E H2O H2,O2 (b) (a) A H B FIGURE 13.7 (a) A physical analogy illustrating the differ-ence between thermodynamic and kinetic stabilities. The boulder is thermodynami-cally more stable (lower potential energy) in position B than in position A but cannot get over the hump H. (b) The reactants and have a strong tendency to form That is, has lower energy than and However, the large activation energy Ea prevents the reaction at In other words, the magnitude of K for the reaction depends on but the reaction rate depends on Ea.
¢E, 25°C.
O2.
H2 H2O H2O.
O2 H2 Sample Exercise 13.7 594 Chapter Thirteen Chemical Equilibrium Solution a. First we calculate the value of Q: Since K Q is much greater than K. To attain equilibrium, the concen-trations of the products must be decreased and the concentrations of the reactants in-creased. The system will shift to the left: b. We calculate the value of Q: In this case so the system is at equilibrium. No shift will occur.
c. The value of Q is Here Q is less than K, so the system will shift to the right to attain equilibrium by in-creasing the concentration of the product and decreasing the reactant concentrations: See Exercises 13.33 through 13.36.
Calculating Equilibrium Pressures and Concentrations A typical equilibrium problem involves finding the equilibrium concentrations (or pres-sures) of reactants and products, given the value of the equilibrium constant and the initial concentrations (or pressures). However, since such problems sometimes become complicated mathematically, we will develop useful strategies for solving them by con-sidering cases for which we know one or more of the equilibrium concentrations (or pressures).
Calculating Equilibrium Pressures I Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions. In the gas phase it decomposes to gaseous nitrogen dioxide: Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2(g).
Kp 0.133.
N2O41g2 ∆2NO21g2 N2 3H2 ¡ 2NH3 2.0 103 Q 3NH340 2 3N2403H240 3 11.0 10422 15.0211.0 10223 Q K, 6.01 102 Q 3NH340 2 3N2403H240 3 12.00 10422 11.50 105213.54 10123 N2 3H2 — 2NH3 6.0 102, 1.3 107 Q 3NH340 2 3N2403H240 3 11.0 10322 11.0 105212.0 10323 Sample Exercise 13.8 13.5 Applications of the Equilibrium Constant 595 Solution We know that the equilibrium pressures of the gases NO2 and N2O4 must satisfy the relationship Since we know we can simply solve for Therefore, See Exercises 13.37 and 13.38.
Calculating Equilibrium Pressures II At a certain temperature a 1.00-L flask initially contained 0.298 mol PCl3(g) and mol PCl5(g). After the system had reached equilibrium, mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the reaction Calculate the equilibrium concentrations of all species and the value of K.
Solution The equilibrium expression for this reaction is To find the value of K, we must calculate the equilibrium concentrations of all species and then substitute these quantities into the equilibrium expression. The best method for finding the equilibrium concentrations is to begin with the initial concentrations, which we will define as the concentrations before any shift toward equilibrium has occurred.
We will then modify these initial concentrations appropriately to find the equilibrium concentrations.
The initial concentrations are Next we find the change required to reach equilibrium. Since no Cl2 was initially pres-ent but Cl2 is present at equilibrium, mol PCl5 must have 2.00 103 2.00 103 M 3PCl54 0 8.70 103 mol 1.00 L 8.70 103 M 3PCl34 0 0.298 mol 1.00 L 0.298 M 3Cl24 0 0 K 3Cl24 3PCl34 3PCl54 + PCl51g2 ∆PCl31g2 Cl21g2 2.00 103 8.70 103 PNO2 20.360 0.600 PNO2 2 Kp1PN2O42 10.133212.712 0.360 PNO2: PN2O4, Kp PNO2 2 PN2O4 0.133 Apollo II lunar landing module at Tranquility Base, 1969.
Sample Exercise 13.9 596 Chapter Thirteen Chemical Equilibrium Sample Exercise 13.10 + + decomposed to form mol Cl2 and mol PCl3. In other words, to reach equilibrium, the reaction shifted to the right: h r p Net amount of PCl5 Net amounts of decomposed products formed Now we apply this change to the initial concentrations to obtain the equilibrium concentrations: These equilibrium concentrations can now be used to find K: See Exercises 13.39 through 13.42.
Sometimes we are not given any of the equilibrium concentrations (or pressures), only the initial values. Then we must use the stoichiometry of the reaction to express concen-trations (or pressures) at equilibrium in terms of the initial values. This is illustrated in Sample Exercise 13.10.
Calculating Equilibrium Concentrations I Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000-L flask.
Solution The balanced equation for the reaction is and Next we calculate the initial concentrations: 3CO40 3H2O4 0 3CO24 0 3H24 0 1.000 mol 1.000 L 1.000 M K 3CO24 3H24 3CO4 3H2O4 5.10 CO1g2 H2O1g2 ∆CO21g2 H21g2 8.96 102 K 3Cl24 3PCl34 3PCl54 12.00 103210.3002 6.70 103 3PCl54 0 3PCl54 8.70 103 M 2.00 103 mol 1.00 L 6.70 103 M 3PCl34 0 3PCl34 0.298 M 2.00 103 mol 1.00 L 0.300 M 3Cl24 0 3Cl24 0 2.00 103 mol 1.00 L 2.00 103 M 2.00 103 mol ¡ 2.00 103 mol 2.00 103 mol PCl51g2 ¡ PCl31g2 Cl21g2 2.00 103 2.00 103 13.5 Applications of the Equilibrium Constant 597 Is the system at equilibrium, and if not, which way will it shift to reach the equilibrium position? These questions can be answered by calculating Q: Since Q is less than K, the system is not at equilibrium initially but must shift to the right.
What are the equilibrium concentrations? As before, we start with the initial con-centrations and modify them to obtain the equilibrium concentrations. We must ask this question: How much will the system shift to the right to attain the equilibrium condition?
In Sample Exercise 13.9 the change needed for the system to reach equilibrium was given.
However, in this case we do not have this information.
Since the required change in concentrations is unknown at this point, we will define it in terms of x. Let’s assume that x mol/L CO must react for the system to reach equi-librium. This means that the initial concentration of CO will decrease by x mol/L: x h h h Equilibrium Initial Change Since each CO molecule reacts with one molecule, the concentration of water vapor also must decrease by x mol/L: As the reactant concentrations decrease, the product concentrations increase. Since all the coefficients are 1 in the balanced reaction, 1 mol CO reacting with 1 mol will pro-duce 1 mol and 1 mol . Or in the present case, to reach equilibrium, x mol/L CO will react with x mol/L to give an additional x mol/L and x mol/L : Thus the initial concentrations of and will increase by x mol/L: Now we have all the equilibrium concentrations defined in terms of the initial concentra-tions and the change x: 3H24 3H24 0 x 3CO24 3CO24 0 x H2 CO2 xCO xH2O ¡ xCO2 xH2 H2 CO2 H2O H2 CO2 H2O 3H2O4 3H2O4 0 x H2O 3CO40 3CO4 Q 3CO24 03H24 0 3CO4 03H2O4 0 11.000 mol/L211.000 mol/L2 11.000 mol/L211.000 mol/L2 1.000 Initial Equilibrium Concentration (mol/L) Change (mol/L) Concentration (mol/L) 1.000 x x 3H24 0 1.000 1.000 x x 3CO24 0 1.000 1.000 x x 3H2O4 0 1.000 1.000 x x 3CO4 0 1.000 Note that the sign of x is determined by the direction of the shift. In this example, the sys-tem shifts to the right, so the product concentrations increase and the reactant concentra-tions decrease. Also note that because the coefficients in the balanced equation are all 1, the magnitude of the change is the same for all species.
Now since we know that the equilibrium concentrations must satisfy the equilibrium expression, we can find the value of x by substituting these concentrations into the expression K 5.10 3CO24 3H24 3CO4 3H2O4 11.000 x211.000 x2 11.000 x211.000 x2 11.000 x22 11.000 x22 598 Chapter Thirteen Chemical Equilibrium Since the right side of the equation is a perfect square, the solution of the problem can be simplified by taking the square root of both sides: Multiplying and collecting terms gives Thus the system shifts to the right, consuming 0.387 mol/L CO and 0.387 mol/L and forming 0.387 mol/L and 0.387 mol/L .
Now the equilibrium concentrations can be calculated: Reality Check: These values can be checked by substituting them back into the equilibrium expression to make sure they give the correct value for K: This result is the same as the given value of K (5.10) within round-off error, so the answer must be correct.
See Exercise 13.45.
Calculating Equilibrium Concentrations II Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of at a certain temperature. In a par-ticular experiment, 3.000 mol of each component was added to a 1.500-L flask. Calculate the equilibrium concentrations of all species.
Solution The balanced equation for the reaction is The equilibrium expression is We first calculate the initial concentrations: Then we find the value of Q: Since Q is much less than K, the system must shift to the right to reach equilibrium.
Q 3HF40 2 3H2403F240 12.00022 12.000212.0002 1.000 3HF4 0 3H24 0 3F24 0 3.000 mol 1.500 L 2.000 M K 1.15 102 3HF4 2 3H24 3F24 H21g2 F21g2 ∆2HF1g2 1.15 102 K 3CO24 3H24 3CO4 3H2O4 11.38722 10.61322 5.12 3CO24 3H24 1.000 x 1.000 0.387 1.387 M 3CO4 3H2O4 1.000 x 1.000 0.387 0.613 M H2 CO2 H2O x 0.387 mol/L 25.10 2.26 1.000 x 1.000 x Sample Exercise 13.11 13.5 Applications of the Equilibrium Constant 599 What change in the concentrations is necessary? Since this is presently unknown, we will define the change needed in terms of x. Let x equal the number of moles per liter of consumed to reach equilibrium. The stoichiometry of the reaction shows that x mol/L also will be consumed and 2x mol/L HF will be formed: Now the equilibrium concentrations can be expressed in terms of x: x mol/L x mol/L ¡ 2x mol/L H21g2 F21g2 ¡ 2HF1g2 F2 H2 These concentrations can be represented in a shorthand table as follows: Initial Equilibrium Concentration (mol/L) Change (mol/L) Concentration (mol/L) 3HF4 2.000 2x 2x 3HF4 0 2.000 3F24 2.000 x x 3F24 0 2.000 3H24 2.000 x x 3H24 0 2.000 H2(g) F2(g) ∆ 2HF(g) Initial: 2.000 2.000 2.000 Change: x x 2x Equilibrium: 2.000 x 2.000 x 2.000 2x To solve for x, we substitute the equilibrium concentrations into the equilibrium expression: The right side of this equation is a perfect square, so taking the square root of both sides gives which yields The equilibrium concentrations can now be calculated: Reality Check: Checking these values by substituting them into the equilibrium expression gives which agrees with the given value of K.
See Exercise 13.46.
3HF42 3H24 3F24 15.05622 10.47222 1.15 102 3HF4 2.000 M 2x 5.056 M 3H24 3F24 2.000 M x 0.472 M x 1.528.
21.15 102 2.000 2x 2.000 x K 1.15 102 3HF42 3H24 3F24 12.000 2x22 12.000 x22 We often refer to this form as an ICE table (indicated by the first letters of Initial, Change, and Equilibrium).
600 Chapter Thirteen Chemical Equilibrium 13.6 Solving Equilibrium Problems We have already considered most of the strategies needed to solve equilibrium problems.
The typical procedure for analyzing a chemical equilibrium problem can be summarized as follows: Procedure for Solving Equilibrium Problems ➥ 1 Write the balanced equation for the reaction.
➥ 2 Write the equilibrium expression using the law of mass action.
➥ 3 List the initial concentrations.
➥ 4 Calculate Q, and determine the direction of the shift to equilibrium.
➥ 5 Define the change needed to reach equilibrium, and define the equilibrium con-centrations by applying the change to the initial concentrations.
➥ 6 Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown.
➥ 7 Check your calculated equilibrium concentrations by making sure they give the correct value of K.
So far we have been careful to choose systems in which we can solve for the un-known by taking the square root of both sides of the equation. However, this type of sys-tem is not really very common, and we must now consider a more typical problem. Suppose for a synthesis of hydrogen fluoride from hydrogen and fluorine, 3.000 mol and 6.000 mol are mixed in a 3.000-L flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is . We calculate the equilibrium concentration of each component as follows: ➥ 1 We begin, as usual, by writing the balanced equation for the reaction: ➥ 2 The equilibrium expression is ➥ 3 The initial concentrations are ➥ 4 There is no need to calculate Q because no HF is present initially, and we know that the system must shift to the right to reach equilibrium.
➥ 5 If we let x represent the number of moles per liter of consumed to reach equilib-rium, we can represent the equilibrium concentrations as follows: H2 3HF4 0 0 3F24 0 6.000 mol 3.000 L 2.000 M 3H24 0 3.000 mol 3.000 L 1.000 M K 1.15 102 3HF4 2 3H24 3F24 H21g2 F21g2 ∆2HF1g2 1.15 102 F2 H2 H2(g) F2(g) 2HF(g) Initial: 1.000 2.000 0 Change: x x 2x Equilibrium: 1.000 x 2.000 x 2x ∆ 13.6 Solving Equilibrium Problems 601 ➥ 6 Substituting the equilibrium concentrations into the equilibrium expression gives Since the right side of this equation is not a perfect square, we cannot take the square root of both sides, but must use some other procedure.
First, do the indicated multiplication: or and collect terms This is a quadratic equation of the general form where the roots can be obtained from the quadratic formula: In this example, , , and . Substituting these values into the quadratic formula gives two values for x: Both these results cannot be valid (since a given set of initial concentrations leads to only one equilibrium position). How can we choose between them? Since the expression for the equilibrium concentration of is the value of x cannot be 2.14 mol/L (because subtracting 2.14 M from 1.000 M gives a negative concentration of , which is physically impossible). Thus the correct value for x is 0.968 mol/L, and the equilibrium concentrations are as follows: Reality Check: ➥ 7 We can check these concentrations by substituting them into the equilibrium expression: This value is in close agreement with the given value for K ( ), so the calcu-lated equilibrium concentrations are correct.
This procedure is further illustrated for a problem involving pressures in Sample Exercise 13.12.
Calculating Equilibrium Pressures Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is . Suppose HI at atm, at atm, and at atm are mixed in a 5.000-L flask.
Calculate the equilibrium pressures of all species.
5.000 103 I2 1.000 102 H2 5.000 101 1.00 102 1.15 102 3HF4 2 3H24 3F24 11.93622 13.2 102211.0322 1.13 102 3HF4 210.968 M2 1.936 M 3F24 2.000 M 0.968 M 1.032 M 3H24 1.000 M 0.968 M 3.2 102 M H2 3H24 1.000 M x H2 x 2.14 mol/L and x 0.968 mol/L c 2.30 102 b 3.45 102 a 1.11 102 x b 2b2 4ac 2a ax2 bx c 0 11.11 1022x2 13.45 1022x 2.30 102 0 11.15 1022x2 3.00011.15 1022x 2.00011.15 1022 4x2 11.000 x212.000 x211.15 1022 12x22 K 1.15 102 3HF42 3H24 3F24 12x22 11.000 x212.000 x2 Use of the quadratic formula is explained in Appendix 1.4.
Sample Exercise 13.12 Substitution into the equilibrium expression gives Multiplying and collecting terms yield the quadratic equation where , and : From the quadratic formula, the correct value for x is atm. The equi-librium pressures can now be calculated from the expressions involving x: PI2 5.000 103 atm 3.55 102 atm 4.05 102 atm PH2 1.000 102 atm 3.55 102 atm 4.55 102 atm PHI 5.000 101 atm 213.55 1022 atm 4.29 101 atm x 3.55 102 19.60 1012x2 3.5x 12.45 1012 0 c 2.45 101 b 3.5, a 9.60 101 Kp 1PHI22 1PH221PI22 15.000 101 2x22 11.000 102 x215.000 103 x2 602 Chapter Thirteen Chemical Equilibrium Solution The balanced equation for this process is and the equilibrium expression in terms of pressure is The given initial pressures are The value of Q for this system is Since Q is greater than K, the system will shift to the left to reach equilibrium.
So far we have used moles or concentrations in stoichiometric calculations. How-ever, it is equally valid to use pressures for a gas-phase system at constant tempera-ture and volume because in this case pressure is directly proportional to the number of moles: — Constant if constant T and V Thus we can represent the change needed to achieve equilibrium in terms of pressures.
Let x be the change in pressure (in atm) of as the system shifts left toward equi-librium. This leads to the following equilibrium pressures: H2 P naRT V b Q 1PHI 022 1PH2 021PI2 02 15.000 101 atm22 11.000 102 atm215.000 103 atm2 5.000 103 PI2 0 5.000 103 atm PH2 0 1.000 102 atm PHI 0 5.000 101 atm Kp PHI 2 1PH221PI22 1.00 102 H21g2 I21g2 ∆2HI1g2 H2(g) I2(g) 2HI(g) Initial: 1.000 102 5.000 103 5.000 101 Change: x x 2x Equilibrium: 1.000 102 x 5.000 103 x 5.000 101 2x ∆ 13.6 Solving Equilibrium Problems 603 The equilibrium concentrations must satisfy the equilibrium expression Multiplying and collecting terms will give an equation with terms containing , , and x, which requires complicated methods to solve directly. However, we can avoid this sit-uation by recognizing that since K is so small ( ), the system will not proceed far to the right to reach equilibrium. That is, x represents a relatively small number. The consequence of this fact is that the term can be approximated by 0.50. That is, when x is small, 0.50 2x 0.50 10.50 2x2 1.6 105 x2 x3 K 1.6 105 3NO4 23Cl24 3NOCl4 2 12x221x2 10.50 2x22 Reality Check: This agrees with the given value of K ), so the calculated equilibrium con-centrations are correct.
See Exercises 13.47 through 13.50.
Treating Systems That Have Small Equilibrium Constants We have seen that fairly complicated calculations are often necessary to solve equilibrium problems. However, under certain conditions, simplifications are possible that greatly re-duce the mathematical difficulties. For example, gaseous NOCl decomposes to form the gases NO and . At the equilibrium constant is . In an experiment in which 1.0 mol NOCl is placed in a 2.0-L flask, what are the equilibrium concentrations?
The balanced equation is and The initial concentrations are Since there are no products initially, the system will move to the right to reach equi-librium. We will define x as the change in concentration of needed to reach equilib-rium. The changes in the concentrations of NOCl and NO can then be obtained from the balanced equation: The concentrations can be summarized as follows: 2x ¡ 2x x 2NOCl1g2 ¡ 2NO1g2 Cl21g2 Cl2 3NOCl4 0 1.0 mol 2.0 L 0.50 M 3NO40 0 3Cl24 0 0 K 3NO423Cl24 3NOCl4 2 1.6 105 2NOCl1g2 ∆2NO1g2 Cl21g2 1.6 105 35°C Cl2 11.00 102 PHI 2 PH2 PI2 14.29 10122 14.55 102214.05 1022 99.9 2NOCl(g) 2NO(g) Cl2(g) Initial: 0.50 0 0 Change: 2x 2x x Equilibrium: 0.50 2x 2x x ∆ 604 Chapter Thirteen Chemical Equilibrium Making this approximation allows us to simplify the equilibrium expression: Solving for gives and .
How valid is this approximation? If , then The difference between 0.50 and 0.48 is 0.02, or 4% of the initial concentration of NOCl, a relatively small discrepancy that will have little effect on the outcome. That is, since 2x is very small compared with 0.50, the value of x obtained in the approximate solution should be very close to the exact value. We use this approximate value of x to calculate the equilibrium concentrations: Reality Check: Since the given value of K is , these calculated concentrations are correct.
This problem was much easier to solve than it appeared at first because the small value of K and the resulting small shift to the right to reach equilibrium allowed simplification.
13.7 Le Châtelier’s Principle It is important to understand the factors that control the position of a chemical equilib-rium. For example, when a chemical is manufactured, the chemists and chemical engi-neers in charge of production want to choose conditions that favor the desired product as much as possible. That is, they want the equilibrium to lie far to the right. When Fritz Haber was developing the process for the synthesis of ammonia, he did extensive studies on how temperature and pressure affect the equilibrium concentration of ammonia. Some of his results are given in Table 13.2. Note that the equilibrium amount of increases NH3 1.6 105 3NO423Cl24 3NOCl4 2 12.0 1022211.0 1022 10.5022 1.6 105 3Cl24 x 1.0 102 M 3NO4 2x 211.0 102 M2 2.0 102 M 3NOCl4 0.50 2x 0.50 M 0.50 2x 0.50 211.0 1022 0.48 x 1.0 102 x 1.0 102 x3 11.6 105210.5022 4 1.0 106 x3 1.6 105 12x221x2 10.50 2x22 12x221x2 10.5022 4x3 10.5022 Approximations can simplify complicated math, but their validity should be checked carefully.
TABLE 13.2 The Percent by Mass of at Equilibrium in a Mixture of as a Function of Temperature and Total Pressure Total Pressure Temperature (°C) 300 atm 400 atm 500 atm 400 48% 55% 61% 500 26% 32% 38% 600 13% 17% 21% Each experiment was begun with a 3:1 mixture of and N2.
H2 NH3 NH3 NH3 NH3 NH3 NH3 NH3 NH3 NH3 N2, H2, and NH3 NH3 13.7 Le Châtelier’s Principle 605 with an increase in pressure but decreases as the temperature is increased. Thus the amount of present at equilibrium is favored by conditions of low temperature and high pressure.
However, this is not the whole story. Carrying out the process at low temperatures is not feasible because then the reaction is too slow. Even though the equilibrium tends to shift to the right as the temperature is lowered, the attainment of equilibrium would be much too slow at low temperatures to be practical. This emphasizes once again that we must study both the thermodynamics and the kinetics of a reaction before we really understand the factors that control it.
We can qualitatively predict the effects of changes in concentration, pressure, and temperature on a system at equilibrium by using Le Châtelier’s principle, which states that if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Although this rule sometimes over-simplifies the situation, it works remarkably well.
The Effect of a Change in Concentration To see how we can predict the effect of change in concentration on a system at equilib-rium, we will consider the ammonia synthesis reaction. Suppose there is an equilibrium position described by these concentrations: What will happen if 1.000 mol/L is suddenly injected into the system? We can answer this question by calculating the value of Q. The concentrations before the system adjusts are h Added N2 Note we are labeling these as “initial concentrations” because the system is no longer at equilibrium. Then Since we are not given the value of K, we must calculate it from the first set of equilib-rium concentrations: As expected, Q is less than K because the concentration of was increased.
The system will shift to the right to come to the new equilibrium position. Rather than do the calculations, we simply summarize the results: N2 K 3NH34 2 3N24 3H24 3 10.20222 10.399211.19723 5.96 102 Q 3NH340 2 3N2403H240 3 10.20222 11.399211.19723 1.70 102 3NH34 0 0.202 M 3H24 0 1.197 M 3N24 0 0.399 M 1.000 M 1.399 M N2 3N24 0.399 M 3H24 1.197 M 3NH34 0.202 M NH3 Visualization: Le Châtelier’s Principle Equilibrium Position I Equilibrium Position II [N2] 0.399 M [N2] 1.348 M [H2] 1.197 M [H2] 1.044 M [NH3] 0.202 M [NH3] 0.304 M 1.000 mol/L of N2 added 606 Chapter Thirteen Chemical Equilibrium Note from these data that the equilibrium position does in fact shift to the right: The concentration of decreases, the concentration of increases, and of course, since nitrogen is added, the concentration of shows an increase relative to the amount present in the original equilibrium position. (However, notice that the nitrogen showed a decrease relative to the amount present immediately after addition of the 1.000 mol .) We can understand this shift by thinking about reaction rates. When we add mol-ecules to the system, the number of collisions between and will increase, thus in-creasing the rate of the forward reaction and in turn increasing the rate of formation of molecules. More molecules will in turn lead to a higher rate for the reverse re-action. Eventually, the forward and reverse reaction rates will again become equal, and the system will reach its new equilibrium position.
We can predict this shift qualitatively by using Le Châtelier’s principle. Since the change imposed is the addition of nitrogen, Le Châtelier’s principle predicts that the sys-tem will shift in a direction that consumes nitrogen. This reduces the effect of the addi-tion. Thus Le Châtelier’s principle correctly predicts that adding nitrogen will cause the equilibrium to shift to the right (see Fig. 13.8).
If ammonia had been added instead of nitrogen, the system would have shifted to the left to consume ammonia. So another way of stating Le Châtelier’s principle is to say that if a component (reactant or product) is added to a reaction system at equilibrium (at con-stant T and P or constant T and V), the equilibrium position will shift in the direction that lowers the concentration of that component. If a component is removed, the opposite effect occurs.
Using Le Châtelier’s Principle I Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid , which is then reduced using carbon: Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions.
a. Addition of carbon monoxide b. Addition or removal of carbon or tetraarsenic hexoxide c. Removal of gaseous arsenic (As4) (As4O6) As4O61s2 6C1s2 ∆As41g2 6CO1g2 As4O6 NH3 NH3 H2 N2 N2 N2 N2 NH3 H2 The system shifts in the direction that compensates for the imposed change.
N2 added (a) (b) (c) N2 H2 NH3 FIGURE 13.8 (a) The initial equilibrium mixture of and (b) Addition of (c) The new equilibrium position for the system containing more (due to addition of ), less and more than in (a).
NH3 H2, N2 N2 N2.
NH3.
H2, N2, Sample Exercise 13.13 13.7 Le Châtelier’s Principle 607 Solution a. Le Châtelier’s principle predicts that the shift will be away from the substance whose concentration is increased. The equilibrium position will shift to the left when carbon monoxide is added.
b. Since the amount of a pure solid has no effect on the equilibrium position, changing the amount of carbon or tetraarsenic hexoxide will have no effect.
c. If gaseous arsenic is removed, the equilibrium position will shift to the right to form more products. In industrial processes, the desired product is often continuously removed from the reaction system to increase the yield.
See Exercise 13.57.
The Effect of a Change in Pressure Basically, there are three ways to change the pressure of a reaction system involving gaseous components: 1. Add or remove a gaseous reactant or product.
2. Add an inert gas (one not involved in the reaction).
3. Change the volume of the container.
We have already considered the addition or removal of a reactant or product. When an inert gas is added, there is no effect on the equilibrium position. The addition of an inert gas increases the total pressure but has no effect on the concentrations or partial pres-sures of the reactants or products. That is, in this case the added molecules do not participate in the reaction in any way and thus cannot affect the equilibrium in any way.
Thus the system remains at the original equilibrium position.
When the volume of the container is changed, the concentrations (and thus the par-tial pressures) of both reactants and products are changed. We could calculate Q and pre-dict the direction of the shift. However, for systems involving gaseous components, there is an easier way: We focus on the volume. The central idea is that when the volume of the Visualization: Equilibrium Decomposition of N2O4 (a) Brown NO2( g) and colorless N2O4( g) in equilibrium in a syringe. (b) The vol-ume is suddenly decreased, giving a greater concentration of both N2O4 and NO2 (indicated by the darker brown color). (c) A few seconds after the sudden volume decrease, the color is much lighter brown as the equilibrium shifts the brown NO2( g) to colorless N2O4( g) as predicted by Le Châtelier’s principle, since in the equilibrium 2NO2( g) ∆ N2O4( g) the product side has the smaller number of molecules.
(a) (b) (c) 608 Chapter Thirteen Chemical Equilibrium container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system.
To see that this is true, we can rearrange the ideal gas law to give or at constant T and P, That is, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas present.
Suppose we have a mixture of the gases nitrogen, hydrogen, and ammonia at equi-librium (Fig. 13.9). If we suddenly reduce the volume, what will happen to the equilib-rium position? The reaction system can reduce its volume by reducing the number of mol-ecules present. This means that the reaction will shift to the right, since in this direction four molecules (one of nitrogen and three of hydrogen) react to produce two molecules (of ammonia), thus reducing the total number of gaseous molecules present. The new equilibrium position will be farther to the right than the original one. That is, the equilibrium position will shift toward the side of the reaction involving the smaller number of gaseous molecules in the balanced equation.
The opposite is also true. When the container volume is increased, the system will shift so as to increase its volume. An increase in volume in the ammonia synthesis system will produce a shift to the left to increase the total number of gaseous molecules present.
Using Le Châtelier’s Principle II Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.
a. The preparation of liquid phosphorus trichloride by the reaction b. The preparation of gaseous phosphorus pentachloride according to the equation c. The reaction of phosphorus trichloride with ammonia: Solution a. Since and are a pure solid and a pure liquid, respectively, we need to consider only the effect of the change in volume on The volume is decreased, so the position Cl2.
PCl3 P4 PCl31g2 3NH31g2 ∆P1NH2231g2 3HCl1g2 PCl31g2 Cl21g2 ∆PCl51g2 P41s2 6Cl21g2 ∆4PCl31l2 N21g2 3H21g2 ∆2NH31g2 V r n V aRT P bn FIGURE 13.9 (a) A mixture of and at equilibrium. (b) The volume is suddenly decreased. (c) The new equilibrium position for the system containing more and less and The reaction shifts to the right (toward the side with fewer molecules) when the container volume is decreased.
3H2(g) ∆2NH3(g) N2(g) H2.
N2 NH3 H2(g) NH3(g), N21g2, (a) N2 H2 NH3 (b) (c) Sample Exercise 13.14 13.7 Le Châtelier’s Principle 609 of the equilibrium will shift to the right, since the reactant side contains six gaseous molecules and the product side has none.
b. Decreasing the volume will shift the given reaction to the right, since the product side contains only one gaseous molecule while the reactant side has two.
c. Both sides of the balanced reaction equation have four gaseous molecules. A change in volume will have no effect on the equilibrium position. There is no shift in this case.
See Exercise 13.58.
The Effect of a Change in Temperature It is important to realize that although the changes we have just discussed may alter the equilibrium position, they do not alter the equilibrium constant. For example, the addi-tion of a reactant shifts the equilibrium position to the right but has no effect on the value of the equilibrium constant; the new equilibrium concentrations satisfy the original equi-librium constant.
The effect of temperature on equilibrium is different, however, because the value of K changes with temperature. We can use Le Châtelier’s principle to predict the direction of the change.
The synthesis of ammonia from nitrogen and hydrogen is exothermic. We can repre-sent this by treating energy as a product: If energy is added to this system at equilibrium by heating it, Le Châtelier’s principle pre-dicts that the shift will be in the direction that consumes energy, that is, to the left. Note that this shift decreases the concentration of and increases the concentrations of N2 NH3 N21g2 3H21g2 ∆2NH31g2 92 kJ Shifting the N2O4( g) n 2NO2( g) equilibrium by changing the temperature. (a) At 100C the flask is definitely red-dish brown due to a large amount of NO2 present. (b) At 0C the equilibrium is shifted toward colorless N2O4( g).
Of course, energy is not a chemical product of this reaction, but thinking of it in this way makes it easy to apply Le Châtelier’s principle.
610 Chapter Thirteen Chemical Equilibrium and thus decreasing the value of K. The experimentally observed change in K with temperature for this reaction is indicated in Table 13.3. The value of K decreases with increased temperature, as predicted.
On the other hand, for an endothermic reaction, such as the decomposition of calcium carbonate, an increase in temperature will cause the equilibrium to shift to the right and the value of K to increase.
In summary, to use Le Châtelier’s principle to describe the effect of a temperature change on a system at equilibrium, treat energy as a reactant (in an endothermic process) or as a product (in an exothermic process), and predict the direction of the shift in the same way as when an actual reactant or product is added or removed. Although Le Châtelier’s principle cannot predict the size of the change in K, it does correctly predict the direction of the change.
Using Le Châtelier’s Principle III For each of the following reactions, predict how the value of K changes as the tempera-ture is increased.
a.
b.
Solution a. This is an endothermic reaction, as indicated by the positive value for Energy can be viewed as a reactant, and K increases (the equilibrium shifts to the right) as the temperature is increased.
b. This is an exothermic reaction (energy can be regarded as a product). As the temper-ature is increased, the value of K decreases (the equilibrium shifts to the left).
See Exercises 13.63 and 13.64.
We have seen how Le Châtelier’s principle can be used to predict the effect of several types of changes on a system at equilibrium. To summarize these ideas, Table 13.4 shows how various changes affect the equilibrium position of the endothermic reaction N2O41g2 ∆2NO21g2 ¢H° 58 kJ ¢H°.
2SO21g2 O21g2 ∆2SO31g2 ¢H° 198 kJ N21g2 O21g2 ∆2NO1g2 ¢H° 181 kJ 556 kJ CaCO31s2 ∆CaO1s2 CO21g2 H2, Sample Exercise 13.15 TABLE 13.3 Observed Value of K for the Ammonia Synthesis Reaction as a Function of Temperature Temperature (K) K 500 90 600 3 700 0.3 800 0.04 For this exothermic reaction, the value of K decreases as the temperature increases, as predicted by Le Châtelier’s principle.
TABLE 13.4 Shifts in the Equilibrium Position for the Reaction 58 kJ N2O4(g) 2NO2(g) Change Shift Addition of Right Addition of Left Removal of Left Removal of Right Addition of None Decrease container Left volume Increase container Right volume Increase temperature Right Decrease temperature Left He1g2 NO21g2 N2O41g2 NO21g2 N2O41g2 ∆ For Review Chemical equilibrium When a reaction takes place in a closed system, it reaches a condition where the concentrations of the reactants and products remain constant over time Dynamic state: reactants and products are interconverted continually • Forward rate reverse rate The law of mass action: for the reaction K 3C4m3D4 n 3A4 j3B4k equilibrium constant jA kB ∆mC nD Key Terms chemical equilibrium Section 13.2 law of mass action equilibrium expression equilibrium constant equilibrium position Section 13.4 homogeneous equilibria heterogeneous equilibria Section 13.5 reaction quotient, Q Section 13.7 Le Châtelier’s principle For Review 611 • A pure liquid or solid is never included in the equilibrium expression • For a gas-phase reaction the reactants and products can be described in terms of their partial pressures and the equilibrium constant is called where is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants Equilibrium position A set of reactant and product concentrations that satisfies the equilibrium constant expression • There is one value of K for a given system at a given temperature • There are an infinite number of equilibrium positions at a given temperature de-pending on the initial concentrations A small value of K means the equilibrium lies to the left; a large value of K means the equilibrium lies to the right • The size of K has no relationship to the speed at which equilibrium is achieved Q, the reaction quotient, applies the law of mass action to initial concentrations rather than equilibrium concentrations • If the system will shift to the left to achieve equilibrium • If the system will shift to the right to achieve equilibrium Finding the concentrations that characterize a given equilibrium position: 1. Start with the given initial concentrations (pressures) 2. Define the change needed to reach equilibrium 3. Apply the change to the initial concentrations (pressures) and solve for the equilibrium concentrations (pressures) Le Châtelier’s principle Enables qualitative prediction of the effects of changes in concentration, pressure, and temperature on a system at equilibrium If a change in conditions is imposed on a system at equilibrium, the system will shift in a direction that compensates for the imposed change • In other words, when a stress is placed on a system at equilibrium, the system shifts in the direction that relieves the stress REVIEW QUESTIONS 1. Characterize a system at chemical equilibrium with respect to each of the following.
a. the rates of the forward and reverse reactions b. the overall composition of the reaction mixture For a general reaction if one starts an experiment with only reactants present, show what the plot of concentrations of A, B, and C versus time would look like. Also sketch the plot illustrating the rate of the forward reaction and rate of the reverse reaction versus time.
2. What is the law of mass action? Is it true that the value of K depends on the amounts of reactants and products mixed together initially? Explain. Is it true that reactions with large equilibrium constant values are very fast? Explain. There is only one value of the equilibrium constant for a particular system at a particular temperature, but there is an infinite number of equilibrium positions. Explain.
3. Consider the following reactions at some temperature: 2NO1g2 ∆N21g2 O21g2 K 1 1031 2NOCl1g2 ∆2NO1g2 Cl21g2 K 1.6 105 3A(g) B(g) ¡ 2C(g), Q 6 K, Q 7 K, ¢n Kp K1RT2¢n Kp: 612 Chapter Thirteen Chemical Equilibrium For each reaction, assume some quantities of the reactants were placed in sepa-rate containers and allowed to come to equilibrium. Describe the relative amounts of reactants and products that would be present at equilibrium. At equilibrium, which is faster, the forward or reverse reaction in each case?
4. What is the difference between K and ? When does for a reaction?
When does for a reaction? If the coefficients in a reaction equation are tripled, how is the new value of K related to the initial value of K? If a reaction is reversed, how is the value of for the reversed reaction related to the value of for the initial reaction?
5. What are homogeneous equilibria? Heterogeneous equilibria? What is the difference in writing K expressions for homogeneous versus heterogeneous re-actions? Summarize which species are included in the K expression and which species are not included.
6. Distinguish between the terms equilibrium constant and reaction quotient.
When what does this say about a reaction? When what does this say about a reaction? When what does this say about a reaction?
7. Summarize the steps for solving equilibrium problems (see the beginning of Section 13.6). In general, when solving an equilibrium problem, you should always set up an ICE table. What is an ICE table?
8. A common type of reaction we will study is that having a very small K value Solving for equilibrium concentrations in an equilibrium problem usu-ally requires many mathematical operations to be performed. However, the math involved when solving equilibrium problems for reactions having small K values is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the “5% rule” is used to check the validity of assuming x (or 2x, 3x, and so on) is very small compared to some number. When x (or 2x, 3x, and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
9. What is Le Châtelier’s principle? Consider the reaction 2NOCl(g) If this reaction is at equilibrium, what happens when the following changes occur?
a. NOCl(g) is added.
b. NO(g) is added.
c. NOCl(g) is removed.
d. Cl2(g) is removed.
e. The container volume is decreased.
For each of these changes, what happens to the value of K for the reaction as equilibrium is reached again? Give an example of a reaction for which the addition or removal of one of the reactants or products has no effect on the equilibrium position.
In general, how will the equilibrium position of a gas-phase reaction be affected if the volume of the reaction vessel changes? Are there reactions that will not have their equilibria shifted by a change in volume? Explain. Why does changing the pressure in a rigid container by adding an inert gas not shift the equilibrium position for a gas-phase reaction?
10. The only “stress” (change) that also changes the value of K is a change in tem-perature. For an exothermic reaction, how does the equilibrium position change as temperature increases, and what happens to the value of K? Answer the same questions for an endothermic reaction. If the value of K increases with a decrease in temperature, is the reaction exothermic or endothermic? Explain.
2NO(g) Cl2(g).
∆ (K V 1) (K V 1).
Q 7 K, Q 6 K, Q K, Kp Kp K Kp K Kp Kp Active Learning Questions 613 order in terms of increasing equilibrium concentration of B and explain.
5. Consider the reaction in a 1.0-L rigid flask. Answer the following questions for each situa-tion (a–d): i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between 95 M and 100 M.
ii. Explain how you decided on the limits for the estimated range.
iii. Indicate what other information would enable you to nar-row your estimated range.
iv. Compare the estimated concentrations for a through d, and explain any differences.
a. If at equilibrium M, and then 1 mol C is added, estimate the value for [A] once equilibrium is reestablished.
b. If at equilibrium M, and then 1 mol C is added, esti-mate the value for [B] once equilibrium is reestablished.
c. If at equilibrium M, and then 1 mol C is added, estimate the value for [C] once equilibrium is reestablished.
d. If at equilibrium M, and then 1 mol C is added, esti-mate the value for [D] once equilibrium is reestablished.
6. Consider the reaction A friend asks the following: “I know we have been told that if a mixture of A, B, C, and D is at equilibrium and more of A is added, more C and D will form. But how can more C and D form if we do not add more B?” What do you tell your friend?
7. Consider the following statements: “Consider the reaction for which at equilibrium M, M, and M. To a 1-L container of the system at equilibrium you add 3 moles of B. A possible equilibrium condi-tion is M, M, and M because in both cases ” Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.
8. Le Châtelier’s principle is stated (Section 13.7) as follows: “If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.” The system is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in concentration and an increase in concentra-tion. In the experiment the volume is assumed to be constant. On the other hand, if is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of actually could decrease and the concentration of would increase as equilibrium is reestablished. Explain how this can hap-pen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does af-fect the equilibrium position. Explain why the addition of an in-ert gas to this system in a rigid container does not affect the equi-librium position.
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
H2 NH3 N2 NH3 H2 N2 3H2 ∆2NH3 K 2.
[C] 6 [B] 3 [A] 1 [C] 4 [B] 1 [A] 2 A(g) B(g) ∆C(g), A(g) B(g) ∆C(g) D(g).
[D] 1 [C] 1 [B] 1 [A] 1 A(g) 2B(g) ∆C(g) D(g) Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equi-librium is reestablished? Justify your answer.
b. You have the original setup at equilibrium, and add more D to the flask. How does the concentration of each chemical com-pare to its original concentration after equilibrium is reestab-lished? Justify your answer.
2. The boxes shown below represent a set of initial conditions for the reaction: Draw a quantitative molecular picture that shows what this sys-tem looks like after the reactants are mixed in one of the boxes and the system reaches equilibrium. Support your answer with calculations.
+ + K = 25 A B ∆C D + 3. For the reaction consider two pos-sibilities: (a) you mix 0.5 mol of each reactant, allow the system to come to equilibrium, and then add another mole of and al-low the system to reach equilibrium again, or (b) you mix 1.5 mol and 0.5 mol and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures?
Explain.
4. Given the reaction consider the following situations: i. You have 1.3 M A and 0.8 M B initially.
ii. You have 1.3 M A, 0.8 M B, and 0.2 M C initially.
iii. You have 2.0 M A and 0.8 M B initially.
Order the preceding situations in terms of increasing equilib-rium concentration of D. Explain your order. Then give the A(g) B(g) ∆C(g) D(g), I2 H2 H2 H2(g) I2(g) ∆2HI(g), 614 Chapter Thirteen Chemical Equilibrium 14. Consider the following reactions.
List two property differences between these two reactions that relate to equilibrium.
15. For a typical equilibrium problem, the value of K and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?
16. Which of the following statements is(are) true? Correct the false statement(s).
a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium.
b. When a product is added to a system at equilibrium at a given temperature, the value of K for the reaction will increase when equilibrium is reestablished.
c. When temperature is increased for a reaction at equilibrium, the value of K for the reaction will increase.
d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium.
e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.
Exercises In this section similar exercises are paired.
The Equilibrium Constant 17. Write the equilibrium expression (K) for each of the following gas-phase reactions.
a.
b.
c.
d.
18. Write the equilibrium expression ( ) for each reaction in Exercise 17.
19. At a given temperature, for the reaction Calculate values of K for the following reactions at this temper-ature.
a.
b.
c.
d.
20. For the reaction at 1495 K. What is the value of for the following reactions at 1495 K?
a.
b.
c.
1 2H21g2 1 2Br21g2 ∆HBr1g2 2HBr1g2 ∆H21g2 Br21g2 HBr1g2 ∆1 2H21g2 1 2Br21g2 Kp Kp 3.5 104 H21g2 Br21g2 ∆2HBr1g2 2N21g2 6H21g2 ∆4NH31g2 NH31g2 ∆1 2N21g2 3 2H21g2 2NH31g2 ∆N21g2 3H21g2 1 2N21g2 3 2H21g2 ∆NH31g2 N21g2 3H21g2 ∆2NH31g2 K 1.3 102 Kp 2PBr31g2 3Cl21g2 ∆2PCl31g2 3Br21g2 SiH41g2 2Cl21g2 ∆SiCl41g2 2H21g2 N2O41g2 ∆2NO21g2 N21g2 O21g2 ∆2NO1g2 H21g2 I21g2 ¡ 2HI1g2 and H21g2 I21s2 ¡ 2HI1g2 Questions 9. Consider the following reaction: Amounts of and are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive 14C, will 14C be found only in CO molecules for an indefinite period of time?
Explain.
10. Consider the same reaction as in Exercise 9. In one experiment 1.0 mol and 1.0 mol are put into a flask and heated to In a second experiment 1.0 mol and 1.0 mol are put into another flask with the same volume as the first. This mixture is also heated to After equilibrium is reached, will there be any difference in the composition of the mixtures in the two flasks?
11. Consider the following reaction at some temperature: Some molecules of and CO are placed in a 1.0-L container as shown below.
When equilibrium is reached, how many molecules of and are present? Do this problem by trial and error—that is, if two molecules of CO react, is this equilibrium; if three molecules of CO react, is this equilibrium; and so on.
12. Consider the following generic reaction: Some molecules of are placed in a 1.0-L container. As time passes, several snapshots of the reaction mixture are taken as il-lustrated below.
Which illustration is the first to represent an equilibrium mixture?
Explain. How many molecules of reacted initially?
13. Explain the difference between K, Kp, and Q.
A2B A2B 2A2B1g2 ∆2A21g2 B21g2 CO2 H2O, CO, H2 H2O H2O1g2 CO1g2 ∆H21g2 CO21g2 K 2.0 350°C.
CO2(g) H2(g) 350°C.
CO(g) H2O(g) CO2 H2O, CO, H2, H2O1g2 CO1g2 ∆H21g2 CO21g2 Exercises 615 28. At 1100 K, for the reaction What is the value of K at this temperature?
29. Write expressions for K and for the following reactions.
a.
b.
c.
d.
30. For which reactions in Exercise 29 is equal to K?
31. Consider the following reaction at a certain temperature: An equilibrium mixture contains 1.0 mol Fe, mol and 2.0 mol all in a 2.0-L container. Calculate the value of K for this reaction.
32. In a study of the reaction at 1200 K it was observed that when the equilibrium partial pres-sure of water vapor is 15.0 torr, that total pressure at equilibrium is 36.3 torr. Calculate the value of for this reaction at 1200 K.
Hint: Apply Dalton’s law of partial pressures.
Equilibrium Calculations 33. The equilibrium constant, K, is at a certain temperature for the reaction For which of the following sets of conditions is the system at equi-librium? For those that are not at equilibrium, in which direction will the system shift?
a. A 1.0-L flask contains 0.024 mol NO, 2.0 mol and 2.6 mol b. A 2.0-L flask contains 0.032 mol NO, 0.62 mol and 4.0 mol c. A 3.0-L flask contains 0.060 mol NO, 2.4 mol and 1.7 mol 34. The equilibrium constant, , is at a certain tempera-ture for the reaction For which of the following sets of conditions is the system at equi-librium? For those that are not at equilibrium, in which direction will the system shift?
a.
atm, atm, atm b.
atm, atm, atm c.
atm, atm, atm 35. At for the reaction At a low temperature, dry ice (solid ), calcium oxide, and calcium carbonate are introduced into a 50.0-L reaction cham-ber. The temperature is raised to resulting in the dry ice 900°C, CO2 CaCO31s2 ∆CaO1s2 CO21g2 Kp 1.04 900°C, PO2 0.18 PN2 0.51 PNO 0.0062 PO2 0.67 PN2 0.36 PNO 0.0078 PO2 2.0 PN2 0.11 PNO 0.010 2NO1g2 ∆N21g2 O21g2 2.4 103 Kp O2.
N2, O2.
N2, O2.
N2, 2NO1g2 ∆N21g2 O21g2 2.4 103 Kp 3Fe1s2 4H2O1g2 ∆Fe3O41s2 4H21g2 Fe2O3 O2, 1.0 103 4Fe1s2 3O21g2 ∆2Fe2O31s2 Kp CuO1s2 H21g2 ∆Cu1l2 H2O1g2 2KClO31s2 ∆2KCl1s2 3O21g2 2NBr31s2 ∆N21g2 3Br21g2 2NH31g2 CO21g2 ∆N2CH4O1s2 H2O1g2 Kp 2SO21g2 O21g2 ∆2SO31g2 Kp 0.25 21. For the reaction it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: and Calculate the value of K for the reaction at this temperature.
22. For the reaction an analysis of an equilibrium mixture is performed at a certain tem-perature. It is found that M, M, and M. Calculate K for the reaction at this temperature.
23. At a particular temperature, a 3.0-L flask contains 2.4 mol 1.0 mol NOCl, and mol NO. Calculate K at this tem-perature for the following reaction: 24. At a particular temperature a 2.00-L flask at equilibrium contains mol mol and mol Calculate K at this temperature for the reaction If M, M, and M, does this represent a system at equilibrium?
25. The following equilibrium pressures at a certain temperature were observed for the reaction Calculate the value for the equilibrium constant at this tem-perature.
26. The following equilibrium pressures were observed at a certain temperature for the reaction Calculate the value for the equilibrium constant at this tem-perature.
If atm, atm, and atm, does this represent a system at equilibrium?
27. At the equilibrium concentrations are for the reaction Calculate at this temperature.
Kp CH3OH1g2 ∆CO1g2 2H21g2 0.15 M, [CO] 0.24 M, and [H2] 1.1 M [CH3OH] 327°, 0.00761 PH2 PNH3 0.0167 PN2 0.525 Kp PH2 3.1 103 atm PN2 8.5 101 atm PNH3 3.1 102 atm N21g2 3H21g2 ∆2NH31g2 Kp PO2 4.5 105 atm PNO 6.5 105 atm PNO2 0.55 atm 2NO21g2 ∆2NO1g2 O21g2 3O24 0.00245 3N2O4 0.200 3N24 2.00 104 2N21g2 O21g2 ∆2N2O1g2 N2O.
2.00 102 O2, 2.50 105 N2, 2.80 104 2NOCl1g2 ∆2NO1g2 Cl21g2 4.5 103 Cl2, [Cl2(g)] 4.3 104 1.4 103 [N2(g)] [NCl3(g)] 1.9 101 N21g2 3Cl21g2 ∆2NCl31g2 2.9 103 M.
[H2O(g)4 [N2(g)] 5.3 102 M, 4.1 105 M, [H2(g)] [NO(g) 4 8.1103M, 2NO1g2 2H21g2 ∆N21g2 2H2O1g2 616 Chapter Thirteen Chemical Equilibrium 41. At a particular temperature, 12.0 mol of is placed into a 3.0-L rigid container, and the dissociates by the reaction At equilibrium, 3.0 mol of is present. Calculate K for this reaction.
42. At a certain temperature, 4.0 mol is introduced into a 2.0-L container, and the partially dissociates by the reaction At equilibrium, 2.0 mol remains. What is the value of K for this reaction?
43. An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At equilibrium, the concentrations are M, M, and M. What were the concentrations of nitro-gen gas and hydrogen gas that were reacted initially?
44. Nitrogen gas reacts with hydrogen gas to form ammo-nia At in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium, the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium.
45. At a particular temperature, for the reaction If all four gases had initial concentrations of 0.800 M, calculate the equilibrium concentrations of the gases.
46. At a particular temperature, for the reaction In an experiment, 1.00 mol 1.00 mol and 1.00 mol HI are introduced into a 1.00-L container. Calculate the concentrations of all species when equilibrium is reached.
47. At for the reaction What is the partial pressure of NO in equilibrium with and that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?
48. At for the reaction Calculate the concentrations of all species at equilibrium for each of the following cases.
a. 1.0 g and 2.0 g are mixed in a 1.0-L flask.
b. 1.0 mol pure HOCl is placed in a 2.0-L flask.
49. At 1100 K, for the reaction Calculate the equilibrium partial pressures of and produced from an initial mixture in which atm and (Hint: If you don’t have a graphing calculator, then PSO3 0.
PSO2 PO2 0.50 SO3 SO2, O2, 2SO21g2 O21g2 ∆2SO31g2 Kp 0.25 Cl2O H2O H2O1g2 Cl2O1g2 ∆2HOCl1g2 K 0.090 25°C, O2 N2 N21g2 O21g2 ∆2NO1g2 Kp 0.050 2200°C, I2, H2, H21g2 I21g2 ∆2HI1g2 K 1.00 102 SO21g2 NO21g2 ∆SO31g2 NO1g2 K 3.75 200°C (NH3).
(H2) (N2) [NH3] 4.0 [N2] 8.0 [H2] 5.0 3H21g2 N21g2 ∆2NH31g2 NH3 2NH31g2 ∆N21g2 3H21g2 NH3 NH3 SO2 2SO31g2 ∆2SO21g2 O21g2 SO3 SO3 converting to gaseous . For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at a. 655 g 95.0 g atm b. 780 g 1.00 g atm c. 0.14 g 5000 g atm d. 715 g 813 g atm 36. Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: Acetic acid Ethanol Ethyl acetate For the following mixtures (a–d), will the concentration of increase, decrease, or remain the same as equilibrium is estab-lished?
a.
b.
c.
d.
e. What must the concentration of water be for a mixture with to be at equilibrium?
f. Why is water included in the equilibrium expression for this reaction?
37. For the reaction at a given temperature. At equilibrium it is found that and What is the concentration of under these conditions?
38. The reaction has at If the equilibrium partial pressure of is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.
39. A 1.00-L flask was filled with 2.00 mol gaseous and 2.00 mol gaseous and heated. After equilibrium was reached, it was found that 1.30 mol gaseous NO was present. Assume that the reaction occurs under these conditions. Calculate the value of the equilib-rium constant, K, for this reaction.
40. A sample of is placed in an otherwise empty rigid container at 1325 K at an initial pressure of 1.00 atm, where it decomposes to by the reaction At equilibrium, the partial pressure of is 0.25 atm. Calculate for this reaction at 1325 K.
Kp S8 S81g2 ∆4S21g2 S21g2 S81g2 SO21g2 NO21g2 ∆SO31g2 NO1g2 NO2 SO2 Br2 25°C.
Kp 109 2NO1g2 Br21g2 ∆2NOBr1g2 O2(g) [H2(g)] 1.9 102 M.
[H2O(g)] 1.1 101 M K 2.4 103 2H2O1g2 ∆2H21g2 O21g2 5.0 M [C2H5OH] [CH3CO2H] 0.10 M, [CH3CO2C2H5] 2.0 M, [C2H5OH] 10.0 M [CH3CO2H] 0.88 M, [H2O] 4.4 M, [CH3CO2C2H5] 4.4 M, [C2H5OH] 6.0 M [CH3CO2H] 0.044 M, [H2O] 0.12 M, [CH3CO2C2H5] 0.88 M, [C2H5OH] 0.10 M [CH3CO2H] 0.0020 M, [H2O] 0.0020 M, [CH3CO2C2H5] 0.22 M, [C2H5OH] 0.010 M [CH3CO2H] 0.010 M, [H2O] 0.10 M, [CH3CO2C2H5] 0.22 M, H2O K 2.2 CH3CO2H C2H5OH ∆CH3CO2C2H5 H2O CaO, PCO2 0.211 CaCO3, CaO, PCO2 1.04 CaCO3, CaO, PCO2 1.04 CaCO3, CaO, PCO2 2.55 CaCO3, 900°C?
CO2 Exercises 617 Le Châtelier’s Principle 57. Suppose the reaction system has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected.
a. Additional is added to the system.
b. The reaction is performed in a glass reaction vessel; attacks and reacts with glass.
c. Water vapor is removed.
58. Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased.
a.
b.
c.
d.
e.
59. An important reaction in the commercial production of hydrogen is How will this system at equilibrium shift in each of the five following cases?
a. Gaseous carbon dioxide is removed.
b. Water vapor is added.
c. The pressure is increased by adding helium gas.
d. The temperature is increased (the reaction is exothermic).
e. The pressure is increased by decreasing the volume of the reaction container.
60. What will happen to the number of moles of in equilibrium with and in the reaction in each of the following cases?
a. Oxygen gas is added.
b. The pressure is increased by decreasing the volume of the re-action container.
c. The pressure is increased by adding argon gas.
d. The temperature is decreased.
e. Gaseous sulfur dioxide is removed.
61. In which direction will the position of the equilibrium be shifted for each of the following changes?
a.
is added.
b.
is removed.
c.
is removed.
d. Some is added.
e. The volume of the container is doubled.
f. The temperature is decreased (the reaction is exothermic).
62. Hydrogen for use in ammonia production is produced by the reaction CO1g2 3H21g2 CH41g2 H2O1g2 Ar1g2 HI1g2 I21g2 H21g2 2HI1g2 ∆H21g2 I21g2 2SO31g2 ∆2SO21g2 O21g2 ¢H° 197 kJ O2 SO2 SO3 CO1g2 H2O1g2 ∆H21g2 CO21g2 CaCO31s2 ∆CaO1s2 CO21g2 COCl21g2 ∆CO1g2 Cl21g2 H21g2 F21g2 ∆2HF1g2 PCl51g2 ∆PCl31g2 Cl21g2 N21g2 3H21g2 ∆2NH31g2 HF1g2 UO2(s) UO21s2 4HF1g2 ∆UF41g2 2H2O1g2 use the method of successive approximations to solve, as discussed in Appendix 1.4.) 50. At a particular temperature, for the reaction a. A flask containing only at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium par-tial pressures of the gases.
b. A flask containing only at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium par-tial pressures of the gases.
c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?
51. At for the reaction Calculate the concentrations of all species at equilibrium for each of the following original mixtures.
a. 2.0 mol pure NOCl in a 2.0-L flask b. 1.0 mol NOCl and 1.0 mol NO in a 1.0-L flask c. 2.0 mol NOCl and 1.0 mol in a 1.0-L flask 52. At a particular temperature, for the reaction In an experiment, 1.0 mol is placed in a 10.0-L vessel. Cal-culate the concentrations of and when this reaction reaches equilibrium.
53. At a particular temperature, for the reaction If 2.0 mol is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species.
54. Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet-proof glass. One of the compounds used to make Lexan is phosgene an extremely poisonous gas. Phosgene decomposes by the reaction for which at 100C. If pure phosgene at an ini-tial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.
55. At for the reaction In an experiment carried out at a certain amount of is placed in an evacuated rigid container and al-lowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.
56. A sample of solid ammonium chloride was placed in an evacu-ated container and then heated so that it decomposed to ammo-nia gas and hydrogen chloride gas. After heating, the total pres-sure in the container was found to be 4.4 atm. Calculate at this temperature for the decomposition reaction NH4Cl1s2 ∆NH31g2 HCl1g2 Kp NH4OCONH2 25°C, NH4OCONH21s2 ∆2NH31g2 CO21g2 Kp 2.9 103 25°C, Kp 6.8 109 COCl21g2 ∆CO1g2 Cl21g2 (COCl22, CO2 2CO21g2 ∆2CO1g2 O21g2 K 2.0 106 NO2 N2O4 N2O4 N2O41g2 ∆2NO21g2 K 4.0 107 Cl2 2NOCl1g2 ∆2NO1g2 Cl21g2 K 1.6 105 35°C, NO2 N2O4 N2O41g2 ∆2NO21g2 Kp 0.25 Ni catalyst w88888888 88888888x 750C 618 Chapter Thirteen Chemical Equilibrium 68 At a certain temperature, for the reaction Calculate the concentrations of in a solution that is initially 2.0 M 69. For the reaction at 600. K, the equilibrium constant, , is 11.5. Suppose that 2.450 g of is placed in an evacuated 500.-mL bulb, which is then heated to 600. K.
a. What would be the pressure of if it did not dissociate?
b. What is the partial pressure of at equilibrium?
c. What is the total pressure in the bulb at equilibrium?
d. What is the degree of dissociation of at equilibrium?
70. At gaseous decomposes to and to the extent that 12.5% of the original (by moles) has de-composed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of for this system.
71. For the following reaction at a certain temperature it is found that the equilibrium concentrations in a 5.00-L rigid container are M, M, and M. If 0.200 mol of is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.
72. Consider the reaction How will the equilibrium position shift if a. water is added, doubling the volume?
b.
is added? (AgSCN is insoluble.) c.
is added? is insoluble.] d.
is added?
73. Chromium(VI) forms two different oxyanions, the orange dichro-mate ion, and the yellow chromate ion, (See the following photos.) The equilibrium reaction between the two ions is Explain why orange dichromate solutions turn yellow when sodium hydroxide is added.
Cr2O7 21aq2 H2O1l2 ∆2CrO4 21aq2 2H1aq2 CrO4 2.
Cr2O7 2, Fe1NO3231aq2 3Fe1OH23 NaOH1aq2 AgNO31aq2 Fe31aq2 SCN1aq2 ∆FeSCN21aq2 F2 [HF] 0.400 [F2] 0.0100 [H2] 0.0500 H21g2 F21g2 ∆2HF1g2 Kp SO2Cl2 Cl2(g) SO2(g) SO2Cl2 25°C, PCl5 PCl5 PCl5 PCl5 Kp PCl51g2 ∆PCl31g2 Cl21g2 FeSCN2.
Fe3, SCN, and FeSCN2 FeSCN21aq2 ∆Fe31aq2 SCN1aq2 K 9.1 104 What will happen to a reaction mixture at equilibrium if a.
is removed?
b. the temperature is increased (the reaction is endothermic)?
c. an inert gas is added?
d.
is removed?
e. the volume of the container is tripled?
63. Old-fashioned “smelling salts” consist of ammonium carbonate, The reaction for the decomposition of ammonium carbonate is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?
64. Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction equilibrium constants ( values) as a function of temperature are Is the reaction exothermic or endothermic?
Additional Exercises 65. Calculate a value for the equilibrium constant for the reaction given Hint: When reactions are added together, the equilibrium expres-sions are multiplied.
66. At for the reaction a. Calculate the concentration of NO, in molecules/cm3, that can exist in equilibrium in air at In air, atm and atm.
b. Typical concentrations of NO in relatively pristine environ-ments range from to molecules/cm3. Why is there a discrepancy between these values and your answer to part a?
67. The gas arsine, decomposes as follows: In an experiment at a certain temperature, pure was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr.
After 48 hours the pressure in the flask was observed to be con-stant at 488.0 torr.
a. Calculate the equilibrium pressure of b. Calculate for this reaction.
Kp H2(g) AsH3(g) 2AsH31g2 ∆2As1s2 3H21g2 AsH3, 1010 108 PO2 0.2 PN2 0.8 25°C.
N21g2 O21g2 ∆2NO1g2 Kp 1 1031 25°C, K 5.8 1034 O31g2 NO1g2 ∆NO21g2 O21g2 NO21g2 ∆ hv NO1g2 O1g2 K 6.8 1049 O21g2 O1g2 ∆O31g2 600°C, 2.25 106 500°C, 1.45 105 300°C, 4.34 103 Kp N21g2 3H21g2 ∆2NH31g2 1NH422CO31s2 ∆2NH31g2 CO21g2 H2O1g2 (NH4)2CO3.
CO1g2 H2O1g2 Challenge Problems 619 50.0% of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?
82. Consider the reaction where at 1325 K. In an experiment where is placed into a container at 1325 K, the equilibrium mixture of and has a total pressure of 1.00 atm.
Calculate the equilibrium pressures of and Calcu-late the fraction (by moles) of that has dissociated to reach equilibrium.
83. The partial pressures of an equilibrium mixture of and are atm and atm at a certain temperature. The volume of the container is doubled. Find the partial pressures of the two gases when a new equilibrium is established.
84. At 125C, for the reaction A 1.00-L flask containing 10.0 g is evacuated and heated to a. Calculate the partial pressures of and after equilib-rium is established.
b. Calculate the masses of and present at equi-librium.
c. Calculate the minimum container volume necessary for all of the to decompose.
85. An 8.00-g sample of was placed in an evacuated container, where it decomposed at according to the following reaction: At equilibrium the total pressure and the density of the gaseous mixture were 1.80 atm and 1.60 g/L, respectively. Calculate for this reaction.
86. A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred: After equilibrium was reached, the total pressure was 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate for each of these reactions.
87. At 5000 K and 1.000 atm, 83.00% of the oxygen molecules in a sample have dissociated to atomic oxygen. At what pressure will 95.0% of the molecules dissociate at this temperature?
88. A sample of is placed in an empty cylinder at After equilibrium is reached the total pressure is 1.5 atm and 16% (by moles) of the original has dissociated to a. Calculate the value of for this dissociation reaction at b. If the volume of the cylinder is increased until the total pres-sure is 1.0 atm (the temperature of the system remains con-stant), calculate the equilibrium pressure of and c. What percentage (by moles) of the original is disso-ciated at the new equilibrium position (total pressure 1.00 atm)?
N2O4(g) NO2(g).
N2O4(g) 25°C.
Kp NO2(g).
N2O4(g) 25°C.
N2O4(g) Kp SO31g2 ∆SO21g2 1 2O21g2 2FeSO41s2 ∆Fe2O31s2 SO31g2 SO21g2 Kp SO31g2 ∆SO21g2 1 2O21g2 600°C SO3 NaHCO3 Na2CO3 NaHCO3 H2O CO2 125°C.
NaHCO3 2NaHCO31s2 ∆Na2CO31s2 CO21g2 H2O1g2 Kp 0.25 PNO2 1.20 PN2O4 0.34 NO2(g) N2O4(g) P4(g) P2(g).
P4(g) P2(g2 P4(g) P4(g) Kp 1.00 101 P41g2 ¡ 2P21g2 74. The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction eco-nomically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia.
a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure 75. Suppose at a certain temperature for the reaction If it is found that the concentration of is twice the concen-tration of what must be the concentration of under these conditions?
76. For the reaction below, If a 20.0-g sample of is put into a 10.0-L container and heated to what percentage by mass of the will react to reach equilibrium?
77. A 2.4156-g sample of was placed in an empty 2.000-L flask and allowed to decompose to and at At equilibrium the total pressure inside the flask was observed to be 358.7 torr. Calculate the partial pressure of each gas at equi-librium and the value of at 78. Consider the decomposition of the compound as follows: When a 5.63-g sample of pure was sealed into an otherwise empty 2.50-L flask and heated to the pressure in the flask gradually rose to 1.63 atm and remained at that value.
Calculate K for this reaction.
Challenge Problems 79. At for the reaction If 2.0 mol NO and 1.0 mol are placed into a 1.0-L flask, cal-culate the equilibrium concentrations of all species.
80. Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300. K. At equi-librium the total pressure was 110.5 torr. The reaction is a. Calculate the value of .
b. What would be the partial pressures of all species if NO and both at an initial partial pressure of 0.30 atm, were al-lowed to come to equilibrium at this temperature?
81. At for the reaction When a certain partial pressure of is put into an other-wise empty rigid vessel at equilibrium is reached when 25°C, NH31g2 N21g2 3H21g2 ∆2NH31g2 Kp 5.3 105 25°C, Br2, Kp 2NO1g2 Br21g2 ∆2NOBr1g2 Cl2 2NOCl1g2 ∆2NO1g2 Cl21g2 K 1.6 105 35°C, 200.°C, C5H6O31g2 C5H6O31g2 ¡ C2H61g2 3CO1g2 C5H6O3 250.0°C.
Kp PCl51g2 ∆PCl31g2 Cl21g2 250.0°C: Cl2 PCl3 PCl5 CaCO3 800.°C, CaCO3 CaCO31s2 ∆CaO1s2 CO21g2 Kp 1.16 at 800.°C.
Cl2 PCl3, PCl5 PCl51g2 ∆PCl31g2 Cl21g2 K 4.5 103 620 Chapter Thirteen Chemical Equilibrium sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation If 3.00 g of solid naphthalene is placed into an enclosed space with a volume of 5.00 L at what percentage of the naph-thalene will have sublimed once equilibrium has been established?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
94. Consider the reaction for which Assume that 0.406 mol C(g) is placed in the cylinder represented below. The temperature is 300.0 K, and the barometric pressure on the piston (which is assumed to be massless and frictionless) is constant at 1.00 atm. The original volume (before the 0.406 mol C(g) begins to decompose) is 10.00 L. What is the volume in the cylinder at equilibrium?
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
P = 1.00 atm Original volume = 10.00 L T = 300.0 K 0.406 mole of pure C(g) (initially) K 1.30 102.
A1g2 B1g2 ∆C1g2 25°C, naphthalene1s2 ÷ naphthalene1g2 K 4.29 106 1at 298 K2 89. A sample of gaseous nitrosyl bromide, NOBr, was placed in a rigid flask, where it decomposed at according to the fol-lowing reaction: At equilibrium the total pressure and the density of the gaseous mixture were found to be 0.0515 atm and 0.1861 g/L, respec-tively. Calculate the value of for this reaction.
90. The equilibrium constant for the reaction at is 0.76 atm. Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at Integrative Problems These problems require the integration of multiple concepts to find the solutions.
91. For the reaction If 2.00 mol each of and are placed in a 5.00-L vessel, what mass of will be pres-ent at equilibrium? What is the pressure of at equilibrium?
92. Given at for the reaction and at for the reaction what is the value of K at the same temperature for the reaction What is the value of at for the reaction? Starting with 1.50 atm partial pressures of both C and D, what is the mole frac-tion of B once equilibrium is reached?
93. The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is 93.71% carbon by mass and a 0.256-mol sample of naphthalene has a mass of 32.8 g. What is the molecular formula of naphtha-lene? This compound works as a pesticide in mothballs by 45°C Kp C1g2 D1g2 ÷ 2B1g2 2A1g2 D1g2 ÷ C1g2 45°C K 7.10 A1g2 B1g2 ÷ C1g2 45°C K 3.50 H2S NH4HS NH4HS H2S, NH3, K 400. at 35.0°C.
NH31g2 H2S1g2 ∆NH4HS1s2 700°C.
700°C CCl41g2 ∆C1s2 2Cl21g2 Kp Kp 2NOBr1g2 ∆2NO1g2 Br21g2 25°C Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
622 14 Acids and Bases Contents 14.1 The Nature of Acids and Bases 14.2 Acid Strength • Water as an Acid and a Base 14.3 The pH Scale 14.4 Calculating the pH of Strong Acid Solutions 14.5 Calculating the pH of Weak Acid Solutions • The pH of a Mixture of Weak Acids • Percent Dissociation 14.6 Bases 14.7 Polyprotic Acids • Phosphoric Acid • Sulfuric Acid 14.8 Acid–Base Properties of Salts • Salts That Produce Neutral Solutions • Salts That Produce Basic Solutions • Base Strength in Aqueous Solutions • Salts That Produce Acidic Solutions 14.9 The Effect of Structure on Acid–Base Properties 14.10 Acid–Base Properties of Oxides 14.11 The Lewis Acid–Base Model 14.12 Strategy for Solving Acid–Base Problems: A Summary This grass pink orchid thrives in the acidic soil of a bog meadow at Illinois Beach State Park.
In this chapter we reencounter two very important classes of compounds, acids and bases. We will explore their interactions and apply the fundamentals of chemical equilibria discussed in Chapter 13 to systems involving proton-transfer reactions.
Acid–base chemistry is important in a wide variety of everyday applications. There are complex systems in our bodies that carefully control the acidity of our blood, since even small deviations may lead to serious illness and death. The same sensitivity exists in other life forms. If you have ever had tropical fish or goldfish, you know how important it is to monitor and control the acidity of the water in the aquarium.
Acids and bases are also important in industry. For example, the vast quantity of sul-furic acid manufactured in the United States each year is needed to produce fertilizers, polymers, steel, and many other materials.
The influence of acids on living things has assumed special importance in the United States, Canada, and Europe in recent years as a result of the phenomenon of acid rain (see the Chemical Impact in Chapter 5). This problem is complex and has diplomatic and eco-nomic overtones that make it all the more difficult to solve.
14.1 The Nature of Acids and Bases Acids were first recognized as a class of substances that taste sour. Vinegar tastes sour be-cause it is a dilute solution of acetic acid; citric acid is responsible for the sour taste of a lemon. Bases, sometimes called alkalis, are characterized by their bitter taste and slippery feel. Commercial preparations for unclogging drains are highly basic.
The first person to recognize the essential nature of acids and bases was Svante Arrhenius. Based on his experiments with electrolytes, Arrhenius postulated that acids produce hydrogen ions in aqueous solution, while bases produce hydroxide ions. At the time, the Arrhenius concept of acids and bases was a major step forward in quantifying acid–base chemistry, but this concept is limited because it applies only to aqueous solu-tions and allows for only one kind of base—the hydroxide ion. A more general definition of acids and bases was suggested by the Danish chemist Johannes Brønsted (1879–1947) and the English chemist Thomas Lowry (1874–1936). In terms of the Brønsted–Lowry model, an acid is a proton (H) donor, and a base is a proton acceptor. For example, when gaseous HCl dissolves in water, each HCl molecule donates a proton to a water mol-ecule and so qualifies as a Brønsted–Lowry acid. The molecule that accepts the proton, in this case water, is a Brønsted–Lowry base.
To understand how water can act as a base, we need to remember that the oxygen of the water molecule has two unshared electron pairs, either of which can form a covalent bond with an H ion. When gaseous HCl dissolves, the following reaction occurs: Note that the proton is transferred from the HCl molecule to the water molecule to form H3O, which is called the hydronium ion. This reaction is represented in Fig. 14.1 us-ing molecular models.
OA OS H O H O Q ClS H O A O H O O O H H SOS Cl Q 623 Don’t taste chemicals!
Acids and bases were first discussed in Section 4.2.
624 Chapter Fourteen Acids and Bases The general reaction that occurs when an acid is dissolved in water can best be rep-resented as (14.1) Acid Base Conjugate Conjugate acid base This representation emphasizes the significant role of the polar water molecule in pulling the proton from the acid. Note that the conjugate base is everything that remains of the acid molecule after a proton is lost. The conjugate acid is formed when the proton is transferred to the base. A conjugate acid–base pair consists of two substances related to each other by the donating and accepting of a single proton. In Equation (14.1) there are two conjugate acid–base pairs: HA and A and H2O and H3O. This reaction is repre-sented by molecular models in Fig. 14.2.
It is important to note that Equation (14.1) really represents a competition for the pro-ton between the two bases H2O and . If H2O is a much stronger base than that is, if H2O has a much greater affinity for H than does the equilibrium position will be far to the right; most of the acid dissolved will be in the ionized form. Conversely, if is a much stronger base than H2O, the equilibrium position will lie far to the left. In this case most of the acid dissolved will be present at equilibrium as HA.
The equilibrium expression for the reaction given in Equation (14.1) is (14.2) where Ka is called the acid dissociation constant. Both H3O(aq) and H(aq) are com-monly used to represent the hydrated proton. In this book we will often use simply H, but you should remember that it is hydrated in aqueous solutions.
In Chapter 13 we saw that the concentration of a pure solid or a pure liquid is always omitted from the equilibrium expression. In a dilute solution we can assume that the Ka 3H3O4 3A4 3HA4 3H4 3A4 3HA4 A A, A, A HA1aq2 H2O1l2 ∆ H3O1aq2 A1aq2 Common household substances that contain acids and bases. Vinegar is a dilute solution of acetic acid. Drain cleaners contain strong bases such as sodium hydroxide.
+ + + – FIGURE 14.1 The reaction of HCl and H2O.
Recall that (aq) means the substance is hydrated.
In this chapter we will always represent an acid as simply dissociating. This does not mean we are using the Arrhenius model for acids. Since water does not affect the equilibrium position, it is simply easier to leave it out of the acid dissociation reaction.
Visualization: Acid Ionization Equilibrium 14.1 The Nature of Acids and Bases 625 concentration of liquid water remains essentially constant when an acid is dissolved. Thus the term [H2O] is not included in Equation (14.2), and the equilibrium expression for Ka has the same form as that for the simple dissociation into ions: You should not forget, however, that water plays an important role in causing the acid to ionize.
Note that Ka is the equilibrium constant for the reaction in which a proton is removed from HA to form the conjugate base We use Ka to represent only this type of reac-tion. Knowing this, you can write the Ka expression for any acid, even one that is totally unfamiliar to you. As you do Sample Exercise 14.1, focus on the definition of the reac-tion corresponding to Ka.
Acid Dissociation (Ionization) Reactions Write the simple dissociation (ionization) reaction (omitting water) for each of the fol-lowing acids.
a. Hydrochloric acid (HCl) b. Acetic acid (HC2H3O2) c. The ammonium ion (NH4 ) d. The anilinium ion (C6H5NH3 ) e. The hydrated aluminum(III) ion [Al(H2O)6]3 Solution a.
b.
c.
d.
e. Although this formula looks complicated, writing the reaction is simple if you con-centrate on the meaning of Ka. Removing a proton, which can come only from one of the water molecules, leaves one and five H2O molecules attached to the Al3 ion.
So the reaction is See Exercises 14.27 and 14.28.
The Brønsted–Lowry model is not limited to aqueous solutions; it can be extended to reactions in the gas phase. For example, we discussed the reaction between gaseous hydrogen chloride and ammonia when we studied diffusion (Section 5.7): In this reaction, a proton is donated by the hydrogen chloride to the ammonia, as shown by these Lewis structures: O A Q Cl H S H O O O A H H O S A SOS Cl Q A H N N H H H NH31g2 HCl1g2 ∆NH4Cl1s2 Al1H2O26 31aq2 ∆H1aq2 Al1H2O25OH21aq2 OH C6H5NH3 1aq2 ∆H1aq2 C6H5NH21aq2 NH4 1aq2 ∆H1aq2 NH31aq2 HC2H3O21aq2 ∆H1aq2 C2H3O2 1aq2 HCl1aq2 ∆H1aq2 Cl1aq2 A.
HA1aq2 ∆H1aq2 A1aq2 + + + – FIGURE 14.2 The reaction of an acid HA with water to form H3O and a conjugate base A.
Sample Exercise 14.1 When HCl(g) and NH3(g) meet in a tube, a white ring of NH4Cl(s) forms.
626 Chapter Fourteen Acids and Bases Note that this is not considered an acid–base reaction according to the Arrhenius concept.
Figure 14.3 shows a molecular representation of this reaction.
14.2 Acid Strength The strength of an acid is defined by the equilibrium position of its dissociation (ioniza-tion) reaction: A strong acid is one for which this equilibrium lies far to the right. This means that almost all the original HA is dissociated (ionized) at equilibrium [see Fig. 14.4(a)]. There is an important connection between the strength of an acid and that of its conjugate base.
A strong acid yields a weak conjugate base—one that has a low affinity for a proton. A strong acid also can be described as an acid whose conjugate base is a much weaker base than water (see Fig. 14.5). In this case the water molecules win the competition for the H ions.
HA1aq2 H2O1l2 ∆H3O1aq2 A1aq2 + + + – FIGURE 14.3 The reaction of NH3 with HCl to form and Cl.
NH4 HA (a) H+ A– HA (b) H+ A– Before dissociation After dissociation, at equilibrium HA Relative acid strength Very strong Strong Weak Very weak Very strong Strong Weak Very weak Relative conjugate base strength FIGURE 14.4 Graphic representation of the behavior of acids of different strengths in aqueous solution. (a) A strong acid.
(b) A weak acid.
FIGURE 14.5 The relationship of acid strength and conju-gate base strength for the reaction HA(aq) H2O(l) H3O(aq) A(aq) Acid Conjugate base ∆ A strong acid has a weak conjugate base.
14.2 Acid Strength 627 TABLE 14.1 Various Ways to Describe Acid Strength Property Strong Acid Weak Acid Ka value Ka is large Ka is small Position of the dissociation Far to the right Far to the left (ionization) equilibrium Equilibrium concentration of compared with original concentration of HA Strength of conjugate base much weaker much stronger compared with that of water base than H2O base than H2O A A H 3H4 3HA40 3H4 3HA4 0 Sulfuric acid (H2SO4) Nitric acid (HNO3) Perchloric acid (HClO4) Perchloric acid can explode if handled improperly. means much less than.
means much greater than.
Conversely, a weak acid is one for which the equilibrium lies far to the left. Most of the acid originally placed in the solution is still present as HA at equilibrium. That is, a weak acid dissociates only to a very small extent in aqueous solution [see Fig. 14.4(b)].
In contrast to a strong acid, a weak acid has a conjugate base that is a much stronger base than water. In this case a water molecule is not very successful in pulling an H ion from the conjugate base. The weaker the acid, the stronger its conjugate base.
The various ways of describing the strength of an acid are summarized in Table 14.1.
Strong and weak acids are represented pictorially in Fig. 14.6.
The common strong acids are sulfuric acid , hydrochloric acid nitric acid , and perchloric acid . Sulfuric acid is actually a diprotic acid, an acid having two acidic protons. The acid H2SO4 is a strong acid, virtu-ally 100% dissociated (ionized) in water: The ion, however, is a weak acid: Most acids are oxyacids, in which the acidic proton is attached to an oxygen atom.
The strong acids mentioned above, except hydrochloric acid, are typical examples. Many common weak acids, such as phosphoric acid (H3PO4), nitrous acid (HNO2), and HSO4 1aq2 ∆H1aq2 SO4 21aq2 HSO4 H2SO41aq2 ¡ H1aq2 HSO4 1aq2 [HClO4(aq)] [HNO3(aq)] [HCl(aq)], [H2SO4(aq)] H+ A– B– FIGURE 14.6 (a) A strong acid HA is completely ionized in water. (b) A weak acid HB exists mostly as undissociated HB molecules in water. Note that the water molecules are not shown in this figure.
628 Chapter Fourteen Acids and Bases hypochlorous acid (HOCl), are also oxyacids. Organic acids, those with a carbon atom backbone, commonly contain the carboxyl group: Acids of this type are usually weak. Examples are acetic acid (CH3COOH), often written HC2H3O2, and benzoic acid (C6H5COOH). Note that the remainder of the hydrogens in these molecules are not acidic—they do not form H in water.
There are some important acids in which the acidic proton is attached to an atom other than oxygen. The most significant of these are the hydrohalic acids HX, where X represents a halogen atom.
Table 14.2 contains a list of common monoprotic acids (those having one acidic proton) and their Ka values. Note that the strong acids are not listed. When a strong acid molecule such as HCl, for example, is placed in water, the position of the dissociation equilibrium lies so far to the right that [HCl] cannot be measured accurately. This prevents an accu-rate calculation of Ka: Very small and highly uncertain Relative Base Strength Using Table 14.2, arrange the following species according to their strengths as bases: H2O, , , NO2 , and .
Solution Remember that water is a stronger base than the conjugate base of a strong acid but a weaker base than the conjugate base of a weak acid. This leads to the following order: Weakest bases Strongest bases ¡ Cl 6 H2O 6 conjugate bases of weak acids CN Cl F Ka 3H4 3Cl4 3HCl4 HCl1aq2 H2O1l2 ∆H1aq2 Cl1aq2 O O C H The units of Ka are customarily omitted.
TABLE 14.2 Values of Ka for Some Common Monoprotic Acids Formula Name Value of Ka Hydrogen sulfate ion Chlorous acid Monochloracetic acid HF Hydrofluoric acid Nitrous acid Acetic acid Hydrated aluminum(III) ion HOCl Hypochlorous acid HCN Hydrocyanic acid NH4 Ammonium ion Phenol 1.6 1010 HOC6H5 5.6 1010 6.2 1010 3.5 108 1.4 105 [Al(H2O)6]3 1.8 105 HC2H3O2 4.0 104 HNO2 7.2 104 1.35 103 HC2H2ClO2 1.2 102 HClO2 1.2 102 HSO4 h 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 Increasing acid strength 8n Sample Exercise 14.2 Phosphoric acid (H3PO4) Hypochlorous acid (HOCl) Nitrous acid (HNO2) Acetic acid (CH3CO2H) Acidic H Benzoic acid (C6H5CO2H) Acidic H Appendix 5.1 contains a table of Ka values.
14.2 Acid Strength 629 We can order the remaining conjugate bases by recognizing that the strength of an acid is inversely related to the strength of its conjugate base. Since from Table 14.2 we have the base strengths increase as follows: The combined order of increasing base strength is See Exercises 14.33 through 14.36.
Water as an Acid and a Base A substance is said to be amphoteric if it can behave either as an acid or as a base. Water is the most common amphoteric substance. We can see this clearly in the autoionization of water, which involves the transfer of a proton from one water molecule to another to produce a hydroxide ion and a hydronium ion: In this reaction, also illustrated in Fig. 14.7, one water molecule acts as an acid by fur-nishing a proton, and the other acts as a base by accepting the proton.
Autoionization can occur in other liquids besides water. For example, in liquid am-monia the autoionization reaction is The autoionization reaction for water leads to the equilibrium expression where Kw, called the ion-product constant (or the dissociation constant for water), always refers to the autoionization of water.
Experiment shows that at in pure water, which means that at 1.0 1014 Kw 3H4 3OH4 11.0 107211.0 1072 25°C 3H4 3OH4 1.0 107 M 25°C Kw 3H3O4 3OH4 3H4 3OH4 2H2O1l2 ∆H3O1aq2 OH1aq2 0/ ?
S S H H H H O N 0/ ?
H H H 0/ ?
H H H N O G D H H N A N S S Q S HG D H S S O O O H 0/ ?O H H H O O G D H H O Cl 6 H2O 6 F 6 NO2 6 CN F 6 NO2 6 CN Ka for HF 7 Ka for HNO2 7 Ka for HCN + + + – FIGURE 14.7 Two water molecules react to form and .
OH H3O acid(1) base(1) acid(2) base(2) H2O H2O ∆H3O OH 1.0 1014 Kw 3H 4 3OH4 630 Chapter Fourteen Acids and Bases It is important to recognize the meaning of Kw. In any aqueous solution at , no matter what it contains, the product of [H] and must always equal There are three possible situations: 1. A neutral solution, where 2. An acidic solution, where 3. A basic solution, where In each case, however, at , Calculating [H+] and [OH–] Calculate [H] or as required for each of the following solutions at , and state whether the solution is neutral, acidic, or basic.
a.
b.
c.
Solution a.
Since is , solving for gives Since , the solution is basic.
b. As in part a, solving for gives Since , the solution is neutral.
c. Solving for gives Since , the solution is acidic.
See Exercises 14.37 and 14.38.
Since Kw is an equilibrium constant, it varies with temperature. The effect of tem-perature is considered in Sample Exercise 14.4.
Autoionization of Water At , the value of Kw is .
a. Using Le Châtelier’s principle, predict whether the reaction is exothermic or endothermic.
b. Calculate [H] and in a neutral solution at .
60°C [OH] 2H2O1l2 ∆H3O1aq2 OH1aq2 1 1013 60°C [H] 7 [OH] 3OH4 1.0 1014 3H4 1.0 1014 10.0 1.0 1015 M [OH] [H] [OH] 3H4 1.0 1014 3OH4 1.0 1014 1.0 107 1.0 107 M [H] [OH] 7 [H] 3H 4 1.0 1014 3OH4 1.0 1014 1.0 105 1.0 109 M [H] 1.0 105 M [OH] Kw [H][OH] 1.0 1014.
10.0 M H 1.0 107 M OH 1.0 105 M OH 25°C [OH] Kw 3H4 3OH4 1.0 1014 25°C [OH] 7 [H].
[H] 7 [OH].
[H] [OH].
1.0 1014.
[OH] 25°C Sample Exercise 14.3 Sample Exercise 14.4 Visualization: Self-Ionization of Water 14.3 The pH Scale 631 Solution a. Kw increases from at to at . Le Châtelier’s principle states that if a system at equilibrium is heated, it will adjust to consume energy. Since the value of Kw increases with temperature, we must think of energy as a reactant, and so the process must be endothermic.
b. At , For a neutral solution, See Exercise 14.39.
14.3 The pH Scale Because [H] in an aqueous solution is typically quite small, the pH scale provides a con-venient way to represent solution acidity. The pH is a log scale based on 10, where Thus for a solution where At this point we need to discuss significant figures for logarithms. The rule is that the number of decimal places in the log is equal to the number of significant figures in the original number. Thus 2 significant figures 2 decimal places Similar log scales are used for representing other quantities; for example, Since pH is a log scale based on 10, the pH changes by 1 for every power of 10 change in [H]. For example, a solution of pH 3 has an H concentration 10 times that of a solution of pH 4 and 100 times that of a solution of pH 5. Also note that because pH is defined as , the pH decreases as [H] increases. The pH scale and the pH values for several common substances are shown in Fig. 14.8.
The pH of a solution is usually measured using a pH meter, an electronic device with a probe that can be inserted into a solution of unknown pH. The probe contains an acidic aqueous solution enclosed by a special glass membrane that allows migration of H ions.
If the unknown solution has a different pH from the solution in the probe, an electric po-tential results, which is registered on the meter (see Fig. 14.9).
Calculating pH and pOH Calculate pH and pOH for each of the following solutions at .
a.
b. 1.0 M OH 1.0 103 M OH 25°C log[H] pK log K pOH log3OH4 pH 9.00 3H4 1.0 109 M pH 17.002 7.00 3H4 1.0 107 M pH log3H4 3H4 3OH4 21 1013 3 107 M 3H4 3OH4 1 1013 60°C 60°C 1 1013 25°C 1 1014 The pH scale is a compact way to represent solution acidity.
Appendix 1.2 has a review of logs.
Visualization: The pH Scale Sample Exercise 14.5 The pH meter is discussed more fully in Section 17.4.
632 Chapter Fourteen Acids and Bases Solution a.
b.
See Exercise 14.41.
It is useful to consider the log form of the expression Kw 3H4 3OH4 pOH log 3OH4 log 11.0 10142 14.00 pH log 3H4 log 11.02 0.00 3OH4 Kw 3H4 1.0 1014 1.0 1.0 1014 M pOH log 3OH4 log 11.0 1032 3.00 pH log 3H4 log 11.0 10112 11.00 3H4 Kw 3OH4 1.0 1014 1.0 103 1.0 1011 M A rnold Beckman died at age 104 in May 2004. Beckman’s leadership of science and business spans virtually the entire twentieth century. He was born in 1900 in Cullom, Illinois, a town of 500 people that had no electricity or tele-phones. Beckman says, “In Cullom we were forced to im-provise. I think it was a good thing.” The son of a blacksmith, Beckman had his interest in science awakened at age nine. At that time, in the attic of his house he discovered J. Dorman Steele’s Fourteen Weeks in Chemistry, a book containing instructions for doing chem-istry experiments. Beckman became so fascinated with chemistry that his father built him a small “chemistry shed” in the back yard for his tenth birthday.
Beckman’s interest in chemistry was fostered by his high school teachers, and he eventually attended the Uni-versity of Illinois, Urbana–Champaign. He graduated with Arnold Beckman.
CHEMICAL IMPACT Arnold Beckman, Man of Science FIGURE 14.9 pH meters are used to measure acidity.
a bachelor’s degree in chemical engineering in 1922 and stayed one more year to get a master’s degree. He then went to Caltech, where he earned a Ph.D. and became a faculty member.
FIGURE 14.8 The pH scale and pH values of some common substances.
[H+] pH 10 –14 14 10 –13 13 10 –12 12 10 –11 11 10 –10 10 10 –9 9 10 –8 8 10 –6 6 10 –5 5 10 –4 4 10 –3 3 10 –2 2 10 –1 1 1 0 Acidic Neutral Basic 1 M NaOH Ammonia (Household cleaner) Blood Pure water Milk Vinegar Lemon juice Stomach acid 1 M HCl 10 –7 7 14.3 The pH Scale 633 That is, or Thus (14.3) Since , Thus, for any aqueous solution at , pH and pOH add up to 14.00: (14.4) Calculating pH The pH of a sample of human blood was measured to be 7.41 at . Calculate pOH, [H], and for the sample.
Solution Since To find [H] we must go back to the definition of pH: Thus 7.41 log3H4 or log3H4 7.41 pH log3H4 pOH 14.00 pH 14.00 7.41 6.59 pH pOH 14.00, [OH] 25°C pH pOH 14.00 25°C pKw log11.0 10142 14.00 Kw 1.0 1014 pKw pH pOH log Kw log3H4 log3OH4 log Kw log3H4 log3OH4 Beckman was always known for his inventiveness. As a youth he designed a pressurized fuel system for his Model T Ford to overcome problems with its normal gravity feed fuel system—you had to back it up steep hills to keep it from starving for fuel. In 1927 he applied for his first patent: a buzzer to alert drivers that they were speeding.
In 1935 Beckman invented something that would cause a revolution in scientific instrumentation. A college friend who worked in a laboratory in the California citrus industry needed an accurate, convenient way to measure the acidity of orange juice. In response, Beckman invented the pH meter, which he initially called the acidimeter. This compact, sturdy device was an immediate hit. It signaled a new era in scientific instrumentation. In fact, business was so good that Beckman left Caltech to head his own company.
Over the years Beckman invented many other devices, including an improved potentiometer and an instrument for measuring the light absorbed by molecules. At age 65 he re-tired as president of Beckman Instruments (headquartered in Fullerton, California). After a merger the company be-came Beckman Coulter; it had sales of more than $2 billion in 2003.
After stepping down as president of Beckman Instru-ments, Beckman began a new career—donating his wealth for the improvement of science. In 1984 he and Mabel, his wife of 58 years, donated $40 million to his alma mater— the University of Illinois—to fund the Beckman Institute.
The Beckmans have also funded many other research insti-tutes, including one at Caltech, and formed a foundation that currently gives $20 million each year to various scientific endeavors.
Arnold Beckman was a man known for his incredible creativity but even more he was recognized as a man of ab-solute integrity. Mr. Beckman has important words for us: “Whatever you do, be enthusiastic about it.” Note: You can see Arnold Beckman’s biography at the Chemical Heritage Foundation Web site (http;//www.chemheritage.org).
Sample Exercise 14.6 634 Chapter Fourteen Acids and Bases We need to know the antilog of 7.41. As shown in Appendix 1.2, taking the antilog is the same as exponentiation; that is, Since , and [H] can be calculated by taking the antilog of pH: In the present case, Similarly, , and See Exercises 14.42 through 14.46.
Now that we have considered all the fundamental definitions relevant to acid–base so-lutions, we can proceed to a quantitative description of the equilibria present in these solutions.
The main reason that acid–base problems sometimes seem difficult is that a typical aqueous solution contains many components, so the problems tend to be complicated. However, you can deal with these problems successfully if you use the following general strategies: • Think chemistry. Focus on the solution components and their reactions. It will almost always be possible to choose one reaction that is the most important.
• Be systematic. Acid–base problems require a step-by-step approach.
• Be flexible. Although all acid–base problems are similar in many ways, important dif-ferences do occur. Treat each problem as a separate entity. Do not try to force a given problem into matching any you have solved before. Look for both the similarities and the differences.
• Be patient. The complete solution to a complicated problem cannot be seen immedi-ately in all its detail. Pick the problem apart into its workable steps.
• Be confident. Look within the problem for the solution, and let the problem guide you. As-sume that you can think it out. Do not rely on memorizing solutions to problems. In fact, memorizing solutions is usually detrimental because you tend to try to force a new problem to be the same as one you have seen before. Understand and think; don’t just memorize.
14.4 Calculating the pH of Strong Acid Solutions When we deal with acid–base equilibria, we must focus on the solution components and their chemistry. For example, what species are present in a 1.0 M solution of HCl? Since hydrochloric acid is a strong acid, we assume that it is completely dissociated. Thus, al-though the label on the bottle says 1.0 M HCl, the solution contains virtually no HCl mol-ecules. Typically, container labels indicate the substance(s) used to make up the solution but do not necessarily describe the solution components after dissolution. Thus a 1.0 M HCl solution contains H and ions rather than HCl molecules.
The next step in dealing with aqueous solutions is to determine which components are significant and which can be ignored. We need to focus on the major species, those solution components present in relatively large amounts. In 1.0 M HCl, for example, the major species are H, , and H2O. Since this is a very acidic solution, is present OH Cl Cl 3OH4 antilog16.592 106.59 2.6 107 M [OH] antilog(pOH) 3H4 antilog1pH2 antilog17.412 107.41 3.9 108 3H4 antilog1pH2 pH log3H4 pH log[H] antilog1n2 10n antilog1n2 log11n2 Common Strong Acids HCl(aq) HNO3(aq) H2SO4(aq) HClO4(aq) 14.5 Calculating the pH of Weak Acid Solutions 635 only in tiny amounts and is classified as a minor species. In attacking acid–base prob-lems, the importance of writing the major species in the solution as the first step cannot be overemphasized. This single step is the key to solving these problems successfully.
To illustrate the main ideas involved, let us calculate the pH of 1.0 M HCl. We first list the major species: H, , and H2O. Since we want to calculate the pH, we will fo-cus on those major species that can furnish H. Obviously, we must consider H from the dissociation of HCl. However, H2O also furnishes H by autoionization, which is of-ten represented by the simple dissociation reaction However, is autoionization an important source of H ions? In pure water at , [H] is M. In 1.0 M HCl solution, the water will produce even less than 1027 M H, since by Le Châtelier’s principle the H from the dissociated HCl will drive the position of the water equilibrium to the left. Thus the amount of H contributed by water is negligible compared with the 1.0 M H from the dissociation of HCl. Therefore, we can say that [H] in the solution is 1.0 M. The pH is then pH of Strong Acids a. Calculate the pH of 0.10 M HNO3.
b. Calculate the pH of .
Solution a. Since HNO3 is a strong acid, the major species in solution are The concentration of HNO3 is virtually zero, since the acid completely dissociates in water. Also, will be very small because the H ions from the acid will drive the equilibrium to the left. That is, this is an acidic solution where , so The sources of H are 1. H from HNO3 (0.10 M) 2. H from H2O The number of H ions contributed by the autoionization of water will be very small compared with the 0.10 M contributed by the HNO3 and can be neglected. Since the dissolved HNO3 is the only important source of H ions in this solution, b. Normally, in an aqueous solution of HCl the major species are H, , and H2O.
However, in this case the amount of HCl in solution is so small that it has no effect; the only major species is H2O. Thus the pH will be that of pure water, or pH 7.00.
See Exercises 14.47 and 14.48.
14.5 Calculating the pH of Weak Acid Solutions Since a weak acid dissolved in water can be viewed as a prototype of almost any equi-librium occurring in aqueous solution, we will proceed carefully and systematically.
Although some of the procedures we develop here may seem unnecessary, they will become Cl 3H4 0.10 M and pH log10.102 1.00 [OH] 107 M.
[H] [OH] H2O1l2 ∆H1aq2 OH1aq2 [OH] H, NO3 , and H2O 1.0 1010 M HCl pH log3H4 log11.02 0 107 25°C H2O1l2 ∆H1aq2 OH1aq2 Cl Sample Exercise 14.7 Always write the major species present in the solution.
The from the strong acid drives the equilibrium to the left.
H2O ∆H OH H In pure water, only M is produced.
H 107 H+ NO3 – H2O Major Species 636 Chapter Fourteen Acids and Bases essential as the problems become more complicated. We will develop the necessary strate-gies by calculating the pH of a 1.00 M solution of HF ( ).
The first step, as always, is to write the major species in the solution. From its small Ka value, we know that hydrofluoric acid is a weak acid and will be dissociated only to a slight extent. Thus, when we write the major species, the hydrofluoric acid will be represented in its dominant form, as HF. The major species in solution are HF and H2O.
The next step (since this is a pH problem) is to decide which of the major species can furnish H ions. Actually, both major species can do so: In aqueous solutions, however, typically one source of H can be singled out as domi-nant. By comparing Ka for HF with Kw for H2O, we see that hydrofluoric acid, although weak, is still a much stronger acid than water. Thus we will assume that hydrofluoric acid will be the dominant source of H. We will ignore the tiny contribution by water.
Therefore, it is the dissociation of HF that will determine the equilibrium concentra-tion of H and hence the pH: The equilibrium expression is To solve the equilibrium problem, we follow the procedures developed in Chapter 13 for gas-phase equilibria. First, we list the initial concentrations, the concentrations before the reaction of interest has proceeded to equilibrium. Before any HF dissociates, the con-centrations of the species in the equilibrium are (Note that the zero value for is an approximation, since we are neglecting the H ions from the autoionization of water.) The next step is to determine the change required to reach equilibrium. Since some HF will dissociate to come to equilibrium (but this amount is presently unknown), we let x be the change in the concentration of HF that is required to achieve equilibrium. That is, we assume that x mol/L HF will dissociate to produce x mol/L H and x mol/L as the system adjusts to its equilibrium position. Now the equilibrium concentrations can be defined in terms of x: Substituting these equilibrium concentrations into the equilibrium expression gives This expression produces a quadratic equation that can be solved using the quadratic for-mula, as for the gas-phase systems in Chapter 13. However, since Ka for HF is so small, HF will dissociate only slightly, and x is expected to be small. This will allow us to sim-plify the calculation. If x is very small compared to 1.00, the term in the denominator can be approximated as follows: 1.00 x 1.00 Ka 7.2 104 3H4 3F4 3HF4 1x21x2 1.00 x 3H4 3H4 0 x 0 x x 3F4 3F4 0 x 0 x x 3HF4 3HF4 0 x 1.00 x F [H]0 3HF4 0 1.00 M 3F4 0 0 3H4 0 107 M 0 Ka 7.2 104 3H4 3F4 3HF4 HF1aq2 ∆H1aq2 F1aq2 H2O1l2 ∆H1aq2 OH1aq2 Kw 1.0 1014 HF1aq2 ∆H1aq2 F1aq2 Ka 7.2 104 Ka 7.2 104 First, always write the major species present in the solution.
HF Major Species H2O 14.5 Calculating the pH of Weak Acid Solutions 637 The equilibrium expression then becomes which yields How valid is the approximation that [HF] 1.00 M? Because this question will arise often in connection with acid–base equilibrium calculations, we will consider it carefully.
The validity of the approximation depends on how much accuracy we demand for the cal-culated value of [H]. Typically, the Ka values for acids are known to an accuracy of only about . It is reasonable therefore to apply this figure when determining the validity of the approximation We will use the following test. First, we calculate the value of x by making the approximation where We then compare the sizes of x and . If the expression is less than or equal to 5%, the value of x is small enough that the approximation will be considered valid.
In our example, and The approximation we made is considered valid, and the value of x calculated using that approximation is acceptable. Thus This problem illustrates all the important steps for solving a typical equilibrium prob-lem involving a weak acid. These steps are summarized as follows: Solving Weak Acid Equilibrium Problems ➥ 1 List the major species in the solution.
➥ 2 Choose the species that can produce H, and write balanced equations for the reactions producing H.
➥ 3 Using the values of the equilibrium constants for the reactions you have writ-ten, decide which equilibrium will dominate in producing H.
➥ 4 Write the equilibrium expression for the dominant equilibrium.
x 3H4 2.7 102 M and pH log 12.7 1022 1.57 x 3HA40 100 2.7 102 1.00 100% 2.7% 3HA40 3HF4 0 1.00 mol/L x 2.7 102 mol/L 3HA4 0 x 3HA40 x 3HA40 100% [HA]0 x2 Ka3HA40 and x 2Ka3HA40 Ka x2 3HA40 x x2 3HA40 3HA40 x 3HA40 5% x 27.2 104 2.7 102 x2 17.2 104211.002 7.2 104 7.2 104 1x21x2 1.00 x 1x21x2 1.00 The validity of an approximation should always be checked.
638 Chapter Fourteen Acids and Bases ➥ 5 List the initial concentrations of the species participating in the dominant equilibrium.
➥ 6 Define the change needed to achieve equilibrium; that is, define x.
➥ 7 Write the equilibrium concentrations in terms of x.
➥ 8 Substitute the equilibrium concentrations into the equilibrium expression.
➥ 9 Solve for x the “easy” way, that is, by assuming that [HA]0 x [HA]0.
➥10 Use the 5% rule to verify whether the approximation is valid.
➥11 Calculate [H] and pH.
We use this systematic approach in Sample Exercise 14.8.
The pH of Weak Acids The hypochlorite ion is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than , for example) and forms the weakly acidic hypochlorous acid ( ). Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid.
Solution ➥ 1 We list the major species. Since HOCl is a weak acid and remains mostly undisso-ciated, the major species in a 0.100 M HOCl solution are ➥ 2 Both species can produce H: ➥ 3 Since HOCl is a significantly stronger acid than H2O, it will dominate in the pro-duction of H.
➥ 4 We therefore use the following equilibrium expression: ➥ 5 The initial concentrations appropriate for this equilibrium are ➥ 6 Since the system will reach equilibrium by the dissociation of HOCl, let x be the amount of HOCl (in mol/L) that dissociates in reaching equilibrium.
➥ 7 The equilibrium concentrations in terms of x are ➥ 8 Substituting these concentrations into the equilibrium expression gives Ka 3.5 108 1x21x2 0.100 x 3H4 3H4 0 x 0 x x 3OCl4 3OCl4 0 x 0 x x 3HOCl4 3HOCl4 0 x 0.100 x 3H4 0 0 (We neglect the contribution from H2O.) 3OCl4 0 0 3HOCl40 0.100 M Ka 3.5 108 3H4 3OCl4 3HOCl4 H2O1l2 ∆H1aq2 OH1aq2 Kw 1.0 1014 HOCl1aq2 ∆H1aq2 OCl1aq2 Ka 3.5 108 HOCl and H2O HOCl, Ka 3.5 108 Cl (OCl) The Ka values for various weak acids are given in Table 14.2 and in Appendix 5.1.
Sample Exercise 14.8 Swimming pool water must be frequently tested for pH and chlorine content.
H2O HOCl Major Species 14.5 Calculating the pH of Weak Acid Solutions 639 ➥ 9 Since Ka is so small, we can expect a small value for x. Thus we make the approxi-mation which leads to the expression Solving for x gives ➥10 The approximation must be validated. To do this, we compare x to [HOCl]0: Since this value is much less than 5%, the approximation is considered valid.
➥11 We calculate [H] and pH: See Exercises 14.53 through 14.55.
The pH of a Mixture of Weak Acids Sometimes a solution contains two weak acids of very different strengths. This case is considered in Sample Exercise 14.9. Note that the steps are again followed (though not labeled).
The pH of Weak Acid Mixtures Calculate the pH of a solution that contains and Also calculate the concentration of cyanide ion in this solution at equilibrium.
Solution Since HCN and HNO2 are both weak acids and are largely undissociated, the major species in the solution are All three of these components produce H: A mixture of three acids might lead to a very complicated problem. However, the situa-tion is greatly simplified by the fact that even though HNO2 is a weak acid, it is much stronger than the other two acids present (as revealed by the K values). Thus HNO2 can be assumed to be the dominant producer of H, and we will focus on the equilibrium expression Ka 4.0 104 3H4 3NO2 4 3HNO24 H2O1l2 ∆H1aq2 OH1aq2 Kw 1.0 1014 HNO21aq2 ∆H1aq2 NO2 1aq2 Ka 4.0 104 HCN1aq2 ∆H1aq2 CN1aq2 Ka 6.2 1010 HCN, HNO2, and H2O (CN) 5.00 M HNO2 (Ka 4.0 104).
1.00 M HCN (Ka 6.2 1010) 3H4 x 5.9 105 M and pH 4.23 x 3HA40 100 x 3HOCl4 0 100 5.9 105 0.100 100 0.059% 0.100 x 0.100 x 5.9 105 Ka 3.5 108 x2 0.100 x x2 0.100 [HA]0 x [HA]0, or 0.100 x 0.100, The same systematic approach applies to all solution equilibria.
Sample Exercise 14.9 H2O HCN HNO2 Major Species 640 Chapter Fourteen Acids and Bases The initial concentrations, the definition of x, and the equilibrium concentrations are as follows: Initial Equilibrium Concentration (mol/L) Concentration (mol/L) 3H 4 x 3H 4 0 0 3NO2 4 x 3NO2 4 0 0 3HNO24 5.00 x 3HNO24 0 5.00 x mol/L HNO2 ¬ ¬ ¬ ¬ ¬ ¬ ¬ ¬ ¡ dissociates Initial: 0 0 Change: Equilibrium: x x 5.00 x x x x 5.00 NO2 1aq2 H1aq2 ∆ HNO21aq2 It is convenient to represent these concentrations in the following shorthand form (called an ICE table): Substituting the equilibrium concentrations in the equilibrium expression and making the approximation that give We solve for x: Using the 5% rule, we show that the approximation is valid: Therefore, We also want to calculate the equilibrium concentration of cyanide ion in this solu-tion. The ions in this solution come from the dissociation of HCN: Although the position of this equilibrium lies far to the left and does not contribute sig-nificantly to [H], HCN is the only source of . Thus we must consider the extent of the dissociation of HCN to calculate . The equilibrium expression for the preced-ing reaction is We know [H] for this solution from the results of the first part of the problem. It is impor-tant to understand that there is only one kind of H in this solution. It does not matter from which acid the H ions originate. The equilibrium [H] we need to insert into the HCN equi-librium expression is even though the H was contributed almost entirely from the dissociation of HNO2. What is [HCN] at equilibrium? We know and since Ka for HCN is so small, a negligible amount of HCN will dissociate. Thus 3HCN4 3HCN40 amount of HCN dissociated 3HCN40 1.00 M [HCN]0 1.00 M, 4.5 102 M, Ka 6.2 1010 3H4 3CN4 3HCN4 [CN] CN HCN1aq2 ∆H1aq2 CN1aq2 CN 3H4 x 4.5 102 M and pH 1.35 x 3HNO24 0 100 4.5 102 5.00 100 0.90% x 4.5 102 Ka 4.0 104 1x21x2 5.00 x x2 5.00 5.00 x 5.00 To avoid clutter we do not show the units of concentration in the ICE tables. All terms have units of mol/L.
14.5 Calculating the pH of Weak Acid Solutions 641 Since [H] and [HCN] are known, we can find from the equilibrium expression: Note the significance of this result. Since and HCN is the only source of this means that only mol/L of HCN dissociated. This is a very small amount compared with the initial concentration of HCN, which is exactly what we would expect from its very small Ka value, and M as assumed.
See Exercises 14.61 and 14.62.
Percent Dissociation It is often useful to specify the amount of weak acid that has dissociated in achieving equilibrium in an aqueous solution. The percent dissociation is defined as follows: (14.5) For example, we found earlier that in a 1.00 M solution of To reach equilibrium, of the original 1.00 M HF dissociates, so For a given weak acid, the percent dissociation increases as the acid becomes more dilute. For example, the percent dissociation of acetic acid ( ) is significantly greater in a 0.10 M solution than in a 1.0 M solution, as demonstrated in Sample Exercise 14.10.
Calculating Percent Dissociation Calculate the percent dissociation of acetic acid in each of the fol-lowing solutions.
a.
b.
Solution a. Since acetic acid is a weak acid, the major species in this solution are HC2H3O2 and H2O. Both species are weak acids, but acetic acid is a much stronger acid than water.
Thus the dominant equilibrium will be and the equilibrium expression is The initial concentrations, definition of x, and equilibrium concentrations are: Ka 1.8 105 3H4 3C2H3O2 4 3HC2H3O24 HC2H3O21aq2 ∆H 1aq2 C2H3O2 1aq2 0.100 M HC2H3O2 1.00 M HC2H3O2 (Ka 1.8 105) HC2H3O2, Ka 1.8 105 Percent dissociation 2.7 102 mol/L 1.00 mol/L 100% 2.7% 2.7 102 mol/L HF, 3H4 2.7 102 M.
Percent dissociation amount dissociated 1mol/L2 initial concentration 1mol/L2 100% [HCN] 1.00 1.4 108 CN, [CN] 1.4 108 M 3CN4 16.2 1010211.002 4.5 102 1.4 108 M Ka 6.2 1010 3H4 3CN4 3HCN4 14.5 1022 3CN4 1.00 [CN] Percent dissociation is also known as percent ionization.
Sample Exercise 14.10 H2O HC2H3O2 Major Species Initial: 1.00 0 0 Change: x x Equilibrium: x x 1.00 x x C2H3O2 (aq) H(aq) ∆ HC2H3O2(aq) 642 Chapter Fourteen Acids and Bases Inserting the equilibrium concentrations into the equilibrium expression and making the usual approximation that x is small compared with [HA]0 give Thus The approximation is valid by the 5% rule (check this yourself), so The percent dissociation is b. This is a similar problem, except that in this case M. Analysis of the problem leads to the expression Thus and See Exercises 14.63 and 14.64.
The results in Sample Exercise 14.10 show two important facts. The concentration of H ion at equilibrium is smaller in the 0.10 M acetic acid solution than in the 1.0 M acetic acid solution, as we would expect. However, the percent dissociation is significantly greater in the 0.10 M solution than in the 1.0 M solution. This is a general result. For solutions of any weak acid HA, [H] decreases as [HA]0 decreases, but the percent dissociation increases as [HA]0 decreases. This phenomenon can be explained as follows.
Consider the weak acid HA with the initial concentration [HA]0, where at equilibrium Thus Now suppose enough water is added suddenly to dilute the solution by a factor of 10. The new concentrations before any adjustment occurs are and Q, the reaction quotient, is Since Q is less than Ka, the system must adjust to the right to reach the new equilibrium position. Thus the percent dissociation increases when the acid is diluted. This behavior Q a x 10b a x 10b 3HA40 10 11x21x2 103HA4 0 1 10Ka 3HA4new 3HA40 10 3A4 new 3H 4 new x 10 Ka 3H4 3A4 3HA4 1x21x2 3HA40 3H4 3A4 x 3HA4 3HA4 0 x 3HA4 0 Percent dissociation 1.3 103 0.10 100% 1.3% x 3H4 1.3 103 M Ka 1.8 105 3H4 3C2H3O2 4 3HC2H3O24 1x21x2 0.100 x x2 0.100 [HC2H3O2] 0.100 3H4 3HC2H3O24 0 100 4.2 103 1.00 100% 0.42% 3H4 x 4.2 103 M 1.00 x 1.00 x2 1.8 105 and x 4.2 103 Ka 1.8 105 3H4 3C2H3O2 4 3HC2H3O24 1x21x2 1.00 x x2 1.00 An acetic acid solution, which is a weak electrolyte, contains only a few ions and does not conduct as much current as a strong electrolyte. The bulb is only dimly lit.
The more dilute the weak acid solution, the greater is the percent dissociation.
14.5 Calculating the pH of Weak Acid Solutions 643 is summarized in Fig. 14.10. In Sample Exercise 14.11 we see how the percent dissocia-tion can be used to calculate the Ka value for a weak acid.
Calculating Ka from Percent Dissociation Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exer-tion, leading to pain and a feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.
Solution From the small value for the percent dissociation, it is clear that HC3H5O3 is a weak acid.
Thus the major species in the solution are the undissociated acid and water: However, even though HC3H5O3 is a weak acid, it is a much stronger acid than water and will be the dominant source of H in the solution. The dissociation reaction is and the equilibrium expression is Ka 3H4 3C3H5O3 4 3HC3H5O34 HC3H5O31aq2 ∆H1aq2 C3H5O3 1aq2 HC3H5O3 and H2O CHEMICAL IMPACT Household Chemistry C ommon household bleach is an aqueous solution con-taining approximately 5% sodium hypochlorite, a potent oxidizing agent that can react with and decolorize chemi-cals that cause stains. Bleaching solutions are manufactured by dissolving chlorine gas in a sodium hydroxide solution to give the reaction As long as the pH of this solution is maintained above 8, the OCl ion is the predominant chlorine-containing species.
However, if the solution is made acidic (the [OH] lowered), elemental chlorine (Cl2) is favored, and since Cl2 is much less soluble in water than is sodium hypochlorite, Cl2 gas is suddenly evolved from the solution. This is why labels on bottles of bleach carry warnings about mixing the bleach with other cleaning solutions. For example, toilet bowl cleaners usually contain acids such as H3PO4 or HSO4 and have pH values around 2. Mixing toilet bowl cleaner with bleach can lead to a very dangerous evolution of chlorine gas.
In addition, if bleach is mixed with a cleaning agent containing ammonia, the chlorine and ammonia can react to produce chloramines, such as NH2Cl, NHCl2, and NCl3.
These compounds produce acrid fumes that can cause res-piratory distress.
Cl21g2 2OH1aq2 ∆OCl1aq2 Cl1aq2 H2O1l2 The label on this bleach bottle warns of the hazards of mixing cleaning solutions.
Sample Exercise 14.11 Acid concentration Percent dissociation H+ concentration More concentrated More dilute FIGURE 14.10 The effect of dilution on the percent dissociation and [H] of a weak acid solution.
644 Chapter Fourteen Acids and Bases The initial and equilibrium concentrations are as follows: Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [H] x [H]0 0 [C3H5O3 ] x [C3H5O3 ]0 0 [HC3H5O3] 0.10 x [HC3H5O3]0 0.10 x mol/L ¬ ¬¬¡ HC3H5O3 dissociates The change needed to reach equilibrium can be obtained from the percent dissociation and Equation (14.5). For this acid, and Now we can calculate the equilibrium concentrations: These concentrations can now be used to calculate the value of Ka for lactic acid: See Exercises 14.65 and 14.66.
14.6 Bases According to the Arrhenius concept, a base is a substance that produces OH ions in aque-ous solution. According to the Brønsted–Lowry model, a base is a proton acceptor. The bases sodium hydroxide (NaOH) and potassium hydroxide (KOH) fulfill both criteria.
They contain ions in the solid lattice and, behaving as strong electrolytes, dissociate completely when dissolved in aqueous solution: leaving virtually no undissociated NaOH. Thus a 1.0 M NaOH solution really contains 1.0 M Na and 1.0 M . Because of their complete dissociation, NaOH and KOH are called strong bases in the same sense as we defined strong acids.
All the hydroxides of the Group 1A elements (LiOH, NaOH, KOH, RbOH, and CsOH) are strong bases, but only NaOH and KOH are common laboratory reagents, because the lithium, rubidium, and cesium compounds are expensive. The alkaline earth (Group 2A) hydroxides—Ca(OH)2, Ba(OH)2, and Sr(OH)2—are also strong bases. For these com-pounds, two moles of hydroxide ion are produced for every mole of metal hydroxide dis-solved in aqueous solution.
The alkaline earth hydroxides are not very soluble and are used only when the sol-ubility factor is not important. In fact, the low solubility of these bases can sometimes be an advantage. For example, many antacids are suspensions of metal hydroxides, such as aluminum hydroxide and magnesium hydroxide. The low solubility of these compounds prevents a large hydroxide ion concentration that would harm the tissues of the mouth, esophagus, and stomach. Yet these suspensions furnish plenty of OH NaOH1s2 ¡ Na 1aq2 OH1aq2 OH Ka 3H4 3C3H5O3 4 3HC3H5O34 13.7 103213.7 1032 0.10 1.4 104 3C3H5O3 4 3H4 x 3.7 103 M 1to the correct number of significant figures2 3HC3H5O34 0.10 x 0.10 M x 3.7 10010.102 3.7 103 mol L Percent dissociation 3.7% x 3HC3H5O34 0 100% x 0.10 100% Strenuous exercise causes a buildup of lactic acid in muscle tissues.
In a basic solution at 25C, pH > 7.
H2O HC3H5O3 Major Species Visualization: Limewater and Carbon Dioxide 14.6 Bases 645 hydroxide ion to react with the stomach acid, since the salts dissolve as this reaction proceeds.
Calcium hydroxide, Ca(OH)2, often called slaked lime, is widely used in industry be-cause it is inexpensive and plentiful. For example, slaked lime is used in scrubbing stack gases to remove sulfur dioxide from the exhaust of power plants and factories. In the scrubbing process a suspension of slaked lime is sprayed into the stack gases to react with sulfur dioxide gas according to the following steps: Slaked lime is also widely used in water treatment plants for softening hard water, which involves the removal of ions, such as Ca2 and Mg2, that hamper the action of detergents. The softening method most often employed in water treatment plants is the lime–soda process, in which lime (CaO) and soda ash (Na2CO3) are added to the water.
As we will see in more detail later in this chapter, the ion reacts with water to produce the ion. When the lime is added to the water, it forms slaked lime, that is, which then reacts with the ion from the added soda ash and the Ca2 ion in the hard water to produce calcium carbonate: p From hard water Thus, for every mole of Ca(OH)2 consumed, 1 mole of Ca2 is removed from the hard water, thereby softening it. Some hard water naturally contains bicarbonate ions. In this case, no soda ash is needed—simply adding the lime produces the softening.
Calculating the pH of a strong base solution is relatively simple, as illustrated in Sam-ple Exercise 14.12.
The pH of Strong Bases Calculate the pH of a NaOH solution.
Solution The major species in this solution are From NaOH Although autoionization of water also produces ions, the pH will be dominated by the ions from the dissolved NaOH. Thus, in the solution, and the concentration of H can be calculated from Kw: Note that this is a basic solution for which 3OH4 7 3H4 and pH 7 7 pH 12.70 3H4 Kw 3OH4 1.0 1014 5.0 102 2.0 1013 M 3OH4 5.0 102 M OH OH ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Na, OH, and H2O 5.0 102 M Ca1OH221aq2 Ca21aq2 2HCO3 1aq2 ¡ 2CaCO31s2 2H2O1l2 HCO3 CaO1s2 H2O1l2 ¡ Ca1OH221aq2 HCO3 CO3 2 Ca1OH2 21aq2 H2SO31aq2 ∆CaSO31s2 2H2O1l2 SO21g2 H2O1l2 ∆H2SO31aq2 An antacid containing aluminum and magnesium hydroxides.
Calcium carbonate is also used in scrub-bing, as discussed in Section 5.10.
Sample Exercise 14.12 H2O Na+ OH– Major Species 646 Chapter Fourteen Acids and Bases The added from the salt has shifted the water autoionization equilibrium to the left, significantly lowering [H] compared with that in pure water.
See Exercises 14.77 through 14.80.
Many types of proton acceptors (bases) do not contain the hydroxide ion. However, when dissolved in water, these substances increase the concentration of hydroxide ion be-cause of their reaction with water. For example, ammonia reacts with water as follows: The ammonia molecule accepts a proton and thus functions as a base. Water is the acid in this reaction. Note that even though the base ammonia contains no hydroxide ion, it still increases the concentration of hydroxide ion to yield a basic solution.
Bases such as ammonia typically have at least one unshared pair of electrons that is capable of forming a bond with a proton. The reaction of an ammonia molecule with a water molecule can be represented as follows: There are many bases like ammonia that produce hydroxide ion by reaction with water.
In most of these bases, the lone pair is located on a nitrogen atom. Some examples are Note that the first four bases can be thought of as substituted ammonia molecules with hydrogen atoms replaced by methyl (CH3) or ethyl (C2H5) groups. The pyridine molecule is like benzene except that a nitrogen atom replaces one of the carbon atoms in the ring. The general reaction between a base B and water is given by (14.6) Base Acid Conjugate Conjugate acid base The equilibrium constant for this general reaction is where Kb always refers to the reaction of a base with water to form the conjugate acid and the hydroxide ion.
Bases of the type represented by B in Equation (14.6) compete with OH, a very strong base, for the H ion. Thus their Kb values tend to be small (for example, for am-monia, Kb 1.8 105), and they are called weak bases. The values of Kb for some common weak bases are listed in Table 14.3.
Kb 3BH 4 3OH4 3B4 B1aq2 H2O1l2 ∆BH1aq2 OH1aq2 N N H H H3C Pyridine Methylamine N H H3C CH3 Dimethylamine N H3C CH3 CH3 Trimethylamine N H C2H5 H Ethylamine H N O O H H H H H H H N H H NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 H2O1l2 ∆H1aq2 OH1aq2 OH A base does not have to contain hydroxide ion.
Appendix 5.3 contains a table of Kb values.
14.6 Bases 647 Typically, pH calculations for solutions of weak bases are very similar to those for weak acids, as illustrated by Sample Exercises 14.13 and 14.14.
The pH of Weak Bases I Calculate the pH for a 15.0 M solution of Solution Since ammonia is a weak base, as can be seen from its small Kb value, most of the dis-solved NH3 will remain as NH3. Thus the major species in solution are Both these substances can produce according to the reactions However, the contribution from water can be neglected, since Kb Kw. The equilibrium for NH3 will dominate, and the equilibrium expression to be used is The appropriate concentrations are Kb 1.8 105 3NH4 4 3OH4 3NH34 H2O1l2 ∆H1aq2 OH1aq2 Kw 1.0 1014 NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 Kb 1.8 105 OH NH3 and H2O NH3 (Kb 1.8 105).
TABLE 14.3 Values of Kb for Some Common Weak Bases Conjugate Name Formula Acid Kb Ammonia NH3 NH4 Methylamine CH3NH2 CH3NH3 Ethylamine C2H5NH2 C2H5NH3 Aniline C6H5NH2 C6H5NH3 Pyridine C5H5N C5H5NH 1.7 109 3.8 1010 5.6 104 4.38 104 1.8 105 Sample Exercise 14.13 H2O NH3 Major Species Visualization: Brønsted–Lowry Reaction Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [OH] x [OH]0 0 [NH4 ] x [NH4 ]0 0 [NH3] 15.0 x [NH3]0 15.0 Refer to the steps for solving weak acid equilibrium problems. Use the same systematic approach for weak base equilibrium problems.
x mol/L NH3 reacts with ¬ ¬¬¬¬¬¬¡ H2O to reach equilibrium Initial: 15.0 — 0 0 Change: — Equilibrium: — x x 15.0 x x x x OH(aq) NH4 (aq) ∆ H2O(l) NH3(aq) In terms of an ICE table, these concentrations are: 648 Chapter Fourteen Acids and Bases Substituting the equilibrium concentrations into the equilibrium expression and making the usual approximation gives Thus The 5% rule validates the approximation (check it yourself), so Since we know that Kw must be satisfied for this solution, we can calculate [H] as follows: Therefore, See Exercises 14.83 and 14.84.
Sample Exercise 14.13 illustrates how a typical weak base equilibrium problem should be solved. Note two additional important points: 1. We calculated [H] from Kw and then calculated the pH, but another method is available.
The pOH could have been calculated from and then used in Equation (14.3): pH 14.00 pOH pKw 14.00 pH pOH [OH] pH log16.3 10132 12.20 3H4 Kw 3OH4 1.0 1014 1.6 102 6.3 1013 M 3OH4 1.6 102 M x 1.6 102 Kb 1.8 105 3NH4 4 3OH4 3NH34 1x21x2 15.0 x x2 15.0 CHEMICAL IMPACT Amines W e have seen that many bases have nitrogen atoms with one lone pair and can be viewed as substituted am-monia molecules, with the general formula Com-pounds of this type are called amines. Amines are widely distributed in animals and plants, and complex amines of-ten serve as messengers or regulators. For example, in the human nervous system, there are two amine stimulants, nor-epinephrine and adrenaline.
HO HO OH CHCH2NHCH3 Adrenaline HO HO OH CHCH2NH2 Norepinephrine RxNH13x2.
Ephedrine, widely used as a decongestant, was a known drug in China over 2000 years ago. Indians in Mexico and the Southwest have used the hallucinogen mescaline, ex-tracted from the peyote cactus, for centuries.
Mescaline CH3 CH3 CH3 O O O CH2CH2NH2 O C H H H H N C CH3 CH3 Ephedrine A table of Kb values for bases is also given in Appendix 5.3.
14.6 Bases 649 Many other drugs, such as codeine and quinine, are amines, but they are usually not used in their pure amine forms. Instead, they are treated with an acid to become acid salts. An example of an acid salt is ammonium chloride, ob-tained by the reaction Amines also can be protonated in this way. The resulting acid salt, written as AHCl (where A represents the amine), contains AH and . In general, the acid salts are more stable and more soluble in water than the parent amines. For instance, the parent amine of the well-known local anesthetic novocaine is insoluble in water, whereas the acid salt is much more soluble.
O O C H CH2CH3 CH2CH3 H H H H Cl N C C Novocaine hydrochloride N H H Cl NH3 HCl ¡ NH4Cl Peyote cactus growing on a rock.
2. In a 15.0 M NH3 solution, the equilibrium concentrations of NH4 and are each Only a small percentage, of the ammonia reacts with water. Bottles containing 15.0 M NH3 solution are often la-beled 15.0 M NH4OH, but as you can see from these results, 15.0 M NH3 is actually a much more accurate description of the solution contents.
The pH of Weak Bases II Calculate the pH of a 1.0 M solution of methylamine Solution Since methylamine (CH3NH2) is a weak base, the major species in solution are Both are bases; however, water can be neglected as a source of OH, so the dominant equilibrium is and Kb 4.38 104 3CH3NH3 4 3OH4 3CH3NH24 CH3NH21aq2 H2O1l2 ∆CH3NH3 1aq2 OH1aq2 CH3NH2 and H2O (Kb 4.38 104).
1.6 102 15.0 100% 0.11% 1.6 102 M.
OH Sample Exercise 14.14 H2O CH3NH2 Major Species 650 Chapter Fourteen Acids and Bases The ICE table is: Initial: 1.0 — 0 0 Change: — Equilibrium: — x x 1.0 x x x x OH(aq) CH3NH3 (aq) ∆ H2O(l) CH3NH2(aq) Substituting the equilibrium concentrations in the equilibrium expression and making the usual approximation give and The approximation is valid by the 5% rule, so See Exercises 14.85 and 14.86.
14.7 Polyprotic Acids Some important acids, such as sulfuric acid (H2SO4) and phosphoric acid (H3PO4), can furnish more than one proton and are called polyprotic acids. A polyprotic acid always dissociates in a stepwise manner, one proton at a time. For example, the diprotic (two-proton) acid carbonic acid (H2CO3), which is so important in maintaining a constant pH in human blood, dissociates in the following steps: The successive Ka values for the dissociation equilibria are designated and . Note that the conjugate base of the first dissociation equilibrium becomes the acid in the second step.
Carbonic acid is formed when carbon dioxide gas is dissolved in water. In fact, the first dissociation step for carbonic acid is best represented by the reaction since relatively little H2CO3 actually exists in solution. However, it is convenient to con-sider CO2 in water as H2CO3 so that we can treat such solutions using the familiar dissociation reactions for weak acids.
Phosphoric acid is a triprotic acid (three protons) that dissociates in the following steps: HPO4 21aq2 ∆H1aq2 PO4 31aq2 Ka3 3H4 3PO4 34 3HPO4 24 4.8 1013 H2PO4 1aq2 ∆H1aq2 HPO4 21aq2 Ka2 3H4 3HPO4 24 3H2PO4 4 6.2 108 H3PO41aq2 ∆H1aq2 H2PO4 1aq2 Ka1 3H4 3H2PO4 4 3H3PO44 7.5 103 CO21aq2 H2O1l2 ∆H1aq2 HCO3 1aq2 HCO3 Ka2 Ka1 HCO3 1aq2 ∆H1aq2 CO3 21aq2 Ka2 3H4 3CO3 24 3HCO3 4 5.6 1011 H2CO31aq2 ∆H1aq2 HCO3 1aq2 Ka1 3H4 3HCO3 4 3H2CO34 4.3 107 pH 14.00 1.68 12.32 pOH 1.68 3OH4 x 2.1 102 M x 2.1 102 Kb 4.38 104 3CH3NH3 4 3OH4 3CH3NH24 1x21x2 1.0 x x2 1.0 For a typical weak polyprotic acid, That is, the acid involved in each step of the dissociation is successively weaker, as shown by the stepwise dissociation constants given in Table 14.4. These values indicate that the loss of a second or third proton occurs less readily than loss of the first proton. This is not surprising; as the negative charge on the acid increases, it becomes more difficult to remove the positively charged proton.
Although we might expect the pH calculations for solutions of polyprotic acids to be complicated, the most common cases are surprisingly straightforward. To illustrate, we will consider a typical case, phosphoric acid, and a unique case, sulfuric acid.
Phosphoric Acid Phosphoric acid is typical of most polyprotic acids in that the successive Ka values are very different. For example, the ratios of successive Ka values (from Table 14.4) are Thus the relative acid strengths are This means that in a solution prepared by dissolving H3PO4 in water, only the first disso-ciation step makes an important contribution to [H]. This greatly simplifies the pH cal-culations for phosphoric acid solutions, as is illustrated in Sample Exercise 14.15.
The pH of a Polyprotic Acid Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of the species Solution The major species in solution are H3PO4 and H2O H3PO4, H2PO4 , HPO4 2, and PO4 3.
H3PO4 H2PO4 HPO4 2 Ka2 Ka3 6.2 108 4.8 1013 1.3 105 Ka1 Ka2 7.5 103 6.2 108 1.2 105 Ka1 7 Ka2 7 Ka3 14.7 Polyprotic Acids 651 TABLE 14.4 Stepwise Dissociation Constants for Several Common Polyprotic Acids Name Formula Phosphoric acid H3PO4 Arsenic acid H3AsO4 Carbonic acid H2CO3 Sulfuric acid H2SO4 Large Sulfurous acid H2SO3 Hydrosulfuric acid H2S Oxalic acid H2C2O4 Ascorbic acid (vitamin C) H2C6H6O6 The Ka2 value for H2S is very uncertain. Because it is so small, the Ka2 value is very difficult to measure accurately.
1.6 1012 7.9 105 6.1 105 6.5 102 1019 1.0 107 1.0 107 1.5 102 1.2 102 5.6 1011 4.3 107 6 1010 8 108 5 103 4.8 1013 6.2 108 7.5 103 Ka3 Ka2 Ka1 A table of Ka values for polyprotic acids is also given in Appendix 5.2.
For a typical polyprotic acid in water, only the first dissociation step is important in determining the pH.
Sample Exercise 14.15 H2O H3PO4 Major Species 652 Chapter Fourteen Acids and Bases None of the dissociation products of H3PO4 is written, since the Ka values are all so small that they will be minor species. The dominant equilibrium is the dissociation of H3PO4: where The ICE table is: Ka1 7.5 103 3H4 3H2PO4 4 3H3PO44 H3PO41aq2 ∆H1aq2 H2PO4 1aq2 Substituting the equilibrium concentrations into the expression for and making the usual approximation give Thus Since is less than 5% of 5.0, the approximation is acceptable, and So far we have determined that and The concentration of can be obtained by using the expression for where Thus To calculate we use the expression for and the values of [H] and calculated previously: These results show that the second and third dissociation steps do not make an im-portant contribution to [H]. This is apparent from the fact that is 6.2 10–8 M, which means that only has dissociated. The value of shows that the dissociation of is even smaller. We must, however, use the second and third dissociation steps to calculate and , since these steps are the only sources of these ions.
See Exercises 14.95 and 14.96.
[PO4 3] [HPO4 2] HPO4 2 [PO4 3] 6.2 108 mol L H2PO4 [HPO4 2] 3PO4 34 14.8 1013216.2 1082 0.19 1.6 1019 M Ka3 3H4 3PO4 34 3HPO4 24 4.8 1013 0.193PO4 34 16.2 1082 [HPO4 2] Ka3 [PO4 3], 3HPO4 24 Ka2 6.2 108 M 3H4 3H2PO4 4 0.19 M Ka2 6.2 108 3H4 3HPO4 24 3H2PO4 4 Ka2: HPO4 2 3H3PO44 5.0 x 4.8 M 3H4 3H2PO4 4 0.19 M pH 0.72 3H4 x 0.19 M 1.9 101 x 1.9 101 Ka1 7.5 103 3H4 3H2PO4 4 3H3PO44 1x21x2 5.0 x x2 5.0 Ka1 Initial: 5.0 0 0 Change: Equilibrium: x x 5.0 x x x x H2PO4 (aq) H(aq) ∆ H3PO4(aq) 14.7 Polyprotic Acids 653 Sulfuric Acid Sulfuric acid is unique among the common acids in that it is a strong acid in its first dis-sociation step and a weak acid in its second step: Sample Exercise 14.16 illustrates how to calculate the pH for sulfuric acid solutions.
The pH of Sulfuric Acid Calculate the pH of a 1.0 M H2SO4 solution.
Solution The major species in the solution are where the first two ions are produced by the complete first dissociation step of H2SO4.
The concentration of H in this solution will be at least 1.0 M, since this amount is pro-duced by the first dissociation step of H2SO4. We must now answer this question: Does the ion dissociate enough to produce a significant contribution to the concentra-tion of H? This question can be answered by calculating the equilibrium concentrations for the dissociation reactions of where The ICE table is: Ka2 1.2 102 3H4 3SO4 24 3HSO4 4 HSO4 1aq2 ∆H1aq2 SO4 21aq2 HSO4 : HSO4 H, HSO4 , and H2O HSO4 1aq2 ∆H1aq2 SO4 21aq2 Ka2 1.2 102 H2SO41aq2 ¡ H1aq2 HSO4 1aq2 Ka1 is very large Note that [H]0 is not equal to zero, as it usually is for a weak acid, because the first dis-sociation step has already occurred. Substituting the equilibrium concentrations into the expression for and making the usual approximation give Thus Since is 1.2% of 1.0, the approximation is valid according to the 5% rule.
Note that x is not equal to [H] in this case. Instead, Thus the dissociation of does not make a significant contribution to the concen-tration of H, and See Exercise 14.97.
3H4 1.0 M and pH 0.00 HSO4 1.0 M 1to the correct number of significant figures2 3H4 1.0 M x 1.0 M 11.2 1022 M 1.2 102 x 1.2 102 Ka2 1.2 102 3H4 3SO4 24 3HSO4 4 11.0 x21x2 1.0 x 11.021x2 11.02 Ka2 Sample Exercise 14.16 A bottle of sulfuric acid.
Initial: 1.0 1.0 0 Change: Equilibrium: x 1.0 x 1.0 x x x x SO4 2(aq) H(aq) ∆ HSO4 (aq) H2O HSO4 – H+ Major Species 654 Chapter Fourteen Acids and Bases Sample Exercise 14.16 illustrates the most common case for sulfuric acid in which only the first dissociation makes an important contribution to the concentration of H. In solutions more dilute than 1.0 M (for example, 0.10 M H2SO4), the dissociation of HSO4 is important, and solving the problem requires use of the quadratic formula, as shown in Sample Exercise 14.17.
The pH of Sulfuric Acid Calculate the pH of a solution.
Solution The major species in solution are Proceeding as in Sample Exercise 14.16, we consider the dissociation of which leads to the following ICE table: HSO4 , H, HSO4 , and H2O 1.00 102 M H2SO4 Only in dilute H2SO4 solutions does the second dissociation step contribute significantly to [H].
Sample Exercise 14.17 Initial: 0.0100 0.0100 0 From dissociation of H2SO4 Change: Equilibrium: x 0.0100 x 0.0100 x x x x SO4 2(aq) H(aq) ∆ HSO4 (aq) Substituting the equilibrium concentrations into the expression for gives If we make the usual approximation, then 0.0100 x 0.0100 and 0.0100 x 0.0100, and we have The calculated value of x is This value is larger than 0.010, clearly a ridiculous result. Thus we cannot make the usual approximation and must instead solve the quadratic equation. The expression leads to This equation can be solved using the quadratic formula where a 1, b 2.2 102, and c 1.2 104. Use of the quadratic formula gives one negative root (which cannot be correct) and one positive root, x 4.5 103 x b 2b2 4ac 2a x2 12.2 1022x 11.2 1042 0 11.2 1042 11.2 1022x 11.0 1022x x2 11.2 102210.0100 x2 10.0100 x21x2 1.2 102 10.0100 x21x2 10.0100 x2 x 1.2 102 0.012 1.2 102 10.0100 x21x2 10.0100 x2 10.01002x 10.01002 1.2 102 Ka2 3H4 3SO4 24 3HSO4 4 10.0100 x21x2 10.0100 x2 Ka2 H2O HSO4 – H+ Major Species 14.8 Acid–Base Properties of Salts 655 Thus and Note that in this case the second dissociation step produces about half as many H ions as the initial step does.
This problem also can be solved by successive approximations, a method illustrated in Appendix 1.4.
See Exercise 14.98. pH 1.84 3H4 0.0100 x 0.0100 0.0045 0.0145 Characteristics of Weak Polyprotic Acids 1. Typically, successive Ka values are so much smaller than the first value that only the first dissociation step makes a significant contribution to the equilibrium concentra-tion of H. This means that the calculation of the pH for a solution of a typical weak polyprotic acid is identical to that for a solution of a weak monoprotic acid.
2. Sulfuric acid is unique in being a strong acid in its first dissociation step and a weak acid in its second step. For relatively concentrated solutions of sulfuric acid (1.0 M or higher), the large concentration of H from the first dissociation step represses the second step, which can be neglected as a contributor of H ions. For dilute solutions of sulfuric acid, the second step does make a significant contribution, and the quadratic equation must be used to obtain the total H concentration.
14.8 Acid–Base Properties of Salts Salt is simply another name for ionic compound. When a salt dissolves in water, we as-sume that it breaks up into its ions, which move about independently, at least in dilute so-lutions. Under certain conditions, these ions can behave as acids or bases. In this section we explore such reactions.
Salts That Produce Neutral Solutions Recall that the conjugate base of a strong acid has virtually no affinity for protons in wa-ter. This is why strong acids completely dissociate in aqueous solution. Thus, when an-ions such as and are placed in water, they do not combine with H and have no effect on the pH. Cations such as K and Na from strong bases have no affinity for H, nor can they produce H, so they too have no effect on the pH of an aqueous solu-tion. Salts that consist of the cations of strong bases and the anions of strong acids have no effect on [H] when dissolved in water. This means that aqueous solutions of salts such as KCl, NaCl, NaNO3, and KNO3 are neutral (have a pH of 7).
Salts That Produce Basic Solutions In an aqueous solution of sodium acetate (NaC2H3O2), the major species are What are the acid–base properties of each component? The Na ion has neither acid nor base properties. The ion is the conjugate base of acetic acid, a weak acid. This means that has a significant affinity for a proton and is a base. Finally, water is a weakly amphoteric substance.
C2H3O2 C2H3O2 Na, C2H3O2 , and H2O NO3 Cl The salt of a strong acid and a strong base gives a neutral solution.
H2O Na+ C2H3O2 – Major Species 656 Chapter Fourteen Acids and Bases The pH of this solution will be determined by the ion. Since is a base, it will react with the best proton donor available. In this case, water is the only source of protons, and the reaction between the acetate ion and water is (14.7) Note that this reaction, which yields a base solution, involves a base reacting with water to produce hydroxide ion and a conjugate acid. We have defined Kb as the equilibrium constant for such a reaction. In this case, The value of Ka for acetic acid is well known But how can we obtain the Kb value for the acetate ion? The answer lies in the relationships among Ka, Kb, and Kw. Note that when the expression for Ka for acetic acid is multiplied by the expression for Kb for the acetate ion, the result is Kw: This is a very important result. For any weak acid and its conjugate base, Thus, when either Ka or Kb is known, the other can be calculated. For the acetate ion, This is the Kb value for the reaction described by Equation (14.7). Note that it is obtained from the Ka value of the parent weak acid, in this case acetic acid. The sodium acetate solution is an example of an important general case. For any salt whose cation has neu-tral properties (such as Naor K) and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic. The Kb value for the anion can be obtained from the relationship Equilibrium calculations of this type are illustrated in Sample Exercise 14.18.
Salts as Weak Bases Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is .
Solution The major species in solution are Since HF is a weak acid, the F ion must have a significant affinity for protons, and the dominant reaction will be which yields the Kb expression The value of Kb can be calculated from Kw and the Ka value for HF: Kb Kw Ka 1for HF2 1.0 1014 7.2 104 1.4 1011 Kb 3HF4 3OH4 3F4 F1aq2 H2O1l2 ∆ HF1aq2 OH1aq2 Na, F, and H2O 7.2 104 Kb Kw Ka.
Kb Kw Ka 1for HC2H3O22 1.0 1014 1.8 105 5.6 1010 Ka Kb Kw Ka Kb 3H4 3C2H3O2 4 3HC2H3O24 3HC2H3O24 3OH4 3C2H3O2 4 3H4 3OH4 Kw (1.8 105).
Kb 3HC2H3O24 3OH4 3C2H3O2 4 C2H3O2 1aq2 H2O1l2 ∆HC2H3O21aq2 OH1aq2 C2H3O2 C2H3O2 A basic solution is formed if the anion of the salt is the conjugate base of a weak acid.
Sample Exercise 14.18 Na+ F– H2O Major Species 14.8 Acid–Base Properties of Salts 657 The corresponding ICE table is: Initial: 0.30 — 0 Change: — Equilibrium: — x x 0.30 x x x x 0 OH(aq) HF(aq) ∆ H2O(l) F(aq) Thus and The approximation is valid by the 5% rule, so As expected, the solution is basic.
See Exercise 14.103.
Base Strength in Aqueous Solutions To emphasize the concept of base strength, let us consider the basic properties of the cyanide ion. One relevant reaction is the dissociation of hydrocyanic acid in water: Since HCN is such a weak acid, CN appears to be a strong base, showing a very high affinity for H compared to H2O, with which it is competing. However, we also need to look at the reaction in which cyanide ion reacts with water: where In this reaction appears to be a weak base; the Kb value is only What accounts for this apparent difference in base strength? The key idea is that in the reaction of with H2O, is competing with for H, instead of competing with H2O, as it does in the HCN dissociation reaction. These equilibria show the following relative base strengths: Similar arguments can be made for other “weak” bases, such as ammonia, the acetate ion, the fluoride ion, and so on.
Salts That Produce Acidic Solutions Some salts produce acidic solutions when dissolved in water. For example, when solid NH4Cl is dissolved in water, NH4 and ions are present, with NH4 behaving as a weak acid: NH4 1aq2 ∆ NH31aq2 H1aq2 Cl OH 7 CN 7 H2O OH CN CN 1.6 105.
CN Kb Kw Ka 1.0 1014 6.2 1010 1.6 105 CN1aq2 H2O1l2 ∆ HCN1aq2 OH1aq2 HCN1aq2 H2O1l2 ∆ H3O1aq2 CN1aq2 Ka 6.2 1010 pH 14.00 5.69 8.31 pOH 5.69 3OH4 x 2.0 106 M x 2.0 106 Kb 1.4 1011 3HF4 3OH4 3F4 1x21x2 0.30 x x2 0.30 658 Chapter Fourteen Acids and Bases The ion, having virtually no affinity for H in water, does not affect the pH of the solution.
In general, salts in which the anion is not a base and the cation is the conjugate acid of a weak base produce acidic solutions.
Salts as Weak Acids I Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is Solution The major species in solution are Note that both NH4 and H2O can produce H. The dissociation reaction for the NH4 ion is for which Note that although the Kb value for NH3 is given, the reaction corresponding to Kb is not appropriate here, since NH3 is not a major species in the solution. Instead, the given value of Kb is used to calculate Ka for NH4 from the relationship Thus Although NH4 is a very weak acid, as indicated by its Ka value, it is stronger than H2O and will dominate in the production of H. Thus we will focus on the dissociation reac-tion of NH4 to calculate the pH in this solution.
We solve the weak acid problem in the usual way: Ka 1for NH4 2 Kw Kb 1for NH32 1.0 1014 1.8 105 5.6 1010 Ka Kb Kw Ka 3NH34 3H4 3NH4 4 NH4 1aq2 ∆ NH31aq2 H1aq2 NH4 , Cl, and H2O 1.8 105.
Cl Sample Exercise 14.19 Initial: 0.10 0 Change: Equilibrium: x x 0.10 x x x x 0 NH3(aq) H(aq) ∆ NH4 (aq) Thus The approximation is valid by the 5% rule, so See Exercise 14.104.
A second type of salt that produces an acidic solution is one that contains a highly charged metal ion. For example, when solid aluminum chloride (AlCl3) is dissolved in water, the resulting solution is significantly acidic. Although the Al3 ion is not 3H4 x 7.5 106 M and pH 5.13 x 7.5 106 5.6 1010 Ka 3H4 3NH34 3NH4 4 1x21x2 0.10 x x2 0.10 H2O Cl– NH4 + Major Species 14.8 Acid–Base Properties of Salts 659 itself a Brønsted–Lowry acid, the hydrated ion Al(H2O)6 3 formed in water is a weak acid: The high charge on the metal ion polarizes the bonds in the attached water mole-cules, making the hydrogens in these water molecules more acidic than those in free water molecules. Typically, the higher the charge on the metal ion, the stronger the acidity of the hydrated ion.
Salts as Weak Acids II Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for Al(H2O)6 3 is Solution The major species in solution are Since the Al(H2O)6 3 ion is a stronger acid than water, the dominant equilibrium is and This is a typical weak acid problem, which we can solve with the usual procedure: 1.4 105 Ka 3Al1OH21H2O25 24 3H4 3Al1H2O26 34 Al1H2O26 31aq2 ∆ Al1OH21H2O25 21aq2 H1aq2 Al1H2O26 3, Cl, and H2O 1.4 105.
O¬H Al1H2O26 31aq2 ∆ Al1OH21H2O25 21aq2 H1aq2 Sample Exercise 14.20 Initial: 0.010 0 Change: Equilibrium: x x 0.010 x x x x 0 H1aq2 Al1OH21H2O25 21aq2 ∆ Al1H2O26 31aq2 Thus Since the approximation is valid by the 5% rule, See Exercises 14.109 and 14.110.
So far we have considered salts in which only one of the ions has acidic or basic properties. For many salts, such as ammonium acetate (NH4C2H3O2), both ions can affect the pH of the aqueous solution. Because the equilibrium calculations for these cases can be quite complicated, we will consider only the qualitative aspects of such problems. We can predict whether the solution will be basic, acidic, or neutral by comparing the Ka value for the acidic ion with the Kb value for the basic ion. If the Ka value for the acidic ion is larger than the Kb value for the basic ion, the solution will be acidic. If the Kb value is larger than the Ka value, the solution will be basic. Equal Ka and Kb values mean a neutral solution. These facts are summarized in Table 14.5.
3H4 x 3.7 104 M and pH 3.43 x 3.7 104 1.4 105 Ka 3Al1OH21H2O25 24 3H4 3Al1H2O26 34 1x21x2 0.010 x x2 0.010 Section 14.9 contains a further discus-sion of the acidity of hydrated ions.
TABLE 14.5 Qualitative Prediction of pH for Solutions of Salts for Which Both Cation and Anion Have Acidic or Basic Properties (acidic) (basic) (neutral) pH 7 Ka Kb pH 7 7 Kb 7 Ka pH 6 7 Ka 7 Kb Cl– Al(H2O)6 3+ H2O Major Species 660 Chapter Fourteen Acids and Bases The Acid–Base Properties of Salts Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral.
a. NH4C2H3O2 b. NH4CN c. Al2(SO4)3 Solution a. The ions in solution are NH4 and C2H3O2 . As we mentioned previously, Ka for NH4 is 5.6 1010 and Kb for C2H3O2 is 5.6 1010. Thus Ka for NH4 is equal to Kb for C2H3O2 , and the solution will be neutral (pH 7).
b. The solution will contain NH4 and CN ions. The Ka value for NH4 is 5.6 1010 and Since Kb for CN is much larger than Ka for NH4 , CN is a much stronger base than NH4 is an acid. This solution will be basic.
c. The solution will contain Al(H2O)6 3 and SO4 2 ions. The Ka value for Al(H2O)6 3 is 1.4 105, as given in Sample Exercise 14.20. We must calculate Kb for SO4 2. The HSO4 ion is the conjugate acid of SO4 2, and its Ka value is for sulfuric acid, or 1.2 102. Therefore, This solution will be acidic, since Ka for Al(H2O)6 3 is much greater than Kb for SO4 2.
See Exercises 14.111 and 14.112.
The acid–base properties of aqueous solutions of various salts are summarized in Table 14.6.
1.0 1014 1.2 102 8.3 1013 Kb 1for SO4 22 Kw Ka2 1for sulfuric acid2 Ka2 Kb1for CN2 Kw Ka 1for HCN2 1.6 105 Sample Exercise 14.21 TABLE 14.6 Acid–Base Properties of Various Types of Salts Type of Salt Examples Comment pH of Solution Cation is from strong base; anion is KCl, KNO3, Neither acts as an Neutral from strong acid NaCl, NaNO3 acid or a base Cation is from strong base; anion is NaC2H3O2, Anion acts as a base; Basic from weak acid KCN, NaF cation has no effect on pH Cation is conjugate acid of weak base; NH4Cl, Cation acts as acid; Acidic anion is from strong acid NH4NO3 anion has no effect on pH Cation is conjugate acid of weak base; NH4C2H3O2, Cation acts as an acid; Acidic if Ka Kb, anion is conjugate base of weak acid NH4CN anion acts as a base basic if Kb Ka, neutral if Ka Kb Cation is highly charged metal ion; Al(NO3)3, Hydrated cation acts as Acidic anion is from strong acid FeCl3 an acid; anion has no effect on pH 14.9 The Effect of Structure on Acid–Base Properties 661 14.9 The Effect of Structure on Acid–Base Properties We have seen that when a substance is dissolved in water, it produces an acidic solution if it can donate protons and produces a basic solution if it can accept protons. What struc-tural properties of a molecule cause it to behave as an acid or as a base?
Any molecule containing a hydrogen atom is potentially an acid. However, many such molecules show no acidic properties. For example, molecules containing COH bonds, such as chloroform (CHCl3) and nitromethane (CH3NO2), do not produce acidic aqueous solutions because a COH bond is both strong and nonpolar and thus there is no tendency to donate protons. On the other hand, although the HOCl bond in gaseous hydrogen chloride is slightly stronger than a COH bond, it is much more polar, and this molecule readily dissociates when dissolved in water.
Thus there are two main factors that determine whether a molecule containing an XOH bond will behave as a Brønsted–Lowry acid: the strength of the bond and the polarity of the bond.
To explore these factors let’s consider the relative acid strengths of the hydrogen halides. The bond polarities vary as shown h h Most polar Least polar because electronegativity decreases going down the group. Based on the high polarity of the HOF bond, we might expect hydrogen fluoride to be a very strong acid. In fact, among HX molecules, HF is the only weak acid when dissolved in water.
The HOF bond is unusually strong, as shown in Table 14.7, and thus is difficult to break.
This contributes significantly to the reluctance of the HF molecules to dissociate in water.
Another important class of acids are the oxyacids, which as we saw in Section 14.2 characteristically contain the grouping HOOOX. Several series of oxyacids are listed with their Ka values in Table 14.8. Note from these data that for a given series the acid strength increases with an increase in the number of oxygen atoms attached to the cen-tral atom. For example, in the series containing chlorine and a varying number of oxygen atoms, HOCl is a weak acid, but the acid strength is successively greater as the number of oxygen atoms increases. This happens because the very electronegative oxygen atoms are able to draw electrons away from the chlorine atom and the OOH bond, as shown in Fig. 14.11. The net effect is to both polarize and weaken the OOH bond; this effect be-comes more important as the number of attached oxygen atoms increases. This means that a proton is most readily produced by the molecule with the largest number of attached oxygen atoms (HClO4).
This type of behavior is also observed for hydrated metal ions. Earlier in this chapter we saw that highly charged metal ions such as Al3 produce acidic solutions. The acidity of the water molecules attached to the metal ion is increased by the attraction of electrons to the positive metal ion: The greater the charge on the metal ion, the more acidic the hydrated ion becomes.
For acids containing the H—O—X grouping, the greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule. Since the electronegativ-ity of X reflects its ability to attract the electrons involved in bonding, we might expect acid strength to depend on the electronegativity of X. In fact, there is an excellent cor-relation between the electronegativity of X and the acid strength for oxyacids, as shown in Table 14.9.
G D O H O H Al3 (Ka 7.2 104) H¬F 7 H¬Cl 7 H¬Br 7 H¬I Further aspects of acid strengths are discussed in Section 20.7.
TABLE 14.7 Bond Strengths and Acid Strengths for Hydrogen Halides Bond Acid Strength Strength Bond (kJ/mol) in Water 565 Weak 427 Strong 363 Strong 295 Strong H¬I H¬Br H¬Cl H¬F H¬X 662 Chapter Fourteen Acids and Bases 14.10 Acid–Base Properties of Oxides We have just seen that molecules containing the grouping H—O—X can behave as acids and that the acid strength depends on the electron-withdrawing ability of X. But sub-stances with this grouping also can behave as bases if the hydroxide ion instead of a pro-ton is produced. What determines which behavior will occur? The answer lies mainly in the nature of the O—X bond. If X has a relatively high electronegativity, the O—X bond will be covalent and strong. When the compound containing the H—O—X grouping is dissolved in water, the O—X bond will remain intact. It will be the polar and relatively weak H—O bond that will tend to break, releasing a proton. On the other hand, if X has Cl O H Electron density Cl O H Electron density O Cl O H Electron density O O Cl O H Electron density O O O FIGURE 14.11 The effect of the number of attached oxygens on the OOH bond in a series of chlorine oxyacids. As the number of oxygen atoms attached to the chlorine atom increases, they become more effec-tive at withdrawing electron density from the OOH bond, thereby weakening and polarizing it. This increases the tendency for the molecule to produce a proton, and so its acid strength increases.
TABLE 14.9 Comparison of Electronegativity of X and Ka Value for a Series of Oxyacids Electronegativity Acid X of X Ka for Acid HOCl Cl 3.0 HOBr Br 2.8 HOI I 2.5 HOCH3 CH3 2.3 (for carbon in CH3) 1015 2 1011 2 109 4 108 A compound containing the group will produce an acidic solution in water if the bond is strong and covalent. If the bond is ionic, the compound will produce a basic solution in water.
O¬X O¬X H¬O¬X TABLE 14.8 Several Series of Oxyacids and Their Ka Values Oxyacid Structure Ka Value HClO4 Large HClO3 HClO2 H—O—Cl—O HClO H—O—Cl H2SO4 Large H2SO3 HNO3 Large HNO2 H—O—N—O 4.0 104 HOO G D O O O N 1.5 102 O O G D O H O O O H S O O G D O O H O O O O H S 3.5 108 1.2 102 1 O G D O H O Cl O O (1072 O G D O O H O Cl O O O 14.11 The Lewis Acid–Base Model 663 a very low electronegativity, the O—X bond will be ionic and subject to being broken in polar water. Examples are the ionic substances NaOH and KOH that dissolve in water to give the metal cation and the hydroxide ion.
We can use these principles to explain the acid–base behavior of oxides when they are dissolved in water. For example, when a covalent oxide such as sulfur trioxide is dis-solved in water, an acidic solution results because sulfuric acid is formed: The structure of H2SO4 is shown in the margin. In this case, the strong, covalent O—S bonds remain intact and the H—O bonds break to produce protons. Other common co-valent oxides that react with water to form acidic solutions are sulfur dioxide, carbon diox-ide, and nitrogen dioxide, as shown by the following reactions: Thus, when a covalent oxide dissolves in water, an acidic solution forms. These oxides are called acidic oxides.
On the other hand, when an ionic oxide dissolves in water, a basic solution results, as shown by the following reactions: These reactions can be explained by recognizing that the oxide ion has a high affinity for protons and reacts with water to produce hydroxide ions: Thus the most ionic oxides, such as those of the Group 1A and 2A metals, produce basic solutions when they are dissolved in water. As a result, these oxides are called basic oxides.
14.11 The Lewis Acid–Base Model We have seen that the first successful conceptualization of acid–base behavior was proposed by Arrhenius. This useful but limited model was replaced by the more general Brønsted–Lowry model. An even more general model for acid–base behavior was sug-gested by G. N. Lewis in the early 1920s. A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. Another way of saying this is that a Lewis acid has an empty atomic orbital that it can use to accept (share) an electron pair from a mol-ecule that has a lone pair of electrons (Lewis base). The three models for acids and bases are summarized in Table 14.10.
O21aq2 H2O1l2 ¡ 2OH1aq2 K2O1s2 H2O1l2 ¡ 2KOH1aq2 CaO1s2 H2O1l2 ¡ Ca1OH221aq2 2NO21g2 H2O1l2 ¡ HNO31aq2 HNO21aq2 CO21g2 H2O1l2 ¡ H2CO31aq2 SO21g2 H2O1l2 ¡ H2SO31aq2 SO31g2 H2O1l2 ¡ H2SO41aq2 TABLE 14.10 Three Models for Acids and Bases Model Definition of Acid Definition of Base Arrhenius H producer producer Brønsted–Lowry H donor H acceptor Lewis Electron-pair acceptor Electron-pair donor OH 664 Chapter Fourteen Acids and Bases Note that Brønsted–Lowry acid–base reactions (proton donor–proton acceptor reac-tions) are encompassed by the Lewis model. For example, the reaction between a proton and an ammonia molecule, that is, can be represented as a reaction between an electron-pair acceptor (H) and an electron-pair donor (NH3). The same holds true for a reaction between a proton and a hydroxide ion: The real value of the Lewis model for acids and bases is that it covers many reactions that do not involve Brønsted–Lowry acids. For example, consider the gas-phase reaction between boron trifluoride and ammonia.
Here the electron-deficient BF3 molecule (there are only six electrons around the boron) completes its octet by reacting with NH3, which has a lone pair of electrons. (see Fig. 14.12.) In fact, as mentioned in Chapter 8, the electron deficiency of boron trifluoride makes it very reactive toward any electron-pair donor. That is, it is a strong Lewis acid.
The hydration of a metal ion, such as Al3, also can be viewed as a Lewis acid–base reaction: O H H Lewis acid Lewis base 6 Al3 3 O H H Al 6 H N H H F B F F H N H H F B F F Lewis acid Lewis base H] H Lewis acid Lewis base O O H H [ H N H H H H H N H H Lewis acid Lewis base + FIGURE 14.12 Reaction of BF3 with NH3.
The Lewis model encompasses the Brønsted–Lowry model, but the reverse is not true.
14.11 The Lewis Acid–Base Model 665 Here the Al3 ion accepts one electron pair from each of six water molecules to form (see Fig. 14.13).
In addition, the reaction between a covalent oxide and water to form a Brønsted–Lowry acid can be defined as a Lewis acid–base reaction. An example is the reaction between sulfur trioxide and water: Note that as the water molecule attaches to sulfur trioxide, a proton shift occurs to form sulfuric acid.
Lewis Acids and Bases For each reaction, identify the Lewis acid and base.
a.
b.
Solution a. Each NH3 molecule donates an electron pair to the Ni2 ion: The nickel(II) ion is the Lewis acid, and ammonia is the Lewis base.
b. The proton is the Lewis acid and the water molecule is the Lewis base: See Exercises 14.119 and 14.120.
Lewis acid Lewis base H O H H H O H H Lewis acid Lewis base 6 Ni2 2 N H H Ni 6 H N H N H H H(aq) H2O(aq) ∆H3O(aq) Ni2(aq) 6NH3(aq) ¡ Ni(NH3)6 2(aq) O O O O H H S S H O O O Lewis acid Lewis base H O Al(H2O)6 3 Al3+ Al3+ FIGURE 14.13 The ion.
Al(H2O)6 3 Sample Exercise 14.22 666 Chapter Fourteen Acids and Bases 14.12 Strategy for Solving Acid–Base Problems: A Summary In this chapter we have encountered many different situations involving aqueous solu-tions of acids and bases, and in the next chapter we will encounter still more. In solving for the equilibrium concentrations in these aqueous solutions, it is tempting to create a pigeonhole for each possible situation and to memorize the procedures necessary to deal with that particular case. This approach is just not practical and usually leads to frustration: Too many pigeonholes are required—there seems to be an infinite number of cases. But you can handle any case successfully by taking a systematic, patient, and thoughtful approach. When analyzing an acid–base equilibrium problem, do not ask yourself how a memorized solution can be used to solve the problem. Instead, ask this question: What are the major species in the solution and what is their chemical behavior?
The most important part of doing a complicated acid–base equilibrium problem is the analysis you do at the beginning of a problem.
What major species are present?
Does a reaction occur that can be assumed to go to completion?
What equilibrium dominates the solution?
Let the problem guide you. Be patient.
The following steps outline a general strategy for solving problems involving acid–base equilibria.
T he New York City Public Library has 88 miles of bookshelves, and on 36 miles of these shelves the books are quietly disintegrat-ing between their covers. In fact, an estimated 40% of the books in the major research collec-tions in the United States will soon be too frag-ile to handle.
The problem results from the acidic pa-per widely used in printing books in the past century. Ironically, books from the eighteenth, seventeenth, sixteenth, and even fifteenth cen-tury are in much better shape. Gutenberg Bibles contain paper that is in remarkably good condition. In those days, paper was made by hand from linen or rags, but in the nine-teenth century, the demand for cheap paper skyrocketed. Paper manufacturers found that paper could be made economically, by machine, using wood pulp. To size the paper (that is, fill in microscopic holes to lower absorption of moisture and prevent seeping or spread-ing of inks), alum was added in large amounts.
[Al2(SO4)3] CHEMICAL IMPACT Self-Destructing Paper Because the hydrated aluminum ion is an acid paper manufactured using alum is quite acidic.
(Ka 105), [Al(H2O)6 3] A book ravaged by the decomposition of acidic paper.
14.12 Strategy for Solving Acid–Base Problems: A Summary 667 Solving Acid–Base Problems ➥ 1 List the major species in solution.
➥ 2 Look for reactions that can be assumed to go to completion—for example, a strong acid dissociating or H reacting with ➥ 3 For a reaction that can be assumed to go to completion: a. Determine the concentration of the products.
b. Write down the major species in solution after the reaction.
➥ 4 Look at each major component of the solution and decide if it is an acid or a base.
➥ 5 Pick the equilibrium that will control the pH. Use known values of the dissociation constants for the various species to help decide on the dominant equilibrium.
a. Write the equation for the reaction and the equilibrium expression.
b. Compute the initial concentrations (assuming the dominant equilibrium has not yet occurred, that is, no acid dissociation, and so on).
c. Define x.
d. Compute the equilibrium concentrations in terms of x.
e. Substitute the concentrations into the equilibrium expression, and solve for x.
f. Check the validity of the approximation.
g. Calculate the pH and other concentrations as required.
Although these steps may seem somewhat cumbersome, especially for simpler prob-lems, they will become increasingly helpful as the aqueous solutions become more com-plicated. If you develop the habit of approaching acid–base problems systematically, the more complex cases will be much easier to manage.
OH.
Over time this acidity causes the paper fibers to disintegrate; the pages of books fall apart when they are used.
One could transfer the contents of the threatened books to microfilm, but that would be a very slow and expensive process. Can the books be chemically treated to neutralize the acid and stop the deterioration? Yes. In fact, you know enough chemistry at this point to design the treatment patented in 1936 by Otto Schierholz. He dipped individual pages in solutions of alkaline earth bicarbonate salts [Mg(HCO3)2, Ca(HCO3)2, and so on]. The HCO3 ions present in these solutions react with the H in the paper to give CO2 and H2O.
This treatment works well and is used today to preserve es-pecially important works, but it is slow and labor-intensive.
It would be much more economical if large numbers of books could be treated at one time without disturbing the bindings. However, soaking entire books in an aqueous so-lution is out of the question. A logical question then is: Are there gaseous bases that could be used to neutralize the acid?
Certainly; the organic amines (general formula, RNH2) are bases, and those with low molar masses are gases under normal conditions. Experiments in which books were treated using ammonia, butylamine (CH3CH2CH2CH2NH2), and other amines have shown that the method works, but only for a short time. The amines do enter the paper and neutralize the acid, but being volatile, they gradually evaporate, leaving the paper in its original acidic condition.
A much more effective treatment involves diethylzinc [(CH3CH2)2Zn], which boils at and 1 atm. Diethylzinc (DEZ) reacts with oxygen or water to produce ZnO as follows: The solid zinc oxide produced in these reactions is deposited among the paper fibers, and being a basic oxide, it neutral-izes the acid present as shown in the equation One major problem is that DEZ ignites spontaneously on contact with air. Therefore, this treatment must be carried out in a chamber filled mainly with N2(g), where the amount of O2 present can be rigorously controlled. The pressure in the chamber must be maintained well below one atmosphere both to lower the boiling point of DEZ and to remove ex-cess moisture from the book’s pages. Several major DEZ fires have slowed its implementation as a book preservative.
However, the Library of Congress has designed a new DEZ treatment plant that includes a chamber large enough for ap-proximately 9000 books to be treated at one time.
ZnO 2H ¡ Zn2 H2O 1CH3CH222Zn1g2 H2O1g2 ¡ ZnO1s2 2CH3CH31g2 ZnO1s2 4CO21g2 5H2O1g2 1CH3CH222Zn1g2 7O21g2 ¡ 117°C 668 Chapter Fourteen Acids and Bases Key Terms Section 14.1 Arrhenius concept Brønsted–Lowry model hydronium ion conjugate base conjugate acid conjugate acid–base pair acid dissociation constant Section 14.2 strong acid weak acid diprotic acid oxyacids organic acids carboxyl group monoprotic acids amphoteric substance autoionization ion-product (dissociation) constant Section 14.3 pH scale Section 14.4 major species Section 14.5 percent dissociation Section 14.6 strong bases slaked lime lime–soda process weak bases amine Section 14.7 polyprotic acid triprotic acid Section 14.8 salt Section 14.10 acidic oxides basic oxides Section 14.11 Lewis acid Lewis base Models for acids and bases Arrhenius model • Acids produce H in solution • Bases produce in solution Brønsted–Lowry model • An acid is a proton donor • A base is a proton acceptor • In this model an acid molecule reacts with a water molecule, which behaves as a base: Acid Base Conjugate Conjugate acid base to form a new acid (conjugate acid) and a new base (conjugate base).
Lewis model • A Lewis acid is an electron-pair acceptor • A Lewis base is an electron-pair donor Acid–base equilibrium The equilibrium constant for an acid dissociating (ionizing) in water is called Ka The Ka expression is which is often simplified as • is never included because it is assumed to be constant Acid strength A strong acid has a very large Ka value • The acid completely dissociates (ionizes) in water • The dissociation (ionization) equilibrium position lies all the way to the right • Strong acids have very weak conjugate bases • The common strong acids are nitric acid hydrochloric acid sulfuric acid and perchloric acid A weak acid has a small Ka value • The acid dissociates (ionizes) to only a slight extent • The dissociation (ionization) equilibrium position lies far to the left • Weak acids have relatively strong conjugate bases • Percent dissociation of a weak acid • The smaller the percent dissociation, the weaker the acid • Dilution of a weak acid increases its percent dissociation Autoionization of water Water is an amphoteric substance: it behaves as both an acid and a base Water reacts with itself in an acid–base reaction H2O1l2 H2O1l2 ∆H3O1aq2 OH1aq2 % dissociation amount dissociated 1mol/L2 initial concentration 1mol/L2 100% [HClO4(aq)] [H2SO(aq)] [HCl(aq)], [HNO3(aq)], [H2O] Ka 3H 4 3A4 3HA4 Ka 3H3O 4 3A4 3HA4 HA(aq) H2O(l) ∆H3O(aq) A(aq) OH For Review For Review 669 which leads to the equilibrium expression • Kw is the ion-product constant for water • At in pure water so Acidic solution: Basic solution: Neutral solution: The pH scale Since pH is a log scale, the pH changes by 1 for every 10-fold change in The log scale is also used for and for Ka values Bases Strong bases are hydroxide salts, such as NaOH and KOH Weak bases react with water to produce • The equilibrium constant for this reaction is called Kb where • In water a base B is always competing with for a proton (H), so Kb values tend to be very small, thus making B a weak base (compared to ) Polyprotic acids A polyprotic acid has more than one acidic proton Polyprotic acids dissociate one proton at a time • Each step has a characteristic Ka value • Typically for a weak polyprotic acid, Sulfuric acid is unique • It is a strong acid in the first dissociation step ( is very large) • It is a weak acid in the second step Acid–base properties of salts Can produce acidic, basic, or neutral solutions Salts that contain: • Cations of strong bases and anions of strong acids produce neutral solutions • Cations of strong bases and anions of weak acids produce basic solutions • Cations of weak bases and anions of strong acids produce acidic solutions Acidic solutions are produced by salts containing a highly charged metal cation— for example, Al3 and Fe3 Effect of structure on acid–base properties Many substances that function as acids or bases contain the grouping • Molecules in which the bond is strong and covalent tend to behave as acids • As X becomes more electronegative, the acid becomes stronger • When the bond is ionic, the substance behaves as a base, releasing ions in water OH O¬X O¬X H¬O¬X Ka1 Ka1 7 Ka2 7 Ka3 OH OH Kb 3BH 4 3OH4 3B4 B1aq2 H2O1l2 ∆BH1aq2 OH1aq2 OH pKa logKa pOH log3OH4 [OH] [H] pH log [H] [H] [OH] [OH] 7 [H] [H] 7 [OH] Kw 1.0 1014 [H] [OH] 1.0 107, 25°C Kw 3H3O 4 3OH4 or 3H4 3OH4 Kw 670 Chapter Fourteen Acids and Bases REVIEW QUESTIONS 1. Define each of the following: a. Arrhenius acid b. Brønsted–Lowry acid c. Lewis acid Which of the definitions is most general? Write reactions to justify your answer.
2. Define or illustrate the meaning of the following terms: a. Ka reaction b. Ka equilibrium constant c. Kb reaction d. Kb equilibrium constant e. conjugate acid–base pair 3. Define or illustrate the meaning of the following terms: a. amphoteric b. Kw reaction c. Kw equilibrium constant d. pH e. pOH f. pKw Give the conditions for a neutral solution at in terms of [H], pH, and the relationship between [H] and Do the same for an acidic solution and for a basic solution. As a solution becomes more acidic, what happens to pH, pOH, [H], and As a solution becomes more basic, what happens to pH, pOH, [H], and 4. How is acid strength related to the value of Ka? What is the difference between strong acids versus weak acids (see Table 14.1)? As the strength of an acid in-creases, what happens to the strength of the conjugate base? How is base strength related to the value of Kb? As the strength of a base increases, what happens to the strength of the conjugate acid?
5. Two strategies are followed when solving for the pH of an acid in water. What is the strategy for calculating the pH of a strong acid in water? What major as-sumptions are made when solving strong acid problems? The best way to recog-nize strong acids is to memorize them. List the six common strong acids (the two not listed in the text are HBr and HI).
Most acids, by contrast, are weak acids. When solving for the pH of a weak acid in water, you must have the Ka value. List two places in this text that pro-vide Ka values for weak acids. You can utilize these tables to help you recognize weak acids. What is the strategy for calculating the pH of a weak acid in water?
What assumptions are generally made? What is the 5% rule? If the 5% rule fails, how do you calculate the pH of a weak acid in water?
6. Two strategies are also followed when solving for the pH of a base in water.
What is the strategy for calculating the pH of a strong base in water? List the strong bases mentioned in the text that should be committed to memory. Why is calculating the pH of Ca(OH)2 solutions a little more difficult than calculating the pH of NaOH solutions?
Most bases are weak bases. The presence of what element most commonly results in basic properties for an organic compound? What is present on this ele-ment in compounds that allows it to accept a proton?
Table 14.3 and Appendix 5 of the text list Kb values for some weak bases.
What strategy is used to solve for the pH of a weak base in water? What as-sumptions are made when solving for the pH of weak base solutions? If the 5% rule fails, how do you calculate the pH of a weak base in water?
[OH]?
[OH]?
[OH].
25°C, For Review 671 7. Table 14.4 lists the stepwise Ka values for some polyprotic acids. What is the difference between a monoprotic acid, a diprotic acid, and a triprotic acid? Most polyprotic acids are weak acids; the major exception is To solve for the pH of a solution of you must solve a strong acid problem as well as a weak acid problem. Explain. Write out the reactions that refer to and for For Write out the reactions that refer to the equilibrium constants.
What are the three acids in a solution of Which acid is strongest? What are the three conjugate bases in a solution of H3PO4? Which conjugate base is strongest? Summarize the strategy for calculating the pH of a polyprotic acid in water.
8. For conjugate acid–base pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water where a. Which two bases are competing for the proton?
b. Which is the stronger base?
c. In light of your answer to part b, why do we classify the acetate ion as a weak base? Use an appropriate reaction to justify your answer.
In general, as base strength increases, conjugate acid strength decreases.
Explain why the conjugate acid of the weak base NH3 is a weak acid.
To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids.
To tie it all together, some instructors have students think of Li, K, Rb, Cs, Ca2, Sr2, and Ba2 as the conjugate acids of the strong bases LiOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not techni-cally correct, the conjugate acid strength of these cations is similar to the con-jugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the cor-rect response. The conjugate base of a weak acid is a base.
The conjugate acid of a weak base is a acid. The conjugate base of a strong acid is a base. The conjugate acid of a strong base is a acid. (Hint: Weak gives you weak and strong gives you worthless.) 9. What is a salt? List some anions that behave as weak bases in water. List some anions that have no basic properties in water. List some cations that behave as weak acids in water. List some cations that have no acidic proper-ties in water. Using these lists, give some formulas for salts that have only weak base properties in water. What strategy would you use to solve for the pH of these basic salt solutions? Identify some salts that have only weak acid properties in water. What strategy would you use to solve for the pH of these acidic salt solutions? Identify some salts that have no acidic or basic proper-ties in water (produce neutral solutions). When a salt contains both a weak acid ion and a weak base ion, how do you predict whether the solution pH is acidic, basic, or neutral?
1 106, (CH3CO2 ) Ka 1.8 105.
CH3CO2H1aq2 H2O1l2 ∆CH3CO2 1aq2 H3O1aq2 H3PO4?
Ka1, Ka2, and Ka3 H3PO4, Ka1 7.5 103, Ka2 6.2 108, and Ka3 4.8 1013.
H2SO4.
Ka2 Ka1 H2SO4, H2SO4.
672 Chapter Fourteen Acids and Bases 10. For oxyacids, how does acid strength depend on a. the strength of the bond to the acidic hydrogen atom?
b. the electronegativity of the element bonded to the oxygen atom that bears the acidic hydrogen?
c. the number of oxygen atoms?
How does the strength of a conjugate base depend on these factors?
What type of solution forms when a nonmetal oxide dissolves in water?
Give an example of such an oxide. What type of solution forms when a metal oxide dissolves in water? Give an example of such an oxide.
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Consider two beakers of pure water at different temperatures.
How do their pH values compare? Which is more acidic? more basic? Explain.
2. Differentiate between the terms strength and concentration as they apply to acids and bases. When is HCl strong? Weak? Con-centrated? Dilute? Answer the same questions for ammonia. Is the conjugate base of a weak acid a strong base?
3. Sketch two graphs: (a) percent dissociation for weak acid HA versus the initial concentration of HA ([HA]0) and (b) H con-centration versus [HA]0. Explain both.
4. Consider a solution prepared by mixing a weak acid HA and HCl. What are the major species? Explain what is occurring in solution. How would you calculate the pH? What if you added NaA to this solution? Then added NaOH?
5. Explain why salts can be acidic, basic, or neutral, and show ex-amples. Do this without specific numbers.
6. Consider two separate aqueous solutions: one of a weak acid HA and one of HCl. Assuming you started with 10 molecules of each: a. Draw a picture of what each solution looks like at equilibrium.
b. What are the major species in each beaker?
c. From your pictures, calculate the Ka values of each acid.
d. Order the following from the strongest to the weakest base: Explain your order.
7. You are asked to calculate the H concentration in a solution of NaOH(aq). Because sodium hydroxide is a base, can we say there is no H, since having H would imply that the solution is acidic?
8. Consider a solution prepared by mixing a weak acid HA, HCl, and NaA. Which of the following statements best describes what happens?
H2O, A, Cl.
a. The H from the HCl reacts completely with the from the NaA. Then the HA dissociates somewhat.
b. The H from the HCl reacts somewhat with the from the NaA to make HA, while the HA is dissociating. Eventually you have equal amounts of everything.
c. The H from the HCl reacts somewhat with the from the NaA to make HA while the HA is dissociating. Eventually all the reactions have equal rates.
d. The H from the HCl reacts completely with the from the NaA. Then the HA dissociates somewhat until “too much” H and are formed, so the H and react to form HA, and so on. Eventually equilibrium is reached.
Justify your choice, and for choices you did not pick, explain what is wrong with them.
9. Consider a solution formed by mixing 100.0 mL of 0.10 M HA 100.00 mL of 0.10 M NaA, and 100.0 mL of 0.10 M HCl. In calculating the pH for the final solution, you would make some assumptions about the order in which var-ious reactions occur to simplify the calculations. State these assumptions. Does it matter whether the reactions actually oc-cur in the assumed order? Explain.
10. A certain sodium compound is dissolved in water to liberate Na ions and a certain negative ion. What evidence would you look for to determine whether the anion is behaving as an acid or a base? How could you tell whether the anion is a strong base?
Explain how the anion could behave simultaneously as an acid and a base.
11. Acids and bases can be thought of as chemical opposites (acids are proton donors, and bases are proton acceptors). Therefore, one might think that . Why isn’t this the case? What is the relationship between Ka and Kb? Prove it with a derivation.
12. Consider two solutions of the salts NaX(aq) and NaY(aq) at equal concentrations. What would you need to know to deter-mine which solution has the higher pH? Explain how you would decide (perhaps even provide a sample calculation).
13. What is meant by pH? True or false: A strong acid solution always has a lower pH than a weak acid solution. Explain.
14. Why is the pH of water at equal to 7.00?
15. Can the pH of a solution be negative? Explain.
25°C Ka 1 Kb (Ka 1.0 106), A A A A A A Exercises 673 c. The [H] is between 0.10 M and 0.20 M.
d. The [H] is 0.20 M.
25. Of the hydrogen halides, only HF is a weak acid. Give a possible explanation.
26. Explain why the following are done, both of which are related to acid/base chemistry.
a. Power plants burning coal with high sulfur content use scrub-bers to help eliminate sulfur emissions.
b. A gardener mixes lime (CaO) into the soil of his garden.
Exercises In this section similar exercises are paired.
Nature of Acids and Bases 27. Write balanced equations that describe the following reactions.
a. the dissociation of perchloric acid in water b. the dissociation of propanoic acid (CH3CH2CO2H) in water c. the dissociation of ammonium ion in water 28. Write the dissociation reaction and the corresponding Ka equi-librium expression for each of the following acids in water.
a. HCN b.
c.
29. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid.
a.
b.
c.
30. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid.
a.
b.
c.
31. Classify each of the following as a strong acid or a weak acid.
32. Consider the following illustrations: H+ A– B– c.
d.
Cl S O H b.
a.
HOCl C6H5NH2 ∆OCl C6H5NH3 H2O HONH3 ∆HONH2 H3O Al(H2O)6 3 H2O ∆H3O Al(H2O)5(OH)2 HCO3 C5H5NH ∆H2CO3 C5H5N C5H5NH H2O ∆C5H5N H3O H2O H2CO3 ∆H3O HCO3 C6H5NH3 HOC6H5 A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 16. Why is H3O the strongest acid and the strongest base that can exist in significant amounts in aqueous solutions?
17. How many significant figures are there in the numbers: 10.78, 6.78, 0.78? If these were pH values, to how many significant figures can you express the [H]? Explain any discrepancies between your answers to the two questions.
18. In terms of orbitals and electron arrangements, what must be present for a molecule or an ion to act as a Lewis acid? What must be present for a molecule or an ion to act as a Lewis base?
19. Give three example solutions that fit each of the following descriptions.
a. A strong electrolyte solution that is very acidic.
b. A strong electrolyte solution that is slightly acidic.
c. A strong electrolyte solution that is very basic.
d. A strong electrolyte solution that is slightly basic.
e. A strong electrolyte solution that is neutral.
20. Derive an expression for the relationship between pKa and pKb for a conjugate acid–base pair. 21. Consider the following statements. Write out an example reaction and K expression that is associated with each statement.
a. The autoionization of water.
b. An acid reacts with water to produce the conjugate base of the acid and the hydronium ion.
c. A base reacts with water to produce the conjugate acid of the base and the hydroxide ion.
22. Which of the following statements is(are) true? Correct the false statements.
a. When a base is dissolved in water, the lowest possible pH of the solution is 7.0.
b. When an acid is dissolved in water, the lowest possible pH is 0.
c. A strong acid solution will have a lower pH than a weak acid solution.
d. A 0.0010 M Ba(OH)2 solution has a pOH that is twice the pOH value of a 0.0010 M KOH solution.
23. Consider the following mathematical expressions.
a.
b.
c.
d.
For each expression, give three solutions where the mathemati-cal expression would give a good approximation for the [H] or [OH]. [Hao] and [B] represent initial concentrations of an acid or a base.
24. Consider a 0.10 M H2CO3 solution and a 0.10 M H2SO4 solution.
Without doing any detailed calculations, choose one of the fol-lowing statements that best describes the [H] of each solution and explain your answer.
a. The [H] is less than 0.10 M.
b. The [H] is 0.10 M.
[OH] (Kb [B]o)1 2 [OH] 2[B]o [H] (Ka [HA]o)1 2 [H] [HA]o (pK logK.) OH 674 Chapter Fourteen Acids and Bases 45. The pH of a sample of gastric juice in a person’s stomach is 2.1.
Calculate the pOH, and for this sample. Is gastric juice acidic or basic?
46. The pOH of a sample of baking soda dissolved in water is 5.74 at Calculate the pH, [H], and for this sample. Is the solution acidic or basic?
Solutions of Acids 47. What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions.
a. HClO4 b. HNO3 48. Calculate the pH of each of the following solutions of a strong acid in water.
a. 0.10 M HCl c.
M HCl b. 5.0 M HCl 49. A solution is prepared by adding 50.0 mL of 0.050 M HCl to 150.0 mL of 0.10 M HNO3. Calculate the concentrations of all species in this solution.
50. A solution is prepared by mixing 90.0 mL of 5.00 M HCl and 30.0 mL of 8.00 M HNO3. Water is then added until the final volume is 1.00 L. Calculate [H], , and the pH for this solution.
[OH] 1.0 1011 [OH] 25°C.
[OH] [H], Which beaker best illustrates what happens when the following acids are dissolved in water?
a.
d. HF b.
e.
c. HCl 33. Use Table 14.2 to order the following from the strongest to the weakest acid.
34. Use Table 14.2 to order the following from the strongest to the weakest base.
35. You may need Table 14.2 to answer the following questions.
a. Which is the stronger acid, HCl or H2O?
b. Which is the stronger acid, H2O or HNO2?
c. Which is the stronger acid, HCN or HOC6H5?
36. You may need Table 14.2 to answer the following questions.
a. Which is the stronger base, or H2O?
b. Which is the stronger base, H2O or c. Which is the stronger base, or ?
Autoionization of Water and the pH Scale 37. Calculate the of each of the following solutions at Identify each solution as neutral, acidic, or basic.
a.
c.
b.
d.
38. Calculate the of each of the following solutions at Identify each solution as neutral, acidic, or basic.
a.
c.
b.
d.
39. Values of Kw as a function of temperature are as follows: 3OH4 7.3 104 M 3OH4 3.6 1015 M 3OH4 1.0 107 M 3OH4 1.5 M 25°C.
[H] 3H4 5.4 105 M 3H4 8.3 1016 M 3H4 12 M 3H4 1.0 107 M 25°C.
[OH] OC6H5 CN NO2 ?
Cl ClO2 , H2O, NH3, ClO4 HClO2, H2O, NH4 , HClO4 HC2H3O2 HNO3 HNO2 Acidic, Basic, pH pOH [H] [OH] or Neutral?
Solution a 6.88 Solution b Solution c 3.11 Solution d 1.0 107 M 8.4 1014 M Acidic, Basic, pH pOH [H] [OH] or Neutral?
Solution a 9.63 Solution b Solution c 0.027 M Solution d 12.2 3.9 106 M 44. Fill in the missing information in the following table.
c.
d.
e.
f.
43. Fill in the missing information in the following table.
pOH 9.60 pOH 5.0 pH 3.20 pH 1.0 Temperature (°C) Kw 0 25 35 40.
50.
5.47 1014 2.92 1014 2.09 1014 1.00 1014 1.14 1015 a. Is the autoionization of water exothermic or endothermic?
b. Calculate [H] and in a neutral solution at .
40. At the value of a. Calculate the [H] and in pure water at .
b. What is the pH of pure water at c. If the hydroxide ion concentration in a solution is 0.10 M, what is the pH at 41. Calculate the pH and pOH of the solutions in Exercises 37 and 38.
42. Calculate [H] and for each solution at Identify each solution as neutral, acidic, or basic.
a.
(the normal pH of blood) b. pH 15.3 pH 7.40 25°C.
[OH] 40°C?
40°C?
40°C [OH] Kw is 2.92 1014.
40°C 50.°C [OH] Exercises 675 d. How is percent dissociation of an acid related to the Ka value for the acid (assuming equal initial concentrations of acids)?
65. A 0.15 M solution of a weak acid is 3.0% dissociated. Calcu-late Ka.
66. An acid HX is 25% dissociated in water. If the equilibrium con-centration of HX is 0.30 M, calculate the Ka value for HX.
67. The pH of a M solution of cyanic acid (HOCN) is 2.77 at . Calculate Ka for HOCN from this result.
68. Trichloroacetic acid (CCl3CO2H) is a corrosive acid that is used to precipitate proteins. The pH of a 0.050 M solution of trichloroacetic acid is the same as the pH of a 0.040 M HClO4 solution. Calculate Ka for trichloroacetic acid.
69. A solution of formic acid has a pH of 2.70. Calculate the initial concentration of formic acid in this solution.
70. One mole of a weak acid HA was dissolved in 2.0 L of solution.
After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate Ka for HA.
Solutions of Bases 71. Write the reaction and the corresponding Kb equilibrium ex-pression for each of the following substances acting as bases in water.
a. NH3 b. C5H5N 72. Write the reaction and the corresponding Kb equilibrium ex-pression for each of the following substances acting as bases in water.
a. aniline, C6H5NH2 b. dimethylamine, (CH3)2NH 73. Use Table 14.3 to help order the following bases from strongest to weakest.
74. Use Table 14.3 to help order the following acids from strongest to weakest.
75. Use Table 14.3 to help answer the following questions.
a. Which is the stronger base, or C6H5NH2?
b. Which is the stronger base, H2O or C6H5NH2?
c. Which is the stronger base, or C6H5NH2?
d. Which is the stronger base, C6H5NH2 or CH3NH2?
76. Use Table 14.3 to help answer the following questions.
a. Which is the stronger acid, HClO4 or C6H5NH3 ?
b. Which is the stronger acid, H2O or C6H5NH3 ?
c. Which is the stronger acid, C6H5NH3 or CH3NH3 ?
77. Calculate the pH of the following solutions.
a. 0.10 M NaOH b.
M NaOH c. 2.0 M NaOH 78. Calculate , pOH, and pH for each of the following.
a. 0.00040 M Ca(OH)2 b. a solution containing 25 g of KOH per liter c. a solution containing 150.0 g of NaOH per liter [OH] 1.0 1010 OH ClO4 HNO3, H2O, NH4 , C5H5NH NO3 , H2O, NH3, C5H5N (HCOOH, Ka 1.8 104) 25°C 1.00 102 51. How would you prepare 1600 mL of a solution using concentrated (12 M) HCl?
52. What mass of is present in 250.0 mL of a nitric acid solution having a 53. What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions.
a. HNO2 b. CH3CO2H (HC2H3O2) 54. What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions.
a. HOC6H5 b. HCN 55. A 0.0560-g sample of acetic acid is added to enough water to make 50.00 mL of solution. Calculate and the pH at equilibrium. Ka for acetic acid is 56. For propanoic acid determine the concentration of all species present, the pH, and the percent dissociation of a 0.100 M solution.
57. Calculate the concentration of all species present and the pH of a 0.020 M HF solution.
58. Calculate the pH of a 0.20 M solution of iodic acid 59. Monochloroacetic acid, HC2H2ClO2, is a skin irritant that is used in “chemical peels” intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of Ka for monochloroacetic acid is 1.35 103. Calculate the pH of a 0.10 M solution of monochloroacetic acid.
60. A typical aspirin tablet contains 325 mg of acetylsalicylic acid, Calculate the pH of a solution that is prepared by dissolving two aspirin tablets in one cup (237 mL) of solution.
Assume the aspirin tablets are pure acetylsalicylic acid, 61. Calculate the pH of each of the following.
a. a solution containing 0.10 M HCl and 0.10 M HOCl b. a solution containing 0.050 M HNO3 and 0.50 M HC2H3O2.
62. Calculate the pH of a solution that contains 1.0 M HF and 1.0 M HOC6H5. Also calculate the concentration of in this so-lution at equilibrium.
63. Calculate the percent dissociation of the acid in each of the fol-lowing solutions.
a. 0.50 M acetic acid b. 0.050 M acetic acid c. 0.0050 M acetic acid d. Use Le Châtelier’s principle to explain why percent dissocia-tion increases as the concentration of a weak acid decreases.
e. Even though the percent dissociation increases from solutions a to c, the [H] decreases. Explain.
64. Using the Ka values in Table 14.2, calculate the percent dissoci-ation in a 0.20 M solution of each of the following acids.
a. nitric acid (HNO3) b. nitrous acid (HNO2) c. phenol (HOC6H5) OC6H5 Ka 3.3 104.
HC9H7O4.
(HIO3, Ka 0.17).
(HC3H5O2, Ka 1.3 105), 1.8 105.
[CH3COOH], [H], [CH3COO], pH 5.10?
HNO3 pH 1.50 676 Chapter Fourteen Acids and Bases 96. Arsenic acid (H3AsO4) is a triprotic acid with 5 103, 8 108, and 6 1010. Calculate [H], [OH], [H3AsO4], [H2AsO4 ], [HAsO4 2], and [AsO4 3] in a 0.20 M arsenic acid solution.
97. Calculate the pH of a 2.0 M H2SO4 solution.
98. Calculate the pH of a M solution of H2SO4.
Acid–Base Properties of Salts 99. Arrange the following 0.10 M solutions in order of most acidic to most basic.
100. Arrange the following 0.10 M solutions in order from most acidic to most basic. See Appendix 5 for Ka and Kb values.
101. Given that the Ka value for acetic acid is and the Ka value for hypochlorous acid is which is the stronger base, or ?
102. The Kb values for ammonia and methylamine are and respectively. Which is the stronger acid, or CH3NH3 ?
103. Sodium azide (NaN3) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN3. The Ka value for hydrazoic acid (HN3) is 104. Calculate the concentrations of all species present in a 0.25 M solution of ethylammonium chloride (C2H5NH3Cl).
105. Calculate the pH of each of the following solutions.
a. 0.10 M CH3NH3Cl b. 0.050 M NaCN 106. Calculate the pH of each of the following solutions.
a. 0.12 M KNO2 c. 0.40 M NH4ClO4 b. 0.45 M NaOCl 107. An unknown salt is either NaCN, NaC2H3O2, NaF, NaCl, or NaOCl. When 0.100 mol of the salt is dissolved in 1.00 L of solution, the pH of the solution is 8.07. What is the identity of the salt?
108. Consider a solution of an unknown salt having the general for-mula BHCl, where B is one of the weak bases in Table 14.3. A 0.10 M solution of the unknown salt has a pH of 5.82. What is the actual formula of the salt?
109. Calculate the pH of a 0.050 M Al(NO3)3 solution. The Ka value for Al(H2O)6 3 is 110. Calculate the pH of a 0.10 M CoCl3 solution. The Ka value for Co(H2O)6 3 is 111. Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The rel-evant Ka and Kb values are found in Tables 14.2 and 14.3.
a. NaNO3 c. C5H5NHClO4 e. KOCl b. NaNO2 d. NH4NO2 f. NH4OCl 112. Are solutions of the following salts acidic, basic, or neutral?
For those that are not neutral, write balanced equations for the 1.0 105.
1.4 105.
1.9 105.
NH4 4.4 104, 1.8 105 C2H3O2 OCl 3.5 108, 1.8 105 CaBr2, KNO2, HClO4, HNO2, HONH3ClO4 KOH, KCl, KCN, NH4Cl, HCl 5.0 103 Ka3 Ka2 Ka1 79. What are the major species present in 0.015 M solutions of each of the following bases?
a. KOH b. Ba(OH)2 What is and the pH of each of these solutions?
80. What are the major species present in the following mixtures of bases?
a. 0.050 M NaOH and 0.050 M LiOH b. 0.0010 M Ca(OH)2 and 0.020 M RbOH What is and the pH of each of these solutions?
81. What mass of KOH is necessary to prepare 800.0 mL of a so-lution having a 82. Calculate the concentration of an aqueous Sr(OH)2 that has 83. What are the major species present in a 0.150 M NH3 solution?
Calculate the and the pH of this solution.
84. For the reaction of hydrazine (N2H4) in water, Kb is Calculate the concentrations of all species and the pH of a 2.0 M solution of hydrazine in water.
85. Calculate , [H], and the pH of 0.20 M solutions of each of the following amines.
a. triethylamine b. hydroxylamine 86. Calculate , [H], and the pH of 0.20 M solutions of each of the following amines (the Kb values are found in Table 14.3).
a. aniline b. methylamine 87. Calculate the pH of a 0.20 M C2H5NH2 solution 88. Calculate the pH of a 0.050 M (C2H5)2NH solution 89. Calculate the percent ionization in each of the following solutions.
a. 0.10 M NH3 b. 0.010 M NH3 90. Calculate the percentage of pyridine (C5H5N) that forms pyri-dinium ion, in a 0.10 M aqueous solution of pyridine 91. Codeine (C18H21NO3) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription be-cause of its addictive properties. If the pH of a M solution of codeine is 9.59, calculate Kb.
92. Calculate the mass of HONH2 required to dissolve in enough water to make 250.0 mL of solution having a pH of 10.00.
Polyprotic Acids 93. Write out the stepwise Ka reactions for the diprotic acid H2SO3.
94. Write out the stepwise Ka reactions for citric acid (H3C6H5O7), a triprotic acid.
95. Using the Ka values in Table 14.4 and only the first dissociation step, calculate the pH of 0.10 M solutions of each of the fol-lowing polyprotic acids.
a. H3PO4 b. H2CO3 1.1 108.) (Kb 1.7 103 (Kb 1.7 109).
C5H5NH, 103).
(Kb 1.3 5.6 104).
(Kb [OH] (HONH2, Kb 1.1 108) [(C2H5)3N, Kb 4.0 104] [OH] 3.0 106.
H2NNH21aq2 H2O1l2 ∆H2NNH3 1aq2 OH1aq2 [OH] pH 10.50.
pH 11.56?
[OH] [OH] Additional Exercises 677 reactions causing the solution to be acidic or basic. The relevant Ka and Kb values are found in Tables 14.2 and 14.3.
a. KCl c. CH3NH3Cl e. NH4F b. NH4C2H3O2 d. KF f. CH3NH3CN Relationships Between Structure and Strengths of Acids and Bases 113. Place the species in each of the following groups in order of in-creasing acid strength. Explain the order you chose for each group.
a. HIO3, HBrO3 c. HOCl, HOI b. HNO2, HNO3 d. H3PO4, H3PO3 114. Place the species in each of the following groups in order of in-creasing base strength. Give your reasoning in each case.
a. IO3 , BrO3 b. NO2 , NO3 c. OCl, OI 115. Place the species in each of the following groups in order of in-creasing acid strength.
a. H2O, H2S, H2Se (bond energies: HOO, 467 kJ/mol; HOS, 363 kJ/mol; HOSe, 276 kJ/mol) b. CH3CO2H, FCH2CO2H, F2CHCO2H, F3CCO2H c. NH4 , HONH3 d. NH4 , PH4 (bond energies: NOH, 391 kJ/mol; POH, 322 kJ/mol) Give reasons for the orders you chose.
116. Using your results from Exercise 115, place the species in each of the following groups in order of increasing base strength.
a. OH, SH, SeH b. NH3, PH3 c. NH3, HONH2 117. Will the following oxides give acidic, basic, or neutral solu-tions when dissolved in water? Write reactions to justify your answers.
a. CaO b. SO2 c. Cl2O 118. Will the following oxides give acidic, basic, or neutral solu-tions when dissolved in water? Write reactions to justify your answers.
a. Li2O b. CO2 c. SrO Lewis Acids and Bases 119. Identify the Lewis acid and the Lewis base in each of the fol-lowing reactions.
a.
b.
c.
120. Identify the Lewis acid and the Lewis base in each of the following reactions.
a.
b.
c.
121. Aluminum hydroxide is an amphoteric substance. It can act as either a Brønsted–Lowry base or a Lewis acid. Write a reaction showing Al(OH)3 acting as a base toward H and as an acid toward .
122. Zinc hydroxide is an amphoteric substance. Write equations that describe Zn(OH)2 acting as a Brønsted–Lowry base toward H and as a Lewis acid toward .
OH OH HgI2(s) 2I(aq) ∆HgI4 2(aq) H2O(l) CN(aq) ∆HCN(aq) OH(aq) Fe3(aq) 6H2O(l) ∆Fe(H2O)6 3(aq) BF3(g) F(aq) ∆BF4 (aq) Ag(aq) 2NH3(aq) ∆Ag(NH3)2 (aq) B(OH)3(aq) H2O(l) ∆B(OH)4 (aq) H(aq) The solution contains one of the following substances: HCl, NaOH, NH4Cl, HCN, NH3, HF, or NaCN. If the solute concen-tration is about 1.0 M, what is the identity of the solute?
130. A 0.25-g sample of lime (CaO) is dissolved in enough water to make 1500 mL of solution. Calculate the pH of the solution.
131. At , a saturated solution of benzoic acid has a pH of 2.80. Calculate the water solubility of benzoic acid in moles per liter.
132. Calculate the pH and [S2] in a 0.10 M solution. Assume 133. A typical vitamin C tablet (containing pure ascorbic acid, H2C6H6O6) weighs 500. mg. One vitamin C tablet is dissolved in enough water to make 200.0 mL of solution. Calculate the pH of this solution. Ascorbic acid is a diprotic acid.
134. Calculate the pH of an aqueous solution containing M HCl, M H2SO4, and M HCN.
1.0 102 1.0 102 1.0 102 Ka2 1.0 1019.
1.0 107; Ka1 H2S (Ka 6.4 105) 25°C 123. Would you expect Fe3 or Fe2 to be the stronger Lewis acid?
Explain.
124. Use the Lewis acid–base model to explain the following reaction.
Additional Exercises 125. A 10.0-mL sample of an HCl solution has a pH of 2.000. What volume of water must be added to change the pH to 4.000?
126. Which of the following represent conjugate acid–base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair.
a.
c.
b.
d.
127. A solution is made by adding 50.0 mL of 0.200 M acetic acid to 50.0 mL of HCl.
a. Calculate the pH of the solution.
b. Calculate the acetate ion concentration.
128. You have 100.0 g of saccharin, a sugar substitute, and you want to prepare a solution. What volume of solution can be prepared? For saccharin, 129. A solution is tested for pH and conductivity as pictured below: log Ka).
HC7H4NSO3, pKa 11.70 (pKa pH 5.75 1.00 103 M (Ka 1.8 105) HC2H3O2, C2H3O2 H2SO4, SO4 2 H3PO4, H2PO4 H2O, OH CO21g2 H2O1l2 ¡ H2CO31aq2 678 Chapter Fourteen Acids and Bases 143. Students are often surprised to learn that organic acids, such as acetic acid, contain —OH groups. Actually, all oxyacids contain hydroxyl groups. Sulfuric acid, usually written as H2SO4, has the structural formula SO2(OH)2, where S is the central atom. Iden-tify the acids whose structural formulas are shown below. Why do they behave as acids, while NaOH and KOH are bases?
a. SO(OH)2 b. ClO2(OH) c. HPO(OH)2 Challenge Problems 144. The pH of M hydrochloric acid is not 8.00. The cor-rect pH can be calculated by considering the relationship between the molarities of the three principal ions in the solution ( ). These molarities can be calculated from algebraic equations that can be derived from the considerations given below.
a. The solution is electrically neutral.
b. The hydrochloric acid can be assumed to be 100% ionized.
c. The product of the molarities of the hydronium ions and the hydroxide ions must equal Kw.
Calculate the pH of a HCl solution.
145. Calculate the pH of a M solution of NaOH in water.
146. Calculate in a M solution of Ca(OH)2.
147. Consider 50.0 mL of a solution of weak acid which has a pH of 4.000. What volume of wa-ter must be added to make the 148. Making use of the assumptions we ordinarily make in calculat-ing the pH of an aqueous solution of a weak acid, calculate the pH of a M solution of hypobromous acid . What is wrong with your answer? Why is it wrong? Without trying to solve the problem, tell what has to be included to solve the problem correctly.
149. Calculate the pH of a 0.200 M solution of C5H5NHF. Hint: C5H5NHF is a salt composed of C5H5NH and ions. The prin-cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is 150. Determine the pH of a 0.50 M solution of NH4OCl. See Exer-cise 149.
151. Calculate in a solution obtained by adding 0.0100 mol of solid NaOH to 1.00 L of 15.0 M NH3.
152. What mass of NaOH(s) must be added to 1.0 L of 0.050 M NH3 to ensure that the percent ionization of NH3 is no greater than 0.0010%? Assume no volume change on addition of NaOH.
153. A certain acid, HA, has a vapor density of 5.11 g/L when in the gas phase at a temperature of and a pressure of 1.00 atm.
When 1.50 g of this acid is dissolved in enough water to make 100.0 mL of solution, the pH is found to be 1.80. Calculate Ka for the acid.
154. Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00 M HC2H3O2 to double the pH of the solution (assume that the NaOH does not change the volume of the solution).
155. Consider the species . Each ion can act as a base in water. Determine the Kb value for each of these species. Which species is the strongest base?
PO4 3, HPO4 2, and H2PO4 25°C 3OH4 C5H5N1aq2 HF1aq2 K 8.2 103 C5H5NH1aq2 F1aq2 ∆ F (HBrO, Ka 2 109) 1.0 106 pH 5.000?
Ka(1.00 106), HA 3.0 107 [OH] 1.0 107 1.0 108 Cl, and OH H, 1.0 108 135. Acrylic acid is a precursor for many impor-tant plastics. Ka for acrylic acid is .
a. Calculate the pH of a 0.10 M solution of acrylic acid.
b. Calculate the percent dissociation of a 0.10 M solution of acrylic acid.
c. Calculate the pH of a 0.050 M solution of sodium acrylate (NaC3H3O2).
136. A 0.20 M sodium chlorobenzoate (NaC7H4ClO2) solution has a pH of 8.65. Calculate the pH of a 0.20 M chlorobenzoic acid (HC7H4ClO2) solution.
137. The equilibrium constant Ka for the reaction is 6.0 103.
a. Calculate the pH of a 0.10 M solution of Fe(H2O)6 3.
b. Will a 1.0 M solution of iron(II) nitrate have a higher or lower pH than a 1.0 M solution of iron(III) nitrate? Explain.
138. Rank the following 0.10 M solutions in order of increasing pH.
a. HI, HF, NaF, NaI b. NH4Br, HBr, KBr, NH3 c. C6H5NH3NO3, NaNO3, NaOH, HOC6H5, KOC6H5, C6H5NH2, HNO3 139. Is an aqueous solution of NaHSO4 acidic, basic, or neutral? What reaction occurs with water? Calculate the pH of a 0.10 M solu-tion of NaHSO4.
140. Calculate in a 0.010 M solution of CO2 in water (H2CO3). If all the in this solution comes from the reaction what percentage of the H ions in the solution is a result of the dissociation of ? When acid is added to a solution of sodium hydrogen carbonate (NaHCO3), vigorous bubbling oc-curs. How is this reaction related to the existence of carbonic acid (H2CO3) molecules in aqueous solution?
141. Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemo-globin molecule contains four iron atoms that are the binding sites for O2 molecules. The oxygen binding is pH dependent.
The relevant equilibrium reaction is Use Le Châtelier’s principle to answer the following.
a. What form of hemoglobin, HbH4 4 or Hb(O2)4, is favored in the lungs? What form is favored in the cells?
b. When a person hyperventilates, the concentration of CO2 in the blood is decreased. How does this affect the oxygen-bind-ing equilibrium? How does breathing into a paper bag help to counteract this effect?
c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary?
142. Calculate the value for the equilibrium constant for each of the following aqueous reactions.
a.
b.
c.
d. HNO2 OH ∆H2O NO2 NH4 OH ∆NH3 H2O NO2 H3O ∆HNO2 H2O NH3 H3O ∆NH4 H2O HbH4 41aq2 4O21g2 ∆Hb1O2241aq2 4H1aq2 HCO3 HCO3 1aq2 ∆H1aq2 CO3 21aq2 CO3 2 [CO3 2] Fe1H2O251OH221aq2 H3O1aq2 Fe1H2O26 31aq2 H2O1l2 ∆ 5.6 105 (CH2“CHCO2H) Marathon Problems 679 163. Papaverine hydrochloride (abbreviated molar mass 378.85 g/mol) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; Calculate the pH of a 30.0 mg/mL aqueous dose of prepared at Kw at is Marathon Problems These problems are designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
164. Captain Kirk, of the Starship Enterprise, has been told by his superiors that only a chemist can be trusted with the combina-tion to the safe containing the dilithium crystals that power the ship. The combination is the pH of solution A, described below, followed by the pH of solution C. (Example: If the pH of solu-tion A is 3.47 and that of solution C is 8.15, then the combina-tion to the safe is 3-47-8-15.) The chemist must determine the combination using only the information below (all solutions are at ): Solution A is 50.0 mL of a 0.100 M solution of the weak mono-protic acid HX.
Solution B is a 0.0500 M solution of the salt NaX. It has a pH of 10.02.
Solution C is made by adding 15.0 mL of 0.250 M KOH to so-lution A.
What is the combination to the safe?
165. For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated pH.
Calculate the pH of each solution.
Group I: 0.20 M NH4Cl, 0.20 M HCl, 0.20 M C6H5NH3Cl, 0.20 M (C2H5)3NHCl Group II: 0.20 M KOI, 0.20 M NaCN, 0.20 M KOCl, 0.20 M NaNO2 a. the solution with the lowest pH b. the solution with the highest pH c. the solution with the pH closest to 7.00 Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
25°C 2.1 1014.
35.0°C 35.0°C.
papHCl Kb 8.33 109 at 35.0°C).
papHCl; 156. Calculate the pH of a 0.10 M solution of sodium phosphate. See Exercise 155.
157. Will 0.10 M solutions of the following salts be acidic, basic, or neutral? See Appendix 5 for Ka values.
a. ammonium bicarbonate b. sodium dihydrogen phosphate c. sodium hydrogen phosphate d. ammonium dihydrogen phosphate e. ammonium formate 158. a. The principal equilibrium in a solution of NaHCO3 is Calculate the value of the equilibrium constant for this reaction.
b. At equilibrium, what is the relationship between [H2CO3] and c. Using the equilibrium derive an expression for the pH of the solution in terms of and using the result from part b.
d. What is the pH of a solution of NaHCO3?
159. A 0.100-g sample of the weak acid HA (molar mass 100.0 g/mol) is dissolved in 500.0 g of water. The freezing point of the resulting solution is . Calculate the value of Ka for this acid. Assume molarity equals molarity in this solution.
160. A sample containing 0.0500 mol of Fe2(SO4)3 is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated and Fe(H2O)6 3 ions. The latter behaves as an acid: a. Calculate the expected osmotic pressure of this solution at if the above dissociation is negligible.
b. The actual osmotic pressure of the solution is 6.73 atm at . Calculate Ka for the dissociation reaction of Fe(H2O)6 3. (To do this calculation, you must assume that none of the ions goes through the semipermeable membrane.
Actually, this is not a great assumption for the tiny H ion.) Integrative Problems These problems require the integration of multiple concepts to find the solutions.
161. A 2.14-g sample of sodium hypoiodite is dissolved in water to make 1.25 L of solution. The solution pH is 11.32. What is Kb for the hypoiodite ion?
162. Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation Upon isolating pure HNCO(l), an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water.
What is the pH of a 100.-mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and H2C2O4, assuming all of the HNCO produced is dissolved in solution? HNCO 1.2 104.) (Ka of 2NaOCN1s2 H2C2O41s2 ¡ 2HNCO1l2 Na2C2O41s2 25°C 25°C Fe1H2O26 31aq2 ∆Fe1H2O25OH21aq2 H1aq2 SO4 2 0.0056°C Ka2 Ka1 H2CO31aq2 ∆2H1aq2 CO3 21aq2 [CO3 2]?
HCO3 1aq2 HCO3 1aq2 ∆H2CO31aq2 CO3 21aq2 Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Education, Inc.
680 15 Applications of Aqueous Equilibria Contents Acid–Base Equilibria 15.1 Solutions of Acids or Bases Containing a Common Ion • Equilibrium Calculations 15.2 Buffered Solutions • Buffering: How Does It Work 15.3 Buffering Capacity 15.4 Titrations and pH Curves • Strong Acid–Strong Base Titrations • Titrations of Weak Acids with Strong Bases • Calculation of Ka • Titrations of Weak Bases with Strong Acids 15.5 Acid–Base Indicators Solubility Equilibria 15.6 Solubility Equilibria and the Solubility Product • Relative Solubilities • Common Ion Effect • pH and Solubility 15.7 Precipitation and Qualitative Analysis • Selective Precipitation • Qualitative Analysis Complex Ion Equilibria 15.8 Equilibria Involving Complex Ions • Complex Ions and Solubility Stalactites are formed when carbonate minerals dissolve in ground water acidified by carbon dioxide and then solidify when the water evaporates.
Much important chemistry, including almost all the chemistry of the natural world, occurs in aqueous solution. We have already introduced one very significant class of aque-ous equilibria, acid–base reactions. In this chapter we consider more applications of acid–base chemistry and introduce two additional types of aqueous equilibria, those in-volving the solubility of salts and those involving the formation of complex ions.
The interplay of acid–base, solubility, and complex ion equilibria is often important in natural processes, such as the weathering of minerals, the uptake of nutrients by plants, and tooth decay. For example, limestone (CaCO3) will dissolve in water made acidic by dissolved carbon dioxide: This two-step process and its reverse account for the formation of limestone caves and the stalactites and stalagmites found therein. In the forward direction of the process, the acidic water (containing carbon dioxide) dissolves the underground limestone deposits, thereby forming a cavern. The reverse process occurs as the water drips from the ceiling of the cave, and the carbon dioxide is lost to the air. This causes solid calcium carbonate to form, producing stalactites on the ceiling and stalagmites where the drops hit the cave floor.
Before we consider the other types of aqueous equilibria, we will deal with acid–base equilibria in more detail.
Acid–Base Equilibria 15.1 Solutions of Acids or Bases Containing a Common Ion In Chapter 14 we were concerned with calculating the equilibrium concentrations of species (particularly H ions) in solutions containing an acid or a base. In this section we discuss solutions that contain not only the weak acid HA but also its salt NaA. Although this appears to be a new type of problem, we will see that this case can be handled rather easily using the procedures developed in Chapter 14.
Suppose we have a solution containing the weak acid hydrofluoric acid (HF, and its salt sodium fluoride (NaF). Recall that when a salt dissolves in water, it breaks up completely into its ions—it is a strong electrolyte: Since hydrofluoric acid is a weak acid and only slightly dissociated, the major species in the solution are HF, Na, , and H2O. The common ion in this solution is , since it is produced by both hydrofluoric acid and sodium fluoride. What effect does the pres-ence of the dissolved sodium fluoride have on the dissociation equilibrium of hydrofluoric acid?
To answer this question, we compare the extent of dissociation of hydrofluoric acid in two different solutions, the first containing 1.0 M HF and the second containing 1.0 M HF F F NaF1s2 ¬ ¡ H2O1l2 Na1aq2 F1aq2 Ka 7.2 104) H1aq2 CaCO31s2 ∆Ca21aq2 HCO3 1aq2 CO21aq2 H2O1l2 ∆H1aq2 HCO3 1aq2 681 682 Chapter Fifteen Applications of Aqueous Equilibria and 1.0 M NaF. By Le Châtelier’s principle, we would expect the dissociation equilibrium for HF in the second solution to be driven to the left by the presence of ions from the NaF.
Thus the extent of dissociation of HF will be less in the presence of dissolved NaF: The shift in equilibrium position that occurs because of the addition of an ion already in-volved in the equilibrium reaction is called the common ion effect. This effect makes a solution of NaF and HF less acidic than a solution of HF alone.
The common ion effect is quite general. For example, solid NH4Cl added to a 1.0 M NH3 solution produces additional ammonium ions: and this causes the position of the ammonia–water equilibrium to shift to the left: This reduces the equilibrium concentration of ions.
The common ion effect is also important in solutions of polyprotic acids. The pro-duction of protons by the first dissociation step greatly inhibits the succeeding dissocia-tion steps, which, of course, also produce protons, the common ion in this case. We will see later in this chapter that the common ion effect is also important in dealing with the solubility of salts.
Equilibrium Calculations The procedures for finding the pH of a solution containing a weak acid or base plus a common ion are very similar to the procedures, which we covered in Chapter 14, for so-lutions containing the acids or bases alone. For example, in the case of a weak acid, the only important difference is that the initial concentration of the anion is not zero in a solution that also contains the salt NaA. Sample Exercise 15.1 illustrates a typical exam-ple using the same general approach we developed in Chapter 14.
Acidic Solutions Containing Common Ions In Section 14.5 we found that the equilibrium concentration of H in a 1.0 M HF solu-tion is and the percent dissociation of HF is 2.7%. Calculate [H] and the percent dissociation of HF in a solution containing 1.0 M HF and 1.0 M NaF.
(Ka 7.2 104) 2.7 102 M, A OH NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 NH4Cl1s2 ¡ H2O NH4 1aq2 Cl1aq2 Equilibrium shifts away from added component. Fewer H ions present.
Added F ions from NaF HF1aq2 ∆H1aq2 F1aq2 F HF1aq2 ∆H1aq2 F1aq2 The common ion effect is an application of Le Châtelier’s principle.
Sample Exercise 15.1 15.1 Solutions of Acids or Bases Containing a Common Ion 683 Solution As the aqueous solutions we consider become more complex, it is more important than ever to be systematic and to focus on the chemistry occurring in the solution before think-ing about mathematical procedures. The way to do this is always to write the major species first and consider the chemical properties of each one.
In a solution containing 1.0 M HF and 1.0 M NaF, the major species are We know that Na ions have neither acidic nor basic properties and that water is a very weak acid (or base). Therefore, the important species are HF and , which participate in the acid dissociation equilibrium that controls [H] in this solution. That is, the posi-tion of the equilibrium will determine [H] in the solution. The equilibrium expression is The important concentrations are shown in the following table.
Ka 3H4 3F4 3HF4 7.2 104 HF1aq2 ∆H1aq2 F1aq2 F HF, F, Na, and H2O Initial Equilibrium Concentration (mol/L) Concentration (mol/L) (from dissolved HF) x mol/L HF dissociates (from dissolved NaF) ¬ ¬¬ ¬¬¡ (neglect contribution from H2O) [H] x [H]0 0 [F] 1.0 x [F]0 1.0 [HF] 1.0 x [HF]0 1.0 Note that because of the dissolved sodium fluoride and that at equilib-rium because when the acid dissociates it produces as well as H. Then (since x is expected to be small).
Solving for x gives Noting that x is small compared to 1.0, we conclude that this result is acceptable. Thus The percent dissociation of HF in this solution is Compare these values for [H] and percent dissociation of HF with those for a 1.0 M HF solution, where [H] 2.7 102 M and the percent dissociation is 2.7%. The large difference shows clearly that the presence of the ions from the dissolved NaF greatly inhibits the dissociation of HF. The position of the acid dissociation equilibrium has been shifted to the left by the presence of ions from NaF.
See Exercises 15.25 and 15.26.
F F 3H4 3HF40 100 7.2 104 M 1.0 M 100 0.072% 3H4 x 7.2 104 M 1The pH is 3.14.2 x 1.0 1.0 17.2 1042 7.2 104 Ka 7.2 104 3H4 3F4 3HF4 1x211.0 x2 1.0 x 1x211.02 1.0 F [F] 7 1.0 M [F]0 1.0 M F– Na+ HF H2O Major Species 684 Chapter Fifteen Applications of Aqueous Equilibria 15.2 Buffered Solutions The most important application of acid–base solutions containing a common ion is for buffering. A buffered solution is one that resists a change in its pH when either hydroxide ions or protons are added. The most important practical example of a buffered solution is our blood, which can absorb the acids and bases produced in biologic reactions without changing its pH. A constant pH for blood is vital because cells can survive only in a very narrow pH range.
A buffered solution may contain a weak acid and its salt (for example, HF and NaF) or a weak base and its salt (for example, NH3 and NH4Cl). By choosing the appropriate components, a solution can be buffered at virtually any pH.
In treating buffered solutions in this chapter, we will start by considering the equi-librium calculations. We will then use these results to show how buffering works. That is, we will answer the question: How does a buffered solution resist changes in pH when an acid or a base is added?
As you do the calculations associated with buffered solutions, keep in mind that these are merely solutions containing weak acids or bases, and the procedures required are the same ones we have already developed. Be sure to use the systematic approach introduced in Chapter 14.
The pH of a Buffered Solution I A buffered solution contains 0.50 M acetic acid and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution.
Solution The major species in the solution are h h h h Weak Neither Base Very weak acid acid nor (conjugate acid or base base of base HC2H3O2) Examination of the solution components leads to the conclusion that the acetic acid dis-sociation equilibrium, which involves both HC2H3O2 and will control the pH of the solution: The concentrations are as follows: Ka 1.8 105 3H4 3C2H3O2 4 3HC2H3O24 HC2H3O21aq2 ∆H1aq2 C2H3O2 1aq2 C2H3O2 , HC2H3O2, Na, C2H3O2 , (HC2H3O2, Ka 1.8 105) The most important buffering system in the blood involves and H2CO3.
HCO3 The systematic approach developed in Chapter 14 for weak acids and bases applies to buffered solutions.
Sample Exercise 15.2 Notice as you do this problem that it is exactly like examples you have seen in Chapter 14.
H2O Na+ C2H3O2 – HC2H3O2 Major Species Initial Equilibrium Concentration (mol/L) Concentration (mol/L) [H] x [H]0 0 [C2H3O2 ] 0.50 x [C2H3O2 ]0 0.50 [HC2H3O2] 0.50 x [HC2H3O2]0 0.50 x mol/L of HC2H3O2 dissociates to reach equilibrium ---¡ A digital pH meter shows the pH of the buffered solution to be 4.740.
and H2O 15.2 Buffered Solutions 685 The corresponding ICE table is HC2H3O2 OH ¡ C2H3O2 H2O Before 1.0 L 0.50 M 0.010 mol 1.0 L 0.50 M reaction: 0.50 mol 0.50 mol After 0.50 0.010 0.010 0.010 0.50 0.010 reaction: 0.49 mol 0 mol 0.51 mol HC2H3O2(aq) ∆ H(aq) C2H3O2 (aq) Initial: 0.50 0 0.50 Change: x x x Equilibrium: 0.50 x x 0.50 x Then and The approximations are valid (by the 5% rule), so See Exercises 15.33 and 15.34.
pH Changes in Buffered Solutions Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution described in Sample Exercise 15.2. Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water.
Solution Since the added solid NaOH will completely dissociate, the major species in solution before any reaction occurs are HC2H3O2, Na, C2H3O2 , OH, and H2O. Note that the solution contains a relatively large amount of the very strong base hydroxide ion, which has a great affinity for protons. The best source of protons is the acetic acid, and the reaction that will occur is Although acetic acid is a weak acid, the hydroxide ion is such a strong base that the reaction above will proceed essentially to completion (until the ions are consumed).
The best approach to this problem involves two distinct steps: (1) assume that the reaction goes to completion, and carry out the stoichiometric calculations, and then (2) carry out the equilibrium calculations.
1. The stoichiometry problem. The stoichiometry for the reaction is shown below.
OH OH HC2H3O2 ¡ H2O C2H3O2 3H4 x 1.8 105 M and pH 4.74 x 1.8 105 Ka 1.8 105 3H4 3C2H3O2 4 3HC2H3O24 1x210.50 x2 0.50 x 1x210.502 0.50 Sample Exercise 15.3 Note that 0.010 mol HC2H3O2 has been converted to 0.010 mol C2H3O2 by the added OH.
OH– H2O Na+ C2H3O2 – HC2H3O2 Major Species H2O Na+ C2H3O2 – HC2H3O2 Major Species 686 Chapter Fifteen Applications of Aqueous Equilibria 2. The equilibrium problem. After the reaction between OH and HC2H3O2 is complete, the major species in solution are The dominant equilibrium involves the dissociation of acetic acid.
This problem is then very similar to that in Sample Exercise 15.2. The only differ-ence is that the addition of 0.010 mol OH has consumed some HC2H3O2 and produced some C2H3O2 , yielding the following ICE table: HC2H3O2, Na, C2H3O2 , and H2O Note that the initial concentrations are defined after the reaction with OH is complete but before the system adjusts to equilibrium. Following the usual procedure gives and The approximations are valid (by the 5% rule), so The change in pH produced by the addition of 0.01 mol OH to this buffered solution is then h h New solution Original solution The pH increased by 0.02 pH units.
Now compare this with what happens when 0.01 mol solid NaOH is added to 1.0 L water to give 0.01 M NaOH. In this case [OH] 0.01 M and Thus the change in pH is h h New solution Pure water The increase is 5.00 pH units. Note how well the buffered solution resists a change in pH as compared with pure water.
See Exercises 15.35 and 15.36.
Sample Exercises 15.2 and 15.3 represent typical buffer problems that involve all the concepts that you need to know to handle buffered solutions containing weak acids. Pay special attention to the following points: 1. Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14. This is not a new type of problem.
12.00 7.00 5.00 pH 12.00 3H4 Kw 3OH4 1.0 1014 1.0 102 1.0 1012 4.76 4.74 0.02 3H4 x 1.7 105 and pH 4.76 x 1.7 105 Ka 1.8 105 3H4 3C2H3O2 4 3HC2H3O24 1x210.51 x2 0.49 x 1x210.512 0.49 (top) Pure water at pH 7.000. (bottom) When 0.01 mol NaOH is added to 1.0 L of pure water, the pH jumps to 12.000.
HC2H3O2(aq) ∆ H(aq) C2H3O2 (aq) Initial: 0.49 0 0.51 Change: x x x Equilibrium: 0.49 x x 0.51 x 15.2 Buffered Solutions 687 Buffering: How Does It Work?
Sample Exercises 15.2 and 15.3 demonstrate the ability of a buffered solution to absorb hydroxide ions without a significant change in pH. But how does a buffer work? Suppose a buffered solution contains relatively large quantities of a weak acid HA and its conju-gate base A. When hydroxide ions are added to the solution, since the weak acid repre-sents the best source of protons, the following reaction occurs: The net result is that OH ions are not allowed to accumulate but are replaced by A ions.
OH HA ¡ A H2O (H+/OH– added) Modified pH Step 1 Do stoichiometry calculations to determine new concentrations.
Assume reaction with H+/OH– goes to completion.
Step 2 Do equilibrium calculations.
Original buffered solution pH 2. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be pre-sented as follows: OH– added Added OH– ions replaced by A– ions Final pH of buffer close to original Original buffer pH The stability of the pH under these conditions can be understood by examining the equi-librium expression for the dissociation of HA: or, rearranging, In other words, the equilibrium concentration of H, and thus the pH, is determined by the ratio [HA][A]. When OH ions are added, HA is converted to A, and the ratio [HA][A] decreases. However, if the amounts of HA and A originally present are very large compared with the amount of OH added, the change in the [HA][A] ratio will be small.
3H4 Ka 3HA4 3A4 Ka 3H4 3A4 3HA4 In a buffered solution the pH is governed by the ratio [HA][A].
Visualization: Buffers 688 Chapter Fifteen Applications of Aqueous Equilibria In Sample Exercises 15.2 and 15.3, Initially After adding 0.01 mol/L OH The change in the ratio [HA][A] is very small. Thus the [H] and the pH remain essentially constant.
The essence of buffering, then, is that [HA] and [A] are large compared with the amount of OH added. Thus, when the OH is added, the concentrations of HA and A change, but only by small amounts. Under these conditions, the [HA][A] ratio and thus the [H] remain virtually constant.
3HA4 3A4 0.49 0.51 0.96 3HA4 3A4 0.50 0.50 1.0 OH– added The OH– added changes HA to A– , but [HA] and [A–] are large compared to the [OH–] added. Final [HA] [A–] close to original Original [HA] [A–] Similar reasoning applies when protons are added to a buffered solution of a weak acid and a salt of its conjugate base. Because the A ion has a high affinity for H, the added H ions react with A to form the weak acid: and free H ions do not accumulate. In this case there will be a net change of A to HA.
However, if [A] and [HA] are large compared with the [H] added, little change in the pH will occur.
The form of the acid dissociation equilibrium expression (15.1) is often useful for calculating [H] in a buffered solution, since [HA] and [A] are known.
For example, to calculate [H] in a buffered solution containing 0.10 M HF (Ka 7.2 104) and 0.30 M NaF, we simply substitute into Equation (15.1): [HF] q Kap r[F] Another useful form of Equation (15.1) can be obtained by taking the negative log of both sides: log3H4 log1Ka2 loga 3HA4 3A4 b 3H4 17.2 1042 0.10 0.30 2.4 104 M 3H4 Ka 3HA4 3A4 H A ¡ HA Visualization: Adding an Acid to a Buffer 15.2 Buffered Solutions 689 That is, or, where inverting the log term reverses the sign: (15.2) This log form of the expression for Ka is called the Henderson–Hasselbalch equation and is useful for calculating the pH of solutions when the ratio [HA][A] is known.
For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A][HA] will have the same pH. For example, a buffered solution con-taining 5.0 M HC2H3O2 and 3.0 M NaC2H3O2 will have the same pH as one containing 0.050 M HC2H3O2 and 0.030 M NaC2H3O2. This can be shown as follows: pH pKa loga 3A4 3HA4 b pKa loga 3base4 3acid4 b pH pKa loga 3HA4 3A4 b Therefore, Note that in using this equation we have assumed that the equilibrium concentrations of and HA are equal to the initial concentrations. That is, we are assuming the valid-ity of the approximations where x is the amount of acid that dissociates. Since the initial concentrations of HA and are relatively large in a buffered solution, this assumption is generally acceptable.
The pH of a Buffered Solution II Calculate the pH of a solution containing 0.75 M lactic acid and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue dur-ing exertion.
Solution The major species in solution are Since Na has no acid–base properties and H2O is a weak acid or base, the pH will be controlled by the lactic acid dissociation equilibrium: Ka 3H4 3C3H5O3 4 3HC3H5O34 1.4 104 HC3H5O31aq2 ∆H1aq2 C3H5O3 1aq2 HC3H5O3, Na, C3H5O3 , and H2O (Ka 1.4 104) A 3A4 3A4 0 x 3A4 0 and 3HA4 3HA4 0 x 3HA40 A pH pKa loga 3C2H3O2 4 3HC2H3O24 b 4.74 log10.602 4.74 0.22 4.52 Sample Exercise 15.4 H2O Na+ HC3H5O3 C3H5O3 – Major Species System [A]/[HA] 5.0 M HC2H3O2 and 3.0 M NaC2H3O2 0.050 M HC2H3O2 and 0.030 M NaC2H3O2 0.030 M 0.050 M 0.60 3.0 M 5.0 M 0.60 690 Chapter Fifteen Applications of Aqueous Equilibria Since [HC3H5O3]0 and are relatively large, and Thus, using the rearranged Ka expression, we have and Alternatively, we could use the Henderson–Hasselbalch equation: See Exercises 15.37 and 15.38.
Buffered solutions also can be formed from a weak base and the corresponding con-jugate acid. In these solutions, the weak base B reacts with any H added: and the conjugate acid BH reacts with any added : The approach needed to perform pH calculations for these systems is virtually identical to that used above. This makes sense because, as is true of all buffered solutions, a weak acid (BH) and a weak base (B) are present. A typical case is illustrated in Sample Exercise 15.5.
The pH of a Buffered Solution III A buffered solution contains 0.25 M NH3 and 0.40 M NH4Cl. Calcu-late the pH of this solution.
Solution The major species in solution are From the dissolved NH4Cl Since is such a weak base and water is a weak acid or base, the important equilib-rium is and The appropriate ICE table is: Kb 1.8 105 3NH4 4 3OH4 3NH34 NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 Cl NH3, NH4 , Cl, and H2O (Kb 1.8 105) BH OH ¡ B H2O OH B H ¡ BH pH pKa loga 3C3H5O3 4 3HC3H5O34 b 3.85 loga0.25 M 0.75 Mb 3.38 pH log14.2 1042 3.38 3H4 Ka 3HC3H5O34 3C3H5O3 4 11.4 1042 10.75 M2 10.25 M2 4.2 104 M 3C3H5O3 4 3C3H5O3 4 0 0.25 M 3HC3H5O34 3HC3H5O34 0 0.75 M [C3H5O3 ]0 Sample Exercise 15.5 H2O Cl– NH4 + NH3 Major Species ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ NH3(aq) H2O(l) ∆ NH4 (aq) OH(aq) Initial: 0.25 — 0.40 0 Change: x — x x Equilibrium: 0.25 x — 0.40 x x 15.2 Buffered Solutions 691 Then and The approximations are valid (by the 5% rule), so This case is typical of a buffered solution in that the initial and equilibrium concen-trations of buffering materials are essentially the same.
Alternative Solution There is another way of looking at this problem. Since the solution contains relatively large quantities of both NH4 and NH3, we can use the equilibrium to calculate and then calculate [H] from Kw as we have just done. Or we can use the dissociation equilibrium for NH4 , that is, to calculate [H] directly. Either choice will give the same answer, since the same equi-librium concentrations of NH3 and NH4 must satisfy both equilibria.
We can obtain the Ka value for NH4 from the given Kb value for NH3, since Then, using the Henderson–Hasselbalch equation, we have See Exercises 15.37 and 15.38.
Adding Strong Acid to a Buffered Solution I Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from Sample Exercise 15.5.
Solution Before any reaction occurs, the solution contains the following major species: What reaction can occur? We know that H will not react with to form HCl. In con-trast to , the NH3 molecule has a great affinity for protons (this is demonstrated by the fact that NH4 is such a weak acid ). Thus NH3 will react with H to form NH4 : NH31aq2 H1aq2 ¡ NH4 1aq2 [Ka 5.6 1010] Cl Cl NH3, NH4 , Cl, H, and H2O 9.25 loga0.25 M 0.40 Mb 9.25 0.20 9.05 pH pKa loga 3base4 3acid4 b Ka Kw Kb 1.0 1014 1.8 105 5.6 1010 Ka Kb Kw: NH4 1aq2 ∆NH31aq2 H1aq2 [OH] NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 pH 14.00 4.95 9.05 pOH 4.95 3OH4 x 1.1 105 x 1.1 105 Kb 1.8 105 3NH4 4 3OH4 3NH34 10.40 x21x2 0.25 x 10.4021x2 0.25 Sample Exercise 15.6 H+ H2O Cl– NH4 + NH3 Major Species 692 Chapter Fifteen Applications of Aqueous Equilibria Since this reaction can be assumed to go essentially to completion to form the very weak acid NH4 , we will do the stoichiometry calculations before we consider the equilibrium cal-culations. That is, we will let the reaction run to completion and then consider the equilibrium.
The stoichiometry calculations for this process are shown below.
NH3 H ¡ NH4 Before (1.0 L)(0.25 M) 0.10 mol (1.0 L)(0.40 M) reaction: 0.25 mol h 0.40 mol Limiting reactant After 0.25 0.10 0 0.40 0.10 reaction: 0.15 mol 0.50 mol After the reaction goes to completion, the solution contains the major species and We can use the Henderson–Hasselbalch equation, where Then Note that the addition of HCl only slightly decreases the pH, as we would expect in a buffered solution.
See Exercise 15.39.
9.25 loga0.15 M 0.50 Mb 9.25 0.52 8.73 pH pKa loga 3NH34 3NH4 4 b 3Acid4 3NH4 4 3NH4 4 0 0.50 M 3Base4 3NH34 3NH34 0 0.15 M 3NH4 4 0 0.50 mol 1.0 L 0.50 M 3NH34 0 0.15 mol 1.0 L 0.15 M NH3, NH4 , Cl, and H2O Remember: Think about the chemistry first. Ask yourself if a reaction will occur among the major species.
H2O Cl– NH4 + NH3 Major Species Summary of the Most Important Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and the cor-responding weak base. They can involve a weak acid HA and the conjugate base A or a weak base B and the conjugate acid BH.
When H is added to a buffered solution, it reacts essentially to completion with the weak base present: When OH is added to a buffered solution, it reacts essentially to completion with the weak acid present: OH HA ¡ A H2O or OH BH ¡ B H2O H A ¡ HA or H B ¡ BH 15.3 Buffering Capacity 693 15.3 Buffering Capacity The buffering capacity of a buffered solution represents the amount of protons or hy-droxide ions the buffer can absorb without a significant change in pH. A buffer with a large capacity contains large concentrations of buffering components and so can absorb a relatively large amount of protons or hydroxide ions and show little pH change. The pH of a buffered solution is determined by the ratio [A][HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A].
Adding Strong Acid to a Buffered Solution II Calculate the change in pH that occurs when 0.010 mol gaseous HCl is added to 1.0 L of each of the following solutions: For acetic acid, .
Solution For both solutions the initial pH can be determined from the Henderson–Hasselbalch equation: In each case, Therefore, the initial pH for both A and B is After the addition of HCl to each of these solutions, the major species before any re-action occurs are From the added HCl Will any reactions occur among these species? Note that we have a relatively large quan-tity of H, which will readily react with any effective base. We know that will not react with H to form HCl in water. However, will react with H to form the weak acid HC2H3O2: Because HC2H3O2 is a weak acid, we assume that this reaction runs to completion; the 0.010 mol of added H will convert 0.010 mol to 0.010 mol HC2H3O2.
For solution A (since the solution volume is 1.0 L, the number of moles equals the molarity), the following calculations apply: C2H3O2 H1aq2 C2H3O2 1aq2 ¡ HC2H3O21aq2 C2H3O2 Cl HC2H3O2, Na, C2H3O2 , H, Cl, and H2O pH pKa log112 pKa log11.8 1052 4.74 [C2H3O2 ] [HC2H3O2].
pH pKa loga 3C2H3O2 4 3HC2H3O24 b Ka 1.8 105 Solution B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2 Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 Sample Exercise 15.7 A buffer with a large capacity contains large concentrations of the buffering components.
⎧ ⎪ ⎨ ⎪ ⎩ H2O Na+ C2H3O2 – HC2H3O2 Cl– H+ Major Species H C2H3O2 ¡ HC2H3O2 Before reaction: 0.010 M 5.00 M 5.00 M After reaction: 0 4.99 M 5.01 M The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A or B and BH) are large compared with the amounts of H or OH added.
694 Chapter Fifteen Applications of Aqueous Equilibria The new pH can be obtained by substituting the new concentrations into the Henderson– Hasselbalch equation: There is virtually no change in pH for solution A when 0.010 mol gaseous HCl is added.
For solution B, the following calculations apply: 4.74 loga4.99 5.01b 4.74 0.0017 4.74 pH pKa loga 3C2H3O2 4 3HC2H3O24 b Original solution New solution H [A] 5.00 [A] 4.99 1.0 0 0.996 [HA] 5.00 [HA] 5.01 added H C2H3O2 ¡ HC2H3O2 Before reaction: 0.010 M 0.050 M 0.050 M After reaction: 0 0.040 M 0.060 M The new pH is Although the pH change for solution B is small, a change did occur, which is in contrast to solution A.
These results show that solution A, which contains much larger quantities of buffer-ing components, has a much higher buffering capacity than solution B.
See Exercises 15.39 and 15.40.
We have seen that the pH of a buffered solution depends on the ratio of the concen-trations of buffering components. When this ratio is least affected by added protons or hy-droxide ions, the solution is the most resistant to a change in pH. To find the ratio that gives optimal buffering, let’s suppose we have a buffered solution containing a large concentration of acetate ion and only a small concentration of acetic acid. Addition of protons to form acetic acid will produce a relatively large percent change in the concen-tration of acetic acid and so will produce a relatively large change in the ratio (see Table 15.1). Similarly, if hydroxide ions are added to remove some acetic acid, the percent change in the concentration of acetic acid is again large. The same effects are seen if the initial concentration of acetic acid is large and that of acetate ion is small.
[C2H3O2 ] [HC2H3O2] 4.74 0.18 4.56 pH 4.74 loga0.040 0.060b [A] 0.050 H [A] 0.040 1.0 0.67 [HA] 0.050 [HA] 0.060 added Original solution New solution Solution A H+ [Ac–] [HAc] = 1.00 2% change Original A [Ac–] [HAc] = 0.98 Final A H+ [H+] =1.8 x 10–5 M pH = 4.74 [H+] =1.8 x 10–5 M pH = 4.74 Final A Original A Solution B H+ [Ac–] [HAc] = 100 50.5% change Original B [Ac–] [HAc] = 49.5 Final B H+ [H+] = 1.8 x 10–7 M pH = 6.74 [H+] = 3.6 x 10–7 M pH = 6.44 Final B Original B TABLE 15.1 Change in [C2H3O2 ]/[HC2H3O2 ] for Two Solutions When 0.01 mol H Is Added to 1.0 L of Each Percent Solution Change Change A 1.00 n 0.98 2.00% B 100 n 49.5 50.5% 0.99 M 0.02 M 49.5 1.00 M 0.01 M 100 0.99 M 1.01 M 0.98 1.00 M 1.00 M 1.00 a [C2H3O2 ] [HC2H3O2] b new a [C2H3O2 ] [HC2H3O2] b orig 15.3 Buffering Capacity 695 Because large changes in the ratio will produce large changes in pH, we want to avoid this situation for the most effective buffering. This type of reasoning leads us to the general conclusion that optimal buffering occurs when [HA] is equal to It is for this condition that the ratio is most resistant to change when H or is added to the buffered solution. This means that when choosing the buffer-ing components for a specific application, we want to equal 1. It follows that since the pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH. For example, suppose we need a buffered solution with a pH of 4.00. The most effective buffering will occur when [HA] is equal to [A]. From the Henderson– Hasselbalch equation, 4.00 is O ORatio 1 for most effective buffer wanted That is, Thus the best choice of a weak acid is one that has Preparing a Buffer A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts): a. chloroacetic acid b. propanoic acid c. benzoic acid d. hypochlorous acid Calculate the ratio required for each system to yield a pH of 4.30. Which system will work best?
Solution A pH of 4.30 corresponds to Since Ka values rather than pKa values are given for the various acids, we use Equa-tion (15.1) rather than the Henderson–Hasselbalch equation. We substitute the required [H] and Ka for each acid into Equation (15.1) to calculate the ratio needed in each case.
[HA] [A] 3H4 Ka 3HA4 3A4 3H4 104.30 antilog14.302 5.0 105 M [HA] [A] (Ka 3.5 108) (Ka 6.4 105) (Ka 1.3 105) (Ka 1.35 103) pKa 4.00 or Ka 1.0 104.
4.00 pKa log(1) pKa 0 and pKa 4.00 pH pKa loga 3A4 3HA4 b pH pKa loga 3A4 3HA4 b pKa log112 pKa [A] [HA] OH [A] [HA] [A].
[A] [HA] Sample Exercise 15.8 h A A h A A 696 Chapter Fifteen Applications of Aqueous Equilibria Since for benzoic acid is closest to 1, the system of benzoic acid and its sodium salt will be the best choice among those given for buffering a solution at pH 4.3.
This example demonstrates the principle that the optimal buffering system has a pKa value close to the desired pH. The pKa for benzoic acid is 4.19.
See Exercises 15.45 and 15.46.
15.4 Titrations and pH Curves As we saw in Chapter 4, a titration is commonly used to determine the amount of acid or base in a solution. This process involves a solution of known concentration (the titrant) delivered from a buret into the unknown solution until the substance being analyzed is just consumed. The stoichiometric (equivalence) point is often signaled by the color change of an indicator. In this section we will discuss the pH changes that occur during an acid–base titration. We will use this information later to show how an appropriate indi-cator can be chosen for a particular titration.
The progress of an acid–base titration is often monitored by plotting the pH of the solution being analyzed as a function of the amount of titrant added. Such a plot is called a pH curve or titration curve.
Strong Acid–Strong Base Titrations The net ionic reaction for a strong acid–strong base titration is To compute [H] at a given point in the titration, we must determine the amount of H that remains at that point and divide by the total volume of the solution. Before we proceed, we need to consider a new unit, which is especially convenient for titrations. Since titrations usually involve small quantities (burets are typically graduated in milliliters), the mole is inconveniently large. Therefore, we will use the millimole (abbreviated mmol), which, as the prefix indicates, is a thousandth of a mole: 1 mmol 1 mol 1000 103 mol H1aq2 OH1aq2 ¡ H2O1l2 [HA] [A] Acid [H] Ka a. Chloroacetic b. Propanoic 3.8 c. Benzoic 0.78 d. Hypochlorous 1.4 103 5.0 105 3.5 108a 3HA4 3A4 b 5.0 105 6.4 105a 3HA4 3A4 b 5.0 105 1.3 105a 3HA4 3A4 b 3.7 102 5.0 105 1.35 103a 3HA4 3A4 b [HA] [A] [HA] [A] A setup used to do the pH titration of an acid or a base.
15.4 Titrations and pH Curves 697 So far we have defined molarity only in terms of moles per liter. We can now define it in terms of millimoles per milliliter, as shown below: A 1.0 M solution thus contains 1.0 mole of solute per liter of solution or, equivalently, 1.0 millimole of solute per milliliter of solution. Just as we obtain the number of moles of solute from the product of the volume in liters and the molarity, we obtain the number of millimoles of solute from the product of the volume in milliliters and the molarity: CASE STUDY: Strong Acid–Strong Base Titration We will illustrate the calculations involved in a strong acid–strong base titration by con-sidering the titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH. We will calculate the pH of the solution at selected points during the course of the titration, where specific volumes of 0.100 M NaOH have been added.
A. No NaOH has been added.
Since HNO3 is a strong acid (is completely dissociated), the solution contains the major species and the pH is determined by the H from the nitric acid. Since 0.200 M HNO3 contains 0.200 M H, B. 10.0 mL of 0.100 M NaOH has been added.
In the mixed solution before any reaction occurs, the major species are Note that large quantities of both H and OH are present. The 1.00 mmol (10.0 mL 0.100 M) of added OH will react with 1.00 mmol H to form water: H, NO3 , Na, OH, and H2O 3H4 0.200 M and pH 0.699 H, NO3 , and H2O Number of mmol volume 1in mL2 molarity Molarity mol solute L solution mol solute 1000 L solution 1000 mmol solute mL solution 1 millimole 1 103 mol 1 mL 1 103 L mmol mL mol L M 0 0 Vol NaOH added (mL) 50.0 A 7.0 13.0 pH 100.0 150.0 200.0 0 0 Vol NaOH added (mL) 50.0 A 7.0 13.0 pH 100.0 150.0 200.0 B The final solution volume is the sum of the original volume of HNO3 and the vol-ume of added NaOH.
H OH ¡ H2O Before 50.0 mL 0.200 M 10.0 mL 0.100 M reaction: 10.0 mmol 1.00 mmol After 10.0 1.00 1.00 1.00 reaction: 9.0 mmol 0 After the reaction, the solution contains H, NO3 , Na, and H2O (the OH ions have been consumed) and the pH will be determined by the H remaining: p r Original volume of Volume of HNO3 solution NaOH added pH log10.152 0.82 3H4 mmol H left volume of solution 1mL2 9.0 mmol 150.0 10.02 mL 0.15 M 698 Chapter Fifteen Applications of Aqueous Equilibria C. 20.0 mL (total) of 0.100 M NaOH has been added.
We consider this point from the perspective that a total of 20.0 mL NaOH has been added to the original solution, rather than that 10.0 mL has been added to the solution from point B. It is best to go back to the original solution each time so that a mistake made at an earlier point does not show up in each succeeding calculation. As before, the added OH will react with H to form water: After the reaction (H remaining) D. 50.0 mL (total) of 0.100 M NaOH has been added.
Proceeding exactly as for points B and C, the pH is found to be 1.301.
E. 100.0 mL (total) of 0.100 M NaOH has been added.
At this point the amount of NaOH that has been added is The original amount of nitric acid was Enough OH has been added to react exactly with the H from the nitric acid. This is the stoichiometric point, or equivalence point, of the titration. At this point the major species in solution are Since Na has no acid or base properties and NO3 is the anion of the strong acid HNO3 and is therefore a very weak base, neither NO3 nor Na affects the pH, and the solution is neutral (the pH is 7.00).
F. 150.0 mL (total) of 0.100 M NaOH has been added.
The stoichiometric calculations for the titration reaction are as follows: Na, NO3 , and H2O 50.0 mL 0.200 M 10.0 mmol 100.0 mL 0.100 M 10.0 mmol pH 0.942 3H4 8.00 mmol 150.0 20.02 mL 0.11 M H OH ¡ H2O Before 50.0 mL 0.200 M 20.0 mL 0.100 M reaction: 10.0 mmol 2.00 mmol After 10.0 2.00 2.00 2.00 reaction: 8.00 mmol 0 mmol 0 0 Vol NaOH added (mL) 50.0 A 7.0 13.0 pH 100.0 150.0 200.0 B C D Equivalence (stoichiometric) point: The point in the titration where an amount of base has been added to exactly react with all the acid originally present.
0 0 Vol NaOH added (mL) 50.0 A 7.0 13.0 pH 100.0 150.0 200.0 B C D E 0 0 Vol NaOH added (mL) 50.0 A 7.0 13.0 pH 100.0 150.0 200.0 B C D E F H OH ¡ H2O Before 50.0 mL 0.200 M 150.0 mL 0.100 M reaction: 10.0 mmol 15.0 mmol After 10.0 10.0 15.0 10.0 reaction: 0 mmol 5.0 mmol c Excess OH added 15.4 Titrations and pH Curves 699 Now OH is in excess and will determine the pH: Since [H][OH] 1.0 1014, G. 200.0 mL (total) of 0.100 M NaOH has been added.
Proceeding as for point F, the pH is found to be 12.60.
The results of these calculations are summarized by the pH curve shown in Fig. 15.1.
Note that the pH changes very gradually until the titration is close to the equivalence point, where a dramatic change occurs. This behavior is due to the fact that early in the titration there is a relatively large amount of H in the solution, and the addition of a given amount of OH thus produces a small change in pH. However, near the equivalence point [H] is relatively small, and the addition of a small amount of OH produces a large change.
The pH curve in Fig. 15.1, typical of the titration of a strong acid with a strong base, has the following characteristics: Before the equivalence point, [H] (and hence the pH) can be calculated by divid-ing the number of millimoles of H remaining by the total volume of the solution in millimeters.
At the equivalence point, the pH is 7.00.
After the equivalence point, [OH] can be calculated by dividing the number of mil-limoles of excess OH by the total volume of the solution. Then [H] is obtained from Kw.
The titration of a strong base with a strong acid requires reasoning very similar to that used above, except, of course, that OH is in excess before the equivalence point and H is in excess after the equivalence point. The pH curve for the titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCl is shown in Fig. 15.2.
3H4 1.0 1014 2.5 102 4.0 1013 M and pH 12.40 3OH4 mmol OH in excess volume 1mL2 5.0 mmol 150.0 150.02 mL 5.0 mmol 200.0 mL 0.025 M 0 0 Vol NaOH added (mL) 50.0 A 7.0 13.0 pH 100.0 150.0 200.0 B C D E F G 0 0 Vol NaOH added (mL) 50.0 7.0 13.0 pH 100.0 150.0 Equivalence point 200.0 FIGURE 15.1 The pH curve for the titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH. Note that the equivalence point occurs at 100.0 mL of NaOH added, the point where ex-actly enough OH has been added to react with all the H originally present. The pH of 7 at the equivalence point is characteris-tic of a strong acid–strong base titration.
Vol 1.0 M HCl added 7.0 14.0 50.00 mL Equivalence point pH FIGURE 15.2 The pH curve for the titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCl. The equivalence point occurs at 50.00 mL of HCl added, since at this point 5.0 mmol H has been added to re-act with the original 5.0 mmol OH.
700 Chapter Fifteen Applications of Aqueous Equilibria Titrations of Weak Acids with Strong Bases We have seen that since strong acids and strong bases are completely dissociated, the cal-culations to obtain the pH curves for titrations involving the two are quite straightforward.
However, when the acid being titrated is a weak acid, there is a major difference: To cal-culate [H] after a certain amount of strong base has been added, we must deal with the weak acid dissociation equilibrium. We have dealt with this same situation earlier in this chapter when we treated buffered solutions. Calculation of the pH curve for a titration of a weak acid with a strong base really amounts to a series of buffer problems. In performing these calculations it is very important to remember that even though the acid is weak, it reacts essentially to completion with hydroxide ion, a very strong base.
Calculating the pH curve for a weak acid–strong base titration involves a two-step procedure.
➥ 1 A stoichiometry problem. The reaction of hydroxide ion with the weak acid is as-sumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined.
➥ 2 An equilibrium problem. The position of the weak acid equilibrium is deter-mined, and the pH is calculated.
It is essential to do these steps separately. Note that the procedures necessary to do these calculations have all been used before.
CASE STUDY: Weak Acid–Strong Base Titration As an illustration, we will consider the titration of 50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka 1.8 105) with 0.10 M NaOH. As before, we will calculate the pH at various points representing volumes of added NaOH.
A. No NaOH has been added.
This is a typical weak acid calculation of the type introduced in Chapter 14. The pH is 2.87. (Check this yourself.) B. 10.0 mL of 0.10 M NaOH has been added.
The major species in the mixed solution before any reaction takes place are The strong base OH will react with the strongest proton donor, which in this case is HC2H3O2.
The Stoichiometry Problem HC2H3O2, OH, Na, and H2O Treat the stoichiometry and equilibrium problems separately.
OH– added New pH Stoichiometry calculation Equilibrium calculation pH You are again doing exactly the same type of calculation already considered in Chapter 14.
The Equilibrium Problem We examine the major components left in the solution after the reaction takes place to decide on the dominant equilibrium. The major species are HC2H3O2, C2H3O2 , Na, and H2O OH HC2H3O2 ¡ C2H3O2 H2O Before 10 mL 0.10 M 50.0 mL 0.10 M 0 mmol reaction: 1.0 mmol 5.0 mmol After 1.0 1.0 5.0 1.0 1.0 mmol reaction: 0 mmol 4.0 mmol c c Formed by Limiting reactant the reaction 15.4 Titrations and pH Curves 701 Since HC2H3O2 is a much stronger acid than H2O, and since C2H3O2 is the conjugate base of HC2H3O2, the pH will be determined by the position of the acetic acid dissocia-tion equilibrium: where We follow the usual steps to complete the equilibrium calculations: Ka 3H4 3C2H3O2 4 3HC2H3O24 HC2H3O21aq2 ∆H1aq2 C2H3O2 1aq2 The appropriate ICE table is Therefore, C. 25.0 mL (total) of 0.10 M NaOH has been added.
The procedure here is very similar to that used at point B and will only be summarized briefly. The stoichiometry problem is summarized as follows: x a4.0 1.0b11.8 1052 7.2 105 3H4 and pH 4.14 1.8 105 Ka 3H4 3C2H3O2 4 3HC2H3O24 xa 1.0 60.0 xb 4.0 60.0 x xa 1.0 60.0b 4.0 60.0 a1.0 4.0bx The initial concentrations are defined after the reaction with OH has gone to completion but before any dissociation of HC2H3O2 occurs.
Note that the approximations made are well within the 5% rule.
∆ Initial: 0 Change: x x x Equilibrium: x 1.0 60.0 x 4.0 60.0 x 1.0 60.0 4.0 60.0 C2H3O2 1aq2 H1aq2 HC2H3O21aq2 OH HC2H3O2 ¡ C2H3O2 H2O Before 25.0 mL 0.10 M 50.0 mL 0.10 M 0 mmol reaction: 2.5 mmol 5.0 mmol After 2.5 2.5 0 5.0 2.5 2.5 mmol reaction: 2.5 mmol Initial Equilibrium Concentration Concentration 3H 4 x 3H 4 0 0 3C2H3O2 4 1.0 60.0 x 3C2H3O2 4 0 1.0 mmol 150.0 10.02 mL 1.0 60.0 3HC2H3O24 4.0 60.0 x 3HC2H3O24 0 4.0 mmol 150.0 10.02 mL 4.0 60.0 -HC2H3O2 ---¡ dissociates x mmol/mL 702 Chapter Fifteen Applications of Aqueous Equilibria After the reaction, the major species in solution are The equilibrium that will control the pH is and the pertinent concentrations are as follows: HC2H3O21aq2 ∆H1aq2 C2H3O2 1aq2 HC2H3O2, C2H3O2 , Na, and H2O The corresponding ICE table is Therefore, This is a special point in the titration because it is halfway to the equivalence point. The original solution, 50.0 mL of 0.10 M HC2H3O2, contained 5.0 mmol HC2H3O2. Thus 5.0 mmol OH is required to reach the equivalence point. That is, 50 mL NaOH is required, since After 25.0 mL NaOH has been added, half the original HC2H3O2 has been converted to C2H3O2 . At this point in the titration [HC2H3O2]0 is equal to [C2H3O2 ]0. We can neglect the effect of dissociation; that is, The expression for Ka at the halfway point is Equal at the halfway point Ka 3H4 3C2H3O2 4 3HC2H3O24 3H4 3C2H3O2 4 0 3HC2H3O24 0 3H4 3C2H3O2 4 3C2H3O2 4 0 x 3C2H3O2 4 0 3HC2H3O24 3HC2H3O24 0 x 3HC2H3O24 0 150.0 mL210.10 M2 5.0 mmol x 1.8 105 3H4 and pH 4.74 1.8 105 Ka 3H4 3C2H3O2 4 3HC2H3O24 xa 2.5 75.0 xb 2.5 75.0 x xa 2.5 75.0b 2.5 75.0 888n At this point, half the acid has been used up, so [HC2H3O2] [C2H3O2 ] Initial Equilibrium Concentration Concentration 3H4 x 3H4 0 0 3C2H3O2 4 2.5 75.0 x 3C2H3O2 4 0 2.5 mmol 150.0 25.02 mL 3HC2H3O24 2.5 75.0 x 3HC2H3O24 0 2.5 mmol 150.0 25.02 mL -HC2H3O2 ---¡ dissociates x mmol/mL ∆ Initial: Change: x x x Equilibrium: x 2.5 75.0 x 2.5 75.0 x 2.5 75.0 0 2.5 75.0 C2H3O2 1aq2 H1aq2 HC2H3O21aq2 15.4 Titrations and pH Curves 703 Then, at the halfway point in the titration, D. 40.0 mL (total) of 0.10 M NaOH has been added.
The procedures required here are the same as those used for points B and C. The pH is 5.35. (Check this yourself.) E. 50.0 mL (total) of 0.10 M NaOH has been added.
This is the equivalence point of the titration; 5.0 mmol OH has been added, which will just react with the 5.0 mmol HC2H3O2 originally present. At this point the solution con-tains the major species Note that the solution contains C2H3O2 , which is a base. Remember that a base wants to combine with a proton, and the only source of protons in this solution is water. Thus the reaction will be This is a weak base reaction characterized by Kb: The relevant concentrations are as follows: Kb 3HC2H3O24 3OH4 3C2H3O2 4 Kw Ka 1.0 1014 1.8 105 5.6 1010 C2H3O2 1aq2 H2O1l2 ∆HC2H3O21aq2 OH1aq2 Na, C2H3O2 , and H2O 3H4 Ka and pH pKa The corresponding ICE table is Initial Concentration (before any C2H3O2 reacts Equilibrium with H2O) Concentration 0.050 M 3HC2H3O24 x 3HC2H3O24 0 0 3OH4 x 3OH4 0 0 3C2H3O2 4 0.050 x 3C2H3O2 4 0 5.0 mmol 150.0 50.02 mL C2H3O2 reacts ------¡ with H2O x mmol/mL Therefore, The approximation is valid (by the 5% rule), so and pH 8.72 3H4 1.9 109 M 3H4 3OH4 Kw 1.0 1014 3OH4 5.3 106 M x 5.3 106 5.6 1010 Kb 3HC2H3O24 3OH4 3C2H3O2 4 1x21x2 0.050 x x2 0.050 ∆ Initial: 0.050 — 0 Change: — Equilibrium: — x x 0.050 x x x x 0 OH1aq2 HC2H3O21aq2 H2O1l2 C2H3O2 1aq2 704 Chapter Fifteen Applications of Aqueous Equilibria This is another important result: The pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7. This is so because the anion of the acid, which remains in solution at the equivalence point, is a base. In contrast, for the titration of a strong acid with a strong base, the pH at the equivalence point is 7.0, be-cause the anion remaining in this case is not an effective base.
F. 60.0 mL (total) of 0.10 M NaOH has been added.
At this point, excess has been added. The stoichiometric calculations are as follows: OH After the reaction is complete, the solution contains the major species There are two bases in this solution, and However, is a weak base compared with . Therefore, the amount of produced by reaction of with H2O will be small compared with the excess already in solution. You can ver-ify this conclusion by looking at point E, where only was produced by . The amount in this case will be even smaller, since the excess will push the Kb equilibrium to the left.
Thus the pH is determined by the excess and G. 75.0 mL (total) of 0.10 M NaOH has been added.
The procedure needed here is very similar to that for point F. The pH is 12.30. (Check this yourself.) The pH curve for this titration is shown in Fig. 15.3. It is important to note the dif-ferences between this curve and that in Fig. 15.1. For example, the shapes of the plots are pH 11.96 3H4 1.0 1014 9.1 103 1.1 1012 M 9.1 103 M 3OH4 mmol of OH in excess volume 1in mL2 1.0 mmol 150.0 60.02 mL OH: OH C2H3O2 5.3 106 M OH OH C2H3O2 OH OH C2H3O2 C2H3O2 .
OH Na, C2H3O2 , OH, and H2O The pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7.
OH HC2H3O2 ¡ C2H3O2 H2O Before 60.0 mL 0.10 M 50.0 mL 0.10 M reaction: 6.0 mmol 5.0 mmol 0 mmol After 6.0 5.0 reaction: 1.0 mmol in excess 5.0 5.0 0 5.0 mmol Vol NaOH added (mL) 25.0 50.0 3.0 9.0 12.0 Equivalence point pH FIGURE 15.3 The pH curve for the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.
Note that the equivalence point occurs at 50.0 mL of NaOH added, where the amount of added OH exactly equals the original amount of acid. The pH at the equivalence point is greater than 7.0 be-cause the C2H3O2 ion present at this point is a base and reacts with water to produce OH.
15.4 Titrations and pH Curves 705 quite different before the equivalence point, although they are very similar after that point.
(The shapes of the strong and weak acid curves are the same after the equivalence points because excess controls the pH in this region in both cases.) Near the beginning of the titration of the weak acid, the pH increases more rapidly than it does in the strong acid case. It levels off near the halfway point and then increases rapidly again. The lev-eling off near the halfway point is caused by buffering effects. Earlier in this chapter we saw that optimal buffering occurs when [HA] is equal to [A]. This is exactly the case at the halfway point of the titration. As we can see from the curve, the pH changes least rapidly in this region of the titration.
The other notable difference between the curves for strong and weak acids is the value of the pH at the equivalence point. For the titration of a strong acid, the equivalence point occurs at pH 7. For the titration of a weak acid, the pH at the equivalence point is greater than 7 because of the basicity of the conjugate base of the weak acid.
It is important to understand that the equivalence point in an acid–base titration is defined by the stoichiometry, not by the pH. The equivalence point occurs when enough titrant has been added to react exactly with all the acid or base being titrated.
Titration of a Weak Acid Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka ) when dissolved in water. If a 50.0-mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution a. after 8.00 mL of 0.100 M NaOH has been added.
b. at the halfway point of the titration.
c. at the equivalence point of the titration.
Solution a. The stoichiometry problem. After 8.00 mL of 0.100 M NaOH has been added, the fol-lowing calculations apply: 6.2 1010 OH Vol NaOH Strong acid pH Weak acid The equivalence point is defined by the stoichiometry, not by the pH.
Sample Exercise 15.9 The equilibrium problem. Since the solution contains the major species the position of the acid dissociation equilibrium will determine the pH.
HCN1aq2 ∆H1aq2 CN1aq2 HCN, CN, Na, and H2O HCN OH ¡ CN H2O Before 50.0 mL 0.100 M 8.00 mL 0.100 M 0 mmol reaction: 5.00 mmol 0.800 mmol After 5.00 0.800 reaction: 4.20 mmol 0.800 0.800 0 0.800 mmol Initial Concentration Equilibrium Concentration 3H4 x 3H 4 0 0 3CN4 0.80 58.0 x 3CN4 0 0.800 mmol 150.0 8.02 mL 3HCN4 4.2 58.0 x 3HCN40 4.2 mmol 150.0 8.02 mL HCN ---¡ dissociates x mmol/mL 706 Chapter Fifteen Applications of Aqueous Equilibria Substituting the equilibrium concentrations into the expression for Ka gives b. At the halfway point of the titration. The amount of HCN originally present can be ob-tained from the original volume and molarity: Thus the halfway point will occur when 2.50 mmol OH has been added: or As was pointed out previously, at the halfway point [HCN] is equal to [CN] and pH is equal to pKa. Thus, after 25.0 mL of 0.100 M NaOH has been added, c. At the equivalence point. The equivalence point will occur when a total of 5.00 mmol OH has been added. Since the NaOH solution is 0.100 M, the equivalence point oc-curs when 50.0 mL NaOH has been added. This amount will form 5.00 mmol CN.
The major species in solution at the equivalence point are Thus the reaction that will control the pH involves the basic cyanide ion extracting a pro-ton from water: and Kb Kw Ka 1.0 1014 6.2 1010 1.6 105 3HCN4 3OH4 3CN4 CN1aq2 H2O1l2 ∆HCN1aq2 OH1aq2 CN, Na, and H2O pH pKa log16.2 10102 9.21 Volume of NaOH 25.0 mL Volume of NaOH 1in mL2 0.100 M 2.50 mmol OH 50.0 mL 0.100 M 5.00 mmol x 3.3 109 M 3H4 and pH 8.49 6.2 1010 Ka 3H4 3CN4 3HCN4 xa0.80 58.0 xb 4.2 58.0 x xa0.80 58.0b a 4.2 58.0b xa0.80 4.2 b The corresponding ICE table is The approximations made here are well within the 5% rule.
∆ Initial: Change: x x x Equilibrium: x 0.80 58.0 x 4.2 58.0 x 0.80 58.0 0 4.2 58.0 CN1aq2 H1aq2 HCN1aq2 Initial Concentration Equilibrium Concentration 3OH4 x 3OH4 0 0 3HCN4 x 3HCN4 0 0 5.00 102 M 3CN4 15.00 1022 x 3CN4 0 5.00 mmol 150.0 50.02 mL CN reacts ----¡ with H2O x mmol/mL of 15.4 Titrations and pH Curves 707 Substituting the equilibrium concentrations into the expression for Kb and solving in the usual way gives Then, from Kw, we have See Exercises 15.55, 15.57, and 15.58.
Two important conclusions can be drawn from a comparison of the titration of 50.0 mL of 0.1 M acetic acid covered earlier in this section and that of 50.0 mL of 0.1 M hydrocyanic acid analyzed in Sample Exercise 15.9. First, the same amount of 0.1 M NaOH is required to reach the equivalence point in both cases. The fact that HCN is a much weaker acid than HC2H3O2 has no bearing on the amount of base re-quired. It is the amount of acid, not its strength, that determines the equivalence point. Second, the pH value at the equivalence point is affected by the acid strength. For the titration of acetic acid, the pH at the equivalence point is 8.72; for the titration of hydrocyanic acid, the pH at the equivalence point is 10.96. This difference occurs because the CN ion is a much stronger base than the C2H3O2 ion. Also, the pH at the halfway point of the titration is much higher for HCN than for HC2H3O2, again because of the greater base strength of the CN ion (or equivalently, the smaller acid strength of HCN).
The strength of a weak acid has a significant effect on the shape of its pH curve. Fig-ure 15.4 shows pH curves for 50-mL samples of 0.10 M solutions of various acids titrated with 0.10 M NaOH. Note that the equivalence point occurs in each case when the same volume of 0.10 M NaOH has been added but that the shapes of the curves are dramati-cally different. The weaker the acid, the greater the pH value at the equivalence point. In particular, note that the vertical region that surrounds the equivalence point becomes shorter as the acid being titrated becomes weaker. We will see in the next section that the choice of an indicator is more limited for such a titration.
Besides being used to analyze for the amount of acid or base in a solution, titrations can be used to determine the values of equilibrium constants, as shown in Sample Exer-cise 15.10.
Calculation of Ka Calculating Ka A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value.
To do so, the chemist dissolves 2.00 mmol of the solid acid in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid?
3H4 1.1 1011 and pH 10.96 3OH4 x 8.9 104 The amount of acid present, not its strength, determines the equivalence point.
∆ Initial: 0.050 — 0 0 Change: x — x x Equilibrium: 0.050 x — x x OH1aq2 HCN1aq2 H2O1l2 CN1aq2 The corresponding ICE table is Vol 0.10 M NaOH added (mL) 10 20 30 40 50 60 2.0 4.0 6.0 8.0 10.0 12.0 0 Ka = 10 –2 Ka = 10 –4 Ka = 10 –6 Ka = 10 –8 Ka = 10 –10 Strong acid pH FIGURE 15.4 The pH curves for the titrations of 50.0-mL samples of 0.10 M acids with various Ka values with 0.10 M NaOH.
Sample Exercise 15.10 708 Chapter Fifteen Applications of Aqueous Equilibria Solution The stoichiometry problem. We represent the monoprotic acid as HA. The stoichiometry for the titration reaction is shown below.
The equilibrium problem. After the reaction the solution contains the major species The pH will be determined by the equilibrium for which Ka 3H4 3A4 3HA4 HA1aq2 ∆H1aq2 A1aq2 HA, A, Na, and H2O The corresponding ICE table is add OH 1.00 mmol HA 1.00 mmol A 2.00 mmol HA Initial Concentration Equilibrium Concentration 3H4 x 3H4 0 0 8.33 103 M 3A4 8.33 103 x 3A4 1.00 mmol 1100.0 20.02 mL 8.33 103 M 3HA4 8.33 103 x 3HA40 1.00 mmol 1100.0 20.02 mL HA(aq) ∆ H(aq) A(aq) Initial: Change: Equilibrium: x 8.33 103 x 8.33 103 x x x x 8.33 103 0 8.33 103 Note that x is known here because the pH at this point is known to be 6.00. Thus Substituting the equilibrium concentrations into the expression for Ka allows calculation of the Ka value: 11.0 106218.33 1032 8.33 103 1.0 106 11.0 106218.33 103 1.0 1062 18.33 1032 11.0 1062 Ka 3H4 3A4 3HA4 x 18.33 103 x2 18.33 1032 x x 3H 4 antilog1pH2 1.0 106 M x mmol/mL HA -------¡ dissociates HA OH ¡ A H2O Before 2.00 mmol 20.0 mL 0.0500 M 0 mmol reaction: 1.00 mmol After 2.00 1.00 reaction: 1.00 mmol 1.00 1.00 0 1.00 mmol 15.4 Titrations and pH Curves 709 There is an easier way to think about this problem. The original solution contained 2.00 mmol of HA, and since 20.0 mL of added 0.0500 M NaOH contains 1.0 mmol OH, this is the halfway point in the titration (where [HA] is equal to [A]). Thus See Exercise 15.63.
Titrations of Weak Bases with Strong Acids Titrations of weak bases with strong acids can be treated using the procedures we intro-duced previously. As always, you should think first about the major species in solution and decide whether a reaction occurs that runs essentially to completion. If such a reac-tion does occur, let it run to completion and do the stoichiometric calculations. Finally, choose the dominant equilibrium and calculate the pH.
CASE STUDY: Weak Base–Strong Acid Titration The calculations involved for the titration of a weak base with a strong acid will be il-lustrated by the titration of 100.0 mL of 0.050 M NH3 with 0.10 M HCl.
Before the addition of any HCl.
1. Major species: NH3 is a base and will seek a source of protons. In this case H2O is the only avail-able source.
2. No reactions occur that go to completion, since NH3 cannot readily take a proton from H2O. This is evidenced by the small Kb value for NH3.
3. The equilibrium that controls the pH involves the reaction of ammonia with water: Use Kb to calculate [OH]. Although NH3 is a weak base (compared with OH), it produces much more OH in this reaction than is produced from the autoionization of H2O.
Before the equivalence point.
1. Major species (before any reaction occurs): From added HCl 2. The NH3 will react with H from the added HCl: This reaction proceeds essentially to completion because the NH3 readily reacts with a free proton. This case is much different from the previous case, where H2O was the only source of protons. The stoichiometric calculations are then carried out using the known volume of 0.10 M HCl added.
3. After the reaction of NH3 with H is run to completion, the solution contains the fol-lowing major species: h Formed in titration reaction NH3, NH4 , Cl, and H2O NH31aq2 H1aq2 ∆NH4 1aq2 NH3, H, Cl, and H2O NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 NH3 and H2O 3H4 Ka 1.0 106 ⎧ ⎪ ⎨ ⎪ ⎩ 710 Chapter Fifteen Applications of Aqueous Equilibria Note that the solution contains NH3 and NH4 , and the equilibria involving these species will determine [H]. You can use either the dissociation reaction of NH4 or the reaction of NH3 with H2O At the equivalence point.
1. By definition, the equivalence point occurs when all the original NH3 is converted to NH4 . Thus the major species in solution are 2. No reactions occur that go to completion.
3. The dominant equilibrium (the one that controls the [H]) will be the dissociation of the weak acid NH4 , for which Beyond the equivalence point.
1. Excess HCl has been added, and the major species are 2. No reaction occurs that goes to completion.
3. Although NH4 will dissociate, it is such a weak acid that [H] will be determined simply by the excess H: The results of these calculations are shown in Table 15.2. The pH curve is shown in Fig. 15.5.
3H4 mmol H in excess mL solution H, NH4 , Cl, and H2O Ka Kw Kb1for NH32 NH4 , Cl, and H2O NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 NH4 1aq2 ∆NH31aq2 H1aq2 TABLE 15.2 Summary of Results for the Titration of 100.0 mL 0.050 M NH3 with 0.10 M HCl Volume of 0.10 M HCl Added (mL) [NH3]0 [NH4 ]0 [H] pH 0 0.05 M 0 1.1 1011 M 10.96 10.0 1.4 1010 M 9.85 25.0 5.6 1010 M 9.25 50.0† 0 4.3 106 M 5.36 60.0‡ 0 2.21 6.2 103 M Halfway point †Equivalence point ‡[H] determined by the 1.0 mmol of excess H 1.0 mmol 160 mL 5.0 mmol 1100 602 mL 5.0 mmol 1100 502 mL 2.5 mmol 1100 252 mL 2.5 mmol 1100 252 mL 1.0 mmol 1100 102 mL 4.0 mmol 1100 102 mL 15.5 Acid–Base Indicators 711 15.5 Acid–Base Indicators There are two common methods for determining the equivalence point of an acid–base titration: 1. Use a pH meter (see Fig. 14.9) to monitor the pH and then plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point (for example, see Figs. 15.1 through 15.5).
2. Use an acid–base indicator, which marks the end point of a titration by changing color.
Although the equivalence point of a titration, defined by the stoichiometry, is not necessarily the same as the end point (where the indicator changes color), careful selection of the indicator will ensure that the error is negligible.
The most common acid–base indicators are complex molecules that are themselves weak acids (represented by HIn). They exhibit one color when the proton is attached to the molecule and a different color when the proton is absent. For example, phenolphthalein, a commonly used indicator, is colorless in its HIn form and pink in its In, or basic, form.
The actual structures of the two forms of phenolphthalein are shown in Fig. 15.6.
To see how molecules such as phenolphthalein function as indicators, consider the fol-lowing equilibrium for some hypothetical indicator HIn, a weak acid with Ka 1.0 108.
FIGURE 15.5 The pH curve for the titration of 100.0 mL of 0.050 M NH3 with 0.10 M HCl. Note the pH at the equivalence point is less than 7, since the solution contains the weak acid NH4 .
FIGURE 15.6 The acid and base forms of the indicator phenolphthalein. In the acid form (Hln), the molecule is colorless. When a proton (plus H2O) is removed to give the base form (ln), the color changes to pink.
The indicator phenolphthalein is colorless in acidic solution and pink in basic solution.
0 Vol 0.10 M HCl (mL) 10 2 20 30 40 50 60 70 4 6 8 10 12 Equivalence point pH HO C OH C O– O (Colorless acid form, HIn) –O C O C O– O (Pink base form, In–) OH 712 Chapter Fifteen Applications of Aqueous Equilibria Red Blue By rearranging, we get Suppose we add a few drops of this indicator to an acidic solution whose pH is 1.0 ([H] 1.0 101). Then This ratio shows that the predominant form of the indicator is HIn, resulting in a red so-lution. As OH is added to this solution in a titration, [H] decreases and the equilibrium shifts to the right, changing HIn to In. At some point in a titration, enough of the In form will be present in the solution so that a purple tint will be noticeable. That is, a color change from red to reddish purple will occur.
How much In must be present for the human eye to detect that the color is different from the original one? For most indicators, about a tenth of the initial form must be con-verted to the other form before a new color is apparent. We will assume, then, that in the titration of an acid with a base, the color change will occur at a pH where Indicator Color Change Bromthymol blue, an indicator with a Ka value of 1.0 107, is yellow in its HIn form and blue in its In form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible?
3In4 3HIn4 1 10 Ka 3H4 1.0 108 1.0 101 107 1 10,000,000 3In4 3HIn4 Ka 3H4 3In4 3HIn4 Ka 3H4 3In4 3HIn4 HIn1aq2 ∆H1aq2 In1aq2 The end point is defined by the change in color of the indicator. The equivalence point is defined by the reaction stoichiometry.
Methyl orange indicator is yellow in basic solution and red in acidic solution.
Sample Exercise 15.11 15.5 Acid–Base Indicators 713 Solution For bromthymol blue, We assume that the color change is visible when That is, we assume that we can see the first hint of a greenish tint (yellow plus a little blue) when the solution contains 1 part blue and 10 parts yellow (see Fig. 15.7). Thus The color change is first visible at pH 6.00.
See Exercises 15.65 through 15.68.
The Henderson–Hasselbalch equation is very useful in determining the pH at which an indicator changes color. For example, application of Equation (15.2) to the Ka expres-sion for the general indicator HIn yields where Ka is the dissociation constant for the acid form of the indicator (HIn). Since we assume that the color change is visible when 3In4 3HIn4 1 10 pH pKa loga 3In4 3HIn4 b 3H4 1.0 106 or pH 6.00 Ka 1.0 107 3H4 112 10 3In4 3HIn4 1 10 Ka 1.0 107 3H4 3In4 3HIn4 (a) (a) (b) (c) FIGURE 15.7 (a) Yellow acid form of bromthymol blue; (b) a greenish tint is seen when the solution contains 1 part blue and 10 parts yellow; (c) blue basic form.
714 Chapter Fifteen Applications of Aqueous Equilibria we have the following equation for determining the pH at which the color change occurs: For bromthymol blue (Ka 1 107, or pKa 7), the pH at the color change is as we calculated in Sample Exercise 15.11.
When a basic solution is titrated, the indicator HIn will initially exist as In in solu-tion, but as acid is added, more HIn will be formed. In this case the color change will be visible when there is a mixture of 10 parts In and 1 part HIn. That is, a color change from blue to blue-green will occur (see Fig. 15.7) due to the presence of some of the yellow HIn molecules. This color change will be first visible when Note that this is the reciprocal of the ratio for the titration of an acid. Substituting this ratio into the Henderson–Hasselbalch equation gives For bromthymol blue (pKa 7), we have a color change at In summary, when bromthymol blue is used for the titration of an acid, the starting form will be HIn (yellow), and the color change occurs at a pH of about 6. When bromthy-mol blue is used for the titration of a base, the starting form is In (blue), and the color change occurs at a pH of about 8. Thus the useful pH range for bromthymol blue is or from 6 to 8. This is a general result. For a typical acid–base indicator with dissociation constant Ka, the color transition occurs over a range of pH values given by pKa 1. The useful pH ranges for several common indicators are shown in Fig. 15.8.
When we choose an indicator for a titration, we want the indicator end point (where the color changes) and the titration equivalence point to be as close as possible. Choos-ing an indicator is easier if there is a large change in pH near the equivalence point of the titration. The dramatic change in pH near the equivalence point in a strong acid–strong base titration (Figs. 15.1 and 15.2) produces a sharp end point; that is, the complete color change (from the acid-to-base or base-to-acid colors) usually occurs over one drop of added titrant.
What indicator should we use for the titration of 100.00 mL of 0.100 M HCl with 0.100 M NaOH? We know that the equivalence point occurs at pH 7.00. In the initially acidic solution, the indicator will be predominantly in the HIn form. As OH ions are added, the pH increases rather slowly at first (see Fig. 15.1) and then rises rapidly at the equivalence point. This sharp change causes the indicator dissociation equilibrium to shift suddenly to the right, producing enough In ions to give a color change. Since we are titrating an acid, the indicator is predominantly in the acid form initially. There-fore, the first observable color change will occur at a pH where Thus pH pKa log1 1 102 pKa 1 3In4 3HIn4 1 10 HIn ∆H In pKa 1bromthymol blue2 1 7 1 pH 7 1 8 pH pKa log110 1 2 pKa 1 3In4 3HIn4 10 1 pH 7 1 6 pH pKa log1 1 102 pKa 1 Universal indicator paper can be used to estimate the pH of a solution.
15.5 Acid–Base Indicators 715 FIGURE 15.8 The useful pH ranges for several common indicators. Note that most indicators have a useful range of about two pH units, as predicted by the expression pKa 1.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.
pH Crystal Violet Cresol Red Thymol Blue Erythrosin B 2,4-Dinitrophenol Bromphenol Blue Methyl Orange Bromcresol Green Methyl Red Eriochrome Black T Bromcresol Purple Alizarin Bromthymol Blue Phenol Red m-Nitrophenol o-Cresolphthalein Phenolphthalein Thymolphthalein Alizarin Yellow R Trademark CIBA GEIGY CORP.
716 Chapter Fifteen Applications of Aqueous Equilibria If we want an indicator that changes color at pH 7, we can use this relationship to find the pKa value for a suitable indicator: Thus an indicator with a pKa value of 8 (Ka 1 108) changes color at about pH 7 and is ideal for marking the end point for a strong acid–strong base titration.
How crucial is it for a strong acid–strong base titration that the indicator change color exactly at pH 7? We can answer this question by examining the pH change near the equiv-alence point of the titration of 100 mL of 0.10 M HCl and 0.10 M NaOH. The data for a few points at or near the equivalence point are shown in Table 15.3. Note that in going from 99.99 to 100.01 mL of added NaOH solution (about half of a drop), the pH changes from 5.3 to 8.7—a very dramatic change. This behavior leads to the following general conclusions about indicators for a strong acid–strong base titration: Indicator color changes will be sharp, occurring with the addition of a single drop of titrant.
There is a wide choice of suitable indicators. The results will agree within one drop of titrant, using indicators with end points as far apart as pH 5 and pH 9 (see Fig. 15.9).
The titration of weak acids is somewhat different. Figure 15.4 shows that the weaker the acid being titrated, the smaller the vertical area around the equivalence point. This al-lows much less flexibility in choosing the indicator. We must choose an indicator whose useful pH range has a midpoint as close as possible to the pH at the equivalence point.
For example, we saw earlier that in the titration of 0.1 M HC2H3O2 with 0.1 M NaOH the pH at the equivalence point is 8.7 (see Fig. 15.3). A good indicator choice would be phe-nolphthalein, since its useful pH range is 8 to 10. Thymol blue (changes color, pH 8–9) also would be acceptable, but methyl red would not. The choice of an indicator is illus-trated graphically in Fig. 15.10.
pH 7 pKa 1 or pKa 7 1 8 TABLE 15.3 Selected pH Values Near the Equivalence Point in the Titration of 100.0 mL of 0.10 M HCl with 0.10 M NaOH NaOH Added (mL) pH 99.99 5.3 100.00 7.0 100.01 8.7 FIGURE 15.9 The pH curve for the titration of 100.0 mL of 0.10 M HCl with 0.10 M NaOH. Note that the end points of phenolphthalein and methyl red occur at virtually the same amounts of added NaOH.
FIGURE 15.10 The pH curve for the titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH. Phenolph-thalein will give an end point very close to the equivalence point of the titration. Methyl red would change color well before the equiv-alence point (so the end point would be very different from the equivalence point) and would not be a suitable indicator for this titration.
pH 0 0 Vol 0.100 M NaOH added (mL) 2 4 6 8 10 12 14 20 40 60 80 100 120 Equivalence point Phenolphthalein Methyl red pH 0 0 Vol 0.100 M NaOH added (mL) 2 4 6 8 10 12 14 20 40 60 80 100 120 Phenolphthalein Methyl red Equivalence point 15.6 Solubility Equilibria and the Solubility Product 717 Solubility Equilibria 15.6 Solubility Equilibria and the Solubility Product Solubility is a very important phenomenon. The fact that substances such as sugar and table salt dissolve in water allows us to flavor foods easily. The fact that calcium sulfate is less soluble in hot water than in cold water causes it to coat tubes in boilers, reducing thermal efficiency. Tooth decay involves solubility: When food lodges between the teeth, acids form that dissolve tooth enamel, which contains a mineral called hydroxyapatite, Ca5(PO4)3OH. Tooth decay can be reduced by treating teeth with fluoride (see Chemical Impact, p. 720). Fluoride replaces the hydroxide in hydroxyapatite to produce the corre-sponding fluorapatite, Ca5(PO4)3F, and calcium fluoride, CaF2, both of which are less soluble in acids than the original enamel. Another important consequence of solubility in-volves the use of a suspension of barium sulfate to improve the clarity of X rays of the gastrointestinal tract. The very low solubility of barium sulfate, which contains the toxic ion Ba2, makes ingestion of the compound safe.
In this section we consider the equilibria associated with solids dissolving to form aqueous solutions. We will assume that when a typical ionic solid dissolves in water, it dissociates completely into separate hydrated cations and anions. For example, calcium fluoride dissolves in water as follows: When the solid salt is first added to the water, no Ca2 and ions are present. However, as the dissolution proceeds, the concentrations of Ca2 and increase, making it more and more likely that these ions will collide and re-form the solid phase. Thus two com-peting processes are occurring—the dissolution reaction and its reverse: Ultimately, dynamic equilibrium is reached: At this point no more solid dissolves (the solution is said to be saturated).
We can write an equilibrium expression for this process according to the law of mass action: where [Ca2] and [F] are expressed in mol/L. The constant Ksp is called the solubility product constant or simply the solubility product for the equilibrium expression.
Since CaF2 is a pure solid, it is not included in the equilibrium expression. The fact that the amount of excess solid present does not affect the position of the solubility equi-librium might seem strange at first; more solid means more surface area exposed to the solvent, which would seem to result in greater solubility. This is not the case, however.
When the ions in solution re-form the solid, they do so on the surface of the solid. Thus doubling the surface area of the solid not only doubles the rate of dissolving, but also dou-bles the rate of re-formation of the solid. The amount of excess solid present therefore has no effect on the equilibrium position. Similarly, although either increasing the surface area by grinding up the solid or stirring the solution speeds up the attainment of equilib-rium, neither procedure changes the amount of solid dissolved at equilibrium. Neither the amount of excess solid nor the size of the particles present will shift the position of the solubility equilibrium.
It is very important to distinguish between the solubility of a given solid and its sol-ubility product. The solubility product is an equilibrium constant and has only one value Ksp 3Ca24 3F4 2 CaF21s2 ∆Ca21aq2 2F1aq2 Ca21aq2 2F1aq2 ¡ CaF21s2 F F CaF21s2 ¡ Ca21aq2 2F1aq2 H2O Adding F to drinking water is controver-sial. See Geoff Rayner-Canham, “Fluoride: Trying to Separate Fact from Fallacy,” Chem 13 News, Sept. 2001, pp. 16–19.
For simplicity, we will ignore the effects of ion associations in these solutions.
Pure liquids and pure solids are never included in an equilibrium expression (Section 13.4).
An X ray of the lower gastrointestinal tract using barium sulfate.
718 Chapter Fifteen Applications of Aqueous Equilibria for a given solid at a given temperature. Solubility, on the other hand, is an equilibrium position. In pure water at a specific temperature a given salt has a particular solubility. On the other hand, if a common ion is present in the solution, the solubility varies according to the concentration of the common ion. However, in all cases the product of the ion con-centrations must satisfy the Ksp expression. The Ksp values at for many common ionic solids are listed in Table 15.4. The units are customarily omitted.
Solving solubility equilibria problems requires many of the same procedures we have used to deal with acid–base equilibria, as illustrated in Sample Exercises 15.12 and 15.13.
Calculating Ksp from Solubility I Copper(I) bromide has a measured solubility of mol/L at . Calculate its Ksp value.
Solution In this experiment the solid was placed in contact with water. Thus, before any reaction occurred, the system contained solid CuBr and H2O. The process that occurs is the dissolving of CuBr to form the separated Cu and ions: CuBr1s2 ∆Cu1aq2 Br1aq2 Br 25°C 2.0 104 25°C Ksp is an equilibrium constant; solubility is an equilibrium position.
Sample Exercise 15.12 TABLE 15.4 Ksp Values at 25ºC for Common Ionic Solids Ionic Solid Ksp (at 25ºC) Ionic Solid Ksp (at 25ºC) Ionic Solid Ksp (at 25ºC) Fluorides Hg2CrO4 2 109 Co(OH)2 2.5 1016 BaF2 2.4 105 BaCrO4 8.5 1011 Ni(OH)2 1.6 1016 MgF2 6.4 109 Ag2CrO4 9.0 1012 Zn(OH)2 4.5 1017 PbF2 4 108 PbCrO4 2 1016 Cu(OH)2 1.6 1019 SrF2 7.9 1010 Hg(OH)2 3 1026 CaF2 4.0 1011 Carbonates Sn(OH)2 3 1027 NiCO3 1.4 107 Cr(OH)3 6.7 1031 Chlorides CaCO3 8.7 109 Al(OH)3 2 1032 PbCl2 1.6 105 BaCO3 1.6 109 Fe(OH)3 4 1038 AgCl 1.6 1010 SrCO3 7 1010 Co(OH)3 2.5 1043 Hg2Cl2 1.1 1018 CuCO3 2.5 1010 ZnCO3 2 1010 Sulfides Bromides MnCO3 8.8 1011 MnS 2.3 1013 PbBr2 4.6 106 FeCO3 2.1 1011 FeS 3.7 1019 AgBr 5.0 1013 Ag2CO3 8.1 1012 NiS 3 1021 Hg2Br2 1.3 1022 CdCO3 5.2 1012 CoS 5 1022 PbCO3 1.5 1015 ZnS 2.5 1022 Iodides MgCO3 6.8 106 SnS 1 1026 PbI2 1.4 108 Hg2CO3 9.0 1015 CdS 1.0 1028 AgI 1.5 1016 PbS 7 1029 Hg2I2 4.5 1029 Hydroxides CuS 8.5 1045 Ba(OH)2 5.0 103 Ag2S 1.6 1049 Sulfates Sr(OH)2 3.2 104 HgS 1.6 1054 CaSO4 6.1 105 Ca(OH)2 1.3 106 Ag2SO4 1.2 105 AgOH 2.0 108 Phosphates SrSO4 3.2 107 Mg(OH)2 8.9 1012 Ag3PO4 1.8 1018 PbSO4 1.3 108 Mn(OH)2 2 1013 Sr3(PO4)2 1 1031 BaSO4 1.5 109 Cd(OH)2 5.9 1015 Ca3(PO4)2 1.3 1032 Pb(OH)2 1.2 1015 Ba3(PO4)2 6 1039 Chromates Fe(OH)2 1.8 1015 Pb3(PO4)2 1 1054 SrCrO4 3.6 105 Contains Hg2 2 ions. K [Hg2 2][X]2 for Hg2X2 salts, for example.
Visualization: Supersaturated Sodium Acetate Visualization: Solution Equilibrium 15.6 Solubility Equilibria and the Solubility Product 719 where Initially, the solution contains no Cu or , so the initial concentrations are The equilibrium concentrations can be obtained from the measured solubility of CuBr, which is This means that mol solid CuBr dissolves per 1.0 L of solution to come to equilibrium with the excess solid. The reaction is Thus We can now write the equilibrium concentrations: and These equilibrium concentrations allow us to calculate the value of Ksp for CuBr: The units for Ksp values are usually omitted.
See Exercise 15.77.
Calculating Ksp from Solubility II Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of Solution The system initially contains H2O and solid Bi2S3, which dissolves as follows: Therefore, Since no Bi3 and ions were present in solution before the Bi2S3 dissolved, Thus the equilibrium concentrations of these ions will be determined by the amount of salt that dissolves to reach equilibrium, which in this case is Since each Bi2S3 unit contains 2Bi3 and ions: The equilibrium concentrations are Then See Exercises 15.78 through 15.80.
Ksp 3Bi34 23S24 3 12.0 10152213.0 101523 1.1 1073 3S24 3S24 0 change 0 3.0 1015 mol/L 3Bi34 3Bi34 0 change 0 2.0 1015 mol/L ¡ 211.0 1015 mol/L2 Bi31aq2 311.0 1015 mol/L2 S21aq2 1.0 1015 mol/L Bi2S31s2 3S2 1.0 1015 mol/L.
3Bi34 0 3S24 0 0 S2 Ksp 3Bi34 23S24 3 Bi2S31s2 ∆2Bi31aq2 3S21aq2 mol/L at 25°C.
1.0 1015 4.0 108 mol2/L2 4.0 108 Ksp 3Cu4 3Br4 12.0 104 mol/L212.0 104 mol/L2 0 2.0 104 mol/L 3Br4 3Br4 0 change to reach equilibrium 0 2.0 104 mol/L 3Cu4 3Cu4 0 change to reach equilibrium 2.0 104 mol/L Cu1aq2 2.0 104 mol/L Br1aq2 ¡ 2.0 104 mol/L CuBr1s2 CuBr1s2 ¡ Cu1aq2 Br1aq2 2.0 104 2.0 104 mol/L.
3Cu4 0 3Br4 0 0 Br Ksp 3Cu4 3Br4 Sample Exercise 15.13 Precipitation of bismuth sulfide.
Sulfide is a very basic anion and really exists in water as HS. We will not consider this complication.
Solubilities must be expressed in mol/L in Ksp calculations.
720 Chapter Fifteen Applications of Aqueous Equilibria We have seen that the experimentally determined solubility of an ionic solid can be used to calculate its Ksp value. The reverse is also possible: The solubility of an ionic solid can be calculated if its Ksp value is known.
Calculating Solubility from Ksp The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4 107 at 25C. Calculate its solu-bility at 25C.
Solution The system initially contains H2O and solid Cu(IO3)2, which dissolves according to the following equilibrium: Therefore, Ksp 3Cu24 3IO3 4 2 Cu1IO3221s2 ∆Cu21aq2 2IO3 1aq2 CHEMICAL IMPACT The Chemistry of Teeth I f dental chemistry continues to progress at the present rate, tooth decay may soon be a thing of the past. Cavities are holes that develop in tooth enamel, which is composed of the mineral hydroxyapatite, Ca5(PO4)3OH. Recent research has shown that there is constant dissolving and re-forming of the tooth mineral in the saliva at the tooth’s surface. De-mineralization (dissolving of tooth enamel) is mainly caused by weak acids in the saliva created by bacteria as they metabolize carbohydrates in food. (The solubility of Ca5 (PO4)3OH in acidic saliva should come as no surprise to you if you understand how pH affects the solubility of a salt with basic anions.) In the first stages of tooth decay, parts of the tooth sur-face become porous and spongy and develop swiss-cheese-like holes that, if untreated, eventually turn into cavities (see photo).
However, recent results indicate that if the affected tooth is bathed in a solution containing appropriate amounts of Ca2, and , it remineralizes. Because the replaces in the tooth mineral (Ca5(PO4)3OH is changed to Ca5(PO4)3F), the remineralized area is more resistant to future decay, since fluoride is a weaker base than hydroxide ion. In addition, it has been shown that the presence of Sr2 in the remineraliz-ing fluid significantly increases resistance to decay.
If these results hold up under further study, the work of dentists will change dramatically. Dentists will be much OH F F PO4 3, more involved in preventing damage to teeth than in repair-ing damage that has already occurred. One can picture the routine use of a remineralization rinse that will repair prob-lem areas before they become cavities. Dental drills could join leeches as a medical anachronism.
X-ray photo showing decay (dark area) on the molar (right).
This calculation assumes that all the dissolved solid is present as separated ions. In some cases, such as CaSO4, large numbers of ion pairs exist in solution, so this method yields an incorrect value for Ksp.
Sample Exercise 15.14 15.6 Solubility Equilibria and the Solubility Product 721 To find the solubility of Cu(IO3)2, we must find the equilibrium concentrations of the Cu2 and IO3 ions. We do this in the usual way by specifying the initial concentrations (before any solid has dissolved) and then defining the change required to reach equilibrium. Since in this case we do not know the solubility, we will assume that x mol/L of the solid dissolves to reach equilibrium. The 1:2 stoichiometry of the salt means that The concentrations are as follows: x mol/L Cu1IO3221s2 ¡ x mol/L Cu21aq2 2x mol/L IO3 1aq2 Initial Concentration (mol/L) Equilibrium (before any Cu(IO3)2 dissolves) Concentration (mol/L) 3IO3 4 2x 3IO3 4 0 0 3Cu24 x 3Cu24 0 0 x mol/L dissolves to reach equilibrium -----¡ Substituting the equilibrium concentrations into the expression for Ksp gives Then Thus the solubility of solid Cu(IO3)2 is 3.3 103 mol/L.
See Exercises 15.81 and 15.82.
Relative Solubilities A salt’s Ksp value gives us information about its solubility. However, we must be careful in using Ksp values to predict the relative solubilities of a group of salts. There are two possible cases: 1. The salts being compared produce the same number of ions. For example, consider Each of these solids dissolves to produce two ions: If x is the solubility in mol/L, then at equilibrium Therefore, in this case we can compare the solubilities for these solids by comparing the Ksp values: Most soluble; Least soluble; largest Ksp smallest Ksp CaSO41s2 7 CuI1s2 7 AgI1s2 x 1Ksp solubility Ksp 3cation4 3anion4 x2 3Anion4 x 3Cation4 x Ksp 3cation4 3anion4 Salt ∆cation anion CaSO41s2 Ksp 6.1 105 CuI1s2 Ksp 5.0 1012 AgI1s2 Ksp 1.5 1016 x 2 3 3.5 108 3.3 103 mol/L 1.4 107 Ksp 3Cu24 3IO3 4 2 1x212x22 4x3 722 Chapter Fifteen Applications of Aqueous Equilibria 2. The salts being compared produce different numbers of ions. For example, consider Because these salts produce different numbers of ions when they dissolve, the Ksp values cannot be compared directly to determine relative solubilities. In fact, if we calculate the solubilities (using the procedure in Sample Exercise 15.14), we obtain the results summarized in Table 15.5. The order of solubilities is Most soluble Least soluble which is opposite to the order of the Ksp values.
Remember that relative solubilities can be predicted by comparing Ksp values only for salts that produce the same total number of ions.
Common Ion Effect So far we have considered ionic solids dissolved in pure water. We will now see what happens when the water contains an ion in common with the dissolving salt. For exam-ple, consider the solubility of solid silver chromate in a 0.100 M solution of AgNO3. Before any Ag2CrO4 dissolves, the solution contains the major species Ag, and H2O, with solid Ag2CrO4 on the bottom of the container. Since is not found in Ag2CrO4, we can ignore it. The relevant initial concentrations (before any Ag2CrO4 dissolves) are The system comes to equilibrium as the solid Ag2CrO4 dissolves according to the reaction for which We assume that x mol/L of Ag2CrO4 dissolves to reach equilibrium, which means that Now we can specify the equilibrium concentrations in terms of x: Substituting these concentrations into the expression for Ksp gives 9.0 1012 3Ag4 23CrO4 24 10.100 2x221x2 3CrO4 24 3CrO4 24 0 change 0 x x 3Ag4 3Ag4 0 change 0.100 2x x mol/L Ag2CrO41s2 ¡ 2x mol/L Ag1aq2 x mol/L CrO4 2 Ksp 3Ag4 23CrO4 24 9.0 1012 Ag2CrO41s2 ∆2Ag1aq2 CrO4 21aq2 3CrO4 24 0 0 3Ag4 0 0.100 M 1from the dissolved AgNO32 NO3 NO3 , 1Ag2CrO4, Ksp 9.0 10122 Bi2S31s2 7 Ag2S1s2 7 CuS1s2 Bi2S31s2 Ksp 1.1 1073 Ag2S1s2 Ksp 1.6 1049 CuS1s2 Ksp 8.5 1045 TABLE 15.5 Calculated Solubilities for CuS, Ag2S, and Bi2S3 at 25ºC Calculated Salt Ksp Solubility (mol/L) CuS 8.5 1045 9.2 1023 Ag2S 1.6 1049 3.4 1017 Bi2S3 1.1 1073 1.0 1015 A potassium chromate solution being added to aqueous silver nitrate, forming silver chromate.
15.6 Solubility Equilibria and the Solubility Product 723 The mathematics required here appear to be complicated, since the multiplication of terms on the right-hand side produces an expression that contains an x3 term. However, as is usually the case, we can make simplifying assumptions. Since the Ksp value for Ag2CrO4 is small (the position of the equilibrium lies far to the left), x is expected to be small com-pared with 0.100 M. Therefore, which allows simplification of the expression: Then Since x is much less than 0.100 M, the approximation is valid (by the 5% rule). Thus and the equilibrium concentrations are Now we compare the solubilities of Ag2CrO4 in pure water and in 0.100 M AgNO3: Note that the solubility of Ag2CrO4 is much less in the presence of Ag ions from AgNO3.
This is another example of the common ion effect. The solubility of a solid is lowered if the solution already contains ions common to the solid.
Solubility and Common Ions Calculate the solubility of solid CaF2 in a 0.025 M NaF solution.
Solution Before any CaF2 dissolves, the solution contains the major species Na, , and H2O. The solubility equilibrium for CaF2 is and Ksp 4.0 1011 3Ca24 3F4 2 CaF21s2 ∆Ca21aq2 2F1aq2 F 1Ksp 4.0 10112 Solubility of Ag2CrO4 in 0.100 M AgNO3 9.0 1010 mol/L Solubility of Ag2CrO4 in pure water 1.3 104 mol/L 3CrO4 24 x 9.0 1010 M 3Ag4 0.100 2x 0.100 219.0 10102 0.100 M Solubility of Ag2CrO4 in 0.100 M AgNO3 x 9.0 1010 mol/L x 9.0 1012 10.10022 9.0 1010 mol/L 9.0 1012 10.100 2x22 1x2 10.100221x2 0.100 2x 0.100, Substituting the equilibrium concentrations into the expression for Ksp gives Assuming that 2x is negligible compared with 0.025 (since Ksp is small) gives x 6.4 108 4.0 1011 1x210.02522 Ksp 4.0 1011 3Ca24 3F4 2 1x210.025 2x22 Sample Exercise 15.15 Initial Concentration (mol/L) Equilibrium (before any CaF2 dissolves) Concentration (mol/L) p p p From 0.025 M NaF From NaF From CaF2 3F4 0.025 2x 3F4 0 0.025 M 3Ca24 x 3Ca24 0 0 x mol/L CaF2 dissolves 8888888888n to reach equilibrium 724 Chapter Fifteen Applications of Aqueous Equilibria The approximation is valid (by the 5% rule), and Thus mol solid CaF2 dissolves per liter of the 0.025 M NaF solution.
See Exercises 15.89 through 15.92.
pH and Solubility The pH of a solution can greatly affect a salt’s solubility. For example, magnesium hydroxide dissolves according to the equilibrium Addition of ions (an increase in pH) will, by the common ion effect, force the equi-librium to the left, decreasing the solubility of Mg(OH)2. On the other hand, an addition of H ions (a decrease in pH) increases the solubility, because ions are removed from solution by reacting with the added H ions. In response to the lower concentration of , the equilibrium position moves to the right. This is why a suspension of solid Mg(OH)2, known as milk of magnesia, dissolves as required in the stomach to combat excess acidity.
This idea also applies to salts with other types of anions. For example, the solubility of silver phosphate (Ag3PO4) is greater in acid than in pure water because the ion is a strong base that reacts with H to form the ion. The reaction occurs in acidic solution, thus lowering the concentration of and shifting the solu-bility equilibrium to the right. This, in turn, increases the solubility of silver phosphate.
Silver chloride (AgCl), however, has the same solubility in acid as in pure water.
Why? Since the ion is a very weak base (that is, HCl is a very strong acid), no HCl molecules are formed. Thus the addition of H to a solution containing does not af-fect and has no effect on the solubility of a chloride salt.
The general rule is that if the anion is an effective base—that is, if HX is a weak acid—the salt MX will show increased solubility in an acidic solution. Examples of com-mon anions that are effective bases are Salts con-taining these anions are much more soluble in an acidic solution than in pure water.
As mentioned at the beginning of this chapter, one practical result of the increased solubility of carbonates in acid is the formation of huge limestone caves such as Mam-moth Cave in Kentucky and Carlsbad Caverns in New Mexico. Carbon dioxide dissolved in groundwater makes it acidic, increasing the solubility of calcium carbonate and even-tually producing huge caverns. As the carbon dioxide escapes to the air, the pH of the dripping water goes up and the calcium carbonate precipitates, forming stalactites and stalagmites.
15.7 Precipitation and Qualitative Analysis So far we have considered solids dissolving in solutions. Now we will consider the reverse process—the formation of a solid from solution. When solutions are mixed, various re-actions can occur. We have already considered acid–base reactions in some detail. In this section we show how to predict whether a precipitate will form when two solutions are OH, S2, CO3 2, C2O4 2, and CrO4 2.
X [Cl] Cl Cl Ag3PO41s2 ∆3Ag1aq2 PO4 31aq2 PO4 3 H PO4 3 ¡ HPO4 2 HPO4 2 PO4 3 OH OH OH Mg1OH221s2 ∆Mg21aq2 2OH1aq2 6.4 108 Solubility x 6.4 108 mol/L 15.7 Precipitation and Qualitative Analysis 725 mixed. We will use the ion product, which is defined just like the expression for Ksp for a given solid except that initial concentrations are used instead of equilibrium concen-trations. For solid CaF2, the expression for the ion product Q is written If we add a solution containing Ca2 ions to a solution containing ions, a precip-itate may or may not form, depending on the concentrations of these ions in the resulting mixed solution. To predict whether precipitation will occur, we consider the relationship between Q and Ksp.
If Q is greater than Ksp, precipitation occurs and will continue until the concentra-tions are reduced to the point that they satisfy Ksp.
If Q is less than Ksp, no precipitation occurs.
Determining Precipitation Conditions A solution is prepared by adding 750.0 mL of 4.00 103 M Ce(NO3)3 to 300.0 mL of 2.00 102 M KIO3. Will Ce(IO3)3 (Ksp 1.9 1010) precipitate from this solution?
Solution First, we calculate [Ce3]0 and [IO3 ]0 in the mixed solution before any reaction occurs: The ion product for Ce(IO3)3 is Since Q is greater than Ksp, Ce(IO3)3 will precipitate from the mixed solution.
See Exercises 15.97 and 15.98.
Sometimes we want to do more than simply predict whether precipitation will occur; we may want to calculate the equilibrium concentrations in the solution after precipita-tion occurs. For example, let us calculate the equilibrium concentrations of Pb2 and ions in a solution formed by mixing 100.0 mL of 0.0500 M Pb(NO3)2 and 200.0 mL of 0.100 M NaI. First, we must determine whether solid forms when the solutions are mixed. To do so, we need to calculate [Pb2]0 and before any reaction occurs: The ion product for PbI2 is Since Q is greater than Ksp, a precipitate of PbI2 will form.
Q 3Pb24 03I4 0 2 11.67 102216.67 10222 7.43 105 3I4 0 mmol I mL solution 1200.0 mL210.100 mmol/mL2 300.0 mL 6.67 102 M 3Pb24 0 mmol Pb2 mL solution 1100.0 mL210.0500 mmol/mL2 300.0 mL 1.67 102 M [I]0 PbI2 (Ksp 1.4 108) I Q 3Ce34 03IO3 4 0 3 12.86 103215.71 10323 5.32 1010 3IO3 4 0 1300.0 mL212.00 102 mmol/mL2 1750.0 300.02 mL 5.71 103 M 3Ce34 0 1750.0 mL214.00 103 mmol/mL2 1750.0 300.02 mL 2.86 103 M F Q 3Ca24 03F4 0 2 Q is used here in a very similar way to the use of the reaction quotient in Chapter 13.
Sample Exercise 15.16 No Yes Precipitation of Ce(IO3)3 No precipitation Is Q > Ksp?
Find Q Ion product is [Ce3+]0 [IO3 –]0 3 Determine initial concentrations Ce3+ and IO3 – For Ce(IO3)3(s), Ksp [Ce3][IO3 ]3.
726 Chapter Fifteen Applications of Aqueous Equilibria Since the Ksp for PbI2 is quite small , only very small quantities of Pb2 and can coexist in aqueous solution. In other words, when Pb2 and are mixed, most of these ions will precipitate out as PbI2. That is, the reaction (which is the reverse of the dissolution reaction) goes essentially to completion.
If, when two solutions are mixed, a reaction occurs that goes virtually to completion, it is essential to do the stoichiometry calculations before considering the equilibrium cal-culations. Therefore, in this case we let the system go completely in the direction toward which it tends. Then we will let it adjust back to equilibrium. If we let Pb2 and react to completion, we have the following concentrations: I Pb21aq2 2I1aq2 ¡ PbI21s2 I I (1.4 108) Pb2 2I ¡ PbI2 Before (100.0 mL)(0.0500 M) (200.0 mL)(0.100 M) The amount of reaction: 5.00 mmol 20.0 mmol PbI2formed does After 0 mmol 20.0 2(5.00) not influence reaction: 10.0 mmol the equilibrium.
Initial Equilibrium Concentration (mol/L) Concentration (mol/L) 3I4 3.33 102 2x 3I4 0 3.33 102 3Pb24 x 3Pb24 0 0 Next we must allow the system to adjust to equilibrium. At equilibrium [Pb2] is not actually zero because the reaction does not go quite to completion. The best way to think about this is that once the PbI2 is formed, a very small amount redissolves to reach equi-librium. Since is in excess, the PbI2 is dissolving into a solution that contains 10.0 mmol per 300.0 mL of solution, or M .
We could state this problem as follows: What is the solubility of solid PbI2 in a M NaI solution? The lead iodide dissolves according to the equation The concentrations are as follows: PbI21s2 ∆Pb21aq2 2I1aq2 3.33 102 I 3.33 102 I I In this reaction 10 mmol I is in excess.
x mol/L PbI2(s) 888888n dissolves Substituting into the expression for Ksp gives Then Note that so the approximation is valid. These Pb2 and concen-trations thus represent the equilibrium concentrations present in a solution formed by mixing 100.0 mL of 0.0500 M Pb(NO3)2 and 200.0 mL of 0.100 M NaI.
Precipitation A solution is prepared by mixing 150.0 mL of M Mg(NO3)2 and 250.0 mL of M NaF. Calculate the concentrations of Mg2 and at equilibrium with solid MgF2 (Ksp 6.4 109).
F 1.00 101 1.00 102 I 3.33 102 2x, 3I4 3.33 102 M 3Pb24 x 1.3 105 M Ksp 1.4 108 3Pb24 3I4 2 1x213.33 102 2x22 1x213.33 10222 Sample Exercise 15.17 The equilibrium constant for formation of solid Pbl2 is so this equilibrium lies far to the right.
1 Ksp, or 7 107, 15.7 Precipitation and Qualitative Analysis 727 Solution The first step is to determine whether solid MgF2 forms. To do this, we need to calculate the concentrations of Mg2 and in the mixed solution and find Q: Since Q is greater than Ksp, solid MgF2 will form.
The next step is to run the precipitation reaction to completion: Q 3Mg24 03F4 0 2 13.75 103216.25 10222 1.46 105 3F4 0 mmol F mL solution 1250.0 mL211.00 101 M2 400.0 mL 6.25 102 M 3Mg24 0 mmol Mg2 mL solution 1150.0 mL211.00 102 M2 400.0 mL 3.75 103 M F Run reaction to completion Ksp = [Mg2+][F–]2 6.4 x 10–9 = (x)(5.50 x 10–2 + 2x)2 Mg2+ + 2F– MgF2(s) Find initial concentrations Mg2+, F– Q > Ksp Q = [Mg2+]0 [F–]0 2 = 1.46 x 10–5 Find Q [Mg2+]0 = 3.75 x 10–3 M [F–]0 = 6.25 x 10–2 M Determine [Mg2+] and [F–] at equilibrium Calculate concentrations of excess F– [F–] excess = 5.50 x 10–2 M [Mg2+] = 2.1 x 10–6 M [F–] = 5.50 x 10–2 M Mg2+ is limiting F– is in excess Mg2 2F ¡ MgF2(s) Before (150.0)(1.00 102) (250.0)(1.00 101) reaction: 1.50 mmol 25.0 mmol After 25.0 2(1.50) reaction: 1.50 1.50 0 22.0 mmol Initial Equilibrium Concentration (mol/L) Concentration (mol/L) 3F4 5.50 102 2x 3F4 0 5.50 102 3Mg24 x 3Mg24 0 0 Note that excess remains after the precipitation reaction goes to completion. The concentration is Although we have assumed that the Mg2 is completely consumed, we know that [Mg2] will not be zero at equilibrium. We can compute the equilibrium [Mg2] by let-ting MgF2 redissolve to satisfy the expression for Ksp. How much MgF2 will dissolve in a M NaF solution? We proceed as usual: Ksp 3Mg24 3F4 2 6.4 109 MgF21s2 ∆Mg21aq2 2F1aq2 5.50 102 3F4 excess 22.0 mmol 400.0 mL 5.50 102 M F x mol/L MgF2(s) 888888n dissolves See Exercises 15.99 and 15.100.
Selective Precipitation Mixtures of metal ions in aqueous solution are often separated by selective precipitation, that is, by using a reagent whose anion forms a precipitate with only one or a few of the 3F4 5.50 102 M 3Mg24 x 2.1 106 M 1x215.50 102 2x22 1x215.50 10222 Ksp 6.4 109 3Mg24 3F4 2 The approximations made here fall within the 5% rule.
728 Chapter Fifteen Applications of Aqueous Equilibria metal ions in the mixture. For example, suppose we have a solution containing both Ba2 and ions. If NaCl is added to the solution, AgCl precipitates as a white solid, but since BaCl2 is soluble, the Ba2 ions remain in solution.
Selective Precipitation A solution contains 1.0 104 M Cu and 2.0 103 M Pb2. If a source of I is added gradually to this solution, will PbI2 (Ksp 1.4 108) or CuI (Ksp 5.3 1012) pre-cipitate first? Specify the concentration of I necessary to begin precipitation of each salt.
Solution For PbI2, the Ksp expression is Since [Pb2] in this solution is known to be M, the greatest concentration of that can be present without causing precipitation of PbI2 can be calculated from the Ksp expression: Any in excess of this concentration will cause solid PbI2 to form.
Similarly, for CuI, the Ksp expression is and A concentration of in excess of M will cause formation of solid CuI.
As is added to the mixed solution, CuI will precipitate first, since the required is less. Therefore, Cu would be separated from Pb2 using this reagent.
See Exercises 15.101 and 15.102.
Since metal sulfide salts differ dramatically in their solubilities, the sulfide ion is of-ten used to separate metal ions by selective precipitation. For example, consider a solu-tion containing a mixture of 103 M Fe2 and 103 M Mn2. Since FeS (Ksp 3.7 1019) is much less soluble than MnS (Ksp 2.3 1013), careful addition of S2 to the mixture will precipitate Fe2 as FeS, leaving Mn2 in solution.
One real advantage of the sulfide ion as a precipitating reagent is that because it is basic, its concentration can be controlled by regulating the pH of the solution. H2S is a diprotic acid that dissociates in two steps: Note from the small value that S2 ions have a high affinity for protons. In an acidic solution (large [H]), [S2] will be relatively small, since under these conditions the dis-sociation equilibria will lie far to the left. On the other hand, in basic solutions [S2] will be relatively large, since the very small value of [H] will pull both equilibria to the right, producing S2.
Ka2 HS ∆H S2 Ka2 1019 H2S ∆H HS Ka1 1.0 107 [I] I 5.3 108 I 3I4 5.3 108 M 5.3 1012 Ksp 3Cu4 3I4 11.0 1042 3I4 I 3I4 2.6 103 M 1.4 108 3Pb24 3I4 2 12.0 10323I4 2 I 2.0 103 1.4 108 Ksp 3Pb24 3I4 2 Ag Sample Exercise 15.18 We can compare Ksp values to find relative solubilities because FeS and MnS produce the same number of ions in solution.
15.7 Precipitation and Qualitative Analysis 729 This means that the most insoluble sulfide salts, such as CuS (Ksp 8.5 1045) and HgS (Ksp 1.6 1054), can be precipitated from an acidic solution, leaving the more soluble ones, such as MnS (Ksp 2.3 1013) and NiS (Ksp 3 1021), still dissolved. The manganese and nickel sulfides can then be precipitated by making the solution slightly basic. This procedure is diagramed in Fig. 15.11.
Qualitative Analysis The classic scheme for qualitative analysis of a mixture containing all the common cations (listed in Fig. 15.12) involves first separating them into five major groups based on solu-bilities. (These groups are not directly related to the groups of the periodic table.) Each group is then treated further to separate and identify the individual ions. We will be con-cerned here only with separation of the major groups.
Group I—Insoluble chlorides When dilute aqueous HCl is added to a solution containing a mixture of the common cations, only Ag, Pb2, and Hg2 2 will precipitate out as insoluble chlorides. All other chlorides are soluble and remain in solution. The Group I precipitate is removed, leaving the other ions in solution for treatment with sulfide ion.
Group II—Sulfides insoluble in acid solution After the insoluble chlorides are removed, the solution is still acidic, since HCl was added.
If H2S is added to this solution, only the most insoluble sulfides (those of Hg2, Cd2, Bi3, Cu2, and Sn4) will precipitate, since [S2] is relatively low because of the high concentration of H. The more soluble sulfides will remain dissolved under these condi-tions, and the precipitate of the insoluble salt is removed.
Group III—Sulfides insoluble in basic solution The solution is made basic at this stage, and more H2S is added. As we saw earlier, a basic solution produces a higher [S2], which leads to precipitation of the more soluble sulfides.
The cations precipitated as sulfides at this stage are Co2, Zn2, Mn2, Ni2, and Fe2. If any Cr3 and Al3 ions are present, they also will precipitate, but as insoluble hydroxides (remember the solution is now basic). The precipitate is separated from the solution containing the rest of the ions.
Solution of Mn2+, Ni2+, Cu2+, Hg2+ Add H2S (acidic, pH ≈ 2) Add OH– to bring pH to 8 Precipitate of CuS, HgS Precipitate of MnS, NiS Solution of Mn2+, Ni2+ FIGURE 15.11 The separation of Cu2 and Hg2 from Ni2 and Mn2 using H2S. At a low pH, [S2] is relatively low and only the very insoluble HgS and CuS precipitate. When OH is added to lower [H], the value of [S2] increases, and MnS and NiS precipitate.
Flame test for potassium.
Flame test for sodium.
730 Chapter Fifteen Applications of Aqueous Equilibria Group IV—Insoluble carbonates At this point, all the cations have been precipitated except those from Groups 1A and 2A of the periodic table. The Group 2A cations form insoluble carbonates and can be pre-cipitated by the addition of CO3 2. For example, Ba2, Ca2, and Mg2 form solid carbonates and can be removed from the solution.
Precipitate of Hg2Cl2, AgCl, PbCl2 (Group I) Precipitate of CoS, ZnS, MnS, NiS, FeS, Cr(OH)3, Al(OH)3 (Group III) Precipitate of CaCO3, BaCO3, MgCO3 (Group IV) Precipitate of HgS, CdS, Bi2S3, CuS, SnS2 (Group II) Solution of Groups II–IV Solution of Groups III–V Solution of Groups IV, V Solution of Group V Solution of Hg2 2+, Ag+, Pb2+, Hg2+, Cd2+, Bi3+, Cu2+, Sn4+, Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+, Al3+, Ca2+, Ba2+, Mg2+, NH4 +, Na+, K+ Add H2S (aq) Add HCl (aq) Add NaOH (aq) Add Na2CO3 (aq) FIGURE 15.12 A schematic diagram of the classic method for separating the common cations by selective precipitation.
From left to right, cadmium sulfide, chromium(III) hydroxide, aluminum hydroxide, and nickel(II) hydroxide.
15.8 Equilibria Involving Complex Ions 731 Group V—Alkali metal and ammonium ions The only ions remaining in solution at this point are the Group 1A cations and the NH4 ion, all of which form soluble salts with the common anions. The Group 1A cations are usually identified by the characteristic colors they produce when heated in a flame. These colors are due to the emission spectra of these ions.
The qualitative analysis scheme for cations based on the selective precipitation pro-cedure described above is summarized in Fig. 15.12.
Complex Ion Equilibria 15.8 Equilibria Involving Complex Ions A complex ion is a charged species consisting of a metal ion surrounded by ligands. A ligand is simply a Lewis base—a molecule or ion having a lone electron pair that can be donated to an empty orbital on the metal ion to form a covalent bond. Some common li-gands are H2O, NH3, Cl, and CN. The number of ligands attached to a metal ion is called the coordination number. The most common coordination numbers are 6, for example, in Co(H2O)6 2 and Ni(NH3)6 2; 4, for example, in CoCl4 2 and Cu(NH3)4 2; and 2, for example, in Ag(NH3)2 ; but others are known.
The properties of complex ions will be discussed in more detail in Chapter 21. For now, we will just look at the equilibria involving these species. Metal ions add ligands one at a time in steps characterized by equilibrium constants called formation constants or stability constants. For example, when solutions containing Ag ions and NH3 mole-cules are mixed, the following reactions take place: where K1 and K2 are the formation constants for the two steps. In a solution containing Ag and NH3, all the species NH3, Ag, Ag(NH3), and Ag(NH3)2 exist at equilibrium.
Calculating the concentrations of all these components can be complicated. However, usu-ally the total concentration of the ligand is much larger than the total concentration of the metal ion, and approximations can greatly simplify the problems.
For example, consider a solution prepared by mixing 100.0 mL of 2.0 M NH3 with 100.0 mL of 1.0 103 MAgNO3. Before any reaction occurs, the mixed solution contains the major species Ag, NO3 , NH3, and H2O. What reaction or reactions will occur in this solution? From our discussions of acid–base chemistry, we know that one reaction is However, we are interested in the reaction between NH3 and Ag to form complex ions, and since the position of the preceding equilibrium lies far to the left (Kb for NH3 is 1.8 105), we can neglect the amount of NH3 used up in the reaction with water. Therefore, before any complex ion formation, the concentrations in the mixed solution are Total volume 3NH34 0 1100.0 mL212.0 M2 1200.0 mL2 1.0 M n 3Ag4 0 1100.0 mL211.0 103 M2 1200.0 mL2 5.0 104 M NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 Ag1NH32 NH3 ∆Ag1NH322 K2 8.2 103 Ag NH3 ∆Ag1NH32 K1 2.1 103 CoCl4 2 732 Chapter Fifteen Applications of Aqueous Equilibria As mentioned already, the Ag ion reacts with NH3 in a stepwise fashion to form AgNH3 and then Ag(NH3)2 : Since both K1 and K2 are large, and since there is a large excess of NH3, both reactions can be assumed to go essentially to completion. This is equivalent to writing the net re-action in the solution as follows: The relevant stoichiometric calculations are as follows: Ag 2NH3 ¡ Ag1NH322 Ag1NH32 NH3 ∆Ag1NH322 K2 8.2 103 Ag NH3 ∆Ag1NH32 K1 2.1 103 Ag 2NH3 888n Ag(NH3)2 Before reaction: 5.0 104 M 1.0 M 0 After reaction: 0 1.0 2(5.0 104) 1.0 M 5.0 104 M Twice as much NH3 as Ag is required n Note that in this case we have used molarities when performing the stoichiometry calculations and we have assumed this reaction to be complete, using all the original Ag to form Ag(NH3)2 . In reality, a very small amount of the Ag(NH3)2 formed will disso-ciate to produce small amounts of Ag(NH3) and Ag. However, since the amount of Ag(NH3)2 dissociating will be so small, we can safely assume that [Ag(NH3)2 ] is 5.0 104 M at equilibrium. Also, we know that since so little NH3 has been consumed, [NH3] is 1.0 M at equilibrium. We can use these concentrations to calculate [Ag] and [Ag(NH3)] using the K1 and K2 expressions.
To calculate the equilibrium concentration of Ag(NH3), we use since [Ag(NH3)2 ] and [NH3] are known. Rearranging and solving for [Ag(NH3)] give Now the equilibrium concentration of Ag can be calculated using K1: So far we have assumed that Ag(NH3)2 is the dominant silver-containing species in solution. Is this a valid assumption? The calculated concentrations are These values clearly support the conclusion that 3Ag1NH322 4 3Ag1NH324 3Ag4 3Ag4 2.9 1011 M 3Ag1NH324 6.1 108 M 3Ag1NH322 4 5.0 104 M 3Ag4 6.1 108 12.1 103211.02 2.9 1011 M K1 2.1 103 3Ag1NH324 3Ag4 3NH34 6.1 108 3Ag4 11.02 3Ag1NH324 3Ag1NH322 4 K23NH34 5.0 104 18.2 103211.02 6.1 108 M K2 8.2 103 3Ag1NH322 4 3Ag1NH324 3NH34 A solution containing the blue CoCl4 2 complex ion.
Essentially all the Ag ions originally present end up in Ag(NH3)2 .
Visualization: Nickel(II) Complexes 15.8 Equilibria Involving Complex Ions 733 Thus the assumption that Ag(NH3)2 is the dominant Ag-containing species is valid, and the calculated concentrations are correct.
This analysis shows that although complex ion equilibria have many species present and look complicated, the calculations are actually quite straightforward, especially if the ligand is present in large excess.
Complex Ions Calculate the concentrations of Ag, Ag(S2O3), and Ag(S2O3)2 3 in a solution prepared by mixing 150.0 mL of 1.00 103 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3. The stepwise formation equilibria are Solution The concentrations of the ligand and metal ion in the mixed solution before any reaction occurs are Since [S2O3 2]0 [Ag]0, and since K1 and K2 are large, both formation reactions can be assumed to go to completion, and the net reaction in the solution is as follows: 3S2O3 24 0 1200.0 mL215.00 M2 1150.0 mL 200.0 mL2 2.86 M 3Ag4 0 1150.0 mL211.00 103 M2 1150.0 mL 200.0 mL2 4.29 104 M Ag1S2O32 S2O3 2 ∆Ag1S2O322 3 K2 3.9 104 Ag S2O3 2 ∆Ag1S2O32 K1 7.4 108 Sample Exercise 15.19 Ag (S2O3)2 3– Ag 2S2O3 2 88n Ag(S2O3)2 3 Before 4.29 104 M 2.86 M 0 reaction: After 0 2.86 2(4.29 104) 4.29 104 M reaction: 2.86 M Note that Ag is limiting and that the amount of S2O3 2 consumed is negligible. Also note that since all these species are in the same solution, the molarities can be used to do the stoichiometry problem.
Of course, the concentration of Ag is not zero at equilibrium, and there is some Ag(S2O3) in the solution. To calculate the concentrations of these species, we must use the K1 and K2 expressions. We can calculate the concentration of Ag(S2O3) from K2: We can calculate [Ag] from K1: These results show that [Ag(S2O3)2 3] [Ag(S2O3)] [Ag] 3Ag4 1.8 1018 M 7.4 108 K1 3Ag1S2O324 3Ag4 3S2O3 24 3.8 109 3Ag412.862 3Ag1S2O324 3.8 109 M 3.9 104 K2 3Ag1S2O322 34 3Ag1S2O324 3S2O3 24 4.29 104 3Ag1S2O32412.862 734 Chapter Fifteen Applications of Aqueous Equilibria Thus the assumption is valid that essentially all the original Ag is converted to Ag(S2O3)2 3 at equilibrium.
See Exercises 15.109 and 15.110.
Complex Ions and Solubility Often ionic solids that are very nearly water-insoluble must be dissolved somehow in aque-ous solutions. For example, when the various qualitative analysis groups are precipitated out, the precipitates must be redissolved to separate the ions within each group. Consider a solution of cations that contains Ag, Pb2, and Hg2 2, among others. When dilute aque-ous HCl is added to this solution, the Group I ions will form the insoluble chlorides AgCl, PbCl2, and Hg2Cl2. Once this mixed precipitate is separated from the solution, it must be redissolved to identify the cations individually. How can this be done? We know that some solids are more soluble in acidic than in neutral solutions. What about chloride salts? For example, can AgCl be dissolved by using a strong acid? The answer is no, because Cl ions have virtually no affinity for H ions in aqueous solution. The position of the disso-lution equilibrium is not affected by the presence of H.
How can we pull the dissolution equilibrium to the right, even though Cl is an ex-tremely weak base? The key is to lower the concentration of Ag in solution by forming complex ions. For example, Ag reacts with excess NH3 to form the stable complex ion Ag(NH3)2 . As a result, AgCl is quite soluble in concentrated ammonia solutions. The rel-evant reactions are The Ag ion produced by dissolving solid AgCl combines with NH3 to form Ag(NH3)2 , which causes more AgCl to dissolve, until the point at which Here [Ag] refers only to the Ag ion that is present as a separate species in solution. It is not the total silver content of the solution, which is For reasons discussed in the previous section, virtually all the Ag from the dissolved AgCl ends up in the complex ion Ag(NH3)2 . Thus we can represent the dissolving of solid AgCl in excess NH3 by the equation Since this equation is the sum of the three stepwise reactions given above, the equilibrium constant for the reaction is the product of the constants for the three reactions. (Demon-strate this to yourself by multiplying together the three expressions for Ksp, K1, and K2.) The equilibrium expression is Ksp K1 K2 11.6 1010212.1 103218.2 1032 2.8 103 K 3Ag1NH322 4 3Cl4 3NH34 2 AgCl1s2 2NH31aq2 ∆Ag1NH322 1aq2 Cl1aq2 3Ag4 total dissolved 3Ag4 3Ag1NH324 3Ag1NH322 4 3Ag4 3Cl4 Ksp 1.6 1010 K2 8.2 103 Ag1NH32 NH3 ∆Ag1NH322 K1 2.1 103 Ag NH3 ∆Ag1NH32 Ksp 1.6 1010 AgCl1s2 ∆Ag Cl AgCl1s2 ∆Ag1aq2 Cl1aq2 (top) Aqueous ammonia is added to silver chloride (white). (bottom) Silver chloride, insoluble in water, dissolves to form Ag(NH3)2 (aq) and Cl(aq).
When reactions are added, the equilibrium constant for the overall process is the product of the constants for the individual reactions.
15.8 Equilibria Involving Complex Ions 735 Using this expression, we will now calculate the solubility of solid AgCl in a 10.0 M NH3 solution. If we let x be the solubility (in mol/L) of AgCl in the solution, we can then write the following expressions for the equilibrium concentrations of the pertinent species: Substituting these concentrations into the equilibrium expression gives No approximations are necessary here. Taking the square root of both sides of the equa-tion gives Thus the solubility of AgCl in 10.0 M NH3 is much greater than its solubility in pure wa-ter, which is In this chapter we have considered two strategies for dissolving a water-insoluble ionic solid. If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution. In cases where the anion is not sufficiently basic, the ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.
Sometimes solids are so insoluble that combinations of reactions are needed to dis-solve them. For example, to dissolve the extremely insoluble HgS (Ksp 1054), it is nec-essary to use a mixture of concentrated HCl and concentrated HNO3, called aqua regia.
The H ions in the aqua regia react with the S2 ions to form H2S, and Cl reacts with Hg2 to form various complex ions, including HgCl4 2. In addition, NO3 oxidizes S2 to elemental sulfur. These processes lower the concentrations of Hg2 and S2 and thus promote the solubility of HgS.
Since the solubility of many salts increases with temperature, simple heating is some-times enough to make a salt sufficiently soluble. For example, earlier in this section we considered the mixed chloride precipitates of the Group I ions—PbCl2, AgCl, and Hg2Cl2.
The effect of temperature on the solubility of PbCl2 is such that we can precipitate PbCl2 with cold aqueous HCl and then redissolve it by heating the solution to near boiling. The silver and mercury(I) chlorides remain precipitated, since they are not significantly solu-ble in hot water. However, solid AgCl can be dissolved using aqueous ammonia. The solid Hg2Cl2 reacts with NH3 to form a mixture of elemental mercury and HgNH2Cl: White Black The mixed precipitate appears gray. This is an oxidation–reduction reaction in which one mercury(I) ion in Hg2Cl2 is oxidized to Hg2 in HgNH2Cl and the other mercury(I) ion is reduced to Hg, or elemental mercury.
The treatment of the Group I ions is summarized in Fig. 15.13. Note that the pres-ence of Pb2 is confirmed by adding CrO4 2, which forms bright yellow lead(II) chromate Hg2Cl21s2 2NH31aq2 ¡ HgNH2Cl1s2 Hg1l2 NH4 1aq2 Cl1aq2 1Ksp 1.3 105 mol/L x 0.48 mol/L solubility of AgCl1s2 in 10.0 M NH3 22.8 103 x 10.0 2x K 2.8 103 3Ag1NH322 4 3Cl4 3NH34 2 1x21x2 110.0 2x22 x2 110.0 2x22 3NH34 10.0 2x — 3Ag1NH322 4 x 3Cl4 x x mol/L of AgCl dissolves to produce x mol/L of Cl and x mol/L of Ag(NH3)2 Formation of x mol/L of Ag(NH3)2 requires 2x mol/L of NH3, since each complex ion contains two NH3 ligands m88888888888 m88888888888 736 Chapter Fifteen Applications of Aqueous Equilibria (PbCrO4). Also note that H added to a solution containing Ag(NH3)2 reacts with the NH3 to form NH4 , destroying the Ag(NH3)2 complex. Silver chloride then re-forms: Note that the qualitative analysis of cations by selective precipitation involves all the types of reactions we have discussed and represents an excellent application of the principles of chemical equilibrium.
2H1aq2 Ag1NH322 1aq2 Cl1aq2 ¡ 2NH4 1aq2 AgCl1s2 Add cold HCl (aq) Heat Precipitate of AgCl(s), Hg2Cl2(s), PbCl2(s) Precipitate of PbCrO4(s) (yellow) Solution of Pb2+ Precipitate of AgCl(s), Hg2Cl2(s) Solution of Ag(NH3)2 +, Cl– Precipitate of Hg(l) (black), HgNH2Cl(s) (white) Precipitate of AgCl(s) (white) Solution of Ag+, Hg2 2+, Pb2+ Add CrO4 2– Add NH3 (aq) Add H+ FIGURE 15.13 The separation of the Group I ions in the classic scheme of qualitative analysis.
Key Terms Section 15.1 common ion common ion effect Section 15.2 buffered solution Henderson–Hasselbalch equation Section 15.3 buffering capacity Section 15.4 pH curve (titration curve) millimole (mmol) equivalence point (stoichiometric point) For Review Buffered solutions Contains a weak acid (HA) and its salt (NaA) or a weak base (B) and its salt (BHCl) Resists a change in its pH when H or is added For a buffered solution containing HA and • The Henderson–Hasselbalch equation is useful: • The capacity of the buffered solution depends on the amounts of HA and present A pH pKa loga 3A4 3HA4 b A OH For Review 737 Section 15.5 acid–base indicator phenolphthalein Section 15.6 solubility product constant (solubility product) Section 15.7 ion product selective precipitation qualitative analysis Section 15.8 complex ion formation (stability) constants • The most efficient buffering occurs when the ratio is close to 1 • Buffering works because the amounts of HA (which reacts with added ) and (which reacts with added H) are large enough that the ratio does not change significantly when strong acids or bases are added Acid–base titrations The progress of a titration is represented by plotting the pH of the solution versus the volume of added titrant; the resulting graph is called a pH curve or titration curve Strong acid–strong base titrations show a sharp change in pH near the equiva-lence point The shape of the pH curve for a strong base–strong acid titration is quite dif-ferent before the equivalence point from the shape of the pH curve for a strong base–weak acid titration • The strong base–weak acid pH curve shows the effects of buffering before the equivalence point • For a strong base–weak acid titration, the pH is greater than 7 at the equiva-lence point because of the basic properties of Indicators are sometimes used to mark the equivalence point of an acid–base titration • The end point is where the indicator changes color • The goal is to have the end point and the equivalence point be as close as possible Solids dissolving in water For a slightly soluble salt, an equilibrium is set up between the excess solid (MX) and the ions in solution The corresponding constant is called Ksp: • The solubility of MX(s) is decreased by the presence from another source of either M or this is called the common ion effect Predicting whether precipitation will occur when two solutions are mixed in-volves calculating Q for the initial concentrations • If precipitation occurs • If no precipitation occurs REVIEW QUESTIONS 1. What is meant by the presence of a common ion? How does the presence of a common ion affect an equilibrium such as What is an acid–base solution called that contains a common ion?
2. Define a buffer solution. What makes up a buffer solution? How do buffers ab-sorb added H or with little pH change?
Is it necessary that the concentrations of the weak acid and the weak base in a buffered solution be equal? Explain. What is the pH of a buffer when the weak acid and conjugate base concentrations are equal?
A buffer generally contains a weak acid and its weak conjugate base, or a weak base and its weak conjugate acid, in water. You can solve for the pH by setting up the equilibrium problem using the reaction of the weak acid or the Ka OH HNO21aq2 ∆H1aq2 NO2 1aq2 Q Ksp, Q 7 Ksp, X; Ksp 3M4 3X4 MX1s2 ∆M1aq2 X1aq2 A [A] [HA] A OH [A] [HA] 738 Chapter Fifteen Applications of Aqueous Equilibria reaction of the conjugate base. Both reactions give the same answer for the pH of the solution. Explain.
A third method that can be used to solve for the pH of a buffer solution is the Henderson–Hasselbalch equation. What is the Henderson–Hasselbalch equa-tion? What assumptions are made when using this equation?
3. One of the most challenging parts of solving acid–base problems is writing out the correct reaction. When a strong acid or a strong base is added to solutions, they are great at what they do and we always react them first. If a strong acid is added to a buffer, what reacts with the H from the strong acid and what are the products? If a strong base is added to a buffer, what reacts with the from the strong base and what are the products? Problems involving the reac-tion of a strong acid or strong base are assumed to be stoichiometry problems and not equilibrium problems. What is assumed when a strong acid or strong base reacts to make it a stoichiometry problem?
A good buffer generally contains relatively equal concentrations of weak acid and conjugate base. If you wanted to buffer a solution at or how would you decide which weak acid–conjugate base or weak base–conjugate acid pair to use? The second characteristic of a good buffer is good buffering capacity. What is the capacity of a buffer? How do the following buffers differ in capacity? How do they differ in pH?
0.01 M acetic acid 0.01 M sodium acetate 0.1 M acetic acid 0.1 M sodium acetate 1.0 M acetic acid 1.0 M sodium acetate 4. Draw the general titration curve for a strong acid titrated by a strong base. At the various points in the titration, list the major species present before any reaction takes place and the major species present after any reaction takes place. What re-action takes place in a strong acid–strong base titration? How do you calculate the pH at the various points along the curve? What is the pH at the equivalence point for a strong acid–strong base titration? Why? Answer the same questions for a strong base–strong acid titration. Compare and contrast a strong acid–strong base titration versus a strong base–strong acid titration.
5. Sketch the titration curve for a weak acid titrated by a strong base. When per-forming calculations concerning weak acid–strong base titrations, the general two-step procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoi-chiometry part of the problem? What is assumed about this reaction?
At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH? Why is at the equivalence point of a weak acid–strong base titration? Does the pH at the halfway point to equivalence have to be less than 7.0? What does the pH at the halfway point equal? Compare and contrast the titration curves for a strong acid–strong base titration and a weak acid–strong base titration.
6. Sketch the titration curve for a weak base titrated by a strong acid. Weak base–strong acid titration problems also follow a two-step procedure. What re-action takes place in the stoichiometry part of the problem? What is assumed about this reaction? At the various points in your titration curve, list the major species present after the strong acid (HNO3, for example) reacts to completion with the weak base, B. What equilibrium problem would you solve at the vari-ous points in your titration curve to calculate the pH? Why is at the equivalence point of a weak base–strong acid titration? If at the pH 6.0 pH 6 7.0 pH 7 7.0 pH 10.00, pH 4.00 OH Kb Active Learning Questions 739 halfway point to equivalence, what is the Kb value for the weak base titrated?
Compare and contrast the titration curves for a strong base–strong acid titration and a weak base–strong acid titration.
7. What is an acid–base indicator? Define the equivalence (stoichiometric) point and the end point of a titration. Why should you choose an indicator so that the two points coincide? Do the pH values of the two points have to be within unit of each other? Explain. Why does an indicator change from its acid color to its base color over a range of pH values? In general, when do color changes start to occur for indicators? Can the indicator thymol blue contain only a single group and no other acidic or basic functional group? Explain.
8. To what reaction does the solubility product constant, Ksp, refer? Table 15.4 lists Ksp values for several ionic solids. For any of these ionic compounds, you should be able to calculate the solubility. What is the solubility of a salt, and what procedures do you follow to calculate the solubility of a salt? How would you calculate the Ksp value for a salt given the solubility?
Under what circumstances can you compare the relative solubilities of two salts directly by comparing the values of their solubility products? When can relative solubilities not be compared based on Ksp values? What is a common ion and how does its presence affect the solubility? List some salts whose solu-bility increases as the pH becomes more acidic. What is true about the anions in these salts? List some salts whose solubility remains unaffected by the solu-tion pH. What is true about the anions in these salts?
9. What is the difference between the ion product, Q, and the solubility product, Ksp? What happens when Mixtures of metal ions in aqueous solution can sometimes be separated by selective precipitation. What is selective precipitation? If a solution contained 0.10 M Mg2, 0.10 M Ca2, and 0.10 M Ba2, how could addition of NaF be used to separate the cations out of solution—that is, what would precipitate first, then second, then third?
How could addition of K3PO4 be used to separate out the cations in a solution that 1.0 M Ag, 1.0 M Pb2, and 1.0 M Sr2?
10. What is a complex ion? The stepwise formation constants for the complex ion Cu(NH3)4 2 are Write the reactions that refer to each of these formation constants. Given that the values of the formation constants are large, what can you deduce about the equilib-rium concentration of Cu(NH3)4 2 versus the equilibrium concentration of Cu2?
When 5 M ammonia is added to a solution containing Cu(OH)2(s), the precipitate will eventually dissolve in solution. Why? If 5 M HNO3 is then added, the Cu(OH)2 precipitate re-forms. Why? In general, what effect does the ability of a cation to form a complex ion have on the solubility of salts containing that cation?
K1 1 103, K2 1 104, K3 1 103, and K4 1 103.
Q 7 Ksp? Q 6 Ksp? Q Ksp?
¬CO2H 0.01 pH Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. What are the major species in solution after NaHSO4 is dissolved in water? What happens to the pH of the solution as more NaHSO4 is added? Why? Would the results vary if baking soda (NaHCO3) were used instead?
2. A friend asks the following: “Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the to form . Thus the amount of acid (HA) is decreased, and the amount of base (A) is increased. Analogously, adding HCl to the buffered so-lution forms more of the acid (HA) by reacting with the base (A). Thus how can we claim that a buffered solution resists changes in the pH of the solution?” How would you explain buffering to this friend?
3. Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of A OH 740 Chapter Fifteen Applications of Aqueous Equilibria 16. Consider the following four titrations.
i. 100.0 mL of 0.10 M HCl titrated by 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated by 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated by 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated by 0.10 M NaOH Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point.
b. increasing pH initially before any titrant has been added.
c. increasing pH at the halfway point in equivalence.
d. increasing pH at the equivalence point.
How would the rankings change if C5H5N replaced CH3NH2 and if HOC6H5 replaced HF?
17. Figure 15.4 shows the pH curves for the titrations of six differ-ent acids by NaOH. Make a similar plot for the titration of three different bases by 0.10 M HCl. Assume 50.0 mL of 0.20 M of the bases and assume the three bases are a strong base (KOH), a weak base with Kb 1 105, and another weak base with Kb 1 1010.
18. Acid–base indicators mark the end point of titrations by “magi-cally” turning a different color. Explain the “magic” behind acid–base indicators.
19. The salts in Table 15.4, with the possible exception of the hy-droxide salts, have one of the following mathematical relation-ships between the Ksp value and the molar solubility, s.
i. Ksp s2 iii. Ksp 27s4 ii. Ksp 4s3 iv. Ksp 108s5 For each mathematical relationship, give an example of a salt in Table 15.4 that exhibits that relationship.
20. List some ways one can increase the solubility of a salt in water.
Exercises In this section similar exercises are paired.
Buffers 21. A certain buffer is made by dissolving NaHCO3 and Na2CO3 in some water. Write equations to show how this buffer neutralizes added H and .
22. A buffer is prepared by dissolving HONH2 and HONH3NO3 in some water. Write equations to show how this buffer neutralizes added H and .
23. Calculate the pH of each of the following solutions.
a. 0.100 M propanoic acid b. 0.100 M sodium propanoate (NaC3H5O2) c. pure H2O d. a mixture containing 0.100 M HC3H5O2 and 0.100 M NaC3H5O2 24. Calculate the pH of each of the following solutions.
a. 0.100 M HONH2 b. 0.100 M HONH3Cl c. pure H2O d. a mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl 25. Compare the percent dissociation of the acid in Exercise 23a with the percent dissociation of the acid in Exercise 23d. Explain the large difference in percent dissociation of the acid.
(Kb 1.1 108) (HC3H5O2, Ka 1.3 105) OH OH each solution added change the effectiveness of the buffer?
Would a buffer solution made by mixing HCl and NaOH be effective? Explain.
4. Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the simi-larities and the differences.
5. Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point.
6. Devise as many ways as you can to experimentally determine the Ksp value of a solid. Explain why each of these would work.
7. You are browsing through the Handbook of Hypothetical Chem-istry when you come across a solid that is reported to have a Ksp value of zero in water at . What does this mean?
8. A friend tells you: “The constant Ksp of a salt is called the sol-ubility product constant and is calculated from the concentra-tions of ions in the solution. Thus, if salt A dissolves to a greater extent than salt B, salt A must have a higher Ksp than salt B.” Do you agree with your friend? Explain.
9. Explain the following phenomenon: You have a test tube with about 20 mL of silver nitrate solution. Upon adding a few drops of sodium chromate solution, you notice a red solid forming in a relatively clear solution. Upon adding a few drops of a sodium chloride solution to the same test tube, you no-tice a white solid and a pale yellow solution. Use the Ksp values in the book to support your explanation, and include the balanced reactions.
10. What happens to the Ksp value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.
11. Which is more likely to dissolve in an acidic solution, silver sul-fide or silver chloride? Why?
12. You have two salts, AgX and AgY, with very similar Ksp values.
You know that the Ka value for HX is much greater than the Ka value for HY. Which salt is more soluble in an acidic solution?
Explain.
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 13. The common ion effect for weak acids is to significantly de-crease the dissociation of the acid in water. The common ion effect for ionic solids (salts) is to significantly decrease the solubility of the ionic compound in water. Explain both of these common ion effects.
14. Consider a buffer solution where [weak acid] [conjugate base].
How is the pH of the solution related to the pKa value of the weak acid? If [conjugate base] [weak acid], how is pH related to pKa?
15. A best buffer has about equal quantities of weak acid and con-jugate base present as well as having a large concentration of each species present. Explain.
25°C Exercises 741 44. a. Carbonate buffers are important in regulating the pH of blood at 7.40. What is the concentration ratio of CO2 (usually writ-ten H2CO3) to in blood at b. Phosphate buffers are important in regulating the pH of in-tracellular fluids at pH values generally between 7.1 and 7.2.
What is the concentration ratio of to in in-tracellular fluid at c. Why is a buffer composed of H3PO4 and ineffective in buffering the pH of intracellular fluid?
45. Consider the acids in Table 14.2. Which acid would be the best choice for preparing a buffer? Explain how to make 1.0 L of this buffer.
46. Consider the bases in Table 14.3. Which base would be the best choice for preparing a buffer? Explain how to make 1.0 L of this buffer.
47. Which of the following mixtures would result in buffered solu-tions when 1.0 L of each of the two solutions are mixed?
a. 0.1 M KOH and 0.1 M CH3NH3Cl b. 0.1 M KOH and 0.2 M CH3NH2 c. 0.2 M KOH and 0.1 M CH3NH3Cl d. 0.1 M KOH and 0.2 M CH3NH3Cl 48. Which of the following mixtures would result in a buffered so-lution when 1.0 L of each of the two solutions are mixed?
a. 0.2 M HNO3 and 0.4 M NaNO3 b. 0.2 M HNO3 and 0.4 M HF c. 0.2 M HNO3 and 0.4 M NaF d. 0.2 M HNO3 and 0.4 M NaOH 49. How many moles of NaOH must be added to 1.0 L of 2.0 M HC2H3O2 to produce a solution buffered at each pH?
a. pH pKa b. pH 4.00 c. pH 5.00 50. Calculate the number of moles of HCl(g) that must be added to 1.0 L of 1.0 M NaC2H3O2 to produce a solution buffered at each pH.
a. pH pKa b. pH 4.20 c. pH 5.00 Acid–Base Titrations 51. Consider the titration of a generic weak acid HA with a strong base that gives the following titration curve: 5 pH 10 15 20 25 pH 5.00 pH 7.00 Ka 7.5 103 H3PO41aq2 ∆H2PO4 1aq2 H1aq2 H2PO4 Ka 6.2 108 H2PO4 1aq2 ∆HPO4 21aq2 H1aq2 pH 7.15?
HPO4 2 H2PO4 Ka 4.3 107 H2CO31aq2 ∆HCO3 1aq2 H1aq2 pH 7.40?
HCO3 26. Compare the percent ionization of the base in Exercise 24a with the percent ionization of the base in Exercise 24d. Explain any differences.
27. Calculate the pH after 0.020 mol HCl is added to 1.00 L of each of the four solutions in Exercise 23.
28. Calculate the pH after 0.020 mol HCl is added to 1.00 L of each of the four solutions in Exercise 24.
29. Calculate the pH after 0.020 mol NaOH is added to 1.00 L of each of the four solutions in Exercise 23.
30. Calculate the pH after 0.020 mol NaOH is added to 1.00 L of each of the solutions in Exercise 24.
31. Which of the solutions in Exercise 23 shows the least change in pH upon the addition of acid or base? Explain.
32. Which of the solutions in Exercise 24 is a buffered solution?
33. Calculate the pH of a solution that is 1.00 M HNO2 and 1.00 M NaNO2.
34. Calculate the pH of a solution that is 0.60 M HF and 1.00 M KF.
35. Calculate the pH after 0.10 mol of NaOH is added to 1.00 L of the solution in Exercise 33, and calculate the pH after 0.20 mol of HCl is added to 1.00 L of the solution in Exercise 33.
36. Calculate the pH after 0.10 mol of NaOH is added to 1.00 L of the solution in Exercise 34, and calculate the pH after 0.20 mol of HCl is added to 1.00 L of the solution in Exercise 34.
37. Calculate the pH of a buffer solution prepared by dissolving 21.46 g of benzoic acid (HC7H5O2) and 37.68 g of sodium ben-zoate in 200.0 mL of solution.
38. A buffered solution is made by adding 50.0 g NH4Cl to 1.00 L of a 0.75 M solution of NH3. Calculate the pH of the final solu-tion. (Assume no volume change.) 39. Calculate the pH after 0.010 mol gaseous HCl is added to 250.0 mL of each of the following buffered solutions.
a. 0.050 M NH3 0.15 M NH4Cl b. 0.50 M NH3 1.50 M NH4Cl Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?
40. An aqueous solution contains dissolved C6H5NH3Cl and C6H5NH2.
The concentration of C6H5NH2 is 0.50 M and pH is 4.20.
a. Calculate the concentration of C6H5NH3 in this buffer solution.
b. Calculate the pH after 4.0 g of NaOH(s) is added to 1.0 L of this solution. (Neglect any volume change.) 41. Calculate the mass of sodium acetate that must be added to 500.0 mL of 0.200 M acetic acid to form a buffer solution.
42. What volumes of 0.50 M HNO2 and 0.50 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at 43. Consider a solution that contains both C5H5N and C5H5NHNO3.
Calculate the ratio [C5H5N] [C5H5NH] if the solution has the following pH values.
a.
c.
b.
d. pH 5.50 pH 5.00 pH 5.23 pH 4.50 pH 3.55?
pH 5.00 742 Chapter Fifteen Applications of Aqueous Equilibria 60. Repeat the procedure in Exercise 57, but for the titration of 25.0 mL of 0.100 M pyridine with 0.100 M hydrochloric acid (Kb for pyridine is ). Do not do the points at 24.9 and 25.1 mL.
61. Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.
a. 100.0 mL of 0.10 M HC7H5O2 titrated by 0.10 M NaOH b. 100.0 mL of 0.10 M C2H5NH2 titrated by 0.20 M HNO3 c. 100.0 mL of 0.50 M HCl titrated by 0.25 M NaOH 62. In the titration of 50.0 mL of 1.0 M methylamine, CH3NH2 (Kb 4.4 104), with 0.50 M HCl, calculate the pH under the following conditions.
a. after 50.0 mL of 0.50 M HCl has been added b. at the stoichiometric point 63. You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA?
64. A sample of an ionic compound NaA, where A is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 500.0 mL of HCl was added, the pH was measured and found to be 5.00.
The experimenter found that 1.00 L of 0.100 M HCl was required to reach the stoichiometric point of the titration.
a. What is the Kb value for A?
b. Calculate the pH of the solution at the stoichiometric point of the titration.
Indicators 65. Two drops of indicator HIn (Ka 1.0 109), where HIn is yellow and In is blue, are placed in 100.0 mL of 0.10 M HCl.
a. What color is the solution initially?
b. The solution is titrated with 0.10 M NaOH. At what pH will the color change (yellow to greenish yellow) occur?
c. What color will the solution be after 200.0 mL of NaOH has been added?
66. Methyl red has the following structure: It undergoes a color change from red to yellow as a solution gets more basic. Calculate an approximate pH range for which methyl red is useful. What is the color change and the pH at the color change when a weak acid is titrated with a strong base using methyl red as an indicator? What is the color change and the pH at the color change when a weak base is titrated with a strong acid using methyl red as an indicator? For which of these two types of titrations is methyl red a possible indicator?
67. Potassium hydrogen phthalate, known as KHP can be obtained in high purity and is used to determine the concentration of solutions of strong bases by the reaction HP1aq2 OH1aq2 ¡ H2O1l2 P21aq2 204.22 g mol), (molar mass (Kb 5.6 104) (Ka 6.4 105) 1.7 109 On the curve, indicate the points that correspond to the following: a. the stoichiometric (equivalence) point b. the region with maximum buffering c.
d. pH depends only on [HA] e. pH depends only on f. pH depends only on the amount of excess strong base added 52. Sketch the titration curve for the titration of a generic weak base B with a strong acid. The titration reaction is On this curve, indicate the points that correspond to the following: a. the stoichiometric (equivalence) point b. the region with maximum buffering c.
d. pH depends only on [B] e. pH depends only on f. pH depends only on the amount of excess strong acid added 53. Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the fol-lowing volumes of KOH have been added.
a. 0.0 mL d. 80.0 mL b. 10.0 mL e. 100.0 mL c. 40.0 mL 54. Consider the titration of 80.0 mL of 0.100 M Ba(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.
a. 0.0 mL d. 40.0 mL b. 20.0 mL e. 80.0 mL c. 30.0 mL 55. Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka by 0.100 M KOH. Calculate the pH of the result-ing solution after the following volumes of KOH have been added.
a. 0.0 mL d. 150.0 mL b. 50.0 mL e. 200.0 mL c. 100.0 mL f. 250.0 mL 56. Consider the titration of 100.0 mL of 0.100 M H2NNH2 by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO3 have been added.
a. 0.0 mL d. 40.0 mL b. 20.0 mL e. 50.0 mL c. 25.0 mL f. 100.0 mL 57. A 25.0-mL sample of 0.100 M lactic acid is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calcu-lations as pH versus milliliters of NaOH added.
58. Repeat the procedure in Exercise 57, but for the titration of 25.0 mL of 0.100 M propanoic acid 105) with 0.100 M NaOH.
59. Repeat the procedure in Exercise 57, but for the titration of 25.0 mL of 0.100 M NH3 with 0.100 M HCl.
(Kb 1.8 105) (HC3H5O2, Ka 1.3 (HC3H5O3, pKa 3.86) (Kb 3.0 106) 1.8 105) [BH] pH pKa B H ∆BH [A] pH pKa Exercises 743 84. A solution contains 0.018 mol each of When the solution is mixed with 200. mL of 0.24 M AgNO3, what mass of AgCl(s) precipitates out, and what is the [Ag]? Assume no volume change.
85. Calculate the molar solubility of Co(OH)3, .
86. Calculate the molar solubility of Cd(OH)2, .
87. For each of the following pairs of solids, determine which solid has the smallest molar solubility.
a. CaF2(s), Ksp 4.0 1011, or BaF2(s), Ksp 2.4 105 b. Ca3(PO4)2(s), Ksp 1.31032, or FePO4(s), Ksp1.01022 88. For each of the following pairs of solids, determine which solid has the smallest molar solubility.
a. FeC2O4, Ksp 2.1 107, or Ksp 1.4 107 b. Ag2CO3, Ksp 8.1 1012, or Ksp 2 1013 89. Calculate the solubility (in moles per liter) of Fe(OH)3 (Ksp 4 1038) in each of the following.
a. water b. a solution buffered at pH 5.0 c. a solution buffered at pH 11.0 90. The Ksp for silver sulfate (Ag2SO4) is 1.2 105. Calculate the solubility of silver sulfate in each of the following.
a. water b. 0.10 M AgNO3 c. 0.20 M K2SO4 91. Calculate the solubility of solid Ca3(PO4)2 (Ksp 1.3 1032) in a 0.20 M Na3PO4 solution.
92. The solubility of Ce(IO3)3 in a 0.20 M KIO3 solution is 4.4 108 mol/L. Calculate Ksp for Ce(IO3)3.
93. What mass of ZnS will dissolve in 300.0 mL of 0.050 M Ignore the basic properties of 94. The concentration of Mg2 in seawater is 0.052 M. At what pH will 99% of the Mg2 be precipitated as the hydroxide salt? [Ksp for ] 95. Which of the substances in Exercises 81 and 82 show increased solubility as the pH of the solution becomes more acidic?
Write equations for the reactions that occur to increase the solubility.
96. For which salt in each of the following groups will the solubil-ity depend on pH?
a. AgF, AgCl, AgBr c. Sr(NO3)2, Sr(NO2)2 b. Pb(OH)2, PbCl2 d. Ni(NO3)2, Ni(CN)2 97. Will a precipitate form when 75.0 mL of 0.020 M BaCl2 and 125 mL of 0.040 M Na2SO4 are mixed together?
98. Will a precipitate form when 100.0 mL of 4.0 104 M Mg(NO3)2 is added to 100.0 mL of 2.0 104 M NaOH?
99. Calculate the final concentrations of K(aq), C2O4 2(aq), Ba2(aq), and Br(aq) in a solution prepared by adding 0.100 L of 0.200 M K2C2O4 to 0.150 L of 0.250 M BaBr2. (For BaC2O4, Ksp 2.3 108.) Mg(OH)2 8.9 1012.
S2.
Zn(NO3)2?
(Ksp 2.5 1022) Mn(OH)2, Cu(IO4)2, Ksp 5.9 1011 Ksp 2.5 1043 AgCl, Ksp 1.6 1010 AgBr, Ksp 5.0 1013 AgI, Ksp 1.5 1016 I, Br, and Cl.
If a typical titration experiment begins with approximately 0.5 g of KHP and has a final volume of about 100 mL, what is an appropriate indicator to use? The pKa for 68. A certain indicator HIn has a pKa of 3.00 and a color change be-comes visible when 7.00% of the indicator has been converted to At what pH is this color change visible?
69. Which of the indicators in Fig. 15.8 could be used for the titrations in Exercises 53 and 55?
70. Which of the indicators in Fig. 15.8 could be used for the titra-tions in Exercises 54 and 56?
71. Which of the indicators in Fig. 15.8 could be used for the titrations in Exercises 57 and 59?
72. Which of the indicators in Fig. 15.8 could be used for the titra-tions in Exercises 58 and 60?
73. Estimate the pH of a solution in which bromcresol green is blue and thymol blue is yellow. (See Fig. 15.8.) 74. A solution has a pH of 7.0. What would be the color of the solu-tion if each of the following indicators were added? (See Fig. 15.8.) a. thymol blue c. methyl red b. bromthymol blue d. crystal violet Solubility Equilibria 75. Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the fol-lowing solids.
a. AgC2H3O2 b. Al(OH)3 c. Ca3(PO4)2 76. Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the fol-lowing solids.
a. Ag2CO3 b. Ce(IO3)3 c. BaF2 77. Use the following data to calculate the Ksp value for each solid.
a. The solubility of CaC2O4 is 6.1 103 g/L.
b. The solubility of BiI3 is 1.32 105 mol/L.
78. Use the following data to calculate the Ksp value for each solid.
a. The solubility of Pb3(PO4)2 is 6.2 1012 mol/L.
b. The solubility of Li2CO3 is 7.4 102 mol/L.
79. The concentration of Pb2 in a solution saturated with PbBr2(s) is 2.14 102 M. Calculate Ksp for PbBr2.
80. The concentration of Ag in a solution saturated with Ag2C2O4(s) is 2.2 104 M. Calculate Ksp for Ag2C2O4.
81. Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties.
a. Ag3PO4, Ksp 1.8 1018 b. CaCO3, Ksp 8.7 109 c. Hg2Cl2, Ksp 1.1 1018 (Hg2 2 is the cation in solution.) 82. Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties.
a. PbI2, Ksp 1.4 108 b. CdCO3, Ksp 5.2 1012 c. Sr3(PO4)2, Ksp 1 1031 83. The solubility of the ionic compound M2X3, having a molar mass of 288 g/mol, is g/L. Calculate the Ksp of the compound.
3.60 107 In.
HP is 5.51.
744 Chapter Fifteen Applications of Aqueous Equilibria What is the concentration of in 500.0 mL of a solution that was originally 0.010 M and 0.78 M ? The reaction is 110. A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of 2.0 103 M CuNO3. Assume that Cu(I) forms com-plex ions with X as follows: with an overall reaction Calculate the following concentrations at equilibrium.
a. CuX3 2 b. CuX2 c. Cu 111. a. Calculate the molar solubility of AgI in pure water. Ksp for AgI is 1.5 1016.
b. Calculate the molar solubility of AgI in 3.0 M NH3. The over-all formation constant for Ag(NH3)2 is 1.7 107.
c. Compare the calculated solubilities from parts a and b. Ex-plain any differences.
112. Solutions of sodium thiosulfate are used to dissolve unexposed AgBr (Ksp 5.0 1013) in the developing process for black-and-white film. What mass of AgBr can dissolve in 1.00 L of 0.500 M Na2S2O3? Ag reacts with S2O3 2 to form a complex ion: 113. Kf for the complex ion Ag(NH3)2 is Ksp for AgCl is Calculate the molar solubility of AgCl in 1.0 M NH3.
114. The copper(I) ion forms a chloride salt that has Copper(I) also forms a complex ion with a. Calculate the solubility of copper(I) chloride in pure water.
(Ignore formation for part a.) b. Calculate the solubility of copper(I) chloride in 0.10 M NaCl.
115. A series of chemicals were added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution with the end result shown below in test tube 1, NH3(aq) was then added with the end result shown in test tube 2, and HNO3(aq) was added last with the end result shown in test tube 3.
Explain the results shown in each test tube. Include a balanced equation for the reaction(s) taking place.
1 2 3 CuCl2 Cu1aq2 2Cl1aq2 ∆CuCl2 1aq2 K 8.7 104 Cl: Ksp 1.2 106.
1.6 1010.
1.7 107.
K 2.9 1013 Ag1aq2 2S2O3 21aq2 ∆Ag1S2O322 31aq2 Cu1aq2 3X1aq2 ∆CuX3 21aq2 K 1.0 109 K3 1.0 103 CuX2 1aq2 X1aq2 ∆CuX3 21aq2 K2 1.0 104 CuX1aq2 X1aq2 ∆CuX2 1aq2 K1 1.0 102 Cu1aq2 X1aq2 ∆CuX1aq2 Hg21aq2 4I1aq2 ∆HgI4 21aq2 I Hg2 Hg2 100. A solution is prepared by mixing 50.0 mL of 0.10 M Pb(NO3)2 with 50.0 mL of 1.0 M KCl. Calculate the concentrations of Pb2 and Cl at equilibrium. Ksp for PbCl2(s) is 1.6 105.
101. A solution contains 1.0 105 M Na3PO4. What is the mini-mum concentration of AgNO3 that would cause precipitation of solid Ag3PO4 (Ksp 1.8 1018)?
102. A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that for successful separation 99% of the metal ion must be pre-cipitated before the other metal ion begins to precipitate, and as-sume no volume change on addition of Na2CO3.
Complex Ion Equilibria 103. Write equations for the stepwise formation of each of the fol-lowing complex ions.
a.
b.
104. Write equations for the stepwise formation of each of the fol-lowing complex ions.
a.
b.
105. Given the following data, calculate the value for the overall formation constant for 106. In the presence of Fe3 forms the complex ion The equilibrium concentrations of Fe3 and are and 1.5 103 M, respectively, in a 0.11 M KCN solution. Calculate the value for the overall formation constant of 107. When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: reacts with to form ) 108. As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mix-ture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write reactions for all the changes mentioned above. What conclusions can you draw concerning the sizes of the Ksp values for AgCl, AgBr, and AgI?
109. The overall formation constant for . That is, 1.0 1030 3HgI4 24 3Hg24 3I4 4 HgI4 2 is 1.0 1030 HgI4 2.
I Hg2 Fe31aq2 6CN1aq2 ∆Fe1CN26 3 Koverall ?
Fe(CN)6 3.
8.5 1040 M Fe(CN)6 3 Fe(CN)6 3.
CN, K 3Mn1C2O422 24 3Mn24 3C2O4 24 2 Mn(C2O4)2 2: K2 7.9 101 MnC2O41aq2 C2O4 21aq2 ∆Mn1C2O422 21aq2 K1 7.9 103 Mn21aq2 C2O4 21aq2 ∆MnC2O41aq2 Zn(NH3)4 2 CoF6 3 V(C2O4)3 3 Ni(CN)4 2 Additional Exercises 745 116. The solubility of copper(II) hydroxide in water can be increased by adding either the base NH3 or the acid HNO3. Explain. Would added NH3 or HNO3 have the same effect on the solubility of silver acetate or silver chloride? Explain.
Additional Exercises 117. Derive an equation analogous to the Henderson–Hasselbalch equa-tion but relating pOH and pKb of a buffered solution composed of a weak base and its conjugate acid, such as NH3 and NH4 .
118. a. Calculate the pH of a buffered solution that is 0.100 M in C6H5CO2H (benzoic acid, ) and 0.100 M in C6H5CO2Na.
b. Calculate the pH after 20.0% (by moles) of the benzoic acid is converted to benzoate anion by addition of strong acid. Use the dissociation equilibrium to calculate the pH.
c. Do the same as in part b, but use the following equilibrium to calculate the pH: d. Do your answers in parts b and c agree? Explain.
119. Consider a solution containing 0.10 M ethylamine (C2H5NH2), 0.20 M C2H5NH3 , and 0.20 M a. Calculate the pH of this solution.
b. Calculate the pH after 0.050 mol of KOH(s) is added to 1.00 L of this solution. (Ignore any volume changes.) 120. You make 1.00 L of a buffered solution (pH 4.00) by mixing acetic acid and sodium acetate. You have 1.00 M solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?
121. You have the following reagents on hand: Cl.
C6H5CO2 1aq2 H2O1l2 ∆C6H5CO2H1aq2 OH1aq2 C6H5CO2H1aq2 ∆C6H5CO2 1aq2 H1aq2 Ka 6.4 105 2.0 L. What is the pH of this buffer? What is the pH after 0.50 mL of 12 M HCl is added to a 200.0-mL portion of the buffer?
123. Calculate the value of the equilibrium constant for each of the following reactions in aqueous solution.
a.
b.
c.
124. The following plot shows the pH curves for the titrations of var-ious acids by 0.10 M NaOH (all of the acids were 50.0-mL sam-ples of 0.10 M concentration).
a. Which pH curve corresponds to the weakest acid?
b. Which pH curve corresponds to the strongest acid? Which point on the pH curve would you examine to see if this acid is a strong acid or a weak acid (assuming you did not know the initial concentration of the acid)?
c. Which pH curve corresponds to an acid with 125. Calculate the volume of M NaOH that must be added to 500.0 mL of 0.200 M HCl to give a solution that has 126. Repeat the procedure in Exercise 57, but for the titration of 25.0 mL of 0.100 M HNO3 with 0.100 M NaOH.
127. The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required 27.36 mL of 0.5106 M NaOH for complete reaction. Addition of 13.68 mL of 0.5106 M HCl to the flask containing the aspirin and the sodium hydrox-ide produced a mixture with Find the molar mass of acetylsalicylic acid and its Ka value. State any assumptions you must make to reach your answer.
128. One method for determining the purity of aspirin (empirical for-mula, C9H8O4) is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: Salicylate ion Acetate ion A sample of aspirin with a mass of 1.427 g was boiled in 50.00 mL of 0.500 M NaOH. After the solution was cooled, it took 31.92 mL of 0.289 M HCl to titrate the excess NaOH. Cal-culate the purity of the aspirin. What indicator should be used for this titration? Why?
¬¡ C7H5O3 1aq2 C2H3O2 1aq2 H2O1l2 C9H8O41s2 2OH1aq2 pH 3.48.
pH 2.15.
1.50 102 Ka 1 106?
Vol 0.10 M NaOH added (mL) 10 20 30 40 50 60 2.0 4.0 6.0 8.0 10.0 12.0 0 pH a c d e f b HCl NaOH ∆NaCl H2O C2H3O2 H ∆HC2H3O2 HC2H3O2 OH ∆C2H3O2 H2O Solids (pKa of Acid Form Is Given) Solutions Benzoic acid (4.19) 5.0 M HCl Sodium acetate (4.74) 1.0 M acetic acid (4.74) Potassium fluoride (3.14) 2.6 M NaOH Ammonium chloride (9.26) 1.0 M HOCl (7.46) What combinations of reagents would you use to prepare buffers at the following pH values?
a. 3.0 b. 4.0 c. 5.0 d. 7.0 e. 9.0 122. Tris(hydroxymethyl)aminomethane, commonly called TRIS or Trizma, is often used as a buffer in biochemical studies. Its buffering range is pH 7 to 9, and Kb is 1.19 106 for the aque-ous reaction TRIS TRISH a. What is the optimal pH for TRIS buffers?
b. Calculate the ratio [TRIS][TRISH] at pH 7.00 and at pH 9.00.
c. A buffer is prepared by diluting 50.0 g TRIS base and 65.0 g TRIS hydrochloride (written as TRISHCl) to a total volume of 1HOCH223CNH2 H2O ∆1HOCH223CNH3 OH Boil 10 min Aspirin 746 Chapter Fifteen Applications of Aqueous Equilibria c. Ethylenediaminetetraacetate (EDTA4) is used as a com-plexing agent in chemical analysis and has the following structure: Solutions of EDTA4 are used to treat heavy metal poison-ing by removing the heavy metal in the form of a soluble complex ion. The complex ion virtually eliminates the heavy metal ions from reacting with biochemical systems. The re-action of EDTA4 with Pb2 is Consider a solution with 0.010 mol Pb(NO3)2 added to 1.0 L of an aqueous solution buffered at pH 13.00 and contain-ing 0.050 M Na4EDTA. Does Pb(OH)2 precipitate from this solution?
Challenge Problems 138. Another way to treat data from a pH titration is to graph the ab-solute value of the change in pH per change in milliliters added versus milliliters added ( pH mL versus mL added). Make this graph using your results from Exercise 57. What advantage might this method have over the traditional method for treating titration data?
139. A buffer is made using 45.0 mL of 0.750 M HC3H5O2 (Ka 1.3 105) and 55.0 mL of 0.700 M NaC3H5O2. What volume of 0.10 M NaOH must be added to change the pH of the original buffer solution by 2.5%?
140. A 0.400 M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?
141. What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00?
142. Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.100 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ba(OH)2, and 10.0 mL of 0.150 M KOH.
Calculate the pH of this solution.
143. When a diprotic acid, H2A, is titrated by NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: pH Vol NaOH added K 1.1 1018 Pb2 1aq2 EDTA4 1aq2 ∆PbEDTA21aq2 N CH2 CH2 N O2C CH2 CH2 O2C CO2 CO2 CH2 CH2 Ethylenediaminetetraacetate 129. A certain acetic acid solution has Calculate the vol-ume of 0.0975 M KOH required to reach the equivalence point in the titration of 25.0 mL of the acetic acid solution.
130. A 0.210-g sample of an acid is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?
131. A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 mL of the weak acid solution has been added to 50.0 mL of the 0.100 M NaOH solution, the pH of the resulting solution is 10.50. Calculate the original concentration of the solution of weak acid.
132. A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 mL of 0.100 M NaOH. The student then adds 13.0 mL of 0.100 M HCl. The pH of the re-sulting solution is 4.7. How is the value of pKa for the unknown acid related to 4.7?
133. a. Using the Ksp value for Cu(OH)2 (1.6 1019) and the overall formation constant for Cu(NH3)4 2 (1.0 1013), calculate the value for the equilibrium constant for the following reaction: b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of Cu(OH)2 in 5.0 M NH3. In 5.0 M NH3 the concentration of is 0.0095 M.
134. The solubility rules outlined in Chapter 4 say that Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble hydroxides. Cal-culate the pH of a saturated solution of each of these marginally soluble hydroxides.
135. The Ksp of hydroxyapatite, Ca5(PO4)3OH, is . Cal-culate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, Ca5(PO4)3F, forms. The Ksp of this substance is . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?
136. In the chapter discussion of precipitate formation, we ran the precipitation reaction to completion and then let some of the pre-cipitate redissolve to get back to equilibrium. To see why, redo Sample Exercise 15.17, where 1 1060 6.8 1037 OH Cu1OH221s2 4NH31aq2 ∆Cu1NH324 21aq2 2OH1aq2 (molar mass 192 g/mol) pH 2.68.
Initial Equilibrium Concentration (mol/L) Concentration (mol/L) [Mg2]0 3.75 103 [F]0 6.25 102 [Mg2] 3.75 103 y [F] 6.25 102 2y x mol/Mg2 88888888n reacts to form MgF2 137. Calculate the concentration of Pb2 in each of the following.
a. a saturated solution of Pb(OH)2, Ksp 1.2 1015 b. a saturated solution of Pb(OH)2 buffered at pH 13.00 Marathon Problem 747 a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at 100.0 mL of NaOH added, what volume of NaOH added corresponds to the second equivalence point?
b. For the following volumes of NaOH added, list the major species present after the OH reacts completely.
i. 0 mL NaOH added ii. between 0 and 100.0 mL NaOH added iii. 100.0 mL NaOH added iv. between 100.0 and 200.0 mL NaOH added v. 200.0 mL NaOH added vi. after 200.0 mL NaOH added c. If the pH at 50.0 mL of NaOH added is 4.0 and the pH at 150.0 mL of NaOH added is 8.0, determine the values and for the diprotic acid.
144. The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution as points A–F.
b. Calculate the pH at the halfway points to equivalence, B and D. Hint: Refer to Exercise 143.
145. A few drops of each of the indicators shown in the accompany-ing table were placed in separate portions of a 1.0 M solution of a weak acid, HX. The results are shown in the last column of the table. What is the approximate pH of the solution contain-ing HX? Calculate the approximate value of Ka for HX.
mL HCl pH A B C D E F Ka2 Ka1 148. You add an excess of solid MX in 250 g of water. You measure the freezing point and find it to be What is the Ksp of the solid? Assume the density of the solution is 1.0 g/cm3.
149. a. Calculate the molar solubility of SrF2 in water, ignoring the basic properties of (For SrF2, .) b. Would the measured molar solubility of SrF2 be greater than or less than the value calculated in part a? Explain.
c. Calculate the molar solubility of SrF2 in a solution buffered at .) 150. A solution saturated with a salt of the type M3X2 has an osmotic pressure of atm at . Calculate the Ksp value for the salt, assuming ideal behavior.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
151. A buffer solution is prepared by mixing 75.0 mL of 0.275 M flu-orobenzoic acid (C7H5O2F) with 55.0 mL of 0.472 M sodium fluorobenzoate. The pKa of this weak acid is 2.90. What is the pH of the buffer solution?
152. The Ksp for Q, a slightly soluble ionic compound composed of M2 2 and ions, is The electron configuration of M is The anion has 54 electrons. What is the molar solubility of Q in a solution of NaX prepared by dis-solving 1.98 g of NaX in 150. mL of water?
153. Calculate the pH of a solution prepared by mixing 250. mL of 0.174 m aqueous HF with 38.7 g of an aqueous solution that is 1.50% NaOH by mass (Ka for Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
154. A 225-mg sample of a diprotic acid is dissolved in enough wa-ter to make 250. mL of solution. The pH of this solution is 2.06.
A saturated solution of calcium hydroxide is prepared by adding excess calcium hydroxide to pure water and then removing the undissolved solid by filtration. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the sec-ond equivalence point (as determined by a pH meter) is 7.96.
The first dissociation constant for the acid is Assume that the volumes of the solutions are additive, all so-lutions are at , and that is at least 1000 times greater than .
a. Calculate the molar mass of the acid.
b. Calculate the second dissociation constant for the acid .
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
(Ka2) Ka2 Ka1 25°C 5.90 102.
(Ka1) (Ksp 1.3 106) HF 7.2 104.) 1.02 g/mL).
(density (density 1.10 g/mL) X [Xe]6s14f 145d10.
4.5 1029.
X 25°C 2.64 102 (Ka for HF is 7.2 104 pH 2.00.
Ksp 7.9 1010 F.
0.028°C.
Color Color pKa Indicator of HIn of In of HIn HX Bromphenol Yellow Blue 4.0 Blue blue Bromcresol Yellow Purple 6.0 Yellow purple Bromcresol Yellow Blue 4.8 Green green Alizarin Yellow Red 6.5 Yellow Used with permission from the Journal of Chemical Education, Vol. 68, No. 11, 1991, pp. 919–922; copyright © 1991, Division of Chemical Ed-ucation, Inc.
146. Consider a solution made by mixing 500.0 mL of 4.0 M NH3 and 500.0 mL of 0.40 M AgNO3. Ag reacts with NH3 to form AgNH3 and Ag(NH3)2 : Determine the concentration of all species in solution.
147. What is the maximum possible concentration of Ni2 ion in water at that is saturated with 0.10 M H2S and maintained at pH 3.0 with HCl?
25°C AgNH3 NH3 ∆Ag1NH322 K2 8.2 103 Ag NH3 ∆AgNH3 K1 2.1 103 748 16 Spontaneity, Entropy, and Free Energy Contents 16.1 Spontaneous Processes and Entropy 16.2 Entropy and the Second Law of Thermodynamics 16.3 The Effect of Temperature on Spontaneity 16.4 Free Energy 16.5 Entropy Changes in Chemical Reactions 16.6 Free Energy and Chemical Reactions 16.7 The Dependence of Free Energy on Pressure • The Meaning of for a Chemical Reaction 16.8 Free Energy and Equilibrium • The Temperature Dependence of K 16.9 Free Energy and Work ¢G Solid carbon dioxide (dry ice), when placed in water, causes violent bubbling as gaseous CO2 is released. The “fog” is moisture condensed from the cold air.
T he first law of thermodynamics is a statement of the law of conservation of energy: Energy can be neither created nor destroyed. In other words, the energy of the universe is constant. Although the total energy is constant, the various forms of energy can be in-terchanged in physical and chemical processes. For example, if you drop a book, some of the initial potential energy of the book is changed to kinetic energy, which is then trans-ferred to the atoms in the air and the floor as random motion. The net effect of this process is to change a given quantity of potential energy to exactly the same quantity of thermal energy. Energy has been converted from one form to another, but the same quantity of en-ergy exists before and after the process.
Now let’s consider a chemical example. When methane is burned in excess oxygen, the major reaction is This reaction produces a quantity of energy, which is released as heat. This energy flow results from the lowering of the potential energy stored in the bonds of CH4 and O2 as they react to form CO2 and H2O. This is illustrated in Fig. 16.1. Potential energy has been converted to thermal energy, but the energy content of the universe has remained constant in accordance with the first law of thermodynamics.
The first law of thermodynamics is used mainly for energy bookkeeping, that is, to answer such questions as How much energy is involved in the change?
Does energy flow into or out of the system?
What form does the energy finally assume?
Although the first law of thermodynamics provides the means for accounting for energy, it gives no hint as to why a particular process occurs in a given direction. This is the main question to be considered in this chapter.
16.1 Spontaneous Processes and Entropy A process is said to be spontaneous if it occurs without outside intervention. Spontaneous processes may be fast or slow. As we will see in this chapter, thermodynamics can tell us the direction in which a process will occur but can say nothing about the speed of the process. As we saw in Chapter 12, the rate of a reaction depends on many factors, such as activation energy, temperature, concentration, and catalysts, and we were able to ex-plain these effects using a simple collision model. In describing a chemical reaction, the discipline of chemical kinetics focuses on the pathway between reactants and products; thermodynamics considers only the initial and final states and does not require knowledge of the pathway between reactants and products (see Fig. 16.2).
CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 energy 749 The first law of thermodynamics: The energy of the universe is constant.
Spontaneous does not mean fast.
CH4, 2O2 CO2, 2H2O Change in energy heat Energy FIGURE 16.1 When methane and oxygen react to form carbon dioxide and water, the products have lower potential energy than the reactants.
This change in potential energy results in energy flow (heat) to the surroundings.
Energy Reaction progress Reactants Products Domain of kinetics (the reaction pathway) Domain of thermodynamics (the initial and final states) 750 Chapter Sixteen Spontaneity, Entropy, and Free Energy In summary, thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process. For example, accord-ing to the principles of thermodynamics, a diamond should change spontaneously to graphite. The fact that we do not observe this process does not mean the prediction is wrong; it simply means the process is very slow. Thus we need both thermodynamics and kinetics to describe reactions fully.
To explore the idea of spontaneity, consider the following physical and chemical processes: A ball rolls down a hill but never spontaneously rolls back up the hill.
If exposed to air and moisture, steel rusts spontaneously. However, the iron oxide in rust does not spontaneously change back to iron metal and oxygen gas.
A gas fills its container uniformly. It never spontaneously collects at one end of the container.
Heat flow always occurs from a hot object to a cooler one. The reverse process never occurs spontaneously.
Wood burns spontaneously in an exothermic reaction to form carbon dioxide and wa-ter, but wood is not formed when carbon dioxide and water are heated together.
At temperatures below C, water spontaneously freezes, and at temperatures above C, ice spontaneously melts.
What thermodynamic principle will provide an explanation of why, under a given set of conditions, each of these diverse processes occurs in one direction and never in the re-verse? In searching for an answer, we could explain the behavior of a ball on a hill in terms of gravity. But what does gravity have to do with the rusting of a nail or the freez-ing of water? Early developers of thermodynamics thought that exothermicity might be the key—that a process would be spontaneous if it were exothermic. Although this factor 0° 0° FIGURE 16.2 The rate of a reaction depends on the path-way from reactants to products; this is the domain of kinetics. Thermodynamics tells us whether a reaction is spontaneous based only on the properties of the reactants and products. The predictions of thermodynam-ics do not require knowledge of the path-way between reactants and products.
A disordered pile of playing cards.
Plant materials burn to form carbon dioxide and water.
16.1 Spontaneous Processes and Entropy 751 does appear to be important, since many spontaneous processes are exothermic, it is not the total answer. For example, the melting of ice, which occurs spontaneously at temper-atures greater than C, is an endothermic process.
What common characteristic causes the processes listed above to be spontaneous in one direction only? After many years of observation, scientists have concluded that the characteristic common to all spontaneous processes is an increase in a property called entropy, denoted by the symbol S. The driving force for a spontaneous process is an in-crease in the entropy of the universe.
What is entropy? Although there is no simple definition that is completely accurate, entropy can be viewed as a measure of molecular randomness or disorder. The natural progression of things is from order to disorder, from lower entropy to higher entropy. To illustrate the natural tendency toward disorder, you only have to think about the condition of your room. Your room naturally tends to get messy (disordered), because an ordered room requires everything to be in its place. There are simply many more ways for things to be out of place than for them to be in their places.
As another example, suppose you have a deck of playing cards ordered in some particular way. You throw these cards into the air and pick them all up at random.
Looking at the new sequence of the cards, you would be very surprised to find that it matched the original order. Such an event would be possible, but very improbable. There are billions of ways for the deck to be disordered, but only one way to be ordered ac-cording to your definition. Thus the chances of picking the cards up out of order are much greater than the chance of picking them up in order. It is natural for disorder to increase.
Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a given state.
Entropy is closely associated with probability. The key concept is that the more ways a particular state can be achieved, the greater is the likelihood (probability) of finding that state. In other words, nature spontaneously proceeds toward the states that have the high-est probabilities of existing. This conclusion is not surprising at all. The difficulty comes in connecting this concept to real-life processes. For example, what does the spontaneous rusting of steel have to do with probability? Understanding the connection between en-tropy and spontaneity will allow us to answer such questions. We will begin to explore this connection by considering a very simple process, the expansion of an ideal gas into 0° Probability refers to likelihood.
Visualization: Entropy FIGURE 16.4 Possible arrangements (states) of four molecules in a two-bulbed flask.
Arrangement I Arrangement II Arrangement III Arrangement IV Arrangement V FIGURE 16.3 The expansion of an ideal gas into an evacuated bulb.
Ideal gas Vacuum 752 Chapter Sixteen Spontaneity, Entropy, and Free Energy a vacuum, as represented in Fig. 16.3. Why is this process spontaneous? The driving force is probability. Because there are more ways of having the gas evenly spread throughout the container than there are ways for it to be in any other possible state, the gas sponta-neously attains the uniform distribution.
To understand this conclusion, we will greatly simplify the system and consider the possible arrangements of only four gas molecules in the two-bulbed container (Fig. 16.4).
How many ways can each arrangement (state) be achieved? Arrangements I and V can be achieved in only one way—all the molecules must be in one end. Arrangements II and V can be achieved in four ways, as shown in Table 16.1. Each configuration that gives a particular arrangement is called a microstate. Arrangement I has one microstate, and arrangement II has four microstates. Arrangement III can be achieved in six ways (six microstates), as shown in Table 16.1. Which arrangement is most likely to occur?
The one that can be achieved in the greatest number of ways. Thus arrangement III is most probable. The relative probabilities of arrangements III, II, and I are 6 : 4 : 1. We have discovered an important principle: The probability of occurrence of a particular arrangement (state) depends on the number of ways (microstates) in which that arrange-ment can be achieved.
The consequences of this principle are dramatic for large numbers of molecules. One gas molecule in the flask in Fig. 16.4 has one chance in two of being in the left bulb. We say that the probability of finding the molecule in the left bulb is For two molecules in the flask, there is one chance in two of finding each molecule in the left bulb, so there is one chance in four ( ) that both molecules will be in the left bulb. As the num-ber of molecules increases, the relative probability of finding all of them in the left bulb decreases, as shown in Table 16.2. For 1 mole of gas, the probability of finding all the molecules in the left bulb is so small that this arrangement would “never” occur.
Thus a gas placed in one end of a container will spontaneously expand to fill the en-tire vessel evenly because, for a large number of gas molecules, there is a huge number of microstates in which equal numbers of molecules are in both ends. On the other hand, the opposite process, 1 2 1 2 1 4 1 2.
TABLE 16.2 Probability of Finding All the Molecules in the Left Bulb as a Function of the Total Number of Molecules Relative Probability of Finding All Molecules Number of Molecules in the Left Bulb 1 2 3 5 10 n a1 2b 61023 101210232 6 1023 (1 mole) 1 2n a1 2b n 1 210 1 1024 1 2 1 2 1 2 1 2 1 2 1 25 1 32 1 2 1 2 1 2 1 23 1 8 1 2 1 2 1 22 1 4 1 2 For two molecules in the flask, there are four possible microstates: Thus there is one chance in four of finding B A B A B A B A A B A TABLE 16.1 The Microstates That Give a Particular Arrangement (State) Arrangement Microstates I II III IV V A B C D A B D C A A D B C C A D B D C B A D C B A C A D B B A D C B A D C B C D A B C D D D A A B B C C D D D A A A B B B C C C 16.1 Spontaneous Processes and Entropy 753 Sample Exercise 16.1 Sgas Sliquid Ssolid 754 Chapter Sixteen Spontaneity, Entropy, and Free Energy although not impossible, is highly improbable, since only one microstate leads to this arrangement. Therefore, this process does not occur spontaneously.
The type of probability we have been considering in this example is called positional probability because it depends on the number of configurations in space (positional mi-crostates) that yield a particular state. A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability, that is, the largest entropy, of the states available to the system.
Positional probability is also illustrated by changes of state. In general, positional en-tropy increases in going from solid to liquid to gas. A mole of a substance has a much smaller volume in the solid state than it does in the gaseous state. In the solid state, the molecules are close together, with relatively few positions available to them; in the gaseous state, the molecules are far apart, with many more positions available to them. The liquid state is closer to the solid state than it is to the gaseous state in these terms. We can sum-marize these comparisons as follows: Solid, liquid, and gaseous states were compared in Chapter 10.
Positional entropy is also very important in the formation of solutions. In Chapter 11 we saw that solution formation is favored by the natural tendency for substances to mix. We can now be more precise. The entropy change associated with the mixing of two pure substances is expected to be positive. An increase in entropy is expected be-cause there are many more microstates for the mixed condition than for the separated condition. This effect is due principally to the increased volume available to a given “par-ticle” after mixing occurs. For example, when two liquids are mixed to form a solution, the molecules of each liquid have more available volume and thus more available posi-tions. Therefore, the increase in positional entropy associated with mixing favors the for-mation of solutions.
Positional Entropy For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature.
a. Solid CO2 and gaseous CO2 b. N2 gas at 1 atm and N2 gas at atm Solution a. Since a mole of gaseous CO2 has the greater volume by far, the molecules have many more available positions than in a mole of solid CO2. Thus gaseous CO2 has the higher positional entropy.
b. A mole of N2 gas at atm has a volume 100 times that (at a given temperature) of a mole of N2 gas at 1 atm. Thus N2 gas at atm has the higher positional entropy.
See Exercise 16.23.
1 102 1 102 1.0 102 The tendency to mix is due to the in-creased volume available to the particles of each component of the mixture. For exam-ple, when two liquids are mixed, the mole-cules of each liquid have more available volume and thus more available positions.
Solids are more ordered than liquids or gases and thus have lower entropy.
16.2 Entropy and the Second Law of Thermodynamics 755 Predicting Entropy Changes Predict the sign of the entropy change for each of the following processes.
a. Solid sugar is added to water to form a solution.
b. Iodine vapor condenses on a cold surface to form crystals.
Solution a. The sugar molecules become randomly dispersed in the water when the solution forms and thus have access to a larger volume and a larger number of possible positions. The positional disorder is increased, and there will be an increase in entropy. is positive, since the final state has a larger entropy than the initial state, and b. Gaseous iodine is forming a solid. This process involves a change from a relatively large volume to a much smaller volume, which results in lower positional disorder. For this process is negative (the entropy decreases).
See Exercise 16.24.
16.2 Entropy and the Second Law of Thermodynamics We have seen that processes are spontaneous when they result in an increase in disorder.
Nature always moves toward the most probable state available to it. We can state this prin-ciple in terms of entropy: In any spontaneous process there is always an increase in the entropy of the universe. This is the second law of thermodynamics. Contrast this with the first law of thermodynamics, which tells us that the energy of the universe is constant.
Energy is conserved in the universe, but entropy is not. In fact, the second law can be paraphrased as follows: The entropy of the universe is increasing.
As in Chapter 6, we find it convenient to divide the universe into a system and the surroundings. Thus we can represent the change in the entropy of the universe as where and represent the changes in entropy that occur in the system and sur-roundings, respectively.
To predict whether a given process will be spontaneous, we must know the sign of If is positive, the entropy of the universe increases, and the process is spontaneous in the direction written. If is negative, the process is spontaneous in the opposite direction. If is zero, the process has no tendency to occur, and the system is at equilibrium. To predict whether a process is spontaneous, we must consider the entropy changes that occur both in the system and in the surroundings and then take their sum.
The Second Law In a living cell, large molecules are assembled from simple ones. Is this process consis-tent with the second law of thermodynamics?
Solution To reconcile the operation of an order-producing cell with the second law of thermody-namics, we must remember that not must be positive for a process to be spontaneous. A process for which is negative can be spontaneous if the associated is both larger and positive. The operation of a cell is such a process.
See Questions 16.7 and 16.8.
¢Ssurr ¢Ssys ¢Ssys, ¢Suniv, ¢Suniv ¢Suniv ¢Suniv ¢Suniv.
¢Ssurr ¢Ssys ¢Suniv ¢Ssys ¢Ssurr ¢S ¢S Sfinal Sinitial.
¢S Sample Exercise 16.3 The total energy of the universe is constant, but the entropy is increasing.
Sample Exercise 16.2 756 Chapter Sixteen Spontaneity, Entropy, and Free Energy 16.3 The Effect of Temperature on Spontaneity To explore the interplay of and in determining the sign of we will first discuss the change of state for one mole of water from liquid to gas, considering the water to be the system and everything else the surroundings.
What happens to the entropy of water in this process? A mole of liquid water (18 grams) has a volume of approximately 18 mL. A mole of gaseous water at 1 atmos-phere and C occupies a volume of approximately 31 liters. Clearly, there are many more positions available to the water molecules in a volume of 31 L than in 18 mL, and the vaporization of water is favored by this increase in positional probability. That is, for this process the entropy of the system increases; has a positive sign.
What about the entropy change in the surroundings? Although we will not prove it here, entropy changes in the surroundings are determined primarily by the flow of energy ¢Ssys 100° H2O1l2 ¡ H2O1g2 ¢Suniv, ¢Ssurr ¢Ssys I n this text we have emphasized the meaning of the second law of thermodynamics—that the entropy of the universe is always increasing. Although the results of all our exper-iments support this conclusion, this does not mean that order cannot appear spontaneously in a given part of the universe. The best example of this phenomenon involves the assembly of cells in living organisms. Of course, when a process that creates an ordered system is examined in detail, it is found that other parts of the process involve an increase in disorder such that the sum of all the entropy changes is positive. In fact, scientists are now finding that the search for maximum entropy in one part of a system can be a powerful force for organization in another part of the system.
To understand how entropy can be an organizing force, look at the accompanying figure. In a system containing large and small “balls” as shown in the figure, the small balls can “herd” the large balls into clumps in the corners and near the walls. This clears out the maximum space for the small balls so that they can move more freely, thus maxi-mizing the entropy of the system, as demanded by the second law of thermodynamics.
In essence, the ability to maximize entropy by sorting different-sized objects creates a kind of attractive force, called a depletion, or excluded-volume, force. These “en-tropic forces” operate for objects in the size range of approximately to approximately m. For entropy-induced ordering to occur, the particles must be constantly jostling each other and must be constantly agitated by sol-vent molecules, thus making gravity unimportant.
106 108 CHEMICAL IMPACT Entropy: An Organizing Force?
There is increasing evidence that entropic ordering is important in many biological systems. For example, this phenomenon seems to be responsible for the clumping of sickle-cell hemoglobin in the presence of much smaller pro-teins that act as the “smaller balls.” Entropic forces also have been linked to the clustering of DNA in cells without nu-clei, and Allen Minton of the National Institutes of Health in Bethesda, Maryland, is studying the role of entropic forces in the binding of proteins to cell membranes.
Entropic ordering also appears in nonbiological set-tings, especially in the ways polymer molecules clump to-gether. For example, polymers added to paint to improve the flow characteristics of the paint actually caused it to coag-ulate because of depletion forces.
Thus, as you probably have concluded already, entropy is a complex issue. As entropy drives the universe to its ul-timate death of maximum chaos, it provides some order along the way.
Boiling water to form steam increases its volume and thus its entropy.
16.3 The Effect of Temperature on Spontaneity 757 into or out of the system as heat. To understand this, suppose an exothermic process trans-fers 50 J of energy as heat to the surroundings, where it becomes thermal energy, that is, kinetic energy associated with the random motions of atoms. Thus this flow of energy into the surroundings increases the random motions of atoms there and thereby increases the entropy of the surroundings. The sign of is positive. When an endothermic process occurs in the system, it produces the opposite effect. Heat flows from the surroundings to the system, and the random motions of the atoms in the surroundings decrease, decreas-ing the entropy of the surroundings. The vaporization of water is an endothermic process.
Thus, for this change of state, is negative.
Remember it is the sign of that tells us whether the vaporization of water is spontaneous. We have seen that is positive and favors the process and that is negative and unfavorable. Thus the components of are in opposition. Which one controls the situation? The answer depends on the temperature. We know that at a pres-sure of 1 atmosphere, water changes spontaneously from liquid to gas at all temperatures above C. Below 100ºC, the opposite process (condensation) is spontaneous.
Since and are in opposition for the vaporization of water, the temperature must have an effect on the relative importance of these two terms. To understand why this is so, we must discuss in more detail the factors that control the entropy changes in the surroundings. The central idea is that the entropy changes in the surroundings are primarily determined by heat flow. An exothermic process in the system increases the entropy of the surroundings, because the resulting energy flow increases the random motions in the sur-roundings. This means that exothermicity is an important driving force for spontaneity. In earlier chapters we have seen that a system tends to undergo changes that lower its energy.
We now understand the reason for this tendency. When a system at constant temperature moves to a lower energy, the energy it gives up is transferred to the surroundings, lead-ing to an increase in entropy there.
The significance of exothermicity as a driving force depends on the temperature at which the process occurs. That is, the magnitude of depends on the temperature at which the heat is transferred. We will not attempt to prove this fact here. Instead, we of-fer an analogy. Suppose that you have $50 to give away. Giving it to a millionaire would not create much of an impression—a millionaire has money to spare. However, to a poor college student, $50 would represent a significant sum and would be received with con-siderable joy. The same principle can be applied to energy transfer via the flow of heat.
If 50 J of energy is transferred to the surroundings, the impact of that event depends greatly on the temperature. If the temperature of the surroundings is very high, the atoms there are in rapid motion. The 50 J of energy will not make a large percent change in these mo-tions. On the other hand, if 50 J of energy is transferred to the surroundings at a very low temperature, where atomic motion is slow, the energy will cause a large percent change in these motions. The impact of the transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures.
For our purposes, there are two important characteristics of the entropy changes that occur in the surroundings: 1. The sign of depends on the direction of the heat flow. At constant temperature, an exothermic process in the system causes heat to flow into the surroundings, in-creasing the random motions and thus the entropy of the surroundings. For this case, is positive. The opposite is true for an endothermic process in a system at con-stant temperature. Note that although the driving force described here really results from the change in entropy, it is often described in terms of energy: Nature tends to seek the lowest possible energy.
2. The magnitude of depends on the temperature. The transfer of a given quantity of energy as heat produces a much greater percent change in the randomness of the surroundings at a low temperature than it does at a high temperature. Thus de-pends directly on the quantity of heat transferred and inversely on temperature. In ¢Ssurr ¢Ssurr ¢Ssurr ¢Ssurr ¢Ssurr ¢Ssurr ¢Ssys 100° ¢Suniv ¢Ssurr ¢Ssys ¢Suniv ¢Ssurr ¢Ssurr In an endothermic process, heat flows from the surroundings into the system.
In an exothermic process, heat flows into the surroundings from the system.
In a process occurring at constant tem-perature, the tendency for the system to lower its energy results from the positive value of ¢Ssurr.
758 Chapter Sixteen Spontaneity, Entropy, and Free Energy other words, the tendency for the system to lower its energy becomes a more impor-tant driving force at lower temperatures.
These ideas are summarized as follows: We can express in terms of the change in enthalpy for a process occurring at constant pressure, since Recall that consists of two parts: a sign and a number. The sign indicates the direc-tion of flow, where a plus sign means into the system (endothermic) and a minus sign means out of the system (exothermic). The number indicates the quantity of energy.
Combining all these concepts produces the following definition of for a reac-tion that takes place under conditions of constant temperature (in kelvins) and pressure: The minus sign is necessary because the sign of is determined with respect to the re-action system, and this equation expresses a property of the surroundings. This means that if the reaction is exothermic, has a negative sign, but since heat flows into the sur-roundings, is positive.
Determining In the metallurgy of antimony, the pure metal is recovered via different reactions, de-pending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores: Carbon is used as the reducing agent for oxide ores: Calculate for each of these reactions at C and 1 atm.
Solution We use where For the sulfide ore reaction, ¢Ssurr 125 kJ 298 K 0.419 kJ/K 419 J/K T 25 273 298 K ¢Ssurr ¢H T 25° ¢Ssurr Sb4O61s2 6C1s2 ¡ 4Sb1s2 6CO1g2 ¢H 778 kJ Sb2S31s2 3Fe1s2 ¡ 2Sb1s2 3FeS1s2 ¢H 125 kJ ¢Ssurr ¢Ssurr ¢H ¢H ¢Ssurr ¢H T ¢Ssurr ¢H Heat flow 1constant P2 change in enthalpy ¢H ¢H ¢Ssurr Endothermic process: ¢Ssurr quantity of heat 1J2 temperature 1K2 Exothermic process: ¢Ssurr quantity of heat 1J2 temperature 1K2 Driving force provided by the energy flow 1heat2 magnitude of the entropy change of the surroundings quantity of heat 1J2 temperature 1K2 Exothermic process: Ssurr positive Endothermic process: Ssurr negative When no subscript is present, the quantity (for example, ) refers to the system.
¢H The minus sign changes the point of view from the system to the surroundings.
Sample Exercise 16.4 The mineral stibnite contains Sb2S3.
16.4 Free Energy 759 Note that is positive, as it should be, since this reaction is exothermic and heat flow occurs to the surroundings, increasing the randomness of the surroundings.
For the oxide ore reaction, In this case is negative because heat flow occurs from the surroundings to the system.
See Exercises 16.25 and 16.26.
We have seen that the spontaneity of a process is determined by the entropy change it produces in the universe. We also have seen that has two components, and If for some process both and are positive, then is positive, and the process is spontaneous. If, on the other hand, both and are negative, the process does not occur in the direction indicated but is spontaneous in the opposite di-rection. Finally, if and have opposite signs, the spontaneity of the process de-pends on the sizes of the opposing terms. These cases are summarized in Table 16.3.
We can now understand why spontaneity is often dependent on temperature and thus why water spontaneously freezes below C and melts above C. The term is temperature-dependent. Since at constant pressure, the value of changes markedly with temperature. The magni-tude of will be very small at high temperatures and will increase as the tem-perature decreases. That is, exothermicity is most important as a driving force at low temperatures.
16.4 Free Energy So far we have used to predict the spontaneity of a process. However, another ther-modynamic function is also related to spontaneity and is especially useful in dealing with the temperature dependence of spontaneity. This function is called the free energy, which is symbolized by G and defined by the relationship where H is the enthalpy, T is the Kelvin temperature, and S is the entropy.
G H TS ¢Suniv ¢Ssurr ¢Ssurr ¢Ssurr ¢H T ¢Ssurr 0° 0° ¢Ssurr ¢Ssys ¢Ssurr ¢Ssys ¢Suniv ¢Ssurr ¢Ssys ¢Ssurr.
¢Ssys ¢Suniv ¢Ssurr ¢Ssurr 778 kJ 298 2.61 kJ/K 2.61 103 J/K ¢Ssurr TABLE 16.3 Interplay of Ssys and Ssurr in Determining the Sign of Suniv Signs of Entropy Changes Ssys Ssurr Suniv Process Spontaneous?
Yes No (reaction will occur in opposite direction) ?
Yes, if Ssys has a larger magnitude than Ssurr ?
Yes, if Ssurr has a larger magnitude than Ssys The symbol G for free energy honors Josiah Willard Gibbs (1839–1903), who was professor of mathematical physics at Yale University from 1871 to 1903. He laid the foundations of many areas of thermo-dynamics, particularly as they apply to chemistry.
760 Chapter Sixteen Spontaneity, Entropy, and Free Energy For a process that occurs at constant temperature, the change in free energy ( ) is given by the equation Note that all quantities here refer to the system. From this point on we will follow the usual convention that when no subscript is included, the quantity refers to the system.
To see how this equation relates to spontaneity, we divide both sides of the equation by to produce Remember that at constant temperature and pressure So we can write We have shown that This result is very important. It means that a process carried out at constant temperature and pressure will be spontaneous only if is negative. That is, a process (at constant T and P) is spontaneous in the direction in which the free energy decreases ( means ).
Now we have two functions that can be used to predict spontaneity: the entropy of the universe, which applies to all processes, and free energy, which can be used for processes carried out at constant temperature and pressure. Since so many chemical re-actions occur under the latter conditions, free energy is the more useful to chemists.
Let’s use the free energy equation to predict the spontaneity of the melting of ice: for which Results of the calculations of and at are shown in Table 16.4. These data predict that the process is spontaneous at C; that is, ice melts at this temperature because is positive and is negative. The opposite is true at where water freezes spontaneously.
Why is this so? The answer lies in the fact that ( ) and oppose each other.
The term favors the melting of ice because of the increase in positional entropy, and favors the freezing of water because it is an exothermic process. At temperatures below ¢Ssurr ¢S° ¢Ssurr ¢S° ¢Ssys 10°C, ¢G° ¢Suniv 10° 10°C, 0°C, and 10°C ¢G° ¢Suniv ¢H° 6.03 103 J/mol and ¢S° 22.1 J/K mol H2O1s2 ¡ H2O1l2 ¢Suniv ¢G ¢G ¢Suniv ¢G T at constant T and P ¢G T ¢H T ¢S ¢Ssurr ¢S ¢Suniv ¢Ssurr ¢H T ¢G T ¢H T ¢S T ¢G ¢H T¢S ¢G The superscript degree symbol ( ) indicates all substances are in their standard states.
° To review the definitions of standard states, see page 246.
TABLE 16.4 Results of the Calculation of Suniv and G for the Process H2O(s) n H2O(l) at 10C, 0C, and 10C Suniv T T H S S Ssurr TS G H TS (°C) (K) (J/mol) (J/K mol) (J/K mol) (J/K mol) (J/mol) (J/mol) 10 263 6.03 103 22.1 22.9 0.8 5.81 103 2.2 102 0 273 6.03 103 22.1 22.1 0 6.03 103 0 10 283 6.03 103 22.1 21.3 0.8 6.25 103 2.2 102 Note that at ( ) controls, and the process occurs even though it is endothermic. At the magnitude of Ssurr is larger than that of , so the process is spontaneous in the opposite (exothermic) direction.
¢S° ¢ 10°C, ¢Ssys ¢S° 10°C, Ssurr H T Visualization: Spontaneous Reactions 16.4 Free Energy 761 C, the change of state occurs in the exothermic direction because is larger in mag-nitude than But above C the change occurs in the direction in which is favorable, since in this case is larger in magnitude than At the opposing tendencies just balance, and the two states coexist; there is no driving force in either direction. An equi-librium exists between the two states of water. Note that is equal to 0 at C.
We can reach the same conclusions by examining At is positive be-cause the term is larger than the term. The opposite is true at C. At C, is equal to and is equal to 0. This means that solid H2O and liquid H2O have the same free energy at and the system is at equilibrium.
We can understand the temperature dependence of spontaneity by examining the behavior of For a process occurring at constant temperature and pressure, If and favor opposite processes, spontaneity will depend on temperature in such a way that the exothermic direction will be favored at low temperatures. For exam-ple, for the process is positive and is positive. The natural tendency for this system to lower its en-ergy is in opposition to its natural tendency to increase its positional randomness. At low temperatures, dominates, and at high temperatures, dominates. The various pos-sible cases are summarized in Table 16.5.
Free Energy and Spontaneity At what temperatures is the following process spontaneous at 1 atm?
What is the normal boiling point of liquid Br2?
Solution The vaporization process will be spontaneous at all temperatures where G is negative. Note that S favors the vaporization process because of the increase in positional entropy, and H favors the opposite process, which is exothermic. These opposite tendencies will exactly balance at the boiling point of liquid Br2, since at this temperature liquid and gaseous Br2 are in equilibrium ( G 0). We can find this temperature by setting G 0 in the equation ¢H° T¢S° 0 ¢H° T¢S° ¢G° ¢H° T¢S° ¢H° 31.0 kJ/mol and ¢S° 93.0 J/K mol Br21l2 ¡ Br21g2 ¢S ¢H ¢S ¢H H2O1s2 ¡ H2O1l2 ¢S ¢H ¢G ¢H T¢S ¢G.
0°C (¢G° Gliquid Gsolid), ¢G° T¢S° ¢H° 0° 10° T¢S° ¢H° 10°, ¢G° ¢G°.
0° ¢Suniv 0°C ¢Ssurr.
¢Ssys ¢Ssys 0° ¢Ssys.
¢Ssurr 0° Sample Exercise 16.5 TABLE 16.5 Various Possible Combinations of H and S for a Process and the Resulting Dependence of Spontaneity on Temperature Case Result S positive, H negative Spontaneous at all temperatures S positive, H positive Spontaneous at high temperatures (where exothermicity is relatively unimportant) S negative, H negative Spontaneous at low temperatures (where exothermicity is dominant) S negative, H positive Process not spontaneous at any temperature (reverse process is spontaneous at all temperatures) Note that although H and S are some-what temperature-dependent, it is a good approximation to assume they are con-stant over a relatively small temperature range.
762 Chapter Sixteen Spontaneity, Entropy, and Free Energy Then At temperatures above 333 K, T S has a larger magnitude than H, and G (or H T S) is negative. Above 333 K, the vaporization process is spontaneous; the opposite process occurs spontaneously below this temperature. At 333 K, liquid and gaseous Br2 coexist in equilibrium. These observations can be summarized as follows (the pressure is 1 atm in each case): 1. T 333 K. The term S controls. The increase in entropy when liquid Br2 is va-porized is dominant.
2. T 333 K. The process is spontaneous in the direction in which it is exothermic.
The term H controls.
3. T 333 K. The opposing driving forces are just balanced ( G 0), and the liquid and gaseous phases of bromine coexist. This is the normal boiling point.
See Exercises 16.29 through 16.31.
16.5 Entropy Changes in Chemical Reactions The second law of thermodynamics tells us that a process will be spontaneous if the en-tropy of the universe increases when the process occurs. We saw in Section 16.4 that for a process at constant temperature and pressure, we can use the change in free energy of the system to predict the sign of and thus the direction in which it is spontaneous.
So far we have applied these ideas only to physical processes, such as changes of state and the formation of solutions. However, the main business of chemistry is studying chem-ical reactions, and, therefore, we want to apply the second law to reactions.
First, we will consider the entropy changes accompanying chemical reactions that oc-cur under conditions of constant temperature and pressure. As for the other types of processes we have considered, the entropy changes in the surroundings are determined by the heat flow that occurs as the reaction takes place. However, the entropy changes in the system (the reactants and products of the reaction) are determined by positional probability.
For example, in the ammonia synthesis reaction four reactant molecules become two product molecules, lowering the number of inde-pendent units in the system, which leads to less positional disorder.
N21g2 3H21g2 ¡ 2NH31g2 ¢Suniv T ¢H° ¢S° 3.10 104 J/mol 93.0 J/K mol 333 K N H Less entropy Greater entropy 16.5 Entropy Changes in Chemical Reactions 763 Fewer molecules mean fewer possible configurations. To help clarify this idea, consider a special container with a million compartments, each large enough to hold a hydrogen molecule. Thus there are a million ways one H2 molecule can be placed in this container.
But suppose we break the HOH bond and place the two independent H atoms in the same container. A little thought will convince you that there are many more than a mil-lion ways to place the two separate atoms. The number of arrangements possible for the two independent atoms is much greater than the number for the molecule. Thus for the process positional entropy increases.
Does positional entropy increase or decrease when the following reaction takes place?
In this case 9 gaseous molecules are changed to 10 gaseous molecules, and the positional entropy increases. There are more independent units as products than as reactants. In general, when a reaction involves gaseous molecules, the change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products. If the number of molecules of the gaseous products is greater than the number of molecules of the gaseous reactants, positional entropy typically increases, and will be positive for the reaction.
Predicting the Sign of S Predict the sign of S for each of the following reactions.
a. The thermal decomposition of solid calcium carbonate: b. The oxidation of SO2 in air: Solution a. Since in this reaction a gas is produced from a solid reactant, the positional entropy increases, and S is positive.
b. Here three molecules of gaseous reactants become two molecules of gaseous products.
Since the number of gas molecules decreases, positional entropy decreases, and S is negative.
See Exercises 16.33 and 16.34.
In thermodynamics it is the change in a certain function that is usually important.
The change in enthalpy determines if a reaction is exothermic or endothermic at constant pressure. The change in free energy determines if a process is spontaneous at constant temperature and pressure. It is fortunate that changes in thermodynamic functions are sufficient for most purposes, since absolute values for many ther modynamic characteristics of a system, such as enthalpy or free energy, cannot be determined.
However, we can assign absolute entropy values. Consider a solid at 0 K, where molecular motion virtually ceases. If the substance is a perfect crystal, its internal arrange-ment is absolutely regular (see Fig. 16.5(a)). There is only one way to achieve this perfect order: Every particle must be in its place. For example, with N coins there is only one 2SO21g2 O21g2 ¡ 2SO31g2 CaCO31s2 ¡ CaO1s2 CO21g2 ¢S 4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 H2 ¡ 2H Sample Exercise 16.6 764 Chapter Sixteen Spontaneity, Entropy, and Free Energy way to achieve the state of all heads. Thus a perfect crystal represents the lowest possible entropy; that is, the entropy of a perfect crystal at 0 K is zero. This is a statement of the third law of thermodynamics.
As the temperature of a perfect crystal is increased, the random vibrational mo-tions increase, and disorder increases within the crystal [see Fig. 16.5(b)]. Thus the entropy of a substance increases with temperature. Since S is zero for a perfect crys-tal at 0 K, the entropy value for a substance at a particular temperature can be calcu-lated by knowing the temperature dependence of entropy. (We will not show such cal-culations here.) The standard entropy values of many common substances at 298 K and 1 atm are listed in Appendix 4. From these values you will see that the entropy of a substance does indeed increase in going from solid to liquid to gas. One especially interesting feature of this table is the very low value for diamond. The structure of diamond is highly or-dered, with each carbon strongly bound to a tetrahedral arrangement of four other carbon atoms (see Section 10.5, Fig. 10.22). This type of structure allows very little disorder and has a very low entropy, even at 298 K. Graphite has a slightly higher entropy because its layered structure allows for a little more disorder.
Because entropy is a state function of the system (it is not pathway-dependent), the entropy change for a given chemical reaction can be calculated by taking the difference between the standard entropy values of products and those of the reactants: where, as usual, represents the sum of the terms. It is important to note that entropy is an extensive property (it depends on the amount of substance present). This means that the number of moles of a given reactant (nr) or product (np) must be taken into account.
Calculating S Calculate at C for the reaction given the following standard entropy values: 2NiS1s2 3O21g2 ¡ 2SO21g2 2NiO1s2 25° ¢S° © ¢S° reaction ©npS° products ©nrS° reactants S° S° FIGURE 16.5 (a) A perfect crystal of hydrogen chloride at 0 K; the dipolar HCl molecules are repre-sented by . The entropy is zero (S 0) for this perfect crystal at 0 K. (b) As the temperature rises above 0 K, lat-tice vibrations allow some dipoles to change their orientations, producing some disorder and an increase in entropy (S 0).
7 – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + (a) (b) The standard entropy values represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.
Sample Exercise 16.7 A perfect crystal at 0 K is an unattainable ideal, taken as a standard but never ac-tually observed.
Substance S° (J/K mol) SO2(g) 248 NiO(s) 38 O2(g) 205 NiS(s) 53 16.5 Entropy Changes in Chemical Reactions 765 Sample Exercise 16.8 Solution Since We would expect to be negative because the number of gaseous molecules decreases in this reaction.
See Exercise 16.37.
Calculating S° Calculate for the reduction of aluminum oxide by hydrogen gas: Use the following standard entropy values: Al2O31s2 3H21g2 ¡ 2Al1s2 3H2O1g2 ¢S° ¢S° 149 J/K 496 J/K 76 J/K 106 J/K 615 J/K 2 mola53 J K molb 3 mola205 J K molb 2 mol a248 J K molb 2 mol a38 J K molb 2S° SO21g2 2S° NiO1s2 2S° NiS1s2 3S° O21s2 ¢S° ©npS° products ©nrS° reactants Solution See Exercises 16.38 through 16.40.
The reaction considered in Sample Exercise 16.8 involves 3 moles of hydrogen gas on the reactant side and 3 moles of water vapor on the product side. Would you expect to be large or small for such a case? We have assumed that depends on the relative numbers of molecules of gaseous reactants and products. Based on this assumption, should be near zero for this reaction. However, is large and ¢S ¢S ¢S ¢S 179 J/K 56 J/K 567 J/K 393 J/K 51 J/K 3 mola131 J K molb 1 mola51 J K molb 2 mola28 J K molb 3 mola189 J K molb 2S° Al1s2 3S° H2O1g2 3S° H21g2 S° Al2O31s2 ¢S° ©npS° products ©nrS° reactants Substance S° (J/K mol) Al2O3(s) 51 H2(g) 131 Al(s) 28 H2O(g) 189 766 Chapter Sixteen Spontaneity, Entropy, and Free Energy positive. Why is this so? The large value for results from the difference in the en-tropy values for hydrogen gas and water vapor. The reason for this difference can be traced to the difference in molecular structure. Because it is a nonlinear, triatomic mo-lecule, H2O has more rotational and vibrational motions (see Fig. 16.6) than does the diatomic H2 molecule. Thus the standard entropy value for H2O(g) is greater than that for H2(g). Generally, the more complex the molecule, the higher the standard entropy value.
16.6 Free Energy and Chemical Reactions For chemical reactions we are often interested in the standard free energy change the change in free energy that will occur if the reactants in their standard states are con-verted to the products in their standard states. For example, for the ammonia synthesis reaction at C, (16.1) This value represents the change in free energy when 1 mol nitrogen gas at 1 atm reacts with 3 mol hydrogen gas at 1 atm to produce 2 mol gaseous NH3 at 1 atm.
It is important to recognize that the standard free energy change for a reaction is not measured directly. For example, we can measure heat flow in a calorimeter to determine but we cannot measure this way. The value of for the ammonia synthe-sis in Equation (16.1) was not obtained by mixing 1 mol N2 and 3 mol H2 in a flask and measuring the change in free energy as 2 mol NH3 formed. For one thing, if we mixed 1 mol N2 and 3 mol H2 in a flask, the system would go to equilibrium rather than to completion. Also, we have no instrument that measures free energy. However, while we cannot directly measure for a reaction, we can calculate it from other measured quantities, as we will see later in this section.
Why is it useful to know for a reaction? As we will see in more detail later in this chapter, knowing the values for several reactions allows us to compare the rel-ative tendency of these reactions to occur. The more negative the value of the fur-ther a reaction will go to the right to reach equilibrium. We must use standard-state free energies to make this comparison because free energy varies with pressure or concentra-tion. Thus, to get an accurate comparison of reaction tendencies, we must compare all re-actions under the same pressure or concentration conditions. We will have more to say about the significance of later.
There are several ways to calculate One common method uses the equation which applies to a reaction carried out at constant temperature. For example, for the reaction the values of and are known to be kJ and 3.05 J/K, respectively, and can be calculated at 298 K as follows: 394.4 kJ1per mole of CO22 3.944 105 J 3.935 105 J 1298 K2 13.05 J/K2 ¢G° ¢H° T¢S° ¢G° 393.5 ¢S° ¢H° C1s2 O21g2 ¡ CO21g2 ¢G° ¢H° T¢S° ¢G°.
¢G° ¢G°, ¢G° ¢G° ¢G° ¢G° ¢G° ¢H°, ¢G° N21g2 3H21g2 ∆2NH31g2 ¢G° 33.3 kJ 25° (¢G°), ¢S H O H O H O H O H Rotation Vibrations H H H FIGURE 16.6 The H2O molecule can vibrate and rotate in several ways, some of which are shown here. This freedom of motion leads to a higher entropy for water than for a sub-stance like hydrogen, a simple diatomic molecule with fewer possible motions.
The value of G tells us nothing about the rate of a reaction, only its eventual equilibrium position.
° 16.6 Free Energy and Chemical Reactions 767 Calculating H°, S°, and G° Consider the reaction carried out at C and 1 atm. Calculate and using the following data: ¢G° ¢H°, ¢S°, 25° 2SO21g2 O21g2 ¡ 2SO31g2 Sample Exercise 16.9 Solution The value of H can be calculated from the enthalpies of formation using the equation we discussed in Section 6.4: Then The value of S can be calculated using the standard entropy values and the equa-tion discussed in Section 16.5: Thus We would expect S to be negative because three molecules of gaseous reactants give two molecules of gaseous products.
The value of G can now be calculated from the equation See Exercises 16.45 through 16.47.
A second method for calculating for a reaction takes advantage of the fact that, like enthalpy, free energy is a state function. Therefore, we can use procedures for find-ing that are similar to those for finding using Hess’s law.
¢H ¢G ¢G 198 kJ 55.7 kJ 142 kJ 198 kJ 1298 K2a187 J Kba 1 kJ 1000 Jb ¢G° ¢H° T¢S° 187 J/K 514 J/K 496 J/K 205 J/K 2 mol 1257 J/K mol2 2 mol 1248 J/K mol2 1 mol1205 J/K mol2 ¢S° 2S° SO31g2 2S° SO21g2 S° O21g2 ¢S° ©npS° products ©nrS° reactants 198 kJ 792 kJ 594 kJ 2 mol 1396 kJ/mol2 2 mol 1297 kJ/mol2 0 ¢H° 2¢Hf °1SO31g22 2¢Hf °1SO21g22 ¢Hf °1O21g22 ¢H° ©np¢Hf ° 1products2 ©nr¢Hf ° 1reactants2 Substance (kJ/mol) S° (J/K mol) SO2(g) 297 248 SO3(g) 396 257 O2(g) 0 205 Hf 768 Chapter Sixteen Spontaneity, Entropy, and Free Energy To illustrate this method for calculating the free energy change, we will obtain for the reaction (16.2) from the following data: (16.3) (16.4) Note that CO(g) is a reactant in Equation (16.2). This means that Equation (16.3) must be reversed, since CO(g) is a product in that reaction as written. When a reaction is re-versed, the sign of is also reversed. In Equation (16.4), CO2(g) is a product, as it is in Equation (16.2), but only one molecule of CO2 is formed. Thus Equation (16.4) must be multiplied by 2, which means the value for Equation (16.4) also must be multi-plied by 2. Free energy is an extensive property, since it is defined by two extensive prop-erties, H and S.
Reversed Equation (16.3) 2 Equation (16.4) This example shows that the values for reactions are manipulated in exactly the same way as the values.
Calculating G° Using the following data (at C) (16.5) (16.6) calculate for the reaction Solution We reverse Equation (16.6) to make graphite a product, as required, and then add the new equation to Equation (16.5): Reversed Equation (16.6) Since is negative for this process, diamond should spontaneously change to graphite at C and 1 atm. However, the reaction is so slow under these conditions that we do not observe the process. This is another example of kinetic rather than thermody-namic control of a reaction. We can say that diamond is kinetically stable with respect to graphite even though it is thermodynamically unstable.
See Exercises 16.51 and 16.52.
25° ¢G° 3 kJ Cdiamond1s2 ¡ Cgraphite1s2 ¢G° 397 kJ 394 kJ CO21g2 ¡ Cgraphite1s2 O21g2 ¢G° 1394 kJ2 Cdiamond1s2 O21g2 ¡ CO21g2 ¢G° 397 kJ Cdiamond1s2 ¡ Cgraphite1s2 ¢G° Cgraphite1s2 O21g2 ¡ CO21g2 ¢G° 394 kJ Cdiamond1s2 O21g2 ¡ CO21g2 ¢G° 397 kJ 25° ¢H ¢G 514 kJ 21801 kJ2 2CO1g2 O21g2 ¡ 2CO21g2 ¢G° 11088 kJ2 2CH41g2 4O21g2 ¡ 2CO21g2 4H2O1g2 ¢G° 21801 kJ2 2CO1g2 4H2O1g2 ¡ 2CH41g2 3O21g2 ¢G° 11088 kJ2 ¢G° ¢G° CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 ¢G° 801 kJ 2CH41g2 3O21g2 ¡ 2CO1g2 4H2O1g2 ¢G° 1088 kJ 2CO1g2 O21g2 ¡ 2CO21g2 ¢G° Sample Exercise 16.10 Graphite Diamond 16.6 Free Energy and Chemical Reactions 769 In Sample Exercise 16.10 we saw that the process is spontaneous but very slow at C and 1 atm. The reverse process can be made to oc-cur at high temperatures and pressures. Diamond has a more compact structure and thus a higher density than graphite, so exerting very high pressure causes it to become ther-modynamically favored. If high temperatures are also used to make the process fast enough to be feasible, diamonds can be made from graphite. The conditions typically used in-volve temperatures greater than C and pressures of about 105 atm. About half of all industrial diamonds are made this way.
A third method for calculating the free energy change for a reaction uses standard free energies of formation. The standard free energy of formation ( G f) of a substance is defined as the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states. For the formation of glucose (C6H12O6), the appropriate reaction is The standard free energy associated with this process is called the free energy of forma-tion of glucose. Values of the standard free energy of formation are useful in calculating for specific chemical reactions using the equation Values of for many common substances are listed in Appendix 4. Note that, analogous to the enthalpy of formation, the standard free energy of formation of an ele-ment in its standard state is zero. Also note that the number of moles of each reactant (nr) and product (np) must be used when calculating for a reaction.
Calculating G° Methanol is a high-octane fuel used in high-performance racing engines. Calculate for the reaction given the following free energies of formation: 2CH3OH1g2 3O21g2 ¡ 2CO21g2 4H2O1g2 ¢G° ¢G° ¢Gf ° ¢G° ©np¢Gf °1products2 ©nr¢Gf °1reactants2 ¢G° 6C1s2 6H21g2 3O21g2 ¡ C6H12O61s2 1000° 25° Cdiamond1s2 ¡ Cgraphite1s2 Sample Exercise 16.11 Substance G° f (kJ/mol) CH3OH(g) 163 O2(g) 0 CO2(g) 394 H2O(g) 229 Solution We use the equation 1378 kJ 2 mol1163 kJ/mol2 2 mol1394 kJ/mol2 4 mol1229 kJ/mol2 3102 2¢Gf °1CO21g22 4¢Gf ° 1H2O1g22 3¢Gf °1O21g22 2¢Gf °1CH3OH1g22 ¢G° ©np¢Gf °1products2 ©nr¢Gf °1reactants2 The standard state of an element is its most stable state of 25°C and 1 atm.
Calculating G° is very similar to calcu-lating H°, as shown in Section 6.4.
770 Chapter Sixteen Spontaneity, Entropy, and Free Energy The large magnitude and the negative sign of indicate that this reaction is very fa-vorable thermodynamically.
See Exercises 16.53 through 16.55.
Free Energy and Spontaneity A chemical engineer wants to determine the feasibility of making ethanol (C2H5OH) by reacting water with ethylene (C2H4) according to the equation Is this reaction spontaneous under standard conditions?
Solution To determine the spontaneity of this reaction under standard conditions, we must deter-mine for the reaction. We can do this using standard free energies of formation at from Appendix 4: Then Thus the process is spontaneous under standard conditions at 25°C.
See Exercise 16.56.
Although the reaction considered in Sample Exercise 16.12 is spontaneous, other features of the reaction must be studied to see if the process is feasible. For example, the chemical engineer will need to study the kinetics of the reaction to determine whether it is fast enough to be useful and, if it is not, whether a catalyst can be found to enhance the rate. In doing these studies, the engineer must remember that depends on temperature: Thus, if the process must be carried out at high temperatures to be fast enough to be fea-sible, must be recalculated at that temperature from the and values for the reaction.
16.7 The Dependence of Free Energy on Pressure In this chapter we have seen that a system at constant temperature and pressure will proceed spontaneously in the direction that lowers its free energy. This is why reactions proceed until they reach equilibrium. As we will see later in this section, the equilib-rium position represents the lowest free energy value available to a particular reaction system. The free energy of a reaction system changes as the reaction proceeds, because free energy is dependent on the pressure of a gas or on the concentration of species in solution. We will deal only with the pressure dependence of the free energy of an ideal gas. The dependence of free energy on concentration can be developed using similar reasoning.
¢S° ¢H° ¢G° ¢G° ¢H° T¢S° ¢G° 6 kJ 175 kJ 1237 kJ2 68 kJ ¢G° ¢Gf °1C2H5OH1l22 ¢Gf °1H2O1l22 ¢Gf °1C2H41g22 ¢Gf °1C2H41g22 68 kJ/mol ¢Gf °1H2O1l22 237 kJ/mol ¢Gf ° 1C2H5OH1l22 175 kJ/mol 25° ¢G° C2H41g2 H2O1l2 ¡ C2H5OH1l2 ¢G° Sample Exercise 16.12 Ethylene Ethanol 16.7 The Dependence of Free Energy on Pressure 771 To understand the pressure dependence of free energy, we need to know how pres-sure affects the thermodynamic functions that comprise free energy, that is, enthalpy and entropy (recall that ). For an ideal gas, enthalpy is not pressure-dependent.
However, entropy does depend on pressure because of its dependence on volume. Con-sider 1 mole of an ideal gas at a given temperature. At a volume of 10.0 L, the gas has many more positions available for its molecules than if its volume is 1.0 L. The positional entropy is greater in the larger volume. In summary, at a given temperature for 1 mole of ideal gas or, since pressure and volume are inversely related, We have shown qualitatively that the entropy and therefore the free energy of an ideal gas depend on its pressure. Using a more detailed argument, which we will not consider here, it can be shown that where is the free energy of the gas at a pressure of 1 atm, G is the free energy of the gas at a pressure of P atm, R is the universal gas constant, and T is the Kelvin temperature.
To see how the change in free energy for a reaction depends on pressure, we will con-sider the ammonia synthesis reaction In general, For this reaction where Substituting these values into the equation gives G reaction The first term (in parentheses) is for the reaction. Thus we have and since the equation becomes ¢G ¢G° RT lna PNH3 2 1PN22 1PH2 32 b 3 ln1PH22 lna 1 PH2 3b ln 1PN22 lna 1 PN2 b 2 ln 1PNH32 ln 1PNH3 22 ¢G ¢G° reaction RT 32 ln1PNH32 ln 1PN22 3 ln1PH22 4 ¢G° 12G° NH3 G° N2 3G° H22 RT 32 ln1PNH32 ln1PN22 3 ln1PH22 4 2G° NH3 G° N2 3G° H2 2RT ln1PNH32 RT ln1PN22 3RT ln1PH22 ¢G 23G° NH3 RT ln1PNH32 4 3G° N2 RT ln1PN224 33G° H2 RT ln1PH224 GH2 G° H2 RT ln1PH22 GN2 G° N2 RT ln1PN22 GNH3 G° NH3 RT ln1PNH32 ¢G 2GNH3 GN2 3GH2 ¢G ©npGproducts ©nrGreactants N21g2 3H21g2 ¡ 2NH31g2 G° G G° RT ln1P2 Slow pressure 7 Shigh pressure Slarge volume 7 Ssmall volume G H TS ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ See Appendix 1.2 to review logarithms.
772 Chapter Sixteen Spontaneity, Entropy, and Free Energy But the term is the reaction quotient Q discussed in Section 13.5. Therefore, we have where Q is the reaction quotient (from the law of mass action), T is the temperature (K), R is the gas law constant and is equal to 8.3145 J/K mol, is the free energy change for the reaction with all reactants and products at a pressure of 1 atm, and is the free energy change for the reaction for the specified pressures of reactants and products.
Calculating G° One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases: Calculate at C for this reaction where carbon monoxide gas at 5.0 atm and hy-drogen gas at 3.0 atm are converted to liquid methanol.
Solution To calculate for this process, we use the equation We must first compute from standard free energies of formation (see Appendix 4).
Since Note that this is the value of for the reaction of 1 mol CO with 2 mol H2 to produce 1 mol CH3OH. We might call this the value of for one “round” of the re-action or for one mole of the reaction. Thus the value might better be written as J/mol of reaction, or J/mol rxn.
We can now calculate using Note that the pure liquid methanol is not included in the calculation of Q. Then 38 kJ/mol rxn 12.9 104 J/mol rxn2 19.4 103 J/mol rxn2 3.8 104 J/mol rxn 12.9 104 J/mol rxn2 18.3145 J/K mol rxn2 1298 K2 ln 12.2 1022 ¢G ¢G° RT ln 1Q2 Q 1 1PCO2 1PH2 22 1 15.02 13.022 2.2 102 T 273 25 298 K R 8.3145 J/K mol ¢G° 2.9 104 J/mol rxn ¢G 2.9 104 2.9 104 ¢G° ¢G° ¢G° ¢G° 166 kJ 1137 kJ2 0 29 kJ 2.9 104 J ¢Gf ° 1CO1g22 137 kJ ¢Gf ° 1H21g22 0 ¢Gf ° 1CH3OH1l22 166 kJ ¢G° ¢G ¢G° RT ln 1Q2 ¢G 25° ¢G CO1g2 2H21g2 ¡ CH3OH1l2 ¢G ¢G° ¢G ¢G° RT ln 1Q2 PNH3 2 1PN221PH2 32 Sample Exercise 16.13 Note in this case that G is defined for “one mole of the reaction,” that is, for 1 mol CO(g) reacting with 2 mol H2(g) to form 1 mol CH3OH(l). Thus G, Gº, and RT ln(Q) all have units of J/mol of reac-tion. In this case the units of R are actu-ally J/K mol of reaction, although they are usually not written this way.
16.7 The Dependence of Free Energy on Pressure 773 Note that is significantly more negative than implying that the reaction is more spontaneous at reactant pressures greater than 1 atm. We might expect this result from Le Châtelier’s principle.
See Exercises 16.57 and 16.58.
The Meaning of G for a Chemical Reaction In this section we have learned to calculate for chemical reactions under various con-ditions. For example, in Sample Exercise 16.13 the calculations show that the formation of CH3OH(l) from CO(g) at 5.0 atm reacting with H2(g) at 3.0 atm is spontaneous. What does this result mean? Does it mean that if we mixed 1.0 mol CO(g) and 2.0 mol H2(g) together at pressures of 5.0 and 3.0 atm, respectively, that 1.0 mol CH3OH(l) would form in the reaction flask? The answer is no. This answer may surprise you in view of what has been said in this section. It is true that 1.0 mol CH3OH(l) has a lower free energy than 1.0 mol CO(g) at 5.0 atm plus 2.0 mol H2(g) at 3.0 atm. However, when CO(g) and H2(g) are mixed under these conditions, there is an even lower free energy available to this sys-tem than 1.0 mol pure CH3OH(l). For reasons we will discuss shortly, the system can achieve the lowest possible free energy by going to equilibrium, not by going to comple-tion. At the equilibrium position, some of the CO(g) and H2(g) will remain in the reac-tion flask. So even though 1.0 mol pure CH3OH(l) is at a lower free energy than 1.0 mol CO(g) and 2.0 mol H2(g) at 5.0 and 3.0 atm, respectively, the reaction system will stop short of forming 1.0 mol CH3OH(l). The reaction stops short of completion because the equilibrium mixture of CH3OH(l), CO(g), and H2(g) exists at the lowest possible free energy available to the system.
To illustrate this point, we will explore a mechanical example. Consider balls rolling down the two hills shown in Fig. 16.7. Note that in both cases point B has a lower po-tential energy than point A.
In Fig. 16.7(a) the ball will roll to point B. This diagram is analogous to a phase change. For example, at C ice will spontaneously change completely to liquid water, because the latter has the lowest free energy. In this case liquid water is the only choice.
There is no intermediate mixture of ice and water with lower free energy.
The situation is different for a chemical reaction system, as illustrated in Fig. 16.7(b).
In Fig. 16.7(b) the ball will not get to point B because there is a lower potential energy at point C. Like the ball, a chemical system will seek the lowest possible free energy, which, for reasons we will discuss below, is the equilibrium position.
Therefore, although the value of for a given reaction system tells us whether the products or reactants are favored under a given set of conditions, it does not mean that the system will proceed to pure products (if is negative) or remain at pure reactants (if is positive). Instead, the system will spontaneously go to the equilibrium position, ¢G ¢G ¢G 25° ¢G ¢G°, ¢G A B A B C (a) (b) FIGURE 16.7 Schematic representations of balls rolling down two types of hills.
774 Chapter Sixteen Spontaneity, Entropy, and Free Energy the lowest possible free energy available to it. In the next section we will see that the value of for a particular reaction tells us exactly where this position will be.
16.8 Free Energy and Equilibrium When the components of a given chemical reaction are mixed, they will proceed, rapidly or slowly depending on the kinetics of the process, to the equilibrium position. In Chap-ter 13 we defined the equilibrium position as the point at which the forward and reverse reaction rates are equal. In this chapter we look at equilibrium from a thermodynamic point of view, and we find that the equilibrium point occurs at the lowest value of free energy available to the reaction system. As it turns out, the two definitions give the same equilibrium state, which must be the case for both the kinetic and thermodynamic mod-els to be valid.
To understand the relationship of free energy to equilibrium, let’s consider the fol-lowing simple hypothetical reaction: where 1.0 mole of gaseous A is initially placed in a reaction vessel at a pressure of 2.0 atm. The free energies for A and B are diagramed as shown in Fig. 16.8(a). As A reacts to form B, the total free energy of the system changes, yielding the following results: As A changes to B, GA will decrease because PA is decreasing [Fig. 16.8(b)]. In contrast, GB will increase because PB is increasing. The reaction will proceed to the right as long as the total free energy of the system decreases (as long as GB is less than GA). At some point the pressures of A and B reach the values Pe A and Pe B that make GA equal to GB.
The system has reached equilibrium [Fig. 16.8(c)]. Since A at pressure Pe A and B at pres-sure Pe B have the same free energy (GA equals GB), G is zero for A at pressure Pe A chang-ing to B at pressure Pe B. The system has reached minimum free energy. There is no longer any driving force to change A to B or B to A, so the system remains at this position (the pressures of A and B remain constant).
Suppose that for the experiment described above the plot of free energy versus the mole fraction of A reacted is defined as shown in Fig. 16.9(a). In this experiment, mini-mum free energy is reached when 75% of A has been changed to B. At this point, the pressure of A is 0.25 times the original pressure, or The pressure of B is 10.75212.0 atm2 1.5 atm 10.25212.0 atm2 0.50 atm Total free energy of system G GA GB Free energy of B GB G° B RT ln 1PB2 Free energy of A GA G° A RT ln 1PA2 A 1g2 ∆B 1g2 ¢G° GA GB (a) GA GB (b) (PA decreasing) (PB increasing) GA GB (c) G G G FIGURE 16.8 (a) The initial free energies of A and B.
(b) As A(g) changes to B(g), the free energy of A decreases and that of B increases. (c) Eventually, pressures of A and B are achieved such that GA GB, the equilibrium position.
0 Fraction of A reacted 0.5 1.0 G 0 Fraction of A reacted 0.5 1.0 G 0 Fraction of A reacted 0.5 1.0 G (a) (b) (c) Equilibrium occurs here Equilibrium occurs here Equilibrium occurs here FIGURE 16.9 (a) The change in free energy to reach equi-librium, beginning with 1.0 mol A(g) at PA 2.0 atm. (b) The change in free energy to reach equilibrium, beginning with 1.0 mol B(g) at PB 2.0 atm. (c) The free energy profile for A(g) B(g) in a system containing 1.0 mol (A plus B) at PTOTAL 2.0 atm. Each point on the curve corresponds to the total free energy of the system for a given combination of A and B.
∆ 16.8 Free Energy and Equilibrium 775 Since this is the equilibrium position, we can use the equilibrium pressures to calculate a value for K for the reaction in which A is converted to B at this temperature: Exactly the same equilibrium point would be achieved if we placed 1.0 mol pure B(g) in the flask at a pressure of 2.0 atm. In this case B would change to A until equilibrium is reached. This is shown in Fig. 16.9(b).
The overall free energy curve for this system is shown in Fig. 16.9(c). Note that any mixture of A(g) and B(g) containing 1.0 mol (A plus B) at a total pressure of 2.0 atm will react until it reaches the minimum in the curve.
In summary, when substances undergo a chemical reaction, the reaction proceeds to the minimum free energy (equilibrium), which corresponds to the point where We can now establish a quantitative relationship between free energy and the value of the equilibrium constant. We have seen that and at equilibrium equals 0 and Q equals K.
So or We must note the following characteristics of this very important equation.
Case 1: G 0. When G equals zero for a particular reaction, the free energies of the reactants and products are equal when all components are in the standard states (1 atm for gases). The system is at equilibrium when the pressures of all reactants and products are 1 atm, which means that K equals 1.
Case 2: G 0. In this case G (G products G reactants) is negative, which means that If a flask contains the reactants and products, all at 1 atm, the system will not be at equilibrium. Since is less than , the system will adjust to the right to reach equilibrium. In this case K will be greater than 1, since the pressures of the products at equilibrium will be greater than 1 atm and the pressures of the reactants at equilibrium will be less than 1 atm.
Case 3: Since is positive, If a flask contains the reactants and products, all at 1 atm, the system will not be at equilibrium. In this case the system will adjust to the left (toward the reactants, which have a lower free energy) to reach equilibrium. The value of K will be less than 1, since at equilibrium the pressures of the reactants will be greater than 1 atm and the pressures of the products will be less than 1 atm.
These results are summarized in Table 16.6. The value of K for a specific reaction can be calculated from the equation as is shown in Sample Exercises 16.14 and 16.15.
¢G° RT ln 1K2 G° reactants 6 G° products 1G° products G° reactants2 ¢G° ¢G° 7 0.
G° reactants G° products G° products 6 G° reactants 6 ¢G° RT ln 1K2 ¢G 0 ¢G° RT ln 1K2 ¢G ¢G ¢G° RT ln1Q2 Gproducts Greactants or ¢G Gproducts Greactants 0 (GB GA) K Pe B P e A 1.5 atm 0.50 atm 3.0 TABLE 16.6 Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction G° K K 6 1 ¢G° 7 0 K 7 1 ¢G° 6 0 K 1 ¢G° 0 For the reaction A(g) B(g), the pressure is constant during the reaction, since the same number of gas molecules is always present.
∆ 776 Chapter Sixteen Spontaneity, Entropy, and Free Energy Free Energy and Equilibrium I Consider the ammonia synthesis reaction where kJ per mole of N2 consumed at . For each of the following mixtures of reactants and products at , predict the direction in which the system will shift to reach equilibrium.
a.
b.
Solution a. We can predict the direction of reaction to equilibrium by calculating the value of using the equation where and Then Since the reactants and products have the same free energies at these partial pressures. The system is already at equilibrium, and no shift will occur.
b. The partial pressures given here are all 1.00 atm, which means that the system is in the standard state. That is, For this reaction at , The negative value for means that in their standard states the products have a lower free energy than the reactants. Thus the system will move to the right to reach equilibrium. That is, K is greater than 1.
See Exercise 16.59.
Free Energy and Equilibrium II The overall reaction for the corrosion (rusting) of iron by oxygen is 4Fe1s2 3O21g2 ∆2Fe2O31s2 ¢G° ¢G° 33.3 kJ/mol 25°C ¢G° RT ln 11.002 ¢G° 0 ¢G° ¢G ¢G° RT ln 1Q2 ¢G° RT ln 11.0022 11.002 11.0023 ¢G 0, 13.33 104 J/mol2 13.33 104 J/mol2 0 ¢G 13.33 104 J/mol2 18.3145 J/K mol2 1298 K2 ln 16.8 1052 ¢G° 33.3 kJ/mol 3.33 104 J/mol R 8.3145 J/K mol T 25 273 298 K Q PNH3 2 1PN221PH2 32 11.0022 11.47211.00 10223 6.80 105 ¢G ¢G° RT ln 1Q2 ¢G PNH3 1.00 atm, PN2 1.00 atm, PH2 1.00 atm PNH3 1.00 atm, PN2 1.47 atm, PH2 1.00 102 atm 25°C 25°C ¢G° 33.3 N21g2 3H2 1g2 ∆2NH31g2 Sample Exercise 16.14 Sample Exercise 16.15 The units of G, G°, and RT ln (Q) all refer to the balanced reaction with all amounts expressed in moles. We might say that the units are joules per “mole of reaction,” although only the “per mole” is indicated for R (as is customary).
16.8 Free Energy and Equilibrium 777 Using the following data, calculate the equilibrium constant for this reaction at .
25°C Substance S° (J/K mol) Fe2O3(s) 826 90 Fe(s) 0 27 O2(g) 0 205 Hf (kJ/mol) Solution To calculate K for this reaction, we will use the equation We must first calculate from where and Then and Thus and This is a very large equilibrium constant. The rusting of iron is clearly very favorable from a thermodynamic point of view.
See Exercise 16.62.
The Temperature Dependence of K In Chapter 13 we used Le Châtelier’s principle to predict qualitatively how the value of K for a given reaction would change with a change in temperature. Now we can specify the quantitative dependence of the equilibrium constant on temperature from the relationship We can rearrange this equation to give ln1K2 ¢H° RT ¢S° R ¢H° R a1 Tb ¢S° R ¢G° RT ln1K2 ¢H° T¢S° K e601 ln1K2 1.490 106 2.48 103 601 ¢G° RT ln1K2 1.490 106 J 18.3145 J/K mol21298 K2 ln1K2 1.490 106 J ¢G° ¢H° T¢S° 11.652 106 J2 1298 K21543 J/K2 T 273 25 298 K 543 J/K 2 mol190 J/K mol2 3 mol1205 J/K mol2 4 mol127 J/K mol2 ¢S° 2S° Fe2O3 3S° O2 4S° Fe 1652 kJ 1.652 106 J 2 mol1826 kJ/mol2 0 0 ¢H° 2¢Hf ° 1Fe2O31s22 3¢Hf ° 1O21g22 4¢Hf ° 1Fe1s22 ¢G° ¢H° T¢S° ¢G° ¢G° RT ln1K2 Formation of rust on bare steel is a sponta-neous process.
778 Chapter Sixteen Spontaneity, Entropy, and Free Energy Note that this is a linear equation of the form y mx b, where y ln(K), m HR slope, x 1T, and b SR intercept. This means that if values of K for a given re-action are determined at various temperatures, a plot of ln(K) versus 1T will be linear, with slope HR and intercept S R. This result assumes that both H and S are independent of temperature over the temperature range considered. This assumption is a good approximation over a relatively small temperature range.
16.9 Free Energy and Work One of the main reasons we are interested in physical and chemical processes is that we want to use them to do work for us, and we want this work done as efficiently and eco-nomically as possible. We have already seen that at constant temperature and pressure, the sign of the change in free energy tells us whether a given process is spontaneous. This is very useful information because it prevents us from wasting effort on a process that has no inherent tendency to occur. Although a thermodynamically favorable chemical reac-tion may not occur to any appreciable extent because it is too slow, it makes sense in this case to try to find a catalyst to speed up the reaction. On the other hand, if the reaction is prevented from occurring by its thermodynamic characteristics, we would be wasting our time looking for a catalyst.
In addition to its qualitative usefulness (telling us whether a process is spontaneous), the change in free energy is important quantitatively because it can tell us how much work can be done with a given process. In fact, the maximum possible useful work ob-tainable from a process at constant temperature and pressure is equal to the change in free energy: This relationship explains why this function is called the free energy. Under certain con-ditions, for a spontaneous process represents the energy that is free to do useful work.
On the other hand, for a process that is not spontaneous, the value of tells us the min-imum amount of work that must be expended to make the process occur.
Knowing the value of for a process thus gives us valuable information about how close the process is to 100% efficiency. For example, when gasoline is burned in a car’s engine, the work produced is about 20% of the maximum work available.
For reasons we will only briefly introduce in this book, the amount of work we actually obtain from a spontaneous process is always less than the maximum possible amount.
To explore this idea more fully, let’s consider an electric current flowing through the starter motor of a car. The current is generated from a chemical change in a battery, and we can calculate for the battery reaction and so determine the energy available to do work. Can we use all this energy to do work? No, because a current flowing through a wire causes frictional heating, and the greater the current, the greater the heat.
This heat represents wasted energy—it is not useful for running the starter motor. We can minimize this energy waste by running very low currents through the motor circuit.
However, zero current flow would be necessary to eliminate frictional heating entirely, and we cannot derive any work from the motor if no current flows. This represents the difficulty in which nature places us. Using a process to do work requires that some of the energy be wasted, and usually the faster we run the process, the more energy we waste.
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. If we could discharge the battery infinitely slowly by an infinitesimal current flow, we would achieve the maximum useful work. Also, if we could then recharge the battery using an infinitesimally small ¢G ¢G ¢G ¢G wmax ¢G Note that “PV work” is not counted as useful work here.
16.9 Free Energy and Work 779 current, exactly the same amount of energy would be used to return the battery to its original state. After we cycle the battery in this way, the universe (the system and surroundings) is exactly the same as it was before the cyclic process. This is a reversible process (see Fig. 16.10).
However, if the battery is discharged to run the starter motor and then recharged us-ing a finite current flow, as is the case in reality, more work will always be required to recharge the battery than the battery produces as it discharges. This means that even though the battery (the system) has returned to its original state, the surroundings have not, be-cause the surroundings had to furnish a net amount of work as the battery was cycled.
The universe is different after this cyclic process is performed, and this function is called an irreversible process. All real processes are irreversible.
In general, after any real cyclic process is carried out in a system, the surroundings have less ability to do work and contain more thermal energy. In other words, in any real cyclic process in the system, work is changed to heat in the surroundings, and the entropy of the universe increases. This is another way of stating the second law of thermody-namics.
Thus thermodynamics tells us the work potential of a process and then tells us that we can never achieve this potential. In this spirit, thermodynamicist Henry Bent has para-phrased the first two laws of thermodynamics as follows: First law: You can’t win, you can only break even.
Second law: You can’t break even.
The ideas we have discussed in this section are applicable to the energy crisis that will probably increase in severity over the next 25 years. The crisis is obviously not one of supply; the first law tells us that the universe contains a constant supply of energy. The problem is the availability of useful energy. As we use energy, we degrade its usefulness.
For example, when gasoline reacts with oxygen in the combustion reaction, the change in potential energy results in heat flow. Thus the energy concentrated in the bonds of the gasoline and oxygen molecules ends up spread over the surroundings as thermal energy, where it is much more difficult to harness for useful work. This is a way in which the entropy of the universe increases: Concentrated energy becomes spread out—more disor-dered and less useful. Thus the crux of the energy problem is that we are rapidly consuming the concentrated energy found in fossil fuels. It took millions of years to concentrate the sun’s energy in these fuels, and we will consume these same fuels in a few hundred years.
Thus we must use these energy sources as wisely as possible.
State 1: Fully charged State 1: Fully charged State 2: Discharged Work (w1) flows to starter Work (w2) flows to battery Alternator Starter motor Battery Battery is charged (energy flows into the battery) Discharge of battery (energy flows out of the battery) w2 w1 FIGURE 16.10 A battery can do work by sending current to a starter motor. The battery can then be recharged by forcing current through it in the opposite direction. If the current flow in both processes is infinitesimally small, w1 w2. This is a reversible process. But if the current flow is finite, as it would be in any real case, w2 w1. This is an irre-versible process (the universe is different after the cyclic process occurs). All real processes are irreversible.
7 When energy is used to do work, it be-comes less organized and less concen-trated and thus less useful.
780 Chapter Sixteen Spontaneity, Entropy, and Free Energy Key Terms Section 16.1 spontaneous process entropy positional probability Section 16.2 second law of thermodynamics Section 16.4 free energy Section 16.5 third law of thermodynamics Section 16.6 standard free energy change standard free energy of formation Section 16.8 equilibrium point (thermodynamic definition) Section 16.9 reversible process irreversible process For Review First law of thermodynamics States that the energy of the universe is constant Provides a way to keep track of energy as it changes form Gives no information about why a particular process occurs in a given direction Second law of thermodynamics States that for any spontaneous process there is always an increase in the entropy of the universe Entropy(S) is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) available to a system existing in a given state • Nature spontaneously proceeds toward states that have the highest probability of occurring • Using entropy, thermodynamics can predict the direction in which a process will occur spontaneously • For a spontaneous process, must be positive • For a process at constant temperature and pressure: • is dominated by “positional” entropy For a chemical reaction, is dominated by changes in the number of gaseous molecules • is determined by heat: is positive for an exothermic process ( is negative) Because depends inversely on T, exothermicity becomes a more impor-tant driving force at low temperatures Thermodynamics cannot predict the rate at which a system will spontaneously change; the principles of kinetics are necessary to do this Third law of thermodynamics States that the entropy of a perfect crystal at 0 K is zero Free energy (G) Free energy is a state function: A process occurring at constant temperature and pressure is spontaneous in the direction in which its free energy decreases For a reaction the standard free energy change is the change in free en-ergy that occurs when reactants in their standard states are converted to products in their standard states The standard free energy change for a reaction can be determined from the stan-dard free energies of formation of the reactants and products: Free energy depends on temperature and pressure: G G° RT ln P ¢G° ©np¢G° f1products2 ©nr ¢G° f1reactants2 (¢G° f) (¢G°) (¢G 6 0) G H TS ¢Ssurr ¢H ¢Ssurr ¢Ssurr ¢H T ¢Ssurr ¢Ssys ¢Ssys ¢Suniv ¢Suniv ¢Ssys ¢Ssurr For Review 781 • This relationship can be used to derive the relationship between for a reaction and the value of its equilibrium constant K: • • • The maximum possible useful work obtainable from a process at constant temper-ature and pressure is equal to the change in free energy: • In any real process, • When energy is used to do work in a real process, the energy of the universe remains constant but the usefulness of the energy decreases • Concentrated energy is spread out in the surroundings as thermal energy REVIEW QUESTIONS 1. Define the following: a. spontaneous process b. entropy c. positional probability d. system e. surroundings f. universe 2. What is the second law of thermodynamics? For any process, there are four pos-sible sign combinations for and . Which sign combination(s) always give a spontaneous process? Which sign combination(s) always give a nonspon-taneous process? Which sign combination(s) may or may not give a spontaneous process?
3. What determines for a process? To calculate at constant pressure and temperature, we use the following equation: . Why does a minus sign appear in the equation, and why is inversely proportional to temperature?
4. The free energy change, , for a process at constant temperature and pressure is related to and reflects the spontaneity of the process. How is re-lated to ? When is a process spontaneous? Nonspontaneous? At equilib-rium? is a composite term composed of , T, and . What is the equation? Give the four possible sign combinations for and . What tem-peratures are required for each sign combination to yield a spontaneous process? If is positive, what does it say about the reverse process? How does the equation reduce when at the melting-point temper-ature of a solid-to-liquid phase change or at the boiling-point temperature of a liquid-to-gas phase change? What is the sign of for the solid-to-liquid phase change at temperatures above the freezing point? What is the sign of for the liquid-to-gas phase change at temperatures below the boiling point?
5. What is the third law of thermodynamics? What are standard entropy values, and how are these values (listed in Appendix 4) used to calculate for a reaction? How would you use Hess’s law to calculate for a reaction? What does the superscript indicate?
Predicting the sign of for a reaction is an important skill to master. For a gas-phase reaction, what do you concentrate on to predict the sign of For a phase change, what do you concentrate on to predict the sign of ¢S°?
¢S°?
¢S° ° ¢S° ¢S° S° S°, ¢G ¢G ¢G ¢H T¢S ¢G ¢S ¢H ¢G ¢S ¢H ¢G ¢Suniv ¢G ¢Suniv ¢G ¢Ssurr ¢Ssurr ¢H T ¢Ssurr ¢Ssurr ¢Ssurr ¢Ssys w 6 wmax wmax ¢G For ¢G° 7 0, K 6 1 For ¢G° 6 0, K 7 1 For ¢G° 0, K 1 ¢G° RT ln K ¢G° 782 Chapter Sixteen Spontaneity, Entropy, and Free Energy That is, how are related to one another? When a solute dissolves in water, what is usually the sign of for this process?
6. What is the standard free energy change, for a reaction? What is the stan-dard free energy of formation, for a substance? How are values used to calculate How can you use Hess’s law to calculate How can you use and values to calculate Of the functions and which depends most strongly on temperature? When is calcu-lated at temperatures other than , what assumptions are generally made concerning and ?
7. If you calculate a value for for a reaction using the values of in Ap-pendix 4 and get a negative number, is it correct to say that the reaction is always spontaneous? Why or why not? Free energy changes also depend on concentration. For gases, how is G related to the pressure of the gas? What are standard pressures for gases and standard concentrations for solutes? How do you calculate for a reaction at nonstandard conditions? The equation to determine at nonstandard conditions has Q in it: What is Q? A reaction is spontaneous as long as is negative; that is, reactions always proceed as long as the products have a lower free energy than the reactants. What is so special about equilibrium? Why don’t reactions move away from equilibrium?
8. Consider the equation . What is the value of for a reaction at equilibrium? What does Q equal at equilibrium? At equilibrium, the previous equation reduces to . When what does it indicate about K? When what does it indicate about K? When what does it indicate about K? predicts spontaneity for a reaction, whereas predicts the equilibrium position. Explain what this statement means. Under what conditions can you use to determine the spontaneity of a reaction?
9. Even if is negative, the reaction may not occur. Explain the interplay be-tween the thermodynamics and the kinetics of a reaction. High temperatures are favorable to a reaction kinetically but may be unfavorable to a reaction thermo-dynamically. Explain.
10. Discuss the relationship between and the magnitude and sign of the free energy change for a reaction. Also discuss for real processes. What is a reversible process?
wmax wmax ¢G ¢G° ¢G° ¢G ¢G° 0, ¢G° 6 0, ¢G° 7 0, ¢G° RT ln(K) ¢G ¢G ¢G° RT ln(Q) ¢G ¢G ¢G ¢G° f ¢G° ¢S° ¢H° 25°C ¢G° ¢G°, ¢S°, ¢H°, ¢G° rxn?
¢S° ¢H° ¢G° rxn?
¢G° rxn?
¢G° f ¢G° f, ¢G°, ¢S° S° solid, S° liquid, and S° gas Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. For the process , which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.
2. For a liquid, which would you expect to be larger, Sfusion or Sevaporation? Why?
3. Gas A2 reacts with gas B2 to form gas AB at a constant tempera-ture. The bond energy of AB is much greater than that of either ¢ ¢ A(l) ¡ A(g) reactant. What can be said about the sign of H?
Ssurr?
S? Ex-plain how potential energy changes for this process. Explain how random kinetic energy changes during the process.
4. What types of experiments can be carried out to determine whether a reaction is spontaneous? Does spontaneity have any relationship to the final equilibrium position of a reaction? Explain.
5. A friend tells you, “Free energy G and pressure P are related by the equation . Also, G is related to the equi-librium constant K in that when the system is at equilibrium. Therefore, it must be true that a system is at equi-librium when all the pressures are equal.” Do you agree with this friend? Explain.
6. You remember that is related to RT ln(K) but cannot re-member if it’s RT ln(K) or RT ln(K). Realizing what and K mean, how can you figure out the correct sign?
¢G° ¢G° Gproducts Greactants, G G° RT ln(P) ¢ ¢ ¢ Exercises 783 17. List three different ways to calculate the standard free energy change, G°, for a reaction at 25°C? How is G° estimated at temperatures other than 25°C? What assumptions are made?
18. What information can be determined from G for a reaction? Does one get the same information from G°, the standard free energy change? G° allows determination of the equilibrium constant K for a reaction. How? How can one estimate the value of K at tem-peratures other than 25°C for a reaction? How can one estimate the temperature where K 1 for a reaction? Do all reactions have a specific temperature where K 1?
Exercises In this section similar exercises are paired.
Spontaneity, Entropy, and the Second Law of Thermodynamics: Free Energy 19. Which of the following processes are spontaneous?
a. Salt dissolves in H2O.
b. A clear solution becomes a uniform color after a few drops of dye are added.
c. Iron rusts.
d. You clean your bedroom.
20. Which of the following processes are spontaneous?
a. A house is built.
b. A satellite is launched into orbit.
c. A satellite falls back to earth.
d. The kitchen gets cluttered.
21. Consider the following energy levels, each capable of holding two objects: _ _ Draw all the possible arrangements of the two identical particles (represented by X) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? As-sume that the particles are indistinguishable from each other.
22. Redo Exercise 21 with two particles A and B, which can be dis-tinguished from each other.
23. Choose the compound with the greatest positional probability in each case.
a. 1 mol H2 (at STP) or 1 mol H2 (at , 0.5 atm) b. 1 mol N2 (at STP) or 1 mol N2 (at 100 K, 2.0 atm) c. 1 mol H2O(s) (at ) or 1 mol H2O(l) (at ) 24. Which of the following involve an increase in the entropy of the system?
a. melting of a solid b. sublimation c. freezing d. mixing e. separation f. boiling 25. Predict the sign of Ssurr for the following processes.
a.
b. CO2(g) ¡ CO2(s) H2O(l) ¡ H2O(g) ¢ 20°C 0°C 100°C XX E 0 E 1 kJ E 2 kJ A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 7. The synthesis of glucose directly from CO2 and H2O and the syn-thesis of proteins directly from amino acids are both nonsponta-neous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of ther-modynamics, how can life exist?
8. When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consis-tent with the second law of thermodynamics? In terms of the sec-ond law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contami-nation before it occurs? Explain.
9. A green plant synthesizes glucose by photosynthesis, as shown in the reaction Animals use glucose as a source of energy: If we were to assume that both these processes occur to the same extent in a cyclic process, what thermodynamic property must have a nonzero value?
10. Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Like-wise, the digital data sent from Voyager II contained one re-dundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of in-formation? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?
11. Entropy has been described as “time’s arrow.” Interpret this view of entropy.
12. A mixture of hydrogen gas and chlorine gas remains unreacted until it is exposed to ultraviolet light from a burning magnesium strip. Then the following reaction occurs very rapidly: Explain.
13. Table 16.1 shows the possible arrangements of four molecules in a two-bulbed flask. What are the possible arrangements if there is one molecule in this two-bulbed flask or two molecules or three molecules? For each, what arrangement is most likely?
14. Ssurr is sometime called the energy disorder term. Explain.
15. The third law of thermodynamics states that the entropy of a per-fect crystal at 0 K is zero. In Appendix 4, F(aq), OH(aq), and S2(aq) all have negative standard entropy values. How can S° val-ues be less than zero?
16. The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?
H21g2 Cl21g2 ¡ 2HCl1g2 C6H12O61s2 6O21g2 ¡ 6CO21g2 6H2O1l2 6CO21g2 6H2O1l2 ¡ C6H12O61s2 6O21g2 784 Chapter Sixteen Spontaneity, Entropy, and Free Energy 36. For each of the following pairs, which substance has the greater value of S?
a. N2O (at 0 K) or He (at 10 K) b. N2O(g) (at 1 atm, 25°C) or He(g) (at 1 atm, 25°C) c. H2O(s) (at 0°C) or H2O(l) (at 0°C) 37. Predict the sign of S and then calculate S for each of the fol-lowing reactions.
a. 2H2S(g) SO2(g) 8n 3Srhombic(s) 2H2O(g) b. 2SO3(g) 8n 2SO2(g) O2(g) c. Fe2O3(s) 3H2(g) 8n 2Fe(s) 3H2O(g) 38. Predict the sign of S and then calculate S for each of the fol-lowing reactions.
a. H2(g) O2(g) 8n H2O(l) b. 2CH3OH(g) 3O2(g) 8n 2CO2(g) 4H2O(g) c. HCl(g) 8n H(aq) Cl(aq) 39. For the reaction S is equal to 358 J/K. Use this value and data from Appendix 4 to calculate the value of S for CF4(g).
40. For the reaction S is equal to 144 J/K. Use this value and data from Appendix 4 to calculate the value of S for solid aluminum bromide.
41. It is quite common for a solid to change from one structure to an-other at a temperature below its melting point. For example, sul-fur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above 95C.
a. Predict the signs of H and S for the process Srhombic 8n Smonoclinic.
b. Which form of sulfur has the more ordered crystalline structure?
42. When most biologic enzymes are heated they lose their catalytic activity. The change that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered?
Explain.
43. Consider the reaction a. Predict the signs of H and S.
b. Would the reaction be more spontaneous at high or low temperatures?
44. Hydrogen cyanide is produced industrially by the following exothermic reaction: Is the high temperature needed for thermodynamic or kinetic reasons?
45. From data in Appendix 4, calculate H, S, and G for each of the following reactions at 25C.
2HCN1g2 6H2O1g2 ¬ ¡ 2NH31g2 3O21g2 2CH41g2 2O1g2 ¡ O21g2 original enzyme ¡ new form 2Al1s2 3Br21l2 ¡ 2AlBr31s2 C2H21g2 4F21g2 ¡ 2CF41g2 H21g2 1 2 26. Calculate Ssurr for the following reactions at and 1 atm.
a. C3H8(g) 5O2(g) 8n 3CO2(g) 4H2O(l) H° 2221 kJ b. 2NO2(g) 8n 2NO(g) O2(g) H° 112 kJ 27. Given the values of and S, which of the following changes will be spontaneous at constant T and P?
a.
b.
c.
d.
28. At what temperatures will the following processes be sponta-neous?
a.
b.
c.
d.
29. Ethanethiol (C2H5SH; also called ethyl mercaptan) is commonly added to natural gas to provide the “rotten egg” smell of a gas leak. The boiling point of ethanethiol is and its heat of va-porization is 27.5 kJ/mol. What is the entropy of vaporization for this substance?
30. For mercury, the enthalpy of vaporization is 58.51 kJ/mol and the entropy of vaporization is 92.92 J/K mol. What is the normal boiling point of mercury?
31. For ammonia (NH3), the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 J/K mol.
a. Will NH3(s) spontaneously melt at 200. K?
b. What is the approximate melting point of ammonia?
32. The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boil-ing point Determine when 1.00 mol of ethanol is vaporized at and 1.00 atm.
Chemical Reactions: Entropy Changes and Free Energy 33. Predict the sign of S for each of the following changes.
a.
b. AgCl(s) 8n Ag(aq) Cl(aq) c. 2H2(g) O2(g) 8n 2H2O(l) d. H2O(l) 8n H2O(g) 34. Predict the sign of S for each of the following changes.
a. Na(s) Cl2(g) 8n NaCl(s) b. N2(g) 3H2(g) 8n 2NH3(g) c. NaCl(s) 8n Na(aq) Cl(aq) d. NaCl(s) 8n NaCl(l) 35. For each of the following pairs of substances, which substance has the greater value of S?
a. Cgraphite(s) or Cdiamond(s) b. C2H5OH(l) or C2H5OH(g) c. CO2(s) or CO2(g) 1 2 78°C ¢Ssys, ¢Ssurr, and ¢Suniv (78°C).
35°C ¢H 18 kJ and ¢S 60. J/K ¢H 18 kJ and ¢S 60. J/K ¢H 18 kJ and ¢S 60. J/K ¢H 18 kJ and ¢S 60. J/K ¢H 10. kJ, ¢S 40. J/K, T 200. K ¢H 10. kJ, ¢S 5.0 J/K, T 298 K ¢H 25 kJ, ¢S 100. J/K, T 300. K ¢H 25 kJ, ¢S 5.0 J/K, T 300. K ¢ ¢H 25°C ¢ 1000ºC Pt-Rh Exercises 785 calculate G for the reaction 53. For the reaction the value of is 374 kJ. Use this value and data from Ap-pendix 4 to calculate the value of for SF4(g).
54. The value of for the reaction is 5490. kJ. Use this value and data from Appendix 4 to calcu-late the standard free energy of formation for C4H10(g).
55. Assuming standard conditions, can the following reaction take place at room temperature?
56. Consider the reaction a. Calculate for this reaction. The values for POCl3(g) and PCl3(g) are 502 kJ/mol and 270. kJ/mol, respectively.
b. Is this reaction spontaneous under standard conditions at 298 K?
c. The value of for this reaction is 179 J/K. At what tem-peratures is this reaction spontaneous at standard conditions?
Assume that and do not depend on temperature.
Free Energy: Pressure Dependence and Equilibrium 57. Using data from Appendix 4, calculate G for the reaction for these conditions: 58. Using data from Appendix 4, calculate G for the reaction for the following conditions at 25C: 59. Consider the reaction For each of the following mixtures of reactants and products at 25C, predict the direction in which the reaction will shift to reach equilibrium.
a. PNO2 PN2O4 1.0 atm b. PNO2 0.21 atm, PN2O4 0.50 atm c. PNO2 0.29 atm, PN2O4 1.6 atm 2NO21g2 ∆N2O41g2 PH2O 3.0 102 atm PSO2 1.0 102 atm PH2S 1.0 104 atm 2H2S1g2 SO21g2 ∆3Srhombic1s2 2H2O1g2 PNO2 1.00 107 atm, PO2 1.00 103 atm PNO 1.00 106 atm, PO3 2.00 106 atm T 298 K NO1g2 O31g2 ¡ NO21g2 O21g2 ¢S° ¢H° ¢S° ¢Gf ° ¢G° 2POCl31g2 ¡ 2PCl31g2 O21g2 ¢G° 1HCl2 95.30 kJ/mol ¢G° 1CH2Cl22 68.85 kJ/mol ¢G° 1CH3Cl2 57.37 kJ/mol ¢G° 1CH42 50.72 kJ/mol 3Cl21g2 2CH41g2 ¡ CH3Cl1g2 CH2Cl21g2 3HCl1g2 2C4H101g2 13O21g2 ¡ 8CO21g2 10H2O1l2 ¢G° ¢G° f ¢G° SF41g2 F21g2 ¡ SF61g2 6C1s2 3H21g2 ¡ C6H61l2 a. CH4(g) 2O2(g) 8n CO2(g) 2H2O(g) b. 6CO2(g) 6H2O(l) 8n C6H12O6(s) 6O2(g) Glucose c. P4O10(s) 6H2O(l) 8n 4H3PO4(s) d. HCl(g) NH3(g) 8n NH4Cl(s) 46. The decomposition of ammonium dichromate [(NH4)2Cr2O7] is called the “volcano” demonstration for its fiery display. The de-composition reaction involves breaking down ammonium dichromate into nitrogen gas, water vapor, and solid chromium(III) oxide. From the data in Appendix 4 and given kJ/mol and J/K mol for (NH4)2Cr2O7, calculate G° for the “volcano” reaction and calculate for ammonium dichromate.
47. For the reaction at 298 K, the values of H and S are 58.03 kJ and 176.6 J/K, re-spectively. What is the value of G at 298 K? Assuming that H and S do not depend on temperature, at what temperature is G 0? Is G negative above or below this temperature?
48. At and 1.00 atm, kJ/mol for the vaporization of water. Estimate for the vaporization of water at and Assume and at and 1.00 atm do not de-pend on temperature.
49. Using data from Appendix 4, calculate and for the following reactions that produce acetic acid: Which reaction would you choose as a commercial method for producing acetic acid (CH3CO2H) at standard conditions? What temperature conditions would you choose for the reaction? As-sume and do not depend on temperature.
50. Consider two reactions for the production of ethanol: Which would be the more thermodynamically feasible at standard conditions? Why?
51. Given the following data: Calculate for 52. Given the following data: ¢G° 237 kJ H21g2 1 2O21g2 ¡ H2O1l2 ¢G° 394 kJ C1s2 O21g2 ¡ CO21g2 ¢G° 6399 kJ 2C6H61l2 15O21g2 ¡ 12CO21g2 6H2O1l2 CH41g2 2O21g2 S CO21g2 2H2O1l2.
¢G° ¢G° 394 kJ C1s2 O21g2 ¡ CO21g2 ¢G° 474 kJ 2H21g2 O21g2 ¡ 2H2O1l2 ¢G° 51 kJ 2H21g2 C1s2 ¡ CH41g2 C2H61g2 H2O1g2 ¡ CH3CH2OH1l2 H21g2 C2H41g2 H2O1g2 ¡ CH3CH2OH1l2 ¢S° ¢H° ¢G° ¢S°, ¢H°, 100.°C ¢S° ¢H° 110.°C.
90.°C ¢G° ¢H° 40.6 100.°C 2NO21g2 ∆N2O41g2 ¢G° f ¢S° 114 ¢H° f 23 786 Chapter Sixteen Spontaneity, Entropy, and Free Energy 67. Consider the relationship: The equilibrium constant for some hypothetical process was de-termined as a function of temperature (in Kelvin) with the results plotted below.
ln1K2 ¢H° RT ¢S° R 60. Consider the following reaction: Calculate G for this reaction under the following conditions (as-sume an uncertainty of 1 in all quantities): a. T 298 K, PN2 PH2 200 atm, PNH3 50 atm b. T 298 K, PN2 200 atm, PH2 600 atm, PNH3 200 atm 61. Consider the following reaction at 25.0C: The values of H and S are 58.03 kJ/mol and 176.6 J/K mol, respectively. Calculate the value of K at 25.0C.
Assuming H and S are temperature independent, estimate the value of K at 100.0C.
62. Consider the reaction a. Calculate H, S, G, and K (at 298 K) using data in Ap-pendix 4.
b. If H2(g), Cl2(g), and HCl(g) are placed in a flask such that the pressure of each gas is 1 atm, in which direction will the sys-tem shift to reach equilibrium at 25C?
63. Calculate for at 600. K, us-ing the following data: 64. The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate H, S, G, and K (at 298 K) for each of the three steps in the Ostwald process (see Appendix 4).
b. Calculate the equilibrium constant for the first step at 825C, assuming H and S do not depend on temperature.
c. Is there a thermodynamic reason for the high temperature in the first step assuming standard conditions?
65. Consider the following reaction at 800. K: An equilibrium mixture contains the following partial pressures: PN2 0.021 atm, PF2 0.063 atm, PNF3 0.48 atm. Calculate G for the reaction at 800. K.
66. Consider the following reaction at 298 K: An equilibrium mixture contains O2(g) and SO3(g) at partial pres-sures of 0.50 atm and 2.0 atm, respectively. Using data from Ap-pendix 4, determine the equilibrium partial pressure of SO2 in the mixture. Will this reaction be most favored at a high or a low tem-perature, assuming standard conditions?
2SO21g2 O21g2 ¡ 2SO31g2 N21g2 3F21g2 ¡ 2NF31g2 3NO21g2 H2O1l2 ¡ 2HNO31l2 NO1g2 2NO1g2 O21g2 ¡ 2NO21g2 4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 K 1.8 1037 at 600. K 2H21g2 O21g2 ∆2H2O1g2 K 2.3 106 at 600. K H21g2 O21g2 ∆H2O21g2 H2O1g2 1 2O21g2 ∆H2O21g2 ¢G° H21g2 Cl21g2 ∆2HCl1g2 2NO21g2 ∆N2O41g2 N21g2 3H21g2 ∆2NH31g2 1000 T(K) ln (K) 10.
1.0 2.0 3.0 20.
30.
40.
Pt 825ºC From the plot, determine the values of H and S for this process. What would be the major difference in the ln(K) versus 1T plot for an endothermic process as compared to an exothermic process?
68. The equilibrium constant K for the reaction was measured as a function of temperature (Kelvin). A graph of ln K versus for this reaction gives a straight line with a slope of K and a of Deter-mine the values of and for this reaction. Reference Exercise 67.
Additional Exercises 69. Using Appendix 4 and the following data, determine for Fe(CO)5(g).
70. Some water is placed in a coffee-cup calorimeter. When 1.0 g of an ionic solid is added, the temperature of the solution increases from to as the solid dissolves. For the dissolving process, what are the signs for Ssys, Ssurr, and Suniv?
71. Consider the following system at equilibrium at What will happen to the ratio of partial pressure of PCl5 to par-tial pressure of PCl3 if the temperature is raised? Explain com-pletely.
72. Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data.
¢G° 92.50 kJ PCl31g2 Cl21g2 ∆PCl51g2 25°C: 24.2°C 21.5°C ¢S° 677 J/K Fe1s25CO1g2 ¡ Fe1CO251l2 ¢S° 107 J/K Fe1CO251l2 ¡ Fe1CO251g2 ¢S° ?
Fe1s2 5CO1g2 ¡ Fe1CO251g2 S° ¢S° ¢H° 14.51.
y-intercept 1.352 104 1 T 2Cl1g2 ∆Cl21g2 Additional Exercises 787 77. In the text the equation was derived for gaseous reactions where the quantities in Q were expressed in units of pressure. We also can use units of mol/L for the quantities in Q, specifically for aqueous reactions. With this in mind, consider the reaction for which at . Calculate G for the reac-tion under the following conditions at .
a.
b.
c.
d.
e.
Based on the calculated values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?
78. Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion (K). The concentration of K in muscle cells is about 0.15 M. The concentration of K in blood plasma is about 0.0050 M. The high internal concen-tration in cells is maintained by pumping K from the plasma.
How much work must be done to transport 1.0 mol K from the blood to the inside of a muscle cell at , normal body temperature? When 1.0 mol K is transferred from blood to the cells, do any other ions have to be transported? Why or why not?
79. Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be writ-ten as where ADP represents adenosine diphosphate. For this reaction, .
a. Calculate K at .
b. If all the free energy from the metabolism of glucose goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?
c. Much of the ATP formed from metabolic processes is used to provide energy for transport of cellular components. What amount (mol) of ATP must be hydrolyzed to provide the energy for the transport of 1.0 mol of K from the blood to the inside of a muscle cell at as described in Exer-cise 78?
80. One reaction that occurs in human metabolism is Glutamic acid O Glutamine NH2 H2NC ‘ CH2CH2C ƒ HCO2H1aq2 H2O1l2 NH2 HO2CCH2CH2C ƒ HCO2H1aq2 NH31aq2 ∆ 37°C C6H12O61s2 6O21g2 ¡ 6CO21g2 6H2O1l2 25°C ¢G° 30.5 kJ/mol ATP1aq2 H2O1l2 ¡ ADP1aq2 H2PO4 1aq2 37°C ¢G 3HF4 0.52 M, 3F4 0.67 M, 3H4 1.0 103 M 3HF4 3F4 0.27 M, 3H4 7.2 104 M 3HF4 3H4 3F4 1.0 105 M 3HF4 0.98 M, 3H4 3F4 2.7 102 M 3HF4 3H4 3F4 1.0 M 25°C ¢ 25°C Ka 7.2 104 HF1aq2 ∆H1aq2 F1aq2 ¢G ¢G° RT ln1Q2 Compare the molar volume of gaseous methane at 112 K with that of gaseous hexane at 342 K. How do the differences in mo-lar volume affect the values of Svap for these liquids?
73. As O2(l) is cooled at 1 atm, it freezes at 54.5 K to form solid I.
At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that for the phase transition is J/mol, and for the same transition is J/K mol. At what temperature are solids I and II in equilibrium?
74. Consider the following reaction: For Cl2O(g), a. Calculate for the reaction using the equation ln(K).
b. Use bond energy values (Table 8.4) to estimate for the reaction.
c. Use the results from parts a and b to estimate for the re-action.
d. Estimate and for HOCl(g).
e. Estimate the value of K at 500. K.
f. Calculate at when and .
75. Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O2. Consider the following reactions and approximate standard free energy changes: Using these data, estimate the equilibrium constant value at for the following reaction: 76. Using the following data, calculate the value of Ksp for Ba(NO3)2, one of the least soluble of the common nitrate salts.
HgbO2 CO ∆HgbCO O2 25°C Hgb CO ¡ HgbCO ¢G° 80 kJ Hgb O2 ¡ HgbO2 ¢G° 70 kJ PHOCl 0.10 torr PCl2O 2.0 torr, PH2O 18 torr, 25°C ¢G S° ¢H° f ¢S° ¢H° RT ¢G° ¢G° S° 266.1 J/K mol ¢H° f 80.3 kJ/mol ¢G° f 97.9 kJ/mol H2O1g2 Cl2O1g2 ∆2HOCl1g2 K298 0.090 17.0 ¢S 743.1 I S II ¢H ¢ Boiling Point (1 atm) Hvap Methane 112 K 8.20 kJ/mol Hexane 342 K 28.9 kJ/mol Species G f 797 kJ/mol Ba1NO3221s2 109 kJ/mol NO3 1aq2 561 kJ/mol Ba21aq2 788 Chapter Sixteen Spontaneity, Entropy, and Free Energy where J/K and W is the number of ways a particular state can be obtained. (This equation is engraved on Boltzmann’s tombstone.) Calculate S for the three arrangements of particles in Table 16.1.
87. a. Using the free energy profile for a simple one-step reaction, show that at equilibrium where kf and kr are the rate constants for the forward and reverse reactions. Hint: Use the relationship ln(K) and represent kf and kr using the Arrhenius equation (k AeEa RT).
¢G° RT K kf kr, k 1.38 1023 For this reaction .
a. Calculate K for this reaction at .
b. In a living cell this reaction is coupled with the hydrolysis of ATP. (See Exercise 79.) Calculate and K at for the following reaction: 81. Consider the reactions (1) (2) where The H values for the two reactions are quite similar, yet Kreaction 2 Kreaction 1. Explain.
82. Use the equation in Exercise 67 to determine H and S for the autoionization of water: T(C) Kw 0 1.14 1015 25 1.00 1014 35 2.09 1014 40.
2.92 1014 50.
5.47 1014 83. Consider the reaction Assuming and do not depend on temperature, calculate the temperature where for this reaction.
Challenge Problems 84. Liquid water at is introduced into an evacuated, insulated ves-sel. Identify the signs of the following thermodynamic functions for the process that occurs: 85. Using data from Appendix 4, calculate and K (at 298 K) for the production of ozone from oxygen: At 30 km above the surface of the earth, the temperature is about 230. K and the partial pressure of oxygen is about atm.
Estimate the partial pressure of ozone in equilibrium with oxygen at 30 km above the earth’s surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.
86. Entropy can be calculated by a relationship proposed by Ludwig Boltzmann: S k ln 1W2 1.0 103 3O21g2 ∆2O31g2 ¢G°, ¢H°, ¢H, ¢S, ¢Twater, ¢Ssurr, ¢Suniv.
25°C K 1.00 ¢S° ¢H° Fe2O31s2 3H21g2 ¡ 2Fe1s2 3H2O1g2 H2O1l2 ∆H1aq2 OH1aq2 en H2N¬CH2¬CH2¬NH2 Ni21aq2 3en1aq2 ¡ Ni1en23 21aq2 Ni21aq2 6NH31aq2 ¡ Ni1NH326 21aq2 Glutamine1aq2 ADP1aq2 H2PO4 1aq2 Glutamic acid1aq2 ATP1aq2 NH31aq2 ∆ 25°C ¢G° 25°C ¢G° 14 kJ at 25°C Reaction coordinate G Reactants Ea (forward) ∆G° Ea (reverse) Products b. Why is the following statement false? “A catalyst can increase the rate of a forward reaction but not the rate of the reverse re-action.” 88. Consider the reaction where kJ/mol. In a particular experiment, equal moles of H2(g) at 1.00 atm and Br2(g) at 1.00 atm were mixed in a 1.00-L flask at and allowed to reach equilibrium. Then the molecules of H2 at equilibrium were counted using a very sensi-tive technique, and molecules were found. For this reaction, calculate the values of K, and .
89. Consider the system at .
a. Assuming that and , cal-culate the value of the equilibrium constant for this reaction.
b. Calculate the equilibrium pressures that result if 1.00 mol A(g) at 1.00 atm and 1.00 mol B(g) at 1.00 atm are mixed at .
c. Show by calculations that at equilibrium.
90. The equilibrium constant for a certain reaction decreases from 8.84 to when the temperature increases from to . Estimate the temperature where for this reac-tion. Estimate the value of for this reaction. Hint: Manipu-late the equation in Exercise 67.
91. If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent sil-ver carbonate from decomposing by the reaction for this reaction is 79.14 kJ/mol in the temperature range of 25 to . Given that the partial pressure of carbon dioxide in 125°C ¢H° Ag2CO3 1s2 ∆Ag2O 1s2 CO21g2 ¢S° K 1.00 75°C 25°C 3.25 102 ¢G 0 25°C G° B 11,718 J/mol G° A 8996 J/mol 25°C A1g2 ¡ B1g2 ¢S° ¢G°, 1.10 1013 25°C ¢H° 103.8 H21g2 Br21g2 ∆2HBr1g2 Marathon Problem 789 98. What is the pH of a 0.125 M solution of the weak base B if kJ and J/K for the following equi-librium reaction at Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
99. Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is a. Without referring to Appendix 4, predict the sign of for the above reaction. Explain.
b. The spontaneity of the above reaction is temperature depen-dent. Predict the sign of Ssurr for this reaction. Explain.
c. For Ni(CO)4(g), kJ/mol and mol at 298 K. Using these values and data in Appendix 4, cal-culate and for the above reaction.
d. Calculate the temperature at which for the above reaction, assuming that and do not depend on temperature.
e. The first step of the Mond process involves equilibrating im-pure nickel with CO(g) and Ni(CO)4(g) at about . The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at .
f. In the second step of the Mond process, the gaseous Ni(CO)4 is isolated and heated to . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the above reaction). Calculate the equilibrium constant for the above reaction at .
g. Why is temperature increased for the second step of the Mond process?
h. The Mond process relies on the volatility of Ni(CO)4 for its success. Only pressures and temperatures at which Ni(CO)4 is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a tem-perature of . Estimate the maximum pressure of Ni(CO)4(g) that can be attained before the gas will liquefy at . The boiling point for Ni(CO)4 is and the enthalpy of vaporization is 29.0 kJ/mol.
[Hint: The phase change reaction and the corresponding equi-librium expression are Ni(CO)4(g) will liquefy when the pressure of Ni(CO)4 is greater than the K value.] Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
Ni1CO24 1l2 ∆Ni1CO241g2 K PNi1CO24 42°C 152°C 152°C 227°C 227°C 50.°C 50°C ¢S° H° ¢G° 0 (K 1) ¢S° ¢H° S° 417 J K ¢H° f 607 ¢ ¢S° Ni 1s2 4CO1g2 ∆Ni1CO241g2 B1aq2 H2O1l2 ÷ BH1aq2 OH1aq2 25°C?
¢S° 175 ¢H° 28.0 equilibrium with pure solid silver carbonate is torr at , calculate the partial pressure of CO2 necessary to pre-vent decomposition of Ag2CO3 at . Hint: Manipulate the equation in Exercise 67.
92. Carbon tetrachloride (CCl4) and benzene (C6H6) form ideal solu-tions. Consider an equimolar solution of CCl4 and C6H6 at The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.
Substance G f C6H6(l) 124.50 kJ/mol C6H6(g) 129.66 kJ/mol CCl4(l) 65.21 kJ/mol CCl4(g) 60.59 kJ/mol 93. Some nonelectrolyte solute was dis-solved in 150. mL of a solvent . The el-evated boiling point of the solution was 355.4 K. What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 kJ/mol, the entropy of vaporization is 95.95 J/K mol, and the boiling-point elevation constant is 2.5 K kg/mol.
94. You have a l.00-L sample of hot water sitting open in a room. Eventually the water cools to while the tem-perature of the room remains unchanged. Calculate for this process. Assume the density of water is 1.00 g/cm3 over this tem-perature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 95. Consider a weak acid, HX. If a 0.10 M solution of HX has a pH of 5.83 at what is for the acid’s dissociation reaction at 96. Sodium chloride is added to water (at ) until it is saturated.
Calculate the Cl concentration in such a solution.
Species G(kJ/mol) NaCl(s) Na+(aq) Integrative Problems These problems require the integration of multiple concepts to find the solutions.
97. For the equilibrium the initial concentrations are atm.
Once equilibrium has been established, it is found that atm. What is for this reaction at 25°C?
¢G° 0.040 [C] [A] [B] [C] 0.100 A1g2 2B1g2 ∆C1g2 131 Cl1aq2 262 384 25°C 25°C?
¢G° 25°C, J/K mol.
¢Ssurr 250°C 25.0°C (90.0°C) (density 0.879 g/cm3) (molar mass 142 g/mol) 25°C.
110.°C 25°C 6.23 103 790 17 Electrochemistry Contents 17.1 Galvanic Cells • Cell Potential 17.2 Standard Reduction Potentials • Line Notation • Complete Description of a Galvanic Cell 17.3 Cell Potential, Electrical Work, and Free Energy 17.4 Dependence of Cell Potential on Concentration • Concentration Cells • The Nernst Equation • Ion-Selective Electrodes • Calculation of Equilibrium Constants for Redox Reactions 17.5 Batteries • Lead Storage Battery • Other Batteries • Fuel Cells 17.6 Corrosion • Corrosion of Iron • Prevention of Corrosion 17.7 Electrolysis • Electrolysis of Water • Electrolysis of Mixtures of Ions 17.8 Commercial Electrolytic Processes • Production of Aluminum • Electrorefining of Metals • Metal Plating • Electrolysis of Sodium Chrloride A nickel half-electroplated with copper.
Electrochemistry constitutes one of the most important interfaces between chemistry and everyday life. Every time you start your car, turn on your calculator, look at your digital watch, or listen to a radio at the beach, you are depending on electrochemical reactions. Our society sometimes seems to run almost entirely on batteries. Certainly the advent of small, dependable batteries along with silicon-chip technology has made possi-ble the tiny calculators, tape recorders, and clocks that we take for granted.
Electrochemistry is important in other less obvious ways. For example, the corro-sion of iron, which has tremendous economic implications, is an electrochemical process. In addition, many important industrial materials such as aluminum, chlorine, and sodium hydroxide are prepared by electrolytic processes. In analytical chemistry, electrochemical techniques employ electrodes that are specific for a given molecule or ion, such as H (pH meters), F, Cl, and many others. These increasingly important methods are used to analyze for trace pollutants in natural waters or for the tiny quantities of chemicals in human blood that may signal the development of a specific disease.
Electrochemistry is best defined as the study of the interchange of chemical and electrical energy. It is primarily concerned with two processes that involve oxidation– reduction reactions: the generation of an electric current from a spontaneous chemical reaction and the opposite process, the use of a current to produce chemical change.
17.1 Galvanic Cells As we discussed in detail in Section 4.9, an oxidation–reduction (redox) reaction in-volves a transfer of electrons from the reducing agent to the oxidizing agent. Recall that oxidation involves a loss of electrons (an increase in oxidation number) and that reduc-tion involves a gain of electrons (a decrease in oxidation number).
To understand how a redox reaction can be used to generate a current, let’s consider the reaction between MnO4 and Fe2: In this reaction, Fe2 is oxidized and MnO4 is reduced; electrons are transferred from Fe2 (the reducing agent) to MnO4 (the oxidizing agent).
It is useful to break a redox reaction into half-reactions, one involving oxidation and one involving reduction. For the reaction above, the half-reactions are The multiplication of the second half-reaction by 5 indicates that this reaction must occur five times for each time the first reaction occurs. The balanced overall reaction is the sum of the half-reactions.
When MnO4 and Fe2 are present in the same solution, the electrons are transferred directly when the reactants collide. Under these conditions, no useful work is obtained from the chemical energy involved in the reaction, which instead is released as heat. How can we harness this energy? The key is to separate the oxidizing agent from the reducing agent, thus requiring the electron transfer to occur through a wire. The current produced in the wire by the electron flow can then be directed through a device, such as an electric motor, to provide useful work.
Oxidation: 51Fe2 ¡ Fe3 e2 Reduction: 8H MnO4 5e ¡ Mn2 4H2O 8H1aq2 MnO4 1aq2 5Fe21aq2 ¡ Mn21aq2 5Fe31aq2 4H2O1l2 791 Balancing half-reactions is discussed in Section 4.10.
792 Chapter Seventeen Electrochemistry For example, consider the system illustrated in Fig. 17.1. If our reasoning has been correct, electrons should flow through the wire from Fe2 to MnO4 . However, when we construct the apparatus as shown, no flow of electrons is apparent. Why? Careful obser-vation shows that when we connect the wires from the two compartments, current flows for an instant and then ceases. The current stops flowing because of charge buildups in the two compartments. If electrons flowed from the right to the left compartment in the apparatus as shown, the left compartment (receiving electrons) would become negatively charged, and the right compartment (losing electrons) would become positively charged.
Creating a charge separation of this type requires a large amount of energy. Thus sus-tained electron flow cannot occur under these conditions.
However, we can solve this problem very simply. The solutions must be connected so that ions can flow to keep the net charge in each compartment zero. This connec-tion might involve a salt bridge (a U-tube filled with an electrolyte) or a porous disk in a tube connecting the two solutions (see Fig. 17.2). Either of these devices allows ions to flow without extensive mixing of the solutions. When we make the provision for ion flow, the circuit is complete. Electrons flow through the wire from reducing agent to oxidizing agent, and ions flow from one compartment to the other to keep the net charge zero.
We now have covered all the essential characteristics of a galvanic cell, a device in which chemical energy is changed to electrical energy. (The opposite process is called electrolysis and will be considered in Section 17.7.) The reaction in an electrochemical cell occurs at the interface between the electrode and the solution where the electron transfer occurs. The electrode compartment in which oxidation occurs is called the anode; the electrode compartment in which reduction oc-curs is called the cathode (see Fig. 17.3).
MnO4 –(aq) Fe2+(aq) H+(aq) Wire FIGURE 17.1 Schematic of a method to separate the oxidizing and reducing agents of a redox reaction. (The solutions also contain counterions to balance the charge.) (a) Salt bridge Porous disk (b) FIGURE 17.2 Galvanic cells can contain a salt bridge as in (a) or a porous-disk connection as in (b). A salt bridge contains a strong elec-trolyte held in a Jello-like matrix. A porous disk contains tiny passages that allow hindered flow of ions.
A galvanic cell uses a spontaneous redox reaction to produce a current that can be used to do work.
Oxidation occurs at the anode. Reduction occurs at the cathode.
17.1 Galvanic Cells 793 Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The “pull,” or driving force, on the electrons is called the cell potential or the electromotive force (emf) of the cell. The unit of electrical potential is the volt (abbreviated V), which is defined as 1 joule of work per coulomb of charge transferred.
How can we measure the cell potential? One possible instrument is a crude voltmeter, which works by drawing current through a known resistance. However, when current flows through a wire, the frictional heating that occurs wastes some of the potentially useful en-ergy of the cell. A traditional voltmeter will therefore measure a potential that is less than the maximum cell potential. The key to determining the maximum potential is to do the measurement under conditions of zero current so that no energy is wasted. Traditionally, this has been accomplished by inserting a variable-voltage device (powered from an ex-ternal source) in opposition to the cell potential. The voltage on this instrument (called a potentiometer) is adjusted until no current flows in the cell circuit. Under such condi-tions, the cell potential is equal in magnitude and opposite in sign to the voltage setting of the potentiometer. This value represents the maximum cell potential, since no energy is wasted heating the wire. More recently, advances in electronic technology have allowed the design of digital voltmeters that draw only a negligible amount of current (see Fig. 17.4).
Since these instruments are more convenient to use, they have replaced potentiometers in the modern laboratory.
1ecell2, FIGURE 17.4 Digital voltmeters draw only a negligible current and are convenient to use.
Porous disk e– e– e– e– e– e– (b) Cathode (a) Anode Reducing agent Oxidizing agent FIGURE 17.3 An electrochemical process involves elec-tron transfer at the interface between the electrode and the solution. (a) The species in the solution acting as the reducing agent supplies electrons to the anode. (b) The species in the solution acting as the oxidizing agent receives electrons from the cathode.
Digital voltmeter A volt is 1 joule of work per coulomb of charge transferred: 1 V 1 J/C.
Visualization: Voltaic Cell: Cathode Reaction Visualization: Voltaic Cell: Anode Reaction 794 Chapter Seventeen Electrochemistry 17.2 Standard Reduction Potentials The reaction in a galvanic cell is always an oxidation–reduction reaction that can be broken down into two half-reactions. It would be convenient to assign a potential to each half-reaction so that when we construct a cell from a given pair of half-reactions we can obtain the cell potential by summing the half-cell potentials. For example, the observed potential for the cell shown in Fig. 17.5(a) is 0.76 V, and the cell reac-tion is For this cell, the anode compartment contains a zinc metal electrode with Zn2 and SO4 2 ions in aqueous solution. The anode reaction is the oxidation half-reaction: The zinc metal, in producing Zn2 ions that go into solution, is giving up electrons, which flow through the wire. For now, we will assume that all cell components are in their stan-dard states, so in this case the solution in the anode compartment will contain 1 M Zn2.
The cathode reaction of this cell is the reduction half-reaction: The cathode consists of a platinum electrode (used because it is a chemically inert conductor) in contact with 1 M H ions and bathed by hydrogen gas at 1 atm. Such an electrode, called the standard hydrogen electrode, is shown in Fig. 17.5(b).
Although we can measure the total potential of this cell (0.76 V), there is no way to measure the potentials of the individual electrode processes. Thus, if we want potentials for the half-reactions (half-cells), we must arbitrarily divide the total cell potential. For example, if we assign the reaction 2H 2e ¡ H2 2H 2e ¡ H2 Zn ¡ Zn2 2e 2H1aq2 Zn1s2 ¡ Zn21aq2 H21g2 The name galvanic cell honors Luigi Galvani (1737–1798), an Italian scientist generally credited with the discovery of electricity. These cells are sometimes called voltaic cells after Alessandro Volta (1745–1827), another Italian, who first constructed cells of this type around 1800.
In this text we will follow the convention of indicating the physical states of the reactants and products only in the overall redox reaction. For simplicity, half-reactions will not include the physical states.
Digital voltmeter e– e– e– e– Zn2+ SO4 2– Zn metal Anode Cathode 1.0 M ZnSO4 solution 1.0 M H+ ions (for example, 1 M HCl) H+ Cl– H2(g) in Pt (electrode) 1 M H + ions ° = 0.000 volt (a) (b) H2(g) FIGURE 17.5 (a) A galvanic cell involving the reactions (at the anode) and (at the cathode) has a potential of 0.76 V. (b) The standard hydrogen electrode where H2(g) at 1 atm is passed over a platinum electrode in contact with 1 M H ions. This electrode process (assuming ideal behavior) is arbitrarily assigned a value of exactly zero volts.
2H 2e S H2 Zn S Zn2 2e An electrochemical cell with a measured potential of 1.10 V.
where a potential of exactly zero volts, then the reaction will have a potential of 0.76 V because h h h 0.76 V 0 V 0.76 V where the superscript indicates that standard states are employed. In fact, by setting the standard potential for the half-reaction 2H 2e S H2 equal to zero, we can assign values to all other half-reactions.
For example, the measured potential for the cell shown in Fig. 17.6 is 1.10 V. The cell reaction is which can be divided into the half-reactions Then e° cell e° Zn S Zn2 e° Cu2 S Cu Cathode: Cu2 2e ¡ Cu Anode: Zn ¡ Zn2 2e Zn1s2 Cu21aq2 ¡ Zn21aq2 Cu1s2 e° cell e° H S H2 e° Zn S Zn2 Zn ¡ Zn2 2e Standard states were discussed in Section 6.4.
Digital voltmeter e– e– e– e– Zn2+ SO4 2– Zn(s) 1.0 M Zn2+ solution Anode 1.0 M Cu2+ solution Cathode Cu2+ SO4 2– Cu(s) 2e– 2e– Zn 2+ Zn 2+ Zn Cu 2+ Cu 2+ Cu FIGURE 17.6 A galvanic cell involving the half-reactions (anode) and (cathode), with e° cell 1.10 V.
2e S Cu Cu2 Zn S Zn2 2e Visualization: Electrochemical Half-Reactions in a Galvanic Cell Visualization: Zinc/Copper Cells with Lemons 3H4 1 M and PH2 1 atm 17.2 Standard Reduction Potentials 795 Since was earlier assigned a value of 0.76 V, the value of must be 0.34 V because The scientific community has universally accepted the half-reaction potentials based on the assignment of zero volts to the process 2H 2e S H2 (under standard conditions where ideal behavior is assumed). However, before we can use these values to calculate cell potentials, we need to understand several essential characteristics of half-cell potentials.
The accepted convention is to give the potentials of half-reactions as reduction processes. For example: The values corresponding to reduction half-reactions with all solutes at 1 M and all gases at 1 atm are called standard reduction potentials. Standard reduction potentials for the most common half-reactions are given in Table 17.1 and Appendix 5.5.
Combining two half-reactions to obtain a balanced oxidation–reduction reaction often requires two manipulations: 1. One of the reduction half-reactions must be reversed (since redox reactions must involve a substance being oxidized and a substance being reduced). The half-reaction with the largest positive potential will run as written (as a reduction), and the other e° Zn2 2e ¡ Zn Cu2 2e ¡ Cu 2H 2e ¡ H2 1.10 V 0.76 V 0.34 V e° Cu2 S Cu e° Zn S Zn2 The standard hydrogen potential is the reference potential against which all half-reaction potentials are assigned.
All half-reactions are given as reduction processes in standard tables.
TABLE 17.1 Standard Reduction Potentials at 25ºC (298 K) for Many Common Half-Reactions Half-Reaction (V) Half-Reaction (V) 2.87 0.40 1.99 0.34 1.82 0.27 1.78 0.22 1.70 0.20 1.69 0.16 1.68 0.00 1.60 0.036 1.51 0.13 1.50 0.14 1.46 0.23 1.36 0.35 1.33 0.40 1.23 0.44 1.21 0.50 1.20 0.73 1.09 0.76 1.00 0.83 0.99 1.18 0.96 1.66 0.954 2.23 0.91 2.37 0.80 2.37 0.80 2.71 0.77 2.76 0.68 2.90 0.56 2.92 0.54 3.05 0.52 Cu e S Cu Li e S Li I2 2e S 2I K e S K MnO4 e S MnO4 2 Ba2 2e S Ba O2 2H 2e S H2O2 Ca2 2e S Ca Fe3 e S Fe2 Na e S Na Hg2 2 2e S 2Hg La3 3e S La Ag e S Ag Mg2 2e S Mg 2Hg2 2e S Hg2 2 H2 2e S 2H ClO2 e S ClO2 Al3 3e S Al NO3 4H 3e S NO 2H2O Mn2 2e S Mn AuCl4 3e S Au 4Cl 2H2O 2e S H2 2OH VO2 2H e S VO2 H2O Zn2 2e S Zn Br2 2e S 2Br Cr3 3e S Cr IO3 6H 5e S 1 2I2 3H2O Cr3 e S Cr2 MnO2 4H 2e S Mn2 2H2O Fe2 2e S Fe O2 4H 4e S 2H2O Cd2 2e S Cd Cr2O7 2 14H 6e S 2Cr3 7H2O PbSO4 2e S Pb SO4 2 Cl2 2e S 2Cl Ni2 2e S Ni PbO2 4H 2e S Pb2 2H2O Sn2 2e S Sn Au3 3e S Au Pb2 2e S Pb MnO4 8H 5e S Mn2 4H2O Fe3 3e S Fe 2e 2H IO4 S IO3 H2O 2H 2e S H2 MnO4 4H 3e S MnO2 2H2O Cu2 e S Cu PbO2 4H SO4 2 2e S PbSO4 2H2O SO4 2 4H 2e S H2SO3 H2O Ce4 e S Ce3 AgCl e S Ag Cl H2O2 2H 2e S 2H2O Hg2Cl2 2e S 2Hg 2Cl Co3 e S Co2 Cu2 2e S Cu Ag2 e S Ag O2 2H2O 4e S 4OH F2 2e S 2F e e Visualization: Galvanic (Voltaic) Cells 796 Chapter Seventeen Electrochemistry 17.2 Standard Reduction Potentials 797 half-reaction will be forced to run in reverse (will be the oxidation reaction). The net potential of the cell will be the difference between the two. Since the reduction process occurs at the cathode and the oxidation process occurs at the anode, we can write Because subtraction means “change the sign and add,” in the examples done here we will change the sign of the oxidation (anode) reaction when we reverse it and add it to the reduction (cathode) reaction.
2. Since the number of electrons lost must equal the number gained, the half-reactions must be multiplied by integers as necessary to achieve the balanced equation. How-ever, the value of is not changed when a half-reaction is multiplied by an integer.
Since a standard reduction potential is an intensive property (it does not depend on how many times the reaction occurs), the potential is not multiplied by the integer required to balance the cell reaction.
Consider a galvanic cell based on the redox reaction The pertinent half-reactions are (1) (2) To balance the cell reaction and calculate the standard cell potential, reaction (2) must be reversed: Note the change in sign for the ° value. Now, since each Cu atom produces two elec-trons but each Fe3 ion accepts only one electron, reaction (1) must be multiplied by 2: Note that is not changed in this case.
Now we can obtain the balanced cell reaction by summing the appropriately modi-fied half-reactions: e° 2Fe3 2e ¡ 2Fe2 e° 0.77 V e Cu ¡ Cu2 2e e° 0.34 V Cu2 2e ¡ Cu e° 0.34 V Fe3 e ¡ Fe2 e° 0.77 V Fe31aq2 Cu1s2 ¡ Cu21aq2 Fe21aq2 e° e° cell e° 1cathode2 e° 1anode2 When a half-reaction is reversed, the sign of is reversed.
e° When a half-reaction is multiplied by an integer, remains the same.
e° Visualization: Copper Metal in Water Visualization: Copper Metal in Sulfuric Acid Visualization: Copper Metal in Hydrochloric Acid Visualization: Copper Metal in Nitric Acid 0.77 V 0.34 V 0.43 V Cell reaction: Cu1s2 2Fe31aq2 ¡ Cu21aq2 2Fe21aq2 e° cell e° 1cathode2 e° 1anode2 Cu ¡ Cu2 2e e 1anode2 0.34 V 2Fe3 2e ¡ 2Fe2 e 1cathode2 0.77 V Galvanic Cells a. Consider a galvanic cell based on the reaction The half-reactions are (1) (2) Give the balanced cell reaction and calculate for the cell.
b. A galvanic cell is based on the reaction MnO4 1aq2 H1aq2 ClO3 1aq2 ¡ ClO4 1aq2 Mn21aq2 H2O1l2 e° Mg2 2e ¡ Mg e° 2.37 V Al3 3e ¡ Al e° 1.66 V Al31aq2 Mg1s2 ¡ Al1s2 Mg21aq2 Sample Exercise 17.1 798 Chapter Seventeen Electrochemistry The half-reactions are (1) (2) Give the balanced cell reaction and calculate for the cell.
Solution a. The half-reaction involving magnesium must be reversed and since this is the oxida-tion process, it is the anode: Also, since the two half-reactions involve different numbers of electrons, they must be multiplied by integers as follows: Mg ¡ Mg2 2e e° 1anode2 12.37 V2 2.37 V e° ClO4 2H 2e ¡ ClO3 H2O e° 1.19 V MnO4 5e 8H ¡ Mn2 4H2O e° 1.51 V 2MnO4 1aq2 6H1aq2 5ClO3 1aq2 ¡ e° cell e° 1cathode2 e° 1anode2 51ClO3 H2O ¡ ClO4 2H 2e2 e° 1anode2 1.19 V 21MnO4 5e 8H ¡ Mn2 4H2O2 e° 1cathode2 1.51 V See Exercises 17.27 and 17.28.
Line Notation We now will introduce a handy line notation used to describe electrochemical cells. In this notation the anode components are listed on the left and the cathode components are listed on the right, separated by double vertical lines (indicating the salt bridge or porous disk). For example, the line notation for the cell described in Sample Exercise 17.1(a) is In this notation a phase difference (boundary) is indicated by a single vertical line. Thus, in this case, vertical lines occur between the solid Mg metal and the Mg2 in aqueous so-lution and between solid Al and Al3 in aqueous solution. Also note that the substance constituting the anode is listed at the far left and the substance constituting the cathode is listed at the far right.
For the cell described in Sample Exercise 17.1(b), all the components involved in the oxidation–reduction reaction are ions. Since none of these dissolved ions can serve as an electrode, a nonreacting (inert) conductor must be used. The usual choice is platinum.
Thus, for the cell described in Sample Exercise 17.1(b), the line notation is Complete Description of a Galvanic Cell Next we want to consider how to describe a galvanic cell fully, given just its half-reactions. This description will include the cell reaction, the cell potential, and the Pt1s2 0 ClO3 1aq2, ClO4 1aq2, H1aq2 0 0 H1aq2, MnO4 1aq2, Mn21aq2 0 Pt1s2 Mg1s2 0Mg21aq2 0 0Al31aq2 0 Al1s2 1.66 V 2.37 V 0.71 V 2Al31aq2 3Mg1s2 ¡ 2Al1s2 3Mg21aq2 e° cell e° 1cathode2 e° 1anode2 31Mg ¡ Mg2 2e2 e° 1anode2 2.37 V 21Al3 3e ¡ Al2 e° 1cathode2 1.66 V b. Half-reaction (2) must be reversed (it is the anode), and both half-reactions must be multiplied by integers to make the number of electrons equal: 1.51 V 1.19 V 0.32 V 2Mn21aq2 3H2O1l2 5ClO4 1aq2 17.2 Standard Reduction Potentials 799 physical setup of the cell. Let’s consider a galvanic cell based on the following half-reactions: In a working galvanic cell, one of these reactions must run in reverse. Which one?
We can answer this question by considering the sign of the potential of a working cell: A cell will always run spontaneously in the direction that produces a positive cell potential. Thus, in the present case, it is clear that the half-reaction involving iron must be reversed, since this choice leads to a positive cell potential: where The balanced cell reaction is obtained as follows: Now consider the physical setup of the cell, shown schematically in Fig. 17.7. In the left compartment the active components in their standard states are pure metallic iron (Fe) and 1.0 M Fe2. The anion present depends on the iron salt used. In this compartment the anion does not participate in the reaction but simply balances the charge. The half-reaction that takes place at this electrode is which is an oxidation reaction, so this compartment is the anode. The electrode consists of pure iron metal.
In the right compartment the active components in their standard states are 1.0 M MnO4 , 1.0 M H, and 1.0 M Mn2, with appropriate unreacting ions (often called coun-terions) to balance the charge. The half-reaction in this compartment is which is a reduction reaction, so this compartment is the cathode. Since neither MnO4 nor Mn2 ions can serve as the electrode, a nonreacting conductor such as platinum must be employed.
The next step is to determine the direction of electron flow. In the left compartment the half-reaction involves the oxidation of iron: In the right compartment the half-reaction is the reduction of MnO4 : Thus the electrons flow from Fe to MnO4 in this cell, or from the anode to the cathode, as is always the case. The line notation for this cell is A complete description of a galvanic cell usually includes four items: • The cell potential (always positive for a galvanic cell where and the balanced cell reaction.
• The direction of electron flow, obtained by inspecting the half-reactions and using the direction that gives a positive ecell.
e° 1anode2 e° cell e° 1cathode2 Fe1s2 0 Fe21aq2 0 0 MnO4 1aq2, Mn21aq2 0 Pt1s2 MnO4 5e 8H ¡ Mn2 4H2O Fe ¡ Fe2 2e MnO4 5e 8H ¡ Mn2 4H2O Fe ¡ Fe2 2e 2MnO4 1aq2 5Fe1s2 16H1aq2 ¡ 5Fe21aq2 2Mn21aq2 8H2O1l2 21MnO4 5e 8H ¡ Mn2 4H2O2 51Fe ¡ Fe2 2e2 e° cell e° 1cathode2 e° 1anode2 1.51 V 0.44 V 1.95 V MnO4 5e 8H ¡ Mn2 4H2O e° 1.51 V Fe ¡ Fe2 2e e° 0.44 V MnO4 5e 8H ¡ Mn2 4H2O e° 1.51 V Fe2 2e ¡ Fe e° 0.44 V A galvanic cell runs spontaneously in the direction that gives a positive value for ecell.
Porous disk e– e– e– e– Fe 1 M Fe2+ Anode 1 M MnO4– 1 M H+ 1 M Mn2+ Cathode Pt FIGURE 17.7 The schematic of a galvanic cell based on the half-reactions: MnO4 5e 8H ¡ Mn2 4H2O Fe ¡ Fe2 2e Anode reaction Cathode reaction 800 Chapter Seventeen Electrochemistry • Designation of the anode and cathode.
• The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half-reaction is a conducting solid.
Description of a Galvanic Cell Describe completely the galvanic cell based on the following half-reactions under standard conditions: (1) (2) Solution Item 1 Since a positive value is required, reaction (2) must run in reverse: Item 2 Since Ag receives electrons and Fe2 loses electrons in the cell reaction, the electrons will flow from the compartment containing Fe2 to the compartment containing Ag.
Item 3 Oxidation occurs in the compartment containing Fe2 (electrons flow from Fe2 to Ag). Hence this compartment functions as the anode. Reduction occurs in the com-partment containing Ag, so this compartment functions as the cathode.
Item 4 The electrode in the AgAg compartment is silver metal, and an inert conductor, such as platinum, must be used in the Fe2Fe3 compartment. Appropriate counterions are assumed to be present. The diagram for this cell is shown in Fig. 17.8. The line notation for this cell is See Exercises 17.29 and 17.30.
17.3 Cell Potential, Electrical Work, and Free Energy So far we have considered electrochemical cells in a very practical fashion without much theoretical background. The next step will be to explore the relationship between ther-modynamics and electrochemistry.
The work that can be accomplished when electrons are transferred through a wire depends on the “push” (the thermodynamic driving force) behind the electrons. This driving force (the emf) is defined in terms of a potential difference (in volts) between two points in the circuit. Recall that a volt represents a joule of work per coulomb of charge transferred: Thus 1 joule of work is produced or required (depending on the direction) when 1 coulomb of charge is transferred between two points in the circuit that differ by a potential of 1 volt.
In this book, work is viewed from the point of view of the system. Thus work flowing out of the system is indicated by a minus sign. When a cell produces a current, the cell emf potential difference 1V2 work 1J2 charge 1C2 Pt1s2 0 Fe21aq2, Fe31aq2 0 0 Ag1aq2 0 Ag1s2 Cell reaction: Ag1aq2 Fe21aq2 ¡ Fe31aq2 Ag1s2 e° cell 0.03 V Fe2 ¡ Fe3 e e° 1anode2 0.77 V Ag e ¡ Ag e° 1cathode2 0.80 V e° cell Fe3 e ¡ Fe2 e° 0.77 V Ag e ¡ Ag e° 0.80 V Sample Exercise 17.2 Porous disk e– e– e– e– Pt Anode Cathode Ag 1 M Fe2+ 1 M Fe3+ 1 M Ag+ ° = 0.03 V FIGURE 17.8 Schematic diagram for the galvanic cell based on the half-reactions Fe 2 ¡ Fe 3 e Ag e ¡ Ag Using a battery-powered drill to insert a screw.
17.3 Cell Potential, Electrical Work, and Free Energy 801 potential is positive, and the current can be used to do work—to run a motor, for instance.
Thus the cell potential and the work w have opposite signs: Therefore, From this equation it can be seen that the maximum work in a cell would be obtained at the maximum cell potential: However, there is a problem. To obtain electrical work, current must flow. When cur-rent flows, some energy is inevitably wasted through frictional heating, and the maximum work is not obtained. This reflects the important general principle introduced in Section 16.9: In any real, spontaneous process some energy is always wasted—the actual work realized is always less than the calculated maximum. This is a consequence of the fact that the entropy of the universe must increase in any spontaneous process. Recall from Section 16.9 that the only process from which maximum work could be realized is the hypothetical reversible process. For a galvanic cell this would involve an infinitesimally small current flow and thus an infinite amount of time to do the work. Even though we can never achieve the maximum work through the actual discharge of a galvanic cell, we can measure the maximum potential. There is negligible current flow when a cell poten-tial is measured with a potentiometer or an efficient digital voltmeter. No current flow im-plies no waste of energy, so the potential measured is the maximum.
Although we can never actually realize the maximum work from a cell reaction, the value for it is still useful in evaluating the efficiency of a real process based on the cell reaction. For example, suppose a certain galvanic cell has a maximum potential (at zero current) of 2.50 V. In a particular experiment 1.33 moles of electrons were passed through this cell at an average actual potential of 2.10 V. The actual work done is where represents the actual potential difference at which the current flowed (2.10 V or 2.10 J/C) and q is the quantity of charge in coulombs transferred. The charge on 1 mole of electrons is a constant called the faraday (abbreviated F), which has the value 96,485 coulombs of charge per mole of electrons. Thus q equals the number of moles of elec-trons times the charge per mole of electrons: Then, for the preceding experiment, the actual work is For the maximum possible work, the calculation is similar, except that the maximum potential is used: Thus, in its actual operation, the efficiency of this cell is w wmax 100% 2.69 105 J 3.21 105 J 100% 83.8% 3.21 105 J a1.33 mol e 96,485 C mol eba2.50 J Cb wmax qe 2.69 105 J w qe 11.33 mol e 96,485 C/mol e2 12.10 J/C2 q nF 1.33 mol e 96,485 C/mol e e w qe wmax qemax or wmax qemax w qe e w q e Work is never the maximum possible if any current is flowing.
Michael Faraday lecturing at the Royal Institution before Prince Albert and others (1855). The faraday was named in honor of Michael Faraday (1791–1867), an English-man who may have been the greatest experimental scientist of the nineteenth century. Among his many achievements were the invention of the electric motor and generator and the development of the principles of electrolysis.
d Work d Charge 802 Chapter Seventeen Electrochemistry Next we want to relate the potential of a galvanic cell to free energy. In Section 16.9 we saw that for a process carried out at constant temperature and pressure, the change in free energy equals the maximum useful work obtainable from that process: For a galvanic cell, Since we have From now on the subscript on will be deleted, with the understanding that any po-tential given in this book is the maximum potential. Thus For standard conditions, This equation states that the maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell. This relationship is impor-tant because it provides an experimental means to obtain G for a reaction. It also confirms that a galvanic cell will run in the direction that gives a positive value for a positive value corresponds to a negative G value, which is the condition for spontaneity.
Calculating for a Cell Reaction Using the data in Table 17.1, calculate G for the reaction Is this reaction spontaneous?
Solution The half-reactions are We can calculate G from the equation Since two electrons are transferred per atom in the reaction, 2 moles of electrons are re-quired per mole of reactants and products. Thus n 2 mol e, F 96,485 C/mol e, and Therefore, The process is spontaneous, as indicated by both the negative sign of G and the posi-tive sign of This reaction is used industrially to deposit copper metal from solutions resulting from the dissolving of copper ores.
See Exercises 17.37 and 17.38.
e° cell.
1.5 105 J ¢G° 12 mol e2a96,485 C mol eba0.78 J Cb e° 0.78 V 0.78 J/C.
¢G° nF e° Cu2 Fe ¡ Fe2 Cu e° cell 0.78 V Fe ¡ Fe2 2e e° 1anode2 0.44 V Cu2 2e ¡ Cu e° 1cathode2 0.34 V Cu21aq2 Fe1s2 ¡ Cu1s2 Fe21aq2 G ecell ecell; ¢G° nFe° ¢G nFe emax ¢G qemax nF emax q nF wmax qemax ¢G wmax ¢G Sample Exercise 17.3 17.4 Dependence of Cell Potential on Concentration 803 Predicting Spontaneity Using the data from Table 17.1, predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3 solution.
Solution The half-reaction for HNO3 acting as an oxidizing agent is The reaction for the oxidation of solid gold to Au3 ions is The sum of these half-reactions gives the required reaction: and Since the value is negative, the process will not occur under standard conditions. That is, gold will not dissolve in 1 M HNO3 to give 1 M Au3. In fact, a mixture (1:3 by volume) of concentrated nitric and hydrochloric acids, called aqua regia, is required to dissolve gold.
See Exercises 17.37 and 17.38.
17.4 Dependence of Cell Potential on Concentration So far we have described cells under standard conditions. In this section we consider the dependence of the cell potential on concentration. Under standard conditions (all con-centrations 1 M), the cell with the reaction has a potential of 1.36 V. What will the cell potential be if [Ce4] is greater than 1.0 M?
This question can be answered qualitatively in terms of Le Châtelier’s principle. An in-crease in the concentration of Ce4 will favor the forward reaction and thus increase the driving force on the electrons. The cell potential will increase. On the other hand, an in-crease in the concentration of a product will oppose the forward reaction, thus decreasing the cell potential.
These ideas are illustrated in Sample Exercise 17.5.
The Effects of Concentration on For the cell reaction predict whether is larger or smaller than for the following cases.
a.
b.
Solution a. A product concentration has been raised above 1.0 M. This will oppose the cell reac-tion and will cause ecell to be less than e° cell (ecell 0.48 V).
b. A reactant concentration has been increased above 1.0 M, and ecell will be greater than e° cell (ecell 0.48 V).
See Exercise 17.51.
3Al34 1.0 M, 3Mn24 3.0 M 3Al34 2.0 M, 3Mn24 1.0 M e° cell ecell 2Al1s2 3Mn21aq2 ¡ 2Al31aq2 3Mn1s2 e° cell 0.48 V e (Cu2 or Ce3) Cu1s2 2Ce41aq2 ¡ Cu 21aq2 2Ce 31aq2 e° e° cell e° 1cathode2 e° 1anode2 0.96 V 1.50 V 0.54 V Au1s2 NO3 1aq2 4H1aq2 ¡ Au31aq2 NO1g2 2H2O1l2 Au ¡ Au3 3e e° 1anode2 1.50 V NO3 4H 3e ¡ NO 2H2O e° 1cathode2 0.96 V Sample Exercise 17.4 A gold ring does not dissolve in nitric acid.
Sample Exercise 17.5 A concentration cell with 1.0 M Cu2 on the right and 0.010 M Cu2 on the left.
804 Chapter Seventeen Electrochemistry Concentration Cells Because cell potentials depend on concentration, we can construct galvanic cells where both compartments contain the same components but at different concentrations. For ex-ample, in the cell in Fig. 17.9, both compartments contain aqueous AgNO3, but with dif-ferent molarities. Let’s consider the potential of this cell and the direction of electron flow.
The half-reaction relevant to both compartments of this cell is If the cell had 1 M Ag in both compartments, However, in the cell described here, the concentrations of Ag in the two compartments are 1 M and 0.1 M. Because the concentrations of Ag are unequal, the half-cell potentials will not be identical, and the cell will exhibit a positive voltage. In which direction will the electrons flow in this cell? The best way to think about this question is to recognize that nature will try to equalize the concentrations of Ag in the two compartments. This can be done by transferring electrons from the compartment containing 0.1 M Ag to the one containing 1 M Ag (left to right in Fig. 17.9). This electron transfer will produce more Ag in the left compartment and consume Ag (to form Ag) in the right compartment.
A cell in which both compartments have the same components but at different con-centrations is called a concentration cell. The difference in concentration is the only factor that produces a cell potential in this case, and the voltages are typically small.
Concentration Cells Determine the direction of electron flow and designate the anode and cathode for the cell represented in Fig. 17.10.
Solution The concentrations of ion in the two compartments can (eventually) be equalized by transferring electrons from the left compartment to the right. This will cause to be formed in the left compartment, and iron metal will be deposited (by reducing ions to Fe) on the right electrode. Since electron flow is from left to right, oxida-tion occurs in the left compartment (the anode) and reduction occurs in the right (the cathode).
See Exercise 17.52.
The Nernst Equation The dependence of the cell potential on concentration results directly from the depen-dence of free energy on concentration. Recall from Chapter 16 that the equation where Q is the reaction quotient, was used to calculate the effect of concentration on G.
Since and the equation becomes Dividing each side of the equation by nF gives (17.1) e e° RT nF ln1Q2 nFe nFe° RT ln1Q2 ¢G° nF e°, ¢G nF e ¢G ¢G° RT ln1Q2 Fe 2 Fe 2 Fe 2 e° cell 0.80 V 0.80 V 0 V Ag e ¡ Ag e° 0.80 V Porous disk e– e– e– e– Ag Anode Cathode Ag 1 M Ag+ 1 M NO3– 0.1 M Ag+ 0.1 M NO3– FIGURE 17.9 A concentration cell that contains a silver electrode and aqueous silver nitrate in both compartments. Because the right compartment contains 1 M Ag and the left compartment contains 0.1 M Ag, there will be a driving force to transfer electrons from left to right. Silver will be deposited on the right electrode, thus lowering the concentration of Ag in the right compartment. In the left compart-ment the silver electrode dissolves (producing Ag ions) to raise the concentration of Ag in solution.
Sample Exercise 17.6 Porous disk Fe Fe 0.01 M Fe2+ 0.1 M Fe2+ FIGURE 17.10 A concentration cell containing iron electrodes and different concentrations of Fe2 ion in the two compartments.
17.4 Dependence of Cell Potential on Concentration 805 Equation (17.1), which gives the relationship between the cell potential and the concen-trations of the cell components, is commonly called the Nernst equation, after the German chemist Walther Hermann Nernst (1864–1941).
The Nernst equation is often given in a form that is valid at 25C: Using this relationship, we can calculate the potential of a cell in which some or all of the components are not in their standard states.
For example, is 0.48 V for the galvanic cell based on the reaction Consider a cell in which The cell potential at 25C for these concentrations can be calculated using the Nernst equation: We know that and Since the half-reactions are we know that Thus Note that the cell voltage decreases slightly because of the nonstandard concentrations.
This change is consistent with the predictions of Le Châtelier’s principle (see Sample Exercise 17.5). In this case, since the reactant concentration is lower than 1.0 M and the product concentration is higher than 1.0 M, is less than The potential calculated from the Nernst equation is the maximum potential before any current flow has occurred. As the cell discharges and current flows from anode to cathode, the concentrations will change, and as a result, will change. In fact, the cell will spontaneously discharge until it reaches equilibrium, at which point A “dead” battery is one in which the cell reaction has reached equilibrium, and there is no longer any chemical driving force to push electrons through the wire. In other words, at equilibrium, the components in the two cell compartments have the same free energy, and G 0 for the cell reaction at the equilibrium concentrations. The cell no longer has the ability to do work.
Q K 1the equilibrium constant2 and ecell 0 ecell e° cell.
ecell 0.48 0.0591 6 11.262 0.48 0.01 0.47 V ecell 0.48 0.0591 6 log1182 n 6 Reduction: 3Mn2 6e ¡ 3Mn Oxidation: 2Al ¡ 2Al3 6e Q 3Al34 2 3Mn24 3 11.5022 10.5023 18 e° cell 0.48 V ecell e° cell 0.0591 n log1Q2 3Mn24 0.50 M and 3Al34 1.50 M 2Al1s2 3Mn21aq2 ¡ 2Al31aq2 3Mn1s2 e° cell e e° 0.0591 n log1Q2 Nernst was one of the pioneers in the development of electrochemical theory and is generally given credit for first stating the third law of thermodynamics.
He won the Nobel Prize in chemistry in 1920.
806 Chapter Seventeen Electrochemistry The Nernst Equation Describe the cell based on the following half-reactions: (1) (2) where Solution The balanced cell reaction is obtained by reversing reaction (2) and multiplying reaction (1) by 2: 3Zn24 1.0 101 M 3VO24 1.0 102 M 3H4 0.50 M 3VO2 4 2.0 M T 25°C Zn2 2e ¡ Zn e° 0.76 V VO2 2H e ¡ VO2 H2O e° 1.00 V Sample Exercise 17.7 Cell reaction: 2VO2 1aq2 4H1aq2 Zn1s2 ¡ 2VO21aq2 2H2O1l2 Zn21aq2 e° cell 1.76 V Zn ¡ Zn2 2e e° 1anode2 0.76 V Reaction 122 reversed 2VO2 4H 2e ¡ 2VO2 2H2O e° 1cathode2 1.00 V 2 reaction 112 Since the cell contains components at concentrations other than 1 M, we must use the Nernst equation, where n 2 (since two electrons are transferred), to calculate the cell potential. At 25C we can use the equation The cell diagram is given in Fig. 17.11.
1.76 0.0591 2 log 14 1052 1.76 0.13 1.89 V 1.76 0.0591 2 log a 11.0 101211.0 10222 12.02210.5024 b 1.76 0.0591 2 log a 3Zn24 3VO24 2 3VO2 4 23H4 4 b e e° cell 0.0591 n log 1Q2 e– e– e– e– Zn Pt cell = 1.89 V Anode Cathode [VO2+] = 1.0 × 10 –2M [VO2+] = 2.0 M [H+] = 0.50 M [Zn2+] = 0.1 M FIGURE 17.11 Schematic diagram of the cell described in Sample Exercise 17.7.
See Exercises 17.55 through 17.58.
17.4 Dependence of Cell Potential on Concentration 807 Ion-Selective Electrodes Because the cell potential is sensitive to the concentrations of the reactants and products involved in the cell reaction, measured potentials can be used to determine the concen-tration of an ion. A pH meter (see Fig. 14.9) is a familiar example of an instrument that measures concentration using an observed potential. The pH meter has three main com-ponents: a standard electrode of known potential, a special glass electrode that changes potential depending on the concentration of H ions in the solution into which it is dipped, and a potentiometer that measures the potential between the electrodes. The potentiome-ter reading is automatically converted electronically to a direct reading of the pH of the solution being tested.
The glass electrode (see Fig. 17.12) contains a reference solution of dilute hydrochloric acid in contact with a thin glass membrane. The electrical potential of the glass electrode depends on the difference in [H] between the reference solution and the solution into which the electrode is dipped. Thus the electrical potential varies with the pH of the so-lution being tested.
Electrodes that are sensitive to the concentration of a particular ion are called ion-selective electrodes, of which the glass electrode for pH measurement is just one example.
Glass electrodes can be made sensitive to such ions as Na, K, or NH4 by changing the composition of the glass. Other ions can be detected if an appropriate crystal replaces the glass membrane. For example, a crystal of lanthanum(III) fluoride (LaF3) can be used in an electrode to measure [F]. Solid silver sulfide (Ag2S) can be used to measure [Ag] and [S2]. Some of the ions that can be detected by ion-selective electrodes are listed in Table 17.2.
Calculation of Equilibrium Constants for Redox Reactions The quantitative relationship between and G allows calculation of equilibrium con-stants for redox reactions. For a cell at equilibrium, Applying these conditions to the Nernst equation valid at 25C, e e° 0.0591 n log1Q2 ecell 0 and Q K e° Reference solution of dilute hydrochloric acid Silver wire coated with silver chloride Thin-walled membrane FIGURE 17.12 A glass electrode contains a reference solution of dilute hydrochloric acid in contact with a thin glass membrane in which a silver wire coated with silver chloride has been embedded. When the electrode is dipped into a solution containing H ions, the electrode potential is determined by the difference in [H] between the two solutions.
TABLE 17.2 Some Ions Whose Concentrations Can Be Detected by Ion-Selective Electrodes Cations Anions H Br Cd2 Cl Ca2 CN Cu2 F K NO3 Ag S2 Na 808 Chapter Seventeen Electrochemistry gives or Equilibrium Constants from Cell Potentials For the oxidation–reduction reaction the appropriate half-reactions are (1) (2) Balance the redox reaction, and calculate and K (at 25C).
Solution To obtain the balanced reaction, we must reverse reaction (2), multiply it by 2, and add it to reaction (1): e° Cr3 e ¡ Cr2 e° 0.50 V S4O6 2 2e ¡ 2S2O3 2 e° 0.17 V S4O6 21aq2 Cr21aq2 ¡ Cr31aq2 S2O3 21aq2 log1K2 ne° 0.0591 at 25°C 0 e° 0.0591 n log1K2 Sample Exercise 17.8 In this reaction, 2 moles of electrons are transferred for every unit of reaction, that is, for every 2 mol Cr2 reacting with 1 mol S4O6 2 to form 2 mol Cr3 and 2 mol S2O3 2.
Thus n 2. Then The value of K is found by taking the antilog of 22.6: This very large equilibrium constant is not unusual for a redox reaction.
See Exercises 17.65, 17.66, 17.69, and 17.70.
17.5 Batteries A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series, where the potentials of the individual cells add to give the total battery poten-tial. Batteries are a source of direct current and have become an essential source of portable power in our society. In this section we examine the most common types of batteries. Some new batteries currently being developed are described at the end of the chapter.
Lead Storage Battery Since about 1915 when self-starters were first used in automobiles, the lead storage bat-tery has been a major factor in making the automobile a practical means of transporta-tion. This type of battery can function for several years under temperature extremes from 30F to 120F and under incessant punishment from rough roads.
K 1022.6 4 1022 log1K2 ne° 0.0591 210.672 0.0591 22.6 The blue solution contains Cr2 ions, and the green solution contains Cr3 ions. Cell reaction: 2Cr21aq2 S4O6 21aq2 ¡ 2Cr31aq2 2S2O3 21aq2 e° 0.67 V 21Cr2 ¡ Cr3 e2 e° 1anode2 10.502 V S4O6 2 2e ¡ 2S2O3 2 e° 1cathode2 0.17 V Reaction 112 2 reaction 122 reversed 17.5 Batteries 809 The typical automobile lead storage battery has six cells connected in series. Each cell contains multiple electrodes in the form of grids (Fig. 17.13) and produces approximately 2 V, to give a total battery potential of about 12 V. Note from the cell reaction that sulfuric acid is consumed as the battery discharges. This lowers the den-sity of the electrolyte solution from its initial value of about 1.28 g/cm3 in the fully charged battery. As a result, the condition of the battery can be monitored by measuring the density of the sulfuric acid solution. The solid lead sulfate formed in the cell reaction during discharge adheres to the grid surfaces of the electrodes. The bat-tery is recharged by forcing current through it in the opposite direction to reverse the cell reaction. A car’s battery is continuously charged by an alternator driven by the automobile engine.
An automobile with a dead battery can be “jump-started” by connecting its battery to the battery in a running automobile. This process can be dangerous, however, be-cause the resulting flow of current causes electrolysis of water in the dead battery, pro-ducing hydrogen and oxygen gases (see Section 17.7 for details). Disconnecting the jumper cables after the disabled car starts causes an arc that can ignite the gaseous CHEMICAL IMPACT Printed Batteries S oon you may reach for a compact disc in your lo-cal record store and, as you touch it, the package will start playing one of the songs on the disc. Or you may stop to look at a product because the package begins to glow as you pass it in the store. These effects could happen soon thanks to the invention of a flexi-ble, superthin battery that can actually be printed onto the package. This battery was developed by Power Paper, Ltd., a company founded by Baruch Levanon and several colleagues.
The battery developed by Power Paper consists of five layers of zinc (anode) and manganese dioxide (cathode) and is only 0.5 millimeter thick. The battery can be printed onto paper with a regular printing press and appears to present no environmental hazards.
The new battery has been licensed by Interna-tional Paper Company, which intends to use it to bring light, sound, and other special effects to packaging to entice potential customers. You might see talking, singing, or glowing packages on the shelves within a year or two.
A CD case with an ultrathin battery that can be “printed” on packages like ink.
Cell reaction: Pb1s2 PbO21s2 2H1aq2 2HSO4 1aq2 ¡ 2PbSO41s2 2H2O1l2 Cathode reaction: PbO2 HSO4 3H 2e ¡ PbSO4 2H2O Anode reaction: Pb HSO4 ¡ PbSO4 H 2e In this battery, lead serves as the anode, and lead coated with lead dioxide serves as the cathode. Both electrodes dip into an electrolyte solution of sulfuric acid. The elec-trode reactions are H2SO4 electrolyte solution Anode (lead grid filled with spongy lead) Cathode (lead grid filled with spongy PbO2) + – FIGURE 17.13 One of the six cells in a 12-V lead storage battery. The anode consists of a lead grid filled with spongy lead, and the cathode is a lead grid filled with lead dioxide. The cell also contains 38% (by mass) sulfuric acid.
810 Chapter Seventeen Electrochemistry mixture. If this happens, the battery may explode, ejecting corrosive sulfuric acid. This problem can be avoided by connecting the ground jumper cable to a part of the engine remote from the battery. Any arc produced when this cable is disconnected will then be harmless.
Traditional types of storage batteries require periodic “topping off” because the wa-ter in the electrolyte solution is depleted by the electrolysis that accompanies the charg-ing process. Recent types of batteries have electrodes made of an alloy of calcium and lead that inhibits the electrolysis of water. These batteries can be sealed, since they re-quire no addition of water.
It is rather amazing that in the 85 years in which lead storage batteries have been used, no better system has been found. Although a lead storage battery does provide ex-cellent service, it has a useful lifetime of 3 to 5 years in an automobile. While it might seem that the battery could undergo an indefinite number of discharge/charge cycles, phys-ical damage from road shock and chemical side-reactions eventually cause it to fail.
Other Batteries The calculators, electronic games, digital watches, and portable CD players that are so fa-miliar to us are all powered by small, efficient batteries. The common dry cell battery was invented more than 100 years ago by George Leclanché (1839–1882), a French chemist. In its acid version, the dry cell battery contains a zinc inner case that acts as the anode and a carbon rod in contact with a moist paste of solid MnO2, solid NH4Cl, and carbon that acts as the cathode (Fig. 17.14). The half-reactions are complex but can be Anode (zinc inner case) Cathode (graphite rod) Paste of MnO2, NH4Cl, and carbon FIGURE 17.14 A common dry cell battery.
CHEMICAL IMPACT Thermophotovoltaics: Electricity from Heat A photovoltaic cell transforms the energy of sunlight into an electric current. These devices are used to power cal-culators, electric signs in rural areas, experimental cars, and an increasing number of other devices. But what happens at night or on cloudy days? Usually photovoltaic power sources employ a battery as a reserve energy source when light lev-els are low.
Now there is an emerging technology, called ther-mophotovoltaics (TPV), that uses a heat source instead of the sun for energy. These devices can operate at night or on an overcast day without a battery. Although TPV devices could use many different sources of heat, the examples cur-rently under development use a propane burner. To produce an electric current, the radiant heat from the burner is used to excite a “radiator,” a device that emits infrared (IR) radi-ation when heated. The emitted IR radiation then falls on a “converter,” which is a semiconductor that contains p–n junctions. The IR radiation excites electrons from valence bands to conduction bands in the semiconductor so that the electrons can flow as a current. A schematic of a TPV gen-erator is illustrated in the diagram.
TPV technology has advanced recently because re-searchers have found that it is possible to use radiators such as silicon carbide, which can operate at relatively low tem-peratures (approximately ), with III–V semiconduc-tor converters such as gallium antimonide (GSb) or gallium arsenide (GaAs). While development work continues on many fronts, the first commercial TPV product is being mar-keted by JX Crystals of Issaquah, Washington. The prod-uct—Midnight Sun—is a propane-powered TPV generator that can produce 30 watts of electricity and is intended for use on boats to charge the batteries that power navigation and other essential equipment. Although at $3000 the TPV generator is more expensive than a conventional diesel-powered generator, Midnight Sun is silent and more reliable because it has no moving parts.
Although TPV technology is still in its infancy, it has many possible uses. The utilization of industrial waste heat—generated by glass and steel manufacturing and other industries—could establish a huge market for TPV. For ex-ample, two-thirds of the energy used in the manufacture of glass ends up as waste heat. If a significant quantity of this 1000°C 17.5 Batteries 811 approximated as follows: This cell produces a potential of about 1.5 V.
In the alkaline version of the dry cell battery, the solid NH4Cl is replaced with KOH or NaOH. In this case the half-reactions can be approximated as follows: The alkaline dry cell lasts longer mainly because the zinc anode corrodes less rapidly under basic conditions than under acidic conditions.
Other types of useful batteries include the silver cell, which has a Zn anode and a cathode that employs Ag2O as the oxidizing agent in a basic environment. Mercury cells, often used in calculators, have a Zn anode and a cathode involving HgO as the oxidizing agent in a basic medium (see Fig. 17.15).
An especially important type of battery is the nickel–cadmium battery, in which the electrode reactions are As in the lead storage battery, the products adhere to the electrodes. Therefore, a nickel– cadmium battery can be recharged an indefinite number of times.
Cathode reaction: NiO2 2H2O 2e ¡ Ni1OH22 2OH Anode reaction: Cd 2OH ¡ Cd1OH22 2e Cathode reaction: 2MnO2 H2O 2e ¡ Mn2O3 2OH Anode reaction: Zn 2OH ¡ ZnO H2O 2e Cathode reaction: 2NH4 2MnO2 2e ¡ Mn2O3 2NH3 H2O Anode reaction: Zn ¡ Zn2 2e now wasted energy could be used to produce electricity, this would have tremendous fiscal implications.
Another promising application of TPV technology is for cars with hybrid energy sources. For example, an exper-imental electric car built at Western Washington University uses a 10-kW TPV generator to supplement the bat-teries that serve as the main power source.
Projections indicate that TPV de-vices could account for $500 million in sales by 2005, mainly by substitut-ing TPV generators for small diesel-powered generators used on boats and by the military in the field. It appears that this technology has a hot future.
Exhaust Quartz shield Window Burner Propane intake Cooling air Room heat Radiator Photovoltaic converter cells Cooling fins Combustion air Cooling fan Diagram of a TPV generator.
Batteries for electronic watches are, by necessity, very tiny.
812 Chapter Seventeen Electrochemistry Fuel Cells A fuel cell is a galvanic cell for which the reactants are continuously supplied. To illus-trate the principles of fuel cells, let’s consider the exothermic redox reaction of methane with oxygen: CH41g2 2O21g2 ¡ CO21g2 2H2O1g2 energy Cathode (steel) Insulation Anode (zinc container) Paste of HgO (oxidizing agent) in a basic medium of KOH and Zn(OH)2 FIGURE 17.15 A mercury battery of the type used in calculators.
CHEMICAL IMPACT Fuel Cells for Cars Y our next car may be powered by a fuel cell. Until recently only affordable to NASA, fuel cells are now ready to become practical power plants in cars. Many car companies are testing vehicles that should be commercially available by 2004 or 2005.
All of these vehicles are powered by hy-drogen–oxygen fuel cells (see Fig. 17.16).
One of the most common types of fuel cells for automobiles uses a proton-exchange membrane (PEM). When H2 re-leases electrons at the anode, H+ ions form and then travel through the membrane to the cathode, where they combine with O2 and electrons to form water. This cell gen-erates about 0.7 V of power. To achieve the desired power level, several cells are stacked in series. Fuel cells of this type have appeared in several prototype vehi-cles, such as Nissan’s Xterra FCV, Ford’s Focus FCV, and DaimlerChrysler’s Mercedes Benz NECAR 5 (see photo).
The main question yet to be answered deals with whether the fuel cells in these cars will be fueled by H2 stored on board or by H2 made from gasoline or methanol as it is needed. The latter systems include an onboard re-former that uses catalysts to produce H2 from other fuels.
The on-board storage of hydrogen could take place in a tank at high pressure (approximately 5000 psi) or it could utilize a metal-hydride–based solid. Energy Conversion Devices (ECD) of Troy, Michigan, is developing a storage system based on a magnesium alloy that absorbs H2 to form a mag-nesium hydride. The H2 gas can be released from this solid by heating it to . According to ECD, the alloy can be fully charged with H2 in about 5 minutes, achieving a hy-drogen density of 103 g/L. This density compares to 71 g/L for liquid hydrogen and 31 g/L for gaseous hydrogen at 5000 psi. ECD claims its storage system furnishes enough H2 to power a fuel-cell car for 300 miles of driving.
Clearly, fuel-cell–powered cars are on the near horizon.
300°C A gathering of several cars powered by fuel cells at Los Angeles Memorial Coliseum.
17.6 Corrosion 813 Usually the energy from this reaction is released as heat to warm homes and to run ma-chines. However, in a fuel cell designed to use this reaction, the energy is used to pro-duce an electric current: The electrons flow from the reducing agent (CH4) to the oxidiz-ing agent (O2) through a conductor.
The U.S. space program has supported extensive research to develop fuel cells. The space shuttle uses a fuel cell based on the reaction of hydrogen and oxygen to form water: A schematic of a fuel cell that employs this reaction is shown in Fig. 17.16. The half-reactions are A cell of this type weighing about 500 pounds has been designed for space vehicles, but this fuel cell is not practical enough for general use as a source of portable power. How-ever, current research on portable electrochemical power is now proceeding at a rapid pace. In fact, cars powered by fuel cells are now being tested on the streets.
Fuel cells are also finding use as permanent power sources. For example, a power plant built in New York City contains stacks of hydrogen–oxygen fuel cells, which can be rapidly put on-line in response to fluctuating power demands. The hydrogen gas is ob-tained by decomposing the methane in natural gas. A plant of this type also has been con-structed in Tokyo.
In addition, new fuel cells are under development that can use fuels such as methane and diesel directly without having to produce hydrogen first.
17.6 Corrosion Corrosion can be viewed as the process of returning metals to their natural state—the ores from which they were originally obtained. Corrosion involves oxidation of the metal.
Since corroded metal often loses its structural integrity and attractiveness, this sponta-neous process has great economic impact. Approximately one-fifth of the iron and steel produced annually is used to replace rusted metal.
Metals corrode because they oxidize easily. Table 17.1 shows that, with the excep-tion of gold, those metals commonly used for structural and decorative purposes all have standard reduction potentials less positive than that of oxygen gas. When any of these half-reactions is reversed (to show oxidation of the metal) and combined with the reduc-tion half-reaction for oxygen, the result is a positive value. Thus the oxidation of most metals by oxygen is spontaneous (although we cannot tell from the potential how fast it will occur).
In view of the large difference in reduction potentials between oxygen and most met-als, it is surprising that the problem of corrosion does not completely prevent the use of metals in air. However, most metals develop a thin oxide coating, which tends to protect their internal atoms against further oxidation. The metal that best demonstrates this phe-nomenon is aluminum. With a reduction potential of 1.7 V, aluminum should be easily oxidized by O2. According to the apparent thermodynamics of the reaction, an aluminum airplane could dissolve in a rainstorm. The fact that this very active metal can be used as a structural material is due to the formation of a thin, adherent layer of aluminum oxide (Al2O3), more properly represented as Al2(OH)6, which greatly inhibits further corrosion.
The potential of the “passive,” oxide-coated aluminum is 0.6 V, a value that causes it to behave much like a noble metal.
Iron also can form a protective oxide coating. This coating is not an infallible shield against corrosion, however; when steel is exposed to oxygen in moist air, the oxide that forms tends to scale off and expose new metal surfaces to corrosion.
e° Cathode reaction: 4e O2 2H2O ¡ 4OH Anode reaction: 2H2 4OH ¡ 4H2O 4e 2H21g2 O21g2 ¡ 2H2O1l2 H2(g) O2(g) K+ OH– OH– H2O Steam Porous carbon electrodes containing catalysts e– e– e– FIGURE 17.16 Schematic of the hydrogen–oxygen fuel cell.
Some metals, such as copper, gold, silver, and platinum, are relatively difficult to oxidize. These are often called noble metals.
814 Chapter Seventeen Electrochemistry The corrosion products of noble metals such as copper and silver are complex and affect the use of these metals as decorative materials. Under normal atmospheric condi-tions, copper forms an external layer of greenish copper carbonate called patina. Silver tarnish is silver sulfide (Ag2S), which in thin layers gives the silver surface a richer ap-pearance. Gold, with a positive standard reduction potential of 1.50 V, significantly larger than that for oxygen (1.23 V), shows no appreciable corrosion in air.
Corrosion of Iron Since steel is the main structural material for bridges, buildings, and automobiles, con-trolling its corrosion is extremely important. To do this, we must understand the corro-sion mechanism. Instead of being a direct oxidation process as we might expect, the corrosion of iron is an electrochemical reaction, as shown in Fig. 17.17.
FIGURE 17.17 The electrochemical corrosion of iron.
Water droplet Anodic area Iron dissolves forming a pit Rust O2 Cathodic area (Anode reaction: Fe Fe2+ + 2e– ) (Cathode reaction: O2 + 2H2O + 4e– 4OH–) e– Fe2+ CHEMICAL IMPACT Paint That Stops Rust—Completely T raditionally, paint has provided the most economical method for protecting steel against corrosion. However, as people who live in the Midwest know well, paint cannot pre-vent a car from rusting indefinitely. Eventually, flaws develop in the paint that allow the ravages of rusting to take place.
This situation may soon change. Chemists at Glidden Research Center in Ohio have developed a paint called Rust-master Pro that worked so well to prevent rusting in its ini-tial tests that the scientists did not believe their results. Steel coated with the new paint showed no signs of rusting after an astonishing 10,000 hours of exposure in a salt spray chamber at .
Rustmaster is a water-based polymer formulation that prevents corrosion in two different ways. First, the polymer layer that cures in air forms a barrier impenetrable to both oxygen and water vapor. Second, the chemicals in the coat-ing react with the steel surface to produce an interlayer between the metal and the polymer coating. This interlayer is a complex mineral called pyroaurite that contains cations 38°C of the form [M1xZx(OH)2]x, where M is a 2 ion (Mg2, Fe2, Zn2, Co2, or Ni2), Z is a 3 ion (Al3, Fe3, Mn3, Co3, or Ni3), and x is a number between 0 and 1. The an-ions in pyroaurite are typically CO3 2, Cl, and/or SO4 2.
This pyroaurite interlayer is the real secret of the paint’s effectiveness. Because the corrosion of steel has an electro-chemical mechanism, motion of ions must be possible be-tween the cathodic and anodic areas on the surface of the steel for rusting to occur. However, the pyroaurite interlayer grows into the neighboring polymer layer, thus preventing this crucial movement of ions. In effect, this layer prevents corrosion in the same way that removing the salt bridge prevents current from flowing in a galvanic cell.
In addition to having an extraordinary corrosion-fighting ability, Rustmaster yields an unusually small quantity of volatile solvents as it dries. A typical paint can produce from 1 to 5 kg of volatiles per gallon; Rustmaster produces only 0.05 kg. This paint may signal a new era in corrosion prevention.
17.6 Corrosion 815 Steel has a nonuniform surface because the chemical composition is not completely homogeneous. Also, physical strains leave stress points in the metal. These nonuniformi-ties cause areas where the iron is more easily oxidized (anodic regions) than it is at oth-ers (cathodic regions). In the anodic regions each iron atom gives up two electrons to form the ion: The electrons that are released flow through the steel, as they do through the wire of a galvanic cell, to a cathodic region, where they react with oxygen: The ions formed in the anodic regions travel to the cathodic regions through the moisture on the surface of the steel, just as ions travel through a salt bridge in a galvanic cell. In the cathodic regions ions react with oxygen to form rust, which is hydrated iron(III) oxide of variable composition: Rust Because of the migration of ions and electrons, rust often forms at sites that are re-mote from those where the iron dissolved to form pits in the steel. The degree of hydra-tion of the iron oxide affects the color of the rust, which may vary from black to yellow to the familiar reddish brown.
The electrochemical nature of the rusting of iron explains the importance of mois-ture in the corrosion process. Moisture must be present to act as a kind of salt bridge between anodic and cathodic regions. Steel does not rust in dry air, a fact that explains why cars last much longer in the arid Southwest than in the relatively humid Midwest.
Salt also accelerates rusting, a fact all too easily recognized by car owners in the colder parts of the United States, where salt is used on roads to melt snow and ice. The severity of rusting is greatly increased because the dissolved salt on the moist steel surface in-creases the conductivity of the aqueous solution formed there and thus accelerates the electrochemical corrosion process. Chloride ions also form very stable complex ions with Fe3, and this factor tends to encourage the dissolving of the iron, again accelerating the corrosion.
Prevention of Corrosion Prevention of corrosion is an important way of conserving our natural resources of en-ergy and metals. The primary means of protection is the application of a coating, most commonly paint or metal plating, to protect the metal from oxygen and moisture.
Chromium and tin are often used to plate steel (see Section 17.8) because they oxidize to form a durable, effective oxide coating. Zinc, also used to coat steel in a process called galvanizing, forms a mixed oxide–carbonate coating. Since zinc is a more active metal than iron, as the potentials for the oxidation half-reactions show, any oxidation that occurs dissolves zinc rather than iron. Recall that the reaction with the most positive standard potential has the greatest thermodynamic tendency to occur. Thus zinc acts as a “sacrificial” coating on steel.
Alloying is also used to prevent corrosion. Stainless steel contains chromium and nickel, both of which form oxide coatings that change steel’s reduction potential to one characteristic of the noble metals. In addition, a new technology is now being developed Zn ¡ Zn2 2e e° 0.76 V Fe ¡ Fe2 2e e° 0.44 V 4Fe 2 1aq2 O2 1g2 14 2n2 H2O 1l2 ¡ 2Fe2O3 nH2O 1s2 8H 1aq2 Fe 2 Fe 2 O2 2H2O 4e ¡ 4OH Fe ¡ Fe 2 2e Fe 2 816 Chapter Seventeen Electrochemistry to create surface alloys. That is, instead of forming a metal alloy such as stainless steel, which has the same composition throughout, a cheaper carbon steel is treated by ion bom-bardment to produce a thin layer of stainless steel or other desirable alloy on the surface.
In this process, a “plasma” or “ion gas” of the alloying ions is formed at high tempera-tures and is then directed onto the surface of the metal.
Cathodic protection is a method most often employed to protect steel in buried fuel tanks and pipelines. An active metal, such as magnesium, is connected by a wire to the pipeline or tank to be protected (Fig. 17.18). Because the magnesium is a better reducing agent than iron, electrons are furnished by the magnesium rather than by the iron, keep-ing the iron from being oxidized. As oxidation occurs, the magnesium anode dissolves, and so it must be replaced periodically. Ships’ hulls are protected in a similar way by at-taching bars of titanium metal to the steel hull (Fig. 17.18). In salt water the titanium acts as the anode and is oxidized instead of the steel hull (the cathode).
17.7 Electrolysis A galvanic cell produces current when an oxidation–reduction reaction proceeds sponta-neously. A similar apparatus, an electrolytic cell, uses electrical energy to produce chemical change. The process of electrolysis involves forcing a current through a cell to produce a chemical change for which the cell potential is negative; that is, electrical work causes an otherwise nonspontaneous chemical reaction to occur. Electrolysis has great practical importance; for example, charging a battery, producing aluminum metal, and chrome plating an object are all done electrolytically.
To illustrate the difference between a galvanic cell and an electrolytic cell, consider the cell shown in Fig. 17.19(a) as it runs spontaneously to produce 1.10 V. In this galvanic cell the reaction at the anode is whereas at the cathode the reaction is Figure 17.19(b) shows an external power source forcing electrons through the cell in the opposite direction to that in (a). This requires an external potential greater than 1.10 V, which must be applied in opposition to the natural cell potential. This device is an elec-trolytic cell. Notice that since electron flow is opposite in the two cases, the anode and cathode are reversed between (a) and (b). Also, ion flow through the salt bridge is oppo-site in the two cells.
Now we will consider the stoichiometry of electrolytic processes, that is, how much chemical change occurs with the flow of a given current for a specified time. Suppose we wish to determine the mass of copper that is plated out when a current of 10.0 amps Cu2 2e ¡ Cu Zn ¡ Zn2 2e FIGURE 17.18 Cathodic protection of an underground pipe.
Ground level Electrolyte (moist soil) Cathode (buried iron pipe) Connecting insulated wire Anode (magnesium) An electrolytic cell uses electrical energy to produce a chemical change that would otherwise not occur spontaneously.
1 A 1 C/s 17.7 Electrolysis 817 (an ampere [amp], abbreviated A, is 1 coulomb of charge per second) is passed for 30.0 minutes through a solution containing . Plating means depositing the neutral metal on the electrode by reducing the metal ions in solution. In this case each ion re-quires two electrons to become an atom of copper metal: This reduction process will occur at the cathode of the electrolytic cell.
To solve this stoichiometry problem, we need the following steps: current quantity of moles moles grams and S charge in S of S of S of time coulombs electrons copper copper ➥1 Since an amp is a coulomb of charge per second, we multiply the current by the time in seconds to obtain the total coulombs of charge passed into the Cu2 solution at the cathode: ➥2 Since 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can calculate the number of moles of electrons required to carry 1.80 104 coulombs of charge: This means that 0.187 mole of electrons flowed into the Cu2 solution.
1.80 104 C 1 mol e 96,485 C 1.87 101 mol e 1.80 104 C 10.0 C s 30.0 min 60.0 s min Coulombs of charge amps seconds C s s Cu 2 1aq2 2e ¡ Cu 1s2 Cu 2 Cu 2 Zn(s) Cu(s) Anode Cathode (a) (b) Power source greater than 1.10 V e– e– e– e– Cations Anions 1.0 M Zn2+ solution 1.0 M Cu2+ solution Zn2+ SO4 2– Cu2+ SO4 2– Zn(s) Cu(s) Cathode Anode e– e– e– e– Cations Anions 1.0 M Zn2+ solution 1.0 M Cu2+ solution Zn2+ SO4 2– Cu2+ SO4 2– FIGURE 17.19 (a) A standard galvanic cell based on the spontaneous reaction (b) A standard electrolytic cell. A power source forces the opposite reaction Cu Zn 2 ¡ Cu 2 Zn Zn Cu 2 ¡ Zn 2 Cu 1 2 3 4 818 Chapter Seventeen Electrochemistry ➥3 Each Cu2 ion requires two electrons to become a copper atom. Thus each mole of electrons produces mole of copper metal: ➥4 We now know the moles of copper metal plated onto the cathode, and we can calculate the mass of copper formed: Electroplating How long must a current of 5.00 A be applied to a solution of Ag to produce 10.5 g silver metal?
Solution In this case, we must use the steps given earlier in reverse: grams moles moles of coulombs time of S of S electrons S of charge S required silver silver required required for plating Each Ag ion requires one electron to become a silver atom: Thus 9.73 102 mol of electrons is required, and we can calculate the quantity of charge carried by these electrons: The 5.00 A (5.00 C/s) of current must produce 9.39 103 C of charge. Thus See Exercises 17.77 through 17.80.
Electrolysis of Water We have seen that hydrogen and oxygen combine spontaneously to form water and that the accompanying decrease in free energy can be used to run a fuel cell to produce electricity.
The reverse process, which is of course nonspontaneous, can be forced by electrolysis: Time 9.39 103 5.00 s 1.88 103 s 31.3 min a5.00 C s b 1time, in s2 9.39 103 C 9.73 102 mol e 96,485 C mol e 9.39 103 C Ag e ¡ Ag 10.5 g Ag 1 mol Ag 107.868 g Ag 9.73 102 mol Ag 9.35 102 mol Cu 63.546 g mol Cu 5.94 g Cu 1.87 101 mol e 1 mol Cu 2 mol e 9.35 102 mol Cu 1 2 Sample Exercise 17.9 or 2H2O ¡ 2H2 O2 4H2O Net reaction: 6H2O ¡ 2H2 O2 41H OH2 e° 2.06 V Cathode reaction: 4H2O 4e ¡ 2H2 4OH e° 0.83 V Anode reaction: 2H2O ¡ O2 4H 4e e° 1.23 V Sample Exercise 17.9 describes only the half-cell of interest. There also must be an anode at which oxidation is occurring.
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 17.7 Electrolysis 819 Note that these potentials assume an anode chamber with 1 M H and a cathode cham-ber with 1 M . In pure water, where [H] [OH] 107 M, the potential for the overall process is .
In practice, however, if platinum electrodes connected to a 6-V battery are dipped into pure water, no reaction is observed because pure water contains so few ions that only a negligible current can flow. However, addition of even a small amount of a solu-ble salt causes an immediate evolution of bubbles of hydrogen and oxygen, as illustrated in Fig. 17.20.
Electrolysis of Mixtures of Ions Suppose a solution in an electrolytic cell contains the ions Ag, and If the voltage is initially very low and is gradually turned up, in which order will the metals be plated out onto the cathode? This question can be answered by looking at the standard reduction potentials of these ions: Remember that the more positive the value, the more the reaction has a tendency to proceed in the direction indicated. Of the three reactions listed, the reduction of Ag occurs most easily, and the order of oxidizing ability is This means that silver will plate out first as the potential is increased, followed by cop-per, and finally zinc.
Relative Oxidizing Abilities An acidic solution contains the ions Ce4, VO2 , and Fe3. Using the values listed in Table 17.1, give the order of oxidizing ability of these species and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage.
Solution The half-reactions and values are The order of oxidizing ability is therefore The Ce4 ion will be reduced at the lowest voltage in an electrolytic cell.
See Exercise 17.89.
The principle described in this section is very useful, but it must be applied with some caution. For example, in the electrolysis of an aqueous solution of sodium chloride, we should be able to use values to predict the products. Of the major species in the solution e° Ce 4 7 VO2 7 Fe 3 Fe3 e ¡ Fe2 e° 0.77 V VO2 2H e ¡ VO2 H2O e° 1.00 V Ce4 e ¡ Ce3 e° 1.70 V e° e° Ag 7 Cu2 7 Zn2 e° Zn2 2e ¡ Zn e° 0.76 V Cu2 2e ¡ Cu e° 0.34 V Ag e ¡ Ag e° 0.80 V Zn2.
Cu2, 1.23 V OH Sample Exercise 17.10 FIGURE 17.20 The electrolysis of water produces hydrogen gas at the cathode (on the right) and oxy-gen gas at the anode (on the left).
Visualization: Electrolysis of Water 820 Chapter Seventeen Electrochemistry (Na, and H2O), only and H2O can be readily oxidized. The half-reactions (writ-ten as oxidization processes) are Since water has the more positive potential, we would expect to see O2 produced at the anode because it is easier (thermodynamically) to oxidize H2O than . Actually, this does not happen. As the voltage is increased in the cell, the ion is the first to be oxidized. A much higher potential than expected is required to oxidize water.
The voltage required in excess of the expected value (called the overvoltage) is much greater for the production of O2 than for Cl2, which explains why chlorine is produced first.
The causes of overvoltage are very complex. Basically, the phenomenon is caused by difficulties in transferring electrons from the species in the solution to the atoms on the electrode across the electrode–solution interface. Because of this situation, values must be used cautiously in predicting the actual order of oxidation or reduction of species in an electrolytic cell.
e° Cl Cl 2H2O ¡ O2 4H 4e e° 1.23 V 2Cl ¡ Cl2 2e e° 1.36 V Cl Cl, CHEMICAL IMPACT The Chemistry of Sunken Treasure W hen the galleon Atocha was destroyed on a reef by a hurricane in 1622, it was bound for Spain carrying ap-proximately 47 tons of copper, gold, and silver from the New World. The bulk of the treasure was silver bars and coins packed in wooden chests. When treasure hunter Mel Fisher salvaged the silver in 1985, corrosion and marine growth had transformed the shiny metal into something that looked like coral. Restoring the silver to its original condition re-quired an understanding of the chemical changes that had occurred in 350 years of being submerged in the ocean.
Much of this chemistry we have already considered at var-ious places in this text.
As the wooden chests containing the silver decayed, the oxygen supply was depleted, favoring the growth of certain bacteria that use the sulfate ion rather than oxygen as an ox-idizing agent to generate energy. As these bacteria consume sulfate ions, they release hydrogen sulfide gas that reacts with silver to form black silver sulfide: Thus, over the years, the surface of the silver became cov-ered with a tightly adhering layer of corrosion, which for-tunately protected the silver underneath and thus prevented total conversion of the silver to silver sulfide.
2Ag1s2 H2S1aq2 ¡ Ag2S1s2 H21g2 Silver coins and tankards salvaged from the wreck of the Atocha.
17.8 Commercial Electrolytic Processes 821 17.8 Commercial Electrolytic Processes The chemistry of metals is characterized by their ability to donate electrons to form ions.
Because metals are typically such good reducing agents, most are found in nature in ores, mixtures of ionic compounds often containing oxide, sulfide, and silicate anions. The no-ble metals, such as gold, silver, and platinum, are more difficult to oxidize and are often found as pure metals.
Production of Aluminum Aluminum is one of the most abundant elements on earth, ranking third behind oxygen and silicon. Since aluminum is a very active metal, it is found in nature as its oxide in an ore called bauxite (named after Les Baux, France, where it was discovered in 1821).
Production of aluminum metal from its ore proved to be more difficult than produc-tion of most other metals. In 1782 Lavoisier recognized aluminum to be a metal “whose affinity for oxygen is so strong that it cannot be overcome by any known re-ducing agent.” As a result, pure aluminum metal remained unknown. Finally, in 1854 Another change that took place as the wood decom-posed was the formation of carbon dioxide. This shifted the equilibrium that is present in the ocean, to the right, producing higher concentrations of . In turn, the reacted with ions present in the sea-water to form calcium carbonate: Calcium carbonate is the main component of limestone.
Thus, over time, the corroded silver coins and bars became encased in limestone.
Both the limestone formation and the corrosion had to be dealt with. Since CaCO3 contains the basic anion acid dissolves limestone: Soaking the mass of coins in a buffered acidic bath for sev-eral hours allowed the individual pieces to be separated, and the black Ag2S on the surfaces was revealed. An abrasive could not be used to remove this corrosion; it would have de-stroyed the details of the engraving—a very valuable feature of the coins to a historian or a collector—and it would have washed away some of the silver. Instead, the corrosion reac-tion was reversed through electrolytic reduction. The coins were connected to the cathode of an electrolytic cell in a di-lute sodium hydroxide solution as represented in the figure.
2H1aq2 CaCO31s2 ¡ Ca21aq2 CO21g2 H2O1l2 CO3 2, Ca 21aq2 HCO3 1aq2 ∆CaCO31s2 H1aq2 Ca 2 HCO3 HCO3 CO21aq2 H2O1l2 ∆HCO3 1aq2 H1aq2 As electrons flow, the Ag ions in the silver sulfide are reduced to silver metal: As a by-product, bubbles of hydrogen gas from the reduc-tion of water form on the surface of the coins: The agitation caused by the bubbles loosens the flakes of metal sulfide and helps clean the coins.
These procedures have made it possible to restore the treasure to very nearly its condition when the Atocha sailed many years ago.
2H2O 2e ¡ H21g2 2OH Ag2S 2e ¡ Ag S2 Na+ OH– H2O Power source Coin coated with Ag2S Cathode Anode e– e– 822 Chapter Seventeen Electrochemistry a process was found for producing metallic aluminum using sodium, but aluminum remained a very expensive rarity. In fact, it is said that Napoleon III served his most honored guests with aluminum forks and spoons, while the others had to settle for gold and silver utensils.
The breakthrough came in 1886 when two men, Charles M. Hall in the United States and Paul Heroult in France, almost simultaneously discovered a practical electrolytic process for producing aluminum (see Fig. 17.21). The key factor in the Hall–Heroult process is the use of molten cryolite (Na3AlF6) as the solvent for the aluminum oxide.
Electrolysis is possible only if ions can move to the electrodes. A common method for producing ion mobility is dissolving the substance to be electrolyzed in water. This is not possible in the case of aluminum because water is more easily reduced than Al3, as the following standard reduction potentials show: Thus aluminum metal cannot be plated out of an aqueous solution of Al3.
Ion mobility also can be produced by melting the salt. But the melting point of solid Al2O3 is much too high to allow practical electrolysis of the molten oxide. A mixture of Al2O3 and Na3AlF6, however, has a melting point of and the result-ing molten mixture can be used to obtain aluminum metal electrolytically. Because of this discovery by Hall and Heroult, the price of aluminum plunged (see Table 17.3), and its use became economically feasible.
Bauxite is not pure aluminum oxide (called alumina); it also contains the oxides of iron, silicon, and titanium, and various silicate materials. To obtain the pure hydrated alu-mina the crude bauxite is treated with aqueous sodium hydroxide. Being amphoteric, alumina dissolves in the basic solution: The other metal oxides, which are basic, remain as solids. The solution containing the aluminate ion is separated from the sludge of the other oxides and is acidified with carbon dioxide gas, causing the hydrated alumina to reprecipitate: The purified alumina is then mixed with cryolite and melted, and the aluminum ion is reduced to aluminum metal in an electrolytic cell of the type shown in Fig. 17.22. Because the electrolyte solution contains a large number of aluminum-containing ions, the chemistry is not completely clear. However, the alumina probably reacts with the cryolite anion as follows: The electrode reactions are thought to be The overall cell reaction can be written as The aluminum produced in this electrolytic process is 99.5% pure. To be useful as a structural material, aluminum is alloyed with metals such as zinc (used for trailer and air-craft construction) and manganese (used for cooking utensils, storage tanks, and highway signs). The production of aluminum consumes about 5% of all the electricity used in the United States.
2Al2O3 3C ¡ 4Al 3CO2 Anode reaction: 2Al2OF6 2 12F C ¡ 4AlF6 3 CO2 4e Cathode reaction: AlF6 3 3e ¡ Al 6F Al2O3 4AlF6 3 ¡ 3Al2OF6 2 6F 2CO21g2 2AlO2 1aq2 1n 12H2O1l2 ¡ 2HCO3 1aq2 Al2O3 nH2O1s2 (AlO2 ) Al2O31s2 2OH1aq2 ¡ 2AlO2 1aq2 H2O1l2 1Al2O3 nH2O2, 1000°C, 12050°C2 2H2O 2e ¡ H2 2OH e° 0.83 V Al3 3e ¡ Al e° 1.66 V FIGURE 17.21 Charles Martin Hall (1863–1914) was a stu-dent at Oberlin College in Ohio when he first became interested in aluminum. One of his professors commented that anyone who could manufacture aluminum cheaply would make a fortune, and Hall decided to give it a try. The 21-year-old Hall worked in a wooden shed near his house with an iron frying pan as a container, a blacksmith’s forge as a heat source, and galvanic cells constructed from fruit jars. Using these crude galvanic cells, Hall found that he could produce aluminum by passing a current through a molten Al2O3/Na3AlF6 mixture. By a strange coincidence, Paul Heroult, a Frenchman who was born and died in the same years as Hall, made the same discovery at about the same time.
TABLE 17.3 The Price of Aluminum over the Past Century Date Price of Aluminum ($/lb) 1855 100,000 1885 100 1890 2 1895 0.50 1970 0.30 1980 0.80 1990 0.74 Note the precipitous drop in price after the discovery of the Hall–Heroult process.
17.8 Commercial Electrolytic Processes 823 Electrorefining of Metals Purification of metals is another important application of electrolysis. For example, im-pure copper from the chemical reduction of copper ore is cast into large slabs that serve as the anodes for electrolytic cells. Aqueous copper sulfate is the electrolyte, and thin sheets of ultrapure copper function as the cathodes (see Fig. 17.23).
The main reaction at the anode is Other metals such as iron and zinc are also oxidized from the impure anode: Fe ¡ Fe2 2e Zn ¡ Zn2 2e Cu ¡ Cu2 2e Carbon dioxide formed at the anodes Carbon-lined iron tank Plug Molten Al2O3/Na3AlF6 mixture To external power source Electrodes of graphite rods Molten aluminum FIGURE 17.22 A schematic diagram of an electrolytic cell for producing aluminum by the Hall–Heroult process. Because molten aluminum is more dense than the mixture of molten cryolite and alumina, it settles to the bottom of the cell and is drawn off periodically. The graphite elec-trodes are gradually eaten away and must be replaced from time to time. The cell operates at a current flow of up to 250,000 A.
FIGURE 17.23 Ultrapure copper sheets that serve as the cathodes are lowered between slabs of im-pure copper that serve as the anodes into a tank containing an aqueous solution of copper sulfate (CuSO4). It takes about four weeks for the anodes to dissolve and for the pure copper to be deposited on the cathodes.
824 Chapter Seventeen Electrochemistry Noble metal impurities in the anode are not oxidized at the voltage used; they fall to the bottom of the cell to form a sludge, which is processed to remove the valuable silver, gold, and platinum.
The ions from the solution are deposited onto the cathode producing copper that is 99.95% pure.
Metal Plating Metals that readily corrode can often be protected by the application of a thin coating of a metal that resists corrosion. Examples are “tin” cans, which are actually steel cans with a thin coating of tin, and chrome-plated steel bumpers for automobiles.
An object can be plated by making it the cathode in a tank containing ions of the plating metal. The silver plating of a spoon is shown schematically in Fig. 17.24(b). In an actual plating process, the solution also contains ligands that form complexes with the silver ion. By lowering the concentration of Ag in this way, a smooth, even coating of silver is obtained.
Electrolysis of Sodium Chloride Sodium metal is mainly produced by the electrolysis of molten sodium chloride. Because solid NaCl has a rather high melting point it is usually mixed with solid CaCl2 to lower the melting point to about . The mixture is then electrolyzed in a Downs cell, as illustrated in Fig. 17.25, where the reactions are Cathode reaction: Na e ¡ Na Anode reaction: 2Cl ¡ Cl2 2e 1600°C2 1800°C2, Cu2 2e ¡ Cu Cu2 Cathode Anode e– e– e– e– (b) Ag+ Ag Ag+ e– Ag e– Power source FIGURE 17.24 (a) A silver-plated teapot. Silver plating is often used to beautify and protect cutlery and items of table service.
(b) Schematic of the electroplating of a spoon. The item to be plated is the cathode, and the anode is a silver bar.
Silver is plated out at the cathode: Note that a salt bridge is not needed here because Ag ions are involved at both electrodes.
Ag e S Ag.
Addition of a nonvolatile solute lowers the melting point of the solvent, molten NaCl in this case.
(a) 17.8 Commercial Electrolytic Processes 825 At the temperatures in the Downs cell, the sodium is liquid and is drained off, then cooled, and cast into blocks. Because it is so reactive, sodium must be stored in an inert solvent, such as mineral oil, to prevent its oxidation.
Electrolysis of aqueous sodium chloride (brine) is an important industrial process for the production of chlorine and sodium hydroxide. In fact, this process is the second largest consumer of electricity in the United States, after the production of aluminum. Sodium is not produced in this process under normal circumstances because H2O is more easily re-duced than Na, as the standard reduction potentials show: Hydrogen, not sodium, is produced at the cathode.
For the reasons we discussed in Section 17.7, chlorine gas is produced at the anode.
Thus the electrolysis of brine produces hydrogen and chlorine: It leaves a solution containing dissolved NaOH and NaCl.
The contamination of the sodium hydroxide by NaCl can be virtually eliminated using a special mercury cell for electrolyzing brine (see Fig. 17.26). In this cell, mercury is the conductor at the cathode, and because hydrogen gas has an extremely high overvoltage with a mercury electrode, Na is reduced instead of H2O. The resulting sodium metal dis-solves in the mercury, forming a liquid alloy, which is then pumped to a chamber where the dissolved sodium reacts with water to produce hydrogen: Relatively pure solid NaOH can be recovered from the aqueous solution, and the re-generated mercury is then pumped back to the electrolysis cell. This process, called the chlor–alkali process, was the main method for producing chlorine and sodium hydroxide in the United States for many years. However, because of the environmen-tal problems associated with the mercury cell, it has been largely displaced in the 2Na1s2 2H2O1l2 ¡ 2Na1aq2 2OH1aq2 H21g2 Cathode reaction: 2H2O 2e ¡ H2 2OH Anode reaction: 2Cl ¡ Cl2 2e 2H2O 2e ¡ H2 2OH e° 0.83 V Na e ¡ Na e° 2.71 V Molten mixture of NaCl and CaCl2 Cathode Cathode Anode Iron screen Liquid sodium metal Cl2 gas Iron screen Anode FIGURE 17.25 The Downs cell for the electrolysis of molten sodium chloride. The cell is de-signed so that the sodium and chlorine produced cannot come into contact with each other to re-form NaCl.
826 Chapter Seventeen Electrochemistry chlor–alkali industry by other technologies. In the United States, nearly 75% of the chlor–alkali production is now carried out in diaphragm cells. In a diaphragm cell the cathode and anode are separated by a diaphragm that allows passage of H2O mol-ecules, Na ions, and, to a limited extent, ions. The diaphragm does not allow ions to pass through it. Thus the H2 and formed at the cathode are kept separate from the Cl2 formed at the anode. The major disadvantage of this process is that the aqueous effluent pumped from the cathode compartment contains a mixture of sodium hydroxide and unreacted sodium chloride, which must be separated if pure sodium hydroxide is a desired product.
In the past 30 years, a new process has been developed in the chlor–alkali industry that employs a membrane to separate the anode and cathode compartments in brine elec-trolysis cells. The membrane is superior to a diaphragm because the membrane is imper-meable to anions. Only cations can flow through the membrane. Because neither nor ions can pass through the membrane separating the anode and cathode compart-ments, NaCl contamination of the NaOH formed at the cathode does not occur. Although membrane technology is now just becoming prominent in the United States, it is the dominant method for chlor–alkali production in Japan.
OH Cl OH OH Cl Key Terms electrochemistry Section 17.1 oxidation–reduction (redox) reaction reducing agent oxidizing agent oxidation reduction half-reactions salt bridge porous disk galvanic cell anode cathode Electrochemistry The study of the interchange of chemical and electrical energy Employs oxidation–reduction reactions Galvanic cell: chemical energy is transformed into electrical energy by separating the oxidizing and reducing agents and forcing the electrons to travel through a wire Electrolytic cell: electrical energy is used to produce a chemical change Galvanic cell Anode: the electrode where oxidation occurs Cathode: the electrode where reduction occurs For Review NaOH solution H2O Brine Brine Hg Na in Hg(l) Hg Cl2 gas H2 gas Anode Mercury cathode Hg/Na FIGURE 17.26 The mercury cell for production of chlorine and sodium hydroxide. The large overvolt-age required to produce hydrogen at a mer-cury electrode means that Na ions are reduced rather than water. The sodium formed dissolves in the liquid mercury and is pumped to a chamber where it reacts with water.
For Review 827 cell potential (electromotive force) volt voltmeter potentiometer Section 17.2 standard hydrogen electrode standard reduction potentials Section 17.3 faraday Section 17.4 concentration cell Nernst equation glass electrode ion-selective electrode Section 17.5 battery lead storage battery dry cell battery fuel cell Section 17.6 corrosion galvanizing cathodic protection Section 17.7 electrolytic cell electrolysis ampere Section 17.8 Downs cell mercury cell chlor–alkali process The driving force behind the electron transfer is called the cell potential ( ) • The potential is measured in units of volts (V), defined as a joule of work per coulomb of charge: • A system of half-reactions, called standard reduction potentials, can be used to calculate the potentials of various cells • The half-reaction is arbitrarily assigned a potential of 0 V Free energy and work The maximum work that a cell can perform is where represents the cell potential when no current is flowing The actual work obtained from a cell is always less than the maximum because energy is lost through frictional heating of the wire when current flows For a process carried out at constant temperature and pressure, the change in free energy equals the maximum useful work obtainable from that process: where F (faraday) equals 96,485 C and n is the number of moles of electrons transferred in the process Concentration cell A galvanic cell in which both compartments have the same components but at different concentrations The electrons flow in the direction that tends to equalize the concentrations Nernst equation Shows how the cell potential depends on the concentrations of the cell components: When a galvanic cell is at equilibrium, and Batteries A battery consists of a galvanic cell or group of cells connected in series that serve as a source of direct current Lead storage battery • Anode: lead • Cathode: lead coated with PbO2 • Electrolyte: H2SO4(aq) Dry cell battery • Contains a moist paste instead of a liquid electrolyte • Anode: usually Zn • Cathode: carbon rod in contact with an oxidizing agent (which varies depend-ing on the application) Fuel cells Galvanic cells in which the reactants are continuously supplied The H2/O2 fuel cell is based on the reaction between H2 and O2 to form water Corrosion Involves the oxidation of metals to form mainly oxides and sulfides Q K e 0 e e0 0.0591 n log Q at 25°C ¢G wmax qemax nFe emax wmax qemax 2H 2e ¡ H2 e1V2 work 1J2 charge 1C2 w q ecell 828 Chapter Seventeen Electrochemistry Some metals, such as aluminum and chromium, form a thin protective oxide coat-ing that prevents further corrosion The corrosion of iron to form rust is an electrochemical process • The Fe2 ions formed at anodic areas of the surface migrate through the mois-ture layer to cathodic regions, where they react with oxygen from the air • Iron can be protected from corrosion by coating it with paint or with a thin layer of metal such as chromium, tin, or zinc; by alloying; and by cathodic protection Electrolysis Used to place a thin coating of metal onto steel Used to produce pure metals such as aluminum and copper REVIEW QUESTIONS 1. What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.
2. Galvanic cells harness spontaneous oxidation–reduction reactions to produce work by producing a current. They do so by controlling the flow of electrons from the species oxidized to the species reduced. How is a galvanic cell de-signed? What is in the cathode compartment? The anode compartment? What purpose do electrodes serve? Which way do electrons always flow in the wire connecting the two electrodes in a galvanic cell? Why is it necessary to use a salt bridge or a porous disk in a galvanic cell? Which way do cations flow in the salt bridge? Which way do the anions flow? What is a cell potential and what is a volt?
3. Table 17.1 lists common half-reactions along with the standard reduction poten-tial associated with each half-reaction. These standard reduction potentials are all relative to some standard. What is the standard (zero point)? If is positive for a half-reaction, what does it mean? If is negative for a half-reaction, what does it mean? Which species in Table 17.1 is most easily reduced? Least easily reduced? The reverse of the half-reactions in Table 17.1 are the oxida-tion half-reactions. How are standard oxidation potentials determined? In Table 17.1, which species is the best reducing agent? The worst reducing agent?
To determine the standard cell potential for a redox reaction, the standard reduction potential is added to the standard oxidation potential. What must be true about this sum if the cell is to be spontaneous (produce a galvanic cell)?
Standard reduction and oxidation potentials are intensive. What does this mean?
Summarize how line notation is used to describe galvanic cells.
4. Consider the equation What are the four terms in this equation?
Why does a minus sign appear in the equation? What does the superscript indicate?
5. The Nernst equation allows determination of the cell potential for a galvanic cell at nonstandard conditions. Write out the Nernst equation. What are non-standard conditions? What do , , n, and Q stand for in the Nernst equation?
What does the Nernst equation reduce to when a redox reaction is at equilib-rium? What are the signs of and when When When Explain the following statement: determines spontaneity, while determines the equilibrium position. Under what conditions can you use to predict spontaneity?
6. What are concentration cells? What is in a concentration cell? What is the driving force for a concentration cell to produce a voltage? Is the higher or the lower ion concentration solution present at the anode? When the anode ion con-centration is decreased and/or the cathode ion concentration is increased, both e° e° e° e K 1?
K 7 1?
K 6 1?
e° ¢G° e° e ° ¢G° nFe°.
e° e° Active Learning Questions 829 give rise to larger cell potentials. Why? Concentration cells are commonly used to calculate the value of equilibrium constants for various reactions. For exam-ple, the silver concentration cell illustrated in Fig. 17.9 can be used to determine the Ksp value for AgCl(s). To do so, NaCl is added to the anode compartment until no more precipitate forms. The in solution is then determined some-how. What happens to when NaCl is added to the anode compartment? To calculate the Ksp value, [Ag] must be calculated. Given the value of , how is [Ag] determined at the anode?
7. Batteries are galvanic cells. What happens to as a battery discharges? Does a battery represent a system at equilibrium? Explain. What is when a bat-tery reaches equilibrium? How are batteries and fuel cells alike? How are they different? The U.S. space program utilizes hydrogen–oxygen fuel cells to pro-duce power for its spacecraft. What is a hydrogen–oxygen fuel cell?
8. Not all spontaneous redox reactions produce wonderful results. Corrosion is an example of a spontaneous redox process that has negative effects. What hap-pens in the corrosion of a metal such as iron? What must be present for the cor-rosion of iron to take place? How can moisture and salt increase the severity of corrosion? Explain how the following protect metals from corrosion: a. paint b. durable oxide coatings c. galvanizing d. sacrificial metal e. alloying f. cathodic protection 9. What characterizes an electrolytic cell? What is an ampere? When the current applied to an electrolytic cell is multiplied by the time in seconds, what quan-tity is determined? How is this quantity converted to moles of electrons required? How are moles of electrons required converted to moles of metal plated out? What does plating mean? How do you predict the cathode and the anode half-reactions in an electrolytic cell? Why is the electrolysis of molten salts much easier to predict in terms of what occurs at the anode and cathode than the electrolysis of aqueous dissolved salts? What is overvoltage?
10. Electrolysis has many important industrial applications. What are some of these applications? The electrolysis of molten NaCl is the major process by which sodium metal is produced. However, the electrolysis of aqueous NaCl does not produce sodium metal under normal circumstances. Why? What is purification of a metal by electrolysis?
ecell ecell ecell ecell [Cl ] Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. Sketch a galvanic cell, and explain how it works. Look at Figs. 17.1 and 17.2. Explain what is occurring in each container and why the cell in Fig. 17.2 “works” but the one in Fig. 17.1 does not.
2. In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.
3. You want to “plate out” nickel metal from a nickel nitrate solu-tion onto a piece of metal inserted into the solution. Should you use copper or zinc? Explain.
4. A copper penny can be dissolved in nitric acid but not in hy-drochloric acid. Using reduction potentials from the book, show why this is so. What are the products of the reaction? Newer pennies contain a mixture of zinc and copper. What happens to the zinc in the penny when the coin is placed in nitric acid? Hy-drochloric acid? Support your explanations with data from the book, and include balanced equations for all reactions.
5. Sketch a cell that forms iron metal from iron(II) while changing chromium metal to chromium(III). Calculate the voltage, show 830 Chapter Seventeen Electrochemistry a. HNO3 e. C6H12O6 i. Na2C2O4 b. CuCl2 f. Ag j. CO2 c. O2 g. PbSO4 k. (NH4)2Ce(SO4)3 d. H2O2 h. PbO2 l. Cr2O3 15. Specify which of the following equations represent oxidation– reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced.
a.
b.
c.
d.
16. Balance each of the following equations by the half-reaction method for the pH conditions specified.
a.
b.
c.
d.
e.
f.
Questions 17. When magnesium metal is added to a beaker of HCl(aq), a gas is produced. Knowing that magnesium is oxidized and that hy-drogen is reduced, write the balanced equation for the reaction.
How many electrons are transferred in the balanced equation?
What quantity of useful work can be obtained when Mg is added directly to the beaker of HCl? How can you harness this reac-tion to do useful work?
18. How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?
19. The free energy change for a reaction, G, is an extensive property.
What is an extensive property? Surprisingly, one can calculate G from the cell potential, , for the reaction. This is surprising because is an intensive property. How can the extensive property G be calculated from the intensive property ?
20. What is wrong with the following statement: The best concen-tration cell will consist of the substance having the most posi-tive standard reduction potential. What drives a concentration cell to produce a large voltage?
21. When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block. Why?
22. In theory, most metals should easily corrode in air. Why? A group of metals called the noble metals are relatively difficult to corrode in air. Some noble metals include: gold, platinum, and silver.
Reference Table 17.1 to come up with a possible reason why the noble metals are relatively difficult to corrode.
23. Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?
24. Although aluminum is one of the most abundant elements on earth, production of pure Al proved very difficult until the late 1800s. At this time, the Hall–Heroult process made it relatively easy to produce pure Al. Why was pure Al so difficult to produce and what was the key discovery behind the Hall–Heroult process?
e e e HCO3 1aq2 Ag1s2 NH31aq2 1basic2 H2CO1aq2 Ag1NH322 1aq2 S Mg1s2 OCl 1aq2 S Mg1OH221s2 Cl 1aq2 1basic2 PO4 31aq2 MnO21s2 1basic2 PO3 31aq2 MnO4 1aq2 S CH3OH1aq2 Ce 41aq2 S CO21aq2 Ce 31aq2 1acidic2 Al1s2 MnO4 1aq2 S Al 31aq2 Mn 21aq2 1acidic2 Cr1s2 NO3 1aq2 S Cr 31aq2 NO1g2 1acidic2 2H 1aq2 2CrO4 21aq2 S Cr2O7 21aq2 H2O1l2 Zn1s2 2HCl1aq2 S ZnCl21aq2 H21g2 2AgNO31aq2 Cu1s2 S Cu1NO3221aq2 2Ag1s2 CH41g2 H2O1g2 S CO1g2 3H21g2 the electron flow, label the anode and cathode, and balance the overall cell equation.
6. Which of the following is the best reducing agent: F2, H2, Na, Na, ? Explain. Order as many of these species as possible from the best to the worst oxidizing agent. Why can’t you order all of them? From Table 17.1 choose the species that is the best oxidizing agent. Choose the best reducing agent. Explain.
7. You are told that metal A is a better reducing agent than metal B.
What, if anything, can be said about A and B? Explain.
8. Explain the following relationships: and w, cell potential and w, cell potential and cell potential and Q. Using these relationships, explain how you could make a cell in which both electrodes are the same metal and both solutions contain the same compound, but at different concentrations. Why does such a cell run spontaneously?
9. Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between and cell potential to do this.) 10. What is the difference between and ? When is equal to zero? When is equal to zero? (Consider “regular” galvanic cells as well as concentration cells.) 11. Consider the following galvanic cell: What happens to as the concentration of is increased?
As the concentration of Ag is increased? What happens to in these cases?
12. Look up the reduction potential for Fe3 to . Look up the reduction potential for to Fe. Finally, look up the reduc-tion potential for Fe3 to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two po-tentials to calculate the third potential.
A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Review of Oxidation–Reduction Reactions If you have trouble with these exercises, you should review Sections 4.9 and 4.10.
13. Define oxidation and reduction in terms of both change in oxi-dation number and electron loss or gain.
14. Assign oxidation numbers to all the atoms in each of the fol-lowing.
Fe2 Fe 2 e° Zn 2 e Zn Ag 1.0 M Zn2+ 1.0 M Ag+ e° e e° e ¢G ¢G, ¢G F Exercises 831 Exercises In this section similar exercises are paired.
Galvanic Cells, Cell Potentials, Standard Reduction Potentials, and Free Energy 25. Sketch the galvanic cells based on the following overall reactions.
Show the direction of electron flow and identify the cathode and anode. Give the overall balanced reaction. Assume that all con-centrations are 1.0 M and that all partial pressures are 1.0 atm.
a.
b.
26. Sketch the galvanic cells based on the following overall reac-tions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction. Assume that all con-centrations are 1.0 M and that all partial pressures are 1.0 atm.
a.
b.
27. Calculate values for the galvanic cells in Exercise 25.
28. Calculate values for the galvanic cells in Exercise 26.
29. Sketch the galvanic cells based on the following half-reactions.
Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm.
a.
b.
30. Sketch the galvanic cells based on the following half-reactions.
Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced reaction, and determine for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm.
a.
b.
31. Give the standard line notation for each cell in Exercises 25 and 29.
32. Give the standard line notation for each cell in Exercises 26 and 30.
33. Consider the following galvanic cells: e° 0.036 V Fe 3 3e S Fe e° 1.18 V Mn 2 2e S Mn e° 0.68 V O2 2H 2e S H2O2 e° 1.78 V H2O2 2H 2e S 2H2O e° e° 1.60 V IO4 2H 2e S IO3 H2O e° 1.51 V MnO4 8H 5e S Mn 2 4H2O Br2 2e S 2Br e° 1.09 V Cl2 2e S 2Cl e° 1.36 V e° e° e° Zn1s2 Ag1aq2 ∆Zn21aq2 Ag1s2 IO3 1aq2 Fe21aq2 ∆Fe31aq2 I21aq2 Cu 21aq2 Mg1s2 ∆Mg21aq2 Cu1s2 Cr31aq2 Cl21g2 ∆Cr2O7 21aq2 Cl 1aq2 For each galvanic cell, give the balanced cell reaction and de-termine . Standard reduction potentials are found in Table 17.1.
34. Give the balanced cell reaction and determine for the gal-vanic cells based on the following half-reactions. Standard re-duction potentials are found in Table 17.1.
a.
b.
35. Calculate values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Bal-ance the reactions. Standard reduction potentials are found in Table 17.1.
a.
b.
36. Calculate values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the reactions that are not already balanced. Standard reduction po-tentials are found in Table 17.1.
a.
b.
37. Chlorine dioxide (ClO2), which is produced by the reaction has been tested as a disinfectant for municipal water treatment.
Using data from Table 17.1, calculate and at for the production of ClO2.
38. The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing ions. These ions are then oxidized to the deeply colored ions by periodate ion in acid solution.
a. Complete and balance an equation describing each of the above reactions.
b. Calculate and at for each reaction.
39. Calculate the maximum amount of work that can be obtained from the galvanic cells at standard conditions in Exercise 33.
40. Calculate the maximum amount of work that can be obtained from the galvanic cells at standard conditions in Exercise 34.
41. Calculate for the reaction using values of in Appendix 4.
42. The equation also can be applied to half-reactions. Use standard reduction potentials to estimate for (aq) and Fe3(aq). .) 43. Using data from Table 17.1, place the following in order of increasing strength as oxidizing agents (all under standard con-ditions).
44. Using data from Table 17.1, place the following in order of increasing strength as reducing agents (all under standard con-ditions).
Cu , F , H , H2O, I2, K Cd 2, IO3 , K, H2O, AuCl4 , I2 (¢G° f for e 0 Fe 2 ¢G° f ¢G° nF e° ¢G° f CH3OH1l2 3 2O21g2 ¡ CO21g2 2H2O1l2 e° 25°C ¢G° e° (IO4 ) MnO4 Mn 2 25°C ¢G° e° 2NaClO21aq2 Cl21g2 ¡ 2ClO21g2 2NaCl1aq2 Au31aq2 Ag1s2 ∆Ag 1aq2 Au1s2 H21g2 ∆H 1aq2 H 1aq2 e° MnO4 1aq2 F 1aq2 ∆F21g2 Mn 21aq2 MnO4 1aq2 I1aq2 ∆I21aq2 Mn 21aq2 e° Al3 3e S Al 2H 2e S H2 H2O2 2H 2e S 2H2O Cr2O7 2 14H 6e S 2Cr 3 7H2O e° e° Pt Au 1.0 M Au3+ 1.0 M Cu+ 1.0 M Cu2+ Pt Cd 1.0 M Cd2+ 1.0 M VO2 + 1.0 M VO2+ 1.0 M H+ (a) (b) 832 Chapter Seventeen Electrochemistry 53. Consider the concentration cell shown below. Calculate the cell potential at when the concentration of Ag in the com-partment on the right is the following.
a. 1.0 M b. 2.0 M c. 0.10 M d.
M e. Calculate the potential when both solutions are 0.10 M in Ag.
For each case, also identify the cathode, the anode, and the di-rection in which electrons flow.
54. Consider a concentration cell similar to the one shown in Exer-cise 53, except that both electrodes are made of Ni and in the left-hand compartment M. Calculate the cell po-tential at when the concentration of in the compart-ment on the right has each of the following values.
a. 1.0 M b. 2.0 M c. 0.10 M d.
M e. Calculate the potential when both solutions are 2.5 M in For each case, also identify the cathode, anode, and the direc-tion in which electrons flow.
55. The overall reaction in the lead storage battery is Calculate at for this battery when that is, At for the lead storage battery.
56. Calculate the pH of the cathode compartment for the follow-ing reaction given when and 57. Consider the cell described below: Calculate the cell potential after the reaction has operated long enough for the to have changed by 0.20 mol/L.
58. Consider the cell described below: Al 0 Al 3 11.00 M2 0 0 Pb2 11.00 M2 0 Pb (Assume T 25°C.) [Zn2] Zn 0 Zn 211.00 M2 0 0Cu 211.00 M2 0 Cu 2Al 31aq2 2Cr 31aq2 7H2O1l2 2Al1s2 Cr2O7 21aq2 14H 1aq2 S [Cr2O7 2] 0.55 M.
[Al3] 0.30 M, [Cr 3] 0.15 M, ecell 3.01 V e° 2.04 V 25°C, [H ] [HSO4 ] 4.5 M.
[H2SO4] 4.5 M, 25°C e 2PbSO41s2 2H2O1l2 Pb1s2 PbO21s2 2H 1aq2 2HSO4 1aq2 ¡ Ni2.
4.0 105 Ni 2 25°C [Ni 2] 1.0 Ag Ag [Ag+] = 1.0 M 4.0 105 25°C 45. Answer the following questions using data from Table 17.1 (all under standard conditions).
a. Is H(aq) capable of oxidizing Cu(s) to (aq)?
b. Is Fe3(aq) capable of oxidizing (aq)?
c. Is H2(g) capable of reducing Ag(aq)?
d. Is (aq) capable of reducing Cr3(aq) to (aq)?
46. Consider only the species (at standard conditions) in answering the following questions. Give reasons for your answers. (Use data from Table 17.1.) a. Which is the strongest oxidizing agent?
b. Which is the strongest reducing agent?
c. Which species can be oxidized by (aq) in acid?
d. Which species can be reduced by Al(s)?
47. Use the table of standard reduction potentials (Table 17.1) to pick a reagent that is capable of each of the following oxidations (under standard conditions in acidic solution).
a. Oxidize to Br2 but not oxidize to Cl2 b. Oxidize Mn to but not oxidize Ni to 48. Use the table of standard reduction potentials (Table 17.1) to pick a reagent that is capable of each of the following reductions (under standard conditions in acidic solution).
a. Reduce to Cu but not reduce to Cu.
b. Reduce Br2 to but not reduce I2 to .
49. Hydrazine is somewhat toxic. Use the half-reactions shown be-low to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia.
50. The compound with the formula TlI3 is a black solid. Given the following standard reduction potentials, would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?
The Nernst Equation 51. A galvanic cell is based on the following half-reactions at Predict whether is larger or smaller than for the fol-lowing cases.
a.
b.
52. Consider the concentration cell in Fig. 17.10. If the con-centration in the right compartment is changed from 0.1 M to predict the direction of electron flow, and des-ignate the anode and cathode compartments.
1 107 M Fe2, Fe 2 [Ag ] 2.0 M, [H2O2] 1.0 M, [H ] 1.0 107 M [Ag ] 1.0 M, [H2O2] 2.0 M, [H ] 2.0 M e° cell ecell H2O2 2H 2e ¡ 2H2O Ag e ¡ Ag 25°C: e° 0.55 V I3 2e ¡ 3I e° 1.25 V Tl 3 2e ¡ Tl e° 0.10 V N2H4 2H2O 2e ¡ 2NH3 2OH e° 0.90 V ClO H2O 2e ¡ 2OH Cl I Br Cu 2 Cu 2 Ni 2 Mn 2 Cl Br SO4 2 Na, Cl , Ag , Ag, Zn 2, Zn, Pb Cr2 Fe 2 I Cu 2 Exercises 833 Calculate the cell potential after the reaction has operated long enough for the [Al3] to have changed by 0.60 mol/L.
59. An electrochemical cell consists of a standard hydrogen elec-trode and a copper metal electrode.
a. What is the potential of the cell at if the copper electrode is placed in a solution in which M?
b. The copper electrode is placed in a solution of unknown . The measured potential at is 0.195 V. What is ? (Assume is reduced.) 60. An electrochemical cell consists of a nickel metal electrode im-mersed in a solution with M separated by a porous disk from an aluminum metal electrode.
a. What is the potential of this cell at if the aluminum electrode is placed in a solution in which M?
b. When the aluminum electrode is placed in a certain solution in which [Al3] is unknown, the measured cell potential at is 1.62 V. Calculate [Al3] in the unknown solution.
(Assume Al is oxidized.) 61. An electrochemical cell consists of a standard hydrogen elec-trode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10 M NaOH that is saturated with Cu(OH)2, what is the cell potential at ? (For Cu(OH)2, 62. An electrochemical cell consists of a nickel metal electrode im-mersed in a solution with separated by a porous disk from an aluminum metal electrode immersed in a solution with . Sodium hydroxide is added to the alu-minum compartment, causing Al(OH)3(s) to precipitate. After precipitation of Al(OH)3 has ceased, the concentration of is and the measured cell potential is 1.82 V. Cal-culate the Ksp value for Al(OH)3.
63. Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell con-tains 1.0 M M2. Solution B in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mol of M(NO3)2 and 0.0100 mol of Na2SO4 are dissolved in so-lution B (ignore volume changes), where the reaction occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at Assume that the process has a standard reduction potential of V and that no other redox process occurs in the cell. Calculate the value of Ksp for MSO4(s) at 64. You have a concentration cell in which the cathode has a silver electrode with 0.10 M Ag. The anode also has a silver electrode with Ag(aq), 0.050 M , and You read the voltage to be 0.76 V.
Ag(S2O3)2 3.
1.0 10 3 M S2O3 2 25°C.
0.31 M 2 2e ¡ M 25°C.
M 21aq2 SO4 21aq2 ∆MSO41s2 Al1OH231s2 ∆Al 31aq2 3OH1aq2 Ksp ?
1.0 104 M OH [Al 3] 1.0 M [Ni 2] 1.0 M Ksp 1.6 1019.) 25°C 25°C 103 [Al 3] 7.2 25°C 3Ni 24 1.0 Cu2 [Cu 2] 25°C [Cu2] [Cu2] 2.5 104 25°C (Assume T 25°C.) a. Calculate the concentration of Ag at the cathode.
b. Determine the value of the equilibrium constant for the for-mation of 65. Calculate and K at for the reactions in Exercises 25 and 29.
66. Calculate and K at for the reactions in Exercises 26 and 30.
67. An excess of finely divided iron is stirred up with a solution that contains Cu2 ion, and the system is allowed to come to equilibrium. The solid materials are then filtered off, and electrodes of solid copper and solid iron are inserted into the remaining solution. What is the value of the ratio 68. Consider the following reaction: Determine the minimum ratio of necessary to make this reaction spontaneous as written.
69. Under standard conditions, what reaction occurs, if any, when each of the following operations is performed?
a. Crystals of I2 are added to a solution of NaCl.
b. Cl2 gas is bubbled into a solution of NaI.
c. A silver wire is placed in a solution of CuCl2.
d. An acidic solution of FeSO4 is exposed to air.
For the reactions that occur, write a balanced equation and cal-culate and K at .
70. A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products.
Which of the following disproportionation reactions are spon-taneous under standard conditions? Calculate and K at for those reactions that are spontaneous under standard conditions.
a.
b.
c.
Use the half-reactions: 71. Consider the galvanic cell based on the following half-reactions: a. Determine the overall cell reaction and calculate b. Calculate and K for the cell reaction at .
c. Calculate at when and .
72. Consider the following galvanic cell at Pt0 Cr 210.30 M2, Cr 312.0 M2 0 0 Co210.20 M2 0 Co 25°C: [Tl] 1.0 104 M [Au 3] 1.0 102 M 25°C ecell 25°C ¢G° e° cell.
e° 0.34 V Tl e ¡ Tl e° 1.50 V Au 3 3e ¡ Au e° 1.65 V HClO2 2H 2e ¡ HClO H2O e° 1.21 V ClO3 3H 2e ¡ HClO2 H2O HClO21aq2 S ClO3 1aq2 HClO1aq2 1unbalanced2 3Fe 21aq2 S 2Fe 31aq2 Fe1s2 2Cu1aq2 S Cu 21aq2 Cu1s2 25°C ¢G° 25°C ¢G°, e°, [Sn2] [Ni 2] Ni21aq2 Sn1s2 S Ni1s2 Sn 21aq2 at 25°C?
[Fe2] [Cu 2] 25°C ¢G° 25°C ¢G° Ag1aq2 2S2O3 21aq2 ∆Ag1S2O322 31aq2 K ?
Ag(S2O3)2 3.
834 Chapter Seventeen Electrochemistry 85. One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is The is then chemically reduced using hy-drogen gas to which is used in the pro-duction of nylon. What current must be used to produce 150. kg of per hour?
86. A single Hall–Heroult cell (as shown in Fig. 17.22) produces about 1 ton of aluminum in 24 hours. What current must be used to accomplish this?
87. It took 2.30 min using a current of 2.00 A to plate out all the sil-ver from 0.250 L of a solution containing Ag. What was the original concentration of Ag in the solution?
88. A solution containing Pt4 is electrolyzed with a current of 4.00 A. How long will it take to plate out 99% of the platinum in 0.50 L of a 0.010 M solution of Pt4?
89. A solution at contains 1.0 M 1.0 M Ag, 1.0 M Au3, and 1.0 M in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.
90. Consider the following half-reactions: A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 M in chloride ion and 0.020 M in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis?
(Assume that 99% of a metal must be plated out before another metal begins to plate out.) 91. What reactions take place at the cathode and the anode when each of the following is electrolyzed?
a. molten NiBr2 b. molten AlF3 c. molten MnI2 92. What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard condi-tions.) a. 1.0 M NiBr2 solution b. 1.0 M AlF3 solution c. 1.0 M MnI2 solution Additional Exercises 93. The saturated calomel electrode, abbreviated SCE, is often used as a reference electrode in making electrochemical measure-ments. The SCE is composed of mercury in contact with a sat-urated solution of calomel (Hg2Cl2). The electrolyte solution is saturated KCl. relative to the standard hydro-gen electrode. Calculate the potential for each of the following galvanic cells containing a saturated calomel electrode and the given half-cell components at standard conditions. In each case, indicate whether the SCE is the cathode or the anode. Standard reduction potentials are found in Table 17.1.
eSCE is 0.242 V e° 0.62 V PdCl4 2 2e ¡ Pd 4Cl e° 0.73 V PtCl4 2 2e ¡ Pt 4Cl e° 0.77 V IrCl6 3 3e ¡ Ir 6Cl Ni 2 Cd 2, 25°C NC¬(CH2)4¬CN H2N¬1CH226¬NH2, NC¬(CH2)4¬CN 2CH2“CHCN 2H 2e ¡ NC¬1CH224¬CN The overall reaction and equilibrium constant value are Calculate the cell potential, for this galvanic cell and for the cell reaction at these conditions.
73. Calculate Ksp for iron(II) sulfide given the following data: 74. For the following half-reaction, Using data from Table 17.1, calculate the equilibrium constant at for the reaction 75. Calculate the value of the equilibrium constant for the reaction of zinc metal in a solution of silver nitrate at 76. The solubility product for CuI(s) is . Calculate the value of for the half-reaction Electrolysis 77. How long will it take to plate out each of the following with a current of 100.0 A?
a. 1.0 kg Al from aqueous Al3 b. 1.0 g Ni from aqueous c. 5.0 mol Ag from aqueous Ag 78. The electrolysis of BiO produces pure bismuth. How long would it take to produce 10.0 g of Bi by the electrolysis of a BiO solution using a current of 25.0 A?
79. What mass of each of the following substances can be produced in 1.0 h with a current of 15 A?
a. Co from aqueous c. I2 from aqueous KI b. Hf from aqueous Hf 4 d. Cr from molten CrO3 80. Aluminum is produced commercially by the electrolysis of Al2O3 in the presence of a molten salt. If a plant has a continuous ca-pacity of 1.00 million amp, what mass of aluminum can be pro-duced in 2.00 h?
81. An unknown metal M is electrolyzed. It took 74.1 s for a cur-rent of 2.00 amp to plate out 0.107 g of the metal from a solu-tion containing M(NO3)3. Identify the metal.
82. Electrolysis of an alkaline earth metal chloride using a current of 5.00 A for 748 s deposits 0.471 g of metal at the cathode.
What is the identity of the alkaline earth metal chloride?
83. What volume of F2 gas, at and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 h?
What mass of potassium metal is produced? At which electrode does each reaction occur?
84. What volumes of H2(g) and O2(g) at STP are produced from the electrolysis of water by a current of 2.50 A in 15.0 min?
25°C Co2 Ni 2 CuI e ¡ Cu I e° 1.1 1012 25°C.
Al 31aq2 6F 1aq2 ∆AlF6 31aq2 K ?
25°C AlF6 3 3e ¡ Al 6F e° 2.07 V: Fe 21aq2 2e S Fe1s2 e° 0.44 V FeS1s2 2e S Fe1s2 S 21aq2 e° 1.01 V ¢G e, 2Cr31aq2 Co1s2 K 2.79 107 2Cr 2 1aq2 Co 21aq2 ¡ Additional Exercises 835 a.
b.
c.
d.
e.
94. Consider the following half-reactions: Explain why platinum metal will dissolve in aqua regia (a mix-ture of hydrochloric and nitric acids) but not in either concen-trated nitric or concentrated hydrochloric acid individually.
95. Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are Ag(s) and Cu(s). Does the cell po-tential increase, decrease, or remain the same when the follow-ing changes occur to the standard cell?
a. CuSO4(s) is added to the copper half-cell compartment (as-sume no volume change).
b. NH3(aq) is added to the copper half-cell compartment. Hint: reacts with NH3 to form (aq).
c. NaCl(s) is added to the silver half-cell compartment. Hint: Ag reacts with to form AgCl(s).
d. Water is added to both half-cell compartments until the vol-ume of solution is doubled.
e. The silver electrode is replaced with a platinum electrode.
96. A standard galvanic cell is constructed so that the overall cell reaction is where M is an unknown metal. If for the over-all cell reaction, identify the metal used to construct the stan-dard cell.
97. The black silver sulfide discoloration of silverware can be re-moved by heating the silver article in a sodium carbonate solu-tion in an aluminum pan. The reaction is a. Using data in Appendix 4, calculate K, and for the above reaction at . (For Al3(aq), b. Calculate the value of the standard reduction potential for the following half-reaction: 98. In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. (The Monitor and the CSS Virginia [formerly the USS Merrimack] fought the first battle between iron-armored ships.) In 1987 in-vestigations were begun to see if the ship could be salvaged. It was reported in Time (June 22, 1987) that scientists were considering adding sacrificial anodes of zinc to the rapidly 2e Ag2S1s2 ¡ 2Ag1s2 S 21aq2 ¢G° f 480. kJ/mol.) 25°C e° ¢G°, 3Ag2S1s2 2Al1s2 ∆6Ag1s2 3S 21aq2 2Al 31aq2 ¢G° 411 kJ 2Al31aq2 3M1s2 ¡ 3M 21aq2 2Al1s2 e° 1.19 V Pt 2 2e ¡ Pt Cl Cu(NH3)4 2 Cu 2 Ag e ¡ Ag Cu2 2e ¡ Cu e° 0.96 V NO3 4H 3e ¡ NO 2H2O e° 0.755 V PtCl4 2 2e ¡ Pt 4Cl e° 1.188 V Pt2 2e ¡ Pt Ni2 2e ¡ Ni Al 3 3e ¡ Al AgCl e ¡ Ag Cl Fe 3 e ¡ Fe 2 Cu 2 2e ¡ Cu corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.
99. When aluminum foil is placed in hydrochloric acid, nothing hap-pens for the first 30 seconds or so. This is followed by vigorous bubbling and the eventual disappearance of the foil. Explain these observations.
100. Which of the following statements concerning corrosion is/are true? For the false statements, correct them.
a. Corrosion is an example of an electrolytic process.
b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen.
c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states.
d. Salting roads in the winter has the added benefit of hinder-ing the corrosion of steel.
e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.
101. A patent attorney has asked for your advice concerning the mer-its of a patent application that describes a single aqueous gal-vanic cell capable of producing a 12-V potential. Comment.
102. The overall reaction and equilibrium constant value for a hydrogen–oxygen fuel cell at 298 K is a. Calculate and at 298 K for the fuel cell reaction.
b. Predict the signs of and for the fuel cell reaction.
c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or re-main the same? Explain.
103. What is the maximum work that can be obtained from a hydrogen–oxygen fuel cell at standard conditions that pro-duces 1.00 kg of water at ? Why do we say that this is the maximum work that can be obtained? What are the ad-vantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?
104. The overall reaction and standard cell potential at for the rechargeable nickel–cadmium alkaline battery is For every mole of Cd consumed in the cell, what is the maxi-mum useful work that can be obtained at standard conditions?
105. An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is The two half-cell reactions are The two half-reactions are carried out in separate compartments connected with a solid mixture of CeO2 and Gd2O3. Oxide ions can move through this solid at high temperatures (about for the overall reaction at under certain concentra-tion conditions is kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.
380 800°C ¢G 800°C). O2 4e ¡ 2O 2 CO O 2 ¡ CO2 2e 2CO1g2 O21g2 ¡ 2CO21g2 e° 1.10 V Ni1OH221s2 Cd1OH221s2 Cd1s2 NiO21s2 2H2O1l2 ¡ 25°C 25°C ¢S° ¢H° ¢G° e° 2H21g2 O21g2 ¡ 2H2O1l2 K 1.28 1083 836 Chapter Seventeen Electrochemistry Calculate the Ksp value for Ag2SO4(s). Note that to obtain silver ions in the right compartment (the cathode compartment), ex-cess solid Ag2SO4 was added and some of the salt dissolved.
114. A zinc–copper battery is constructed as follows at The mass of each electrode is 200. g.
a. Calculate the cell potential when this battery is first con-nected.
b. Calculate the cell potential after 10.0 A of current has flowed for 10.0 h. (Assume each half-cell contains 1.00 L of solution.) c. Calculate the mass of each electrode after 10.0 h.
d. How long can this battery deliver a current of 10.0 A before it goes dead?
115. A galvanic cell is based on the following half-reactions: where the iron compartment contains an iron electrode and and the hydrogen compartment con-tains a platinum electrode, atm, and a weak acid, HA, at an initial concentration of 1.00 M. If the observed cell potential is 0.333 V at calculate the Ka value for the weak acid HA.
116. Consider a cell based on the following half-reactions: a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concen-trations, as appropriate.
b. When enough NaCl(s) is added to the compartment contain-ing gold to make the the cell potential is observed to be 0.31 V. Assume that Au3 is reduced and as-sume that the reaction in the compartment containing gold is Calculate the value of K for this reaction at .
117. The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at is given by the equation where contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that and that a. What is the uncertainty in the values of pH and [H] if the uncertainty in the measured potential is b. To what precision must the potential be measured for the un-certainty in pH to be pH unit?
118. Zirconium is one of the few metals that retains its structural in-tegrity upon exposure to radiation. For this reason, the fuel rods in most nuclear reactors are made of zirconium. Answer the fol-lowing questions about the redox properties of zirconium based on the half-reaction ZrO2 H2O H2O 4e ¡ Zr 4OH e° 2.36 V 0.02 1 mV (0.001 V)?
emeas 0.480 V.
eref 0.250 V eref emeas eref 0.05916 pH 25.00°C 25°C Au31aq2 4Cl 1aq2 ∆AuCl4 1aq2 [Cl ] 0.10 M, Fe 3 e ¡ Fe2 e° 0.77 V Au 3 3e ¡ Au e° 1.50 V 25°C, PH2 1.00 [Fe 2] 1.00 103 M e° 0.000 V 2H 2e ¡ H21g2 e° 0.440 V Fe2 2e ¡ Fe1s2 Zn 0 Zn 210.10 M2 0 0 Cu212.50 M2 0 Cu 25°C: 106. A fuel cell designed to react grain alcohol with oxygen has the following net reaction: The maximum work 1 mol of alcohol can yield by this process is 1320 kJ. What is the theoretical maximum voltage this cell can achieve?
107. Gold is produced electrochemically from an aqueous solution of containing an excess of Gold metal and oxy-gen gas are produced at the electrodes. What amount (moles) of O2 will be produced during the production of 1.00 mol of gold?
108. In the electrolysis of a sodium chloride solution, what volume of H2(g) is produced in the same time it takes to produce 257 L of Cl2(g), with both volumes measured at and 2.50 atm?
109. An aqueous solution of an unknown salt of ruthenium is elec-trolyzed by a current of 2.50 A passing for 50.0 min. If 2.618 g Ru is produced at the cathode, what is the charge on the ruthe-nium ions in solution?
110. It takes 15 kWh (kilowatt-hours) of electrical energy to pro-duce 1.0 kg of aluminum metal from aluminum oxide by the Hall–Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg of aluminum metal. Why is it eco-nomically feasible to recycle aluminum cans? (The enthalpy of fusion for aluminum metal is 10.7 kJ/mol ) Challenge Problems 111. Combine the equations to derive an expression for as a function of temperature. De-scribe how one can graphically determine and from measurements of at different temperatures, assuming that and do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?
112. The overall reaction in the lead storage battery is a. For the cell reaction and Calculate . Assume and do not depend on temperature.
b. Calculate when c. Consider your answer to Exercise 55. Why does it seem that batteries fail more often on cold days than on warm days?
113. Consider the following galvanic cell: Pb Ag 1.8 M Pb2+ ? M Ag+ ? M SO4 2-0.83V Ag2SO4 (s) [HSO4 ] [H] 4.5 M.
e at 20.°C ¢S° ¢H° e° at 20.°C ¢S° 263.5 J/K.
¢H° 315.9 kJ 2PbSO41s2 2H2O1l2 Pb1s2 PbO21s2 2H 1aq2 2HSO4 1aq2 ¡ ¢S° ¢H° e° ¢S° ¢H° e° ¢G° nFe° and ¢G° ¢H° T¢S° [1 watt 1 J/s].
50.°C CN .
Au(CN)2 C2H5OH1l2 3O21g2 ¡ 2CO21g2 3H2O1l2 Integrative Problems 837 a. Is zirconium metal capable of reducing water to form hydro-gen gas at standard conditions?
b. Write a balanced equation for the reduction of water by zirconium metal.
c. Calculate and K for the reduction of water by zir-conium metal.
d. The reduction of water by zirconium occurred during the ac-cident at Three Mile Island, Pennsylvania, in 1979. The hy-drogen produced was successfully vented and no chemical explosion occurred. If kg of Zr reacts, what mass of H2 is produced? What volume of H2 at 1.0 atm and is produced?
e. At Chernobyl, USSR, in 1986, hydrogen was produced by the reaction of superheated steam with the graphite reactor core: A chemical explosion involving the hydrogen gas did occur at Chernobyl. In light of this fact, do you think it was a cor-rect decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.
119. A galvanic cell is based on the following half-reactions: In this cell, the silver compartment contains a silver electrode and excess AgCl(s) and the copper com-partment contains a copper electrode and a. Calculate the potential for this cell at .
b. Assuming 1.0 L of 2.0 M in the copper compartment, calculate the moles of NH3 that would have to be added to give a cell potential of 0.52 V at (assume no volume change on addition of NH3).
120. Given the following two standard reduction potentials, solve for the standard reduction potential of the half-reaction (Hint: You must use the extensive property to determine the standard reduction potential.) 121. You make a galvanic cell with a piece of nickel, 1.0 M Ni2(aq), a piece of silver, and 1.0 M Ag(aq). Calculate the concentra-tions of Ag(aq) and Ni2(aq) once the cell is “dead.” 122. A chemist wishes to determine the concentration of electrochemically. A cell is constructed consisting of a satu-rated calomel electrode (SCE; see Exercise 93) and a silver wire coated with Ag2CrO4. The value for the following half-reaction is 0.446 V relative to the standard hydrogen electrode: a. Calculate and at for the cell reaction when b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant.
[CrO4 2] 1.00 mol/L.
25°C ¢G ecell Ag2CrO4 2e ¡ 2Ag CrO4 2 e° CrO4 2 ¢G° M 3 e S M 2 e° 0.50 V M2 2e S M e° 0.10 V M3 3e S M Cu1NH324 21aq2 K 1.0 1013 Cu 21aq2 4NH31aq2 ∆ 25°C Cu 2 25°C [Cu 2] 2.0 M.
(Ksp 1.6 1010), Cu 2 2e ¡ Cu1s2 e° 0.34 V Ag e ¡ Ag1s2 e° 0.80 V C1s2 H2O1g2 ¡ CO1g2 H21g2 1000.°C 1.00 103 e°, ¢G°, c. If the coated silver wire is placed in a solution (at ) in which what is the expected cell potential?
d. The measured cell potential at is 0.504 V when the coated wire is dipped into a solution of unknown What is for this solution?
e. Using data from this problem and from Table 17.1, calculate the solubility product (Ksp) for Ag2CrO4.
123. You have a concentration cell with Cu electrodes and (right side) and (left side).
a. Calculate the potential for this cell at .
b. The Cu2 ion reacts with NH3 to form Cu(NH3)4 2 where the stepwise formation constants are Calculate the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium 124. When copper reacts with nitric acid, a mixture of NO(g) and NO2(g) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium Consider the following standard reduction potentials at a. Calculate the equilibrium constant for the above reaction.
b. What concentration of nitric acid will produce a NO and NO2 mixture with only 0.20% NO2 (by moles) at and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
125. The following standard reduction potentials have been deter-mined for the aqueous chemistry of indium: a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below?
b. What is for (aq) if for In3(aq)?
126. An electrochemical cell is set up using the following balanced reaction: Given the standard reduction potentials are: e° 0.240 V N 2 2e ¡ N e° 0.400 V M a ae ¡ M M a1aq2 N1s2 ¡ N21aq2 M1s2 ¢G° f 97.9 kJ/mol In ¢G° f 3In 1aq2 ¡ 2In1s2 In 31aq2 e° 0.126 V In 1aq2 e ¡ In1s2 e° 0.444 V In 31aq2 2e ¡ In 1aq2 25°C e° 0.775 V e 2H 1aq2 NO3 1aq2 ¡ NO21g2 2H2O1l2 e° 0.957 V 3e 4H 1aq2 NO3 1aq2 ¡ NO1g2 2H2O1l2 25°C: 2H 1aq2 2NO3 1aq2 NO1g2 ∆3NO21g2 H2O1l2 [NH3] 2.0 M.
1.0 104, K3 1.0 103, and K4 1.0 10 3.
K1 1.0 103, K2 25°C 1.0 10 4 M 1.00 M [Cu 2] [CrO4 2] [CrO4 2].
25°C [CrO4 2] 1.00 10 5 M, 25°C 838 Chapter Seventeen Electrochemistry a vanadium electrode and V2 at an unknown concentration. The compartment containing the vanadium (1.00 L of solution) was titrated with 0.0800 M resulting in the reaction The potential of the cell was monitored to determine the stoi-chiometric point for the process, which occurred at a volume of 500.0 mL of solution added. At the stoichiometric point, was observed to be 1.98 V. The solution was buffered at a pH of 10.00.
a. Calculate before the titration was carried out.
b. Calculate the value of the equilibrium constant, K, for the titration reaction.
c. Calculate at the halfway point in the titration.
129. The table below lists the cell potentials for the 10 possible gal-vanic cells assembled from the metals A, B, C, D, and E, and their respective 1.00 M 2 ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table 17.1 in the text. Assign a reduction potential of 0.00 V to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct.
ecell ecell ecell H2EDTA 2 K ?
∆VEDTA 21aq2 2H 1aq2 H2EDTA 21aq2 V 21aq2 H2EDTA2, The cell contains 0.10 M N2 and produces a voltage of 0.180 V.
If the concentration of Ma is such that the value of the reaction quotient Q is 9.32 103, calculate [Ma]. Calculate wmax for this electrochemical cell.
127. Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g of chromium metal was deposited from a chromium(III) nitrate solution. In the second cell, 3.15 g of os-mium was deposited from a solution made of Osn and nitrate ions.
What is the name of the salt? In the third cell, the electrical charge passed through a solution containing X2 ions caused deposition of 2.11 g of metallic X. What is the electron configuration of X?
Marathon Problems These problems are designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
128. A galvanic cell is based on the following half-reactions: In this cell, the copper compartment contains a copper electrode and and the vanadium compartment contains [Cu 2] 1.00 M, e° 1.20 V V 21aq2 2e ¡ V1s2 e° 0.34 V Cu 21aq2 2e ¡ Cu1s2 Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
A(s) in A2(aq) B(s) in B2(aq) C(s) in C2(aq) D(s) in D2(aq) 0.28 V 0.81 V 0.13 V 1.00 V 0.72 V 0.19 V 1.13 V — 0.41 V 0.94 V — — 0.53 V — — — B1s2 in B 21aq2 C1s2 in C 21aq2 D1s2 in D 21aq2 E1s2 in E 21aq2 840 18 The Nucleus: A Chemist’s View Contents 18.1 Nuclear Stability and Radioactive Decay • Types of Radioactive Decay 18.2 The Kinetics of Radioactive Decay • Half-Life 18.3 Nuclear Transformations 18.4 Detection and Uses of Radioactivity • Dating by Radioactivity • Medical Applications of Radioactivity 18.5 Thermodynamic Stability of the Nucleus 18.6 Nuclear Fission and Nuclear Fusion • Nuclear Fission • Nuclear Reactors • Breeder Reactors • Fusion 18.7 Effects of Radiation Workers inside a giant chamber at the National Ignition Facility in California. This chamber will be used to induce nuclear fusion by aiming 192 lasers at a pellet of fuel.
Since the chemistry of an atom is determined by the number and arrangement of its electrons, the properties of the nucleus are not of primary importance to chemists. In the simplest view, the nucleus provides the positive charge to bind the electrons in atoms and molecules. However, a quick reading of any daily newspaper will show you that the nu-cleus and its properties have an important impact on our society. This chapter considers those aspects of the nucleus about which everyone should have some knowledge.
Several aspects of the nucleus are immediately impressive: its very small size, its very large density, and the magnitude of the energy that holds it together. The radius of a typical nucleus appears to be about cm. This can be compared to the radius of a typical atom, which is on the order of cm. A visualization will help you appreciate the small size of the nucleus: If the nucleus of the hydrogen atom were the size of a Ping-Pong ball, the electron in the 1s orbital would be, on average, 0.5 kilometer (0.3 mile) away. The density of the nucleus is equally impressive—approximately g/cm3. A sphere of nuclear material the size of a Ping-Pong ball would have a mass of 2.5 billion tons! In addition, the energies involved in nuclear processes are typically millions of times larger than those associated with normal chemical reactions. This fact makes nuclear processes very attractive for feeding the voracious energy appetite of our civilization.
Atomos, the Greek root of the word atom, means “indivisible.” It was originally be-lieved that the atom was the ultimate indivisible particle of which all matter was com-posed. However, as we discussed in Chapter 2, Lord Rutherford showed in 1911 that the atom is not homogeneous, but rather has a dense, positively charged center surrounded by electrons. Subsequently, scientists have learned that the nucleus of the atom can be subdivided into particles called neutrons and protons. In fact, in the past two decades it has become apparent that even the protons and neutrons are composed of smaller particles called quarks.
For most purposes, the nucleus can be regarded as a collection of nucleons (neutrons and protons), and the internal structures of these particles can be ignored. As we discussed in Chapter 2, the number of protons in a particular nucleus is called the atomic number (Z), and the sum of the neutrons and protons is the mass number (A). Atoms that have identical atomic numbers but different mass number values are called isotopes. However, we usually do not use the singular form isotope to refer to a particular member of a group of isotopes. Rather, we use the term nuclide. A nuclide is a unique atom, represented by the symbol where X represents the symbol for a particular element. For example, the following nuclides constitute the isotopes of carbon: carbon-12 (12 6C), carbon-13 (13 6C), and carbon-14 (14 6C).
18.1 Nuclear Stability and Radioactive Decay Nuclear stability is the central topic of this chapter and forms the basis for all the impor-tant applications related to nuclear processes. Nuclear stability can be considered from both a kinetic and a thermodynamic point of view. Thermodynamic stability, as we will use the term here, refers to the potential energy of a particular nucleus as compared with the sum of the potential energies of its component protons and neutrons. We will use the term kinetic stability to describe the probability that a nucleus will undergo decomposition A ZX 1.6 1014 108 1013 841 The term isotopes refers to a group of nuclides with the same atomic number.
Each individual atom is properly called a nuclide, not an isotope.
The atomic number Z is the number of protons in a nucleus; the mass number A is the sum of protons and neutrons in a nucleus.
842 Chapter Eighteen The Nucleus: A Chemist’s View to form a different nucleus—a process called radioactive decay. We will consider radioactivity in this section.
Many nuclei are radioactive; that is, they decompose, forming another nucleus and producing one or more particles. An example is carbon-14, which decays as follows: where represents an electron, which is called a beta particle, or particle, in nu-clear terminology. This equation is typical of those representing radioactive decay in that both A and Z must be conserved. That is, the Z values must give the same sum on both sides of the equation as must the A values Of the approximately 2000 known nuclides, only 279 are stable with respect to ra-dioactive decay. Tin has the largest number of stable isotopes—10.
It is instructive to examine how the numbers of neutrons and protons in a nucleus are related to its stability with respect to radioactive decay. Figure 18.1 shows a plot of the positions of the stable nuclei as a function of the number of protons (Z) and the number of neutrons (A Z). The stable nuclides are said to reside in the zone of stability.
The following are some important observations concerning radioactive decay: • All nuclides with 84 or more protons are unstable with respect to radioactive decay.
• Light nuclides are stable when Z equals A Z, that is, when the neutron/proton ratio is 1. However, for heavier elements the neutron/proton ratio required for stability is greater than 1 and increases with Z.
(14 14 0).
(6 7 1), B 0 1e 6 14C ¡ 7 14N 1 0e FIGURE 18.1 The zone of stability. The red dots indicate the nuclides that do not undergo radio-active decay. Note that as the number of protons in a nuclide increases, the neutron/proton ratio required for stability also increases.
0 Number of protons (Z) 20 40 60 80 100 20 40 60 80 100 120 140 160 Number of neutrons (A – Z) 0 Unstable region (too many neutrons; spontaneous beta production) 202 80Hg (1.53:1 ratio) Unstable region (too many protons; spontaneous positron production) 110 48 Cd (1.29:1 ratio) 1:1 neutron-to-proton ratio Stable nuclides in the zone of stability 6 3Li (1:1 ratio) 18.1 Nuclear Stability and Radioactive Decay 843 • Certain combinations of protons and neutrons seem to confer special stability. For ex-ample, nuclides with even numbers of protons and neutrons are more often stable than those with odd numbers, as shown by the data in Table 18.1.
• There are also certain specific numbers of protons or neutrons that produce especially stable nuclides. These magic numbers are 2, 8, 20, 28, 50, 82, and 126. This behavior parallels that for atoms in which certain numbers of electrons (2, 10, 18, 36, 54, and 86) produce special chemical stability (the noble gases).
Types of Radioactive Decay Radioactive nuclei can undergo decomposition in various ways. These decay processes fall into two categories: those that involve a change in the mass number of the decaying nucleus and those that do not. We will consider the former type of process first.
An alpha particle, or particle, is a helium nucleus (4 2He). Alpha-particle pro-duction is a very common mode of decay for heavy radioactive nuclides. For example, the predominant (99.3%) isotope of natural uranium, decays by -particle production: Another -particle producer is : Another decay process in which the mass number of the decaying nucleus changes is spontaneous fission, the splitting of a heavy nuclide into two lighter nuclides with similar mass numbers. Although this process occurs at an extremely slow rate for most nuclides, it is important in some cases, such as for where spontaneous fission is the predominant mode of decay.
The most common decay process in which the mass number of the decaying nucleus remains constant is -particle production. For example, the thorium-234 nuclide produces a particle and is converted to protactinium-234: Iodine-131 is also a -particle producer: The particle is assigned the mass number 0, since its mass is tiny compared with that of a proton or neutron. Because the value of Z is for the particle, the atomic num-ber for the new nuclide is greater by 1 than for the original nuclide. Thus the net effect of -particle production is to change a neutron to a proton. We therefore expect nuclides b b 1 b 53 131I ¡ 1 0e 54 131Xe 90 234Th ¡ 91 234Pa 1 0e b B 254 98Cf, 90 230Th ¡ 2 4He 88 226Ra 230 90Th 92 238U ¡ 2 4He 90 234Th 238 92U, A TABLE 18.1 Number of Stable Nuclides Related to Numbers of Protons and Neutrons Number of Number of Number of Protons Neutrons Stable Nuclides Examples Even Even 168 Even Odd 57 Odd Even 50 Odd Odd 4 Note: Even numbers of protons and neutrons seem to favor stability.
2 1H, 6 3Li 19 9F, 23 11Na 13 6C, 47 22Ti 12 6C, 16 8O -particle production involves a change in A for the decaying nucleus; -particle production has no effect on A.
b a Visualization: Nuclear Particles 844 Chapter Eighteen The Nucleus: A Chemist’s View that lie above the zone of stability (those nuclides whose neutron/proton ratios are too high) to be -particle producers.
It should be pointed out that although the particle is an electron, the emitting nucleus does not contain electrons. As we shall see later in this chapter, a given quantity of en-ergy (which is best regarded as a form of matter) can become a particle (another form of matter) under certain circumstances. The unstable nuclide creates an electron as it releases energy in the decay process. The electron thus results from the decay process rather than being present before the decay occurs. Think of this as somewhat like talking: Words are not stored inside us but are formed as we speak. Later in this chapter we will discuss in more detail this very interesting phenomenon where matter in the form of particles and matter in the form of energy can interchange.
A gamma ray, or ray, refers to a high-energy photon. Frequently, -ray produc-tion accompanies nuclear decays and particle reactions, such as in the -particle decay of : where two rays of different energies are produced in addition to the particle. The emission of rays is one way a nucleus with excess energy (in an excited nuclear state) can relax to its ground state.
Positron production occurs for nuclides that are below the zone of stability (those nuclides whose neutron/proton ratios are too small). The positron is a particle with the same mass as the electron but opposite charge. An example of a nuclide that decays by positron production is sodium-22: Note that the net effect is to change a proton to a neutron, causing the product nuclide to have a higher neutron/proton ratio than the original nuclide.
Besides being oppositely charged, the positron shows an even more fundamental dif-ference from the electron: It is the antiparticle of the electron. When a positron collides with an electron, the particulate matter is changed to electromagnetic radiation in the form of high-energy photons: This process, which is characteristic of matter–antimatter collisions, is called annihilation and is another example of the interchange of the forms of matter.
Electron capture is a process in which one of the inner-orbital electrons is captured by the nucleus, as illustrated by the process Inner-orbital electron This reaction would have been of great interest to the alchemists, but unfortunately it does not occur at a rate that would make it a practical means for changing mercury to gold.
Gamma rays are always produced along with electron capture to release excess energy.
The various types of radioactive decay are summarized in Table 18.2.
Nuclear Equations I Write balanced equations for each of the following processes.
a.
produces a positron.
b.
produces a particle.
c.
produces an particle.
a 237 93Np b 241 83Bi 11 6C 80 201Hg 1 0e ¡ 79 201Au 0 0g 1 0e 1 0e ¡ 2 0 0g 11 22Na ¡ 1 0e 10 22Ne g a g 92 238U ¡ 2 4He 90 234Th 2 0 0g 238 92U a g b b ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ Sample Exercise 18.1 18.1 Nuclear Stability and Radioactive Decay 845 Solution a. We must find the product nuclide represented by in the following equation: h Positron We can find the identity of by recognizing that the total of the Z and A values must be the same on both sides of the equation. Thus for X, Z must be and A must be Therefore, is 11 5B. (The fact that Z is 5 tells us that the nu-clide is boron.) Thus the balanced equation is b. Knowing that a particle is represented by and that Z and A are conserved, we can write so must be c. Since an particle is represented by 4 2He, the balanced equation must be See Exercises 18.11 and 18.12.
Nuclear Equations II In each of the following nuclear reactions, supply the missing particle.
a.
b.
Solution a. Since A does not change and Z decreases by 1, the missing particle must be an electron: This is an example of electron capture.
b. To conserve Z and A, the missing particle must be a positron: Thus potassium-38 decays by positron production.
See Exercises 18.13 and 18.14.
38 19K ¡ 38 18Ar 0 1e 79 195Au 1 0e ¡ 78 195Pt 38 19K S 38 18Ar ?
195 79Au ? S 195 78Pt 93 237Np ¡ 2 4He 91 233Pa a 214 84Po.
A ZX 83 214Bi ¡ 1 0e 84 214X 0 1e b 6 11C ¡ 1 0e 5 11B A ZX 11 0 11.
6 1 5 A ZX 6 11C ¡ 1 0e Z AX A ZX TABLE 18.2 Various Types of Radioactive Processes Showing the Changes That Take Place in the Nuclides Change Change Change in Process in A in Z Neutron/Proton Ratio Example -particle (electron) production 0 1 Decrease Positron production 0 1 Increase Electron capture 0 1 Increase -particle production 4 2 Increase -ray production 0 0 — Spontaneous fission — — — 254 98Cf ¡ lighter nuclides neutrons Excited nucleus ¡ ground-state nucleus 0 0g g 210 84Po ¡ 206 82Pb 4 2He a 73 33As 0 1e ¡ 73 32Ge 13 7N ¡ 13 6C 0 1e 227 89Ac ¡ 227 90Th 0 1e b Sample Exercise 18.2 846 Chapter Eighteen The Nucleus: A Chemist’s View Often a radioactive nucleus cannot reach a stable state through a single decay process.
In such a case, a decay series occurs until a stable nuclide is formed. A well-known ex-ample is the decay series that starts with and ends with as shown in Fig. 18.2.
Similar series exist for : and for 18.2 The Kinetics of Radioactive Decay In a sample containing radioactive nuclides of a given type, each nuclide has a certain probability of undergoing decay. Suppose that a sample of 1000 atoms of a certain nu-clide produces 10 decay events per hour. This means that over the span of an hour, 1 out of every 100 nuclides will decay. Given that this probability of decay is characteristic for this type of nuclide, we could predict that a 2000-atom sample would give 20 decay events per hour. Thus, for radioactive nuclides, the rate of decay, which is the negative of the change in the number of nuclides per unit time is directly proportional to the number of nuclides N in a given sample: Rate ¢N ¢t r N a¢N ¢t b 90 232Th ¡ 82 208Pb 90 232Th: 92 235U ¡ 82 207Pb 235 92U 206 82Pb, 238 92U FIGURE 18.2 The decay series from to Each nuclide in the series except is unsta-ble, and the successive transformations (shown by the arrows) continue until is finally formed. Note that horizontal arrows indicate processes where A is unchanged, while diagonal arrows signify that both A and Z change.
206 82Pb 206 82Pb.
206 82Pb.
238 92U Rates of reaction are discussed in Chapter 12.
0 Atomic number (Z) 238 236 234 232 230 228 226 224 222 220 218 216 214 212 210 208 206 204 Mass number (A) 82 83 84 85 86 87 88 89 90 91 92 93 U Th Pa U Th Ra Rn Po Pb Bi Po Pb Bi Po Pb Series of decays Series of decays 18.2 The Kinetics of Radioactive Decay 847 The negative sign is included because the number of nuclides is decreasing. We now insert a proportionality constant k to give This is the rate law for a first-order process, as we saw in Chapter 12. As shown in Sec-tion 12.4, the integrated first-order rate law is where N0 represents the original number of nuclides and N represents the number remaining at time t.
Half-Life The half-life of a radioactive sample is defined as the time required for the number of nuclides to reach half the original value We can use this definition in connection with the integrated first-order rate law (as we did in Section 12.4) to produce the following expression for : Thus, if the half-life of a radioactive nuclide is known, the rate constant can be easily calculated, and vice versa.
Kinetics of Nuclear Decay I Technetium-99m is used to form pictures of internal organs in the body and is often used to assess heart damage. The m for this nuclide indicates an excited nuclear state that de-cays to the ground state by gamma emission. The rate constant for decay of is known to be What is the half-life of this nuclide?
Solution The half-life can be calculated from the expression Thus it will take 5.98 h for a given sample of technetium-99m to decrease to half the orig-inal number of nuclides.
See Exercise 18.21.
As we saw in Section 12.4, the half-life for a first-order process is constant. This is shown for the -particle decay of strontium-90 in Fig. 18.3; it takes 28.8 years for each halving of the amount of Contamination of the environment with poses serious health hazards because of the similar chemistry of strontium and calcium (both are in Group 2A). Strontium-90 in grass and hay is incorporated into cow’s milk along with calcium and is then passed on to humans, where it lodges in the bones. Because of its rel-atively long half-life, it persists for years in humans, causing radiation damage that may lead to cancer.
90 38Sr 90 38Sr.
b 5.98 h t1 2 0.693 k 0.693 1.16 101/h 1.16 101/h.
99m 043Tc t1 2 ln122 k 0.693 k t1 2 (N0 2).
(t1 2) (at t 0) lna N N0 b kt Rate ¢N ¢t kN Sample Exercise 18.3 The image of a bone scan of a normal chest (posterior view). Radioactive technetium-99m is injected into the patient and is then concentrated in bones, allowing a physician to look for abnormalities such as might be caused by cancer.
Visualization: Half-Life of Nuclear Decay The harmful effects of radiation will be discussed in Section 18.7.
848 Chapter Eighteen The Nucleus: A Chemist’s View Kinetics of Nuclear Decay II The half-life of molybdenum-99 is 67.0 h. How much of a 1.000-mg sample of is left after 335 h?
Solution The easiest way to solve this problem is to recognize that 335 h represents five half-lives for : We can sketch the change that occurs, as is shown in Fig. 18.4. Thus, after 335 h, 0.031 mg remains.
See Exercise 18.23.
The half-lives of radioactive nuclides vary over a tremendous range. For example, has a half-life of years, while has a half-life of second. To give you some perspective on this, the half-lives of the nuclides in the decay series are given in Table 18.3.
238 92U 2 104 214 84Po 5 1015 144 60Nd 99 42Mo 335 5 67.0 99 42Mo 99 42Mo FIGURE 18.4 The change in the amount of with time (t1 2 67 h).
99 42Mo Mo (mg) 99 42 67 67 67 67 67 1.000 mg 0.500 mg 0.250 mg 0.125 mg 0.062 mg 0.031 mg Time (h) FIGURE 18.3 The decay of a 10.0-g sample of strontium-90 over time. Note that the half-life is a constant 28.8 years.
1 half-life 2 half-lives 3 half-lives 4 half-lives 10.0 8.0 6.0 4.0 2.0 0 20 40 60 80 100 120 t1/2 = 28.8 t1/2 = 28.8 t1/2 = 28.8 t1/2 = 28.8 Time (yr) 90 38 Mass of Sr (g) Sample Exercise 18.4 18.3 Nuclear Transformations 849 18.3 Nuclear Transformations In 1919 Lord Rutherford observed the first nuclear transformation, the change of one element into another. He found that by bombarding with particles, the nuclide could be produced: Fourteen years later, Irene Curie and her husband Frederick Joliot observed a similar trans-formation from aluminum to phosphorus: where represents a neutron.
Over the years, many other nuclear transformations have been achieved, mostly using particle accelerators, which, as the name reveals, are devices used to give particles very high velocities. Because of the electrostatic repulsion between the target nucleus and a positive ion, accelerators are needed when positive ions are used as bombarding parti-cles. The particle, accelerated to a very high velocity, can overcome the repulsion and penetrate the target nucleus, thus effecting the transformation. A schematic diagram of one type of particle accelerator, the cyclotron, is shown in Fig. 18.5. The ion is intro-duced at the center of the cyclotron and is accelerated in an expanding spiral path by use of alternating electric fields in the presence of a magnetic field. The linear accelerator 1 0n 13 27Al 2 4He ¡ 15 30P 0 1n 7 14N 2 4He ¡ 8 17O 1 1H 17 8O a 14 7N TABLE 18.3 The Half-Lives of Nuclides in the Decay Series Nuclide Particle Produced Half-Life Uranium-238 Thorium-234 24.1 days Protactinium-234 6.75 hours Uranium-234 years Thorium-230 years Radium-226 years Radon-222 3.82 days Polonium-218 3.1 minutes Lead-214 26.8 minutes Bismuth-214 19.7 minutes Polonium-214 second Lead-210 20.4 years Bismuth-210 5.0 days Polonium-210 138.4 days Lead-206 — Stable 1206 82Pb2 T a 1210 84Po2 T b 1210 83Bi2 T b 1210 82Pb2 T 1.6 104 a 1214 84Po2 T b 1214 83Bi2 T b 1214 82Pb2 T a 1218 84Po2 T a 1222 86Rn2 T 1.62 103 a 1226 88Ra2 T 8.0 104 a 1230 90Th2 T 2.48 105 a 1234 92U2 T b 1234 91Pa2 T b 1234 90Th2 T 4.51 109 years a 1238 92U2 238 92U 850 Chapter Eighteen The Nucleus: A Chemist’s View illustrated in Fig. 18.6 employs changing electric fields to achieve high velocities on a linear pathway.
In addition to positive ions, neutrons are often employed as bombarding particles to effect nuclear transformations. Because neutrons are uncharged and thus not repelled elec-trostatically by a target nucleus, they are readily absorbed by many nuclei, leading to new nuclides. The most common source of neutrons for this purpose is a fission reactor (see Section 18.6).
FIGURE 18.5 A schematic diagram of a cyclotron. The ion is intro-duced in the center and is pulled back and forth between the hollow D-shaped electrodes by constant reversals of the electric field. Magnets above and below these electrodes produce a spiral path that ex-pands as the particle velocity increases. When the particle has sufficient speed, it exits the accelerator and is directed at the target nucleus.
H ow did all the matter around us originate? The scien-tific answer to this question is a theory called stellar nucleosynthesis—literally, the formation of nuclei in stars.
Many scientists believe that our universe originated as a cloud of neutrons that became unstable and produced an immense explosion, giving this model its name—the big bang theory. The model postulates that, following the initial explosion, neutrons decomposed into protons and electrons, which eventually recombined to form clouds of hydrogen.
Over the eons, gravitational forces caused many of these hy-drogen clouds to contract and heat up sufficiently to reach temperatures where proton fusion was possible, which released large quantities of energy. When the tendency to 1 0n ¡ 1 1H 0 1e CHEMICAL IMPACT Stellar Nucleosynthesis expand due to the heat from fusion and the tendency to contract due to the forces of gravity are balanced, a stable young star such as our sun can be formed.
Eventually, when the supply of hydrogen is ex-hausted, the core of the star will again contract with fur-ther heating until temperatures are reached where fusion of helium nuclei can occur, leading to the formation of and nuclei. In turn, when the supply of helium nuclei runs out, further contraction and heating will occur, until fusion of heavier nuclei takes place. This process occurs repeatedly, forming heavier and heavier nuclei until iron nuclei are formed. Because the iron nucleus is the most stable of all, energy is required to fuse iron nuclei. This endothermic fusion process cannot furnish energy to sus-tain the star, and therefore it cools to a small, dense white dwarf.
16 8O 12 6C Hollow D-shaped electrodes (±) (±) Oscillating voltage Ion source Exit port Target Direction of magnetic field 18.3 Nuclear Transformations 851 By using neutron and positive-ion bombardment, scientists have been able to extend the periodic table. Prior to 1940, the heaviest known element was uranium but in 1940, neptunium was produced by neutron bombardment of The process initially gives which decays to by -particle production: t12 23 min In the years since 1940, the elements with atomic numbers 93 through 112, called the transuranium elements, have been synthesized. Many of these elements have very short half-lives, as shown in Table 18.4. As a result, only a few atoms of some have ever been formed. This, of course, makes the chemical characterization of these elements extremely difficult.
238 92U 1 0n ¡ 239 92U ¬¬¬¡ 238 92Np 0 1p b 239 93Np 239 92U, 238 92U.
(Z 93) (Z 92), FIGURE 18.6 Schematic diagram of a linear accelerator, which uses a changing electric field to accelerate a positive ion along a linear path. As the ion leaves the source, the odd-numbered tubes are negatively charged, and the even-numbered tubes are positively charged. The positive ion is thus attracted into tube 1. As the ion leaves tube 1, the tube polarities are reversed.
Now tube 1 is positive, repelling the positive ion, and tube 2 is negative, attracting the positive ion. This process continues, eventually producing high particle velocity.
For more information see G. B. Kauffman, “Beyond uranium,” Chem. Eng. News (Nov. 19, 1990): 18.
The evolution just described is characteristic of small and medium-sized stars. Much larger stars, however, become unstable at some time during their evolution and undergo a supernova explosion. In this explosion, some medium-mass nuclei are fused to form heavy elements. Also, some light nuclei capture neutrons. These neutron-rich nuclei then pro-duce particles, increasing their atomic number with each event. This eventually leads to heavy nuclei. In fact, almost all nuclei heavier than iron are thought to originate from su-pernova explosions. The debris of a supernova explosion thus contains a large variety of elements and might eventu-ally form a solar system such as our own.
Although other theories for the origin of matter have been suggested, there is much evidence to support the big bang theory, and it continues to be widely accepted.
b Image of a portion of the Cygnus Loop supernova remnant, taken by the Hubble space telescope.
For more information see V. E. Viola, “Formation of the chemical ele-ments and the evolution of our universe,” J. Chem. Ed. 67 (1990): 723.
Ion source 2 3 5 4 6 Target 1 A physicist works with a small cyclotron at the University of California at Berkeley.
852 Chapter Eighteen The Nucleus: A Chemist’s View 18.4 Detection and Uses of Radioactivity Although various instruments measure radioactivity levels, the most familiar of them is the Geiger–Müller counter, or Geiger counter (see Fig. 18.7). This instrument takes ad-vantage of the fact that the high-energy particles from radioactive decay processes pro-duce ions when they travel through matter. The probe of the Geiger counter is filled with argon gas, which can be ionized by a rapidly moving particle. This reaction is demon-strated by the equation: Normally, a sample of argon gas will not conduct a current when an electrical potential is applied. However, the formation of ions and electrons produced by the passage of the high-energy particle allows a momentary current to flow. Electronic devices detect this current flow, and the number of these events can be counted. Thus the decay rate of the radioactive sample can be determined.
Another instrument often used to detect levels of radioactivity is a scintillation counter, which takes advantage of the fact that certain substances, such as zinc sulfide, Ar1g2 ¬¬¬¡ Ar1g2 e TABLE 18.4 Syntheses of Some of the Transuranium Elements Element Neutron Bombardment Half-Life Neptunium 2.35 days Plutonium 24,400 years Americium 458 years Element Positive-Ion Bombardment Half-Life Curium 163 days Californium 44 minutes or Rutherfordium Dubnium Seaborgium 249 98Cf 18 8O ¡ 263 106Sg 4 1 0n 1Z 1062 249 98Cf 15 7N ¡ 260 105Db 4 1 0n 1Z 1052 249 98Cf 12 6C ¡ 257 104Rf 4 1 0n 1Z 1042 238 92U 12 6C ¡ 246 98Cf 4 1 0n 1245 98Cf2 242 96Cm 4 2He ¡ 245 98Cf 1 0n 1Z 982 1242 96Cm2 239 94Pu 4 2He ¡ 242 96Cm 1 0n 1Z 962 1241 95Am2 239 94Pu 2 1 0n ¡ 241 94Pu ¡ 241 95Am 0 1e 1Z 952 1239 94Pu2 239 93Np ¡ 239 94Pu 0 1e 1Z 942 1239 93Np2 238 92U 1 0n ¡ 239 93Np 0 1e 1Z 932 FIGURE 18.7 A schematic representation of a Geiger–Müller counter. The high-energy radioactive particle enters the window and ionizes argon atoms along its path. The resulting ions and electrons produce a mo-mentary current pulse, which is amplified and counted.
High-energy particle Geiger counters are often called survey meters in the industry.
+ + Window Particle path Amplifier and counter e– e– e– Argon atoms (+) (–) + Visualization: Geiger Counter 18.4 Detection and Uses of Radioactivity 853 give off light when they are struck by high-energy radiation. A photocell senses the flashes of light that occur as the radiation strikes and thus measures the number of decay events per unit of time.
Dating by Radioactivity Archeologists, geologists, and others involved in reconstructing the ancient history of the earth rely heavily on radioactivity to provide accurate dates for artifacts and rocks. A method that has been very important for dating ancient articles made from wood or cloth is ra-diocarbon dating, or carbon-14 dating, a technique originated in the 1940s by Willard Libby, an American chemist who received a Nobel Prize for his efforts in this field.
Radiocarbon dating is based on the radioactivity of the nuclide which decays via -particle production: Carbon-14 is continuously produced in the atmosphere when high-energy neutrons from space collide with nitrogen-14: Thus carbon-14 is continuously produced by this process, and it continuously decomposes through -particle production. Over the years, the rates for these two processes have be-come equal, and like a participant in a chemical reaction at equilibrium, the amount of that is present in the atmosphere remains approximately constant.
Carbon-14 can be used to date wood and cloth artifacts because the along with the other carbon isotopes in the atmosphere, reacts with oxygen to form carbon dioxide.
A living plant consumes carbon dioxide in the photosynthesis process and incorporates the carbon, including into its molecules. As long as the plant lives, the ratio in its molecules remains the same as in the atmosphere because of the continuous uptake of carbon. However, as soon as a tree is cut to make a wooden bowl or a flax plant is har-vested to make linen, the ratio begins to decrease because of the radioactive decay of (the nuclide is stable). Since the half-life of is 5730 years, a wooden bowl found in an archeological dig showing a ratio that is half that found in currently living trees is approximately 5730 years old. This reasoning assumes that the current ratio is the same as that found in ancient times.
Dendrochronologists, scientists who date trees from annual growth rings, have used data collected from long-lived species of trees, such as bristlecone pines and sequoias, to show that the content of the atmosphere has changed significantly over the ages. These data have been used to derive correction factors that allow very accurate dates to be de-termined from the observed ratio in an artifact, especially for artifacts 10,000 years old or younger. Recent measurements of uranium/thorium ratios in ancient coral indicate that dates in the 20,000- to 30,000-year range may have errors as large as 3000 years. As a result, efforts are now being made to recalibrate the dates over this period.
14C Dating The remnants of an ancient fire in a cave in Africa showed a 14 6C decay rate of 3.1 counts per minute per gram of carbon. Assuming that the decay rate of 14 6C in freshly cut wood (corrected for changes in the 14 6C content of the atmosphere) is 13.6 counts per minute per gram of carbon, calculate the age of the remnants. The half-life of 14 6C is 5730 years.
Solution The key to solving this problem is to realize that the decay rates given are directly propor-tional to the number of 14 6C nuclides present. Radioactive decay follows first-order kinetics: Rate kN 14 6C 14 6C 12 6C 14 6C 14 6C 12 6C 14 6C 12 6C 14 6C 12 6C 14 6C 14 6C 12 6C 14 6C 12 6C 14 6C, 14 6C, 14 6C b 14 7N 1 0n ¡ 14 6C 1 1H 14 6C ¡ 0 1e 14 7N b 14 6C, Brigham Young researcher Scott Woodward taking a bone sample for carbon-14 dating at an archeological site in Egypt.
A dendrochronologist cutting a section from a dead tree in South Africa.
Radioactive nuclides are often called radionuclides. Carbon dating is based on the radionuclide 14 6C.
The ratio is the basis for carbon-14 dating.
14 6C 12 6C Sample Exercise 18.5 854 Chapter Eighteen The Nucleus: A Chemist’s View Thus Number of nuclides o present at time t r Number of nuclides present at time 0 We can now use the integrated first-order rate law: where to solve for t, the time elapsed since the campfire: Solving this equation gives t 12,000 years; the campfire in the cave occurred about 12,000 years ago.
See Exercises 18.31 and 18.32.
One drawback of radiocarbon dating is that a fairly large piece of the object (from a half to several grams) must be burned to form carbon dioxide, which is then analyzed for radioactivity. Another method for counting 14 6C nuclides avoids destruction of a signif-icant portion of a valuable artifact. This technique, requiring only about 103 g, uses a mass spectrometer (see Chapter 3), in which the carbon atoms are ionized and accelerated through a magnetic field that deflects their path. Because of their different masses, the var-ious ions are deflected by different amounts and can be counted separately. This allows a very accurate determination of the ratio in the sample.
In their attempts to establish the geologic history of the earth, geologists have made extensive use of radioactivity. For example, since decays to the stable nuclide, the ratio of to in a rock can, under favorable circumstances, be used to esti-mate the age of the rock. The radioactive nuclide , which decays to , has a half-life of 37 billion years (only 186 nuclides out of 10 trillion decay each year!). Thus this nuclide can be used to date very old rocks. With this technique, scientists have estimated that the earth’s crust formed 4.3 billion years ago.
Dating by Radioactivity A rock containing and was examined to determine its approximate age. Analy-sis showed the ratio of atoms to atoms to be 0.115. Assuming that no lead was originally present, that all the formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between and is neg-ligible, calculate the age of the rock. The half-life of is 4.5 109 years.
Solution This problem can be solved using the integrated first-order rate law: lna N N0 b kt a 0.693 4.5 109 yearsbt 238 92U 206 82Pb 238 92U 206 82Pb 238 92U 206 82Pb 206 82Pb 238 92U 176 72Hf 176 71Lu 238 92U 206 82Pb 206 82Pb 238 92U 12 6C 14 6C lna N N0 b ln10.232 a 0.693 5730 yearsbt k 0.693 t1 2 0.693 5730 years lna N N0 b kt N N0 0.23 3.1 counts/min g 13.6 counts/min g rate at time t rate at time 0 kN kN0 Sample Exercise 18.6 Because the half-life of is very long compared with those of the other mem-bers of the decay series (see Table 18.3) to reach the number of nuclides present in intermediate stages of decay is negligible. That is, once a nuclide starts to decay, it reaches relatively fast.
206 82Pb 238 92U 206 82Pb, 238 92U 18.4 Detection and Uses of Radioactivity 855 where NN0 represents the ratio of atoms now found in the rock to the number pres-ent when the rock was formed. We are assuming that each nuclide present must have come from decay of a atom: Thus Think carefully about what this means. For every 1115 atoms originally present in the rock, 115 have been changed to and 1000 remain as . Thus oNow present 23 9 8 2U originally present This is the approximate age of the rock. It was formed sometime in the Cambrian period.
See Exercises 18.33 and 18.34.
Medical Applications of Radioactivity Although the rapid advances of the medical sciences in recent decades are due to many causes, one of the most important has been the discovery and use of radiotracers, radioactive nuclides that can be introduced into organisms in food or drugs and whose pathways can be traced by monitoring their radioactivity. For example, the incorporation of nuclides such as and into nutrients has produced important information about metabolic pathways.
Iodine-131 has proved very useful in the diagnosis and treatment of illnesses of the thyroid gland. Patients drink a solution containing small amounts of and the up-take of the iodine by the thyroid gland is monitored with a scanner (see Fig. 18.8).
Na131I, 32 15P 14 6C t 7.1 108 years lna N N0 b ln10.89692 a 0.693 4.5 109 yearsbt N N0 92 238U 82 206Pb 92 238U 1000 1115 0.8969 238 92U 206 82Pb 238 92U Atoms of 82 206Pb now present Atoms of 92 238U now present 0.115 0.115 1.000 115 1000 number of 92 238U atoms now present number of 82 206Pb atoms now present Number of 92 238U atoms originally present 92 238U ¡ 82 206Pb 238 92U 206 82Pb 238 92U ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ FIGURE 18.8 After consumption of Na131l, the patient’s thyroid is scanned for radioactivity levels to determine the efficiency of iodine absorption. (left) A normal thyroid. (right) An enlarged thyroid.
A pellet containing radioactive 131I.
856 Chapter Eighteen The Nucleus: A Chemist’s View Thallium-201 can be used to assess the damage to the heart muscle in a person who has suffered a heart attack, because thallium is concentrated in healthy muscle tissue.
Technetium-99m is also taken up by normal heart tissue and is used for damage assess-ment in a similar way.
Radiotracers provide sensitive and noninvasive methods for learning about biologic systems, for detection of disease, for monitoring the action and effectiveness of drugs, and for early detection of pregnancy, and their usefulness should continue to grow. Some use-ful radiotracers are listed in Table 18.5.
18.5 Thermodynamic Stability of the Nucleus We can determine the thermodynamic stability of a nucleus by calculating the change in potential energy that would occur if that nucleus were formed from its constituent pro-tons and neutrons. For example, let’s consider the hypothetical process of forming a nucleus from eight neutrons and eight protons: The energy change associated with this process can be calculated by comparing the sum of the masses of eight protons and eight neutrons with that of the oxygen nucleus: h h Mass of 1 0n Mass of 1 1H The difference in mass for one nucleus is The difference in mass for formation of 1 mole of nuclei is therefore Thus 0.1366 g of mass would be lost if 1 mole of oxygen-16 were formed from protons and neutrons. What is the reason for this difference in mass, and how can this informa-tion be used to calculate the energy change that accompanies this process?
The answers to these questions can be found in the work of Albert Einstein. As we discussed in Section 7.2, Einstein’s theory of relativity showed that energy should be con-sidered a form of matter. His famous equation E mc2 12.269 1025 g/nucleus216.022 1023 nuclei/mol2 0.1366 g/mol 16 8O Mass of 8 16O mass of 18 0 1n 8 1 1H2 2.269 1025 g Mass of 8 16O nucleus 2.65535 1023 g 2.67804 1023 g Mass of 18 0 1n 8 1 1H2 811.67493 1024 g2 811.67262 1024 g2 8 0 1n 8 1 1H ¡ 8 16O 16 8O TABLE 18.5 Some Radioactive Nuclides, with Half-Lives and Medical Applications as Radiotracers Nuclide Half-Life Area of the Body Studied 131I 8.1 days Thyroid 59Fe 45.1 days Red blood cells 99Mo 67 hours Metabolism 32P 14.3 days Eyes, liver, tumors 51Cr 27.8 days Red blood cells 87Sr 2.8 hours Bones 99mTc 6.0 hours Heart, bones, liver, and lungs 133Xe 5.3 days Lungs 24Na 14.8 hours Circulatory system Energy is a form of matter.
18.5 Thermodynamic Stability of the Nucleus 857 where c is the speed of light, gives the relationship between a quantity of energy and its mass. When a system gains or loses energy, it also gains or loses a quantity of mass, given by Ec2. Thus the mass of a nucleus is less than that of its component nucleons because the process is so exothermic.
Einstein’s equation in the form where m is the change in mass, or the mass defect, can be used to calculate E for the hypothetical formation of a nucleus from its component nucleons.
Nuclear Binding Energy I Calculate the change in energy if 1 mol nuclei was formed from neutrons and protons.
Solution We have already calculated that 0.1366 g of mass would be lost in the hypothetical process of assembling 1 mol 16 8O nuclei from the component nucleons. We can calculate the change in energy for this process from where Thus The negative sign for the E value indicates that the process is exothermic. Energy, and thus mass, is lost from the system.
See Exercises 18.35 through 18.37.
The energy changes observed for nuclear processes are extremely large compared with those observed for chemical and physical changes. Thus nuclear processes constitute a potentially valuable energy resource.
The thermodynamic stability of a particular nucleus is normally represented as en-ergy released per nucleon. To illustrate how this quantity is obtained, we will continue to consider . First, we calculate E per nucleus by dividing the molar value from Sam-ple Exercise 18.7 by Avogadro’s number: In terms of a more convenient energy unit, a million electronvolts (MeV), where Next, we can calculate the value of E per nucleon by dividing by A, the sum of neu-trons and protons: 7.98 MeV/nucleon ¢E per nucleon for 8 16O 1.28 102 MeV/nucleus 16 nucleons/nucleus 1.28 102 MeV/nucleus ¢E per 8 16O nucleus 12.04 1011 J/nucleus2a 1 MeV 1.60 1013 Jb 1 MeV 1.60 1013 J ¢E per 8 16O nucleus 1.23 1013 J/mol 6.022 1023 nuclei/mol 2.04 1011 J/nucleus 16 8O ¢E 11.366 104 kg/mol213.00 108 m/s22 1.23 1013 J/mol c 3.00 108 m/s and ¢m 0.1366 g/mol 1.366 104 kg/mol ¢E ¢mc2 16 8O Energy change ¢E ¢mc2 The energy changes associated with nor-mal chemical reactions are small enough that the corresponding mass changes are not detectable.
Sample Exercise 18.7 858 Chapter Eighteen The Nucleus: A Chemist’s View This means that 7.98 MeV of energy per nucleon would be released if were formed from neutrons and protons. The energy required to decompose this nucleus into its com-ponents has the same numeric value but a positive sign (since energy is required). This is called the binding energy per nucleon for .
The values of the binding energy per nucleon for the various nuclides are shown in Fig. 18.9. Note that the most stable nuclei (those requiring the largest energy per nucleon to decompose the nucleus) occur at the top of the curve. The most stable nucleus known is , which has a binding energy per nucleon of 8.79 MeV.
Nuclear Binding Energy II Calculate the binding energy per nucleon for the nucleus (atomic masses: 4.0026 amu; 1.0078 amu).
Solution First, we must calculate the mass defect ( m) for . Since atomic masses (which include the electrons) are given, we must decide how to account for the electron mass: p Electron mass o Thus, since a nucleus is “synthesized” from two protons and two neutrons, we see that Mass of Mass of Mass of nucleus nucleus (proton) neutron Note that in this case the electron mass cancels out in taking the difference. This will al-ways happen in this type of calculation if the atomic masses are used both for the nuclide of interest and for . Thus 0.0304 amu of mass is lost per nucleus formed.
2 4He 1 1H 0.0304 amu 4.0026 211.00782 211.00872 4.0026 2me 211.00782 2me 211.00872 1 1H 2 4He ¢m 14.0026 2me2 3211.0078 me2 211.00872 4 2 4He 1.0078 mass of 1 1H atom mass of 1 1H nucleus me 4.0026 mass of 2 4He atom mass of 2 4He nucleus 2me 2 4He 1 1H 2 4He 2 4He 26 56Fe 16 8O 16 8O Sample Exercise 18.8 FIGURE 18.9 The binding energy per nucleon as a func-tion of mass number. The most stable nu-clei are at the top of the curve. The most stable nucleus is .
56 26Fe Since atomic masses include the masses of the electrons, to obtain the mass of a given atomic nucleus from its atomic mass, we must subtract the mass of the electrons.
9 8 7 6 5 4 3 2 1 0 Binding energy per nucleon (MeV) 20 40 60 80 100 120 140 160 180 200 220 240 260 2H 3He 3H 6Li 7Li 4He 14N 12C 16O 34S 56Fe 84Kr 119Sn 205Tl 235U 238U Mass number (A) ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ 18.6 Nuclear Fission and Nuclear Fusion 859 The corresponding energy change can be calculated from where and Thus This means that 4.54 1012 J of energy is released per nucleus formed and that 4.54 1012 J would be required to decompose the nucleus into the constituent neutrons and protons. Thus the binding energy (BE) per nucleon is See Exercises 18.38 through 18.40.
18.6 Nuclear Fission and Nuclear Fusion The graph shown in Fig. 18.9 has very important implications for the use of nuclear processes as sources of energy. Recall that energy is released, that is, is negative, when a process goes from a less stable to a more stable state. The higher a nuclide is on the curve, the more stable it is. This means that two types of nuclear processes will be exother-mic (see Fig. 18.10): 1. Combining two light nuclei to form a heavier, more stable nucleus. This process is called fusion.
2. Splitting a heavy nucleus into two nuclei with smaller mass numbers. This process is called fission.
Because of the large binding energies involved in holding the nucleus together, both these processes involve energy changes more than a million times larger than those associated with chemical reactions.
Nuclear Fission Nuclear fission was discovered in the late 1930s when nuclides bombarded with neu-trons were observed to split into two lighter elements: 1 0n 235 92U ¡ 141 56Ba 92 36Kr 3 1 0n 235 92U ¢E 7.13 MeV/nucleon a1.14 1012 J nucleonba 1 MeV 1.60 1013 Jb 1.14 1012 J/nucleon BE per nucleon 4.54 1012 J/nucleus 4 nucleons/nucleus 4.54 1012 J/nucleus ¢E a5.04 1029 kg nucleusba3.00 108 m s b 2 c 3.00 108 m/s 5.04 1029 kg nucleus ¢m 0.0304 amu nucleus a0.0304 amu nucleusba1.66 1027 kg amub ¢E ¢mc2 Visualization: Nuclear Fission 860 Chapter Eighteen The Nucleus: A Chemist’s View This process, shown schematically in Fig. 18.11, releases 3.5 1011 J of energy per event, which translates to 2.1 1013 J per mole of . Compare this figure with that for the combustion of methane, which releases only 8.0 105 J of energy per mole. The fission of produces about 26 million times more energy than the combustion of methane.
The process shown in Fig. 18.11 is only one of the many fission reactions that can undergo. Another is In fact, over 200 different isotopes of 35 different elements have been observed among the fission products of .
In addition to the product nuclides, neutrons are produced in the fission reactions of . This makes it possible to have a self-sustaining fission process—a chain reaction (see Fig. 18.12). For the fission process to be self-sustaining, at least one neutron from each fission event must go on to split another nucleus. If, on average, less than one neutron causes another fission event, the process dies out and the reaction is said to be subcritical.
If exactly one neutron from each fission event causes another fission event, the process sustains itself at the same level and is said to be critical. If more than one neutron from each fission event causes another fission event, the process rapidly escalates and the heat buildup causes a violent explosion. This situation is described as supercritical.
235 92U 235 92U 1 0n 235 92U ¡ 137 52Te 97 40Zr 2 1 0n 235 92U 235 92U 235 92U FIGURE 18.10 Both fission and fusion produce more stable nuclides and are thus exothermic.
FIGURE 18.11 On capturing a neutron, the nucleus undergoes fission to produce two lighter nuclides, free neutrons (typically three), and a large amount of energy.
235 92U 9 8 7 6 5 4 3 2 1 0 Binding energy per nucleon (MeV) 20 40 60 80 100 120 140 160 180 200 220 240 260 Mass number (A) 56 26Fe Fission Fusion 235 92U n 236 92 (Unstable nucleus) U n n n + Energy 141 56Ba 92 36Kr 18.6 Nuclear Fission and Nuclear Fusion 861 To achieve the critical state, a certain mass of fissionable material, called the critical mass, is needed. If the sample is too small, too many neutrons escape before they have a chance to cause a fission event, and the process stops. This is illustrated in Fig. 18.13.
During World War II, an intense research effort called the Manhattan Project was car-ried out by the United States to build a bomb based on the principles of nuclear fission.
This program produced the fission bombs that were used with devastating effects on the cities of Hiroshima and Nagasaki in 1945. Basically, a fission bomb operates by suddenly combining subcritical masses of fissionable material to form a supercritical mass, thereby producing an explosion of incredible intensity.
Nuclear Reactors Because of the tremendous energies involved, it seemed desirable to develop the fission process as an energy source to produce electricity. To accomplish this, reactors were de-signed in which controlled fission can occur. The resulting energy is used to heat water to produce steam to run turbine generators, in much the same way that a coal-burning power plant generates energy. A schematic diagram of a nuclear power plant is shown in Fig. 18.14.
In the reactor core, shown in Fig. 18.15, uranium that has been enriched to approx-imately 3% (natural uranium contains only 0.7% ) is housed in cylinders. A moderator surrounds the cylinders to slow down the neutrons so that the uranium fuel can capture them more efficiently. Control rods, composed of substances that absorb 235 92U 235 92U FIGURE 18.12 Representation of a fission process in which each event produces two neutrons, which can go on to split other nuclei, leading to a self-sustaining chain reaction.
Two neutrons from fission Nucleus Neutron FIGURE 18.13 If the mass of fissionable material is too small, most of the neutrons escape before causing another fission event, and the process dies out.
Nucleus Subcritical mass (too many neutrons escape to keep the reaction sustained) Supercritical mass (most released neutrons interact with nuclides and the chain reaction multiplies) Large proportion of escapes Small proportion of escapes 862 Chapter Eighteen The Nucleus: A Chemist’s View neutrons, are used to regulate the power level of the reactor. The reactor is designed so that should a malfunction occur, the control rods are automatically inserted into the core to stop the reaction. A liquid (usually water) is circulated through the core to extract the heat generated by the energy of fission; the energy can then be passed on via a heat ex-changer to water in the turbine system.
Although the concentration of in the fuel elements is not great enough to allow a supercritical mass to develop in the core, a failure of the cooling system can lead to temper-atures high enough to melt the core. As a result, the building housing the core must be de-signed to contain the core even if meltdown occurs. A great deal of controversy now exists about the efficiency of the safety systems in nuclear power plants. Accidents such as the one at the Three Mile Island facility in Pennsylvania in 1979 and in Chernobyl, Ukraine, in 1986 have led to questions about the wisdom of continuing to build fission-based power plants.
Breeder Reactors One potential problem facing the nuclear power industry is the supply of . Some sci-entists have suggested that we have nearly depleted those uranium deposits rich enough in to make production of fissionable fuel economically feasible. Because of this possibility, breeder reactors have been developed, in which fissionable fuel is actually produced while the reactor runs. In the breeder reactors now being studied, the major component of natural uranium, nonfissionable , is changed to fissionable . The reaction involves absorption of a neutron, followed by production of two particles: 239 93Np ¡ 239 94Pu 0 1e 239 92U ¡ 239 93Np 0 1e 1 0n 238 92U ¡ 239 92U 239 94Pu 238 92U 235 92U 235 92U 235 92U FIGURE 18.14 A schematic diagram of a nuclear power plant.
FIGURE 18.15 A schematic of a reactor core. The posi-tion of the control rods determines the level of energy production by regulating the amount of fission taking place.
Uranium oxide (refined uranium).
See C. A. Atwood, “Chernobyl—What happened?” J. Chem. Ed. 65 (1988): 1037.
Pump Pump Pump Steam turbine Condenser (steam from turbine is condensed) Electrical output Large water source 27°C 38°C Steam generator Water Reactor Control rods Containment shell Steam Incoming coolant Hot coolant Control rods of neutron-absorbing material Uranium fuel cylinders 18.7 Effects of Radiation 863 As the reactor runs and is split, some of the excess neutrons are absorbed by to produce . The is then separated out and used to fuel another reactor. Such a reactor thus “breeds” nuclear fuel as it operates.
Although breeder reactors are now used in France, the United States is proceeding slowly with their development because of their controversial nature. One problem involves the hazards in handling plutonium, which flames on contact with air and is very toxic.
Fusion Large quantities of energy are also produced by the fusion of two light nuclei. In fact, stars produce their energy through nuclear fusion. Our sun, which presently consists of 73% hydrogen, 26% helium, and 1% other elements, gives off vast quantities of energy from the fusion of protons to form helium: Intense research is under way to develop a feasible fusion process because of the ready availability of many light nuclides (deuterium, , in seawater, for example) that can serve as fuel in fusion reactors. The major stumbling block is that high temperatures are required to initiate fusion. The forces that bind nucleons together to form a nucleus are effective only at very small distances (1013 cm). Thus, for two protons to bind to-gether and thereby release energy, they must get very close together. But protons, because they are identically charged, repel each other electrostatically. This means that to get two protons (or two deuterons) close enough to bind together (the nuclear binding force is not electrostatic), they must be “shot” at each other at speeds high enough to overcome the electrostatic repulsion.
The electrostatic repulsion forces between two nuclei are so great that a tem-perature of 4 107 K is required to give them velocities large enough to cause them to collide with sufficient energy that the nuclear forces can bind the particles together and thus release the binding energy. This situation is represented in Fig. 18.16.
Currently, scientists are studying two types of systems to produce the extremely high temperatures required: high-powered lasers and heating by electric currents. At present, many technical problems remain to be solved, and it is not clear which method will prove more useful or when fusion might become a practical energy source.
However, there is still hope that fusion will be a major energy source sometime in the future.
18.7 Effects of Radiation Everyone knows that being hit by a train is very serious. The problem is the energy trans-fer involved. In fact, any source of energy is potentially harmful to organisms. Energy transferred to cells can break chemical bonds and cause malfunctioning of the cell systems.
This fact is behind the concern about the ozone layer in the earth’s upper atmosphere, which screens out high-energy ultraviolet radiation from the sun. Radioactive elements, which are sources of high-energy particles, are also potentially hazardous, although the effects are usually quite subtle. The reason for the subtlety of radiation damage is that even though high-energy particles are involved, the quantity of energy actually deposited in tissues per event is quite small. However, the resulting damage is no less real, although the effects may not be apparent for years.
1 2H 2 1H 3 2He 1 1H ¡ 4 2He 0 1e 3 2He 3 2He ¡ 4 2He 2 1 1H 1 1H 2 1H ¡ 3 2He 1 1H 1 1H ¡ 2 1H 0 1e 239 94Pu 239 94Pu 238 92U 235 92U Visualization: Nuclear Fusion FIGURE 18.16 A plot of energy versus the separation distance for two nuclei. The nuclei must have sufficient velocities to get over the electrostatic repulsion “hill” and get close enough for the nuclear binding forces to become effective, thus “fusing” the particles into a new nucleus and re-leasing large quantities of energy. The binding force is at least 100 times the electrostatic repulsion.
2 1H Distance between the particles Energy of attraction due to the strong nuclear force Electrostatic repulsion E 0 The ozone layer is discussed in Section 20.5.
864 Chapter Eighteen The Nucleus: A Chemist’s View CHEMICAL IMPACT Nuclear Physics: An Introduction N uclear physics is concerned with the fundamental nature of matter. The central focuses of this area of study are the relationship between a quantity of energy and its mass, given by and the fact that matter can be converted from one form (energy) to another (particulate) in particle accelerators. Collisions between high-speed particles have produced a dazzling array of new particles—hundreds of them. These events can best be interpreted as conversions of kinetic energy into particles. For example, a collision of suf-ficient energy between a proton and a neutron can produce four particles: two protons, one antiproton, and a neutron: where is the symbol for an antiproton, which has the same mass as a proton but the opposite charge. This process is a little like throwing one baseball at a very high speed into another and having the energy of the collision converted into two additional baseballs.
The results of such accelerator experiments have led sci-entists to postulate the existence of three types of forces im-portant in the nucleus: the strong force, the weak force, and the electromagnetic force. Along with the gravitational force, these forces are thought to account for all types of in-teractions found in matter. These forces are believed to be generated by the exchange of particles between the inter-acting pieces of matter. For example, gravitational forces are 1 1H 1 1H 1 0n ¡ 2 1 1H 1 1H 1 0n E mc2, thought to be carried by particles called gravitons. The elec-tromagnetic force (the classical electrostatic force between charged particles) is assumed to be exerted through the ex-change of photons. The strong force, not charge-related and effective only at very short distances (1013 cm), is pos-tulated to involve the exchange of particles called gluons.
The weak force is 100 times weaker than the strong force and seems to be exerted through the exchange of two types of large particles, the W (has a mass 70 times the proton mass) and the Z (has a mass 90 times the proton mass).
The particles discovered have been classified into sev-eral categories. Three of the most important classes are as follows: 1. Hadrons are particles that respond to the strong force and have internal structure.
2. Leptons are particles that do not respond to the strong force and have no internal structure.
3. Quarks are particles with no internal structure that are thought to be the fundamental constituents of hadrons.
Neutrons and protons are hadrons that are thought to be composed of three quarks each.
The world of particle physics appears mysterious and complicated. For example, particle physicists have discov-ered new properties of matter they call “color,” “charm,” Radiation damage to organisms can be classified as somatic or genetic damage.
Somatic damage is damage to the organism itself, resulting in sickness or death. The effects may appear almost immediately if a massive dose of radiation is received; for smaller doses, damage may appear years later, usually in the form of cancer. Genetic damage is damage to the genetic machinery, which produces malfunctions in the offspring of the organism.
The biologic effects of a particular source of radiation depend on several factors: 1. The energy of the radiation. The higher the energy content of the radiation, the more damage it can cause. Radiation doses are measured in rads (which is short for radi-ation absorbed dose), where 1 rad corresponds to 102 J of energy deposited per kilo-gram of tissue.
2. The penetrating ability of the radiation. The particles and rays produced in radioac-tive processes vary in their abilities to penetrate human tissue: rays are highly pen-etrating, particles can penetrate approximately 1 cm, and particles are stopped by the skin.
3. The ionizing ability of the radiation. Extraction of electrons from biomolecules to form ions is particularly detrimental to their functions. The ionizing ability of radiation 18.7 Effects of Radiation 865 varies dramatically. For example, rays penetrate very deeply but cause only occa-sional ionization. On the other hand, particles, although not very penetrating, are very effective at causing ionization and produce a dense trail of damage. Thus in-gestion of an -particle producer, such as plutonium, is particularly damaging.
4. The chemical properties of the radiation source. When a radioactive nuclide is in-gested into the body, its effectiveness in causing damage depends on its residence time. For example, and are both -particle producers. However, since kryp-ton is chemically inert, it passes through the body quickly and does not have much time to do damage. Strontium, being chemically similar to calcium, can collect in bones, where it may cause leukemia and bone cancer.
Because of the differences in the behavior of the particles and rays produced by radioactive decay, both the energy dose of the radiation and its effectiveness in causing biologic damage must be taken into account. The rem (which is short for roentgen equiv-alent for man) is defined as follows: where RBE represents the relative effectiveness of the radiation in causing biologic damage.
Number of rems 1number of rads2 RBE 38 90Sr 36 85Kr and “strangeness” and have postulated conservation laws involving these properties. This area of science is ex-tremely important because it should help us to understand the interactions of matter in a more elegant and unified way. For example, the classification of force into four cat-egories is probably necessary only because we do not un-derstand the true nature of forces. All forces may be spe-cial cases of a single, all-pervading force field that governs all of nature. In fact, Einstein spent the last 30 years of his life looking for a way to unify the gravitational and electromagnetic forces—without success. Physicists may now be on the verge of accomplishing what Einstein failed to do.
Although the practical aspects of the work in nuclear physics are not yet totally apparent, a more fundamental un-derstanding of the way nature operates could lead to presently undreamed-of devices for energy production and communication, which could revolutionize our lives.
(left) An aerial view of Fermilab, a high-energy particle accelerator in Batavia, Illinois.
(right) The accelerator tunnel at Fermilab.
866 Chapter Eighteen The Nucleus: A Chemist’s View FIGURE 18.17 The two models for radiation damage. In the linear model, even a small dosage causes a proportional risk. In the threshold model, risk begins only after a certain dosage.
Exposure level 0 Radiation damage in humans Linear model Threshold model Threshold dosage TABLE 18.7 Typical Radiation Exposures for a Person Living in the United States (1 millirem 103 rem) Exposure (millirems/year) Cosmic radiation 50 From the earth 47 From building materials 3 In human tissues 21 Inhalation of air 5 Total from natural sources 126 X-ray diagnosis 50 Radiotherapy 10 Internal diagnosis/ therapy 1 Nuclear power industry 0.2 TV tubes, industrial wastes, etc.
2 Radioactive fallout 4 Total from human activities 67 Total 193 TABLE 18.6 Effects of Short-Term Exposures to Radiation Dose (rem) Clinical Effect 0–25 Nondetectable 25–50 Temporary decrease in white blood cell counts 100–200 Strong decrease in white blood cell counts 500 Death of half the exposed population within 30 days after exposure Table 18.6 shows the physical effects of short-term exposure to various doses of ra-diation, and Table 18.7 gives the sources and amounts of radiation exposure for a typical person in the United States. Note that natural sources contribute about twice as much as human activities to the total exposure. However, although the nuclear industry contributes only a small percentage of the total exposure, the major controversy associated with nu-clear power plants is the potential for radiation hazards. These arise mainly from two sources: accidents allowing the release of radioactive materials and improper disposal of the radioactive products in spent fuel elements. The radioactive products of the fission of , although only a small percentage of the total products, have half-lives of several hundred years and remain dangerous for a long time. Various schemes have been advanced for the disposal of these wastes. The one that seems to hold the most promise is the in-corporation of the wastes into ceramic blocks and the burial of these blocks in geologi-cally stable formations. At present, however, no disposal method has been accepted, and nuclear wastes continue to accumulate in temporary storage facilities.
Even if a satisfactory method for permanent disposal of nuclear wastes is found, there will continue to be concern about the effects of exposure to low levels of radiation. Ex-posure is inevitable from natural sources such as cosmic rays and radioactive minerals, and many people are also exposed to low levels of radiation from reactors, radioactive tracers, or diagnostic X rays. Currently, we have little reliable information on the long-term effects of low-level exposure to radiation.
Two models of radiation damage, illustrated in Fig. 18.17, have been proposed: the linear model and the threshold model. The linear model postulates that damage from ra-diation is proportional to the dose, even at low levels of exposure. Thus any exposure is dangerous. The threshold model, on the other hand, assumes that no significant damage occurs below a certain exposure, called the threshold exposure. Note that if the linear model is correct, radiation exposure should be limited to a bare minimum (ideally at the natural levels). If the threshold model is correct, a certain level of radiation exposure be-yond natural levels can be tolerated. Most scientists feel that since there is little evidence available to evaluate these models, it is safest to assume that the linear hypothesis is cor-rect and to minimize radiation exposure.
235 92U For Review 867 Key Terms neutron proton nucleon atomic number mass number isotopes nuclide Section 18.1 thermodynamic stability kinetic stability radioactive decay beta ( ) particle zone of stability alpha ( ) particle -particle production spontaneous fission -particle production gamma ( ) ray positron production electron capture decay series Section 18.2 rate of decay half-life Section 18.3 nuclear transformation particle accelerator cyclotron linear accelerator transuranium elements Section 18.4 Geiger–Müller counter (Geiger counter) scintillation counter radiocarbon dating (carbon-14 dating) radiotracers Section 18.5 mass defect binding energy Section 18.6 fusion fission chain reaction subcritical reaction critical reaction supercritical reaction critical mass reactor core moderator control rods breeder reactor Section 18.7 somatic damage genetic damage rad rem g b a a b For Review Radioactivity Certain nuclei decay spontaneously into more stable nuclei Types of radioactive decay: • -particle production • -particle ( 0 1e) production • Positron production • rays are usually produced in a radioactive decay event A decay series involves several radioactive decays to finally reach a stable nuclide Radioactive decay follows first-order kinetics • Half-life of a radioactive sample: the time required for half of the nuclides to decay The transuranium elements (those beyond uranium in the periodic table) can be synthesized by particle bombardment of uranium or heavier elements Radiocarbon dating employs the 14 6C12 6C ratio in an object to establish its date of origin Thermodynamic stability of a nucleus Compares the mass of a nucleus to the sum of the masses of its component nucleons When a system gains or loses energy, it also gains or loses mass as described by the relationship The difference between the sum of the masses of the component nucleons and the actual mass of a nucleus (called the mass defect) can be used to calculate the nu-clear binding energy Nuclear energy production Fusion: the process of combining two light nuclei to form a heavier, more stable nucleus Fission: the process of splitting a heavy nucleus into two lighter, more stable nuclei • Current nuclear power reactors employ controlled fission to produce energy Radiation damage Radiation can cause direct (somatic) damage to a living organism or genetic damage to the organism’s offspring The biologic effects of radiation depend on the energy, the penetrating ability, the ionizing ability of the radiation, and the chemical properties of the nuclide producing the radiation REVIEW QUESTIONS 1. Define or illustrate the following terms: a. thermodynamic stability b. kinetic stability c. radioactive decay d. beta-particle production e. alpha-particle production f. positron production g. electron capture h. gamma-ray emissions In radioactive decay processes, A and Z are conserved. What does this mean?
2. Figure 18.1 illustrates the zone of stability. What is the zone of stability? Stable light nuclides have about equal numbers of neutrons and protons. What happens to the neutron-to-proton ratio for stable nuclides as the number of protons E mc2 g (0 1e) b (4 2He) a 868 Chapter Eighteen The Nucleus: A Chemist’s View increases? Nuclides that are not already in the zone of stability undergo ra-dioactive processes to get to the zone of stability. If a nuclide has too many neutrons, which process(es) can the nuclide undergo to become more stable?
Answer the same question for a nuclide having too many protons.
3. All radioactive decay processes follow first-order kinetics. What does this mean?
What happens to the rate of radioactive decay as the number of nuclides is halved? Write the first-order rate law and the integrated first-order rate law.
Define the terms in each equation. What is the half-life equation for radioactive decay processes? How does the half-life depend on how many nuclides are pres-ent? Are the half-life and rate constant k directly related or inversely related?
4. What is a nuclear transformation? How do you balance nuclear transformation reactions? Particle accelerators are used to perform nuclear transformations.
What is a particle accelerator?
5. What is a Geiger counter and how does it work? What is a scintillation counter and how does it work? Radiotracers are used in the medical sciences to learn about metabolic pathways. What are radiotracers? Explain why 14C and 32P radioactive nuclides would be very helpful in learning about metabolic pathways.
Why is I-131 useful for diagnosis of diseases of the thyroid? How could you use a radioactive nuclide to demonstrate that chemical equilibrium is a dynamic process?
6. Explain the theory behind carbon-14 dating. What assumptions must be made and what problems arise when using carbon-14 dating?
The decay of uranium-238 to lead-206 is also used to estimate the age of objects. Specifically, 206Pb-to-238U ratios allow dating of rocks. Why is the 238U decay to 206Pb useful for dating rocks but worthless for dating objects 10,000 years old or younger? Similarly, why is carbon-14 dating useful for dating objects 10,000 years old or younger but worthless for dating rocks?
7. Define mass defect and binding energy. How do you determine the mass defect for a nuclide? How do you convert the mass defect into the binding energy for a nuclide? Iron-56 has the largest binding energy per nucleon among all known nuclides. Is this good or bad for iron-56? Explain.
8. Define fission and fusion. How does the energy associated with fission or fusion processes compare to the energy changes associated with chemical reactions?
Fusion processes are more likely to occur for lighter elements, whereas fission processes are more likely to occur for heavier elements. Why? (Hint: Reference Figure 18.10.) The major stumbling block for turning fusion reactions into a feasible source of power is the high temperature required to initiate a fusion re-action. Why are elevated temperatures necessary to initiate fusion reactions but not fission reactions?
9. The fission of U-235 is used exclusively in nuclear power plants located in the United States. There are many different fission reactions of U-235, but all the fission reactions are self-sustaining chain reactions. Explain. Differentiate between the terms critical, subcritical, and supercritical. What is the critical mass? How does a nuclear power plant produce electricity? What are the purposes of the moderator and the control rods in a fission reactor? What are some problems associated with nuclear reactors? What are breeder reactors?
What are some problems associated with breeder reactors?
10. The biological effects of a particular source of radiation depend on several factors.
List some of these factors. Even though 85Kr and 90Sr are both beta-particle emit-ters, the dangers associated with the decay of 90Sr are much greater than those linked to 85Kr. Why? Although gamma rays are far more penetrating than alpha particles, the latter are more likely to cause damage to an organism. Why? Which type of radiation is more effective at promoting the ionization of biomolecules?
Exercises 869 A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 1. When nuclei undergo nuclear transformations, rays of charac-teristic frequencies are observed. How does this fact, along with other information in the chapter on nuclear stability, suggest that a quantum mechanical model may apply to the nucleus?
2. There is a trend in the United States toward using coal-fired power plants to generate electricity rather than building new nuclear fission power plants. Is the use of coal-fired power plants without risk? Make a list of the risks to society from the use of each type of power plant.
3. Which type of radioactive decay has the net effect of changing a neutron into a proton? Which type of decay has the net effect of turning a proton into a neutron?
4. What is annihilation in terms of nuclear processes?
5. What are transuranium elements and how are they synthesized?
6. Scientists have estimated that the earth’s crust was formed 4.3 bil-lion years ago. The radioactive nuclide 176Lu, which decays to 176Hf, was used to estimate this age. The half-life of 176Lu is 37 billion years. How are ratios of 176Lu to 176Hf utilized to date very old rocks?
7. Why are the observed energy changes for nuclear processes so much larger than the energy changes for chemical and physical processes?
8. Natural uranium is mostly nonfissionable 238U; it contains only about 0.7% of fissionable 235U. For uranium to be useful as a nuclear fuel, the relative amount of 235U must be increased to about 3%. This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of 238UF6(g) and 235UF6(g). The fluoride mixture is then enriched through a multistage diffusion process to produce a 3% 235U nuclear fuel. The diffusion process utilizes Graham’s law of effusion (see Chapter 5, Section 5.7). Explain how Graham’s law of effusion al-lows natural uranium to be enriched by the gaseous diffusion process.
9. Strontium-90 and radon-222 both pose serious health risks. 90Sr decays by -particle production and has a relatively long half-life (28.8 yr). Radon-222 decays by alpha-particle production and has a relatively short half-life (3.82 days). Explain why each decay process poses health risks.
10. A recent study concluded that any amount of radiation exposure can cause biological damage. Explain the differences between the two models of radiation damage, the linear model and the thresh-old model.
Exercises In this section similar exercises are paired.
Radioactive Decay and Nuclear Transformations 11. Write balanced equations for each of the processes described below.
a. Chromium-51, which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture.
g b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a particle.
12. Write balanced equations for each of the processes described below.
a. Phosphorus-32, which accumulates in the liver, decays by -particle production.
b. Uranium-235, which is used in atomic bombs, decays initially by -particle production.
13. Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parenthe-ses, except for electron capture, where an electron is a reactant.) a. 68Ga (electron capture) c. 212Fr () b. 62Cu (positron) d. 129Sb () 14. In each of the following nuclear reactions, supply the missing particle.
a. 73Ga S 73Ge ?
c. 205Bi S 205Pb ?
b. 192Pt S 188Os ?
d. 241Cm ? S 241Am 15. The radioactive isotope 247Bk decays by a series of -particle and -particle productions, taking 247Bk through many transformations to end up as 207Pb. In the complete decay series, how many par-ticles and particles are produced?
16. One type of commercial smoke detector contains a minute amount of radioactive americium-241 (241Am), which decays by -parti-cle production. The particles ionize molecules in the air, al-lowing it to conduct an electric current. When smoke particles en-ter, the conductivity of the air is changed and the alarm buzzes.
a. Write the equation for the decay of 241 95Am by -particle production.
b. The complete decay of 241Am involves successively , , , , , , , , , , , and production. What is the final sta-ble nucleus produced in this decay series?
c. Identify the 11 intermediate nuclides.
17. There are four stable isotopes of iron with mass numbers 54, 56, 57, and 58. There are also two radioactive isotopes: iron-53 and iron-59. Predict modes of decay for these two isotopes. (See Table 18.2.) 18. The only stable isotope of fluorine is fluorine-19. Predict possi-ble modes of decay for fluorine-21, fluorine-18, and fluorine-17.
19. In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, Sg, in honor of Glenn T. Seaborg, dis-coverer of the transuranium elements.
a. 263Sg was produced by the bombardment of 249Cf with a beam of 18O nuclei. Complete and balance an equation for this reaction.
b. 263Sg decays by emission. What is the other product result-ing from the decay of 263Sg?
20. Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators.
Complete the following nuclear reactions, which have been used to synthesize elements.
a. _ 4 2He S 243 97Bk 1 0n b.
238 92U 12 6C S _ 6 1 0n c.
249 98Cf _ S 260 105Db 4 1 0n d. 249 98Cf 10 5B S 257 103Lr _ a b b a a b b 870 Chapter Eighteen The Nucleus: A Chemist’s View mass of NaBr should you order (assuming all of the Br in the NaBr was Br-82)?
30. Fresh rainwater or surface water contains enough tritium ( ) to show 5.5 decay events per minute per 100. g of water. Tritium has a half-life of 12.3 years. You are asked to check a vintage wine that is claimed to have been produced in 1946. How many decay events per minute should you expect to observe in 100. g of that wine?
31. A living plant contains approximately the same fraction of carbon-14 as in atmospheric carbon dioxide. Assuming that the observed rate of decay of carbon-14 from a living plant is 13.6 counts per minute per gram of carbon, how many counts per minute per gram of carbon will be measured from a 15,000-year-old sample? Will radiocarbon dating work well for small samples of 10 mg or less?
(For14C, years.) 32. Assume a constant ratio of 13.6 counts per minute per gram of living matter. A sample of a petrified tree was found to give 1.2 counts per minute per gram. How old is the tree?
33. A rock contains 0.688 mg of 206Pb for every 1.000 mg of 238U present. Assuming that no lead was originally present, that all the 206Pb formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between 238U and 206Pb is negligible, calculate the age of the rock. (For 238U, years.) 34. The mass ratios of 40Ar to 40K also can be used to date geologic materials. Potassium-40 decays by two processes: 4 1 0 9K 0 1e 88n 4 1 0 8Ar (10.7%) t12 1.27 109 years 4 1 0 9K 88n 4 2 0 0Ca 0 1e (89.3%) a. Why are 40Ar40K ratios used to date materials rather than 40Ca40K ratios?
b. What assumptions must be made using this technique?
c. A sedimentary rock has an 40Ar40K ratio of 0.95. Calculate the age of the rock.
d. How will the measured age of a rock compare to the actual age if some 40Ar escaped from the sample?
Energy Changes in Nuclear Reactions 35. The sun radiates J of energy into space every second.
What is the rate at which mass is lost from the sun?
36. The earth receives kJ/s of solar energy. What mass of solar material is converted to energy over a 24-h period to provide the daily amount of solar energy to the earth? What mass of coal would have to be burned to provide the same amount of energy? (Coal releases 32 kJ of energy per gram when burned.) 37. Many transuranium elements, such as plutonium-232, have very short half-lives. (For 232Pu, the half-life is 36 minutes.) However, some, like protactinium-231 have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mol of 232Pu nu-clei and 1 mol of 231Pa nuclei are each formed from their respec-tive number of protons and neutrons.
1half-life 3.34 104 years2, 1.8 1014 3.9 1023 t1 2 4.5 109 (t1 2 14C 5730 years.) 14C 12C t1 2 5730 3 1H Kinetics of Radioactive Decay 21. The rate constant for a certain radioactive nuclide is 1.0 What is the half-life of this nuclide?
22. Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of 241Am is 432.2 years, and it decays by emitting alpha particles. How many alpha particles are emitted each second by a 5.00-g sample of 241Am?
23. Krypton consists of several radioactive isotopes, some of which are listed in the following table.
Half-life Kr-73 27 s Kr-74 11.5 min Kr-76 14.8 h Kr-81 2.1 105 yr Which of these isotopes is most stable and which isotope is “hottest”? How long does it take for 87.5% of each isotope to decay?
24. Radioactive copper-64 decays with a half-life of 12.8 days.
a. What is the value of k in b. A sample contains 28.0 mg 64Cu. How many decay events will be produced in the first second? Assume the atomic mass of 64Cu is 64.0.
c. A chemist obtains a fresh sample of 64Cu and measures its ra-dioactivity. She then determines that to do an experiment, the radioactivity cannot fall below 25% of the initial measured value. How long does she have to do the experiment?
25. Phosphorus-32 is a commonly used radioactive nuclide in bio-chemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phospho-rus-32 is left from an original sample of 175 mg of Na3 32PO4 after 35.0 days? Assume the atomic mass of 32P is 32.0.
26. The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to decay events per second (the number of decay events from 1 g of radium in 1 s).
a. What mass of Na2 38SO4 has an activity of 10.0 mCi? Sulfur-38 has an atomic mass of 38.0 and a half-life of 2.87 h.
b. How long does it take for 99.99% of a sample of sulfur-38 to decay?
27. The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July 16, 1945. What fraction of the strontium-90 ( years) originally produced by that ex-plosion still remains as of July 16, 2006?
28. Iodine-131 is used in the diagnosis and treatment of thyroid dis-ease and has a half-life of 8.1 days. If a patient with thyroid dis-ease consumes a sample of Na131I containing 10 of 131I, how long will it take for the amount of 131I to decrease to of the original amount?
29. The Br-82 nucleus has a half-life of min. If you wanted 1.0 g of Br-82 and the delivery time was 3.0 days, what 1.0 103 1 100 mg t1 2 28.8 3.7 1010 s1?
103 h1.
Additional Exercises 871 46. When using a Geiger–Müller counter to measure radioactivity, it is necessary to maintain the same geometrical orientation between the sample and the Geiger–Müller tube to compare different meas-urements. Why?
47. Photosynthesis in plants can be represented by the following overall reaction: Algae grown in water containing some 18O (in H2 18O) evolve oxy-gen gas with the same isotopic composition as the oxygen in the water. When algae growing in water containing only 16O were fur-nished carbon dioxide containing 18O, no 18O was found to be evolved from the oxygen gas produced. What conclusions about photosynthesis can be drawn from these experiments?
48. Consider the following reaction to produce methyl acetate: Methyl acetate When this reaction is carried out with CH3OH containing oxygen-18, the water produced does not contain oxygen-18. Explain.
49. U-235 undergoes many different fission reactions. For one such reaction, when U-235 is struck with a neutron, Ce-144 and Sr-90 are produced along with some neutrons and electrons. How many neutrons and -particles are produced in this fission reaction?
50. Breeder reactors are used to convert the nonfissionable nuclide to a fissionable product. Neutron capture of the is fol-lowed by two successive beta decays. What is the final fissionable product?
51. Which do you think would be the greater health hazard: the re-lease of a radioactive nuclide of Sr or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chem-ical properties of Sr and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?
52. Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most -particle radiation.
ii. Plutonium is an -particle producer.
iii. The chemistry of Pu4 is similar to that of Fe3.
iv. Pu oxidizes readily to Pu4.
Why is plutonium one of the most toxic substances known?
Additional Exercises 53. Predict whether each of the following nuclides is stable or un-stable (radioactive). If the nuclide is unstable, predict the type of radioactivity you would expect it to exhibit.
a. 45 19K b. 56 26Fe c. 20 11Na d. 194 81Tl 54. At a flea market, you’ve found a very interesting painting done in the style of Rembrandt’s “dark period” (1642–1672). You suspect that you really do not have a genuine Rembrandt, but you take it to the local university for testing. Living wood shows a carbon-14 238 092U 238 092U CH3OH CH3COH CH3COCH3 H2O O 8n O 6CO21g2 6H2O1l2 ¡ C6H12O61s2 6O21g2 Atom or Atomic Particle Mass Neutron 1.67493 1024 g Proton 1.67262 1024 g Electron 9.10939 1028 g Pu-232 3.85285 1022 g Pa-231 3.83616 1022 g (Since the masses of 232Pu and 231Pa are atomic masses, they each include the mass of the electrons present. The mass of the nucleus will be the atomic mass minus the mass of the electrons.) 38. The most stable nucleus in terms of binding energy per nucleon is 56Fe. If the atomic mass of 56Fe is 55.9349 amu, calculate the binding energy per nucleon for 56Fe.
39. Calculate the binding energy in J/nucleon for carbon-12 (atomic mass 12.0000) and uranium-235 (atomic mass 235.0439). The atomic mass of 1 1H is 1.00782 amu and the mass of a neutron is 1.00866 amu. The most stable nucleus known is 56Fe (see Exer-cise 38). Would the binding energy per nucleon for 56Fe be larger or smaller than that of 12C or 235U? Explain.
40. Calculate the binding energy per nucleon for 2 1H and 3 1H. The atomic masses are 2 1H, 2.01410, and 3 1H, 3.01605.
41. The mass defect for a Li-6 nucleus is g/mol. Calculate the atomic mass of Li-6.
42. The binding energy per nucleon for Mg-27 is J/nucleon. Calculate the atomic mass of Mg-27.
43. Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are 1 1H, 1.00782 amu, 2 1H, 2.01410 amu, and an electron, amu. (Hint: Think carefully about how to account for the electron mass.) 44. The easiest fusion reaction to initiate is Calculate the energy released per 4 2He nucleus produced and per mole of 4 2He produced. The atomic masses are 2 1H, 2.01410; 3 1H, 3.01605; and 4 2He, 4.00260. The masses of the electron and neu-tron are and 1.00866 amu, respectively.
Detection, Uses, and Health Effects of Radiation 45. The typical response of a Geiger–Müller tube is shown below.
Explain the shape of this curve.
Disintegrations/s from sample Counts/s 5.4858 104 2 1H 3 1H ¡ 4 2He 0 1n 1 1H 1 1H ¡ 2 1H 0 1e 104 5.4858 1.326 1012 0.03434 Light 872 Chapter Eighteen The Nucleus: A Chemist’s View 64. Zirconium is one of the few metals that retains its structural in-tegrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the fol-lowing questions about the redox properties of zirconium based on the half-reaction a. Is zirconium metal capable of reducing water to form hydro-gen gas at standard conditions?
b. Write a balanced equation for the reduction of water by zirconium.
c. Calculate e°, G°, and K for the reduction of water by zirco-nium metal.
d. The reduction of water by zirconium occurred during the ac-cidents at Three Mile Island in 1979. The hydrogen produced was successfully vented and no chemical explosion occurred.
If 1.00 103 kg of Zr reacts, what mass of H2 is produced?
What volume of H2 at 1.0 atm and 1000.°C is produced?
e. At Chernobyl in 1986, hydrogen was produced by the reaction of superheated steam with the graphite reactor core: It was not possible to prevent a chemical explosion at Cher-nobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the at-mosphere at Three Mile Island? Explain.
65. In addition to the process described in the text, a second process called the carbon–nitrogen cycle occurs in the sun: Overall reaction: a. What is the catalyst in this process?
b. What nucleons are intermediates?
c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of 1 1H and 4 2He are 1.00782 and 4.00260, respectively.) 66. The most significant source of natural radiation is radon-222.
222Rn, a decay product of 238U, is continuously generated in the earth’s crust, allowing gaseous Rn to seep into the basements of buildings. Because 222Rn is an -particle producer with a rela-tively short half-life of 3.82 days, it can cause biological damage when inhaled.
a. How many particles and particles are produced when 238U decays to 222Rn? What nuclei are produced when 222Rn decays?
b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling 222Rn?
c. Another problem associated with 222Rn is that the decay of 222Rn produces a more potent -particle producer ( 3.11min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by -particle production. Why is the solid a more potent -particle producer?
a a t1 2 a b a a 4 1 1H ¡ 4 2He 2 0 1e 1 1H 15 7N ¡ 12 6C 4 2He 0 0g 15 8O ¡ 15 7N 0 1e 1 1H 14 7N ¡ 15 8O 0 0g 1 1H 13 6C ¡ 14 7N 0 0g 13 7N ¡ 13 6C 0 1e 1 1H 12 6C ¡ 13 7N 0 0g C1s2 H2O1g2 ¡ CO1g2 H21g2 e° 2.36 V ZrO2 H2O H2O 4e ¡ Zr 4OH activity of 15.3 counts per minute per gram. Your painting showed a carbon-14 activity of 15.1 counts per minute per gram. Could it be a genuine Rembrandt?
55. Define “third-life” in a similar way to “half-life” and determine the “third-life” for a nuclide that has a half-life of 31.4 years.
56. A proposed system for storing nuclear wastes involves storing the radioactive material in caves or deep mine shafts. One of the most toxic nuclides that must be disposed of is plutonium-239, which is produced in breeder reactors and has a half-life of 24,100 years.
A suitable storage place must be geologically stable long enough for the activity of plutonium-239 to decrease to 0.1% of its orig-inal value. How long is this for plutonium-239?
57. During World War II, tritium (3H) was a component of fluores-cent watch dials and hands. Assume you have such a watch that was made in January 1944. If 17% or more of the original tritium was needed to read the dial in dark places, until what year could you read the time at night? (For 3H, yr.) 58. A positron and an electron can annihilate each other on colliding, producing energy as photons: Assuming that both rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.
59. A small atomic bomb releases energy equivalent to the detona-tion of 20,000 tons of TNT; a ton of TNT releases J of energy when exploded. Using J/mol as the energy re-leased by fission of 235U, approximately what mass of 235U un-dergoes fission in this atomic bomb?
60. During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were in-vestigated. In one type of bomb, a “gun” shot one piece of fis-sionable material into a cavity containing another piece of fis-sionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when deto-nated, compressed the fissionable material into a smaller volume.
Discuss what is meant by critical mass, and explain why the abil-ity to achieve a critical mass is essential to sustaining a nuclear reaction.
61. Using the kinetic molecular theory (Section 5.6), calculate the root mean square velocity and the average kinetic energy of nuclei at a temperature of K. (See Exercise 44 for the appro-priate mass values.) Challenge Problems 62. A 0.20-mL sample of a solution containing that produces cps is injected into the bloodstream of an animal. Af-ter allowing circulatory equilibrium to be established, a 0.20-mL sample of blood is found to have an activity of 20. cps. Calculate the blood volume of the animal.
63. A 0.10-cm3 sample of a solution containing a radioactive nuclide ( counts per minute per milliliter) is injected into a rat.
Several minutes later 1.0 cm3 of blood is removed. The blood shows 48 counts per minute of radioactivity. Calculate the vol-ume of blood in the rat. What assumptions must be made in per-forming this calculation?
5.0 103 3.7 103 3 1H 4 107 2 1H 2 1013 4 109 g 0 1e 0 1e ¡ 2 0 0g t1 2 12.3 Integrative Problems 873 neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is 15.0 seconds. If 199 atoms of bohrium-267 could be synthesized, how much time would elapse before only 11 atoms of bohrium-267 remain? What is the expected electron configuration of elemental bohrium?
70. Radioactive cobalt-60 is used to study defects in vitamin B12 ab-sorption because cobalt is the metallic atom at the center of the vitamin B12 molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: ). What is the de Broglie wavelength of the emitted particle if it has a velocity equal to 0.90c where c is the speed of light?
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
1H 1.00782 amu 60Co 59.9338 amu; d. The U.S. Environmental Protection Agency (EPA) recom-mends that 222Rn levels not exceed 4 pCi per liter of air ( decay events per second; 1 pCi ). Convert 4.0 pCi per liter of air into concentra-tions units of 222Rn atoms per liter of air and moles of 222Rn per liter of air.
67. To determine the Ksp value of Hg2I2, a chemist obtained a solid sample of Hg2I2 in which some of the iodine is present as ra-dioactive 131I. The count rate of the Hg2I2 sample is counts per minute per mole of I. An excess amount of Hg2I2(s) is placed into some water, and the solid is allowed to come to equi-librium with its respective ions. A 150.0-mL sample of the satu-rated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the Ksp value for Hg2I2.
68. Estimate the temperature needed to achieve the fusion of deu-terium to make an alpha particle. The energy required can be estimated from Coulomb’s law [use the form (Q1Q2 r), using C for a proton, and m for the helium nucleus; the unit for the proportionality constant in Coloumb’s law is ] Integrative Problems These problems require the integration of multiple concepts to find the solutions.
69. A recently reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with J m C2.
1015 r 2 Q 1.6 1019 E 9.0 109 Hg2I21s2 ∆Hg2 21aq2 2I1aq2 Ksp 3Hg2 24 3I4 2 5.0 1011 1 1012 Ci 1 Ci 1 curie 3.7 1010 19 The Representative Elements: Groups 1A Through 4A Contents 19.1 A Survey of the Representative Elements • Atomic Size and Group Anomalies • Abundance and Preparation 19.2 The Group 1A Elements 19.3 Hydrogen 19.4 The Group 2A Elements 19.5 The Group 3A Elements 19.6 The Group 4A Elements Scanning electron micrograph of calcium crystals, a representative element in Group 2A.
So far in this book we have covered the major principles and explored the most im-portant models of chemistry. In particular, we have seen that the chemical properties of the elements can be explained very successfully by the quantum mechanical model of the atom. In fact, the most convincing evidence of that model’s validity is its ability to relate the observed periodic properties of the elements to the number of valence electrons in their atoms.
We have learned many properties of the elements and their compounds, but we have not discussed in detail the relationship between the chemical properties of a particular element and its position on the periodic table. In this chapter and the next we will explore the chemical similarities and differences among the elements in the several groups of the periodic table and will try to interpret these data using the quantum mechanical model of the atom. In the process we will illustrate a great variety of chemical properties and further demonstrate the practical importance of chemistry.
19.1 A Survey of the Representative Elements The traditional form of the periodic table is shown in Fig. 19.1. Recall that the repre-sentative elements, whose chemical properties are determined by the valence-level s and p electrons, are designated Groups 1A through 8A. The transition metals, in the center of the table, result from the filling of d orbitals. The elements that correspond to the filling of the 4f and 5f orbitals are listed separately as the lanthanides and actinides, respectively.
The heavy black line in Fig. 19.1 divides the metals from the nonmetals. Some ele-ments just on either side of this line, such as silicon and germanium, exhibit both metal-lic and nonmetallic properties and are often called metalloids or semimetals. The fundamental chemical difference between a metal and a nonmetal is that metals tend to lose their valence electrons to form cations, usually with the valence-electron configura-tion of the noble gas from the preceding period, and nonmetals tend to gain electrons to 875 H Li Na K Rb Cs Fr Be Mg Ca Sr Ba Ra Ga In Tl B Al Ge Sn Pb C Si 2A 3A 4A 1A 876 Chapter Nineteen The Representative Elements: Groups 1A Through 4A form anions that exhibit the electron configuration of the noble gas in the same period.
Metallic character is observed to increase going down a given group, which is consistent with the trends in ionization energy, electron affinity, and electronegativity discussed earlier (see Sections 7.13 and 8.2).
Atomic Size and Group Anomalies Although the chemical properties of the members of a group show many similarities, there are also important differences. In fact, the relatively large increase in atomic radius in going from the first to the second member of a group causes the first element to show properties that are often quite different from the others. Consequently, hydrogen, beryl-lium, boron, carbon, nitrogen, oxygen, and fluorine all have some properties that distin-guish them from the other members of their groups. For example, in Group 1A hydro-gen is a nonmetal and lithium is a very active metal. This extreme difference results primarily from the very large difference in the atomic radii of hydrogen and lithium, as shown in Fig. 19.2. The small hydrogen atom has a much greater attraction for electrons than do the larger members of Group 1A and forms covalent bonds with nonmetals; the other members of Group 1A lose their valence electrons to nonmetals to form 1 cations in ionic compounds.
This effect of size is also evident in other groups. For example, the oxides of the met-als in Group 2A are all quite basic except for the first member of the series; beryllium oxide (BeO) is amphoteric. Recall from Section 14.10 that the basicity of an oxide de-pends on its ionic character. Ionic oxides contain the ion, which reacts with water to form two ions. All the oxides of the Group 2A metals are highly ionic except for beryllium oxide, which has considerable covalent character. The small ion can effectively polarize the electron “cloud” of the ion, producing significant electron sharing. We see the same pattern in Group 3A, where only the small boron atom behaves as a nonmetal, or sometimes as a semimetal, and aluminum and the other members are active metals.
In Group 4A the effect of size is reflected in the dramatic differences between the chemistry of carbon and that of silicon. The chemistry of carbon is dominated by mole-cules containing chains of bonds, but silicon compounds mainly contain bonds rather than bonds. Carbon forms a wide variety of stable compounds with strong single bonds. Silicon also forms compounds with chains of bonds, Si¬Si C¬C Si¬Si Si¬O C¬C O2 Be2 OH O2 FIGURE 19.1 The periodic table. The elements in the A groups are the representative elements. The elements shown in pink are called transi-tion metals. The dark line approximately di-vides the nonmetals from the metals. The elements that have both metallic and non-metallic properties (semimetals) are shaded in blue.
H Li Na K Rb Cs Fr Be Mg Ca Sr Ba Ra Sc Y La Ac Ti Zr Hf Rf V Nb Ta Ha Cr Mo W Unh Mn Tc Re Uns Uno Fe Ru Os Co Rh Ir Une Ds Rg Uub Ni Pd Pt Cu Ag Au Zn Cd Hg Ga In Tl B Al Ge Sn Pb C Si As Sb Bi Uut Uuq Uup N P Se Te Po O S F Cl Br I At Ne Ar Kr Xe Rn He Ce Th Pr Pa Nd U Pm Np Sm Pu Eu Am Gd Cm Tb Bk Dy Cf Ho Es Er Fm Tm Md Yb No Lu Lr 1A 2A 3A 4A 5A 6A 7A 8A Lanthanides Actinides Metallic character increases going down a group in the periodic table.
19.1 A Survey of the Representative Elements 877 but these compounds are much more reactive than the corresponding carbon compounds.
The reasons for the difference in reactivity between the carbon and silicon compounds are quite complex but are likely related to the differences in the sizes of the carbon and silicon atoms.
Carbon and silicon differ markedly in their abilities to form bonds. As we discussed in Section 9.1, carbon dioxide is composed of discrete molecules with the Lewis structure where the carbon and oxygen atoms achieve the [Ne] configuration by forming bonds.
In contrast, the structure of silica (empirical formula SiO2) is based on SiO4 tetrahedra with bridges, as shown in Fig. 19.3. The silicon 3p valence orbitals do not overlap very effectively with the smaller oxygen 2p orbitals to form bonds; therefore, discrete SiO2 molecules with the Lewis structure p Si¬O¬Si p CO2 p FIGURE 19.2 The atomic radii of some atoms in picometers.
H 37 He 32 Li 152 Be 113 B 88 C 77 N 70 O 66 F 64 Ne 69 Na 186 Mg 160 Al 143 Si 117 P 110 S 104 Cl 99 Ar 97 K 227 Ca 197 Ga 122 Ge 122 As 121 Se 117 Br 114 Kr 110 Rb 247 Sr 215 In 163 Sn 140 Sb 141 Te 143 I 133 Xe 130 Cs 265 Ba 217 Tl 170 Pb 175 Bi 155 Po 167 At 140 Rn 145 8A 7A 6A 5A 4A 3A 2A 1A Atomic radius increases Atomic radius decreases 878 Chapter Nineteen The Representative Elements: Groups 1A Through 4A are not stable. Instead, the silicon atoms achieve a noble gas configuration by forming several single bonds.
The importance of bonding for the relatively small elements of the second period also explains the different elemental forms of the members of Groups 5A and 6A. For ex-ample, elemental nitrogen consists of very stable N2 molecules with the Lewis structure Elemental phosphorus forms larger aggregates of atoms, the simplest being the tetrahedral P4 molecules found in white phosphorus (see Fig. 20.12). Like silicon atoms, the rela-tively large phosphorus atoms do not form strong bonds and prefer to achieve a noble gas configuration by forming single bonds to several other phosphorus atoms. In contrast, its very strong bonds make the N2 molecule the most stable form of elemental nitro-gen. Similarly, in Group 6A the most stable form of elemental oxygen is the O2 molecule with a double bond, but the larger sulfur atom forms bigger aggregates, such as the cyclic S8 molecule (see Fig. 20.16), which contains only single bonds.
The relatively large change in size in going from the first to second member of a group also has important consequences for the Group 7A elements. For example, fluorine has a smaller electron affinity than chlorine. This reversal of the expected trend can be at-tributed to the small fluorine 2p orbitals, which result in unusually large electron repul-sions. The relative weakness of the bond in the F2 molecule can be explained in terms of the repulsions among the lone pairs, shown in the Lewis structure: The small size of the fluorine atoms allows close approach of the lone pairs, which leads to much greater repulsions than are found in the Cl2 molecule with its much larger atoms.
Abundance and Preparation Table 19.1 shows the distribution of elements in the earth’s crust, oceans, and atmosphere.
The major element is, of course, oxygen, which is found in the atmosphere as O2, in the oceans in H2O, and in the earth’s crust primarily in silicate and carbonate minerals. The second most abundant element, silicon, is found throughout the earth’s crust in the silica and silicate minerals that form the basis of most sand, rocks, and soil. The most abundant metals, aluminum and iron, are found in ores in which they are combined with nonmetals, most commonly oxygen. One notable fact revealed by Table 19.1 is the small incidence of most transition metals. Since many of these relatively rare elements are assuming p p p Si¬O FIGURE 19.3 The structure of quartz, which has the em-pirical formula SiO2. Note that the structure is based on interlocking SiO4 tetrahedra, where each oxygen atom is shared by two silicon atoms.
Sand dunes in Monument Valley, Arizona.
TABLE 19.1 Distribution (Mass Percent) of the 18 Most Abundant Elements in the Earth’s Crust, Oceans, and Atmosphere Element Mass Percent Element Mass Percent Oxygen 49.2 Chlorine 0.19 Silicon 25.7 Phosphorus 0.11 Aluminum 7.50 Manganese 0.09 Iron 4.71 Carbon 0.08 Calcium 3.39 Sulfur 0.06 Sodium 2.63 Barium 0.04 Potassium 2.40 Nitrogen 0.03 Magnesium 1.93 Fluorine 0.03 Hydrogen 0.87 All others 0.49 Titanium 0.58 O O O O Si 19.1 A Survey of the Representative Elements 879 increasing importance in our high-technology society, it is possible that the control of transition metal ores may ultimately have more significance for world politics than the control of petroleum supplies.
The distribution of elements in living materials is very different from that found in the earth’s crust. Table 19.2 shows the relative abundance of elements in the human body.
Oxygen, carbon, hydrogen, and nitrogen form the basis for all biologically important mol-ecules. The other elements, even though they are found in relatively small amounts, are often crucial for life. For example, zinc is found in over 150 different biomolecules in the human body.
Only about one-fourth of the elements occur naturally in the free state. Most are found in a combined state. The process of obtaining a metal from its ore is called metallurgy.
Since the metals in ores are found in the form of cations, the chemistry of metallurgy always involves reduction of the ions to the elemental metal (with an oxidation state of zero). A variety of reducing agents can be used, but carbon is the usual choice because of its wide availability and relatively low cost. As we will see in Chapter 21, carbon is the primary reducing agent in the production of steel. Carbon also can be used to produce tin and lead from their oxides: Hydrogen gas also can be used as a reducing agent for metal oxides, as in the production of tin: Electrolysis is often used to reduce the most active metals. In Chapter 17 we con-sidered the electrolytic production of aluminum metal. The alkali metals are also produced by electrolysis, usually of their molten halide salts.
The preparation of nonmetals varies widely. Elemental nitrogen and oxygen are usu-ally obtained from the liquefaction of air, which is based on the principle that a real gas cools when it expands. After each expansion, part of the cooler gas is compressed, while the rest is used to carry away the heat of the compression. The compressed gas is then al-lowed to expand again. This cycle is repeated many times. Eventually, the remaining gas becomes cold enough to form the liquid state. Because liquid nitrogen and liquid oxygen have different boiling points, they can be separated by the distillation of liquid air. Both SnO1s2 H21g2 ¡ Heat Sn1s2 H2O1g2 2PbO1s2 C1s2 ¡ Heat 2Pb1s2 CO21g2 2SnO1s2 C1s2 ¡ Heat 2Sn1s2 CO21g2 TABLE 19.2 Abundance of Elements in the Human Body Trace Elements Major Elements Mass Percent (in alphabetical order) Oxygen 65.0 Arsenic Carbon 18.0 Chromium Hydrogen 10.0 Cobalt Nitrogen 3.0 Copper Calcium 1.4 Fluorine Phosphorus 1.0 Iodine Magnesium 0.50 Manganese Potassium 0.34 Molybdenum Sulfur 0.26 Nickel Sodium 0.14 Selenium Chlorine 0.14 Silicon Iron 0.004 Vanadium Zinc 0.003 Metallurgy is discussed in more detail in Chapter 21.
Carbon is the cheapest and most readily available industrial reducing agent for metallic ions.
880 Chapter Nineteen The Representative Elements: Groups 1A Through 4A substances are important industrial chemicals, ranking in the top five in terms of the amounts manufactured in the United States. Hydrogen can be obtained from the electrolysis of water, but more commonly it is obtained from the decomposition of the methane in natural gas. Sulfur is found underground in its elemental form and is recovered using the Frasch process (see Section 20.6). The halogens are obtained by oxidation of the anions from halide salts (see Section 20.7).
19.2 The Group 1A Elements The Group 1A elements with their valence-electron configurations are all very active metals (they lose their valence electrons very readily), except for hydrogen, which behaves as a nonmetal. We will discuss the chemistry of hydrogen in the next section. Many of the properties of the alkali metals have been described previously (see Section 7.13). The sources and methods of preparation of pure alkali metals are given in Table 19.3. The ion-ization energies, standard reduction potentials, ionic radii, and melting points for the alkali metals are listed in Table 19.4. Lepidolite, shown in Fig. 19.4, contains several pure alkali metals.
In Section 7.13 we saw that the alkali metals all react vigorously with water to release hydrogen gas: We will reconsider this process briefly because it illustrates several important concepts.
Based on the ionization energies, we might expect lithium to be the weakest of the alkali metals as a reducing agent in water. However, the standard reduction potentials indicate that it is the strongest. This reversal results mainly from the very large energy of hydration 2M1s2 2H2O1l2 ¡ 2M1aq2 2OH1aq2 H21g2 ns1 The preparation of sulfur and the halogens is discussed in Chapter 20.
H Li Na K Rb Cs Fr 1A Several properties of the alkali metals are given in Table 7.8.
TABLE 19.3 Sources and Methods of Preparation of the Pure Alkali Metals Element Source Method of Preparation Lithium Silicate minerals such as Electrolysis of molten LiCl spodumene, LiAl(Si2O6) Sodium NaCl Electrolysis of molten NaCl Potassium KCl Electrolysis of molten KCl Rubidium Impurity in lepidolite, Reduction of RbOH with Mg Li2(F,OH)2Al2(SiO3)3 and H2 Cesium Pollucite (Cs4Al4Si9O26 H2O) and Reduction of CsOH with Mg an impurity in lepidolite and H2 (see Fig. 19.4) TABLE 19.4 Selected Physical Properties of the Alkali Metals Ionization Standard Reduction Radius Melting Energy Potential (V) for of M Point Element (kJ/mol) M e M (pm) (°C) Lithium 520 3.05 60 180 Sodium 495 2.71 95 98 Potassium 419 2.92 133 63 Rubidium 409 2.99 148 39 Cesium 382 3.02 169 29 S FIGURE 19.4 Lepidolite is composed mainly of lithium, aluminum, silicon, and oxygen, but it also contains significant amounts of rubidium and cesium.
19.2 The Group 1A Elements 881 of the small ion. Because of its relatively high charge density, the ion very ef-fectively attracts water molecules. A large quantity of energy is released in the process, favoring the formation of the ion and making lithium a strong reducing agent in aque-ous solution.
We also saw in Section 7.13 that lithium, although it is the strongest reducing agent, reacts more slowly with water than does sodium or potassium. From the discussions in Chapters 12 and 16, we know that the equilibrium position for a reaction (in this case indicated by the values) is controlled by thermodynamic factors but that the rate of a reaction is controlled by kinetic factors. There is no direct connection between these fac-tors. Lithium reacts more slowly with water than does sodium or potassium because as a solid it has a higher melting point than either of them. It does not become molten from the heat of reaction with water as sodium and potassium do, and thus it has a smaller area of contact with the water.
The relative ease with which the alkali metals lose electrons to form cations means that they react with nonmetals to form ionic compounds. Although we might expect the alkali metals to react with oxygen to form regular oxides of the general formula M2O, lithium is the only one that does so in the presence of excess oxygen gas: Sodium forms solid Na2O if the oxygen supply is limited, but in excess oxygen it forms sodium peroxide: Sodium peroxide contains the basic anion and reacts with water to form hydrogen peroxide and hydroxide ions: Hydrogen peroxide is a strong oxidizing agent often used as a bleach for hair and as a disinfectant.
Potassium, rubidium, and cesium react with oxygen to produce superoxides of the general formula MO2, which contains the anion. For example, potassium reacts with oxygen as follows: The superoxides release oxygen gas in reactions with water or carbon dioxide: This chemistry makes superoxides very useful in the self-contained breathing apparatuses used by firefighters. These “airpacks” are also used as emergency equipment in labs and production facilities in case toxic fumes are released.
The types of compounds formed by the alkali metals with oxygen are summarized in Table 19.5. Table 19.6 summarizes some important reactions of the alkali metals.
4MO21s2 2CO21g2 ¡ 2M2CO31s2 3O21g2 2MO21s2 2H2O1l2 ¡ 2M1aq2 2OH1aq2 O21g2 H2O21aq2 K1s2 O21g2 ¡ KO21s2 O2 Na2O21s2 2H2O1l2 ¡ 2Na1aq2 H2O21aq2 2OH1aq2 O2 2 2Na1s2 O21g2 ¡ Na2O21s2 4Li1s2 O21g2 ¡ 2Li2O1s2 M e° Li Li Li TABLE 19.5 Types of Compounds Formed by the Alkali Metals with Oxygen General Formula Name Examples M2O Oxide Li2O, Na2O M2O2 Peroxide Na2O2 MO2 Superoxide KO2, RbO2, CsO2 O H O H Sodium reacting with water.
Airpacks are an essential source of oxygen for firefighters.
Hydrogen peroxide has the Lewis structure 882 Chapter Nineteen The Representative Elements: Groups 1A Through 4A The alkali metal ions are very important for the proper functioning of biologic sys-tems, such as nerves and muscles, and and ions are present in all body cells and fluids. In human blood plasma the concentrations are For the fluids inside the cells the concentrations are reversed: Since the concentrations are so different inside and outside the cells, an elaborate mech-anism is needed to transport and ions through the cell membranes.
Recently, studies have been carried out concerning the role of the Li ion in the human brain, and lithium carbonate has been used extensively in the treatment of manic-depressive patients. The ion apparently affects the levels of neurotransmitters, molecules that assist the transmission of messages along the nerve networks. Incorrect concentrations of these molecules can lead to depression or mania.
Predicting Reaction Products Predict the products formed by the following reactants.
a. Li3N(s) and H2O(l) b. KO2(s) and H2O(l) Solution a. Solid Li3N contains the anion, which has a strong attraction for three H ions to form NH3. Thus the reaction is b. Solid KO2 is a superoxide that characteristically reacts with water to produce O2, H2O2, and : See Exercises 19.17 and 19.18.
2KO21s2 2H2O1l2 ¡ 2K1aq2 2OH1aq2 O21g2 H2O21aq2 OH Li3N1s2 3H2O1l2 ¡ NH31g2 3Li1aq2 3OH1aq2 N3 Li K Na 3Na4 0.005 M and 3K4 0.16 M 3Na4 0.15 M and 3K4 0.005 M K Na TABLE 19.6 Selected Reactions of the Alkali Metals Reaction Comment Excess oxygen M K, Rb, or Cs Li only Violent reaction!
2M 2H S 2M H2 2M 2H2O S 2MOH H2 2M H2 S 2MH 12M P4 S 4M3P 6Li N2 S 2Li3N 2M S S M2S M O2 S MO2 2Na O2 S Na2O2 4Li O2 S 2Li2O X2 any halogen molecule 2M X2 S 2MX Sodium reacts violently with chlorine.
Sample Exercise 19.1 19.3 Hydrogen 883 19.3 Hydrogen Under ordinary conditions of temperature and pressure, hydrogen is a colorless, odorless gas composed of H2 molecules. Because of its low molar mass and nonpolarity, hydrogen has a very low boiling point and melting point . Hydrogen gas is highly flammable, and mixtures of air containing from 18% to 60% hydrogen by volume are explosive. In a common lecture demonstration hydrogen and oxygen gases are bub-bled into soapy water. The resulting bubbles can be ignited with a candle on a long stick, giving a loud explosion.
The major industrial source of hydrogen gas is the reaction of methane and water at high temperatures and high pressures with a metallic cata-lyst, often nickel: Large quantities of hydrogen are also formed as a by-product of gasoline production, in which hydrocarbons with high molar masses are broken down (cracked) to produce smaller molecules more suitable for use as a motor fuel.
Very pure hydrogen can be produced by electrolysis of water (see Section 17.7), but this method is currently not economically feasible for large-scale production because of the relatively high cost of electricity.
The major industrial use of hydrogen is in the production of ammonia by the Haber process. Large quantities of hydrogen are also used for hydrogenating unsaturated veg-etable oils (those containing carbon–carbon double bonds) to produce solid shortenings that are saturated (containing carbon–carbon single bonds): The catalysis of this process was discussed in Section 12.8.
Chemically, hydrogen behaves as a typical nonmetal, forming covalent compounds with other nonmetals and forming salts with very active metals. Binary compounds containing hydrogen are called hydrides, of which there are three classes. The ionic CO1g2 3H21g2 CH41g2 H2O1g2 (1050 atm) (8001000°C) (260°C) (253°C) Heat, pressure ¬ ¬¬ ¬¬¬¡ Catalyst (left) Hydrogen gas being used to blow soap bubbles. (right) As the bubbles float upward, they are lighted using a candle on a long pole. The orange flame is due to the heat from the reaction of hydrogen with the oxygen in the air that excites sodium ions in the soap solution.
884 Chapter Nineteen The Representative Elements: Groups 1A Through 4A (or saltlike) hydrides are formed when hydrogen combines with the most active met-als, those from Groups 1A and 2A. Examples are LiH and CaH2, which can best be characterized as containing hydride ions and metal cations. Because the presence of two electrons in the small 1s orbital produces large electron–electron repulsions and the nucleus has only a 1 charge, the hydride ion is a strong reducing agent (easily loses electrons). For example, when ionic hydrides are placed in water, a violent re-action takes place. This reaction results in the formation of hydrogen gas, as seen in the equation Covalent hydrides are formed when hydrogen combines with other nonmetals.
We have encountered many of these compounds already: HCl, CH4, NH3, H2O, and so on. The most important covalent hydride is water. The polarity of the H2O molecule leads to many of water’s unusual properties. Water has a much higher boiling point than is expected from its molar mass. It has a large heat of vaporization and a large heat capacity, both of which make it a very useful coolant. Water has a higher density as a liquid than as a solid. This is due to the open structure of ice, which results from maximizing the hydrogen bonding (see Fig. 19.5). Because water is an excellent solvent for ionic and polar substances, it provides an effective medium for life processes. In fact, water is one of the few covalent hydrides that is nontoxic to organisms.
The third class of hydrides, the metallic (or interstitial) hydrides, are formed when transition metal crystals are treated with hydrogen gas. The hydrogen molecules dissoci-ate at the metal’s surface, and the small hydrogen atoms migrate into the crystal structure to occupy holes, or interstices. These metal–hydrogen mixtures are more like solid solu-tions than true compounds. Palladium can absorb about 900 times its own volume of hydrogen gas. In fact, hydrogen can be purified by placing it under slight pressure in a vessel containing a thin wall of palladium. The hydrogen diffuses into and through the metal wall, leaving the impurities behind.
Although hydrogen can react with transition metals to form compounds of constant composition, most of the interstitial hydrides have variable compositions (often called non-stoichiometric compositions) with formulas such as LaH2.76 and VH0.56. The compositions of the nonstoichiometric hydrides vary with the length of exposure of the metal to hydrogen gas and other factors.
LiH1s2 H2O1l2 ¡ H21g2 Li1aq2 OH1aq2 (H) Boiling points of covalent hydrides are discussed in Section 10.1. H2O Ice FIGURE 19.5 The structure of ice, showing the hydrogen bonding.
19.4 The Group 2A Elements 885 When interstitial hydrides are heated, much of the absorbed hydrogen is expelled as hydrogen gas. Because of this behavior, these materials offer possibilities for storing hy-drogen for use as a portable fuel. The internal combustion engines in current automobiles can burn hydrogen gas with little modification, but storage of enough hydrogen to pro-vide an acceptable mileage range remains a problem. One possible solution might be to use a fuel tank containing a porous solid that includes a transition metal into which the hydrogen gas could be pumped to form the interstitial hydride. The hydrogen gas could then be released as required by the engine.
19.4 The Group 2A Elements The Group 2A elements (with valence-electron configuration ns2) are very reactive, losing their two valence electrons to nonmetals to form ionic compounds containing M2 cations.
These elements are commonly called the alkaline earth metals because of the basicity of their oxides: Only beryllium oxide (BeO) also shows some acidic properties, such as dissolving in aque-ous solutions containing hydroxide ions: The more active alkaline earth metals react with water as the alkali metals do, pro-ducing hydrogen gas: Calcium, strontium, and barium react vigorously at . The less easily oxidized beryl-lium and magnesium show no observable reaction with water at , although magne-sium reacts with boiling water. Table 19.7 summarizes various properties, sources, and preparations of the alkaline earth metals.
25°C 25°C M1s2 2H2O1l2 ¡ M21aq2 2OH1aq2 H21g2 BeO1s2 2OH1aq2 H2O1l2 ¡ Be1OH24 21aq2 MO1s2 H2O1l2 ¡ M21aq2 2OH1aq2 See Section 6.6 for a discussion of the feasibility of using hydrogen gas as a fuel.
An amphoteric oxide displays both acidic and basic properties.
TABLE 19.7 Selected Physical Properties, Sources, and Methods of Preparation for the Group 2A Elements Radius of M2 e (V) for Method of Element (pm) First Second M2 2e M Source Preparation Beryllium 30 900 1760 1.70 Beryl Electrolysis of (Be3Al2Si6O18) molten BeCl2 Magnesium 65 738 1450 2.37 Magnesite (MgCO3), Electrolysis of dolomite (MgCO3 CaCO3), molten MgCl2 carnallite (MgCl2 KCl 6H2O) Calcium 99 590 1146 2.76 Various minerals containing Electrolysis of CaCO3 molten CaCl2 Strontium 113 549 1064 2.89 Celestite (SrSO4), Electrolysis of strontianite (SrCO3) molten SrCl2 Barium 135 503 965 2.90 Baryte (BaSO4), Electrolysis of witherite (BaCO3) molten BaCl2 Radium 140 509 979 2.92 Pitchblende Electrolysis of (1 g of Ra/7 tons of ore) molten RaCl2 S Be Mg Ca Sr Ba Ra 2A Ionization Energy (kJ/mol) 886 Chapter Nineteen The Representative Elements: Groups 1A Through 4A As we saw in Section 19.1, the small size and relatively high electronegativity of the beryllium atom cause its bonds to be more covalent than is usual for a metal. For exam-ple, beryllium chloride with the Lewis structure exists as a linear molecule, as predicted by the VSEPR model. The bonds are covalent, and beryllium is best described as being sp hybridized. As a solid, BeCl2 achieves an octet of electrons around each beryllium atom by forming an extended structure containing Be in a tetrahedral environment, as shown in Fig. 19.6. The lone pairs on the chlorine atoms are used to form bonds.
The alkaline earth metals have great practical importance. Calcium and magnesium ions are essential for human life. Calcium is found primarily in the structural minerals constituting bones and teeth; magnesium (as the Mg2 ion) plays a vital role in metabo-lism and muscle functions. Also, magnesium was formerly used to produce the bright light for photographic flash bulbs from its reaction with oxygen: Because magnesium metal has a relatively low density and moderate strength, it is a useful structural material, especially if alloyed with aluminum.
Table 19.8 summarizes some important reactions of the alkaline earth metals.
2Mg1s2 O21g2 ¡ 2MgO1s2 light Be¬Cl Be¬Cl Calcium metal reacting with water to form bubbles of hydrogen gas.
Cl Be Cl Be Cl Be Cl Cl Cl (a) Cl Be Cl Be Be Cl Cl (b) (c) FIGURE 19.6 (a) Solid BeCl2 can be visualized as being formed from many BeCl2 molecules, where lone pairs on the chlorine atoms are used to bond to the beryllium atoms in adjacent BeCl2 molecules. (b) The extended structure of solid BeCl2.
(c) The ball-and-stick model of the extended structure.
TABLE 19.8 Selected Reactions of the Group 2A Elements Reaction Comment Ba gives BaO2 as well High temperatures High temperatures M Ca, Sr, or Ba; high temperatures; Mg at high pressure M Ca, Sr, or Ba Be 2OH 2H2O S Be(OH)4 2 H2 M 2H S M2 H2 M 2H2O S M(OH)2 H2 M H2 S MH2 6M P4 S 2M3P2 3M N2 S M3N2 M S S MS 2M O2 S 2MO X2 any halogen molecule M X2 S MX2 Bones contain large amounts of calcium.
19.4 The Group 2A Elements 887 Relatively large concentrations of Ca2 and Mg2 ions are often found in natural water supplies. These ions in this hard water interfere with the action of detergents and form precipitates with soap. In Section 14.6 we saw that Ca2 is often removed by precipitation as CaCO3 in large-scale water softening. In individual homes Ca2, Mg2, and other cations are removed by ion exchange. An ion-exchange resin con-sists of large molecules (polymers) that have many ionic sites. A cation-exchange resin is represented schematically in Fig. 19.7(a), showing Na ions bound ionically to the groups that are covalently attached to the resin polymer. When hard water is passed over the resin, Ca2 and Mg2 bind to the resin in place of Na, which is re-leased into the solution [Fig. 19.7(b)]. Replacing Mg2 and Ca2 by Na [Fig. 19.7(c)] “softens” the water because the sodium ions interfere much less with the action of soaps and detergents.
Electrolytic Production of Magnesium Calculate the amount of time required to produce kg of magnesium metal by the electrolysis of molten MgCl2 using a current of A.
Solution The reaction for plating magnesium is which means that 2 moles of electrons are required for each mole of Mg produced. The number of moles of magnesium in kg is Thus 2 mol e 1 mol Mg 4.11 104 mol Mg 8.22 104 mol e 1.00 103 kg 1000 g kg 1 mol Mg 24.31 g 4.11 104 mol Mg 1.00 103 Mg2 2e ¡ Mg 1.00 102 1.00 103 SO3 The Dolomite mountains in Italy. Dolomite is a source of magnesium.
SO3– Na+ SO3– Na+ Mg2+ Ca2+ SO3– Na+ SO3– Na+ SO3– Na+ Hard water Resin Resin polymer (a) SO3– Na+ Mg2+ Hard water Resin (b) Soft water Resin (c) SO3– Na+ SO3– Na+ Ca2+ SO3– SO3– SO3– SO3– Na+ Na+ Na+ Na+ Ca2+ Mg2+ FIGURE 19.7 (a) A schematic representation of a typical cation-exchange resin. (b) and (c) When hard water is passed over the cation-exchange resin, the Ca2 and Mg2 bind to the resin.
Sample Exercise 19.2 888 Chapter Nineteen The Representative Elements: Groups 1A Through 4A Using the faraday , we can calculate the coulombs of charge: Since an ampere is a coulomb of charge per second, we can now calculate the time required: See Exercises 19.29 and 19.30.
19.5 The Group 3A Elements The Group 3A elements (valence-electron configuration ns2np1) generally show the in-crease in metallic character in going down the group that is characteristic of the repre-sentative elements. Some physical properties, sources, and methods of preparation for the Group 3A elements are summarized in Table 19.9.
Boron is a nonmetal, and most of its compounds are covalent. The most interesting compounds of boron are the covalent hydrides called boranes. We might expect BH3 to be the simplest hydride, since boron has three valence electrons to share with three hydrogen atoms. However, this compound is unstable, and the simplest known member of the series is diborane (B2H6), with the structure shown in Fig. 19.8(a). In this molecule the terminal bonds are normal covalent bonds, each involving one electron pair. The bridging bonds are three-center bonds using a single pair of electrons to bond all three atoms. An-other interesting borane contains the square pyramidal B5H9 molecule [Fig. 19.8(b)], which has four three-center bonds around the base of the pyramid. Because the boranes are ex-tremely electron-deficient, they are highly reactive. The boranes react very exothermi-cally with oxygen and were once evaluated as potential fuels for rockets in the U.S. space program.
Aluminum, the most abundant metal on earth, has metallic physical properties, such as high thermal and electrical conductivities and a lustrous appearance, but its bonds to nonmetals are significantly covalent. This covalency is responsible for the amphoteric B¬H 7.93 109 C 1.00 102 C/s 7.93 107 s or 918 days 8.22 104 mol e 96,485 C mol e 7.93 109 C (96,485 C/mol e) Ga In Tl B Al 3A TABLE 19.9 Selected Physical Properties, Sources, and Methods of Preparation for the Group 3A Elements Radius Ionization of M3 Energy e (V) for Method of Element (pm) (kJ/mol) M3 3e S M Sources Preparation Boron 20 798 — Kernite, a form of Reduction by Mg or H2 borax (Na2B4O7 4H2O) Aluminum 51 581 1.71 Bauxite (Al2O3) Electrolysis of Al2O3 in molten Na3AlF6 Gallium 62 577 0.53 Traces in various minerals Reduction with H2 or electrolysis Indium 81 556 0.34 Traces in various minerals Reduction with H2 or electrolysis Thallium 95 589 0.72 Traces in various minerals Electrolysis 19.5 The Group 3A Elements 889 nature of Al2O3, which dissolves in acidic or basic solution, and for the acidity of Al(H2O)6 3 (see Section 14.8): One especially interesting property of gallium is its unusually low melting point at , which is in contrast to the melting point of aluminum. Gallium’s boiling point is approximately . This gives gallium the largest liquid range of any metal, which makes it useful for thermometers, especially to measure high temperatures. Gallium, like water, expands when it freezes. The chemistry of gallium is quite similar to that of aluminum. For example, Ga2O3 is amphoteric.
Table 19.10 summarizes some important reactions of the Group 3A elements.
The practical importance of the Group 3A elements centers on aluminum. Since the discovery of the electrolytic production process by Hall and Heroult (see Section 17.8), aluminum has become a highly important structural material in a wide variety of appli-cations from aircraft bodies to bicycle components. Aluminum is especially valuable be-cause it has a high strength-to-weight ratio and because it protects itself from corrosion by developing a tough, adherent oxide coating.
2400°C 660°C 29.8°C Al1H2O26 31aq2 ∆Al1OH21H2O25 21aq2 H1aq2 CHEMICAL IMPACT Boost Your Boron E veryone realizes that the body needs protein, carbohy-drates, vitamins, and even fat. The importance of several trace elements in our diet, however, is often poorly under-stood. An example is the element boron. People in the United States have a relatively low intake of boron. For example, the U.S. population consumes a little more than 1 mg of boron per day, which is about 10% less than people living in Great Britain and Egypt and about 35% less than people in Germany and Mexico.
To study the importance of boron intake, Zuo-Fen Zhang and his colleagues in the School of Public Health at the University of California–Los Angeles examined nutri-tion data collected from thousands of men and women who filled out the National Health and Nutrition Examination Survey (NHANES). Zhang and his coworkers learned that boron seems to protect against prostate cancer. Comparing the diets of men with prostate cancer to those without the disease indicated a strong correlation between boron con-sumption and the absence of the disease. The prostate can-cer risk for men eating at least 1.8 mg boron per day was only one-third that of men who consumed less than 0.9 mg boron per day. The data show that boron offers no apparent protection for other types of cancer, just very specific pro-tection for prostate cancer. Other studies involving animals indicate that boron consumption can furnish protection against autoimmune diseases such as rheumatoid arthritis.
Although boron intake in the neighborhood of 3.0 mg/day seems beneficial, large amounts of boron can be toxic. The best way to obtain extra boron in your diet is by consuming foods such as nuts and noncitrus fruits.
(a) (b) Boron Hydrogen B2H6 B5H9 FIGURE 19.8 (a) The structure of B2H6 with its two three-center B H B bridging bonds and four “normal” B H bonds. (b) The structure of B5H9. There are five “normal” B H bonds to terminal hydro-gens and four three-center bridging bonds around the base.
¬ ¬ ¬ ¬ Gallium metal has such a low melting point (30°C) that it melts from the heat of a hand.
890 Chapter Nineteen The Representative Elements: Groups 1A Through 4A 19.6 The Group 4A Elements Group 4A (with the valence-electron configuration ns2np2) contains two of the most im-portant elements on earth: carbon, the fundamental constituent of the molecules neces-sary for life, and silicon, which forms the basis of the geologic world. The change from nonmetallic to metallic properties seen in Group 3A is also apparent in going down Group 4A from carbon, a typical nonmetal, to silicon and germanium, usually considered semi-metals, to the metals tin and lead. Table 19.11 summarizes some physical properties, sources, and methods of preparation for the elements in this group.
All the Group 4A elements can form four covalent bonds to nonmetals—for example, CH4, SiF4, GeBr4, SnCl4, and PbCl4. In each of these tetrahedral molecules, the central atom is described as sp3 hybridized by the localized electron model. All these compounds, except those of carbon, can react with Lewis bases to form two additional covalent bonds. For example, SnCl4, which is a fuming liquid , can add two chloride ions: Carbon compounds cannot react in this way because of the small atomic size of carbon and because there are no d orbitals on carbon to accommodate the extra electrons, as there are on the other elements in the group.
SnCl4 2Cl ¡ SnCl6 2 (bp 114°C) Aluminum is used in airplane construction.
TABLE 19.10 Selected Reactions of the Group 3A Elements Reaction Comment X2 any halogen molecule; Tl gives TlX as well, but no TlI3 High temperatures; Tl gives Tl2O as well High temperatures; Tl gives Tl2S as well M Al only M Al, Ga, or In; Tl gives Tl M Al or Ga 2M 2OH 6H2O S 2M(OH)4 3H2 2M 6H S 2M3 3H2 2M N2 S 2MN 2M 3S S M2S3 4M 3O2 S 2M2O3 2M 3X2 S 2MX3 Ge Sn Pb C Si 4A Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
19.6 The Group 4A Elements 891 We have seen that carbon also differs markedly from the other members of Group 4A in its ability to form bonds. This accounts for the completely different structures and properties of CO2 and SiO2. Note from Table 19.12 that bonds and bonds are stronger than bonds. This partly explains why the chemistry of carbon is dom-inated by bonds, whereas that of silicon is dominated by bonds.
Carbon occurs in the earth’s crust mainly in two allotropic forms—graphite and di-amond. In addition, new forms of elemental carbon, including buckminsterfullerene (C60) and other related substances, have recently been characterized. The structures of graphite and diamond are given in Section 10.5.
Carbon monoxide (CO), one of three oxides of carbon, is an odorless, colorless, and toxic gas formed as a by-product of the combustion of carbon-containing compounds when there is a limited oxygen supply. Incidents of carbon monoxide poisoning are especially common in the winter in cold areas of the world when blocked furnace vents limit the availability of oxygen. The bonding in carbon monoxide, which has the Lewis structure is described in terms of sp hybridized carbon and oxygen atoms that interact to form one and two bonds.
Carbon dioxide, a linear molecule with the Lewis structure has an sp hybridized carbon atom, and is a product of human and animal respiration and of the combustion of fossil fuels. It is also produced by fermentation, a process by which the sugar in fruits and grains is changed to ethanol (C2H5OH) and carbon dioxide (see Section 22.4): Glucose Carbon dioxide dissolves in water to produce an acidic solution: Carbon suboxide, the third carbon oxide, is a linear molecule with the Lewis structure which contains sp hybridized carbon atoms.
P P P P S S S S C O O C C CO21aq2 H2O1l2 ∆H1aq2 HCO3 1aq2 C6H12O61aq2 ¡ 2C2H5OH1aq2 2CO21g2 p s Si¬O C¬C Si¬Si Si¬O C¬C p TABLE 19.11 Selected Physical Properties, Sources, and Methods of Preparation for the Group 4A Elements Melting Boiling Method of Element Electronegativity Point ( C) Point ( C) Sources Preparation Carbon 2.5 3727 — Graphite, diamond, — (sublimes) petroleum, coal Silicon 1.8 1410 2355 Silicate minerals, silica Reduction of K2SiF6 with Al, or reduction of SiO2 with Mg Germanium 1.8 937 2830 Germanite (mixture Reduction of GeO2 with of copper, iron, and H2 or C germanium sulfides) Tin 1.8 232 2270 Cassiterite (SnO2) Reduction of SnO2 with C Lead 1.9 327 1740 Galena (PbS) Roasting of PbS with O2 to form PbO2 and then reduction with C Although graphite is thermodynamically more stable than diamond, the transfor-mation of diamond to graphite is not observed under normal conditions.
Fullerenes have been discovered recently by geologists in ancient rocks in Russia.
TABLE 19.12 Strengths of C C, Si Si, and Si O Bonds Bond Energy Bond (kJ/mol) 347 340 452 Si¬O Si¬Si C¬C ¬ ¬ ¬ Enzymes (top) A processed silicon wafer with (bottom) a silicon microchip.
892 Chapter Nineteen The Representative Elements: Groups 1A Through 4A Silicon, the second most abundant element in the earth’s crust, is a semimetal found widely distributed in silica and silicates (see Section 10.5). Approximately 85% of the earth’s crust is composed of these substances. Although silicon is found in some steel and aluminum alloys, its major use is in semiconductors for electronic devices (see the Chemical Impact in Section 10.5).
Germanium, a relatively rare element, is a semimetal used mainly in the manufacture of semiconductors for transistors and similar electronic devices.
Tin is a soft silvery metal that can be rolled into thin sheets (tin foil) and has been used for centuries in various alloys such as bronze , solder , and pewter .
Tin exists as three allotropes: white tin, stable at normal temperatures; gray tin, stable at temperatures below ; and brittle tin, found at temperatures above . When tin is exposed to low temperatures, it gradually changes to the powdery gray tin and crumbles away; this process is known as tin disease.
161°C 13.2°C 2% Sb) (85% Sn, 7% Cu, 6% Bi, and (33% Sn and 67% Pb) (20% Sn and 80% Cu) The compositions of these alloys vary significantly. For example, the tin content of bronze varies from 5% to 30%, and the tin content of pewter is often as high as 95%.
CHEMICAL IMPACT Concrete Learning C oncrete has literally paved the way for civilization over the past 5000 years, tracing its roots to the an-cient Egyptians. At a cost of about a penny per pound, concrete is ubiquitous in today’s world—used in houses, factories, roads, dams, cooling towers, pipes, skyscrapers, and countless other places. In the United States alone there are an estimated $6 trillion worth of concrete-based structures.
Most concretes are based on Portland cement, patented in 1824 by an English bricklayer named J. Aspdin and so named because it forms a product that resembles the natural limestone on the Isle of Portland in England. Portland cement is a pow-der containing a mixture of calcium silicates [Ca2SiO4 (26%) and Ca3SiO5 (51%)], calcium aluminate [Ca3Al2O6 (11%)], and calcium iron aluminate [Ca4Al2Fe2O10 (1%)]. Portland ce-ment is made from a mixture of limestone, sand, shale, clay, and gypsum . When the cement is mixed with sand, gravel, and water, it turns into a muddy substance that eventually hardens into the familiar concrete that finds so many uses in our world. The hardening of concrete occurs not through drying but through hydration. The material becomes dry and hard as the water is consumed in building the complex silicate structures present in cured concrete.
Although many of the details of this process are poorly understood, the main “glue” that holds the components of concrete together is calcium silicate hydrate, which forms a three-dimensional network mainly responsible for concrete’s strength.
Despite its strength when newly produced, concrete contains pockets of air and water dispersed throughout, (CaSO4 2H2O) making it porous and subject to deterioration. Thus, despite all the advantages of concrete, it cracks and deteriorates seriously over time.
Much research is now being carried out to improve the durability of concrete. Most of these efforts are directed to-ward lowering the porosity of concrete and making it less brittle. One group of additives aimed at solving this prob-lem consists of molecules with carbon atom backbones that have sulfate groups attached. These so-called superplasti-cizers allow the formation of concrete using much less water, and these chemicals have doubled the strength of ready-mix concrete over the past 20 years.
Researchers also have found that the properties of con-crete can be greatly improved by adding fibers of various kinds, including those made of steel, glass, and carbon-based polymers. One type of fiber concrete—called slurry infil-trated fiber concrete (SIFCON), which is tough enough to be used to make missile silos and can be formed into com-plex shapes—may be especially useful for structures in earthquake-prone areas.
Other efforts to improve concrete center on replacing Portland cement with other binders such as carbon-based polymers. Although these polymer-based concretes will burn and do lose their shapes at high temperatures, they are much more resistant to the effects of water, acids, and salts than those made with Portland cement.
Despite the fact that most concrete now used is very similar to that used by the Romans to build the Pantheon, progress is being made, and revolutionary improvements may be just around the corner.
19.6 The Group 4A Elements 893 Roman baths such as these in Bath, England, used lead pipes for water.
The major current use for tin is as a protective coating for steel, especially for cans used as food containers. The thin layer of tin, applied electrolytically, forms a protective oxide coating that prevents further corrosion.
Lead is easily obtained from its ore, galena (PbS). Because lead melts at such a low temperature, it may have been the first pure metal obtained from its ore. We know that lead was used as early as 3000 B.C. by the Egyptians and was later used by the Romans to make eating utensils, glazes on pottery, and even intricate plumbing systems. Lead is very toxic, however. In fact, the Romans had so much contact with lead that it may have contributed to the demise of their civilization. Analysis of bones from that era shows significant levels of lead.
Although lead poisoning has been known since at least the second century B.C., the incidences of this problem have been relatively isolated. However, the widespread use of tetraethyl lead, (C2H5)4Pb, as an antiknock agent in gasoline increased the lead levels in our environment in the twentieth century. Concern about the effects of this lead pollution has caused the U.S. government to require the gradual replacement of the lead in gaso-line with other antiknock agents. The largest commercial use of lead (about 1.3 million tons annually) is for electrodes in the lead storage batteries used in automobiles (see Section 17.5).
Table 19.13 summarizes some important reactions of the Group 4A elements.
CHEMICAL IMPACT Beethoven: Hair Is the Story L udwig van Beethoven, arguably the greatest composer who ever lived, led a troubled life fraught with sickness, deafness, and personality aberrations. Now we may know the source of these difficulties: lead poisoning. Scientists have recently reached this conclusion through analysis of Beethoven’s hair. When Beethoven died in 1827 at age 56, many mourners took samples of the great man’s hair. In fact, it was said at the time that he was practically bald by the time he was buried. The hair that was recently analyzed consisted of 582 strands—3 to 6 inches long—bought for the Center of Beethoven Studies for $7300 in 1994 from Sotheby’s auction house in London.
According to William Walsh of the Health Research In-stitute (HRI) in suburban Chicago, Beethoven’s hair showed a lead concentration 100 times the normal levels. The sci-entists concluded that Beethoven’s exposure to lead came as an adult, possibly from the mineral water he drank and swam in when he visited spas.
The lead poisoning may well explain Beethoven’s volatile temper—the composer was subject to towering rages and sometimes had the look of a wild animal. In rare cases lead poisoning has been known to cause deafness, but the researchers remain unsure if this problem led to Beethoven’s hearing loss.
According to Walsh, the scientists at HRI were origi-nally looking for mercury, a common treatment for syphilis in the early nineteenth century, in Beethoven’s hair. The absence of mercury supports the consensus of scholars that Beethoven did not have this disease. Not surprisingly, Beethoven himself wanted to know what made him so ill.
In a letter to his brothers in 1802, he asked them to have doctors find the cause of his frequent abdominal pain after his death.
Portrait of Beethoven by Josef Kari Stieler.
Key Terms Section 19.1 representative elements transition metals lanthanides actinides metalloids (semimetals) metallurgy liquefaction Section 19.2 alkali metals superoxide Section 19.3 hydride ionic (saltlike) hydride covalent hydride metallic (interstitial) hydride Section 19.4 alkaline earth metals hard water ion exchange ion-exchange resin cation-exchange resin Section 19.5 boranes 894 Chapter Nineteen The Representative Elements: Groups 1A Through 4A Representative elements Chemical properties are determined by their s and p valence-electron configurations Metallic character increases going down the group The properties of the first element in a group usually differ most from the proper-ties of the other elements in the group due to a significant difference in size • In Group 1A, hydrogen is a nonmetal and the other members of the group are active metals • The first member of a group forms the strongest bonds, causing nitrogen and oxygen to exist as N2 and O2 molecules Elemental abundances on earth Oxygen is the most abundant element, followed by silicon The most abundant metals are aluminum and iron, which are found as ores Group 1A elements (alkali metals) Have valence configuration ns1 Except for hydrogen, readily lose one electron to form M ions in their com-pounds with nonmetals React vigorously with water to form M and ions and hydrogen gas Form a series of oxides of the types M2O (oxide), M2O2 (peroxide), and MO2 (superoxide) • Not all metals form all types of oxide compounds Hydrogen forms covalent compounds with nonmetals With very active metals, hydrogen forms hydrides that contain the ion Group 2A (alkaline earth metals) Have valence configuration ns2 React less violently with water than alkali metals The heavier alkaline earths form nitrides and hydrides Hard water contains Ca2 and Mg2 ions • Form precipitates with soap • Usually removed by ion-exchange resins that replace the Ca2 and Mg2 ions with Na Group 3A Have valence configuration ns2np1 Show increasing metallic character going down the group Boron is a nonmetal that forms many types of covalent compounds, including bo-ranes, which are highly electron-deficient and thus are very reactive The metals aluminum, gallium, and indium show some covalent tendencies H OH p For Review TABLE 19.13 Selected Reactions of the Group 4A Elements Reaction Comment M Ge or Sn; Pb gives PbX2 M Ge or Sn; high temperatures; Pb gives PbO or M Sn or Pb M 2H S M2 H2 Pb3O4 M O2 S MO2 X2 any halogen molecule; M 2X2 S MX4 Lead(II) oxide, known as litharge.
Group 4A Have valence configuration ns2np2 Lighter members are nonmetals; heavier members are metals • All group members can form covalent bonds to nonmetals Carbon forms a huge variety of compounds, most of which are classified as organic compounds REVIEW QUESTIONS 1. What are the two most abundant elements by mass in the earth’s crust, oceans, and atmosphere? Does this make sense? Why? What are the four most abundant elements by mass in the human body? Does this make sense? Why?
2. What evidence supports putting hydrogen in Group 1A of the periodic table? In some periodic tables hydrogen is listed separately from any of the groups. In what ways is hydrogen unlike a typical Group 1A element?
3. What is the valence electron configuration for the alkali metals? List some com-mon properties of alkali metals. How are the pure metals prepared? Predicting formulas for the compound formed when an alkali metal reacts with oxygen can be difficult. Why? What is the difference between an oxide, a peroxide, and a superoxide? Describe how potassium superoxide is used in a self-contained breathing apparatus. Predict the formulas of the compounds formed when an alkali metal reacts with F2, S, N2, H2, and H2O.
4. List two major industrial uses of hydrogen. Name the three types of hydrides.
How do they differ from one another?
5. What is the valence electron configuration for alkaline earth metals? List some com-mon properties of alkaline earth metals. How are alkaline earth metals prepared?
What ions are found in hard water? What happens when water is “softened”?
6. Predict the formulas of the compounds formed when an alkaline earth metal re-acts with F2, O2, S, N2, H2, and H2O.
7. What is the valence electron configuration for the Group 3A elements? How does metallic character change as one goes down this group? How are boron and aluminum different? A12O3 is amphoteric. What does this mean?
8. Predict the formulas of the compounds formed when aluminum reacts with F2, O2, S, and N2.
9. What is the valence electron configuration for Group 4A elements? Group 4A con-tains two of the most important elements on earth. What are they, and why are they so important? How does metallic character change as one goes down Group 4A?
Why is the chemistry of carbon dominated by bonds, whereas that of silicon is dominated by bonds? What are the two allotropic forms of carbon?
10. List some properties of germanium, tin, and lead. Predict the formulas of the compounds formed when Ge reacts with F2 and O2.
Si¬O C¬C Questions 895 A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 1. Although the earth was formed from the same interstellar mate-rial as the sun, there is little elemental hydrogen (H2) in the earth’s atmosphere. Explain.
2. Many lithium salts are hygroscopic (absorb water), whereas the corresponding salts of the other alkali metals are not. Explain.
3. How do the acidities of the aqueous solutions of the alkaline earth metal ions (M2) change in going down the group?
4. What are three-centered bonds?
5. Why is graphite a good lubricant? What advantages does it have over grease- or oil-based lubricants?
6. What are some of the structural differences between quartz and amorphous SiO2?
896 Chapter Nineteen The Representative Elements: Groups 1A Through 4A 7. What type of semiconductor is formed when a Group 3A element is added as an impurity to Si or Ge?
8. Diagonal relationships in the periodic table exist as well as the vertical relationships. For example, Be and Al are similar in some of their properties as are B and Si. Rationalize why these diago-nal relationships hold for properties such as size, ionization en-ergy, and electron affinity.
9. Atomic size seems to play an important role in explaining some of the differences between the first element in a group and the subsequent group elements. Explain.
10. What will be the atomic number of the next alkali metal to be dis-covered? How would you expect the physical properties of the next alkali metal to compare with the properties of the other al-kali metals summarized in Table 19.4?
11. In most compounds, the solid phase is denser than the liquid phase.
Why isn’t this true for water?
12. Beryllium shows some covalent characteristics in some of its com-pounds, unlike the other alkaline earth halides. Give a possible explanation for this phenomenon.
Exercises In this section similar exercises are paired.
Group 1A Elements 13. Hydrogen is produced commercially by the reaction of methane with steam: a. Calculate and for this reaction (use the data in Ap-pendix 4).
b. What temperatures will favor product formation at standard con-ditions? Assume and do not depend on temperature.
14. The major industrial use of hydrogen is in the production of am-monia by the Haber process: a. Using data from Appendix 4, calculate , , and for the Haber process reaction.
b. Is the reaction spontaneous at standard conditions?
c. At what temperatures is the reaction spontaneous at standard con-ditions? Assume and do not depend on temperature.
15. Name each of the following compounds.
a. Li2O b. KO2 c. Na2O2 16. Write the formula for each of the following compounds.
a. lithium nitride c. rubidium hydroxide b. potassium carbonate d. sodium hydride 17. Complete and balance the following reactions.
a.
b.
c.
d.
18. Write balanced equations describing the reaction of lithium metal with each of the following: O2, S8, Cl2, P4, H2, H2O, and HCl.
19. Lithium reacts with acetylene in liquid ammonia to produce LiC2H (lithium acetylide, ) and hydrogen gas. Write a balanced equation for this reaction. What type of reaction is this?
LiC‚CH KO2(s) H2O(l) ¡ LiH(s) H2O(l) ¡ Na2O2(s) H2O(l) ¡ Li2O(s) H2O(l) ¡ ¢S° ¢H° ¢G° ¢S° ¢H° 3H21g2 N21g2 S 2NH31g2 ¢S° ¢H° ¢S° ¢H° CH41g2 H2O1g2 ∆CO1g2 3H21g2 20. The electrolysis of aqueous sodium chloride (brine) is an impor-tant industrial process for the production of chlorine and sodium hydroxide. In fact, this process is the second largest consumer of electricity in the United States, after the production of aluminum.
Write a balanced equation for the electrolysis of aqueous sodium chloride (hydrogen gas is also produced).
Group 2A Elements 21. Name each of the following compounds.
a. MgCO3 b. BaSO4 c. Sr(OH)2 22. Write the formula for each of the following compounds.
a. calcium nitride b. beryllium chloride c. barium hydride 23. One harmful effect of acid rain is the deterioration of structures and statues made of marble or limestone, both of which are es-sentially calcium carbonate. The reaction of calcium carbonate with sulfuric acid yields carbon dioxide, water, and calcium sul-fate. Because calcium sulfate is marginally soluble in water, part of the object is washed away by the rain. Write a balanced chemical equation for the reaction of sulfuric acid with calcium carbonate.
24. Write balanced equations describing the reaction of Sr with each of the following: O2, S8, Cl2, P4, H2, H2O, and HCl.
25. Predict the structure of BeF2 in the gas phase. What structure would you predict for BeF2(s)?
26. The beryllium atom in BeCl2 is electron-deficient (only four va-lence electrons surround it), which makes it very reactive toward electron-pair donors such as ammonia. Draw a Lewis structure for the expected product when BeCl2 reacts with excess ammonia.
27. The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 mg per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the max-imum molarity of calcium ions in hard water if the fluoride con-centration is at the USPHS recommended level? 28. Slaked lime, Ca(OH)2, is used to soften hard water by removing calcium ions from hard water through the reaction Although CaCO3(s) is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of CaCO3 in water .
29. What mass of barium is produced when molten BaCl2 is elec-trolyzed by a current of A for 6.00 h?
30. Electrolysis of an alkaline earth metal chloride using a current of 5.00 A for 748 s deposits 0.471 g of metal at the cathode. What is the identity of the alkaline earth metal chloride?
Group 3A Elements 31. Write the formula for each of the following compounds.
a. aluminum nitride b. gallium fluoride c. gallium sulfide 32. Thallium and indium form 1 and 3 oxidation states when in compounds. Predict the formulas of the possible compounds be-tween thallium and oxygen and between indium and chlorine.
Name the compounds.
2.50 105 (Ksp 8.7 109) 2CaCO31s2 2H2O1l2 Ca1OH221aq2 Ca21aq2 2HCO3 1aq2 S 4.0 1011) (Ksp for CaF2 F Additional Exercises 897 Additional Exercises 49. A 0.250-g chunk of sodium metal is cautiously dropped into a mix-ture of 50.0 g of water and 50.0 g of ice, both at C. The reaction is Will the ice melt? Assuming the final mixture has a specific heat capacity of calculate the final temperature. The en-thalpy of fusion for ice is 6.02 kJ/mol.
50. One of the chemical controversies of the nineteenth century con-cerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming Be3 ions) and that it gave an oxide with the formula Be2O3. This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming Be2 ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of 9.0. In 1894, A.
Combes (Comptes Rendus 1894, p. 1221) reacted beryllium with the anion and measured the density of the gaseous prod-uct. Combes’s data for two different experiments are as follows: C5H7O2 4.18 J/g °C, 2Na1s2 2H2O1l2 ¡ 2NaOH1aq2 H21g2 ¢H 368 kJ 0° 33. Boron hydrides were once evaluated for possible use as rocket fuels. Complete and balance the following equation for the com-bustion of diborane.
34. Elemental boron is produced by reduction of boron oxide with magnesium to give boron and magnesium oxide. Write a balanced equation for this reaction.
35. Ga2O3 is an amphoteric oxide, and In2O3 is a basic oxide. Write equations for the reactions that illustrate these properties.
36. Aluminum hydroxide is amphoteric and will dissolve in both acidic and basic solutions. Write balanced chemical equations representing each process.
37. Write equations describing the reactions of Ga with each of the following: F2, O2, S8, N2, and HCl.
38. Write a balanced equation describing the reaction of aluminum metal with concentrated aqueous sodium hydroxide.
Group 4A Elements 39. Draw Lewis structures for CF4, GeF4, and . Predict the molecular structure (including bond angles), and give the expected hybridization of the central atom in these three substances. Ex-plain why does not form.
40. Carbon and sulfur form compounds with the formulas CS2 and C3S2. Draw Lewis structures and predict the shapes of these two compounds.
41. Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction.
a.
b. Silicon tetrachloride is reacted with very pure magnesium, pro-ducing silicon and magnesium chloride.
c.
42. Write equations describing the reactions of Sn with each of the following: Cl2, O2, and HCl.
43. Why are people advised not to drink hot tap water if their plumb-ing contains lead solder?
44. Calculate the solubility of Pb(OH)2 in wa-ter. Is Pb(OH)2 more or less soluble in acidic solutions? Explain.
45. The fermentation of glucose produces ethanol and carbon diox-ide. Write a balanced equation for this reaction.
46. Tin forms compounds in the 2 and 4 oxidation states. There-fore, when tin reacts with fluorine, two products are possible.
Write balanced equations for the production of the two tin halide compounds and name them.
47. The resistivity (a measure of electrical resistance) of graphite is in the basal plane. (The basal plane is the plane of the six-membered rings of carbon atoms.) The re-sistivity is along the axis perpendicular to the plane. The resistivity of diamond is 1014 to 1016 and is independent of direction. How can you account for this behavior in terms of the structures of graphite and diamond?
48. Silicon carbide (SiC) is an extremely hard substance. Propose a structure for SiC.
ohm cm 0.2 to 1.0 ohm cm (0.4 to 5.0) 104 ohm cm (Ksp 1.2 1015) Na2SiF6(s) Na(s) S Si(s) NaF(s) SiO2(s) C(s) S Si(s) CO(g) CF6 2 GeF6 2 B2H6 O2 ¡ B1OH23 I II Mass 0.2022 g 0.2224 g Volume 22.6 cm3 26.0 cm3 Temperature Pressure 765.2 mm Hg 764.6 mm 17°C 13°C If beryllium is a divalent metal, the molecular formula of the product will be Be(C5H7O2)2; if it is trivalent, the formula will be Be(C5H7O2)3. Show how Combes’s data help to confirm that beryllium is a divalent metal.
51. It takes 15 kWh (kilowatt-hours) of electrical energy to produce 1.0 kg of aluminum metal from aluminum oxide by the Hall–Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg of aluminum metal. Why is it economically feasible to recy-cle aluminum cans? (The enthalpy of fusion for aluminum metal is 10.7 kJ/mol [1 watt 1 J/s].) 52. Borazine (B3N3H6) has often been called “inorganic” benzene.
Write Lewis structures for borazine. Borazine contains a six-membered ring of alternating boron and nitrogen atoms with one hydrogen bonded to each boron and nitrogen.
53. Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O2. Consider the fol-lowing reactions and approximate standard free energy changes: Using these data, estimate the equilibrium constant value at for the following reaction: 54. The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide (CO2), and carbon suboxide (C3O2). The space-filling models for these three compounds are HgbO2 CO ∆HgbCO O2 25°C Hgb CO ¡ HgbCO ¢G° 80 kJ Hgb O2 ¡ HgbO2 ¢G° 70 kJ 898 Chapter Nineteen The Representative Elements: Groups 1A Through 4A 65. In Exercise 107 in Chapter 5, the pressure of CO2 in a bottle of sparkling wine was calculated assuming that the CO2 was insolu-ble in water. This was a bad assumption. Redo this problem by as-suming CO2 obeys Henry’s law. Use the data given in that problem to calculate the partial pressure of CO2 in the gas phase and the sol-ubility of CO2 in the wine at . The Henry’s law constant for CO2 is at with Henry’s law in the form where C is the concentration of the gas in mol/L.
66. The compound Pb3O4 (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in Pb3O4?
67. Lead hydrogen arsenate, an inorganic insecticide used against the potato beetle, is produced by the following reaction: Balance this equation.
Challenge Problems 68. Provide a reasonable estimate for the number of atoms in an average adult human. Explain your answer. Use the information given in Table 19.2.
69. Suppose 10.00 g of an alkaline earth metal reacts with 10.0 L of water to produce 6.10 L of hydrogen gas at 1.00 atm and .
Identify the metal and determine the pH of the solution.
70. Gallium arsenide, GaAs, has gained widespread use in semi-conductor devices that convert light and electrical signals in fiber-optic communications systems. Gallium consists of 60.% 69Ga and 40.% 71Ga. Arsenic has only one naturally occurring isotope, 75As.
Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and Ga2As2. What would the distribution of peaks look like for these two fragments?
71. Consider dissolving 0.50 mol of CO2(g) to enough water to make a 1.0-L solution. Determine the pH of this solution, and Use data from Appendix 5, Table 5.2.
72. a. Many biochemical reactions that occur in cells require rela-tively high concentrations of potassium ion (K). The concen-tration of K in muscle cells is about 0.15 M. The concentra-tion of K in blood plasma is about 0.0050 M. The high internal concentration in cells is maintained by pumping K from the plasma. How much work must be done to transport 1.0 mol of K from the blood to the inside of a muscle cell at (nor-mal body temperature)?
b. When 1.0 mol of K is transferred from blood to the cells, do any other ions have to be transported? Why or why not?
c. Cells use the hydrolysis of adenosine triphosphate, abbreviated ATP, as a source of energy. Symbolically, this reaction can be represented as where ADP represents adenosine diphosphate. For this reaction at . How many moles of ATP must be hy-drolyzed to provide the energy for the transport of 1.0 mol of K? Assume standard conditions for the ATP hydrolysis reaction.
73. EDTA is used as a complexing agent in chemical analysis. Solu-tions of EDTA, usually containing the disodium salt Na2H2EDTA, are also used to treat heavy metal poisoning. The equilibrium con-stant for the following reaction is Pb21aq2 H2EDTA21aq2 ∆PbEDTA21aq2 2H1aq2 1.0 1023: 37°C, K 1.7 105 ATP1aq2 H2O1l2 ¡ ADP1aq2 H2PO4 1aq2 37°C [CO3 2].
25°C Pb1NO3221aq2 H3AsO41aq2 S PbHAsO41s2 HNO31aq2 C kP, 25°C 3.1 102 mol/L atm 25°C For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).
55. The overall reaction in the lead storage battery is Calculate at for this battery when that is, At for the lead storage battery.
56. The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the fre-quency and the energy of a photon of light at each of these wave-lengths? What are the energies in kJ/mol?
57. In the 1950s and 1960s, several nations conducted tests of nuclear warheads in the atmosphere. It was customary following each test to monitor the concentration of strontium-90 (a radioactive isotope of strontium) in milk. Why would strontium-90 tend to accumulate in milk?
58. The compound cannot be dehydrated easily by heating. It dissolves in water to give an acidic solution. Explain these observations.
59. The inert-pair effect is sometimes used to explain the tendency of heavier members of group 3A to exhibit 1 and 3 oxidation states. What does the inert-pair effect reference? Hint: Consider the valence electron configuration for group 3A elements.
60. Assume that element 113 is produced. What is the expected elec-tron configuration for element 113?
61. Calculate the pH of a 0.050 M Al(NO3)3 solution. The Ka value for Al(H2O)6 3 is 62. The compound with the formula TlI3 is a black solid. Given the following standard reduction potentials: would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?
63. How could you determine experimentally whether the compound Ga2Cl4 contains two gallium(II) ions or one gallium(I) and one gallium(III) ion? (Hint: Consider the electron configurations of the three possible ions.) 64. Tricalcium aluminate, an important component of Portland cement, is 44.4% calcium and 20.0% aluminum by mass. The remainder is oxygen.
a. Calculate the empirical formula of tricalcium aluminate.
b. The structure of tricalcium aluminate was not determined until 1975.
The aluminate anion has the following structure: What is the molecular formula of tricalcium aluminate?
c. How would you describe the bonding in the anion?
Al6O18 18 = Al = O (Al6O18 18) I3 2e ¡ 3I e° 0.55 V Tl3 2e ¡ Tl e° 1.25 V 1.4 105.
BeSO4 4H2O e° 2.04 V 25°C, [H] [HSO4 ] 4.5 M.
[H2SO4] 4.5 M, 25°C e 2PbSO41s2 2H2O1l2 Pb1s2 PbO21s2 2H1aq2 2HSO 4 1aq2 ¡ Marathon Problem 899 fiber-optic devices. This material can be synthesized at 900. K according to the following reaction: a. If 2.56 L of In(CH3)3 at 2.00 atm is allowed to react with 1.38 L of PH3 at 3.00 atm, what mass of InP(s) will be produced assuming the reaction is 87% efficient?
b. When an electric current is passed through an optoelectronic device containing InP, the light emitted has an energy of J. What is the wavelength of this light and is it visible to the human eye? c. The semiconducting properties of InP can be altered by doping.
If a small number of phosphorus atoms are replaced by atoms with an electron configuration of [Kr]5s24d105p4, is this n-type or p-type doping?
80. The chemistry of tin(II) fluoride is particularly complex and demonstrates a wide range of reactivities. For example, in aque-ous solutions of tin(II) fluoride containing sodium fluoride, the predominant species is a. What is the molecular geometry of and the hybridiza-tion of the tin atom?
b. When tin(II) fluoride is crystallized from aqueous solutions containing sodium fluoride, one of the products is the poly-atomic cluster Na4Sn3F10. Write a balanced chemical reaction for the formation of Na4Sn3F10 from tin(II) fluoride and NaF.
c. Assuming complete conversion, what mass of Na4Sn3F10 can be prepared by mixing 15.5 mL of 1.48 M tin(II) fluoride with 35.0 mL of 1.25 M NaF?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
81. Use the symbols of the elements described in the following clues to fill in the blanks that spell out the name of a famous American scientist. Although this scientist was better known as a physicist than as a chemist, the Philadelphia institute that bears his name does include a biochemistry research facility.
__ ______ (1) (2) (3) (4) (5) (6) (7) (1) The oxide of this alkaline earth metal is amphoteric.
(2) You might be surprised to learn that a binary compound of sodium with this element has the formula NaX3, a compound used in airbags.
(3) This alkali metal is radioactive.
(4) This element is the alkali metal with the least negative stan-dard reduction potential. Write its symbol in reverse order.
(5) Potash is an oxide of this alkali metal.
(6) This is the only alkali metal that reacts directly with nitrogen to make a binary compound with formula M3N.
(7) This element is the first in Group 3A for which the 1 oxi-dation state is exhibited in stable compounds. Use only the second letter of its symbol.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
SnF3 SnF3 .
2.03 1019 In1CH3231g2 PH31g2 ¡ InP1s2 3CH41g2 Calculate [Pb2] at equilibrium in a solution originally 0.0010 M in Pb2, 0.050 M in H2EDTA2, and buffered at .
74. The compounds CCl4 and H2O do not react with each other. On the other hand, silicon tetrachloride reacts with water according to the equation Discuss the importance of thermodynamics and kinetics in the reactivity of water with SiCl4 as compared with its lack of reac-tivity with CCl4.
75. One reason suggested to account for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is 210 kJ/mol, which is less than either the energy. Why would a similar mechanism not be expected to be very important in the decom-position of long carbon chains?
76. From the information on the temperature stability of white and gray tin given in this chapter, which form would you expect to have the more ordered structure?
77. Lead forms compounds in the 2 and 4 oxidation states. All lead(II) halides are known (and are known to be ionic). Only PbF4 and PbCl4 are known among the possible lead(IV) halides. Pre-sumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: Suppose 25.00 g of a lead(IV) halide reacts to form 16.12 g of a lead(II) halide and the free halogen. Identify the halogen.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
78. The heaviest member of the alkaline earth metals is radium (Ra), a naturally radioactive element discovered by Pierre and Marie Curie in 1898. Radium was initially isolated from the uranium ore pitchblende, in which it is present as approximately 1.0 g per 7.0 metric tons of pitchblende. How many atoms of radium can be isolated from g of pitchblende (1 metric ton 1000 kg)? One of the early uses of radium was as an additive to paint so that watch dials coated with this paint would glow in the dark.
The longest-lived isotope of radium has a half-life of years. If an antique watch, manufactured in 1925, contains 15.0 mg of radium, how many atoms of radium will remain in 2025?
79. Indium(III) phosphide is a semiconducting material that has been frequently used in lasers, light-emitting diodes (LED) and 1.60 103 1.75 108 PbX4 ¡ PbX2 X2 Si¬Si or Si¬H SiCl41l2 2H2O1l2 ¡ SiO21s2 4HCl1aq2 pH 6.00 CH2 CH2 CH2 O2C CH2 O2C CH2 CO2 EDTA4 N N CH2 CO2 Ethylenediaminetetraacetate 20 The Representative Elements: Groups 5A Through 8A Contents 20.1 The Group 5A Elements 20.2 The Chemistry of Nitrogen • Nitrogen Hydrides • Nitrogen Oxides • Oxyacids of Nitrogen 20.3 The Chemistry of Phosphorus • Phosphorus Oxides and Oxyacids • Phosphorus in Fertilizers • Phosphorus Halides 20.4 The Group 6A Elements 20.5 The Chemistry of Oxygen 20.6 The Chemistry of Sulfur • Sulfur Oxides • Oxyacids of Sulfur • Other Compounds of Sulfur 20.7 The Group 7A Elements • Hydrogen Halides • Oxyacids and Oxyanions • Other Halogen Compounds 20.8 The Group 8A Elements The carnivorous pitcher plant “eats” insects to utilize the nitrogen held in the insect tissue.
I n Chapter 19 we saw that vertical groups of elements tend to show similar chemical characteristics because they have identical valence-electron configurations. Generally, metallic character increases going down a group, as the valence electrons are found farther from the nucleus. Also, recall that the most dramatic change in properties occurs after the first group member, mainly because the most dramatic change in size occurs between the first and second group members.
As we proceed from Group 1A to Group 7A, the elements change from active metals (electron donors) to strong nonmetals (electron acceptors). Thus it is not sur-prising that the middle groups show the most varied chemistry: Some group members behave principally as metals, others behave mainly as nonmetals, and some show both tendencies. The elements in Groups 5A and 6A show great chemical variety and form many compounds of considerable practical value. The halogens (Group 7A) are non-metals that are also found in many everyday substances such as household bleach, photographic films, and “automatic” sunglasses. The elements in Group 8A (the noble gases) are most useful in their elemental forms, but their ability to form compounds, discovered only within the past 40 years, has provided important tests for the theories of chemical bonding.
In this chapter we give an overview of the elements in Groups 5A through 8A, con-centrating on the chemistry of the most important elements in these groups: nitrogen, phos-phorus, oxygen, sulfur, and the halogens.
As Sb Bi N P Se Te Po O S 5A 6A Br I At F Cl Kr Xe Rn Ne Ar He 7A 8A 20.1 The Group 5A Elements The Group 5A elements (with the valence-electron configuration ns2np3) are prepared as shown in Table 20.1, and they show remarkably varied chemical properties. As usual, metallic character increases going down the group, as is apparent from the electronega-tivity values (Table 20.1). Nitrogen and phosphorus are nonmetals that can gain three elec-trons to form 3 anions in salts with active metals—for example, magnesium nitride (Mg3N2) and beryllium phosphide (Be3P2). The chemistry of these two important elements is discussed in the next two sections.
As Sb Bi N P 5A 901 902 Chapter Twenty The Representative Elements: Groups 5A Through 8A Bismuth and antimony tend to be metallic, readily losing electrons to form cations.
Although these elements have five valence electrons, so much energy is required to re-move all five that no ionic compounds containing Bi5 or Sb5 ions are known. Three pentahalides (BiF5, SbCl5, and SbF5) are known, but these are molecular rather than ionic compounds.
The Group 5A elements can form molecules or ions that involve three, five, or six covalent bonds to the Group 5A atom. Examples involving three single bonds are NH3, PH3, NF3, and AsCl3. Each of these molecules has a lone pair of electrons (and thus can behave as a Lewis base) and a pyramidal shape as predicted by the VSEPR model (see Fig. 20.1).
All the Group 5A elements except nitrogen can form molecules with five cova-lent bonds (of general formula MX5). Nitrogen cannot form such molecules because of its small size and lack of available d orbitals. The MX5 molecules have a trigonal bipyra-midal shape (see Fig. 20.1) as predicted by the VSEPR model, and the central atom is de-scribed as dsp3 hybridized. The MX5 molecules can accept an additional electron pair to form ionic species containing six covalent bonds. An example is where the anion has an octahedral shape (see Fig. 20.1) and the phosphorus atom is described as d2sp3 hybridized.
Although the MX5 molecules have a trigonal bipyramidal structure in the gas phase, the solids of many of these compounds contain the ions MX4 and (Fig. 20.2), where the MX4 cation is tetrahedral (the atom represented by M is described as sp3 hybridized) and the anion is octahedral (the atom repre-sented by M is described as d2sp3 hybridized). Examples are PCl5, which in the solid state contains PCl4 and and AsF3Cl2, which in the solid state contains AsCl4 and As discussed in Section 19.1, the ability of the Group 5A elements to form bonds decreases dramatically after nitrogen. This explains why elemental nitrogen exists as N2 molecules, whereas the other elements in the group exist as larger aggregates containing single bonds. For example, in the gas phase, the elements phosphorus, arsenic, and anti-mony consist of P4, As4, and Sb4 molecules, respectively.
p AsF6 .
PCl6 , MX6 MX6 PF6 PF5 F ¡ PF6 TABLE 20.1 Selected Physical Properties, Sources, and Methods of Preparation for the Group 5A Elements Electro-Method of Element negativity Sources Preparation Nitrogen 3.0 Air Liquefaction of air Phosphorus 2.1 Phosphate rock 2Ca3(PO4)2 6SiO2 S (Ca3(PO4)2) 6CaSiO3 P4O10 Fluorapatite P4O10 10C S (Ca5(PO4)3F) 4P 10CO Arsenic 2.0 Arsenopyrite Heating arsenopyrite in (Fe3As2, FeS) the absence of air Antimony 1.9 Stibnite Roasting Sb2S3 in air (Sb2S3) to form Sb2O3 and then reduction with carbon Bismuth 1.9 Bismite (Bi2O3), Roasting Bi2S3 in air bismuth glance to form Bi2O3 and then (Bi2S3) reduction with carbon 20.2 The Chemistry of Nitrogen 903 20.2 The Chemistry of Nitrogen At the earth’s surface, virtually all elemental nitrogen exists as the N2 molecule with its very strong triple bond (941 kJ/mol). Because of this high bond strength, the N2 mole-cule is so unreactive that it can coexist with most other elements under normal condi-tions without undergoing any appreciable reaction. This property makes nitrogen gas very useful as a medium for experiments involving substances that react with oxygen or wa-ter. Such experiments can be done using an inert-atmosphere box of the type shown in Fig. 20.3.
The strength of the triple bond in the N2 molecule is important both thermodynami-cally and kinetically. Thermodynamically, the great stability of the bond means that most binary compounds containing nitrogen decompose exothermically to the elements.
For example: Of these compounds, only ammonia is thermodynamically more stable than its component elements. That is, only for ammonia is energy required (endothermic process, positive NH31g2 ¡ 1 2N21g2 3 2H21g2 ¢H° 46 kJ N2H41g2 ¡ N21g2 2H21g2 ¢H° 95 kJ NO21g2 ¡ 1 2N21g2 O21g2 ¢H° 34 kJ NO1g2 ¡ 1 2N21g2 1 2O21g2 ¢H° 90 kJ N2O1g2 ¡ N21g2 1 2O21g2 ¢H° 82 kJ N‚N Molecule Type Molecular structure Hybridization of M MX3 MX5 MX6 X Pyramidal Trigonal bipyramidal Octahedral M X X X M X X X X X X X X X X M M Lone pair sp3 M M dsp3 d2sp3 X X X X M + M – X X X X X X FIGURE 20.1 The molecules of the types MX3, MX5, and MX6 formed by Group 5A elements.
FIGURE 20.2 The structures of the tetrahedral MX4 and octahedral MX6 ions.
904 Chapter Twenty The Representative Elements: Groups 5A Through 8A value of ) to decompose the molecule to its elements. For the remaining molecules, energy is released when decomposition to the elements occurs as a result of the great sta-bility of N2.
The importance of the thermodynamic stability of N2 can be seen clearly in the power of nitrogen-based explosives, such as nitroglycerin (C3H5N3O9), which has the structure ¢H° At the Centers for Disease Control and Prevention in Atlanta, Georgia, a worker checks samples stored in a liquid nitrogen tank.
FIGURE 20.3 An inert-atmosphere box used when work-ing with oxygen- or water-sensitive materi-als. The box is filled with an inert gas such as nitrogen, and work is done through the ports fitted with large rubber gloves.
The decomposition of nitroglycerin is a complex process that occurs in many steps. This equation only summarizes the stoichiometry of the reaction.
20.2 The Chemistry of Nitrogen 905 When ignited or subjected to sudden impact, nitroglycerin decomposes very rapidly and exothermically: An explosion occurs; that is, large volumes of gas are produced in a fast, highly exother-mic reaction. Note that 4 moles of liquid nitroglycerin produce 29 moles of gaseous products. This alone produces a large increase in volume. However, also note that the products, including N2, are very stable molecules with strong bonds. Their formation is therefore accompanied by the release of large quantities of energy as heat, which increases the gaseous volume. The hot, rapidly expanding gases produce a pressure surge and damaging shock wave.
Pure nitroglycerin is quite dangerous because it explodes with little provocation.
However, in 1867 the Swedish inventor Alfred Nobel found that if nitroglycerin is ab-sorbed in porous silica, it can be handled quite safely. This tremendously important explosive (see Fig. 20.4), which he called dynamite, earned Nobel a great fortune, which he subsequently used to establish the Nobel Prizes.
Most high explosives are organic compounds that, like nitroglycerin, contain nitro groups and produce nitrogen and other gases as products. Another example is trinitrotoluene, or TNT, a solid at normal temperatures, which decomposes as follows: Note that 2 moles of solid TNT produce 20 moles of gaseous products plus energy.
Decomposition of NH4NO2 When ammonium nitrite is heated, it decomposes to nitrogen gas and water. Calculate the volume of N2 gas produced from 1.00 g of solid NH4NO2 at and 1.00 atm.
Solution The decomposition reaction is Using the molar mass of NH4NO2 (64.05 g/mol), we first calculate the moles of NH4NO2: Since 1 mol N2 is produced for each mole of NH4NO2, mol N2 will be pro-duced in the given experiment. We can calculate the volume of N2 from the ideal gas law: In this case we have and the volume of N2 is See Exercises 20.17 and 20.18.
0.670 L V nRT P 11.56 102 mol2 a0.08206 L atm K molb1523 K2 1.00 atm T 250 273 523 K R 0.08206 L atm/K mol n 1.56 102 mol P 1.00 atm PV nRT 1.56 10 2 1.00 g NH4NO2 1 mol NH4NO2 64.05 g NH4NO2 1.56 102 mol NH4NO2 NH4NO21s2 ¡ Heat N21g2 2H2O1g2 250°C 2C7H5N3O61s2 ¡ 12CO1g2 5H21g2 3N21g2 2C1s2 energy (¬NO2) (6 12 10 1) 4C3H5N3O91l2 ¡ 6N21g2 12CO21g2 10H2O1g2 O21g2 energy NO2 NO2 CH3 NO2 TNT Sample Exercise 20.1 FIGURE 20.4 Chemical explosives are used to demolish a building in Miami, Florida.
906 Chapter Twenty The Representative Elements: Groups 5A Through 8A The effect of bond strength on the kinetics of reactions involving the N2 molecule is il-lustrated by the synthesis of ammonia from nitrogen and hydrogen, a reaction we have dis-cussed many times before. Because a large quantity of energy is required to disrupt the bond, the ammonia synthesis reaction has a negligible rate at room temperature, even though the equilibrium constant is very large at Of course, the most direct way to increase the rate is to raise the temperature, but since the reaction is very exothermic, that is, the value of K decreases significantly with a temperature increase Obviously, the kinetics and the thermodynamics of this reaction are in opposition. A compromise must be reached: high pressure to force the equilibrium to the right and high temperature to produce a reasonable rate. The Haber process for manufacturing ammo-nia illustrates this compromise (see Fig. 20.5). The process is carried out at a pressure of about 250 atm and a temperature of approximately Even higher temperatures would be required except that a catalyst, consisting of a solid iron oxide mixed with small amounts of potassium oxide and aluminum oxide, is used to facilitate the reaction.
Nitrogen is essential to living systems. The problem with nitrogen is not one of supply— we are surrounded by it—but of changing it from inert N2 molecules to a form usable by plants and animals. The process of transforming N2 to other nitrogen-containing compounds is called nitrogen fixation. The Haber process is one example of nitrogen fixation. The am-monia produced can be applied to the soil as a fertilizer, since plants can readily employ the nitrogen in ammonia to make the nitrogen-containing biomolecules essential for their growth.
Nitrogen fixation also results from the high-temperature combustion process in auto-mobile engines. The nitrogen in the air drawn into the engine reacts at a significant rate with oxygen to form nitric oxide (NO), which further reacts with oxygen from the air to form nitrogen dioxide (NO2). This nitrogen dioxide, which is an important contributor to photochemical smog in many urban areas (see Section 12.8), eventually reacts with mois-ture in the air and reaches the soil to form nitrate salts, which are plant nutrients.
Nitrogen fixation also occurs naturally. For example, lightning provides the energy to disrupt N2 and O2 molecules in the air, producing highly reactive nitrogen and oxygen atoms that attack other N2 and O2 molecules to form nitrogen oxides that eventually be-come nitrates. Although lightning traditionally has been credited with forming about 10% of the total fixed nitrogen, recent studies indicate that lightning may account for as much as half the fixed nitrogen available on earth. Another natural nitrogen fixation process is provided by bacteria that reside in the root nodules of plants such as beans, peas, and al-falfa. These nitrogen-fixing bacteria readily allow the conversion of nitrogen to ammonia and other nitrogen-containing compounds useful to plants. The efficiency of these bacte-ria is intriguing: They produce ammonia at soil temperature and 1 atm of pressure, whereas the Haber process requires severe conditions of and 250 atm. For obvious reasons, researchers are studying these bacteria intensively.
When plants and animals die, they decompose, and the elements they consist of are returned to the environment. In the case of nitrogen, the return of the element to the at-mosphere as nitrogen gas, called denitrification, is carried out by bacteria that change nitrates to nitrogen. The complex nitrogen cycle is summarized in Fig. 20.6. It has been estimated that as much as 10 million tons per year more nitrogen is currently being fixed by natural and human processes than is being returned to the atmosphere. This fixed ni-trogen is accumulating in the soil, lakes, rivers, and oceans, where it can promote the growth of algae and other undesirable organisms.
Nitrogen Hydrides By far the most important hydride of nitrogen is ammonia (NH3). A toxic, colorless gas with a pungent odor, ammonia is manufactured in huge quantities mainly for use in fertilizers.
(30 billion pounds per year), 400°C 400°C.
(at 500°C, K 102).
N21g2 3H21g2 ¡ 2NH31g2 ¢H° 92 kJ 25°C.
(K 10 6) N‚N Unreacted N2, H2 Impure N2, H2 Unwanted trace gases removed Pure N2, H2 Catalytic reactors NH3 Cooling chamber Liquid NH3 (yield 20% on each cycle) FIGURE 20.5 A schematic diagram of the Haber process for the manufacture of ammonia.
Nodules on the roots of pea plants contain nitrogen-fixing bacteria.
20.2 The Chemistry of Nitrogen 907 The pyramidal ammonia molecule (see Fig. 20.1) has a lone pair of electrons on the nitrogen atom and polar bonds. This structure leads to a high degree of intermo-lecular interaction by hydrogen bonding in the liquid state and produces an unusually high boiling point for a substance of such low molar mass. Note, however, that the hydrogen bonding in liquid ammonia is clearly not as important as that in liquid water, which has about the same molar mass but a much higher boiling point. The water mole-cule has two polar bonds involving hydrogen and two lone pairs—the right combination for optimal hydrogen bonding—in contrast to the one lone pair and three polar bonds of the ammonia molecule.
As we saw in Chapter 14, ammonia behaves as a base and reacts with acids to pro-duce ammonium salts. For example, A second nitrogen hydride of major importance is hydrazine (N2H4). The Lewis struc-ture of hydrazine is indicating that each nitrogen atom should be sp3 hybridized with bond angles close to 109.5 degrees (the tetrahedral angle), since the nitrogen atom is surrounded by four elec-tron pairs. The observed structure with bond angles of 112 degrees (see Fig. 20.7) agrees reasonably well with these predictions. Hydrazine, a colorless liquid with an ammonia-like odor, freezes at and boils at This boiling point is quite high for a com-pound with a molar mass of 32 g/mol; this suggests that considerable hydrogen bonding must occur among the polar hydrazine molecules.
Hydrazine is a powerful reducing agent that has been used widely as a rocket pro-pellant. For example, its reaction with oxygen is highly exothermic: Since hydrazine also reacts vigorously with the halogens, fluorine is often used instead of oxygen as the oxidizer in rocket engines. Substituted hydrazines, where one or more of the hydrogen atoms are replaced by other groups, are also useful rocket fuels. For exam-ple, monomethylhydrazine, N2H41l2 O21g2 ¡ N21g2 2H2O1g2 ¢H° 622 kJ 113.5°C.
2°C NH31g2 HCl1g2 ¡ NH4Cl1s2 (33.4°C) N¬H Nitrates Plant and animal protein Ammonia Nitrites N2 in the atmosphere N-fixing bacteria Bacteria Bacteria Decay processes Denitrifying bacteria Lightning FIGURE 20.6 The nitrogen cycle. To be used by plants and animals, nitrogen must be converted from N2 to nitrogen-containing compounds, such as nitrates, ammonia, and proteins.
The nitrogen is returned to the atmosphere by natural decay processes.
H H N 112° H N H FIGURE 20.7 The molecular structure of hydrazine (N2H4). This arrangement minimizes the repulsion between the lone pairs on the nitrogen atoms by placing them on oppo-site sides.
908 Chapter Twenty The Representative Elements: Groups 5A Through 8A is used with the oxidizer dinitrogen tetroxide (N2O4) to power the U.S. space shuttle or-biter. The reaction is Because of the large number of gaseous molecules produced and the exothermic nature of this reaction, a very high thrust per mass of fuel is achieved. The reaction is also self-starting—it begins immediately when the fuels are mixed—which is a useful property for rocket engines that must be started and stopped frequently.
Heats of Reaction from Bond Energies Using the bond energies in Table 8.4, calculate the approximate value of for the reaction between gaseous monomethylhydrazine and dinitrogen tetroxide: The bonding in N2O4 is described by resonance structures that predict that the bonds are intermediate in strength between single and double bonds (assume an average bond energy of 440 kJ/mol).
Solution To calculate H for this reaction, we must compare the energy necessary to break the bonds of the reactants and the energy released by formation of the bonds in the products: Breaking bonds requires energy (positive sign), and forming bonds releases energy (neg-ative sign). As summarized in the following table, The reaction is highly exothermic.
¢H 121.1 103 kJ2 126.1 103 kJ2 5.0 103 kJ N¬O N¬O 5N2O41g2 4N2H31CH321g2 ¡ 12H2O1g2 9N21g2 4CO21g2 ¢H 5N2O41l2 4N2H31CH321l2 ¡ 12H2O1g2 9N21g2 4CO21g2 Sample Exercise 20.2 Bonds Broken Energy Required (kJ/mol) Bonds Formed Energy Released (kJ/mol) 5 4 20 20 440 8.8 103 12 2 24 24 467 1.12 104 5 4 9 9 160 1.4 103 9 9 941 8.5 103 4 3 12 12 391 4.7 103 4 2 8 8 799 6.4 103 4 3 12 12 413 5.0 103 4 1 4 4 305 1.2 103 Total 21.1 103 Total 26.1 103 C¬N C¬H C“O N¬H N‚N N¬N O¬H N“O See Exercise 20.19.
The use of hydrazine as a rocket propellant is a rather specialized application. The main industrial use of hydrazine is as a “blowing” agent in the manufacture of plastics.
Hydrazine decomposes to form nitrogen gas, which causes foaming in the liquid plastic, which results in a porous texture. Another major use of hydrazine is in the production of 20.2 The Chemistry of Nitrogen 909 agricultural pesticides. Of the many hundreds of hydrazine derivatives (substituted hy-drazines) that have been tested, 40 are used as fungicides, herbicides, insecticides, and plant growth regulators.
The manufacture of hydrazine involves the oxidation of ammonia by the hypochlo-rite ion in basic solution: Although this reaction looks straightforward, the actual process involves many steps and requires high pressure, high temperature, and catalysis to optimize the yield of hydrazine in the face of many competing reactions.
Nitrogen Oxides Nitrogen forms a series of oxides in which it has an oxidation state from 1 to 5, as shown in Table 20.2.
Dinitrogen monoxide (N2O), more commonly called nitrous oxide or “laughing gas,” has an inebriating effect and has been used as a mild anesthetic by dentists. Because of its high solubility in fats, nitrous oxide is used widely as a propellant in aerosol cans of whipped cream. It is dissolved in the liquid in the can at high pressure and forms bubbles that produce foaming as the liquid is released from the can. A significant amount of N2O exists in the atmosphere, mostly produced by soil microorganisms, and its concentration 2NH31aq2 OCl1aq2 ¡ N2H41aq2 Cl 1aq2 H2O1l2 Blowing agents such as hydrazine, which forms nitrogen gas on decomposition, are used to produce porous plastics like these styrofoam products.
TABLE 20.2 Some Common Nitrogen Compounds Oxidation State of Lewis Nitrogen Compound Formula Structure 3 Ammonia NH3 2 Hydrazine N2H4 1 Hydroxylamine NH2OH 0 Nitrogen N2 1 Dinitrogen monoxide N2O (nitrous oxide) 2 Nitrogen monoxide NO (nitric oxide) 3 Dinitrogen trioxide N2O3 4 Nitrogen dioxide NO2 5 Nitric acid HNO3 In some cases, additional resonance structures are needed to fully describe the electron distribution.
910 Chapter Twenty The Representative Elements: Groups 5A Through 8A appears to be gradually increasing. Because it can strongly absorb infrared radiation, nitrous oxide plays a small but probably significant role in controlling the earth’s tem-perature in the same way that atmospheric carbon dioxide and water vapor do (see the discussion of the greenhouse effect in Section 6.5). Some scientists fear that the rapid de-crease of tropical rain forests resulting from development in countries such as Brazil will significantly affect the rate of production of N2O by soil organisms and thus will have im-portant effects on the earth’s temperature.
In the laboratory, nitrous oxide is prepared by the thermal decomposition of ammo-nium nitrate: This experiment must be done carefully because ammonium nitrate can explode. In fact, one of the greatest industrial disasters in U.S. history occurred in 1947 in Texas, when a ship loaded with ammonium nitrate for use as fertilizer exploded and killed nearly 600 people.
Nitrogen monoxide (NO), commonly called nitric oxide, is a colorless gas under nor-mal conditions that can be produced in the laboratory by reaction of 6 M nitric acid with copper metal: When this reaction is run in the air, the nitric oxide is immediately oxidized to brown ni-trogen dioxide (NO2).
Although nitric oxide is toxic when inhaled, it has been shown to be produced in cer-tain tissues of the human body, where it behaves as a neurotransmitter. Current research indicates that nitric oxide plays a role in regulating blood pressure, blood clotting, and the muscle changes that allow erection of the penis in males.
Since the NO molecule has an odd number of electrons, it is most conveniently de-scribed in terms of the molecular orbital model. The molecular orbital energy-level dia-gram is shown in Fig. 20.8. Note that the NO molecule is paramagnetic and should have a bond order of 2.5, a prediction that is supported by experimental observations. Since the NO molecule has one high-energy electron, it is not surprising that it can be rather eas-ily oxidized to form NO, the nitrosyl ion. Because an antibonding electron is removed in going from NO to NO, the ion should have a stronger bond (the predicted bond or-der is 3) than the molecule. This is borne out by experiment. The bond lengths and bond 8H 1aq2 2NO3 1aq2 3Cu1s2 ¡ 3Cu21aq2 4H2O1l2 2NO1g2 NH4NO31s2 ¡ Heat N2O1g2 2H2O1g2 Do not attempt this experiment unless you have the proper safety equipment.
A copper penny reacts with nitric acid to produce NO gas, which is immediately oxidized in air to brown NO2.
FIGURE 20.8 The molecular orbital energy-level diagram for nitric oxide (NO). The bond order is 2.5, or (8 3)2.
σ2p 2p Molecular orbitals in NO π2p π2p 2p Atomic orbitals of oxygen Atomic orbitals of nitrogen σ2p π2p π2p 2s 2s σ2s σ2s E 20.2 The Chemistry of Nitrogen 911 energies for nitric oxide and the nitrosyl ion are shown in Table 20.3. The nitrosyl ion is formed when nitric oxide and nitrogen dioxide are dissolved in concentrated sulfuric acid: The ionic compound can be isolated from this solution.
Nitric oxide is thermodynamically unstable and decomposes to nitrous oxide and ni-trogen dioxide: Nitrogen dioxide (NO2) is also an odd-electron molecule and has a V-shaped structure.
The brown, paramagnetic NO2 molecule readily dimerizes to form dinitrogen tetroxide, which is diamagnetic and colorless. The value of the equilibrium constant is for this process at and since the dimerization is exothermic, K decreases as the temperature increases.
Molecular Orbital Description of NO Use the molecular orbital model to predict the bond order and magnetism of the ion.
Solution Using the energy-level diagram for the NO molecule in Fig. 20.8, we can see that has one more antibonding electron than NO. Thus there will be unpaired electrons in the two orbitals, and will be paramagnetic with a bond order of or 2.
Note that the bond in the ion is weaker than that in the NO molecule.
See Exercises 20.23 and 20.24.
The least common of the nitrogen oxides are dinitrogen trioxide (N2O3), a blue liq-uid that readily dissociates into gaseous nitric oxide and nitrogen dioxide, and dinitrogen pentoxide (N2O5), which under normal conditions is a solid that is best viewed as a mix-ture of NO2 and ions. Although N2O5 molecules do exist in the gas phase, they readily dissociate to nitrogen dioxide and oxygen: This reaction follows first-order kinetics, as was discussed in Section 12.4.
Oxyacids of Nitrogen Nitric acid is an important industrial chemical (almost 10 million tons is produced an-nually) used in the manufacture of many products, such as nitrogen-based explosives and ammonium nitrate for use as fertilizer.
2N2O51g2 ∆4NO21g2 O21g2 NO3 NO (8 4) 2, NO p2p NO NO 55°C, 1 2NO21g2 ∆N2O41g2 3NO1g2 ¡ N2O1g2 NO21g2 NOHSO4 NO1g2 NO21g2 3H2SO41aq2 ¡ 2NO 1aq2 3HSO4 1aq2 H3O 1aq2 TABLE 20.3 Comparison of the Bond Lengths and Bond Energies for Nitric Oxide and the Nitrosyl Ion NO NO Bond length (pm) 115 109 Bond energy (kJ/mol) 630 1020 Bond order (predicted by MO model) 2.5 3 Production of NO2 by power plants and automobiles leads to smog (see Section 12.8).
Sample Exercise 20.3 912 Chapter Twenty The Representative Elements: Groups 5A Through 8A Nitric acid is produced commercially by the oxidation of ammonia in the Ostwald process (see Fig. 20.9). In the first step of this process, ammonia is oxidized to nitric oxide: Although this reaction is highly exothermic, it is very slow at There is also a side reaction between nitric oxide and ammonia: which is particularly undesirable because it traps the nitrogen as very unreactive N2 mol-ecules. To speed up the desired reaction and minimize the effects of the competing reac-tion, the ammonia oxidation is carried out using a catalyst of a platinum–rhodium alloy heated to Under these conditions, there is a 97% conversion of the ammonia to nitric oxide.
In the second step, nitric oxide reacts with oxygen to produce nitrogen dioxide: This oxidation reaction has a rate that decreases with increasing temperature. Because of this very unusual behavior, the reaction is carried out at and is kept at this tem-perature by cooling with water.
The third step in the Ostwald process is the absorption of nitrogen dioxide by water: The gaseous NO produced in this reaction is recycled to be oxidized to NO2. The aqueous nitric acid from this process is about 50% HNO3 by mass, which can be increased to 68% by distillation to remove some of the water. The maximum concentration attainable this way is 68% because nitric acid and water form an azeotrope at this concentration. The solution 3NO21g2 H2O1l2 ¡ 2HNO31aq2 NO1g2 ¢H° 139 kJ 25°C 2NO1g2 O21g2 ¡ 2NO21g2 ¢H° 113 kJ 900°C.
4NH31g2 6NO1g2 ¡ 5N21g2 6H2O1g2 25°C.
4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 ¢H° 905 kJ CHEMICAL IMPACT Nitrous Oxide: Laughing Gas That Propels Whipped Cream and Cars N itrous oxide (N2O), more properly called dinitrogen monoxide, is a compound with many interesting uses.
It was discovered in 1772 by Joseph Priestley (who is also given credit for discovering oxygen gas), and its intoxicating effects were noted almost immediately. In 1798, the 20-year-old Humphry Davy became director of the Pneumatic Institute, which was set up to investigate the medical effects of various gases. Davy tested the ef-fects of N2O on himself, reporting that after inhaling 16 quarts of the gas in 7 minutes, he became “absolutely intoxicated.” Over the next century “laughing gas,” as nitrous oxide became known, was developed as an anesthetic, particularly for dental procedures. Nitrous oxide is still used as an anesthetic, although it has been largely replaced by more modern drugs.
One major use of nitrous oxide today is as the propellant in cans of “instant” whipped cream. The high solubility of N2O in the whipped cream mixture makes it an excellent candidate for pressurizing the cans of whipped cream.
Another current use of nitrous oxide is to produce “in-stant horsepower” for hot rods and street racers. Because the reaction of N2O with O2 to form NO actually absorbs heat, this reaction has a cooling effect when placed in the fuel mixture in an automobile engine. This cooling effect lowers combustion temperatures, thus allowing the fuel–air mixture to be significantly more dense (the density of a gas is in-versely proportional to temperature). This effect can produce a burst of additional power in excess of 200 horsepower. Be-cause engines are not designed to run steadily at such high power levels, the nitrous oxide is injected from a tank when extra power is desired.
NO byproduct NH3 Oxidation at 900°C with Pt – Rh catalyst NO Oxidation with O2 at 25°C NO2 Dissolved in H2O HNO3 FIGURE 20.9 The Ostwald process.
20.3 The Chemistry of Phosphorus 913 can be further concentrated to 95% HNO3 by treatment with concentrated sulfuric acid, which strongly absorbs water; H2SO4 is often used as a dehydrating (water-removing) agent.
Nitric acid is a colorless, fuming liquid with a pungent odor; it decom-poses in sunlight via the following reaction: As a result, nitric acid turns yellow as it ages because of the dissolved nitrogen dioxide.
The common laboratory reagent known as concentrated nitric acid is 15.9 M HNO3 (70.4% HNO3 by mass) and is a very strong oxidizing agent. The resonance structures and mo-lecular structure of HNO3 are shown in Fig. 20.10. Note that the hydrogen is bound to an oxygen atom rather than to nitrogen, as the formula suggests.
Nitric acid reacts with metal oxides, hydroxides, and carbonates and with other ionic compounds containing basic anions to form nitrate salts. For example, Nitrate salts are generally very soluble in water.
Nitrous acid (HNO2) is a weak acid, that forms pale yellow nitrite salts. In contrast to nitrates, which are often used as explosives, nitrites are quite stable even at high temperatures. Nitrites are usually pre-pared by bubbling equal numbers of moles of nitric oxide and nitrogen dioxide into the appropriate aqueous solution of a metal hydroxide. For example, 20.3 The Chemistry of Phosphorus Although phosphorus lies directly below nitrogen in Group 5A of the periodic table, its chemical properties are significantly different from those of nitrogen. The differences arise mainly from four factors: nitrogen’s ability to form much stronger bonds, the greater electronegativity of nitrogen, the larger size of the phosphorus atom, and the availability of empty valence d orbitals on phosphorus.
The chemical differences are apparent in the elemental forms of nitrogen and phosphorus. In contrast to the diatomic form of elemental nitrogen, which is stabilized by strong bonds, there are several solid forms of phosphorus, all containing aggregates of atoms. White phosphorus, which contains discrete tetrahedral P4 molecules [see Fig. 20.11(a)], is very reactive and bursts into flames on contact with air (it is said to be pyrophoric). To prevent this, white phosphorus is commonly stored under water. White phosphorus is quite toxic, and the P4 molecules are very damaging to tissue, particularly the cartilage and bones of the nose and jaw. The much less reactive forms known as black phosphorus and red phosphorus are network solids (see Section 10.5). Black phosphorus has a regular crystalline structure [Fig. 20.11(b)], but red phosphorus is amorphous and is thought to consist of chains of P4 units [Fig. 20.11(c)]. Red phosphorus can be obtained p p NO1g2 NO21g2 2NaOH1aq2 ¡ 2NaNO21aq2 H2O1l2 (NO2 ) HNO21aq2 ∆H 1aq2 NO2 1aq2 Ka 4.0 104 Ca1OH221s2 2HNO31aq2 ¡ Ca1NO3221aq2 2H2O1l2 4HNO31l2 ¡ hv 4NO21g2 2H2O1l2 O21g2 1bp 83°C2 O O O H ~105° 120° N O O N O H O O N O H (a) (b) FIGURE 20.10 (a) The molecular structure of HNO3.
(b) The resonance structures of HNO3.
An azeotrope is a solution that, like a pure liquid, distills at a constant tempera-ture without a change in composition.
Visualization: Barking Dogs: Reaction of Phosphorus 914 Chapter Twenty The Representative Elements: Groups 5A Through 8A by heating white phosphorus in the absence of oxygen at 1 atm. Black phosphorus is obtained from either white or red phosphorus by heating at high pressures.
Even though phosphorus has a lower electronegativity than nitrogen, it will form phos-phides (ionic substances containing the anion) such as Na3P and Ca3P2. Phosphide salts react vigorously with water to produce phosphine (PH3), a toxic, colorless gas: Phosphine is analogous to ammonia, although it is a much weaker base and much less soluble in water. Because phosphine has a relatively small affinity for protons, phosphonium (PH4 ) salts are very uncommon and not very stable—only PH4I, PH4Cl, and PH4Br are known.
Phosphine has the Lewis structure and a pyramidal molecular structure, as we would predict from the VSEPR model. How-ever, it has bond angles of 94 degrees, rather than 107 degrees, as found in the ammonia (Kb 1026) 2Na3P1s2 6H2O1l2 ¡ 2PH31g2 6Na 1aq2 6OH 1aq2 P 3 (a) (b) (c) FIGURE 20.11 (a) The P4 molecule found in white phos-phorus. (b) The crystalline network struc-ture of black phosphorus. (c) The chain structure of red phosphorus.
CHEMICAL IMPACT Phosphorus: An Illuminating Element T he elemental form of phosphorus was discovered by ac-cident in 1669 by German alchemist Henning Brand when he heated dried urine with sand (alchemists often investigated the chemistry of body fluids in an attempt to better understand the “stuff of life”). When Brand passed the resulting vapors through water, he was able to isolate the form of elemental phosphorus known as white phosphorus (contains P4 molecules). The name phosphorus is derived from the Latin phos, meaning “light,” and phorus, meaning “bearing.” It seems that when Brand stored the solid white phosphorus in a sealed bottle, it glowed in the dark! This effect—a glow that persists even after the light source has been removed—came to be called phosphorescence. Inter-estingly, the term phosphorescence is derived from the name of an element that really does not phosphoresce. The glow that Brand saw actually was the result of a reaction of oxygen from the air on the surface of white phosphorus. If isolated completely from air, phosphorus does not glow in the dark after being irradiated.
After its discovery, phosphorus became quite a novelty in the seventeenth century. People would deposit a film of phosphorus on their faces and hands so that they would glow in the dark. This fascination was short-lived—painful, slow-healing burns result from the spontaneous reaction of phosphorus with oxygen from the air.
The greatest consumer use of phosphorus compounds concerns the chemistry of matches. Two kinds of matches are currently available—strike-anywhere matches and safety An interesting reference to white phosphorus can be found in the Sherlock Holmes mystery, The Hound of the Baskervilles, where a large dog was coated with white phosphorus to scare Baskerville family members to death.
20.3 The Chemistry of Phosphorus 915 molecule. The reasons for this are complex, and we will simply regard phosphine as an exception to the simple version of the VSEPR model that we use.
Phosphorus Oxides and Oxyacids Phosphorus reacts with oxygen to form oxides in which it has oxidation states of 5 and 3. The oxide P4O6 is formed when elemental phosphorus is burned in a limited supply of oxygen, and P4O10 is produced when the oxygen is in excess. These oxides, as shown in Fig. 20.12, can be pictured as being constructed by adding oxygen atoms to the fun-damental P4 structure. The intermediate states, P4O7, P4O8, and P4O9, which contain one, two, and three terminal oxygen atoms, respectively, are also known.
matches. Both types of matches use phosphorus (in different forms) to help initiate a flame at the match head. The chemistry of matches is quite interesting. The tip of a strike-anywhere match is made from a mixture of powdered glass, binder, and tetraphosphorus trisulfide (P4S3). When the match is struck, friction ignites the combustion reaction of P4S3: The heat from this reaction causes an oxidizing agent such as potassium chlorate to decompose: which in turn causes solid sulfur to melt and react with oxygen, producing sulfur dioxide and more heat. This then ignites a paraffin wax that helps to “light” the wooden stem of the match.
The chemistry of a safety match is quite similar, but the location of the reactants is different. The phosphorus needed to initiate all the reactions is found on the striking surface of the box. Thus, in theory, a safety match is able to ignite 2KClO31s2 ¡ 2KCl1s2 3O21g2 P4S31s2 6O21g2 ¡ P4O61g2 3SO21g2 only when used with the box. For a safety match, the strik-ing surface contains red phosphorus, which is easily con-verted to white phosphorus by the friction of the match head on the striking surface. White phosphorus ignites sponta-neously in air and generates enough heat to initiate all the other reactions to ignite the match stem.
P41s21white2 5O21g2 ¡ P4O101s2 heat 4P 1red2 energy 1friction2 ¡ The phosphorus in safety matches helps ignite the flame in the match.
The terminal oxygens are the nonbridging oxygen atoms.
FIGURE 20.12 The structures of P4O6 and P4O10.
Limited O2 Excess O2 P4O6 P4O10 P P P P O P 916 Chapter Twenty The Representative Elements: Groups 5A Through 8A Tetraphosphorus decoxide (P4O10), which was formerly represented as P2O5, has a great affinity for water and thus is a powerful dehydrating agent. For example, it can be used to convert HNO3 and H2SO4 to their parent oxides, N2O5 and SO3, respectively.
When tetraphosphorus decoxide dissolves in water, phosphoric acid (H3PO4), also called orthophosphoric acid, is produced: Pure phosphoric acid is a white solid that melts at Aqueous phosphoric acid is a much weaker acid than nitric acid or sulfuric acid and is a poor oxidizing agent.
Phosphate minerals are the main source of phosphoric acid. Unlike nitrogen, phos-phorus is found in nature exclusively in a combined state, principally as the ion in phosphate rock, which is mainly calcium phosphate, Ca3(PO4)2, and fluorapatite, Ca5(PO4)3F. Fluorapatite can be converted to phosphoric acid by grinding up the phos-phate rock and forming a slurry with sulfuric acid: (A similar reaction can be written for the conversion of calcium phosphate.) The solid product called gypsum, is used to manufacture wallboard for the con-struction of buildings.
The process just described, called the wet process, produces only impure phosphoric acid. In another procedure, phosphate rock, sand (SiO2), and coke are heated in an elec-tric furnace to form white phosphorus: The white phosphorus obtained is burned in air to form tetraphosphorus decoxide, which is then combined with water to give phosphoric acid.
Phosphoric acid easily undergoes condensation reactions, where a molecule of wa-ter is eliminated in the joining of two molecules of acid: The product (H4P2O7) is called pyrophosphoric acid. Further heating produces polymers, such as tripolyphosphoric acid (H5P3O10), which has the structure The sodium salt of tripolyphosphoric acid is widely used in detergents because the anion can form complexes with metal ions such as Mg2 and Ca2, which would other-wise interfere with detergent action.
Structure of Phosphoric Acid What are the molecular structure and the hybridization of the central atom of the phosphoric acid molecule?
P3O10 5 O O O P O H H H O O P O H O O P O H 12Ca51PO423F 43SiO2 90C ¡ 9P4 90CO 2013CaO 2SiO22 3SiF4 CaSO4 2H2O, ¡ HF1aq2 5CaSO4 2H2O1s2 3H3PO41aq2 Ca51PO423F1s2 5H2SO41aq2 10H2O1l2 PO4 3 (Ka1 102) 42°C.
P4O101s2 6H2O1l2 ¡ 4H3PO41aq2 Sample Exercise 20.4 The mineral hydroxyapatite, Ca5(PO4)3OH, the principal component of tooth enamel, can be converted to fluorapatite by reac-tion with fluoride. Fluoride ions added to drinking water and toothpaste help pre-vent tooth decay because fluorapatite is less soluble in the acids of the mouth than hydroxyapatite.
20.3 The Chemistry of Phosphorus 917 Solution In the phosphoric acid molecule, the hydrogen atoms are attached to oxygens, and the Lewis structure is as shown below. Thus the phosphorus atom is surrounded by four ef-fective pairs, which are arranged tetrahedrally. The atom is described as sp3 hybridized.
See Exercises 20.27 and 20.28.
When P4O6 is placed in water, phosphorous acid (H3PO3) is formed [Fig. 20.13(a)].
Although the formula suggests a triprotic acid, phosphorous acid is a diprotic acid. The hydrogen atom bonded directly to the phosphorus atom is not acidic in aqueous solution; only those hydrogen atoms bonded to the oxygen atoms in H3PO3 can be released as protons.
A third oxyacid of phosphorus is hypophosphorous acid (H3PO2) [Fig. 20.13(b)], which is a monoprotic acid.
Phosphorus in Fertilizers Phosphorus is essential for plant growth. Although most soil contains large amounts of phosphorus, it is often present as insoluble minerals, which makes it inaccessible to plants.
Soluble phosphate fertilizers are manufactured by treating phosphate rock with sulfuric acid to make superphosphate of lime, a mixture of and If phosphate rock is treated with phosphoric acid, Ca(H2PO4)2, or triple phosphate, is pro-duced. The reaction of ammonia and phosphoric acid gives ammonium dihydrogen phos-phate (NH4H2PO4), a very efficient fertilizer because it furnishes both phosphorus and nitrogen.
Phosphorus Halides Phosphorus forms all possible halides of the general formulas PX3 and PX5, with the ex-ception of PI5. The PX3 molecule has the expected pyramidal structure [Fig. 20.14(a)].
Under normal conditions of temperature and pressure, PF3 is a colorless gas, PCl3 is a liq-uid PBr3 is a liquid and PI3 is an unstable red solid All the PX3 compounds react with water to produce phosphorous acid: In the gaseous and liquid states, the PX5 compounds have molecules with a trigonal bipyramidal structure [Fig. 20.14(b)]. However, PCl5 and PBr5 form ionic solids: Solid PX3 3H2O1l2 ¡ H3PO31aq2 3HX1aq2 61°C).
(mp (bp 175°C), (bp 74°C), Ca(H2PO4)2 H2O.
CaSO4 2H2O H H H O O O O P O O O P O H H H Lewis structure P O O H H O H (a) P O H H O H (b) FIGURE 20.13 (a) The structure of phosphorous acid (H3PO3). (b) The structure of hypophospho-rous acid (H3PO2).
FIGURE 20.14 Structures of the phosphorus halides.
(a) The PX3 compounds have pyramidal molecules. (b) The gaseous and liquid phases of the PX5 compounds are com-posed of trigonal bipyramidal molecules.
X X X P P (a) (b) X X X X X Unshared electron pair 918 Chapter Twenty The Representative Elements: Groups 5A Through 8A PCl5 contains a mixture of octahedral ions and tetrahedral PCl4 ions, and solid PBr5 appears to consist of PBr4 and ions.
The PX5 compounds react with water to form phosphoric acid: 20.4 The Group 6A Elements Although in Group 6A (Table 20.4) there is the usual tendency for metallic properties to increase going down the group, none of the Group 6A elements (valence-electron con-figuration n2np4) behaves as a typical metal. The most common chemical behavior of a Group 6A atom is to achieve a noble gas electron configuration by adding two electrons to become a anion in ionic compounds with metals. In fact, for most metals, the ox-ides and sulfides constitute the most common minerals.
The Group 6A elements can form covalent bonds with other nonmetals. For exam-ple, they combine with hydrogen to form a series of covalent hydrides of the general for-mula H2X. Those members of the group that have valence d orbitals available (all except oxygen) commonly form molecules in which they are surrounded by more than eight elec-trons. Examples are SF4, SF6, TeI4, and SeBr4.
In recent years there has been a growing interest in the chemistry of selenium, an el-ement found throughout the environment in trace amounts. Selenium’s toxicity has long been known, but recent medical studies have shown an inverse relationship between the incidence of cancer and the selenium levels in soil. It has been suggested that the result-ing greater dietary intake of selenium by people living in areas of relatively high levels of selenium somehow furnishes protection from cancer. These studies are only prelimi-nary, but selenium is known to be physiologically important (it is involved in the activity of vitamin E and certain enzymes) and selenium deficiency has been shown to be con-nected to the occurrence of congestive heart failure. Also of importance is the fact that selenium (along with tellurium) is a semiconductor and therefore finds some application in the electronics industry.
Polonium was discovered in 1898 by Marie and Pierre Curie in their search for the sources of radioactivity in pitchblende. Polonium has 27 isotopes and is highly toxic and very radioactive. It has been suggested that the isotope 210Po, a natural contaminant of to-bacco and an -particle emitter (see Section 18.1), might be at least partly responsible for the incidence of cancer in smokers.
a 2 PX5 4H2O1l2 ¡ H3PO41aq2 5HX1aq2 Br PCl6 Se Te Po O S 6A TABLE 20.4 Selected Physical Properties, Sources, and Methods of Preparation for the Group 6A Elements Electro-Radius Method of Element negativity of X2 (pm) Source Preparation Oxygen 3.5 140 Air Distillation from liquid air Sulfur 2.5 184 Sulfur deposits Melted with hot water and pumped to the surface Selenium 2.4 198 Impurity in Reduction of sulfide ores H2SeO4 with SO2 Tellurium 2.1 221 Nagyagite (mixed Reduction of ore sulfide and with SO2 telluride) Polonium 2.0 230 Pitchblende 20.5 The Chemistry of Oxygen 919 20.5 The Chemistry of Oxygen It is hard to overstate the importance of oxygen, the most abundant element in and near the earth’s crust. Oxygen is present in the atmosphere in oxygen gas and ozone; in soil and rocks in oxide, silicate, and carbonate minerals; in the oceans in water; and in our bodies in water and in a myriad of molecules. In addition, most of the energy we need to live and to run our civilization comes from the exothermic reactions of oxygen and carbon-containing molecules.
The most common elemental form of oxygen (O2) constitutes 21% of the volume of the earth’s atmosphere. Since nitrogen has a lower boiling point than oxygen, nitrogen can be boiled away from liquid air, leaving oxygen and small amounts of argon, another component of air. Liquid oxygen is a pale blue liquid that freezes at and boils at The paramagnetism of the O2 molecule can be demonstrated by pouring liquid oxygen between the poles of a strong magnet, where it “sticks” as it boils away (see Fig. 9.40). The paramagnetism of the O2 molecule can be accounted for by the molecu-lar orbital model (see Fig. 9.39), which also explains its bond strength.
The other form of elemental oxygen is ozone (O3), a molecule that can be represented by the resonance structures The bond angle in the O3 molecule is 117 degrees, in reasonable agreement with the pre-diction of the VSEPR model (three effective pairs require a trigonal planar arrangement).
That the bond angle is slightly less than 120 degrees can be explained by concluding that more space is required for the lone pair than for the bonding pairs.
Ozone can be prepared by passing an electric discharge through pure oxygen gas. The electrical energy disrupts the bonds in some O2 molecules to give oxygen atoms, which react with other O2 molecules to form O3. Ozone is much less stable than oxygen at and 1 atm. For example, for the equilibrium A pale blue, highly toxic gas, ozone is a much more powerful oxidizing agent than oxygen. Because of its oxidizing ability, ozone is being considered as a replacement for chlorine in municipal water purification. Chlorine leaves residues of chloro compounds, such as chloroform (CHCl3), which may cause cancer after long-term exposure. Although ozone effectively kills the bacteria in water, one problem with ozonolysis is that the wa-ter supply is not protected against recontamination, since virtually no ozone remains af-ter the initial treatment. In contrast, for chlorination, significant residual chlorine remains after treatment.
The oxidizing ability of ozone can be highly detrimental, especially when it is formed in the pollution from automobile exhausts (see Section 5.10).
Ozone exists naturally in the upper atmosphere of the earth. The ozone layer is espe-cially important because it absorbs ultraviolet light and thus acts as a screen to prevent this radiation, which can cause skin cancer, from penetrating to the earth’s surface. When an ozone molecule absorbs this energy, it splits into an oxygen molecule and an oxygen atom: If the oxygen molecule and atom collide, they will not stay together as ozone unless a “third body,” such as a nitrogen molecule, is present to help absorb the energy released in the bond formation. The third body absorbs the energy as kinetic energy; its temperature is increased. Therefore, the energy originally absorbed as ultraviolet radiation is eventually changed to thermal energy. Thus the ozone prevents the harmful high-energy ultraviolet light from reaching the earth.
O3 ¡ hv O2 O 3O21g2 ∆2O31g2 K 10 57 25°C 183°C.
219°C 920 Chapter Twenty The Representative Elements: Groups 5A Through 8A 20.6 The Chemistry of Sulfur Sulfur is found in nature both in large deposits of the free element and in widely distrib-uted ores, such as galena (PbS), cinnabar (HgS), pyrite (FeS2), gypsum epsomite and glauberite About 60% of the sulfur produced in the United States comes from the underground deposits of elemental sulfur found in Texas and Louisiana. This sulfur is recovered using the Frasch process developed by Herman Frasch in the 1890s. Superheated water is pumped into the deposit to melt the sulfur which is then forced to the surface by air pressure (see Fig. 20.15). The remaining 40% of sulfur produced in the United States is either a by-product of the purification of fossil fuels before combustion to prevent pollution or comes from the sulfur dioxide (SO2) scrubbed from the exhaust gases when sulfur-containing fuels are burned.
In contrast to oxygen, elemental sulfur exists as S2 molecules only in the gas phase at high temperatures. Because sulfur atoms form much stronger bonds than bonds, S2 is less stable at than larger aggregates such as S6 and S8 rings and Sn chains (Fig. 20.16). The most stable form of sulfur at and 1 atm is called rhombic sulfur [see Fig. 20.17(a)], which contains stacked S8 rings. If rhombic sulfur is melted and heated to it forms monoclinic sulfur as it cools slowly [Fig. 20.17(b)]. This form also contains S8 rings, but the rings are stacked differently than they are in rhombic sulfur.
As sulfur is heated beyond its melting point, a relatively nonviscous liquid containing S8 rings forms initially. With continued heating, the liquid becomes highly viscous as the rings first break and then link up to form long chains. Further heating lowers the viscosity 120°C, 25°C 25°C p s (mp 113°C), (Na2SO4 CaSO4).
(MgSO4 7H2O), (CaSO4 2H2O), The scrubbing of sulfur dioxide from exhaust gases was discussed in Section 5.10.
Air Superheated water Molten sulfur Molten sulfur (a) (b) FIGURE 20.15 The Frasch process for recovering sulfur from underground deposits.
FIGURE 20.16 (a) The S8 molecule. (b) Chains of sulfur atoms in viscous liquid sulfur. The chains may contain as many as 10,000 atoms.
20.6 The Chemistry of Sulfur 921 because the long chains are broken down as the energetic sulfur atoms break loose. If the liquid is suddenly cooled, a substance called plastic sulfur, which contains Sn chains and has rubberlike qualities, is formed. Eventually, this form reverts back to the more stable S8 rings.
Sulfur Oxides From its position below oxygen in the periodic table, we might expect the simplest stable oxide of sulfur to have the formula SO. However, sulfur monoxide, which can be produced in small amounts when gaseous sulfur dioxide (SO2) is subjected to an electrical discharge, is very unstable. The difference in the stabilities of the O2 and SO molecules probably re-flects the stronger bonding between oxygen atoms than between sulfur and oxygen atoms.
Sulfur burns in air with a bright blue flame to give sulfur dioxide (SO2), a colorless gas with a pungent odor, which condenses to a liquid at and 1 atm. Sulfur diox-ide is a very effective antibacterial agent and is often used to preserve stored fruit. Its structure is given in Fig. 20.18.
Sulfur dioxide reacts with oxygen to produce sulfur trioxide (SO3): However, this reaction is very slow in the absence of a catalyst. One of the mysteries during early research on air pollution was how the sulfur dioxide produced from the combustion of sulfur-containing fuels is so rapidly converted to sulfur trioxide. It is now known that dust and other particles can act as heterogeneous catalysts for this process (see Section 12.8). In the preparation of sulfur trioxide for the manufacture of sulfuric acid, a platinum or vana-dium(V) oxide (V2O5) catalyst is used, and the reaction is carried out at even though this temperature decreases the value of the equilibrium constant for this exothermic reaction.
The bonding in the SO3 molecule is usually described in terms of the resonance struc-tures shown in Fig. 20.19. The molecule is trigonal planar, as predicted by the VSEPR 500°C, 2SO21g2 O21g2 ¡ 2SO31g2 10°C p (a) (b) FIGURE 20.17 (a) Crystals of rhombic sulfur. (b) Crystals of monoclinic sulfur.
Pouring liquid sulfur into water to produce plastic sulfur.
FIGURE 20.18 (a) Two of the resonance structures for SO2. (b) SO2 is a bent molecule with a 119-degree bond angle, as predicted by the VSEPR model.
FIGURE 20.19 (a) Three of the resonance structures of SO3. (b) A resonance structure with three double bonds. (c) SO3 is a planar molecule with 120-degree bond angles.
922 Chapter Twenty The Representative Elements: Groups 5A Through 8A model. Sulfur trioxide is a corrosive gas with a choking odor that forms white fumes of sulfuric acid when it reacts with moisture in the air. Thus sulfur trioxide and nitrogen dioxide, which reacts with water to form a mixture of nitrous and nitric acids, are the major culprits in the formation of acid rain.
Sulfur trioxide condenses to a colorless liquid at and freezes at to give three solid forms, one containing S3O9 rings and the other two containing (SO3)x chains (Fig. 20.20).
Oxyacids of Sulfur Sulfur dioxide dissolves in water to form an acidic solution. The reaction is often repre-sented as where H2SO3 is called sulfurous acid. However, very little H2SO3 actually exists in the solution. The major form of sulfur dioxide in water is SO2, and the acid dissociation equi-libria are best represented as This situation is analogous to the behavior of carbon dioxide in water (see Section 14.7).
Although H2SO3 cannot be isolated, salts of (sulfites) and (hydrogen sulfites) are well known.
Sulfur trioxide reacts violently with water to produce the diprotic acid sulfuric acid: Manufactured in greater amounts than any other chemical, sulfuric acid is usually pro-duced by the contact process, which is described at the end of Chapter 3. About 60% of the sulfuric acid manufactured in the United States is used to produce fertilizers from phosphate rock (see Section 20.3). The other 40% is used in lead storage batteries and in petroleum refining, in steel manufacturing, and for various purposes in most of the chem-ical industries.
Because sulfuric acid has a high affinity for water, it is often used as a dehydrating agent. Gases that do not react with sulfuric acid, such as oxygen, nitrogen, and carbon dioxide, are often dried by bubbling them through concentrated solutions of the acid.
SO31g2 H2O1l2 ¡ H2SO41aq2 HSO3 SO3 2 HSO3 1aq2 ∆H 1aq2 SO3 21aq2 Ka2 1.0 107 SO21aq2 H2O1l2 ∆H 1aq2 HSO3 1aq2 Ka1 1.5 102 SO21g2 H2O1l2 ¡ H2SO31aq2 16.8°C 44.5°C (left) A sulfur deposit. (right) Melted sulfur obtained from underground deposits by the Frasch process.
(a) (b) O S FIGURE 20.20 Different structures for solid SO3. (a) S3O9 rings. (b) (SO3)x chains. In both cases the sulfur atoms are surrounded by a tetrahe-dral arrangement of oxygen atoms.
20.6 The Chemistry of Sulfur 923 Sulfuric acid is such a powerful dehydrating agent that it will remove hydrogen and oxygen from a substance in a 2:1 ratio even when the substance contains no molecular water. For example, concentrated sulfuric acid reacts vigorously with common table sugar (sucrose), leaving a charred mass of carbon (see Fig. 20.21): Sucrose Sulfuric acid is a moderately strong oxidizing agent, especially at high temperatures.
Hot concentrated sulfuric acid oxidizes bromide or iodide ions to elemental bromine or iodine. For example, Hot sulfuric acid attacks copper metal: The cold acid does not react with copper.
Other Compounds of Sulfur Sulfur reacts with both metals and nonmetals to form a wide variety of compounds in which it has a 6, 4, 2, 0, or oxidation state (see Table 20.5). The oxidation state occurs in the metal sulfides and in hydrogen sulfide (H2S), a toxic, foul-smelling gas that acts as a diprotic acid when dissolved in water. Hydrogen sulfide is a strong reduc-ing agent in aqueous solution, producing a milky-looking suspension of finely divided sul-fur as one of the products. For example, hydrogen sulfide reacts with chlorine in aqueous solution as follows: p Milky suspension of sulfur H2S1g2 Cl21aq2 ¡ 2H 1aq2 2Cl 1aq2 S1s2 2 2 Cu1s2 2H2SO41aq2 ¡ CuSO41aq2 2H2O1l2 SO21aq2 2I1aq2 3H2SO41aq2 ¡ I21aq2 SO21aq2 2H2O1l2 2HSO4 1aq2 C12H22O111s2 11H2SO41conc2 ¡ 12C1s2 11H2SO4 H2O1l2 FIGURE 20.21 (a) A beaker of sucrose (table sugar). (b) Concentrated sulfuric acid reacts with the sucrose to produce a column of carbon (c), accompanied by an intense burnt-sugar odor.
(a) (b) (c) The acidic properties of sulfuric acid solutions are discussed in Section 14.7.
924 Chapter Twenty The Representative Elements: Groups 5A Through 8A Sulfur also forms the thiosulfate ion which has the Lewis structure Note that this anion can be viewed as a sulfate ion in which one of the oxygen atoms has been replaced by sulfur, which is reflected in the name thiosulfate. The thiosulfate ion can be formed by heating sulfur with a sulfite salt in aqueous solution: One important use of thiosulfate ion is in photography, where dissolves solid sil-ver bromide by forming a complex with the Ag ion (see the Chemical Impact in Section 20.7).
Thiosulfate ion is also a good reducing agent and is often used to analyze for iodine: where is called the tetrathionate ion.
Sulfur reacts with the halogens to form a variety of compounds, such as S2Cl2, SF4, SF6, and S2F10. The structures of these molecules are shown in Fig. 20.22.
20.7 The Group 7A Elements In our coverage of the representative elements, we have progressed from groups of metal-lic elements (Groups 1A and 2A), through groups in which the lighter members are non-metals and the heavier members are metals (Groups 3A, 4A, and 5A), to a group of all nonmetals (Group 6A—although some might prefer to call polonium a metal). The S2O6 2 2S2O3 21aq2 I21aq2 ¡ S4O6 21aq2 2I 1aq2 S2O3 2 S1s2 SO3 21aq2 ¡ S2O3 21aq2 (S2O3 2), TABLE 20.5 Common Compounds of Sulfur with Various Oxidation States Oxidation State of Sulfur Compounds 6 SO3, H2SO4, SF6 4 SO2, 2 SCl2 S8 and all other forms of elemental sulfur H2S, S 2 2 0 HSO3 , SO3 2, SF4 SO4 2, The preparation of sulfur trioxide provides an example of the compromise that often must be made between thermodynamics and kinetics.
The prefix thio means “sulfur.” FIGURE 20.22 The structures of (a) SF4, (b) SF6, (c) S2F10, and (d) S2Cl2.
(a) (b) (c) (d) Br I At F Cl 7A 20.7 The Group 7A Elements 925 Group 7A elements, the halogens (with the valence-electron configuration ns2np5), are also all nonmetals whose properties generally vary smoothly going down the group. The only notable exceptions are the unexpectedly low values for the electron affinity of fluor-ine and the bond energy of the F2 molecule (see Section 19.1). Table 20.6 summarizes the trends in some physical properties of the halogens.
Because of their high reactivities, the halogens are not found as the free elements in nature. Instead, they are found as halide ions in various minerals and in seawater (see Table 20.7).
Although astatine is a member of Group 7A, its chemistry is of no practical impor-tance because all its known isotopes are radioactive. The longest-lived isotope, 210At, has a half-life of only 8.3 hours.
The halogens, particularly fluorine, have very high electronegativity values (Table 20.6).
They tend to form polar covalent bonds with other nonmetals and ionic bonds with met-als in their lower oxidation states. When a metal ion is in a higher oxidation state, such as 3 or 4, the metal–halogen bonds are polar covalent ones. For example, TiCl4 and SnCl4 are both covalent compounds that are liquids under normal conditions.
Hydrogen Halides The hydrogen halides can be prepared by a reaction of the elements: This reaction occurs with explosive vigor when fluorine and hydrogen are mixed. On the other hand, hydrogen and chlorine can coexist with little apparent reaction for relatively H21g2 X21g2 ¡ 2HX1g2 1X 2 TABLE 20.6 Trends in Selected Physical Properties of the Group 7A Elements Electro-Radius ° (V) for Bond Energy of Element negativity of X (pm) X2 2e S 2X X2(kJ/mol) Fluorine 4.0 136 2.87 154 Chlorine 3.0 181 1.36 239 Bromine 2.8 185 1.09 193 Iodine 2.5 216 0.54 149 Astatine 2.2 — — — e TABLE 20.7 Some Physical Properties, Sources, and Methods of Preparation for the Group 7A Elements Color and Percentage of Melting Boiling Method of Element State Earth’s Crust Point (°C) Point (°C) Sources Preparation Fluorine Pale yellow 0.07 Fluorospar (CaF2), Electrolysis of gas cryolite (Na3AlF6), molten KHF2 fluorapatite (Ca5(PO4)3F) Chlorine Yellow-green 0.14 Rock salt (NaCl), Electrolysis of gas halite (NaCl), aqueous NaCl sylvite (KCl) Bromine Red-brown liquid 59 Seawater, brine wells Oxidation of by Iodine Violet-black 113 184 Seaweed, brine wells Oxidation of solid by electrolysis or MnO2 I 3 10 5 Cl2 Br 7.3 2.5 10 4 34 101 188 220 Chlorine, bromine, and iodine.
926 Chapter Twenty The Representative Elements: Groups 5A Through 8A long periods in the dark. However, ultraviolet light causes an explosively fast reaction, and this is the basis of a popular lecture demonstration, the “hydrogen–chlorine cannon.” Bromine and iodine also react with hydrogen, but more slowly.
The hydrogen halides also can be prepared by treating halide salts with acid. For ex-ample, hydrogen fluoride and hydrogen chloride can be prepared as follows: Sulfuric acid is capable of oxidizing to Br2 and to I2 and so cannot be used to prepare hydrogen bromide and hydrogen iodide. However, phosphoric acid, a nonoxi-dizing acid, can be used to treat bromides and iodides to form the corresponding hydro-gen halides.
Some physical properties of the hydrogen halides are listed in Table 20.8. Note the very high boiling point for hydrogen fluoride, which results from extensive hydrogen bond-ing among the very polar HF molecules (Fig. 20.23). Fluoride ion has such a high affin-ity for protons that in concentrated aqueous solutions of hydrogen fluoride, the ion exists, in which an H ion is centered between two ions.
When dissolved in water, the hydrogen halides behave as acids, and all except hy-drogen fluoride are completely dissociated. Because water is a much stronger base than F [F¬H¬F] I Br 2NaCl1s2 H2SO41aq2 ¡ Na2SO41s2 2HCl1g2 CaF21s2 H2SO41aq2 ¡ CaSO41s2 2HF1g2 CHEMICAL IMPACT Photography I n black-and-white photography, light from an object is fo-cused onto a special paper containing an emulsion of solid silver bromide. Silver salts turn dark when exposed to light because the radiant energy stimulates the transfer of an elec-tron to the Ag ion, forming an atom of elemental silver.
When photographic paper (film) is exposed to light, the ar-eas exposed to the brightest light form the most silver atoms.
The next step is the application of a chemical reducing agent to the film, a process called developing. The real advantage of silver-based films is that the silver atoms already present from exposure to light catalyze the reduction of millions of Ag ions in the immediate vicinity in the developing process. Thus the effect of exposure to light is greatly in-tensified in this chemical reduction process. Once the image has been developed, the unchanged solid silver bromide must be removed so that the film is no longer light-sensitive and the image is fixed. A solution of sodium thiosulfate (called hypo) is used in this fixing process: After the excess silver bromide is dissolved and washed away, the fixed image (the negative) is ready to produce the positive print. By shining light through the negative onto a fresh sheet of film and repeating the developing and fixing processes, a black-and-white photograph can be produced.
AgBr1s2 2S2O3 21aq2 ¡ Ag1S2O322 31aq2 Br 1aq2 Positive and negative images.
A candle burning in an atmosphere of Cl2(g).
The exothermic reaction, which involves breaking COC and COH bonds in the wax and forming COCl bonds in their places, produces enough heat to make the gases in the region incandescent (a flame results).
20.7 The Group 7A Elements 927 or ion, the acid strengths of HCl, HBr, and HI cannot be differentiated in water. However, in a less basic solvent, such as glacial (pure) acetic acid, the acids show different strengths of the order Strongest Weakest acid acid To see why hydrogen fluoride is the only weak acid in water among the HX mole-cules, let’s consider the dissociation equilibrium where from a thermodynamic point of view. Recall that acid strength is reflected by the magni-tude of small Ka value means a weak acid. Also recall that the value of an equi-librium constant depends on the standard free energy change for the reaction as follows: As becomes more negative, K becomes larger; a decrease in free energy favors a given reaction. As we saw in Chapter 16, free energy depends on enthalpy, entropy, and temperature. For a process at constant temperature, Thus, to explain the various acid strengths of the hydrogen halides, we must focus on the factors that determine and for the acid dissociation reaction.
What energy terms are important in determining for the dissociation of HX in water? (Keep in mind that large positive contributions to the value of will tend to make more highly positive, Ka smaller, and the acid weaker.) One important factor is certainly the bond strength. Note from Table 20.8 that the bond is much stronger than the other bonds. This factor tends to make HF a weaker acid than the others.
Another important contribution to is the enthalpy of hydration (see Section 11.2) of (see Table 20.9). As we would expect, the smallest of the halide ions, has the most negative value—its hydration is the most exothermic. This term favors the dissoci-ation of HF into its ions more so than it does for the other HX molecules.
So far we have two conflicting factors: The large HF bond energy tends to make HF a weaker acid than the other hydrogen halides, but the enthalpy of hydration favors the dissociation of HF more than that of the others. In comparing data for HF and HCl, the difference in bond energy (138 kJ/mol) is slightly smaller than the difference in the en-thalpies of hydration for the anions (144 kJ/mol). If these were the only important fac-tors, HF should be a stronger acid than HCl because the large enthalpy of hydration of more than makes up for the large HF bond strength.
F F, X ¢H° H¬X H¬F H¬X ¢G° ¢H° ¢H° ¢S° ¢H° ¢G° ¢H°T¢S° ¢G° ¢G° RT ln1K2 Ka¬a Ka 3H 4 3X 4 3HX4 HX1aq2 ∆H 1aq2 X 1aq2 H¬I 7 H¬Br 7 H¬Cl H¬F I Br , Cl, TABLE 20.8 Some Physical Properties of the Hydrogen Halides Melting Boiling HOX Bond HX Point (°C) Point (°C) Energy (kJ/mol) HF 20 565 HCl 427 HBr 363 HI 295 35 51 67 87 85 114 83 FIGURE 20.23 The hydrogen bonding among HF mole-cules in liquid hydrogen fluoride.
When H2O molecules cluster around an ion, an ordering effect occurs, and is negative.
¢S° hyd 928 Chapter Twenty The Representative Elements: Groups 5A Through 8A As it turns out, the deciding factor is entropy. Note from Table 20.9 that the entropy of hydration for is much more negative than for the other halides because of the high degree of ordering that occurs as the water molecules associate with the small ion.
Remember that a negative change in entropy is unfavorable. Thus, although the enthalpy of hydration favors dissociation of HF, the entropy of hydration strongly opposes it.
When all these factors are taken into account, for the dissociation of HF in wa-ter is positive; that is, Ka is small. In contrast, for dissociation of the other HX mol-ecules in water is negative (Ka is large). This example illustrates the complexity of the processes that occur in aqueous solutions and the importance of entropy effects in that medium.
In practical terms, hydrochloric acid is the most important of the hydrohalic acids, the aqueous solutions of the hydrogen halides. About 3 million tons of hydrochloric acid is produced annually for use in cleaning steel before galvanizing and in the manufacture of many other chemicals.
Hydrofluoric acid is used to etch glass by reacting with the silica in glass to form the volatile gas SiF4: Oxyacids and Oxyanions All the halogens except fluorine combine with various numbers of oxygen atoms to form a series of oxyacids, as shown in Table 20.10. The strengths of these acids vary in direct proportion to the number of oxygen atoms attached to the halogen, as we discussed in Section 14.9.
The only member of the chlorine series that has been obtained in the pure state is perchloric acid (HOClO3), a strong acid and powerful oxidizing agent. Because perchloric acid reacts explosively with many organic materials, it must be handled with great caution.
The other oxyacids of chlorine are known only in solution, although salts containing their anions are well known (Fig. 20.24).
SiO21s2 4HF1aq2 ¡ SiF41g2 2H2O1l2 ¢G° ¢G° F F TABLE 20.9 The Enthalpies and Entropies of Hydration for the Halide Ions X (kJ/mol) (J/K mol) 64 291 I 81 334 Br 96 366 Cl 159 510 F S H X(g) ¡ H2O X(aq) TABLE 20.10 The Known Oxyacids of the Halogens Oxidation State General Name General Name of Halogen Fluorine Chlorine Bromine Iodine of Acids of Salts 1 HOF HOCl HOBr HOI Hypohalous acid Hypohalites, MOX 3 HOClO Halous acid Halites, MXO2 5 HOClO2 HOBrO2 HOIO2 Halic acid Halates, MXO3 7 HOClO3 HOBrO3 HOIO3 Perhalic acid Perhalates, MXO4 Iodine also forms H4I2O9 (mesodiperiodic acid) and H5IO6 (paraperiodic acid).
Compound is unknown.
Hydration becomes more exothermic as the charge density of an ion icreases.
Thus, for ions of a given charge, the smallest is most strongly hydrated.
Stomach acid is 0.1 M HCl.
This Steuben glass design was etched using hydrofluoric acid.
20.7 The Group 7A Elements 929 Hypochlorous acid (HOCl) is formed when chlorine gas is dissolved in cold water: Note that in this reaction chlorine is both oxidized (from 0 in Cl2 to 1 in HOCl) and reduced (from 0 in Cl2 to in ). Such a reaction, where a given element is both ox-idized and reduced, is called a disproportionation reaction. Hypochlorous acid and its salts are strong oxidizing agents, and solutions of them are widely used as household bleaches and disinfectants.
Chlorate salts, such as KClO3, are also strong oxidizing agents and are used as weed killers and as oxidizers in fireworks (see Chapter 7) and explosives.
Fluorine forms only one oxyacid, hypofluorous acid (HOF), but at least two oxides.
When fluorine gas is bubbled into a dilute solution of sodium hydroxide, the compound oxygen difluoride (OF2) is formed: Oxygen difluoride is a pale yellow gas which is a strong oxidizing agent.
The oxide dioxygen difluoride (O2F2) is an orange solid that can be prepared by passing an electric discharge through an equimolar mixture of fluorine and oxygen gases: Bonding Description of OF2 Give the Lewis structure, molecular structure, and hybridization of the oxygen atom for OF2.
Solution The OF2 molecule has 20 valence electrons and the Lewis structure is The four effective pairs around the oxygen are arranged tetrahedrally. Therefore, the oxy-gen atom is sp3 hybridized. The molecule is bent (V-shaped), and the bond angle is pre-dicted to be smaller than 109.5 degrees because of the lone-pair repulsions.
See Exercise 20.37.
F21g2 O21g2 ¡ O2F21s2 1bp 145°C2, 4F21g2 3H2O1l2 ¡ 6HF1aq2 OF21g2 O21g2 Cl 1 Cl21aq2 H2O1l2 ∆HOCl1aq2 H 1aq2 Cl 1aq2 FIGURE 20.24 The structures of the oxychloro anions.
The name for OF2 is oxygen difluoride rather than difluorine oxide because fluorine has a higher electronegativity than oxygen and thus is named as if it were an anion.
Hypochlorite ion, OCl– Chlorate ion, ClO3– Chlorite ion, ClO2– Perchlorate ion, ClO4– O Cl Cl O O O O Cl O O O Cl O O Cl O Electric discharge Sample Exercise 20.5 930 Chapter Twenty The Representative Elements: Groups 5A Through 8A Other Halogen Compounds The halogens react readily with most nonmetals to form a variety of compounds, some of which are shown in Table 20.11.
Halogens react with each other to form interhalogen compounds with the general formula ABn, where n is typically 1, 3, 5, or 7 and A is the larger of the two halogens. The structures of these compounds (see Fig. 20.25) are predicted accurately by the VSEPR model. The interhalogens are volatile, highly reactive compounds that act as strong oxidizing agents. They react readily with water, forming the halide ion of the more electronegative halogen and the hypohalous acid of the less electronegative halogen. For example, ICl1s2 H2O1l2 ¡ H 1aq2 Cl 1aq2 HOI1aq2 FIGURE 20.25 The idealized structures of the interhalo-gens ClF3 and lF5. In reality, the lone pairs cause the bond angles to be slightly less than 90 degrees.
F Cl F F F Unshared electron pairs F I F F F F Unshared electron pair (b) IF5 is square pyramidal (a) ClF3 is “T-shaped” TABLE 20.11 Some Compounds of the Halogens with Nonmetals Compounds with Compounds with Compounds with Compounds with Compounds with Group 3A Nonmetals Group 4A Nonmetals Group 5A Nonmetals Group 6A Nonmetals Group 7A Nonmetals BX3 (X F, Cl, Br, I) CX4 (X F, Cl, Br, I) NX3 (X F, Cl, Br, I) OF2 ICl N2F4 O2F2 IBr SiF4 OCl2 BrF PX3 (X F, Cl, Br, I) OBr2 BrCl SiCl4 PF5 ClF PCl5 SF2 GeF4 PBr5 SCl2 ClF3 S2F2 BrF3 GeCl4 AsF3 S2Cl2 ICl3 AsF5 SF4 IF3 SCl4 SbF3 SF6 ClF5 SbF5 BrF5 SeF4 IF5 SeF6 SeCl2 IF7 SeCl4 SeBr4 TeF4 TeF6 TeCl4 TeBr4 TeI4 GeF6 2 SiF6 2 BF4 20.8 The Group 8A Elements 931 20.8 The Group 8A Elements The Group 8A elements, the noble gases, are characterized by filled s and p valence or-bitals (electron configuration of 2s2 for helium and ns2np6 for the others). Because of this, these elements are very unreactive. In fact, no noble gas compounds were known 50 years ago. Selected properties of the Group 8A elements are summarized in Table 20.12.
Helium was identified by its characteristic emission spectrum as a component of the sun before it was found on earth. The major sources of helium on earth are natural gas deposits, where helium was formed from -particle decay of radioactive elements. The particle is a helium nucleus that can easily pick up electrons from the environment to form a helium atom. Although helium forms no compounds, it is an important substance that is used as a coolant, as a pressurizing gas for rocket fuels, as a diluent in the gases used for deep-sea diving and spaceship atmospheres, and as the gas in lighter-than-air airships (blimps).
a a CHEMICAL IMPACT Automatic Sunglasses S unglasses can be troublesome. It seems they are always getting lost or sat on. One solution to this problem for people who wear glasses is photochromic glass—glass that darkens in response to intense light. Recall that glass is a complex, noncrystalline material that is composed of poly-meric silicates (see Chapter 10). Of course, glass transmits visible light—its transparency is its most useful property.
Glass can be made photochromic by adding tiny silver chloride crystals that get trapped in the glass matrix as the glass solidifies. Silver chloride has the unusual property of darkening when struck by light—the property that makes the silver halide salts so useful for photographic films. This darkening occurs because light causes an electron transfer from to Ag in the silver chloride crystal, forming a sil-ver atom and a chlorine atom. The silver atoms formed in this way tend to migrate to the surface of the silver chloride crystal, where they aggregate to form a tiny crystal of sil-ver metal, which is opaque to light.
In photography the image defined by the grains of sil-ver is fixed by chemical treatment so that it remains per-manent. However, in photochromic glass this process must be reversible—the glass must become fully transparent again when the person goes back indoors. The secret to the re-versibility of photochromic glass is the presence of Cu ions.
The added Cu ions serve two important functions. First, they reduce the Cl atoms formed in the light-induced reac-tion. This prevents them from escaping from the crystal: Cl Cu ¡ Cu 2 Cl Ag Cl ¡ Ag Cl Cl Second, when the exposure to intense light ends (the person goes indoors), the Cu2 ions migrate to the surface of the sil-ver chloride crystal, where they accept electrons from silver atoms as the tiny crystal of silver atoms disintegrates: The Ag ions are re-formed in this way, then return to their places in the silver chloride crystal, making the glass trans-parent once again.
Typical photochromic glass decreases to about 20% transmittance (transmits 20% of the light that strikes it) in strong sunlight, and then over a period of a few minutes re-turns to about 80% transmittance indoors (normal glass has 92% transmittance).
Cu 2 Ag ¡ Cu Ag Glasses with photosensitive lenses. The right lens has been exposed to light and the left one has not.
Kr Xe Rn Ne Ar He 8A hv 932 Chapter Twenty The Representative Elements: Groups 5A Through 8A Like helium, neon and argon form no compounds but are used extensively. For ex-ample, neon is employed in luminescent lighting (neon signs), and argon is used to pro-vide the noncorrosive atmosphere in incandescent light bulbs, which prolongs the life of the tungsten filament.
Of the Group 8A elements, only krypton and xenon have been observed to form chem-ical compounds. The first of these was prepared in 1962 by Neil Bartlett, an English chemist who used Xe(g) and PtF6(g) to make the ionic compound with the empirical for-mula XePtF6.
Less than a year after Bartlett’s report of XePtF6, a group at Argonne National Lab-oratory near Chicago prepared xenon tetrafluoride by reaction of xenon and fluorine gases in a nickel reaction vessel at and 6 atm: Xenon tetrafluoride forms stable colorless crystals. Two other xenon fluorides, XeF2 and XeF6, were synthesized by the group at Argonne, and a highly explosive xenon oxide (XeO3) also was found. The xenon fluorides react with water to form hydrogen fluoride and oxycompounds. For example, In the past 40 years, other xenon compounds have been prepared—for example, XeO4 (explosive), XeO2F4, XeO2F2, and XeO3F2. These compounds contain discrete molecules with covalent bonds between the xenon and the other atoms. The structures of some of these xenon compounds are summarized in Fig. 20.26. A few compounds of krypton, such as KrF2 and KrF4, also have been observed. There is evidence that radon also reacts with fluorine, but the radioactivity of radon makes its chemistry very difficult to study.
The Structure of XeF6 Use the VSEPR model to predict whether XeF6 has an octahedral structure.
Solution The XeF6 molecule contains 50 valence electrons and the Lewis structure is The xenon atom has seven pairs of electrons around it (one lone pair and six bonding pairs), one more pair than can be accommodated in an octahedral arrangement. Thus XeF6 [8 6(7)] XeF61s2 H2O1l2 ¡ XeOF41aq2 2HF1g2 XeF61s2 3H2O1l2 ¡ XeO31aq2 6HF1aq2 Xe1g2 2F21g2 ¡ XeF41s2 400°C TABLE 20.12 Selected Properties of Group 8A Elements Atmospheric Melting Boiling Abundance Examples of Element Point (°C) Point (°C) (% by Volume) Compounds Helium None Neon None Argon None Krypton KrF2 Xenon XeF4, XeO3, XeF6 9 10 6 107 112 1 10 4 153 157 9 10 1 186 189 1 10 3 246 249 5 10 4 269 270 Neon, a noble gas, is used in luminescent lighting (neon signs).
Sample Exercise 20.6 For Review 933 will not have an octahedral structure, but should be distorted from this geometry by the extra electron pair. There is experimental evidence that the structure of XeF6 is not octahedral.
See Exercises 20.43 and 20.44.
FIGURE 20.26 The structures of several known xenon compounds.
F F XeF2 Linear XeO2F2 Distorted tetrahedron XeO3 Pyramidal XeO4 Tetrahedral XeO2F4 Octahedral XeO3F2 Trigonal bipyramid XeF4 Square planar F F For Review Group 5A Elements show a wide variety of chemical properties • Nitrogen and phosphorus are nonmetals • Antimony and bismuth tend to be metallic, although no ionic compounds con-taining Sb5 and Bi5 are known; the compounds containing Sb(V) and Bi(V) are molecular rather than ionic • All group members except N form molecules with five covalent bonds • The ability to form bonds decreases dramatically after N Chemistry of nitrogen • Most nitrogen-containing compounds decompose exothermically, forming the very stable N2 molecule, which explains the power of nitrogen-based explosives • The nitrogen cycle, which consists of a series of steps, shows how nitrogen is cycled in the natural environment • Nitrogen fixation changes the N2 in air into compounds useful to plants • The Haber process is a synthetic method of nitrogen fixation • In the natural world, nitrogen fixation occurs through nitrogen-fixing bacteria in the root nodules of certain plants and through lightning in the atmosphere • Ammonia is the most important hydride of nitrogen • Contains pyramidal NH3 molecules • Widely used as a fertilizer • Hydrazine (N2H4) is a powerful reducing agent p Key Terms Section 20.2 Haber process nitrogen fixation nitrogen-fixing bacteria denitrification nitrogen cycle ammonia hydrazine nitric acid Ostwald process Section 20.3 phosphoric (orthophosphoric) acid condensation reactions phosphorous acid superphosphate of lime Section 20.5 ozone ozonolysis Section 20.6 Frasch process sulfuric acid thiosulfate ion 934 Chapter Twenty The Representative Elements: Groups 5A Through 8A • Nitrogen forms a series of oxides including N2O, NO, NO2, and N2O5 • Nitric acid (HNO3) is a very important strong acid manufactured by the Ostwald process Chemistry of phosphorus • Exists in three elemental forms: white (contains P4 molecules), red, and black • Phosphine (PH3) has bond angles close to 90 degrees • Phosphorus forms oxides including P4O6 and P4O10 (which dissolves in water to form phosphoric acid, H3PO4) Group 6A Metallic character increases going down the group but no element behaves as a typical metal The lighter members tend to gain two electrons to form ions in compounds with metals Chemistry of oxygen • Elemental forms are O2 and O3 • Oxygen forms a wide variety of oxides • O2 and especially O3 are powerful oxidizing agents Chemistry of sulfur • The elemental forms are called rhombic and monoclinic sulfur, both of which contain S8 molecules • The most important oxides are SO2 (which forms H2SO3 in water) and SO3 (which forms H2SO4 in water) • Sulfur forms a wide variety of compounds in which it shows the oxidation states 6, 4, 2, 0, and Group 7A (halogens) All nonmetals Form hydrides of the type HX that behave as strong acids in water except for HF, which is a weak acid The oxyacids of the halogens become stronger as more oxygen atoms are present The interhalogens contain two or more different halogens Group 8A (noble gases) All elements are monatomic gases and are generally very unreactive The heavier elements form compounds with electronegative elements such as fluorine and oxygen REVIEW QUESTIONS 1. What is the valence electron configuration for Group 5A elements? Metallic character increases when going down a group. Give some examples illustrating how Bi and Sb have metallic characteristics not associated with N, P, and As.
The Group 5A elements can form molecules or ions that involve three, five, or six covalent bonds; NH3, AsCl5, and are examples. Draw the Lewis struc-ture for each of these substances and predict the molecular structure and hy-bridization for each. Why doesn’t NF5 or form?
2. Table 20.2 lists some common nitrogen compounds having oxidation states rang-ing from to 5. Rationalize this spread in oxidation states. For each sub-stance listed in Table 20.2, list some of its special properties.
3. Ammonia forms hydrogen-bonding intermolecular forces resulting in an unusu-ally high boiling point for a substance with the small size of NH3. Can hydra-zine, N2H4, also form hydrogen-bonding interactions?
The synthesis of ammonia gas from nitrogen gas and hydrogen gas is a classic case in which a knowledge of kinetics and equilibrium was exploited to 3 NCl6 PF6 2 X2 Section 20.7 halogens hydrochloric acid hydrohalic acids disproportionation reaction interhalogen compounds Section 20.8 noble gases For Review 935 make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia: a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure In many natural waters, nitrogen and phosphorus are the least abundant nutrients available for plant life. Some waters that become polluted from agri-cultural runoff or municipal sewage become infested with algae. The algae con-sume most of the dissolved oxygen in the water, and fish life dies off as a result.
Describe how these events are chemically related.
4. White phosphorus is much more reactive than black or red phosphorus. Explain.
How is phosphine’s (PH3) structure different from that of ammonia? Phosphoric acid (H3PO4) is a triprotic acid, phosphorous acid (H3PO3) is a diprotic acid, and hypophosphorous acid (H3PO2) is a monoprotic acid. Explain this phenomenon.
5. What is the valence electron configuration of Group 6A elements? What are some property differences between oxygen and polonium? What are the Lewis structures for the two allotropic forms of oxygen? How can the paramagnetism of O2 be explained using the molecular orbital model? What is the molecular structure and the bond angle in ozone? Ozone is desirable in the upper atmos-phere and undesirable in the lower atmosphere. A dictionary states that ozone has the scent of a fresh spring day. How can these seemingly conflicting state-ments be reconciled in terms of the chemical properties of ozone?
6. The most stable form of solid sulfur is the rhombic form; however, a solid form called monoclinic sulfur can also form. What is the difference between rhombic and monoclinic sulfur? Explain why O2 is much more stable than S2 or SO. When SO2(s) or SO3(g) reacts with water, an acidic solution forms. Explain. What are the molecular structures and bond angles in SO2 and SO3? Explain the bonding in SO2 and SO3. H2SO4 is a powerful dehydrating agent: What does this mean?
7. What is the valence electron configuration of the halogens? Why do the boiling points and melting points of the halogens increase steadily from F2 to I2? Give two reasons why F2 is the most reactive of the halogens. Explain why HF is a weak acid, whereas HCl, HBr, and HI are all strong acids.
Explain why the boiling point of HF is much higher than the boiling points of HCl, HBr, and HI. In nature, the halogens are generally found as halide ions in various minerals and seawater. What is a halide ion, and why are halide salts so stable? The oxidation states of the halogens vary from to 7. Identify compounds of chlorine that have 1, 1, 3, 5, and 7 oxidation states.
How does the oxyacid strength of the halogens vary as the number of oxygens in the formula increases?
8. Table 20.11 lists many compounds or ions that halogens form with other non-metals. For each compound or ion, give the molecular structure, including bond angles, and give the hybridization of the central atom in each species (ignore IF7). Why does ICl3 form but not FCl3?
9. What special property of the noble gases makes them unreactive? The boiling points and melting points of the noble gases increase steadily from He to Xe.
Explain. Although He is the second most abundant element in the universe, it is very rare on earth. Why? The noble gases were among the last elements discov-ered; their existence was not predicted by Mendeleev when he published his first periodic table. Explain. In chemistry textbooks written before 1962, the noble gases were referred to as the inert gases. Why do we no longer use this term?
10. For the structures of the xenon compounds in Figure 20.26, give the bond angles exhibited and the hybridization of the central atom in each compound.
1 936 Chapter Twenty The Representative Elements: Groups 5A Through 8A 11. Several important compounds contain only nitrogen and oxygen.
Place the following compounds in order of increasing mass per-cent of nitrogen.
a. NO, a gas formed by the reaction of N2 with O2 in internal combustion engines b. NO2, a brown gas mainly responsible for the brownish color of photochemical smog c. N2O4, a colorless liquid used as fuel in space shuttles d. N2O, a colorless gas sometimes used as an anesthetic by den-tists (known as laughing gas) 12. Nitric acid is produced commercially by the Ostwald process, rep-resented by the following equations: What mass of NH3 must be used to produce kg HNO3 by the Ostwald process? Assume 100% yield in each reaction and assume that the NO produced in the third step is not recycled.
13. Complete and balance each of the following reactions.
a. the decomposition of solid ammonium nitrate b. the decomposition of gaseous dinitrogen pentoxide c. the reaction between solid potassium phosphide and water d. the reaction between liquid phosphorus tribromide and water e. the reaction between aqueous ammonia and aqueous sodium hypochlorite 14. Arsenic reacts with oxygen to form oxides that react with water in a manner analogous to that of the phosphorus oxides. Write balanced chemical equations describing the reaction of arsenic with oxygen and the reaction of the resulting oxide with water.
15. Phosphorus occurs naturally in the form of fluorapatite, CaF2 3Ca3(PO4)2, where the dot indicates 1 part CaF2 to 3 parts Ca3(PO4)2. This mineral is reacted with an aqueous solution of sulfuric acid in the preparation of a fertilizer. The products are phosphoric acid, hydrogen fluoride, and gypsum, CaSO4 2H2O. Write and balance the chemical equation describing this process.
16. Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structure for the reactants and products in the reactions described below.
a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide.
b. Boron trihydride accepts a pair of electrons from ammonia, forming BH3NH3.
Give a possible explanation for why these two reactions occur.
17. Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively according to the following reaction: How many moles of NaN3(s) must be reacted to inflate an air bag to 70.0 L at STP?
2NaN31s2 ¡ 2Na1s2 3N21g2 1.0 10 6 3NO21g2 H2O1l2 ¡ 2HNO31aq2 NO1g2 2NO1g2 O21g2 ¡ 2NO21g2 4NH31g2 5O21g2 ¡ 4NO1g2 6H2O1g2 A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 1. Elemental nitrogen exists as N2, whereas in the gas phase the el-ements phosphorus, arsenic, and antimony consist of P4, As4, and Sb4 molecules, respectively. Give a possible reason for this dif-ference between N2 and the other group 5A elements.
2. What is nitrogen fixation? Give some examples of nitrogen fixation.
3. In large doses, selenium is toxic. However, in moderate intake, selenium is a physiologically important element. How is selenium physiologically important?
4. Ozone is a possible replacement for chlorine in municipal water purification. Unlike chlorine, virtually no ozone remains after treatment. This has good and bad consequences. Explain.
5. Sulfur forms a wide variety of compounds in which it has 6, 4, 2, 0, and 2 oxidation states. Give examples of sulfur com-pounds having each of these oxidation states.
6. When a halogen is a central atom in a compound, the compound typically is sp3, dsp3, or d2sp3 hybridized. Using bromine as your central atom, give example compounds for each type of hybridiza-tion. What is the molecular structure for each of your examples?
7. Explain the following observations regarding reactions of halogens.
a. In the hydrogen-chlorine cannon lecture demonstration, a lit magnesium strip is held to a mixture of H2 and Cl2, resulting in a reaction that sends the stopper to the cannon flying across the lecture room.
b. When a brown bromine solution is added dropwise to an or-ganic compound called an alkene, the brown color disappears, resulting in a colorless reaction mixture.
c. When aluminum is reacted with iodine, the reaction container emits sparks and a deep purple colored smoke.
8. There is evidence that radon reacts with fluorine to form com-pounds similar to those formed by xenon and fluorine. Predict the formulas of these RaFx compounds. Why is the chemistry of radon difficult to study?
Exercises In this section similar exercises are paired.
Group 5A Elements 9. The oxyanion of nitrogen in which it has the highest oxidation state is the nitrate ion The corresponding oxyanion of phosphorus is The ion is known but not very sta-ble. The ion is not known. Account for these differences in terms of the bonding in the four anions.
10. In each of the following pairs of substances, one is stable and known, and the other is unstable. For each pair, choose the stable substance, and explain why the other is unstable.
a. NF5 or PF5 b. AsF5 or AsI5 c. NF3 or NBr3 PO3 NO4 3 PO4 3.
(NO3 ).
Exercises 937 a. Draw Lewis structures for all the species shown in the pre-ceding reactions. Predict the hybridization of the central Sb and S atoms in each structure.
b. This superacid solution is capable of protonating (adding H to) virtually every known organic compound. What is the ac-tive protonating agent in the superacid solution?
Group 6A Elements 29. Use bond energies to estimate the maximum wavelength of light that will cause the reaction 30. The xerographic (dry writing) process was invented in 1938 by C.
Carlson. In xerography, an image is produced on a photoconduc-tor by exposing it to light. Selenium is commonly used, since its conductivity increases three orders of magnitude upon exposure to light in the range from 400 to 500 nm. What color light should be used to cause selenium to become conductive? (See Figure 7.2.) 31. Complete and balance each of the following reactions.
a. the reaction between sulfur dioxide gas and oxygen gas b. the reaction between sulfur trioxide gas and water c. the reaction between aqueous sodium thiosulfate and aqueous iodine d. the reaction between copper metal and aqueous hot sulfuric acid 32. Write a balanced equation describing the reduction of H2SeO4 by SO2 to produce selenium.
33. For each of the following, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybridization of the central atom.
a.
c. SCl2 e. TeF6 b. O3 d. SeBr4 34. Disulfur dinitride (S2N2) exists as a ring of alternating sulfur and nitrogen atoms. S2N2 will polymerize to polythiazyl, which acts as a metallic conductor of electricity along the polymer chain.
Write a Lewis structure for S2N2.
35. Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes upon con-tact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams upon con-tact with blood, which provides a cleansing action. In the labora-tory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: What mass of hydrogen peroxide should result when 1.50 g of barium peroxide is treated with 25.0 mL of hydrochloric acid so-lution containing 0.0272 g of HCl per mL? What mass of which reagent is left unreacted?
36. During the developing process of black-and-white film, silver bro-mide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosulfate. The net ionic equa-tion for the reaction is AgBr1s2 2S2O3 21aq2 ¡ Ag1S2O322 31aq2 Br 1aq2 BaO21s2 2HCl1aq2 ¡ H2O21aq2 BaCl21aq2 SO3 2 O3 ¡ O2 O 18. Urea (H2NCONH2) is used extensively as a nitrogen source in fer-tilizers. It is produced commercially from the reaction of ammo-nia and carbon dioxide: Ammonia gas at and 90. atm flows into a reactor at a rate of 500. L/min. Carbon dioxide at and 45 atm flows into the reactor at a rate of 600. L/min. What mass of urea is produced per minute by this reaction assuming a 100% yield?
19. Hydrazine (N2H4) is used as a fuel in liquid-fueled rockets. When hydrazine reacts with oxygen gas, nitrogen gas and water vapor are produced. Write a balanced equation and use bond energies from Table 8.4 to estimate for this reaction.
20. The space shuttle orbiter utilizes the oxidation of methylhydrazine by dinitrogen tetroxide for propulsion: Calculate for this reaction using data in Appendix 4. Com-pare your answer to the value determined in Sample Exercise 20.2. Explain any discrepancies.
21. Many oxides of nitrogen have positive values for the standard free energy of formation. Using NO as an example, explain why this is the case.
22. Using data from Appendix 4 calculate and for the reaction Why does NO form in an automobile engine but then does not readily decompose back to N2 and O2 in the atmosphere?
23. Compare the Lewis structures with the molecular orbital view of the bonding in NO, NO, and Account for any discrepan-cies between the two models.
24. The energy to break a particular bond is not always constant. It takes about 200 kJ/mol less energy to break the bond in NOCl as compared with NCl3: Why is there such a great discrepancy in the apparent bond energies? Hint: Consider what happens to the nitrogen–oxygen bond in the first reaction.
25. Predict the relative acid strengths of the following compounds.
a. H3PO4 and H3PO3 b. H3PO4, and 26. Trisodium phosphate (TSP) is an effective grease remover. Like many cleaners, TSP acts as a base in water. Write a balanced equa-tion to account for this behavior.
27. Isohypophosphonic acid (H4P2O6) and diphosphonic acid (H4P2O5) are tri- and diprotic acids, respectively. Draw Lewis structures for these acids that are consistent with these facts.
28. One of the most strongly acidic solutions known is a mixture of antimony pentafluoride (SbF5) and fluorosulfonic acid (HSO3F).
The dominant equilibria are F5SbOSO2FH HSO3F ∆H2SO3F F5SbOSO2F SbF5 HSO3F ∆F5SbOSO2FH HPO4 2 H2PO4 , N¬Cl NCl3 ¡ NCl2 Cl ¢H° 375 kJ/mol NOCl ¡ NO Cl ¢H° 158 kJ/mol N¬Cl NO.
N21g2 O21g2 ¡ 2NO1g2 ¢G° ¢H°, ¢S°, ¢H ¢H° 4N2H3CH31l2 5N2O41l2 ¡ 12H2O1g2 9N21g2 4CO21g2 ¢H 223°C 223°C 2NH31g2 CO21g2 ¡ H2NCONH21s2 H2O1g2 hv 938 Chapter Twenty The Representative Elements: Groups 5A Through 8A 45. Xenon difluoride has proven to be a versatile fluorinating agent.
For example, in the reaction the by-products Xe and HF are easily removed, leaving pure C6H5F. Xenon difluoride is stored in an inert atmosphere free from oxygen and water. Why is this necessary?
46. Using the data in Table 20.12, calculate the mass of argon at and 1.0 atm in a room How many Ar atoms are in this room? How many Ar atoms do you inhale in one breath (approximately 2 L) of air at and 1.0 atm? Argon gas is inert, so it poses no serious health risks. However, if significant amounts of radon were inhaled into the lungs, lung cancer is a possible result. Explain the health-risk differences between argon gas and radon gas.
47. Which do you think would be the greater health hazard, the re-lease of a radioactive nuclide of Sr or a radioactive nuclide of Xe into the environment? Assume the amount of radioactivity is the same in each case. Explain your answer on the basis of the chem-ical properties of Sr and Xe. Why are the chemical properties of a radioactive substance important in assessing its potential health hazards?
48. The most significant source of natural radiation is radon-222.
222Rn, a decay product of 238U, is continuously generated in the earth’s crust, allowing gaseous Rn to seep into the basements of buildings. Because 222Rn is an -particle producer with a rela-tively short half-life of 3.82 days, it can cause biological damage when inhaled.
a. How many particles and particles are produced when 238U decays to 222Rn? What nucleus is produced when 222Rn decays?
b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling 222Rn?
Additional Exercises 49. The compound NF3 is quite stable, but NCl3 is very unstable (NCl3 was first synthesized in 1811 by P. L. Dulong, who lost three fin-gers and an eye studying its properties). The compounds NBr3 and NI3 are rare, although the explosive compound NI3 NH3 is known. Account for the instability of these halides of nitrogen.
50. The N2O molecule is linear and polar.
a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer.
b. On the basis of your answer in part a, write the Lewis struc-ture of N2O (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom.
c. How would the multiple bonding in SNqNOO O QS be described in terms of orbitals?
51. Oxidation of the cyanide ion produces the stable cyanate ion, The fulminate ion, on the other hand, is very un-stable. Fulminate salts explode when struck; Hg(CNO)2 is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable?
52. Sodium bismuthate (NaBiO3) is used to test for the presence of Mn2 in solution by the following reaction: Mn 21aq2 NaBiO31s2 ¡ MnO4 1aq2 BiO3 31aq2 CNO , OCN.
b a a 25°C 10.0 m 10.0 m 10.0 m.
25°C C6H61l2 XeF21g2 ¡ C6H5F1l2 Xe1g2 HF1g2 What mass of AgBr can be dissolved by 1.00 L of 0.200 M Na2S2O3? (Assume the reaction goes to completion.) Group 7A Elements 37. Write the Lewis structure for O2F2. Predict the bond angles and hybridization of the two central oxygen atoms. Assign oxidation states and formal charges to the atoms in O2F2. The compound O2F2 is a vigorous and potent oxidizing and fluorinating agent.
Are oxidation states or formal charges more useful in accounting for these properties of O2F2?
38. For each of the following, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybridization of the central atom.
a. Freon-12 (CCl2F2) c. iodine trichloride b. perchloric acid d. bromine pentafluoride 39. Complete and balance each of the following reactions.
a.
b.
c.
40. Hypofluorous acid is the most recently prepared of the halogen oxyacids. Weighable amounts were first obtained in 1971 by M. H.
Studies and E. N. Appelman using the fluorination of ice. Hypo-fluorous acid is exceedingly unstable, decomposing spontaneously (with a half-life of 30 min) to HF and O2 in a Teflon container at room temperature. It reacts rapidly with water to produce HF, H2O2, and O2. In dilute acid, H2O2 is the major product; in dilute base, O2 is the major product.
a. Write balanced equations for the reactions described above.
b. Assign oxidation states to the elements in hypofluorous acid.
Does this suggest why hypofluorous acid is so unstable?
41. Hydrazine is somewhat toxic. Use the following half-reactions to explain why household bleach (highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia.
42. What is a disproportionation reaction? Use the following reduc-tion potentials to predict whether HClO2 will disproportionate.
Group 8A Elements 43. The xenon halides and oxides are isoelectronic with many other compounds and ions containing halogens. Give a molecule or ion in which iodine is the central atom that is isoelectronic with each of the following.
a. xenon tetroxide d. xenon tetrafluoride b. xenon trioxide e. xenon hexafluoride c. xenon difluoride 44. For each of the following, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybridization of the central atom.
a. KrF2 b. KrF4 c. XeO2F2 d. XeO2F4 HClO2 2H 2e ¡ HClO H2O e° 1.65 V ClO3 3H 2e ¡ HClO2 H2O e° 1.21 V N2H4 2H2O 2e ¡ 2NH3 2OH e° 0.10 V ClO H2O 2e ¡ 2OH Cl e° 0.90 V SiO21s2 HF1aq2 ¡ BrF1s2 H2O1l2 ¡ BaCl21s2 H2SO41aq2 ¡ Additional Exercises 939 58. Phosphate buffers are important in regulating the pH of intracel-lular fluids at pH values generally between 7.1 and 7.2. What is the concentration ratio of to in intracellular fluid at pH 7.15?
Why is a buffer composed of H3PO4 and ineffective in buffering the pH of intracellular fluid?
59. Commercial cold packs and hot packs are available for treating athletic injuries. Both types contain a pouch of water and a dry chemical. When the pack is struck, the pouch of water breaks, dis-solving the chemical, and the solution becomes either hot or cold.
Many hot packs use magnesium sulfate, and many cold packs use ammonium nitrate. Write reactions to show how these strong elec-trolytes break apart when they dissolve in water.
60. Classify each of the following as a strong acid or a weak acid.
61. Consider the following Lewis structure where E is an unknown element: What are some possible identities for element E? Predict the mo-lecular structure (including bond angles) for this ion.
62. Consider the following Lewis structure where E is an unknown element: What are some possible identities for element E? Predict the mo-lecular structure (including bond angles) for this ion.
63. The unit cell for a pure xenon fluoride compound is shown be-low. What is the formula of the compound?
Xenon Fluorine F D G ð š O 2– ½ k F E O ð – O E O O O O A ð ð ð ð c.
d.
Cl S O H b.
a.
Ka 7.5 103 H3PO41aq2 ∆H2PO4 1aq2 H 1aq2 H2PO4 Ka 6.2 108 H2PO4 1aq2 ∆HPO4 21aq2 H 1aq2 HPO4 2 H2PO4 a. Balance this equation.
b. Given that bismuth does not form double bonds with oxygen in and that NaBiO3 is relatively insoluble in water, what type of structure must NaBiO3 have to account for this behavior?
53. Bacterial digestion is an economical method of sewage treatment.
The reaction Bacterial tissue is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. How much bacterial tissue is pro-duced in a treatment plant for every kg of wastewater containing 3.0% NH4 ions by mass? Assume that 95% of the ammonium ions are consumed by the bacteria.
54. An unknown element is a nonmetal and has a valence electron configuration of ns2np4.
a. How many valence electrons does this element have?
b. What are some possible identities for this element?
c. What is the formula of the compound this element would form with lithium?
d. Would this element have a larger or smaller radius than barium?
e. Would this element have a greater or smaller ionization energy than fluorine?
55. The structure of is Draw a complete Lewis structure for and explain the dis-tortion from the ideal square pyramidal structure.
56. Photogray lenses contain small embedded crystals of solid silver chloride. Silver chloride is light-sensitive because of the reaction Small particles of metallic silver cause the lenses to darken. In the lenses this process is reversible. When the light is removed, the reverse reaction occurs. However, when pure white silver chlo-ride is exposed to sunlight it darkens; the reverse reaction does not occur in the dark.
a. How do you explain this difference?
b. Photogray lenses do become permanently dark in time. How do you account for this?
57. Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction equilibrium constants (Kp values) as a function of temperature are Is the reaction exothermic or endothermic?
600°C, 2.25 106 500°C, 1.45 105 300°C, 4.34 103 N21g2 3H21g2 ∆2NH31g2 AgCl1s2 ¡ Ag1s2 Cl TeF5 , F F F F Te 79˚ F TeF5 1.0 10 4 C5H7O2N1s2 54NO2 1aq2 52H2O1l2 109H 1aq2 5CO21g2 55NH4 1aq2 76O21g2 ¬¡ Bacteria BiO3 hv 940 Chapter Twenty The Representative Elements: Groups 5A Through 8A Assuming a two-step mechanism, propose the second step in the mechanism and give the overall balanced equation.
e. The activation energy for Cl-catalyzed destruction of ozone is 2.1 kJ/mol. Estimate the efficiency with which Cl atoms de-stroy ozone as compared with NO molecules at Assume that the frequency factor A is the same for each catalyzed reaction and assume similar rate laws for each catalyzed reaction.
68. Using data from Appendix 4, calculate and Kp (at 298 K) for the production of ozone from oxygen: At 30 km above the surface of the earth, the temperature is about 230. K, and the partial pressure of oxygen is about atm. Estimate the partial pressure of ozone in equilibrium with oxygen at 30 km above the earth’s surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.
69. You travel to a distant, cold planet where the ammonia flows like water. In fact, the inhabitants of this planet use ammonia (an abun-dant liquid on their planet) much as earthlings use water. Ammo-nia is also similar to water in that it is amphoteric and undergoes autoionization. The K value for the autoionization of ammonia is at the standard temperature of the planet. What is the pH of ammonia at this temperature?
70. Nitrogen gas reacts with hydrogen gas to form ammonia gas. You have an equimolar mixture of nitrogen and hydrogen gases in a 15.0-L container fitted with a piston in a room with a pressure of 1.00 atm. The piston apparatus allows the container volume to change in order to keep the pressure constant at 1.00 atm. Assume ideal behavior, constant temperature, and complete reaction.
a. What is the partial pressure of ammonia in the container when the reaction is complete?
b. What is the mole fraction of ammonia in the container when the reaction is complete?
c. What is the volume of the container when the reaction is com-plete?
71. A cylinder fitted with a movable piston initially contains 2.00 mol O2(g) and an unknown amount of SO2(g). The oxygen is known to be in excess. The density of the mixture is 0.8000 g/L at some T and P.After the reaction has gone to completion, forming SO3(g), the density of the resulting gaseous mixture is 0.8471 g/L at the same T and P. Calculate the mass of SO3 formed in the reaction.
72. One way to determine Ksp for the salt Ca(IO3)2 is to titrate it with sodium thiosulfate (Na2S2O3). First, make a saturated solution of calcium iodate. Then, add KI and a strong acid (hydrochloric acid and sulfuric acid are generally used). The iodate ion will react ac-cording to the equation Note that molecular iodine is a product of this reaction. Adding a starch indicator will turn the solution of I2 a dark blue-black color. A solution of sodium thiosulfate is added through a buret, which reacts with iodine as follows: The dark blue-black color disappears, when all of the I2 has re-acted. This is the endpoint of the titration.
I2 2S2O3 2 ¡ 2I S4O6 2 IO3 5I 6H ¡ 3I2 3H2O 1.8 1012 1.0 103 3O21g2 ∆2O31g2 ¢H°, ¢G°, 25°C.
Challenge Problems 64. Many structures of phosphorus-containing compounds are drawn with some bonds. These bonds are not the typical bonds we’ve considered, which involve the overlap of two p orbitals. In-stead, they result from the overlap of a d orbital on the phospho-rus atom with a p orbital on oxygen. This type of bonding is sometimes used as an explanation for why H3PO3 has the first structure below rather than the second: Draw a picture showing how a d orbital and a p orbital overlap to form a bond.
65. Use bond energies (Table 8.4) to show that the preferred products for the decomposition of N2O3 are NO2 and NO rather than O2 and N2O. (The single bond energy is 201 kJ/mol.) Hint: Consider the reaction kinetics.
66. Sodium tripolyphosphate (Na5P3O10) is used in many synthetic de-tergents to soften the water by complexing Mg2 and Ca2 ions.
It also increases the efficiency of surfactants (wetting agents) that lower a liquid’s surface tension. The K value for the formation of is The reaction is Calculate the concentration of Mg2 in a solution that was origi-nally 50. ppm of Mg2 (50. mg/L of solution) after 40. g Na5P3O10 is added to 1.0 L of the solution.
67. One pathway for the destruction of ozone in the upper atmosphere is Slow Fast a. Which species is a catalyst?
b. Which species is an intermediate?
c. The activation energy Ea for the uncatalyzed reaction is 14.0 kJ. Ea for the same reaction when catalyzed by the pres-ence of NO is 11.9 kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at Assume that the frequency factor A is the same for each reaction.
d. One of the concerns about the use of Freons is that they will migrate to the upper atmosphere, where chlorine atoms can be generated by the reaction Freon-12 Chlorine atoms also can act as a catalyst for the destruction of ozone. The first step of a proposed mechanism for chlorine-catalyzed ozone destruction is Slow Cl1g2 O31g2 ¡ ClO1g2 O21g2 CCl2F2 ¡ hv CF2Cl Cl 25°C?
O31g2 O1g2 ¡ 2O21g2 Overall reaction: O31g2 O1g2 ¡ 2O21g2 NO21g2 O1g2 ¡ NO1g2 O21g2 O31g2 NO1g2 ¡ NO21g2 O21g2 Mg 2 P3O10 5 ÷ MgP3O10 3 4.0 108.
MgP3O10 3 N¬O p OH O P OH H HO P OH OH p p P“O Marathon Problem 941 open. He calls this the “fishhook” strategy. Mr. Spock has sent a coded message to the chemists on the fighters to tell the ships what to do next. The outline of the message is _ _ _ _ _ (1) (2) (3) (4) (5) (6) _ _ _ _ _ ____ (7) (8) (9) (10) (11) (12) (10) (11) Fill in the blanks of the message using the following clues.
(1) Symbol of the halogen whose hydride has the second highest boiling point in the series of HX compounds that are hydrogen halides.
(2) Symbol of the halogen that is the only hydrogen halide, HX, that is a weak acid in aqueous solution.
(3) Symbol of the element whose existence on the sun was known before its existence on earth was discovered.
(4) Symbol of the element whose presence can interfere with the qualitative analysis for Pb2, Hg2 2, and Ag. When chloride ions are added to an aqueous solution of this metal ion, a white precipitate forms with formula MOCl.
(5) Symbol of the Group 6A element that, like selenium, is a semiconductor.
(6) Symbol for the element known in rhombic and monoclinic forms.
(7) Symbol for the element that exists as diatomic molecules in a yellow-green gas when not combined with another ele-ment; its silver, lead, and mercury(I) salts are white and insoluble in water.
(8) Symbol for the most abundant element in and near the earth’s crust.
(9) Symbol for the element that seems to give some protection against cancer when a diet rich in this element is consumed.
(10) Symbol for the only noble gas besides xenon that has been shown to form compounds under some circumstances (write the symbol backward and split the letters as shown).
(11) Symbol for the toxic element that, like phosphorus and an-timony, forms tetrameric molecules when uncombined with other elements (split the letters of the symbol as shown).
(12) Symbol for the element that occurs as an inert component of air but is a very prominent part of fertilizers and explosives.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
Consider starting with a 10.0-mL sample of a saturated cal-cium iodate solution. Upon titrating, you find that 14.9 mL of 0.100 M Na2S2O3 is required to reach the end point of the titra-tion. Calculate Ksp for Ca(IO3)2.
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
73. Although nitrogen trifluoride (NF3) is a thermally stable com-pound, nitrogen triiodide (NI3) is known to be a highly explosive material. NI3 can be synthesized according to the equation a. What is the enthalpy of formation for NI3(s) given the enthalpy of reaction ( kJ) and the enthalpies of formation for BN(s) ( kJ/mol), IF(g) ( kJ/mol), and BF3(g) ( kJ/mol)?
b. It is reported that when the synthesis of NI3 is conducted us-ing 4 mol IF for every 1 mol BN, one of the by-products iso-lated is What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?
74. While selenic acid has the formula H2SeO4 and thus is directly related to sulfuric acid, telluric acid is best visualized as H6TeO6 or Te(OH)6.
a. What is the oxidation state of tellurium in Te(OH)6?
b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with pKa1 7.68 and pKa2 11.29. Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas: If a cubic block of tellurium (density 6.240 g/cm3) measuring 0.545 cm on edge is allowed to react with 2.34 L of fluorine gas at 1.06 atm and what is the pH of a solution of Te(OH)6 formed by dissolving the isolated TeF6(g) in 115 mL of water?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
75. Captain Kirk has set a trap for the Klingons who are threatening an innocent planet. He has sent small groups of fighter rockets to sites that are invisible to Klingon radar and put a decoy in the 25°C, Te1s2 3F21g2 ¡ TeF61g2 TeF61g2 6H2O1l2 ¡ Te1OH261aq2 6HF1aq2 [IF2][BF4].
1136 96 254 307 BN1s2 3IF1g2 ¡ BF31g2 NI31g2 942 21 Transition Metals and Coordination Chemistry Contents 21.1 The Transition Metals: A Survey • General Properties • Electron Configurations • Oxidation States and Ionization Energies • Standard Reduction Potentials • The 4d and 5d Transition Series 21.2 The First-Row Transition Metals 21.3 Coordination Compounds • Coordination Number • Ligands • Nomenclature 21.4 Isomerism • Structural Isomerism • Stereoisomerism 21.5 Bonding in Complex Ions: The Localized Electron Model 21.6 The Crystal Field Model • Octahedral Complexes • Other Coordination Geometries 21.7 The Biologic Importance of Coordination Complexes 21.8 Metallurgy and Iron and Steel Production • Hydrometallurgy • The Metallurgy of Iron • Production of Steel • Heat Treatment of Steel Copper ore deposit in Namibia.
T ransition metals have many uses in our society. Iron is used for steel; copper for electrical wiring and water pipes; titanium for paint; silver for photographic paper; man-ganese, chromium, vanadium, and cobalt as additives to steel; platinum for industrial and automotive catalysts; and so on.
One indication of the importance of transition metals is the great concern shown by the U.S. government for continuing the supply of these elements. In recent years the United States has been a net importer of about 60 “strategic and critical” minerals, including cobalt, manganese, platinum, palladium, and chromium. All these metals play a vital role in the U.S. economy and defense, and approximately 90% of the required amounts must be imported (see Table 21.1).
In addition to being important in industry, transition metal ions play a vital role in living organisms. For example, complexes of iron provide for the transport and storage of oxygen, molybdenum and iron compounds are catalysts in nitrogen fixation, zinc is found in more than 150 biomolecules in humans, copper and iron play a crucial role in the respiratory cycle, and cobalt is found in essential biomolecules such as vitamin B12.
In this chapter we explore the general properties of transition metals, paying particular attention to the bonding, structure, and properties of the complex ions of these metals.
21.1 The Transition Metals: A Survey General Properties One striking characteristic of the representative elements is that their chemistry changes markedly across a given period as the number of valence electrons changes. The chemi-cal similarities occur mainly within the vertical groups. In contrast, the transition metals show great similarities within a given period as well as within a given vertical group.
This difference occurs because the last electrons added for transition metals are inner elec-trons: d electrons for the d-block transition metals and f electrons for the lanthanides and actinides. These inner d and f electrons cannot participate as easily in bonding as can the valence s and p electrons. Thus the chemistry of the transition elements is not affected as greatly by the gradual change in the number of electrons as is the chemistry of the representative elements.
Group designations are traditionally given on the periodic table for the d-block tran-sition metals (see Fig. 21.1). However, these designations do not relate as directly to the 943 TABLE 21.1 Some Transition Metals Important to the U.S. Economy and Defense Metal Uses Percentage Imported Chromium Stainless steel (especially for parts exposed 91% to corrosive gases and high temperatures) Cobalt High-temperature alloys in jet engines, 93% magnets, catalysts, drill bits Manganese Steelmaking 97% Platinum and Catalysts 87% palladium 944 Chapter Twenty-One Transition Metals and Coordination Chemistry chemical behavior of these elements as they do for the representative elements (the A groups), so we will not use them.
As a class, the transition metals behave as typical metals, possessing metallic luster and relatively high electrical and thermal conductivities. Silver is the best conductor of heat and electric current. However, copper is a close second, which explains copper’s wide use in the electrical systems of homes and factories.
Ce Th Pr Pa Nd U Pm Sm Pu Eu Am Gd Cm Tb Bk Dy Cf Ho Es Er Fm Tm Md Yb No Lu Lr Sc Y La Ac Ti Zr Hf Rf V Nb Ta Db Cr Mo W Sg Mn Tc Re Bh Fe Ru Os Co Rh Ir Ni Pd Pt Cu Ag Au Zn Cd Hg Hs Mt Ds Rg Uub Np FIGURE 21.1 The position of the transition elements on the periodic table. The d-block elements correspond to filling the 3d, 4d, 5d, or 6d orbitals. The inner transition metals corre-spond to filling the 4f (lanthanides) or 5f (actinides) orbitals.
Sc Y La Ac† Ti Zr Hf Rf V Nb Ta Db Cr Mo W Sg Mn Tc Re Fe Ru Os Co Rh Ir Ni Pd Pt Cu Ag Au Zn Cd Hg Ce Th Pr Pa Nd U Pm Np Sm Pu Eu Am Gd Cm Tb Bk Dy Cf Ho Es Er Fm Tm Md Yb No Lu Lr d-block transition elements Lanthanides †Actinides f-block transition elements Bh Hs Mt Ds Rg Uub 21.1 The Transition Metals: A Survey 945 Despite their many similarities, the transition metals do vary considerably in certain properties. For example, tungsten has a melting point of and is used for filaments in light bulbs; mercury is a liquid at Some transition metals such as iron and titanium are hard and strong and make very useful structural materials; others such as copper, gold, and silver are relatively soft. The chemical reactivity of the transition metals also varies significantly. Some react readily with oxygen to form oxides. Of these metals, some, such as chromium, nickel, and cobalt, form oxides that adhere tightly to the metallic surface, protecting the metal from further oxidation. Others, such as iron, form oxides that scale off, constantly exposing new metal to the corrosion process. On the other hand, the no-ble metals—primarily gold, silver, platinum, and palladium—do not readily form oxides.
In forming ionic compounds with nonmetals, the transition metals exhibit several typ-ical characteristics: More than one oxidation state is often found. For example, iron combines with chlo-rine to form FeCl2 and FeCl3.
The cations are often complex ions, species where the transition metal ion is surrounded by a certain number of ligands (molecules or ions that behave as Lewis bases). For ex-ample, the compound [Co(NH3)6]Cl3 contains the Co(NH3)6 3 cation and anions.
NH3 NH3 NH3 NH3 Co The Co(NH3)6 3+ ion 3+ NH3 NH3 Cl 25°C.
3400°C Sterling silver candlesticks and bowl from Japan.
946 Chapter Twenty-One Transition Metals and Coordination Chemistry (from left to right) Aqueous solutions con-taining the metal ions Co2, Mn2, Cr3, Fe3, and Ni2.
Most compounds are colored, because the transition metal ion in the complex ion can absorb visible light of specific wavelengths.
Many compounds are paramagnetic (they contain unpaired electrons).
In this chapter we will concentrate on the first-row transition metals (scandium through zinc) because they are representative of the other transition series and because they have great practical significance. Some important properties of these elements are summarized in Table 21.2 and are discussed in the next section.
Electron Configurations The electron configurations of the first-row transition metals were discussed in Sec-tion 7.11. The 3d orbitals begin to fill after the 4s orbital is complete, that is, after cal-cium ([Ar]4s2). The first transition metal, scandium, has one electron in the 3d orbitals; the second, titanium, has two; and the third, vanadium, has three. We would expect chromium, the fourth transition metal, to have the electron configuration [Ar]4s23d 4. How-ever, the actual configuration is [Ar]4s13d5, which shows a half-filled 4s orbital and a half-filled set of 3d orbitals (one electron in each of the five 3d orbitals). It is tempting to say that the configuration results because half-filled “shells” are especially stable. Although there are some reasons to think that this explanation might be valid, it is an oversimplifi-cation. For instance, tungsten, which is in the same vertical group as chromium, has the configuration [Xe]6s24f 145d4, where half-filled s and d shells are not found. There are several similar cases.
Basically, the chromium configuration occurs because the energies of the 3d and 4s orbitals are very similar for the first-row transition elements. We saw in Section 7.11 that when electrons are placed in a set of degenerate orbitals, they first occupy each orbital (clockwise from upper left) Calcite stalac-tites colored by traces of iron. Quartz is often colored by the presence of transition metals such as Mn, Fe, and Ni. Wulfenite contains PbMoO4. Rhodochrosite is a mineral containing MnCO3.
21.1 The Transition Metals: A Survey 947 singly to minimize electron repulsions. Since the 4s and 3d orbitals are virtually degenerate in the chromium atom, we would expect the configuration 4s __ 3d __ __ __ __ __ rather than 4s __ 3d __ __ __ __ __ since the second arrangement has greater electron–electron repulsions and thus a higher energy.
The only other unexpected configuration among the first-row transition metals is that of copper, which is [Ar]4s13d10 rather than the expected [Ar]4s23d 9.
In contrast to the neutral transition metals, where the 3d and 4s orbitals have very similar energies, the energy of the 3d orbitals in transition metal ions is significantly less than that of the 4s orbital. This means that the electrons remaining after the ion is formed occupy the 3d orbitals, since they are lower in energy. First-row transition metal ions do not have 4s electrons. For example, manganese has the configuration [Ar]4s23d5, while that of Mn2 is [Ar]3d5. The neutral titanium atom has the configuration [Ar]4s23d2, while that of Ti3 is [Ar]3d1.
Oxidation States and Ionization Energies The transition metals can form a variety of ions by losing one or more electrons. The common oxidation states of these elements are shown in Table 21.2. Note that for the TABLE 21.2 Selected Properties of the First-Row Transition Metals Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Atomic 21 22 23 24 25 26 27 28 29 30 number Electron 4s23d1 4s23d2 4s23d3 4s13d5 4s23d5 4s23d6 4s23d 7 4s23d8 4s13d10 4s23d10 configuration Atomic 162 147 134 130 135 126 125 124 128 138 radius (pm) Ionization energies (eV/atom) First 6.54 6.82 6.74 6.77 7.44 7.87 7.86 7.64 7.73 9.39 Second 12.80 13.58 14.65 16.50 15.64 16.18 17.06 18.17 20.29 17.96 Third 24.76 27.49 29.31 30.96 33.67 30.65 33.50 35.17 36.83 39.72 Reduction 2.08 1.63 1.2 0.91 1.18 0.44 0.28 0.23 0.34 0.76 potential† (V) Common 3 2,3, 2,3, 2,3, 2,3, 2,3 2,3 2 1,2 2 oxidation 4 4,5 6 4,7 states Melting 1397 1672 1710 1900 1244 1530 1495 1455 1083 419 point (°C) Density 2.99 4.49 5.96 7.20 7.43 7.86 8.9 8.90 8.92 7.14 (g/cm3) Electrical — 2 3 10 2 17 24 24 97 27 conductivity‡ Each atom has an argon inner-core configuration.
†For the reduction process M2 2e n M (except for scandium, where the ion is Sc3).
‡Compared with an arbitrarily assigned value of 100 for silver.
Chromium has the electron configuration [Ar]4s13d 5.
A set of orbitals with the same energy is said to be degenerate.
Copper has the electron configuration [Ar]4s13d 10.
In transition metal ions, the 3d orbitals are lower in energy than the 4s orbitals.
948 Chapter Twenty-One Transition Metals and Coordination Chemistry first five metals the maximum possible oxidation state corresponds to the loss of all the 4s and 3d electrons. For example, the maximum oxidation state of chromium ([Ar]4s13d 5) is 6. Toward the right end of the period, the maximum oxidation states are not observed; in fact, the 2 ions are the most common. The higher oxidation states are not seen for these metals because the 3d orbitals become lower in energy as the nuclear charge increases, and the electrons become increasingly difficult to remove.
From Table 21.2 we see that ionization energy increases gradually going from left to right across the period. However, the third ionization energy (when an electron is removed from a 3d orbital) increases faster than the first ionization energy, clear evidence of the significant decrease in the energy of the 3d orbitals going across the period (see Fig. 21.2).
Standard Reduction Potentials When a metal acts as a reducing agent, the half-reaction is This is the reverse of the conventional listing for half-reactions in tables. Thus, to rank the transition metals in order of reducing ability, it is most convenient to reverse the re-actions and the signs given in Table 21.2. The metal with the most positive potential is then the best reducing agent. The transition metals are listed in order of reducing ability in Table 21.3.
Since is zero for the process all the metals except copper can reduce H ions to hydrogen gas in 1 M aqueous solu-tions of strong acid: As Table 21.3 shows, the reducing abilities of the first-row transition metals generally de-crease going from left to right across the period. Only chromium and zinc do not follow this trend.
The 4d and 5d Transition Series In comparing the 3d, 4d, and 5d transition series, it is instructive to consider the atomic radii of these elements (Fig. 21.3). Note that there is a general, although not regular, M1s2 2H1aq2 ¡ H21g2 M21aq2 2H 2e ¡ H2 e° M ¡ Mn ne FIGURE 21.2 Plots of the first (red dots) and third (blue dots) ionization energies for the first-row transition metals.
40 35 30 25 20 15 10 5 Sc Ti V Cr Mn Fe Co Ni Cu Zn Ionization energy (eV/atom) TABLE 21.3 Relative Reducing Abilities of the First-Row Transition Metals in Aqueous Solution Reaction Potential (V) Sc n Sc3 3e 2.08 Ti n Ti2 2e 1.63 V n V2 2e 1.2 Mn n Mn2 2e 1.18 Cr n Cr2 2e 0.91 Zn n Zn2 2e 0.76 Fe n Fe2 2e 0.44 Co n Co2 2e 0.28 Ni n Ni2 2e 0.23 Cu n Cu2 2e 0.34 88888888888888n Reducing ability 21.2 The First-Row Transition Metals 949 decrease in size going from left to right for each of the series. Also note that although there is a significant increase in radius in going from the 3d to the 4d metals, the 4d and 5d metals are remarkably similar in size. This latter phenomenon is the result of the lanthanide contraction. In the lanthanide series, consisting of the elements between lan-thanum and hafnium (see Fig. 21.1), electrons are filling the 4f orbitals. Since the 4f or-bitals are buried in the interior of these atoms, the additional electrons do not add to the atomic size. In fact, the increasing nuclear charge (remember that a proton is added to the nucleus for each electron) causes the radii of the lanthanide elements to decrease signif-icantly going from left to right. This lanthanide contraction just offsets the normal increase in size due to going from one principal quantum level to another. Thus the 5d elements, instead of being significantly larger than the 4d elements, are almost identical to them in size. This leads to a great similarity in the chemistry of the 4d and 5d elements in a given vertical group. For example, the chemical properties of hafnium and zirconium are remarkably similar, and they always occur together in nature. Their separation, which is probably more difficult than the separation of any other pair of elements, often requires fractional distillation of their compounds.
In general, the differences between the 4d and 5d elements in a group increase grad-ually going from left to right. For example, niobium and tantalum are also quite similar, but less so than zirconium and hafnium.
Although generally less well known than the 3d elements, the 4d and 5d transition metals have certain very useful properties. For example, zirconium and zirconium oxide (ZrO2) have great resistance to high temperatures and are used, along with niobium and molybdenum alloys, for space vehicle parts that are exposed to high temperatures during reentry into the earth’s atmosphere. Niobium and molybdenum are also important alloying materials for certain types of steel. Tantalum, which has a high resistance to attack by body fluids, is often used for replacement of bones. The platinum group metals—ruthenium, osmium, rhodium, iridium, palladium, and platinum—are all quite similar and are widely used as catalysts for many types of industrial processes.
21.2 The First-Row Transition Metals We have seen that the transition metals are similar in many ways but also show impor-tant differences. We will now explore some of the specific properties of each of the 3d transition metals.
Scandium is a rare element that exists in compounds mainly in the 3 oxidation state—for example, in ScCl3, Sc2O3, and Sc2(SO4)3. The chemistry of scandium strongly resembles that of the lanthanides, with most of its compounds being colorless and FIGURE 21.3 Atomic radii of the 3d, 4d, and 5d transi-tion series.
0.2 Atomic radii (nm) Atomic number 0.1 0.15 La Hf Ta W Re Os Ir Pt Au Zr Y Nb Mo Tc Ru Rh Pd Ag Sc Ti V Cr Mn Fe Co Ni Cu 1st series (3d) 2nd series (4d) 3rd series (5d) Niobium was originally called columbium and is still occasionally referred to by that name.
950 Chapter Twenty-One Transition Metals and Coordination Chemistry diamagnetic. This is not surprising; as we will see in Section 21.6, the color and mag-netism of transition metal compounds usually arise from the d electrons on the metal ion, and Sc3 has no d electrons. Scandium metal, which can be prepared by electrolysis of molten ScCl3, is not widely used because of its rarity, but it is found in some electronic devices, such as high-intensity lamps.
Titanium is widely distributed in the earth’s crust (0.6% by mass). Because of its rel-atively low density and high strength, titanium is an excellent structural material, espe-cially in jet engines, where light weight and stability at high temperatures are required.
Nearly 5000 kg of titanium alloys is used in each engine of a Boeing 747 jetliner. In ad-dition, the resistance of titanium to chemical attack makes it a useful material for pipes, pumps, and reaction vessels in the chemical industry.
The most familiar compound of titanium is no doubt responsible for the white color of this paper. Titanium dioxide, or more correctly, titanium(IV) oxide (TiO2), is a highly opaque substance used as the white pigment in paper, paint, linoleum, plastics, syn-thetic fibers, whitewall tires, and cosmetics (sunscreens, for example). Approximately 700,000 tons is used annually in these and other products. Titanium(IV) oxide is widely dispersed in nature, but the main ores are rutile (impure TiO2) and ilmenite (FeTiO3).
Rutile is processed by treatment with chlorine to form volatile TiCl4, which is separated from the impurities and burned to form TiO2: Ilmenite is treated with sulfuric acid to form a soluble sulfate: When this aqueous mixture is allowed to stand, under vacuum, solid forms first and is removed. The mixture is then heated, and the insoluble titanium(IV) oxide hydrate forms. The water of hydration is driven off by heating to form pure TiO2: In its compounds, titanium is most often found in the 4 oxidation state. Examples are TiO2 and TiCl4, the latter a colorless liquid that fumes in moist air to produce TiO2: Titanium(III) compounds can be produced by reduction of the 4 state. In aqueous so-lution, Ti3 exists as the purple Ti(H2O)6 3 ion, which is slowly oxidized to titanium(IV) by air. Titanium(II) is not stable in aqueous solution but does exist in the solid state in compounds such as TiO and the dihalides of general formula TiX2.
Vanadium is widely spread throughout the earth’s crust (0.02% by mass). It is used mostly in alloys with other metals such as iron (80% of vanadium is used in steel) and ti-tanium. Vanadium(V) oxide (V2O5) is used as an industrial catalyst in the production of materials such as sulfuric acid.
Pure vanadium can be obtained from the electrolytic reduction of fused salts, such as VCl2, to produce a metal similar to titanium that is steel gray, hard, and corrosion resistant. Often the pure element is not required for alloying. For example, ferrovanadium, produced by the reduction of a mixture of V2O5 and Fe2O3 with aluminum, is added to iron to form vanadium steel, a hard steel used for engine parts and axles.
The principal oxidation state of vanadium is 5, found in compounds such as the or-ange V2O5 and the colorless VF5 The oxidation states from 5 to 2 all exist in aqueous solution (see Table 21.4). The higher oxidation states, (mp 19.5°C).
(mp 650°C) TiCl41l2 2H2O1l2 ¡ TiO21s2 4HCl1g2 (bp 137°C) TiO2 H2O1s2 ¡ Heat TiO21s2 H2O1g2 (TiO2 H2O) FeSO4 7H2O FeTiO31s2 2H2SO41aq2 ¡ Fe21aq2 TiO21aq2 2SO4 21aq2 2H2O1l2 TiCl41g2 O21g2 ¡ TiO21s2 2Cl21g2 An X ray of a patient who has had a hip re-placement. The normal hip joint is on the left; the hip joint constructed from tanta-lum metal is on the right.
Ti(H2O)6 3 is purple in solution.
The manufacture of sulfuric acid was discussed at the end of Chapter 3.
The most common oxidation state for vanadium is 5.
21.2 The First-Row Transition Metals 951 5 and 4, do not exist as hydrated ions of the type Vn(aq) because the highly charged ion causes the attached water molecules to be very acidic. The H ions are lost to give the oxycations VO2 and VO2. The hydrated V3 and V2 ions are easily oxidized and thus can function as reducing agents in aqueous solution.
Although chromium is relatively rare, it is a very important industrial material. The chief ore of chromium is chromite (FeCr2O4), which can be reduced by carbon to give ferrochrome, Ferrochrome which can be added directly to iron in the steelmaking process. Chromium metal, which is often used to plate steel, is hard and brittle and maintains a bright surface by developing a tough invisible oxide coating.
Chromium commonly forms compounds in which it has the oxidation state 2, 3, or 6, as shown in Table 21.5. The Cr2 (chromous) ion is a powerful reducing agent in aqueous solution. In fact, traces of O2 in other gases can be removed by bubbling through a Cr2 solution: The chromium(VI) species are excellent oxidizing agents, especially in acidic solu-tion, where chromium(VI) as the dichromate ion is reduced to the Cr3 ion: Cr2O7 21aq2 14H1aq2 6e ¡ 2Cr31aq2 7H2O1l2 e° 1.33 V (Cr2O7 2) 4Cr21aq2 O21g2 4H1aq2 ¡ 4Cr31aq2 2H2O1l2 FeCr2O41s2 4C1s2 ¡ Fe1s2 2Cr1s2 4CO1g2 CHEMICAL IMPACT Titanium Dioxide—Miracle Coating T itanium dioxide, more properly called titanium(IV) ox-ide, is a very important material. Approximately 1.5 mil-lion tons of the substance is produced each year in the United States for use as a pigment in paper and paints and as a component of sunscreens.
In recent years, however, scientists have found a new use for TiO2. When surfaces are coated with titanium dioxide, they become resistant to dirt and bacteria. For example, the Pilkington Glass Company is now making glass coated with TiO2 that cleans itself. All the glass needs is sun and rain to keep itself clean. The self-cleaning action arises from two ef-fects. First, the coating of TiO2 acts as a catalyst in the pres-ence of ultraviolet (UV) light to break down carbon-based pollutants to carbon dioxide and water. Second, because TiO2 reduces surface tension, rainwater “sheets” instead of form-ing droplets on the glass, thereby washing away the grime on the surface of the glass. Although this self-cleaning glass is bad news for window washers, it could save millions of dol-lars in maintenance costs for owners of commercial buildings.
Because the TiO2-treated glass requires UV light for its action, it does not work well for interior surfaces where UV light is present only in small amounts. However, a team of Japanese researchers has found that if the TiO2 coating is doped with nitrogen atoms, it will catalyze the breakdown of dirt in the presence of visible light as well as UV light.
Studies also show that this N-doped TiO2 surface coating kills many types of bacteria in the presence of visible or ultraviolet light. This discovery could lead to products such as self-sterilizing bathroom tiles, counters, and toilets. In addition, because the TiO2 on the surface of glass has such a strong attraction for water molecules (greatly lowering the surface tension), water does not bead up to form droplets.
Just as this effect produces sheeting action on exterior glass, so it prevents interior windows and mirrors from “fogging up.” Titanium dioxide, a cheap and plentiful material, may prove to be worth its weight in gold as a surface coating.
TABLE 21.4 Oxidation States and Species for Vanadium in Aqueous Solution Oxidation State of Species in Aqueous Vanadium Solution 5 VO2 (yellow) 4 VO2 (blue) 3 V3(aq) (blue-green) 2 V2(aq) (violet) TABLE 21.5 Typical Chromium Compounds Oxidation Examples of State of Compounds Chromium (X halogen) 2 CrX2 3 CrX3 Cr2O3 (green) Cr(OH)3 (blue-green) 6 K2Cr2O7 (orange) Na2CrO4 (yellow) CrO3 (red) ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 952 Chapter Twenty-One Transition Metals and Coordination Chemistry The oxidizing ability of the dichromate ion is strongly pH-dependent, increasing as [H] increases, as predicted by Le Châtelier’s principle. In basic solution, chromium(VI) exists as the chromate ion, a much less powerful oxidizing agent: The structures of the and ions are shown in Fig. 21.4.
Red chromium(VI) oxide (CrO3) dissolves in water to give a strongly acidic, red-orange solution: 2CrO31s2 H2O1l2 ¡ 2H1aq2 Cr2O7 21aq2 CrO4 2 Cr2O7 2 e° 0.13 V CrO4 21aq2 4H2O1l2 3e ¡ Cr1OH231s2 5OH1aq2 FIGURE 21.4 The structures of the chromium(VI) anions: (a) which exists in acidic solution, and (b) which exists in basic solution.
CrO4 2, Cr2O7 2, F O O O O O O 115° O 2– O O O O Cr 2– (b) (a) Cr Cr CHEMICAL IMPACT Titanium Makes Great Bicycles O ne of the most interesting characteristics of the world of bicycling is the competition among various frame ma-terials. Bicycle frames are now built from steel, aluminum, carbon fiber composites, and titanium, with each material having advantages and disadvantages. Steel is strong, eco-nomical, adaptable, and (unfortunately) “rustable.”Aluminum is light and stiff but has relatively low fatigue limits (resis-tance to repeated stresses). Carbon fiber composites have amaz-ing strength-to-mass ratios and have shock- and vibration-dampening properties superior to any metal; however, they are very expensive. Titanium has a density approximately 43% less than that of steel, a yield strength (when alloyed with metals such as aluminum and tin) that is 30% greater than that of steel, an extraordinary resistance to fatigue, and a high resistance to corrosion, but it is expensive and diffi-cult to work.
Of all these materials, titanium gives the bicycle that fa-natics seem to love the most. After their first ride on a bi-cycle with a titanium frame, most experienced cyclists find themselves shaking their heads and searching hard for the right words to describe the experience. Typically, the word “magic” is used a great deal in the ensuing description.
The magic of titanium results from its combination of toughness, stretchability, and resilience. A bicycle that is built stiff to resist pedaling loads usually responds by giv-ing a harsh, uncomfortable ride. A titanium bike is very stiff against high pedaling torques, but it seems to transmit much less road shock than bikes made of competitive materials.
Why titanium excels in dampening vibrations is not entirely clear. Despite titanium’s significantly lower density than steel, shock waves travel more slowly in titanium than in steel. Whatever the explanation for its shock-absorbing abil-ities, titanium provides three things that cyclists find crucial: light weight, stiffness, and a smooth ride—magic.
Titanium is quite abundant in the earth’s crust, ranking ninth of all the elements and second among the transition elements. The metallurgy of titanium presents special chal-lenges. Carbon, the reducing agent most commonly used to obtain metals from their oxide ores, cannot be used because it forms intractable interstitial carbides with titanium. These 21.2 The First-Row Transition Metals 953 It is possible to precipitate bright orange dichromate salts, such as K2Cr2O7, from these solutions. When made basic, the solution turns yellow, and chromate salts such as Na2CrO4 can be obtained. A mixture of chromium(VI) oxide and concentrated sulfuric acid, com-monly called cleaning solution, is a powerful oxidizing medium that can remove organic materials from analytical glassware, yielding a very clean surface.
Manganese is relatively abundant (0.1% of the earth’s crust), although no significant sources are found in the United States. The most common use of manganese is in the pro-duction of an especially hard steel used for rock crushers, bank vaults, and armor plate.
One interesting source of manganese is from manganese nodules found on the ocean floor.
These roughly spherical “rocks” contain mixtures of manganese and iron oxides as well as smaller amounts of other metals such as cobalt, nickel, and copper. Apparently, the nodules were formed at least partly by the action of marine organisms. Because of the abundance of these nodules, there is much interest in developing economical methods for their recovery and processing.
Manganese can exist in all oxidation states from 2 to 7, although 2 and 7 are the most common. Manganese(II) forms an extensive series of salts with all the common anions. In aqueous solution Mn2 forms Mn(H2O)6 2, which has a light pink color. Man-ganese(VII) is found in the intensely purple permanganate ion (MnO4 ). Widely used as an analytical reagent in acidic solution, the MnO4 ion behaves as a strong oxidizing agent, with the manganese becoming Mn2: Several typical compounds of manganese are shown in Table 21.6.
MnO4 1aq2 8H1aq2 5e ¡ Mn21aq2 4H2O1l2 e° 1.51 V TABLE 21.6 Some Compounds of Manganese in Its Most Common Oxidation States Oxidation State of Examples of Manganese Compounds 2 Mn(OH)2 (pink) MnS (salmon) MnSO4 (reddish) MnCl2 (pink) 4 MnO2 (dark brown) 7 KMnO4 (purple) carbides are extraordinarily hard and have melting points close to However, if chlorine gas is used in con-junction with carbon to treat the ore, volatile TiCl4 is formed, which can be distilled off and then reduced with magnesium or sodium at approximately to form a titanium “sponge.” This sponge is then ground up, cleaned with aqua regia (a 1:3 mixture of concentrated HNO3 and concentrated HCl), melted under a blanket of inert gas (to prevent reac-tion with oxygen), and cast into ingots. Titanium, a lustrous, silvery metal with a high melting point crystal-lizes in a hexagonal closest packed structure. Because tita-nium tends to become quite brittle when trace impurities such as C, N, and O are present, it must be fabricated with great care.
Titanium’s unusual ability to stretch makes it hard to machine. It tends to push away even from a very sharp cut-ting blade, giving a rather unpredictable final dimension.
Also, because titanium is embrittled by reaction with oxy-gen, all welding operations must be carried out under a shielding gas such as argon.
However, the bicycle that results is worth all these dif-ficulties. One woman described a titanium bicycle as “the one God rides on Sunday.” (1667°C), 1000°C 3000°C.
A titanium bicycle.
954 Chapter Twenty-One Transition Metals and Coordination Chemistry Iron is the most abundant heavy metal (4.7% of the earth’s crust) and the most im-portant to our civilization. It is a white, lustrous, not particularly hard metal that is very reactive toward oxidizing agents. For example, in moist air it is rapidly oxidized by oxy-gen to form rust, a mixture of iron oxides.
The chemistry of iron mainly involves its 2 and 3 oxidation states. Typical compounds are shown in Table 21.7. In aqueous solutions iron(II) salts are generally light green because of the presence of Fe(H2O)6 2. Although the Fe(H2O)6 3 ion is col-orless, aqueous solutions of iron(III) salts are usually yellow to brown in color due to the presence of Fe(OH)(H2O)5 2, which results from the acidity of Fe(H2O)6 3 (Ka 6 103): Although cobalt is relatively rare, it is found in ores such as smaltite (CoAs2) and cobaltite (CoAsS) in large enough concentrations to make its production econom-ically feasible. Cobalt is a hard, bluish white metal mainly used in alloys such as stain-less steel and stellite, an alloy of iron, copper, and tungsten that is used in surgical instruments.
The chemistry of cobalt involves mainly its 2 and 3 oxidation states, although compounds containing cobalt in the 0, 1, or 4 oxidation state are known. Aqueous solutions of cobalt(II) salts contain the Co(H2O)6 2 ion, which has a characteristic rose color. Cobalt forms a wide variety of coordination compounds, many of which will be discussed in later sections of this chapter. Some typical cobalt compounds are shown in Table 21.8.
Nickel, which ranks twenty-fourth in elemental abundance in the earth’s crust, is found in ores, where it is combined mainly with arsenic, antimony, and sulfur. Nickel metal, a silvery white substance with high electrical and thermal conductivities, is quite resistant to corrosion and is often used for plating more active metals. Nickel is also widely used in the production of alloys such as steel.
Nickel in compounds is almost exclusively in the 2 oxidation state. Aqueous solu-tions of nickel(II) salts contain the Ni(H2O)6 2 ion, which has a characteristic emerald green color. Coordination compounds of nickel(II) will be discussed later in this chapter.
Some typical nickel compounds are shown in Table 21.9.
Fe1H2O26 31aq2 ∆Fe1OH21H2O25 21aq2 H1aq2 TABLE 21.7 Typical Compounds of Iron Oxidation State of Examples of Iron Compounds 2 FeO (black) FeS (brownish black) FeSO4 7H2O (green) K4Fe(CN)6 (yellow) 3 FeCl3 (brownish black) Fe2O3 (reddish brown) K3Fe(CN)6 (red) Fe(SCN)3 (red) 2, 3 Fe3O4 (black) (mixture) KFe[Fe(CN)6] (deep blue, “Prussian blue”) TABLE 21.8 Typical Compounds of Cobalt Oxidation Examples of State Compounds 2 CoSO4 (dark blue) [Co(H2O)6]Cl2 (pink) Co(H2O)62 (red) CoS (black) CoO (greenish brown) 3 CoF3 (brown) Co2O3 (charcoal) K3[Co(CN)6] (yellow) [Co(NH3)6]Cl3 (yellow) TABLE 21.9 Typical Compounds of Nickel Oxidation State of Examples of Nickel Compounds 2 NiCl2 (yellow) [Ni(H2O)6]Cl2 (green) NiO (greenish black) NiS (black) [Ni(H2O)6]SO4 (green) Ni(NH3)62 (blue) An aqueous solution containing the Ni2 ion.
21.3 Coordination Compounds 955 Copper, widely distributed in nature in ores containing sulfides, arsenides, chlorides, and carbonates, is valued for its high electrical conductivity and its resistance to corro-sion. It is widely used for plumbing, and 50% of all copper produced annually is used for electrical applications. Copper is a major constituent in several well-known alloys (see Table 21.10).
Although copper is not highly reactive (it will not reduce H to H2, for example), the reddish metal does slowly corrode in air, producing the characteristic green patina con-sisting of basic copper sulfate Basic copper sulfate and other similar compounds.
The chemistry of copper principally involves the 2 oxidation state, but many com-pounds containing copper(I) are also known. Aqueous solutions of copper(II) salts are a characteristic bright blue color due to the presence of the Cu(H2O)6 2 ion. Table 21.11 lists some typical copper compounds.
Although trace amounts of copper are essential for life, copper in large amounts is quite toxic; copper salts are used to kill bacteria, fungi, and algae. For example, paints containing copper are used on ship hulls to prevent fouling by marine organisms.
Widely dispersed in the earth’s crust, zinc is mainly refined from sphalerite (ZnS), which often occurs with galena (PbS). Zinc is a white, lustrous, very active metal that behaves as an excellent reducing agent and tarnishes rapidly. About 90% of the zinc produced is used for galvanizing steel. Zinc forms colorless salts in the 2 oxidation state.
21.3 Coordination Compounds Transition metal ions characteristically form coordination compounds, which are usu-ally colored and often paramagnetic. A coordination compound typically consists of a complex ion, a transition metal ion with its attached ligands (see Section 15.8), and counterions, anions or cations as needed to produce a compound with no net charge.
The substance [Co(NH3)5Cl]Cl2 is a typical coordination compound. The brackets in-dicate the composition of the complex ion, in this case Co(NH3)5Cl2, and the two Cl counterions are shown outside the brackets. Note that in this compound one Cl acts as a ligand along with the five NH3 molecules. In the solid state this compound consists of the large Co(NH3)5Cl2 cations and twice as many Cl anions, all packed together as efficiently as possible. When dissolved in water, the solid behaves like any ionic solid; the cations and anions are assumed to separate and move about independently: 3Co1NH325Cl4Cl21s2 ¡ H2O Co1NH325Cl21aq2 2Cl1aq2 3Cu1s2 2H2O1l2 SO21g2 2O21g2 ¡ Cu31OH24SO4 TABLE 21.11 Typical Compounds of Copper Oxidation State of Examples of Copper Compounds 1 Cu2O (red) Cu2S (black) CuCl (white) 2 CuO (black) CuSO4 5H2O (blue) CuCl2 2H2O (green) Cu(H2O)62 (blue) TABLE 21.10 Alloys Containing Copper Alloy Composition (% by mass in parentheses) Brass Cu (20–97), Zn (2–80), Sn (0–14), Pb (0–12), Mn (0–25) Bronze Cu (50–98), Sn (0–35), Zn (0–29), Pb (0–50), P (0–3) Sterling silver Cu (7.5), Ag (92.5) Gold (18-karat) Cu (5–15), Au (75), Ag (10–20) Gold (14-karat) Cu (12–28), Au (58), Ag (4–30) Copper roofs and bronze statues, such as the Statue of Liberty, turn green in air because Cu3(OH)4SO4 and Cu4(OH)6SO4 form.
956 Chapter Twenty-One Transition Metals and Coordination Chemistry Coordination compounds have been known since about 1700, but their true nature was not understood until the 1890s when a young Swiss chemist named Alfred Werner (1866–1919) proposed that transition metal ions have two types of valence (combin-ing ability). One type of valence, which Werner called the secondary valence, refers to the ability of a metal ion to bind to Lewis bases (ligands) to form complex ions.
The other type, the primary valence, refers to the ability of the metal ion to form ionic bonds with oppositely charged ions. Thus Werner explained that the compound, orig-inally written as CoCl3 5NH3, was really [Co(NH3)5Cl]Cl2, where the Co3 ion has a primary valence of 3, satisfied by the three Cl ions, and a secondary valence of 6, satisfied by the six ligands (five NH3 and one Cl). We now call the primary valence the oxidation state and the secondary valence the coordination number, which re-flects the number of bonds formed between the metal ion and the ligands in the com-plex ion.
Coordination Number The number of bonds formed by metal ions to ligands in complex ions varies from two to eight depending on the size, charge, and electron configuration of the transition metal ion. As shown in Table 21.12, 6 is the most common coordination number, followed closely by 4, with a few metal ions showing a coordination number of 2. Many metal ions show more than one coordination number, and there is really no simple way to predict what the coordination number will be in a particular case. The typical geome-tries for the various common coordination numbers are shown in Fig. 21.5. Note that six ligands produce an octahedral arrangement around the metal ion. Four ligands can form either a tetrahedral or a square planar arrangement, and two ligands give a linear structure.
Ligands A ligand is a neutral molecule or ion having a lone electron pair that can be used to form a bond to a metal ion. The formation of a metal–ligand bond therefore can be described as the interaction between a Lewis base (the ligand) and a Lewis acid (the metal ion). The resulting bond is often called a coordinate covalent bond.
A ligand that can form one bond to a metal ion is called a monodentate ligand, or a unidentate ligand (from root words meaning “one tooth”). Examples of unidentate li-gands are shown in Table 21.13.
Some ligands have more than one atom with a lone electron pair that can be used to bond to a metal ion. Such ligands are said to be chelating ligands, or chelates (from the TABLE 21.12 Typical Coordination Numbers for Some Common Metal Ions Coordination Coordination Coordination M Numbers M2 Numbers M3 Numbers Cu 2, 4 Mn2 4, 6 Sc3 6 Ag 2 Fe2 6 Cr3 6 Au 2, 4 Co2 4, 6 Co3 6 Ni2 4, 6 Cu2 4, 6 Au3 4 Zn2 4, 6 Coordination number 2 4 6 Geometry Linear Tetrahedral Square planar Octahedral FIGURE 21.5 The ligand arrangements for coordination numbers 2, 4, and 6.
TABLE 21.13 Some Common Ligands Type Examples Unidentate/monodentate H2O CN SCN (thiocyanate) X (halides) NH3 NO2 (nitrite) OH Bidentate Oxalate Ethylenediamine (en) Polydentate Diethylenetriamine (dien) Ethylenediaminetetraacetate (EDTA) C O O O H2C CH2 (CH2)2 C O O CH2 C N N C O O H2C O () () () () Six coordinating atoms (CH2)2 NH2 H2N NH (CH2)2 Three coordinating atoms C O C O M () () O O H2C H2N CH2 NH2 M Mn+ H H N H (b) (a) Mn+ N N H H H H C C H H H H C O B O B B O O N O O N O O B O O O C H2 C CH2 CH2 CH2 2 CH2 CH2 C C M FIGURE 21.6 (a) The bidentate ligand ethylenediamine can bond to the metal ion through the lone pair on each nitrogen atom, thus forming two coordinate covalent bonds. (b) Ammonia is a monodentate ligand.
FIGURE 21.7 The coordination of EDTA with a 2 metal ion.
Sample Exercise 21.1 In an older system the negatively charged ligands were named first, then neutral ligands, with positively charged ligands named last. We will follow the newer convention in this text.
TABLE 21.14 Names of Some Common Unidentate Ligands Neutral Molecules Aqua H2O Ammine NH3 Methylamine CH3NH2 Carbonyl CO Nitrosyl NO Anions Fluoro Chloro Bromo Iodo Hydroxo Cyano CN OH I Br Cl F TABLE 21.15 Latin Names Used for Some Metal Ions in Anionic Complex Ions Name in an Metal Anionic Complex Iron Ferrate Copper Cuprate Lead Plumbate Silver Argentate Gold Aurate Tin Stannate 21.3 Coordination Compounds 959 disregarding the prefix. Since the counterions are chloride ions, the compound is named as a chloride salt: Pentaamminechlorocobalt(III) chloride Cation Anion b. First, we determine the oxidation state of the iron by considering the other charged species. The compound contains three K ions and six CN ions. Therefore, the iron must carry a charge of 3, giving a total of six positive charges to balance the six negative charges. The complex ion present is thus Fe(CN)6 3. The cyanide ligands are each designated cyano, and the prefix hexa- indicates that six are present. Since the complex ion is an anion, we use the Latin name ferrate. The oxidation state is indicated by (III) at the end of the name. The anion name is therefore hexacyanoferrate(III). The cations are K ions, which are simply named potassium. Putting this together gives the name Potassium hexacyanoferrate(III) Cation Anion (The common name of this compound is potassium ferricyanide.) c. We first determine the oxidation state of the iron by looking at the other charged species: four NO2 ions and one SO4 2 ion. The ethylenediamine is neutral. Thus the two iron ions must carry a total positive charge of 6 to balance the six negative charges. This means that each iron has a 3 oxidation state and is designated as iron(III).
Since the name ethylenediamine already contains di, we use bis- instead of di- to indicate the two en ligands. The name for NO2 as a ligand is nitro, and the prefix di-indicates the presence of two NO2 ligands. Since the anion is sulfate, the compound’s name is Cation Anion Because the complex ion is a cation, the Latin name for iron is not used.
See Exercises 21.29 through 21.32.
Naming Coordination Compounds II Given the following systematic names, give the formula of each coordination compound.
a. Triamminebromoplatinum(II) chloride b. Potassium hexafluorocobaltate(III) Solution a. Triammine signifies three ammonia ligands, and bromo indicates one bromide ion as a ligand. The oxidation state of platinum is 2, as indicated by the Roman numeral II.
Thus the complex ion is [Pt(NH3)3Br]. One chloride ion is needed to balance the 1 charge of this cation. The formula of the compound is [Pt(NH3)3Br]Cl. Note that brackets enclose the complex ion.
b. The complex ion contains six fluoride ligands attached to a Co3 ion to give CoF6 3.
Note that the -ate ending indicates that the complex ion is an anion. The cations are K ions, and three are required to balance the 3 charge on the complex ion. Thus the formula is K3[CoF6].
See Exercises 21.33 and 21.34.
Bis1ethylenediamine2dinitroiron1III2 sulfate ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎧ ⎨ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Sample Exercise 21.2 (top) An aqueous solution of [Co(NH3)5Cl]Cl2. (bottom) Solid K3Fe(CN)6.
960 Chapter Twenty-One Transition Metals and Coordination Chemistry 21.4 Isomerism When two or more species have the same formula but different properties, they are said to be isomers. Although isomers contain exactly the same types and numbers of atoms, the arrangements of the atoms differ, and this leads to different properties. We will con-sider two main types of isomerism: structural isomerism, where the isomers contain the same atoms but one or more bonds differ, and stereoisomerism, where all the bonds in the isomers are the same but the spatial arrangements of the atoms are different. Each of these classes also has subclasses (see Fig. 21.8), which we will now consider.
Structural Isomerism The first type of structural isomerism we will consider is coordination isomerism, in which the composition of the complex ion varies. For example, [Cr(NH3)5SO4]Br and [Cr(NH3)5Br]SO4 are coordination isomers. In the first case, SO4 2 is coordinated to Cr3, and Br is the counterion; in the second case, the roles of these ions are reversed.
CHEMICAL IMPACT Alfred Werner: Coordination Chemist D uring the early and middle parts of the nineteenth cen-tury, chemists prepared a large number of colored com-pounds containing transition metals and other substances such as ammonia, chloride ion, cyanide ion, and water. These compounds were very interesting to chemists who were try-ing to understand the nature of bonding (Dalton’s atomic theory of 1808 was very new at this time), and many theo-ries were suggested to explain these substances. The most widely accepted early theory was the chain theory, champi-oned by Sophus Mads Jorgensen (1837–1914), professor of chemistry at the University of Copenhagen. The chain theory got its name from the postulate that metal ammine com-plexes contain chains of NH3 molecules. For example, Jorgensen proposed the structure explanation for the constitution of these compounds. Writ-ing furiously the rest of that night and into the late after-noon of the following day, he constructed a scientific paper containing his now famous coordination theory. This model postulates an octahedral arrangement of ligands around the Co3 ion, producing the Co(NH3)6 3 complex ion with three ions as counterions. Thus Werner’s picture of Co(NH3)6Cl3 varied greatly from the chain theory.
In his paper on the coordination theory, Werner ex-plained not only the metal–ammine compounds but also most of the other known transition metal compounds, and the importance of his contribution was recognized immedi-ately. He was appointed professor at the University of Zurich, where he spent the rest of his life studying coordi-nation compounds and refining his theory. Alfred Werner was a confident, impulsive man of seemingly boundless en-ergy, who was known for his inspiring lectures, his intoler-ance of incompetence (he once threw a chair at a student who performed poorly on an oral exam), and his intuitive scientific brilliance. For example, he was the first to show that stereochemistry is a general phenomenon, not one ex-hibited only by carbon, as was previously thought. He also recognized and named many types of isomerism.
In 1913, for his work on coordination chemistry and stereochemistry, Werner became the fourteenth Nobel Prize winner in chemistry and the first Swiss chemist to be so honored. Werner’s work is even more remarkable when one realizes that his ideas preceded by many years any real understanding of the nature of covalent bonds.
Cl Co CI NH3 NH3 NH3 NH3 CI NH3 CI NH3 for the compound Co(NH3)6Cl3. In the late nineteenth cen-tury this theory was used in classrooms around the world to explain the nature of metal–ammine compounds.
However, in 1890, a young Swiss chemist named Al-fred Werner, who had just obtained a Ph.D. in the field of organic chemistry, became so interested in these compounds that he apparently even dreamed about them. In the middle of one night Werner awoke realizing that he had the correct Ammine is the name for NH3 as a ligand.
21.4 Isomerism 961 Another example of coordination isomerism is the [Co(en)3][Cr(ox)3] and [Cr(en)3] [Co(ox)3] pair, where ox represents the oxalate ion, a bidentate ligand shown in Table 21.13.
In a second type of structural isomerism, linkage isomerism, the composition of the complex ion is the same, but the point of attachment of at least one of the ligands differs.
Two ligands that can attach to metal ions in different ways are thiocyanate (SCN), which can bond through lone electron pairs on the nitrogen or the sulfur atom, and the nitrite ion (NO2 ), which can bond through lone electron pairs on the nitrogen or the oxygen atom. For example, the following two compounds are linkage isomers: Tetraamminechloronitrocobalt(III) chloride (yellow) Tetraamminechloronitritocobalt(III) chloride (red) In the first case, the ligand is called nitro and is attached to Co3 through the ni-trogen atom; in the second case, the ligand is called nitrito and is attached to Co3 through an oxygen atom (see Fig. 21.9).
Stereoisomerism Stereoisomers have the same bonds but different spatial arrangements of the atoms. One type, geometrical isomerism, or cis–trans isomerism, occurs when atoms or groups of atoms can assume different positions around a rigid ring or bond. An important example is the compound Pt(NH3)2Cl2, which has a square planar structure. The two possible arrangements of the ligands are shown in Fig. 21.10. In the trans isomer, the ammonia molecules are across (trans) from each other. In the cis isomer, the ammonia molecules are next (cis) to each other.
Geometrical isomerism also occurs in octahedral complex ions. For example, the com-pound [Co(NH3)4Cl2]Cl has cis and trans isomers (Fig. 21.11).
A second type of stereoisomerism is called optical isomerism because the isomers have opposite effects on plane-polarized light. When light is emitted from a source such as a glowing filament, the oscillating electric fields of the photons in the beam are ori-ented randomly, as shown in Fig. 21.12. If this light is passed through a polarizer, only the photons with electric fields oscillating in a single plane remain, constituting plane-polarized light.
In 1815, a French physicist, Jean Biot (1774–1862), showed that certain crystals could rotate the plane of polarization of light. Later it was found that solutions of certain NO2 NO2 3Co1NH3241ONO2Cl4Cl 3Co1NH3241NO22Cl4Cl FIGURE 21.8 Some classes of isomers.
FIGURE 21.9 As a ligand, NO2 can bond to a metal ion (a) through a lone pair on the nitrogen atom or (b) through a lone pair on one of the oxygen atoms.
Isomers (same formula but different properties) Stereoisomers (same bonds, different spatial arrangements) Structural isomers (different bonds) Optical isomerism Geometric (cis-trans) isomerism Linkage isomerism Coordination isomerism Co (a) N O O Co (b) N O O H3N H3N Cl Cl Pt (a) NH3 H3N Cl Cl Pt (b) FIGURE 21.10 (a) The cis isomer of Pt(NH3)2Cl2 (yellow).
(b) The trans isomer of Pt(NH3)2Cl2 (pale yellow).
962 Chapter Twenty-One Transition Metals and Coordination Chemistry FIGURE 21.11 (a) The trans isomer of [Co(NH3)4Cl2].
The chloride ligands are directly across from each other. (b) The cis isomer of [Co(NH3)4Cl2]. The chloride ligands in this case share an edge of the octahedron. Be-cause of their different structures, the trans isomer of [Co(NH3)4Cl2]Cl is green and the cis isomer is violet.
FIGURE 21.12 Unpolarized light consists of waves vibrat-ing in many different planes (indicated by the arrows). The polarizing filter blocks all waves except those vibrating in a given plane.
FIGURE 21.13 The rotation of the plane of polarized light by an optically active substance. The angle of rotation is called theta ( ).
u H3N Co H3N NH3 NH3 Cl Cl H3N Co H3N NH3 Cl Cl NH3 Cl Cl Co Cl Cl Co Light source Unpolarized light Polarizing filter Plane polarized light Unpolarized light Polarizing filter Polarized light Tube containing sample θ Rotated polarized light compounds could do the same thing (see Fig. 21.13). Louis Pasteur (1822–1895) was the first to understand this behavior. In 1848 he noted that solid sodium ammonium tartrate (NaNH4C4H4O4) existed as a mixture of two types of crystals, which he painstakingly separated with tweezers. Separate solutions of these two types of crystals rotated plane-polarized light in exactly opposite directions. This led to a connection between optical activity and molecular structure.
21.4 Isomerism 963 We now realize that optical activity is exhibited by molecules that have nonsuperim-posable mirror images. Your hands are nonsuperimposable mirror images (Fig. 21.15).
The two hands are related like an object and its mirror image; one hand cannot be turned to make it identical to the other. Many molecules show this same feature, such as the CHEMICAL IMPACT The Importance of Being cis S ome of the most important advancements of science are the results of accidental discoveries—for example, peni-cillin, Teflon, and the sugar substitutes cyclamate and aspar-tame. Another important chance discovery occurred in 1964, when a group of scientists using platinum electrodes to ap-ply an electric field to a colony of E. coli bacteria noticed that the bacteria failed to divide but continued to grow, form-ing long fibrous cells. Further study revealed that cell divi-sion was inhibited by small concentrations of cis-Pt(NH3)2Cl2 and cis-Pt(NH3)2Cl4 formed electrolytically in the solution.
Cancerous cells multiply very rapidly because cell division is uncontrolled. Thus these and similar platinum complexes were evaluated as antitumor agents, which in-hibit the division of cancer cells. The results showed that cis-Pt(NH3)2Cl2 was active against a wide variety of tumors, including testicular and ovarian tumors, which are very re-sistant to treatment by more traditional methods. However, although the cis complex showed significant antitumor activity, the corresponding trans complex had no effect on tumors. This shows the importance of isomerism in biologic systems. When drugs are synthesized, great care must be taken to obtain the correct isomer.
Although cis-Pt(NH3)2Cl2 has proven to be a valuable drug, unfortunately it has some troublesome side effects, the most serious being kidney damage. As a result, the search continues for even more effective antitumor agents. Promis-ing candidates are shown in Fig. 21.14. Note that they are all cis complexes.
FIGURE 21.14 Some cis complexes of platinum and palladium that show significant antitumor activity. It is thought that the cis complexes work by losing two adjacent ligands and forming coordinate covalent bonds to adjacent bases on a DNA molecule.
Left hand Right hand Mirror image of right hand FIGURE 21.15 A human hand exhibits a nonsuperimposable mirror image. Note that the mirror image of the right hand (while identical to the left hand) cannot be turned in any way to make it identical to (superimposable on) the actual right hand.
964 Chapter Twenty-One Transition Metals and Coordination Chemistry complex ion [Co(en)3]3 shown in Fig. 21.16. Objects that have nonsuperimposable mir-ror images are said to be chiral (from the Greek word cheir, meaning “hand”).
The isomers of [Co(en)3]3 (Fig. 21.17) are nonsuperimposable mirror images called enantiomers, which rotate plane-polarized light in opposite directions and are thus optical isomers. The isomer that rotates the plane of light to the right (when viewed down the beam of oncoming light) is said to be dextrorotatory, designated by d. The isomer that rotates the plane of light to the left is levorotatory (l). An equal mixture of the d and FIGURE 21.16 Isomers I and II of Co(en)3 3 are mirror im-ages (the image of I is identical to II) that cannot be superimposed. That is, there is no way that I can be turned in space so that it is the same as II.
N N N N N N Co N N N N N N Co Mirror image of Isomer I Isomer I Isomer II N N N N N N Co Cl Cl N N N N Co Cl Cl N N N N Co Cl Cl N N N N Co Cl Cl N N N N Co Cl Cl N N N N Co Isomer II Isomer I cis trans Isomer II cannot be superimposed exactly on isomer I. They are not identical structures.
The trans isomer and its mirror image are identical. They are not isomers of each other.
Isomer II has the same structure as the mirror image of isomer I.
(b) (a) FIGURE 21.17 (a) The trans isomer of Co(en)2Cl2 and its mirror image are identical (superimposable). (b) The cis isomer of Co(en)2Cl2 and its mirror image are not superimposable and are thus a pair of optical isomers.
21.5 Bonding in Complex Ions: The Localized Electron Model 965 l forms in solution, called a racemic mixture, does not rotate the plane of the polarized light at all because the two opposite effects cancel each other.
Geometrical isomers are not necessarily optical isomers. For instance, the trans isomer of [Co(en)2Cl2] shown in Fig. 21.17 is identical to its mirror image. Since this isomer is superimposable on its mirror image, it does not exhibit optical isomerism and is not chiral.
On the other hand, cis-[Co(en)2Cl2] is not superimposable on its mirror image; a pair of enantiomers exists for this complex ion (the cis isomer is chiral).
Most important biomolecules are chiral, and their reactions are highly structure de-pendent. For example, a drug can have a particular effect because its molecules can bind to chiral molecules in the body. To bind correctly, however, the correct optical isomer of the drug must be administered. Just as the right hand of one person requires the right hand of another to perform a handshake, a given isomer in the body requires a specific isomer of the drug to bind together. Because of this, the syntheses of drugs, which are usually very complicated molecules, must be carried out in a way that produces the correct “handedness,” a requirement that greatly adds to the synthetic difficulties.
Geometrical and Optical Isomerism Does the complex ion [Co(NH3)Br(en)2]2 exhibit geometrical isomerism? Does it exhibit optical isomerism?
Solution The complex ion exhibits geometrical isomerism because the ethylenediamine ligands can be across from or next to each other: The cis isomer of the complex ion also exhibits optical isomerism because its mirror images cannot be turned in any way to make them superimposable. Thus these mirror-image iso-mers of the cis complex are shown to be enantiomers that will rotate plane-polarized light in opposite directions.
See Exercises 21.41 and 21.42.
21.5 Bonding in Complex Ions: The Localized Electron Model In Chapters 8 and 9 we considered the localized electron model, a very useful model for describing the bonding in molecules. Recall that a central feature of this model is the formation of hybrid atomic orbitals that are used to share electron pairs to form bonds s Sample Exercise 21.3 966 Chapter Twenty-One Transition Metals and Coordination Chemistry between atoms. This same model can be used to account for the bonding in complex ions, but there are two important points to keep in mind: 1. The VSEPR model for predicting structure generally does not work for complex ions.
However, we can safely assume that a complex ion with a coordination number of 6 will have an octahedral arrangement of ligands, and complexes with two ligands will be linear. On the other hand, complex ions with a coordination number of 4 can be either tetrahedral or square planar, and there is no completely reliable way to predict which will occur in a particular case.
2. The interaction between a metal ion and a ligand can be viewed as a Lewis acid–base reaction with the ligand donating a lone pair of electrons to an empty orbital of the metal ion to form a coordinate covalent bond: The hybrid orbitals used by the metal ion depend on the number and arrangement of the ligands. For example, accommodating the lone pairs from the six ammonia mol-ecules in the octahedral Co(NH3)6 3 ion requires a set of six empty hybrid atomic or-bitals in an octahedral arrangement. As we discussed in Section 9.1, an octahedral set of orbitals is formed by the hybridization of two d, one s, and three p orbitals to give a set of six d2sp3 orbitals (see Fig. 21.18).
The hybrid orbitals required on a metal ion in a four-coordinate complex depend on whether the structure is tetrahedral or square planar. For a tetrahedral arrangement of li-gands, an sp3 hybrid set is required (see Fig. 21.19). For example, in the tetrahedral ion, the Co2 can be described as sp3 hybridized. A square planar arrangement of ligands requires a dsp2 hybrid orbital set on the metal ion (see Fig. 21.19). For example, in the square planar ion, the Ni2 is described as dsp2 hybridized.
Ni(CN)4 2 CoCl4 2 FIGURE 21.18 A set of six d2sp3 hybrid orbitals on Co3 can accept an electron pair from each of six NH3 ligands to form the Co(NH3)6 3 ion.
Co N H H H N H H H N H H H N H H H N H H H N H H H 21.6 The Crystal Field Model 967 A linear complex requires two hybrid orbitals 180 degrees from each other. This arrangement is given by an sp hybrid set (see Fig. 21.19). Thus, in the linear Ag(NH3)2 ion, the Ag can be described as sp hybridized.
Although the localized electron model can account in a general way for metal–ligand bonds, it is rarely used today because it cannot readily account for important properties of complex ions, such as magnetism and color. Thus we will not pursue the model any further.
21.6 The Crystal Field Model The main reason the localized electron model cannot fully account for the properties of complex ions is that it gives no information about how the energies of the d orbitals are affected by complex ion formation. This is critical because, as we will see, the color and magnetism of complex ions result from changes in the energies of the metal ion d orbitals caused by the metal–ligand interactions.
The crystal field model focuses on the energies of the d orbitals. In fact, this model is not so much a bonding model as it is an attempt to account for the colors and magnetic properties of complex ions. In its simplest form, the crystal field model assumes that the ligands can be approximated by negative point charges and that metal–ligand bonding is entirely ionic.
Octahedral Complexes We will illustrate the fundamental principles of the crystal field model by applying it to an octahedral complex. Figure 21.20 shows the orientation of the 3d orbitals relative to an octahedral arrangement of point-charge ligands. The important thing to note is that two of the orbitals, dz2 and dx2 y2, point their lobes directly at the point-charge ligands and three of the orbitals, dxz, dyz, and dxy, point their lobes between the point charges.
To understand the effect of this difference, we need to consider which type of orbital is lower in energy. Because the negative point-charge ligands repel negatively charged electrons, the electrons will first fill the d orbitals farthest from the ligands to minimize repulsions. In other words, the dxz, dyz, and dxy orbitals (known as the t2g set) are at a lower FIGURE 21.19 The hybrid orbitals required for tetrahedral, square planar, and linear complex ions.
The metal ion hybrid orbitals are empty, and the metal ion bonds to the ligands by accepting lone pairs.
FIGURE 21.20 An octahedral arrangement of point-charge ligands and the orientation of the 3d orbitals.
Tetrahedral ligand arrangement; sp3 hybridization Square planar ligand arrangement; dsp2 hybridization Linear ligand arrangement; sp hybridization M M M z x y – – – – – – dz2 dx2 – y2 dxy dyz dxz 968 Chapter Twenty-One Transition Metals and Coordination Chemistry energy in the octahedral complex than are the dz2 and dx2 y2 orbitals (the eg set). This is shown in Fig. 21.21. The negative point-charge ligands increase the energies of all the d orbitals. However, the orbitals that point at the ligands are raised in energy more than those that point between the ligands.
It is this splitting of the 3d orbital energies (symbolized by ) that explains the color and magnetism of complex ions of the first-row transition metal ions. For example, in an octahedral complex of Co3 (a metal ion with six 3d electrons), there are two pos-sible ways to place the electrons in the split 3d orbitals (Fig. 21.22). If the splitting pro-duced by the ligands is very large, a situation called the strong-field case, the electrons will pair in the lower-energy t2g orbitals. This gives a diamagnetic complex in which all the electrons are paired. On the other hand, if the splitting is small (the weak-field case), the electrons will occupy all five orbitals before pairing occurs. In this case the complex has four unpaired electrons and is paramagnetic.
The crystal field model allows us to account for the differences in the magnetic prop-erties of Co(NH3)6 3 and The Co(NH3)6 3 ion is known to be diamagnetic and thus corresponds to the strong-field case, also called the low-spin case, since it yields the minimum number of unpaired electrons. In contrast, the ion, which is known to have four unpaired electrons, corresponds to the weak-field case, also known as the high-spin case, since it gives the maximum number of unpaired electrons.
Crystal Field Model I The ion is known to have one unpaired electron. Does the ligand produce a strong or weak field?
Solution Since the ligand is and the overall complex ion charge is the metal ion must be Fe3, which has a 3d5 electron configuration. The two possible arrangements of the five electrons in the d orbitals split by the octahedrally arranged ligands are The strong-field case gives one unpaired electron, which agrees with the experimental observation. The ion is a strong-field ligand toward the Fe3 ion.
See Exercises 21.45 and 21.46.
From studies of many octahedral complexes, we can arrange ligands in order of their ability to produce d-orbital splitting. A partial listing of ligands in this spectrochemical series is Strong-field Weak-field ligands ligands (large ) (small ) The ligands are arranged in order of decreasing values toward a given metal ion.
It also has been observed that the magnitude of for a given ligand increases as the charge on the metal ion increases. For example, NH3 is a weak-field ligand toward Co2 but acts as a strong-field ligand toward Co3. This makes sense; as the metal ion charge increases, the ligands will be drawn closer to the metal ion because of the increased charge density. As the ligands move closer, they cause greater splitting of the d orbitals and produce a larger value.
¢ ¢ ¢ CN 7 NO2 7 en 7 NH3 7 H2O 7 OH 7 F 7 Cl 7 Br 7 I CN eg t2g Large E E eg t2g Small 3, CN CN Fe(CN)6 3 CoF6 3 CoF6 3.
¢ FIGURE 21.21 The energies of the 3d orbitals for a metal ion in an octahedral complex. The 3d or-bitals are degenerate (all have the same energy) in the free metal ion. In the octa-hedral complex the orbitals are split into two sets as shown. The difference in energy between the two sets is designated as (delta).
¢ FIGURE 21.22 Possible electron arrangements in the split 3d orbitals in an octahedral complex of Co3 (electron configuration 3d6). (a) In a strong field (large value), the electrons fill the t2g set first, giving a diamagnetic complex. (b) In a weak field (small value), the electrons occupy all five orbitals before any pairing occurs.
¢ ¢ ∆ eg(dz2, dx2 – y2) t2g(dxz, dyz, dxy) Free metal ion 3d orbital energies E eg eg t2g t2g Large E E Strong field (a) Weak field (b) Small Sample Exercise 21.4 21.6 The Crystal Field Model 969 Crystal Field Model II Predict the number of unpaired electrons in the complex ion Solution The net charge of means that the metal ion present must be Cr2 which has a 3d4 electron configuration. Since is a strong-field ligand (see the spec-trochemical series), the correct crystal field diagram for is The complex ion will have two unpaired electrons. Note that the ligand produces such a large splitting that all four electrons will occupy the t2g set even though two of the electrons must be paired in the same orbital.
See Exercises 21.47 and 21.48.
We have seen how the crystal field model can account for the magnetic properties of octahedral complexes. The same model also can explain the colors of these complex ions.
For example, Ti(H2O)6 3, an octahedral complex of Ti3, which has a 3d1 electron con-figuration, is violet because it absorbs light in the middle of the visible region of the spectrum (see Fig. 21.23). When a substance absorbs certain wavelengths of light in the visible region, the color of the substance is determined by the wavelengths of visible light that remain. We say that the substance exhibits the color complementary to those absorbed.
The Ti(H2O)6 3 ion is violet because it absorbs light in the yellow-green region, thus letting red light and blue light pass, which gives the observed violet color. This is shown schemat-ically in Fig. 21.24. Table 21.16 shows the general relationship between the wavelengths of visible light absorbed and the approximate color observed.
CN E eg t2g Large [Cr(CN)6]4 CN (6 2 4), 4 [Cr(CN)6]4.
FIGURE 21.24 (a) When white light shines on a filter that absorbs in the yellow-green region, the emerging light is violet. (b) Because the complex ion Ti(H2O)6 3 absorbs yellow-green light, a solution of it is violet.
FIGURE 21.23 The visible spectrum.
Sample Exercise 21.5 Wavelength (nm) 500 600 700 400 Violet Blue Green Yellow Orange Red Filter absorbs yellow-green light (a) (b) Ti(H2O)6 3+ TABLE 21.16 Approximate Relationship of Wavelength of Visible Light Absorbed to Color Observed Absorbed Wavelength in nm (Color) Observed Color 400 (violet) Greenish yellow 450 (blue) Yellow 490 (blue-green) Red 570 (yellow-green) Violet 580 (yellow) Dark blue 600 (orange) Blue 650 (red) Green 970 Chapter Twenty-One Transition Metals and Coordination Chemistry The reason that the Ti(H2O)6 3 ion absorbs a specific wavelength of visible light can be traced to the transfer of the lone d electron between the split d orbitals, as shown in Fig. 21.25. A given photon of light can be absorbed by a molecule only if the wavelength of the light provides exactly the energy needed by the molecule. In other words, the wave-length absorbed is determined by the relationship where represents the energy spacing in the molecule (we have used simply in this chapter) and represents the wavelength of light needed. Because the d-orbital splitting in most octahedral complexes corresponds to the energies of photons in the visible region, octahedral complex ions are usually colored.
Since the ligands coordinated to a given metal ion determine the size of the d-orbital splitting, the color changes as the ligands are changed. This occurs because a change in means a change in the wavelength of light needed to transfer electrons between the t2g and eg orbitals. Several octahedral complexes of Cr3 and their colors are listed in Table 21.17.
Other Coordination Geometries Using the same principles developed for octahedral complexes, we will now consider com-plexes with other geometries. For example, Fig. 21.26 shows a tetrahedral arrangement of ¢ l ¢ ¢E ¢E hc l FIGURE 21.25 The complex ion Ti(H2O)6 3 can absorb visible light in the yellow-green region to transfer the lone d electron from the t2g to the eg set.
TABLE 21.17 Several Octahedral Complexes of Cr3 and Their Colors Isomer Color [Cr(H2O)6]Cl3 Violet [Cr(H2O)5Cl]Cl2 Blue-green [Cr(H2O)4Cl2]Cl Green [Cr(NH3)6]Cl3 Yellow [Cr(NH3)5Cl]Cl2 Purple [Cr(NH3)4Cl2]Cl Violet CHEMICAL IMPACT Transition Metal Ions Lend Color to Gems T he beautiful pure color of gems, so valued by cultures everywhere, arises from trace transition metal ion im-purities in minerals that would otherwise be colorless. For example, the stunning red of a ruby, the most valuable of all gemstones, is caused by Cr3 ions, which replace about 1% of the Al3 ions in the mineral corundum, which is a form of aluminum oxide (Al2O3) that is nearly as hard as dia-mond. In the corundum structure the Cr3 ions are sur-rounded by six oxide ions at the vertices of an octahedron.
This leads to the characteristic octahedral splitting of chromium’s 3d orbitals, such that the Cr3 ions absorb strongly in the blue-violet and yellow-green regions of the visible spectrum but transmit red light to give the charac-teristic ruby color. (On the other hand, if some of the Al3 ions in corundum are replaced by a mixture of Fe2, Fe3, and Ti4 ions, the gem is a sapphire with its brilliant blue color; or if some of the Al3 ions are replaced by Fe3 ions, the stone is a yellow topaz.) Emeralds are derived from the mineral beryl, a beryllium aluminum silicate (empirical formula 3BeO Al2O3 6SiO2).
When some of the Al3 ions in beryl are replaced by Cr3 ions, the characteristic green color of emerald results. In this envi-ronment the splitting of the Cr3 3d orbitals causes it to strongly absorb yellow and blue-violet light and to transmit green light.
A gem closely related to ruby and emerald is alexan-drite, named after Alexander II of Russia. This gem is based on the mineral chrysoberyl, a beryllium aluminate with the empirical formula in which approximately 1% of the Al3 ions are replaced by Cr3 ions. In the chrysoberyl environment Cr3 absorbs strongly in the yellow region of BeO Al2O3 21.6 The Crystal Field Model 971 point charges in relation to the 3d orbitals of a metal ion. There are two important facts to note: 1. None of the 3d orbitals “point at the ligands” in the tetrahedral arrangement, as the dx2 y2 and dz2 orbitals do in the octahedral case. Thus the tetrahedrally arranged li-gands do not differentiate the d orbitals as much in the tetrahedral case as in the octahedral case. That is, the difference in energy between the split d orbitals will be sig-nificantly less in tetrahedral complexes. Although we will not derive it here, the tetra-hedral splitting is that of the octahedral splitting for a given ligand and metal ion: 2. Although not exactly pointing at the ligands, the dxy, dxz, and dyz orbitals are closer to the point charges than are the dz2 and dx2 y2 orbitals. This means that the tetrahedral d-orbital splitting will be opposite to that for the octahedral arrangement. The two arrangements are contrasted in Fig. 21.27. Because the d-orbital splitting is relatively small for the tetrahedral case, the weak-field case (high-spin case) always applies.
There are no known ligands powerful enough to produce the strong-field case in a tetrahedral complex.
¢tet 4 9¢oct 4 9 Solutions of [Cr(NH3)6]Cl3 (yellow) and [Cr(NH3)5Cl]Cl2 (purple).
FIGURE 21.26 (a) Tetrahedral and octahedral arrange-ments of ligands shown inscribed in cubes.
Note that in the two types of arrangements, the point charges occupy opposite parts of the cube; the octahedral point charges are at the centers of the cube faces, and the tetrahedral point charges occupy opposite corners of the cube. (b) The orientations of the 3d orbitals relative to the tetrahedral set of point charges.
– – – – – – – – – – dz2 dx2 – y2 dxy dyz dxz (a) (b) the spectrum. Alexandrite has the interesting property of changing colors depending on the light source. When the first alexandrite stone was discovered deep in a mine in the Russian Ural Mountains in 1831, it appeared to be a deep red color in the firelight of the miners’ lamps. However, when the stone was brought to the surface, its color was blue. This seemingly magical color change occurs because the firelight of a miner’s helmet is rich in the yellow and red wavelengths of the visible spectrum but does not contain much blue. Absorption of the yellow by the stone produces a reddish color. However, daylight has much more intensity in the blue region than firelight. Thus the extra blue in the light transmitted by the stone gives it bluish color in daylight.
Once the structure of a natural gem is known, it is usually not very difficult to make the gem artificially. For example, ru-bies and sapphires are made on a large scale by fusing Al(OH)3 with the appropriate transition metal salts at approximately to make the “doped” corundum. With these techniques 1200°C gems of astonishing size can be manufactured: Rubies as large as 10 lb and sapphires up to 100 lb have been synthesized.
Smaller synthetic stones produced for jewelry are virtually identical to the corresponding natural stones, and it takes great skill for a gemologist to tell the difference.
Alexandrite, a gem closely related to ruby and emerald.
972 Chapter Twenty-One Transition Metals and Coordination Chemistry Crystal Field Model III Give the crystal field diagram for the tetrahedral complex ion Solution The complex ion contains Co2, which has a 3d7 electron configuration. The splitting of the d orbitals will be small, since this is a tetrahedral complex, giving the high-spin case with three unpaired electrons.
See Exercises 21.53 through 21.56.
The crystal field model also applies to square planar and linear complexes. The crys-tal field diagrams for these cases are shown in Fig. 21.28. The ranking of orbitals in these diagrams can be explained by considering the relative orientations of the point charges and the orbitals. The diagram in Fig. 21.27 for the octahedral arrangement can be used to ob-tain these orientations. We can obtain the square planar complex by removing the two point charges along the z axis. This will greatly lower the energy of dz2, leaving only dx2 y2, E dx2y2 dyz Small dxy dxz dz2 CoCl4 2.
FIGURE 21.28 (a) The crystal field diagram for a square planar complex oriented in the xy plane with ligands along the x and y axes. The position of the orbital is higher than those of the dxz and dyz orbitals because of the “doughnut” of electron density in the xy plane. The actual position of is some-what uncertain and varies in different square planar complexes. (b) The crystal field diagram for a linear complex where the ligands lie along the z axis.
dz2 dz2 FIGURE 21.27 The crystal field diagrams for octahedral and tetrahedral complexes. The relative en-ergies of the sets of d orbitals are reversed.
For a given type of ligand, the splitting is much larger for the octahedral complex ( ) because in this arrangement the and orbitals point their lobes directly at the point charges and are thus relatively high in energy.
dx2y2 dz2 ¢oct 7 ¢tet Sample Exercise 21.6 Free metal ion Complex dx2 – y2 dxy dz2 dxz dyz M (a) x y E E Free metal ion Complex dz2 dxy dxz dyz dx2 – y2 M z (b) 21.7 The Biologic Importance of Coordination Complexes 973 which points at the four remaining ligands as the highest-energy orbital. We can obtain the linear complex from the octahedral arrangement by leaving the two ligands along the z axis and removing the four in the xy plane. This means that only the dz2 points at the ligands and is highest in energy.
21.7 The Biologic Importance of Coordination Complexes The ability of metal ions to coordinate with and release ligands and to easily undergo ox-idation and reduction makes them ideal for use in biologic systems. For example, metal ion complexes are used in humans for the transport and storage of oxygen, as electron-transfer agents, as catalysts, and as drugs. Most of the first-row transition metals are es-sential for human health, as summarized in Table 21.18. We will concentrate on iron’s role in biologic systems, since several of its coordination complexes have been studied extensively.
Iron plays a central role in almost all living cells. In mammals, the principal source of energy comes from the oxidation of carbohydrates, proteins, and fats. Although oxygen is the oxidizing agent for these processes, it does not react directly with these molecules.
Instead, the electrons from the breakdown of these nutrients are passed along a complex chain of molecules, called the respiratory chain, eventually reaching the O2 molecule. The principal electron-transfer molecules in the respiratory chain are iron-containing species called cytochromes, consisting of two main parts: an iron complex called heme and a protein. The structure of the heme complex is shown in Fig. 21.29. Note that it contains an iron ion (it can be either Fe2 or Fe3) coordinated to a rather complicated planar ligand called a porphyrin. As a class, porphyrins all contain the same central ring structure but have different substituent groups at the edges of the rings. The various porphyrin molecules act as tetradentate ligands for many metal ions, including iron, cobalt, and magnesium. In fact, chlorophyll, a substance essential to the process of photosynthesis, is a magnesium–porphyrin complex of the type shown in Fig. 21.30.
In addition to participating in the transfer of electrons from nutrients to oxygen, iron plays a principal role in the transport and storage of oxygen in mammalian blood and tissues. Oxygen is stored in a molecule called myoglobin, which consists of a heme A protein is a large molecule assembled from amino acids, which have the gen-eral structure in which R represents various groups.
R H C COOH O A A H2NO TABLE 21.18 The First-Row Transition Metals and Their Biologic Significance First-Row Transition Metal Biologic Function(s) Scandium None known.
Titanium None known.
Vanadium None known in humans.
Chromium Assists insulin in the control of blood sugar; may also be involved in the control of cholesterol.
Manganese Necessary for a number of enzymatic reactions.
Iron Component of hemoglobin and myoglobin; involved in the electron-transport chain.
Cobalt Component of vitamin B12, which is essential for the metabolism of carbohydrates, fats, and proteins.
Nickel Component of the enzymes urease and hydrogenase.
Copper Component of several enzymes; assists in iron storage; involved in the production of color pigments of hair, skin, and eyes.
Zinc Component of insulin and many enzymes.
974 Chapter Twenty-One Transition Metals and Coordination Chemistry complex and a protein in a structure very similar to that of the cytochromes. In myoglobin, the Fe2 ion is coordinated to four nitrogen atoms of the porphyrin ring and to a nitrogen atom of the protein chain, as shown in Fig. 21.31. Since Fe2 is normally six-coordinate, this leaves one position open for attachment of an O2 molecule.
One especially interesting feature of myoglobin is that it involves an O2 molecule at-taching directly to Fe2. However, if gaseous O2 is bubbled into an aqueous solution con-taining heme, the Fe2 is immediately oxidized to Fe3. This oxidation of the Fe2 in heme does not happen in myoglobin. This fact is of crucial importance because Fe3 does FIGURE 21.29 The heme complex, in which an Fe2 ion is coordinated to four nitrogen atoms of a planar porphyrin ligand.
FIGURE 21.30 Chlorophyll is a porphyrin complex of Mg2. There are two similar forms of chlorophyll, one of which is shown here.
N Fe N N HC CH CH HC N CH3 CH3 CH3 CH2 CH3 CH CH2 CH CH2 CH2 CH2 COOH CH2COOH N N N Mg N 21.7 The Biologic Importance of Coordination Complexes 975 not form a coordinate covalent bond with O2, and myoglobin would not function if the bound Fe2 could be oxidized. Since the Fe2 in the “bare” heme complex can be oxi-dized, it must be the protein that somehow prevents the oxidation. How? Based on much research, the answer seems to be that the oxidation of Fe2 to Fe3 involves an oxygen bridge between two iron ions (the circles indicate the ligands): The bulky protein around the heme group in myoglobin prevents two molecules from get-ting close enough to form the oxygen bridge, and so oxidation of the Fe2 is prevented.
The transport of O2 in the blood is carried out by hemoglobin, a molecule consist-ing of four myoglobin-like units, as shown in Fig. 21.32. Each hemoglobin can therefore CHEMICAL IMPACT The Danger of Mercury I n August 1989, four family members from Lincoln Park, Michigan, died under mysterious circumstances. All four victims had severe tissue damage to the esophagus and lungs, which led doctors to suspect exposure to some type of caustic chemical. Subsequently, local police and fire in-vestigators discovered a crude laboratory in the basement that was used to recover valuable silver from stolen dental amalgams. A dental amalgam is a metal alloy that a dentist uses to fill a tooth. The alloy typically contains silver, tin, copper, and zinc dissolved in liquid mercury. The mixture is placed in a cavity, where it hardens, resulting in a stan-dard “filling.” One of the victims worked at a manufacturing facility for amalgams and was stealing some of the products. At home in his crude lab he heated the amalgam to drive off the mercury (which vaporized at relatively low temperatures) so that he could recover the silver that was left behind. In the process, the colorless, odorless, tasteless mercury vapor entered the ductwork of the home, which was contaminated with mercury at levels 1500 times normal—levels certain to result in death to those exposed. In fact, postmortem analy-sis revealed extreme levels of mercury in the vital organs of all four victims. Mercury vapor was the silent killer.
The toxicity of mercury varies significantly depending on the route of entry into the body. Inhalation is the most dangerous route because mercury vapor entering the lungs quickly passes across the lung–blood barrier and into the bloodstream, where it can interfere with normal blood chem-istry. One of the reactions that takes place in the blood is the decomposition of hydrogen peroxide (a metabolic waste product) by the enzyme peroxidase: When elemental mercury enters the bloodstream, it reacts with hydrogen peroxide in the presence of peroxidase to pro-duce mercury(II) oxide and water: If this conversion to mercury(II) occurs within vital tissues, the mercury(II) cation can denature proteins, inhibit enzyme activity, and disrupt cell membranes. Death often results from respiratory or kidney failure.
If mercury is so toxic, how can it be used in dental fill-ings? Surprisingly, unlike elemental mercury in the vapor form, mercury bound as a solid in a dental amalgam pre-sents little, if any, risk. Because the mercury is not mobile, even in the harsh environment of the human mouth, the American Dental Association has determined it to be a min-imal health risk to dental patients. Even if a filling loosens and is accidentally swallowed, it is passed through the di-gestive system and excreted before it can pose any risk. The mercury that the four victims in this story were exposed to resulted from heating the amalgam in a smelting furnace, thus vaporizing the mercury and exposing the occupants of the house to the most hazardous route of entry—inhalation.
Printed with permission, ChemMatters magazine. Copyright © 1999, American Chemical Society.
HgO H2O Hg H2O2 2H2O O2 2H2O2 Peroxidase ¡ Peroxidase ¡ Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
976 Chapter Twenty-One Transition Metals and Coordination Chemistry bind four O2 molecules to form a bright red diamagnetic complex. The diamagnetism occurs because oxygen is a strong-field ligand toward Fe2, which has a 3d6 electron configuration. When the oxygen molecule is released, water molecules occupy the sixth coordination position around each Fe2, giving a bluish paramagnetic complex (H2O is a weak-field ligand toward Fe2) that gives venous blood its characteristic bluish tint.
FIGURE 21.31 A representation of the myoglobin mole-cule. The Fe2 ion is coordinated to four ni-trogen atoms in the porphyrin of the heme (represented by the disk in the figure) and on nitrogen from the protein chain. This leaves a sixth coordination position (indi-cated by the W) available for an oxygen molecule.
FIGURE 21.32 A representation of the hemoglobin struc-ture. There are two slightly different types of protein chains ( and ). Each hemoglo-bin has two chains and two chains, each with a heme complex near the center.
Thus each hemoglobin molecule can com-plex with four O2 molecules.
HC H NA GH A EF FG W F G E D CD C B AB F E B H EF G H′ G′ G B A EF F E C F1 A1 H23 F8 FG3 FG5 FG5 G1 G2 G3 C7 G9 CD5 B14 C6 C3 EF1 E7 A12 B1 G19 H9 E1 D1 E1 E7 EF1 F1 H15 H23 F8 H9 G9 G4 G3 FG4 FG3 G1 C6 C3 C2 C5 CD5 B11 B1 A9 A G19 C F ′ A′ E′ H′ F ′ A′ E′ β1 α1 β2 α2 H 21.7 The Biologic Importance of Coordination Complexes 977 Hemoglobin dramatically demonstrates how sensitive the function of a biomolecule is to its structure. In certain people, in the synthesis of the proteins needed for hemoglo-bin, an improper amino acid is inserted into the protein in two places. This may not seem very serious, since there are several hundred amino acids present. However, because the incorrect amino acid has a nonpolar substituent instead of the polar one found on the proper amino acid, the hemoglobin drastically changes its shape. The red blood cells are then sickle-shaped rather than disk-shaped, as shown in Fig. 21.33. The misshapen cells can aggregate, causing clogging of tiny capillaries. This condition, known as sickle cell anemia, is the subject of intense research.
Our knowledge of the workings of hemoglobin allows us to understand the effects of high altitudes on humans. The reaction between hemoglobin and oxygen can be repre-sented by the following equilibrium: Hemoglobin Oxyhemoglobin At high altitudes, where the oxygen content of the air is lower, the position of this equilibrium will shift to the left, according to Le Châtelier’s principle. Because less oxyhemoglobin is formed, fatigue, dizziness, and even a serious illness called high-altitude sickness can result.
One way to combat this problem is to use supplemental oxygen, as most high-altitude moun-tain climbers do. However, this is impractical for people who live at high elevations. In fact, the human body adapts to the lower oxygen concentrations by making more hemoglobin, causing the equilibrium to shift back to the right. Someone moving from Chicago to Boulder, Colorado (5300 feet), would notice the effects of the new altitude for a couple of weeks, but as the hemoglobin level increased, the effects would disappear. This change is called high-altitude acclimatization, which explains why athletes who want to compete at high el-evations should practice there for several weeks prior to the event.
Our understanding of the biologic role of iron also allows us to explain the toxicities of substances such as carbon monoxide and the cyanide ion. Both CO and are very good ligands toward iron and so can interfere with the normal workings of the iron complexes in the body. For example, carbon monoxide has about 200 times the affinity for the Fe2 in hemoglobin as oxygen does. The resulting stable complex, carboxyhemoglobin, prevents the normal uptake of O2, thus depriving the body of needed oxygen. Asphyxiation can result if enough carbon monoxide is present in the air. The mechanism for the toxicity of the cyanide ion is somewhat different. Cyanide coordinates strongly to cytochrome oxidase, an iron-containing cytochrome enzyme that catalyzes the oxidation–reduction reactions of certain cytochromes. The coordinated cyanide thus prevents the electron-transfer process and rapid death results. Because of its behavior, cyanide is called a respiratory inhibitor.
CN Hb1aq2 4O21g2 ∆Hb1O2241aq2 FIGURE 21.33 A normal red blood cell (right) and a sickle cell (left), both magnified 18,000 times.
Sherpa and Balti porters are acclimatized to high elevations such as those around the K2 mountain peak in Pakistan.
978 Chapter Twenty-One Transition Metals and Coordination Chemistry 21.8 Metallurgy and Iron and Steel Production In the preceding section we saw the importance of iron in biologic systems. Of course, iron is also very important in many other ways in our world. In this section we will discuss the isolation of metals from their natural sources and the formulation of metals into useful materials, with special emphasis on the role of iron.
Metals are very important for structural applications, electrical wires, cooking utensils, tools, decorative items, and many other purposes. However, because the main chemical characteristic of a metal is its ability to give up electrons, almost all metals in nature are found in ores, combined with nonmetals such as oxygen, sulfur, and the halogens. To recover and use these metals, we must separate them from their ores and reduce the metal ions. Then, because most metals are unsuitable for use in the pure state, we must form alloys that have the desired properties. The process of separating a metal from its ore and preparing it for use is known as metallurgy. The steps in this process are typically 1. Mining 2. Pretreatment of the ore 3. Reduction to the free metal 4. Purification of the metal (refining) 5. Alloying An ore can be viewed as a mixture containing minerals (relatively pure metal com-pounds) and gangue (sand, clay, and rock). Some typical minerals are listed in Table 21.19.
Although silicate minerals are the most common in the earth’s crust, they are typically A steel mill in Brazil.
CHEMICAL IMPACT Supercharged Blood D uring the 1964 Winter Olympics in Innsbruck, Austria, a Finn named Eero Maentyranta won three gold medals in cross-country skiing and immediately became a national hero. His success was due in no small part to the fact that his blood carried 25% to 50% more hemoglobin than the av-erage man’s blood. Maentyranta suffered from a rare genetic disorder that results in unusually elevated levels of red blood cells. This extra oxygen-carrying capacity lends itself to in-creased stamina and endurance, certainly an advantage in the rigorous sport of cross-country skiing. Several years later, geneticists at the University of Helsinki determined that the disorder was due to a mutation of a protein respon-sible for red blood cell production called the erythropoietin receptor (EPO-R). The mutation, which was common in the Maentyranta family, resulted in a protein that was missing 70 amino acids (out of 550). This mutation deleted the por-tion of the protein that contained the “off switch” for red blood cell production.
Since the discovery of EPO’s role in red blood cell pro-duction, genetic engineers have been able to synthesize EPO by bioengineering techniques. The protein, marketed in the United States by Amgen of Thousand Oaks, California, is used by kidney dialysis, AIDS, and cancer patients to boost red blood cell production. It has become biotechnology’s biggest revenue producer, with over $1 billion in annual sales.
While the potential benefits of this protein are obvious, the potential for abuse may be even greater. The Tour de France, thought by many to be the world’s greatest endurance race, was rocked in 1998 by controversy when several riders (including three race favorites) were expelled from the race over allegations of blood doping with the red blood cell– enchancing hormone EPO. EPO has become the drug of choice for many endurance athletes trying to gain an unfair advantage because it is metabolized quickly and, being iden-tical to the body’s own EPO, is almost impossible to detect by blood or urine analysis. But the hazards are great. En-durance athletes are especially at risk from abuse of EPO due to extreme water loss during the athletic competition, which, when coupled with elevated levels of EPO, results in an 21.8 Metallurgy and Iron and Steel Production 979 extreme thickening of the blood. This puts an obvious strain on the heart. Since EPO became available in 1986, several world-class athletes (mostly bicyclists and distance runners) have died under mysterious circumstances thought to be associated with abuse of EPO. Because of the widespread abuse of EPO, new methods of detection have been put in place to limit its impact on organized sports and its participants.
The advantage that Eero Maen-tyranta gained in the 1964 Olympics was the result of a disorder that he had lived with for his entire life. His body had become accustomed to operating with elevated levels of hemoglobin, and it became a natural advantage. Synthetic EPO holds much promise for those suffering conditions that result in hemo-globin deficiency. But as a performance-enhancing drug, EPO’s advantages come with enormous risk.
The 1998 Tour de France.
TABLE 21.19 Common Minerals Found in Ores Anion Examples None (free metal) Au, Ag, Pt, Pd, Rh, Ir, Ru Oxide Fe2O3 (hematite) Fe3O4 (magnetite) Al2O3 (bauxite) SnO2 (cassiterite) Sulfide PbS (galena) ZnS (sphalerite) FeS2 (pyrite) HgS (cinnabar) Cu2S (chalcocite) Chloride NaCl (rock salt) KCl (sylvite) KCl MgCl2 (carnalite) Carbonate FeCO3 (siderite) CaCO3 (limestone) MgCO3 (magnesite) MgCO3 CaCO3 (dolomite) Sulfate CaSO4 2H2O (gypsum) BaSO4 (barite) Silicate Be3Al2Si6O18 (beryl) Al2(Si2O8)(OH)4 (kaolinite) LiAl(SiO3)2 (spondumene) 980 Chapter Twenty-One Transition Metals and Coordination Chemistry very hard and difficult to process, making metal extraction relatively expensive. There-fore, other ores are used when available.
After mining, an ore must be treated to remove the gangue and to concentrate the mineral. The ore is first pulverized and then processed in a variety of devices, including cyclone separators (see Fig. 21.34), inclined vibrating tables, and flotation tanks.
In the flotation process, the crushed ore is fed into a tank containing a water– oil–detergent mixture. Because of the difference in the surface characteristics of the mineral particles and the silicate rock particles, the oil wets the mineral particles. A stream of air blown through the mixture causes tiny bubbles to form on the oil-covered pieces, which then float to the surface, where they can be skimmed off.
After the mineral has been concentrated, it is often chemically altered in prepara-tion for the reduction step. For example, nonoxide minerals are often converted to oxides before reduction. Carbonates and hydroxides can be converted by simple heating: Sulfide minerals can be converted to oxides by heating in air at temperatures below their melting points, a process called roasting: As we have seen earlier, sulfur dioxide causes severe problems if released into the at-mosphere, and modern roasting operations collect this gas and use it in the manufacture of sulfuric acid.
The method chosen to reduce the metal ion to the free metal, a process called smelt-ing, depends on the affinity of the metal ion for electrons. Some metals are good enough oxidizing agents that the free metal is produced in the roasting process. For example, the roasting reaction for cinnabar is where the Hg2 is reduced by electrons donated by the ion, which is then further ox-idized by O2 to form SO2.
The roasting of a more active metal produces the metal oxide, which must be reduced to obtain the free metal. The most common reducing agents are coke (impure carbon), carbon monoxide, and hydrogen. The following are some common examples of the re-duction process: The most active metals, such as aluminum and the alkali metals, must be reduced electrolytically, usually from molten salts (see Section 17.8).
The metal obtained in the reduction step is invariably impure and must be refined.
The methods of refining include electrolytic refining (see Section 17.8), oxidation of impurities (as for iron, see below), and distillation of low-boiling metals such as mer-cury and zinc. One process used when highly pure metals are needed is zone refining.
In this process a bar of the impure metal travels through a heater (see Fig. 21.35), which causes melting and recrystallizing of the metal as the bar cools. Purification of the metal occurs because as the crystal re-forms, the metal ions are likely to fit much better in the crystal lattice than are the atoms of impurities. Thus the impurities tend to be excluded and carried to the end of the bar. Several repetitions of this process give a very pure metal bar.
ZnO1s2 C1s2 ¡ Heat Zn1l2 CO1g2 WO31s2 3H21g2 ¡ Heat W1l2 3H2O1g2 Fe2O31s2 3CO1g2 ¡ Heat 2Fe1l2 3CO21g2 S2 HgS1s2 O21g2 ¡ Heat Hg1l2 SO21g2 2ZnS1s2 3O21g2 ¡ Heat 2ZnO1s2 2SO21g2 Mg1OH221s2 ¡ Heat MgO1s2 H2O1g2 CaCO31s2 ¡ Heat CaO1s2 CO21g2 FIGURE 21.34 A schematic diagram of a cyclone separator.
The ore is pulverized and blown into the separator. The more dense mineral particles are thrown toward the walls by centrifugal force and fall down the funnel. The lighter particles (gangue) tend to stay closer to the center and are drawn out through the top by the stream of air.
21.8 Metallurgy and Iron and Steel Production 981 Hydrometallurgy The metallurgical processes we have considered so far are usually called pyrometallurgy (pyro means “at high temperatures”). These traditional methods require large quantities of energy and have two other serious problems: atmospheric pollution (mainly by sulfur dioxide) and relatively high costs that make treatment of low-grade ores economically unfeasible.
In the past hundred years, a different process, hydrometallurgy (hydro means “water”), has been employed to extract metals from ores by use of aqueous chemical solutions, a process called leaching. The first two uses of hydrometallurgy were for the extraction of gold from low-grade ores and for the production of aluminum oxide, or alumina, from bauxite, an aluminum-bearing ore.
Gold is sometimes found in ores in the elemental state, but it usually occurs in rela-tively small concentrations. A process called cyanidation treats the crushed ore with an aqueous cyanide solution in the presence of air to dissolve the gold by forming the com-plex ion Pure gold is then recovered by reaction of the solution of Au(CN)2 with zinc powder to reduce Au to Au: The extraction of alumina from bauxite (the Bayer process) leaches the ore with sodium hydroxide at high temperatures and pressures to dissolve the amphoteric aluminum oxide: This process leaves behind solid impurities such as SiO2, Fe2O3, and TiO2, which are not appreciably soluble in basic solution. After the solid impurities are removed, the pH of the solution is lowered, causing the pure aluminum oxide to re-form. It is then electrolyzed to produce aluminum metal (see Section 17.8).
As illustrated by these processes, hydrometallurgy involves two distinct steps: selec-tive leaching of a given metal ion from the ore and recovery of the metal ion from the so-lution by selective precipitation as an ionic compound.
The leaching agent can simply be water if the metal-containing compound is a water-soluble chloride or sulfate. However, most commonly, the metal is present in a water-insoluble substance that must somehow be dissolved. The leaching agents used in such cases are usually aqueous solutions containing acids, bases, oxidizing agents, salts, or some combination of these. Often the dissolving process involves the forma-tion of complex ions. For example, when an ore containing water-insoluble lead sulfate is treated with an aqueous sodium chloride solution, the soluble complex ion is formed: PbSO41s2 4Na1aq2 4Cl1aq2 ¡ 4Na1aq2 PbCl4 21aq2 SO4 21aq2 PbCl4 2 Al2O31s2 2OH1aq2 ¡ 2AlO2 1aq2 H2O1l2 2Au1CN22 1aq2 Zn1s2 ¡ 2Au1s2 Zn1CN24 21aq2 4Au1s2 8CN1aq2 O21g2 2H2O1l2 ¡ 4Au1CN22 1aq2 4OH1aq2 Au(CN)2 : Precipitation reactions are discussed in Section 15.7.
FIGURE 21.35 A schematic representation of zone refining.
Impure solid Impurities are concentrated here Molten zone Purified solid 982 Chapter Twenty-One Transition Metals and Coordination Chemistry Formation of a complex ion also occurs in the cyanidation process for the recovery of gold. However, since the gold is present in the ore as particles of metal, it must first be oxidized by oxygen to produce Au, which then reacts with CN to form the soluble Au(CN)2 species. Thus, in this case, the leaching process involves a combination of oxidation and complexation.
Sometimes just oxidation is used. For example, insoluble zinc sulfide can be con-verted to soluble zinc sulfate by pulverizing the ore and suspending it in water to form a slurry through which oxygen is bubbled: One advantage of hydrometallurgy over the traditional processes is that sometimes the leaching agent can be pumped directly into the ore deposits in the earth. For exam-ple, aqueous sodium carbonate (Na2CO3) can be injected into uranium-bearing ores to form water-soluble complex carbonate ions.
Recovering the metal ions from the leaching solution involves forming an insoluble solid containing the metal ion to be recovered. This step may involve addition of an an-ion to form an insoluble salt, reduction to the solid metal, or a combination of reduction and precipitation of a salt. Examples of these processes are shown in Table 21.20. Because of its suitability for treating low-grade ores economically and without significant pollu-tion, hydrometallurgy is becoming more popular for recovering many important metals such as copper, nickel, zinc, and uranium.
The Metallurgy of Iron Iron is present in the earth’s crust in many types of minerals. Iron pyrite (FeS2) is widely distributed but is not suitable for production of metallic iron and steel because it is almost impossible to remove the last traces of sulfur. The presence of sulfur makes the resulting steel too brittle to be useful. Siderite (FeCO3) is a valuable iron mineral that can be converted to iron oxide by heating. The iron oxide minerals are hematite (Fe2O3), the more abundant, and magnetite (Fe3O4, really ). Taconite ores contain iron oxides mixed with silicates and are more difficult to process than the others. However, taconite ores are being increasingly used as the more desirable ores are consumed.
To concentrate the iron in iron ores, advantage is taken of the natural magnetism of Fe3O4 (hence its name, magnetite). The Fe3O4 particles can be separated from the gangue by magnets. The ores that are not magnetic are often converted to Fe3O4; hematite is partially reduced to magnetite, while siderite is first converted to FeO thermally, then FeO Fe2O3 ZnS1s2 2O21g2 ¡ Zn21aq2 SO4 21aq2 TABLE 21.20 Examples of Methods for Recovery of Metal Ions from Leaching Solutions Method Examples Precipitation of a salt Cu2(aq) S2(aq) ¡ CuS(s) Cu(aq) HCN(aq) ¡ CuCN(s) H(aq) Au(aq) Fe2(aq) ¡ Au(s) Fe3(aq) Chemical Cu2(aq) Fe(s) ¡ Cu(s) Fe2(aq) Reduction Ni2(aq) H2(g) ¡ Ni(s) 2H(aq) Cu2(aq) 2e ¡ Cu(s) Electrolytic Al3(aq) 3e ¡ Al(s) Reduction plus precipitation 2Cu2(aq) 2Cl(aq) H2SO3(aq) H2O(l) ¡ 2CuCl(s) 3H(aq) HSO4 (aq) ⎧ ⎪ ⎨ ⎪ ⎩ 21.8 Metallurgy and Iron and Steel Production 983 oxidized to Fe2O3, and then reduced to Fe3O4: Sometimes the nonmagnetic ores are concentrated by flotation processes.
The most commonly used reduction process for iron takes place in the blast furnace (Fig. 21.36). The raw materials required are concentrated iron ore, coke, and limestone (which serves as a flux to trap impurities). The furnace, which is approximately 25 feet in diameter, is charged from the top with a mixture of iron ore, coke, and limestone. A very strong blast (350 mi/h) of hot air is injected at the bottom, where the oxygen re-acts with the carbon in the coke to form carbon monoxide, the reducing agent for the iron.
The temperature of the charge increases as it travels down the furnace, with reduction of the iron to iron metal occurring in steps: Iron can reduce carbon dioxide, so complete reduction of the iron occurs only if the carbon dioxide is destroyed by adding excess coke: CO2 C ¡ 2CO Fe CO2 ¡ FeO CO FeO CO ¡ Fe CO2 Fe3O4 CO ¡ 3FeO CO2 3Fe2O3 CO ¡ 2Fe3O4 CO2 3Fe2O31s2 C1s2 ¡ 2Fe3O41s2 CO1g2 4FeO1s2 O21g2 ¡ 2Fe2O31s2 FeCO31s2 ¡ Heat FeO1s2 CO21g2 FIGURE 21.36 The blast furnace used in the production of iron.
Exhaust gases 1900°C 1300°C 1000°C 800°C 200°C Iron ore, limestone, and coke 3Fe2O3 + CO 2Fe3O4 + CO2 Fe3O4 + CO 3FeO + CO2 FeO + CO Fe + CO2 Fe + CO2 FeO + CO C + CO2 2CO CaCO3 CaO + CO2 Slag formation CaO + SiO2 CaSiO3 C + CO2 2CO C + O2 CO2 Pig iron Slag Oxygen-enriched air 984 Chapter Twenty-One Transition Metals and Coordination Chemistry The limestone (CaCO3) in the charge loses carbon dioxide, or calcines, in the hot furnace and combines with silica and other impurities to form slag, which is mostly molten cal-cium silicate, CaSiO3, and alumina (Al2O3). The slag floats on the molten iron and is skimmed off. The gas that escapes from the top of the furnace contains carbon monoxide, which is combined with air to form carbon dioxide. The energy released in this exothermic reaction is collected in a heat exchanger and used in heating the furnace.
The iron collected from the blast furnace, called pig iron, is quite impure. It contains 90% iron, 5% carbon, 2% manganese, 1% silicon, 0.3% phosphorus, and 0.04% sulfur (from impurities in the coke). The production of 1 ton of pig iron requires approximately 1.7 tons of iron ore, 0.5 ton of coke, 0.25 ton of limestone, and 2 tons of air.
Iron oxide also can be reduced in a direct reduction furnace, which operates at much lower temperatures (1300–2000F) than a blast furnace and produces a solid “sponge iron” rather than molten iron. Because of the milder reaction conditions, the direct reduction furnace requires a higher grade of iron ore (with fewer impurities) than that used in a blast furnace. The iron from the direct reduction furnace is called DRI (directly reduced iron) and contains 95% iron, with the balance mainly silica and alumina.
Production of Steel Steel is an alloy and can be classified as carbon steel, which contains up to about 1.5% carbon, or alloy steel, which contains carbon plus other metals such as Cr, Co, Mn, and Mo. The wide range of mechanical properties associated with steel is determined by its chemical composition and by the heat treatment of the final product.
The production of iron from its ore is fundamentally a reduction process, but the con-version of iron to steel is basically an oxidation process in which unwanted impurities are eliminated. Oxidation is carried out in various ways, but the two most common are the open hearth process and the basic oxygen process.
In the oxidation reactions of steelmaking, the manganese, phosphorus, and silicon in the impure iron react with oxygen to form oxides, which in turn react with appro-priate fluxes to form slag. Sulfur enters the slag primarily as sulfides, and excess carbon forms carbon monoxide or carbon dioxide. The flux chosen depends on the major impurities present. If manganese is the chief impurity, an acidic flux of silica is used: If the main impurity is silicon or phosphorus, a basic flux, usually lime (CaO) or mag-nesia (MgO), is needed to give reactions such as Whether an acidic or a basic slag will be needed is a factor that must be considered when a furnace is constructed so that its refractory linings will be compatible with the slag. Sil-ica bricks would deteriorate quickly in the presence of basic slag, and magnesia or lime bricks would dissolve in acid slag.
The open hearth process (Fig. 21.37) uses a dishlike container that holds 100 to 200 tons of molten iron. An external heat source is required to keep the iron molten, and a concave roof over the container reflects heat back toward the iron surface. A blast of air or oxygen is passed over the surface of the iron to react with impurities. Silicon P4O101s2 6CaO1s2 ¡ Heat 2Ca31PO4221l2 SiO21s2 MgO1s2 ¡ Heat MgSiO31l2 MnO1s2 SiO21s2 ¡ Heat MnSiO31l2 CaO SiO2 ¡ CaSiO3 21.8 Metallurgy and Iron and Steel Production 985 and manganese are oxidized first and enter the slag, followed by oxidation of carbon to carbon monoxide, which causes agitation and foaming of the molten bath. The exothermic oxidation of carbon raises the temperature of the bath, causing the limestone flux to calcine: The resulting lime floats to the top of the molten mixture (an event called the lime boil), where it combines with phosphates, sulfates, and silicates. Next comes the refining process, which involves continued oxidation of carbon and other impurities. Because the melting point increases as the carbon content decreases, the bath temperatures must be increased during this phase of the operation. If the carbon content falls below that desired in the fi-nal product, coke or pig iron may be added.
The final composition of the steel is “fine-tuned” after the charge is poured. For ex-ample, aluminum is sometimes added at this stage to remove trace amounts of oxygen via the reaction Alloying metals such as vanadium, chromium, titanium, manganese, and nickel are also added to give the steel the properties needed for a specific application.
The processing of a batch of steel by the open hearth process is quite slow, taking up to 8 hours. The basic oxygen process is much faster. Molten pig iron and scrap iron are placed in a barrel-shaped container (Fig. 21.38) that can hold as much as 300 tons of material. A high-pressure blast of oxygen is directed at the surface of the molten iron, oxidizing impurities in a manner very similar to that used in the open hearth process.
Fluxes are added after the oxygen blow begins. One advantage of this process is that the exothermic oxidation reactions proceed so rapidly that they produce enough heat to raise the temperature nearly to the boiling point of iron without an external heat source. Also, at these high temperatures only about an hour is needed to complete the oxidation processes.
The electric arc method, which was once used only for small batches of specialty steels, is being utilized more and more in the steel industry. In this process an electric arc between carbon electrodes is used to melt the charge. This means that no fuel-borne im-purities are added to the steel, since no fuel is needed. Also, higher temperatures are pos-sible than in the open hearth or basic oxygen processes, and this leads to more effective removal of sulfur and phosphorus impurities. Oxygen is added in this process so that the oxide impurities in the steel can be controlled effectively.
4Al 3O2 ¡ 2Al2O3 CaCO3 ¡ Heat CaO CO2 FIGURE 21.37 A schematic diagram of the open hearth process for steelmaking. The checker cham-bers contain bricks that absorb heat from gases passing over the molten charge. The flow of air and gases is reversed periodically.
Checker chamber Gas or liquid fuel Burner Alternate burner Air Checker chamber Burned gases Steel ladle Slag pot Hearth Tap hole Molten metal FIGURE 21.38 The basic oxygen process for steelmaking.
Oxygen Molten metal Flux 986 Chapter Twenty-One Transition Metals and Coordination Chemistry Heat Treatment of Steel One way of producing the desired physical properties in steel is by controlling the chem-ical composition (see Table 21.21). Another method for tailoring the properties of steel involves heat treatment. Pure iron exists in two different crystalline forms, depending on the temperature. At any temperature below iron has a body-centered cubic structure and is called -iron. Between and iron has a face-centered cubic structure called austentite, or -iron. At iron changes to -iron, a body-centered cubic structure identical to -iron.
When iron is alloyed with carbon, which fits into holes among the iron atoms to form the interstitial alloy carbon steel, the situation becomes even more complex. For exam-ple, the temperature at which -iron changes to austentite is lowered by about Also, at high temperatures iron and carbon react by an endothermic reversible reaction to form an iron carbide called cementite: (Heat) Cementite By Le Châtelier’s principle, we can predict that cementite will become more stable rela-tive to iron and carbon as the temperature is increased. This is the observed result.
Thus steel is really a mixture of iron metal in one of its crystal forms, carbon, and cementite. The proportions of these components are very important in determining the physical properties of steel.
When steel is heated to temperatures in the region of much of the carbon is converted to cementite. If the steel is then allowed to cool slowly, the equilibrium shown above shifts to the left, and small crystals of carbon precipitate, giving a steel that is relatively ductile. If the cooling is very rapid, the equilibrium does not have time to adjust. The cementite is trapped, and the steel has a high cementite content, making it quite brittle. The proportions of carbon crystals and cementite can be “fine-tuned” to give the desired properties by heating to intermediate temperatures followed by rapid cooling, a process called tempering. The rate of heating and cooling determines not only the amounts of cementite present but also the size of its crystals and the form of crystalline iron present.
1000°C, 3Fe C energy ∆ Fe3C 200°C.
a a d 1394°C, g 1394°C, 912°C a 912°C, TABLE 21.21 Percent Composition and Uses of Various Types of Steel Type of % % % % % % % % Steel Carbon Manganese Phosphorus Sulfur Silicon Nickel Chromium Other Uses Plain carbon 1.35 1.65 0.04 0.05 0.60 — — — Sheet steel, tools High-strength 0.25 1.65 0.04 0.05 0.15– 0.4–1 0.3–1.3 Cu (0.2–0.6) Transportation (low-alloy) 0.9 Sb (0.01–0.08) equipment, V (0.01–0.08) structural beams Alloy 1.00 3.50 0.04 0.05 0.15– 0.25– 0.25–4.0 Mo (0.08–4.0) Automobile 2.0 10.0 V (0–0.2) and aircraft W (0–18) engine parts Co (0–5) Stainless 0.03– 1.0–10 0.04– 0.03 1–3 1–22 4.0–27 — Engine parts, 1.2 0.06 steam turbine parts, kitchen utensils Silicon — — — — 0.5– — — — Electric 5.0 motors and transformers Refer to Section 10.3 for a review of packing and crystal lattices.
For Review 987 Key Terms Section 21.1 complex ion first-row transition metals lanthanide contraction lanthanide series Section 21.3 coordination compound counterion oxidation state coordination number ligand coordinate covalent bond monodentate (unidentate) ligand chelating ligand (chelate) bidentate ligand Section 21.4 isomers structural isomerism stereoisomerism coordination isomerism linkage isomerism geometrical (cis–trans) isomerism trans isomer cis isomer optical isomerism chiral enantiomers Section 21.6 crystal field model d-orbital splitting strong-field (low-spin) case weak-field (high-spin) case spectrochemical series Section 21.7 cytochromes heme porphyrin myoglobin hemoglobin carboxyhemoglobin Section 21.8 metallurgy minerals gangue flotation process roasting smelting zone refining pyrometallurgy hydrometallurgy leaching cyanidation blast furnace slag pig iron For Review First-row transition metals (scandium–zinc) All have one or more electrons in the 4s orbital and various numbers of 3d electrons All exhibit metallic properties • A particular element often shows more than one oxidation state in its compounds Most compounds are colored, and many are paramagnetic Most commonly form coordination compounds containing a complex ion involv-ing ligands (Lewis bases) attached to a central transition metal ion • The number of attached ligands (called the coordination number) can vary from 2 to 8, with 4 and 6 being most common Many transition metal ions have major biologic importance in molecules such as enzymes and those that transport and store oxygen • Chelating ligands form more than one bond to the transition metal ion Isomerism Isomers: two or more compounds with the same formula but different properties • Coordination isomerism: the composition of the coordination sphere varies • Linkage isomerism: the point of attachment of one or more ligands varies • Stereoisomerism: isomers have identical bonds but different spatial arrangements • Geometric isomerism: ligands assume different relative positions in the coordi-nation sphere; examples are cis and trans isomers • Optical isomerism: molecules with nonsuperimposable mirror images rotate plane-polarized light in opposite directions Spectral and magnetic properties Usually explained in terms of the crystal field model Model assumes the ligands are point charges that split the energies of the 3d orbitals Color and magnetism are explained in terms of how the 3d electrons occupy the split 3d energy levels • Strong-field case: relatively large orbital splitting • Weak-field case: relatively small orbital splitting Metallurgy The processes connected with separating a metal from its ore • The minerals in ores are often converted to oxides (roasting) before being re-duced to the metal (smelting) The metallurgy of iron: most common method for reduction uses a blast furnace; process involves iron ore, coke, and limestone • Impure product (90% iron) is called pig iron Steel is manufactured by oxidizing the impurities in pig iron REVIEW QUESTIONS 1. What two first-row transition metals have unexpected electron configurations? A statement in the text says that first-row transition metal ions do not have 4s electrons. Why not? Why do transition metal ions often have several oxidation states, whereas representative metals generally have only one?
2. Define each of the following terms: a. coordination compound b. complex ion c. counterions d. coordination number 988 Chapter Twenty-One Transition Metals and Coordination Chemistry direct reduction furnace carbon steel alloy steel open hearth process basic oxygen process electric arc method tempering e. ligand f. chelate g. bidentate How would transition metal ions be classified using the Lewis definition of acids and bases? What must a ligand have to bond to a metal? What do we mean when we say that a bond is a “coordinate covalent bond”?
3. When a metal ion has a coordination number of 2, 4, or 6, what are the ob-served geometries and associated bond angles? For each of the following, give the correct formulas for the complex ions.
a. linear Ag complex ions having ligands b. tetrahedral Cu complex ions having H2O ligands c. tetrahedral Mn2 complex ions having oxalate ligands d. square planar Pt2 complex ions having NH3 ligands e. octahedral Fe3 complex ions having EDTA ligands f. octahedral Co2 complex ions having ligands g. octahedral Cr3 complex ions having ethylenediamine ligands What is the electron configuration for the metal ion in each of the complex ions in a–g?
4. What is wrong with the following formula–name combinations? Give the cor-rect names for each.
a. [Cu(NH3)4]Cl2 copperammine chloride b. [Ni(en)2]SO4 bis(ethylenediamine)nickel(IV) sulfate c. K[Cr(H2O)2Cl4] potassium tetrachlorodiaquachromium(III) d. Na4[Co(CN)4C2O4] tetrasodium tetracyanooxalatocobaltate(II) 5. Define each of the following and give examples of each.
a. isomer b. structural isomer c. stereoisomer d. coordination isomer e. linkage isomer f. geometrical isomer g. optical isomer Consider the cis and trans forms of the octahedral complex Cr(en)2Cl2. Are both of these isomers optically active? Explain.
Another way to determine whether a substance is optically active is to look for a plane of symmetry in the molecule. If a substance has a plane of symme-try, then it will not exhibit optical activity (the mirror image will be superim-posable). Show the plane of symmetry in the trans isomer and prove to yourself that the cis isomer does not have a plane of symmetry.
6. What is the major focus of the crystal field model? Why are the d orbitals split into two sets for an octahedral complex? What are the two sets of orbitals?
Define each of the following.
a. weak-field ligand b. strong-field ligand c. low-spin complex d. high-spin complex Why is Co(NH3)6 3 diamagnetic whereas is paramagnetic? Some octa-hedral complex ions have the same d-orbital splitting diagrams whether they are high-spin or low-spin. For which of the following is this true?
a. V3 b. Ni2 c. Ru2 CoF6 3 Cl CN Questions 989 7. The crystal field model predicts magnetic properties of complex ions and ex-plains the colors of these complex ions. How? Solutions of [Cr(NH3)6]Cl3 are yellow, but Cr(NH3)6 3 does not absorb yellow light. Why? What color light is absorbed by Cr(NH3)6 3? What is the spectrochemical series, and how can the study of light absorbed by various complex ions be used to develop this series?
Would you expect Co(NH3)6 2 to absorb light of a longer or shorter wavelength than Co(NH3)6 3? Explain. Compounds of copper(II) are generally colored, but compounds of copper(I) are not colored. Explain. Would you expect Cd(NH3)4Cl2 to be colored? Explain. Compounds of Sc3 are not colored, but those of Ti3 and V3 are colored. Explain.
8. Why do tetrahedral complex ions have a different crystal field diagram than octahedral complex ions? What is the tetrahedral crystal field diagram? Why are virtually all tetrahedral complex ions “high spin”?
Explain the crystal field diagram for square planar complex ions and for linear complex ions.
9. Review Table 21.18, which lists some important biological functions associated with different first-row transition metals. The transport of O2 in the blood is carried out by hemoglobin. Briefly explain how hemoglobin transports O2 in the blood. Why are and CO toxic to humans and animals?
10. Define and give an example of each of the following.
a. roasting b. smelting c. flotation d. leaching e. gangue What are the advantages and disadvantages of hydrometallurgy? Describe the process by which a metal is purified by zone refining.
CN A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 5. Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?
6. Four different octahedral chromium coordination compounds ex-ist that all have the same oxidation state for chromium and have H2O and Cl as the ligands and counterions. When 1 mol of each of the four compounds is dissolved in water, how many mol of silver chloride will precipitate upon addition of excess AgNO3?
7. Figure 21.17 shows that the cis isomer of Co(en)2Cl2 is optically active while the trans isomer is not optically active. Is the same true for Co(NH3)4Cl2 ? Explain.
8. A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each hav-ing the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four un-paired electrons and the other two are diamagnetic. What can be de-duced from this information about the four coordination compounds?
Active Learning Questions These questions are designed to be used by groups of students in class. The questions allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the learning that occurs while students talk to each other about chemical concepts.
1. You isolate a compound with the formula PtCl4 2KCl. From elec-trical conductance tests of an aqueous solution of the compound, you find that three ions per formula unit are present, and you also notice that addition of AgNO3 does not cause a precipitate. Give the formula for this compound that shows the complex ion present. Ex-plain your findings. Name this compound.
2. Both Ni(NH3)4 2 and Ni(CN)4 2 have four ligands. The first is paramagnetic, and the second is diamagnetic. Are the complex ions tetrahedral or square planar? Explain.
3. Which is more likely to be paramagnetic, Fe(CN)6 4 or Fe(H2O)6 2?
Explain.
4. A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low-spin octahe-dral complex. Name some possible metal ions for which this would be true.
990 Chapter Twenty-One Transition Metals and Coordination Chemistry 23. Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium: Purple Pink What color will such an indicator be if rain is imminent?
24. Chromium(VI) forms two different oxyanions, the orange dichro-mate ion, and the yellow chromate ion, The equi-librium reaction between the two ions is Cr2O7 21aq2 H2O1l2 ∆2CrO4 21aq2 2H1aq2 CrO4 2.
Cr2O7 2, CoCl21s2 6H2O1g2 ∆CoCl2 6H2O1s2 9. CoCl4 2 forms a tetrahedral complex ion and Co(CN)6 3 forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the d orbital split-ting diagrams?
a. CoCl4 2 is an example of a strong-field case having two un-paired electrons.
b. Because CN is a weak field ligand, Co(CN)6 3 will be a low spin case having four unpaired electrons.
10. The following statements discuss some coordination compounds.
For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion.
a.
is a compound used in novelty devices that predict rain.
b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosulfate. The equa-tion for the reaction is: c. The compound cisplatin, Pt(NH3)2Cl2, has been studied exten-sively as an antitumor agent. The reaction for the synthesis of cisplatin is: Assume these platinum complex ions have a square planar geometry.
d. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically re-sistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: Assume these copper complex ions have tetrahedral geometry.
11. What causes high-altitude sickness and what is high-altitude acclimatization?
12. Almost all metals in nature are found as ionic compounds in ores instead of being in the pure state. Why? What must be done to a sam-ple of ore to obtain a metal substance that has desirable properties?
Exercises In this section similar exercises are paired.
Transition Metals and Coordination Compounds 13. Write electron configurations for the following metals.
a. Ni b. Cd c. Zr d. Os 14. Write electron configurations for the following ions.
a. Ni2 c. Zr3 and Zr4 b. Cd2 d. Os2 and Os3 15. Write electron configurations for each of the following.
a. Ti, Ti2, Ti4 b. Re, Re2, Re3 c. Ir, Ir2, Ir3 Cu1NH324Cl21aq2 4NH31aq2 Cu1s2 ¡ 2Cu1NH324Cl1aq2 K2PtCl41aq2 2NH31aq2 ¡ Pt1NH322Cl21s2 2KCl1aq2 NaBr1aq2 AgBr1s2 2Na2S2O31aq2 ¡ Na33Ag1S2O32241aq2 CoCl2 6H2O Atomic Atomic Element Radius (Å) Element Radius (Å) Sc 1.57 Ti 1.477 Y 1.693 Zr 1.593 La 1.915 Hf 1.476 16. Write electron configurations for each of the following.
a. Cr, Cr2, Cr3 b. Cu, Cu, Cu2 c. V, V2, V3 17. What is the electron configuration for the transition metal ion in each of the following compounds?
a. K3[Fe(CN)6] b. [Ag(NH3)2]Cl c. [Ni(H2O)6]Br2 d. [Cr(H2O)4(NO2)2]I 18. What is the electron configuration for the transition metal ion(s) in each of the following compounds?
a. (NH4)2[Fe(H2O)2Cl4] b. [Co(NH3)2(NH2CH2CH2NH2)2]I2 c. Na2[TaF7] d. [Pt(NH3)4I2][PtI4] Pt forms 2 and 4 oxidation states in compounds.
19. Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as MoS2, which is then converted to MoO3. The MoO3 can be used directly in the production of stainless steel for high-speed tools (which accounts for about 85% of the molybdenum used). Molybdenum can be purified by dissolving MoO3 in aqueous ammonia and crys-tallizing ammonium molybdate. Depending on conditions, either (NH4)2Mo2O7 or is obtained.
a. Give names for MoS2 and MoO3.
b. What is the oxidation state of Mo in each of the compounds mentioned above?
20. Iron is present in the earth’s crust in many types of minerals. The iron oxide minerals are hematite (Fe2O3) and magnetite (Fe3O4).
What is the oxidation state of iron in each mineral? The iron ions in magnetite are a mixture of Fe2 and Fe3 ions. What is the ratio of Fe3 to Fe2 ions in magnetite? The formula for magnetite is usually written as Does this make sense? Explain.
21. What is the lanthanide contraction? How does the lanthanide con-traction affect the properties of the 4d and 5d transition metals?
22. We expect the atomic radius to increase down a group in the pe-riodic table. Can you suggest why the atomic radius of hafnium breaks this rule? (See the following data.) FeO Fe2O3.
(NH4)6Mo7O24 4H2O Exercises 991 31. Name the following coordination compounds.
a. [Co(NH3)6]Cl2 d. K4[PtCl6] b. [Co(H2O)6]I3 e. [Co(NH3)5Cl]Cl2 c. K2[PtCl4] f. [Co(NH3)3(NO2)3] 32. Name the following coordination compounds.
a. [Cr(H2O)5Br]Br2 c. [Fe(NH2CH2CH2NH2)2(NO2)2]Cl b. Na3[Co(CN)6] d. [Pt(NH3)4I2][PtI4] 33. Give formulas for the following.
a. potassium tetrachlorocobaltate(II) b. aquatricarbonylplatinum(II) bromide c. sodium dicyanobis(oxalato)ferrate(III) d. triamminechloroethylenediaminechromium(III) iodide 34. Give formulas for the following complex ions.
a. tetrachloroferrate(III) ion b. pentaammineaquaruthenium(III) ion c. tetracarbonyldihydroxochromium(III) ion d. amminetrichloroplatinate(II) ion 35. Draw geometrical isomers of each of the following complex ions.
a. [Co(C2O4)2(H2O)2] c. [Ir(NH3)3Cl3] b. [Pt(NH3)4I2]2 d. [Cr(en)(NH3)2I2] 36. Draw structures of each of the following.
a. cis-dichloroethylenediamineplatinum(II) b. trans-dichlorobis(ethylenediamine)cobalt(II) c. cis-tetraamminechloronitrocobalt(III) ion d. trans-tetraamminechloronitritocobalt(III) ion e. trans-diaquabis(ethylenediamine)copper(II) ion 37. Amino acids can act as ligands toward transition metal ions. The simplest amino acid is glycine (NH2CH2CO2H). Draw a structure of the glycinate anion acting as a bidentate li-gand. Draw the structural isomers of the square planar complex Cu(NH2CH2CO2)2.
38. BAL is a chelating agent used in treating heavy metal poisoning.
It acts as a bidentate ligand. What type of linkage isomers are pos-sible when BAL coordinates to a metal ion?
39. Which of the following ligands are capable of linkage isomerism?
Explain your answer.
40. Draw all geometrical and linkage isomers of Co(NH3)4(NO2)2.
41. Acetylacetone, abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as as shown below, where M is a transition metal: P O G D J M O C CH3 CH G D G CH3 O C acac, SCN, N3 , NO2 , NH2CH2CH2NH2, OCN, I CH2 CH2 CH SH O SH O A A OH O BAL (NH2CH2CO2 ) The following pictures show what happens when sodium hy-droxide is added to a dichromate solution.
Explain what happened.
25. A series of chemicals was added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution with the end result shown below in test tube 1, NH3(aq) was then added with the end result shown in test tube 2, and HNO3(aq) was added last with the end result shown in test tube 3.
Explain the results shown in each test tube. Include a balanced equation for the reaction(s) taking place.
26. When an aqueous solution of KCN is added to a solution con-taining Ni2 ions, a precipitate forms, which redissolves on addi-tion of more KCN solution. Write reactions describing what happens in this solution. (Hint: is a Brønsted–Lowry base and a Lewis base.) 27. Consider aqueous solutions of the following coordination com-pounds: Co(NH3)6I3, Pt(NH3)4I4, Na2PtI6, and Cr(NH3)4I3. If aque-ous AgNO3 is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will pre-cipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.
28. A coordination compound of cobalt(III) contains four ammonia molecules, one sulfate ion, and one chloride ion. Addition of aque-ous BaCl2 solution to an aqueous solution of the compound gives no precipitate. Addition of aqueous AgNO3 to an aqueous solu-tion of the compound produces a white precipitate. Propose a structure for this coordination compound.
29. Name the following complex ions.
a. Ru(NH3)5Cl2 c. Mn(NH2CH2CH2NH2)3 2 b. Fe(CN)6 4 d. Co(NH3)5NO2 2 30. Name the following complex ions.
a. Ni(CN)4 2 c. Fe(C2O4)3 3 b. Cr(NH3)4Cl2 d. Co(SCN)2(H2O)4 [Kb 105] CN 1 2 3 992 Chapter Twenty-One Transition Metals and Coordination Chemistry For each compound, predict the predominant color of light ab-sorbed. If the complex ions are Cr(NH3)6 3, Cr(H2O)6 3, and Cr(H2O)4Cl2 , what is the identity of the complex ion in each test tube? Hint: Reference the spectrochemical series.
52. Consider the complex ions Co(NH3)6 3, Co(CN)6 3, and CoF6 3.
The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are 770 nm, 440 nm, and 290 nm.
Match the complex ion to the wavelength of absorbed electro-magnetic radiation.
53. The wavelength of absorbed electromagnetic radiation for is m. Will the complex ion absorb electro-magnetic radiation having a wavelength longer or shorter than m? Explain.
54. Tetrahedral complexes of Co2 are quite common. Use a d-orbital splitting diagram to rationalize the stability of Co2 tetrahedral complex ions.
55. How many unpaired electrons are present in the tetrahedral ion ?
56. The complex ion is diamagnetic. Propose a structure for Metallurgy 57. A blast furnace is used to reduce iron oxides to elemental iron.
The reducing agent for this reduction process is carbon monoxide.
a. Given the following data: determine H for the reaction b. The CO2 produced in a blast furnace during the reduction process actually can oxidize iron into FeO. To eliminate this reaction, excess coke is added to convert CO2 into CO by the reaction Using data from Appendix 4, determine and for this reaction. Assuming and do not depend on tempera-ture, at what temperature is the conversion reaction of CO2 into CO spontaneous at standard conditions?
58. What roles do kinetics and thermodynamics play in the effect that the following reaction has on the properties of steel?
59. Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. Cyanide ion is often used to extract the silver by the following reaction that occurs in basic solution: Balance this equation by using the half-reaction method.
Ag1s2 CN1aq2 O21g2 ¡ Basic Ag1CN22 1aq2 3Fe C ∆Fe3C ¢S° ¢H° ¢S° ¢H° CO21g2 C1s2 ¡ 2CO1g2 FeO1s2 CO1g2 ¡ Fe1s2 CO21g2 ¢H° 18 kJ Fe3O41s2 CO1g2 ¡ 3FeO1s2 CO21g2 ¢H° 39 kJ 3Fe2O31s2 CO1g2 ¡ 2Fe3O41s2 CO21g2 ¢H° 23 kJ Fe2O31s2 3CO1g2 ¡ 2Fe1s2 3CO21g2 PdCl4 2.
PdCl4 2 FeCl4 3.4 106 CoBr6 4 3.4 106 CoBr4 2 Which of the following complexes are optically active: cis-Cr(acac)2(H2O)2, trans-Cr(acac)2(H2O)2, and Cr(acac)3?
42. Draw all geometrical isomers of Pt(CN)2Br2(H2O)2. Which of these isomers has an optical isomer? Draw the various optical isomers.
Bonding, Color, and Magnetism in Coordination Compounds 43. Draw the d-orbital splitting diagrams for the octahedral complex ions of each of the following.
a. Fe2 (high and low spin) b. Fe3 (high spin) c. Ni2 44. Draw the d-orbital splitting diagrams for the octahedral complex ions of each of the following.
a. Zn2 b. Co2 (high and low spin) c. Ti3 45. The ion is known to have four unpaired electrons. Does the ligand produce a strong or weak field?
46. The Co(NH3)6 3 ion is diamagnetic, but Fe(H2O)6 2 is paramag-netic. Explain.
47. How many unpaired electrons are in the following complex ions?
a. Ru(NH3)6 2 (low-spin case) b. Ni(H2O)6 2 c. V(en)3 3 48. The complex ion is paramagnetic with one unpaired electron. The complex ion has five unpaired electrons.
Where does lie in the spectrochemical series relative to 49. Rank the following complex ions in order of increasing wave-length of light absorbed.
50. The complex ion has an absorption maximum at around 800 nm. When four ammonias replace water, [Cu(NH3)4(H2O)2]2, the absorption maximum shifts to around 600 nm. What do these results signify in terms of the relative field splittings of NH3 and H2O? Explain.
51. The following test tubes each contain a different chromium com-plex ion.
[Cu(H2O)6]2 3Co1H2O264 3, 3Co1CN264 3, 3Co1I62 43, 3Co1en234 3 CN?
SCN Fe1SCN26 3 Fe1CN26 3 F CrF6 4 Additional Exercises 993 a. acetylacetone(acacH) b. diethylenetriamine c. salen d. porphine 66. Until the discoveries of Werner, it was thought that carbon had to be present in a compound for it to be optically active. Werner pre-pared the following compound containing ions as bridging groups and separated the optical isomers.
a. Draw structures of the two optically active isomers of this compound.
b. What are the oxidation states of the cobalt ions?
c. How many unpaired electrons are present if the complex is the low-spin case?
67. The complex ion Ru(phen)3 2 has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in Ru(phen)3 2?
b. Ru(phen)3 2 is diamagnetic (as are all complex ions of Ru2).
Draw the crystal field diagram for the d orbitals in this com-plex ion.
Phen 1,10-phenanthroline N N 3 H Co Co( NH3 H )4 A A O OD G Cl6 OH NH HN N N HO OH N N NH2¬CH2¬CH2¬NH¬CH2¬CH2¬NH2 C B O O CH3 CH2 C O B O O CH3 O 60. One of the classic methods for the determination of the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorp-tion of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.
Additional Exercises 61. Ammonia and potassium iodide solutions are added to an aque-ous solution of Cr(NO3)3. A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess O2, 0.0203 g of CrO3 was formed.
ii. In a second experiment it took 32.93 mL of 0.100 M HCl to titrate completely the NH3 present in 0.341 g of compound A.
iii. Compound A was found to contain 73.53% iodine by mass.
iv. The freezing point of water was lowered by 0.64ºC when 0.601 g of compound A was dissolved in 10.00 g of H2O What is the formula of the compound? What is the structure of the complex ion present? (Hints: Cr3 is expected to be six-coordinate, with NH3 and possibly as ligands. The ions will be the counterions if needed.) 62. A transition metal compound contains a cobalt ion, chloride ions, and water molecules. The H2O molecules are the ligands in the complex ion and the ions are the counterions. A 0.256-g sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g. What is the oxidation state of cobalt in the complex ion and what is the formula of the compound?
63. When aqueous KI is added gradually to mercury(II) nitrate, an or-ange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: Hg2 reacts with to form Would you expect to form colored solutions? Explain.
64. In the production of printed circuit boards for the electronics in-dustry, a 0.60-mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resis-tant polymer is printed on the board. The unwanted copper is re-moved by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is a. Is this reaction an oxidation–reduction process? Explain.
b. A plant needs to manufacture 10,000 printed circuit boards, each 8.0 16.0 cm in area. An average of 80.% of the cop-per is removed from each board (density of copper 8.96 g/cm3). What masses of [Cu(NH3)4]Cl2 and NH3 are needed to do this? Assume 100% yield.
65. How many bonds could each of the following chelates form with a metal ion?
¡ 23Cu1NH3244Cl1aq2 3Cu1NH3244Cl21aq2 4NH31aq2 Cu1s2 HgI4 2 HgI4 2.
I Cl I I 1.86°C kg/mol).
(Kf 994 Chapter Twenty-One Transition Metals and Coordination Chemistry 73. Carbon monoxide is toxic because it binds more strongly to iron in hemoglobin (Hb) than does O2. Consider the following reac-tions and approximate standard free energy changes: Using these data, estimate the equilibrium constant value at 25C for the following reaction: 74. For the process what would be the expected ratio of cis to trans isomers in the product?
Challenge Problems 75. The complex trans-[NiA2B4]2, where A and B represent neutral ligands, is known to be diamagnetic. Do A and B produce very similar or very different crystal fields? Explain.
76. Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is a. Without referring to Appendix 4, predict the sign of S for the preceding reaction. Explain.
b. The spontaneity of the preceding reaction is temperature de-pendent. Predict the sign of Ssurr for this reaction. Explain.
c. For Ni(CO)4(g), Hf 607 kJ/mol and S 417 J/K mol at 298 K. Using these values and data in Appendix 4, calcu-late H and S for the preceding reaction.
d. Calculate the temperature at which G 0 (K 1) for the preceding reaction, assuming that H and S do not depend on temperature.
e. The first step of the Mond process involves equilibrating im-pure nickel with CO(g) and Ni(CO)4(g) at about 50C. The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at 50.C.
f. In the second step of the Mond process, the gaseous Ni(CO)4 is isolated and heated at 227C. The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the above reaction at 227C.
g. Why is temperature increased for the second step of the Mond process?
77. Consider the following data: where en ethylenediamine.
a. Calculate e for the half-reaction Co1en23 3 e ¡ Co1en23 2 Co3 3en ¡ Co1en23 3 K 2.0 1047 Co2 3en ¡ Co1en23 2 K 1.5 1012 Co3 e ¡ Co2 e° 1.82 V Ni1s2 4CO1g2 ∆Ni1CO241g2 Co1NH325Cl2 Cl ¡ Co1NH324Cl2 NH3 HbO2 CO ∆HbCO O2 Hb CO ¡ HbCO ¢G° 80 kJ Hb O2 ¡ HbO2 ¢G° 70 kJ 68. A compound related to acetylacetone is 1,1,1-trifluoroacetyl-acetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise 41.) Both Be2 and Cu2 form complexes with tfa hav-ing the formula M(tfa)2. Two isomers are formed for each metal complex.
a. The Be2 complexes are tetrahedral. Draw the two isomers of Be(tfa)2. What type of isomerism is exhibited by Be(tfa)2?
b. The Cu2 complexes are square planar. Draw the two isomers of Cu(tfa)2. What type of isomerism is exhibited by Cu(tfa)2?
69. Would it be better to use octahedral complexes or octahe-dral complexes to determine whether a given ligand is a strong-field or weak-field ligand by measuring the number of un-paired electrons? How else could the relative ligand field strengths be determined?
70. The equilibrium constant, Ka, for the reaction is 6.0 103.
a. Calculate the pH of a 0.10 M solution of Fe(H2O)6 3.
b. Will a 1.0 M solution of iron(II) nitrate have a higher or lower pH than a 1.0 M solution of iron(III) nitrate? Explain.
71. Ethylenediaminetetraacetate is used as a complexing agent in chemical analysis with the structure shown in Figure 21.7.
Solutions of are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion.
The complex ion virtually eliminates the heavy metal ions from reacting with biochemical systems. The reaction of with is Consider a solution with 0.010 mol Pb(NO3)2 added to 1.0 L of an aqueous solution buffered at pH 13.00 and containing 0.050 M Na4EDTA. Does Pb(OH)2 precipitate from this solution? (Ksp for Pb(OH)2 1.2 1015.) 72. Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemo-globin molecule contains four iron atoms that serve as the bind-ing sites for O2 molecules. The oxygen binding is pH dependent.
The relevant equilibrium reaction is Use Le Châtelier’s principle to answer the following.
a. What form of hemoglobin, or Hb(O2)4, is favored in the lungs? What form is favored in the cells?
b. When a person hyperventilates, the concentration of CO2 in the blood decreases. How does this affect the oxygen-binding equilibrium? How does breathing into a paper bag help to coun-teract this effect?
c. When a person has suffered a cardiac arrest, an injection of a sodium bicarbonate solution is given. Why is this step necessary?
HbH4 4 HbH4 41aq2 4O21g2 ∆Hb1O2241aq2 4H1aq2 K 1.1 1018 Pb21aq2 EDTA41aq2 ∆PbEDTA21aq2 Pb2 EDTA4 EDTA4 (EDTA4) Fe1H2O251OH221aq2 H3O1aq2 Fe1H2O26 31aq2 H2O1l2 ∆ Cr2 Ni2 CF3CCH2CCH3 O O Marathon Problem 995 d. What mass of AgBr will dissolve in 250.0 mL of 3.0 M NH3?
e. What effect does adding HNO3 have on the solubilities calcu-lated in parts a and b?
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
85. The ferrate ion, is such a powerful oxidizing agent that in acidic solution, aqueous ammonia is reduced to elemental ni-trogen along with the formation of the iron(III) ion.
a. What is the oxidation state of iron in and what is the electron configuration of iron in this polyatomic ion?
b. If 25.0 mL of a 0.243 M solution is allowed to react with 55.0 mL of 1.45 M aqueous ammonia, what volume of nitrogen gas can form at and 1.50 atm?
86. a. In the absorption spectrum of the complex ion there is a band corresponding to the absorption of a photon of light with an energy of Given what is the wavelength of this photon?
b. The bond angle in is predicted to be What is the hybridization of the N atom in the ligand when a Lewis acid–base reaction occurs between Cr3 and that would give a bond angle?
undergoes substitution by ethylenediammine (en) according to the equation Does [Cr(NCS)2(en)2] exhibit geometric isomerism? Does [Cr(NCS)2(en)2] exhibit optical isomerism?
Marathon Problem This problem is designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
87. There are three salts that contain complex ions of chromium and have the molecular formula Treating 0.27 g of the first salt with a strong dehydrating agent resulted in a mass loss of 0.036 g. Treating 270 mg of the second salt with the same de-hydrating agent resulted in a mass loss of 18 mg. The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to 100.0-mL por-tions of 0.100 M solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 mg AgCl; another, 2870 mg AgCl; the third, 4300 mg AgCl. Two of the salts are green and one is violet.
Suggest probable structural formulas for these salts, defend-ing your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the mag-netic properties of the salts be helpful in determining the struc-tural formulas? Explain.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
CrCl3 6H2O.
3Cr1NCS264 3 2en ¡ 3Cr1NCS221en224 4NCS [Cr(NCS)6]3 Cr¬N¬C 180° NCS NCS 180°.
[Cr(NCS)6]3 Cr¬N¬C 1023 J, 1.986 1 cm1 1.75 104 cm1.
[Cr(NCS)6]3, 25°C FeO4 2 FeO4 2, FeO4 2, b. Based on your answer to part a, which is the stronger oxidizing agent, Co3 or Co(en)3 3?
c. Use the crystal field model to rationalize the result in part b.
78. Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation–reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment.
Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is Cr(H2O)5Cl2. Is this consistent with the reaction proceeding through formation of (H2O)5CrOClOCo(NH3)5 as an intermediate? Explain.
79. Chelating ligands often form more stable complex ions than the corresponding monodentate ligands with the same donor atoms.
For example, where en is ethylenediamine and penten is This increased stability is called the chelate effect. Based on bond energies, would you expect the enthalpy changes for the above reactions to be very different? What is the order (from least fa-vorable to most favorable) of the entropy changes for the above reactions? How do the values of the formation constants correlate with S? How can this be used to explain the chelate effect?
80. Qualitatively draw the crystal field splitting of the d orbitals in a trigonal planar complex ion. (Let the z axis be perpendicular to the plane of the complex.) 81. Qualitatively draw the crystal field splitting for a trigonal bi-pyramidal complex ion. (Let the z axis be perpendicular to the trigonal plane.) 82. Sketch a d-orbital energy diagram for the following.
a. a linear complex with ligands on the x axis b. a linear complex with ligands on the y axis 83. Sketch and explain the most likely pattern for the crystal field di-agram for the complex ion trans-diamminetetracyanonickelate(II), where produces a much stronger crystal field than NH3. Ex-plain completely and label the d orbitals in your diagram. Assume the NH3 ligands lie on the axis.
84. a. Calculate the molar solubility of AgBr in pure water. Ksp for AgBr is 5.0 1013.
b. Calculate the molar solubility of AgBr in 3.0 M NH3. The over-all formation constant for Ag(NH3)2 is 1.7 107, that is, c. Compare the calculated solubilities from parts a and b. Explain any differences.
K 1.7 107.
Ag 1aq2 2NH31aq2 ¡ Ag1NH322 1aq2 CN O G D G D N CH2 OCH2O N NH2CH2CH2 NH2CH2CH2 NH2 CH2CH2 NH2 CH2CH2 K 2.0 1019 Ni2 1aq2 penten1aq2 ∆Ni1penten22 1aq2 K 1.6 1018 Ni21aq2 3en1aq2 ∆Ni1en23 2 1aq2 K 3.2 108 Ni21aq2 6NH31aq2 ∆Ni1NH326 2 1aq2 ¡ Cr1III2 complexes Co1II2 complexes Cr1H2O26 2 1aq2 Co1NH325Cl21aq2 996 22 Organic and Biological Molecules Contents 22.1 Alkanes: Saturated Hydrocarbons • Isomerism in Alkanes • Nomenclature • Reactions of Alkanes • Cyclic Alkanes 22.2 Alkenes and Alkynes • Reactions of Alkenes and Alkynes 22.3 Aromatic Hydrocarbons 22.4 Hydrocarbon Derivatives • Alcohols • Aldehydes and Ketones • Carboxylic Acids and Esters • Amines 22.5 Polymers • The Development and Properties of Polymers • Types of Polymers • Polymers Based on Ethylene 22.6 Natural Polymers • Proteins • Carbohydrates • Nucleic Acids Polarized light micrograph of crystals of phenylalanine, one of the essential amino acids that the body cannot synthesize.
T wo Group 4A elements, carbon and silicon, form the basis of most natural substances.
Silicon, with its great affinity for oxygen, forms chains and rings containing bridges to produce the silica and silicates that form the basis for most rocks, sands, and soils. What silicon is to the geological world, carbon is to the biological world. Carbon has the unusual ability of bonding strongly to itself to form long chains or rings of carbon atoms. In addition, carbon forms strong bonds to other nonmetals such as hydrogen, nitrogen, oxygen, sulfur, and the halogens. Because of these bonding properties, there are a myriad of carbon compounds; several million are now known, and the number continues to grow rapidly. Among these many compounds are the biomolecules, those responsible for maintaining and reproducing life.
The study of carbon-containing compounds and their properties is called organic chemistry. Although a few compounds involving carbon, such as its oxides and carbon-ates, are considered to be inorganic substances, the vast majority are organic compounds that typically contain chains or rings of carbon atoms.
Originally, the distinction between inorganic and organic substances was based on whether a compound was produced by living systems. For example, until the early nine-teenth century it was believed that organic compounds had some sort of “life force” and could be synthesized only by living organisms. This view was dispelled in 1828 when the German chemist Friedrich Wöhler (1800–1882) prepared urea from the inorganic salt ammonium cyanate by simple heating: Urea is a component of urine, so it is clearly an organic material; yet here was clear evidence that it could be produced in the laboratory as well as by living things.
Organic chemistry plays a vital role in our quest to understand living systems. Beyond that, the synthetic fibers, plastics, artificial sweeteners, and drugs that are such an accepted part of modern life are products of industrial organic chemistry. In addition, the energy on which we rely so heavily to power our civilization is based mostly on the organic materials found in coal and petroleum.
Because organic chemistry is such a vast subject, we can provide only a brief intro-duction to it in this book. We will begin with the simplest class of organic compounds, the hydrocarbons, and then show how most other organic compounds can be considered to be derivatives of hydrocarbons.
22.1 Alkanes: Saturated Hydrocarbons As the name indicates, hydrocarbons are compounds composed of carbon and hydrogen.
Those compounds whose carbon–carbon bonds are all single bonds are said to be saturated, because each carbon is bound to four atoms, the maximum number. Hydro-carbons containing carbon–carbon multiple bonds are described as being unsaturated, O Ammonium cyanate Urea Heat O B NH4OCN H2N C O NH2 Si¬O¬Si 997 998 Chapter Twenty-Two Organic and Biological Molecules since the carbon atoms involved in a multiple bond can react with additional atoms, as shown by the addition of hydrogen to ethylene: Note that each carbon in ethylene is bonded to three atoms (one carbon and two hydro-gens) but that each can bond to one additional atom if one bond of the carbon–carbon double bond is broken.
The simplest member of the saturated hydrocarbons, which are also called the alkanes, is methane (CH4). As discussed in Section 9.1, methane has a tetrahedral structure and can be described in terms of a carbon atom using an sp3 hybrid set of orbitals to bond to the four hydrogen atoms (see Fig. 22.1). The next alkane, the one containing two carbon atoms, is ethane (C2H6), as shown in Fig. 22.2. Each carbon in ethane is surrounded by four atoms and thus adopts a tetrahedral arrangement and sp3 hybridization, as predicted by the localized electron model.
The next two members of the series are propane (C3H8) and butane (C4H10), shown in Fig. 22.3. Again, each carbon is bonded to four atoms and is described as sp3 hybridized.
Alkanes in which the carbon atoms form long “strings” or chains are called normal, straight-chain, or unbranched hydrocarbons. As can be seen from Fig. 22.3, the chains in normal alkanes are not really straight but zig-zag, since the tetrahedral COCOC angle is 109.5. The normal alkanes can be represented by the structure where n is an integer. Note that each member is obtained from the previous one by in-serting a methylene (CH2) group. We can condense the structural formulas by omitting some of the COH bonds. For example, the general formula for normal alkanes shown above can be condensed to The first ten normal alkanes and some of their properties are listed in Table 22.1. Note that all alkanes can be represented by the general formula CnH2n 2. For example, nonane, which has nine carbon atoms, is represented by C9H(2 9) 2, or C9H20. Also note from Table 22.1 that the melting points and boiling points increase as the molar masses in-crease, as we would expect.
Isomerism in Alkanes Butane and all succeeding members of the alkanes exhibit structural isomerism. Recall from Section 21.4 that structural isomerism occurs when two molecules have the same atoms CH3¬ 1 CH22n ¬CH3 O G DH C H2 A G D P H H H H C C O O A A A H H H H H C Unsaturated Saturated FIGURE 22.1 The COH bonds in methane.
FIGURE 22.2 (a) The Lewis structure of ethane (C2H6).
(b) The molecular structure of ethane repre-sented by space-filling and ball-and-stick models.
FIGURE 22.3 The structures of (a) propane (CH3CH2CH3) and (b) butane (CH3CH2CH2CH3). Each angle shown in red is 109.5°.
C sp3 sp3 sp3 sp3 H1s H1s H1s H1s C H H C H H H H (a) (b) (a) (b) 22.1 Alkanes: Saturated Hydrocarbons 999 but different bonds. For example, butane can exist as a straight-chain molecule (normal butane, or n-butane) or with a branched-chain structure (called isobutane), as shown in Fig. 22.4. Because of their different structures, these molecules exhibit different properties.
For example, the boiling point of n-butane is 0.5C, whereas that of isobutane is 12C.
Structural Isomerism Draw the isomers of pentane.
Solution Pentane (C5H12) has the following isomeric structures: 1.
CH3 CH2 CH2 CH2 CH3 n-Pentane FIGURE 22.4 (a) Normal butane (abbreviated n-butane).
(b) The branched isomer of butane (called isobutane).
TABLE 22.1 Selected Properties of the First Ten Normal Alkanes Number of Melting Boiling Structural Name Formula Molar Mass Point (°C) Point (°C) Isomers Methane CH4 16 182 162 1 Ethane C2H6 30 183 89 1 Propane C3H8 44 187 42 1 Butane C4H10 58 138 0 2 Pentane C5H12 72 130 36 3 Hexane C6H14 86 95 68 5 Heptane C7H16 100 91 98 9 Octane C8H18 114 57 126 18 Nonane C9H20 128 54 151 35 Decane C10H22 142 30 174 75 (a) (b) Sample Exercise 22.1 1000 Chapter Twenty-Two Organic and Biological Molecules 2.
3.
Note that the structures which might appear to be other isomers, are actually identical to structure 2.
See Exercise 22.13.
Nomenclature Because there are literally millions of organic compounds, it would be impossible to re-member common names for all of them. We must have a systematic method for naming them. The following rules are used in naming alkanes.
Rules for Naming Alkanes 1. The names of the alkanes beyond butane are obtained by adding the suffix -ane to the Greek root for the number of carbon atoms (pent- for five, hex- for six, and so on). For a branched hydrocarbon, the longest continuous chain of carbon atoms determines the root name for the hydrocarbon. For example, in the alkane the longest chain contains six carbon atoms, and this compound is named as a hexane.
2. When alkane groups appear as substituents, they are named by dropping the -ane and adding -yl. For example, OCH3 is obtained by removing a hydrogen from methane and is called methyl, OC2H5 is called ethyl, OC3H7 is called propyl, and so on. The compound above is therefore an ethylhexane. (See Table 22.2.) CH3 A CH2 Six carbons O A CH CH2 O CH2 O O CH3 A CH3 CH2 CH3 CH3 CH3 C CH3 Neopentane CH3 CH3 CH CH2 CH3 Isopentane TABLE 22.2 The Most Common Alkyl Substituents and Their Names Structure Name† OCH3 Methyl OCH2CH3 Ethyl OCH2CH2CH3 Propyl Isopropyl OCH2CH2CH2CH3 Butyl sec-Butyl Isobutyl tert-Butyl The bond with one end open shows the point of attachment of the substituent to the carbon chain.
†For the butyl groups, sec- indicates attach-ment to the chain through a secondary car-bon, a carbon atom attached to two other carbon atoms. The designation tert- signi-fies attachment through a tertiary carbon, a carbon attached to three other carbon atoms.
CO A CH3 O A CH3 CH3 H CH2 CH3 O O A O A CH3 C CH3CHCH2CH3 A CH3CHCH3 A 22.1 Alkanes: Saturated Hydrocarbons 1001 3. The positions of substituent groups are specified by numbering the longest chain of carbon atoms sequentially, starting at the end closest to the branching. For example, the compound is called 3-methylhexane. Note that the top set of numbers is correct since the left end of the molecule is closest to the branching, and this gives the smallest number for the position of the substituent. Also, note that a hyphen is written between the number and the substituent name.
4. The location and name of each substituent are followed by the root alkane name.
The substituents are listed in alphabetical order, and the prefixes di-, tri-, and so on, are used to indicate multiple, identical substituents.
Isomerism and Nomenclature Draw the structural isomers for the alkane and give the systematic name for each one.
Solution We will proceed systematically, starting with the longest chain and then rearranging the carbons to form the shorter, branched chains.
1. CH3CH2CH2CH2CH2CH3 Hexane Note that although a structure such as may look different it is still hexane, since the longest carbon chain has six atoms.
2. We now take one carbon out of the chain and make it a methyl substituent.
Since the longest chain consists of five carbons, this is a substituted pentane: 2-methylpentane. The 2 indicates the position of the methyl group on the chain. Note that if we numbered the chain from the right end, the methyl group would be on carbon 4.
Because we want the smallest possible number, the numbering shown is correct.
3. The methyl substituent can also be on carbon 3 to give Note that we have now exhausted all possibilities for placing a single methyl group on pentane.
A CH3 CH3CH2CHCH2CH3 5 3-Methylpentane 1 2 3 4 CH3CHCH2CH2CH3 A CH3 2-Methylpentane 1 2 3 4 5 A CH3 CH2CH2CH2CH2 A CH3 Six carbon atoms C6H14 O A CH CH3 O CH3 CH2 CH2 O O CH3 O CH2 1 2 3 4 5 6 Correct numbering Incorrect numbering 6 5 4 3 2 1 Sample Exercise 22.2 1002 Chapter Twenty-Two Organic and Biological Molecules 4. Next, we can take two carbons out of the original six-member chain: Since the longest chain now has four carbons, the root name is butane. Since there are two methyl groups, we use the prefix di-. The numbers denote that the two methyl groups are positioned on the second and third carbons in the butane chain. Note that when two or more numbers are used, they are separated by a comma.
5. The two methyl groups can also be attached to the same carbon atom as shown here: We might also try ethyl-substituted butanes, such as However, note that this is instead a pentane (3-methylpentane), since the longest chain has five carbon atoms. Thus it is not a new isomer. Trying to reduce the chain to three atoms provides no further isomers either. For example, the structure is actually 2,2-dimethylbutane.
Thus there are only five distinct structural isomers of C6H14: hexane, 2-methylpentane, 3-methylpentane, 2,3-dimethylbutane, and 2,2-dimethylbutane.
See Exercises 22.15 and 22.16.
Structures from Names Determine the structure for each of the following compounds.
a. 4-ethyl-3,5-dimethylnonane b. 4-tert-butylheptane Solution a. The root name nonane signifies a nine-carbon chain. Thus we have A CH3 O CH3 CH2 A CH3CH2CH CH3 CH CHCH2CH2CH2CH3 A A 7 8 9 O 1 2 3 4 5 6 CH3 CH3 AO A CH2 O A C CH3 CH3 A CH3 CHCH2CH3 O A CH3 CH2 Pentane A CH2CH3 O CH3 1 2 3 4 2,2-Dimethylbutane O A CH3 C CH3 1 2 3 4 2,3-Dimethylbutane CH3 CH3CHO A A CHCH3 CH3 Sample Exercise 22.3 22.1 Alkanes: Saturated Hydrocarbons 1003 b. Heptane signifies a seven-carbon chain, and the tert-butyl group is Thus we have See Exercises 22.19 and 22.20.
Reactions of Alkanes Because they are saturated compounds and because the and bonds are rela-tively strong, the alkanes are fairly unreactive. For example, at they do not react with acids, bases, or strong oxidizing agents. This chemical inertness makes them valu-able as lubricating materials and as the backbone for structural materials such as plastics.
At a sufficiently high temperature alkanes do react vigorously and exothermically with oxygen, and these combustion reactions are the basis for their widespread use as fuels. For example, the reaction of butane with oxygen is The alkanes can also undergo substitution reactions, primarily where halogen atoms replace hydrogen atoms. For example, methane can be successively chlorinated as follows: Note that the products of the last two reactions have two names; the systematic name is given first, followed by the common name in parentheses. (This format will be used throughout this chapter for compounds that have common names.) Also, note that ul-traviolet light (hv) furnishes the energy to break the bond to produce chlorine atoms: A chlorine atom has an unpaired electron, as indicated by the dot, which makes it very reactive and able to attack the bond.
Substituted methanes with the general formula containing both chlorine and fluorine as substituents are called chlorofluorocarbons (CFCs) and are also known as Fre-ons. These substances are very unreactive and have been extensively used as coolant flu-ids in refrigerators and air conditioners. Unfortunately, their chemical inertness allows CFxCl4x C¬H Cl2 ¡ Cl Cl Cl¬Cl (carbon tetrachloride) Tetrachloromethane HCl C Cl4 CHCl3 Cl2 ¡ hv (chloroform) Trichloromethane HCl CH Cl3 CH2Cl2 Cl2 ¡ hv Dichloromethane HCl CH2Cl2 CH3Cl Cl2 ¡ hv Chloromethane HCl CH3Cl CH4 Cl2 ¡ hv 2C4H101g2 13O21g2 ¡ 8CO21g2 10H2O1g2 25°C C¬H C¬C CH3 CH3 A CH3CH2CH2CHCH2CH2CH3 A 5 6 7 O H3C CO 1 2 3 4 A A OC CH3 H3C OCH3 The hv above the arrow represents ultraviolet light.
A butane lighter used for camping.
FIGURE 22.6 The (a) chair and (b) boat forms of cyclohexane.
1004 Chapter Twenty-Two Organic and Biological Molecules Freons to remain in the atmosphere so long that they eventually reach altitudes where they are a threat to the protective ozone layer (see Section 12.8), and the use of these com-pounds is being rapidly phased out.
Alkanes can also undergo dehydrogenation reactions in which hydrogen atoms are removed and the product is an unsaturated hydrocarbon. For example, in the presence of chromium(III) oxide at high temperatures, ethane can be dehydrogenated, yielding ethylene: Ethylene Cyclic Alkanes Besides forming chains, carbon atoms also form rings. The simplest of the cyclic alkanes (general formula CnH2n) is cyclopropane (C3H6), shown in Fig. 22.5(a). Since the carbon atoms in cyclopropane form an equilateral triangle with 60 bond angles, their sp3 hybrid orbitals do not overlap head-on as in normal alkanes [Fig. 22.5(b)]. This results in unusually weak, or strained, COC bonds; thus the cyclopropane molecule is much more reactive than straight-chain propane. The carbon atoms in cyclobutane (C4H8) form a square with 88 bond angles, and cyclobutane is also quite reactive.
The next two members of the series, cyclopentane (C5H10) and cyclohexane (C6H12), are quite stable, because their rings have bond angles very close to the tetrahedral angles, which allows the sp3 hybrid orbitals on adjacent carbon atoms to overlap head-on and form normal bonds, which are quite strong. To attain the tetrahedral angles, the cy-clohexane ring must “pucker”—that is, become nonplanar. Cyclohexane can exist in two forms, the chair and the boat forms, as shown in Fig. 22.6. The two hydrogen atoms above the ring in the boat form are quite close to each other, and the resulting repulsion between these atoms causes the chair form to be preferred. At more than 99% of cyclohexane exists in the chair form.
For simplicity, the cyclic alkanes are often represented by the following structures: Thus the structure represents methylcyclopropane.
25°C C¬C CH3CH3¬¡ Cr2O3 500°C CH2“CH2 H2 C C C 109.5° 60° No “head-on” overlap of atomic orbitals (a) (b) FIGURE 22.5 (a) The molecular structure of cyclopropane (C3H6). (b) The overlap of the sp3 orbitals that form the COC bonds in cyclopropane.
Chair (a) Boat (b) These two H atoms repel each other 22.2 Alkenes and Alkynes 1005 The nomenclature for cycloalkanes follows the same rules as for the other alkanes except that the root name is preceded by the prefix cyclo-. The ring is numbered to yield the smallest substituent numbers possible.
Naming Cyclic Alkanes Name each of the following cyclic alkanes.
a.
b.
Solution a. The six-carbon cyclohexane ring is numbered as follows: There is an isopropyl group at carbon 1 and a methyl group at carbon 3. The name is 1-isopropyl-3-methylcyclohexane, since the alkyl groups are named in alphabetical order.
b. This is a cyclobutane ring, which is numbered as follows: The name is 1-ethyl-2-propylcyclobutane.
See Exercise 22.22.
22.2 Alkenes and Alkynes Multiple carbon–carbon bonds result when hydrogen atoms are removed from alkanes.
Hydrocarbons that contain at least one carbon–carbon double bond are called alkenes and have the general formula CnH2n. The simplest alkene (C2H4), commonly known as ethylene, has the Lewis structure As discussed in Section 9.1, each carbon in ethylene can be described as sp2 hybridized.
The COC bond is formed by sharing an electron pair between sp2 orbitals, and the bond is formed by sharing a pair of electrons between p orbitals (Fig. 22.7).
The systematic nomenclature for alkenes is quite similar to that for alkanes.
1. The root hydrocarbon name ends in -ene rather than -ane. Thus the systematic name for C2H4 is ethene and the name for C3H6 is propene.
2. In alkenes containing more than three carbon atoms, the location of the double bond is indicated by the lowest-numbered carbon atom involved in the bond. Thus CH2PCHCH2CH3 is called 1-butene, and CH3CHPCHCH3 is called 2-butene.
1 2 3 4 CH2CH2CH3 CH2CH3 O 1 2 3 4 5 6 CH3 O CH3 CH3 CH CH2CH3 CH2CH2CH3 CH3 CH3 CH CH3 Sample Exercise 22.4 FIGURE 22.10 The bonding in acetylene.
1006 Chapter Twenty-Two Organic and Biological Molecules Note from Fig. 22.7 that the p orbitals on the two carbon atoms in ethylene must be lined up (parallel) to allow formation of the bond. This prevents rotation of the two CH2 groups relative to each other at ordinary temperatures, in contrast to alkanes, where free rotation is possible (see Fig. 22.8). The restricted rotation around doubly bonded carbon atoms means that alkenes exhibit cis–trans isomerism. For example, there are two stereoiso-mers of 2-butene (Fig. 22.9). Identical substituents on the same side of the double bond are designated cis and those on opposite sides are labeled trans.
Alkynes are unsaturated hydrocarbons containing at least one triple carbon–carbon bond. The simplest alkyne is C2H2 (commonly called acetylene), which has the systematic name ethyne. As discussed in Section 9.1, the triple bond in acetylene can be described as one bond between two sp hybrid orbitals on the two carbon atoms and two bonds involving two 2p orbitals on each carbon atom (Fig. 22.10).
The nomenclature for alkynes involves the use of -yne as a suffix to replace the -ane of the parent alkane. Thus the molecule CH3CH2CqCCH3 has the name 2-pentyne.
Like alkanes, unsaturated hydrocarbons can exist as ringed structures, for example, FIGURE 22.7 The bonding in ethylene.
FIGURE 22.9 The two stereoisomers of 2-butene: (a) cis-2-butene and (b) trans-2-butene.
FIGURE 22.8 The bonding in ethane.
For cyclic alkenes, number through the double bond toward the substituent.
H1s sp2 sp2 sp2 sp2 H1s 2p C C sp2 sp2 H H H H H H C C 2p 2p H1s H sp 2p 2p C sp 2p 2p 2p C H H1s 2p sp 22.2 Alkenes and Alkynes 1007 Naming Alkenes and Alkynes Name each of the following molecules.
a.
b.
Solution a. The longest chain, which contains six carbon atoms, is numbered as follows: Thus the hydrocarbon is a 2-hexene. Since the hydrogen atoms are located on oppo-site sides of the double bond, this molecule corresponds to the trans isomer. The name is 4-methyl-trans-2-hexene.
b. The longest chain, consisting of seven carbon atoms, is numbered as shown (giving the triple bond the lowest possible number): The hydrocarbon is a 3-heptyne. The full name is 5-ethyl-3-heptyne, where the posi-tion of the triple bond is indicated by the lower-numbered carbon atom involved in this bond.
See Exercises 22.25 through 22.28 and 22.44.
Reactions of Alkenes and Alkynes Because alkenes and alkynes are unsaturated, their most important reactions are addition reactions. In these reactions bonds, which are weaker than the bonds, are bro-ken, and new bonds are formed to the atoms being added. For example, hydrogena-tion reactions involve the addition of hydrogen atoms: 1-Propene Propane For this reaction to proceed rapidly at normal temperatures, a catalyst of platinum, pal-ladium, or nickel is used. The catalyst serves to help break the relatively strong bond, as was discussed in Section 12.8. Hydrogenation of alkenes is an important in-dustrial process, particularly in the manufacture of solid shortenings where unsaturated fats (fats containing double bonds), which are generally liquid, are converted to solid saturated fats.
H¬H CH2“CHCH3 H2 ¬¡ Catalyst CH3CH2CH3 C¬C s p qCCHCH2CH3 A 1 2 3 4 5 6 7 CH2 CH3CH2C A CH3 G D A CH3CH2CH CH3 CG DCH3 C H H 1 2 3 4 5 6 P Sample Exercise 22.5 A worker using an oxyacetylene torch.
1008 Chapter Twenty-Two Organic and Biological Molecules Halogenation of unsaturated hydrocarbons involves addition of halogen atoms. For example, 1-Pentene 1,2-Dibromopentane Another important reaction involving certain unsaturated hydrocarbons is polymer-ization, a process in which many small molecules are joined together to form a large mol-ecule. Polymerization will be discussed in Section 22.5.
22.3 Aromatic Hydrocarbons A special class of cyclic unsaturated hydrocarbons is known as the aromatic hydrocar-bons. The simplest of these is benzene , which has a planar ring structure, as shown in Fig. 22.11(a). In the localized electron model of the bonding in benzene, resonance structures of the type shown in Fig. 22.11(b) are used to account for the known equiva-lence of all the carbon–carbon bonds. But as we discussed in Section 9.5, the best description of the benzene molecule assumes that sp2 hybrid orbitals on each carbon are used to form the and bonds, while the remaining 2p orbital on each carbon is used to form molecular orbitals. The delocalization of these electrons is usually indicated by a circle inside the ring [Fig. 22.11(c)].
The delocalization of the electrons makes the benzene ring behave quite differently from a typical unsaturated hydrocarbon. As we have seen previously, unsaturated hydro-carbons generally undergo rapid addition reactions. However, benzene does not. Instead, it undergoes substitution reactions in which hydrogen atoms are replaced by other atoms.
For example, p p C¬H s C¬C (C6H6) CH2“CHCH2CH2CH3 Br2 ¡ CH2BrCHBrCH2CH2CH3 In each case the substance shown over the arrow is needed to catalyze these substitution reactions.
Substitution reactions are characteristic of saturated hydrocarbons, and addition reactions are characteristic of unsaturated ones. The fact that benzene reacts more like a saturated hydrocarbon indicates the great stability of the delocalized electron system.
22.3 Aromatic Hydrocarbons 1009 The nomenclature of benzene derivatives is similar to the nomenclature for saturated ring systems. If there is more than one substituent present, numbers are used to indicate substituent positions. For example, the compound is named 1,2-dichlorobenzene. Another nomenclature system uses the prefix ortho- (o-) for two adjacent substituents, meta- (m-) for two substituents with one carbon between them, and para- (p-) for two substituents opposite each other. When benzene is used as a substituent, it is called the phenyl group. Examples of some aromatic compounds are shown in Fig. 22.12.
Benzene is the simplest aromatic molecule. More complex aromatic systems can be viewed as consisting of a number of “fused” benzene rings. Some examples are given in Table 22.3.
1 2 3 4 5 6 Cl Cl FIGURE 22.11 (a) The structure of benzene, a planar ring system in which all bond angles are 120°.
(b) Two of the resonance structures of benzene. (c) The usual representation of benzene. The circle represents the electrons in the delocalized system. All COC bonds in benzene are equivalent.
FIGURE 22.12 Some selected substituted benzenes and their names. Common names are given in parentheses.
1010 Chapter Twenty-Two Organic and Biological Molecules 22.4 Hydrocarbon Derivatives The vast majority of organic molecules contain elements in addition to carbon and hy-drogen. However, most of these substances can be viewed as hydrocarbon derivatives, molecules that are fundamentally hydrocarbons but that have additional atoms or groups of atoms called functional groups. The common functional groups are listed in Table 22.4. Because each functional group exhibits characteristic chemistry, we will consider the groups separately.
Alcohols Alcohols are characterized by the presence of the hydroxyl group (OOH). Some com-mon alcohols are shown in Table 22.5. The systematic name for an alcohol is obtained by replacing the final -e of the parent hydrocarbon with -ol. The position of the OOH group is specified by a number (where necessary) chosen so that it is the smallest of the substituent numbers. Alcohols are classified according to the number of hydrocar-bon fragments bonded to the carbon where the OOH group is attached (see mar-gin), where R, R , and R (which may be the same or different) represent hydrocarbon fragments.
Alcohols usually have much higher boiling points than might be expected from their molar masses. For example, both methanol and ethane have a molar mass of 30, but the boiling point for methanol is 65C while that for ethane is 89C. This difference can be understood if we consider the types of intermolecular attractions that occur in these liquids. Ethane molecules are nonpolar and exhibit only weak London dispersion inter-actions. However, the polar OOH group of methanol produces extensive hydrogen bond-ing similar to that found in water (see Section 10.1), which results in the relatively high boiling point.
Although there are many important alcohols, the simplest ones, methanol and ethanol, have the greatest commercial value. Methanol, also known as wood alcohol because it was formerly obtained by heating wood in the absence of air, is prepared industrially TABLE 22.3 More Complex Aromatic Systems Structural Formula Name Use of Effect Naphthalene Formerly used in mothballs Anthracene Dyes Phenanthrene Dyes, explosives, and synthesis of drugs 3,4-Benzpyrene Active carcinogen found in smoke and smog 22.4 Hydrocarbon Derivatives 1011 (approximately 4 million tons annually in the United States) by the hydrogenation of car-bon monoxide: Methanol is used as a starting material for the synthesis of acetic acid and for many types of adhesives, fibers, and plastics. It is also used (and such use may increase) as a motor fuel. Methanol is highly toxic to humans and can lead to blindness and death if ingested.
Ethanol is the alcohol found in beverages such as beer, wine, and whiskey; it is pro-duced by the fermentation of glucose in corn, barley, grapes, and so on: Glucose Ethanol C6H12O6 ¬ ¡ Yeast 2CH3CH2OH 2CO2 CO 2H2 ¬ ¬ ¡ 400°C ZnO/Cr2O3 CH3OH TABLE 22.4 The Common Functional Groups Functional General Class Group Formula Example Halohydrocarbons OX (F, Cl, Br, I) ROX CH3I Iodomethane (methyl iodide) Alcohols OOH ROOH CH3OH Methanol (methyl alcohol) Ethers OOO ROOOR CH3OCH3 Dimethyl ether CH2O Aldehydes Methanal (formaldehyde) CH3COCH3 Ketones Propanone (dimethyl ketone or acetone) CH3COOH Carboxylic acids Ethanoic acid (acetic acid) CH3COOCH2CH3 Esters Ethyl acetate Amines ONH2 RONH2 CH3NH2 Aminomethane (methylamine) R and represent hydrocarbon fragments.
R¿ TABLE 22.5 Some Common Alcohols Formula Systematic Name Common Name CH3OH Methanol Methyl alcohol CH3CH2OH Ethanol Ethyl alcohol CH3CH2CH2OH 1-Propanol n-Propyl alcohol CH3CHCH3 2-Propanol Isopropyl alcohol !
OH A winemaker drawing off a glass of wine in a modern wine cellar.
1012 Chapter Twenty-Two Organic and Biological Molecules The reaction is catalyzed by the enzymes found in yeast. This reaction can proceed only until the alcohol content reaches about 13% (the percentage found in most wines), at which point the yeast can no longer survive. Beverages with higher alcohol content are made by distilling the fermentation mixture.
Ethanol, like methanol, can be burned in the internal combustion engines of auto-mobiles and is now commonly added to gasoline to form gasohol (see Section 6.6). It is also used in industry as a solvent and for the preparation of acetic acid. The commercial production of ethanol (500,000 tons per year in the United States) is carried out by reac-tion of water with ethylene: Many polyhydroxyl (more than one OOH group) alcohols are known, the most im-portant being 1,2-ethanediol (ethylene glycol), a toxic substance that is the major constituent of most automobile antifreeze solutions.
The simplest aromatic alcohol is which is commonly called phenol. Most of the 1 million tons of phenol produced annu-ally in the United States is used to make polymers for adhesives and plastics.
Naming and Classifying Alcohols For each of the following alcohols, give the systematic name and specify whether the al-cohol is primary, secondary, or tertiary.
a.
c.
b.
Solution a. The chain is numbered as follows: The compound is called 2-butanol, since the OOH group is located at the number 2 position of a four-carbon chain. Note that the carbon to which the OOH is attached also has OCH3 and OCH2CH3 groups attached: Therefore, this is a secondary alcohol.
O CH3 AO A H C R R OH CH2CH3 A CH3CHCH2CH3 1 2 3 4 OH ClCH2CH2CH2OH CH2“CH2 H2O ¬ ¬ ¡ Acid Catalyst CH3CH2OH Ethanol is being tested in selected areas as a fuel for automobiles.
Sample Exercise 22.6 22.4 Hydrocarbon Derivatives 1013 b. The chain is numbered as follows: 3 2 1 The name is 3-chloro-1-propanol. This is a primary alcohol: c. The chain is numbered as follows: The name is 6-bromo-2-methyl-2-hexanol. This is a tertiary alcohol since the carbon where the OOH is attached also has three other R groups attached.
See Exercises 22.51 and 22.52.
Aldehydes and Ketones Aldehydes and ketones contain the carbonyl group, In ketones this group is bonded to two carbon atoms, as in acetone, In aldehydes the carbonyl group is bonded to at least one hydrogen atom, as in formalde-hyde, or acetaldehyde, The systematic name for an aldehyde is obtained from the parent alkane by remov-ing the final -e and adding -al. For ketones the final -e is replaced by -one, and a number indicates the position of the carbonyl group where necessary. Examples of common alde-hydes and ketones are shown in Fig. 22.13. Note that since the aldehyde functional group always occurs at the end of the carbon chain, the aldehyde carbon is assigned the num-ber 1 when substituent positions are listed in the name.
Ketones often have useful solvent properties (acetone is found in nail polish remover, for example) and are frequently used in industry for this purpose. Aldehydes typically have strong odors. Vanillin is responsible for the pleasant odor in vanilla beans; cinnamaldehyde A O 1 2 6 OH A CO CH2Br O CH2 CH2 CH2 O O A CH3 3 4 5 CH3 CH2CH2 ClO H H OH One R group attached to the carbon with the OOH group A O A CO Cl¬CH2¬CH2¬CH2¬OH Cinnamaldehyde produces the characteristic odor of cinnamon.
1014 Chapter Twenty-Two Organic and Biological Molecules produces the characteristic odor of cinnamon. On the other hand, the unpleasant odor in rancid butter arises from the presence of butyraldehyde.
Aldehydes and ketones are most often produced commercially by the oxidation of alcohols. For example, oxidation of a primary alcohol yields the corresponding aldehyde: Oxidation of a secondary alcohol results in a ketone: Carboxylic Acids and Esters Carboxylic acids are characterized by the presence of the carboxyl group that gives an acid of the general formula RCOOH. Typically, these molecules are weak acids in aqueous solution (see Section 14.5). Organic acids are named from the parent alkane by dropping the final -e and adding -oic. Thus CH3COOH, commonly called acetic acid, has the systematic name ethanoic acid, since the parent alkane is ethane. Other examples of carboxylic acids are shown in Fig. 22.14.
Many carboxylic acids are synthesized by oxidizing primary alcohols with a strong oxidizing agent. For example, ethanol can be oxidized to acetic acid by using potassium permanganate: CH3CH2OH ¬¬¡ KMnO4(aq) CH3COOH A B CH3CCH3 Oxidation O OH CH3CHCH3 G J CH3CH2OH Oxidation CH3C O H FIGURE 22.13 Some common ketones and aldehydes. Note that since the aldehyde functional group always appears at the end of a carbon chain, carbon is assigned the number 1 when the compound is named.
FIGURE 22.14 Some carboxylic acids.
22.4 Hydrocarbon Derivatives 1015 A carboxylic acid reacts with an alcohol to form an ester and a water molecule. For example, the reaction of acetic acid with ethanol produces ethyl acetate and water: Esters often have a sweet, fruity odor that is in contrast to the often pungent odors of the parent carboxylic acids. For example, the odor of bananas is caused by n-amyl acetate, and that of oranges is caused by n-octyl acetate, The systematic name for an ester is formed by changing the -oic ending of the parent acid to -oate. The parent alcohol chain is named first with a -yl ending. For example, the sys-tematic name for n-octyl acetate is n-octylethanoate (from ethanoic acid).
A very important ester is formed from the reaction of salicylic acid and acetic acid: The product is acetylsalicylic acid, commonly known as aspirin, which is used in huge quantities as an analgesic (painkiller).
Amines Amines are probably best viewed as derivatives of ammonia in which one or more NOH bonds are replaced by NOC bonds. The resulting amines are classified as primary if one NOC bond is present, secondary if two NOC bonds are present, and tertiary if all three NOH bonds in NH3 have been replaced by NOC bonds (Fig. 22.15). Examples of some common amines are given in Table 22.6.
Common names are often used for simple amines; the systematic nomenclature for more complex molecules uses the name amino- for the ONH2 functional group. For ex-ample, the molecule is named 2-aminobutane.
Many amines have unpleasant “fishlike” odors. For example, the odors associated with decaying animal and human tissues are caused by amines such as putrescine (H2NCH2CH2CH2NH2) and cadaverine (H2NCH2CH2CH2CH2CH2NH2).
Aromatic amines are primarily used to make dyes. Since many of them are carcino-genic, they must be handled with great care.
Aspirin tablets.
Computer-generated space-filling model of acetylsalicylic acid (aspirin).
1016 Chapter Twenty-Two Organic and Biological Molecules 22.5 Polymers Polymers are large, usually chainlike molecules that are built from small molecules called monomers. Polymers form the basis for synthetic fibers, rubbers, and plastics and have played a leading role in the revolution that has been brought about in daily life by chem-istry. It has been estimated that about 50% of the industrial chemists in the United States work in some area of polymer chemistry, a fact that illustrates just how important poly-mers are to our economy and standard of living.
The Development and Properties of Polymers The development of the polymer industry provides a striking example of the importance of serendipity in the progress of science. Many discoveries in polymer chemistry arose from accidental observations that scientists followed up.
The age of plastics might be traced to a day in 1846 when Christian Schoenbein, a chemistry professor at the University of Basel in Switzerland, spilled a flask containing nitric and sulfuric acids. In his hurry to clean up the spill, he grabbed his wife’s cotton apron, which he then rinsed out and hung up in front of a hot stove to dry. Instead of dry-ing, the apron flared and burned.
Very interested in this event, Schoenbein repeated the reaction under more controlled conditions and found that the new material, which he correctly concluded to be nitrated cellulose, had some surprising properties. As he had experienced, the nitrated cellulose is extremely flammable and, under certain circumstances, highly explosive. In addition, he found that it could be molded at moderate temperatures to give objects that were, upon cooling, tough but elastic. Predictably, the explosive nature of the substance was initially of more interest than its other properties, and cellulose nitrate rapidly became the basis for smokeless gun powder. Although Schoenbein’s discovery cannot be described as a truly synthetic polymer (because he simply found a way to modify the natural polymer cellulose), it formed the basis for a large number of industries that grew up to produce photographic films, artificial fibers, and molded objects of all types.
The first synthetic polymers were produced as by-products of various organic reac-tions and were regarded as unwanted contaminants. Thus the first preparations of many of the polymers now regarded as essential to our modern lifestyle were thrown away in TABLE 22.6 Some Common Amines Formula Common Name Type CH3NH2 Methylamine Primary CH3CH2NH2 Ethylamine Primary (CH3)2NH Dimethylamine Secondary (CH3)3N Trimethylamine Tertiary Aniline Primary Diphenylamine Secondary FIGURE 22.15 The general formulas for primary, sec-ondary, and tertiary amines. R, R , and R represent carbon-containing substituents.
The soybeans on the left are coated with a red acrylic polymer to delay soybean emergence. This allows farmers to plant their crops more efficiently.
22.5 Polymers 1017 disgust. One chemist who refused to be defeated by the “tarry” products obtained when he reacted phenol with formaldehyde was the Belgian-American chemist Leo H. Baekeland (1863–1944). Baekeland’s work resulted in the first completely synthetic plastic (called Bakelite), a substance that when molded to a certain shape under high pressure and temperature cannot be softened again or dissolved. Bakelite is a thermoset polymer. In contrast, cellulose nitrate is a thermoplastic polymer; that is, it can be remelted after it has been molded.
The discovery of Bakelite in 1907 spawned a large plastics industry, producing tele-phones, billiard balls, and insulators for electrical devices. During the early days of poly-mer chemistry, there was a great deal of controversy over the nature of these materials.
Although the German chemist Hermann Staudinger speculated in 1920 that polymers were very large molecules held together by strong chemical bonds, most chemists of the time assumed that these materials were much like colloids, in which small molecules are ag-gregated into large units by forces weaker than chemical bonds.
One chemist who contributed greatly to the understanding of polymers as giant mol-ecules was Wallace H. Carothers of the DuPont Chemical Company. Among his accom-plishments was the preparation of nylon. The nylon story further illustrates the importance of serendipity in scientific research. When nylon is first prepared, the resulting product is a sticky material with little structural integrity. Because of this, it was initially put aside as having no apparently useful characteristics. However, Julian Hill, a chemist in the Carothers research group, one day put a small ball of this nylon on the end of a stirring rod and drew it away from the remaining sticky mass, forming a string. He noticed the silky appearance and strength of this thread and realized that nylon could be drawn into useful fibers.
The reason for this behavior of nylon is now understood. When nylon is first formed, the individual polymer chains are oriented randomly, like cooked spaghetti, and the sub-stance is highly amorphous. However, when drawn out into a thread, the chains tend to line up (the nylon becomes more crystalline), which leads to increased hydrogen bonding between adjacent chains. This increase in crystallinity, along with the resulting increase in hydrogen-bonding interactions, leads to strong fibers and thus to a highly useful mate-rial. Commercially, nylon is produced by forcing the raw material through a spinneret, a plate containing small holes, which forces the polymer chains to line up.
Another property that adds strength to polymers is crosslinking, the existence of co-valent bonds between adjacent chains. The structure of Bakelite is highly crosslinked, which accounts for the strength and toughness of this polymer. Another example of crosslinking occurs in the manufacture of rubber. Raw natural rubber consists of chains of the type and is a soft, sticky material unsuitable for tires. However, in 1839 Charles Goodyear (1800–1860), an American chemist, accidentally found that if sulfur is added to rubber and the resulting mixture is heated (a process called vulcanization), the resulting rubber is still elastic (reversibly stretchable) but is much stronger. This change in character occurs because sulfur atoms become bonded between carbon atoms on different chains. These sulfur atoms form bridges between the polymer chains, thus linking the chains together.
Types of Polymers The simplest and one of the best-known polymers is polyethylene, which is constructed from ethylene monomers: OA P H C CH2 nCH2 Catalyst CO O A A A n H H H A radio from the 1930s made of Bakelite.
Nylon netting magnified 62 times.
Charles Goodyear tried for many years to change natural rubber into a useful product. In 1839 he accidentally dropped some rubber containing sulfur on a hot stove. Noting that the rubber did not melt as expected, Goodyear pursued this lead and developed vulcanization.
1018 Chapter Twenty-Two Organic and Biological Molecules where n represents a large number (usually several thousand). Polyethylene is a tough, flexible plastic used for piping, bottles, electrical insulation, packaging films, garbage bags, and many other purposes. Its properties can be varied by using substituted ethylene monomers. For example, when tetrafluoroethylene is the monomer, the polymer Teflon is obtained: The discovery of Teflon, a very important substituted polyethylene, is another illus-tration of the role of chance in chemical research. In 1938 a DuPont chemist named Roy Plunkett was studying the chemistry of gaseous tetrafluoroethylene. He synthesized about 100 pounds of the chemical and stored it in steel cylinders. When one of the cylinders failed to produce perfluoroethylene gas when the valve was opened, the cylinder was cut open to reveal a white powder. This powder turned out to be a polymer of perfluoroeth-ylene, which was eventually developed into Teflon. Because of the resistance of the strong COF bonds to chemical attack, Teflon is an inert, tough, and nonflammable material widely used for electrical insulation, nonstick coatings on cooking utensils, and bearings for low-temperature applications.
Other polyethylene-type polymers are made from monomers containing chloro, methyl, cyano, and phenyl substituents, as summarized in Table 22.7 on page 1020. In each case the double carbon–carbon bond in the substituted ethylene monomer becomes a single bond in the polymer. The different substituents lead to a wide variety of properties.
CHEMICAL IMPACT Heal Thyself O ne major problem with structural materials is that they crack and weaken as they age. The human body has mechanisms for healing itself if the skin is cut or a bone is broken. However, inanimate materials have had no such mechanisms—until now. Scientists at the University of Illinois at Urbana–Champaign (UIUC) have invented a plas-tic that automatically heals microscopic cracks before they can develop into large cracks that would degrade the use-fulness of the material. This accomplishment was achieved by an interdisciplinary team of scientists including aero-nautical engineering professors Scott White and Philippe Geubelle, applied mechanics professor Nancy Sottos, and chemistry professor Jeffrey Moore.
The self-healing system is based on microcapsules con-taining liquid dicyclopentadiene Dicyclopentadiene that are blended into the plastic. When a microscopic crack develops, it encounters and breaks a microcapsule. The di-cyclopentadiene then leaks out, where it encounters a cata-lyst (blended into the plastic when it was formulated) that mediates a repair polymerization process. This process in-volves opening the cyclopentadiene rings, which leads to a highly cross-linked repair of the crack.
The trickiest part of the repair mechanism is to get the microcapsules to be the correct size and to have the appro-priate wall strength. They must be small enough not to de-grade the strength of the plastic. The walls must also be thick enough to survive the molding of the plastic but thin enough to burst as the lengthening crack reaches them.
Self-healing materials should have many applications.
The U.S. Air Force, which partially funded the research at UIUC, is interested in using the materials in tanks that hold gases and liquids under pressure. The current materials used for these tanks are subject to microcracks that eventually grow, causing the tanks to leak. Self-healing materials would Cross-linking gives the rubber in these tires strength and toughness.
22.5 Polymers 1019 The polyethylene polymers illustrate one of the major types of polymerization reac-tions, called addition polymerization, in which the monomers simply “add together” to produce the polymer. No other products are formed. The polymerization process is initi-ated by a free radical (a species with an unpaired electron) such as the hydroxyl radical (HO). The free radical attacks and breaks the bond of an ethylene molecule to form a new free radical, which is then available to attack another ethylene molecule: Repetition of this process thousands of times creates a long-chain polymer. Termination of the growth of the chain occurs when two radicals react to form a bond, a process that consumes two radicals without producing any others.
D G A O D G H C C H H H H O A A OH C H O O A O A A C C C H H A A H H H H A C O A O A A H H H H HO P C also be valuable in situations where re-pair is impossible or impractical, such as electronic circuit boards, components of deep space probes, and implanted med-ical devices.
A scanning electron microscope image showing the fractured plane of a self-healing material with a ruptured microcapsule in a thermosetting matrix.
1020 Chapter Twenty-Two Organic and Biological Molecules Another common type of polymerization is condensation polymerization, in which a small molecule, such as water, is formed for each extension of the polymer chain. The most familiar polymer produced by condensation is nylon. Nylon is a copolymer, since two different types of monomers combine to form the chain; a homopolymer is the re-sult of polymerizing a single type of monomer. One common form of nylon is produced when hexamethylenediamine and adipic acid react by splitting out a water molecule to form a CON bond: TABLE 22.7 Some Common Synthetic Polymers, Their Monomers and Applications Monomer Polymer Name Formula Name Formula Uses Ethylene H2CPCH2 Polyethylene O (CH2OCH2) O n Plastic piping, bottles, electrical insulation, toys Propylene Polypropylene Film for packaging, carpets, lab wares, toys Vinyl Polyvinyl chloride Piping, siding, floor chloride (PVC) tile, clothing, toys Acrylonitrile Polyacrylonitrile Carpets, fabrics (PAN) Tetrafluoro F2CPCF2 Teflon O (CF2OCF2) O n Cooking utensils, ethylene electrical insulation, bearings Styrene Polystyrene Containers, thermal insulation, toys Butadiene Polybutadiene O (CH2CHPCHCH2) O n Tire tread, coating resin Butadiene (See above.) Styrene-butadiene Synthetic rubber and styrene rubber Visualization: Synthesis of Nylon 22.5 Polymers 1021 The molecule formed, called a dimer (two monomers joined), can undergo further con-densation reactions since it has an amino group at one end and a carboxyl group at the other. Thus both ends are free to react with another monomer. Repetition of this process leads to a long chain of the type which is the basic structure of nylon. The reaction to form nylon occurs quite readily and is often used as a lecture demonstration (see Fig. 22.16). The properties of nylon can be varied by changing the number of carbon atoms in the chain of the acid or amine monomer.
More than 1 million tons of nylon is produced annually in the United States for use in clothing, carpets, rope, and so on. Many other types of condensation polymers are also produced. For example, Dacron is a copolymer formed from the condensation reaction of ethylene glycol (a dialcohol) and p-terephthalic acid (a dicarboxylic acid): The repeating unit of Dacron is Note that this polymerization involves a carboxylic acid and an alcohol forming an ester group: Thus Dacron is called a polyester. By itself or blended with cotton, Dacron is widely used in fibers for the manufacture of clothing.
Polymers Based on Ethylene A large section of the polymer industry involves the production of macromolecules from ethylene or substituted ethylenes. As discussed previously, ethylene molecules polymer-ize by addition after the double bond has been broken by some initiator: This process continues by adding new ethylene molecules to eventually give polyethyl-ene, a thermoplastic material.
There are two forms of polyethylene: low-density polyethylene (LDPE) and high-density polyethylene (HDPE). The chains in LDPE contain many branches and thus do not pack as tightly as those in HDPE, which consist of mostly straight-chain molecules.
Traditionally, LDPE has been manufactured under conditions of high pressure (20,000 psi) and high temperature (500C). These severe reaction conditions require specially designed equipment, and for safety reasons the reaction usually has been run behind a reinforced concrete barrier. More recently, lower reaction pressures and FIGURE 22.16 The reaction to form nylon can be carried out at the interface of two immiscible liq-uid layers in a beaker. The bottom layer contains adipoyl chloride, dissolved in CCl4, and the top layer con-tains hexamethylenediamine, H2NO (CH2)6 ONH2 dissolved in water. A molecule of HCl is formed as each bond forms.
C¬N psi is the abbreviation for pounds per square inch: 15 psi 1 atm.
1022 Chapter Twenty-Two Organic and Biological Molecules temperatures have become possible through the use of catalysts. One catalytic system us-ing triethylaluminum, Al(C2H5)3, and titanium(IV) chloride was developed by Karl Ziegler in Germany and Giulio Natta in Italy. Although this catalyst is very efficient, it catches fire on contact with air and must be handled very carefully. A safer catalytic system was developed at Phillips Petroleum Company. It uses a chromium(III) oxide (Cr2O3) and alu-minosilicate catalyst and has mainly taken over in the United States. The product of the catalyzed reaction is highly linear (unbranched) and is often called linear low-density poly-ethylene. It is very similar to HDPE.
The major use of LDPE is in the manufacture of the tough transparent film that is used in packaging so many consumer goods. Two-thirds of the approximately 10 billion pounds of LDPE produced annually in the United States are used for this purpose. The major use of HDPE is for blow-molded products, such as bottles for consumer products (see Fig. 22.17).
The useful properties of polyethylene are due primarily to its high molecular weight (molar mass). Although the strengths of the interactions between specific points on the nonpolar chains are quite small, the chains are so long that these small attractions accu-mulate to a very significant value, so that the chains stick together very tenaciously. There is also a great deal of physical tangling of the lengthy chains. The combination of these interactions gives the polymer strength and toughness. However, a material like polyeth-ylene can be melted and formed into a new shape (thermoplastic behavior), because in the melted state the molecules can readily flow past one another.
CHEMICAL IMPACT Wallace Hume Carothers W allace H. Carothers. a brilliant organic chemist who was principally responsible for the development of nylon and the first synthetic rubber (Neoprene), was born in 1896 in Burlington, Iowa. As a youth, Carothers was fascinated by tools and mechanical devices and spent many hours experimenting. In 1915 he entered Tarkio College in Missouri. Carothers so excelled in chemistry that even before his graduation, he was made a chemistry instructor.
Carothers eventually moved to the University of Illinois at Urbana–Champaign, where he was appointed to the faculty when he completed his Ph.D. in organic chemistry in 1924.
He moved to Harvard University in 1926, and then to DuPont in 1928 to participate in a new program in fundamental re-search. At DuPont, Carothers headed the organic chemistry division, and during his ten years there played a prominent role in laying the foundations of polymer chemistry.
By the age of 33, Carothers had become a world-famous chemist whose advice was sought by almost everyone work-ing in polymers. He was the first industrial chemist to be elected to the prestigious National Academy of Sciences.
Carothers was an avid reader of poetry and a lover of classical music. Unfortunately, he also suffered from severe bouts of depression that finally led to his suicide in 1937 in a Philadelphia hotel room, where he drank a cyanide solution. He was 41 years old. Despite the brevity of his ca-reer, Carothers was truly one of the finest American chemists of all time. His great intellect, his love of chemistry, and his insistence on perfection produced his special genius.
Wallace H. Carothers.
Molecular weight (not molar mass) is the common terminology in the polymer industry.
22.5 Polymers 1023 Since a high molecular weight gives a polymer useful properties, one might think that the goal would be to produce polymers with chains as long as possible. However, this is not the case—polymers become much more difficult to process as the molecular weights increase. Most industrial operations require that the polymer flow through pipes as it is processed. But as the chain lengths increase, viscosity also increases. In practice, the up-per limit of a polymer’s molecular weight is set by the flow requirements of the manu-facturing process. Thus the final product often reflects a compromise between the optimal properties for the application and those needed for ease of processing.
Although many polymer properties are greatly influenced by molecular weight, some other important properties are not. For example, chain length does not affect a polymer’s resistance to chemical attack. Physical properties such as color, refractive index, hardness, density, and electrical conductivity are also not greatly influenced by molecular weight.
We have already seen that one way of altering the strength of a polymeric materials is to vary the chain length. Another method for modifying polymer behavior involves vary-ing the substituents. For example, if we use a monomer of the type the properties of the resulting polymer depend on the identity of X. The simplest exam-ple is polypropylene, whose monomer is and that has the form The CH3 groups can be arranged on the same side of the chain (called an isotactic chain) as shown above, can alternate (called a syndiotactic chain) as shown below, or can be randomly distributed (called an atactic chain).
Open die Compressed air Blow-molded bottle (a) (b) (c) (d) HDPE tube FIGURE 22.17 A major use of HDPE is for blow-molded objects such as bottles for soft drinks, shampoos, bleaches, and so on. (a) A tube composed of HDPE is inserted into the mold (die). (b) The die closes, sealing the bottom of the tube. (c) Compressed air is forced into the warm HDPE tube, which then expands to take the shape of the die. (d) The molded bottle is removed from the die.
1024 Chapter Twenty-Two Organic and Biological Molecules The chain arrangement has a significant effect on the polymer’s properties. Most polypropylene is made using the Ziegler-Natta catalyst, Al(C2H5)3 TiCl4, which produces highly isotactic chains that pack together quite closely. As a result, polypropylene is more crystalline, and therefore stronger and harder, than polyethylene. The major uses of polypropylene are for molded parts (40%), fibers (35%), and packaging films (10%).
Polypropylene fibers are especially useful for athletic wear because they do not absorb water from perspiration, as cotton does. Rather, the moisture is drawn away from the skin to the surface of the polypropylene garment, where it can evaporate. The annual U.S. pro-duction of polypropylene is about 7 billion pounds.
Another related polymer, polystyrene, is constructed from the monomer styrene, Pure polystyrene is too brittle for many uses, so most polystyrene-based polymers are ac-tually copolymers of styrene and butadiene, CHEMICAL IMPACT Plastic That Talks and Listens I magine a plastic so “smart” that it can be used to sense a baby’s breath, measure the force of a karate punch, sense the presence of a person 100 feet away, or make a balloon that sings. There is a plastic film capable of doing all these things. It’s called polyvinylidene difluoride (PVDF), which has the structure When this polymer is processed in a particular way, it be-comes piezoelectric and pyroelectric. A piezoelectric sub-stance produces an electric current when it is physically de-formed or alternatively undergoes a deformation caused by the application of a current. A pyroelectric material is one that develops an electrical potential in response to a change in its temperature.
Because PVDF is piezoelectric, it can be used to con-struct a paper-thin microphone; it responds to sound by pro-ducing a current proportional to the deformation caused by the sound waves. A ribbon of PVDF plastic one-quarter of F F F F C C C C H H H H an inch wide could be strung along a hallway and used to listen to all the conversations going on as people walk through. On the other hand, electric pulses can be applied to the PVDF film to produce a speaker. A strip of PVDF film glued to the inside of a balloon can play any song stored on a microchip attached to the film—hence a balloon that can sing “Happy Birthday” at a party. The PVDF film can also be used to construct a sleep apnea monitor, which, when placed beside the mouth of a sleeping infant, will set off an alarm if the breathing stops, thus helping to prevent sudden infant death syndrome (SIDS). The same type of film is used by the U.S. Olympic karate team to measure the force of kicks and punches as the team trains. Also, gluing two strips of film together gives a material that curls in response to a current, creating an artificial muscle. In addition, because the PVDF film is pyroelectric, it responds to the infrared (heat) radiation emitted by a human as far away as 100 feet, making it useful for burglar alarm systems.
Making the PVDF polymer piezoelectric and pyroelec-tric requires some very special processing, which makes it costly ($10 per square foot). This expense seems a small price to pay for its near-magical properties.
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22.6 Natural Polymers 1025 thus incorporating bits of butadiene rubber into the polystyrene matrix. The resulting poly-mer is very tough and is often used as a substitute for wood in furniture.
Another polystyrene-based product is acrylonitrile-butadiene-styrene (ABS), a tough, hard, and chemically resistant plastic used for pipes and for items such as ra-dio housings, telephone cases, and golf club heads, for which shock resistance is an essential property. Originally, ABS was produced by copolymerization of the three monomers: It is now prepared by a special process called grafting, in which butadiene is polymerized first, and then the cyanide and phenyl substituents are added chemically.
Another high-volume polymer, polyvinyl chloride (PVC), is constructed from the monomer vinyl chloride, 22.6 Natural Polymers Proteins We have seen that many useful synthetic materials are polymers. Thus it should not be surprising that a great many natural materials are also polymers: starch, hair, silicate chains in soil and rocks, silk and cotton fibers, and the cellulose in woody plants, to name only a few.
In this section we consider a class of natural polymers, the proteins, which make up about 15% of our bodies and have molecular weights (molar masses) that range from about 6000 to over 1,000,000 grams per mole. Proteins perform many functions in the hu-man body. Fibrous proteins provide structural integrity and strength for many types of tissue and are the main components of muscle, hair, and cartilage. Other proteins, usually called globular proteins because of their roughly spherical shape, are the “worker” mol-ecules of the body. These proteins transport and store oxygen and nutrients, act as cata-lysts for the thousands of reactions that make life possible, fight invasion by foreign objects, participate in the body’s many regulatory systems, and transport electrons in the complex process of metabolizing nutrients.
The building blocks of all proteins are the -amino acids, where R may represent H, CH3, or a more complex substituent. These molecules are called -amino acids be-cause the amino group (ONH2) is always attached to the -carbon, the one next to the carboxyl group (OCO2H). The 20 amino acids most commonly found in proteins are shown in Fig. 22.18.
PVC pipe is widely used in industry.
The protein in muscles enables them to contract.
H O OH R C NH2 A O G J O A -Carbon C 1026 Chapter Twenty-Two Organic and Biological Molecules H Glycine (Gly) H2N H C O OH A O G J O A C H2N A O O A C O OH G J H C Alanine (Ala) CH3 H C O OH G D J OA CH2 O C N H2C H O D G H2 C Proline (Pro) CH2 Phenylalanine (Phe) H2N H C O OH A O G J O A C G D CH3 H3C G Isoleucine (Ile) H2N H C O OH A O G J O A C CH2 CH C NH Tryptophan (Trp) H2N H O OH A O G J P O A C A CH2 C CH A A CH2 S Methionine (Met) H2N H O OH A O G J OA C A C CH2 CH3 Leucine (Leu) H2N H O OH A O G J O A C A C CH2 CH D G CH3 H3C H2N H O OH A O G J O A C CH D G H3C C CH3 Valine (Val) Serine (Ser) H2N H C O OH A O G J OA C A OH CH2 H2N H C O OH A O G J O A C A A CH2 CH2 Glutamine (Gln) M DNH2 O C H2N H C OH A O G J O A CH2 C O OH Tyrosine (Tyr) H2N H C O OH A O G J OA C CH2 O B f A r B O B HC N CH C NH Histidine (His) Asparagine (Asn) C O NH2 H2N H C O OH C CH2 H2N H C O OH A O G J O A C CH3G D OH CH Threonine (Thr) C O H2N H C O OH C CH2 Aspartic acid (Asp) OH Cysteine (Cys) SH H2N H C O OH A O G J OA C A CH2 C O H2N H C O OH C CH2 CH2 Glutamic acid (Glu) OH H2N H C O OH A O G J OA C A A A A CH2 CH2 CH2 CH2 NH2 Lysine (Lys) H2N H C O OH A O G J OA C A A A A CH2 CH2 CH2 NH CJ G NH H2N Arginine (Arg) Nonpolar R groups Polar R groups FIGURE 22.18 The 20 -amino acids found in most proteins. The R group is shown in color.
a 22.6 Natural Polymers 1027 Note from Fig. 22.18 that the amino acids are grouped into polar and nonpolar classes, determined by the R groups, or side chains. Nonpolar side chains contain mostly carbon and hydrogen atoms, whereas polar side chains contain large numbers of nitrogen and oxygen atoms. This difference is important, because polar side chains are hydrophilic (water-loving), but nonpolar side chains are hydrophobic (water-fearing), and this characteristic greatly affects the three-dimensional structure of the resulting protein.
The protein polymer is built by condensation reactions between amino acids. For example, The product shown above is called a dipeptide. This name is used because the structure is called a peptide linkage by biochemists. (The same grouping is called an amide by or-ganic chemists.) Additional condensation reactions lengthen the chain to produce a polypeptide, eventually yielding a protein.
You can imagine that with 20 amino acids, which can be assembled in any order, there is essentially an infinite variety possible in the construction of proteins. This flexi-bility allows an organism to tailor proteins for the many types of functions that must be carried out.
The order, or sequence, of amino acids in the protein chain is called the primary structure, conveniently indicated by using three-letter codes for the amino acids (see Fig.
22.18), where it is understood that the terminal carboxyl group is on the right and the ter-minal amino group is on the left. For example, one possible sequence for a tripeptide con-taining the amino acids lysine, alanine, and leucine is which is represented in the shorthand notation by Note from Sample Exercise 22.7 that there are six sequences possible for a polypep-tide with three given amino acids. There are three possibilities for the first amino acid (any one of the three given amino acids), there are two possibilities for the second amino acid (one has already been accounted for), but there is only one possibility left for the third amino acid. Thus the number of sequences is 3 2 1 6. The product 3 2 1 is often written 3! (and is called 3 factorial). Similar reasoning shows that for a polypeptide with four amino acids, there are 4!, or 4 3 2 1 24, possible sequences.
lys-ala-leu A A O O O H A H A H (CH2)4 A CH3 A H A H A NH2 A CH2 A HC(CH3)2 C B O B O C O N C O O N C OC OCOOH H2N Lysine Alanine Leucine C O H N At the pH in biological fluids, the amino acids shown in Fig. 22.18 exist in a dif-ferent form, with the proton of the group transferred to the group. For example, glycine would be in the form H3 NCH2COO.
¬NH2 ¬COOH The peptide linkage is also found in nylon (see Section 22.5).
A tripeptide containing glycine, cysteine, and alanine.
Tripeptide Sequences Write the sequences of all possible tripeptides composed of the amino acids tyrosine, his-tidine, and cysteine.
Solution There are six possible sequences: See Exercise 22.89.
Polypeptide Sequences What number of possible sequences exists for a polypeptide composed of 20 different amino acids?
Solution The answer is 20!, or See Exercise 22.90.
A striking example of the importance of the primary structure of polypeptides can be seen in the differences between oxytocin and vasopressin. Both of these molecules are nine-unit polypeptides that differ by only two amino acids (Fig. 22.19), yet they perform completely different functions in the human body. Oxytocin is a hormone that triggers contraction of the uterus and milk secretion. Vasopressin raises blood pressure levels and regulates kidney function.
A second level of structure in proteins, beyond the sequence of amino acids, is the arrangement of the chain of the long molecule. The secondary structure is determined to a large extent by hydrogen bonding between lone pairs on an oxygen atom in the car-bonyl group of an amino acid and a hydrogen atom attached to a nitrogen of another amino acid: Such interactions can occur within the chain coils to form a spiral structure called an -helix, as shown in Fig. 22.20 and Fig. 22.21. This type of secondary structure gives the protein elasticity (springiness) and is found in the fibrous proteins in wool, hair, and tendons. Hydrogen bonding can also occur between different protein chains, joining them together in an arrangement called a pleated sheet, as shown in Fig. 22.22. Silk contains this arrangement of proteins, making its fibers flexible yet very strong and resistant to stretching. The pleated sheet is also found in muscle fibers. The hydrogen bonds in the -helical protein are called intrachain (within a given protein chain), and those in the pleated sheet are said to be interchain (between protein chains).
As you might imagine, a molecule as large as a protein has a great deal of flexibil-ity and can assume a variety of overall shapes. The specific shape that a protein assumes depends on its function. For long, thin structures, such as hair, wool and silk fibers, and tendons, an elongated shape is required. This may involve an -helical secondary struc-ture, as found in the protein -keratin in hair and wool or in the collagen found in ten-dons [Fig. 22.23(a)], or it may involve a pleated-sheet secondary structure, as found in 20 19 18 17 16 p 5 4 3 2 1 2.43 1018 tyr-cys-his his-cys-tyr cys-his-tyr tyr-his-cys his-tyr-cys cys-tyr-his 1028 Chapter Twenty-Two Organic and Biological Molecules Sample Exercise 22.8 cys–tyr–ile–gln–asn–cys–pro–leu–gly cys–tyr–phe–gln–asn–cys–pro–arg–gly (a) (b) FIGURE 22.19 The amino acid sequences in (a) oxytocin and (b) vasopressin. The differing amino acids are boxed.
Carbon Nitrogen FIGURE 22.20 Hydrogen bonding within a protein chain causes it to form a stable helical structure called the -helix. Only the main atoms in the helical backbone are shown here. The hydrogen bonds are not shown.
a Sample Exercise 22.7 22.6 Natural Polymers 1029 silk [Fig. 22.23(b)]. Many of the proteins in the body having nonstructural functions are globular, such as myoglobin (see Fig. 21.31). Note that the secondary structure of myoglobin is basically -helical. However, in the areas where the chain bends to give the protein its compact globular structure, the -helix breaks down to give a secondary configuration known as the random-coil arrangement.
R Carbon Nitrogen Hydrogen Oxygen Hydrogen bonding Hydrogen bond Side chain R Imaginary axis of "-helix R R R R R R R R R FIGURE 22.21 Ball-and-stick model of a portion of a pro-tein chain in the -helical arrangement, showing the hydrogen-bonding interactions.
a Carbon Hydrogen Nitrogen Hydrogen bond Oxygen R group R FIGURE 22.22 When hydrogen bonding occurs between protein chains rather than within them, a stable struc-ture (the pleated sheet) results. This structure contains many protein chains and is found in natural fibers, such as silk, and in muscles.
(a) (b) FIGURE 22.23 (a) Collagen, a protein found in tendons, consists of three protein chains (each with a helical structure) twisted together to form a superhelix. The result is a long, relatively narrow protein. (b) The pleated-sheet arrangement of many proteins bound together to form the elongated protein found in silk fibers.
The overall shape of the protein, long and narrow or globular, is called its tertiary structure and is maintained by several different types of interactions: hydrogen bonding, dipole–dipole interactions, ionic bonds, covalent bonds, and London dispersion forces be-tween nonpolar groups. These bonds, which represent all the bonding types discussed in this text, are summarized in Fig. 22.24.
The amino acid cysteine plays a special role in stabilizing the tertiary structure of many proteins because the OSH groups on two cysteines can react in the presence of an oxidizing agent to form a SOS bond called a disulfide linkage: A practical application of the chemistry of disulfide bonds is permanent waving of hair, as summarized in Fig. 22.25. The SOS linkages in the protein of hair are bro-ken by treatment with a reducing agent. The hair is then set in curlers to change the tertiary protein structure to the desired shape. Then treatment with an oxidizing agent causes new SOS bonds to form, which allow the hair protein to retain the new structure.
The three-dimensional structure of a protein is crucial to its function. The process of breaking down this structure is called denaturation (Fig. 22.26). For example, the dena-turation of egg proteins occurs when an egg is cooked. Any source of energy can cause denaturation of proteins and is thus potentially dangerous to living organisms. For example, ultraviolet and X-ray radiation or nuclear radioactivity can disrupt protein structure, which may lead to cancer or genetic damage. Protein damage is also caused by chemicals like ben-zene, trichloroethane, and 1,2-dibromoethane. The metals lead and mercury, which have a very high affinity for sulfur, cause protein denaturation by disrupting disulfide bonds between protein chains.
The tremendous flexibility in the various levels of protein structure allows the tailor-ing of proteins for a wide range of specific functions. Proteins are the “workhorse” mol-ecules of living organisms.
1030 Chapter Twenty-Two Organic and Biological Molecules (a) N H H H C O O– (b) O H C O S S (c) (d) C H H H H H H C (e) C O H H H C O H H H + FIGURE 22.24 Summary of the various types of interactions that stabilize the tertiary structure of a protein: (a) ionic, (b) hydrogen bonding, (c) covalent, (d) London dispersion, and (e) dipole–dipole.
S S Natural cysteine linkages in hair Reduction S S S S S S S S S S S S S S SH SH SH SH SH SH SH SH Chains shift Oxidation Hair set in curlers alters tertiary structures New cysteine linkages in waved hair HS HS HS HS HS HS HS HS FIGURE 22.25 The permanent waving of hair.
22.6 Natural Polymers 1031 Carbohydrates Carbohydrates form another class of biologically important molecules. They serve as a food source for most organisms and as a structural material for plants. Because many car-bohydrates have the empirical formula CH2O, it was originally believed that these sub-stances were hydrates of carbon, thus accounting for the name.
Most important carbohydrates, such as starch and cellulose, are polymers composed of monomers called monosaccharides, or simple sugars. The monosaccharides are poly-hydroxy ketones and aldehydes. The most important contain five carbon atoms (pentoses) or six carbon atoms (hexoses). One important hexose is fructose, a sugar found in honey and fruit. Its structure is where the asterisks indicate chiral carbon atoms. In Section 21.4 we saw that molecules with nonsuperimposable mirror images exhibit optical isomerism. A carbon atom with four different groups bonded to it in a tetrahedral arrangement always has a nonsuperim-posable mirror image (see Fig. 22.27), which gives rise to a pair of optical isomers. For example, the simplest sugar, glyceraldehyde, which has one chiral carbon, has two optical isomers, as shown in Fig. 22.28.
HO H O A A A A O O O C H OH CH2OH CH2OH C C O A O O H OH C Fructose Energy c d d b Mirror b c c b C C d C c C a a d a a b FIGURE 22.26 A schematic representation of the thermal denaturation of a protein.
FIGURE 22.27 When a tetrahedral carbon atom has four different substituents, there is no way that its mirror image can be superimposed. The lower two forms show other possible orien-tations of the molecule. Compare these with the mirror image and note that they cannot be superimposed.
FIGURE 22.28 The mirror image optical isomers of glyceraldehyde. Note that these mirror images cannot be superimposed.
1032 Chapter Twenty-Two Organic and Biological Molecules Sample Exercise 22.9 CHEMICAL IMPACT Tanning in the Shade A mong today’s best-selling cosmetics are self-tanning lo-tions. Many light-skinned people want to look like they have just spent a vacation in the Caribbean, but they recog-nize the dangers of too much sun—it causes premature aging and may lead to skin cancer. Chemistry has come to the res-cue in the form of lotions that produce an authentic-looking tan. All of these lotions have the same active ingredient: dihydroxyacetone (DHA). DHA, which has the structure is a nontoxic, simple sugar that occurs as an intermediate in carbohydrate metabolism in higher-order plants and animals.
The DHA used in self-tanners is prepared by bacterial fer-mentation of glycerine, H H H H H H H O O H O C C C H H H H H O O C C C OH The tanning effects of DHA were discovered by acci-dent in the 1950s at Children’s Hospital at the University of Cincinnati, where DHA was being used to treat children with glycogen storage disease. When the DHA was accidentally spilled on the skin, it produced brown spots.
The mechanism of the browning process involves the Maillard reaction, which was discovered by Louis-Camille Maillard in 1912. In this process amino acids react with sug-ars to create brown or golden brown products. The same re-action is responsible for much of the browning that occurs during the manufacture and storage of foods. It is also the reason that beer is golden brown.
The browning of skin occurs in the stratum corneum— the outermost, dead layer—where the DHA reacts with free amino groups of the proteins found there.
DHA is present in most tanning lotions at concentra-tions between 2% and 5%, although some products designed to give a deeper tan are more concentrated. Because the lo-tions themselves turn brown above pH 7, the tanning lotions are buffered at pH 5.
Thanks to these new products, tanning is now both safe and easy.
1¬NH22 In fructose each of the three chiral carbon atoms satisfies the requirement of being surrounded by four different groups. This leads to a total of 23, or 8, isomers that differ in their ability to rotate polarized light. The particular isomer whose structure is shown in Table 22.8 is called D-fructose. Generally, monosaccharides have one isomer that is more common in nature than the others. The most important pentoses and hexoses are shown in Table 22.8.
Chiral Carbons in Carbohydrates Determine the number of chiral carbon atoms in the following pentose: Solution We must look for carbon atoms that have four different substituents. The top carbon has only three substituents and thus cannot be chiral. The three carbon atoms shown in blue J G A H C O OH CH2OH A A A O O O O O O OH OH H H H C C C 22.6 Natural Polymers 1033 Self-tanning products and a close-up of a label showing the contents.
TABLE 22.8 Some Important Monosaccharides Pentoses D-Ribose D-Arabinose D-Ribulose Hexoses D-Glucose D-Mannose D-Galactose D-Fructose H OH O A A A A O O O O O CH2OH C H OH CH2OH C C O A O O H C HO H OH O A A A A O O O O O CH2OH C H OH CH2OH C C O A O O H C HO HO H O A A A A O O O O O CHO C H OH CH2OH C C A O O H OH C H HO HO H O A A A A O O O O O CHO C H OH CH2OH C C OH H A O O H OH C O A A A A O O O C H OH CH2OH C C O OH H CH2OH P H O A A A A O O O O O CHO C H OH CH2OH C C OH H HO O A A A A O O O O O CHO C H OH CH2OH C C OH OH H H General Name Number of of Sugar Carbon Atoms Triose 3 Tetrose 4 Pentose 5 Hexose 6 Heptose 7 Octose 8 Nonose 9 1034 Chapter Twenty-Two Organic and Biological Molecules each have four different groups attached to them: Since the fifth carbon atom has only three types of substituents (it has two hydrogen atoms), it is not chiral.
Thus the three chiral carbon atoms in this pentose are those shown in blue: Note that D-ribose and D-arabinose, shown in Table 22.8, are two of the eight isomers of this pentose.
See Exercises 22.108 and 22.113 through 22.116.
Although we have so far represented the monosaccharides as straight-chain molecules, they usually cyclize, or form a ring structure, in aqueous solution. Figure 22.29 shows this reaction for fructose. Note that a new bond is formed between the oxygen of the ter-minal hydroxyl group and the carbon of the ketone group. In the cyclic form fructose is a five-membered ring containing a COOOC bond. The same type of reaction can occur between a hydroxyl group and an aldehyde group, as shown for D-glucose in Fig. 22.30.
In this case a six-membered ring is formed.
More complex carbohydrates are formed by combining monosaccharides. For example, sucrose, common table sugar, is a disaccharide formed from glucose and fructose by elimination of water to form a COOOC bond between the rings, which is called a glycoside linkage (Fig. 22.31). When sucrose is consumed in food, the above reaction is reversed. An enzyme in saliva catalyzes the breakdown of this disaccharide.
Large polymers consisting of many monosaccharide units, called polysaccharides, can form when each ring forms two glycoside linkages, as shown in Fig. 22.32. Three of the most important of these polymers are starch, cellulose, and glycogen. All these substances are polymers of glucose, differing from each other in the nature of the glycoside linkage, the amount of branching, and molecular weight (molar mass).
Starch, a polymer of -D-glucose, consists of two parts: amylose, a straight-chain polymer of -glucose [see Fig. 22.32(a)], and amylopectin, a highly branched poly-mer of -glucose with a molecular weight that is 10 to 20 times that of amylose. Branch-ing occurs when a third glycoside linkage attaches a branch to the main polymer chain.
Starch, the carbohydrate reservoir in plants, is the form in which glucose is stored by the plant for later use as cellular fuel. Glucose is stored in this high-molecular-weight form because it results in less stress on the plant’s internal structure by osmotic pressure. Recall J G A H C O OH CH2OH A A A O O O O O O OH OH H H H C C C A O P H C O OH CH2OH A A A O O O O O O OH OH H H H C C C A O P H C O OH CH2OH A A A O O O O O O OH OH H H H C C C A H C OH CH2OH A A A O O O O O O OH OH H H C C C J G O H FIGURE 22.29 The cyclization of D-fructose.
22.6 Natural Polymers 1035 from Section 11.6 that it is the concentration of solute molecules (or ions) that determines the osmotic pressure. Combining the individual glucose molecules into one large chain keeps the concentration of solute molecules relatively low, minimizing the osmotic pressure.
Cellulose, the major structural component of woody plants and natural fibers (such as cotton), is a polymer of -D-glucose and has the structure shown in Fig. 22.32(b). Note that the -glycoside linkages in cellulose give the glucose rings a different relative ori-entation than is found in starch. Although this difference may seem minor, it has very im-portant consequences. The human digestive system contains -glycosidases, enzymes that can catalyze breakage of the -glycoside bonds in starch. These enzymes are not effec-tive on the -glycoside bonds of cellulose, presumably because the different structure C O C C H OH H C C CH2OH OH H HO H H HO C O C C H OH H C C CH2OH OH " H HO H HO H C O C C H OH H C C CH2OH H OH H HO H H O HO H CH2OH O H OH H H H OH OH HO H CH2OH O OH H H H " # # H OH OH FIGURE 22.30 The cyclization of glucose. Two different rings are possible; they differ in the orien-tation of the hydroxy group and hydrogen on one carbon, as indicated. The two forms are designated and and are shown here in two representations.
CH2OH H C C O C H OH C C CH2OH H Fructose OH HO CH2OH H H OH C C H OH HO H CH2OH O H OH H H C C C H OH OH "-D-glucose C O C CH2OH Sucrose Glycoside linkage CH2OH H O O H H C C C H OH OH –H2O +H2O C C HO H C FIGURE 22.31 Sucrose is a disaccharide formed from -D-glucose and fructose.
Bowl of sugar cubes.
1036 Chapter Twenty-Two Organic and Biological Molecules results in a poor fit between the enzyme’s active site and the carbohydrate. The enzymes necessary to cleave -glycoside linkages, the -glycosidases, are found in bacteria that exist in the digestive tracts of termites, cows, deer, and many other animals. Thus, unlike humans, these animals can derive nutrition from cellulose.
Glycogen, the main carbohydrate reservoir in animals, has a structure similar to that of amylopectin but with more branching. It is this branching that is thought to facilitate the rapid breakdown of glycogen into glucose when energy is required.
Nucleic Acids Life is possible only because each cell, when it divides, can transmit the vital information about how it works to the next generation. It has been known for a long time that this process involves the chromosomes in the nucleus of the cell. Only since 1953, however, have scientists understood the molecular basis of this intriguing cellu-lar “talent.” The substance that stores and transmits the genetic information is a polymer called deoxyribonucleic acid (DNA), a huge molecule with a molecular weight as high as sev-eral billion grams per mole. Together with other similar nucleic acids called the ribonu-cleic acids (RNA), DNA is also responsible for the synthesis of the various proteins needed by the cell to carry out its life functions. The RNA molecules, which are found in the cy-toplasm outside the nucleus, are much smaller than DNA polymers, with molecular weights of only 20,000 to 40,000 grams per mole.
The monomers of the nucleic acids, called nucleotides, are composed of three dis-tinct parts: 1. A five-carbon sugar, deoxyribose in DNA and ribose in RNA (Fig. 22.33) 2. A nitrogen-containing organic base of the type shown in Fig. 22.34 3. A phosphoric acid molecule (H3PO4) HO H CH2OH O H O H H H OH OH O O H CH2OH O H H O H H H OH OH OH H O H O H H CH2OH HO H CH2OH O H H O H H H OH OH OH H O H O H H CH2OH HO H CH2OH O H H H (a) (b) H OH OH O H CH2OH O H H H H OH OH O H CH2OH O H H H H OH OH FIGURE 22.32 (a) The polymer amylose is a major component of starch and is made up of -D-glucose monomers. (b) The polymer cellulose, which consists of -D-glucose monomers.
a C C OH OH OH O C C CH2OH CH2OH H H H H H OH C C C C H H H H OH Deoxyribose Ribose (a) (b) FIGURE 22.33 The structure of the pentoses (a) deoxyri-bose and (b) ribose. Deoxyribose is the sugar molecule present in DNA; ribose is found in RNA.
22.6 Natural Polymers 1037 The base and the sugar combine as shown in Fig. 22.35(a) to form a unit that in turn re-acts with phosphoric acid to create the nucleotide, which is an ester [see Fig. 22.35(b)].
The nucleotides become connected through condensation reactions that eliminate water to give a polymer of the type represented in Fig. 22.36; such a polymer can contain a bil-lion units.
N H N H Uracil (U) RNA N H Cytosine (C) DNA RNA N H Thymine (T) DNA O O O O O O HN N N N N N N H N HN HN CH3 NH2 H2N NH2 Adenine (A) DNA RNA Guanine (G) DNA RNA FIGURE 22.34 The organic bases found in DNA and RNA.
A computer image of the base pairs of DNA. The blue lines represent the sugar–phosphate backbone and the colored bars represent the hydrogen bonding between the base pairs.
H2O C O C H C C HOCH2 OH Ribose Adenine OH H H OH H H N N N N H2O C O C H C C HOCH2 OH Adenosine OH H H NH2 NH2 H N N N N NH2 N N N N C O C H C C CH2 O OH Adenosine OH H H H O H HO P H C O C H C C OH Adenosine 5-phosphoric acid OH H H NH2 OH O Phosphoric acid CH2 O HO P OH O H N N N N (a) (b) FIGURE 22.35 (a) Adenosine is formed by the reaction of adenine with ribose. (b) The reaction of phosphoric acid with adenosine to form the ester adenosine 5-phosphoric acid, a nucleotide. (At biological pH, the phosphoric acid would not be fully protonated as is shown here.) 1038 Chapter Twenty-Two Organic and Biological Molecules The key to DNA’s functioning is its double-helical structure with complementary bases on the two strands. The bases form hydrogen bonds to each other, as shown in Fig.
22.37. Note that the structures of cytosine and guanine make them perfect partners for hydrogen bonding, and they are always found as pairs on the two strands of DNA. Thymine and adenine form similar hydrogen-bonding pairs.
There is much evidence to suggest that the two strands of DNA unwind during cell division and that new complementary strands are constructed on the unraveled strands (Fig. 22.38). Because the bases on the strands always pair in the same way—cytosine with guanine and thymine with adenine—each unraveled strand serves as a template for at-taching the complementary bases (along with the rest of the nucleotide). This process re-sults in two double-helix DNA structures that are identical to the original one. Each new double strand contains one strand from the original DNA double helix and one newly syn-thesized strand. This replication of DNA allows for the transmission of genetic informa-tion as the cells divide.
The other major function of DNA is protein synthesis. A given segment of the DNA, called a gene, contains the code for a specific protein. These codes transmit the primary structure of the protein (the sequence of amino acids) to the construction “machinery” of the cell. There is a specific code for each amino acid in the protein, which ensures that the correct amino acid will be inserted as the protein chain grows. A code consists of a set of three bases called a codon.
DNA stores the genetic information, while RNA molecules are responsible for trans-mitting this information to the ribosomes, where protein synthesis actually occurs. This complex process involves, first, the construction of a special RNA molecule called mes-senger RNA (mRNA). The mRNA is built in the cell nucleus on the appropriate section of DNA (the gene); the double helix is “unzipped,” and the complementarity of the bases O O C C C C H H H H O O O H Base CH2 CH2 P P HO O C C C C H H H H O O H Base P HO O CH2 O C C C C H H H H O O H Base P HO O CH2 O C C C C H H H H O O O H Base P HO Repeating unit along DNA chain A T T A G C A T C G A T A T G C C G T A T A G C N N H N H H N N H H O N N CH3 Deoxyribose Adenine Thymine (a) (b) O N N N H H N N Guanine Cytosine O H H H H N (c) H O H Deoxyribose Deoxyribose Deoxyribose N N H FIGURE 22.36 A portion of a typical nucleic acid chain.
Note that the backbone consists of sugar–phosphate esters.
FIGURE 22.37 (a) The DNA double helix contains two sugar–phosphate backbones, with the bases from the two strands hydrogen-bonded to each other. The complementarity of the (b) thymine-adenine and (c) cytosine-guanine pairs.
22.6 Natural Polymers 1039 is used in a process similar to that used in DNA replication. The mRNA then migrates into the cytoplasm of the cell where, with the assistance of the ribosomes, the protein is synthesized.
Small RNA fragments, called transfer RNA (tRNA), are tailored to find specific amino acids and then to attach them to the growing protein chain as dictated by the codons in the mRNA. Transfer RNA has a lower molecular weight than messenger RNA. It consists of a chain of 75 to 80 nucleotides, including the bases adenine, cy-tosine, guanine, and uracil, among others. The chain folds back onto itself in various places as the complementary bases along the chain form hydrogen bonds. The tRNA decodes the genetic message from the mRNA, using a complementary triplet of bases called an anticodon. The nature of the anticodon governs which amino acid will be brought to the protein under construction.
The protein is built in several steps. First, a tRNA molecule brings an amino acid to the mRNA [the anticodon of the tRNA must complement the codon of the mRNA (see Fig. 22.39)]. Once this amino acid is in place, another tRNA moves to the second codon site of the mRNA with its specific amino acid. The two amino acids link via a peptide bond, and the tRNA on the first codon breaks away. The process is repeated down the chain, always matching the tRNA anticodon with the mRNA codon.
G C T A A A T T C A T G C C G T A G G C T A C G A A T G C C G T A A A T C G A T A T C G T T G C C G T A A Old New New Old T C G New New A T T G C C G T A A A T G C T A A T C G A T Old Old FIGURE 22.38 During cell division the original DNA dou-ble helix unwinds and new complementary strands are constructed on each original strand.
1040 Chapter Twenty-Two Organic and Biological Molecules Nucleus DNA RNA tRNA Amino acid + mRNA + Ribosome moves along mRNA Growing polypeptide chain Ribosome Completed protein (as specified by mRNA) FIGURE 22.39 The mRNA molecule, constructed from a specific gene on the DNA, is used as the pattern to construct a given protein with the assistance of ribosomes. The tRNA molecules attach to specific amino acids and put them in place as called for by the codons on the mRNA.
For Review Hydrocarbons Compounds composed of mostly carbon and hydrogen atoms that typically contain chains or rings of carbon atoms Alkanes • Contain compounds with only single bonds • Can be represented by the formula CnH2n2 • Are said to be saturated because each carbon present is bonded to the maximum number of atoms (4) • The carbon atoms are described as being sp3 hybridized • Their structural isomerism involves the formation of branched chains • React with O2 to form CO2 and H2O (called a combustion reaction) • Undergo substitution reactions Alkenes • Contain one or more double bonds • Simplest alkene is C2H4 (ethylene) which is described as containing sp2 hybri-dized carbon atoms • Restricted rotation about the bonds in alkenes can lead to cis–trans isomerism • Undergo addition reactions Alkynes • Contain one or more triple bonds • Simplest example is C2H2 (acetylene), described as containing sp-hybridized car-bon atoms • Undergo addition reactions Aromatic hydrocarbons • Contain rings of carbon atoms with delocalized electrons • Undergo substitution reactions rather than addition reactions p C‚C C“C C“C C¬C Key Terms biomolecule organic chemistry Section 22.1 hydrocarbons saturated unsaturated alkanes normal (straight-chain or unbranched) hydrocarbons structural isomerism combustion reaction substitution reaction dehydrogenation reaction cyclic alkanes Section 22.2 alkenes cis–trans isomerism alkynes addition reaction hydrogenation reaction halogenation polymerization Section 22.3 aromatic hydrocarbons phenyl group Section 22.4 hydrocarbon derivatives functional group alcohols phenol carbonyl group For Review 1041 ketones aldehydes carboxylic acids carboxyl group ester amines Section 22.5 polymers thermoset polymer thermoplastic polymer crosslinking vulcanization addition polymerization free radical condensation polymerization copolymer homopolymer dimer polyester isotactic chain syndiotactic chain atactic chain polystyrene polyvinyl chloride (PVC) Section 22.6 proteins fibrous proteins globular proteins -amino acids side chains dipeptide peptide linkage polypeptide primary structure secondary structure -helix pleated sheet random-coil arrangement tertiary structure disulfide linkage denaturation carbohydrates monosaccharides (simple sugars) pentoses hexoses sucrose disaccharide glycoside linkage starch cellulose glycogen deoxyribonucleic acid (DNA) ribonucleic acid (RNA) nucleotides protein synthesis gene codon Hydrocarbon derivatives Contain one or more functional groups Alcohols: contain the group Aldehydes: contain a H group Ketones: contain the group Carboxylic acids: contain the group Polymers Large molecules formed from many small molecules (called monomers) • Addition polymerization: monomers add together by a free radical mechanism • Condensation polymerization: monomers connect by splitting out a small mole-cule, such as water Proteins A class of natural polymers with molar masses ranging from 600 to 1,000,000 Fibrous proteins form the structural basis of muscle, hair, and cartilage Globular proteins perform many biologic functions, including transport and stor-age of oxygen, catalysis of biologic reactions, and regulation of biologic systems Building blocks of proteins (monomers) are -amino acids, which connect by a condensation reaction to form a peptide linkage Protein structure • Primary: the order of amino acids in the chain • Secondary: the arrangement of the protein chain • -helix • pleated sheet • Tertiary structure: the overall shape of the protein Carbohydrates Contain carbon, hydrogen, and oxygen Serve as food sources for most organisms Monosaccharides are most commonly five-carbon and six-carbon polyhydroxy ketones and aldehydes • Monosaccharides combine to form more complex carbohydrates, such as su-crose, starch, and cellulose Genetic processes When a cell divides, the genetic information is transmitted via deoxyribonucleic acid (DNA), which has a double helical structure • During cell division, the double helix unravels and a new polymer forms along each strand of the original DNA • The genetic code is carried by organic bases that hydrogen-bond to each other in specific pairs in the interior of the DNA double helix REVIEW QUESTIONS 1. What is a hydrocarbon? What is the difference between a saturated hydrocarbon and an unsaturated hydrocarbon? Distinguish between normal and branched hydro-carbons. What is an alkane? What is a cyclic alkane? What are the two general for-mulas for alkanes? What is the hybridization of carbon atoms in alkanes? What are the bond angles in alkanes? Why are cyclopropane and cyclobutane so reactive?
a a C O C O ¬OH 1042 Chapter Twenty-Two Organic and Biological Molecules messenger RNA (mRNA) transfer RNA (tRNA) anticodon The normal (unbranched) hydrocarbons are often referred to as straight-chain hydrocarbons. What does this name refer to? Does it mean that the carbon atoms in a straight-chain hydrocarbon really have a linear arrangement? Explain. In the shorthand notation for cyclic alkanes, the hydrogens are usually omitted. How do you determine the number of hydrogens bonded to each carbon in a ring structure?
2. What is an alkene? What is an alkyne? What are the general formulas for alkenes and alkynes, assuming one multiple bond in each? What are the bond angles in alkenes and alkynes? Describe the bonding in alkenes and alkynes us-ing C2H4 and C2H2 as your examples. Why is there restricted rotation in alkenes and alkynes? Is the general formula for a cyclic alkene CnH2n? If not, what is the general formula, assuming one multiple bond?
3. What are aromatic hydrocarbons? Benzene exhibits resonance. Explain. What are the bond angles in benzene? Give a detailed description of the bonding in benzene. The electrons in benzene are delocalized, while the electrons in alkenes and alkynes are localized. Explain the difference.
4. Summarize the nomenclature rules for alkanes, alkenes, alkynes, and aromatic compounds. Correct the following false statements regarding nomenclature of hydrocarbons.
a. The root name for a hydrocarbon is based on the shortest continuous chain of carbon atoms.
b. The suffix used to name all hydrocarbons is -ane.
c. Substituent groups are numbered so as to give the largest numbers possible.
d. No number is required to indicate the positions of double or triple bonds in alkenes and alkynes.
e. Substituent groups get the lowest number possible in alkenes and alkynes.
f. The ortho- term in aromatic hydrocarbons indicates the presence of two substituent groups bonded to carbon-1 and carbon-3 in benzene.
5. What functional group distinguishes each of the following hydrocarbon derivatives?
a. halohydrocarbons b. alcohols c. ethers d. aldehydes e. ketones f. carboxylic acids g. esters h. amines Give examples of each functional group. What prefix or suffix is used to name each functional group? What are the bond angles in each? Describe the bonding in each functional group. What is the difference between a primary, secondary, and tertiary alcohol? For the functional groups in a–h, when is a number required to indicate the position of the functional group? Carboxylic acids are often written as RCOOH. What does indicate and what does R indicate?
Aldehydes are sometimes written as RCHO. What does indicate?
6. Distinguish between isomerism and resonance. Distinguish between structural and geometric isomerism. When writing the various structural isomers, the most difficult task is identifying which are different isomers and which are identical to a previously written structure—that is, which are compounds that differ only by the rotation of a carbon single bond. How do you distinguish between struc-tural isomers and those that are identical?
Alkenes and cycloalkanes are structural isomers of each other. Give an example of each using C4H10. Another common feature of alkenes and cycloalkanes ¬CHO ¬COOH p p For Review 1043 is that both have restricted rotation about one or more bonds in the compound, so both can exhibit cis–trans isomerism. What is required for an alkene or cy-cloalkane to exhibit cis–trans isomerism? Explain the difference between cis and trans isomers.
Alcohols and ethers are structural isomers of each other, as are aldehydes and ketones. Give an example of each to illustrate. Which functional group in Table 22.4 can be structural isomers of carboxylic acids?
What is optical isomerism? What do you look for to determine whether an organic compound exhibits optical isomerism? 1-Bromo-1-chloroethane is opti-cally active whereas 1-bromo-2-chloroethane is not optically active. Explain.
7. What type of intermolecular forces do hydrocarbons exhibit? Explain why the boiling point of n-heptane is greater than that of n-butane. A general rule for a group of hydrocarbon isomers is that as the amount of branching increases, the boiling point decreases. Explain why this would be true.
The functional groups listed in Table 22.4 all exhibit London dispersion forces, but they also usually exhibit additional dipole–dipole forces. Explain why this is the case for each functional group. Although alcohols and ethers are structural isomers of each other, alcohols always boil at significantly higher temperatures than similar-size ethers. Explain. What would you expect when comparing the boiling points of similar-size carboxylic acids to esters? CH3CH2CH3, CH3CH2OH, CH3CHO, and HCOOH all have about the same molar mass, but they boil at very different tem-peratures. Why? Place these compounds in order by increasing boiling point.
8. Distinguish between substitution and addition reactions. Give an example of each type of reaction. Alkanes and aromatics are fairly stable compounds. To make them react, a special catalyst must be present. What catalyst must be present when reacting Cl2 with an alkane or with benzene? Adding Cl2 to an alkene or alkyne does not require a special catalyst. Why are alkenes and alkynes more re-active than alkanes and aromatic compounds? All organic compounds can be combusted. What is the other reactant in a combustion reaction, and what are the products, assuming the organic compound contains only C, H, and perhaps O?
The following are some other organic reactions covered in Section 22.4.
Give an example to illustrate each type of reaction.
a. Adding H2O to an alkene (in the presence of H) yields an alcohol.
b. Primary alcohols are oxidized to aldehydes, which can be further oxidized to carboxylic acids.
c. Secondary alcohols are oxidized to ketones.
d. Reacting an alcohol with a carboxylic acid (in the presence of H) produces an ester.
9. Define and give an example of each of the following.
a. addition polymer b. condensation polymer c. copolymer d. homopolymer e. polyester f. polyamide Distinguish between a thermoset polymer and a thermoplastic polymer. How do the physical properties of polymers depend on chain length and extent of chain branching? Explain how crosslinking agents are used to change the physical properties of polymers. Isotactic polypropylene makes stronger fibers than atactic polypropylene. Explain. In which polymer, polyethylene or polyvinyl chloride, would you expect to find the stronger intermolecular forces (assuming the average chain lengths are equal)?
c. cis-4-methyl-3-pentene d. 2-bromo-3-butanol 4. The following organic compounds cannot exist. Why?
a. 2-chloro-2-butyne b. 2-methyl-2-propanone c. 1,l-dimethylbenzene d. 2-pentanal e. 3-hexanoic acid f. 5,5-dibromo-1-cyclobutanol 5. If you had a group of hydrocarbons, what structural features would you look at to rank the hydrocarbons in order of increas-ing boiling point?
6. Which of the functional groups in Table 22.4 can exhibit hydrogen bonding intermolecular forces? Can CH2CF2 exhibit hydrogen bonding? Explain.
7. A polypeptide is also called a polyamide. What is a polyamide?
Consider a polyhydrocarbon, a polyester, and a polyamide. As-suming average chain lengths are equal, which polymer would you expect to make the strongest fibers and which polymer would you expect to make the weakest fibers? Explain.
8. Give an example reaction that would yield the following products. Name the organic reactant and product in each reaction.
a. alkane b. monohalogenated alkane A blue question or exercise number indicates that the answer to that question or exercise appears at the back of this book and a solution appears in the Solutions Guide.
Questions 1. A confused student was doing an isomer problem and listed the following six names as different structural isomers of C7H16.
a. 1-sec-butylpropane b. 4-methylhexane c. 2-ethylpentane d. 1-ethyl-1-methylbutane e. 3-methylhexane f. 4-ethylpentane How many different structural isomers are actually present in these six names?
2. For the following formulas, what types of isomerism could be exhibited? For each formula, give an example that illustrates the specific type of isomerism. The types of isomerism are struc-tural, geometric, and optical.
a. C6H12 b. C5H12O c. C6H4Br2 3. What is wrong with the following names? Give the correct name for each compound.
a. 2-ethylpropane b. 5-iodo-5, 6-dimethylhexane 1044 Chapter Twenty-Two Organic and Biological Molecules 10. Give the general formula for an amino acid. Some amino acids are labeled hy-drophilic and some are labeled hydrophobic. What do these terms refer to?
Aqueous solutions of amino acids are buffered solutions. Explain. Most of the amino acids in Fig. 22.18 are optically active. Explain. What is a peptide bond?
Show how glycine, serine, and alanine react to form a tripeptide. What is a pro-tein, and what are the monomers in proteins? Distinguish between the primary, secondary, and tertiary structures of a protein. Give examples of the types of forces that maintain each type of structure. Describe how denaturation affects the function of a protein.
What are carbohydrates, and what are the monomers in carbohydrates? The monosaccharides in Table 22.8 are all optically active. Explain. What is a disac-charide? Which monosaccharide units make up the disaccharide sucrose? What do you call the bond that forms between the monosaccharide units? What forces are responsible for the solubility of starch in water? What is the difference be-tween starch, cellulose, and glycogen?
Describe the structural differences between DNA and RNA. The monomers in nucleic acids are called nucleotides. What are the three parts of a nucleotide?
The compounds adenine, guanine, cytosine, and thymine are called the nucleic acid bases. What structural features in these compounds make them bases? DNA exhibits a double-helical structure. Explain. Describe how the complementary base pairing between the two individual strands of DNA forms the overall double-helical structure. How is complementary base pairing involved in the replication of DNA molecule during cell division? Describe how protein synthe-sis occurs. What is a codon, and what is a gene? The deletion of a single base from a DNA molecule can constitute a fatal mutation, whereas substitution of one base for another is often not as serious a mutation. Explain.
Exercises 1045 19. Draw the structural formula for each of the following.
a. 3-isobutylhexane b. 2,2,4-trimethylpentane, also called isooctane. This substance is the reference (100 level) for octane ratings.
c. 2-tert-butylpentane d. The names given in parts a and c are incorrect. Give the cor-rect names for these hydrocarbons.
20. Draw the structure for 4-ethyl-2,3-diisopropylpentane. This name is incorrect. Give the correct systematic name.
21. Name each of the following: a.
b.
c.
d.
22. Name each of the following cyclic alkanes, and indicate the for-mula of the compound.
a.
b.
c.
23. Give two examples of saturated hydrocarbons. How many other atoms are bonded to each carbon in a saturated hydrocarbon?
24. Draw the structures for two examples of unsaturated hydrocar-bons. What structural feature makes a hydrocarbon unsaturated?
25. Name each of the following alkenes.
a. CH2PCHOCH2OCH3 b.
c.
26. Name each of the following alkenes or alkynes.
a.
CH C 3 CH3 CH3 C CH3 O A CH CH OCHCH3 CH3 A CH3 CH3CH2CH B CH2CH3 CH CH CH CH 3 CH3 CH3 CH3 CH2CH2CH3 CH3 CH3 CH3 CH3 C CH3 CH3 CH CH3 CH2 C CH2 CH2 CH3 CH2 CH3 CH2 CH3 CH2 CH3 CH3 CH3 CH2 C CH3 CH3 CH3 C CH2 CH3 CH2 CH2 CH2 CH2 CH2 CH CH3 CH3 CH3 CH3 CH3 CH3 CH2 C CH2 CH CH3 c. dihalogenated alkane d. tetrahalogenated alkane e. monohalogenated benzene f. alkene 9. Give an example reaction that would yield the following prod-ucts as major organic products. See Exercises 22.62 and 22.65 for some hints. For oxidation reactions, just write oxidation over the arrow and don’t worry about the actual reagent.
a. primary alcohol b. secondary alcohol c. tertiary alcohol d. aldehyde e. ketone f. carboxylic acid g. ester 10. What is polystyrene? The following processes result in a stronger polystyrene polymer. Explain why in each case.
a. addition of catalyst to form syndiotactic polystyrene b. addition of 1,3-butadiene and sulfur c. producing long chains of polystyrene d. addition of a catalyst to make linear polystyrene 11. Answer the following questions regarding the formation of poly-mers.
a. What structural features must be present in a monomer in or-der to form a homopolymer polyester?
b. What structural features must be present in the monomers in order to form a copolymer polyamide?
c. What structural features must be present in a monomer that can form both an addition polymer and a condensation polymer?
12. In Section 22.6, three important classes of biologically impor-tant natural polymers are discussed. What are the three classes, what are the monomers used to form the polymers, and why are they biologically important?
Exercises In this section similar exercises are paired.
Hydrocarbons 13. Draw the five structural isomers of hexane (C6H14).
14. Name the structural isomers in Exercise 13.
15. Draw all the structural isomers for C8H18 that have the following root name (longest carbon chain). Name the structural isomers.
a. heptane b. butane 16. Draw all the structural isomers for C8H18 that have the following root name (longest carbon chain). Name the structural isomers.
a. hexane b. pentane 17. Draw a structural formula for each of the following compounds.
a. 2-methylpropane b. 2-methylbutane c. 2-methylpentane d. 2-methylhexane 18. Draw a structural formula for each of the following compounds.
a. 2,2-dimethylheptane c. 3,3-dimethylheptane b. 2,3-dimethylheptane d. 2,4-dimethylheptane f.
g.
Isomerism 33. There is only one compound that is named 1,2-dichloroethane, but there are two distinct compounds that can be named 1,2-dichloroethene. Why?
34. Consider the following four structures: a. Which of these compounds would have the same physical properties (melting point, boiling point, density, and so on)?
b. Which of these compounds is (are) trans isomers?
c. Which of these compounds do not exhibit cis–trans isomerism?
35. Which of the compounds in Exercises 25 and 27 exhibit cis–trans isomerism?
36. Which of the compounds in Exercises 26 and 28 exhibit cis–trans isomerism?
37. Draw all the structural isomers of C5H10. Ignore any cyclic isomers.
38. Which of the structural isomers in Exercise 37 exhibit cis–trans isomerism?
39. Draw all the structural and geometrical (cis–trans) isomers of C3H5Cl.
40. Draw all the structural and geometrical (cis–trans) isomers of bromochloropropene.
41. Draw all structural and geometrical (cis–trans) isomers of C4H7F.
Ignore any cyclic isomers.
42. Cis–trans isomerism is also possible in molecules with rings.
Draw the cis and trans isomers of 1,2-dimethylcyclohexane. In Exercise 41, you drew all of the noncyclic structural and geometric isomers of C4H7F. Now draw the cyclic structural and geometric isomers of C4H7F.
43. Draw the following.
a. cis-2-hexene b. trans-2-butene c. cis-2,3-dichloro-2-pentene C H3C H C P D G D G H C H H H C P D G D C H3C H C P D G D G H C H H H C P D G D C H3C H H H H H C P D G D G C C P D G G C H3C H H H H H C P D G D G C C P D G G (i) (ii) (iii) (iv) CH3 Br CH3 Br b.
c.
27. Give the structure for each of the following.
a. 3-hexene b. 2,4-heptadiene c. 2-methyl-3-octene 28. Give the structure for each of the following.
a. 4-methyl-1-pentyne b. 2,3,3-trimethyl-1-hexene c. 3-ethyl-4-decene 29. Give the structure of each of the following aromatic hydrocarbons.
a. o-ethyltoluene b. p-di-tert-butylbenzene c. m-diethylbenzene d. 1-phenyl-2-butene 30. Cumene is the starting material for the industrial production of acetone and phenol. The structure of cumene is Give the systematic name for cumene.
31. Name each of the following.
a.
b. CH3CH2CH2CCl3 c.
d. CH2FCH2F 32. Name each of the following compounds.
a.
b.
c.
d.
e. CH3 Br CH3 CH3 CH3 CH2 CH2 CH3 Cl CH2CH3 CH3 CH3CHCH Cl CH2 O G D O A A CCl Cl CH CH OCH3 OCH3 CH2 CH3 CH3 O A O Cl CH2 CH2O Cl O CH CH3 A CH CH3 A CH3 CH3 CH CH2 CH3 C CH3 CH2 CH C C CH3 CH3 CH3 CH2 1046 Chapter Twenty-Two Organic and Biological Molecules Exercises 1047 48. Identify the functional groups present in the following com-pounds.
a.
b.
c.
49. Mimosine is a natural product found in large quantities in the seeds and foliage of some legume plants and has been shown to cause inhibition of hair growth and hair loss in mice.
a. What functional groups are present in mimosine?
b. Give the hybridization of the eight carbon atoms in mimosine.
c. How many and bonds are found in mimosine?
50. Minoxidil (C9H15N5O) is a compound produced by Pharmacia Company that has been approved as a treatment of some types of male pattern baldness.
a. Would minoxidil be more soluble in acidic or basic aqueous solution? Explain.
b. Give the hybridization of the five nitrogen atoms in minoxidil.
c. Give the hybridization of each of the nine carbon atoms in minoxidil.
d. Give approximate values of the bond angles marked a, b, c, d, and e.
N N N N N G D D G G H O H H H a b c d e O Q S S S S O p s NH CH CHCH2 OCH3 H2N CH2 Aspartame C C OH C O O O CH O CH3O HO Vanillin O CH3 CH3 OH Testosterone 44. Name the following compounds.
a.
b.
c.
45. If one hydrogen in a hydrocarbon is replaced by a halogen atom, the number of isomers that exist for the substituted compound depends on the number of types of hydrogen in the original hydrocarbon. Thus there is only one form of chloroethane (all hydrogens in ethane are equivalent), but there are two isomers of propane that arise from the substitution of a methyl hydrogen or a methylene hydrogen. How many isomers can be obtained when one hydrogen in each of the compounds named below is replaced by a chlorine atom?
a. n-pentane c. 2,4-dimethylpentane b. 2-methylbutane d. methylcyclobutane 46. There are three isomers of dichlorobenzene, one of which has now replaced naphthalene as the main constituent of mothballs.
a. Identify the ortho, the meta, and the para isomers of dichlo-robenzene.
b. Predict the number of isomers for trichlorobenzene.
c. It turns out that the presence of one chlorine atom on a ben-zene ring will cause the next substituent to add ortho or para to the first chlorine atom on the benzene ring. What does this tell you about the synthesis of m-dichlorobenzene?
d. Which of the isomers of trichlorobenzene will be the hardest to prepare?
Functional Groups 47. Identify each of the following compounds as a carboxylic acid, ester, ketone, aldehyde, or amine.
a. Anthraquinone, an important starting material in the manu-facture of dyes: b.
c.
d.
A O B O O C HO CH2CHCH3 CH3 CH3CHCH2 H I I CH3CH2CH2 C C CH3 CH2CH3 CH3CH2 CH2CH2CH3 C C CH3 Br H H C C 58. Draw a structural formula for each of the following.
a. 3-methylpentanoic acid b. ethyl methanoate c. methyl benzoate d. 3-chloro-2,4-dimethylhexanoic acid 59. Which of the following statements is (are) false? Explain why the statement(s) is (are) false.
O B a.
is a structural isomer of pentanoic acid.
O CH3 B A b.
is a structural isomer of 2-methyl-3-pentanone.
c.
is a structural isomer of 2-pentanol.
d.
is a structural isomer of 2-butenal.
A OH e. Trimethylamine is a structural isomer of 60. Draw the isomer(s) specified. There may be more than one pos-sible isomer for each part.
a. a cyclic compound that is an isomer of trans-2-butene b. an ester that is an isomer of propanoic acid c. a ketone that is an isomer of butanal d. a secondary amine that is an isomer of butylamine e. a tertiary amine that is an isomer of butylamine f. an ether that is an isomer of 2-methyl-2-propanol g. a secondary alcohol that is an isomer of 2-methyl-2-propanol Reactions of Organic Compounds 61. Complete the following reactions.
a.
b.
c.
d.
62. Reagents such as HCl, HBr, and HOH (H2O) can add across car-bon–carbon double and triple bonds, with H forming a bond to one of the carbon atoms in the multiple bond and Cl, Br, or OH forming a bond to the other carbon atom in the multiple bond.
In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that al-ready has the greater number of hydrogen atoms bonded to it.
With this rule in mind, draw the structure of the major product in each of the following reactions.
a.
b.
c.
d.
CH3 H2O CH3CH2C‚CH 2HBr ¡ CH3CH2CH“CH2 HBr ¡ CH3CH2CH“CH2 H2O ¡ CH2 O2 CH3 CH3C Spark Cl2 FeCl3 CH CH2 2Cl2 CH3 CHCHCH CH3 CH3CH“CHCH3 H2 ¡ Pt CH3CH2CH2NH2.
CH2“CHCHCH3 CH3CH2OCH2CH2CH3 HCCH2CH2CHCH3 CH3CH2CH2COCH3 e. Including all the hydrogen atoms, how many bonds exist in minoxidil?
f. How many bonds exist in minoxidil?
51. For each of the following alcohols, give the systematic name and specify whether the alcohol is primary, secondary, or tertiary.
a.
b.
c.
52. Draw structural formulas for each of the following alcohols. In-dicate whether the alcohol is primary, secondary, or tertiary.
a. 1-butanol c. 2-methyl-1-butanol b. 2-butanol d. 2-methyl-2-butanol 53. Name all the alcohols that have the formula C5H12O. How many ethers have the formula C5H12O?
54. Name all the aldehydes and ketones that have the formula C5H10O.
55. Name the following compounds.
a.
b.
c.
56. Draw the structural formula for each of the following.
a. formaldehyde (methanal) b. 4-heptanone c. 3-chlorobutanal d. 5,5-dimethyl-2-hexanone 57. Name the following compounds.
a.
b.
c. HCOOH CH3 C O OH CH2CH2CH3 CH3CH2CHCH C O OH Cl CH O CH3 CH2CH3 O CH3 HCCHCHCH3 Cl O CH3 CH3CHCHCCH2 Cl CH3 OH CH2CH2CH3 CH3CCH2CH3 OH Cl CH3CHCH2CH2 OH p s 1048 Chapter Twenty-Two Organic and Biological Molecules Exercises 1049 c.
d. CH3CH2NH2, CH3OCH3 69. How would you synthesize the following esters?
a. n-octylacetate b.
70. Salicylic acid has the following structure: Since salicylic acid has both an alcohol functional group and a car-boxylic acid functional group, it can undergo two different esteri-fication reactions depending on which functional group reacts. For example, when treated with ethanoic acid (acetic acid), salicylic acid behaves as an alcohol and the ester produced is acetylsalicylic acid (aspirin). On the other hand, when reacted with methanol, sal-icylic acid behaves as an acid and the ester methyl salicylate (oil of wintergreen) is produced. Methyl salicylate is also an analgesic and part of the formulation of many liniments for sore muscles. What are the structures of acetylsalicylic acid and methyl salicylate?
Polymers 71. Kel-F is a polymer with the structure What is the monomer for Kel-F?
72. What monomer(s) must be used to produce the following polymers?
a.
b.
c.
d.
e.
CH3 CH3 CH CH CH CH n CH2 C CH3 CH2 C CH3 CH2 C CH3 n CH2 CH2 CH2 CH2 C C O N H N H O n CH2 O CH2 CH2 O CH2 C C O O n F F F n CH2 CH CH CH2 CH CH2 Cl F C F F C Cl F C F F C Cl F C F F C n CO2H OH CCH2CH3 O CH3CH2CH2CH2CH2CH2O CH3CH2CH2 CH3 CH3 OH, C O e.
63. When toluene (C6H5CH3) reacts with chlorine gas in the pres-ence of iron(III) catalyst, the product is a mixture of the ortho and para isomers of C6H4ClCH3. However, when the reaction is light-catalyzed with no Fe3 catalyst present, the product is C6H5CH2Cl. Explain.
64. Why is it preferable to produce chloroethane by the reaction of HCl(g) with ethene than by the reaction of Cl2(g) with ethane?
(See Exercise 62.) 65. Using appropriate reactants, alcohols can be oxidized into alde-hydes, ketones, and/or carboxylic acids. Primary alcohols can be oxidized into aldehydes, which can then be oxidized into car-boxylic acids. Secondary alcohols can be oxidized into ketones, while tertiary alcohols do not undergo this type of oxidation.
Give the structure of the product(s) resulting from the oxidation of each of the following alcohols.
a. 3-methyl-1-butanol b. 3-methyl-2-butanol c. 2-methyl-2-butanol d.
e.
f.
66. Oxidation of an aldehyde yields a carboxylic acid: Draw the structures for the products of the following oxidation reactions.
a. propanal b. 2,3-dimethylpentanal c. 3-ethylbenzaldehyde 67. How would you synthesize each of the following?
a. 1,2-dibromopropane from propene b. acetone (2-propanone) from an alcohol c. tert-butyl alcohol (2-methyl-2-propanol) from an alkene (See Exercise 62.) d. propanoic acid from an alcohol 68. What tests could you perform to distinguish between the following pairs of compounds?
a. CH3CH2CH2CH3, CH2PCHCH2CH3 b.
CH3CH2CH2 CH2 CH3 COOH, CH3 C O ¡ 3ox4 ¡ 3ox4 ¡ 3ox4 CH O R R C OH O [ox] CH2 CH3 OH OH HO CH3 OH CH2 OH CH3CH2 CH3 CH3 H C C HCl 77. Polyimides are polymers that are tough and stable at tempera-tures of up to . They are used as a protective coating on the quartz fibers used in fiber optics. What monomers were used to make the following polyimide?
78. The Amoco Chemical Company has successfully raced a car with a plastic engine. Many of the engine parts, including piston skirts, connecting rods, and valve-train components, were made of a polymer called Torlon: What monomers are used to make this polymer?
79. Polystyrene can be made more rigid by copolymerizing styrene with divinylbenzene: How does the divinylbenzene make the copolymer more rigid?
80. Polyesters containing double bonds are often crosslinked by re-acting the polymer with styrene.
a. Draw the structure of the copolymer of b. Draw the structure of the crosslinked polymer (after the polyester has been reacted with styrene).
81. Which of the following polymers would be stronger or more rigid? Explain your choices.
a. The copolymer of ethylene glycol and terephthalic acid or the copolymer of 1,2-diaminoethane and terephthalic acid (1,2-diaminoethane b. The polymer of or that of c. Polyacetylene or polyethylene (The monomer in polyacety-lene is ethyne.) HO¬(CH2)6¬CO2H NH2CH2CH2NH2) HO¬CH2CH2¬OH and HO2C¬CH“CH¬CO2H n N C O B H N C O B O C B N n C O B O C B N C O B O C B 400°C f.
g.
(This polymer is Kodel, used to make fibers of stain-resistant carpeting.) Classify these polymers as condensation or addition polymers.
Which are copolymers?
73. “Super glue” contains methyl cyanoacrylate, which readily polymerizes upon exposure to traces of water or al-cohols on the surfaces to be bonded together. The polymer provides a strong bond between the two surfaces. Draw the structure of the polymer formed by methyl cyanoacrylate.
74. Isoprene is the repeating unit in natural rubber. The structure of isoprene is a. Give a systematic name for isoprene.
b. When isoprene is polymerized, two polymers of the form are possible. In natural rubber, the cis configuration is found. The polymer with the trans configuration about the double bond is called gutta percha and was once used in the manufacture of golf balls. Draw the structure of nat-ural rubber and gutta percha showing three repeating units and the configuration about the carbon–carbon double bonds.
75. Kevlar, used in bulletproof vests, is made by the condensation copolymerization of the monomers Draw the structure of a portion of the Kevlar chain.
76. The polyester formed from lactic acid, is used for tissue implants and surgical sutures that will dissolve in the body. Draw the structure of a portion of this polymer.
H2N NH2 and HO2C CO2H CH2 CH CH2 C CH3 n CH2 CH CH2 C CH3 NC CH3 O O C C CH2 COC O H COC OC H H H H H COC C O H O O H n ¬ 1 CClFCF2CClFCF2CClFCF2 2 ¬ n 1050 Chapter Twenty-Two Organic and Biological Molecules Exercises 1051 91. Give an example of amino acids that could give rise to the in-teractions pictured in Fig. 22.24 that maintain the tertiary struc-tures of proteins.
92. What types of interactions can occur between the side chains of the following amino acids that would help maintain the tertiary structure of a protein?
a. cysteine and cysteine c. glutamic acid and lysine b. glutamine and serine d. proline and leucine 93. Oxygen is carried from the lungs to tissues by the protein he-moglobin in red blood cells. Sickle cell anemia is a disease re-sulting from abnormal hemoglobin molecules in which a valine is substituted for a single glutamic acid in normal hemoglobin.
How might this substitution affect the structure of hemoglobin?
94. Over 100 different kinds of mutant hemoglobin molecules have been detected in humans. Unlike sickle cell anemia (see Exercise 93), not all of these mutations are as serious. In one nonlethal mutation, glutamine substitutes for a single glutamic acid in nor-mal hemoglobin. Rationalize why this substitution is nonlethal.
95. Draw cyclic structures for D-ribose and D-mannose.
96. Indicate the chiral carbon atoms found in the monosaccharides D-ribose and D-mannose.
97. In addition to using numerical prefixes in the general names of sugars to indicate how many carbon atoms are present, we often use the prefixes keto- and aldo- to indicate whether the sugar is a ketone or an aldehyde. For example, the monosaccharide fruc-tose is frequently called a ketohexose to emphasize that it con-tains six carbons as well as the ketone functional group. For each of the monosaccharides shown in Table 22.8 classify the sugars as aldohexoses, aldopentoses, ketohexoses, or ketopentoses.
98. Glucose can occur in three forms: two cyclic forms and one open-chain structure. In aqueous solution, only a tiny fraction of the glucose is in the open-chain form. Yet tests for the pres-ence of glucose depend on reaction with the aldehyde group, which is found only in the open-chain form. Explain why these tests work.
99. What are the structural differences between - and -glucose?
These two cyclic forms of glucose are the building blocks to form two different polymers. Explain.
100. Cows can digest cellulose, but humans can’t. Why not?
101. Which of the amino acids in Fig. 22.18 contain more than one chiral carbon atom? Draw the structures of these amino acids and indicate all chiral carbon atoms.
102. Why is glycine not optically active?
103. Which of the noncyclic isomers of bromochloropropene are op-tically active?
104. How many chiral carbon atoms does the following structure have?
O OH OH CH3CHOH b a 82. Poly(lauryl methacrylate) is used as an additive in motor oils to counter the loss of viscosity at high temperature. The structure is The long hydrocarbon chain of poly(lauryl methacrylate) makes the polymer soluble in oil (a mixture of hydrocarbons with mostly 12 or more carbon atoms). At low temperatures the poly-mer is coiled into balls. At higher temperatures the balls uncoil and the polymer exists as long chains. Explain how this helps control the viscosity of oil.
Natural Polymers 83. Which of the amino acids in Fig. 22.18 contain the following functional groups in their R group?
a. alcohol c. amine b. carboxylic acid d. amide 84. When pure crystalline amino acids are heated, decomposition gen-erally occurs before the solid melts. Account for this observation.
(Hint: Crystalline amino acids exist as called zwitterions.) 85. Aspartame, the artificial sweetner marketed under the name Nutra-Sweet, is a methyl ester of a dipeptide: a. What two amino acids are used to prepare aspartame?
b. There is concern that methanol may be produced by the decomposition of aspartame. From what portion of the molecule can methanol be produced? Write an equation for this reaction.
86. Glutathione, a tripeptide found in virtually all cells, functions as a reducing agent. The structure of glutathione is O O B B OOCCHCH2CH2CNHCHCNHCH2COO A A NH3 CH2SH What amino acids make up glutathione?
87. Draw the structures of the two dipeptides that can be formed from serine and alanine.
88. Draw the structures of the tripeptides gly–ala–ser and ser–ala– gly. How many other tripeptides are possible using these three amino acids?
89. Write the sequence of all possible tetrapeptides composed of the following amino acids.
a. two phenylalanines and two glycines b. two phenylalanines, glycine, and alanine 90. How many different pentapeptides can be formed using five dif-ferent amino acids?
H3N CRHCOO, A O J C O CH2 (CH2)11 CH3 G A O O O C O CH3 O n b. 1,3-dimethylcyclobutane Cl2 c. 2,3-dimethylbutane Cl2 113. Polychlorinated dibenzo-p-dioxins (PCDDs) are highly toxic substances that are present in trace amounts as by-products of some chemical manufacturing processes. They have been impli-cated in a number of environmental incidents—for example, the chemical contamination at Love Canal and the herbicide spray-ing in Vietnam. The structure of dibenzo-p-dioxin, along with the customary numbering convention, is The most toxic PCCD is 2,3,7,8-tetrachloro-dibenzo-p-dioxin.
Draw the structure of this compound. Also draw the structures of two other isomers containing four chlorine atoms.
114. Consider the following five compounds.
a. CH3CH2CH2CH2CH3 b.
c. CH3CH2CH2CH2CH2CH3 d.
e.
The boiling points of these five compounds are . Which compound boils at Explain.
115. The two isomers having the formula C2H6O boil at and Draw the structure of the isomer that boils at and of the isomer that boils at 116. Ignoring ring compounds, which isomer of C2H4O2 should boil at the lowest temperature?
117. Explain why methyl alcohol is soluble in water in all propor-tions, while stearyl alcohol [CH3(CH2)16OH] is a waxy solid that is not soluble in water.
118. Is octanoic acid more soluble in 1 M HCl, 1 M NaOH, or pure water? Explain. Drugs such as morphine (C17H19NO3) are often treated with strong acids. The most commonly used form of morphine is morphine hydrochloride (C17H20ClNO3).
Why is morphine treated in this way? (Hint: Morphine is an amine.) 119. Consider the compounds butanoic acid, pentanal, n-hexane, and 1-pentanol. The boiling points of these compounds (in no spe-cific order) are , , , and . Match the boil-ing points to the correct compound.
120. Consider the reaction to produce the ester methyl acetate: O CH3OH CH3COH O H2O CH3COCH3 Methyl acetate 164°C 137°C 103°C 69°C 78.5°C.
23°C 78.5°C.
23°C 36°C?
69°C, 76°C, and 117°C 9.5°C, 36°C, CH3CCH3 CH3 CH3 CH3CH2CH2CH O CH3CH2CH2CH2 OH O O 1 9 2 3 4 6 7 8 ¡ hv ¡ hv 105. Part of a certain DNA sequence is G–G–T–C–T–A–T–A–C.
What is the complementary sequence?
106. The codons (words) in DNA (that specify which amino acid should be at a particular point in a protein) are three bases long.
How many such three-letter words can be made from the four bases adenine, cytosine, guanine, and thymine?
107. Which base will hydrogen-bond with uracil within an RNA mol-ecule? Draw the structure of this base pair.
108. Tautomers are molecules that differ in the position of a hydro-gen atom. A tautomeric form of thymine has the structure If the tautomer above, rather than the stable form of thymine were present in a strand of DNA during replication, what would be the result?
109. The base sequences in mRNA that code for certain amino acids are Glu: GAA, GAG Val: GUU, GUC, GUA, GUG Met: AUG Trp: UGG Phe: UUU, UUC Asp: GAU, GAC These sequences are complementary to the sequences in DNA.
a. Give the corresponding sequences in DNA for the amino acids listed above.
b. Give a DNA sequence that would code for the peptide trp–glu–phe–met.
c. How many different DNA sequences can code for the butapeptide in part b?
d. What is the peptide that is produced from the DNA sequence T–A–C–C–T–G–A–A–G?
e. What other DNA sequences would yield the same tripeptide as in part d?
110. The change of a single base in the DNA sequence for normal he-moglobin can encode for the abnormal hemoglobin giving rise to sickle cell anemia. Which base in the codon for glu in DNA is replaced to give the codon(s) for val? (See Exercises 93 and 109.) Additional Exercises 111. Draw the following incorrectly named compounds and name them correctly.
a. 2-ethyl-3-methyl-5-isopropylhexane b. 2-ethyl-4-tert-butylpentane c. 3-methyl-4-isopropylpentane d. 2-ethyl-3-butyne 112. In the presence of light, chlorine can substitute for one (or more) of the hydrogens in an alkane. For the following reactions, draw the possible monochlorination products.
a. 2,2-dimethylpropane Cl2 ¡ hv 1052 Chapter Twenty-Two Organic and Biological Molecules Additional Exercises 1053 129. Ethylene oxide, is an important industrial chemical. Although most ethers are un-reactive, ethylene oxide is quite reactive. It resembles C2H4 in its reactions in that addition reactions occur across the bond in ethylene oxide.
a. Why is ethylene oxide so reactive? (Hint: Consider the bond angles in ethylene oxide as compared with those predicted by the VSEPR model.) b. Ethylene oxide undergoes addition polymerization, forming a polymer used in many applications requiring a nonionic sur-factant. Draw the structure of this polymer.
130. Another way of producing highly crosslinked polyesters is to use glycerol. Alkyd resins are a polymer of this type.
The polymer forms very tough coatings when baked onto a surface and is used in paints for automobiles and large appliances. Draw the structure of the polymer formed from the condensation of Explain how crosslinking occurs in this polymer.
131. Monosodium glutamate (MSG) is commonly used as a flavoring in foods. Draw the structure of MSG.
132. a. Use bond energies (Table 8.4) to estimate for the reac-tion of two molecules of glycine to form a peptide linkage.
b. Would you predict to favor the formation of peptide link-ages between two molecules of glycine?
c. Would you predict the formation of proteins to be a sponta-neous process?
133. The reaction to form a phosphate ester linkage between two nu-cleotides can be approximated as follows: Would you predict the formation of a dinucleotide from two nu-cleotides to be a spontaneous process?
134. Considering your answers to Exercises 132 and 133, how can you justify the existence of proteins and nucleic acids in light of the second law of thermodynamics?
135. All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has and the amino group has .
Three ions of alanine are possible when alanine is dissolved in Kb 7.4 105 Ka 4.5 103 O P O A A O O O O O O CH2 H2O O Sugar OH HO CH2 sugar O O O P O A A O OO O ¢S ¢H OH OH OH CH2 CH CH2 CO2H CO2H Glyercol Phthalic acid and C¬O O CH2 CH2 When this reaction is carried out with CH3OH containing radioac-tive oxygen-18, the water produced does not contain oxygen-18.
Explain the results of this radioisotope tracer experiment.
121. A compound containing only carbon and hydrogen is 85.63% C by mass. Reaction of this compound with H2O produces a sec-ondary alcohol as the major product and a primary alcohol as the minor product (see Exercise 62). If the molar mass of the hydrocarbon is between 50 and 60 g/mol, name the compound.
122. Diborane, B2H6, is a highly unstable compound that reacts ex-plosively with oxygen. Ethane, C2H6, combines with oxygen only at elevated temperatures. Explain the differences in these two compounds.
123. Three different organic compounds have the formula C3H8O.
Only two of these isomers react with KMnO4 (a strong oxidiz-ing agent). What are the names of the products when these iso-mers react with excess KMnO4?
124. Consider the following polymer: Is this polymer a homopolymer or a copolymer, and is it formed by addition polymerization or condensation polymerization?
What is (are) the monomer(s) for this polymer?
125. Nylon is named according to the number of C atoms between the N atoms in the chain. Nylon-46 has 4 C atoms then 6 C atoms, and this pattern repeats. Nylon-6 always has 6 atoms in a row. Speculate as to why nylon-46 is stronger than nylon-6.
(Hint: Consider the strengths of interchain forces.) 126. The polymer nitrile is a copolymer made from acrylonitrile and butadiene; it is used to make automotive hoses and gaskets. Draw the structure of nitrile. (Hint: See Table 22.7.) 127. Polyaramid is a term applied to polyamides containing aromatic groups. These polymers were originally made for use as tire cords but have since found many other uses.
a. Kevlar is used in bulletproof vests and many high-strength O O O n C O C CH2 CH2 CN OA n H B A H O N O NC H B A O NC H B B A O O C composites. The structure of Kevlar is Which monomers are used to make Kevlar?
b. Nomex is a polyaramid used in fire-resistant clothing. It is a copolymer of Draw the structure of the Nomex polymer. How do Kevlar and Nomex differ in their structures?
128. When acrylic polymers are burned, toxic fumes are produced. For example, in many airplane fires, more passenger deaths have been caused by breathing toxic fumes than by the fire itself. Using poly-acrylonitrile as an example, what would you expect to be one of the most toxic, gaseous combustion products created in the reaction?
H2N and NH2 HO2C CO2H 141. The structure of tartaric acid is a. Is the form of tartaric acid pictured below optically active?
Explain.
Note: The dashed lines show groups behind the plane of the page. The wedges show groups in front of the plane.
b. Draw the optically active forms of tartaric acid.
142. Using one of the Lewis structures for benzene (C6H6), estimate Hf for C6H6(g) using bond energies and given the standard en-thalpy of formation of C(g) is 717 kJ/mol. The experimental Hf value for C6H6(g) is 83 kJ/mol. Explain the discrepancy between the experimental value and the calculated Hf value for C6H6(g).
143. Mycomycin, a naturally occurring antibiotic produced by the fungus Nocardia acidophilus, has the molecular formula and the systematic name 3,5,7,8-tridecatetraene-10,12-diynoic acid. Draw the structure of mycomycin.
144. Sorbic acid is used to prevent mold and fungus growth in some food products, especially cheeses. The systematic name for sor-bic acid is 2,4-hexadienoic acid. Draw structures for the four geometrical isomers of sorbic acid.
145. Consider the following reactions. For parts b–d, see Exercise 62.
a. When is reacted with in the presence of ultra-violet light, four different monochlorination products form.
What is the structure of in this reaction?
b. When is reacted with , a tertiary alcohol is pro-duced as the major product. What is the structure of in this reaction?
c. When is reacted with HCl, 1-chloro-1-methylcyclo-hexane is produced as the major product. What are the two possible structures for in this reaction?
d. When a hydrocarbon is reacted with and the major prod-uct of this reaction is then oxidized, acetone (2-propanone) is produced. What is the structure of the hydrocarbon in this reaction?
e. When is oxidized, a carboxylic acid is produced.
What are the possible structures for in this reaction?
146. Polycarbonates are a class of thermoplastic polymers that are used in the plastic lenses of eyeglasses and in the shells of bi-cycle helmets. A polycarbonate is made from the reaction of bisphenol A (BPA) with phosgene : ¡ polycarbonate 2nHCl HOO OOH nCOCl2 OCO CH3 CH3 n A A BPA (COCl2) C5H12O C5H12O H2O C7H12 C7H12 C4H8 H2O C4H8 C5H12 Cl2(g) C5H12 C13H10O2 water. Which of these ions would predominate in a solution with In a solution with 136. The average molar mass of one base pair of nucleotides in DNA is approximately 600 g/mol. The spacing between successive base pairs is about 0.34 nm, and a complete turn in the helical structure of DNA occurs about every 3.4 nm. If a DNA molecule has a molar mass of g/mol, approx-imately how many complete turns exist in the DNA -helix structure?
137. When heat is added to proteins, the hydrogen bonding in the sec-ondary structure is disrupted. What are the algebraic signs of and for the denaturation process?
138. In glycine, the carboxylic acid group has and the amino group has . Use these equilibrium constant values to calculate the equilibrium constants for the fol-lowing.
a.
b.
c.
Challenge Problems 139. The isoelectric point of an amino acid is the pH at which the molecule has no net charge. For glycine, that point would be the pH at which virtually all glycine molecules are in the form . This form of glycine is amphoteric since it can act as both an acid and a base. If we assume that the principal equilibrium at the isoelectric point has the best acid reacting with the best base present, then the reaction is (i) Assuming this reaction is the principal equilibrium, then the fol-lowing relationship must hold true: (ii) Use this result and your answer to part c of Exercise 138 to calculate the pH at which equation (ii) is true. It will be the isoelectric point of glycine.
140. In 1994 chemists at Texas A & M University reported the syn-thesis of a non-naturally occurring amino acid (C & E News, April 18, 1994, pp. 26–27): a. To which naturally occurring amino acid is this compound most similar?
b. A tetrapeptide, phe–met–arg–phe—NH2, is synthesized in the brains of rats addicted to morphine and heroin. (The indicates that the peptide ends in instead of ) The TAMU scientists synthesized a similar tetrapeptide, with the synthetic amino acid above replacing one of the original amino acids. Draw a structure for the tetrapeptide containing the synthetic amino acid.
c. Indicate the chiral carbon atoms in the synthetic amino acid.
¬CO2H.
¬C ƒ ƒ O ¬NH2 ¬NH2 G D CH2SCH3 H G C CH2 H2N CO2H D C G D 3H2NCH2CO2 4 3 H3NCH2CO2H4 2H3NCH2CO2 ∆H2NCH2CO2 H3NCH2CO2H H3NCH2CO2 H3NCH2CO2H ∆2H H2NCH2CO2 H2NCH2CO2 H2O ∆H2NCH2CO2H OH H3NCH2CO2 H2O ∆H2NCH2CO2 H3O Kb 6.0 105 Ka 4.3 103 ¢S ¢H 4.5 109 [OH] 1.0 M?
[H] 1.0 M?
1054 Chapter Twenty-Two Organic and Biological Molecules Catalyst ¬ ¬¬¡ Integrative Problems 1055 151. A chemical “breathalyzer” test works because ethyl alcohol in the breath is oxidized by the dichromate ion (orange) to form acetic acid and chromium(III) ion (green). The balanced reaction is You analyze a breathalyzer test in which 4.2 mg of K2Cr2O7 was reduced. Assuming the volume of the breath was 0.500 L at and 750. mm Hg, what was the mole percent alcohol of the breath?
152. Consider a sample of a hydrocarbon at 0.959 atm and 298 K.
Upon combusting the entire sample in oxygen, you collect a mix-ture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 K. This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon.
Determine the molecular formula of the hydrocarbon and name it.
153. Estradiol is a female hormone with the following structure: How many chiral carbon atoms are in estradiol?
Integrative Problems These problems require the integration of multiple concepts to find the solutions.
154. Helicenes are extended fused polyaromatic hydrocarbons that have a helical or screw-shaped structure.
a. A 0.1450-g sample of solid helicene is combusted in air to give 0.5063 g of CO2. What is the empirical formula of this helicene?
b. If a 0.0938-g sample of this helicene is dissolved in 12.5 g of solvent to give a 0.0175 m solution, what is the molecular formula of this helicene?
c. What is the balanced reaction for the combustion of this helicene?
155. An organometallic compound is one containing at least one metal–carbon bond. An example of an organometallic species is (CH3CH2)MBr, which contains a metal–ethyl bond.
a. If M2 has the electron configuration [Ar]3d10, what is the percent by mass of M in (CH3CH2)MBr?
b. One of the reactions in which (CH3CH2)MBr becomes involved is the conversion of a ketone to an alcohol as illustrated here: How does the hybridization of the starred carbon atom change, if at all, in going from reactants to products?
c. What is the systematic name of the product?
Hint: In this shorthand notation, all the bonds have been eliminated and the lines represent bonds, unless shown differently. As is typical of most organic compounds, each carbon atom has four bonds to it and the oxygen atoms have only two bonds.
C¬C C¬H O OH CH3 HO OH 30.°C 3HC2H3O21aq2 4Cr31aq2 11H2O1l2 3C2H5OH1aq2 2Cr2O7 21aq2 2H1aq2 ¡ Phenol is used to terminate the polymer (stop its growth).
a. Draw the structure of the polycarbonate chain formed from the above reaction.
b. Is this reaction a condensation or addition polymerization?
147. A urethane linkage occurs when an alcohol adds across the carbon–nitrogen double bond in an isocyanate: Polyurethanes (formed from the copolymerization of a diol with a diisocyanate) are used in foamed insulation and a variety of other construction materials. What is the structure of the polyurethane formed by the following reaction?
148. ABS plastic is a tough, hard plastic used in applications requir-ing shock resistance. The polymer consists of three monomer units: acrylonitrile (C3H3N), butadiene (C4H6), and styrene (C8H8).
a. Draw two repeating units of ABS plastic assuming that the three monomer units react in a 1:1:1 mole ratio and react in the same order as the monomers listed above.
b. A sample of ABS plastic contains 8.80% N by mass. It took 0.605 g of to react completely with a 1.20-g sample of ABS plastic. What is the percent by mass of acrylonitrile, butadiene, and styrene in this polymer sample?
c. ABS plastic does not react in a 1:1:1 mole ratio among the three monomer units. Using the results from part b, determine the relative numbers of the monomer units in this sample of ABS plastic.
149. Stretch a rubber band while holding it gently to your lips. Then slowly let it relax while still in contact with your lips.
a. What happens to the temperature of the rubber band on stretching?
b. Is the stretching an exothermic or endothermic process?
c. Explain the above result in terms of intermolecular forces.
d. What is the sign of and for stretching the rubber band?
e. Give the molecular explanation for the sign of for stretching.
150. Alcohols are very useful starting materials for the production of many different compounds. The following conversions, starting with 1-butanol, can be carried out in two or more steps. Show the steps (reactants/catalysts) you would follow to carry out the conversions, drawing the formula for the organic product in each step. For each step, a major product must be produced.
See Exercise 62. Hint: in the presence of H, an alcohol is con-verted into an alkene and water. This is the exact reverse of the reaction of adding water to an alkene to form an alcohol.
a. 1-butanol b. 1-butanol -butanone ¡ 2 ¡ butane ¢S ¢G ¢S Br2 ROOOH OPCPNOR ROOCONOR H O A B Alcohol Isocyanate A urethane (C6H5OH) u. The typically sweet-smelling compounds called result from the condensation reaction of an organic acid with an .
157. Choose one of the following terms to match the description given in statements (1)–(17). All of the following pertain to proteins or carbohydrates.
a. aldohexose g. disaccharides m. ketohexoses b. saliva h. disulfide n. oxytocin c. cellulose i. globular o.
pleated sheet d. CH2O j. glycogen p. polypeptide e. cysteine k. glycoside linkage q. primary structure f. denaturation l. hydrophobic (1) polymer consisting of many amino acids (2) linkage that forms between two cysteine species (3) peptide hormone that triggers milk secretion (4) proteins with roughly spherical shape (5) sequence of amino acids in a protein (6) silk protein secondary structure (7) water-repelling amino acid side chain (8) amino acid responsible for permanent wave in hair (9) breakdown of a protein’s tertiary and/or secondary structure (10) animal polymer of glucose (11) bond between rings in disaccharide sugars (12) empirical formula leading to the name carbohydrate (13) where enzymes catalyzing the breakdown of glycoside link-ages are found (14) six-carbon ketone sugars (15) structural component of plants, polymer of glucose (16) sugars consisting of two monomer units (17) six-carbon aldehyde sugars 158. For each of the following, fill in the blank with the correct re-sponse(s). All of the following pertain to nucleic acids.
a. The substance in the nucleus of the cell that stores and transmits genetic information is DNA, which stands for .
b. The basic repeating monomer units of DNA and RNA are called .
c. The pentose deoxyribose is found in DNA, whereas is found in RNA.
d. The basic linkage in DNA or RNA between the sugar mole-cule and phosphoric acid is a phosphate linkage.
e. The bases on opposite strands of DNA are said to be to each other, which means the bases fit together specifically by hydrogen bonding to one another.
f. In a strand of normal DNA, the base is always found paired with the base adenine, whereas is always found paired with cytosine.
g. A given segment of the DNA molecule, which contains the molecular coding for a specific protein to be synthesized, is referred to as a .
h. During protein synthesis, RNA molecules attach to and transport specific amino acids to the appropriate posi-tion on the pattern provided by RNA molecules.
i. The codes specified by are responsible for assem-bling the correct primary structure of proteins.
Get help understanding core concepts and visualizing molecular-level interactions, and practice problem solving, by visiting the Online Study Center at college.hmco.com/ PIC/zumdahl7e.
¬C¬O¬C¬ Marathon Problems These problems are designed to incorporate several concepts and techniques into one situation. Marathon Problems can be used in class by groups of students to help facilitate problem-solving skills.
156. For each of the following, fill in the blank with the correct response. All of these fill-in-the-blank problems pertain to ma-terial covered in the sections on alkanes, alkenes and alkynes, aromatic hydrocarbons, and hydrocarbon derivatives.
a.
The first “organic” compound to be synthesized in the labora-tory, rather than being isolated from nature, was , which was prepared from .
b.
An organic compound whose carbon–carbon bonds are all single bonds is said to be .
c.
The general orientation of the four pairs of electrons around the carbon atoms in alkanes is .
d. Alkanes in which the carbon atoms form a single unbranched chain are said to be alkanes.
e.
Structural isomerism occurs when two molecules have the same number of each type of atom but exhibit different arrangements of the between those atoms.
f.
The systematic names of all saturated hydrocarbons have the ending added to a root name that indicates the number of carbon atoms in the molecule.
g.
For a branched hydrocarbon, the root name for the hydro-carbon comes from the number of carbon atoms in the continuous chain in the molecule.
h. The positions of substituents along the hydrocarbon frame-work of a molecule are indicated by the of the carbon atom to which the substituents are attached.
i.
The major use of alkanes has been in reactions, as a source of heat and light.
j.
With very reactive agents, such as, the halogen elements, alkanes undergo reactions, whereby a new atom replaces one or more hydrogen atoms of the alkane.
k. Alkenes and alkynes are characterized by their ability to un-dergo rapid, complete reactions, by which other atoms attach themselves to the carbon atoms of the double or triple bond.
l.
Unsaturated fats may be converted to saturated fats by the process of .
m. Benzene is the parent member of the group of hydrocarbons called hydrocarbons.
n. An atom or group of atoms that imparts new and characteristic properties to an organic molecule is called a group.
o.
A alcohol is one in which there is only one hy-drocarbon group attached to the carbon atom holding the hydroxyl group.
p. The simplest alcohol, methanol, is prepared industrially by the hydrogenation of .
q. Ethanol is commonly prepared by the of certain sugars by yeast.
r.
Both aldehydes and ketones contain the group, but they differ in where this group occurs along the hydro-carbon chain.
s.
Aldehydes and ketones can be prepared by of the corresponding alcohol.
t.
Organic acids, which contain the group, are typ-ically weak acids.
1056 Chapter Twenty-Two Organic and Biological Molecules A1 Appendixes Appendix One Mathematical Procedures A1.1 Exponential Notation The numbers characteristic of scientific measurements are often very large or very small; thus it is convenient to express them using powers of 10. For example, the number 1,300,000 can be expressed as 1.3 106, which means multiply 1.3 by 10 six times, or Note that each multiplication by 10 moves the decimal point one place to the right: Thus the easiest way to interpret the notation 1.3 106 is that it means move the deci-mal point in 1.3 to the right six times: 1 2 3 4 5 6 Using this notation, the number 1985 can be expressed as 1.985 103. Note that the usual convention is to write the number that appears before the power of 10 as a number between 1 and 10. To end up with the number 1.985, which is between 1 and 10, we had to move the decimal point three places to the left. To compensate for that, we must mul-tiply by 103, which says that to get the intended number we start with 1.985 and move the decimal point three places to the right; that is: 1 2 3 Some other examples are given below.
Number Exponential Notation 5.6 39 943 1126 So far we have considered numbers greater than 1. How do we represent a number such as 0.0034 in exponential notation? We start with a number between 1 and 10 and divide by the appropriate power of 10: 0.0034 3.4 10 10 10 3.4 103 3.4 103 1.126 103 9.43 102 3.9 101 5.6 100 or 5.6 1 1.985 103 1 9 8 5.
1.3 106 1 3 0 0 0 0 0 1,300,000 o 130 10 1300. 13 10 130. 1.3 10 13.
106 1 million 1.3 106 1.3 10 10 10 10 10 10 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ A2 Appendixes Division by 10 moves the decimal point one place to the left. Thus the number 0. 0 0 0 0 0 0 1 4 7 6 5 4 3 2 1 can be written as 1.4 107.
To summarize, we can write any number in the form where N is between 1 and 10 and the exponent n is an integer. If the sign preceding n is positive, it means the decimal point in N should be moved n places to the right. If a neg-ative sign precedes n, the decimal point in N should be moved n places to the left.
Multiplication and Division When two numbers expressed in exponential notation are multiplied, the initial numbers are multiplied and the exponents of 10 are added: For example (to two significant figures, as required), When the numbers are multiplied, if a result greater than 10 is obtained for the initial num-ber, the decimal point is moved one place to the left and the exponent of 10 is increased by 1: (two significant figures) Division of two numbers expressed in exponential notation involves normal division of the initial numbers and subtraction of the exponent of the divisor from that of the div-idend. For example, Divisor If the initial number resulting from the division is less than 1, the decimal point is moved one place to the right and the exponent of 10 is decreased by 1. For example, Addition and Subtraction To add or subtract numbers expressed in exponential notation, the exponents of the num-bers must be the same. For example, to add 1.31 105 and 4.2 104, we must rewrite one number so that the exponents of both are the same. The number 1.31 105 can be written 13.1 104, since moving the decimal point one place to the right can be com-pensated for by decreasing the exponent by 1. Now we can add the numbers: 13.1 104 4.2 104 17.3 104 7.7 103 6.4 103 8.3 105 6.4 8.3 101352 0.77 102 4.8 108 2.1 103 4.8 2.1 101832 2.3 105 2.5 1011 2.49 1011 15.8 102214.3 1082 24.9 1010 13.2 104212.8 1032 9.0 107 1M 10m21N 10n2 1MN2 10m n N 10n ⎧ ⎪ ⎨ ⎪ ⎩ Appendixes A3 In correct exponential notation the result is expressed as 1.73 105.
To perform addition or subtraction with numbers expressed in exponential notation, only the initial numbers are added or subtracted. The exponent of the result is the same as those of the numbers being added or subtracted. To subtract 1.8 102 from 8.99 103, we write 8.99 103 0.18 103 8.81 103 Powers and Roots When a number expressed in exponential notation is taken to some power, the initial num-ber is taken to the appropriate power and the exponent of 10 is multiplied by that power: For example, (two significant figures) When a root is taken of a number expressed in exponential notation, the root of the initial number is taken and the exponent of 10 is divided by the number representing the root: For example, Because the exponent of the result must be an integer, we may sometimes have to change the form of the number so that the power divided by the root equals an integer. For example, In this case, we moved the decimal point one place to the left and increased the exponent from 3 to 4 to make n2 an integer.
The same procedure is followed for roots other than square roots. For example, and 3.6 103 1 3 46 103 2 3 4.6 1010 14.6 101021 3 146 10921 3 8.8 101 0.88 102 1 3 0.69 102 2 3 6.9 105 16.9 10521 3 10.69 10621 3 4.4 101 0.44 102 10.19 102 21.9 103 11.9 10321 2 10.19 10421 2 1.7 103 12.9 10621 2 12.9 106 2 1N 10n 1n 10n21 2 1N 10n 2 4.2 108 4.22 108 422 106 17.5 10223 7.53 1032 1N 10n2m N m 10mn Refer to the instruction booklet for your calculator for directions concerning how to take roots and powers of numbers.
A4 Appendixes A1.2 Logarithms A logarithm is an exponent. Any number N can be expressed as follows: For example, The common, or base 10, logarithm of a number is the power to which 10 must be taken to yield the number. Thus, since 1000 103, Similarly, For a number between 10 and 100, the required exponent of 10 will be between 1 and 2. For example, 65 101.8129; that is, log 65 1.8129. For a number between 100 and 1000, the exponent of 10 will be between 2 and 3. For example, 650 102.8129 and log 650 2.8129.
A number N greater than 0 and less than 1 can be expressed as follows: For example, Thus Although common logs are often tabulated, the most convenient method for obtain-ing such logs is to use an electronic calculator. On most calculators the number is first entered and then the log key is punched. The log of the number then appears in the display. Some examples are given below. You should reproduce these results on your calculator to be sure that you can find common logs correctly.
Number Common Log 36 1.56 1849 3.2669 0.156 0.807 4.775 1.68 105 log 0.1 1 log 0.01 2 log 0.001 3 0.1 1 10 1 101 101 0.01 1 100 1 102 102 0.001 1 1000 1 103 103 N 10x 1 10x log 1 0 log 10 1 log 100 2 log 1000 3 1 100 10 101 100 102 1000 103 N 10x Refer to the instruction booklet for your calculator for the exact sequence to obtain logarithms.
Appendixes A5 Note that the number of digits after the decimal point in a common log is equal to the number of significant figures in the original number.
Since logs are simply exponents, they are manipulated according to the rules for ex-ponents. For example, if A 10 x and B 10y, then their product is and For division, we have and For a number raised to a power, we have and It follows that or, for n 1, When a common log is given, to find the number it represents, we must carry out the process of exponentiation. For example, if the log is 2.673, then N 102.673. The process of exponentiation is also called taking the antilog, or the inverse logarithm. This operation is usually carried out on calculators in one of two ways. The majority of calculators require that the log be entered first and then the keys INV and LOG pressed in succession. For ex-ample, to find N 102.673 we enter 2.673 and then press INV and LOG . The number 471 will be displayed; that is, N 471. Some calculators have a 10x key. In that case, the log is entered first and then the 10x key is pressed. Again, the number 471 will be displayed.
Natural logarithms, another type of logarithm, are based on the number 2.7183, which is referred to as e. In this case, a number is represented as N ex 2.7183x. For example, To find the natural log of a number using a calculator, the number is entered and then the ln key is pressed. Use the following examples to check your technique for finding nat-ural logs with your calculator: Number (e x) Natural Log(x) 784 6.664 7.384 16.118 1.00 0 1.00 107 1.61 103 ln 7.15 x 1.967 N 7.15 ex log 1 A log A log 1 An log An n log A log An nx n log A An 110x2n 10nx log A B x y log A log B A B 10x 10y 10xy log AB x y log A log B A B 10x 10y 10xy A6 Appendixes If a natural logarithm is given, to find the number it represents, exponentiation to the base e (2.7183) must be carried out. With many calculators this is done using a key marked ex (the natural log is entered, with the correct sign, and then the ex key is pressed).
The other common method for exponentiation to base e is to enter the natural log and then press the INV and ln keys in succession. The following examples will help you check your technique: ln N(x) N(e x) 3.256 25.9 5.169 13.112 Since natural logarithms are simply exponents, they are also manipulated according to the mathematical rules for exponents given earlier for common logs.
A1.3 Graphing Functions In interpreting the results of a scientific experiment, it is often useful to make a graph. If possible, the function to be graphed should be in a form that gives a straight line. The equation for a straight line (a linear equation) can be represented by the general form where y is the dependent variable, x is the independent variable, m is the slope, and b is the intercept with the y axis.
To illustrate the characteristics of a linear equation, the function y 3x 4 is plotted in Fig. A.1. For this equation m 3 and b 4. Note that the y intercept occurs when x 0. In this case the intercept is 4, as can be seen from the equation (b 4).
The slope of a straight line is defined as the ratio of the rate of change in y to that in x: For the equation y 3x 4, y changes three times as fast as x (since x has a coefficient of 3). Thus the slope in this case is 3. This can be verified from the graph. For the trian-gle shown in Fig. A.1, ¢y 34 10 24 and ¢x 10 2 8 m slope ¢y ¢x y mx b 4.95 105 5.69 103 Intercept 60 50 40 30 20 10 0 y x y3x4 y x 10 20 30 40 FIGURE A.1 Graph of the linear equation y 3x 4.
Appendixes A7 Thus The preceding example illustrates a general method for obtaining the slope of a line from the graph of that line. Simply draw a triangle with one side parallel to the y axis and the other parallel to the x axis as shown in Fig. A.1. Then determine the lengths of the sides to give y and x, respectively, and compute the ratio y x.
Sometimes an equation that is not in standard form can be changed to the form y mx b by rearrangement or mathematical manipulation. An example is the equation k AeEaRT described in Section 12.7, where A, Ea, and R are constants; k is the dependent variable; and 1T is the independent variable. This equation can be changed to standard form by taking the natural logarithm of both sides, noting that the log of a product is equal to the sum of the logs of the individual terms and that the natural log of eEaRT is simply the exponent EaRT. Thus, in standard form, the equation k AeEaRT is written h h h h y m x b A plot of ln k versus 1T (see Fig. A.2) gives a straight line with slope EaR and in-tercept ln A.
Other linear equations that are useful in the study of chemistry are listed in standard form in Table A.1.
A1.4 Solving Quadratic Equations A quadratic equation, a polynomial in which the highest power of x is 2, can be written as One method for finding the two values of x that satisfy a quadratic equation is to use the quadratic formula: x b 2b2 4ac 2a ax2 bx c 0 ln k Ea R a1 Tb ln A ln k ln AeEa RT ln A ln eEa RT ln A Ea RT Slope ¢y ¢x 24 8 3 Slope = −Ea R Intercept = ln A 1 T ln k FIGURE A.2 Graph of ln k versus 1/T.
TABLE A.1 Some Useful Linear Equations in Standard Form What Is Equation Plotted Slope Intercept Section (y mx b) ( y vs. x) (m) (b) in Text k 12.4 k 12.4 k 12.4 C 10.8 ¢Hvap R ln Pvap vs. 1 T ln Pvap ¢Hvap R a1 Tb C 1 3A40 1 3A4 vs. t 1 3A4 kt 1 3A4 0 ln3A4 0 ln3A4 vs. t ln3A4 kt ln3A4 0 3A4 0 3A4 vs. t 3A4 kt 3A4 0 ⎧ ⎨ ⎩ { { { A8 Appendixes where a, b, and c are the coefficients of x2 and x and the constant, respectively. For example, in determining [H] in a solution of 1.0 104 M acetic acid the following expression arises: which yields where a 1, b 1.8 105, and c 1.8 109. Using the quadratic formula, we have Thus and Note that there are two roots, as there always will be, for a polynomial in x2. In this case x represents a concentration of H (see Section 14.3). Thus the positive root is the one that solves the problem, since a concentration cannot be a negative number.
A second method for solving quadratic equations is by successive approximations, a systematic method of trial and error. A value of x is guessed and substituted into the equa-tion everywhere x (or x2) appears, except for one place. For example, for the equation we might guess x 2 105. Substituting that value into the equation gives or Thus Note that the guessed value of x(2 105) is not the same as the value of x that is cal-culated (3.7 105) after inserting the estimated value. This means that x 2 105 is not the correct solution, and we must try another guess. We take the calculated value (3.7 105) as our next guess: x2 1.8 109 6.7 1010 1.1 109 x2 11.8 105213.7 1052 1.8 109 0 x 3.7 105 x2 1.8 109 3.6 1010 1.4 109 x2 11.8 105212 1052 1.8 109 0 x2 11.8 1052x 1.8 109 0 x 10.5 105 2 5.2 105 x 6.9 105 2 3.5 105 1.8 105 8.7 105 2 1.8 105 27.5 109 2 1.8 105 23.24 1010 7.2 109 2 1.8 105 23.24 1010 14211211.8 1092 2112 x b 2b2 4ac 2a x2 11.8 1052x 1.8 109 0 1.8 105 x2 1.0 104 x Appendixes A9 Thus Now we compare the two values of x again: These values are closer but not close enough. Next we try 3.3 105 as our guess: Thus Again we compare: Next we guess x 3.5 105 to give Thus Now the guessed value and the calculated value are the same; we have found the correct solution. Note that this agrees with one of the roots found with the quadratic formula in the first method.
To further illustrate the method of successive approximations, we will solve Sample Exercise 14.17 using this procedure. In solving for [H] for 0.010 M H2SO4, we obtain the following expression: which can be rearranged to give We will guess a value for x, substitute it into the right side of the equation, and then cal-culate a value for x. In guessing a value for x, we know it must be less than 0.010, since a larger value will make the calculated value for x negative and the guessed and calcu-lated values will never match. We start by guessing x 0.005.
The results of the successive approximations are shown in the following table: Guessed Calculated Trial Value for x Value for x 1 0.0050 0.0040 2 0.0040 0.0051 3 0.00450 0.00455 4 0.00452 0.00453 Note that the first guess was close to the actual value and that there was oscillation be-tween 0.004 and 0.005 for the guessed and calculated values. For trial 3, an average of x 11.2 1022a0.010 x 0.010 xb 1.2 102 x10.010 x2 0.010 x x 3.5 105 x2 1.8 109 6.3 1010 1.2 109 x2 11.8 105213.5 1052 1.8 109 0 Calculated: x 3.5 105 Guessed: x 3.3 105 x 3.5 105 x2 1.8 109 5.9 1010 1.2 109 x2 11.8 105213.3 1052 1.8 109 0 Calculated: x 3.3 105 Guessed: x 3.7 105 x 3.3 105 A10 Appendixes these values was used as the guess, and this led rapidly to the correct value (0.0045 to the correct number of significant figures). Also, note that it is useful to carry extra digits un-til the correct value is obtained. That value can then be rounded off to the correct num-ber of significant figures.
The method of successive approximations is especially useful for solving polynomi-als containing x to a power of 3 or higher. The procedure is the same as for quadratic equations: Substitute a guessed value for x into the equation for every x term but one, and then solve for x. Continue this process until the guessed and calculated values agree.
A1.5 Uncertainties in Measurements Like all the physical sciences, chemistry is based on the results of measurements. Every measurement has an inherent uncertainty, so if we are to use the results of measurements to reach conclusions, we must be able to estimate the sizes of these uncertainties.
For example, the specification for a commercial 500-mg acetaminophen (the active painkiller in Tylenol) tablet is that each batch of tablets must contain 450 to 550 mg of acet-aminophen per tablet. Suppose that chemical analysis gave the following results for a batch of acetaminophen tablets: 428 mg, 479 mg, 442 mg, and 435 mg. How can we use these results to decide if the batch of tablets meets the specification? Although the details of how to draw such conclusions from measured data are beyond the scope of this text, we will con-sider some aspects of how this is done. We will focus here on the types of experimental uncertainty, the expression of experimental results, and a simplified method for estimating experimental uncertainty when several types of measurement contribute to the final result.
Types of Experimental Error There are two types of experimental uncertainty (error). A variety of names are applied to these types of errors: Precision random error indeterminate error Accuracy systematic error determinate error The difference between the two types of error is well illustrated by the attempts to hit a target shown in Fig. 1.7 in Chapter 1.
Random error is associated with every measurement. To obtain the last significant fig-ure for any measurement, we must always make an estimate. For example, we interpolate between the marks on a meter stick, a buret, or a balance. The precision of replicate meas-urements (repeated measurements of the same type) reflects the size of the random er-rors. Precision refers to the reproducibility of replicate measurements.
The accuracy of a measurement refers to how close it is to the true value. An inac-curate result occurs as a result of some flaw (systematic error) in the measurement: the presence of an interfering substance, incorrect calibration of an instrument, operator er-ror, and so on. The goal of chemical analysis is to eliminate systematic error, but random errors can only be minimized. In practice, an experiment is almost always done to find an unknown value (the true value is not known—someone is trying to obtain that value by doing the experiment). In this case the precision of several replicate determinations is used to assess the accuracy of the result. The results of the replicate experiments are expressed as an average (which we assume is close to the true value) with an error limit that gives some indication of how close the average value may be to the true value. The error limit represents the uncertainty of the experimental result.
Expression of Experimental Results If we perform several measurements, such as for the analysis for acetaminophen in painkiller tablets, the results should express two things: the average of the measurements and the size of the uncertainty.
K · K · Appendixes A11 There are two common ways of expressing an average: the mean and the median. The mean (x ) is the arithmetic average of the results, or where means take the sum of the values. The mean is equal to the sum of all the meas-urements divided by the number of measurements. For the acetaminophen results given previously, the mean is The median is the value that lies in the middle among the results. Half the meas-urements are above the median and half are below the median. For results of 465 mg, 485 mg, and 492 mg, the median is 485 mg. When there is an even number of results, the median is the average of the two middle results. For the acetaminophen results, the median is There are several advantages to using the median. If a small number of measurements is made, one value can greatly affect the mean. Consider the results for the analysis of acetaminophen: 428 mg, 479 mg, 442 mg, and 435 mg. The mean is 446 mg, which is larger than three of the four weights. The median is 438 mg, which lies near the three val-ues that are relatively close to one another.
In addition to expressing an average value for a series of results, we must express the uncertainty. This usually means expressing either the precision of the measurements or the observed range of the measurements. The range of a series of measurements is de-fined by the smallest value and the largest value. For the analytical results on the aceta-minophen tablets, the range is from 428 mg to 479 mg. Using this range, we can express the results by saying that the true value lies between 428 mg and 479 mg. That is, we can express the amount of acetaminophen in a typical tablet as 446 33 mg, where the er-ror limit is chosen to give the observed range (approximately).
The most common way to specify precision is by the standard deviation, s, which for a small number of measurements is given by the formula where xi is an individual result, x is the average (either mean or median), and n is the to-tal number of measurements. For the acetaminophen example, we have Thus we can say the amount of acetaminophen in the typical tablet in the batch of tablets is 446 mg with a sample standard deviation of 23 mg. Statistically this means that any additional measurement has a 68% probability (68 chances out of 100) of being between 423 mg (446 23) and 469 mg (446 23). Thus the standard deviation is a measure of the precision of a given type of determination.
The standard deviation gives us a means of describing the precision of a given type of determination using a series of replicate results. However, it is also useful to be able to estimate the precision of a procedure that involves several measurements by combining the precisions of the individual steps. That is, we want to answer the following question: s c 1428 44622 1479 44622 1442 44622 1435 44622 4 1 d 1 2 23 s £ a n i1 1xi x22 n 1 § 1 2 442 435 2 438 mg x 428 479 442 435 4 446 mg Mean x a n i1 xi n x1 x2 p xn n A12 Appendixes How do the uncertainties propagate when we combine the results of several different types of measurements? There are many ways to deal with the propagation of uncertainty. We will discuss only one simple method here.
A Simplified Method for Estimating Experimental Uncertainty To illustrate this method, we will consider the determination of the density of an irregu-larly shaped solid. In this determination we make three measurements. First, we measure the mass of the object on a balance. Next, we must obtain the volume of the solid. The easiest method for doing this is to partially fill a graduated cylinder with a liquid and record the volume. Then we add the solid and record the volume again. The difference in the measured volumes is the volume of the solid. We can then calculate the density of the solid from the equation where M is the mass of the solid, V1 is the initial volume of liquid in the graduated cylinder, and V2 is the volume of liquid plus solid. Suppose we get the following results: The calculated density is Now suppose that the precision of the balance used is 0.02 g and that the volume measurements are precise to 0.05 mL. How do we estimate the uncertainty of the den-sity? We can do this by assuming a worst case. That is, we assume the largest uncertain-ties in all measurements, and see what combinations of measurements will give the largest and smallest possible results (the greatest range). Since the density is the mass divided by the volume, the largest value of the density will be that obtained using the largest possi-ble mass and the smallest possible volume: Largest possible mass 23.06 .02 o p r Smallest possible V2 Largest possible V1 The smallest value of the density is Smallest possible mass o p r Largest possible V2 Smallest possible V1 Thus the calculated range is from 7.20 to 7.69 and the average of these values is 7.44.
The error limit is the number that gives the high and low range values when added and subtracted from the average. Therefore, we can express the density as 7.44 0.25 g/mL, which is the average value plus or minus the quantity that gives the range calculated by assuming the largest uncertainties.
Dmin 23.04 13.35 10.35 7.20 g/mL Dmax 23.08 13.45 10.45 7.69 g/mL 23.06 g 13.5 mL 10.4 mL 7.44 g/mL V2 13.5 mL V1 10.4 mL M 23.06 g D M V2 V1 Appendixes A13 Analysis of the propagation of uncertainties is useful in drawing qualitative conclu-sions from the analysis of measurements. For example, suppose that we obtained the pre-ceding results for the density of an unknown alloy and we want to know if it is one of the following alloys: Alloy A: D 7.58 g/mL Alloy B: D 7.42 g/mL Alloy C: D 8.56 g/mL We can safely conclude that the alloy is not C. But the values of the densities for alloys A and B are both within the inherent uncertainty of our method. To distinguish between A and B, we need to improve the precision of our determination: The obvious choice is to improve the precision of the volume measurement.
The worst-case method is very useful in estimating uncertainties when the results of several measurements are combined to calculate a result. We assume the maximum un-certainty in each measurement and calculate the minimum and maximum possible result.
These extreme values describe the range and thus the error limit.
Appendix Two The Quantitative Kinetic Molecular Model We have seen that the kinetic molecular model successfully accounts for the properties of an ideal gas. This appendix will show in some detail how the postulates of the kinetic molecular model lead to an equation corresponding to the experimentally obtained ideal gas equation.
Recall that the particles of an ideal gas are assumed to be volumeless, to have no at-traction for each other, and to produce pressure on their container by colliding with the container walls.
Suppose there are n moles of an ideal gas in a cubical container with sides each of length L. Assume each gas particle has a mass m and that it is in rapid, random, straight-line motion colliding with the walls, as shown in Fig. A.3. The collisions will be assumed to be elastic—no loss of kinetic energy occurs. We want to compute the force on the walls from the colliding gas particles and then, since pressure is force per unit area, to obtain an expression for the pressure of the gas.
Before we can derive the expression for the pressure of a gas, we must first discuss some characteristics of velocity. Each particle in the gas has a particular velocity u that can be divided into components ux, uy, and uz, as shown in Fig. A.4. First, using ux and uy and the Pythagorean theorem, we can obtain uxy as shown in Fig. A.4(c): p r r Hypotenuse of Sides of right triangle right triangle Then, constructing another triangle as shown in Fig. A.4(c), we find h or Now let’s consider how an “average” gas particle moves. For example, how often does this particle strike the two walls of the box that are perpendicular to the x axis? It is impor-tant to realize that only the x component of the velocity affects the particle’s impacts on these two walls, as shown in Fig. A.5(a). The larger the x component of the velocity, the faster the particle travels between these two walls, and the more impacts per unit of time it will make on these walls. Remember, the pressure of the gas is due to these collisions with the walls.
u2 ux 2 uy 2 uz 2 u2 uxy 2 uz 2 uxy 2 ux 2 uy 2 L L L FIGURE A.3 An ideal gas particle in a cube whose sides are of length L. The particle collides elastically with the walls in a random, straight-line motion.
⎧ ⎪ ⎨ ⎪ ⎩ A14 Appendixes The collision frequency (collisions per unit of time) with the two walls that are perpendicular to the x axis is given by Next, what is the force of a collision? Force is defined as mass times acceleration (change in velocity per unit of time): where F represents force, a represents acceleration, u represents a change in velocity, and t represents a given length of time.
Since we assume that the particle has constant mass, we can write The quantity mu is the momentum of the particle (momentum is the product of mass and velocity), and the expression F (mu) t implies that force is the change in momen-tum per unit of time. When a particle hits a wall perpendicular to the x axis, as shown in Fig. A.5(b), an elastic collision results in an exact reversal of the x component of veloc-ity. That is, the sign, or direction, of ux reverses when the particle collides with one of the walls perpendicular to the x axis. Thus the final momentum is the negative, or opposite, of the initial momentum. Remember that an elastic collision means that there is no change in the magnitude of the velocity. The change in momentum in the x direction is then p r Final Initial momentum momentum in x direction in x direction 2mux mux mux Change in momentum ¢1mux2 final momentum initial momentum F m¢u ¢t ¢1mu2 ¢t F ma ma ¢u ¢t b ux L 1Collision frequency2x velocity in the x direction distance between the walls (a) z x y L (b) L u uz L uy ux (c) uz uz uy uy uxy u ux FIGURE A.4 (a) The Cartesian coordinate axes.
(b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, ux, uy, and uz. This can be represented as a rectangular solid with sides ux, uy, and uz and body diagonal u.
(c) In the xy plane, by the Pythagorean theorem. Since uxy and ux are also perpendicular, u2 uxy 2 uz 2 ux 2 uy 2 uz 2 ux 2 uy 2 uxy 2 L x z (a) (b) L L ux u x z ux −ux u −u FIGURE A.5 (a) Only the x component of the gas particle’s velocity affects the frequency of impacts on the shaded walls, the walls that are perpendicular to the x axis.
(b) For an elastic collision, there is an exact reversal of the x component of the velocity and of the total velocity. The change in momentum (final initial) is then mux mux 2mux Appendixes A15 But we are interested in the force the gas particle exerts on the walls of the box. Since we know that every action produces an equal but opposite reaction, the change in mo-mentum with respect to the wall on impact is (2mux), or 2mux.
Recall that since force is the change in momentum per unit of time, for the walls perpendicular to the x axis.
This expression can be obtained by multiplying the change in momentum per im-pact by the number of impacts per unit of time: p r Change in momentum Impacts per per impact unit of time That is, So far we have considered only the two walls of the box perpendicular to the x axis.
We can assume that the force on the two walls perpendicular to the y axis is given by and that on the two walls perpendicular to the z axis by Since we have shown that the total force on the box is Now since we want the average force, we use the average of the square of the veloc-ity to obtain Next, we need to compute the pressure (force per unit of area) p r The 6 sides Area of of the cube each side 2mu2 L 6L2 mu2 3L3 Pressure due to “average” particle force TOTAL areaTOTAL Force TOTAL 2m L 1u22 1u22 2m L 1ux 2 uy 2 uz 22 2m L 1u22 2mux 2 L 2muy 2 L 2muz 2 L Force TOTAL forcex forcey forcez u2 ux 2 uy 2 uz 2 Forcez 2muz 2 L Forcey 2muy 2 L Forcex 2mux 2 L Forcex 12mux2aux L b change in momentum per unit of time Forcex ¢1mux2 ¢t A16 Appendixes Since the volume V of the cube is equal to L3, we can write So far we have considered the pressure on the walls due to a single, “average” par-ticle. Of course, we want the pressure due to the entire gas sample. The number of parti-cles in a given gas sample can be expressed as follows: where n is the number of moles and NA is Avogadro’s number.
The total pressure on the box due to n moles of a gas is therefore Next we want to express the pressure in terms of the kinetic energy of the gas mol-ecules. Kinetic energy (the energy due to motion) is given by where m is the mass and u is the velocity. Since we are using the average of the velocity squared and since we have or Thus, based on the postulates of the kinetic molecular model, we have been able to derive an equation that has the same form as the ideal gas equation, This agreement between experiment and theory supports the validity of the assumptions made in the kinetic molecular model about the behavior of gas particles, at least for the limiting case of an ideal gas.
PV n RT PV n a2 3bNA11 2mu22 P a2 3b nNA11 2mu22 V mu2 211 2mu22, 1u22, 1 2mu2, P nNA mu2 3V Number of gas particles nNA Pressure P mu2 3V Appendix Three Spectral Analysis Although volumetric and gravimetric analyses are still very commonly used, spectroscopy is the technique most often used for modern chemical analysis. Spectroscopy is the study of elec-tromagnetic radiation emitted or absorbed by a given chemical species. Since the quantity of radiation absorbed or emitted can be related to the quantity of the absorbing or emitting species present, this technique can be used for quantitative analysis. There are many spectroscopic techniques, as electromagnetic radiation spans a wide range of energies to include X rays, ul-traviolet, infrared, and visible light, and microwaves, to name a few of its familiar forms. We will consider here only one procedure, which is based on the absorption of visible light.
If a liquid is colored, it is because some component of the liquid absorbs visible light.
In a solution the greater the concentration of the light-absorbing substance, the more light absorbed, and the more intense the color of the solution.
The quantity of light absorbed by a substance can be measured by a spectropho-tometer, shown schematically in Fig. A.6. This instrument consists of a source that emits all wavelengths of light in the visible region (wavelengths of 400 to 700 nm); a mono-chromator, which selects a given wavelength of light; a sample holder for the solution Appendixes A17 being measured; and a detector, which compares the intensity of incident light I0 to the intensity of light after it has passed through the sample I. The ratio II0, called the trans-mittance, is a measure of the fraction of light that passes through the sample. The amount of light absorbed is given by the absorbance A, where The absorbance can be expressed by the Beer–Lambert law: where is the molar absorptivity or the molar extinction coefficient (in L/mol cm), l is the distance the light travels through the solution (in cm), and c is the concentration of the absorbing species (in mol/L). The Beer–Lambert law is the basis for using spectroscopy in quantitative analysis. If and l are known, measuring A for a solution allows us to cal-culate the concentration of the absorbing species in the solution.
Suppose we have a pink solution containing an unknown concentration of Co2(aq) ions. A sample of this solution is placed in a spectrophotometer, and the absorbance is measured at a wavelength where for Co2(aq) is known to be 12 L/mol cm. The ab-sorbance A is found to be 0.60. The width of the sample tube is 1.0 cm. We want to de-termine the concentration of Co2(aq) in the solution. This problem can be solved by a straightforward application of the Beer–Lambert law, where Solving for the concentration gives To obtain the unknown concentration of an absorbing species from the measured ab-sorbance, we must know the product l, since c A l c A l 0.60 a12 L mol cmb11.0 cm2 5.0 102 mol/L l light path 1.0 cm 12 L mol cm A 0.60 A lc A lc A log I I0 Source Monochromator Sample Detector I0 I l FIGURE A.6 A schematic diagram of a simple spectrophotometer. The source emits all wavelengths of visible light, which are dispersed using a prism or grating and then focused, one wavelength at a time, onto the sample. The detector compares the intensity of the incident light (I0) to the intensity of the light after it has passed through the sample (l).
A18 Appendixes We can obtain the product l by measuring the absorbance of a solution of known con-centration, since Measured using a o spectrophotometer r Known from making up the solution However, a more accurate value of the product l can be obtained by plotting A versus c for a series of solutions. Note that the equation A lc gives a straight line with slope l when A is plotted against c.
For example, consider the following typical spectroscopic analysis. A sample of steel from a bicycle frame is to be analyzed to determine its manganese content. The proce-dure involves weighing out a sample of the steel, dissolving it in strong acid, treating the resulting solution with a very strong oxidizing agent to convert all the manganese to per-manganate ion (MnO4 ), and then using spectroscopy to determine the concentration of the intensely purple MnO4 ions in the solution. To do this, however, the value of l for MnO4 must be determined at an appropriate wavelength. The absorbance values for four solutions with known MnO4 concentrations were measured to give the following data: Concentration of Solution MnO4 (mol/L) Absorbance 1 7.00 105 0.175 2 1.00 104 0.250 3 2.00 104 0.500 4 3.50 104 0.875 A plot of absorbance versus concentration for the solutions of known concentration is shown in Fig. A.7. The slope of this line (change in Achange in c) is 2.48 103 L/mol. This quan-tity represents the product l.
A sample of the steel weighing 0.1523 g was dissolved and the unknown amount of manganese was converted to MnO4 ions. Water was then added to give a solution with a final volume of 100.0 mL. A portion of this solution was placed in a spectrophotometer, l A c 1.0 × 10– 4 2.0 × 10– 4 Concentration (mol/L) 3.0 × 10– 4 0.10 0 0.20 0.30 0.40 Absorbance 0.50 0.60 0.70 0.80 0.780 Slope = 0.90 1.00 3.15 × 10– 4 0.558 2.25 × 10– 4 = 2.48 × 103 C = 2.25 × 10– 4 A = 0.558 FIGURE A.7 A plot of absorbance versus concentration of MnO4 in a series of solutions of known concentration.
Appendixes A19 and its absorbance was found to be 0.780. Using these data, we want to calculate the per-cent manganese in the steel. The MnO4 ions from the manganese in the dissolved steel sample show an absorbance of 0.780. Using the Beer–Lambert law, we calculate the con-centration of MnO4 in this solution: There is a more direct way for finding c. Using a graph such as that in Fig. A.7 (of-ten called a Beer’s law plot), we can read the concentration that corresponds to A 0.780. This interpolation is shown by dashed lines on the graph. By this method, c 3.15 104 mol/L, which agrees with the value obtained above.
Recall that the original 0.1523-g steel sample was dissolved, the manganese was con-verted to permanganate, and the volume was adjusted to 100.0 mL. We now know that [MnO4 ] in that solution is 3.15 104 M. Using this concentration, we can calculate the total number of moles of MnO4 in that solution: Since each mole of manganese in the original steel sample yields a mole of MnO4 , that is, the original steel sample must have contained 3.15 105 mol of manganese. The mass of manganese present in the sample is Since the steel sample weighed 0.1523 g, the present manganese in the steel is This example illustrates a typical use of spectroscopy in quantitative analysis. The steps commonly involved are as follows: 1. Preparation of a calibration plot (a Beer’s law plot) by measuring the absorbance values of a series of solutions with known concentrations.
2. Measurement of the absorbance of the solution of unknown concentration.
3. Use of the calibration plot to determine the unknown concentration.
1.73 103 g of Mn 1.523 101 g of sample 100 1.14% 3.15 105 mol of Mn 54.938 g of Mn 1 mol of Mn 1.73 103 g of Mn 1 mol of MnO4 1 mol of Mn 3.15 105mol mol of MnO4 100.0 mL 1 L 1000 mL 3.15 104 mol L c A l 0.780 2.48 103 L/mol 3.15 104 mol/L Oxidation 8 8888888n Appendix Four Selected Thermodynamic Data Note: All values are assumed precise to at least 1.
Substance and State Aluminum Al(s) 0 0 28 Al2O3(s) 1676 1582 51 Al(OH)3(s) 1277 AlCl3(s) 704 629 111 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 Substance and State Barium Ba(s) 0 0 67 BaCO3(s) 1219 1139 112 BaO(s) 582 552 70 Ba(OH)2(s) 946 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 (continued) A20 Appendixes Appendix Four (continued) Substance and State Barium, continued BaSO4(s) 1465 1353 132 Beryllium Be(s) 0 0 10 BeO(s) 599 569 14 Be(OH)2(s) 904 815 47 Bromine Br2(l) 0 0 152 Br2(g) 31 3 245 Br2(aq) 3 4 130 Br(aq) 121 104 82 HBr(g) 36 53 199 Cadmium Cd(s) 0 0 52 CdO(s) 258 228 55 Cd(OH)2(s) 561 474 96 CdS(s) 162 156 65 CdSO4(s) 935 823 123 Calcium Ca(s) 0 0 41 CaC2(s) 63 68 70 CaCO3(s) 1207 1129 93 CaO(s) 635 604 40 Ca(OH)2(s) 987 899 83 Ca3(PO4)2(s) 4126 3890 241 CaSO4(s) 1433 1320 107 CaSiO3(s) 1630 1550 84 Carbon C(s) (graphite) 0 0 6 C(s) (diamond) 2 3 2 CO(g) 110.5 137 198 CO2(g) 393.5 394 214 CH4(g) 75 51 186 CH3OH(g) 201 163 240 CH3OH(l) 239 166 127 H2CO(g) 116 110 219 HCOOH(g) 363 351 249 HCN(g) 135.1 125 202 C2H2(g) 227 209 201 C2H4(g) 52 68 219 CH3CHO(g) 166 129 250 C2H5OH(l) 278 175 161 C2H6(g) 84.7 32.9 229.5 C3H6(g) 20.9 62.7 266.9 C3H8(g) 104 24 270 C2H4O(g) (ethylene oxide) 53 13 242 CH2PCHCN(g) 185.0 195.4 274 CH3COOH(l) 484 389 160 C6H12O6(s) 1275 911 212 CCl4 135 65 216 Chlorine Cl2(g) 0 0 223 Cl2(aq) 23 7 121 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 Substance and State Chlorine, continued Cl(aq) 167 131 57 HCl(g) 92 95 187 Chromium Cr(s) 0 0 24 Cr2O3(s) 1128 1047 81 CrO3(s) 579 502 72 Copper Cu(s) 0 0 33 CuCO3(s) 595 518 88 Cu2O(s) 170 148 93 CuO(s) 156 128 43 Cu(OH)2(s) 450 372 108 CuS(s) 49 49 67 Fluorine F2(g) 0 0 203 F(aq) 333 279 14 HF(g) 271 273 174 Hydrogen H2(g) 0 0 131 H(g) 217 203 115 H(aq) 0 0 0 OH(aq) 230 157 11 H2O(l) 286 237 70 H2O(g) 242 229 189 Iodine I2(s) 0 0 116 I2(g) 62 19 261 I2(aq) 23 16 137 I(aq) 55 52 106 Iron Fe(s) 0 0 27 Fe3C(s) 21 15 108 Fe0.95O(s) (wustite) 264 240 59 FeO 272 255 61 Fe3O4(s) (magnetite) 1117 1013 146 Fe2O3(s) (hematite) 826 740 90 FeS(s) 95 97 67 FeS2(s) 178 166 53 FeSO4(s) 929 825 121 Lead Pb(s) 0 0 65 PbO2(s) 277 217 69 PbS(s) 100 99 91 PbSO4(s) 920 813 149 Magnesium Mg(s) 0 0 33 MgCO3(s) 1113 1029 66 MgO(s) 602 569 27 Mg(OH)2(s) 925 834 64 Manganese Mn(s) 0 0 32 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 Appendixes A21 Appendix Four (continued) Substance and State Manganese, continued MnO(s) 385 363 60 Mn3O4(s) 1387 1280 149 Mn2O3(s) 971 893 110 MnO2(s) 521 466 53 MnO4 (aq) 543 449 190 Mercury Hg(l) 0 0 76 Hg2Cl2(s) 265 211 196 HgCl2(s) 230 184 144 HgO(s) 90 59 70 HgS(s) 58 49 78 Nickel Ni(s) 0 0 30 NiCl2(s) 316 272 107 NiO(s) 241 213 38 Ni(OH)2(s) 538 453 79 NiS(s) 93 90 53 Nitrogen N2(g) 0 0 192 NH3(g) 46 17 193 NH3(aq) 80 27 111 NH4 (aq) 132 79 113 NO(g) 90 87 211 NO2(g) 34 52 240 N2O(g) 82 104 220 N2O4(g) 10 98 304 N2O4(l) 20 97 209 N2O5(s) 42 134 178 N2H4(l) 51 149 121 N2H3CH3(l) 54 180 166 HNO3(aq) 207 111 146 HNO3(l) 174 81 156 NH4ClO4(s) 295 89 186 NH4Cl(s) 314 203 96 Oxygen O2(g) 0 0 205 O(g) 249 232 161 O3(g) 143 163 239 Phosphorus P(s) (white) 0 0 41 P(s) (red) 18 12 23 P(s) (black) 39 33 23 P4(g) 59 24 280 PF5(g) 1578 1509 296 PH3(g) 5 13 210 H3PO4(s) 1279 1119 110 H3PO4(l) 1267 — — H3PO4(aq) 1288 1143 158 P4O10(s) 2984 2698 229 Potassium K(s) 0 0 64 KCl(s) 436 408 83 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 Substance and State Potassium, continued KClO3(s) 391 290 143 KClO4(s) 433 304 151 K2O(s) 361 322 98 K2O2(s) 496 430 113 KO2(s) 283 238 117 KOH(s) 425 379 79 KOH(aq) 481 440 9.20 Silicon SiO2(s) (quartz) 911 856 42 SiCl4(l) 687 620 240 Silver Ag(s) 0 0 43 Ag(aq) 105 77 73 AgBr(s) 100 97 107 AgCN(s) 146 164 84 AgCl(s) 127 110 96 Ag2CrO4(s) 712 622 217 AgI(s) 62 66 115 Ag2O(s) 31 11 122 Ag2S(s) 32 40 146 Sodium Na(s) 0 0 51 Na(aq) 240 262 59 NaBr(s) 360 347 84 Na2CO3(s) 1131 1048 136 NaHCO3(s) 948 852 102 NaCl(s) 411 384 72 NaH(s) 56 33 40 NaI(s) 288 282 91 NaNO2(s) 359 NaNO3(s) 467 366 116 Na2O(s) 416 377 73 Na2O2(s) 515 451 95 NaOH(s) 427 381 64 NaOH(aq) 470 419 50 Sulfur S(s) (rhombic) 0 0 32 S(s) (monoclinic) 0.3 0.1 33 S2(aq) 33 86 15 S8(g) 102 50 431 SF6(g) 1209 1105 292 H2S(g) 21 34 206 SO2(g) 297 300 248 SO3(g) 396 371 257 SO4 2(aq) 909 745 20 H2SO4(l) 814 690 157 H2SO4(aq) 909 745 20 Tin Sn(s) (white) 0 0 52 Sn(s) (gray) 2 0.1 44 SnO(s) 285 257 56 SnO2(s) 581 520 52 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 (continued) Equilibrium Constants and Appendix Five Reduction Potentials A22 Appendixes Appendix Four (continued) Substance and State Tin, continued Sn(OH)2(s) 561 492 155 Titanium TiCl4(g) 763 727 355 TiO2(s) 945 890 50 Uranium U(s) 0 0 50 UF6(s) 2137 2008 228 UF6(g) 2113 2029 380 UO2(s) 1084 1029 78 U3O8(s) 3575 3393 282 UO3(s) 1230 1150 99 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 Substance and State Xenon Xe(g) 0 0 170 XeF2(g) 108 48 254 XeF4(s) 251 121 146 XeF6(g) 294 XeO3(s) 402 Zinc Zn(s) 0 0 42 ZnO(s) 348 318 44 Zn(OH)2(s) 642 ZnS(s) (wurtzite) 193 ZnS(s) (zinc blende) 206 201 58 ZnSO4(s) 983 874 120 S 1J/K mol2 ¢G f 1kJ/mol2 ¢H f 1kJ/mol2 A5.1 Values of Ka for Some Common Monoprotic Acids Name Formula Value of Ka Hydrogen sulfate ion HSO4 1.2 102 Chlorous acid HClO2 1.2 102 Monochloracetic acid HC2H2ClO2 1.35 103 Hydrofluoric acid HF 7.2 104 Nitrous acid HNO2 4.0 104 Formic acid HCO2H 1.8 104 Lactic acid HC3H5O3 1.38 104 Benzoic acid HC7H5O2 6.4 105 Acetic acid HC2H3O2 1.8 105 Hydrated aluminum(III) ion [Al(H2O)6]3 1.4 105 Propanoic acid HC3H5O2 1.3 105 Hypochlorous acid HOCl 3.5 108 Hypobromous acid HOBr 2 109 Hydrocyanic acid HCN 6.2 1010 Boric acid H3BO3 5.8 1010 Ammonium ion NH4 5.6 1010 Phenol HOC6H5 1.6 1010 Hypoiodous acid HOI 2 1011 Appendixes A23 A5.2 Stepwise Dissociation Constants for Several Common Polyprotic Acids Name Formula Ka1 Ka2 Ka3 Phosphoric acid H3PO4 7.5 103 6.2 108 4.8 1013 Arsenic acid H3AsO4 5 103 8 108 6 1010 Carbonic acid H2CO3 4.3 107 5.6 1011 Sulfuric acid H2SO4 Large 1.2 102 Sulfurous acid H2SO3 1.5 102 1.0 107 Hydrosulfuric acid H2S 1.0 107 1019 Oxalic acid H2C2O4 6.5 102 6.1 105 Ascorbic acid H2C6H6O6 7.9 105 1.6 1012 (vitamin C) Citric acid H3C6H5O7 8.4 104 1.8 105 4.0 106 A5.3 Values of Kb for Some Common Weak Bases Conjugate Name Formula Acid Kb Ammonia NH3 NH4 1.8 105 Methylamine CH3NH2 CH3NH3 4.38 104 Ethylamine C2H5NH2 C2H5NH3 5.6 104 Diethylamine (C2H5)2NH (C2H5)2NH2 1.3 103 Triethylamine (C2H5)3N (C2H5)3NH 4.0 104 Hydroxylamine HONH2 HONH3 1.1 108 Hydrazine H2NNH2 H2NNH3 3.0 106 Aniline C6H5NH2 C6H5NH3 3.8 1010 Pyridine C5H5N C5H5NH 1.7 109 A24 Appendixes A5.4 Ksp Values at 25°C for Common Ionic Solids Ionic Solid Ksp (at 25°C) Ionic Solid Ksp (at 25°C) Ionic Solid Ksp (at 25°C) Fluorides Hg2CrO4 2 109 Co(OH)2 2.5 1016 BaF2 2.4 105 BaCrO4 8.5 1011 Ni(OH)2 1.6 1016 MgF2 6.4 109 Ag2CrO4 9.0 1012 Zn(OH)2 4.5 1017 PbF2 4 108 PbCrO4 2 1016 Cu(OH)2 1.6 1019 SrF2 7.9 1010 Hg(OH)2 3 1026 CaF2 4.0 1011 Carbonates Sn(OH)2 3 1027 NiCO3 1.4 107 Cr(OH)3 6.7 1031 Chlorides CaCO3 8.7 109 Al(OH)3 2 1032 PbCl2 1.6 105 BaCO3 1.6 109 Fe(OH)3 4 1038 AgCl 1.6 1010 SrCO3 7 1010 Co(OH)3 2.5 1043 Hg2Cl2 1.1 1018 CuCO3 2.5 1010 ZnCO3 2 1010 Sulfides Bromides MnCO3 8.8 1011 MnS 2.3 1013 PbBr2 4.6 106 FeCO3 2.1 1011 FeS 3.7 1019 AgBr 5.0 1013 Ag2CO3 8.1 1012 NiS 3 1021 Hg2Br2 1.3 1022 CdCO3 5.2 1012 CoS 5 1022 Iodides PbCO3 1.5 1015 ZnS 2.5 1022 PbI2 1.4 108 MgCO3 1 105 SnS 1 1026 AgI 1.5 1016 Hg2CO3 9.0 1015 CdS 1.0 1028 Hg2I2 4.5 1029 PbS 7 1029 Hydroxides CuS 8.5 1045 Sulfates Ba(OH)2 5.0 103 Ag2S 1.6 1049 CaSO4 6.1 105 Sr(OH)2 3.2 104 HgS 1.6 1054 Ag2SO4 1.2 105 Ca(OH)2 1.3 106 SrSO4 3.2 107 AgOH 2.0 108 Phosphates PbSO4 1.3 108 Mg(OH)2 8.9 1012 Ag3PO4 1.8 1018 BaSO4 1.5 109 Mn(OH)2 2 1013 Sr3(PO4)2 1 1031 Cd(OH)2 5.9 1015 Ca3(PO4)2 1.3 1032 Chromates Pb(OH)2 1.2 1015 Ba3(PO4)2 6 1039 SrCrO4 3.6 105 Fe(OH)2 1.8 1015 Pb3(PO4)2 1 1054 Contains Hg2 2 ions. Ksp [Hg2 2][X]2 for Hg2X2 salts.
Appendixes A25 A5.5 Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions Half-Reaction e° (V) Half-Reaction e° (V) F2 2e n 2F 2.87 O2 2H2O 4e n 4OH 0.40 Ag2 e n Ag 1.99 Cu2 2e n Cu 0.34 Co3 e n Co2 1.82 Hg2Cl2 2e n 2Hg 2Cl 0.34 H2O2 2H 2e n 2H2O 1.78 AgCl e n Ag Cl 0.22 Ce4 e n Ce3 1.70 SO4 2 4H 2e n H2SO3 H2O 0.20 PbO2 4H SO4 2 2e n PbSO4 2H2O 1.69 Cu2 e n Cu 0.16 MnO4 4H 3e n MnO2 2H2O 1.68 2H 2e n H2 0.00 2e 2H IO4 n IO3 H2O 1.60 Fe3 3e n Fe 0.036 MnO4 8H 5e n Mn2 4H2O 1.51 Pb2 2e n Pb 0.13 Au3 3e n Au 1.50 Sn2 2e n Sn 0.14 PbO2 4H 2e n Pb2 2H2O 1.46 Ni2 2e n Ni 0.23 Cl2 2e n 2Cl 1.36 PbSO4 2e n Pb SO4 2 0.35 Cr2O7 2 14H 6e n 2Cr3 7H2O 1.33 Cd2 2e n Cd 0.40 O2 4H 4e n 2H2O 1.23 Fe2 2e n Fe 0.44 MnO2 4H 2e n Mn2 2H2O 1.21 Cr3 e n Cr2 0.50 IO3 6H 5e n 1 2 I2 3H2O 1.20 Cr3 3e n Cr 0.73 Br2 2e n 2Br 1.09 Zn2 2e n Zn 0.76 VO2 2H e n VO2 H2O 1.00 2H2O 2e n H2 2OH 0.83 AuCl4 3e n Au 4Cl 0.99 Mn2 2e n Mn 1.18 NO3 4H 3e n NO 2H2O 0.96 Al3 3e n Al 1.66 ClO2 e n ClO2 0.954 H2 2e n 2H 2.23 2Hg2 2e n Hg2 2 0.91 Mg2 2e n Mg 2.37 Ag e n Ag 0.80 La3 3e n La 2.37 Hg2 2 2e n 2Hg 0.80 Na e n Na 2.71 Fe3 e n Fe2 0.77 Ca2 2e n Ca 2.76 O2 2H 2e n H2O2 0.68 Ba2 2e n Ba 2.90 MnO4 e n MnO4 2 0.56 K e n K 2.92 I2 2e n 2I 0.54 Li e n Li 3.05 Cu e n Cu 0.52 A26 Appendixes Appendix Six SI Units and Conversion Factors Length SI unit: meter (m) 1 meter 1.0936 yards 1 centimeter 0.39370 inch 1 inch 2.54 centimeters (exactly) 1 kilometer 0.62137 mile 1 mile 5280 feet 1.6093 kilometers 1 angstrom 1010 meter 100 picometers Mass SI unit: kilogram (kg) 1 kilogram 1000 grams 2.2046 pounds 1 pound 453.59 grams 0.45359 kilogram 16 ounces 1 ton 2000 pounds 907.185 kilograms 1 metric ton 1000 kilograms 2204.6 pounds 1 atomic mass unit 1.66056 1027 kilograms Volume SI unit: cubic meter (m3) 1 liter 103 m3 1 dm3 1.0567 quarts 1 gallon 4 quarts 8 pints 3.7854 liters 1 quart 32 fluid ounces 0.94633 liter Temperature SI unit: kelvin (K) 0 K 273.15°C 459.67°F K °C 273.15 °C (°F 32) °F (°C) 32 9 5 5 9 Energy SI unit: joule (J) 1 joule 1 kg m2/s2 0.23901 calorie 9.4781 104 btu (British thermal unit) 1 calorie 4.184 joules 3.965 103 btu 1 btu 1055.06 joules 252.2 calories Pressure SI unit: pascal (Pa) 1 pascal 1 N/m2 1 kg/m s2 1 atmosphere 101.325 kilopascals 760 torr (mmHg) 14.70 pounds per square inch 1 bar 105 pascals A27 Glossary Accuracy the agreement of a particular value with the true value. (1.4) Acid a substance that produces hydrogen ions in solution; a proton donor. (2.8; 4.2; 4.8) Acid–base indicator a substance that marks the end point of an acid–base titration by changing color. (15.5) Acid dissociation constant (Ka) the equilibrium constant for a reaction in which a proton is removed from an acid by H2O to form the conjugate base and H3O. (14.1) Acid rain a result of air pollution by sulfur dioxide. (5.10) Acidic oxide a covalent oxide that dissolves in water to give an acidic solution. (14.10) Actinide series a group of 14 elements following actinium in the periodic table, in which the 5f orbitals are being filled. (7.11; 19.1) Activated complex (transition state) the arrangement of atoms found at the top of the potential energy barrier as a reaction pro-ceeds from reactants to products. (12.7) Activation energy the threshold energy that must be overcome to produce a chemical reaction. (12.7) Addition polymerization a type of polymerization in which the monomers simply add together to form the polymer, with no other products. (22.5) Addition reaction a reaction in which atoms add to a carbon– carbon multiple bond. (22.2) Adsorption the collection of one substance on the surface of another.
(12.8) Air pollution contamination of the atmosphere, mainly by the gaseous products of transportation and production of electricity. (5.10) Alcohol an organic compound in which the hydroxyl group is a sub-stituent on a hydrocarbon. (22.4) Aldehyde an organic compound containing the carbonyl group bonded to at least one hydrogen atom. (22.4) Alkali metal a Group 1A metal. (2.7; 19.2) Alkaline earth metal a Group 2A metal. (2.7; 19.4) Alkane a saturated hydrocarbon with the general formula CnH2n 2.
(22.1) Alkene an unsaturated hydrocarbon containing a carbon–carbon double bond. The general formula is CnH2n. (22.2) Alkyne an unsaturated hydrocarbon containing a triple carbon– carbon bond. The general formula is CnH2n2. (22.2) Alloy a substance that contains a mixture of elements and has metal-lic properties. (10.4) Alloy steel a form of steel containing carbon plus other metals such as chromium, cobalt, manganese, and molybdenum. (21.8) Alpha () particle a helium nucleus. (18.1) Alpha-particle production a common mode of decay for radio-active nuclides in which the mass number changes. (18.1) Amine an organic base derived from ammonia in which one or more of the hydrogen atoms are replaced by organic groups. (14.6; 22.4) -Amino acid an organic acid in which an amino group and an R group are attached to the carbon atom next to the carboxyl group. (22.6) Amorphous solid a solid with considerable disorder in its structure.
(10.3) Ampere the unit of electric current equal to one coulomb of charge per second. (17.7) Amphoteric substance a substance that can behave either as an acid or as a base. (14.2) Angular momentum quantum number ( ) the quantum number relating to the shape of an atomic orbital, which can assume any integral value from 0 to n 1 for each value of n. (7.6) Anion a negative ion. (2.6) Anode the electrode in a galvanic cell at which oxidation occurs.
(17.1) Antibonding molecular orbital an orbital higher in energy than the atomic orbitals of which it is composed. (9.2) Aqueous solution a solution in which water is the dissolving medium or solvent. (4) Aromatic hydrocarbon one of a special class of cyclic unsaturated hydrocarbons, the simplest of which is benzene. (22.3) Arrhenius concept a concept postulating that acids produce hydro-gen ions in aqueous solution, while bases produce hydroxide ions.
(14.1) Arrhenius equation the equation representing the rate constant as k AeEaRT, where A represents the product of the collision fre-quency and the steric factor, and eEaRT is the fraction of collisions with sufficient energy to produce a reaction. (12.7) Atactic chain a polymer chain in which the substituent groups such as CH3 are randomly distributed along the chain. (22.5) Atmosphere the mixture of gases that surrounds the earth’s surface.
(5.10) Atomic number the number of protons in the nucleus of an atom.
(2.5; 18) Atomic radius half the distance between the nuclei in a molecule consisting of identical atoms. (7.12) Atomic solid a solid that contains atoms at the lattice points. (10.3) Atomic weight the weighted average mass of the atoms in a naturally occurring element. (2.3) Aufbau principle the principle stating that as protons are added one by one to the nucleus to build up the elements, electrons are simi-larly added to hydrogen-like orbitals. (7.11) Autoionization the transfer of a proton from one molecule to another of the same substance. (14.2) Avogadro’s law equal volumes of gases at the same temperature and pressure contain the same number of particles. (5.2) Avogadro’s number the number of atoms in exactly 12 grams of pure 12C, equal to 6.022 1023. (3.3) Ball-and-stick model a molecular model that distorts the sizes of atoms but shows bond relationships clearly. (2.6) Band model a molecular model for metals in which the electrons are assumed to travel around the metal crystal in molecular orbitals formed from the valence atomic orbitals of the metal atoms. (10.4) / A28 Glossary Carboxyhemoglobin a stable complex of hemoglobin and carbon monoxide that prevents normal oxygen uptake in the blood. (21.7) Carboxyl group the OCOOH group in an organic acid. (14.2; 22.4) Carboxylic acid an organic compound containing the carboxyl group; an acid with the general formula RCOOH. (22.4) Catalyst a substance that speeds up a reaction without being con-sumed. (12.8) Cathode the electrode in a galvanic cell at which reduction occurs.
(17.1) Cathode rays the “rays” emanating from the negative electrode (cath-ode) in a partially evacuated tube; a stream of electrons. (2.4) Cathodic protection a method in which an active metal, such as mag-nesium, is connected to steel to protect it from corrosion. (17.6) Cation a positive ion. (2.6) Cell potential (electromotive force) the driving force in a galvanic cell that pulls electrons from the reducing agent in one compart-ment to the oxidizing agent in the other. (17.1) Ceramic a nonmetallic material made from clay and hardened by firing at high temperature; it contains minute silicate crystals suspended in a glassy cement. (10.5) Chain reaction (nuclear) a self-sustaining fission process caused by the production of neutrons that proceed to split other nuclei. (18.6) Charles’s law the volume of a given sample of gas at constant pres-sure is directly proportional to the temperature in kelvins. (5.2) Chelating ligand (chelate) a ligand having more than one atom with a lone pair that can be used to bond to a metal ion. (21.3) Chemical bond the force or, more accurately, the energy, that holds two atoms together in a compound. (2.6) Chemical change the change of substances into other substances through a reorganization of the atoms; a chemical reaction. (1.9) Chemical equation a representation of a chemical reaction show-ing the relative numbers of reactant and product molecules. (3.7) Chemical equilibrium a dynamic reaction system in which the con-centrations of all reactants and products remain constant as a func-tion of time. (13) Chemical formula the representation of a molecule in which the symbols for the elements are used to indicate the types of atoms present and subscripts are used to show the relative numbers of atoms. (2.6) Chemical kinetics the area of chemistry that concerns reaction rates.
(12) Chemical stoichiometry the calculation of the quantities of material consumed and produced in chemical reactions. (3) Chirality the quality of having nonsuperimposable mirror images.
(21.4) Chlor–alkali process the process for producing chlorine and sodium hydroxide by electrolyzing brine in a mercury cell. (17.8) Chromatography the general name for a series of methods for sep-arating mixtures by employing a system with a mobile phase and a stationary phase. (1.9) Coagulation the destruction of a colloid by causing particles to ag-gregate and settle out. (11.8) Codons organic bases in sets of three that form the genetic code. (22.6) Colligative properties properties of a solution that depend only on the number, and not on the identity, of the solute particles. (11.5) Collision model a model based on the idea that molecules must collide to react; used to account for the observed characteristics of reaction rates. (12.7) Barometer a device for measuring atmospheric pressure. (5.1) Base a substance that produces hydroxide ions in aqueous solution, a proton acceptor. (4.8) Basic oxide an ionic oxide that dissolves in water to produce a basic solution. (14.10) Basic oxygen process a process for producing steel by oxidizing and removing the impurities in iron using a high-pressure blast of oxy-gen. (21.8) Battery a group of galvanic cells connected in series. (17.5) Beta () particle an electron produced in radioactive decay. (18.1) Beta-particle production a decay process for radioactive nuclides in which the mass number remains constant and the atomic number changes. The net effect is to change a neutron to a proton. (18.1) Bidentate ligand a ligand that can form two bonds to a metal ion.
(21.3) Bimolecular step a reaction involving the collision of two mol-ecules. (12.6) Binary compound a two-element compound. (2.8) Binding energy (nuclear) the energy required to decompose a nucleus into its component nucleons. (18.5) Biomolecule a molecule responsible for maintaining and/or repro-ducing life. (22) Blast furnace a furnace in which iron oxide is reduced to iron metal by using a very strong blast of hot air to produce carbon monoxide from coke, and then using this gas as a reducing agent for the iron.
(21.8) Bond energy the energy required to break a given chemical bond.
(8.1) Bond length the distance between the nuclei of the two atoms con-nected by a bond; the distance where the total energy of a diatomic molecule is minimal. (8.1) Bond order the difference between the number of bonding electrons and the number of antibonding electrons, divided by two. It is an index of bond strength. (9.2) Bonding molecular orbital an orbital lower in energy than the atomic orbitals of which it is composed. (9.2) Bonding pair an electron pair found in the space between two atoms.
(8.9) Borane a covalent hydride of boron. (19.5) Boyle’s law the volume of a given sample of gas at constant temper-ature varies inversely with the pressure. (5.2) Breeder reactor a nuclear reactor in which fissionable fuel is produced while the reactor runs. (18.6) Brønsted–Lowry model a model proposing that an acid is a proton donor, and a base is a proton acceptor. (14.1) Buffered solution a solution that resists a change in its pH when either hydroxide ions or protons are added. (15.2) Buffering capacity the ability of a buffered solution to absorb pro-tons or hydroxide ions without a significant change in pH; deter-mined by the magnitudes of [HA] and [A] in the solution. (15.3) Calorimetry the science of measuring heat flow. (6.2) Capillary action the spontaneous rising of a liquid in a narrow tube.
(10.2) Carbohydrate a polyhydroxyl ketone or polyhydroxyl aldehyde or a polymer composed of these. (22.6) Carbon steel an alloy of iron containing up to about 1.5% carbon.
(21.8) Glossary A29 Critical mass the mass of fissionable material required to produce a self-sustaining chain reaction. (18.6) Critical point the point on a phase diagram at which the tempera-ture and pressure have their critical values; the end point of the liquid–vapor line. (10.9) Critical pressure the minimum pressure required to produce lique-faction of a substance at the critical temperature. (10.9) Critical reaction (nuclear) a reaction in which exactly one neutron from each fission event causes another fission event, thus sustain-ing the chain reaction. (18.6) Critical temperature the temperature above which vapor cannot be liquefied no matter what pressure is applied. (10.9) Crosslinking the existence of bonds between adjacent chains in a polymer, thus adding strength to the material. (22.5) Crystal field model a model used to explain the magnetism and col-ors of coordination complexes through the splitting of the d orbital energies. (21.6) Crystalline solid a solid with a regular arrangement of its components. (10.3) Cubic closest packed (ccp) structure a solid modeled by the clos-est packing of spheres with an abcabc arrangement of layers; the unit cell is face-centered cubic. (10.4) Cyanidation a process in which crushed gold ore is treated with an aqueous cyanide solution in the presence of air to dissolve the gold.
Pure gold is recovered by reduction of the ion to the metal. (21.8) Cyclotron a type of particle accelerator in which an ion introduced at the center is accelerated in an expanding spiral path by the use of alternating electrical fields in the presence of a magnetic field.
(18.3) Cytochromes a series of iron-containing species composed of heme and a protein. Cytochromes are the principal electron-transfer mol-ecules in the respiratory chain. (21.7) Dalton’s law of partial pressures for a mixture of gases in a con-tainer, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. (5.5) Degenerate orbitals a group of orbitals with the same energy. (7.7) Dehydrogenation reaction a reaction in which two hydrogen atoms are removed from adjacent carbons of a saturated hydrocarbon, giv-ing an unsaturated hydrocarbon. (22.1) Denaturation the breaking down of the three-dimensional structure of a protein resulting in the loss of its function. (22.6) Denitrification the return of nitrogen from decomposed matter to the atmosphere by bacteria that change nitrates to nitrogen gas. (20.2) Density a property of matter representing the mass per unit volume. (1.8) Deoxyribonucleic acid (DNA) a huge nucleotide polymer having a double-helical structure with complementary bases on the two strands. Its major functions are protein synthesis and the storage and transport of genetic information. (22.6) Desalination the removal of dissolved salts from an aqueous solution. (11.6) Dialysis a phenomenon in which a semipermeable membrane allows transfer of both solvent molecules and small solute mole-cules and ions. (11.6) Diamagnetism a type of magnetism, associated with paired electrons, that causes a substance to be repelled from the inducing magnetic field. (9.3) Differential rate law an expression that gives the rate of a reaction as a function of concentrations; often called the rate law. (12.2) Colloid (colloidal dispersion) a suspension of particles in a dispers-ing medium. (11.8) Combustion reaction the vigorous and exothermic reaction that takes place between certain substances, particularly organic compounds, and oxygen. (22.1) Common ion effect the shift in an equilibrium position caused by the addition or presence of an ion involved in the equilibrium reaction. (15.1) Complete ionic equation an equation that shows all substances that are strong electrolytes as ions. (4.6) Complex ion a charged species consisting of a metal ion surrounded by ligands. (15.8; 21.1) Compound a substance with constant composition that can be broken down into elements by chemical processes. (1.9) Concentration cell a galvanic cell in which both compartments con-tain the same components, but at different concentrations. (17.4) Condensation the process by which vapor molecules reform a liquid. (10.8) Condensation polymerization a type of polymerization in which the formation of a small molecule, such as water, accompanies the ex-tension of the polymer chain. (22.5) Condensation reaction a reaction in which two molecules are joined, accompanied by the elimination of a water molecule. (20.3) Condensed states of matter liquids and solids. (10.1) Conjugate acid the species formed when a proton is added to a base.
(14.1) Conjugate acid–base pair two species related to each other by the donating and accepting of a single proton. (14.1) Conjugate base what remains of an acid molecule after a proton is lost. (14.1) Continuous spectrum a spectrum that exhibits all the wavelengths of visible light. (7.3) Control rods rods in a nuclear reactor composed of substances that ab-sorb neutrons. These rods regulate the power level of the reactor. (18.6) Coordinate covalent bond a metal–ligand bond resulting from the interaction of a Lewis base (the ligand) and a Lewis acid (the metal ion). (21.3) Coordination compound a compound composed of a complex ion and counter ions sufficient to give no net charge. (21.3) Coordination isomerism isomerism in a coordination compound in which the composition of the coordination sphere of the metal ion varies. (21.4) Coordination number the number of bonds formed between the metal ion and the ligands in a complex ion. (21.3) Copolymer a polymer formed from the polymerization of more than one type of monomer. (22.5) Core electron an inner electron in an atom; one not in the outermost (valence) principal quantum level. (7.11) Corrosion the process by which metals are oxidized in the atmos-phere. (17.6) Coulomb’s law E 2.31 1019 where E is the energy of interaction between a pair of ions, expressed in joules; r is the distance between the ion centers in nm; and Q1 and Q2 are the nu-merical ion charges. (8.1) Counterions anions or cations that balance the charge on the com-plex ion in a coordination compound. (21.3) Covalent bonding a type of bonding in which electrons are shared by atoms. (2.6; 8.1) aQ1Q2 r b, A30 Glossary Element a substance that cannot be decomposed into simpler substances by chemical or physical means. (1.9) Elementary step a reaction whose rate law can be written from its molecularity. (12.6) E mc2 Einstein’s equation proposing that energy has mass; E is energy, m is mass, and c is the speed of light. (7.2) Empirical formula the simplest whole number ratio of atoms in a compound. (3.6) Enantiomers isomers that are nonsuperimposable mirror images of each other. (21.4) Endpoint the point in a titration at which the indicator changes color.
(4.8) Endothermic refers to a reaction where energy (as heat) flows into the system. (6.1) Energy the capacity to do work or to cause heat flow. (6.1) Enthalpy a property of a system equal to E PV, where E is the internal energy of the system, P is the pressure of the system, and V is the volume of the system. At constant pressure the change in enthalpy equals the energy flow as heat. (6.2) Enthalpy (heat) of fusion the enthalpy change that occurs to melt a solid at its melting point. (10.8) Entropy a thermodynamic function that measures randomness or dis-order. (16.1) Enzyme a large molecule, usually a protein, that catalyzes biological reactions. (12.8) Equilibrium constant the value obtained when equilibrium concen-trations of the chemical species are substituted in the equilibrium expression. (13.2) Equilibrium expression the expression (from the law of mass action) obtained by multiplying the product concentrations and di-viding by the multiplied reactant concentrations, with each concentration raised to a power represented by the coefficient in the balanced equation. (13.2) Equilibrium point (thermodynamic definition) the position where the free energy of a reaction system has its lowest possible value. (16.8) Equilibrium position a particular set of equilibrium concentrations.
(13.2) Equivalence point (stoichiometric point) the point in a titration when enough titrant has been added to react exactly with the sub-stance in solution being titrated. (4.9; 15.4) Ester an organic compound produced by the reaction between a car-boxylic acid and an alcohol. (22.4) Exothermic refers to a reaction where energy (as heat) flows out of the system. (6.1) Exponential notation expresses a number as N 10M, a convenient method for representing a very large or very small number and for easily indicating the number of significant figures. (1.5) Faraday a constant representing the charge on one mole of electrons; 96,485 coulombs. (17.3) Filtration a method for separating the components of a mixture con-taining a solid and a liquid. (1.9) First law of thermodynamics the energy of the universe is constant; same as the law of conservation of energy. (6.1) Fission the process of using a neutron to split a heavy nucleus into two nuclei with smaller mass numbers. (18.6) Flotation process a method of separating the mineral particles in an ore from the gangue that depends on the greater wettability of the mineral pieces. (21.8) Diffraction the scattering of light from a regular array of points or lines, producing constructive and destructive interference. (7.2) Diffusion the mixing of gases. (5.7) Dilution the process of adding solvent to lower the concentration of solute in a solution. (4.3) Dimer a molecule formed by the joining of two identical monomers.
(22.5) Dipole–dipole attraction the attractive force resulting when polar molecules line up so that the positive and negative ends are close to each other. (10.1) Dipole moment a property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge. (8.3) Direct reduction furnace a furnace in which iron oxide is reduced to iron metal using milder reaction conditions than in a blast furnace. (21.8) Disaccharide a sugar formed from two monosaccharides joined by a glycoside linkage. (22.6) Disproportionation reaction a reaction in which a given element is both oxidized and reduced. (20.7) Distillation a method for separating the components of a liquid mix-ture that depends on differences in the ease of vaporization of the components. (1.9) Disulfide linkage an SOS bond that stabilizes the tertiary structure of many proteins. (22.6) Double bond a bond in which two pairs of electrons are shared by two atoms. (8.8) Downs cell a cell used for electrolyzing molten sodium chloride.
(17.8) Dry cell battery a common battery used in calculators, watches, ra-dios, and tape players. (17.5) Dual nature of light the statement that light exhibits both wave and particulate properties. (7.2) Effusion the passage of a gas through a tiny orifice into an evacuated chamber. (5.7) Electrical conductivity the ability to conduct an electric current. (4.2) Electrochemistry the study of the interchange of chemical and elec-trical energy. (17) Electrolysis a process that involves forcing a current through a cell to cause a nonspontaneous chemical reaction to occur. (17.7) Electrolyte a material that dissolves in water to give a solution that conducts an electric current. (4.2) Electrolytic cell a cell that uses electrical energy to produce a chem-ical change that would otherwise not occur spontaneously. (17.7) Electromagnetic radiation radiant energy that exhibits wavelike behavior and travels through space at the speed of light in a vacuum. (7.1) Electron a negatively charged particle that moves around the nucleus of an atom. (2.4) Electron affinity the energy change associated with the addition of an electron to a gaseous atom. (7.12) Electron capture a process in which one of the inner-orbital electrons in an atom is captured by the nucleus. (18.1) Electron spin quantum number a quantum number representing one of the two possible values for the electron spin; either or (7.8) Electronegativity the tendency of an atom in a molecule to attract shared electrons to itself. (8.2) 1 2.
1 2 Glossary A31 Ground state the lowest possible energy state of an atom or mole-cule. (7.4) Group (of the periodic table) a vertical column of elements having the same valence electron configuration and showing similar properties. (2.7) Haber process the manufacture of ammonia from nitrogen and hy-drogen, carried out at high pressure and high temperature with the aid of a catalyst. (3.10; 20.2) Half-life (of a radioactive sample) the time required for the number of nuclides in a radioactive sample to reach half of the original value. (18.2) Half-life (of a reactant) the time required for a reactant to reach half of its original concentration. (12.4) Half-reactions the two parts of an oxidation–reduction reaction, one representing oxidation, the other reduction. (4.10; 17.1) Halogen a Group 7A element. (2.7; 20.7) Halogenation the addition of halogen atoms to unsaturated hydro-carbons. (22.2) Hard water water from natural sources that contains relatively large concentrations of calcium and magnesium ions. (19.4) Heat energy transferred between two objects due to a temperature difference between them. (6.1) Heat capacity the amount of energy required to raise the tempera-ture of an object by one degree Celsius. (6.2) Heat of fusion the enthalpy change that occurs to melt a solid at its melting point. (10.8) Heat of hydration the enthalpy change associated with placing gaseous molecules or ions in water; the sum of the energy needed to expand the solvent and the energy released from the solvent–solute interactions. (11.2) Heat of solution the enthalpy change associated with dissolving a solute in a solvent; the sum of the energies needed to expand both solvent and solute in a solution and the energy released from the solvent–solute interactions. (11.2) Heat of vaporization the energy required to vaporize one mole of a liquid at a pressure of one atmosphere. (10.8) Heating curve a plot of temperature versus time for a substance where energy is added at a constant rate. (10.8) Heisenberg uncertainty principle a principle stating that there is a fundamental limitation to how precisely both the position and mo-mentum of a particle can be known at a given time. (7.5) Heme an iron complex. (21.7) Hemoglobin a biomolecule composed of four myoglobin-like units (proteins plus heme) that can bind and transport four oxygen mol-ecules in the blood. (21.7) Henderson–Hasselbalch equation an equation giving the relation-ship between the pH of an acid–base system and the concentrations of base and acid: pH pKa (15.2) Henry’s law the amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. (11.3) Hess’s law in going from a particular set of reactants to a particular set of products, the enthalpy change is the same whether the reac-tion takes place in one step or in a series of steps; in summary, en-thalpy is a state function. (6.3) Heterogeneous equilibrium an equilibrium involving reactants and/or products in more than one phase. (13.4) log a 3base4 3acid4 b.
Formal charge the charge assigned to an atom in a molecule or poly-atomic ion derived from a specific set of rules. (8.12) Formation constant (stability constant) the equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solu-tion. (15.8) Formula equation an equation representing a reaction in solution showing the reactants and products in undissociated form, whether they are strong or weak electrolytes. (4.6) Fossil fuel coal, petroleum, or natural gas; consists of carbon-based molecules derived from decomposition of once-living organisms.
(6.5) Frasch process the recovery of sulfur from underground deposits by melting it with hot water and forcing it to the surface by air pres-sure. (20.6) Free energy a thermodynamic function equal to the enthalpy (H) minus the product of the entropy (S) and the Kelvin temperature (T); G H TS. Under certain conditions the change in free en-ergy for a process is equal to the maximum useful work. (16.4) Free radical a species with an unpaired electron. (22.5) Frequency the number of waves (cycles) per second that pass a given point in space. (7.1) Fuel cell a galvanic cell for which the reactants are continuously sup-plied. (17.5) Functional group an atom or group of atoms in hydrocarbon derivatives that contains elements in addition to carbon and hydrogen. (22.4) Fusion the process of combining two light nuclei to form a heavier, more stable nucleus. (18.6) Galvanic cell a device in which chemical energy from a spontaneous redox reaction is changed to electrical energy that can be used to do work. (17.1) Galvanizing a process in which steel is coated with zinc to prevent corrosion. (17.6) Gamma () ray a high-energy photon. (18.1) Gangue the impurities (such as clay or sand) in an ore. (21.8) Geiger–Müller counter (Geiger counter) an instrument that mea-sures the rate of radioactive decay based on the ions and electrons produced as a radioactive particle passes through a gas-filled cham-ber. (18.4) Gene a given segment of the DNA molecule that contains the code for a specific protein. (22.6) Geometrical (cis–trans) isomerism isomerism in which atoms or groups of atoms can assume different positions around a rigid ring or bond. (21.4; 22.2) Glass an amorphous solid obtained when silica is mixed with other compounds, heated above its melting point, and then cooled rapidly.
(10.5) Glass electrode an electrode for measuring pH from the potential dif-ference that develops when it is dipped into an aqueous solution containing H ions. (17.4) Glycoside linkage a COOOC bond formed between the rings of two cyclic monosaccharides by the elimination of water. (22.6) Graham’s law of effusion the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. (5.7) Greenhouse effect a warming effect exerted by the earth’s atmos-phere (particularly CO2 and H2O) due to thermal energy retained by absorption of infrared radiation. (6.5) A32 Glossary Ion exchange (water softening) the process in which an ion-exchange resin removes unwanted ions (for example, Ca2 and Mg2) and replaces them with Na ions, which do not interfere with soap and detergent action. (19.4) Ion pairing a phenomenon occurring in solution when oppositely charged ions aggregate and behave as a single particle. (11.7) Ion-product (dissociation) constant (Kw) the equilibrium constant for the auto-ionization of water; Kw [H][OH]. At 25°C, Kw equals 1.0 1014. (14.2) Ion-selective electrode an electrode sensitive to the concentration of a particular ion in solution. (17.4) Ionic bonding the electrostatic attraction between oppositely charged ions. (2.6; 8.1) Ionic compound (binary) a compound that results when a metal re-acts with a nonmetal to form a cation and an anion. (8.1) Ionic solid (salt) a solid containing cations and anions that dissolves in water to give a solution containing the separated ions which are mobile and thus free to conduct electrical current. (2.6; 10.3) Irreversible process any real process. When a system undergoes the changes State 1 n State 2 n State 1 by any real pathway, the uni-verse is different than before the cyclic process took place in the system. (16.9) Isoelectronic ions ions containing the same number of electrons. (8.4) Isomers species with the same formula but different properties. (21.4) Isotactic chain a polymer chain in which the substituent groups such as CH3 are all arranged on the same side of the chain. (22.5) Isotonic solutions solutions having identical osmotic pressures.
(11.6) Isotopes atoms of the same element (the same number of protons) with different numbers of neutrons. They have identical atomic numbers but different mass numbers. (2.5; 18) Ketone an organic compound containing the carbonyl group bonded to two carbon atoms. (22.4) Kinetic energy ( mv2) energy due to the motion of an object; depen-dent on the mass of the object and the square of its velocity. (6.1) Kinetic molecular theory (KMT) a model that assumes that an ideal gas is composed of tiny particles (molecules) in constant motion.
(5.6) Lanthanide contraction the decrease in the atomic radii of the lan-thanide series elements, going from left to right in the periodic table.
(21.1) Lanthanide series a group of 14 elements following lanthanum in the periodic table, in which the 4f orbitals are being filled. (7.11; 19.1; 21.1) Lattice a three-dimensional system of points designating the positions of the centers of the components of a solid (atoms, ions, or mole-cules). (10.3) Lattice energy the energy change occurring when separated gaseous ions are packed together to form an ionic solid. (8.5) Law of conservation of energy energy can be converted from one form to another but can be neither created nor destroyed. (6.1) Law of conservation of mass mass is neither created nor destroyed.
(1.2; 2.2) 1 2 Hexagonal closest packed (hcp) structure a structure composed of closest packed spheres with an ababab arrangement of layers; the unit cell is hexagonal. (10.4) Homogeneous equilibrium an equilibrium system where all reac-tants and products are in the same phase. (13.4) Homopolymer a polymer formed from the polymerization of only one type of monomer. (22.5) Hund’s rule the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli exclusion principle in a particular set of degenerate orbitals, with all unpaired electrons having parallel spins. (7.11) Hybrid orbitals a set of atomic orbitals adopted by an atom in a mol-ecule different from those of the atom in the free state. (9.1) Hybridization a mixing of the native orbitals on a given atom to form special atomic orbitals for bonding. (9.1) Hydration the interaction between solute particles and water mole-cules. (4.1) Hydride a binary compound containing hydrogen. The hydride ion, H, exists in ionic hydrides. The three classes of hydrides are co-valent, interstitial, and ionic. (19.3) Hydrocarbon a compound composed of carbon and hydrogen. (22.1) Hydrocarbon derivative an organic molecule that contains one or more elements in addition to carbon and hydrogen. (22.4) Hydrogen bonding unusually strong dipole–dipole attractions that occur among molecules in which hydrogen is bonded to a highly electronegative atom. (10.1) Hydrogenation reaction a reaction in which hydrogen is added, with a catalyst present, to a carbon–carbon multiple bond. (22.2) Hydrohalic acid an aqueous solution of a hydrogen halide. (20.7) Hydrometallurgy a process for extracting metals from ores by use of aqueous chemical solutions. Two steps are involved: selective leaching and selective precipitation. (21.8) Hydronium ion the H3O ion; a hydrated proton. (14.1) Hypothesis one or more assumptions put forth to explain the observed behavior of nature. (1.2) Ideal gas law an equation of state for a gas, where the state of the gas is its condition at a given time; expressed by PV nRT, where P pressure, V volume, n moles of the gas, R the universal gas constant, and T absolute temperature. This equa-tion expresses behavior approached by real gases at high T and low P. (5.3) Ideal solution a solution whose vapor pressure is directly propor-tional to the mole fraction of solvent present. (11.4) Indicator a chemical that changes color and is used to mark the end point of a titration. (4.8; 15.5) Integrated rate law an expression that shows the concentration of a reactant as a function of time. (12.2) Interhalogen compound a compound formed by the reaction of one halogen with another. (20.7) Intermediate a species that is neither a reactant nor a product but that is formed and consumed in the reaction sequence. (12.6) Intermolecular forces relatively weak interactions that occur between molecules. (10.1) Internal energy a property of a system that can be changed by a flow of work, heat or both; E q w, where E is the change in the internal energy of the system, q is heat, and w is work. (6.1) Ion an atom or a group of atoms that has a net positive or negative charge. (2.6) Glossary A33 Mass number the total number of protons and neutrons in the atomic nucleus of an atom. (2.5; 18) Mass percent the percent by mass of a component of a mixture (11.1) or of a given element in a compound. (3.5) Mass spectrometer an instrument used to determine the relative masses of atoms by the deflection of their ions on a magnetic field.
(3.2) Matter the material of the universe. (1.9) Messenger RNA (mRNA) a special RNA molecule built in the cell nucleus that migrates into the cytoplasm and participates in protein synthesis. (22.6) Metal an element that gives up electrons relatively easily and is lus-trous, malleable, and a good conductor of heat and electricity. (2.7) Metalloids (semimetals) elements along the division line in the pe-riodic table between metals and nonmetals. These elements exhibit both metallic and nonmetallic properties. (7.13; 19.1) Metallurgy the process of separating a metal from its ore and prepar-ing it for use. (19.1; 21.8) Millimeters of mercury (mmHg) a unit of pressure, also called a torr, 760 mm Hg 760 torr 101,325 Pa 1 standard atmosphere. (5.1) Mineral a relatively pure compound as found in nature. (21.8) Model (theory) a set of assumptions put forth to explain the observed behavior of matter. The models of chemistry usually involve assumptions about the behavior of individual atoms or molecules.
(1.2) Moderator a substance used in a nuclear reactor to slow down the neutrons. (18.6) Molal boiling-point elevation constant a constant characteristic of a particular solvent that gives the change in boiling point as a func-tion of solution molality; used in molecular weight determinations.
(11.5) Molal freezing-point depression constant a constant characteristic of a particular solvent that gives the change in freezing point as a function of the solution molality; used in molecular weight deter-minations (11.5) Molality the number of moles of solute per kilogram of solvent in a solution. (11.1) Molar heat capacity the energy required to raise the temperature of one mole of a substance by one degree Celsius. (6.2) Molar mass the mass in grams of one mole of molecules or formula units of a substance; also called molecular weight. (3.4) Molar volume the volume of one mole of an ideal gas; equal to 22.42 liters at STP. (5.4) Molarity moles of solute per volume of solution in liters. (4.3; 11.1) Mole (mol) the number equal to the number of carbon atoms in ex-actly 12 grams of pure 12C: Avogadro’s number. One mole repre-sents 6.022 1023 units. (3.3) Mole fraction the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. (5.5; 11.1) Mole ratio (stoichiometry) the ratio of moles of one substance to moles of another substance in a balanced chemical equation. (3.9) Molecular formula the exact formula of a molecule, giving the types of atoms and the number of each type. (3.6) Molecular orbital (MO) model a model that regards a molecule as a collection of nuclei and electrons, where the electrons are assumed to occupy orbitals much as they do in atoms, but having the orbitals extend over the entire molecule. In this model the electrons are as-sumed to be delocalized rather than always located between a given pair of atoms. (9.2; 10.4) Law of definite proportion a given compound always contains ex-actly the same proportion of elements by mass. (2.2) Law of mass action a general description of the equilibrium condi-tion; it defines the equilibrium constant expression. (13.2) Law of multiple proportions a law stating that when two elements form a series of compounds, the ratios of the masses of the second element that combine with one gram of the first element can always be reduced to small whole numbers. (2.2) Leaching the extraction of metals from ores using aqueous chemical solutions. (21.8) Lead storage battery a battery (used in cars) in which the anode is lead, the cathode is lead coated with lead dioxide, and the elec-trolyte is a sulfuric acid solution. (17.5) Le Châtelier’s principle if a change is imposed on a system at equi-librium, the position of the equilibrium will shift in a direction that tends to reduce the effect of that change. (13.7) Lewis acid an electron-pair acceptor. (14.11) Lewis base an electron-pair donor. (14.11) Lewis structure a diagram of a molecule showing how the valence electrons are arranged among the atoms in the molecule. (8.10) Ligand a neutral molecule or ion having a lone pair of electrons that can be used to form a bond to a metal ion; a Lewis base. (21.3) Lime–soda process a water-softening method in which lime and soda ash are added to water to remove calcium and magnesium ions by precipitation. (14.6) Limiting reactant (limiting reagent) the reactant that is completely consumed when a reaction is run to completion. (3.10) Line spectrum a spectrum showing only certain discrete wave-lengths. (7.3) Linear accelerator a type of particle accelerator in which a chang-ing electrical field is used to accelerate a positive ion along a lin-ear path. (18.3) Linkage isomerism isomerism involving a complex ion where the ligands are all the same but the point of attachment of at least one of the ligands differs. (21.4) Liquefaction the transformation of a gas into a liquid. (19.1) Localized electron (LE) model a model which assumes that a mol-ecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. (8.9) London dispersion forces the forces, existing among noble gas atoms and nonpolar molecules, that involve an accidental dipole that in-duces a momentary dipole in a neighbor. (10.1) Lone pair an electron pair that is localized on a given atom; an elec-tron pair not involved in bonding. (8.9) Magnetic quantum number m, the quantum number relating to the orientation of an orbital in space relative to the other orbitals with the same quantum number. It can have integral values between and , including zero. (7.6) Main-group (representative) elements elements in the groups labeled 1A, 2A, 3A, 4A, 5A, 6A, 7A, and 8A in the periodic table. The group number gives the sum of the valence s and p electrons. (7.11; 18.1) Major species the components present in relatively large amounts in a solution. (14.3) Manometer a device for measuring the pressure of a gas in a con-tainer. (5.1) Mass the quantity of matter in an object. (1.3) Mass defect the change in mass occurring when a nucleus is formed from its component nucleons. (18.5) A34 Glossary Nucleon a particle in an atomic nucleus, either a neutron or a proton. (18) Nucleotide a monomer of the nucleic acids composed of a five-carbon sugar, a nitrogen-containing base, and phosphoric acid. (22.6) Nucleus the small, dense center of positive charge in an atom. (2.4) Nuclide the general term applied to each unique atom; represented by A ZX, where X is the symbol for a particular element. (18) Octet rule the observation that atoms of nonmetals tend to form the most stable molecules when they are surrounded by eight electrons (to fill their valence orbitals). (8.10) Open hearth process a process for producing steel by oxidizing and removing the impurities in molten iron using external heat and a blast of air or oxygen. (21.8) Optical isomerism isomerism in which the isomers have opposite ef-fects on plane-polarized light. (21.4) Orbital a specific wave function for an electron in an atom. The square of this function gives the probability distribution for the elec-tron. (7.5) d-Orbital splitting a splitting of the d orbitals of the metal ion in a complex such that the orbitals pointing at the ligands have higher energies than those pointing between the ligands. (21.6) Order (of reactant) the positive or negative exponent, determined by experiment, of the reactant concentration in a rate law. (12.2) Organic acid an acid with a carbon-atom backbone; often contains the carboxyl group. (14.2) Organic chemistry the study of carbon-containing compounds (typ-ically chains of carbon atoms) and their properties. (22) Osmosis the flow of solvent into a solution through a semipermeable membrane. (11.6) Osmotic pressure () the pressure that must be applied to a solution to stop osmosis; MRT. (11.6) Ostwald process a commercial process for producing nitric acid by the oxidation of ammonia. (20.2) Oxidation an increase in oxidation state (a loss of electrons). (4.9; 17.1) Oxidation–reduction (redox) reaction a reaction in which one or more electrons are transferred. (4.9; 17.1) Oxidation states a concept that provides a way to keep track of elec-trons in oxidation–reduction reactions according to certain rules.
(4.9; 21.3) Oxidizing agent (electron acceptor) a reactant that accepts electrons from another reactant. (4.9; 17.1) Oxyacid an acid in which the acidic proton is attached to an oxygen atom. (14.2) Ozone O3, the form of elemental oxygen in addition to the much more common O2. (20.5) Paramagnetism a type of induced magnetism, associated with unpaired electrons, that causes a substance to be attracted into the inducing magnetic field. (9.3) Partial pressures the independent pressures exerted by different gases in a mixture. (5.5) Particle accelerator a device used to accelerate nuclear particles to very high speeds. (18.3) Pascal the SI unit of pressure; equal to newtons per meter squared.
(5.1) Pauli exclusion principle in a given atom no two electrons can have the same set of four quantum numbers. (7.8) Molecular orientations (kinetics) orientations of molecules during col-lisions, some of which can lead to reaction while others cannot. (12.7) Molecular solid a solid composed of neutral molecules at the lattice points. (10.3) Molecular structure the three-dimensional arrangement of atoms in a molecule. (8.13) Molecularity the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism.
(12.6) Molecule a bonded collection of two or more atoms of the same or different elements. (2.6) Monodentate (unidentate) ligand a ligand that can form one bond to a metal ion. (21.3) Monoprotic acid an acid with one acidic proton. (14.2) Monosaccharide (simple sugar) a polyhydroxy ketone or aldehyde containing from three to nine carbon atoms. (22.6) Myoglobin an oxygen-storing biomolecule consisting of a heme com-plex and a proton. (21.7) Natural law a statement that expresses generally observed behavior.
(1.2) Nernst equation an equation relating the potential of an electro-chemical cell to the concentrations of the cell components: (17.4) Net ionic equation an equation for a reaction in solution, where strong electrolytes are written as ions, showing only those compo-nents that are directly involved in the chemical change. (4.6) Network solid an atomic solid containing strong directional covalent bonds. (10.5) Neutralization reaction an acid–base reaction. (4.8) Neutron a particle in the atomic nucleus with mass virtually equal to the proton’s but with no charge. (2.5; 18) Nitrogen cycle the conversion of N2 to nitrogen-containing com-pounds, followed by the return of nitrogen gas to the atmosphere by natural decay processes. (20.2) Nitrogen fixation the process of transforming N2 to nitrogen-containing compounds useful to plants. (20.2) Nitrogen-fixing bacteria bacteria in the root nodules of plants that can convert atmospheric nitrogen to ammonia and other nitrogen-containing compounds useful to plants. (20.2) Noble gas a Group 8A element. (2.7; 20.8) Node an area of an orbital having zero electron probability. (7.7) Nonelectrolyte a substance that, when dissolved in water, gives a non-conducting solution. (4.2) Nonmetal an element not exhibiting metallic characteristics. Chem-ically, a typical nonmetal accepts electrons from a metal. (2.7) Normal boiling point the temperature at which the vapor pressure of a liquid is exactly one atmosphere. (10.8) Normal melting point the temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure on the system is one atmosphere. (10.8) Normality the number of equivalents of a substance dissolved in a liter of solution. (11.1) Nuclear atom an atom having a dense center of positive charge (the nucleus) with electrons moving around the outside. (2.4) Nuclear transformation the change of one element into another.
(18.3) e e° 0.0591 n log1Q2 at 25°C Glossary A35 Precipitation reaction a reaction in which an insoluble substance forms and separates from the solution. (4.5) Precision the degree of agreement among several measurements of the same quantity; the reproducibility of a measurement. (1.4) Primary structure (of a protein) the order (sequence) of amino acids in the protein chain. (22.6) Principal quantum number (n) the quantum number relating to the size and energy of an orbital; it can have any positive integer value.
(7.6) Probability distribution the square of the wave function indicating the probability of finding an electron at a particular point in space. (7.5) Product a substance resulting from a chemical reaction. It is shown to the right of the arrow in a chemical equation. (3.7) Protein a natural high-molecular-weight polymer formed by conden-sation reactions between amino acids. (22.6) Proton a positively charged particle in an atomic nucleus. (2.5; 18) Pure substance a substance with constant composition. (1.9) Pyrometallurgy recovery of a metal from its ore by treatment at high temperatures. (21.8) Quantization the concept that energy can occur only in discrete units called quanta. (7.2) Rad a unit of radiation dosage corresponding to 102 J of energy de-posited per kilogram of tissue (from radiation absorbed dose). (18.7) Radioactive decay (radioactivity) the spontaneous decomposition of a nucleus to form a different nucleus. (2.4; 18.1) Radiocarbon dating (carbon-14 dating) a method for dating ancient wood or cloth based on the rate of radioactive decay of the nuclide 14 6C. (18.4) Radiotracer a radioactive nuclide, introduced into an organism for diagnostic purposes, whose pathway can be traced by monitoring its radioactivity. (18.4) Random error an error that has an equal probability of being high or low. (1.4) Raoult’s law the vapor pressure of a solution is directly proportional to the mole fraction of solvent present. (11.4) Rate constant the proportionality constant in the relationship between reaction rate and reactant concentrations. (12.2) Rate of decay the change in the number of radioactive nuclides in a sample per unit time. (18.2) Rate-determining step the slowest step in a reaction mechanism, the one determining the overall rate. (12.6) Rate law (differential rate law) an expression that shows how the rate of reaction depends on the concentration of reactants. (12.2) Reactant a starting substance in a chemical reaction. It appears to the left of the arrow in a chemical equation. (3.7) Reaction mechanism the series of elementary steps involved in a chemical reaction. (12.6) Reaction quotient, Q a quotient obtained by applying the law of mass action to initial concentrations rather than to equilibrium concen-trations. (13.5) Reaction rate the change in concentration of a reactant or product per unit time. (12.1) Reactor core the part of a nuclear reactor where the fission reaction takes place. (18.6) Reducing agent (electron donor) a reactant that donates electrons to another substance to reduce the oxidation state of one of its atoms.
(4.9; 17.1) Peptide linkage the bond resulting from the condensation reaction between amino acids; represented by: (22.6) Percent dissociation the ratio of the amount of a substance that is dissociated at equilibrium to the initial concentration of the sub-stance in a solution, multiplied by 100. (14.5) Percent yield the actual yield of a product as a percentage of the the-oretical yield. (3.10) Periodic table a chart showing all the elements arranged in columns with similar chemical properties. (2.7) pH curve (titration curve) a plot showing the pH of a solution be-ing analyzed as a function of the amount of titrant added. (15.4) pH scale a log scale based on 10 and equal to log[H]; a conve-nient way to represent solution acidity. (14.3) Phase diagram a convenient way of representing the phases of a sub-stance in a closed system as a function of temperature and pressure.
(10.9) Phenyl group the benzene molecule minus one hydrogen atom. (22.3) Photochemical smog air pollution produced by the action of light on oxygen, nitrogen oxides, and unburned fuel from auto exhaust to form ozone and other pollutants. (5.10) Photon a quantum of electromagnetic radiation. (7.2) Physical change a change in the form of a substance, but not in its chemical composition; chemical bonds are not broken in a physi-cal change. (1.9) Pi () bond a covalent bond in which parallel p orbitals share an electron pair occupying the space above and below the line joining the atoms. (9.1) Planck’s constant the constant relating the change in energy for a system to the frequency of the electromagnetic radiation absorbed or emitted; equal to 6.626 1034 J s. (7.2) Polar covalent bond a covalent bond in which the electrons are not shared equally because one atom attracts them more strongly than the other. (8.1) Polar molecule a molecule that has a permanent dipole moment. (4.1) Polyatomic ion an ion containing a number of atoms. (2.6) Polyelectronic atom an atom with more than one electron. (7.9) Polymer a large, usually chainlike molecule built from many small molecules (monomers). (22.5) Polymerization a process in which many small molecules (monomers) are joined together to form a large molecule. (22.2) Polypeptide a polymer formed from amino acids joined together by peptide linkages. (22.6) Polyprotic acid an acid with more than one acidic proton. It disso-ciates in a stepwise manner, one proton at a time. (14.7) Porous disk a disk in a tube connecting two different solutions in a galvanic cell that allows ion flow without extensive mixing of the solutions. (17.1) Porphyrin a planar ligand with a central ring structure and various substituent groups at the edges of the ring. (21.7) Positional probability a type of probability that depends on the num-ber of arrangements in space that yield a particular state. (16.1) Positron production a mode of nuclear decay in which a particle is formed having the same mass as an electron but opposite charge.
The net effect is to change a proton to a neutron. (18.1) Potential energy energy due to position or composition. (6.1) A36 Glossary Single bond a bond in which one pair of electrons is shared by two atoms. (8.8) Smelting a metallurgical process that involves reducing metal ions to the free metal. (21.8) Solubility the amount of a substance that dissolves in a given volume of solvent at a given temperature. (4.2) Solubility product constant the constant for the equilibrium expres-sion representing the dissolving of an ionic solid in water. (15.6) Solute a substance dissolved in a liquid to form a solution. (4.2; 11.1) Solution a homogeneous mixture. (1.9) Solvent the dissolving medium in a solution. (4.2) Somatic damage radioactive damage to an organism resulting in its sickness or death. (18.7) Space-filling model a model of a molecule showing the relative sizes of the atoms and their relative orientations. (2.6) Specific heat capacity the energy required to raise the temperature of one gram of a substance by one degree Celsius. (6.2) Spectator ions ions present in solution that do not participate directly in a reaction. (4.6) Spectrochemical series a listing of ligands in order based on their ability to produce d-orbital splitting. (21.6) Spontaneous fission the spontaneous splitting of a heavy nuclide into two lighter nuclides. (18.1) Spontaneous process a process that occurs without outside intervention. (16.1) Standard atmosphere a unit of pressure equal to 760 mm Hg. (5.1) Standard enthalpy of formation the enthalpy change that accompa-nies the formation of one mole of a compound at 25°C from its elements, with all substances in their standard states at that tem-perature. (6.4) Standard free energy change the change in free energy that will occur for one unit of reaction if the reactants in their standard states are converted to products in their standard states. (16.6) Standard free energy of formation the change in free energy that accompanies the formation of one mole of a substance from its con-stituent elements with all reactants and products in their standard states. (16.6) Standard hydrogen electrode a platinum conductor in contact with 1 M H ions and bathed by hydrogen gas at one atmosphere. (17.2) Standard reduction potential the potential of a half-reaction under standard state conditions, as measured against the potential of the standard hydrogen electrode. (17.2) Standard solution a solution whose concentration is accurately known. (4.3) Standard state a reference state for a specific substance defined ac-cording to a set of conventional definitions. (6.4) Standard temperature and pressure (STP) the condition 0°C and 1 atmosphere of pressure. (5.4) Standing wave a stationary wave as on a string of a musical instrument; in the wave mechanical model, the electron in the hy-drogen atom is considered to be a standing wave. (7.5) State function (property) a property that is independent of the path-way. (6.1) States of matter the three different forms in which matter can exist; solid, liquid, and gas. (1.9) Stereoisomerism isomerism in which all the bonds in the isomers are the same but the spatial arrangements of the atoms are different. (21.4) Steric factor the factor (always less than 1) that reflects the fraction of collisions with orientations that can produce a chemical reaction. (12.7) Reduction a decrease in oxidation state (a gain of electrons). (4.9; 17.1) Rem a unit of radiation dosage that accounts for both the energy of the dose and its effectiveness in causing biological damage (from roentgen equivalent for man). (18.7) Resonance a condition occurring when more than one valid Lewis structure can be written for a particular molecule. The actual elec-tronic structure is not represented by any one of the Lewis struc-tures but by the average of all of them. (8.12) Reverse osmosis the process occurring when the external pressure on a solution causes a net flow of solvent through a semipermeable membrane from the solution to the solvent. (11.6) Reversible process a cyclic process carried out by a hypothetical pathway, which leaves the universe exactly the same as it was be-fore the process. No real process is reversible. (16.9) Ribonucleic acid (RNA) a nucleotide polymer that transmits the ge-netic information stored in DNA to the ribosomes for protein syn-thesis. (22.6) Roasting a process of converting sulfide minerals to oxides by heat-ing in air at temperatures below their melting points. (21.8) Root mean square velocity the square root of the average of the squares of the individual velocities of gas particles. (5.6) Salt an ionic compound. (14.8) Salt bridge a U-tube containing an electrolyte that connects the two compartments of a galvanic cell, allowing ion flow without exten-sive mixing of the different solutions. (17.1) Scientific method the process of studying natural phenomena, involving observations, forming laws and theories, and testing of theories by experimentation. (1.2) Scintillation counter an instrument that measures radioactive decay by sensing the flashes of light produced in a substance by the radiation. (18.4) Second law of thermodynamics in any spontaneous process, there is always an increase in the entropy of the universe. (16.2) Secondary structure (of a protein) the three-dimensional structure of the protein chain (for example, -helix, random coil, or pleated sheet). (22.6) Selective precipitation a method of separating metal ions from an aqueous mixture by using a reagent whose anion forms a precipi-tate with only one or a few of the ions in the mixture. (4.7; 15.7) Semiconductor a substance conducting only a slight electrical cur-rent at room temperature, but showing increased conductivity at higher temperatures. (10.5) Semipermeable membrane a membrane that allows solvent but not solute molecules to pass through. (11.6) SI system International System of units based on the metric system and units derived from the metric system. (1.3) Side chain (of amino acid) the hydrocarbon group on an amino acid represented by H, CH3, or a more complex substituent. (22.6) Sigma () bond a covalent bond in which the electron pair is shared in an area centered on a line running between the atoms. (9.1) Significant figures the certain digits and the first uncertain digit of a measurement. (1.4) Silica the fundamental silicon–oxygen compound, which has the em-pirical formula SiO2, and forms the basis of quartz and certain types of sand. (10.5) Silicates salts that contain metal cations and polyatomic silicon– oxygen anions that are usually polymeric. (10.5) Glossary A37 Thermodynamic stability (nuclear) the potential energy of a par-ticular nucleus as compared to the sum of the potential energies of its component protons and neutrons. (18.1) Thermodynamics the study of energy and its interconversions. (6.1) Thermoplastic polymer a substance that when molded to a certain shape under appropriate conditions can later be remelted. (22.5) Thermoset polymer a substance that when molded to a certain shape under pressure and high temperatures cannot be softened again or dissolved. (22.5) Third law of thermodynamics the entropy of a perfect crystal at 0 K is zero. (16.5) Titration a technique in which one solution is used to analyze another. (4.8) Torr another name for millimeter of mercury (mm Hg). (5.1) Transfer RNA (tRNA) a small RNA fragment that finds specific amino acids and attaches them to the protein chain as dictated by the codons in mRNA. (22.6) Transition metals several series of elements in which inner orbitals (d or f orbitals) are being filled. (7.11; 19.1) Transuranium elements the elements beyond uranium that are made artificially by particle bombardment. (18.3) Triple bond a bond in which three pairs of electrons are shared by two atoms. (8.8) Triple point the point on a phase diagram at which all three states of a substance are present. (10.9) Tyndall effect the scattering of light by particles in a suspension.
(11.8) Uncertainty (in measurement) the characteristic that any measure-ment involves estimates and cannot be exactly reproduced. (1.4) Unimolecular step a reaction step involving only one molecule.
(12.6) Unit cell the smallest repeating unit of a lattice. (10.3) Unit factor method an equivalence statement between units used for converting from one unit to another. (1.6) Universal gas constant the combined proportionality constant in the ideal gas law; 0.08206 L atm/K mol or 8.3145 J/K mol. (5.3) Valence electrons the electrons in the outermost principal quantum level of an atom. (7.11) Valence shell electron-pair repulsion (VSEPR) model a model whose main postulate is that the structure around a given atom in a molecule is determined principally by minimizing electron-pair repulsions. (8.13) Van der Waals equation a mathematical expression for describing the behavior of real gases. (5.8) Van’t Hoff factor the ratio of moles of particles in solution to moles of solute dissolved. (11.7) Vapor pressure the pressure of the vapor over a liquid at equilib-rium. (10.8) Vaporization (evaporization) the change in state that occurs when a liquid evaporates to form a gas. (10.8) Viscosity the resistance of a liquid to flow. (10.2) Volt the unit of electrical potential defined as one joule of work per coulomb of charge transferred. (17.1) Voltmeter an instrument that measures cell potential by drawing elec-tric current through a known resistance. (17.1) Volumetric analysis a process involving titration of one solution with another. (4.8) Stoichiometric quantities quantities of reactants mixed in exactly the correct amounts so that all are used up at the same time. (3.10) Strong acid an acid that completely dissociates to produce an H ion and the conjugate base. (4.2; 14.2) Strong base a metal hydroxide salt that completely dissociates into its ions in water. (4.2; 14.6) Strong electrolyte a material that, when dissolved in water, gives a solution that conducts an electric current very efficiently. (4.2) Structural formula the representation of a molecule in which the relative positions of the atoms are shown and the bonds are in-dicated by lines. (2.6) Structural isomerism isomerism in which the isomers contain the same atoms but one or more bonds differ. (21.4; 22.1) Subcritical reaction (nuclear) a reaction in which less than one neu-tron causes another fission event and the process dies out. (18.6) Sublimation the process by which a substance goes directly from the solid to the gaseous state without passing through the liquid state.
(10.8) Subshell a set of orbitals with a given azimuthal quantum number.
(7.6) Substitution reaction (hydrocarbons) a reaction in which an atom, usually a halogen, replaces a hydrogen atom in a hydrocarbon.
(22.1) Supercooling the process of cooling a liquid below its freezing point without its changing to a solid. (10.8) Supercritical reaction (nuclear) a reaction in which more than one neutron from each fission event causes another fission event. The process rapidly escalates to a violent explosion. (18.6) Superheating the process of heating a liquid above its boiling point without its boiling. (10.8) Superoxide a compound containing the O2 anion. (19.2) Surface tension the resistance of a liquid to an increase in its surface area. (10.2) Surroundings everything in the universe surrounding a thermody-namic system. (6.1) Syndiotactic chain a polymer chain in which the substituent groups such as CH3 are arranged on alternate sides of the chain. (22.5) Syngas synthetic gas, a mixture of carbon monoxide and hydrogen, obtained by coal gasification. (6.6) System (thermodynamic) that part of the universe on which atten-tion is to be focused. (6.1) Systematic error an error that always occurs in the same direction.
(1.4) Tempering a process in steel production that fine-tunes the propor-tions of carbon crystals and cementite by heating to intermediate temperatures followed by rapid cooling. (21.8) Termolecular step a reaction involving the simultaneous collision of three molecules. (12.6) Tertiary structure (of a protein) the overall shape of a protein, long and narrow or globular, maintained by different types of intramolecular interactions. (22.6) Theoretical yield the maximum amount of a given product that can be formed when the limiting reactant is completely consumed. (3.10) Theory a set of assumptions put forth to explain some aspect of the observed behavior of matter. (1.2) Thermal pollution the oxygen-depleting effect on lakes and rivers of using water for industrial cooling and returning it to its natural source at a higher temperature. (11.3) A38 Glossary Weight the force exerted on an object by gravity. (1.3) Work force acting over a distance. (6.1) X-ray diffraction a technique for establishing the structure of crys-talline solids by directing X rays of a single wavelength at a crys-tal and obtaining a diffraction pattern from which interatomic spaces can be determined. (10.3) Zone of nuclear stability the area encompassing the stable nuclides on a plot of their positions as a function of the number of protons and the number of neutrons in the nucleus. (18.1) Zone refining a metallurgical process for obtaining a highly pure metal that depends on continuously melting the impure material and recrystallizing the pure metal. (21.8) Vulcanization a process in which sulfur is added to rubber and the mixture is heated, causing crosslinking of the polymer chains and thus adding strength to the rubber. (22.5) Wave function a function of the coordinates of an electron’s position in three-dimensional space that describes the properties of the electron. (7.5) Wave mechanical model a model for the hydrogen atom in which the electron is assumed to behave as a standing wave. (7.5) Wavelength the distance between two consecutive peaks or troughs in a wave. (7.1) Weak acid an acid that dissociates only slightly in aqueous solution.
(4.2; 14.2) Weak base a base that reacts with water to produce hydroxide ions to only a slight extent in aqueous solution. (4.2; 14.6) Weak electrolyte a material which, when dissolved in water, gives a solution that conducts only a small electric current. (4.2) Photo Credits Chapter 1 p. xxiv: Ed Reschke. p. 2: (top left), DI/Veeco Metrology. p. 2: (top right), Courtesy, Dr. Randall M. Feenstra. p. 2: (bottom), Courtesy of IBM Research, Almaden Research Center. All rights reserved. p. 3: (top left), Cour-tesy, David Wineland. p. 3: (inset photo), Dr. Jeremy Burgess/Science Photo Library/Photo Researchers, Inc. p. 3: (bottom right), Chuck Place. p. 4: Dennis Brack/Estock/National Gallery of Art, Washington DC, Gallery Archives. p. 6, Corbis/Bettmann. p. 8: (top), NASA. p. 8: (bottom), Steve Borick/American Color. p. 10: Courtesy, Mettler Toledo, p. 16: Steve Borick/American Color.
p. 21: Peter Steiner/The Stock Market/Corbis. p. 23: (top) Courtesy, Briton Engineering Developments, (bottom), Richard Megna/Fundamental Photo-graphs. p 27: Ken O’Donoghue. p. 29: (bottom), Kristen Brockmann/Funda-mental Photographs.
Chapter 2 p. 38: (c) UNEP/Peter Arnold, Inc. p. 40: (top), Courtesy, Roald Hoffman/Cornell University. p. 40: (bottom), Paul Soud-ers/ Stone/Getty Images. p. 41: (Detail) Antoine Laurent Lavoisier and His Wife by Jacques Louis David, The Metropolitan Museum of Art, Purchase, Mr.
and Mrs. Charles Wrightsman. Gift, in honor of Everett Fahy, 1977. p. 42: Re-produced by permission, Manchester Literary and Philosophical Society.
p. 44: The Granger Collection, NY. p. 45: The Cavendish Laboratory. p. 46: Richard Megna/Fundamental Photographs. p. 47: Roger Du Buisson/The Stock Market/Corbis. p. 48: (top), Bob Daemmrich/Stock Boston. p. 48: (bottom), Topham Picture Library/The Image Works. p. 50: Steve Borick/American Color. p. 53: (left), Frank Cox. p. 53: (right), Ken O’Donoghue. p. 54: (far left), Ken O’Donoghue. p. 54: (center left), Sean Brady/American Color. p. 54: (center right), Sean Brady/American Color. p. 54: (far right), Charles D.
Winters/Photo Researchers, Inc. p. 55: (top), Ken O’Donoghue. p. 55: (bot-tom), Steve Borick/American Color. p. 60: Sean Brady/American Color. p. 61: Richard Megna/Fundamental Photographs.
Chapter 3 p. 76: Ken Karp. p. 79: Siede Preis/Photodisc/Getty Images. p. 78: Geoff Tompkinson/Science Library/Photo Researchers, Inc. p. 79: Sean Brady/American Color. p. 80: Tom Pantages. p. 81: (top), Courtesy, Joseph Wilmnoff/University of Washington and Luann Becker/University of Hawaii. p. 81: (bottom), Jeff J. Daly/Visuals Unlimited. p. 82: Ken O’Donoghue. p. 84: (left), Sean Brady/American Color.
p. 84: (right), Tom Pantages. p. 85: Russ Lappa/Science Source/Photo Re-searchers, Inc. p. 87: Grace Davies/Photo Network. p. 88: Kenneth Lorenzen.
p. 90: Phil Degginger/Stone/Getty Images. p. 94: (both), Ken O’Donoghue.
p. 95: Frank Cox. p. 97: Steve Borick/American Color. p. 98: Sean Brady/ American Color. p. 101: (both), Ken O’Donoghue. p. 105: NASA. p. 106: Steve Borick/American Color. p. 108: Grant Heilman Photography. p. 112: AP Photo/Marc Matheny.
Chapter 4 p. 126: Richard Megna/Fundamental Photographs. p. 130: (all), Ken O’Donoghue. p. 132: The Swedish Royal Academy of Sciences. p. 135: Sean Brady/American Color. p. 136: Ken O’Donoghue. p. 138: Image courtesy of Caliper Technologies Corporation. p. 140: Richard Megna/Fundamental Photographs. p. 141: (left), Steve Borick/American Color, p. 141: (right), Steve Borick/American Color. p. 142: Sean Brady/American Color. p. 143: (all), © Houghton Mifflin Company.
p. 144: Sean Brady/American Color. p. 152: Richard Megna/Fundamental Photographs. p. 155: (left), Sean Brady. p. 155: (center left), Ken O’Donoghue.
p. 155: (center right), Ken O’Donoghue. p. 155: (far right), Sean Brady/American Color. p. 156: Richard Megna/Fundamental Photographs. p. 158: Sean Brady/ American Color. p. 160: Sean Brady/American Color. p. 161: C Squared Studios/PhotoDisc/PictureQuest. p. 165: (all) © Houghton Mifflin Company; Chapter 5 p. 178: James Sparshatt/Corbis. p. 179: (both), Sean Brady/American Color. p. 181: Steve Borick/American Color. p. 184: Nick Nicholson/The Image Bank/Getty Images. p. 187: Ken O’Donoghue. p. 190: Runk/Schoenberger/ Grant Heilman Photography. p. 191: Steve Borick/American Color. p. 197: Courtesy, Ford Motor Corporation. p. 200: (both), Steve Borick/American Color. p. 203: (all), Ken O’Donoghue. p. 207: Ken O’Donoghue. p. 213: (both), The Field Museum, Chicago. p. 214: David Woodfall/Stone/Getty Images.
Chapter 6 p. 228: (c) UNEP/Peter Arnold Inc. p. 230: Courtesy, Sierra Pacific Innovations. p. 235: Mark E. Gibson/Visuals Unlimited. p. 239: Neil Lucas/Nature Picture Library. p. 240: E.R. Degginger. p. 241: AP Photo/Itsuo Inouye. p. 243: Argonne National Laboratory. p. 244: (left), Rich Treptow/ Visuals Unlimited. p. 244: (right), Comstock Images. p. 247: Sean Brady/ American Color. p. 251: Steve Borick/American Color. p. 253: (c) Helga Lade/Peter Arnold Inc. p. 255: Tony Freeman/PhotoEdit. p. 257: NASA. p. 259: Courtesy, FPL Energy LLC. p. 262: Courtesy, National Biodiesel Board, Inc.
p. 263: Arthur C. Smith III/Grant Heilman Photography.
Chapter 7 p. 274: Philip Habib/Stone/Getty Images. p. 276: David B. Fleetham/Stone/Getty Images. p. 277: Ken O’Donoghue. p. 278: Donald Clegg. p. 280: AFP/Corbis.
p. 279: Corbis/Bettmann. p. 281: (top), Dr. David Wexler/SPL/Photo Researchers, Inc. p. 281: (bottom), Science VU/Visuals Unlimited. p. 283: Sony. p. 284: PictureQuest. p. 286: Emilio Segre Visuals Archives. p. 289: Pho-toDisc/Getty Images. p. 290: Dave Blackburn. p. 300: Image Select/Art Re-source, NY. p. 301: Annalen der Chemie und Pharmacie, VIII, Supplementary Volume for 1872. p. 302: Courtesy, Lawrence Livermore National Laboratory: p. 304: (both), Sean Brady/American Color. p. 305: (top), Ken O’Donoghue.
p. 305: (bottom), Leslie Zumdahl. p. 309: The Granger Collection, NY. p. 317: (top), Gabe McDonald/Visuals Unlimited. p. 317: (bottom), Sean Brady/American Color.
Chapter 8 p. 328: Ken Eward/Photo Researchers, Inc. p. 329: Sean Brady/American Color. p. 337: (top, center), Ken O’Donoghue. p. 337: (bottom), Sean Brady/American Color. p. 340: Courtesy, Aluminum Company of America. p. 343: Sean Brady/American Color. p. 348: Ken O’Donoghue. p. 349: Will & Deni McIntyre/Photo Researchers, Inc.
p. 354: Courtesy, The Bancroft Library/University of California, Berkeley.
p. 358: Carnegie Institution of Washington. Photo by Ho-Kwang Mao and J.
Shu. p. 369: Steve Borick/American Color. p. 371: (all), Ken O’Donoghue.
p. 378: Kenneth Lorenzen. p. 379: Steve Borick/American Color.
Chapter 9 p. 390: Dr. Jeremy Burgess/Science Photo Library/Photo Researchers, Inc.
p. 399: Ken O’Donoghue. p. 407: Sean Brady/American Color. p. 411: Donald Clegg. Chapter 10 p. 424: Robert Fried Photography. p. 430: (left, center), Sean Brady/American Color. p. 430: (right), Ray Massey/Stone/Getty Images.
p. 431: (both) Courtesy, Lord Corporation. p. 432: (both), Brian Parker/Tom Stack & Associates. p. 434: Bruce Fox/Michigan State University Chemistry Photo Gallery. p. 435: (left), Comstock Images. p. 435: (center), Alfred Pasieka/Peter Arnold Inc. p. 435: (right), Richard C. Walters/Visuals Unlim-ited. p. 439: Steve Borick/American Color. p. 440: Denise Applewhite/Princeton University. p. 441: Sean Brady/American Color. p. 442: Chip Clark. p. 444: Paul Silvermann/Fundamental Photographs. p. 445: Courtesy, RMS Titanic, Inc. All rights reserved. p. 446: (top), Ken Eward/Photo Researchers, Inc. 446: (bottom), Courtesy, IBM Corporation. p. 448: (top), Richard Pasley/Stock Boston. p. 448: (bottom), Mathias Oppersdorff/Photo Researchers, Inc. p. 449: Courtesy, LiquidMetal Golf. p. 454: (left), Sean Brady/American Color. p. 454: (center), Ken O’Donoghue. p. 454: (right), Richard Megna/Fundamental Photographs. p. 455: Dr. Thomas Thundat/Oak Ridge National Laboratory.
p. 466: Steve Borick/American Color. p. 469: Chris Noble/Stone/Getty Images.
p. 470: Courtesy, Badger Fire Protection, Inc.
Chapter 11 p. 484: Charles Derby/Photo Researchers, Inc. p. 487: Sean Brady/American Color. p. 489: Photo Courtesy of E Ink. p. 491: E.R. Degginger. p. 493: (top left, right), Frank Cox. p. 493: (bottom) Sean Brady/American Color. p. 496: T. Orban/Sygma/ Corbis. p. 497: Betz/Visuals Unlimited. p. 500: David Young-Wolfe/PhotoEdit.
p. 505: Steve Borick/American Color. p. 506: (top), Craig Newbauer/Peter Arnold, Inc. p. 507: Leslie Zumdahl. p. 510: Visuals Unlimited. p. 511: (left), A39 Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
A40 Photo Credits Photographs. p. 862: Marvin Lazarus/Photo Researchers, Inc. p. 865 (both): Courtesy Fermilab Visual Media Services.
Chapter 19 p. 874: Oliver Meckes/Nicole Ottawa/Photo Researchers, Inc. p. 878: Wendell Metzen/Bruce Coleman, Inc. p. 880: E.R. Degginger. p. 881 (top): Steve Borick/American Color. p. 881 (bottom): Peter Skinner/Photo Researchers, Inc. p. 882: Yoav Levy/Phototake. p. 883: Richard Megna/Fundamental Photographs. p. 886 (top): Sean Brady/American Color. p. 886 (bottom): Hoa Qui/Index Stock Imagery/PictureQuest. p. 887: Spencer Swanger/Tom Stack & Associates.
p. 889: Sean Brady/American Color. p. 890: Andrea Pistolesi/The Image Bank/Getty Images. p. 891 (top): M. Kulyk/Photo Researchers, Inc. p. 893 (top): The Granger Collection, NY. p. 893 (bottom): Michael Holford. p. 894: Sean Brady/American Color.
Chapter 20 p. 900: Kjell B. Sandved/Visuals Unlimted. p. 904 (top): Ron Sherman/Stone/Getty Images. p. 904 (bottom): Courtesy Inert Systems. p. 905: Randy G. Taylor/Estock Photo. p. 906: Hugh Spencer/Photo Researchers, Inc. p. 909: Sean Brady/American Color. p. 910: Sean Brady/American Color. p. 915: R.H. Laron/PhotoEdit. p. 921 (top left, bottom): Ken O’Donoghue. p. 921 (top right): E.R. Degginger. p. 922 (left): E.R. Degginger. p. 922 (right): Farrell Greham/Photo Researchers, Inc. p. 923 (all): Steve Borick/American Color. p. 925: Tom Pantages. p. 926 (top): E.R.
Degginger. p. 926 (bottom): Yoav Levy/Phototake. p. 928: James L. Amos/ Peter Arnold, Inc. p. 931: E.R. Degginger. p. 932: Deborah Davis/PhotoEdit.
Chapter 21 p. 942: Bios/Peter Arnold, Inc. p. 945: Peter Harholdt/Superstock.
p. 946 (top two): Paul Silverman/Fundamental Photographs. p. 946 (center left): Sean Brady/American Color, Courtesy, Charles Lewis. p. 946 (center right): Ken O’Donoghue. p. 946 (bottom left): Sean Brady/American Color. p. 950 (top): RNHRD NHS Trust/Stone/Getty Images. p. 950 (bottom): Ken O’Donoghue. p. 953: Jodi Jacobson/Peter Arnold, Inc. p. 954: Sean Brady/ American Color. p. 959 (both): Sean Brady/American Color. p. 963: Martin Bough/Fundamental Photographs. p. 971 (top): Tom Pantages. p. 971 (bottom): Ken O’Donoghue. p. 977: p. 977 (top): Stanley Flegler/Visual Unlimited.
p. 977 (bottom): Galen Rowell/Peter Arnold, Inc. p. 979: Victoire de Marco Pantani/Liaison/Getty News Images. p. 978: Luis Veiga/The Image Bank/Getty Images. p. 991: Sean Brady/American Color. p. 992: Sean Brady/American Color.
Chapter 22 p. 996: Alfred Pasieka/Science Photo Library/Photo Researchers, Inc. p. 1003: Steve Borick/American Color. p. 1007: Chuck Keeler Jr./The Stock Market/Corbis. p. 1011: Ian Shaw/Stone/Getty Images. p. 1012: AP Photo/Nati Harnik. p. 1013: Inga Spence/Visuals Unlimited. p. 1015 (top): Steve Borick/American Color. p. 1015 (bottom): Laguna Design/SPL/Photo Researchers, Inc. p. 1016: AP Photo/Indianapolis Star, Karen Ducey. p. 1017 (top): Phil Nelson. p. 1017 (bottom): Ron Boardman/Frank Lane Picture Agency/Corbis. p. 1018: Scott White/University of Illinois at Urbana-Champaign.
p. 1019: Wendell Metzen/Index Stock Imagery/PictureQuest. p. 1021: Sean Brady/American Color. p. 1022: Courtesy Dupont Company. p. 1025 (top): Bob Gomel/The Stock Market/Corbis. p. 1025 (bottom): Michael Abbey/ Science Source/Photo Researchers, Inc. p. 1027: Ken Eward/Biografx/Science Source/Photo Researchers, Inc. p. 1033 (both): Sean Brady/American Color.
p. 1035: Steve Borick/American Color. p. 1037: Alfred Pasieka/Science Source/Photo Researchers, Inc. Stanley Flagler/Visuals Unlimited. p. 511: (center, right), David M. Phillips/ Visuals Unlimited. p. 512: Courtesy, Southern California Electric. p. 514: Stephen Frisch. p. 515: Sean Brady/American Color. p. 516: Doug Allan/Animals, Animals.
Chapter 12 p. 526: Vandystadt/TIPS Images. p. 528: (top left), Bob Daemmrich/Stock Boston. p. 528: (top right), Thomas H. Brakefield/The Stock Market/Corbis. p. 528: (bottom left), Tom Stillo/Omni-Photo Communications.
p. 528: (bottom right), E. Bordis/Estock Photo. p. 531: (left), Chad Ehlers/Photo Network. p. 531: (right), Stephen L. Saks/Photo Network. p. 555: Alvin E.
Staffan/Photo Researchers, Inc. p. 558: Steve Borick/American Color. p. 567: NASA/Goddard Space Flight Center. p. 563: Ken Eward/Photo Researchers, Inc.
Chapter 13 p. 578: Richard Megna/Fundamental Photographs. p. 585: Grant Heilman Photography. p. 589: Martin Bond/SPL/Photo Researchers, Inc.
p. 590: Richard Megna/Fundamental Photographs. p. 595: NASA. p. 608 (all): Ken O’Donoghue. p. 609 (both): Sean Brady/American Color. p. 618 (all): Sean Brady/American Color.
Chapter 14 p. 622: Ed Reschke/Peter Arnold, Inc. p. 624: Eric Fordham. p. 625: Ken O’Donoghue. p. 632 (top): Copyright Los Angeles Times Syndicate, photo from Chemical Heritage Foundation.
p. 632 (bottom): Ken O’Donoghue. p. 638: Spencer Grant/PhotoEdit. p. 642: Ken O’Donoghue. p. 643: Steve Borick/American Color. p. 644: Bob Daemmrich/Stock Boston. p. 645: Steve Borick/American Color. p. 649: E.R.
Degginger. p. 653: Steve Borick/American Color. p. 666: E.R. Degginger.
Chapter 15 p. 680: Demetrio Carrasco/Stone/Getty Images. p. 684: Ken O’Donoghue. p. 686 (both): Ken O’Donoghue. p. 696: Steve Borick/American Color. p. 711 (both): Sean Brady/American Color. p. 712: Sean Brady/Amer-ican Color. p. 713: Ken O’Donoghue. p. 714: Steve Borick/American Color.
p. 717: CNRI/Science Photo Library/Photo Researchers, Inc. p. 719: Ken O’Donoghue. p. 720: Max A. Listgarten/Visuals Unlimited. p. 722: Ken O’Donoghue. p. 729 (both): Ken O’Donoghue. p. 730: Sean Brady/American Color. p. 732: Steve Borick/American Color. p. 734 (both): Richard Megna/Fundamental Photographs.
Chapter 16 p. 748: Barry L. Runk/Grant Heilman Photography. p. 751 (top): Kent & Donna Dennen/Photo Researchers, Inc. p. 751 (bottom): Sean Brady/American Color. p. 757: Tony Freeman/Pho-toEdit. p. 759: Sean Brady/American Color. p. 777: Hank Morgan/Rainbow.
Chapter 17 p. 790: Jeff J. Daly/Visuals Unlimited. p. 794: Steve Borick/Amer-ican Color. p. 800: Steve Borick/American Color. p. 801: Courtesy, the Royal Institution of Great Britain. p. 803 (top): Sean Brady/American Color. p. 803 (bottom): Steve Borick/American Color. p. 807: Ken O’Donoghue. p. 808: Ken O’Donoghue. p. 809: AP Photo/Ed Bailey. p. 812 (top): AutoWeek/Crain Com-munications, Inc. p. 812 (bottom): Paul Silverman/Fundamental Photographs.
p. 819: Charles D. Winters/Photo Researchers, Inc. p. 820: Courtesy, Mel Fisher’s Motivation, Inc. p. 822: Courtesy, Oberlin College Archives, Oberlin, Ohio. p. 823: Tom Hollyman/Photo Researchers, Inc. p. 824: Bill Gallery/Stock Boston.
Chapter 18 p. 840: Lawrence Livermore National Laboratory.
p. 847: Simon Fraser/Medical Physics: RVI, Newcastle-Upon-Tyne/SPL/Photo Researchers, Inc. p. 851 (top): Arizona State University and NASA. p. 851 (bottom): James A. Sugar/Corbis. p. 853 (top): Mark A. Philbrick/Brigham Young University. p. 853 (bottom): Peter Dunwiddle/Visuals Unlimited. p. 855 (left, center): SIU/Visuals Unlimited. p. 855 (right): Richard Megna/Fundamental A41 The answers listed here are from the Complete Solutions Guide, in which round-ing is carried out at each intermediate step in a calculation in order to show the correct number of significant figures for that step. Therefore, an answer given here may differ in the last digit from the result obtained by carrying ex-tra digits throughout the entire calculation and rounding at the end (the pro-cedure you should follow).
Chapter 1 17. A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction or the ideal gas law, A theory (model) is an at-tempt to explain why something happens. Dalton’s atomic theory explains why mass is conserved in a chemical reaction. The kinetic molecular theory explains why pressure and volume are inversely related at constant tempera-ture and moles of gas.
19. A qualitative observation expresses what makes something what it is; it does not involve a number; e.g., the air we breathe is a mixture of gases, ice is less dense than water, rotten milk stinks. The SI units are mass in grams, length in meters, and volume in the derived units of The assumed uncertainty in a number is in the last significant figure of the number. The precision of an instrument is related to the number of significant figures associated with an experimental reading on that instrument. Different instruments for measuring mass, length, or volume have varying degrees of precision. Some instruments only give a few significant figures for a mea-surement while others will give more significant figures.
21. Significant fig-ures are the digits we associate with a number. They contain all of the certain digits and the first uncertain digit (the first estimated digit). What follows is one thousand indicated to varying numbers of significant figures: 1000 or (1 S.F.); (2 S.F.); (3 S.F.); 1000. or (4 S.F.). To perform the calculation, the addition/subtraction significant rule is applied to 1.5 1.0. The result of this is the one significant figure answer of 0.5. Next, the multiplication/division rule is applied to A one signif-icant number divided by a two significant number yields an answer with one significant figure 23. The slope of the vs. plot is and the y-intercept is 32F. The slope of vs. plot is one and the y-intercept is 273C.
25. a. exact; b. inexact; c. exact; d. inex-act 27. a. 3; b. 4; c. 4; d. 1; e. 7; f. 1; g. 3; h. 3 29. a. 3.42 104; b. 1.034 104; c. 1.7992 101; d. 3.37 105 31. a. 641.0; b. 1.327; c. 77.34; d. 3215; e. 0.420 33. a. 188.1; b. 12.; c. 4 107; d. 6.3 1026; e. 4.9; Uncertainty appears in the first decimal place. The average of several numbers can be only as precise as the least precise number. Averages can be exceptions to the significant figure rules. f. 0.22 35. a. 84.3 mm; b. 2.41 m; c. 2.945 105 cm; d. 14.45 km; e. 2.353 105 mm; f. 0.9033 m 37. a. 8 lb and 9.9 oz; 20 1 4 in; b. 4.0 104 km, 4.0 107 m; c. 1.2 102 m3, 12 L, 730 in3, 0.42 ft3 39. a. 4.00 102 rods; 10.0 fur-longs; 2.01 103 m; 2.01 km; b. 8390.0 rods; 209.75 furlongs; 42,195 m; 42.195 km 41. a. 0.373 kg, 0.822 lb; b. 31.1 g, 156 carats; c. 19.3 cm3 43. 2.95 109 knots; 3.36 109 mi/h 45. To the proper number of signif-icant figures, the car is traveling at 40. mi/h, which would not violate the speed limit.
47. 0.68 Canadian/L 49. a. 273C, 0 K; b. 40.C, 233 K; c. 20.C, 293 K; d. 4 107 C, 4 107 K 51. a. 312.4 K; 102.6F; b. 248 K; 13F; c. 0 K; 459F; d. 1074 K; 1470F 53. It will float (density 0.80 g/cm3).
55. 1 106 g/cm3 57. 0.28 cm3 59. 3.8 g/cm3 61. a. Both are the same mass; b. 1.0 mL mercury; c. Both are the same mass; d. 1.0 L benzene 63. a. 1.0 kg feather; b. 100 g water; c. same 65. 2.77 cm 67. a. Picture iv represents a gaseous compound. Pictures ii and iii also contain a gaseous compound but have a gaseous element present.
b. Picture vi represents a mixture of two elemental gases. c. Picture v represents a solid element. d. Pictures ii and iii both represent a mixture of a gaseous element and a gaseous compound.
69. a. heterogeneous; TK TC 1.8 1 9 52 TC TF 1answer 12.
0.5 0.50.
1.000 103 1.00 103 1.0 103 1 103 1 m3.
PV nRT.
b. homogeneous (hopefully); c. homogeneous; d. homogeneous (hopefully); e. heterogeneous; f. heterogeneous 71. a. physical; b. chemical; c. physical; d. chemical 73. 24 capsules 75. 1.0 105 bags 77. 56.56°C 79. a. Volume density mass; the orange block is more dense. Since mass (orange) mass (blue) and volume (orange) volume (blue), then the den-sity of the orange block must be greater to account for the larger mass of the orange block. b. Which block is more dense cannot be determined. Since mass (orange) mass (blue) and volume (orange) volume (blue), then the den-sity of the orange block may or may not be larger than the blue block. If the blue block is more dense, then its density cannot be so large that the mass of the smaller blue block becomes larger than the orange block mass. c. The blue block is more dense. Since mass (blue) mass (orange) and volume (blue) volume (orange), then the density of the blue block must be larger to equate the masses. d. The blue block is more dense. Since mass (blue) mass (orange) and the volumes are equal, then the density of the blue block must be larger to give the blue block the larger mass.
81. 8.5 0.5 g/cm3 83. a. 2%; b. 2.2%; c. 0.2% 85. dold 8.8 g/cm3, dnew 7.17 g/cm3; dnew/dold massnew/massold 0.81; The difference in mass is accounted for by the difference in the alloy used (if the assumptions are correct).
87. 7.0% 89. a. One possibility is that rope B is not attached to anything and rope A and rope C are connected via a pair of pulleys and/or gears; b. Try to pull rope B out of the box. Measure the distance moved by C for a given movement of A. Hold either A or C firmly while pulling on the other rope.
91. $160/person; 3.20 103 nickels/person; 85.6 £/per-son 93. 200.0F 93.33C; 100.0F 73.3C; 93.33C 366.48 K; 73.3C 199.9 K; difference of temperatures in C 166.6; difference of temperatures in K 166.6; No, there is not a difference of 300 degrees in C or K.
Chapter 2 15. A compound will always contain the same numbers (and types) of atoms.
A given amount of hydrogen will react only with a specific amount of oxygen.
Any excess oxygen will remain unreacted.
17. Law of conservation of mass: mass is neither created nor destroyed. The mass before a chemical reaction always equals the mass after a chemical reaction. Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass. Water is always 1 g H for every 8 g oxygen. Law of multiple propor-tions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with one gram of the first element can always be reduced to small whole numbers. For and CO discussed in section 2.2, the mass ratios of oxygen that react with 1 g of carbon in each compound are in a 2:1 ratio.
19. J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negatively charged particles which we now call electrons. Ernest Rutherford and his alpha bombardment of metal foil experiments led him to postulate the nuclear atom—an atom with a tiny dense center of positive charge (the nucleus) with electrons moving about the nucleus at relatively large distances away; the distance is so large that an atom is mostly empty space.
21. The number and arrangement of electrons in an atom de-termines how the atom will react with other atoms. The electrons determine the chemical properties of an atom. The number of neutrons present determine the isotope identity.
23. Statements a and b are true. Counting over in the periodic table, element 118 will be the next noble gas (a nonmetal). For state-ment c, hydrogen has mostly nonmetallic properties. For statement d, a fam-ily of elements is also known as a group of elements. For statement e, two items are incorrect. When a metal reacts with a nonmetal, an ionic compound is produced and the formula of the compound would be (since alkaline earth metals for ions and halogens form 1 ions in ionic compounds). The correct statement would be: When alkaline earth metal, A reacts with a 2 AX2 CO2 Answers to Selected Exercises A42 Answers to Selected Exercises most of the solid material is converted to gases, which escape. c. Atoms are not an indivisible particle. Atoms are composed of electrons, neutrons, and protons. d. The two hydride samples contain different isotopes of either hy-drogen and/or lithium. Isotopes may have different masses but have similar chemical properties.
91. tantalum(V) oxide; the formula would have the same subscripts, Ta2S5; 40 protons 93. Ge4; 99Tc Chapter 3 19. From the relative abundances, there would be 9889 atoms of and 111 atoms of in the 10,000-atom sample. The average mass of carbon is inde-pendent of the sample size; it will always be 12.01 amu. The total mass would be For one mol of carbon the av-erage mass would still be 12.01 amu. There would be of and of The total mass would be The total mass in grams is 12.01 g/mol.
21. Each person would have 100 tril-lion dollars.
23. The mass percent of a compound is a constant no matter what amount of substance is present. Compounds always have constant composition.
25. The information needed is mostly the coefficients in the balanced equation and the molar masses of the reactants and products. For percent yield, we would need the actual yield of the reaction and the amounts of reactants used.
a. mass of CB produced molecules b. atoms of A produced molecules c. mol of C reacted molecules d.
The theoretical mass of CB produced was calculated in part a. If the actual mass of CB produced is given, then the percent yield can be determined for the reaction using the percent yield equa-tion.
27. 207.2 amu, Pb 29. 185 amu 31. There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes, differing in mass by two mass units.
33. 4.64 1020 g Fe 35. 1.00 1022 atoms C 37. Al2O3, 101.96 g mol; Na3AlF6, 209.95 g mol 39. a. 17.03 g mol; b. 32.05 g/mol; c. 252.08 g/mol 41. a. 0.0587 mol NH3; b. 0.0312 mol N2H4; c. 3.97 103 mol (NH4)2Cr2O7 43. a. 85.2 g NH3; b. 160. g N2H4; c. 1260 g (NH4)2Cr2O7 45. a. 70.1 g N; b. 140. g N; c. 140. g N 47. a. 3.54 1022 molecules NH3; b. 1.88 1022 molecules N2H4; c. 2.39 1021 formula units (NH4)2Cr2O7 49. a. 3.54 1022 atoms N; b. 3.76 1022 atoms N; c. 4.78 1021 atoms N 51. 176.12 g/mol; 2.839 103 mol; 1.710 1021 molecules 53. a. 0.9393 mol; b. 2.17 104 mol; c.
2.5 108 mol 55. a. 4.01 1022 atoms N; b. 5.97 1022 atoms N; c. 3.67 1022 atoms N; d. 6.54 1022 atoms N 57. a. 294.30 g/mol; b. 3.40 102 mol; c. 459 g; d. 1.0 1019 mole-cules; e. 4.9 1021 atoms; f. 4.9 1013 g; g. 4.887 1022 g 59. a. 50.00% C, 5.595% H, 44.41% O; b. 55.80% C, 7.025% H, 37.18% O; c. 67.90% C, 5.699% H, 26.40% N 61. NO2 N2O4 NO N2O 63. 1360 g/mol 65. a. 39.99% C, 6.713% H, 53.30% O; b. 40.00% C, 6.714% H, 53.29% O; c. 40.00% C, 6.714% H, 53.29% O (all the same ex-cept for rounding differences) 67. a. NO2; b. CH2; c. P2O5; d. CH2O 69. C8H11O3N 71. compound I: HgO; compound II: Hg2O 73. SN; S4N4 75. C3H5O2; C6H10O4 77. C3H8 79. C3H4, C9H12 81. a. C6H12O6(s) 6O2(g) n 6CO2(g) 6H2O(g); b. Fe2S3(s) 6HCl(g) n 2FeCl3(s) 3H2S(g); c. CS2(l) 2NH3(g) n H2S(g) NH4SCN(s) 83. a. 3Ca(OH)2(aq) 2H3PO4(aq) n 6H2O(l) Ca3(PO4)2(s); b. Al(OH)3(s) 3HCl(aq) n AlCl3(aq) 3H2O(l); c. 2AgNO3(aq) H2SO4(aq) n Ag2SO4(s) 2HNO3(aq) 85. a. 2C6H6(l) 15O2(g) n 12CO2(g) 6H2O(g); b. 2C4H10(g) 13O2(g) n 8CO2(g) 10H2O(g); c. C12H22O11(s) %yield actual mass theoretical mass 100; 1 mol A2B2 6.022 1023 molecules A2B2 2 mol C 1 mol A2B2 A2B2 1.00 104 A2B2 2 atoms A 1 molecule A2B2 1.00 104 A2B2 1 mol A2B2 6.022 1023 molecules A2B2 2 mol CB 1 mol A2B2 molar mass of CB mol CB 1.00 104 7.232 1024 amu.
13C.
6.68 1021 atoms 12C 5.955 1023 atoms 16.022 1023 atoms C2, 1.201 105 amu.
13C 12C halogen, X, the formula of the ionic compound formed should be 25. a. The composition of a substance depends on the number of atoms of each element making up the compound (depends on the formula of the com-pound) and not on the composition of the mixture from which it was formed.
b. Avogadro’s hypothesis implies that volume ratios are equal to molecule ratios at constant temperature and pressure. H2(g) Cl2(g) n 2 HCl(g); from the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted.
27. All the masses of hydrogen in these three compounds can be expressed as simple whole-number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6:9:1.
29. O, 7.94; Na, 22.8; Mg, 11.9; O and Mg are incorrect by a factor of 2; correct formulas are H2O, Na2O, and MgO.
31. Using r 5 1014 cm, dnucleus 3 1015 g cm3; using r 1 108 cm, datom 0.4 g cm3 33. 37 35. sodium, Na; radium, Ra; iron, Fe; gold, Au; manganese, Mn; lead, Pb 37. Sn, tin; Pt, plat-inum; Hg, mercury; Mg, magnesium; K, potassium; Ag, silver 39. a. Metals: Mg, Ti, Au, Bi, Ge, Eu, Am; nonmetals: Si, B, At, Rn, Br; b. metalloids: Si, Ge, B, At. The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At. These elements are all considered met-alloids. Aluminum has mostly properties of metals.
41. a. 6; b. 5; c. 4; d. 6 43. a. 35 p, 44 n, 35 e; b. 35 p, 46 n, 35 e; c. 94 p, 145 n, 94 e; d. 55 p, 78 n, 55 e; e. 1 p, 2 n, 1 e; f. 26 p, 30 n, 26 e 45. a. 17 8O; b. 37 17Cl; c. 60 27Co; d. 57 26Fe; e. 131 53I; f. 7 3Li 47. 151 63Eu3; 118 50Sn2 49. 238 92U, 92 p, 146 n, 92 e, 0; 40 20Ca2, 20 p, 20 n, 18 e, 2; 51 23V3, 23 p, 28 n, 20 e, 3; 89 39Y, 39 p, 50 n, 39 e, 0; 79 35Br, 35 p, 44 n, 36 e, 1; 31 15P3, 15 p, 16 n, 18 e, 3 51. a. transition metals; b. alkaline earth metals; c. alkali metals; d. noble gases; e. halogens 53. a. lose 2 e to form Ra2; b. lose 3 e to form In3; c. gain 3 e to form P3; d. gain 2 e to form Te2; e. gain 1 e to form Br; f. lose 1 e to form Rb 55. a. sodium bromide; b. rubidium ox-ide; c. calcium sulfide; d. aluminum iodide; e. SrF2; f. Al2Se3; g. K3N; h. Mg3P2 57. a. cesium fluoride; b. lithium nitride; c. silver sulfide (Sil-ver forms only 1 ions so no Roman numerals are needed); d. manganese(IV) oxide; e. titanium(IV) oxide; f. strontium phosphide 59. a. barium sulfite; b. sodium nitrite; c. potassium permanganate; d. potassium dichromate 61. a. dinitrogen tetroxide; b. iodine trichloride; c. sulfur dioxide; d. diphosphorus pentasulfide 63. a. copper(I) iodide; b. copper(II) iodide; c. cobalt(II) iodide; d. sodium carbonate; e. sodium hydrogen carbonate or sodium bicarbonate; f. tetrasulfur tetranitride; g. sulfur hexafluoride; h. sodium hypochlorite; i. barium chromate; j. ammonium nitrate 65. a. SF2; b. SF6; c. NaH2PO4; d. Li3N; e. Cr2(CO3)3; f. SnF2; g. NH4C2H3O2; h. NH4HSO4; i. Co(NO3)3; j. Hg2Cl2; Mercury(I) exists as Hg2 2 ions.
k. KClO3; l. NaH 67. a. Na2O; b. Na2O2; c. KCN; d. Cu(NO3)2; e. SeBr4; f. HIO2; g. PbS2; h. CuCl; i. GaAs (from the positions in the pe-riodic table, Ga3 and As3 are the predicted ions); j. CdSe; k. ZnS; l. HNO2; m. P2O5 69. a. nitric acid, HNO3; b. perchloric acid, HClO4; c. acetic acid, HC2H3O2; d. sulfuric acid, H2SO4; e. phosphoric acid, H3PO4 71. Yes, 1.0 g H would react with 37.0 g 37Cl and 1.0 g H would react with 35.0 g 35Cl.
No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37Cl and 1 g H/35 g Cl for 35Cl. As long as we had pure 35Cl or pure 37Cl, these ratios will always hold. If we have a mixture (such as the natural abundance of chlorine), the ratio will also be constant as long as the composition of the mixture of the two isotopes does not change.
73. Only statement a is true. For statement b, X has 34 protons. For statement c, X has 45 neutrons. For statement d, X is selenium.
75. a. lead(II) acetate; b. copper(II) sulfate; c. calcium oxide; d. magnesium sulfate; e. magnesium hydroxide; f. calcium sulfate; g. dini-trogen monoxide or nitrous oxide (common) 77. X Ra, 142 neutrons 79. a. Ca3N2; calcium nitride; b. K2O; potassium oxide; c. RbF; rubidium fluoride; d. MgS; magnesium sulfide; e. BaI2; barium iodide; f. Al2Se3; alu-minum selenide; g. Cs3P; cesium phosphide; h. InBr3; indium(III) bromide (In forms compounds with 1 and 3 ions. You would predict a 3 ion from the position of In in the periodic table.) 81. 116 g S; 230. g O 83. Cu, Ag, and Au 85. C:H 8:18 or 4:9 87. The ratio of the masses of R that com-bine with 1.00 g Q is 3:1, as expected by the law of multiple proportions. R3Q is the formula of the first compound.
89. a. The compounds are isomers of each other. Isomers are compounds with the same formula but the atoms are attached differently, resulting in different properties. b. When wood burns, AX2.
Answers to Selected Exercises A43 12O2(g) n 12CO2(g) 11H2O(g); d. 4Fe(s) 3O2(g) n 2Fe2O3(s); e. 4FeO(s) O2(g) n 2Fe2O3(s) 87. a. SiO2(s) 2C(s) n Si(s) 2CO(g); b. SiCl4(l) 2Mg(s) n Si(s) 2MgCl2(s); c. Na2SiF6(s) 4Na(s) n Si(s) 6NaF(s) 89. 7.26 g Al; 21.5 g Fe2O3; 13.7 g Al2O3 91. 4.355 kg 93. 32 kg 95. a. 73.9 g; b. 1.30 102 g 97. 2NO(g) O2(g) n 2NO2(g); NO is limiting.
99. 0.301 g H2O2; 3.6 102 g HCl in excess 101. 2.81 106 g HCN; 5.63 106 g H2O 103. 1.96 g (theoretical yield); 76.5% 105. 1.20 103 kg 1.20 metric tons 107. 6 109. a. stoichiometric mixture; b. O2; c. H2; d. H2; e. H2; f. stoichiometric mixture; g. H2 111. 9.25 1022 H atoms 113. 4.30 102 mol; 2.50 g 115. 5 117. 42.8% 119. 81.1 g 121. 86.2% 123. C20H30O 125. C7H5N3O6 127. 38.7% 129. 40.08% 131. N4H6 133. 1.8 106 g Cu(NH3)4Cl2; 5.9 105 g NH3 135. 207 amu, Pb 137. Al2Se3 139. 0.48 mol 141. a. 113 Fe atoms; b. mass 9.071 1020 g; c. 540 Ru atoms 143. M Y, X Cl, yttrium(III) chloride; 1.84 g Chapter 4 9. a. Polarity is a term applied to covalent compounds. Polar covalent com-pounds have an unequal sharing of electrons in bonds that results in an unequal charge distribution in the overall molecule. Polar molecules have a partial neg-ative end and a partial positive end. These are not full charges like in ionic compounds, but are charges much less in magnitude. Water is a polar mole-cule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial posi-tive end of the polar solute while the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charged attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in wa-ter and do not dissolve. b. KF is a soluble ionic compound so it is a strong electrolyte. KF(aq) actually exists as separate hydrated ions and hydrated ions in solution: is a polar covalent molecule that is a nonelec-trolyte. is hydrated as described in part a. c. RbCl is a soluble ionic compound so it exists as separate hydrated ions and hydrated ions in solution. AgCl is an insoluble ionic compound so the ions stay together in solution and fall to the bottom of the container as a precipitate. d.
is a strong acid and exists as separate hydrated ions and hydrated ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a.
11. Bromides: NaBr, KBr, and HgBr (and others) would be soluble and AgBr, and would be insoluble. Sulfates: and (and others) would be soluble and and (or would be insoluble. Hydroxides: NaOH, KOH, (and others) would be soluble and and (and others) would be insoluble. Phosphates: (and others) would be soluble and (and others) would be insolu-ble. Lead: and PbS (and others) would be insoluble. would be a soluble salt.
13. The Brønsted-Lowry definitions are best for our purposes. An acid is a proton donor and a base is a proton acceptor. A proton is an ion. Neutral hydrogen has 1 electron and 1 proton, so an ion is just a proton. An acid–base reaction is the transfer of an ion (a proton) from an acid to a base.
15. a. The species reduced is the element that gains electrons. The reducing agent causes reduction to oc-cur by itself being oxidized. The reducing agent is generally listed as the en-tire formula of the compound/ion that contains the element oxidized. b. The species oxidized is the element that loses electrons. The oxidizing agent causes oxidation to occur by itself being reduced. The oxidizing agent is generally listed as the entire formula of the compound/ion that contains the element re-duced. c. For simple binary ionic compounds, the actual charge on the ions are the oxidation states. For covalent compounds, nonzero oxidation states are pretend charges the elements would have if they were held together by ionic bonds (assuming the bond is between two different nonmetals). Nonzero oxidation states for elements in covalent compounds are not actual charges.
Oxidation states for covalent compounds are a bookkeeping method to keep track of electrons in a reaction.
H H H Pb2 Pb1NO322 PbCl2, PbBr2, PbI2, Pb1OH22, PbSO4, Ag3PO4, Ca31PO422, and FePO4 1NH423PO4 K3PO4, Na3PO4, Cu1OH22 Fe1OH23, Al1OH23, Ca1OH22 Hg2SO42 PbSO4 BaSO4, CaSO4, 1NH422SO4 K2SO4, Na2SO4, Hg2Br2 PbBr2, NO3 H HNO3 Cl Rb C6H12O6 C6H12O6 F K 17.
c. For answers c–i, we will describe what should be in each solution. For c, the drawing should have three times as many NO3 anions as Al3 cations.
d. The drawing should have twice as many NH4 cations as SO4 2 anions.
e. The drawing should have equal numbers of Na cations and OH anions.
f. The drawing should have equal numbers of Fe2 cations and SO4 2 anions.
g. The drawing should have equal numbers of K cations and MnO4 anions.
h. The drawing should have equal numbers of H cations and ClO4 anions.
i. The drawing should have equal numbers of NH4 cations and C2H3O2 anions.
19. CaCl2(s) n Ca2(aq) 2Cl(aq) 21. a. 0.2677 M; b. 1.255 103 M; c. 8.065 103 M 23. a.
1.00 M, 2.00 M; b.
4.0 M, 2.0 M; c.
0.187 M; d.
0.0564 M, 0.0188 M 25. 100.0 mL of 0.30 M AlCl3 27. 4.00 g 29. a. Place 20.0 g NaOH in a 2-L volumetric flask; add water to dissolve the NaOH and fill to the mark. b. Add 500. mL of the 1.00 M NaOH stock solu-tion to a 2-L volumetric flask; fill to the mark with water. c. As in a, instead using 38.8 g K2CrO4. d. As in b, instead using 114 mL of 1.75 M K2CrO4 stock solution.
31. MNH4 0.272 M, MSO42 0.136 M 33. 5.94 108 M steroid 35. Aluminum nitrate, magnesium chloride, and rubidium sulfate are soluble.
37. a. no precipitate forms; b. Al(OH)3(s); c. CaSO4(s); d. NiS(s) 39. a. No reaction occurs because all possible products are soluble salts.
b. 2Al(NO3)3(aq) 3Ba(OH)2(aq) n 2Al(OH)3(s) 3Ba(NO3)2(aq); 2Al3(aq) 6NO3 (aq) 3Ba2(aq) 6OH(aq) n 2Al(OH)3(s) 3Ba2(aq) 6NO3 (aq); Al3(aq) 3OH(aq) n Al(OH)3(s); c. CaCl2(aq) Na2SO4(aq) n CaSO4(s) 2NaCl(aq); Ca2(aq) 2Cl(aq) 2Na(aq) SO4 2(aq) n CaSO4(s) 2Na(aq) 2Cl(aq); Ca2(aq) SO4 2(aq) n CaSO4(s); d. K2S(aq) Ni(NO3)2(aq) n 2KNO3(aq) NiS(s); 2K(aq) S2(aq) Ni2(aq) 2NO3 (aq) n 2K(aq) 2NO3 (aq) NiS(s); Ni2(aq) S2(aq) n NiS(s) 41. a. CuSO4(aq) Na2S(aq) n CuS(s) Na2SO4(aq); Cu2(aq) S2(aq) n CuS(s); the grey spheres are the Na spec-tator ions and the blue-green spheres are the SO4 2 spectator ions; b. CoCl2(aq) 2NaOH(aq) n Co(OH)2(s) 2NaCl(aq); Co2(aq) 2 OH(aq) n Co(OH)2(s); the grey spheres are the Na spectator ions and the green spheres are the Cl spectator ions; c. AgNO3(aq) KI(aq) n AgI(s) KNO3(aq); Ag(aq) I(aq) n AgI(s); the red spheres are the K spectator ions and the blue spheres are the NO3 spectator ions 43. a. Ba2(aq) SO4 2(aq) n BaSO4(s); b. Pb2(aq) 2Cl(aq) n PbCl2(s); c. No reaction; d. No reaction; e. Cu2(aq) 2OH(aq) n Cu(OH)2(s) 45. Ca2, Sr2, or Ba2 could all be present.
47. 0.607 g 49. 0.520 g Al(OH)3 51. 2.9 g AgCl; 0 M Ag; 0.10 M NO3 ; 0.075 M Ca2; 0.050 M Cl 53. 23 amu; Na 55. a. 2HClO4(aq) Mg(OH)2(s) n Mg(ClO4)2(aq) 2H2O(l); 2H(aq) 2ClO4 (aq) Mg(OH)2(s) n Mg2(aq) 2ClO4 (aq) 2H2O(l); 2H(aq) Mg(OH)2(s) n Mg2(aq) 2H2O(l); b. HCN(aq) NaOH(aq) n NaCN(aq) H2O(l); HCN(aq) Na(aq) OH(aq) n Na(aq) CN(aq) H2O(l); HCN(aq) OH(aq) n H2O(l) CN(aq); c. HCl(aq) NaOH(aq) n NaCl(aq) H2O(l); H(aq) Cl(aq) Na(aq) OH(aq) n Na(aq) Cl(aq) H2O(l); H(aq) OH(aq) n H2O(l) 57. a. KOH(aq) HNO3(aq) n H2O(l) KNO3(aq); K(aq) OH(aq) H(aq) NO3 (aq) n H2O(l) K(aq) NO3 (aq); OH(aq) H(aq) n H2O(l); b. Ba(OH)2(aq) 2HCl(aq) n 2H2O(l) BaCl2(aq); Ba2(aq) 2OH(aq) 2H(aq) 2Cl(aq) n Ba2(aq) 2Cl(aq) 2H2O(l); OH(aq) H(aq) n H2O(l); c. 3HClO4(aq) Fe(OH)3(s) n 3H2O(l) Fe(ClO4)3(aq); 3H(aq) 3ClO4 (aq) Fe(OH)3(s) n 3H2O(l) Fe3(aq) 3ClO4 (aq); 3H(aq) Fe(OH)3(s) n 3H2O(l) Fe3(aq) 59. a. 100. mL; b. 66.7 mL; c. 50.0 mL 61. 2.0 102 M excess OH 63. 0.102 M 65. 0.4178 g 67. a. K, 1; O, 2; Mn, 7; b. Ni, 4; O, 2; c. Na, 1; Fe, 2; O, MPO4 3 MK MCl MNH4 MSO4 2 MNa MNO3 MCa2 Mg2+ Mg2+ Mg2+ Cl– Cl– Cl– Cl– Cl– Cl– Na+ Na+ Na+ Br– Br– Br– a.
b.
A44 Answers to Selected Exercises mercury, the column of water must be 13.6 times longer than that of mercury to match the force exerted by the columns of liquid standing on the surface.
19. The P versus 1V plot is incorrect. The plot should be linear with positive slope and a y-intercept of zero. PV k so P k (1/V), which is in the form of the straight-line equation y mx b.
21. d (molar mass) PRT; Be-cause d is directly proportional to the molar mass of the gas. Helium, which has the smallest molar mass of all the noble gases, will have the smallest den-sity.
23. No; At any nonzero Kelvin temperature, there is a distribution of kinetic energies. Similarily, there is a distribution of velocities at any nonzero Kelvin temperature.
25.
At constant P and T, volume is directly proportional to the moles of gas present. In the reaction, the moles of gas doubles as reactants are converted to products, so the volume of the container should double. At constant V and T, P is directly proportional to the moles of gas present. As the moles of gas doubles, the pressure will double. The partial pressure of will be 12 the initial pressure of and the partial pressure of will be 3/2 the initial pressure of The partial pressure of will be three times the partial pressure of 27 a. 3.6 103 mm Hg; b. 3.6 103 torr; c. 4.9 105 Pa; d. 71 psi 29. 65 torr, 8.7 103 Pa, 8.6 102 atm 31. a. 642 torr, 0.845 atm; 8.56 104 Pa; b. 975 torr; 1.28 atm; 1.30 105 Pa; c. 517 torr; 850. torr 33. The balloon will burst.
35. 0.89 mol 37. a. 14.0 L; b. 4.72 102 mol; c. 678 K; d. 133 atm 39. 4.44 103 g He; 2.24 103 g H2 41. a. 69.6 K; b. 32.3 atm 43. 1.27 mol 45. PB 2PA 47. 5.1 104 torr 49. The volume of the balloon increases from 1.00 L to 2.82 L, so the change in volume is 1.82 L.
51. 3.21 g Al 53. 135 g NaN3 55. 1.5 107 g Fe, 2.6 107 g 98% H2SO4 57. 2.47 mol H2O 59. a. 2CH4(g) 2NH3(g) 3O2(g) n 2HCN(g) 6H2O(g); b. 13.3 L 61. Cl2 63. 12.6 g/L 65. 1.1 atm, 1.1 atm, PTOTAL 2.1 atm 67.
317 torr, 50.7 torr, PTOTAL 368 torr 69. a.
0.412, 0.588; b. 0.161 mol; c. 1.06 g CH4, 3.03 g O2 71. 0.990 atm; 0.625 g 73. 18.0% 75. Ptot 6.0 atm; 1.5 atm; 4.5 atm 77. Both CH4(g) and N2(g) have the same average kinetic energy at the various tempera-tures. 273 K, 5.65 1021 J/molecule; 546 K, 1.13 1020 J/molecule 79. CH4: 652 m/s (273 K); 921 m/s (546 K); N2: 493 m/s (273 K); 697 m/s (546 K) 81.
Wall-Collision Average KE Average Velocity Frequency a. Increase Increase Increase b. Decrease Decrease Decrease c. Same Same Increase d. Same Same Increase 83. a. All the same; b. Flask C 85. CF2Cl2 87. The relative rates of ef-fusion of 12C16O, 12C17O, and 12C18O are 1.04, 1.02, and 1.00. Advantage: CO2 isn't as toxic as CO. Disadvantages: Can get a mixture of oxygen isotopes in CO2, so some species would effuse at about the same rate.
89. a. 12.24 atm; b. 12.13 atm; c. The ideal gas law is high by 0.91%.
91. 5.2 106 atm; 1.3 1014 atoms He/cm3 93. 2NO2(g) H2O(l) n HNO3(aq) HNO2(aq); SO3(g) H2O(l) n H2SO4 (aq) 95. a.
PH2 PN2 xO2 xCH4 PN2 PH2 PCO2 N2.
H2 NH3.
H2 NH3 N2 2NH31g2 S N21g2 3H21g2; 2; H, 1 d. H, 1; O, 2; N, 3; P, 5; e. P, 3; O, 2; f. O, 2; Fe, g. O, 2; F, 1; Xe, 6; h. S, 4; F, 1; i. C, 2; O, 2; j. C, 0; H, 1; O, 2 69. a. 3; b. 3; c. 2; d. 2; e. 1; f. 4; g. 3; h. 5; i. 0 71.
Oxidizing Reducing Substance Substance Redox?
Agent Agent Oxidized Reduced a. Yes Ag Cu Cu Ag b. No — — — — c. No — — — — d. Yes SiCl4 Mg Mg SiCl4 (Si) e. No — — — — In b, c, and e, no oxidation numbers change from reactants to products.
73. a. Zn 2HCl n Zn2 H2 2Cl; b. 2H 3I ClO n I3 Cl H2O; c. 7H2O 4H 3As2O3 4NO3 n 4NO 6H3AsO4; d. 16H 2MnO4 10Br n 5Br2 2Mn2 8H2O; e. 8H 3CH3OH Cr2O7 2 n 2Cr3 3CH2O 7H2O 75. a. 2H2O Al MnO4 n Al(OH)4 MnO2; b. 2OH Cl2 n Cl ClO H2O; c. OH H2O NO2 2Al n NH3 2AlO2 77. 4NaCl 2H2SO4 MnO2 n 2Na2SO4 MnCl2 Cl2 2H2O 79. Only statement b is true. a. A non-electrolyte solute can make a concentrated solution. c. Weak acids like acetic acid, HC2H3O2, are weak electrolytes. d. Some ionic compounds do not dis-solve in water (are insoluble). These compounds are not strong electrolytes (nor any type of electrolyte). The electrolyte designation refers to solutes that are soluble in water.
81. a. AgNO3, Pb(NO3)2, and Hg2(NO3)2 would form precipitates with the Cl ion; Ag(aq) Cl(aq) n AgCl(s); Pb2(aq) 2Cl(aq) n PbCl2(s); Hg2 2(aq) 2Cl(aq) n Hg2Cl2(s); b. Na2SO4, Na2CO3, and Na3PO4 would form precipitates with the Ca2 ion; Ca2(aq) SO4 2(aq) n CaSO4(s); Ca2(aq) CO3 2(aq) n CaCO3(s); 3Ca2(aq) 2PO4 3(aq) n Ca3(PO4)2(s); c. NaOH, Na2S, and Na2CO3 would form pre-cipitates with the Fe3 ion; Fe3(aq) 3OH(aq) n Fe(OH)3(s); 2Fe3(aq) 3S2(aq) n Fe2S3(s); 2Fe3(aq) 3CO3 2(aq) n Fe2(CO3)3(s); d. BaCl2, Pb(NO3)2, and Ca(NO3)2 would form precipitates with the SO4 2 ion; Ba2(aq) SO4 2(aq) n BaSO4(s); Pb2(aq) SO4 2(aq) n PbSO4(s); Ca2(aq) SO4 2(aq) n CaSO4(s); e. Na2SO4, NaCl, and NaI would form precipitates with the Hg2 2 ion; Hg2 2(aq) SO4 2(aq) n Hg2SO4(s); Hg2 2(aq) 2Cl(aq) n Hg2Cl2(s); Hg2 2(aq) 2I(aq) n Hg2I2(s); f. NaBr, Na2CrO4, and Na3PO4 would form precipitates with the Ag ion; Ag(aq) Br(aq) n AgBr(s); 2Ag(aq) CrO4 2(aq) n Ag2CrO4(s); 3Ag(aq) PO4 3(aq) n Ag3PO4(s) 83. Ba 85. 39.49 mg/tablet; 67.00% 87. 2.00 M 89. a. 0.8393 M; b. 5.010% 91. C6H8O6 93. Ca(OH)2, Sr(OH)2, and Ba(OH)2 are possibilities for the base.
95.
2H(aq) Mn(s) 2HNO3(aq) n Mn2(aq) 2NO2(g) 2H2O(l); 3H2O(l) 2Mn2(aq) 5IO4 (aq) n 2MnO4 (aq) 5IO3 (aq) 6H(aq) 97. a. 24.8% Co, 29.7% Cl, 5.09% H, 40.4% O; b. CoCl2 6H2O; c. CoCl2 6H2O(aq) 2AgNO3(aq) n 2AgCl(s) Co(NO3)2(aq) 6H2O(l), CoCl2 6H2O(aq) 2NaOH(aq) n Co(OH)2(s) 2NaCl(aq) 6H2O(l), 4Co(OH)2(s) O2(g) n 2Co2O3(s) 4H2O(l) 99. a. 7.000 M K; b. 0.750 M CrO4 2 101. 0.123 g SO4 2, 60.0% SO4 2; 61% K2SO4 and 39% Na2SO4 103. 4.90 M 105. Y, 2.06 mL/min; Z, 4.20 mL/min 107. 57.6 mL 109. a. MgO(s) 2HCl(aq) n MgCl2(aq) H2O(l), Mg(OH)2(s) 2HCl(aq) n MgCl2(aq) 2H2O(l), Al(OH)3(s) 3HCl(aq) n AlCl3(aq) 3H2O(l); b. MgO 111. Citric acid has three acidic hydrogens per citric acid molecule.
113. 0.07849 0.00016 M or 0.0785 0.0002 M 115. 3(NH4)2CrO4(aq) 2Cr(NO2)3(aq) n 6NH4NO2 (aq) Cr2(CrO4)3(s); 7.34 g 117. X Se; H2Se is hydroselenic acid; 0.252 g Chapter 5 17. higher than; 13.6 times taller; When the pressure of the column of liquid standing on the surface of the liquid is equal to the pressure exerted by air on the rest of the surface of the liquid, then the height of the column of liquid is a measure of atmospheric pressure. Because water is 13.6 times less dense than 8 3; PV V P T b.
Answers to Selected Exercises A45 97. 0.772 atm L; In Sample Exercise 5.3, 1.0 mol of gas was present at 0C.
The moles of gas and/or the temperature must have been different for Boyle’s data.
99. MnCl4 101. 1490 103. 24 torr 105. 4.1 106 L air; 7.42 105 L H2 107. 490 atm 109. 13.3% N 111. C12H21NO; C24H42N2O2 113. C3H8 should have the largest b constant. Since CO2 has the largest a con-stant, CO2 will have the strongest intermolecular attractions.
115. 13.4% CaO, 86.6% BaO 117. C2H6 119. a. 8.7 103 L air/min; b.
0.0017, 0.032, 0.13, 0.77, 0.067 121. a. 1.01 104 g; b. 6.65 104 g; c. 8.7 103 g 123. a. Due to air’s larger aver-age molar mass, a given volume of air at a given set of conditions has a higher density than helium. We need to heat the air to greater than 25C to lower the air density (by driving air out of the hot-air balloon) until the density is the same as that for helium (at 25C and 1.00 atm). b. 2150 K 125. a. 0.19 torr; b. 6.6 1015 molecules CO/cm3 127. 0.023 mol 129. 4.8 g/L; UF3 will effuse 1.02 times faster.
Chapter 6 9. Path-dependent functions for a trip from Chicago to Denver are those quan-tities that depend on the route taken. One can fly directly from Chicago to Denver or one could fly from Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc. State functions are path-independent; they only depend on the initial and final states. Some state functions for an airplane trip from Chicago to Denver would be longitude change, latitude change, elevation change, and overall time zone change.
11. Both q and w are negative.
13.
15. Fossil fuels contain carbon; the incomplete combustion of fossil fuels pro-duces CO(g) instead of This occurs when the amount of oxygen reacting is not sufficient to convert all of the carbon in fossil fuels to Carbon monoxide is a poisonous gas to humans.
17. 150 J 19. 1.0 kg ob-ject with velocity of 2.0 m/s.
21. a. 41 kJ; b. 35 kJ; c. 47 kJ; d. part a only 23. 375 J heat transferred to the system 25. 13.2 kJ 27. 11.0 L 29. q 30.9 kJ, w 12.4 kJ, E 18.5 kJ 31. This is an endothermic reaction, so heat must be absorbed to convert reactants into products. The high-temperature environment of internal combustion engines provides the heat.
33. a. endothermic; b. exothermic; c. exothermic; d. endothermic 35. a. 1650 kJ; b. 826 kJ; c. 7.39 kJ; d. 34.4 kJ 37. 4400 g C3H8 39. When a liquid is converted into a gas, there is an increase in volume.
CO2.
CO21g2.
¢H ¢H1 ¢H2 44 kJ H2O1l2 S H2O1g2 ¢H2 1 2 1803 kJ2 1 2CH41g2 O21g2 S 1 2CO21g2 H2O1g2 ¢H1 1 2 1891 kJ2 H2O1l2 1 2CO21g2 S 1 2CH41g2 O21g2 xH2O xN2 xO2 xCO2 xCO The 2.5 kJ/mol quantity can be considered as the work done by the vapor-ization process in pushing back the atmosphere.
41. H2O(l); 2.30 103 J; Hg(l); 140C 43. Al(s) 45. 311 K 47. 23.7C 49. 0.25 J/g C 51. 66 kJ/mol 53. 39.2C 55. a. 31.5 kJ/C; b. 1.10 103 kJ/mol 57. 220.8 kJ 59. 1268 kJ; No, because this reaction is very endothermic, it would not be a practical way of making ammonia due to the high energy costs.
61. 233 kJ 63. 713 kJ 65. The enthalpy change for the for-mation of one mole of a compound from its elements, with all substances in their standard states. Na(s) Cl2(g) n NaCl(s); H2(g) O2(g) n H2O(l); 6Cgraphite(s) 6H2(g) 3O2(g) n C6H12O6(s); Pb(s) S(s) 2O2(g) n PbSO4(s) 67. a. 940. kJ; b. 265 kJ; c. 176 kJ 69. a. 908 kJ, 112 kJ, 140. kJ; b. 12NH3(g) 21O2(g) n 8HNO3(aq) 4NO(g) 14H2O(g), exothermic 71. 2677 kJ 73. 169 kJ/mol 75. 29.67 kJ/g 77. For C3H8(g), 50.37 kJ/g vs. 47.7 kJ/g for octane. Because of the low boiling point of propane, there are extra safety hazards associated with using the necessary high-pressure compressed gas tanks.
79. 1.05 105 L 81. a. 2SO2(g) O2(g) n 2SO3(g); w 0; b. COCl2(g) n CO(g) Cl2(g); w 0; c. N2(g) O2(g) n 2NO(g); w 0; Compare the sum of the coefficients of all the product gases in the balanced equation to the sum of the coefficients of all the reactant gases. When a balanced reaction has more mol of product gases than mol of reactant gases, then the reaction will expand in volume ( V is positive) and the system does work on the surroundings (w 0). When a balanced reaction has a decrease in the mol of gas from reactants to products, then the reaction will contract in volume ( V is negative) and the surroundings does compression work on the system (w 0). When there is no change in the mol of gas from reactants to prod-ucts, then V 0 and w 0.
83. 25 J 85. 4.2 kJ heat released 87. The calculated H value will be less positive (smaller) than it should be.
89. 25.91C 91. a. 632 kJ; b. C2H2(g) 93. a. 361 kJ; b. 199 kJ; c. 227 kJ; d. 112 kJ 95. a. C12H22O11(s) 12O2(g) n 12CO2(g) 11H2O(l); b. 5630 kJ; c. 5630 kJ 97. 37 m2 99. 1 104 steps 101.
56.9 kJ 103. 1.74 kJ 105. 3.3 cm Chapter 7 15. The equations relating the terms are and From the equations, wavelength and frequency are inversely related, photon energy and frequency are directly related, and photon energy and wavelength are inversely related. The unit of 1 Joule (J) This is why you must change mass units to kg when using the deBroglie equation.
17. Sample Exercise 7.3 calculates the deBroglie wavelength of a ball and of an electron.
The ball has a wavelength on the order of This is incredibly short and, as far as the wave-particle duality is concerned, the wave properties of large objects are insignificant. The electron, with its tiny mass, also has a short wavelength; on the order of However, this wavelength is significant as it is on the same order as the spacing between atoms in a typical crystal.
For very tiny objects like electrons, the wave properties are important. The wave properties must be considered, along with the particle properties, when hypothesizing about the electron motion in an atom.
19. For the radial prob-ability distribution, the space around the hydrogen nucleus is cut up into a se-ries of thin spherical shells. When the total probability of finding the electron in each spherical shell is plotted versus the distance from the nucleus, we get the radial probability distribution graph. The plot shows a steady increase with distance from the nucleus, maximizes at a certain distance from the nucleus, then shows a steady decrease. Even though it is likely to find an electron near the nucleus, the volume of the spherical shell close to the nucleus is tiny, re-sulting in a low radial probability. The maximum radial probability distribu-tion occurs at a distance of from the nucleus; the electron is most likely to be found in the volume of the shell centered at this distance from the nucleus. The distance is the exact radius of innermost orbit in the Bohr model.
21. If one more electron is added to a half-filled subshell, electron–electron repulsions will increase because two electrons must now occupy the same atomic orbital. This may slightly decrease the stability of the atom. Hence, half-filled subshells minimize electron–electron 1n 12 5.29 102 nm 5.29 102 nm 1010 m.
1034 m.
1 kg m2/s2.
E hc l.
E hn, nl c, 1 2 1 2 c.
d.
T V P V P 1/V T PV P e.
f.
A46 Answers to Selected Exercises 69. Si: 1s22s22p63s23p2 or [Ne]3s23p2; Ga: 1s22s22p63s23p64s23d104p1 or [Ar]4s23d104p1; As: [Ar]4s23d104p3; Ge: [Ar]4s23d104p2; Al: [Ne]3s23p1; Cd: [Kr]5s24d10; S: [Ne]3s23p4; Se: [Ar]4s23d104p4 71. Sc: 1s22s22p63s23p6 4s23d1; Fe: 1s22s22p63s23p64s23d 6; P:1s22s22p63s23p3; Cs: 1s22s22p63s23p6 4s23d104p65s24d105p66s1; Eu: 1s22s22p63s23p64s23d104p65s24d105p66s24f 65d1 (Actual: [Xe]6s24f 7); Pt: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d 8 (Actual: [Xe]6s14f 145d9); Xe: 1s22s22p63s23p64s23d104p65s24d105p6; Br: 1s22s22p63s23p64s23d104p5; 73. a. I: [Kr]5s24d105p5; b. element 120: [Rn]7s25f 146d107p68s2; c. Rn:[Xe]6s24f 145d106p6; d. Cr:[Ar]4s13d5 75. a. 18; b. 30; c. 10; d. 40 77. B: 1s22s22p1 n m ms 1s 1 0 0 1 2 1s 1 0 0 1 2 2s 2 0 0 1 2 2s 2 0 0 1 2 2p 2 1 1 1 2 For boron, there are six possibilities for the 2p electron. For nitrogen, all the 2p electrons could have ms 1 2 .
79. none; an excited state; energy re-leased 81. C, O, Si, S, Ti, Ni, Ge, Se 83. Li (1 unpaired electron), N (3 un-paired electrons), Ni (2 unpaired electrons), and Te (2 unpaired electrons) are all expected to be paramagnetic because they have unpaired electrons.
85. a. S Se Te; b. Br Ni; K; c. F Si Ba 87. a. Te Se S; b. K Ni Br; c. Ba Si F 89. a. He; b. Cl; c. element 117; d. Si; e. Na 91. a. [Rn]7s25f146d4; b. W; c. SgO3 and SgO4 2 probably would form (similar to Cr).
93. Se is an exception to the general ionization trend. There are extra electron–electron repulsions in Se because two electrons are in the same 4p orbital, resulting in a lower ionization energy than expected.
95. a. C, Br; b. N, Ar; c. C, Br 97. Al (44), Si (120), P (74), S (200.4), Cl (348.7); Based on the increasing nuclear charge, we would expect the electron affinity (EA) values to become more exothermic as we go from left to right in the period. Phosphorus is out of line. The reaction for the EA of P is [Ne]3s23p3 [Ne]3s23p4 The additional electron in P will have to go into an orbital that already has one electron. There will be greater repulsions between the paired elec-trons in P, causing the EA of P to be less favorable than predicted based solely on attractions to the nucleus.
99. a. I Br F Cl; b. N O F 101. a.
b.
c.
; d.
103. potassium peroxide, K2O2; K2 unstable 105. 6.582 1014 s1; 4.361 1019J 107. Yes; the ionization energy general trend decreases down a group and the atomic radius trend increases down a group. The data in Table 7.8 confirm both of these general trends.
109. a.
; b.
111. 386 nm 113. 200 s 115. 4.104 105 cm so violet light is emitted.
117. a. true for H only; b. true for all atoms; c. true for all atoms 119. When the p and d orbital functions are evaluated at various points in space, the results sometimes have positive values and sometimes have nega-tive values. The term phase is often associated with the and signs. For example, a sine wave has alternating positive and negative phases. This is analogous to the positive and negative values (phases) in the p and d orbitals.
121. The element with the smallest first ionization energy (I1) is Al (the green plot), the next highest I1 belongs to Mg (the blue plot), and Si has the largest I1 (the red plot). Mg is the element with the huge jump between I2 and I3; it has two valence electrons so the third electron removed is an inner-core electron.
Inner-core electrons are always much more difficult to remove as compared to valence electrons because they are closer to the nucleus, on average, as S1s2 S Rb2S1s2 2Rb1s2 6Li1s2 N21g2 S 2Li3N1s2 Mg1g2 S Mg1g2 e e S Fe21g2 Fe31g2 S1g2 e S S21g2; Se31g2 S Se41g2 e; P1g2 e S P1g2 N: 1s22s22p3 n m ms 1s 1 0 0 1 2 1s 1 0 0 1 2 2s 2 0 0 1 2 2s 2 0 0 1 2 2p 2 1 1 1 2 2p 2 1 0 1 2 2p 2 1 1 1 2 repulsions.
23. The valence electrons are strongly attracted to the nucleus for elements with large ionization energies. One would expect these species to readily accept another electron and have very exothermic electron affinities.
The noble gases are an exception; they have a large ionization energy but an endothermic electron affinity. Noble gases have a stable arrangement of elec-trons. Adding an electron disrupts this stable arrangement, resulting in unfa-vorable electron affinities.
25. For hydrogen, all orbitals with the same value of n have the same energy. For polyatomic atoms/ions, the energy of the or-bitals also depends on . Because there are more nondegenerate energy levels for polyatomic atoms/ions as compared with hydrogen, there are many more possible electronic transitions resulting in more complicated line spec-tra.
27. Yes, the maximum number of unpaired electrons in any configura-tion corresponds to a minimum in electron–electron repulsions.
29. Ioniza-tion energy applies to the removal of the electron from the atom in the gas phase. The work function applies to the removal of an electron from the solid.
31. 4.5 1014 s1 33. 3.0 1010 s1, 2.0 1023 J/photon, 12 J/mol 35. Wave a has the longer wavelength (4.0 104 m). Wave b has the higher fre-quency (1.5 1012 s1) and larger photon energy (9.9 1022 J). Since both of these waves represent electromagnetic radiation, they both should travel at the same speed, c, the speed of light. Both waves represent infrared radiation.
37. 1.50 1023 atoms 39. 427.7 nm 41. a. 2.4 1011m; b. 3.4 1034 m 43. 1.6 1027 kg 45. a. 656.7 nm (visible); b. 486.4 nm (visible); c. 121.6 nm (ultraviolet) 47.
49. n 1 n n 5, 95.00 nm; n 2 n n 6, 410.4 nm; visible light has sufficient energy for the n 2 n n 6 transition but does not have sufficient energy for the n 1 n n 5 transition.
51. n 1, 91.20 nm; n 2, 364.8 nm 53. n 2 55. a. 5.79 104 m; b. 3.64 1033 m; c. The diameter of an H atom is roughly 1.0 108 cm. The uncertainty in position is much larger than the size of the atom. d. The uncertainty is insignificant compared to the size of a baseball.
57. n 1, 2, 3, . . . ; 0, 1, 2, . . . (n 1); m , . . . , 2, 1, 0, 1, 2, . . . , .
59. b. For can range from 3 to 3; thus 4 is not allowed.
c. n cannot equal zero.
d.
cannot be a negative number.
61. 2 gives the probability of finding the electron at that point.
63. 3; 1; 5; 25; 16 65. a. 32; b. 8; c. 25; d. 10; e. 6 67. a.
b.
c.
1s 2s 2p 3p 4s 3d 4p 3s 3 1s 2s 2p 3p or 4s 3d 3s 3 1s 2s 2p 3s 3s or / / 3, m/ 1 E a b c 3 2 4 Answers to Selected Exercises A47 compared to the valence electrons.
123. Valence electrons are easier to remove than inner-core electrons. The large difference in energy between I2 and I3 in-dicates that this element has two valence electrons. The element is most likely an alkaline earth metal because alkaline earth metals have two valence electrons.
125. a. 146 kJ; b. 407 kJ; c. 1117 kJ; d. 1524 kJ 127. a. line A, n 6 to n 3; line B, n 5 to n 3; b. 121.6 nm 129. For r a0 and 0, 2 2.46 1028. For r a0 and 90, 2 0.
As expected, the xy plane is a node for the 2pz, atomic orbital.
131. a.
b. 2, 4, 12, and 20; c. There are many possibilities. One example of each for-mula is XY 1 11, XY2 6 11, X2Y 1 10, XY3 7 11, and X2Y3 7 10; d. 6; e. 0; f. 18 133. The ratios for Mg, Si, P, Cl, and Ar are about the same. However, the ratios for Na, Al, and S are higher. For Na, the second IE is extremely high because the electron is taken from n 2 (the first electron is taken from n 3). For Al, the first electron requires a bit less energy than expected by the trend due to the fact it is a 3p electron. For S, the first electron requires a bit less energy than expected by the trend due to elec-trons being paired in one of the p orbitals.
135. a. As we remove succeed-ing electrons, the electron being removed is closer to the nucleus and there are fewer electrons left repelling it. The remaining electrons are more strongly attracted to the nucleus and it takes more energy to remove these electrons.
b. For I4, we begin removing an electron with n 2. For I3, we removed an elec-tron with n 3. In going from n 3 to n 2 there is a big jump in ionization energy because the n 2 electrons (inner core) are much closer to the nucleus on the average than the n 3 electrons (valence electrons). Since the n 2 electrons are closer to the nucleus, they are held more tightly and require a much larger amount of energy to remove them. c. Al4; the electron affinity for Al4 is H for the reaction d. The greater the number of electrons, the greater the size. So 137. Solving for the molar mass of the element gives 40.2 g/mol; this is cal-cium.
139. a. Fr [Rn]7s1, Fr [Rn]; b. 7.7 1022 Fr atoms; c. 2.27790 1022 g Chapter 8 13. (NH4)2SO4 and Ca3(PO4)2 are compounds with both ionic and covalent bonds.
15. Electronegativity increases left to right across the periodic table and decreases from top to bottom. Hydrogen has an electronegativity value be-tween B and C in the second row, and identical to P in the third row. Going further down the periodic table, H has an electronegativity value between As and Se (row 4) and identical to Te (row 5). It is important to know where hy-drogen fits into the electronegativity trend, especially for rows 2 and 3. If you know where H fits into the trend, then you can predict bond dipole directions for nonmetals bonded to hydrogen.
17. For ions, concentrate on the number of protons and the number of electrons present. The species whose nucleus holds the electrons most tightly will be smallest. For example, compare the size of an anion to the neutral atom. The anion has more electrons held by the same number of protons in the nucleus. These electrons will not be held as tightly, resulting in a bigger size for the anion as compared to the neutral atom.
For isoelectronic ions, the same number of electrons are held by different numbers of protons in the various ions. The ion with the most protons holds the electrons tightest and has smallest size.
19. Fossil fuels contain a lot of carbon and hydrogen atoms. Combustion of fossil fuels (reaction with produces and Both these compounds have very strong bonds.
Because strong bonds are formed, combustion reactions are very exothermic.
H2O.
CO2 O22 Al4 6 Al3 6 Al2 6 Al 6 Al Al41g2 e ¡ Al31g2 ¢H I4 11,600 kJ/mol 21.
The formal charges are shown above the atoms in the three Lewis structures.
The best Lewis structure for , from a formal charge standpoint, is the first structure with each oxygen double bonded to carbon. This structure has a for-mal charge of zero on all atoms (which is preferred). The other two resonance structures have nonzero formal charges on the oxygens, making them less rea-sonable. For we usually ignore the last two resonance structures and think of the first structure as the true Lewis structure for 23. a. C N O; b. Se S Cl; c. Sn Ge Si; d. Tl Ge S 25. a. GeOF; b. POCl; c. SOF; d. TiOCl 27. Order of electronegativity from Fig. 8.3: a. C(2.5) N (3.0) O (3.5), same; b. Se (2.4) S (2.5) Cl (3.0), same; c. Si Ge Sn (1.8), different; d. Tl (1.8) Ge (1.8) S (2.5), different. Most polar bonds using actual electronegativity values: a. SiOF and GeOF equal polarity (Ge—F predicted); b. POCl (same as predicted); c. SOF (same as predicted); d. TiOCl (same as predicted) 29. Incorrect: b, d, e; b.
; d.
nonpolar bond so no dipole moment; e.
31. FOH OOH NOH COH POH 33. Fr: [Xe]6s24f 145d106p6; Be2: ls2; P3 and Cl: [Ne]3s23p6; Se2: [Ar]4s23d104p6 35. a. Sc3; b. Te2; c. Ce4, Ti4; d. Ba2 37. Sb3, Te2, I, Cs, Ba2, and La3 are some possibilities. La3 Ba2 Cs I Te2 Sb3 39. a. Cu Cu Cu2; b. Pt2 Pd2 Ni2; c. O2 O O; d. La3 Eu3 Gd3 Yb3; e. Te2 I Cs Ba2 La3 41. a. Al2S3, aluminum sulfide; b. K3N1, Potassium nitride; c. MgCl2, Magne-sium chloride; d. CsBr, Cesium bromide 43. a. NaCl, Na smaller than K; b. LiF, F smaller than Cl; c. MgO, O2 greater charge than OH; d. Fe(OH)3, Fe3 greater charge than Fe 2; e. Na2O, O2 greater charge than Cl; f. MgO, Mg2 smaller than Ba2, and O2 smaller than S2.
45. 437 kJ/mol 47. The lattice energy for Mg2O2 will be much more exothermic than for Mg O.
49. 181 kJ/mol 51. Ca2 has greater charge than Na, and Se2 is smaller than Te2. Charge differences affect lattice energy values more than size differences, and we expect the trend from most exothermic to least exothermic to be: CaSe CaTe Na2Se Na2Te (2862) (2721) (2130) (2095) 53. a. 183 kJ; b. 109 kJ 55. 42 kJ 57. 1276 kJ 59. 295 kJ 61. 485 kJ/mol 63. a. Using standard enthalpies of formation, H° 184 kJ vs. 183 kJ from bond energies; b. Using standard enthalpies of formation, H 92 kJ vs. 109 kJ from bond energies. Bond energies give a reasonably good estimate for H, especially when all reactants and products are gases.
65. a. Using SF4 data: DSF 342.5 kJ/mol. Using SF6 data: DSF 327.0 kJ/mol. b. The SOF bond energy in the table is 327 kJ/mol. The value in the table was based on the SOF bond in SF6. c. S(g) and F(g) are not the most stable forms of the elements at 25°C. The most stable forms are S8(s) and F2(g); H° f 0 for these two species.
67. a.
b.
c.
d.
e.
f.
g.
h.
i.
69. HOBeOH H H H B H Br O ð š O O P š š C O O P P š š F Se O OF š ð š šð O C H H B D G ð ð H H O O A N H H A + H Cl C Cl Cl O O A A š ð š ð ð ð H H H P O O A š H N Cq O ð dO¬Pd dCl¬Id CO2.
CO2, CO2 mn mn PO P O C qO O O C P O O O O C 1 1 1 1 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A48 Answers to Selected Exercises 75.
77. With resonance all carbon–carbon bonds are equivalent (we indicate this with a circle in the ring), giving three different structures: Localized double bonds give four unique structures.
79. N2 (triple bond) N2F2 (double bond) N2F4 (single bond) 81. a.–f. and h. all have similar Lewis structures: g.
Formal charges: a. 1; b. 2; c. 3; d. 1; e. 2; f. 4; g. 2; h. 1 83.
Oxidation numbers are more useful. We are forced to assign 1 as the oxida-tion number for oxygen. Oxygen is very electronegative and 1 is not a sta-ble oxidation state for this element.
85.
87. a. linear, 180; b.
trigonal pyramid, 109.5; c. tetrahedral, 109.5; d. tetrahedral, 109.5; e. trigonal planar, 120; f. V-shaped, 109.5; g. linear, 180; h. and i. linear, no bond angle in diatomic molecules; a. NO2 : V-shaped, 120; NO3 : trigonal planar, 120; N2O4: trigonal pla-nar about both N atoms, 120; b. all are linear, 180 89. Br3 : linear; ClF3; T-shaped; SF4: see-saw 91. a. trigonal planar; 120; b. V-shaped; 120 93. a. linear; 180; b. T-shaped; 90; c. see-saw; 90 and 120; d. trigonal bipyramid; 90 and 120 95. SeO2 (bond dipoles do not cancel each other out in SeO2) 97. ICl3 and TeF4 (bond dipoles do not cancel each other out in ICl3 and TeF4) 99. a. OCl2 V-shaped, polar KrF2 Linear, nonpolar Kr F F O O š ð šð ý k Cl O O D G ½ k ½ ý ý k Cl S Cl S F F Formal charge: Oxidation number: O O 0 1 0 1 0 1 0 1 ClO3 – – O Cl O O O O A ð ð ð ð X Y Y Y Y O O A A ð š ð ð š ð š š ð š šð Cl Cl Cl Cl Cl Cl C C C C H H H H H B mn C C A A H A K H H E H E E N C C C C H H H H H B C C A A H A E N H E H E K H 71.
73. a.
N2O4 b.
OCN– – C O q N O š ð ð – C O P N P – C Oq N ; O š ð ð ½ ý k N k SCN– – C S q N O š ð ð – C S P N P – C Sq N ; O š ð ð ½ ý k N3 – – N N q N O š ð ð – N N P P – N Nq N O š ð ð ½ ý f f f f f f f f f f f f N N O O O O N N O N N N N O O O O O O O O O O O – O O O N A G mn J ð ð ð ð ý ½ ý – O O O N A M D ð ð ½ ý ý NO3 – – O O O N B G mn D ð ð ½ ý ½ ý NO2 – – N OP O O š š š šð – N O P O O š ð š mn PF5 SF4 ClF3 Br3 _ P F F F F F S F F F F Br O Br O – Br Cl F F F Cl Answers to Selected Exercises A49 BeH2 HOBeOH Linear, nonpolar SO2 V-shaped, polar (one other resonance structure possible) b. SO3 Trigonal planar, nonpolar (Two other resonance structures possible) NF3 Trigonal pyramid, polar IF3 T-shaped, polar c. CF4 Tetrahedral, nonpolar SeF4 See-saw, polar KrF4 Square planar, nonpolar d. IF5 Square pyramid, polar AsF5 Trigonal bipyramid, nonpolar 101. Element E is a halogen (F, Cl, Br, or I); trigonal pyramid; 109.5 103. The polar bonds are symmetrically arranged about the central atoms, and all the individual bond dipoles cancel to give no net dipole moment for each molecule, i.e., the molecules are nonpolar.
105. a. radius: N N N; I.E. N N N; b. radius: Cl Cl Se Se; I.E.
Se Se Cl Cl; c. radius: Sr2 Rb Br; I.E. Br Rb Sr2; 107. a. 1549 kJ; b. 1390. kJ c. 1312 kJ; d. 1599 kJ 109. a. NaBr: In NaBr2, the sodium ion would have a 2 charge assuming each bromine has a 1 charge. Sodium doesn’t form stable Na2 compounds.
b. ClO4 : ClO4 has 31 valence electrons so it is impossible to satisfy the octet rule for all atoms in ClO4. The extra electron from the 1 charge in ClO4 allows for complete octets for all atoms. c. XeO4: We can’t draw a Lewis F As F O A F F F A D G ð ð ð ð ð ð ð F F F F F I AD G D G ð š ð ð š ðš ð š ðš F F F F Kr Se F F F F O O D G ð ð ð ð ð ð F C A F F F A E H ð š ð š ð šð ð ð F F I F N F F F A E H š ð šð ð ð O S O O D G ½ ½ ý k B ð ð S O O J G š ½ ý k structure that obeys the octet rule for SO4 (30 electrons), unlike XeO4 (32 elec-trons). d. SeF4: Both compounds require the central atom to expand its octet.
O is too small and doesn’t have low-energy d orbitals to expand its octet (which is true for all row 2 elements).
111. a. Both have one or more 180 bond an-gles; both are made up entirely of Xe and Cl; both have the individual bond dipoles arranged so they cancel each other (both are nonpolar); both have lone pairs on the central Xe atom; both have a central Xe atom that has more than 8 electrons around it. b. All have lone pairs on the central atom; all have a net dipole moment (all are polar).
113. Yes, each structure has the same num-ber of effective pairs around the central atom. (We count a multiple bond as a single group of electrons.) 115.
The lone pair of electrons around Te exerts a stronger repulsion than the bond-ing pairs. The stronger repulsion pushes the four square planner F atoms away from the lone pair, reducing the bond angles between the axial F atom and the square planar F atoms.
117. 17 kJ/mol 119. See Fig. 8.11 to see the data supporting MgO as an ionic compound. Note that the lattice energy is large enough to evercome all of the other processes (removing 2 electrons from Mg, and so on). The bond energy for O2 (247 kJ/mol) and electron affinity (737 kJ/mol) are the same when making CO. However, the energy needed to ionize carbon to form a C2 ion must be too large. Figure 7.30 shows that the first ionization energy for carbon is about 400 kJ/mol greater than the first IE for magnesium. If all other numbers were equal, the overall energy change would be 200 kJ/mol (see Fig. 8.11). It is not unreasonable to assume that the second ionization energy for carbon is more than 200 kJ/mol greater than the second ionization energy of magnesium.
121. As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around the small nitrogen atom, and the NX3 molecule becomes less stable.
123. reaction i: 2636 kJ; reaction ii: 3471 kJ; reaction iii: 3543 kJ; Reaction iii yields the most energy per kg (8085 kJ/kg) 125.
127. a. Two possible structures exist; each has a T-shaped molecular structure: b. Three possible structures exist; each has a see-saw molecular structure.
O Xe O F F 90 bond angle between O atoms O Xe F F O 180 bond angle between O atoms F Xe O O F 120 bond angle between O atoms I I Br F 180 bond angle between I atoms I F Br I 90 bond angle between I atoms H C O O C O O O O H H A A O O G N ð ð O O ý ý ý ý k k P mn P G H C O O C O O O O H H A A O O N ð ð O O ý ý k k k G P P P P mn H C O O C O O H H A A A O O N ð ð O O ý ý k k k G Te F F – F F F A G D G D ð ð ð ð ð ð (This form is not important by formal charge arguments.) A50 Answers to Selected Exercises H2O has a tetrahedral arrangement of the electron pairs about the O atom that requires sp3 hybridization. Two of the sp3 hybrid orbitals are used to form bonds to the two hydrogen atoms, and the other two sp3 hybrid orbitals hold the two lone pairs on oxygen.
17.
The central carbon atom has a trigonal planar arrangement of the electron pairs that requires sp2 hybridization. Two of the sp2 hybrid orbitals are used to form the two bonds to hydrogen. The other sp2 hybrid orbital forms the bond to oxygen. The unchanged (unhybridized) p orbital on carbon is used to form the bond between carbon and oxygen.
19. Ethane: The carbon atoms are sp3 hybridized. The six COH bonds are formed from the sp3 hybrid orbitals on C with the 1s atomic orbitals from the hydrogen atoms. The carbon–carbon bond is formed from an sp3 hybrid orbital on each C atom.
Ethanol: The two C atoms and the O atom are all sp3 hybridized. All bonds are formed from these sp3 hybrid orbitals. The COH and OOH bonds form from sp3 hy-brid orbitals and the 1s atomic orbitals from the hydrogen atom. The COC and COO bonds are formed from sp3 hybrid orbitals on each atom.
21. a. sp; b. sp3; c. sp3; d. sp3; e. sp2; f. sp3; g. sp; h. each O is sp2 hy-bridized; i. Br is sp3 hybridized a. NO2 , sp2; NO3 , sp2; N2O4: both N atoms are sp2 hybridized; b. All are sp hybridized.
23. All exhibit dsp3 hybridization.
25. The molecules in Exercise 91 all exhibit sp2 hybridization about the central atom; the molecules in Exercise 92 all exhibit sp3 hybridization about the central atom.
27. a. tetrahedral, 109.5°, sp3, nonpolar b. trigonal pyramid, 109.5°, sp3, polar c. V-shaped, 109.5°, sp3, polar d. trigonal planar, 120°, sp2, nonpolar e. linear, 180°, sp, nonpolar HOBeOH F D G ý F ý k k k k B F A ð ð O F F D G ð ý k ý k k N F F F O D G ð ð ð ð C F F F F A A E H ð ð ð ð ð ð H C O H H H H C H H H C H H H H C H2CO C H H D D ð O B ð ð c. Three possible structures exist; each has a square pyramid molecular structure.
129. H 0 131. X is iodine; square pyramid 133. Cr As P Cl Chapter 9 7. In hybrid orbital theory, some or all of the valence atomic orbitals of the central atom in a molecule are mixed together to form hybrid orbitals; these hybrid orbitals point to where the bonded atoms and lone pairs are oriented.
The bonds are formed from the hybrid orbitals overlapping head to head with an appropriate orbital on the bonded atom. The bonds in hybrid orbital theory are formed from unhybridized p atomic orbitals. The p orbitals overlap side to side to form the bond where the electrons occupy the space above and below a line joining the atoms the internuclear axis). Assuming the z-axis is the internuclear axis, then the pz atomic orbital will always be hybridized whether the hybridization is sp, sp2, sp3, dsp2 or d2sp3. For sp hybridization, the and atomic orbitals are unhybridized; they are used to form two bonds to the bonded atom(s). For hybridization, either the or atomic orbital is hybridized (along with the s and orbitals); the other p orbital is used to form a bond to a bonded atom. For hybridization, the s and all of the p orbitals are hybridized; no unhybridized p atomic orbitals are present, so typical bonds do not form with hybridization. For and hybridization, we just mix in one or two d orbitals into the hybridization process. Which specific d orbitals are used is not important to our discussion.
9. We use d orbitals when we have to; i.e., we use d orbitals when the central atom on a molecule has more than eight electrons around it. The d orbitals are necessary to accommodate the electrons over eight. Row 2 elements never have more than eight electrons around them so they never hybridize d orbitals. We rationalize this by saying there are no d orbitals close in energy to the valence 2s and 2p orbitals (2d orbitals are forbidden energy levels). However, for row 3 and heavier elements, there are 3d, 4d, 5d, etc. orbitals which will be close in energy to the valence s and p orbitals. It is row 3 and heavier nonmetals that hybridize d orbitals when they have to. For sulfur, the valence electrons are in 3s and 3p orbitals. Therefore, 3d orbitals are closest in energy and are avail-able for hybridization. Arsenic would hybridize 4d orbitals to go with the valence 4s and 4p orbitals while iodine would hybridize 5d orbitals since the valence electrons are in 11. Bonding and antibonding molecular orbitals are both solutions to the quantum mechanical treatment of the mole-cule. Bonding orbitals form when in phase orbitals combine to give construc-tive interference. This results in enhanced electron probability located between the two nuclei. The end result is that a bonding MO is lower in energy than the atomic orbitals of which it is composed. Antibonding orbitals form when out-of-phase orbitals combine. The mismatched phases produce destructive interference leading to a node in the electron probability between the two nuclei.
With electron distribution pushed to the outside, the energy of an antibonding orbital is higher than the energy of the atomic orbitals of which it is composed.
13. The localized electron model does not deal effectively with molecules containing unpaired electrons. We can draw all of the possible resonance struc-tures for NO, but still not have a good feel for whether the bond in NO is weaker or stronger than the bond in MO theory can handle odd electron species without any modifications. In addition, hybrid orbital theory does not predict that is paramagnetic. The MO theory correctly makes this prediction.
15. H2O O H H D D ð ý k NO NO.
n 5.
d2sp3 dsp3 sp3 p sp3 p pz py px sp2 p py px p p p Te Cl Cl Cl One F atom is 180 from lone pair.
Cl Te Cl Cl F F F F Both F atoms are 90 from lone pair and 90 from each other.
Cl Te F Cl F Cl Both F atoms are 90 from lone pair and 180 from each other.
Answers to Selected Exercises A51 f.
a 120°, see-saw, dsp3, polar b 90° g.
a 90°, trigonal bipyramid, dsp3, nonpolar b 120° h.
linear, 180°, dsp3, nonpolar i.
square planar, 90°, d2sp3, nonpolar j.
octahedral, 90°, d2sp3, nonpolar k. l.
square pyramid, 90°; T-shaped, 90°; d2sp3, polar dsp3, polar 29. The bond forces all six atoms into the same plane.
31. Biacetyl All CCO angles are 120°. The six atoms are not in the same plane. 11 and 2 Acetoin angle a 120°, angle b 109.5°, 13 and 1 bond 33. To complete the Lewis structure, add lone pairs to complete octets for each atom. a. 6; b. 4; c. The center N in ONPNPN group; d. 33 ; e. 5 ; f. 180°; g. 109.5°; h. sp3 35. a. H2 , H2, H2 ; b. He2 2 and He2 37. a. (
2s)2; B.O. 1; diamagnetic (0 unpaired electrons); b. (
2s)2(
2s)2(2p)4; B.O. 2; diamagnetic (0 unpaired electrons); c. (
3s)2(
3s)2(
3p)2(3p)4(3p)2; B.O. 2; paramagnetic (2 unpaired electrons) 39. When O2 loses an elec-tron, it comes from a pi antibonding orbital, which strengthens the bond from a bond order of 2 to a bond order of 2.5. When N2 loses an electron, C H H H H B O O A A H A C C C O H O A f f f f f f f A E E H H a b sp3 sp3 sp2 E E G h ð ð ý k C C C O O sp2 sp3 H H B K h h h h O A H A HO H A COH A E H ð š ý k F I F F F I F F F F Se F F F F F F O O O O A A D D ð ð ð ð ð ð ð ð Kr F F F O F O D D ð ð ð ð ð ð FOKr F O ð ð ý k As a b F F F F O A F A D f f f f G ð ð ð ð ð ð ð Te F a b A f f f f F F F A E H ð ð ð ð ð ð ð it comes a pi bonding orbital, which changes the bond order from 3 to 2.5 (the bond weakens).
41. N2 and N2 43. a. (
2s)2(
2s)2(2p)4(
2p)2; B.O. 3; diamagnetic; b. (
2s)2(
2s)2 (2p)4(
2p)1; B.O. 2.5; paramag-netic; c. (
2s)2(
2s)2(2p)4; B.O. 2; diamagnetic; bond length: CO CO CO2; bond energy: CO2 CO CO 45. H2; B2; C2 2 47.
49. a. The electrons would be closer to F on the average. The F atom is more electronegative than the H atom, and the 2p orbital of F is lower in energy than the 1s orbital of H; b. The bonding MO would have more flu-orine 2p character because it is closer in energy to the fluorine 2p orbital; c. The antibonding MO would place more electron density closer to H and would have a greater contribution from the higher-energy hydrogen 1s atomic orbital.
51. O3 and NO2 have identical Lewis structures, so we need to discuss only one of them. The Lewis structure for O3 is Localized electron model: The central oxygen atom is sp2 hybridized, which is used to form the two bonds and hold the lone pair of electrons. An unchanged (unhybridized) p atomic orbital forms the bond with the neighboring oxygen atoms. The bond resonates between the two positions. Molecular orbital model: There are two localized bonds and a bond that is delocalized over the entire surface of the molecule. The delocalized bond results from over-lap of a p atomic oribtal on each oxygen atom in O3.
53. a. Trigonal pyramid; sp3 b. Tetrahedral; sp3 c. Square pyramid; d2sp3 d. T-shaped; dsp3 e. Trigonal bipyramid; dsp3 O F F O O O O Xe F O O Xe F F or or O O Xe F F Xe O F Xe O F F or O F F F F Xe F F F F O Xe or O Xe O O O O Xe O O OK O OH mn ½ ½ OK O OH ½ ½ A52 Answers to Selected Exercises The central N is sp hybridized. We can probably ignore the third resonance structure on the basis of formal charge. c. sp hybrid orbitals on the center N overlap with atomic orbitals (or hybrid orbitals) on the other two atoms to form two bonds. The remaining p orbitals on the center N overlap with p orbitals on the other N to form two bonds.
63. N2: (
2s)2(
2s)2(2p)4(
2p)2 in ground state, B.O. 3, diamagnetic; 1st excited state: (
2s)2(
2s)2(2p)4(
2p)1(2p)1, B.O. 2, paramagnetic (two unpaired electrons) 65. F2 (
2s)2(
2s)2(
2p)2(2p)4(2p)4; F2 should have a lower ionization energy than F. The electron removed from F2 is in a 2p antibonding molecular orbital, which is higher in energy than the 2p atomic orbitals from which the electron in atomic fluorine is removed. Since the electron removed from F2 is higher in energy than the electron removed from F, then it should be easier to remove an electron from F2 than from F.
67. molecular orbital 69. 6 sp2; 6 sp3; 0 sp; 25 ; 4 71. a. No, some atoms are attached differently; b. Structure 1: All N sp3, all C sp2; structure 2: All C and N sp2; c. The first structure with the carbon–oxygen double bonds is slightly more stable.
73. a. NCN2: Dicyandiamide: Melamine: b. NCN2: C is sp hybridized. Each resonance structure predicts a different hy-bridization for the N atoms. For the remaining compounds, we will predict hy-brids for the favored resonance structures only.
Melamine: N in NH2 groups are all sp3 hybridized. Atoms in ring are all sp2 hybridized; c. NCN2: 2 and 2 bonds; H2NCN: 4 and 2 bonds; dicyandiamide: 9 and 3 bonds; melamine: 15 and 3 bonds; d. The system forces the ring to be planar just as the benzene ring is planar.
N C H H NH2 NH2 sp3 O m sp sp sp2 sp3 m m N q D G D G ð ð ½ N C q m m m m m m N O C P ½ H N N H H C C N N H B C N A N H H A O O A A mn G D H E K N ð ½ H N N H H C C N N H B C N A N A H A O HO O A A G D H E K N ½ N +1 –1 Favored by formal charge C H H P N P mn D G ð k N 0 0 0 0 C H H O N q D G H2NCN: N C 2– P N P mn ð ð ½ ý ½ ý N C 2– q N O mn ð ð N C 2– q N O 55.
a. 21 bonds; 4 bonds (The electrons in the 3 bonds in the ring are delocalized.) b. angles a, c, and g: 109.5; angles b, d, e, and f: 120; c. 6 sp2 carbons; d. 4 sp3 atoms; e. Yes, the electrons in the ring are delocalized. The atoms in the ring are all sp2 hybridized. This leaves a p or-bital perpendicular to the plane of the ring from each atom. Overlap of all six of these p orbitals results in a molecular orbital system where the electrons are delocalized above and below the plane of the ring (similar to benzene in Fig. 9.48 of the text).
57. 267 kJ/mol; this amount of energy must be supplied to break the bond.
59. a.
Trigonal planar, nonpolar, 120, sp2 b. N2F2 can also be V-shaped about both N atoms, 120°, sp2 These are distinctly different molecules.
c.
d. ICl3 T-shaped, polar, 90°, dsp3 61. a. The NNO structure is correct. From the Lewis structures we would pre-dict both NNO and NON to be linear, but NON would be nonpolar. NNO is polar.
b.
N N O –1 +1 0 P P mn ð ð ð ð ½ ý ½ ý N N O 0 +1 –1 q O mn N N O –2 +1 +1 Formal charges O q I Cl Cl O A Cl A ð ð ð ð ð ½ ý Trigonal planar about all C atoms, nonpolar, 120, sp2 H C H H H H H C C C N F F N Polar P D G ð ð ½ N F F N Nonpolar P D G ð ð ð ð H H H B H H H C O H O C C O C C C N C H C H H H H H H H C O H O C C O C C C N C H C H H H H mn NPC N N N H H P mn C –1 0 +1 0 0 0 H P O A D H D G G ð ð ½ ½ ý NqC N N N H H H C Favored by formal charge 0 0 0 0 0 0 H O O O A D D G N ½ Answers to Selected Exercises A53 e. The structure is the most important because it has three different CN bonds. This structure is also favored on the basis of formal charge.
75. a. 25-nm light has sufficient energy to ionize N and N2 and to break the triple bond in N2. Thus, N2, N2 , N, and N will all be present, assuming excess N2. b. 85.33 nm 127 nm; c. The ionization energy of a substance is the energy it takes to completely remove an electron. N2: (
2s)2(
2s)2(2p)4(
2p)2; the electron removed from N2 is in the 2p molecular orbital, which is lower in energy than the 2p atomic or-bital from which the electron in atomic nitrogen is removed. Since the elec-tron removed from N2 is lower in energy than the electron removed in N, then the ionization energy of N2 is greater than the ionization energy of N.
77. Both reactions apparently involve only the breaking of the NOCl bond.
However, in the reaction ONCl n NO Cl some energy is released in form-ing the stronger NO bond, lowering the value of H. Therefore, the apparent NOCl bond energy is artificially low for this reaction. The first reaction involves only the breaking of the NOCl bond.
79. a. The CO bond is po-lar, with the negative end around the more electronegative oxygen atom. We would expect metal cations to be attracted to and to bond to the oxygen end of CO on the basis of electronegativity. b. The formal charge on C is 1, and the formal charge on O is 1. From formal charge, we would expect metal cations to bond to the carbon (with the negative formal charge.) c. In molec-ular orbital theory, only orbitals with proper symmetry overlap to form bond-ing orbitals. The metals that form bonds to CO are usually transition metals, all of which have outer electrons in the d orbitals. The only molecular orbitals of CO that have proper symmetry to overlap with d orbitals are the 2p or-bitals, whose shape is similar to that of the d orbitals (see Fig. 9.34). Since the antibonding molecular orbitals have more carbon character, one would expect the bond to form through carbon.
81. The species with the smallest ioniza-tion energy has the highest energy electron. O2, N2 2, N2 , and O2 all have at least one electron in the high-energy 2p orbitals. Because N2 2 has the highest ratio of electrons to protons, the 2p electrons are least attracted to the nuclei and easiest to remove, translating into the smallest ionization energy.
83. a. Li2 bond order 1; B2 bond order 1; b. 4 electrons must be removed; c. 4.5 105 kJ 85. T-shaped and dsp3 hybridized Chapter 10 13. Atoms have an approximately spherical shape. It is impossible to pack spheres together without some empty space between the spheres.
15. Evap-oration takes place when some molecules at the surface of a liquid have enough energy to break the intermolecular forces holding them in the liquid phase.
When a liquid evaporates, the molecules that escape have high kinetic ener-gies. The average kinetic energy of the remaining molecules is lower; thus the temperature of the liquid is lower.
17. An alloy is a substance that contains a mixture of elements and has metallic properties. In a substitutional alloy, some of the host metal atoms are replaced by other metal atoms of similar size (e.g., in brass, pewter, plumber’s solder). An interstitial alloy is formed when some of the interstices (holes) in the closest packed metal structure are occu-pied by smaller atoms (e.g., in carbon steels).
19. a. As intermolecular forces increase, the rate of evaporation decreases. b. increase T: increase rate; c. increase surface area: increase rate 21. Sublimation will occur, allowing water to escape as 23. The strength of intermolecular forces deter-mines relative boiling points. The types of intermolecular forces for covalent compounds are London dispersion forces, dipole forces, and hydrogen bond-ing. Because the three compounds are assumed to have similar molar mass and shape, the strength of the London dispersion forces will be about equal be-tween the three compounds. One of the compounds will be nonpolar so it only has London dispersion forces. The other two compounds will be polar so they have additional dipole forces and will boil at a higher temperature than the H2O1g2.
C N C N H H O O Nq N H H O A D G D M ð ð ½ nonpolar compound. One of the polar compounds will have an H covalently bonded to either N, O, or F. This gives rise to the strongest type of covalent intermolecular force, hydrogen bonding. This compound exhibiting hydrogen bonding will have the highest boiling point while the polar compound with no hydrogen bonding will boil at an intermediate temperature.
25. a. Both and are molecular solids. Both have an ordered array of the individual molecules, with the molecular units occupying the lattice points. A difference within each solid lattice is the strength of the intermolecular forces. is nonpolar and only exhibits London dispersion forces. exhibits the rela-tively strong hydrogen bonding interactions. The difference in strength is evi-denced by the solid phase change that occurs at 1 atm. sublimes at a rel-atively low temperature of In sublimation, all of the intermolecular forces are broken. However, doesn’t have a solid phase change until and in this phase change from ice to water, only a fraction of the intermolec-ular forces are broken. The higher temperature and the fact that only a portion of the intermolecular forces are broken are attributed to the strength of the in-termolecular forces in as compared to Related to the intermolec-ular forces is the relative densities of the solid and liquid phases for these two compounds. is denser than while is less dense than For and for most solids, the molecules pack together as close as possible, which is why solids are usually more dense than the liquid phase.
For each molecule has two lone pairs and two bonded hydrogen atoms.
Because of the equal number of lone pairs and bonds, each mole-cule can form two hydrogen bonding interactions to other molecules. To keep this symmetric arrangement (which maximizes the hydrogen bonding in-teractions), the molecules occupy positions that create empty space in the lattice. This translates into smaller density for (less mass per unit volume).
b. Both NaCl and CsCl are ionic compounds with the anions at the lattice points of the unit cell and the cations occupying the empty spaces cre-ated by the anions (called holes). In NaCl, the anions occupy the lattice points of a face-centered unit cell with the cations occupying the octahe-dral holes. Octahedral holes are the empty spaces created by six ions. CsCl has the ions at the lattice points of a simple cubic unit cell with the cations occupying the middle of the cube.
27. In the ln versus 1T plot, the slope of the straight line is equal to Because is always positive, the slope of the line will always be negative.
29. a. LD (London dispersion); b. dipole, LD; c. hydrogen bonding, LD; d. ionic; e. LD; f. dipole, LD; g. ionic 31. a. OCS; b. SeO2; c. H2NCH2CH2NH2; d. H2CO; e. CH3OH 33. a. Neopentane is more compact than n-pentane.
There is less surface area contact among neopentane molecules. This leads to weaker London dispersion forces and a lower boiling point. b. HF is capable of hydrogen bonding; HCl is not. c. LiCl is ionic, and HCl is a molecular solid with only dipole forces and London dispersion forces. Ionic forces are much stronger than the forces for molecular solids. d. n-Hexane is a larger molecule, so it has stronger London dispersion forces.
35. a. HBr has di-pole forces in addition to LD forces; b. NaCl, stronger ionic forces; c. I2, larger molecule so stronger LD forces; d. N2, smallest nonpolar compound present, has weakest LD forces; e. CH4, smallest nonpolar compound present, has weakest LD forces; f. HF, can form relatively strong hydrogen bonding interactions, unlike the other compounds; g. CH3CH2CH2OH, unlike others, has relatively strong hydrogen bonding.
37. H2O is attracted to glass while Hg is not.
39. The structure of H2O2 produces greater hydrogen bonding than wa-ter. Long chains of hydrogen bonded H2O2 molecules then get tangled to-gether.
41. 313 pm 43. 0.704 Å 45. 1.54 g/cm3 47. 174 pm; 11.6 g/cm3 49. edge, 328 pm; radius, 142 pm 51. face-centered cubic unit cell 53. For a cubic closest packed structure, 74.06% of the volume of each unit cell is occu-pied by atoms; in a simple cubic unit cell structure, 52.36% is occupied. The cu-bic (and hexagonal) closest packed structures provide the most efficient means for packing atoms.
55. Doping silicon with phosphorus produces an n-type semiconductor. The phosphorus adds electrons at energies near the conduction band of silicon. Electrons do not need as much energy to move from filled to unfilled energy levels so conduction increases. Doping silicon with gallium produces a p-type semiconductor. Because gallium has fewer valence electrons than silicon, holes (unfilled energy levels) at energies in the previously filled ¢Hvap ¢Hvap/R.
Pvap Cs Cl Cl Na Cl H2O1s2 H2O1s2 H2O H2O O¬H H2O, CO21s2 H2O1l2.
H2O1s2 CO21l2 CO21s2 CO2.
H2O 0°C, H2O 78°C.
CO2 H2O CO2 H2O CO2 A54 Answers to Selected Exercises 109. 46.7 kJ/mol; 90% 111. The solids with high melting points (NaCl, MgCl2, NaF, MgF2, AlF3) are all ionic solids. SiCl4, SiF4, Cl2, F2, PF5, and SF6 are nonpolar covalent molecules with LD forces. PCl3 and SCl2 are polar molecules with LD and dipole forces. In these 8 molecular substances the intermolecular forces are weak and the melting points low. AlCl3 is intermediate. The melting point indicates there are stronger forces present than in the nonmetal halides, but not as strong as for an ionic solid. AlCl3 illus-trates a gradual transition from ionic to covalent bonding; from an ionic solid to discrete molecules.
113. TiO1.182 or Ti0.8462O; 63.7% Ti2, 36.3% Ti3 115. 6.58 g/cm3 117.
As P is lowered, we go from a to b on the phase diagram. The water boils. The evaporation of the water is endothermic and the water is cooled (b n c), forming some ice. If the pump is left on, the ice will sublime until none is left. This is the basis of freeze drying.
119. The volume of the hole is 121. CdS; n-type 123. 2.53 torr; atoms Chapter 11 9. 9.74 M 11. 4.5 M 13. As the temperature increases, the gas molecules will have a greater average kinetic energy. A greater fraction of the gas mole-cules in solution will have kinetic energy greater than the attractive forces be-tween the gas molecules and the solvent molecules. More gas molecules will escape to the vapor phase, and the solubility of the gas will decrease.
15. The levels of the liquids in each beaker will become constant when the con-centration of solute is the same in both beakers. Because the solute is less volatile, the beaker on the right will have a larger volume when the concentra-tions become equal. Water will initially condense in this beaker in a larger amount than solute is evaporating, while the net change occurring initially in the other beaker is for water to evaporate in a larger amount than solute is con-densing. Eventually the rate that solute and H2O leave and return to each beaker will become equal when the concentrations become equal.
17. No. For an ideal solution, Hsoln 0 19. Normality is the number of equivalents per liter of solution. For an acid or a base, an equivalent is the mass of acid or base that can furnish 1 mol of protons (if an acid) or accept 1 mol of protons (if a base). A proton is an ion. Molarity is defined as the moles of solute per liter of solution. When the number of equivalents equals the number of moles of solute, then normality molarity. This is true for acids which only have one acidic proton in them and for bases that accept only one proton per formula unit. Examples of acids where equivalents moles solute are HCl, HF, and Examples of bases where equivalents moles solute are NaOH, KOH, and When equivalents moles solute, then normality molar-ity. This is true for acids that donate more than one proton etc.) and for bases that react with more than one proton per formula unit 21. Only statement b is true. A sub-stance freezes when the vapor pressure of the liquid and solid phases are the same. When a solute is added to water, the vapor pressure of the solution at 0C is less than the vapor pressure of the solid; the net result is for any ice etc.4.
Sr1OH22, Ba1OH22, Ca1OH22, H2CO3, H3PO4, 1H2SO4, NH3.
HC2H3O2.
HNO3, H 6.38 1022 4 3pca223 2 2 br d 3 T c b a P 30.79 kJ/mol ¢H molecular orbitals are created, which induces greater electron movement (greater conductivity).
57. p-type 59. 5.0 102 nm 61. NaCl: 4Na, 4Cl; CsCl: 1Cs, 1Cl; ZnS: 4Zn2, 4S2; TiO2: 2Ti4, 4O2 63. CoF2 65. ZnAl2S4 67. MF2 69.
71. a. CO2: molecular; b. SiO2: covalent network; c. Si: atomic, covalent network; d. CH4: molecular; e. Ru: atomic, metallic; f. I2: molecular; g. KBr: ionic; h. H2O: molecular; i. NaOH: ionic; j. U: atomic, metallic; k. CaCO3: ionic; l. PH3: molecular 73. a. The unit cell consists of Ni at the cube corners and Ti at the body center or Ti at the cube corners and Ni at the body center. b. NiTi; c. Both have a coordination number of 8.
75. CaTiO3; six oxygens around each Ti 77. a. YBa2Cu3O9; b. The structure of this su-perconductor material is based on the second perovskite structure. The YBa2Cu3O9 structure is three of these cubic perovskite unit cells stacked on top of each other. The oxygens are in the same places, Cu takes the place of Ti, two Ca are replaced by two Ba, and one Ca is replaced by Y.
c. YBa2Cu3O7 79. Li, 158 kJ/mol; Mg, 139 kJ/mol. Bonding is stronger in Li.
81. 89°C 83.
85.
87. 1680 kJ 89. 1490 g 91. A: solid; B: liquid; C: vapor; D: solid va-por; E: solid liquid vapor (triple point); F: liquid vapor; G: liquid vapor (critical point); H: vapor; the first dashed line (at the lower temperature) is the normal melting point, and the second dashed line is the normal boiling point. The solid phase is denser.
93. a. two; b. higher pressure triple point: graphite, diamond, and liquid; lower pressure triple point: graphite, liquid and vapor; c. It is converted to diamond (the more dense solid form); d. Diamond is more dense, which is why graphite can be converted to diamond by apply-ing pressure.
95. Because the density of the liquid phase is greater than the density of the solid phase, the slope of the solid–liquid boundary line is neg-ative (as in H2O). With a negative slope, the melting points increase with a de-crease in pressure so the normal melting point of X should be greater than 97. Chalk is composed of the ionic compound calcium carbonate (CaCO3). The electrostatic forces in ionic compounds are much stronger than the intermolecular forces in covalent compounds. Therefore, CaCO3 should have a much higher boiling point than the covalent compounds found in motor oil and in H2O. Motor oil is composed of nonpolar and bonds.
The intermolecular forces in motor oil are therefore London dispersion forces. We generally consider these forces to be weak. However, with com-pounds that have large molar masses, these London dispersion forces add up significantly and can overtake the relatively strong hydrogen-bonding inter-actions in water.
99. A: CH4; B: SiH4 C: NH3 101. If TiO2 conducts electricity as a liquid, then it would be ionic.
103. B2H6, molecular; SiO2, network; CsI, ionic; W, metallic 105. 4.65 kg/h 107.
27.86 kJ/mol; ¢E C¬H C¬C 225°C.
Time –60 –40 –20 0 1 2 3 4 5 Slope 5 > slope 3 > slope 1 Time 4 = 4 × time 2 20 40 60 80 Temp (°C) 77°C rMg2 6.15 109 cm rO2 1.49 108 cm; Answers to Selected Exercises A55 present to convert to liquid in order to try to equalize the vapor pressures (which never can occur at 0C). A lower temperature is needed to equalize the vapor pressure of water and ice, hence the freezing point is depressed. For statement a, the vapor pressure of a solution is directly related to the mole fraction of solvent (not solute) by Raoult’s law. For statement c, colligative properties de-pend on the number of solute particles present and not on the identity of the solute. For statement d, the boiling point of water is increased because the sugar solute decreases the vapor pressure of the water; a higher temperature is required for the vapor pressure of the solution to equal the external pressure so boiling can occur.
23. Isotonic solutions are those which have identical osmotic pressures. Crenation and hemolysis refer to a phenomena that occurs when red blood cells are bathed in solutions having a mismatch in osmotic pressure between the inside and the outside of the cell. When red blood cells are in a solution having a higher osmotic pressure than that of the cells, the cells shrivel as there is a net transfer of water out of the cells. This is called crenation. Hemolysis occurs when the red blood cells are bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells rupture as there is a net transfer of water to inside the red blood cells.
25. 1.06 g/mL; 0.0180 mole fraction H3PO4, 0.9820 mole fraction H2O; 0.981 mol/L; 1.02 mol/kg 27. HCl: 12 M, 17 m, 0.23; HNO3: 16 M, 37 m, 0.39; H2SO4: 18 M, 200 m, 0.76; HC2H3O2: 17 M, 2000 m, 0.96; NH3: 15 M, 23 m, 0.29 29. 35%; 0.39; 7.3 m; 3.1 M 31. 23.9%; 1.6 m, 0.028, 4.11 N 33. NaI(s) S Na(aq) I(aq) Hsoln 8 kJ/mol 35. The attraction of water molecules for Al3 and OH cannot overcome the larger lattice energy of Al(OH)3.
37. a. CCl4; b. H2O; c. H2O; d. CCl4; e. H2O; f. H2O; g. CCl4; 39. Ability to form hy-drogen bonding interactions, ability to break up into ions, and polarity are some factors affecting solute solubility. a. CH3CH2OH; b. CHCl3; c. CH3CH2OH 41. As the length of the hydrocarbon chain increases, the solubility decreases because the nonpolar hydrocarbon chain interacts poorly with the polar water molecules.
43. 1.04 103 mol/L atm; 1.14 103 mol/L 45. 50.0 torr 47. 3.0 102 g/mol 49. a. 290 torr; b. 0.69 51. methanol propanol 0.500 53. solution c 55. Pideal 188.6 torr; acetone 0.512, methanol 0.488; the actual vapor pressure of the solution is smaller than the ideal vapor pressure, so this solution exhibits a negative deviation from Raoult’s law. This occurs when solute–solvent attractions are stronger than for the pure substances.
57.
59. 14.8 g C3H8O3 61. Tf 29.9°C, Tb 108.2°C 63. 776 g/mol 65. a. T 2.0 105°C, 0.20 torr; b. Osmotic pressure is better for determining the molar mass of large mole-cules. A temperature change of 105°C is very difficult to measure. A change in height of a column of mercury by 0.2 mm is not as hard to measure pre-cisely.
67. 0.327 M 69. a. 0.010 m Na3PO4 and 0.020 m KCl; b. 0.020 m HF; c. 0.020 m CaBr2 71. a. Tf 0.28°C; Tb 100.077°C; b. Tf 0.37°C; Tb 100.10°C 73. 2.63 (0.0225 m), 2.60 (0.0910 m), 2.57 (0.278 m); iaverage 2.60 75. a. yes; b. no 77. a. 26.6 kJ/mol; b. 657 kJ/mol 79. a. Water boils when the vapor pressure equals the pressure above the water. In an open pan, Patm 1.0 atm. In a pressure cooker, Pinside 1.0 atm and water boils at a higher temperature. The higher the cooking temperature, the faster the cooking time. b. Salt dissolves in water, forming a solution with a melting point lower than that of pure water ( Tf Kfm). This happens in water on the surface of ice. If it is not too cold, the ice melts. This process won’t occur if the ambient temperature is lower than the depressed freezing point of the salt solution. c. When water freezes from a solution, if freezes as pure water, leaving behind a more concentrated salt solution. d. On the CO2 phase diagram, the triple point is above 1 atm and CO2(g) is the stable phase at 1 atm and room temperature. CO2(l) can’t exist at normal atmos-pheric pressures, which explains why dry ice sublimes rather than boils. In a fire extinguisher, P 1 atm and CO2(l) can exist. When CO2 is released from the fire extinguisher, CO2(g) forms as predicted from the phase dia-gram. e. Adding a solute to a solvent increases the boiling point and decreases the freezing point of the solvent. Thus, the solvent is a liquid over a wider range of temperatures when a solute is dissolved.
81. 0.600 83. C2H4O3; 151 g/mol (exp.); 152.10 g/mol (calc.); C4H8O6 85. 1.97% NaCl 87. a.
b. 23.1 mm Hg; c. Assume an ideal solution; assume no ions form (i 1).
100.77°C; 101.5°C 89.
91. 72.7% sucrose and 27.3% NaCl by mass; 0.2 93. 0.050 95. 44% naph-thalene, 56% anthracene 97. 0.20°C, 100.056C 99. a. 46 L; b. No; A reverse osmosis system that applies 8.0 atm can purify only water with solute concentrations less than 0.32 mol/L. Salt water has a solute concentration of 2(0.60 M) 1.2 M ions. The solute concentration of salt water is much too high for this reverse osmosis unit to work.
101. i 3.00; CdCl2 Chapter 12 9. In a unimolecular reaction, a single reactant molecule decomposes to products. In a bimolecular reaction, two molecules collide to give products.
The probability of the simultaneous collision of three molecules with enough energy and orientation is very small, making termolecular steps very unlikely.
11. All of these choices would affect the rate of the reaction, but only b and c affect the rate by affecting the value of the rate constant k. The value of the rate constant is dependent on temperature. It also depends on the activation energy. A catalyst will change the value of k because the activation energy changes. Increasing the concentration (partial pressure) of either H2 or NO does not affect the value of k, but it does increase the rate of the reaction because both concentrations appear in the rate law.
13. The average rate decreases with time because the reverse reaction occurs more fre-quently as the concentration of products increase. Initially, with no products present, the rate of the forward reaction is at its fastest; but as time goes on, the rate gets slower and slower since products are converting back into reac-tants. The instantaneous rate will also decrease with time. The only rate that is constant is the initial rate. This is the instantaneous rate taken at t At this time, the amount of products is insignificant and the rate of the reaction only depends on the rate of the forward reaction.
15. When the rate doubles as the concentration quadruples, the order is For a reactant that has an order of 1, the rate will decrease by a factor of when the concentra-tions are doubled.
17. Two reasons are: a. the collision must involve enough energy to produce the reaction; i.e., the collision energy must equal or exceed the activation energy. b. the relative orientation of the reactants must allow formation of any new bonds necessary to produce products.
19. P4: 6.0 104 mol/L s; H2: 3.6 103 mol/L s 21. a. average rate of decomposition of rate of production of b. average rate of decomposition of rate of production of 23. a. mol/L s; b. mol/L s; c. s1; d. L/mol s; e. L2/mol2 s 25. a. rate k[NO]2[Cl2]; b. 1.8 102 L2/mol2 min 27. a. rate k[NOCl]2; b. 6.6 1029 cm3/molecules s; c. 4.0 108 L/mol s 29. a. first order in Hb and first order in CO; b. rate k[Hb][CO]; c. 0.280 L/mol s; d. 2.26 mol/L s 31. rate k[H2O2]; ln[H2O2] kt ln[H2O2]0; k 8.3 104 s1; 0.037 M 33. rate k[NO2]2; kt k 2.08 104 L/mol s; 0.131 M 35. a. rate k; [C2H5OH] kt [C2H5OH]0; because slope k, then k 4.00 105 mol/L s; b. 156 s; c. 313 s 37. rate k[C4H6]2; kt 1 3C4H64 1 3NO24 0 ; 1 3NO24 O2 5.80 106 mol/L s 1.16 10 5 mol/L s, H2O2 O2 1.16 105 mol/L s; H2O2 2.31 10 5 mol/L s, 1 2 1 2.
0.
80.% A: xA V 0.80x 0.80x 0.20y, xB V 1 xA V 50.% A: xA V x x y, xB V 1 xA V; 30.% A: xA V 0.30x 0.30x 0.70y, xB V 1 0.30x 0.30x 0.70y; 80.% A: xA 0.80y 0.20x 0.80y, xB 1 xA; 50.% A: xA y x y, xB 1 y x y; 30.% A: xA 0.30y 0.70x 0.30y, xB 1 xA; A56 Answers to Selected Exercises reaction is at equilibrium. When the reaction is not at equilibrium and one can determine what has to be the net charge for the system to get to equilibrium.
15. We always try to make good assumptions that simplify the math. In some problems, we can set up the problem so that the net change, x, that must occur to reach equilibrium is a small number. This comes in handy when you have expressions like or Since x is small, we assume that it makes little difference when subtracted from or added to some relatively big number. When this is true, and If the assumption holds by the 5% rule, then the as-sumption is assumed valid. The 5% rule refers to x (or 2x or 3x, etc.) that was assumed small compared to some number. If x (or 2x or 3x, etc.) is less than 5% of the number the assumption was made against, then the assumption will be assumed valid. If the 5% rule fails to work, one can generally use a math procedure called the method of successive approximations to solve the quadratic or cubic equation.
17. a.
b.
c.
d.
19. a. 0.11; b. 77; c. 8.8; d. 4.6 104 21.
23. 1.7 105 25. 6.3 1013 27.
29. a.
b.
c.
d.
31. 8.0 109 33. a. Q K, so reaction shifts left to reach equilibrium. b. Q K, so reaction is at equilibrium. c. Q K, so reaction shifts right to reach equilibrium.
35. a. decrease; b. no change; c. no change; d. increase 37.
39. 3.4 41. 0.056 43. [N2]0 10.0 M, 45. [SO3] [NO] 1.06 M; [SO2] [NO2] 0.54 M 47. 7.8 102 atm 49. PSO2 0.38 atm; PO2 0.44 atm; PSO3 0.12 atm 51. a. [NO] 0.032 M, [Cl2] 0.016 M, [NOCl] 1.0 M; b. [NO] [NOCl] 1.0 M, [Cl2] 1.6 105 M; c.
[NO] 8.0 103 M, [Cl2] 1.0 M, [NOCl] 2.0 M 53. [CO2] 0.39 M, [CO] 8.6 103 M, [O2] 4.3 103 M 55. 0.27 atm 57. a. no effect; b. shifts left; c. shifts right 59. a. right b. right; c. no effect; d. left; e. no effect 61. a. left; b. right; c. left; d. no effect; e. no effect; f. right 63. increase 65. 2.6 1081 67. a. 0.379 atm; b. 0.786 69. a. 1.16 atm; b. 0.10 atm; c. 2.22 atm; d. 91.4% 71. [H2] [F2] 0.0251 M; [HF] 0.450 M 73. Added OH reacts with H to produce H2O. As H is removed, the reaction shifts right to produce more H and CrO4 2. Because more CrO4 2 is produced, the solution turns yellow. 75.
77.
Kp 1.92 79. [NOCl] 2.0 M, [NO] 0.050 M, [Cl2] 0.025 M 81. 2.1 103 atm 83. PNO2 0.704 atm, PN2O4 0.12 atm 85. 0.63 87. 0.240 atm 89.
91.
93. C10H8; 0.0919% Chapter 14 17. 10.78 (4 significant figures); 6.78 (3 significant figures); 0.78 (2 signifi-cant figures); A pH value is a logarithm. The numbers to the left of the deci-mal place identify the power of 10 to which [H] is expressed in scientific no-tation—for example, The number of decimal places in a pH value identifies the number of significant figures in [H]. In all three pH values, the [H] should be expressed only to two significant figures since these pH values have only two decimal places.
19. a. These would be 0.10 M solutions of strong acids like HCl, HBr, HI, or b. These are salts of the conjugate acids of the bases in Table 14.3. These con-jugate acids are all weak acids. Three examples would be 0.10 M solutions of and Note that the anions used to form these salts are conjugate bases of strong acids; this is because they have no acidic or basic properties in water (with the exception of which has weak acid properties). c. These would be 0.10 M solutions of strong bases like LiOH, HSO4 , C2H5NH3Br.
CH3NH3NO3, NH4Cl, HClO4.
H2SO4 HNO3, 1011, 10 7, 10 1.
192 g; 0.25 atm 9.17 10 3 PPCl5 0.0259 atm; PPCl3 PCl2 0.2230 atm, 9.0 10 3 M [H2]0 11.0 M 8.0 10 2 M K 3H2O4 3H24 , Kp PH2O PH2 K 3O24 3, Kp P3 O2 ; K 3N24 3Br24 3, Kp PN2 P 3 Br2; K 3H2O4 3NH34 23CO24, Kp PH2O P2 NH3 PCO2 ; 1.1 10 3 4.0 10 6 K 3PCl34 23Br24 3 3PBr34 23Cl24 3 K 3SiCl44 3H24 2 3SiH44 3Cl24 2; K 3NO24 2 3N2O44; K 3NO4 2 3N24 3O24; 0.727 2x 0.727.
0.12 x 0.12 0.727 2x.
0.12 x Q K, k 1.4 102 L/mol s 39. second order; 0.1 M 41. a. [A] kt [A]0; b. 1.0 102 s; c. 2.5 104 M 43. a.
for both the first and second half-life; b. 532 s 45. 12.5 s 47. a. 1.1 102 M; b. 0.025 M 49. a. rate k[CH3NC]; b. rate k[O3][NO]; c. rate k[O3]; d. rate k[O3][O] 51. Rate k[C4H9Br]; C4H9Br 2H2O n C4H9OH Br H3O; the intermediates are C4H9 and C4H9OH2 .
53.
55. 341 kJ/mol 57. The graph of lnk versus 1T is linear with slope 1.2 104 K; Ea 1.0 102 kJ/mol 59.
61. 51C 63. H3O(aq) OH(aq) n 2H2O(l) should have the faster rate.
H3O and OH will be electrostatically attracted to each other; Ce4 and Hg2 2 will repel each other (so Ea is much larger).
65. a. NO; b. NO2; c. 2.3 67. CH2DOCH2D should be the product. If the mechanism is possi-ble, then the reaction must be C2H4 D2 n CH2DCH2D. If we got this prod-uct, then we could conclude that this is a possible mechanism. If we got some other product, e.g., CH3CHD2, then we would conclude that the mechanism is wrong. Even though this mechanism correctly predicts the products of the reaction, we cannot say conclusively that this is the correct mechanism; we might be able to conceive of other mechanisms that would give the same prod-uct as our proposed one.
69.
71. 5.68 1018 molecules/cm3 s 73. 1.0 102 kJ/mol 75. At high [S], the enzyme is completely saturated with substrate. Once the enzyme is completely saturated, the rate of decom-position of ES can no longer increase, and the overall rate remains constant.
77. a.
b. 87.0 s; c.
79. rate k 6.0 101 s1 81. a. first order with respect to both reactants; b. rate k[NO][O3]; c. k 1.8 s1; k 3.6 s1; d. k 1.8 1014 cm3/molecules s 83. a. 25 kJ/mol; b. 12 s; c.
T Interval 54 2(Intervals) 21.0C 16.3 s 21C.
27.8C 13.0 s 28C.
30.0C 12 s 0 30.C This rule of thumb gives excellent agreement to two significant figures.
85. a. [B] [A] so that [B] can be considered constant over the experi-ments. (This gives us a pseudo-order rate law equation.) b.
87.
89.
91.
Chapter 13 9. No, equilibrium is a dynamic process. Both the forward and reverse reactions are occurring at equilibrium, just at equal rates. Thus the forward and reverse reactions will distribute 14C atoms between CO and CO2.
11. 4 molecules H2O, 2 molecules CO, 4 molecules H2, and 4 molecules CO2 are present at equilibrium.
13. K and are equilibrium constants as determined by the law of mass action. For K, the units used for concentrations are mol/L, for partial pressures in units of atm are used (generally). Q is called the reaction quotient. Q has the exact same form as K or but instead of equilibrium con-centrations, initial concentrations are used to calculate the Q value. Q is of use when it is compared to the K value. When (or when ), the Qp Kp Q K Kp, Kp, Kp torr 1.3 10 5 s 1; 112 2.20 105 s 1; 5.99 10 21 molecules Rate k[A][B] 2, k 1.4 102 L 2/mol2 s k 0.050 L 2/mol 2 s Rate k[A] 2[B], k3I4 3OCl4 3OH4 ; [A] 1.27 10 5 M, [B] 1.00 M 115 L 3/mol 3 s; 215°C 9.5 105 L/mol s Ea R RC R P E Ea E 160. s t1 2 1 3C4H64 0 ; Answers to Selected Exercises A57 NaOH, KOH, RbOH, CsOH, and d. These are salts of the conjugate bases of the neutrally charged weak acids in Table 14.2.
The conjugate bases of weak acids are weak bases themselves. Three exam-ples would be 0.10 M solutions of and The cations used to form these salts are Rb, and since these cations have no acidic or basic properties in water. Notice that these are the cations in the list of the strong bases listed in part c that you should memorize. e. There are two ways to make a neutral salt. The easiest way is to combine a conjugate base of a strong acid (except for with one of the cations from the strong bases. These ions have no acidic/basic properties in water so salts of these ions are neutral. Three examples would be 0.10 M solutions of NaCl, and Another type of strong electrolyte that can produce neutral solutions are salts that contain an ion with weak acid proper-ties combined with an ion of opposite charge having weak base properties. If the for the weak acid ion is equal to the for the weak base ion, then the salt will produce a neutral solution. The most common example of this type of salt is ammonium acetate, For this salt, for for This salt, at any concentration, produces a neutral solution.
21. a.
or b.
or c.
23. a. This expression holds true for solutions of strong acids having a con-centration greater than M. For example, 0.10 M HCl, 7.8 M and M are solutions where this expression holds true.
b. This expression holds true for solutions of weak acids where the two nor-mal assumptions hold. The two assumptions are that the contribution of from water is negligible and that the acid is less than 5% dissociated in water (from the assumption that x is small compared to some number). This expres-sion will generally hold true for solutions of weak acids having a value less than as long as there is a significant amount of weak acid present.
Three example solutions are 1.5 M 0.10 M HOCl, and 0.72 M c. This expression holds true for strong bases that donate 2 ions per formula unit. As long as the concentration of the base is above M, this expression will hold true. Three examples are M and d. This ex-pression holds true for solutions of weak bases where the two normal as-sumptions hold. The assumptions are that the contribution from water is negligible and that the base is less than 5% ionized in water. For the 5% rule to hold, you generally need bases with and concentrations of weak base greater than 0.10 M. Three examples are 0.10 M 0.54 M and 1.1 M 25. One reason HF is a weak acid is that the bond is unusually strong and thus, is difficult to break. This contributes to the reluctance of the HF molecules to dissociate in water.
27. a. HClO4(aq) H2O(l) S H3O(aq) ClO4 (aq) or HClO4(aq) S H(aq)ClO4 (aq); wa-ter is commonly omitted from Ka reactions. b. CH3CH2CO2H(aq) H(aq) CH3CH2CO2 (aq); c. NH4 (aq) H(aq) NH3(aq) 29. a. H2O, base; H2CO3, acid; H3O, conjugate acid; conjugate base; b. C5H5NH, acid; H2O, base; C5H5N, conjugate base; H3O, conjugate acid; c. HCO3 , base; C5H5NH, acid; H2CO3, conjugate acid; C5H5N, conjugate base 31. a. HClO4, strong acid; b. HOCl, weak acid; c. H2SO4, strong acid; d. H2SO3, weak acid 33.
35. a. HCl; b. HNO2; c. HCN since it has a larger Ka value.
37. a.
M, neutral; b. 12 M, basic; c.
M, acidic; d.
M, acidic 39. a. endothermic; b.
41. a.
b.
c.
d.
a. pH 14.18, pOH pH 4.27, pOH 9.73 pOH 15.08; pH 1.08, pOH 1.08; pH 15.08, pH pOH 7.00; 2.34 107 M [H ] [OH ] 1.9 10 10 8.3 1016 1.0 10 7 NH4 7 H2O HClO4 7 HClO2 7 HCO3 , ∆ ∆ H¬F C5H5N.
C6H5NH2, NH3, Kb 6 1 104 OH Ba1OH22.
9.1 105 M Sr1OH22, 2.1 104 Ca1OH22, 5.0 103 M 5 107 OH NH4NO3.
HC2H3O2, 1 104, Ka H HClO4 3.6 104 HNO3, 1.0 106 K Kb 3C5H5NH4 3OH4 3C5H5N4 C5H5N1aq2 H2O1l2 ∆C5H5NH1aq2 OH1aq2 HF1aq2 ∆H1aq2 F1aq2 K Ka 3H4 3F4 3HF4 H2O1l2 ∆F1aq2 H3O1aq2 HF1aq2 H2O1l2 ∆H1aq2 OH1aq2 K Kw 3H 4 3OH4 H2O1l2 H2O1l2 ∆H3O1aq2 OH1aq2 C2H3O2 5.6 1010.
NH4 Kb Ka NH4C2H3O2.
Kb Ka SrI2.
KNO3, HSO4 2 Ba2 Sr2, Ca2, Cs, K, Na, Li, CaF2.
KC2H3O2, NaClO2, Ba1OH22.
Sr1OH22 Ca1OH22, b.
c.
d.
43. a.
M, acidic; b.
acidic; c.
basic; d. pH pOH 7.00, [H] [OH] 1.0 107 M, neutral 45. pOH 11.9, [H] 8 103 M, [OH] 1 1012 M, acidic 47. a. H, H2O; 0.602; b.
H2O; 0.602 49. [H] 0.088 M, [OH] 1.1 1013 M, [Cl] 0.013 M, [NO3 ] 0.075 M 51. Add 4.2 mL of 12 M HCl to water with mixing; add enough water to bring the solution volume to 1600 mL.
53. a. HNO2 and H2O, 2.00; b. HC2H3O2 and H2O, 2.68 55.
57. [H] [F] 3.5 103 M, [OH] 2.9 1012 M, [HF] 0.017 M, 2.46 59. 1.96 61. a. 1.00; b. 1.30 63. a. 0.60%; b. 1.9%; c. 5.8%; d. Dilution shifts equilibrium to the side with the greater number of particles (% dissociation increases). e. [H] also depends on initial concentration of weak acid.
65. 1.4 104 67. 3.5 104 69. 0.024 M 71. a.
b.
73. NH3 C5H5N H2O NO3 75. a. C6H5NH2; b. C6H5NH2; c. OH; d. CH3NH2 77. a. 13.00; b. 7.00; c. 14.30 79. a. K, OH, and H2O, 0.015 M, 12.18; b. Ba2, OH, and H2O, 0.030 M, 12.48 81. 0.16 g 83. NH3 and H2O, 1.6 103 M, 11.20 85. a. [OH] 8.9 103 M, [H] 1.1 1012 M, 11.96; b. [OH] 4.7 105 M, [H] 2.1 1010 M, 9.68 87. 12.00 89. a. 1.3%; b. 4.2% 91. 9.2 107 93.
95. a. 1.62; b. 3.68 97. 0.30 99. HCl NH4Cl KCl KCN KOH 101. OCl 103. [HN3] [OH] 2.3 106 M, [Na] 0.010 M, [N3 ] 0.010 M, [H] 4.3 109 M 105. a. 5.82; b. 10.95 107. NaF 109. 3.08 111. a. neutral; b. basic; NO2 H2O HNO2 OH; c. acidic; C5H5NH C5H5N H; d. acidic because NH4 is a stronger acid than NO2 is a base; NH4 NH3 H; NO2 H2O HNO2 OH; e. basic; OCl H2O HOCl OH; f. basic because OCl is a stronger base than NH4 is an acid; OCl H2O HOCl OH, NH4 NH3 H 113. a. HIO3 HBrO3; as the electronegativity of the central atom increases, acid strength increases. b. HNO2 HNO3; as the num-ber of oxygen atoms attached to the central atom increases, acid strength increases.
c. HOI HOCl; same reasoning as in part a. d. H3PO3 H3PO4; same rea-soning as in part b.
115. a. H2O H2S H2Se; acid strength increases as bond energy decreases. b. CH3CO2H FCH2CO2H F2CHCO2H F3CCO2H; as the electronegativity of the neighboring atoms increases, acid strength increases.
c. NH4 HONH3 ; same reasoning as in part b. d. NH4 PH4 ; same rea-soning as in part a.
117. a. basic; CaO(s) H2O(l) → Ca(OH)2(aq); b. acidic; SO2(g) H2O(l) → H2SO3(aq); c. acidic; Cl2O(g) H2O(l) → 2HOCl(aq) 119. a. B(OH)3, acid; H2O, base; b. Ag, acid; NH3, base; c. BF3, acid; F, base 121. Al(OH)3(s) 3H(aq) → Al3(aq) 3H2O(l); Al(OH)3(s) OH(aq) → Al(OH)4 (aq) 123. Fe3; because it is smaller with a greater pos-itive charge, Fe3 will be more strongly attracted to a lone pair of electrons from a Lewis base.
125. 990 mL H2O 127. a. 2.80; b.
129. NH4Cl 131.
133. 3.00 135. a. 2.62; b. 2.4%; c. 8.48 137. a. 1.66; b. Fe2 ions will produce a less acidic solution (higher pH) due to the lower charge on Fe2 as compared with Fe3. As the charge on a metal ion increases, acid strength of the hydrated ion increases.
139. acidic; HSO4 SO4 2 H; 1.54 141. a. Hb(O2)4 in lungs, HbH4 4 in cells; b. Decreasing [CO2] will decrease [H], favoring Hb(O2)4 formation. Breathing into a bag raises [CO2]. c. NaHCO3 lowers the acidity from accumulated CO2.
∆ 4.2 10 2 M 1.1 10 3 M ∆ ∆ ∆ ∆ ∆ ∆ ∆ HSO3 1aq2 ∆SO3 2 1aq2 H 1aq2 Ka2 reaction H2SO31aq2 ∆HSO3 1aq2 H1aq2 Ka1 reaction Kb 3C5H5NH4 3OH4 3C5H5N4 C5H5N 1aq2 H2O 1l2 ∆C5H5NH 1aq2 OH1aq2 Kb 3NH4 4 3OH4 3NH34 ; NH31aq2 H2O1l2 ∆NH4 1aq2 OH1aq2 pH 3.24 [CH3COOH] 0.0181 M, [H ] [CH3COO ] 5.8 10 4 M, NO3 , H, ClO4 , 7.8 104 M, [OH ] [H ] 1.3 10 11 M, pH 10.89, pOH 3.11, [OH ] 8.4 1014 M, 0.12 M, pH 0.92, pOH 13.08, [H ] [OH] 7.6 10 8 M, pOH 7.12, [H ] 1.3 10 7 pH 6.88, 10.85, pOH 3.14 pH pH pOH 7.00; pH 0.44, pOH 14.44; 0.18; A58 Answers to Selected Exercises HCO3 (aq); OH(aq) HCO3 (aq) → CO3 2(aq) H2O(l) 23. a. 2.96; b. 8.94; c. 7.00; d. 4.89 25. 1.1% vs. 1.3 102% dissociated; the pres-ence of C3H5O2 in solution 23d greatly inhibits the dissociation of HC3H5O2.
This is called the common ion effect.
27. a. 1.70; b. 5.49; c. 1.70; d. 4.71 29. a. 4.29; b. 12.30; c. 12.30; d. 5.07 31. solution d; solution d is a buffer solution that resists pH changes.
33.
3.40 35. 3.48; 3.22 37. 4.36 39. a. 7.97; b. 8.73; both solutions have an initial pH 8.77. The two solu-tions differ in their buffer capacity. Solution b with the larger concentrations has the greater capacity to resist pH change.
41. 15 g 43. a. 0.19; b. 0.59; c. 1.0; d. 1.9 45. HOCl; there are many possibilities. One possibility is a solution with [HOCl] 1.0 M and [NaOCl] 0.35 M.
47. solution d 49. a. 1.0 mol; b. 0.30 mol; c. 1.3 mol 51. a. 22 mL base added; b. buffer region is from 1 mL to 21 mL base added. The maximum buffer-ing region would be from 5 mL to 17 mL of base added with the halfway point to equivalence (11 mL) as the best buffer point. c. 11 mL base added; d. 0 mL base added; e. 22 mL base added (the stoichiometric point); f. any point after the stoichiometric point (volume base added 22 mL) 53. a. 0.699; b. 0.854; c. 1.301; d. 7.00; e. 12.15 55. a. 2.72; b. 4.26; c. 4.74; d. 5.22; e. 8.79; f. 12.15 57.
Volume (mL) pH 0.0 2.43 4.0 3.14 8.0 3.53 12.5 3.86 20.0 4.46 24.0 5.24 24.5 5.6 24.9 6.3 25.0 8.28 25.1 10.3 26.0 11.29 28.0 11.75 30.0 11.96 See Solutions Guide for pH plot.
59.
Volume (mL) pH 0.0 11.11 4.0 9.97 8.0 9.58 12.5 9.25 20.0 8.65 24.0 7.87 24.5 7.6 24.9 6.9 25.0 5.28 25.1 3.7 26.0 2.71 28.0 2.24 30.0 2.04 See Solutions Guide for pH plot.
61. a. 4.19, 8.45; b. 10.74; 5.96; c. 0.89, 7.00 63.
65. a. yel-low; b. 8.0; c. blue 67. phenolphthalein 69. Phenol red is one possible indicator for the titration in Exercise 53. Phenolphthalein is one possible indicator for the titration in Exercise 55.
71. Phenolphthalein is one possi-ble indicator for Exercise 57. Bromcresol green is one possible indicator for Exercise 59.
73. The pH is between 5 and 8.
75. a. AgC2H3O2(s) Ag(aq) C2H3O2 (aq); Ksp [Ag][C2H3O2 ]; b. Al(OH)3(s) Al3(aq) 3OH(aq); Ksp [Al3][OH]3; c. Ca3(PO4)2(s) 3Ca2(aq) 2PO4 3(aq); Ksp [Ca2]3[PO4 3]2 77. a. 2.3 109; b. 8.20 1019 79. 3.92 105 81. a. 1.6 105 mol/L; b. 9.3 105 mol/L; c. 6.5 107 mol/L 83.
85. 2.5 1022 mol/L 87. a. CaF2; b. FePO4 89. a. 4 1017 mol/L; b. 4 1011 mol/L; c. 4 1029 mol/L 91. 2.3 1011 mol/L 93.
95. If the anion in the salt can act as a base in water, then the solubility of the salt will increase 1.5 10 19 g 3.30 10 43 ∆ ∆ ∆ 2.1 10 6 143. a. H2SO3; b. HClO3; c. H3PO3; NaOH and KOH are ionic compounds composed of either Na or K cations and OH anions. When soluble ionic com-pounds dissolve in water, they form the ions from which they are formed. The acids in this problem are all covalent compounds. When these acids dissolve in water, the covalent bond between oxygen and hydrogen breaks to form H ions.
145. 7.20.
147. 4540 mL 149. 4.17 151. 0.022 M 153.
155. PO4 3, Kb 0.021; HPO4 2, Kb 1.6 107; H2PO4 , Kb 1.3 1012; from the Kb values, PO4 3 is the strongest base.
157. a. basic; b. acidic; c. basic; d. acidic; e. acidic 159. 1.0 103 161. 5.4 104 163. 3.36 Chapter 15 13. When an acid dissociates or when a salt dissolves, ions are produced. A common ion is when one of the product ions in a particular equilibrium is added from an outside source. For a weak acid dissociating to its conjugate base and the common ion would be the conjugate base; this would be added by dissolving a soluble salt of the conjugate base into the acid solution.
The presence of the conjugate base from an outside source shifts the equilib-rium to the left so less acid dissociates. For the reaction of a salt dissolv-ing into its respective ions, the common ion would be if one of the ions in the salt was added from an outside source. When a common ion is present, the equilibrium shifts to the left resulting in less of the salt dissolving into its ions.
15. The more weak acid and conjugate base present, the more and/or that can be absorbed by the buffer without significant pH change. When the concentrations of weak acid and conjugate base are equal (so that the buffer system is equally efficient at absorbing either or If the buffer is overloaded with weak acid or with conjugate base, then the buffer is not equally efficient at absorbing either or 17. The three key points to emphasize in your sketch are the initial pH, pH at the halfway point to equiv-alence, and the pH at the equivalence point. For the two weak bases titrated, at the halfway point to equivalence (50.0 mL HCl added) because at this point. For the initial pH, the strong base has the highest pH (most basic), while the weakest base has the lowest pH (least basic). At the equivalence point, the strong base titration has The weak bases titrated have acidic pHs at the equivalence point because the conjugate acids of the weak bases titrated are the major species present. The weakest base has the strongest conjugate acid so its pH will be lowest (most acidic) at the equivalence point.
19. i. This is the result when you have a salt that breaks up into two ions. Ex-amples of these salts (but not all) would be AgCl, and ZnCO3 ii. This is the result when you have a salt that breaks up into three ions, either two cations and one anion or one cation and two anions. Some examples are and iii. This is the result when you have a salt that breaks up into four ions, either three cations and one anion or one cation and three anions (ignoring the hydroxides, there are no examples of this type of salt in Table 15.4). iv. This is the result when you have a salt that breaks up into five ions, either three cations and two anions or two cations and three anions (no examples of this type of salt are in Table 15.4).
21. When strong acid or strong base is added to a sodium bicarbonate/sodium carbonate buffer mixture, the strong acid/base is neutralized. The reaction goes to completion resulting in the strong acid/base being replaced with a weak acid/base.
This results in a new buffer solution. The reactions are H(aq) CO3 2(aq) → Sr31PO422 1Ag3PO42 Ag2SO4.
Hg2I2, SrF2, BaCrO4, SrSO4, pH 7.0 Volume HCl added (ml) Strong base Kb=10–5 Kb=10–10 pH 7.0.
3weak base4 3conjugate acid4 pH pKa OH.
H OH.
H pH pKa2, OH H Ksp Ksp H, 2.5 10 3 Copyright 2007 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
Answers to Selected Exercises A59 as the solution becomes more acidic. Added H will react with the base, form-ing the conjugate acid. As the basic anion is removed, more of the salt will dissolve to replenish the basic anion. The salts with basic anions are Ag3PO4, CaCO3, CdCO3, and Sr3(PO4)2. Hg2Cl2 and PbI2 do not have any pH depen-dence because Cl and I are terrible bases (the conjugate bases of strong acids).
97. yes; Q 1.9 104 Ksp 99. [K] 0.160 M, [C2O4 2] 3.3 107 M, [Ba2] 0.0700 M, [Br] 0.300 M 101. [Ag] 5.6 105 M 103. a.
K1 K2 K3 K4 b.
K1 K2 K2 105.
107. Hg2(aq) 2I(aq) →HgI2(s) (orange precipitate); HgI2(s) 2I(aq) → HgI4 2(aq) (soluble complex ion) 109. 3.3 1032 M 111. a. 1.2 108 mol/L; b. 1.5 104 mol/L; c. The presence of NH3 in-creases the solubility of AgI. Added NH3 removes Ag from solution by form-ing the complex ion Ag(NH3)2 . As Ag is removed, more AgI will dissolve to replenish the Ag concentration.
113.
115. Test tube 1: added Cl reacts with Ag to form the silver chloride precipitate. The net ionic equation is Ag(aq) Cl(aq) → AgCl(s). Test tube 2: added NH3 re-acts with Ag ions to form the soluble complex ion Ag(NH3)2 . As this com-plex ion forms, Ag is removed from solution, which causes AgCl(s) to dissolve. When enough NH3 is added, then all of the silver chloride precipi-tate will dissolve. The equation is AgCl(s) 2NH3(aq) → Ag(NH3)2 (aq) Cl(aq). Test tube 3: added H reacts with the weak base NH3 to form NH4 .
As NH3 is removed, Ag ions are released to solution, which can then react with Cl to reform AgCl(s). The equations are Ag(NH3)2 (aq) 2H(aq) → Ag(aq) 2NH4 (aq) and Ag(aq) Cl(aq) →AgCl(s).
117. pOH 119. a.
b.
121. a. potassium fluoride HCl; b. benzoic acid NaOH; c. acetic acid sodium acetate; d. HOCl NaOH; e. ammonium chloride NaOH 123. a. 1.8 109; b. 5.6 104; c. 1.0 1014 125. 4.4 L 127. 180. g/mol; 3.3 104; assume acetylsali-cylic acid is a weak monoprotic acid.
129. 65 mL 131. 0.210 M 133. a.
b. 0.056 mol/L 135. 2.7 105 mol L; the solubility of hydroxyapatite will increase as a solution gets more acidic, since both phos-phate and hydroxide can react with H. 6 108 mol L; the hydroxyapatite in the tooth enamel is converted to the less soluble fluorapatite by fluoride-treated water. The less soluble fluorapatite will then be more difficult to dis-solve, making teeth less susceptible to decay. See Chemical Impact on “The Chemistry of Teeth.” 137. a. 6.7 106 mol L; b. 1.2 1013 mol L; c. Pb(OH)2(s) will not form since Q Ksp 139. 49 mL 141. 3.9 L 143. a. 200.0 mL; b. i. H2A, H2O; ii. H2A, HA, H2O, Na; iii. HA, H2O, Na; iv. HA, A2, H2O, Na; v. A2, H2O, Na; vi. A2, H2O, Na, OH; 1.6 10 6; 10.74 10.44; pKb log 3acid4 3base4 4.7 10 2 mol/L 6.2 10 5 V 3 3 C2O4 2 ∆V1C2O423 3 V1C2O422 C2O4 2 ∆V1C2O32 3 VC2O4 C2O4 2 ∆V1C2O422 V3 C2O4 2 ∆VC2O4 Ni 2 4CN ∆Ni1CN24 2 Ni1CN23 CN ∆Ni1CN24 2 Ni1CN22 CN ∆Ni1CN23 NiCN CN ∆Ni1CN22 Ni 2 CN ∆NiCN 3Sr21aq2 2H3PO41aq2 Sr31PO4221s2 2H1aq2 ¡ 3Sr21aq2 2HPO4 21aq2 ¡ excessH Cd21aq2 H2CO31aq2 3H2O1l2 CO21g24 CdCO31s2 H1aq2 ¡ Cd21aq2 HCO3 1aq2 ¡ excessH Ca21aq2 H2CO31aq2 3H2O1l2 CO21g24 CaCO31s2 H1aq2 ¡ Ca21aq2 HCO3 1aq2 ¡ excess H 3Ag1aq2 H3PO41aq2 Ag3PO41s2 H1aq2 ¡ 3Ag1aq2 HPO4 21aq2 ¡ excess H c.
1 104; 1 108 145.
147. 3 M 149. a. 5.8 104 mol L; b. Greater; F is a weak base (Kb 1.4 1011), so some of the F is removed by reaction with water. As F is removed, more SrF2 will dissolve; c. 3.5 103 mol L 151. 3.00 153. 2.78 Chapter 16 7. Living organisms need an external energy source to produce the necessary “ordering.” 9.
11. As any process occurs, will increase; cannot decrease. Time also goes in one direction, just as goes in one direction.
13. Possible arrangements for one molecule: Both are equally probable.
Possible arrangements for two molecules: Possible arrangement for three molecules: 15. Note that these substances are not in the solid state, but are in the aque-ous state; water molecules are also present. There is an apparent increase in ordering when these ions are placed in water. The hydrating water molecules must be in a highly ordered state when surrounding these anions.
17. One can determine and for the reaction using the standard entropies and standard enthalpies of formation in Appendix 4, then use the equation One can also use the standard free energies of forma-tion in Appendix 4. And finally, one can use Hess’s law to calculate Here, reactions having known values are manipulated to determine for a different reaction. For temperatures other than 25(C, is estimated using the equation. The assumptions made are that the and values determined from Appendix 4 data are temperature indepen-dent. We use the same and values as determined when then plug in the new temperature in Kelvin into the equation to estimate at the new temperature.
19. a, b, c 21. We draw all of the possible arrange-ments of the two particles in the three levels.
2 kJ — — — x — — x — xx 1 kJ — — x — — xx — x — 0 kJ — xx — x — x — — — Total E 0 kJ 1 kJ 2 kJ 2 kJ 3 kJ 4 kJ The most likely total energy is 2 kJ.
23. a. H2 at 100C and 0.5 atm; b. N2 at STP; c. H2O(l) 25. a. negative; b. positive 27. G 0 for b, c, d 29. 89.3 J/K mol 31. a. yes ( G 0); b. 196 K 33. a. negative; b. positive; c. negative; d. positive 35. a. Cgraphite(s); b. C2H5OH(g); c. CO2(g) 37. a. negative, 186 J/K; b. positive, 187 J/K; c. hard to predict since n 0.; 138 J/K 39. 262 J/K mol 41. a. H and S are both positive; b. Srhombic 43. a. H and S are both negative; b. low ¢G° T 25°C, ¢S° ¢H° ¢S° ¢H° T¢S ¢G° ¢H° ¢G° ¢G° ¢G° ¢G°.
T¢S°.
¢G° ¢H° ¢H° ¢S° 1 way 3 ways 3 ways 1 way Equally most probable 1 way 2 ways Most probable 1 way 1 way 1 way ¢Suniv ¢Suniv ¢Suniv ¢Suniv pH 5.0; Ka 1 10 10 Ka2 Ka1 A60 Answers to Selected Exercises electrons through a wire. This is accomplished by making a galvanic cell which separates the reduction reaction from the oxidation reaction in order to control the flow of electrons through a wire to produce a voltage.
19. An extensive property is one that depends on the amount of substance. The free energy change for a reaction depends on whether 1 mol of product is produced or 2 mol of product is produced or 1 million mol of product is produced. This is not the case for cell potentials which do not depend on the amount of sub-stance. The equation that relates to E is nFE. It is the n term that converts the intensive property E into the extensive property The n is the number of mol of electrons transferred in the balanced reaction that is associated with.
21. A potential hazard when jump-starting a car is that the electrolysis of can occur. When is electrolyzed, the products are the explosive gas mixture of and A spark produced during jump starting a car could ignite any and produced. Grounding the jumper cable far from the battery minimizes the risk of a spark nearby the battery where and could be collecting.
23. You need to know the iden-tity of the metal so you know which molar mass to use. You need to know the oxidation state of metal ion in the salt so the mol of electrons transferred can be determined. And finally, you need to know the amount of current and the time the current was passed through the electrolytic cell. If you know these four quantities, then the mass of metal plated out can be calculated.
25. See Figure 17.3 of the text for a typical galvanic cell. The anode compartment con-tains the oxidation half-reaction compounds/ions, and the cathode compart-ment contains the reduction half-reaction compounds/ions. The electrons flow from the anode to the cathode. For each of the following answers, all solutes are 1.0 M and all gases are at 1.0 atm. a. 7H2O(l) 2Cr3(aq) 3Cl2(g) n Cr2O7 2(aq) 6Cl(aq) 14H(aq); cathode: Pt electrode; Cl2 bubbled into solution, Cl in solution; anode: Pt electrode; Cr3, H, and Cr2O7 2 in solu-tion; b. Cu2(aq) Mg(s) n Cu(s) Mg2(aq); cathode: Cu electrode; Cu2 in solution; anode: Mg electrode; Mg2 in solution 27. a. 0.03 V; b. 2.71 V 29. See Exercise 25 for a description of a galvanic cell. For each of the fol-lowing answers, all solutes are 1.0 M and all gases are at 1.0 atm. In the salt bridge, cations flow to the cathode and anions flow to the anode. a. Cl2(g) 2Br(aq) n Br2(aq) 2Cl(aq), e 0.27 V; cathode: Pt electrode; Cl2(g) bubbled in, Cl in solution; anode: Pt electrode; Br2 and Br in solution; b. 3H2O(l) 5IO4 (aq) 2Mn2(aq) n 5IO3 (aq) 2MnO4 (aq) 6H(aq), e 0.09 V; cathode: Pt electrode; IO4 , IO3 , and H2SO4 (as a source of H) in solution; anode: Pt electrode; Mn2, MnO4 , and H2SO4 in solution 31. 25a. PtCr3 (1.0 M), H (1.0 M), Cr2O7 2 (1.0 M) 0 0 Cl2 (1.0 atm)Cl (1.0 M) Pt; 25b. Mg 0 Mg2 (1.0 M) 0 0 Cu2 (1.0 M) 0 Cu; 29a. Pt Br (1.0 M), Br2 (1.0 M) 0 0 Cl2 (1.0 atm) 0 Cl (1.0 M) 0 Pt; 29b. Pt 0 Mn2 (1.0 M), MnO4 (1.0 M ), H (1.0 M) 0 0 IO4 (1.0 M), H(1.0 M) IO3 (1.0 M), 0 Pt 33. a. Au3(aq) 3Cu(aq) n 3Cu2(aq) Au(s), e 1.34 V; b. 2VO2 (aq) 4H(aq) Cd(s) n Cd2(aq) 2VO2(aq) 2H2O(l), e 1.40 V 35. a.
spontaneous; b.
37. e 0.41 V, G 79 kJ 39. 33a. 388 kJ; 33b. 270. kJ 41. 1.21 V 43. K H2O Cd2 I2 AuCl4 IO3 45. a. no; b. yes; c. yes; d. no 47. a. Cr2O7 2, O2, MnO2, IO3 ; b. PbSO4, Cd2, Fe2, Cr3, Zn2, H2O 49. ClO(aq) 2NH3(aq) n Cl(aq) N2H4(aq) H2O(l), ecell 1.00 V; Because ecell is positive for this reaction, at standard conditions ClO can spontaneously oxidize NH3 to the somewhat toxic N2H4.
51. a. larger; b. smaller 53. Electron flow is always from the anode to the cathode. For the cells with a nonzero cell potential, we will identify the cathode, which means the other compartment is the anode. a. 0; b. 0.018 V; compartment with [Ag] 2.0 M is cathode; c. 0.059 V; compartment with [Ag] 1.0 M is cathode; d. 0.26 V; compartment with [Ag] 1.0 M is cathode; e. 0 55. 2.12 V 57. 1.09 V 59. a. 0.23 V; b. 1.2 105 M 61. 0.16 V, copper is oxidized.
63.
65. a. G 20 kJ; 1 103; b. G 523 kJ; 5.12 1091; a. G 52 kJ; 1.4 109; b. G 90 kJ; 2 1015 67.
69. a. no reaction; b. Cl2(g) 2I(aq) S I2(s) 2Cl(aq), ecell 0.82 V; G 160 kJ; K 5.6 1027; c. no reaction; d. 4Fe2(aq) 4H(aq) O2(g) S 2.5 1026 1.7 10 30 not spontaneous e°cell 1.36 V, 2Mn2 8H2O, 16H 2MnO4 10F S 5F2 e°cell 0.97 V, 2MnO4 10I S 5I2 2Mn 2 8H2O, 16H O21g2 H21g2 O21g2 H21g2 O21g2.
H21g2 H2O1l2 H2O1l2 ¢G ¢G.
¢G ¢G temperatures 45. a. H 803 kJ, S 4 J/K, G 802 kJ; b. H 2802 kJ, S 262 J/K, G 2880. kJ; c. H 416 kJ, S 209 J/K, G 354 kJ; d. H 176 kJ, S 284 J/K, G 91 kJ 47.
is negative below 328.6 K.
49. CH4(g) CO2(g) n CH3CO2H(l), H 16 kJ, S 240. J/K, G 56 KJ; CH3OH(g) CO(g) n CH3CO2H(l), H 173 kJ, S 278 J/K, G 90. kJ; the second reaction is preferred. It should be run at temperatures below 622 K.
51.
53.
kJ/mol 55. yes 57. 188 kJ 59. a. shifts right; b. no shift since the reaction is at equilib-rium; c. shifts left 61. 8.72; 0.0789 63. 140 kJ 65. 71 kJ/mol 67. H 1.1 105 J/mol; S 330 J/K mol; The major difference in the plot is the slope of the line. An endothermic process has a negative slope for the ln(K) versus 1T plot, whereas an exothermic process has a positive slope.
69.
71. decreases; will be negative since 2 mol of gaseous reactants form 1 mol of gaseous product. For to be negative, must be negative (exothermic). For exothermic reactions, K decreases as T increases, so the ratio of the partial pressure of PCl5 to the partial pressure of PCl3 will decrease.
73. 43.7 K 75. 60 77. a. 1.8 104 J/mol; shifts left; b. 0; no shift since at equilibrium; c. 1.1 104 J/mol; shifts right; d. 0; no shift since at equilibrium; e. 2 103 J/mol; shifts left 79. a. 2.22 105; b. 94.3; c. 0.29 mol ATP 81. S is more favorable for reaction 2 than for reaction 1, resulting in K2 K1. In reaction 1, seven particles in so-lution form one particle. In reaction 2, four particles form one, which results in a smaller decrease in disorder than for reaction 1.
83. 725 K 85. H 286 kJ; G 326 kJ; K 7.22 1058; PO3 3.3 1041 atm; This partial pressure represents one molecule of ozone per 9.5 1017 L of air. Equilib-rium is probably not maintained under the conditions because the concentra-tion of ozone is not large enough to maintain equilibrium 87. a. Because and Because Because then b. A catalyst increases the value of the rate con-stant (increases rate) by lowering the activation energy. For the equilibrium constant K to remain constant, both kf and kr must increase by the same fac-tor. Therefore, a catalyst must increase the rate of both the forward and the re-verse reactions.
89. a. 0.333; b. PA 1.50 atm; PB 0.50 atm; c. G G RT ln(PBPA) 2722 J 2722 J 0 91. at least 7.5 torr 93. 16 g 95. 61 kJ 97.
Chapter 17 13. Oxidation: increase in oxidation number, loss of electrons; reduction: de-crease in oxidation number, gain of electrons 15. Reactions a, b, and c are oxidation–reduction reactions.
Oxidizing Reducing Substance Substance Agent Agent Oxidized Reduced a. H2O CH4 CH4(C) H2O(H) b. AgNO3 Cu Cu AgNO3(Ag) c. HCl Zn Zn HCl(H) 17. Magnesium is an alkaline earth metal; Mg will oxidize to The ox-idation state of hydrogen in HCl is To be reduced, the oxidation state of H must decrease. The obvious choice for the hydrogen product is where hydrogen has a zero oxidation state. The balanced reaction is: Mg goes from the 0 to the 2 ox-idation state by losing two electrons. Each H atom goes from the 0 to the 1 oxidation state by gaining one electron. Since there are two H atoms in the balanced equation, then a total of two electrons are gained by the H atoms.
Hence, two electrons are transferred in the balanced reaction. When the electrons are transferred directly from Mg to no work is obtained. In or-der to harness this reaction to do useful work, we must control the flow of H, Mg1s2 2HCl1aq2 S MgCl21aq2 H21g2.
H21g2 1.
Mg2.
4.1 kJ/mol K kf kr .
exp a¢G° RT b, K kf kr exp a Ea RT 1Ea ¢G°2 RT b exp a¢G° RT b.
kr A exp a 1Ea ¢G°2 RT b, kf A exp a Ea RT b ¢H ¢G ¢S 447 J/K mol 731 817 kJ 5.40 kJ; 328.6 K; ¢G° Answers to Selected Exercises A61 4Fe3(aq) 2H2O(l), ecell 0.46 V; G 180 kJ; K 1.3 1031; 71. a. Au3(aq) 3Tl(s) S Au(s) 3Tl(aq); ecell 1.84 V; b. G 533 kJ; K 2.52 1093; c. 2.04 V 73.
75.
77. a. 30. hours; b. 33 s; c. 1.3 hours 79. a. 16 g; b. 25 g; c. 71 g; d. 4.9 g 81. Bi 83. 9.12 L F2 (anode), 29.2 g K (cathode) 85. 7.44 104 A 87. 1.14 102 M 89. Au followed by Ag followed by Ni fol-lowed by Cd 91. a. cathode: Ni2 2e S Ni; anode: 2Br S Br2 2e; b. cathode: Al3 3e S Al; anode: 2F S F2 2e; c. cathode: Mn2 2e S Mn; anode: 2I S I2 2e 93. a. 0.10 V, SCE is anode; b. 0.53 V, SCE is anode; c. 0.02 V, SCE is cathode; d. 1.90 V, SCE is cath-ode; e. 0.47 V, SCE is cathode 95. a. decrease; b. increase; c. decrease; d. decrease; e. same 97. a. G 582 kJ; K 3.45 10102; e 1.01 V; b. 0.65 V; 99. Aluminum has the ability to form a durable oxide coat-ing over its surface. Once the HCl dissolves this oxide coating, Al is exposed to H and is easily oxidized to Al3. Thus, the Al foil disappears after the ox-ide coating is dissolved.
101. The claim is impossible. The strongest oxi-dizing agent and reducing agent when combined give e of only about 6 V.
103. wmax 13,200 kJ; the work done can be no larger than the free en-ergy change. If the process were reversible all of the free energy released would go into work, but this does not occur in any real process. Fuel cells are more efficient in converting chemical energy to electrical energy; they are also less massive. Major disadvantage: They are expensive.
105. 0.98 V 107. 0.250 mol 109. 3 111.
if we graph e versus T, we should get a straight line (y mx b). The slope of the line is equal to SnF and the y-intercept is equal to H nF. e will have little temperature dependence for cell reactions with S close to zero.
113. 9.8 106 115. 2.39 107 117. a. 0.02 pH units; 6 106 M H; b. 0.001 V 119. a. 0.16 V; b. 8.6 mol 121.
123. a. 0.12 V; b. 0.54 V 125. a.
b.
127. Osmium(IV) nitrate; [Ar]4s13d10 Chapter 18 1. The characteristic frequencies of energies emitted in a nuclear reaction suggest that discrete energy levels exist in the nucleus. The extra stability of certain numbers of nucleons and the predominance of nuclei with even num-bers of nucleons suggest that the nuclear structure might be described by using quantum numbers.
3.
particle production has the net effect of turning a neutron into a proton. Radioactive nuclei having too many neutrons typically undergo particle decay. Positron production has the net effect of turning a proton into a neutron. Nuclei having too many protons typically undergo positron decay.
5. The transuranium elements are the elements having more protons than uranium. They are synthesized by bombarding heavier nuclei with neutrons and positive ions in a particle accelerator.
7.
The key difference is the mass change when going from reactants to products. In chem-ical reactions, the mass change is indiscernible. In nuclear processes, the mass change is discernable. It is the conversion of this discernable mass change into energy that results in the huge energies associated with nuclear processes.
9. Sr-90 is an alkaline earth metal having chemical properties similar to cal-cium. Sr-90 can collect in bones replacing some of the calcium. Once imbed-ded inside the human body, particles can do significant damage. Rn-222 is a noble gas so one would expect Rn to be unreactive and pass through the body quickly; it does. The problem with Rn-222 is the rate at which it produces al-pha particles. With a short half-life, the few moments that Rn-222 is in the lungs, a significant number of decay events can occur; each decay event produces an alpha particle which is very effective at causing ionization and can produce a dense trail of damage.
11. a. 51 24Cr 1 0e S 51 23V; b. 131 53I S 131 54Xe 13. a. 68 31Ga 1 0e S 68 30Zn; b.
62 29Cu S 1 0e 28 62Ni; c. 212 87Fr S 4 2He 208 85At; d. 129 51Sb S 1 0e 129 52Te 15. 10 particles; 5 particles 17. 53 26Fe has too many protons. It will undergo positron production, electron capture, and/or alpha-particle production. 59 26Fe has too many neu-trons and will undergo beta-particle production. (See Table 18.2 of the text.) 0 1e ¢E ¢mc2; b-b-12.2 kJ/mol 5.77 10 10; 1.5 M 10 18 M; 3Ni 24 3Ag 4 4.6 e° T¢S° nF ¢H° nF ; 6.19 1052 5.1 10 20 19. a. 249 98Cf 18 8O S 263 106Sg 40 1n; b. 259 104Rf 21. 690 hours 23. 81Kr is most stable since it has the longest half-life. 73Kr is “hottest” since it decays very rapidly due to its very short half-life. 73Kr, 81s; 74Kr, 34.5 min; 76Kr, 44.4 h; 81Kr, 6.3 105 yr 25. 6.22 mg 32P remains 27. 0.230 29. 26 g 31. 2.3 counts per minute per gram of C. No; for a 10.-mg C sample, it would take roughly 40 min to see a single disintegration. This is too long to wait, and the background radiation would probably be much greater than the 14C activ-ity. Thus 14C dating is not practical for very small samples.
33. 3.8 109 yr 35. 4.3 106 kg/s 37. 232Pu, 1.715 1014 J/mol; 231Pa, 1.714 1014 J/mol 39. 12C: 1.23 1012 J/nucleon; 235U: 1.2154 1012 J/nucleon; since 56Fe is the most stable known nucleus, then the binding energy per nucleon for 56Fe would be larger than that of 12C or 235U. (See Fig. 18.9 of the text.) 41. 6.01513 amu 43. 2.0 1010 J/g of hydrogen nuclei 45. The Geiger–Müller tube has a certain response time. After the gas in the tube ion-izes to produce a “count,” some time must elapse for the gas to return to an electrically neutral state. The response of the tube levels off because, at high activities, radioactive particles are entering the tube faster than the tube can re-spond to them.
47. All evolved O2(g) comes from water.
49. 2 neutrons; 4 particles 51. Strontium. Xe is chemically unreactive and not readily in-corporated into the body. Sr can be easily oxidized to Sr2. Strontium is in the same family as calcium and could be absorbed and concentrated in the body in a fashion similar to Ca2. The chemical properties determine where ra-dioactive material may be concentrated in the body or how easily it may be excreted.
53. a. unstable; beta production; b. stable; c. unstable; positron production or electron capture; d. unstable, positron production, electron capture, or alpha production.
55. 49.7 yr 57. 1975 59. 900 g 235U 61.
; 63. Assuming that (1) the radionu-clide is long lived enough that no significant decay occurs during the time of the experiment, and (2) the total activity is uniformly distributed only in the rat’s blood; V 10. mL.
65. a. 12 6C; b. 13N, 13C, 14N, 15O, and 15N; c. 5.950 1011 J/mol 1H 67. 4.3 1029 69. 249 97Bk 22 10Ne 267 107Bh 40 1n; 62.7s; [Rn]7s25f146d 5 Chapter 19 1. The gravity of the earth cannot keep H2 in the atmosphere.
3. The acid-ity decreases. Solutions of Be2 are acidic, while solutions of the other M2 ions are neutral.
5. The planes of carbon atoms slide easily. Graphite is not volatile so the lubricant will not be lost when used in a high vacuum environ-ment.
7. p-type semiconductor 9. For groups 1A–3A, the small size of H (as compared to Li), Be (as compared to Mg), and B (as compared to Al) seems to be the reason why these elements have nonmetallic properties, while others in the groups 1A–3A are strictly metallic. The small size of H, Be, and B also causes these species to polarize the electron cloud in nonmetals, thus forcing a sharing of electrons when bonding occurs. For groups 4A–6A, a major dif-ference between the first and second members of a group is the ability to form bonds. The smaller elements form stable bonds, while the larger elements do not exhibit good overlap between parallel p orbitals and, in turn, do not form strong bonds. For group 7A, the small size of F as compared to Cl is used to explain the low electron affinity of F and the weakness of the bond.
11. In order to maximize hydrogen bonding interactions in the solid phase, ice is forced into an open structure. This open structure is why is less dense than 13. a. H 207 kJ, S 216 J/K; b. T 958 K 15. a. lithium oxide; b. potassium superoxide; c. sodium peroxide 17. a. Li2O(s) H2O(l) S 2LiOH(aq); b. Na2O2(s) 2H2O(l) S 2NaOH(aq) H2O2(aq); c. LiH(s) H2O(l) S H2(g) LiOH(aq); d. 2KO2(s) 2H2O(l) S 2KOH(aq) O2(g) H2O2(aq) 19. 2Li(s) 2C2H2(g) S 2LiC2H(s) H2(g); oxidation–reduction 21. a. magnesium carbonate; b. barium sulfate; c. strontium hydroxide 23. CaCO3(s) H2SO4(aq) S CaSO4(aq) H2O(l) CO2(g) 25. In the gas phase, linear molecules would exist: F Be F H2O1l2.
H2O1s2 F¬F p p p S 7 105 m/s; 8 1016 J/nuclei A62 Answers to Selected Exercises to the occurrence of congestive heart failure.
5.
oxidation state: ; oxidation state: ; oxidation state: ; 0 oxidation state: and all other elemental forms of sulfur; 2 oxidation state: 7. a.
this reaction produces a lot of energy which can be used in a cannon apparatus to send a stopper flying across the room. To initiate this extremely slow reaction, light of specific wavelengths is needed. This is the purpose of lighting the magnesium strip.
When magnesium is oxidized to MgO, an intense white light is produced.
Some of the wavelengths of this light can break bonds and get the reaction started. b.
is brown. The disappearance of the brown color indicates that all of the has reacted with the alkene (no free re-mains). c.
This is a highly exothermic reaction, hence the sparks that accompany this reaction. The purple smoke is excess being vaporized the purple smoke is 9. Nitrogen’s small size does not provide room for all four oxygen atoms, making NO4 3 un-stable. Phosphorus is larger so PO4 3 is more stable. To form NO3 , a pi bond must form. Phosphorus doesn’t form strong pi bonds as readily as nitrogen.
11.
13. a. NH4NO3(s) N2O(g) 2H2O(g); b. 2N2O5(g) S 4NO2(g) O2(g); c. 2K3P(s) 6H2O(l) S 2PH3(g) 6KOH(aq); d. PBr3(l) 3H2O(l) S H3PO3(aq) 3HBr(aq); e. 2NH3(aq) NaOCl(aq) S N2H4(aq) NaCl(aq) H2O(l) 15. CaF2 3Ca3(PO4)2(s) 10H2SO4(aq) 20H2O(l) S 6H3PO4(aq) 2HF(aq) 10CaSO4 2H2O(s) 17. 2.08 mol 19. N2H4(l) O2(g) S N2(g) 2H2O(g); H 590. kJ 21.
N2(g) O2(g) S NO(g) G G f(NO); NO (and some other oxides of nitrogen) have weaker bonds as compared with the triple bond of N2 and the double bond of O2. Because of this, NO (and some other oxides of nitrogen) has a higher (positive) standard free energy of formation as compared to the relatively stable N2 and O2 molecules.
23.
Bond order # unpaired e M.O.
NO 2.5 1 NO 3 0 NO 2 2 Lewis Lewis structures are not adequate for NO and NO. The M.O. model gives correct results for all three species. For NO, Lewis structures fail for odd-electron species. For NO, Lewis structures fail to predict that NO is paramagnetic.
25. a. H3PO4 H3PO3; b. H3PO4 H2PO4 HPO4 2 27. The acidic protons are attached to oxygen.
29. 821 nm 31. a. 2SO2(g) O2(g) S 2SO3(g); b. SO3(g) H2O(l) S H2SO4(aq); c. 2Na2S2O3(aq) I2(aq) S Na2S4O6(aq) 2NaI(aq); d. Cu(s) 2H2SO4(aq) S CuSO4(aq) 2H2O(l) SO2(aq) 33. a.
b.
O O O V-shaped; ≈120; sp2 O O O O S O O 2 Trigonal pyramid; ≈109.5; sp3 O P H H O O H P O H O O P H O H O O P H O H O H4P2O5 H4P2O6 NO+ + N O q NO N O P mn N O P mn N O P NO – – N O P 1 2 1 2 ¡ Heat NO2 N2O4 6 NO 6 N2O I21g24.
3 I21s2 2Al1s2 3I21s2 S 2AlI31s2; Br2 Br2 Br2 Cl¬Cl H21g2 Cl21g2 S 2HCl1g2; Na2S H2S, S8 SCl2 2 SF4 SO2, SO3 2, 4 SF6 SO3, SO4 2, 6 In the solid state, BeF2 has the following extended structure: 27. 2 102 M 29. 3.84 106 g Ba 31. a. AlN; b. GaF3; c. Ga2S3 33. B2H6(g) 3O2(g) S 2B(OH)3(s) 35. In2O3(s) 6H(aq) S 2In3(aq) 3H2O(l); In2O3(s) OH(aq) S no reaction; Ga2O3(s) 6H(aq) S 2Ga3(aq) 3H2O(l); Ga2O3(s) 2OH(aq) 3H2O(l) S 2Ga(OH)4 (aq) 37. 2Ga(s) 3F2(g) S 2GaF3(s); 4Ga(s) 3O2(g) S 2Ga2O3(s); 16Ga(s) 3S8(s) S 8Ga2S3(s); 2Ga(s) N2(g) S 2GaN(s); 2Ga(s) 6HCl(aq) S 2GaCl3(aq) 3H2(g) 39.
To form CF6 2, carbon would have to expand its octet of electrons. Carbon compounds do not expand their octet because of the small atomic size of car-bon and because there are no low energy d orbitals on carbon to accommodate the extra electrons.
41. a. SiO2(s) 2C(s) S Si(s) 2CO(g); b. SiCl4(l) 2Mg(s) S Si(s) 2MgCl2(s); c. Na2SiF6(s) 4Na(s) S Si(s) 6NaF(s) 43. Lead is very toxic. As the temperature of the water increases, the solubil-ity of Pb increases. Drinking hot tap water from pipes containing lead solder could result in higher Pb concentrations in the body.
45. C6H12O6(aq) S 2C2H5OH(aq) 2CO2(g) 47. The electrons are free to move in graphite, thus giving it a greater conductivity (lower resistance). The electrons have the greatest mobility within the sheets of carbon atoms. Electrons in diamond are not mobile (high resistance). The structure of diamond is uniform in all direc-tions; thus there is no directional dependence of the resistivity.
49. Only some of the ice will melt; 51. It is feasible to recycle Al by melting the metal because, in theory, it takes less than 1% of the energy required to produce the same amount of Al by the Hall–Heroult process.
53. 60 55. 2.12 V 57. Strontium and calcium are both alkaline earth metals, so both have simi-lar chemical properties. Since milk is a good source of calcium, strontium could replace some calcium in milk without much difficulty.
59. The inert-pair ef-fect refers to the difficulty of removing the pair of valence s electrons from some of the elements in the fifth and sixth periods of the periodic table. As a result, multiple oxidation states are exhibited for the heavier elements of Groups 3A (and 4A). In, In3, Tl, and Tl3 oxidation states are all important to the chemistry of In and Tl.
61. 3.08 63. If the compound contained Ga(II) it would be paramagnetic and if the compound contained Ga(I) and Ga(III), it would be diamagnetic. Paramagnetic compounds have an apparent greater mass in a magnetic field.
65. 59 atm in the gas phase; 1.8 mol CO2/L in the wine 67. Pb(NO3)2(aq) H3AsO4(aq) S PbHAsO4(s) 2HNO3(aq) 69. Ca; 12.698 71.
73. 2.0 1037 M 75. Carbon is much smaller than Si and cannot form a fifth bond in the transition state.
77. I 79. a. 7.1 g; b. 979 nm; This electromagnetic radiation is not visible to humans; it is in the infrared region of the electromagnetic radiation spectrum, c. n-type Chapter 20 1. This is due to nitrogen’s ability to form strong bonds whereas heavier group 5A elements do not form strong bonds. Therefore, and do not form since two bonds are required to form these diatomic substances.
3. There are medical studies that have shown an inverse relationship between the incidence of cancer and the selenium levels in soil. The foods grown in these soils and eventually digested are assumed to somehow furnish protection from cancer. Selenium is also involved in the activity of vitamin E and certain enzymes in the human body. In addition, selenium deficiency has been linked p Sb2 As2, P2, p p 3.34, 5.6 1011 M 0°C F F F F C Tetrahedral; 109.5; sp3 F F F F Ge Tetrahedral; 109.5; sp3 F F F F Ge F F Octahedral; 90; d2sp3 2 F F F F F F Be Be Be Answers to Selected Exercises A63 c.
d.
e.
35. 0.301 g H2O2; g excess HCl 37. From the following Lewis structure, each oxygen atom has a tetrahedral arrangement of electron pairs.
Therefore, bond angles 109.5, and each O is sp3 hybridized.
Oxidation states are more useful. We are forced to assign 1 as the oxidation state for oxygen. Oxygen is very electronegative, and 1 is not a stable oxi-dation state for this element.
39. a. BaCl2(s) H2SO4(aq) S BaSO4(s) 2HCl(g); b. BrF(s) H2O(l) S HF(aq) HOBr(aq); c. SiO2(s) 4HF(aq) S SiF4(g) 2H2O(l) 41. ClO can oxide NH3 to the somewhat toxic N2H4.
43. a. IO4 ; b. IO3 ; c. IF2 ; d. IF4 ; e. IF6 45. XeF2 can react with oxygen and water to produce explosive xenon oxides and oxy-fluorides.
47. Release of Sr is probably more harmful. Xe is chemically un-reactive. Strontium is in the same family as calcium and could be absorbed and concentrated in the body in a fashion similar to Ca. This puts the radioactive Sr in the bones, and red blood cells are produced in bone marrow. Xe would not be readily incorporated into the body. The chemical properties determine where a radioactive material may concentrate in the body or how easily it may be excreted. The length of time of exposure and the body part exposed to ra-diation significantly affects the health hazard.
49. As the halogen atoms get larger, it becomes more difficult to fit three halogens around the small N.
51.
All the resonance structures for fulminate involve greater formal charges than in cyanate, making fulminate more reactive (less stable).
53. 32 kg bacterial tissue 55.
The lone pair of electrons around Te exerts a stronger repulsion than the bond-ing pairs, pushing the four square planar F’s away from the lone pair.
57. exothermic 59. MgSO4(s) Mg2(aq) (aq); NH4NO3(s) NH4 (aq) (aq) 61. F, Cl, Br, or I; trigonal pyramid; 63. XeF2 109.5° NO3 S SO4 2 S Te F F – F F F A G D G D F F Formal charge: Oxidation state: O O 0 1 0 1 0 1 0 1 3.6 10 2 F F F F Te F F Octahedral; 90; d2sp3 Br Se Br Br Br See-saw; a ≈ 120, b ≈ 90; dsp3 b a b Cl S Cl V-shaped; ≈109.5; sp3 65. For the reaction the activation energy must in some way involve the breaking of a nitrogen– nitrogen single bond. For the reaction at some point nitrogen–oxygen bonds must be broken. NON single bonds (160 kJ/mol) are weaker than NOO single bonds (201 kJ/mol). In addition, reso-nance structures indicate that there is more double-bond character in the NOO bonds than in the NON bond. Thus NO2 and NO are preferred by kinetics be-cause of the lower activation energy.
67. a. NO; b. NO2; c. kcatkun 2.3; d. ClO(g) O(g) S O2(g) Cl(g); O3(g) O(g) S 2O2(g); e. The Cl-catalyzed reaction is roughly 52 times faster (more efficient) than the NO-catalyzed reaction.
69. 5.89 71. 20. g 73. a. 287 kJ/mol; b. [IF2]: V-shaped; sp3; tetrahedral; sp3 Chapter 21 5.
The oxalate anion forms a soluble complex ion with iron in rust which allows rust stains to be removed.
7. No; both the trans or the cis forms of have mirror images that are superimposable. For the cis form, the mirror image only needs a 90 rotation to produce the original structure.
Hence, neither the trans nor cis forms are optically active.
9. a.
is an example of a weak-field case having three unpaired electrons. b.
is a strong field ligand so will be a low-spin case having zero unpaired electrons.
11. At high altitudes, the oxygen content of air is lower, so less oxyhemoglo-bin is formed which diminishes the transport of oxygen in the blood. A seri-ous illness called high-altitude sickness can result from the decrease of in the blood. High altitude acclimatization is the phenomena that occurs in the human body in response to the lower amounts of oxyhemoglobin in the blood. This response is to produce more hemoglobin, and, hence, increase the oxyhemoglobin in the blood. High-altitude acclimatization takes several weeks to take hold for people moving from lower altitudes to higher altitudes. 13. a. Ni: [Ar]4s23d8; b. Cd: [Kr]5s24d10; c. Zr: [Kr]5s24d2; d. Os: [Xe]6s24f 145d 6 15. a. Ti: [Ar]4s23d2; Ti2: [Ar]3d2; Ti4: [Ne]3s23p6 or [Ar]; b. Re: [Xe]6s24f 145d5; Re2: [Xe]4f 145d5; Re3: [Xe]4f 145d4; c. Ir: [Xe]6s24f 145d7; Ir 2: [Xe]4f145d7; Ir3: [Xe]4f 145d6 17. a. Fe3: [Ar]3d5; b. Ag: [Kr]4d10; c. Ni2: [Ar]3d8; d. Cr3: [Ar]3d3 19. a. molybdenum(IV) sulfide, molybdenum(VI) oxide; b. MoS2, 4; MoO3, 6; (NH4)2 Mo2O7, 6; (NH4)6Mo7O24 4H2O, 6 21. The lan-thanide elements are located just before the 5d transition metals. The lanthanide contraction is the steady decrease in the atomic radii of the lanthanide elements when going from left to right across the periodic table. As a result of the lan-thanide contraction, the sizes of the 4d and 5d elements are very similar. This leads to a greater similarity in the chemistry of the 4d and 5d elements in a given vertical group.
23. If rain is imminent, the large amount of water vapor in the air would cause the reaction to shift to the right. The indicator would take O2 Co1CN26 3 CN O O O O O large ∆ CoCl4 2 O O O O O small ∆ Co1NH324Cl2 1Fe2O32, Fe2O31s2 6 H2C2O41aq2 S 2 Fe1C2O423 31aq2 3 H2O1l2 6 H1aq2; 3BF44 : N N O O2 + N2O O O M D O P mn mn N N O NO2 + NO O O M D O P O C P N – 0 Formal charge: 0 –1 P mn ð ½ ý O C O N – – –1 0 0 q ð ð mn O C q N +1 0 –2 O ð ð C N P O – –2 Formal charge: +1 0 P mn ½ ý ½ ý C N q O – – –1 +1 –1 O ð ð mn C N O O –3 +1 +1 q ð ð A64 Answers to Selected Exercises 39. SCN, NO2 , and OCN can form linkage isomers; all are able to bond to the metal ion in two different ways.
41. Cr(acac)3 and cis-Cr(acac)2(H2O)2 are optically active.
43. a. Fe2 __ h __ h_ ___ ___ _ h _ g _ __ h_ __ h_ _ h _ g _ _ h _ g _ _ h _ g _ High spin Low spin b. Fe3 __ h_ __ h_ c. Ni2 __ h_ __ h_ __ h_ __ h_ __ h_ _ h _ g _ _ h _ g _ _ h _ g _ High spin 45. weak field 47. a. 0; b. 2; c. 2 49.
51. The violet complex ion absorbs yellow-green light ( 570 nm), the yellow complex ion absorbs blue light ( 450 nm), and the green complex ion absorbs red light ( 650 nm). The violet com-plex ion is Cr(H2O)6 3, the yellow complex ion is Cr(NH3)6 3, and the green complex ion is Cr(H2O)4Cl2 .
53. CoBr4 2 is a tetrahedral complex ion, while CoBr6 4 is an octahedral complex ion. Since tetrahedral d-orbital split-ting is less than one-half the octahedral d-orbital splitting, the octahedral com-plex ion (CoBr6 4) will absorb higher-energy light, which will have a shorter wavelength than 3.4 106 m (E hc).
55. 5 57. a. 11 kJ; b. H° 172.5 kJ; S° 176 J/K; T 980. K 59.
61. [Cr(NH3)5I]I2; octahedral 63.
orange precipitate; soluble complex ion 65. a. 2; b. 3; c. 4; d. 4 67. a. op-tical isomerism b.
___ ___ _ h _ g _ _ h _ g _ _ h _ g _ 69. Octahedral Cr2 complexes should be used. Cr2: [Ar]3d 4; High-spin (weak-field) Cr2 complexes have four unpaired electrons and low-spin (strong-field) Cr2 complexes have two unpaired electrons. Ni2: [Ar]3d8; Octahedral Ni2 complexes will always have two unpaired electrons, whether high or low spin. Therefore, Ni2 complexes cannot be used to distinguish weak- from strong-field ligands by examining magnetic properties. Alternatively, the ligand field strengths can be measured using visible spectra. Either Cr2 or Ni2 com-plexes can be used for this method.
71. Pb(OH)2 will not form since Q is less than Ksp.
73. 60 75.
If A and B produced very similar crys-tal fields, the complex trans-[NiA2B4]2 would give an octahedral crystal field diagram: This is paramagnetic. Because the complex ion is diamagnetic, ligands A and B must produce different crystal fields resulting in a unique d-orbital splitting diagram.
77. a. 0.26 V; b. From standard reduction potentials, Co3 (e° 1.82 V) is a much stronger oxidizing agent than Co(en)3 3 (e° 0.26 V); c. In aqueous solution, Co3 forms the hydrated transition metal complex, Co(H2O)6 3. In both complexes, Co(H2O)6 3 and Co(en)3 3, cobalt exists as Co3, which has 6 d electrons. If we assume a strong-field case for each com-plex ion, then the d-orbital splitting diagram for each has the six electrons paired in the lower-energy t2g orbitals. When each complex ion gains an elec-tron, the electron enters the higher-energy eg orbitals. Since en is a stronger-field ligand than H2O, then the d-orbital splitting is larger for Co(en)3 3, and it takes more energy to add an electron to Co(en)3 3 than to Co(H2O)6 3. There-fore, it is more favorable for Co(H2O)6 3 to gain an electron than for Co(en)3 3 to gain an electron.
79. No, since in all three cases six bonds are formed be-tween Ni2 and nitrogen. So H values should be similar. S° for formation of the complex ion is most negative for 6NH3 molecules reacting with a metal ion (7 independent species become 1). For penten reacting with a metal ion, 2 independent species become 1, so S° is the least negative. Thus the chelate effect occurs because the more bonds a chelating agent can form to the metal, O O O O O Ni2 d8; S HgI4 21aq2, HgI21s2 2I1aq2 Hg21aq2 2I1aq2 S HgI21s2, 2H2O1l2 S 4Ag1CN22 1aq2 4OH1aq2 8CN1aq2 4Ag1s2 O21g2 3Co1H2O264 3 6 3CoI64 3 3Co1CN264 3 6 3Co1en234 3 6 on the color of the pink CoCl2 6H2O.
25. Test tube 1: added Cl reacts with Ag to form the silver chloride precipitate. The net ionic equation is Ag(aq) Cl(aq) S AgCl(s). Test tube 2: added NH3 reacts with Ag ions to form the soluble complex ion Ag(NH3)2 . As this complex ion forms, Ag is removed from the solution, which causes the AgCl(s) to dissolve. When enough NH3 is added, then all of the silver chloride precipitate will dissolve.
The equation is AgCl(s) 2NH3(aq) S Ag(NH3)2 (aq) Cl(aq). Test tube 3: added H reacts with the weak base NH3 to form NH4 . As NH3 is removed, Ag ions are released to solution, which can then react with Cl to reform AgCl(s). The equations are Ag(NH3)2 (aq) 2H(aq) S Ag(aq) 2NH4 (aq) and Ag(aq) Cl(aq) S AgCl(s).
27. [Co(NH3)6]I3: 3 mol AgI; [Pt(NH3)4I2]I2: 2 mol AgI; Na2[PtI6]: 0 mol AgI; [Cr(NH3)4I2]I: 1 mol AgI.
29. a. pentaamminechlororuthenium(III) ion; b. hexacyanoferrate(II) ion; c. tris(ethylenediamine)manganese(II) ion; d. pentaamminenitrocobalt(III) ion 31. a. hexaamminecobalt(II) chloride; b. hexaaquacobalt(III) iodide; c. potassiumtetrachloro-platinate(II); d. potassium hexachloroplatinate(II); e. pentaamminechlorocobalt(III) chloride; f. triamminetrinitrocobalt(III) 33. a. K2[CoCl4]; b. [Pt(H2O)(CO)3]Br2; c. Na3[Fe(CN)2(C2O4)2]; d. [Cr(NH3)3Cl(H2NCH2CH2NH2)]I2 35. a.
b.
c.
d.
37.
O O M C CH2 H2N J Cr N I I N + NH3 H3N A A D G D G Cr en = N N NH2CH2CH2NH2 N I I N + NH3 H3N A A D G D G Cr N I I N + NH3 H3N A A D G D G M Ir cis Cl Cl H3N NH3 Cl H3NO O A A D D Ir trans Cl Cl H3N Cl NH3 H3NO O A A D D Pt H3N H3N NH3 I I 2+ O O A NH3 cis A D D Pt H3N H3N NH3 I NH3 2+ O O A I trans A D D Co O H2O H2O O O cis O O O – K K A A Co OH2O O O O O O OH2 trans – A A A E H E H K K A E H K K A E H K K O O A E H E H O O E H E H O O NH2 CH2 Cu and C J M M O O O O H2N O O C H2CO O O O A A A A A D G D G O O NH2 CH2 Cu C M O O O O O H2N O O C H2CO O O O O A A A A A A D G D G Answers to Selected Exercises A65 the more favorable S° is for the formation of the complex ion, and the larger the formation constant.
81.
___ dz 2 ___ ___ dx2 y2, dxy ___ ___ dxz, dyz 83. Place the trans NH3 ligands on the z axis with the ligands oriented as indicated in the following diagram. Since produces a much stronger crystal field, the diagram will most resemble that of a square planar complex: 85. a. 6; [Ar]3d2; b. 0.0496 L N2 Chapter 22 1. a. 1-sec-butylpropane b. 4-methylhexane 3-methylhexane is correct.
3-methylhexane is correct.
c. 2-ethylpentane d. 1-ethyl-1-methylbutane 3-methylhexane is correct.
3-methylhexane is correct.
e. 3-methylhexane f. 4-ethylpentane 3-methylhexane is correct.
All six of these are the same compound. They only differ from each other by rotations about one or more carbon-carbon single bonds. Only one isomer of is present in all of these names, 3-methylhexane.
3. a.
b.
The longest chain is 4 carbons The longest chain is 7 carbons long long. The correct name is and we would start the numbering 2-methylbutane.
system at the other end for lowest possible numbers. The correct name is 3-iodo-3-methylheptane.
CH3CH2CH2CH2C A A CH3 CH3 A I CH2 O CH3CHCH3 A CH2CH3 C7H16 CH3CH2CH2CHCH3 A CH2CH3 CH3CH2CHCH2CH2CH3 A CH3 CHCH2CH2CH3 A A CH2CH3 CH3 CH2CH3 A CH3CHCH2CH2CH3 CH3CH2CH2CHCH2CH3 A CH3 CH3CHCH2CH3 A CH2CH2CH3 O O O O O dx2_ y2 dz2 dxy dxz dyz CN x CN y NH3 NH3 NC NC z Ni CN CN c.
d.
This compound cannot exhibit cis The OH functional group gets the trans isomerism since one of the lowest number. 3-bromo-2-butanol double bonded carbons has the same is correct.
two groups attached. The numbering system should also start at the other end to give the double bond the lowest possible number. 2-methyl-2-pentene is correct.
5. Hydrocarbons are nonpolar substances exhibiting only London dispersion forces. Size and shape are the two most important structural features relating to the strength of London dispersion forces. For size, the bigger the molecule (the larger the molar mass), the stronger the London dispersion forces and the higher the boiling point. For shape, the more branching present in a compound, the weaker the London dispersion forces and the lower the boiling point.
7. The correct order of strength is The difference in strength is related to the types of intermolecular forces present. All of these polymers have London dispersion forces. However, polyhydrocarbons only have London dispersion forces. The polar ester group in polyesters and the polar amide group in polyamides give rise to additional dipole forces. The polyamide has the ability to form relatively strong hydro-gen bonding interactions, hence why it would form the strongest fibers.
9. a.
b.
c.
d.
e.
f.
or g.
11. a. A polyester forms when an alcohol functional group reacts with a car-boxylic acid functional group. The monomer for a homopolymer polyester must have an alcohol functional group and a carboxylic acid functional group present CH3OH HOCCH3 CH3 O P O P O O CCH3 H2O O H ester CH3CH2CH oxidation CH3CH2C O P O P OH O CH3CH2CH2OH carboxylic acid oxidation CH3CH2C O P OH O CH3CHCH3 ketone oxidation CH3CCH3 O P OH A CH3CH2OH aldehyde oxidation CH3CH O P P CH3C CH2 H2O A CH3 3 alcohol H O CH3C A OH A H A CH3 CH2 major product P O CH2CH CH2 H2O CH2CH CH2 A OH A H 2 alcohol H major product P O CH2 CH2 H2O CH2 CH2 A OH A H 1 alcohol H polyamide.
polyhydrocarbon 6 polyester 6 1CH32 CH3CHCHCH3 A Br A OH CH3CH2CHPCOCH3 A CH3 A66 Answers to Selected Exercises c.
d. 4-ethyl-2-methylheptane; 2,2,3-trimethylhexane 21. a. 2,2,4-trimethylhexane: b. 5-methylnonane; c, 2,2,4,4-tetra-methylpentane; d. 3-ethyl-3-methyloctane 23.
Each carbon is bonded to four other atoms.
25. a. 1-butene; b. 4-methyl-2-hexene; c. 2,5-dimethyl-3-heptene 27. a. CH3CH2CHPCHCH2CH3; b. CH3CHPCHCHPCHCH2CH3; c.
29. a.
b.
c.
d.
31. a. 1,3-dichlorobutane; b. 1,1,1-trichlorobutane; c. 2,3-dichloro-2, 4-dimethylhexane; d. 1,2-difluoroethane; 33. CH2Cl CH2Cl,1-2-dichloroethane: There is free rotation about the C C single bond that doesn’t lead to different compounds. CHCl CHCl, 1-2-dichloroethene: There is no ro-tation about the C C double bond. This creates the cis and trans isomers, which are different compounds.
35. , compounds b and c; , all compounds 37. CH2PCHCH2CH2CH3 CH3CHPCHCH2CH3 39.
41.
CH2 CCH2 F P CH3 A A CH2 CHCHCH3 P F A F G C CH3 H3C H C P D G D H G C F CH2CH3 H C P D G D C Cl CH3 H H C P D G D G C H CH3 Cl Cl H C P D G D G Cl CH2CH CH2 P A Cl CH2 CCH3 P A CH3C CH3 CHCH3 P A CH3CHCH CH3 CH2 P A CH2 CH3 CCH2CH3 P A “ “ ¬ ¬ CH2CH CHCH3 P CH2CH3 CH2CH3 C C CH3 H3C CH3 CH3 O O A A CH3 CH3 A A CH3 CH2CH3 CH3 CH3 CHCH CHCH2CH2CH2CH3 P A O C O H COH H H A A H A A A A H O C O H COH CH3¬CH2¬CH2¬CH3; CH3 CH CH3 C CH3 CH3 CH2CH2CH3 in the structure.
b. A polyamide forms when an amine functional group re-acts with a carboxylic acid functional group. For a copolymer polyamide, one monomer would have at least two amine functional groups present and the other monomer would have at least two carboxylic acid functional groups present. For polymerization to occur, each monomer must have two reactive functional groups present.
c. To form an addition polymer, a carbon-carbon double bond must be present. To form a polyester, the monomer would need the alcohol and carboxylic acid functional groups present. To form a polyamide, the monomer would need the amine and carboxylic acid functional groups present. The two possibilities are for the monomer to have a carbon-carbon double bond, an alcohol functional group, and a carboxylic acid functional group all present, or to have a carbon-carbon double bond, an amine functional group, and a carboxylic acid functional group present.
13.
15. a.
b.
17. a.
b.
c.
d.
19. a.
b.
CH3 CH3 C CH3 CH3 CH2 CH3 CH CH3CH2 CH2CH2CH3 CH CH2 CH3 CH CH3 CH3CHCH2CH2CH2CH3 CH3 A CH3CHCH2CH2CH3 CH3 A CH3CHCH2CH3 CH3 A CH3CHCH3 CH3 A CH3 CH3 C CH3 CH3 C CH3 CH3 2,2,3,3-tetramethylbutane CH3 CH3CHCH2CH2CH2CH2CH3 2-methylheptane CH3 CH2 CH3 CHCH2CH2CH2CH3 3-methylheptane CH3 CH2 CH2 CH3 CHCH2CH2CH3 4-methylheptane CH3 CH2 CH2 CH2 CH2 CH3 CH3 CH3 CH CH2 CH2 CH3 CH3 C CH3 CH3 CH CH3 CH CH3 CH2 CH3 CH3 CH3 CH3 CH3 CH2 CH CH2 CH3 Answers to Selected Exercises A67 43. a.
b.
c.
45. a. 3 monochloro isomers of n-pentane; b. 4 monochloro isomers of 2-methylbutane; c. 3 monochloro isomers of 2,4-dimethylpentane; d. 4 mono-chloro isomers of methylcyclobutane 47. a. ketone; b. aldehyde; c. car-boxylic acid; d. amine 49. a.
b. 5 carbons in ring and the carbon in ¬CO2H: sp2; the other two carbons: sp3; c. 24 sigma bonds, 4 pi bonds 51. a. 3-chloro-1-butanol, primary: b. 3-methyl-3-hexanol, tertiary; c. 2-methylcyclopentanol, secondary 53. 1-pentanol; 2-pentanol; 3-pentanol; 2-methyl-1-butanol; 2-methyl-2-butanol; 3-methyl-2-butanol; 3-methyl-1-butanol; 2,2-dimethyl-1-propanol; 6 ethers 55. a. 4,5-dichloro-3-hexanone; b. 2,3-dimethylpentanal; c. 3-methylbenzaldehyde or m-methylbenzaldehyde 57. a. 4-chlorobenzoic acid or p-chlorobenzoic acid; b. 3-ethyl-2-methylhexanoic acid; c. methanoic acid (common name = formic acid) 59. Only statement d is false. 2-butenal: .
The formula of 2-butenal is C4H6O, while the ether has a formula of C4H8O.
61. a. CH3CH2CH2CH3 b.
c.
d. C4H8(g) 6O2(g) ¡ 4CO2(g) 4H2O(g) 63. For the iron-catalyzed reaction, one of the ortho or para hydrogens in ben-zene is replaced by chlorine. When an iron catalyst is not present, then the ben-zene hydrogens are unreactive, which is seen for the light-catalyzed reaction where one of the methyl hydrogens is replaced by chlorine.
Cl HCl Cl Cl Cl Cl CH3 CH3 CH2CHCHCHCH O P P HCCH CHCH3 H C C H C O N B B B B D O H D D C D D D D H D O O O C H H H O O O N H H O O C O C O O O H D C D ketone alcohol amine amine carboxylic acid CH3 CH2CH3 C C Cl Cl B D D D D CH3 H C C H CH3 B D D D D CH3 CH2CH2CH3 C C H H B D D D D CH2 CHCH2CH2 P F A G C CH3 H2CF H C P D G H D F G C CH2CH3 H C P D G H D H G C CH3 H2CF H C P D G D CH CCH3 P F A CH3 A CH2 CCH2CH3 P F A G C F CH3 H3C H C P D G D 65. a.
b.
c. No reaction d.
e.
f.
67. a. CH3CHPCH2 Br2 S CH3CHBrCH2Br; b.
c.
d.
69. a.
b.
71. CFClPCF2 73.
75.
O HN C n NHC B O B O O CN C CH2 B O O A C O OCH3 O A CN n C CH2 B O O O O A C O OCH3 O A H2O CH2(CH2)4CH3 CH3CH2CO O O O B CH3CH2C HOCH2(CH2)4CH3 OH O O B H2O; CH2(CH2)6CH3 CH3C O O O O B CH3C HOCH2(CH2)6CH3 OH O O B CH3CH2CH2OH CH3CH2C KMnO4 OH O O B CH2 CH3 H2O C P O CH3 A A CH2 CH3 H CH3; C H A A OH O O CH3 CH3 CH O O OH A CH3 oxidation CH3; C O O O B CH3 OH O C H O OH CH3 OH O C O CH3 O C H O C OH O CH3 C CHCH3 CH3 O H C CH2CHCH3 CH3 O HO C CH2CHCH3 CH3 O A68 Answers to Selected Exercises 103.
is optically active. The chiral carbon is marked with an asterisk.
105. C–C–A–G–A–T–A–T–G 107. Uracil will H-bond to adenine.
109. a. glu: CTT, CTC; val: CAA, CAG, CAT, CAC; met: TAC; trp: ACC; phe: AAA, AAG; asp: CTA, CTG; b. ACC–CTT–AAA–TAC or ACC–CTC– AAA–TAC or ACC–CTT–AAG–TAC or ACC–CTC–AAG– TAC; c. four (see answer in part b); d. met–asp–phe e. TAC–CTA–AAG; TAC–CTA–AAA; TAC–CTG–AAA 111. a. 2,3,5,6-tetramethyloctane; b. 2,2,3,5-tetra-methylheptane; c. 2,3,4-trimethylhexane; d. 3-methyl-1-pentyne 113.
There are many possibilities for isomers. Any structure with four chlorines replacing four hydrogens in any of the numbered positions would be an isomer; i.e., 1,2,3,4-tetrachloro-dibenzo-p-dioxin is a possible isomer.
115.
117. Alcohols con-sist of two parts, the polar OH group and the nonpolar hydrocarbon chain attached to the OH group. As the length of the nonpolar hydrocarbon chain in-creases, the solubility of the alcohol decreases. In methyl alcohol (methanol), the polar OH group can override the effect of the nonpolar CH3 group, and methyl alcohol is soluble in water. In stearyl alcohol, the molecule consists mostly of the long nonpolar hydrocarbon chain, so it is insoluble in water.
119. n-hexane, 69C; pentanal, 103C; 1-pentanol, 137C; butanoic acid, 164C.
121. 1-butene 123. ethanoic acid 125. In nylon, hydrogen-bond-ing interactions occur due to the presence of bonds in the polymer. For a given polymer chain length, there are more groups in nylon-46 as com-pared to nylon-6. Hence, nylon-46 forms a stronger polymer compared to nylon-6 due to the increased hydrogen-bonding interactions.
127. a.
b. Repeating unit: The two polymers differ in the substitution pattern on the benzene rings. The Kevlar chain is straighter, and there is more efficient hydrogen bonding be-tween Kevlar chains than between Nomex chains.
129. a. The bond angles in the ring are about 60. VSEPR predicts bond angles close to 109. The bond-ing electrons are much closer together than they prefer, resulting in strong electron–electron repulsions. Thus ethylene oxide is unstable (reactive).
b. The ring opens up during polymerization and the monomers link together through the formation of OOC bonds.
131. H2N CH O CO2H CH2CH2CO2 –Na+ O A H2N or CH O CO2 –Na+ CH2CH2CO2H O A O CH2CH2 O O n OO O O CH2CH2 CH2CH2 OO O HN NHC B O C n O B O O H2N NH2 and HO2C CO2H N¬H N¬H 78.5°C: CH3¬CH2¬OH 23°C: CH3¬O¬CH3; O O Cl Cl Cl Cl HN NH N N N Sugar Sugar O O H O O N H Uracil Adenine B B B K , ,, O NH N O A A A A A P N Br CH CH2 C Cl H P O O A A 77.
79. Divinylbenzene inserts itself into two adjacent polymer chains and bonds them together. The chains cannot move past each other because of the crosslinks making the polymer more rigid.
81. a. The polymer from 1,2-diaminoethane and terephthalic acid is stronger because of the possibility of hydrogen bond-ing between chains. b. The polymer of is more rigid because the chains are stiffer due to the rigid benzene rings in the chains. c. Polyacetylene is nHCKCH n O ( CHPCHO )n. Polyacetylene is more rigid because the double bonds in the chains make the chains stiffer.
83. a. serine; tyrosine; threonine; b. aspartic acid; glutamic acid; c. histi-dine; lysine; arginine; tryptophan; d. glutamine; asparagine 85. a. aspartic acid and phenylalanine; b. Aspartame contains the methyl ester of phenylala-nine. This ester can hydrolyze to form methanol, ROCO2CH3 H2O ∆ RCO2H CH3OH.
87.
89. a. Six tetrapeptides are possible. From NH2 to CO2H end: phe–phe–gly–gly, gly–gly–phe–phe, gly–phe–phe–gly, phe–gly–gly–phe, phe–gly–phe–gly, gly–phe–gly–phe b. Twelve tetrapeptides are possible.
From NH2 to CO2H end: phe–phe–gly–ala, phe–phe–ala–gly, phe–gly–phe– ala, phe–gly–ala–phe, phe–ala–phe–gly, phe–ala–gly–phe, gly–phe–phe–ala, gly–phe–ala–phe, gly–ala–phe–phe, ala–phe–phe–gly, ala–phe–gly–phe, ala–gly–phe–phe 91. Ionic: his, lys, or arg with asp or glu; hydrogen bonding: ser, glu, tyr, his, arg, asn, thr, asp, gln, or lys with any amino acid; covalent: cys with cys; London dispersion: all amino acids with nonpolar R groups (gly, ala, pro, phe, ile, trp, met, leu, val); dipole–dipole: tyr, thr, and ser with each other 93. Glutamic acid has a polar R group and valine has a nonpolar R group. The change in polarity of the R groups could affect the tertiary structure of hemo-globin and affect the ability of hemoglobin to bond to oxygen.
95.
97. aldohexose: glucose, mannose, galactose; aldopentose: ribose, arabinose; ketohexose: fructose; ketopentose; ribulose 99. They differ in the orientation of a hydroxy group on a particular carbon. Starch is composed from -D-glucose, and cellulose is composed from -D-glucose.
101. The chiral carbons are marked with asterisks.
H3C CH2CH3 C H H O O H2N CO2H C O O A A A H3C OH C H H Isoleucine Threonine O O H2N CO2H C O O A A A CH2OH CH2OH H OH OH OH HO HO OH OH D-Ribose D-Mannose H H H H H H H H O O O H2NCHC NHCHCO2H CH2 CH3 OH ser–ala O H2NCHC NHCHCO2H CH2 CH3 OH ala–ser HO CO2H H2N NH2 and HO2C HO2C CO2H CO2H Answers to Selected Exercises A69 The first structure is MSG, which is impossible for you to predict.
133. In the reaction, POO and OOH bonds are broken and POO and OOH bonds are formed. Thus H L 0. S 0, since two molecules are going to form one molecule. Thus G 0, not spontaneous.
135.
137. Both and are positive values.
139. 6.07 141. a. No; the mir-ror image is superimposable.
b.
143.
145. a.
b. CH2 CCH3 CH3 CH3 CH3CHCH2CH3 C OH CO2H H A C OH A O O O 0 0 0 0 C OH HO2C CO2H H H e e CO2H H e e A C OH A O O O ¢S ¢H CH O CH3 C H2N 1.0 M OH: O CH O CH3 C H3N 1.0 M H: OH; c.
d. CH2PCHCH3 CH3 A e. CH2CH2CH2CH2CH3 CH2CHCH2CH3 A A OH OH CH3 CH3 A A CH3CHCH2CH2 CH2OCOCH3 A A A OH OH CH3 147.
149. a. The temperature of the rubber band increases when it is stretched; b. exothermic (heat is released); c. As the chains are stretched, they line up more closely resulting in stronger dispersion forces between the chains.
Heat is released as the strength of the intermolecular forces increases.
d. G is positive and S is negative; e. The structure of the stretched poly-mer chains is more ordered than in unstretched rubber. Disorder decreases as the rubber band is stretched.
151. 0.11% 153. 5 155. a. 37.50%; b. The hybridization changes from sp2 to sp3; c. 3,4-dimethyl-3-hexanol CH3 CH2 HC 13 12 11 10 9 8 7 6 5 4 3 2 1 C C q C C O P P B q O C HC CH PCH PCH HC OH CH2 O O O HC O O O O OCH2CH2OCNH B O NHCOCH2CH2OCNH NHC B O B O n B O O A70 A (mass number), 50, 52, 841, 842 Abelson, Phillip H., 302 Absolute zero, 185, 764 Absorbance, A17–A19 Absorption, 558 Accelerators, 57, 83, 302, 849–850, 864 Accuracy, 12–13, A10 Acetic acid, 628, 1014 in buffered solution, 684–686, 689, 693–694 percent dissociation, 641–642 reaction with strong base, 149 titration of, 700–705, 707, 716 as weak electrolyte, 130, 132 Acetone, 502, 504, 1013 Acetylene, 396–397, 1006 Acetylsalicylic acid, 1015 Acid–base equilibrium problems approximations in, 637, 638 with bases, 645–650 with buffered solutions, 684–695 with common ion, 681–683 major species in, 634–635, 636, 637, 666, 667 with polyprotic acids, 650–655 with salts, 655–660 strategies for, 634–635, 666–667 with strong acids, 634–635 with weak acids, 635–644 Acid–base indicators, 151–152, 711–716 Acid–base reactions, 140, 149–154 equivalent mass in, 487 in gas phase, 625–626 Acid–base titrations, 151–153, 696–711. See also Equivalence point; pH curve end point of, 152, 711, 714 indicators for, 151–152, 711–716 Ka derived from, 707–709 strong acid–strong base, 696–699, 705, 716 strong base–strong acid, 699 weak acid–strong base, 153–154, 700–709, 716 weak base–strong acid, 709–711 Acid dissociation constant (Ka), 624–625 calculating from Kb, 656 for hydrogen halides, 927–928 for monoprotic acids, 628, A22 from percent dissociation, 643–644 pH of salt solutions and, 659–660 for polyprotic acids, 650–651, 655, A23 from titration, 707–709 Acidic oxides, 662–663, 665 Acidic solutions with common ions, 681–683 of covalent oxides, 662–663, 665 ion-product constant in, 630 oxidation–reduction reactions in, 162–166 of salts, 657–659, 660 solubility in, 724, 728, 729, 730, 735 Acid rain, 212–213, 214, 559, 922 Acids. See also pH; Strong acids; Weak acids concepts of, 131, 149, 623–626, 663–665 dilution of, 137–139 diprotic, 627, 650, 653–655 monoprotic, 628, A22 names of, 66–67 organic, 628, 1014–1015 polyprotic, 650–655, 682, A23 reaction with water, 623–624 safe dilution of, 139 sour taste of, 131, 623 strength of, 626–631, 661–662 as strong electrolytes, 131 structural properties of, 661–662 triprotic, 650–652 water as, 629, 646 Acid salts, of amines, 649 Acrylonitrile-butadiene-styrene (ABS), 1025 Actinide series, 302, 306, 308, 875, 943 Activated complex, 553 Activation energy, 553–557, 558, 593 Addition in exponential notation, A2–A3 significant figures, 14–15 Addition polymerization, 1019, 1021 Addition reactions, 998, 1007 Adhesive forces, 429 Adrenaline, 648 Adsorption, 558, 559 Aerosol cans, 500–501, 561, 909, 912 Aerosols, 500, 515 Air. See also Atmosphere liquefaction of, 879–880, 919 Air bags, 197 Air pollution, 179, 211–214 acid rain and, 212–213, 214, 559, 922 catalytic converters and, 254, 559, 560 electrostatic precipitator and, 515 metallurgy and, 981 photochemical smog, 213, 906 Alchemy, 39, 40, 46, 47 Alcohols, 1010–1013. See also Ethanol; Methanol Aldehydes, 1013–1014 Alkali metals (Group 1A), 55, 316–318, 880–882 atomic radii, 876 crystal structure, 441 hydrides of, 884 hydroxides of, 318, 644 ionization energies, 310, 316, 317 naming of ions, 60, 61 production of, 824–825, 879, 880 qualitative analysis of, 731 solubility of salts, 144 Alkaline earth metals (Group 2A), 55, 884, 885–888 hydroxides of, 644–645, 724, 885 Alkalis. See Bases Alkanes, 997–1005 Alkenes, 1005–1008. See also Ethylene Alkyl substituents, 1000 Alkynes, 1006, 1007 Allomones, 379 Alloys, 442–443, 978 for corrosion prevention, 815–816 dental amalgam, 975 Alloy steels, 443, 815–816, 984, 985, 986 Alpha () particles, 48–49, 843, 844, 931 biologic effects, 864, 865 Alum (aluminum sulfate), 666 Alumina. See Aluminum oxide Aluminosilicates, 448 Aluminum, 304, 878, 888–890 corrosion of, 813, 889 hydrated ion of, 659, 661, 664–665, 666, 889 production of, 821–822, 981 Aluminum hydroxide, 644–645 Aluminum oxide, 339–340, 813, 889 in aluminum production, 822, 981 in gemstones, 970 in slag, 984 Alvarez, Luis, 1, 80 Alvarez, Walter, 80 Amide linkage, 1027 Amines, 646, 648–649, 1015–1016 Amino acids, 562, 1025–1028, 1032, 1039, 1040 Ammine ligand, 958, 960 Ammonia, 906–907 autoionization of, 629 in buffered solution, 690–692 hydrogen bonding in, 907 as Lewis base, 664 as ligand, 731–733, 734–735, 736, 957, 960, 968 molecular structure, 369, 393 as polar molecule, 336, 907 thermodynamic stability, 903–904 titration of, 709–711 as weak base, 132–133, 646, 647–649 Ammonia synthesis. See also Haber process bacterial, 906 entropy change, 762–763 equilibrium constant, 585, 586, 587, 609–610, 906 equilibrium position, 585, 604–606 free energy change, 766, 771–772, 776 rate, 582, 906 reaction quotient, 593–594 stoichiometry, 108–110 Ammonium chloride, 649, 657–658, 682, 907 Ammonium dihydrogen phosphate, 917 Index Index A71 Ammonium ion, 55, 144 Ammonium nitrate, 910, 911 Ammonium nitrite, 905 Ammonium salts, 907 Amorphous solids, 430, 432. See also Glass red phosphorus, 913–914, 915 Ampere (A), 817 Amphoteric substances, 629–631, 876, 888–889, 981 amu (atomic mass units), 78, 79, 82, 83 Amylopectin, 1034 Amylose, 1034 Analyte, 151, 152 Anderson, Christopher, 40 Angular momentum, 285, 293–294 Anions, 54 common monatomic, 61 as effective bases, 724 electron affinity and, 312–313 names of, 58, 61, 62, 66–67 of nonmetals, 55, 315, 876 oxyanions, 62, 66–67, 928–929 sizes of, 340–342 Anode, 792, 798, 800, 816 Anodic regions, in steel, 815 Antacids, 105–106, 644–645 Antibiotics, 90–91, 379 Antibonding molecular orbital, 404–405 Anticodon, 1039 Antifreeze, 24, 497, 506–507, 1012 Antilog, 634, A5 Antimony, 901, 902 Antioxidants, 160–161 Antiparticles, 844, 864 Antitumor agents, 963 Approximations. See also Uncertainty in measurement in acid–base calculations, 637, 638 of model, 199–200 successive, A8–A10 Aqua ligand, 958 Aqua regia, 735, 803 Aqueous solutions. See also Acidic solutions; Basic solutions; Electrolytes; Hydration; Solutions base strength in, 657 ion-product constant in, 629–631, 632–633, 656 in metallurgy, 981–982 oxidation–reduction reactions in, 162–168 phase diagram for, 504, 506 properties of water in, 127–129 symbol for, 98, 128, 131 temperature effects, 495–496 Argon, 304, 305, 436, 919, 932 Aromatic alcohols, 1012 Aromatic amines, 1015, 1016 Aromatic hydrocarbons, 1008–1010 Arrhenius, Svante, 129, 131, 132–133, 553, 554, 623 Arrhenius acid–base concept, 131, 149, 623, 626, 644, 663 Arrhenius equation, 555–557 Arsenic, 901, 902 Art, chemistry of, 4 Aspdin, J., 892 Aspirin, 238, 1015 Astatine, 924, 925 Atactic chain, 1023 Atmosphere. See also Air carbon dioxide in, 211, 212, 254–256 chemistry in, 179, 211–214 composition of, 211 ozone in, 211, 212–213, 260, 559–561, 1004 Atmospheric pressure, 179–180, 181, 211 Atomic masses, 78–81, 82 early work on, 44–45, 46, 300 Atomic mass units (amu), 78, 79, 82, 83 Atomic number (Z), 50, 52, 55, 841, 842 Atomic orbitals, 291, 292. See also d orbitals; f orbitals; Hybridization; p orbitals; s orbitals filling, in periodic table, 302–309 of hydrogen atom, 292–296 in localized electron bonding model, 353–354 of polyelectronic atoms, 299 Atomic radii, 313–314, 876–878 of transition metals, 948–949 Atomic solids, 435, 436, 458. See also Metals network solids, 436, 443–454, 458 Atomic spectrum, 284–285, 290 Atomic weights, 44, 79 Atoms, 3–4. See also Elements; Hydrogen atom ancient Greek ideas, 39–40, 275 Bohr model, 285–287, 290, 291, 292, 293 calculating the number of, 84–85 Dalton’s theory, 41–44, 275, 960 early physics experiments, 45–49 polyelectronic, 298–299 quantum mechanical model, 275, 290–299 structure of, 49–50 symbols for, 50, 52, 841 viewing with microscope, 2, 438–439 Aufbau principle, 302, 350 Austentite, 986 Autoionization, 629–631, 635, 646 Automobiles. See Cars Average, of measurements, 12, 13, A10–A11 Average atomic mass, 79, 81 Average mass, 77–78 Average rate, 530 Avogadro, Amadeo, 44, 185 Avogadro’s hypothesis, 44 Avogadro’s law, 185–186, 202 Avogadro’s number, 82 Azeotrope, 912, 913 Baekeland, Leo H., 1017 Bakelite, 1017 Baking soda, 105–106 Balance, 10, 11, 12 Balancing chemical equations, 97–102 for oxidation–reduction, 162–168 Ball-and-stick models, 53, 55 Band model, 441–442 Bar, 246 Barium, 885, 886 Barium hydroxide, 144, 644 Barium sulfate, 144, 489, 717 Barometer, 180–181, 460 Bartlett, Neil, 932 Bases. See also Acid–base titrations; pH; Strong bases; Weak bases common anions as, 724 concepts of, 131, 149, 623–624, 625–626, 644, 663–665 in nucleic acids, 1036, 1038, 1039 strength of, 628–629, 657 taste and feel of, 131, 623 water as, 623–624, 626, 627, 629, 657 Basic oxides, 662, 663, 667, 876 Basic oxygen process, 985 Basic solutions of ionic oxides, 663, 876 ion-product constant in, 630 oxidation–reduction reactions in, 166–168 of salts, 655–657, 659, 660 solubility in, 724, 728–729, 730 Batteries, 808–813 charging, 816 dead, 805, 809 definition of, 808 jump starting, 4, 809–810 lead storage, 24, 808–810, 893, 922 printed, 809 work done by, 778–779 Bauer, Georg, 39 Bauxite, 821, 822, 981 Bayer process, 981 Becker, Luann, 80, 81 Beckman, Arnold, 632–633 Becquerel, Henri, 48 Beer–Lambert law, A17–A19 Beethoven, Ludwig van, 893 Bent, Henry, 779 Bent structure, 372 Benzene, 414–415, 1008–1009 Berrie, Barbara, 4 Beryllium, 303, 360, 407, 885, 886 Beryllium chloride, 367, 886 Beryllium oxide, 876, 885 Berzelius, Jöns Jakob, 45, 46–47 Beta () particles, 48, 842, 843–844, 847, 853 biologic effects, 864, 865 in nucleosynthesis, 851 Bicarbonate ion, 62, 63 in water softening, 645 Bicycles, materials for, 443, 889, 952–953 Bidentate ligand, 957, 961 Big bang theory, 850–851 Bimolecular step, 550, 551 Binary covalent compounds, naming of, 63–65 Binary ionic compounds. See also Ionic compounds (salts) crystal structures, 344, 435, 436, 456–458 formation of, 342–346 Lewis structures, 354 naming of, 58–61, 65–66 predicting formulas, 339–340 Binding energy per nucleon, 858–859 Biodiesel, 262–263 Biologic systems. See also Carbohydrates; Human body; Nucleic acids; Plants; Proteins alkali metal ions in, 317, 882 alkaline earth metals in, 886 chiral molecules in, 1031–1032, 1034 communication in, 378–379 coordination complexes in, 973–977 energy in, 154, 229, 973 entropy in, 755, 756 enzymes in, 557, 558, 562–563 ice and, 22–23, 516 mercury toxicity, 893, 975 nitrogen in, 906 radiation and, 847, 863–866 Biot, Jean, 961 Bipyramidal structure, trigonal, 371–372, 401, 402–403, 902, 903, 917 Bismuth, 901, 902 A72 Index Black phosphorus, 913, 914 Blasco, Steve, 444 Blast furnace, 983–984 Bleach, household, 643 Blowing agent, 908 Body-centered cubic (bcc) unit cell, 433, 441 Bogs, lead in, 51 Bohr, Niels, 285, 286 Bohr model, 285–287, 290, 291, 292, 293 Boiler scale, 496 Boiling, 26, 467 Boiling chips, 466 Boiling point, 464, 465, 466 hydrogen bonding and, 427–428 pressure dependence of, 469 Boiling-point elevation, 504–505, 513 Bomb calorimeter, 240–242 Bond angles, 370–371, 375, 376–377, 378 Bond energies average values, 351 bonding concept and, 330–332, 348–349 bond order and, 409–410 in chemical reactions, 351–353, 749 electronegativity and, 334 of homonuclear diatomic molecules, 409–410 of hydrogen halides, 661, 927 as potential energy, 231–232, 331, 332 Bonding molecular orbital, 404, 405, 441 Bonding pairs, 353, 354, 355 in VSEPR model, 367, 369, 371 Bond length, 331, 351, 352 of homonuclear diatomic molecules, 409 Bond order, 405–406 of homonuclear diatomic molecules, 407, 408, 409–410, 411 Bond polarity. See Polar covalent bonds Bonds, 52, 329–330. See also Bond energies; Covalent bonds; Ionic bonds electronegativity and, 333–335, 346 types of, 330–333 Bond strength. See Bond energies; Bond order Boranes, 245, 888 Boron, 303, 888 electron-deficient compounds, 358–359, 360, 664, 888 molecular orbital model, 407–408, 409, 410 nutritional requirement, 889 paramagnetism, 410 Boron tetrafluoride ion, 402 Boron trifluoride, 358–359, 367–368, 664 Borosilicate glass, 448 Boyle, Robert, 6, 39–40, 181 Boyle’s law, 181–184, 186, 189, 200–201 Bragg, William Henry, 433 Bragg, William Lawrence, 433 Bragg equation, 433, 434 Brand, Henning, 914 Brass, 443, 955 Breeder reactors, 862–863 Brine, electrolysis, 825–826 Brittle tin, 892 Bromide salts, solubility, 144 Bromine, 924–927, 928, 929–930 Bromo ligand, 958, 968 Bromthymol blue, 712–713, 714, 715 Br$nsted, Johannes N., 149, 623 Br$nsted-Lowry acid–base concept, 149, 623, 625–626, 644, 661, 663, 664, 665 Bronze, 892, 955 Brooks, Robert, 40 Buckminsterfullerenes, 80–81, 891 Buffered solutions, 684–696 choice of weak acid for, 695–696 how they work, 687–689 pH changes in, 684, 685–686, 693–695 pH of, 684–685, 689–691 summary, 692–693 in titration of weak acid, 700, 705 Buffering capacity, 693–696 Bumping, 466 Buret, 10–13, 151, 152, 696 Burton, William, 253 Butane, 998, 999, 1003 Cadmium, 306, 308 Calcine, 984, 985 Calcite. See Calcium carbonate Calcium, 305, 885, 886–887 Calcium carbonate. See also Limestone as boiler scale, 496 decomposition, 589, 610 on sunken treasure, 821 in water softening, 645, 887 Calcium hydroxide, 144, 644, 645 Calcium oxide, 214, 339, 589, 645, 663 Calcium phosphate, 159, 717, 720, 916 Calcium sulfate, 144, 917 in gypsum, 213, 892, 916, 920 Calcium sulfite, 214 Calculations. See Approximations; Significant figures Calorimetry, 237–242 Cannizzaro, Stanislao, 45 Capillary action, 429 Capsaicin, 414 Captive zeros, 13, 14 Carbohydrates, 1031–1036. See also Sugars Carbon, 303, 890–891. See also Diamond; Graphite; Organic chemistry atomic mass, 78, 79, 82 atomic size, 876–877 as coke, 980, 983, 984, 985 as fullerenes, 80–81, 891 hybrid orbitals of, 391–392, 393–397 isotopes of, 79, 84 as network solid, 443–446 phase diagram, 470–471 as reducing agent, 879, 980 in steel, 443, 816, 879, 984, 985, 986 Carbon-14, 842, 853–854 Carbonate minerals, 979, 980 Carbonate salts, 144, 730 Carbon dioxide, 681, 891 acidic solution of, 650, 663, 724, 891 atmospheric, 211, 212, 254–256 bond polarities, 336 in fire extinguishers, 471 from fossil fuels, 179, 229, 255, 891 liquid, 471 molecular structure, 356, 395–396, 877, 891 phase diagram, 470–471 as real gas, 208, 210–211 solid, 454, 463, 470–471, 748 solubility in water, 493, 495, 497 in space vehicles, 105 Carbon fiber composites, 952 Carbonic acid, 650 Carbon monoxide, 891 bonding in, 401–402, 891 catalytic conversion of, 560 in iron metallurgy, 980, 983, 984 as ligand, 958, 977 in syngas, 256–257 toxicity of, 891, 977 Carbon suboxide, 891 Carbonyl group, 1013 Carbonyl ligand, 958 Carboxyhemoglobin, 977 Carboxyl group, 628, 1014 Carboxylic acids, 1014–1015. See also Acetic acid Carboxypeptidase-A, 562, 563 Carothers, Wallace H., 1017, 1022 Cars. See also Air pollution; Gasoline air bags, 197 antifreeze, 24, 497, 506–507, 1012 batteries, 4, 24, 778–779, 808–810 catalytic converters, 254, 559, 560 electric, 811 ethanol fuel, 263, 1012 fuel cells for, 812, 813 hydrogen fuel, 261–262, 812, 813, 885 magnetorheological shock absorbers, 431 methanol fuel, 257, 263, 1011 nitrous oxide in fuel, 912 Catalysis, 557–563 by enzymes, 557, 558, 562–563 in Haber process, 558, 582, 906 heterogeneous, 558–559 homogeneous, 558, 559–563 Catalytic converters, 254, 559, 560 Cathode, 792, 798, 800, 816 Cathode-ray tubes, 45–46, 282 Cathodic protection, 816 Cathodic regions, 815 Cation-exchange resin, 887 Cations, 53, 54. See also Complex ions common monatomic, 58, 61 of metals, 55, 315, 340, 875 names of, 59, 60, 61, 62, 65 qualitative analysis of, 729–731, 734–736 sizes of, 340–342 Cavendish, Henry, 132 Cell potential, 793. See also Galvanic cells concentration and, 803–808 in electrolysis, 816, 818–820 at equilibrium, 805, 807–808 free energy and, 802, 804, 805 from half-cell potentials, 794–800 work done by, 793, 800–801, 816 Cellulose, 1034, 1035–1036 Cellulose nitrate, 1016, 1017 Celsius scale, 19–23, 185 Cementite, 986 Centimeter (cm), 9, 13, 17–18 Ceramics, 448–449 Certain digits, 11 Cesium, 316, 317, 344, 880, 881, 882 Chain reaction, nuclear, 860–861 Index A73 Chain theory, 960 Changes of state, 425, 426, 459, 463–466 entropy and, 754, 756–757 phase diagrams for, 467–471, 504, 506 Charge current and, 817 on electron, 45, 48, 50 formal, 363–367 on ion, 59, 60, 61, 157 on mole of electrons, 801 oxidation state and, 155–156, 157, 363–364 partial, 128, 332 work and, 793, 800, 801 Charge density, of ion, 318 Charge distribution, of dipolar molecule, 335–336 Charles, Jacques, 184 Charles’s law, 184–185, 186, 189, 201 Chelating ligands, 956–957 Chemical bonds. See Bonds Chemical change, 27–28, 96–97 Chemical energy, 231–232, 749. See also Bond energies Chemical equations, 96–97 balancing, 97–102 meaning of, 98 for oxidation–reduction, 162–168 physical states in, 98 for reactions in solution, 145–146 Chemical equilibrium. See Equilibrium Chemical formula, 52 determination of, 91–96 from name, 59 Chemical kinetics. See Rate laws; Reaction mechanisms; Reaction rates Chemical reactions. See Reactions Chemical vapor deposition (CVD), 471 Chemistry, history of, 39–42, 43–45 Chi (mole fraction), 195–198, 485, 486 Chip laboratories, 138 Chiral isomers, 964–965, 1031–1032, 1034 Chlor-alkali process, 825–826 Chlorate salts, 929 Chloride ion, 54–55, 58 as ligand, 731, 958, 968 Chloride salts. See also Sodium chloride in qualitative analysis, 729, 735–736 solubility in water, 144 Chlorinated alkanes, 1003–1004 Chlorinated organic pollutants, 156–157 Chlorine, 924–930 from bleach, 643 oxyacids and oxyanions of, 62, 67, 928–929 production of, 825–826 in water purification, 919 Chlorite ion, 929 Chlorofluorocarbons (CFCs), 561, 1003–1004 Chloro ligand, 958, 968 Chlorophyll, 973 Chromate salts, 144, 952, 953 Chromatography, 27 Chromium, 947, 948, 951–953 electron configuration, 305, 308, 315, 350, 946 Chromous ion, 951 cis–trans isomerism, 961, 963, 965, 1006 Clausius–Clapeyron equation, 463 Clays, 332–333, 447, 448, 515, 892 Cleaning solution, 953 Closest packing in ionic solids, 456–458 in metals, 436–441 Coagulation, 515 Coal, 213–214, 252–253, 254 Coal gasification, 256–257 Coal slurries, 257 Cobalt, 1, 305, 947, 954 Codons, 1038, 1039 Coefficients, 98 Coffee-cup calorimeter, 237 Cohesive forces, 429 Coke, 980, 983, 984, 985 Colligative properties, 504 boiling-point elevation, 504–505, 513 of electrolyte solutions, 512–513 freezing-point depression, 504, 505, 506–507, 512, 513 osmotic pressure, 508–511, 513 Collision model, of rates, 552–557 Collision of gas molecules. See Kinetic molecular theory Colloids, 514–515 Colors of complex ions, 946, 955, 969–971 of fireworks, 288–289 of gemstones, 970–971 of paintings, 4 of plasma monitor, 282–283 in spectroscopy, A16, A18 Combustion of alkanes, 1003 calorimetry of, 241–242 early studies of, 40–41 of ethanol, 98–99 for formula determination, 91–92 of methane, 812–813 oxidation–reduction in, 154, 158–159 Common ion effect, 681–683, 718, 722–724. See also Buffered solutions Common names, 57, 64 Communication, chemical, 378–379 Complete ionic equation, 145, 146 Complex ions, 945–946. See also Coordination compounds colors of, 946, 955, 969–971 crystal field model, 967–973 equilibria of, 731–736 in hydrometallurgy, 981–982 isomerism in, 960–965 localized electron model, 965–967 naming of, 958–959 paramagnetic, 946, 955, 968, 976 in qualitative analysis, 734–735 Composition mass percent, 89–91, 485, 486, 488 of solutions, 133–140, 485–488 Compounds. See also Names of compounds Dalton’s atomic theory and, 41–44 definition of, 27 enthalpy of formation, 246 formula determination, 91–96 noble gas configurations of, 338–339 percent composition, 89–91 standard states, 246 Compressibility, 25, 210, 211, 425 Compton, Arthur, 280 Computer chips, 47, 452–453 Concentrated solution, 485 Concentration cell potential and, 803–808 equilibrium position and, 585–586, 605–607 free energy and, 770 of a gas, 586, 587, 588 of pure solid or liquid, 589 of solutions, 133–140, 485–488 Concentration cells, 804 Concrete, 892 Condensation, 459 Condensation polymerization, 916, 1027, 1037 Condensed states, 426 Conduction bands, 442, 444, 446, 450 Conductivity. See Electrical conductivity; Thermal conductivity Conjugate acid, 624 of weak base, 646, 658 Conjugate base, 624, 625, 628–629 of strong acid, 626 of weak acid, 627, 656 Conservation of energy, 229, 232, 749 Conservation of mass, 6, 41, 98 Constant-pressure calorimetry, 237–240 Constant-volume calorimetry, 240–242 Constructive interference, 282, 432, 434 Contact potential, 451 Contact process, 922 Conté, Nicolas-Jacques, 332 Continuous spectrum, 284–285 Control rods, 861–862 Conversion of units, 16–19, A26 for pressure, 181 for temperature, 19–23 Coordinate covalent bond, 731, 956, 966 Coordination compounds, 955–960. See also Complex ions biologic, 973–977 Coordination isomerism, 960–961 Coordination number, 731, 956 Copolymers, 1020–1021, 1024–1025 Copper, 55, 955 corrosion of, 814, 955 electron configuration, 305, 308, 315, 350, 947 electroplating of, 816–818, 819 electrorefining of, 823–824 properties, 436, 438–439, 947 Core electrons, 304, 309–310, 312 Corrosion, 813–816 of aluminum, 813, 889 of copper, 814, 955 of iron, 776–777, 813, 814–816, 954 prevention of, 814, 815–816, 824 of silver, 814, 820, 821 of transition metals, 945 Coulomb’s law, 330, 344 Counterions, 955, 958, 959, 960 Covalent atomic radii, 313 Covalent bonds, 52, 332. See also Bond energies; Lewis structures; Localized electron (LE) model; Polar covalent bonds; VSEPR model coordinate, 731, 956, 966 electronegativity and, 335, 346 A74 Index Covalent bonds (continued) as a model, 347–350 noble gas configuration and, 338 in nonmetals, 55, 338 in solids, 436, 443–444, 458 Covalent compounds binary, names of, 63–65 oxidation states in, 155–158 Covalent hydrides, 884 boiling points of, 427–428 of boron, 245, 888 of Group 6A elements, 918 Cracking, 253, 883 Crenation, 510 Critical fission reaction, 860, 861 Critical mass, 861 Critical point, 468, 470 Critical pressure, 468 Critical temperature, 468 Crosslinking, of polymers, 1017, 1018 Cryolite, 822 Crystal field model, 967–973 Crystalline solids, 430, 432–436.
See also Ionic compounds (salts); Metals; Solids entropy at 0 K, 763–764 lattices of, 432, 435, 436 surface properties, 438–439 types of, 435–436 X-ray diffraction by, 281–282, 432–434 Cubic closest packed (ccp) structure, 436, 437, 438–441, 456–458 Cubic unit cells, 433 Curie, Irene, 849 Curie, Marie, 918 Curie, Pierre, 918 Current. See Electric current Cutler, Richard, 160 Cyanidation, 981, 982 Cyanide ion, 356–357, 412, 639–641, 657, 977 Cyano ligand, 958, 959, 968, 977 Cycles per second, 276 Cyclic alkanes, 1004–1005 Cyclic alkenes, 1006 Cyclotron, 849–850 Cysteine, 1030 Cytochromes, 973, 977 Dacron, 1021 Dalton, John, 41–44, 52, 194, 275, 960 Dalton’s law of partial pressures, 194–199, 202 Dating methods, 51, 853–855 Davis, James H., 495 Davisson, C. J., 282 Davy, Humphry, 912 DDT, 489 de Broglie, Louis, 281, 290 de Broglie’s equation, 281, 282 Decay series, 846, 849 Decimal point in exponential notation, A1–A2 significant figures and, 13–16 Definite proportion, law of, 41 Degenerate orbitals atomic, 295, 299, 303 molecular, 408 Degree, of temperature, 19, 20, 21 Degree symbol, on thermodynamic function, 246, 760 Dehydrogenation reactions, 1004 Delocalized electrons, 349 in benzene, 414–415, 1008–1009 combined model of, 413–415 in graphite, 446 in metallic solids, 436 resonance and, 362, 413–414, 415 Delta partial charge, 128, 332 splitting of 3d orbitals, 968–972 Demokritos, 39, 275 Denaturation of proteins, 1030 Denitrification, 906 Density, 24–25 of closest packed solid, 438–441 of gas, 193–194 in group of elements, 316 uncertainty in measurement, A12–A13 of water, 425, 426, 469 Dependent variable, A6 Depletion force, 756 Desalination, 26, 511 Desorption, 559 Destructive interference, 282, 432, 434 Detergents, 645, 887, 916 Determinate error, 12, A10 Dextrorotatory isomer, 964–965, 1031–1034 Dialysis, 509–510 Diamagnetism, 409, 410, 412, 968, 976 Diamond, 436, 443–444, 446, 891 entropy of, 764 graphite and, 244, 246, 446, 768–769, 769 low-pressure production of, 470–471 Diamond anvil pressure cell, 358 Diaphragm cell, 826 Diatomic molecules, 3, 406–413 Diborane, 245, 888 Dichromate ion, 951–953 Dicyclopentadiene, 1018–1019 Diesel, Rudolf, 262 Diethylenetriamine (dien), 957 Diethyl ether, 461–462 Diethylzinc (DEZ), 667 Differential rate law, 533, 534–538, 548 Diffraction, 281–282, 432–434 Diffractometer, 434 Diffusion, 206, 207–208 Dihydroxyacetone (DHA), 1032–1033 Dilute solution, 485 Dilution, 137–140 Dimensional analysis, 16–19 Dimer, 1021 Dinitrogen monoxide, 546, 909–910, 912 Dinitrogen pentoxide, 911 kinetics of decomposition, 534–535, 538–540 Dinitrogen tetroxide, 594–595, 911 Dinitrogen trioxide, 909, 911 Dinosaurs, 1, 51, 80, 81 Dioxygen difluoride, 929 Dipeptide, 1027 Dipole–dipole attraction, 426. See also Hydrogen bonding in molecular solids, 454, 456 in proteins, 1030 Dipole moments, 335–338, 346.
See also Polar molecules induced, 428–429 square planar structure and, 374 of sulfur dioxide, 376 Diprotic acids, 627, 650, 653–655 Direct reduction furnace, 984 Disaccharides, 1034. See also Sucrose Disorder, 751, 764, 779 Dispersion, colloidal, 514–515 Dispersion forces. See London dispersion forces Disproportionation, 929 Dissociation, percent of, 641–644 Dissociation constant. See Acid dissociation constant; Ion-product constant Distillation, 26 Disulfide linkage, 1030 Division in exponential notation, A2 significant figures, 14, 15 DNA (deoxyribonucleic acid), 756, 1036–1039 Dobereiner, Johann, 300 Doping, of semiconductors, 450, 452, 453 d orbitals, 294, 295, 297, 299 in bonding, 359–360, 918 in complex ions, 966, 967–973 filling of, 305–307 of Group 4A elements, 890 of phosphorus, 913 in transition metals, 943, 946–947 Double bonds, 351, 352. See also Ethylene in hydrocarbons, 998, 1005–1008 hydrogenation and, 558–559, 883, 998, 1007 Lewis structures, 356 sigma and pi bonds in, 394–395, 396, 1005, 1006 in VSEPR model, 375–376 Double displacement, 140 Downs cell, 824–825 Drake, Edwin, 253 dsp3 orbitals, 397–399, 401 in complex ion, 966 in Group 5A, 902, 903 d2sp3 orbitals, 399–400, 401 in complex ion, 966 in Group 5A, 902, 903 Dry ice, 454, 463, 471, 748 Ductility, 55, 436, 441, 443, 444 Duet rule, 354, 355 du Pont, Eleuthère, 288 Dyes, 277 Dynamite, 905 Effusion, 206–207 Eiler, John M., 260 Einstein, Albert, 278–280, 856–857, 865 Electrical conductivity of aqueous solutions, 129–130, 132 of graphite, 444, 446 of melted ionic compound, 347, 348 of metals, 436, 441, 442, 944 of silicon, 450 Electrical insulator, 444 Electrical potential. See Cell potential Electric arc method, 985 Electric charge. See Charge Electric current conduction bands and, 442 in electrochemistry, 791 in electrolyte solutions, 129, 435 in electroplating, 816–817 in Geiger counter, 852 in piezoelectric substance, 1024 in semiconductor, 451 units of, 817 work done by, 778–779, 800–801 Electric power. See Energy sources Electrochemistry, 791. See also Cell potential; Electrolysis; Galvanic cells Index A75 Electrodes, 792, 798, 800, 816 glass, 807 ion-selective, 807 overvoltage and, 820 standard hydrogen, 794 Electrolysis, 816–820 commercial uses of, 821–826, 879, 885, 887–888, 980 of mixtures of ions, 819–820 overvoltage in, 820 of water, 3–4, 27, 258, 809–810, 818–819, 883 Electrolytes, 129–133, 145, 435, 512–515 Electromagnetic force, 864, 865 Electromagnetic radiation, 275–277. See also Light; Photons; Ultraviolet radiation gamma rays, 48, 844, 864, 865 infrared, 254–255, 276, 279, 810 quantization of, 278–280, 285 spectroscopy with, A16–A19 Electromotive force (emf). See Cell potential Electron, 48, 50. See also Beta () particles; Delocalized electrons; Valence electrons annihilation of, 844 in Bohr model, 285–287, 290, 291, 292, 293 early experiments, 45–48 paramagnetism and, 409 photoelectric effect and, 279–280 as wave, 282–284, 290 Electron acceptor, 159 Electron affinity, 312–313, 316 Electron capture, 844 Electron configurations of atoms, 302–309, 314–315 of ionic compounds, 339–340 of stable compounds, 338–339 Electron correlation problem, 298, 403 Electron-deficient compounds, 358–359, 360, 664, 888 Electron density map, 292, 293, 296 Electron donor, 159 Electronegativity, 333–335 acid strength and, 661, 662 percent ionic character and, 346, 347 Electron-pair acceptor, 663–665 Electron-pair donor, 663–665 ligand as, 731 Electron-pair repulsion. See VSEPR model Electron sea model, 441 Electron spin, 296, 298, 303 Electron transfer, 154, 158–160, 163, 167 Electroplating, 815, 816–817, 819, 824 Electrorefining, 823–824 Electrostatic forces, 426, 514–515, 863 Electrostatic precipitator, 515 Elementary step, 550 Elements. See also Periodic table; Representative elements abundance, 878–879 definition of, 28 diatomic, 3 early ideas about, 39–40, 41, 43, 46, 47 original names, 56 oxidation state, 156, 879 preparation, 879–880 standard enthalpy of formation, 249 standard free energy of formation, 769 standard state, 246 transformation of, 849–852 emf (electromotive force). See Cell potential Emission spectrum of alkali metal ions, 731 in fireworks, 288 of hydrogen atom, 284–285, 290 Empirical formula, 93–94, 96 Enantiomers, 964–965 Endothermic process, 231, 232, 233 enthalpy change, 236 entropy change, 757 equilibrium constant, 610 End point, of titration, 152, 711, 714 Energy, 229–235. See also Bond energies; Heat; Kinetic energy; Potential energy of activation, 553–557, 558, 593 biologic, 154, 229, 973 chemical, 231–232, 749 conservation, 229, 232, 749 definition of, 229 electrical, 792 of electromagnetic radiation, 276, 278–280 of hydrogen atom, 285–287, 290, 295–296 internal, 232–236 of ionic lattice, 342, 343, 344–346 lower, as driving force, 757, 758 mass associated with, 280, 844, 856–857, 864 of nuclear binding, 858–859 of polyelectronic atoms, 299 quantization of, 278–280, 285–287, 290, 291, 293 radiation damage and, 863, 864, 865 sign of, 232–233 of solution formation, 488–492 as state function, 231, 342 thermal, 757 transfer of, 230 units of, 205, 233, A26 wasted, 778, 779, 801, 810–811 zero point of, 286, 331 Energy crisis, 214, 779 Energy sources. See also Batteries; Fossil fuels alternative, 256–263 conventional, 252–256 fuel cells, 812–813 nuclear power, 861–863, 866 redox reactions in, 154 thermophotovoltaic, 810–811 English system of units, 8, 16–19 Enthalpy, 235–236 bond energies and, 351–353 calorimetry and, 237–240 entropy change and, 758 as extensive property, 238, 243–244 free energy and, 759–762, 766–767 Hess’s law for, 242–246, 247 pressure and, 771 sign of change in, 239, 243, 249 standard, 246–252, 767 Enthalpy of formation, standard, 246–252, 767, A19–A22 Enthalpy of fusion, 425, 464 Enthalpy of hydration, 491, 927, 928 Enthalpy of solution, 489–492, 496, 502, 503 Enthalpy of vaporization, 425, 426, 459, 461–463 Entropy, 751–755. See also Spontaneous processes absolute values, 763–764 in chemical reactions, 762–766 definition of, 758 free energy and, 759–762, 766–767 of hydration, 928 irreversible processes and, 779 molecular structure and, 766 as organizing force, 756 positional, 754–755, 756, 762–764, 771 pressure dependence of, 771 second law of thermodynamics and, 755–756, 762, 779 sign of change in, 758–759 of solution, 754 standard values, 764–765, A19–A22 temperature and, 764 of vaporization, 756–757 Enzymes, 557, 558, 562–563 Ephedrine, 648 Equation of state, 187. See also Ideal gas law Equations. See Chemical equations; Linear equations; Quadratic equations Equilibrium, 579–582 definition of, 579 entropy and, 755, 761 free energy and, 761, 766, 770, 773–778 heterogeneous, 588–590 homogeneous, 588 of liquid and vapor, 459–460 reaction rates and, 581–582, 593, 606 Equilibrium constant (K), 582–586 for complex ion formation, 731–734 extent of reaction and, 592–593 free energy and, 775, 777–778 for heterogeneous equilibria, 588–590 pressures in, 586–588, 594–596, 601–603 reaction quotient and, 593–594, 605 for redox reactions, 807–808 small, 603–604 for sum of reactions, 734 temperature dependence of, 605, 609–610, 777–778 units in, 583, 584 Equilibrium expression, 582–584 for heterogeneous equilibrium, 589 with pressures, 586–588, 594–596, 601–603 Equilibrium point, 774, 775 Equilibrium position, 585–586, 591–592, 774 Le Châtelier’s principle and, 604–610 in solubility equilibrium, 717–718 Equilibrium problems, 600–604. See also Acid–base equilibrium problems; Solubility calculating concentrations, 596–599, 600–601, 603–604 calculating pressures, 594–596, 601–603 pictorial example, 591–592 reaction quotient in, 593–594, 600, 605, 607 with small K, 603–604 summary of procedure, 600 A76 Index Equilibrium vapor pressure, 460, 465. See also Vapor pressure of solution, 497–498 Equivalence point, 151–152, 696 determination of, 711–716 strong acid–strong base, 698, 699, 705 strong base–strong acid, 699 weak acid–strong base, 702, 703, 704, 705, 706, 707 weak base–strong acid, 709, 710, 711 Equivalent mass, 487, 488 Erements, Mikhail, 358 Error, 12–13, A10 Error limit, A10, A11, A12 Erythropoietin (EPO), 978–979 Esters, 1015 Estimation. See Uncertainty in measurement Ethane, 998 Ethanol, 1011–1012, 1014 from fermentation, 258–259, 263, 891, 1011–1012 as fuel, 263, 1012 hydrogen bonding in, 428 as nonelectrolyte, 133 solubility in water, 129 vapor pressure, 461 Ethylene hydrogenation, 558–559, 998 polymers based on, 1017–1019, 1021–1022 structure, 393–395, 1005 synthesis, 1004 Ethylenediamine (en), 957, 959, 968 Ethylenediaminetetraacetate (EDTA), 957 Ethylene glycol, 1012 Ethyl group, 1000 Evaporation, 459. See also Vaporization; Vapor pressure Evaporative cooling, 459, 514 Exact numbers, 13 Excited state of hydrogen atom, 284, 286, 287, 296 of nucleus, 844 Excluded-volume force, 756 Exclusion principle, 298, 302, 303 Exothermic process, 231–232 enthalpy change, 236 entropy change, 757–759, 760–761 equilibrium constant, 609–610 spontaneity and, 750–751, 760–761 Experimental error, 10–13, A10–A13 Experiments, 5, 6, 199, 200 Explosives detector for, 455 in fireworks, 288 nitrogen-based, 7, 358, 410, 455, 582, 904–905, 911, 1016 Exponential notation, 13–14, A1–A3 Exponentiation, A5 Extensive property enthalpy change, 238, 243–244 entropy, 764 Extinction coefficient, A17–A19 Extinctions, 1, 51, 80 Face-centered cubic unit cell, 433, 436, 437, 438–441, 456–457 Fahrenheit scale, 19–23 Families. See Groups Faraday (F), 801 Faraday, Michael, 801 Fats. See Oils Fat-soluble vitamins, 492–493 Feldspar, 448 Fermentation, 258, 263, 891, 1011–1012 Ferrate ion, 959 Ferric ion, 59 Ferrochrome, 951 Ferrous ion, 59 Ferrovanadium, 950 Fertilizers, 7, 906, 910, 911, 917, 922 Filtration, 26 Fire extinguishers, 471 Firewalking, 241 Fireworks, 277, 278, 288–289 First ionization energy, 309, 310, 311–312, 316, 317 First law of thermodynamics, 232, 233, 749 First-order radioactive decay, 847 First-order reaction, 535 half-life, 541–543, 545 integrated rate law, 538–543, 548 pseudo-first-order, 547 Fisher, Mel, 820 Fission, nuclear, 843, 859–863, 866 5% rule, 637, 638 Flame test, 729, 731 Flask, volumetric, 10, 136, 137, 139 Flat-panel display, 282–283 Fleming, Alexander, 90 Flotation process, 980, 983 Fluorapatite, 717, 720, 916 Fluoride, tooth decay and, 717, 720 Fluorine, 304, 924–930. See also Hydrogen fluoride in chlorofluorocarbons, 561, 1003–1004 molecular structure, 355, 409, 410, 878 oxidation state, 156 Fluxes, 983, 984, 985 Foecke, Tim, 445 f orbitals, 294, 295, 297, 299 in lanthanides and actinides, 306, 943, 949 Forces. See also Intermolecular forces bonding and, 332, 347, 348 electromagnetic, 864, 865 electrostatic, 426, 514–515, 863 entropic, 756 fundamental, 864, 865 gravitational, 10, 180, 211, 864, 865 nuclear, 863, 864 potential energy and, 230 pressure and, 181, 233, A14–A16 work and, 230 Formal charge, 363–367 Formaldehyde, 1013 Formation constants, 731–734 Formula, chemical, 52 determination of, 91–96 from name, 59 Formula, structural, 52, 53 Formula equation, 145, 146 Forward bias, 451, 452 Fossil fuels, 229, 253–254. See also Air pollution; Coal; Gasoline; Natural gas; Petroleum carbon dioxide from, 179, 229, 255, 891 Fractional charge. See Partial charge Francium, 316 Frasch process, 920 Free energy, 759–762 cell potential and, 802, 804, 805 in chemical reactions, 766–770, 771–777, 802 equilibrium and, 761, 766, 770, 773–778 pressure dependence of, 770–773, 774–777 spontaneity and, 759–762, 770, 773–774, 778 work and, 778–779, 802, 805 Free energy of formation, standard, 769, 772, A19–A22 Free radical, 1019 Freezing, 26 free energy change, 760 Freezing point, 466, 469. See also Melting point Freezing-point depression, 504, 505, 506–507, 512, 513 Freons, 561, 1003–1004 Frequency, 275–276, 277, 278–280 Frequency factor, 555 Frictional heating, 230 by electric current, 778, 793, 801 of ice, 469 Fructose, 515, 1031, 1032, 1033, 1034 Fry, Art, 7 Fuel cells, 812–813 Fullerenes, 80–81, 891 Functional groups, 1010 Fusion, nuclear, 850–851, 859, 863 Galena, 161, 893, 920, 955 Galileo, 7, 180 Gallium, 305, 888, 889, 890 Gallium arsenide, 279 Galvani, Luigi, 794 Galvanic cells, 791–793. See also Batteries; Cell potential; Fuel cells compared to electrolytic cells, 816 complete description, 798–800 concentration cells, 804 corrosion compared to, 815 efficiency of, 801 line notation for, 798 standard reduction potentials, 794–800, 948, A25 Galvanizing, 815, 928, 955 Gamma rays, 48, 844, 864, 865 Gangue, 978, 980, 982 Gas constant (R), 187, 204–205 Gases. See also Atmosphere; Ideal gas; Kinetic molecular theory; Pressure; Vaporization acid–base reactions in, 625–626 Avogadro’s hypothesis, 44 Avogadro’s law, 185–186, 202 Boyle’s law, 181–184, 186, 189, 200–201 Charles’s law, 184–185, 186, 189, 201 in chemical equations, 98 collected over water, 198–199 compression of, 233, 234 Dalton’s law of partial pressures, 194–199, 202 diffusion of, 206, 207–208 early studies of, 40–41, 44–45 effusion of, 206–207 equilibria with, 586–588, 594–596, 601–603, 607–608 expansion of, 233–235 fundamental properties of, 25, 179 mixing of, 206, 207–208 Index A77 mixtures of, 194–199, 202 molar concentration of, 586, 587, 588 molar mass of, 193–194 molar volume of, 190–191 positional entropy of, 754–755, 756, 762–763, 771 real, 182–184, 187, 191, 200, 205, 208–211, 879 separation of, 196 solubility of, 493–495, 496 solutions of, 485 standard state, 246 standard temperature and pressure, 191 state of, 187 stoichiometry of, 190–194 velocity distribution in, 205–206, 553–554 work performed by, 233–235 Gasohol, 263, 1012 Gasoline, 253 air pollution and, 211–213 combustion, 252, 778, 779 compared to hydrogen, 261 lead in, 253–254, 893 from methanol, 257 octane rating, 103 Gay-Lussac, Joseph, 44–45 Geiger–Müller counter, 852 Gemstones, 970–971 Gene, 1038 Genetic damage, 864 Geometrical isomerism, 961, 963, 965, 1006 Germanium, 300, 301, 890, 891, 892, 894 Germer, L. H., 282 Geubelle, Philippe, 1018 Gibbs, Josiah Willard, 759 Glass capillary action in, 429 cleaning solution for, 953 etched with acid, 928 metallic, 449 photochromic, 931 structure of, 432, 447–448 titanium dioxide on, 951 Glass electrode, 807 Global warming, 179, 229, 255–256, 910 Glucose, 891, 1011–1012, 1033, 1034 Gluons, 864 Glycerol, 429–430 Glycogen, 1034, 1036 Glycoside linkages, 1034, 1035–1036 Gold, 433, 814, 821, 955, 981, 982 in plants, 40 Golf clubs, 449 Goodyear, Charles, 1017 Goudsmit, Samuel, 296 Graduated cylinder, 11, 12–13 Grafting, onto polymer, 1025 Graham, Thomas, 206 Graham’s law, 206–207 Grams, 9, 83 Graphing functions, A6–A7 Graphite, 332–333, 443, 444–446, 891 diamond and, 244, 246, 446, 470–471, 768–769 entropy of, 764 Gravitational force, 10, 864, 865 atmosphere and, 180, 211 Gray tin, 892 Greenhouse effect, 255–256, 495, 910 Ground state, 286, 287, 290, 296 Groundwater clean-up, 156–157 Group 1A. See Alkali metals; Hydrogen Group 2A. See Alkaline earth metals Group 3A, 876, 888–890. See also Aluminum; Boron Group 4A, 876–878, 890–894.
See also Carbon; Lead; Silicon Group 5A, 878, 901–903. See also Nitrogen; Phosphorus Group 6A, 878, 918. See also Oxygen; Selenium; Sulfur Group 7A. See Halogens Group 8A. See Noble gases Groups, 55, 307 alternate designations for, 57, 307–308 atomic radii and, 313, 314 chemical properties of, 308, 314 electron affinities in, 313 ionization energies in, 310 names for, 55, 315 of transition metals, 307, 943–944 valence electrons and, 304–305, 307, 308, 314 Guericke, Otto von, 180 Guldberg, Cato Maximilian, 582 Gutierrez, Sidney M., 105 Gypsum, 213, 892, 916, 920 Haber, Fritz, 582, 604 Haber process, 906. See also Ammonia synthesis equilibrium, 583–584, 604–606 hydrogen for, 106, 883 reaction rate, 527, 558, 582 Hadrons, 864 Hafnium, 308–309, 949 Half-cell potentials, 794–800. See also Cell potential Half-life of radioactive sample, 847–849 of reactant, 541–543, 545–546 of transuranium elements, 851 Half-reactions electrochemical, 791–792, 794–800 oxidation–reduction, 162–168 Hall, Charles M., 822, 889 Hall–Heroult process, 822, 889 Halogenation of hydrocarbons, 1003–1004, 1008 Halogens (Group 7A), 55, 924–930. See also Hydrochloric acid; Hydrogen fluoride atomic radii, 878 electron affinities, 313 hydrogen halides of, 628, 661, 925–928 interhalogen compounds, 930 as ligands, 958, 968 oxyacids of, 67, 928–929 phosphorus compounds with, 917–918 sulfur compounds with, 924 Hassium, 57 Heat. See also Endothermic process; Exothermic process; Frictional heating; Temperature calorimetry and, 237–242 coagulation by, 515 conductivity of, in metals, 436, 441, 442, 944 at constant pressure, 235–236, 237–240, 243 at constant volume, 240–242 electricity from, 810–811 as energy, 229, 232, 749 entropy changes and, 757–759, 762 sign of, 232, 233 state functions and, 231 temperature and, 230 as wasted energy, 778, 779, 801, 810–811 work done by gas and, 234–235 Heat capacity, 237–238, 239, 241 Heating curve, 463–464 Heat of fusion, 425, 464 Heat of hydration, 491, 927, 928 Heat of reaction, 236. See also Enthalpy Heat of solution, 489–492, 496, 502, 503 Heat of vaporization, 425, 426 definition of, 459 vapor pressure and, 461–463 Heat radiation, 254–255 Heisenberg, Werner, 290, 291 Heisenberg uncertainty principle, 291–292, 296 Helium, 302, 931, 932 in buckyballs, 80 Lewis structure, 354 molecular orbital model, 406 Schrödinger equation for, 298 stellar fusion and, 850, 863 Helium nucleus. See Alpha () particles -Helix, 1028, 1029 Hematite, 982 Heme, 973–975 Hemoglobin, 756, 975–979 Hemolysis, 510 Henderson–Hasselbalch equation, 689, 695, 713–714 Henry’s law, 494–495, 497 Heroult, Paul, 822, 889 Hertz (Hz), 276 Hess’s law, 242–246, 247, 767 Heterogeneous catalysis, 558–559 Heterogeneous equilibria, 588–590 Heterogeneous mixture, 25 Heteronuclear diatomic molecules, 412–413 Hexadentate ligand, 957 Hexagonal closest packed (hcp) structure, 436, 437, 438, 441, 442, 456 Hexahydro-1,3,5-triazine (RDX), 455 Hexoses, 1031, 1032 High-altitude sickness, 977 High-spin case, 968, 971, 972 Hill, Julian, 1017 Hole in closest packed structure, 456–458 in semiconductor, 450–451 Homogeneous catalysis, 558, 559–563 Homogeneous equilibria, 588 Homogeneous mixture, 25. See also Solutions Homonuclear diatomic molecules, 406–412 Homopolymer, 1020 Human body. See also Biologic systems aging of, 160–161 boron requirement, 889 elements in, 879 knee prosthesis, 431 radioactivity applications, 847, 855–856 selenium in, 47, 918 transition metals in, 973–977 Hund’s rule, 303 A78 Index Hybridization, 391–403 in alkanes, 998 dsp3 orbitals, 397–399 d2sp3 orbitals, 399–400 in complex ions, 966–967 in Group 4A, 890, 891 in Group 5A, 902, 903 sp orbitals, 395–397 sp2 orbitals, 393–395 sp3 orbitals, 391–393 summary, 401 Hydration of alkali metal ions, 318, 881 of aluminum ions, 659, 661, 664–665, 666, 889 of concrete, 892 enthalpy of, 491, 927, 928 entropy of, 928 of halide ions, 927–928 of ions in solution, 128–129, 130 of protons, 624, 629 of solids, 213 Hydrazine, 907–909 Hydride ion, 340, 884 Hydrides, 427–428, 883–885, 918 of boron, 245, 888 metallic, 262, 884–885 of nitrogen, 906–909 Hydrocarbon derivatives, 1010–1016 Hydrocarbons. See also Benzene; Ethylene; Methane aromatic, 1008–1010 in petroleum, 253 saturated, 997–1005 unsaturated, 997–998, 1004, 1005–1010 Hydrochloric acid, 131, 627, 927, 928. See also Hydrogen chloride in aqua regia, 735, 803 Br$nsted-Lowry model, 625–626 pH calculations, 634–635 Hydrocyanic acid, 657, 705–707 Hydrofluoric acid, 926–928. See also Hydrogen fluoride with common ion, 681–683 pH calculations, 636–637 Hydrogen, 883–885 bonding in, 331–332, 354, 403–405, 406 in early universe, 850 entropy of, 765–766 as fuel, 257–262, 812, 813, 885 as nonmetal, 55, 316, 876 oxidation state, 156 preparation of, 880, 883 as real gas, 208, 210–211 as reducing agent, 879, 980 in syngas, 256–257 in water, 3–4 Hydrogenation, 558–559, 883, 998, 1007 Hydrogen atom atomic radius, 876 Bohr model, 285–287, 290, 291, 292, 293 emission spectrum, 284–285, 290 orbital diagram, 302 quantum mechanical model, 290–296 summary, 296 Hydrogen bonding, 426–428 in alcohols, 428, 1010 in ammonia, 907 boiling points and, 427–428 in hydrazine, 907 in hydrogen fluoride, 926 in molecular solids, 454, 456 in nonideal solution, 502, 504 in nucleic acids, 1038, 1039 in nylon, 1017 in proteins, 1028, 1030 solubility and, 490, 491 viscosity and, 430 in water, 426–427, 454, 456, 459, 461, 464, 490, 491, 884 Hydrogen bromide, 927 Hydrogen chloride. See also Hydrochloric acid dipole moment, 337 Hydrogen–chlorine cannon, 926 Hydrogen cyanide. See also Cyanide ion from millipede, 379 Hydrogen electrode, 794 Hydrogen fluoride, 636–637, 661, 925–928 bonding in, 332–333, 335–336, 413 with common ion, 681–683 Hydrogen halides, 628, 661, 925–928. See also Hydrochloric acid; Hydrogen fluoride Hydrogen iodide, 927 Hydrogen ions. See also pH; Protons from autoionization of water, 629–631 in definition of acid, 131, 149, 623, 664 Hydrogenlike orbitals, 299, 302 Hydrogen peroxide, 159, 881, 975 Hydrogen sulfide, 338, 923 Hydrogen sulfites, 922 Hydrohalic acids, 628, 661, 926–928. See also Hydrochloric acid; Hydrogen fluoride Hydrometallurgy, 981–982 Hydronium ion, 623–624, 626, 629 Hydrophilic side chains, 1027 Hydrophilic substances, 492 Hydrophobic side chains, 1027 Hydrophobic substances, 492 Hydrostatic pressure. See Osmotic pressure Hydroxide ion in Arrhenius base, 131, 149, 623, 644 from autoionization of water, 629–631 buffering and, 685–686, 690–691, 692–693 ionic oxides and, 663, 876 as Lewis base, 664 pOH and, 631–632, 633–634, 648 in salt solutions, 656–657 as strong base, 131, 149 weak bases and, 132–133, 646–650 Hydroxides of alkali metals, 318, 644 of alkaline earth metals, 644–645, 724, 885 solubility in water, 144 Hydroxo ligand, 958, 968 Hydroxyapatite, 159, 717, 720 Hydroxyl group, 1010 Hydroxyl radical, 212, 213, 1019 Hyperbola, 182 Hypochlorite ion, 638, 643, 929 Hypochlorous acid, 628, 638, 929 Hypofluorous acid, 929 Hypophosphorous acid, 917 Hypothesis, 5–6, 7 Ice, 25, 26, 425 biology and, 22–23, 516 density of, 425, 426, 469 freezing-point depression, 506–507 melting of, 425, 426, 463–464, 760–761 on phase diagram, 467–469 structure of, 435, 454, 456, 884 sublimation of, 467–468, 469 supercooled water and, 466, 516 vapor pressure, 465 ICE table, 640 Ideal gas definition of, 183, 187 expansion into vacuum, 751–754 free energy of, 771–772 kinetic molecular theory of, 199–206, 207, 208, A13–A16 molar volume of, 190–191 partial pressure of, 194–196 Ideal gas law, 186–190 derivation of, 202–204, A13–A16 molar mass and, 193–194 molar volume and, 190–191 Ideal solution, 502, 503 Ilmenite, 950 Independent variable, A6 Indeterminate error, 12, A10 Indicators, acid–base, 151–152, 711–716 Indium, 300, 888, 890 Inert-atmosphere box, 903 Inert gases, 373. See also Noble gases equilibrium and, 607 nitrogen as, 903 Infrared radiation, 276 from earth’s surface, 254–255, 910 imaging with, 279 in thermophotovoltaics, 810 Initial rates, method of, 535–538, 548 Ink, electronic, 488–489 Insoluble solids, 143, 735 Instantaneous rate, 530 Insulator, electrical, 444 Integers, in calculations, 13 Integrated rate law, 533, 534, 538–547, 548–549 first-order, 538–543 for radioactive decay, 854–855 second-order, 543–546 with several reactants, 546–547, 549 zero-order, 546 Integration, 539 Intensive property, 238 reduction potential as, 797 Intercept, A6, A7 Interference of waves, 282, 432 Interhalogen compounds, 930 Intermediate, 549 Intermolecular forces, 426–429. See also Hydrogen bonding; London dispersion forces changes of state and, 464 in Group 8A solids, 436 in liquid–liquid solutions, 502, 504 in liquids, 429–430, 460–461 in molecular solids, 454–456 in proteins, 1030 in real gas, 200, 209, 210–211 solubility and, 489 vapor pressure and, 460–461 Internal energy, 232–236 Interstitial alloys, 443, 986 Interstitial (metallic) hydrides, 262, 884–885 Intramolecular bonding, 426 Inverse relationship, 182 Iodide salts, solubility, 144 Iodine, 924–927, 928, 929–930 Index A79 sublimation of, 463 triiodide ion, 361, 375, 398–399 Iodine-131, 855 Ion exchange, 887 Ionic bonds, 53–55, 330, 335 percent ionic character, 346–347 in proteins, 1030 Ionic compounds (salts), 53–55.
See also Binary ionic compounds; Crystalline solids; Solubility acid–base properties, 655–660 in buffered solutions, 684, 688 common ions from, 681–683, 718, 722–724 coordination compounds as, 955 crystal structures, 344, 435, 436, 456–458 as electrolytes, 129–131, 145, 435, 512–513 formation of, 342–346 insoluble, 143, 735 Lewis structures, 354 molten, 347, 348 naming of, 59–63, 65–66 oxidation states in, 156 polyatomic ions in, 55, 62–63, 65, 66–67, 346 predicting formulas of, 339–340 radii of ions in, 340–342 Ionic equation, 145–146, 150 Ionic hydrides, 883–884 Ionic liquids, 494–495 Ionic radii, 340–342 Ionic solids, 55, 339, 435, 436, 456–458. See also Ionic compounds (salts) Ion interchange, 144 Ionization, of strong electrolytes, 130 Ionization energies, 309–312, 315, 316 of transition metals, 948 Ionizing radiation, 864–865 Ion pairing, 512–513 Ion-product constant (Kw), 629–631, 632–633, 656 Ion product (Q), 725, 727 Ions, 53–54. See also Anions; Cations; Complex ions; Electrolytes; Hydration charges on, 59, 60, 61, 157 in colloid, 514, 515 concentrations of, 134–135 energy of interaction, 330 naming of, 59, 60, 61, 62, 66–67 oxidation states of, 156, 157 polyatomic, 55, 62–63, 65, 66–67, 346 selective precipitation of, 727–731, 734–736, 981, 982 sizes of, 340–342 Ion-selective electrodes, 807 Iridium, 1, 51, 80, 81, 949 Iron, 305, 878, 947, 954. See also Steel in biologic systems, 973–977 corrosion of, 776–777, 813, 814–816, 954 crystalline forms, 986 in magnetorheological fluid, 431 metallurgy of, 982–984 for water clean-up, 156–157 Iron oxides, 158, 982–983, 984 Iron pyrite, 982 Irreversible process, 779 Isoelectronic ions, 341–342, 344 Isomerism, 960–965 alkanes, 998–1000, 1001–1002 alkenes, 1006 complex ions, 960–965 sugars, 1031–1032, 1034 Isotactic chains, 1023, 1024 Isotonic solutions, 510–511 Isotopes, 50, 51, 79, 80–81, 84, 841 Jacobson, Joseph M., 488, 489 Johnson, William L., 449 Joliot, Frederick, 849 Jorgensen, Sophus Mads, 960 Joule (J), 205, 233, 801 Juglone, 86, 379 Juhl, Daniel, 258 Junction potential, 451 Junction transistor, 452 K. See Equilibrium constant Ka. See Acid dissociation constant Kb, 646 calculating from Ka, 656 for common weak bases, 646–647, A23 pH of salt solutions and, 659–660 Kp, 586–588, 594–595, 601–603 Ksp, 717–724, 725–729, A24 Kw, 629–631, 632–633, 656 Kairomones, 379 Kaolinite, 448 Kelvin temperature, 19–22 absolute zero of, 185, 764 Charles’s law and, 185, 201 ideal gas law and, 190 kinetic energy and, 200, 203, 204 Kerogen, 262–263 Kerosene, 253 Ketones, 1013–1014 Kinetic energy, 230 bonding and, 331 internal energy and, 232 reaction rates and, 553 temperature and, 200, 203, 204, 206, 461 Kinetic molecular theory (KMT), 199–206 effusion and, 207 quantitative, A13–A16 reaction rates and, 552 real gases and, 200, 208 Kinetics of chemical reactions. See Rate laws; Reaction mechanisms; Reaction rates Kinetics of nuclear decay, 846–849 Kinetic stability, of nucleus, 841–849 Knee prosthesis, 431 Krypton, 305, 932 Kuznick, Steven, 196 Lake Nyos tragedy, 497 Lanthanide contraction, 949 Lanthanide series, 306, 308, 875, 943, 949 Lasers, 87, 863 Lattice, 432, 435, 436 Lattice energy, 342, 343, 344–346 Lavoisier, Antoine, 7, 41, 275, 288, 821 Law of conservation of energy, 229, 232, 749 Law of conservation of mass, 6, 41, 98 Law of definite proportion, 41 Law of mass action, 582, 585, 589, 593, 600 solubility and, 717 Law of multiple proportions, 42–43 Law of nature, 6, 199 Leaching, 981–982 Lead, 890, 891, 893, 894 in gasoline, 253–254, 893 metallurgy of, 51, 161–162, 879, 893, 981 in peat bogs, 51 radioactive decay to, 854–855 Leading zeros, 13, 14 Lead poisoning, 1, 893, 957 Lead salts, solubility of, 144 Lead storage batteries, 24, 808–810, 893, 922 Le Châtelier’s principle, 604–610 common ion effect and, 682 for concentration, 605–607, 803 for pressure, 607–609, 773 for temperature, 609–610 Leclanché, George, 810 Lee, David S., 449 LE model. See Localized electron (LE) model Length, units of, 9, 16–19, A26 LEO says GER, 160 Lepidolite, 880 Leptons, 864 Leucippus, 39, 275 Levanon, Baruch, 809 Levorotatory isomer, 964–965 Lewis, G. N., 348, 354, 663 Lewis acid–base model, 663–665 of complex ions, 731, 945, 956, 966 Group 4A elements and, 890 Group 5A elements and, 902 Lewis structures, 354–358 exceptions to octet rule, 358–361 formal charge and, 363–367 in localized electron model, 354, 358, 360, 400 odd-electron molecules and, 363 resonance of, 362–367, 413 in VSEPR model, 369, 377 Libby, Willard, 853 Ligands, 731, 945, 955, 956–960.
See also Complex ions in biology, 973, 977 coordination number and, 731, 956 definition of, 956 as Lewis bases, 731, 945, 956, 966 naming of, 958 spectrochemical series, 968 Light. See also Colors; Electromagnetic radiation diffraction of, 281 dual nature of, 280–281 dyes and, 277 in photography, 926 polarized, 961–962, 1032 quantization of, 278–280, 285 scattering of, 514 in scintillation counter, 852–853 smog and, 213 spectroscopy with, A16–A19 spectrum of, 284–285 speed of, 275–276, 280 titanium dioxide and, 951 “Like dissolves like,” 129, 489, 491 Lime. See Calcium oxide Lime boil, 985 Lime–soda process, 645 Limestone. See also Calcium carbonate acid rain and, 212–213 caverns in, 681, 724 in iron metallurgy, 983–984 in Portland cement, 892 for scrubbing, 214 in steelmaking, 985 A80 Index Limiting reactant, 106–113 in solution, 147–148 Limiting reagent, 108 Linear complex ions, 956, 966, 973 Linear equations, A6–A7 Linear model, of radiation damage, 866 Linear structure, 367, 371, 372, 375, 401, 402 Line notation, 798 Line spectrum, 285 Linkage isomerism, 961 Liquefaction of air, 879–880, 919 Liquid–liquid solutions, 501–504 Liquids. See also Solutions; Vapor pressure in chemical equations, 98 fundamental properties of, 25, 425, 429–430 in heterogeneous equilibria, 589 intermolecular forces in, 429–430, 460–461 ionic, 494–495 magnetorheological fluids, 431 positional entropy of, 754 standard state, 246 structural model for, 430 superheated, 466 Liter (L), 9 Lithium, 302, 316, 317–318, 880–881, 882. See also Alkali metals atomic radius, 318, 876 manic depressive disease and, 1, 882 molecular orbital model, 406–407 Lobes, of orbitals, 295 Localized electron (LE) model, 353–354 combined with MO model, 413–415 of complex ions, 965–967 hybrid orbitals in, 391–403 Lewis structures and, 354, 358, 360, 400 limitations of, 403 odd-electron molecules and, 363 resonance and, 362–363, 413–415 summary of, 400–403 Logarithms, 631, A4–A6 London dispersion forces, 428–429 in ethane, 1010 in Group 8A solids, 436 in molecular solids, 454 in oil, 490 in proteins, 1030 vapor pressure and, 461 Lone pairs formal charge and, 364, 366 in hydrogen bonding, 427 in localized electron model, 353, 354, 355, 356 in VSEPR model, 370–371, 375, 376–377 Lowry, Thomas M., 149, 623 Low-spin case, 968 Macroscopic world, 2–3 Maentyranta, Eero, 978, 979 Magic numbers, 843 Magnesium, 304, 442, 885, 886–888 for cathodic protection, 816 Magnesium hydroxide, 105–106, 644–645, 724 Magnesium oxide, 344–346 Magnetic moment, 296, 298 Magnetic quantum number, 294 Magnetism. See also Diamagnetism; Paramagnetism in metallurgy, 982–983 Magnetite, 158, 982 Magnetorheological fluid, 431 Maillard, Louis-Camille, 1032 Maillard reaction, 1032 Main-group elements. See Representative elements Major species, 634–635, 636, 637, 666, 667 Malleability, 55, 436, 441, 443 Manganese, 305, 947, 953 Manic depressive disease, 1, 882 Manometer, 180–181 Marble, acid rain and, 212–213 Mass. See also Density; Molar mass atomic, 44–45, 46, 78–81, 82, 300 average, 77–78 compared to weight, 9–10 conservation of, 6, 41, 98 definition of, 10 in definition of matter, 25 effusion of gases and, 206–207 energy associated with, 280, 844, 856–857, 864 equivalent, 487, 488 measurement of, 10, 11, 12 of reactants and products, 102–106 units of, 9, 16, A26 Mass action. See Law of mass action Mass defect, 857, 858–859 Mass number (A), 50, 52, 841, 842 Mass percent, 89–91 of solute, 485, 486, 488 Mass spectrometric measurement of atomic mass, 78–79 of Avogadro’s number, 82 of isotopic composition, 80–81, 84 of molar mass, 87 for radiocarbon dating, 854 Matches, 914–915 Matrix-assisted laser desorption, 87 Matter, 25–28 quantum nature of, 277–284 McMillan, Edwin M., 302 Mean, A11 Mean free path, 205 Measurement of mass, 10, 11, 12 of temperature, 19–23 uncertainty in, 10–13, 16, A10–A13 units of, 5, 8–10, 16–19, A26 of volume, 9, 10–13 Measuring pipet, 137, 139 Mechanisms, reaction, 527, 549–552 Median, A11 Medicine. See Human body Melting, 463–464 of ice, 425, 426, 463–464, 468–469, 760–761 Melting point, 464, 465, 466. See also Freezing point depression; Phase diagrams Mendeleev, Dmitri Ivanovich, 300, 301, 304, 314 Meniscus, 10, 429 Mercury convex meniscus of, 429 salts of, solubility, 144 toxicity of, 893, 975 Mercury barometer, 180–181, 460 Mercury cell, 811, 825–826 Mercury manometer, 180–181 Messenger RNA (mRNA), 1038–1039, 1040 Metal ions. See also Complex ions; Hydration; Ions naming of, 58, 59, 60, 61 selective precipitation of, 727–731, 734–736, 981, 982 Metallic glasses, 449 Metallic hydrides, 262, 884–885 Metallic radii, 313 Metalloids, 316, 875, 890. See also Silicon Metallurgy, 39, 879, 978–982 electrorefining, 823–824 of iron, 982–984 of lead, 51, 161–162, 879, 893, 981 Metal plating. See Electroplating Metals. See also Alloys; Corrosion; Ores; Transition metals bonding in, 441–442, 458 bonding with nonmetals, 339, 945 crystal structures, 433, 435, 436–441 in periodic table, 55, 315, 316, 875, 876 physical properties, 55, 436, 441, 442, 944–945 as reducing agents, 316–318, 881, 948 Meter (m), 9, 18–19 Methane bond energies, 350 bond polarities, 338 from coal gasification, 256 combustion of, 812–813 as greenhouse gas, 255, 256 halogenated, 1003–1004 hydrogen derived from, 883 in natural gas, 196, 253–254, 258 as real gas, 208, 210–211 structure of, 52–53, 347, 349, 368, 391–392, 998 Methanol, 1010–1011 combustion of, 252 as fuel, 257, 263, 1011 hydrogen bonding in, 428, 1010 preparation of, 111–112 VSEPR model, 377 Method of initial rates, 535–538, 548 Methylamine ligand, 958 Methylene group, 998 Methyl group, 1000 Metric system, 8–10, 16–19 Meyer, Julius Lothar, 300 Microchip laboratories, 138 Microelectronics, 47, 452–453 Microencapsulation, 488–489 Microscope, scanning tunneling (STM), 2, 438–439 Microscopic world, 3 Microstates, 752, 753, 754 Microwave radiation, 276 Milk of magnesia, 105–106, 724 Millikan, Robert, 47–48 Milliliter (mL), 9, 10–13 Millimole (mmol), 696 Mineral acids, 131. See also Nitric acid; Sulfuric acid Minerals, 978 Minor species, 635 Minton, Allen, 756 Mirror image isomers, 963–965, 1031–1032, 1034 Mixing, of gases, 206, 207–208 Index A81 Mixtures, 25, 485. See also Colloids; Solutions entropy and, 754 of gases, 194–199, 202 separation of, 25–27, 196 mm Hg, 181 Mobile phase, 27 Models, 6, 199–200, 348–350.
See also Localized electron (LE) model; Molecular orbital (MO) model; VSEPR model ball-and-stick, 53, 55 chemical bond as, 347–350 for chemical kinetics, 552–557 kinetic molecular theory as, 200, 208 space-filling, 52–53 Moderator, 861 Molal boiling-point elevation constant, 505 Molal freezing-point depression constant, 505, 506 Molality (m), 486, 488 Molar concentration of a gas, 586, 587, 588 Molar heat capacity, 237 Molarity (M), 133–140, 485, 486, 487 in mmol per mL, 697 temperature and, 486 Molar mass, 86–88 from boiling-point elevation, 505 from freezing-point depression, 507 of ideal gas, 193–194 from osmotic pressure, 508–509 from Raoult’s law, 499 vapor pressure and, 461 Molar volume, of ideal gas, 190–191 Mole, 82–85 of electrons, 801 of gas, 186, 187, 202 Molecular formula, 93–96 Molecularity, 550 Molecular orbital (MO) model, 403–406 of benzene, 414–415, 1008 combined with LE model, 413–415 of diamond, 443–444 of diatomic molecules, 406–413 of graphite, 444–446 of metals, 441–442 of nitric oxide, 412, 910 paramagnetism and, 409–410, 412 of silicon, 450, 451 Molecular sieve, 196 Molecular solids, 435, 436, 454–456, 458 Molecular structure, 367. See also VSEPR model in biologic systems, 378 standard entropy and, 766 Molecular weight, 86, 193 Molecules, 52–53 Mole fraction, 195–198, 485, 486 Mole ratio, 103–106 with limiting reactant, 110, 111, 112, 113 Molybdenum, 949 Momentum in kinetic theory, A14–A15 uncertainty principle and, 291–292 Monatomic gases, 55 Monatomic ions names of, 58, 59, 60, 61 oxidation states of, 156 Monoclinic sulfur, 920 Monodentate ligand, 956, 958 Monomers, 1016 Monoprotic acids, 628, A22 Monosaccharides, 1031–1034. See also Glucose Moore, Jeffrey, 1018 Multiple bonds, 351, 375–376. See also Double bonds; Triple bonds Multiple proportions, law of, 42–43 Multiplication in exponential notation, A2 significant figures, 14, 15 Myoglobin, 973–975, 1029 Names of compounds, 57–67 acids, 66–67, 1014 alcohols, 1010, 1012–1013 aldehydes, 1013 alkanes, 1000–1003, 1005 alkenes, 1005 alkynes, 1006 amines, 1015 benzene derivatives, 1009 binary covalent, 63–65 binary ionic, 58–61, 65–66 carboxylic acids, 1014 coordination compounds, 958–959 cyclic alkanes, 1005 esters, 1015 formulas from, 59 ketones, 1013 with polyatomic ions, 62–63, 65, 66–67 Names of groups, 55, 315, 1010 Names of ions, 58, 59, 60, 61, 62, 65, 66–67 Nanoscale devices, 439 Natta, Giulio, 1022 Natural gas, 196, 252–254, 258.
See also Methane Natural law, 6, 199 Natural logarithms, A5–A6, A7 Neon, 304, 355, 411, 932 Nernst, Walther Hermann, 805 Nernst equation, 804–806 Net ionic equation, 145–146, 150 Network solids, 436, 443–454, 458 Neutralization analysis, 153–154 Neutralization reactions, 150–151 enthalpy change in, 237–238 Neutralization titration, 153 Neutral solutions, 630 of salts, 655, 659, 660 Neutrons, 49–50, 52, 841, 864 in chain reaction, 860–861 in nuclear transformation, 850–851, 852 in radioactive decay, 842–844 Newlands, John, 300 Newton (unit of force), 181 Newton, Isaac, 275 Nickel, 305, 815, 947, 954 Nickel–cadmium battery, 811 Nicotine, 379 Night vision equipment, 279 Niobium, 1, 949 Nitrate ion, 55, 362–363, 375–376, 413, 415 Nitrate salts, 144, 906, 913 Nitric acid, 131, 627, 911–913 in air pollution, 212, 922 in aqua regia, 735, 803 Nitric oxide, 909, 910–911, 912 air pollution and, 212, 213, 559–560, 906 kinetics of production, 527–533 as ligand, 958 molecular orbitals, 412, 910 as odd-electron molecule, 363, 910 paramagnetism, 412, 910 Nitrite anion, 363, 961 Nitrite salts, 913 Nitrito ligand, 961 Nitrogen, 304, 901–913 in fertilizers, 7, 906, 910, 917 liquid, 904 in living systems, 906 molecular structure, 397, 409, 410, 878, 903 oxyacids of, 911–913 preparation of, 110–111, 879–880 as real gas, 208, 210–211 solid state, 358 Nitrogen cycle, 906 Nitrogen dioxide, 909, 911 acidic solution of, 663 in air pollution, 211–213, 906, 922 dimerization of, 579 as odd-electron molecule, 363 reaction kinetics, 527–533, 549–551 Nitrogen fixation, 906 Nitrogen hydrides, 906–909. See also Ammonia Nitrogen monoxide. See Nitric oxide Nitrogen oxides, 909–911. See also specific compounds in air, 211–213, 906 names of, 64 Nitroglycerin, 904–905 Nitro ligand, 961, 968 Nitrosyl ion, 910–911 Nitrosyl ligand, 958 Nitrous acid, 212, 627–628, 639–640, 913, 922 Nitrous oxide, 909–910, 911, 912.
See also Dinitrogen monoxide Nobel, Alfred, 905 Noble gas electron configuration of covalent compounds, 338 of ionic compounds, 339–340 of ions, 341–342, 875–876 in Lewis structures, 354, 355, 356 Noble gases (Group 8A), 55, 931–933. See also specific gases freezing points, 428 inside buckyballs, 80–81 intermolecular forces, 428–429 ionization energies, 310 solid state, 436, 458 Noble metals, 814, 815, 821, 824, 945 Nodes of orbitals, 295 of standing wave, 291 Nonelectrolytes, 129, 130, 133, 149 Nonideal solutions, 502–504 Nonmetals, 55, 315–316, 875–876 anions of, 55, 315, 876 binary covalent compounds of, 63–65 bonding by, 338–339 oxidation of metals by, 316–317 preparation of, 879–880 transition metal compounds of, 945 Nonpolar molecules, 129, 428, 429 Nonpolar solvent, 489, 491–492 Nonstoichiometric compositions, 884 Norepinephrine, 648 A82 Index Normal boiling point, 466 Normal hydrocarbons, 998, 999 Normality (N), 487, 488 Normal melting point, 466, 467 Novocaine, 649 n-type semiconductor, 450–451, 452, 453 Nuclear atom, 49–50 Nuclear binding energy, 857–859 Nuclear fission, 843, 859–863, 866 Nuclear fusion, 850–851, 859, 863 Nuclear physics, 864–865 Nuclear reactors, 861–863, 866 Nuclear stability, 841–849, 856–859 Nuclear transformations, 849–852 Nucleic acids, 1036–1040 Nucleons, 841. See also Neutrons; Protons Nucleosynthesis, stellar, 850–851, 863 Nucleotides, 1036–1037, 1039 Nucleus, 49–50, 841 Nuclides, 841, 842–844 Nylon, 1017, 1020–1021, 1022 Observations, 5, 6, 7. See also Measurement Octahedral complex ions, 956, 960, 961, 966, 967–970 Octahedral holes, 456, 457, 458 Octahedral hybrid orbitals, 399–400, 401, 902, 903 Octahedral structure, 371, 372, 374, 918 Octane rating, 103 Octaves, 300 Octet rule, 355, 356 exceptions to, 358–361 guidelines for, 360 Odd-electron molecules, 363 OIL RIG mnemonic, 160 Oils. See also Petroleum hydrogenation of, 883, 1007 immiscibility in water, 490 from plants, 262–263 Oil shale, 262–263 Open hearth process, 984–985 Operator, 291 Optical isomerism, 961–965 in sugars, 1031–1032, 1034 Orbital diagrams, 302–304 Orbitals. See Atomic orbitals; Hybridization; Molecular orbital (MO) model Order of reactant, 532 Order of reaction, 535, 536, 538 Ores, 821, 878, 879, 978, 979, 980. See also Metallurgy Organic chemistry, 997. See also Hydrocarbons Orthophosphoric acid. See Phosphoric acid Osmium, 51, 949 Osmotic pressure, 508–511, 513 in plants, 1034–1035 Ostwald process, 101, 912–913 Overvoltage, 820 Oxalate ligand, 957, 961 Oxidation. See also Corrosion aging and, 160–161 definition of, 159 Oxidation half-reaction, 162–168 Oxidation numbers. See Oxidation states Oxidation–reduction (redox) reactions, 140, 154–168. See also Electrolysis; Galvanic cells; Oxidation states in acidic solution, 162–166 balancing equations for, 162–168 in basic solution, 166–168 in biologic systems, 973, 977 characteristics of, 158–161 definition of, 154 disproportionation in, 929 electron transfer in, 154, 158–160, 163, 167 equilibrium constants, 807–808 equivalent mass in, 487 between metal and nonmetal, 316–317 in metallurgy, 161–162, 978, 980, 982, 983 Oxidation states, 155–158 compared to charge, 155–156, 157, 363–364 in complex ions, 956, 958, 959 noninteger, 158 rules for, 156 of transition metals, 945, 947–948 Oxide ion, 313, 339–340, 345–346, 663, 876 Oxide minerals, 979, 980 Oxides, 918, 919. See also Corrosion acid–base properties, 662–663, 665, 667, 876 of Group 2A metals, 876, 885, 886 Oxidizing agents, 159–160 in electrochemistry, 791, 793 equivalent mass of, 487 in fireworks, 288, 289 Oxyacids, 627–628, 661–662. See also specific acids of halogens, 67, 928–929 of nitrogen, 911–913 of phosphorus, 916–917 of sulfur, 922–923 Oxyanions, 62, 66–67, 928–929 Oxygen, 304, 919 abundance, 878 in biologic systems, 973–979 early research on, 40, 41 liquid, 410, 919 molecular orbital model, 410–411 oxidation states, 156 paramagnetism, 410, 919 preparation of, 879–880 superoxides as source of, 881 Oxygen difluoride, 929 Oxytocin, 1028 Ozone, 413, 919 in smog formation, 212–213, 559–560 of upper atmosphere, 211, 260, 559, 561, 919, 1004 Paint for corrosion prevention, 814, 815 entropic forces in, 756 pigments in, 4 titanium dioxide in, 950, 951 Palladium, 884, 949, 963 Paper acid decomposition, 666–667 titanium dioxide in, 950, 951 Paper chromatography, 27 Paracelsus, 39 Paramagnetism, 409–410, 412 of coordination compounds, 946, 955, 968, 976 of nitric oxide, 412, 910 of nitrogen dioxide, 911 of oxygen, 410, 919 Partial charge, 128, 332 Partial ionic character, 346–347 Partial pressures Dalton’s law of, 194–199, 202 equilibrium and, 586–588, 594–596, 601–603 solubility and, 494–495 Particle accelerators, 57, 83, 302, 849–850, 864 Particle physics, 864–865 Particles, 277 wave nature of, 281, 282–284, 290 Pascal (Pa), 181 Pasteur, Louis, 962 Pathway, 230, 749 Patina, 814, 955 Pauli, Wolfgang, 298 Pauli exclusion principle, 298, 302, 303 Pauling, Linus, 334, 348 Peat bogs, lead in, 51 Peker, Atakan, 449 Penetration effect, 299, 306 Penicillin, 90–91 Pentoses, 1031, 1032 Peptide linkage, 1027 Percent composition, 89–91 Percent dissociation, 641–644 Percent ionic character, 346 Percent yield, 111–112 Perchlorate ion, 929 Perchloric acid, 627, 928 Periodic table, 55–57, 875–876. See also Elements; Groups alternate version of, 57, 307–308 filling of orbitals in, 302–309 history of, 299–301 information contained in, 314–316 ionic radii and, 341 predictions based on, 300, 301 transuranium elements in, 302–303, 851–852 valence electrons in, 304–305, 307, 308 Periods, 57 atomic properties in, 310, 311, 312, 313, 314 transition metal properties in, 943 Permanganate ion, 953, A18–A19 Peroxide anion, 881 Peroxides, 156, 159, 881, 975 Petroleum, 252–254, 257. See also Gasoline air pollution and, 211, 213 dwindling supply, 214, 229 immiscibility in water, 490 Pewter, 443, 892 pH, 631–634. See also Buffered solutions at equivalence point, 704, 705, 707 indicators and, 713–716 of polyprotic acids, 651–655 of salt solutions, 655–659 solubility and, 724, 728, 729 of strong acid, 634–635 of strong base, 645 of weak acid, 635–644 of weak base, 647–650 Phase diagrams, 467–471, 504, 506 pH curve, 696 indicators and, 711, 716 strong acid–strong base, 699 strong base–strong acid, 699 weak acid–strong base, 704, 705, 707 weak base–strong acid, 711 Phenol, 1012 Phenolphthalein, 152, 711, 715, 716 Phenyl group, 1009 Pheromones, 1–2, 379 Index A83 Phlogiston, 40, 41 pH meter, 631, 633, 711, 807 Phosphate rock, 916, 917, 922 Phosphate salts, solubility of, 144, 724 Phosphides, 914 Phosphine, 378, 914 Phosphorescence, 914 Phosphoric acid, 627–628, 650–652, 916–917, 918, 926 Phosphorous acid, 917 Phosphorus, 901–902, 913–918 bonding in, 411–412, 454, 455, 878, 913–915 in fertilizers, 917, 922 halides of, 917–918 oxides of, 94, 915–916 Phosphorus pentachloride decomposition, 595–596 structure, 360, 373, 397–398, 902, 917–918 Photochemical smog, 213, 906 Photochromic glass, 931 Photoelectric effect, 279–280 Photography, 924, 926 Photons, 278–280, 286–287, 844, 864. See also Electromagnetic radiation Photosynthesis, 154, 229, 252, 254, 259 alternative pathways, 84 chlorophyll in, 973 radiocarbon dating and, 853 Photovoltaic cell, 810 Physical changes, 26, 464 Physical states. See States of matter Pi () bonds, 394–395, 396–397, 402, 1005, 1006, 1008 atomic size and, 447, 877–878, 902 resonance and, 413–414, 415 Piezoelectric substance, 1024 Pig iron, 984, 985 Pigments, 4 Pi () molecular orbitals, 408, 415, 446, 1008 Pipet, 10, 11, 137, 139 pK, 631 pKa of buffered solution, 689, 690, 691, 692, 693, 694, 695 of indicator, 713–714, 716 Planar structure bond polarities and, 337 square, 374, 956, 961, 966, 972–973 trigonal, 368, 371, 375–376, 393, 401, 445 Planck, Max, 278, 280, 285 Planck’s constant, 278, 292 Plane-polarized light, 961–962, 1032 Plants. See also Photosynthesis chemical messengers of, 379 fuel from, 252, 254, 258–260, 262–263 gold in, 40 polysaccharides of, 1034–1036 thermogenic, 238–239 Plastics. See also Polymers blowing agent for, 908 Plastic sulfur, 921 Plating. See Electroplating Platinum in antitumor agents, 963 as catalyst, 559, 949 as electrode, 794, 819 surface reaction on, 546 Platinum group metals, 949 Pleated sheet, 1028–1029 Plum pudding model, 47, 48–49 Plunkett, Roy, 1018 Plutonium, 302, 862–863 p–n junction, 450–451, 452, 454, 810 pOH, 631–632, 633–634, 648 Polar covalent bonds, 332–333 acid strength and, 661 dipole moments and, 335–338 electronegativity and, 334–335 molecular orbitals and, 413 partial ionic character of, 346–347 Polarizability, 428–429, 461 Polarized light, 961–962, 1032 Polar molecules, 128, 129. See also Dipole moments; Hydration; Hydrogen bonding ammonia as, 336, 907 capillary action and, 429 surface tension and, 429 water as, 128, 332–333, 336 Polar side chains, 1027 Polar solvent, 489, 490, 491–492 Polonium, 433, 918, 924 Polyatomic ions, 55, 62–63, 65, 66–67, 346 Polydentate ligands, 957 Polyelectronic atoms, 298–299 Polyester, 1021 Polyethylene, 1017–1019, 1021–1022 Polymers, 1008, 1016–1025. See also Carbohydrates; Nucleic acids; Proteins entropic forces and, 756 ethylene-based, 1017–1019, 1021–1022 historical development of, 1016–1017 ion-exchange resins, 887 of phosphoric acid, 916 products based on, 7, 23, 449, 814, 892, 1018–1019 properties of, 1022–1025 types of, 1017–1021 Polypeptides, 1027–1028 Polypropylene, 1023–1024 Polyprotic acids, 650–655, 682, A23 Polysaccharides, 1034–1036 Polystyrene, 1024–1025 Polyvinyl chloride (PVC), 1025 Polyvinylidene difluoride (PVDF), 1024 p orbitals, 294, 295, 296, 299. See also Pi () bonds filling of, 303–304, 305 Poreda, Robert J., 80 Porous disk, 792, 798 Porphyrin, 973, 974 Portland cement, 892 Positional entropy, 754–755, 756, 762–764, 771 Positron production, 844 Post-it Notes, 7 Potassium, 305, 316, 317–318, 880, 881, 882. See also Alkali metals Potassium dichromate, 165–166 Potassium ferricyanide, 959 Potassium hydrogen phthalate (KHP), 153 Potassium hydroxide, 131, 144, 644, 663. See also Strong bases Potential difference, 800. See also Cell potential Potential energy, 229–230 in bonding, 331, 332 in chemical reaction, 231–232, 749 as internal energy, 232 of nucleus, 841, 856 in quantum mechanics, 291, 298 reaction rates and, 553 Potentiometer, 793, 801, 807 Power plants. See Energy sources Powers of a number, A3 Precipitate, 140 Precipitation, selective, 727–731, 734–736, 981, 982 Precipitation reactions, 140–145, 724–727 net ionic equation for, 145 stoichiometry of, 147–148 Precipitator, electrostatic, 515 Precision, 12–13, 16, A10, A11–A13 Predictions, 5, 6, 199–200 Pressure. See also Ideal gas law; Partial pressures; Vapor pressure atmospheric, 179–180, 181, 211 Boyle’s law, 181–184, 186, 189, 200–201 conversion of units, 181, A26 definition of, 181 in diamond anvil cell, 358 enthalpy and, 235–236 entropy and, 771 in equilibrium expression, 586–588, 594–596, 601–603 equilibrium position and, 607–609 free energy and, 770–773, 774–777 measurement of, 180–181 osmotic, 508–511, 513 phase diagrams and, 467–471 of real gas, 208–211 solubility and, 493–495, 497 in standard state, 246 temperature and, 201 work and, 233–235 Priestley, Joseph, 40, 912 Primary structure, of protein, 1027–1028 Primary valence, 956 Principal quantum number, 293, 299, 308 Printed circuits, 452–453, 454 Probability entropy and, 751–755, 762–764 mixing and, 491 of nuclear decay, 846 Probability distribution, 292–293, 295, 296, 299 Products, 97–98 calculating mass of, 102–106 in solution, 145–146 Propagation of uncertainties, A12–A13 Propane, 998 Propyl group, 1000 Proteins, 1025–1031 coordination complexes and, 973, 974, 975, 977 entropic forces and, 756 enzymes, 557, 558, 562–563 erythropoietin, 978–979 ice formation and, 516 molar mass determination, 87 Protein synthesis, 1038–1040 Proton acceptor, 149, 623, 624, 644, 646 Proton donor, 149, 623, 624, 661, 662 Proton-exchange membrane (PEM), 812 Protons, 49–50, 52, 841, 842–844, 864. See also Hydrogen ions Proust, Joseph, 41 Pseudo-first-order rate law, 547 p–s mixing, 409 A84 Index p-type semiconductor, 450–451, 452, 453 Pure substance, 25 PV work, 234–235 Pyramidal structure, trigonal, 369, 902, 903, 914, 917 Pyridine, 646 Pyroaurite, 814 Pyroelectric substance, 1024 Pyrolytic cracking, 253 Pyrometallurgy, 981 Pyrophoric substance, 913 Pyrophosphoric acid, 916 Q (ion product), 725, 727 Q (reaction quotient), 593–594, 600, 605, 607 cell potential and, 804–806 free energy change and, 772 Quadratic equations, A7–A10 in acid–base problems, 636–637, 654–655 in equilibrium problems, 601, 602 Qualitative analysis, 729–731, 734–736 Qualitative observations, 5 Quantitative analysis, spectroscopic, A16–A19 Quantitative observations, 5 Quantization of energy, 278–280, 285–287, 290 Quantum mechanics, 275, 290–296 electron spin in, 296, 298, 303 of hydrogen molecule, 403, 404 of polyelectronic atoms, 298–299 Quantum model, of Bohr, 285–287, 290, 291, 292, 293 Quantum numbers, 293–294, 296, 298, 299, 308 Quarks, 841, 864 Quartz, 329, 446, 447, 878, 946 Racemic mixture, 965 Radial probability distribution, 292–293, 299 Radiation, electromagnetic. See Electromagnetic radiation Radiation damage, 847, 863–866 Radii atomic, 313–314, 876–878, 948–949 ionic, 340–342 Radioactive waste, 866 Radioactivity, 48, 842–843 dating with, 51, 853–855 detection of, 852–853 kinetics of decay, 846–849 medical applications, 847, 855–856 of polonium, cancer and, 918 types of, 843–845 Radiotracers, 855–856 Radium, 309, 885 Radon, 932 Rads, 864 Rainbow, 284–285 Random-coil arrangement, 1029 Random error, 12, A10 Randomness, entropy and, 751, 757 Range, of measurements, A11 Raoult, François, 498 Raoult’s law for liquid–liquid solutions, 501–504 for nonvolatile solutes, 498–501 Rate definition of, 528, 530, 533 instantaneous, 530 Rate constant, 532, 533, 537, 538, 548 Arrhenius equation for, 555–557 half-life and, 542, 545–546 for nuclear decay, 847 Rate-determining step, 550–551 Rate laws, 532–534. See also Reaction rates for complex reactions, 546–547, 549 determining form of, 534–538, 548 integrated, 533, 534, 538–547, 546, 548–549 for radioactive decay, 847 reaction mechanisms and, 549–552 summary, 534, 548–549 RBE (relative biological effectiveness), 865 Reactants, 97–98 calculating mass of, 102–106 limiting, 106–113, 147–148 order of, 532 in solution, 145–146 Reaction mechanisms, 527, 549–552 Reaction order, 535, 536, 538 Reaction quotient (Q), 593–594, 600, 605, 607 cell potential and, 804–806 free energy change and, 772 Reaction rates, 527–532. See also Rate laws catalysis and, 557–563 definition of, 528, 530, 533 equilibrium and, 581–582, 593, 606, 774 mechanisms and, 527, 549–552 model for, 552–557 temperature and, 552–558 thermodynamics and, 527, 593, 605, 749–750, 770, 774, 778 Reactions, 96–97. See also Acid–base reactions; Chemical equations; Oxidation–reduction (redox) reactions; Precipitation reactions; Stoichiometry bond energies and, 351–353, 749 energy change, 231–232 entropy change, 762–766 extent of, 592–593 free energy change, 766–770, 771–777, 802 types of, 140 Reactor core, 861, 862 Real gases, 208–211 Boyle’s law and, 182–184 collisions in, 205 cooling on expansion, 879 ideal gas law and, 187, 200 molar volumes of, 191 Rectifier, 451 Redox reactions. See Oxidation– reduction (redox) reactions Red phosphorus, 913–914, 915 Reducing agents, 159–160 alkali metals, 316–318, 881 carbon, 879, 980 in electrochemistry, 791, 793 equivalent mass of, 487 in fireworks, 288 hydrogen, 879, 980 in metallurgy, 980 transition metals, 948 Reduction, 159 at cathode, 792 in metallurgy, 978, 980, 982, 983 Reduction half-reaction, 162–168 Reduction potentials, standard, 794–800, 948, A25 Refining, 978, 980, 985 electrorefining, 823–824 Relative solubilities, 721–722 Relativity, special theory of, 280, 856–857 Rem, 865 Representative elements, 307, 314–315, 875–880 abundance, 878–879 atomic radii, 313–314, 876–878 ionization energies, 311 preparation, 879–880 Residual oil, 257 Resonance, 362–367, 375, 377, 413–415 Respiratory chain, 973, 977 Reverse bias, 451, 452 Reverse osmosis, 511 Reverse reaction, rate of, 532, 534 Reversible process, 779 Rhodium, 949 Rhombic sulfur, 920 Ribonucleic acids (RNA), 1036–1037, 1038–1040 Roasting, 161, 606, 980 Rocket fuels, 907–908. See also Space shuttle Roman numerals in complex ion names, 958 in compound names, 59–61, 63, 65, 66 Root mean square velocity, 204–205, 207, 208 Roots of number, A3 of quadratic equation, A7–A10 Rounding, 15, 16 Rubber, 1017, 1022 Rubidium, 305, 316, 317, 880, 881, 882 Rust, 815, 954. See also Corrosion Ruthenium, 949 Rutherford, Ernest, 48–49, 284, 841, 849 Rutile, 950 Salicylic acid, 238, 1015 Salt bridge, 792, 798, 815, 816 Saltlike hydrides, 883–884 Salts. See Ionic compounds (salts) Sand, 3, 892, 978 Saturated hydrocarbons, 997–1005. See also Methane Saturated solution, 717 Scale. See Balance Scandium, 305, 946, 947, 949–950 Scanning tunneling microscope (STM), 2, 438–439 Scheele, Karl W., 40, 332 Schierholz, Otto, 667 Schmidt, Lanny, 258 Schoenbein, Christian, 1016 Schrödinger, Erwin, 290, 291 Schrödinger equation, 291, 293, 298 Scientific method, 5–7, 199. See also Models Scintillation counter, 852–853 Scoville, Wilbur, 414 Screening electrons. See Shielding electrons Scrubbing, 214, 645, 920 Seaborg, Glenn T., 302, 303 Seawater, 26, 511 Secondary structure, of protein, 1028–1029 Secondary valence, 956 Second ionization energy, 309, 311–312 Index A85 Second law of thermodynamics, 755–756, 762, 779 Second-order reaction, 536, 543–546, 548 Seddon, Kenneth R., 494 Seed oil, 263 See-saw structure, 372 Selective precipitation, 727–731, 734–736 in metallurgy, 981, 982 Selenium, 46–47, 918 Seltzer, 40 Semiconductors, 279, 450–454, 810, 892, 918 Semimetals, 316, 875, 890. See also Silicon Semiochemicals, 378–379 Semipermeable membrane, 508, 509–510, 511 Separation of mixtures, 25–27, 196 Shielding electrons, 299, 310, 311, 312, 313 Shotyk, William, 51 Sickle cell anemia, 756, 977 Side chains, 1027 Side reactions, 111 Siderite, 982 Sieve, molecular, 196 Sigma (
) bonds, 394–395, 396–397, 399, 1005, 1006, 1008 resonance and, 413–414 Sigma (
) molecular orbitals, 404–405 Significant figures, 13–16 for logarithms, 631, A5 in measurement, 11, A10 Signs, of thermodynamic quantities, 232–233 Silica, 446–448, 877–878, 905 Silicates, 447, 448 in concrete, 892 Silicon, 46, 47, 446–454, 890, 891, 892 abundance, 878 atomic size, 876–878 in explosives detector, 455 Silicon chip, 47, 452–453 Silver alloy of, 443, 955 corrosion of, 814, 820, 821 crystal structure, 438–441 naming of compounds, 60 in photochromic glass, 931 in photography, 924, 926 physical properties, 944, 945 plating with, 818, 819, 824 salts of, solubility, 144, 724 Silver, Spencer F., 7 Silver cell, 811 Silver sulfide, 814, 820, 821 Single bonds, 351, 352 SI system of units, 8–9, A26 frequency in, 276 pressure in, 181 Slag, 984, 985 Slaked lime. See Calcium hydroxide Slightly soluble substance, 143, 144 Slope of straight line, A6–A7 of tangent, 530–531 Slurry, 214, 257 Smelting, 980 Smog. See Air pollution Snow, artificial, 22–23, 184 Soda ash, 645 Sodium, 304, 316–318, 880, 881–882. See also Alkali metals production of, 824–825 Sodium carbonate, 645 Sodium chloride crystal structure, 344, 456, 457–458 electrolysis of liquid, 824–825 electrolysis of solution, 819–820, 825–826 as electrolyte, 130–131, 435 formation from elements, 154, 158, 159, 316 ionic bonding in, 53–55, 330 ion pairing in solution of, 512–513 isotonic solution, 510–511 solubility, 491 Sodium chloride structure, 344 Sodium hydroxide, 131, 144, 644, 645–646. See also Strong bases production of, 825–826 standardized solution of, 153 Sodium hypochlorite, 643 Sodium ion, 53, 55, 58 Sodium peroxide, 881 Sodium thiosulfate, 926 Solar energy sources, 810 Solar fusion of hydrogen, 863 Solar radiation, greenhouse effect and, 255–256 Solder, 443, 892 Solids. See also Crystalline solids; Glass; Ionic compounds (salts); Metals in chemical equations, 98 fundamental properties of, 25, 425, 458 in heterogeneous catalysis, 558–559 in heterogeneous equilibria, 589 positional entropy of, 754, 763–764 in solubility equilibria, 717 solutions of, 485 standard state, 246, 764 types of, 430, 432, 435–436, 458–459 vapor pressure of, 463, 465–466, 467–468 Solubility, 717–721. See also Precipitation in acidic solution, 724, 728, 729, 730, 735 in basic solution, 724, 728–729, 730 common ion effect on, 718, 722–724 complex ions and, 734–736 energy considerations in, 488–492 of gases, 493–495, 496 pressure and, 493–495, 497 relative, 721–722 structure effects in, 492–493 temperature and, 495–496, 735 in water, 128–129, 143–144, 490–491 Solubility product (Ksp), 717–724, 725–729, A24 Solubility rules, 143–144 Solute, 129 Solutions. See also Aqueous solutions; Colligative properties; Solubility composition of, 133–140, 485–488 definition of, 25, 485 dilution of, 137–140 energy of formation, 488–492, 496, 502, 503 entropy of formation, 754 equations for reactions in, 145–146 liquid–liquid, 501–504 nonideal, 502–504 preparation of, 136–137 reaction types in, 140 standard, 136, 152–153 standard state in, 246 stock, 137 types of, 485 vapor pressures of, 497–504 Solvents, 127, 129, 489, 490, 491–492 s orbitals, 291, 292–296, 299 filling of, 302–303, 305–306 Sottos, Nancy, 1018 Space-filling model, 52–53 Space shuttle, 105, 257, 813, 908 Special theory of relativity, 280, 856–857 Specific heat capacity, 237–238, 239 Spectator ions, 145, 146, 149 Spectrochemical series, 968 Spectrophotometer, A16–A17, A18–A19 Spectroscopy, A16–A19 liquid structure and, 430 Spectrum. See Emission spectrum Speed. See Rate; Velocity Spheres, packing of, 436–441, 456–458 Spin, electron, 296, 298, 303 Spinneret, 1017 Splitting of 3d energies, 968–973 Spontaneous fission, 843 Spontaneous processes, 749–754.
See also Entropy cell potential and, 799, 802–803 entropy change and, 751, 754–755, 756 free energy and, 759–762, 770, 773–774, 778 rates of, 527 temperature and, 756–762 work and, 778, 801 sp orbitals, 395–397, 401 in alkynes, 1006 in complex ions, 967 sp2 orbitals, 393–395, 401 in alkenes, 1005 in aromatic hydrocarbons, 1008 sp3 orbitals, 391–393, 401 in alkanes, 998 in complex ions, 966 in Group 4A, 890 Sports drinks, 514–515 Square planar structure, 374 of complex ion, 956, 961, 966, 972–973 Square roots, A3 Stability constants, 731–734 Stahl, Georg, 40 Stainless steel, 815–816, 954, 986 Stalactites and stalagmites, 680, 681, 724 Standard atmosphere (atm), 181 Standard deviation, A12 Standard enthalpy of formation, 246–252, 767, A19–A22 Standard free energy change, 766–770, 772–773 equilibrium constant and, 775, 777–778 Standard free energy of formation, 769, 772, A19–A22 Standard hydrogen electrode, 794 Standard reduction potentials, 794–800, 948, A25 Standard solution, 136, 152–153 Standard states, 246 cell potentials and, 795 enthalpy and, 246, 247 entropy and, 764 free energy and, 766 A86 Index Standard temperature and pressure (STP), 191 Standing wave, 290–291, 296 Starch, 1034–1035 Stars, 80, 850–851, 863 State function, 230–231 energy as, 231, 342 enthalpy as, 235, 242, 247 entropy as, 764 free energy as, 767 State property. See State function States of matter, 25, 425, 426 in chemical equation, 98 entropy of, 754 Stationary phase, 27 Staudinger, Hermann, 1017 Steam, 25, 26. See also Water vapor Steel, 443 alloying metals in, 443, 815–816, 954, 984, 985, 986 carbon in, 443, 816, 879, 986 composition of various types, 986 corrosion of, 813, 814–816 galvanized, 815, 928, 955 heat treatment of, 986 plating of, 815, 824, 893 production of, 984–985 from Titanic, 444–445 Stellar nucleosynthesis, 850–851, 863 Stereoisomerism, 960, 961–965 in alkenes, 1006 in sugars, 1031–1032, 1034 Steric factor, 555 Stock solutions, 137 Stoichiometric point. See Equivalence point Stoichiometric quantities, 106 Stoichiometry, 102–106 of acid–base reactions, 149–151 average atomic mass in, 79 definition of, 77 of electrolytic processes, 816–818 of gases, 190–194 with limiting reactant, 106–113, 147–148 of precipitation reactions, 147–148 in solutions, 133 summarized, 104, 113 Straight-chain hydrocarbons, 998, 999 Strong acids, 626, 627, 628 added to buffered solution, 691–692, 693–694 pH calculations, 634–635 reaction with strong base, 149 as strong electrolytes, 131 titration with strong base, 696–699, 705, 716 Strong bases, 644–646 pH calculations, 645 reaction with acid, 149 as strong electrolytes, 131 titration with strong acid, 699 Strong electrolytes, 129, 130–131, 145 Strong-field case, 968, 971 Strong force, 864 Strontium, 305, 885, 886 Strontium-90, 847, 865 Structural formula, 52, 53 Structural isomerism, 960–961 in alkanes, 998–1000, 1001–1002 Subcritical fission reaction, 860 Sublimation, 343, 463, 467–468, 469, 470–471 Subshells, 294 Substitutional alloys, 443 Substitution reactions of alkanes, 1003 of benzene, 1008–1009 Substrate, 562 Subtraction in exponential notation, A2–A3 significant figures, 14–15 Successive approximations, A8–A10 Sucrose, 133, 435, 923, 1034 Sugars, 1031–1034. See also Sucrose fermentation of, 263, 891, 1011–1012 in nucleic acids, 1036, 1037 in sports drinks, 514–515 Sulfate ion, Lewis structures, 364–366 Sulfate minerals, 979 Sulfate salts, solubility, 144 Sulfides, 918, 923 Sulfide salts, 144, 728–730, 735 Sulfites, 922, 924 Sulfur, 308, 920–924 bonding in, 454, 455, 878, 920–921 in steel, 444 Sulfur dioxide, 921, 922 acid rain and, 212–214 catalytic oxidation of, 559 from coal burning, 213–214, 254 dissolved in water, 663 from metallurgy, 980, 981 scrubbing of, 214, 645, 920 VSEPR model, 376–377 Sulfur hexafluoride, 359–360, 399 Sulfuric acid, 922–923 in acid rain, 212–213, 214, 559, 922 in cleaning solution, 953 as dehydrating agent, 913, 922–923 equilibrium calculations, 653–655 in lead storage battery, 809–810 molecular structure, 663 as strong acid, 131, 627, 653 from sulfur trioxide, 663, 665, 922 Sulfur monoxide, 921 Sulfurous acid, 922 Sulfur trioxide, 921–922 air pollution and, 214, 559 bond polarities, 337–338 reaction with water, 663, 665 Sun. See Solar energy sources Sunglasses, automatic, 931 Supercooled water, 466, 516 Supercritical fission reaction, 860, 862 Superheated liquid, 466 Supernova explosion, 851 Superoxide dismutase (SOD), 160–161 Superoxides, 881 Superphosphate of lime, 917 Superplasticizers, 892 Surface alloys, 816 Surfaces, 438–439, 546, 558–559 Surface tension, 429, 951 Surroundings, 231, 755, 757–759, 762, 779 Suspensions, 514–515 Swartzentruber, Brian, 438 Syndiotactic chain, 1023 Syngas, 256–257 System, 231, 755, 762, 779, 800 Systematic error, 12, 13, A10 Taconite, 982 Tangent line, 530–531 Tanning lotions, 1032–1033 Tantalum, 949 Tarnish, 814, 820, 821 Technetium-99m, 847, 856 Teeth, 159, 717, 720, 975 Teflon, 1018 Tellurium, 918 Temperature. See also Heat absolute zero of, 185, 764 altitude and, 211 autoionization and, 631 catalysis and, 557–558 changes of state and, 463–466 Charles’s law and, 184–185, 186, 189, 201 entropy and, 764 equilibrium constant and, 605, 609–610, 777–778 heat and, 230 heat capacity and, 237–238 as intensive property, 238 in kinetic molecular theory, 200–202, 203–206, A16 measurement of, 19–23 phase diagrams and, 467–471, 504, 506 pressure and, 201 pyroelectric material and, 1024 reaction rate and, 552–558 of real gas, 208–210 solubility and, 495–496, 735 spontaneity and, 756–762 vapor pressure and, 461–463, 465–466 Tempering, of steel, 986 Termolecular step, 550 Tertiary structure, of protein, 1030 Tetraethyl lead, 253–254, 893 Tetrahedral complex ions, 956, 966, 970–972 Tetrahedral holes, 456–457 Tetrahedral hybrid orbitals, 391–393, 401, 402 in alkanes, 998 in Group 5A cations, 902, 903, 918 in Group 4A elements, 890 Tetrahedral structure, 368, 369–370, 371 bond polarities and, 337 distorted, 378 Tetrahedron, 347 Tetraphosphorus decoxide, 916 Tetraphosphorus trisulfide, 915 Tetrathionate ion, 924 Thallium, 856, 888, 890 Theoretical yield, 111, 112 Theory, 6, 7, 199. See also Models Thermal conductivity, 436, 441, 442, 944 Thermal energy, 757. See also Heat Thermal pollution, 496 Thermite reaction, 251 Thermodynamics, 232 absolute values of functions in, 763–764 compared to kinetics, 527, 593, 605, 749–750, 770, 774, 778 first law, 232, 233, 749 second law, 755–756, 762, 779 sign convention in, 232–233 third law, 764 Thermodynamic stability, of nucleus, 841, 856–859 Thermogenic plants, 238–239 Thermophotovoltaics (TPV), 810–811 Thermoplastic polymers, 1017, 1021, 1022 Thermoset polymer, 1017 Thiocyanate, 40, 961 Index A87 Thiosulfate ion, 924, 926 Third law of thermodynamics, 764 Thomson, J. J., 45–47, 48, 284 Three-center bonds, 888 Thresh, L. T., 414 Threshold model, of radiation damage, 866 Thundat, Thomas, 455 Thymol blue, 715, 716 Tin, 879, 890, 891, 892–893, 894 plated on steel, 815, 824, 893 Tin disease, 892 Titanic, 444–445 Titanium, 305, 946, 947, 950 in bicycles, 952–953 on ships’ hulls, 816 Titanium dioxide, 950, 951 Titrant, 151, 152, 696 Titration curve. See pH curve Titrations. See Acid–base titrations Tooth. See Teeth Torr, 181 Torricelli, Evangelista, 180, 181 Trailing zeros, 13, 14 Transfer pipet, 137 Transfer RNA (tRNA), 1039, 1040 trans isomer, 961, 963, 965, 1006 Transistors, 452–453, 892 Transition metals, 55, 875. See also Complex ions 3d (first-row), 946–948, 949–955, 973 4d (second-row), 948–949 5d (third-row), 948–949 abundance, 878–879 binary ionic compounds of, 60, 61, 66 in biologic systems, 973–977 electron configurations, 305–306, 308, 315, 946–947 in gems, 970–971 general properties, 943–946 interstitial hydrides of, 884–885 ionization energies, 948 oxidation states, 945, 947–948 reducing ability, 948 strategic importance of, 943 Transition state, 553 Transmittance, A17 Transuranium elements, 302–303, 851–852 Triads, 300 Trigonal bipyramidal structure, 371–372, 401, 402–403, 902, 903, 917 Trigonal holes, 456 Trigonal planar structure, 368, 371, 375–376, 393, 401, 445 Trigonal pyramidal structure, 369, 902, 903, 914–915, 917 Triiodide ion, 361, 375, 398–399 Trinitrotoluene (TNT), 905 Triple bonds, 351, 352, 376, 396–397, 401–402 of nitrogen, 397, 410, 878, 903 Triple phosphate, 917 Triple point, 468, 470 Tripolyphosphoric acid, 916 Triprotic acid, 650–652 Troposphere, 211 Tsapatsis, Michael, 196 T-shaped structure, 372 Tungsten, 946 Tyndall effect, 514 Uhlenback, George, 296 Ultraviolet radiation, 276 chlorination reactions and, 1003 hydrogen–chlorine cannon and, 926 ozone layer and, 211, 561, 919 titanium dioxide and, 951 Unbranched hydrocarbons, 998, 999 Uncertain digit, 11 Uncertainty in measurement, 10–13, 16, A10–A13 Uncertainty principle, 291–292, 296 Unidentate ligand, 956, 958 Unimolecular step, 550 Unit cells, 432 closest packing and, 436–441, 456–458 Unit factor method, 16–19 Units of measurement, 5, 8–10 conversions, 16–19, A26 in equilibrium constant, 583, 584 for pressure, 180–181 for temperature, 19–23 Universal gas constant (R), 187, 204–205 Unsaturated hydrocarbons, 997–998, 1004, 1005–1010.
See also Ethylene hydrogenation of, 558–559, 883, 998, 1007 Uranium fission of, 859–860, 861–862, 866 geologic dating with, 854–855 metallurgy of, 982 radioactive decay of, 48, 843, 844, 845 Urea, 997 Valence electrons, 304–305, 307.
See also Lewis structures chemical properties and, 308, 314 ionization energies and, 309–310 of metal, 441–442 of semiconductor, 450 of stable compound, 338–339 Valence shell electron-pair repulsion model. See VSEPR model Vanadium, 305, 946, 947, 950–951 Vanadium steel, 950 van der Waals, Johannes, 208 van der Waals equation, 210–211 van Gastel, Raoul, 438 van’t Hoff, J. H., 512 van’t Hoff factor (i), 512–513 Vapor, definition of, 459 Vaporization, 459 enthalpy change, 425, 426, 459, 461–463 entropy change, 756–757 free energy change, 761–762 of water, 425, 756–757 Vapor pressure, 459–463 changes of state and, 465–466 phase diagram and, 467–468, 469, 504 of solids, 463, 465–466, 467–468 of solutions, 497–504, 506 of superheated liquid, 466 temperature dependence of, 461–463, 465–466 of water, 198, 461–462, 463, 465 Vasopressin, 1028 Velocity distribution of, in gas, 205–206, 553–554 kinetic energy and, 203, 230 in kinetic molecular theory, 203, 204–206, 207, A13–A16 root mean square, 204–205, 207, 208 Viscosity, 429–430, 431, 1023 Visible light. See Light Vitamin E, 160 Vitamins, solubility, 492–493 Volatile liquids, 460 Volt (V), 793, 800 Voltaic cells. See Galvanic cells Voltmeter, 793, 801 Volume density and, 24 enthalpy and, 235–236 equilibrium position and, 607–609 measurement of, 9, 10–13 molar, of ideal gas, 190–191 molarity and, 133–140 moles of gas and, 186, 187, 202 pressure of gas and, 181–184, 200–201 states of matter and, 25 temperature of gas and, 184–185, 201 units of, 9, 16, A26 work and, 233–235 Volumetric analysis, 151 Volumetric flask, 10, 136, 137, 139 Volumetric pipet, 137 VSEPR model, 354, 367–378 complex ions and, 966 evaluation of, 377–378 Group 5A elements and, 902, 903 Lewis structures in, 369, 377 in localized electron model, 391, 400 lone pairs in, 370–371, 375, 376–377 multiple bonds and, 375–376 names of structures in, 372–373 with no central atom, 377 resonance and, 375, 377 steps in, 369 summary of, 377 V-shaped structure, 371, 372, 376–377 Vulcanization, 1017 Waage, Peter, 582 Walsh, William, 893 Water, 3–4. See also Aqueous solutions; Hydration; Ice; Solubility as acid, 629, 646 as amphoteric substance, 629–631 autoionization of, 629–631, 635, 646 as base, 623–624, 626, 627, 629, 657 boiling point, 427, 464, 465, 466, 469 bonds in, 127–128 capillary action of, 429 changes of state, 26, 425, 426 cooling by, 459, 514 as covalent hydride, 884 density of, 425, 426, 469 desalination of, 26, 511 electrolysis of, 3–4, 27, 258, 809–810, 818–819, 883 entropy of, 765–766 heating curve, 463–464 hydrogen bonding in, 426–427, 454, 456, 459, 461, 464, 490, 491, 884 Lewis structure, 355–356 as ligand, 958, 968 meniscus of, 429 naming of, 64 in natural mixtures, 26 as nonelectrolyte, 149 phase diagram, 467–469, 504, 506 A88 Index Water (continued) as polar molecule, 128, 332–333, 336 purification of, 919 softening of, 645, 887 as solvent, 128–129, 143–144, 490–491 states of, 25, 26, 425, 426 supercooled, 466, 516 vaporization of, 459, 756–757 vapor pressure, 198, 461–462, 463, 465 VSEPR model, 369–370 Water pollution, 156–157 Waters of hydration, 213 Water-soluble vitamins, 492, 493 Water vapor, 425, 426, 464. See also Vapor pressure, of water atmospheric, 255 condensation of, 179–180 on phase diagram, 467–468, 469, 504 Wave function, 291, 292–293. See also Atomic orbitals; Molecular orbital (MO) model of molecular orbital, 404 Wavelength of electromagnetic wave, 275–276, 277 of particle, 281, 282–284 Wave mechanics. See Quantum mechanics Waves, 275–276, 277 diffraction of, 281–282, 432–434 interference of, 282, 432 particles and, 280–281, 282–284, 290 standing, 290–291 Weak acids, 627–628 in buffered solutions, 684–686, 687–690, 692–696 carboxylic, 1014–1015 with common ion, 681–683 conjugate base of, 627, 656 definition of, 627 equilibrium problems, 635–644 indicators, 711–716 Ka calculation, 643–644, 707–709 mixtures of, 639–641 percent dissociation, 641–644 pH calculations, 635–644 reaction with strong base, 149 salts as, 657–659 titration with strong base, 153–154, 700–709, 716 as weak electrolytes, 131–132 Weak bases, 646–650 in buffered solutions, 684, 686, 690–693 with common ion, 682 conjugate acid of, 646, 658 salts as, 656–657 strengths of, 628–629, 657 titration with strong acid, 709–711 as weak electrolytes, 132–133 Weak electrolytes, 129–130, 131–133 Weak-field case, 968, 971 Weak force, 864 Weight, 9–10, 11, 12 atomic, 44, 79 molecular, 86, 193 Weight percent, 89, 485 Wentorf, Robert H., Jr., 470 Werner, Alfred, 956, 960 Wet process, 916 White, Scott, 1018 White phosphorus, 454, 455, 878, 913, 914, 915, 916 White tin, 892 Wind power, 258–259 Wöhler, Friedrich, 997 Wood, as energy source, 252, 253 Work definition of, 230 electrochemical, 791, 793, 800–801, 816 energy and, 229, 230, 231 free energy and, 778–779 by gas, 233–235 internal energy and, 232 sign of, 233, 800–801 Xenon, 932–933 Xenon difluoride, 402–403, 933 Xenon tetrafluoride structure, 374, 400, 933 synthesis, 243–244, 932 Xenon trioxide, 366, 932, 933 X-ray diffraction, 281–282, 432–434 Yield, 111–112 Z (atomic number), 50, 52, 55, 841, 842 Zero-order reaction, 546 Zeros, in calculations, 13, 14 Ziegler, Karl, 1022 Ziegler-Natta catalyst, 1022, 1024 Zinc, 305, 947, 955 in galvanized steel, 815, 955 metallurgy of, 982 naming of compounds, 60 Zirconium, 949 Zone of stability, 842, 844 Zone refining, 980 Zuo-Fen Zhang, 889 58 Ce 140.1 90 Th 232.0 59 Pr 140.9 91 Pa (231) 60 Nd 144.2 92 U 238.0 61 Pm (145) 93 Np (237) 62 Sm 150.4 94 Pu (244) 63 Eu 152.0 95 Am (243) 64 Gd 157.3 96 Cm (247) 65 Tb 158.9 97 Bk (247) 66 Dy 162.5 98 Cf (251) 67 Ho 164.9 99 Es (252) 68 Er 167.3 100 Fm (257) 69 Tm 168.9 101 Md (258) 70 Yb 173.0 102 No (259) 71 Lu 175.0 103 Lr (260) 1 H 1.008 3 Li 6.941 11 Na 22.99 19 K 39.10 37 Rb 85.47 55 Cs 132.9 87 Fr (223) 4 Be 9.012 12 Mg 24.31 20 Ca 40.08 38 Sr 87.62 56 Ba 137.3 88 Ra 226 21 Sc 44.96 39 Y 88.91 57 La 138.9 89 Ac† (227) 22 Ti 47.88 40 Zr 91.22 72 Hf 178.5 104 Rf (261) 23 V 50.94 41 Nb 92.91 73 Ta 180.9 105 Db (262) 24 Cr 52.00 42 Mo 95.94 74 W 183.9 106 Sg (263) 25 Mn 54.94 43 Tc (98) 75 Re 186.2 107 Bh (264) 26 Fe 55.85 44 Ru 101.1 76 Os 190.2 108 Hs (265) 27 Co 58.93 45 Rh 102.9 77 Ir 192.2 109 Mt (268) 110 Ds (271) 111 Rg (272) 112 Uub 113 Uut 115 Uup 114 Uuq 28 Ni 58.69 46 Pd 106.4 78 Pt 195.1 29 Cu 63.55 47 Ag 107.9 79 Au 197.0 30 Zn 65.38 3 4 5 6 7 8 9 10 11 12 48 Cd 112.4 80 Hg 200.6 31 Ga 69.72 49 In 114.8 81 Tl 204.4 5 B 10.81 13 Al 26.98 32 Ge 72.59 50 Sn 118.7 82 Pb 207.2 6 C 12.01 14 Si 28.09 33 As 74.92 51 Sb 121.8 83 Bi 209.0 7 N 14.01 15 P 30.97 34 Se 78.96 52 Te 127.6 84 Po (209) 8 O 16.00 16 S 32.07 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9 85 At (210) 10 Ne 20.18 18 Ar 39.95 36 Kr 83.80 54 Xe 131.3 86 Rn (222) 2 He 4.003 1A 2A Transition metals 3A 4A 5A 6A 7A 8A 1 2 13 14 15 16 17 18 Alkali metals Alkaline earth metals Halogens †Actinides Lanthanides Group numbers 1–18 represent the system recommended by the International Union of Pure and Applied Chemistry.
metals nonmetals Noble gases Periodic Table of the Elements Table of Atomic Masses The values given here are to four significant figures where possible.
§A value given in parentheses denotes the mass of the longest-lived isotope.
Atomic Atomic Element Symbol Number Mass Actinium Ac 89 § Aluminum Al 13 26.98 Americium Am 95 Antimony Sb 51 121.8 Argon Ar 18 39.95 Arsenic As 33 74.92 Astatine At 85 Barium Ba 56 137.3 Berkelium Bk 97 Beryllium Be 4 9.012 Bismuth Bi 83 209.0 Bohrium Bh 107 Boron B 5 10.81 Bromine Br 35 79.90 Cadmium Cd 48 112.4 Calcium Ca 20 40.08 Californium Cf 98 Carbon C 6 12.01 Cerium Ce 58 140.1 Cesium Cs 55 132.90 Chlorine Cl 17 35.45 Chromium Cr 24 52.00 Cobalt Co 27 58.93 Copper Cu 29 63.55 Curium Cm 96 Darmstadtium Ds 110 Dubnium Db 105 Dysprosium Dy 66 162.5 Einsteinium Es 99 Erbium Er 68 167.3 Europium Eu 63 152.0 Fermium Fm 100 Fluorine F 9 19.00 Francium Fr 87 Gadolinium Gd 64 157.3 Gallium Ga 31 69.72 Germanium Ge 32 72.59 Atomic Atomic Element Symbol Number Mass Gold Au 79 197.0 Hafnium Hf 72 178.5 Hassium Hs 108 Helium He 2 4.003 Holmium Ho 67 164.9 Hydrogen H 1 1.008 Indium In 49 114.8 Iodine I 53 126.9 Iridium Ir 77 192.2 Iron Fe 26 55.85 Krypton Kr 36 83.80 Lanthanum La 57 138.9 Lawrencium Lr 103 Lead Pb 82 207.2 Lithium Li 3 6.9419 Lutetium Lu 71 175.0 Magnesium Mg 12 24.31 Manganese Mn 25 54.94 Meitnerium Mt 109 Mendelevium Md 101 Mercury Hg 80 200.6 Molybdenum Mo 42 95.94 Neodymium Nd 60 144.2 Neon Ne 10 20.18 Neptunium Np 93 Nickel Ni 28 58.69 Niobium Nb 41 92.91 Nitrogen N 7 14.01 Nobelium No 102 Osmium Os 76 190.2 Oxygen O 8 16.00 Palladium Pd 46 106.4 Phosphorus P 15 30.97 Platinum Pt 78 195.1 Plutonium Pu 94 Polonium Po 84 Potassium K 19 39.10 Atomic Atomic Element Symbol Number Mass Praseodymium Pr 59 140.9 Promethium Pm 61 Protactinium Pa 91 Radium Ra 88 226 Radon Rn 86 Rhenium Re 75 186.2 Rhodium Rh 45 102.9 Roentgenium Rg 111 Rubidium Rb 37 85.47 Ruthenium Ru 44 101.1 Rutherfordium Rf 104 Samarium Sm 62 150.4 Scandium Sc 21 44.96 Seaborgium Sg 106 Selenium Se 34 78.96 Silicon Si 14 28.09 Silver Ag 47 107.9 Sodium Na 11 22.99 Strontium Sr 38 87.62 Sulfur S 16 32.07 Tantalum Ta 73 180.9 Technetium Tc 43 Tellurium Te 52 127.6 Terbium Tb 65 158.9 Thallium Tl 81 204.4 Thorium Th 90 232.0 Thulium Tm 69 168.9 Tin Sn 50 118.7 Titanium Ti 22 47.88 Tungsten W 74 183.9 Uranium U 92 238.0 Vanadium V 23 50.94 Xenon Xe 54 131.3 Ytterbium Yb 70 173.0 Yttrium Y 39 88.91 Zinc Zn 30 65.38 Zirconium Zr 40 91.22 Bond Energies 351 Electron Configurations of the Elements 307 Ionization Constants of Acids and Bases 628, 647, 651, A24–A25 Reduction Potentials 796, A26 Solubility Products 718, A25 Thermodynamic Data A21–A23 Vapor Pressures of Water 461 Page Numbers of Some Important Tables Constant Symbol Value Atomic mass unit amu Avogadro’s number N Bohr radius Boltzmann constant k Charge of an electron e Faraday constant F Gas constant R Mass of an electron Mass of a neutron Mass of a proton Planck’s constant h Speed of light c 2.99792458 108 m/s 6.62608 1034 J s 1.00728 amu 1.67262 1027 kg mp 1.00866 amu 1.67493 1027 kg mn 5.48580 104 amu 9.10939 1031 kg me 0.08206 L atm/K mol 8.31451 J/K mol 96,485 C/mol 1.60218 1019 C 1.38066 1023 J/K 5.292 1011 m a0 6.02214 1023 mol1 1.66054 1027 kg Physical Constants |
7572 | https://pubmed.ncbi.nlm.nih.gov/6380304/ | Effect of epinephrine on glucose metabolism in humans: contribution of the liver - PubMed
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. 1984 Aug;247(2 Pt 1):E157-65.
doi: 10.1152/ajpendo.1984.247.2.E157.
Effect of epinephrine on glucose metabolism in humans: contribution of the liver
R S Sherwin,L Saccà
PMID: 6380304
DOI: 10.1152/ajpendo.1984.247.2.E157
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Effect of epinephrine on glucose metabolism in humans: contribution of the liver
R S Sherwin et al. Am J Physiol.1984 Aug.
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. 1984 Aug;247(2 Pt 1):E157-65.
doi: 10.1152/ajpendo.1984.247.2.E157.
Authors
R S Sherwin,L Saccà
PMID: 6380304
DOI: 10.1152/ajpendo.1984.247.2.E157
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Abstract
Epinephrine causes a prompt increase in blood glucose concentration in the postabsorptive state. This effect is mediated by a transient increase in hepatic glucose production and an inhibition of glucose disposal by insulin-dependent tissues. Epinephrine augments hepatic glucose production by stimulating glycogenolysis and gluconeogenesis. Although its effect on glycogenolysis rapidly wanes, hyperglycemia continues because the effects of epinephrine on gluconeogenesis and glucose disposal persist. Epinephrine-induced hyperglycemia is markedly accentuated by concomitant elevations of glucagon and cortisol or in patients with diabetes. In both cases, the effect of epinephrine on hepatic glucose production is converted from a transient to a sustained response, thereby accounting for the exaggerated hyperglycemia. During glucose feeding, mild elevations of epinephrine that have little effect on fasting glucose levels cause marked glucose intolerance. This exquisite sensitivity to the diabetogenic effects of epinephrine is accounted for by its capacity to interfere with each of the components of the glucoregulatory response, i.e., stimulation of splanchnic and peripheral glucose uptake and suppression of hepatic glucose production. Our findings suggest that epinephrine is an important contributor to stress-induced hyperglycemia and the susceptibility of diabetics to the adverse metabolic effects of stress.
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7573 | https://www.processon.com/view/652faf546f082c588ffa6f3d | 管理经济学 第8版 贝叶
2023-11-07 19:06:28 2.86k 1 举报
第八章
完全竞争市场、垄断市场
与垄断竞争市场中的管理
导言
首先对某行业中管理者的产量决策进行分析。
最简单的情况是：管理决策对市场价格不会产生显著影响。
先分析完全竞争市场的产量决策，然后分析有市场势力的厂商（垄断和垄断竞争市场）的产量决策。
完全竞争
完全竞争市场（perfetly competitive market）中的产量决策。
完全竞争的基本特征如下：
1、市场上有众多买者和卖着，它们相对于市场来说都很”小“。
2、每个厂商生产同质（相同）的产品。
3、买者和卖着获得完全市场信息。
4、市场中没有交易成本。
5、可以自由进入市场。
总之，上述前四个假设表明没有哪个厂商可以影响产品的价格。
在完全竞争市场中，所有厂商的产品价格相同——该价格取决于市场上所有买者和卖着的相互作用。
自由进出的假设，意味着完全竞争市场中厂商的长期经济利润为零。
市场需求和厂商需求
完全竞争市场中的管理者
制定利润最大化的产量决策时，
首先要了解整个市场的需求
和单个厂商的需求之间的关系。
市场均衡价格Pe由市场供给曲线和市场需求曲线共同作用决定。
对于单个厂商来说，它可以按照价格Pe销售任何数量的产品，
所以单个厂商的需求曲线是水平直线（标记为Df）。
单个厂商需求曲线具有完全弹性，
说明如果厂商价格高于市场价格将销售不出去产品。
单个厂商的价格决策毫无价值：
它的定价定于行业中其他厂商的定价。
所以只能考虑产量决策以实现利润最大化。
短期产量决策
短期指存在固定生产要素的期间。为了实现短期利润最大化，
管理者须在给定的固定投入（即固定成本）下，
改变可变投入并制定产量决策。利润最大化：
在完全竞争市场，
单个厂商的需求曲线就是产品的市场价格（标记为P）。
假如用Q代表厂商产量，产量为Q单位时厂商总收益R=PQ。
由于每单位产量只能以市场价格P销售，
因此每单位产品所增加的收益为P美元。
完全竞争厂商的收益和产量之间存在线性关系。
边际收益（marginal revenue）
是最后一单位产量带来的收益变化，
即收益曲线的斜率。
完全竞争厂商的边际收益就是市场价格。
微积分表达式边际收益是收益函数的导数。
假定收益与产量的函数关系为R=R（Q）
那么MR=dR/dQ
原理 完全竞争厂商的需求
完全竞争厂商的产品需求曲线是一条等于市场价格的水平线。
这个价格也是完全竞争厂商的边际收益。Df=P=MR
微积分表达式边际收益是收益函数的导数。
假设完全竞争厂商的收益函数为：R=PQ
P为均衡价格。因此MR=dR/dQ=P
完全竞争厂商的利润等于收益与成本之差：
Π=PQ-C(Q)
利润由成本曲线与收益曲线之间的垂直距离决定。
原理 完全竞争产量法则
为实现利润最大化，
完全竞争厂商的产量应保证价格等于边际成本：
P=MC（Q）
微积分表达式
完全竞争厂商的利润为：Π=PQ-C(Q)
实现利润最大化的一阶条件是边际利润为零：
亏损最小化
短期经营亏损 存在固定成本的情况下，假设市场价格Pe
处于平均总成本曲线之下，但位于平均变动成本曲线之上。
如果该厂商的产量为Q使得Pe=MC，则亏损为阴影区域。
由于价格高于平均变动成本，
每单位销售产生的收益大于每单位投入的变动成本，
因此短期内厂商应该继续生产（即使出现亏损）。
该厂商即使停止经营也得支付固定成本。
价格大于产量为Q时的平均变动成本，
厂商赚取的单位销售收益足以弥补单位生产成本。通过生产Q
单位产品，厂商可以用超过变动成本的收入弥补固定成本的支出。
简言之，即使厂商遭受短期经营损失，
该损失仍小于完全停产所造成的固定成本亏损。
停产决策 假定市场价格低于平均变动成本，
厂商的产量为Q使得P=MC，
则产生的亏损为两个阴影矩形面积之和。
即每销售一单位产品，厂商将损失ATC(Q)-Pe
每单位损失乘以Q即两个阴影矩形面积之和。
假定厂商停产而不生产Q单位的产品，
则亏损等于固定成本（即使不生产也须支付）。
最上面的矩形表示固定成本，该矩形的面积为：
其等于固定成本。
因此当价格低于平均变动成本时，
厂商停产（生产零单位产品）的损失
小于生产Q单位产品的损失。
原理 完全竞争市场的短期决策
为实现短期利润最大化，
当P≥AVC时，完全竞争厂商应该在边际成本增长阶段生产，此时P=MC
厂商和行业的短期供给曲线
完全竞争厂商的利润最大化产量位于价格等于边际成本处。
当价格在P0~P1之间时，
产量由价格和边际成本的交点决定。
但是当价格位于AVC曲线之下时，
厂商的产量为零，因为边际收益不能弥补变动成本。
所以，完全竞争厂商在任意价格下的产量
就是边际成本等于价格的产量水平。
为了保证产量为正，
价格必须高于平均变动成本曲线。
原理 厂商短期供给曲线
完全竞争厂商的短期供给曲线是AVC最低点之上的边际成本曲线。
完全竞争市场（或行业）
的供给曲线反映了任意价格水平的市场总产量，
这与单个厂商的供给曲线密切相关。
完全竞争行业的市场供给曲线
就是所有厂商各自AVC最低点之上的边际成本曲线的水平加总。
行业供给曲线比单个厂商的供给曲线平坦；
行业中厂商越多，市场供给曲线的位置越靠右。
长期决策
进出自由是完全竞争市场的重要假设。
如果厂商短期能够获取经济利润，
长期内将吸引其他厂商进入并瓜分经济利润。
随着大量厂商进入，行业供给曲线向右移动。
从S0移到S1，市场均衡价格从P0移到P1。
这使单个厂商的需求曲线下移，单个厂商利润降低。
如果完全竞争市场的厂商在短期内遭受亏损，
它们可能因无法弥补机会成本而退出该行业。
随着厂商的退出，
市场供给曲线从S0下降到S2，
导致市场价格从P0上升到P2。
这使单个厂商的需求曲线上移，
行业内留存厂商的利润增加。
上述过程会一直持续到市场价格使市场上所有厂商的经济利润为零。
价格Pe每个厂商的收入正好弥补产品的平均成本
（从长期看无固定成本和变动成本之分），经济利润为零。
如果经济利润大于零，其他厂商将进入市场，导致价格下降；
如果经济利润小于零，厂商退出，
导致市场价格上涨，直至单个厂商的需求曲线与曲线AC相切。
原理 长期竞争均衡
从长期来看，完全竞争厂商的产量满足：
1、P=MC
2、P=AC的最小值
完全竞争市场的长期均衡有两个重要的福利含义：
首先，市场价格等于产品的边际成本。市场价格反映了额外提供单位产量对社会的价值，该价值以市场上所有消费者的偏好为基础。
完全竞争市场的价格等于边际成本，这是最具社会效率的产量水平。
其次，长期完全竞争均衡位于价格等于平均成本即曲线的最低点处。这表明不仅厂商获取零经济利润（收入刚好弥补机会成本），
而且所有规模经济效益被充分发挥，不可能有更低的平均成本了。
零经济利润意味着会计利润刚好弥补生产中的隐性成本。
厂商收益正好等于资源的机会成本。
正因为如此，从长期来看即使经济利润为零，厂商也将继续生产。
垄断（monopoly）
指只有一家厂商服务与整个市场，
且产品没有近似的替代品。
垄断势力
公用事业公司都是地方性垄断企业。
垄断的本质是给定市场中是否有其他厂商提供类似产品。
由于市场上只有一个生产商，
市场需求曲线就是垄断厂商的需求曲线。
若没有法律限制，垄断厂商可以自由定价，
但这并不意味着垄断厂商能销售任意数量的产品。
垄断厂商一旦定价，消费者将决定购买多少。
垄断厂商（受消费者需求限制）
智能选择市场需求曲线上的价格—数量组合。
垄断势力的来源
规模经济
垄断势力的第一个来源是技术。
当长期平均成本随着产量增加而下降时，存在规模经济。
当长期平均成本随着产量增加而增加时，则存在规模不经济。
许多技术都存在规模经济和规模不经济的范围。
如果一家厂商生产QM单位产品，则消费者原因支付的价格为每单位PM。
由于PM＞ATC（QM），该厂商能够以高于产品平均成本的价格出售产品并获得正的经济利润。
假设另一家厂商进入导致两家厂商平分市场（每家生产商生产QM/2）。
市场总量不变，因此价格保持在PM，但是每家厂商仅生产QM/2。
导致平均成本降低。
两家厂商的平均总成本为ATC(QM/2)
——该成本高于单一厂商生产所有产量的平均总成本，
因为市场上存在每家厂商都必须承担的间接成本，
每家厂商所分摊的成本大于单一垄断企业垄断生产的一半。
各厂商的平均成本均高于PM，即高于生产QM单位产品时消费者愿意支付的价格，
所以市场上若存在两家厂商则将亏损，
只有一家厂商才会获得正的经济利润——因为单个厂商的产量更高，
规模经济
由此，规模经济可能导致单个厂商垄断整个市场。
在确定一个厂商是否为垄断厂商时，定义相关市场很重要。
范围经济
范围经济指同一厂商生产两种产品的总成本低于由不同厂商分别生产这些产品的成本，
即联合生产的产品产量为Q1和Q2时成本更低。
当存在范围经济时，
一家厂商生产多种产品是有效率的。
范围经济将促成”更大规模“厂商。高资本成本也成为一种进入壁垒。
成本互补多产品成本函数中的成本互补指当一种产品的产量增加时，
另一种产品的边际成本下降，即产品2的产量增加降低了产品1的边际成本。
专利和其他法律壁垒
政府会授予个人或厂商垄断权。
专利制度给予新产品发明者在一定时期内独家销售产品的权利。
利润最大化
边际收益
随着产量增加至小于Q0，需求富有弹性且产量增加使总收益增加。
当产量增加超过Q0后进入缺乏弹性阶段，进一步增加产量导致总收益减少。
总收益在产量为Q0时达到最大，对应的价格为P0，此时需求为单位弹性。
直线MR表明垄断厂商的边际收益。
边际收益是最后一单位产量所带来的总收益变化，它是总收益曲线的斜率。
垄断厂商的边际收益曲线位于需求曲线下方；
考虑线性需求曲线，边际收益曲线正好位于需求曲线与纵轴中间。
也就是说，垄断厂商的边际收益小于产品价格。
边际收益曲线是总收益曲线R(Q)的斜率。
随着产量从零增加到Q0，总收益曲线的斜率随之下降至Q0处变为零。
当产量超过Q0时，总收益曲线的斜率为负值，
产量进一步增加则负值增大，这说明当产量大于Q0时，边际收益为负值。
产量决策
垄断厂商售出Q单位
获得的收益是R(Q)=Q[P(Q)]，
当成本函数为C(Q)时，
垄断厂商的利润是：
Π=R(Q)-C(Q)
收益函数和成本函数之间的垂直距离
就是垄断厂商在不同产量时的利润。
产量低于点A或高于点B意味着厂商将面临亏损
（因为此时的成本曲线位于收益曲线之上）。
产量位于点A和点B之间，
收益曲线位于成本曲线之上，厂商利润为正。
利润函数，即R和C之间的距离。
产量为QM时利润最大，
此产量下收益曲线和成本曲线之间的垂直距离最大。
利润最大化产量QM的重要特性是：
收益曲线的斜率等于成本曲线的斜率，
即产量为QM时边际收益等于边际成本。
原理 垄断产量法则
垄断厂商为实现利润最大化应该生产QM
单位产品，以保证边际收益等于边际成本：
MR(QM)=MC(QM)
微积分表达式 垄断厂商的利润为：
Π=R(Q)=C(Q)
式中，R(Q)为总收益。
要实现利润最大化，令边际利润等于零：MR=MC或
供给曲线不存在
供给曲线指给定价格下的产量水平。
但垄断厂商根据边际收益（小于价格）决定产量，这是因为垄断厂商产量的变化将影响价格的变化，
由此就无法判断不同价格下垄断厂商的产量决策。事实上，具有垄断势力的厂商所服务的市场是没有供给曲线的。
多工厂决策
垄断厂商
在不同地点会有不同的分厂。
多工厂垄断厂商的管理者
的重要任务之一就是确定每个工厂的产量。
假设垄断厂商有两个工厂。
工厂1生产Q1单位产品的成本为C1(Q1)，
工厂2生产Q2单位产品的成本为C2(Q2)。
假设两个工厂的产品相同，
那么消费者原以为两个工厂的
总产量（即Q=Q1+Q2）支付的价格为P(Q)。
利润最大化发生在垄断厂商的
每个工厂各自产量所对应的边际成本
等于总产量的边际收益处。
原理 多工厂产量法则
令MR(Q)表示总产量的边际收益，总产量Q=Q1+Q2，
假设工厂1生产Q1单位产品的成本为C1(Q1)，
工厂2生产Q2单位产品的成本为C2(Q2)。
垄断厂商要通过两个工厂实现利润最大化，两个工厂的产量分配应该满足：
微积分表达式 假定利润为：
利润最大化一阶条件为：
多工厂利润最大化的条件如下：
MC1(Q1)=MC2(Q2)
进入壁垒
如果一个垄断市场存在较高的进入壁垒，
将有效阻止其他厂商进入市场分割利润。
因此，只要垄断企业能够保持其垄断势力，
垄断利润将一直存在。
但要注意的是，垄断势力并非必然带来正的利润；
垄断利润取决于需求曲线与平均总成本曲线的相对位置。
图中垄断厂商的经济利润为零，
这是应为最优价格恰好等于平均总成本。
此外还要强调的是，垄断厂商并非总是盈利，
它在短期内也可能亏损。
垄断厂商的垄断势力通常意味着给整个社会带来一定的社会成本。
垄断厂商利润最大化的产量为QM，定价为PM。
垄断价格超过了边际成本：PM＞MC。
当价格高于边际成本时，说明垄断厂商的产量小于社会有效水平，
所以社会为额外一单位产量愿意支付的价格高于其生产成本。
但是垄断厂商拒绝增加产量，因为垄断厂商的边际收益位于需求曲线下方，
多供给产品会减少厂商的利润。
阴影部分的面积就是垄断的无谓损失（deadweight loss of monopoly），
这是由于垄断厂商的产量低于完全竞争厂商的产量而造成的社会福利损失。
垄断竞争
垄断竞争（monopolistic competition）是介于垄断和完全竞争两种极端情况之间的一种市场结构。
垄断竞争的条件
当市场结构满足以下条件时，为垄断竞争市场：
1、市场上有许多的买者和卖者。
2、每个厂商提供差异化产品。
3、可以自由进入或退出市场。
垄断竞争与完全竞争的显著差异在于：
垄断竞争市场上每个厂商的产品与其他厂商的产品略有不同，这些产品虽然类似但不能完全替代。
垄断竞争市场的产品呈现不完全替代性，这意味着各厂商面临的是一条向下倾斜的需求曲线。
为了销售更多产品，厂商必须降低产品价格。
从这个意义上讲，垄断竞争厂商的需求曲线更类似于垄断厂商而非完全竞争厂商。
垄断竞争与完全垄断的不同点在于：
首先，垄断竞争厂商的需求曲线虽然向下倾斜，
但市场上存在提供类似产品的厂商，而完全垄断市场上只有一家企业提供产品。
其次，垄断竞争市场不存在进入壁垒。如果在位厂商获得正的经济利润，新厂商将会进入市场。
利润最大化
在垄断竞争市场上，
厂商利润最大化的价格和产量决策与完全垄断市场完全一致。
由于需求曲线向下倾斜，
边际收益曲线位于需求曲线下方。
垄断竞争厂商利润最大化的产量位于边际收益等于边际成本处，
图中的Q。
利润最大化价格是厂商的产量为Q时消费者愿意支付的价格P。
厂商的利润为图中阴影区域。
虽然垄断竞争下与垄断下利润最大化的基本原理一致，
但两者之间有一个重要区别：
垄断竞争厂商面临的需求曲线和边际收益曲线是基于单个厂商产品的需求，
而不是基于产品的市场需求，
但垄断厂商面临的需求曲线是整个市场的需求曲线。
垄断竞争市场的厂商提供差异化产品，使得行业或市场需求曲线难以界定。
原理 垄断竞争的利润最大化法则
为了实现利润最大化，
垄断竞争厂商按照边际收益等于边际成本来确定产量。
利润最大化价格是消费者为利润最大化产量愿意支付的最高价格。
利润最大化产量满足以下等式：MR(Q)=MC(Q)
P=P(Q)
长期均衡
假设某垄断竞争厂商生产X品牌产品，其初始需求曲线为D0，
需求曲线位于ATC曲线上方，该厂商获得正的经济利润，
这将吸引其他厂商进入该行业。
随着新厂商的进入，部分消费者会转而购买新厂商的产品，
旧厂商的需求将下降，其需求曲线下降到D1。
该需求曲线正好与厂商的平均成本曲线相切，
导致厂商的经济利润为零，新厂商进入该行业的激励丧失。
如果行业内厂商遭受损失，亏损厂商将退出该行业，
造成剩余厂商的产品需求增加，
从而增加剩余厂商的利润（或者说减少损失）。
当剩余厂商的经济利润为零时，就不会有厂商退出市场了。
每个厂商的经济利润为零且要超过生产该成本的边际成本。
原理 长期经营和垄断竞争
垄断竞争厂商的长期产量符合以下条件：
与完全垄断一样，
在垄断竞争行业中，均衡价格同样高于边际成本，
这意味着厂商产量低于社会有效水平。
消费者愿意为额外一单位产量支付的价格高于其生产成本，
但厂商因为关注利润而不会增加产量。
当价格等于平均成本时，垄断竞争厂商的经济利润为零。
这说明即使垄断竞争厂商有一定的价格控制力，
厂商之间的竞争也会导致没有一个厂商能够赚取超额经济利润。
最后需要指出的是：
由于长期均衡价格高于平均成本曲线的最低点，
因此厂商在生产中没有充分发挥规模经济优势。
这是因为行业内有太多厂商，
以至于任何单个厂商无法在生产中充分发挥规模经济优势。
也有观点认为，
这是产品多样化带来的社会成本
——如果市场上的厂商数量少一些，
规模经济优势可能被充分利用，
这样的话，产品的多样化将受到影响。
产品差异化的含义
完全竞争和垄断竞争的本质区别在于：垄断竞争厂商提供差异化产品。
由于垄断竞争行业中存在众多厂商，厂商的价格控制力的唯一来源是消费者对产品差异性的认可。
当消费者认为其他厂商的产品不能代替该产品时，该厂商的产品将缺乏弹性。
一个厂商的产品越缺乏弹性，获得利润的潜力就越大。
正因为如此，在垄断竞争市场上，
许多厂商努力让消费者相信自己的产品优于其他厂商的产品。
有的生产厂商甚至会主动引入多种产品。
垄断竞争行业的厂商为了使消费者认可其品牌通常采用两种战略。
其一是花费大量财力做广告站。
这些宣传往往采用比较广告（comparative advertisign），
即广告策略是为了区分本厂商品牌与竞争厂商品牌。
比较广告在快餐业很普遍。
在一定程度上，比较广告越有效，消费者为特殊品牌愿意支付的金额就越高。
这就是品牌价值（brand equity）——品牌贷给产品的附加值。
其二是垄断竞争厂商经常向市场推出新产品以进一步区分自身产品与其他厂商的产品。
新产品不仅包括新的或改良的产品，还包括完全不同的产品线。
垄断竞争厂商会试图开发或宣传产品以满足市场上的特殊需求，
这种战略称为利基营销（niche marketing）
——产品或服务针对特殊的消费者群体。
比如通过绿色营销（green marketing），
厂商开发和宣传环境友好型产品，试图抓住关心环境问题的消费者。
遗憾的是，成功的差异化和品牌战略有时会使管理者失去远见。
品牌短视（brand myopic）的管理者会读现有的产品品牌感到满意，
放慢退出新产品的速度，而且不关心新兴行业的趋势和消费者偏好的变化。
品牌短视的公司会故步自封，
有可能错失提升品牌（进而保护品牌）的机会。
作为一个垄断竞争厂商的管理者，务必记住：
如果短期内靠自身产品获得了经济利润，
从长期来看必然会有新厂商进入该市场。
你可能通过引入新的产品线获得了短期利润，
大其他厂商会仿制或者推出新产品，导致你的利润下降为零。
最优广告决策
完全竞争市场的厂商发现广告是不能带来利润的，
因为消费者掌握完全信息——知道任何一种产品都存在大量替代品。
但拥有市场势力的厂商（例如垄断厂商和垄断竞争厂商）
却发现，将部分收入用于广告会带来更大的利润。
最优广告支出同样取决于边际收益和边际成本相等之时：
要实现利润最大化，管理者应该增加广告支出，
直到广告带来的增量收益等于广告投入的增量成本。
广告投入的增量成本就是做广告所耗费的资源的货币成本，
包括为额外广告空间支付的费用和为广告活动所投入的人力资源的机会成本。
增量收益就是由于广告宣传而给厂商带来的额外收益，
包括由广告宣传所带来的额外销售量及每单位销售量所带来的收益。
通过以下简单公式，管理者可以确定利润最大化的广告投入。
公式 利润最大化的广告支出
与销售额的比率（A/R）：
式中，EQ,P表示厂商产品的需求价格弹性；
EQ,A表示厂商产品的需求广告弹性；
A表示厂商的广告支出；
R=PQ表示厂商的销售额（即厂商收益）。
微积分表达式 厂商利润等于收益减去生产成本和广告支出。
令A表示广告支出，Q=Q(P,A)表示对厂商产品的需求，
C(Q)表示生产成本，则厂商的利润是P和A的函数：
第九章 寡头垄断的基本模型
导言
四种具体的寡头垄断情形下的产量和定价决策：斯维奇寡头垄断、古诺寡头垄断、斯塔克博格寡头垄断、伯川德寡头垄断。
寡头垄断的条件
寡头垄断（oligopoly）
指行业中只有几家大厂商存在的情形。通常为2~10家。厂商提供的产品可能同质，也可能有差别。如果只有两家厂商则称为双寡头垄断。
寡头垄断下的经营是最难的。
原因在于：
寡头垄断市场上只有少数厂商，管理者必须关注其决策对行业内其他厂商可能产生的影响，同时还要关注其他厂商的行为对本企业决策的影响。
信念和战略互动
寡头垄断的厂商之间相互依赖。
假设该厂商起初在点B定价定价为P0。
需求曲线D1假定竞争对手会匹配该厂商的价格变化，
需求曲线D2则假定它们不会匹配该厂商的价格变化。
注意，竞争对手匹配价格变化时的需求较不匹配时更缺乏弹性。
理由很简单，对以一次降价，如果竞争对手没有随之降价（D2），
与竞争对手也随之降价（D1）相比，厂商的销售量更大。
当竞争对手也做出降价反应时，降价带来的需求增量很小。
同理，对于一次提价，与维持现有价格（D2）相比，
当竞争对手也随之提价（D1）时，厂商的销售更大。
上述分析说明，寡头垄断市场中的厂商需求取决于竞争对手对该厂商定价决策的反应。
总之，寡头垄断的利润最大化原则与垄断市场一样，
但最大的困难是，如何确定竞争对手是否会相应其价格变化。
四种寡头垄断的利润最大化
斯威齐寡头垄断
斯威齐(Sweezy oliigopoly)寡头垄断的建立
基于竞争对手对提价和降价的响应。
特征：1、市场中只有几家厂商为众多消费者提供服务。
2、厂商的产品差异化。
3、每家厂商都相信竞争对手会随之降价，但不会随之提价。
4、存在进入壁垒。
厂商的产品需求曲线如图中的ABD。
价格高于P0时的需求曲线为D2（注意对应的边际收益曲线为AC）。
价格低于P0时的需求曲线为（注意对应的边际收益曲线为EF）。
总之，厂商的边际收益曲线（MR）最初是对应于D2的边际收益曲线；
但在Q0之后跳转到对应D1的边际收益曲线。
也就是说，斯威齐寡头通断的边际收益曲线MR是图中的ACEF。
利润最大化的产量位于边际收益等于边际成本处，
利润最大化的价格是消费者在该产量所支付的最高价格。
如果边际成本为MC0，在点C边际收益等于边际成本，
此时利润最大化的产量为Q0、最优价格为P0。
当价格超过边际成本（P0＞MC0），产量低于社会效率水平，
说明出现了无谓损失（消费者剩余和生产者剩余）。
寡头垄断的斯威齐模型的一个重要含义是：
在一个给定范围（CE）内，边际成本的变化不会影响利润最大化的产量。
而在完全竞争、垄断竞争、垄断厂商市场，当边际成本下降时都会增加产量。
斯威齐寡头垄断中，如果厂商的边际成本在给定范围内保持不变，厂商也没有动力改变其定价。
这纯粹是因为竞争对手会随之降价而不会随之提价这一假设。
斯威齐模型有一定的局限性，它无法解释行业的初始价格为何固定为P0，
为何在价格为P0时每个厂商的需求曲线都反生了弯曲。
但是斯威齐模型展示了厂商之间的战略互动和管理者对竞争对手如何反应的预期，这对定价决策有深刻的影响。
实际上，初始价格P0和管理者预期是基于管理者所掌握的给定市场上竞争对手的价格模式。
如果经验告诉你竞争对手会随之降价而不会随之提价，那么斯威齐模型也许是最好的定价决策工具。
古诺寡头垄断
古诺寡头垄断（Cournot oligopoly）
每家厂商必须在其他厂商确定产量的同时决定自己的产量，
或每家厂商都期望自己的产量决策不会影响竞争对手的产量决策
特征：
1、市场中只有几家厂商为众多消费者提供服务。
2、厂商的产品差异化或同质化。
3、每家厂商都相信，当自己改变产量时竞争对手的产量不变。
4、存在较高的进入壁垒。
反应函数与均衡因为市场价格随厂商2的产量提高而降低，导致产商1的边际收益降低。
这意味着厂商1的利润最大化产量受制于产商2的产量。
反应函数（reaction function）
定义了在给定另一家厂商的产量时，本厂商的利润最大化产量。
比如，给定厂商2的产量为 Q2，厂商1的利润最大化产量为：
Q1=r1(Q2)
如果厂商2的产量为零，厂商1的利润最大化产量为QM1，厂商1的反应函数（r1）上的QM1对应的Q2为零。
因此QM1是厂商1为垄断厂商时的最优产量。如果厂商2的产量为Q2，厂商1的利润最大化产量为Q1，
此为反应函数r1所对应的厂商2的产量Q2。
因为厂商1产品的需求依赖于市场上其他厂商的产量。当厂商2增加产量时，厂商1的需求量和边际收益会下降。
厂商1要实现利润最大化就要降低其产量。
假设厂商1的产量为QM1。给定该产量，厂商2的利润最大化产量为r2上的A点。
给定产商2的产量为正，厂商1的利润最大化产量将不再是QM1而是r1上的点B。
给定厂商1的产量减少，点C将成为厂商2的反应函数上的利润最大化产量。
给定厂商2的新产量，厂商1则会将产量减少至其反应函数上的点D。
点E处厂商1的产量为Q1，而厂商2的产量为Q2。
如果两家厂商都相信对方会保持这一产量不变，并且都不愿改变现有产量，
点E为古诺均衡（Cournot equilibrium），即给定其他厂商产量时，没有一家厂商愿意改变自身的产量。
这一情形对应于反应曲线的交点。
公式：古诺寡头垄断的边际收益
假设生产同质产品的古诺双寡头垄断的（反）需求函数为：
P=a-b（Q1+Q2)
公式：古诺双寡头垄断的反应函数
对于线性（反）需求函数: P=a-b（Q1+Q2）
和成本函数 C1(Q1)=C1 Q1
C2(Q2)=C2 Q2
上述公式的推导过程如下，为了使利润最大化，厂商1的产量必须满足：
MR1(Q1,Q2)=MC1
对于线性（反）需求函数和成本函数来说，上述等式变为：
a-bQ2-2bQ1=c1
求解方程，并用Q2来表示Q1，可得：
无论古诺寡头垄断厂商生产同质产品还是差异化产品。
行业产量都是低于社会有效水平的。
低效率出现的原因是均衡价格高于边际成本。
随着厂商数量增加，均衡价格降低并趋向边际成本。
当厂商的数量为任意大时，
生产同质产品的古诺市场中的均衡价格将无限接近边际成本，
行业产量接近完全竞争市场（没有无谓损失）。
等利润线（isoprofit curve），
寡头垄断厂商的利润不仅取决于自身产量，还取决于其他寡头垄断厂商的产量。
它定义了所有厂商不同产量组合下使某给定厂商保持同等利润水平的曲线。
厂商1的反应函数（r1），沿着该曲线有三条等利润线（分别标位
Π0、Π1和Π2）。
有四个方面需要重点关注：
1、等利润线上的每一个点都表示厂商1的利润相同。
2、等利润线越接近厂商1的垄断产量QM1，表示厂商利润越高。
3、厂商1的等利润的最高带你位于与反映函数的相交处。
4、等利润线彼此不相交。
合谋
当市场被几家厂商主导时，
厂商可能通过“协议”限制产量，
或者以损害消费者利益为代价提高定价而获利。
这种行为称为合谋。
点C对应于一个古诺均衡，它是两家厂商反应函数的交点。
厂商1的均衡利润由等利润线ΠC1给出，厂商2的均衡利润由等利润线ΠC2给出。
阴影区囊括了两家厂商的产量，该区域内两家厂商的总利润比古诺均衡所获得的利润高。
如在点D处每家厂商产量更少但利润更高，因为点D处每家厂商的等利润线更接近各自的垄断点。
也就是说，如果每家厂商都同意限制产量，厂商可以定价更高并获得更多利润。
如厂商1的利润在点A处最高（在该点厂商1是垄断厂商）。
厂商2的利润在点B处最高（在该点厂商2是垄断厂商）。
如果各厂商对产量”达成协议“使总产量等于垄断产量，则两家厂商将在直线AB上任一位置进行生产。
换句话说，直线AB上的任意产量组合都将使整个行业的利润最大化。
直线EF上的产量可以使整个行业的利润最大化，
由于它们处于透镜区域内，使两家厂商获得比它们在点C（古诺均衡）处更高的利润。
如果厂商合谋来限制产量并分享垄断利润，它们可以在点D处生产并获得更高利润Π合谋1和Π合谋2。
该点处相应的市场价格和产量与垄断状态一样：合谋导致价格超过边际成本，
其产量在社会最优水平之下且存在无谓损失。
然而合谋的厂商比相互竞争的古诺寡头垄断者获得的利润更高。
但是，厂商达成合谋协议并不容易。
假设厂商同意合谋，
每家厂商按照点D对应的合谋产量进行生产并获取合谋利润。
给定厂商2的产量为Q合谋2，
厂商1有动机采取”欺骗“行为，如将产量扩大到点G，
这样厂商1可以获得比合谋更高的利润，
但以牺牲合谋伙伴的利益为代价。
厂商都知道存在这样的诱因，
因此他们很难快速达成合谋协议。
实施上更严重的是，
厂商2在点G（厂商1有欺骗行为）处获得的利润
比在点C（古诺均衡）处获得的利润更少。
斯塔克博格寡头垄断
斯塔克博格寡头垄断（Stackelberg oligopoly）
中各厂商的决策时间不同。
假设一家厂商（领导者）在其他厂商之前作出产量决策。
其他厂商（跟随者）将领导者的产量视为给定的，
然后选择其利润最大化的产量。
在斯塔克博格寡头垄断中，每个跟随者的行动类似于古诺寡头垄断厂商，
但领导者不会将跟随者的产量视为给定的，
而是相信每个跟随者会根据古诺反应函数对领导者的产量决策作出反应，
在此基础上选择使自己利润最大化的产量。
特征：
1、市场上只有几家厂商为众多消费者提供服务。
2、厂商的产品差异化或同质化。
3、某一厂商（领导者）在其他厂商确定产量之前制定产量决策。
4、其他厂商（跟随者）将领导者的产量视为给定的，
然后在给定领导者产量的情况下确定其利润最大化产量。
5、存在进入壁垒。
考虑只有两家厂商的情形。
厂商1是领导者，具有先动优势，即厂商1在厂商2之前确定产量。
厂商2是跟随者，在给定领导者产量后确定利润最大化产量。
跟随者的生产决策在领导者之后，
所以跟随者的利润最大化产量取决于其对领导者产量的反应函数，该反应函数为r2。
领导者知道跟随者将根据r2作出反应，
因此，领导者在无论自己如何行动跟随者都会作出反应的情况下，决定自己利润最大化的产量。
领导者知道跟随者会沿着r2生产，因此只需要在跟随者的反应曲线上选择使自己利润最大化的点，
所以领导者的等利润线将接近垄断产量，最终领导者选择点S。
此时等利润线记为ΠS1。这条等利润线与厂商2的反应函数相切，领导者的产量为QS1。
跟随者观察到这个产量而生产QS2，即对应与QS1的利润最大化产量。
领导者的利润由ΠS1给出，跟随者的利润由ΠS2给出。
注意，领导者的利润高于其古诺均衡（点C）的利润，而跟随者的利润低于其古诺均衡的利润。
由于先动优势，领导者获得了更高得的利润。
伯川德寡头垄断
伯川德寡头垄断（Bertrand oligopoly）
假设厂商提供同质产品且消费者愿意为产品支付（有限的）垄断价格。
特征：
1、市场中只有几家厂商为众多消费者提供产品或服务。
2、厂商的边际成本固定且产品同质。
3、产商参与价格竞争并对竞争者的价格作出最优反应。
4、消费者获得完全信息且没有交易成本。
5、存在进入壁垒。
从管理者角度看，伯川德寡头垄断不是理想市场，
因为即使市场上只存在两家厂商，也可能导致零利润。
但从消费者角度看，伯川德寡头垄断很受欢迎
——它将导致与完全竞争一样的结果。
由于消费者获得完全信息和零交易成本，
且产品同质，所有消费者都将购买低价产品。
假设厂商1索要垄断价格，厂商2只要定价略低于垄断价格，
就将占据整个市场并获得正利润，而厂商1将卖不出去任何产品。
因此厂商1会以比厂商2更低的价格来报复厂商2，从而重新占据整个市场。
当每家厂商的定价等于其边际成本时：P1=P2=MC。
没有任何一家产商有动力继续降价，因为价格低于边际成本会导致亏损。
同时也没有厂商愿意提高价格，因为提价将导致产品卖不出去。
所以伯川德寡头垄断和同质产品导致每家厂商以边际成本定价和经济利润为零的情形。
由于P=MC，同质产品的伯川德寡头垄断产生一个社会有效产量。
总市场产量等于完全竞争行业的产量且没有无谓损失。
应对”伯川德陷阱“的策略
——在同质产品的伯川德寡头垄断中规避残酷竞争，
关键是增加转换成本或消除产品同质性，
产品差异化使厂商定价高于边际成本而不会失去全部消费者。
寡头垄断模型的比较
古诺模型
为便于比较，在此假定它们具有相同的市场需求和成本条件，如反市场需求函数为：P=10000-（Q1+Q2）
每家厂商的成本函数相同，且给定为：Ci(Qi)=4Qi
给定上述需求函数和成本函数，
单个古诺厂商的利润函数为：Πi=[1000-(Q1+Q2)]Qi-4Qi
古诺寡头垄断厂商的
反应函数为：
求解Q1，Q2这两个反应函数，得到古诺均衡的产量Q1=Q2=332。
因此，市场总产量为664单位，价格为336美元。
将这些数值代入利润函数，得到每家厂商将获得110 224美元的利润。
斯塔克博格模型
根据上述需求函数和成本函数，斯塔克博格领导者的产量为：
跟随者将此产量视为给定的，其反应函数为：
因此，市场总产量为747单位，该产量下的家阿哥为253美元。
斯塔克博格寡头垄断中的市场总产量高于古诺寡头垄断中的市场总产量。
这使得斯塔克博格寡头垄断中的价格低于古诺寡头垄断中的价格。
领导者的利润为124 002美元，跟随者的利润为62 001美元。
由于其先动优势，
斯塔克博格寡头垄断中领导者的利润高于古诺寡头垄断厂商。
但是，斯塔克博格寡头垄断中跟随者的利润低于古诺寡头垄断厂商。
伯川德模型
所有参与伯川德模型竞争的产商的最终定价将等于其边际成本。
在给定的反需求函数和成本函数下，每家厂商的产品价格等于其边际成本（4美元），利润均为零。市场总产量为996单位
合谋
合谋的结果就是厂商选择使整个行业利润最大化的产量。
当厂商进行合谋时，行业总产量就是市场反需求曲线的垄断产量。由于市场反需求曲线是：P=1000-Q
因此边际收益为：MR=1000-2Q
注意：该边际收益函数假定这些厂商作为一个整体利润最大化厂商采取了联合行动（合谋）。
令边际收益等于边际成本（即4美元），可得：1000-2Q=4
则Q=498合谋下整个行业的总产量为498单位，两家厂商各生产一半。
合谋价格为：P=1000-498=502，每家厂商获得的利润为124002美元。
不同的寡头垄断情形下的结果对比如下：市场产量最高的是伯川德寡头垄断厂商，其次是斯塔克博格寡头垄断厂商，再次是古诺寡头垄断产商，最后是合谋厂商。利润最高的是斯塔克博格的领导者和合谋厂商，其次是古诺寡头垄断厂商，再次是斯塔克博格的跟随者，伯川德寡头垄断厂商获得最低利润。
如果你是寡头垄断市场的管理者，
必须认识到最优决策和利润将取决于市场中寡头垄断厂商之间的相互作用的不同类型。
可竞争市场
可自由进入的市场就是经济学家所指的
可竞争市场（contestable market）。
一个可竞争市场
需要满足以下条件：
1、所有的生产者获得了相同的生产技术。
2、消费者读及挨个变化的反应迅速。
3、在位厂商不能通过降价对进入作出快速反应。
4、不存在沉没成本。
如果上述四个条件成立，在位厂商对消费者没有市场控制力。
也就是说，该均衡价格对应于边际成本且厂商的经济利润为零。
即使市场中只有一家产商亦如此。
原因在于：如果在位产商的价格超过边际成本，具有相同技术的新厂商可迅速进入市场且定价比在位厂商低。
由于在位厂商无法迅速降价应对，进入者就可以通过低价获得在位厂商的所有消费者。
在位厂商也明白这一点，所以别无选择，只有以生产成本定价来阻止进入者。
因此，如果一个市场是完全可竞争市场，在位厂商会受到新厂商进入威胁的约束。
可竞争市场的一个重要条件是不存在沉没成本。
沉没成本指新进入厂商必须承担的且退出市场时不能收回的成本。
第十章 博弈论：剖析寡头垄断
导言
运用一般性工具帮助管理者在寡头垄断市场制定各种决策，包括定价、广告投放、引进新产品、是否进入新市场等。
研究这些问题将用到博弈论这个基本工具。博弈论也可以用于分析厂商内部决策，如员工监督和工资议价决策。
博弈与战略性思维
在博弈模型中，博弈方（参与者）是决策个体。博弈方的决策称为策略。博弈方收益就是策略带来的利润或损失。
由于相互影响，一个博弈方的收益不仅取决于自己的策略，而且取决于其他博弈方所采取的策略。
在博弈分析中，博弈方做决策的顺序很重要。
同时行动博弈（simultaneous-move game）指每个博弈方都在不知道其他博弈方决策的前提下制定自己的决策。
在序贯行动博弈（sequential-move game）中，一个博弈方观察到另一个博弈方的行动后才采取行动。
区分一次性博弈和重复性博弈也很重要。
一次性博弈指这个博弈只能进行一次。重复博弈指博弈活动可能进行多次（不止一次）。
同时行动博弈与一次性博弈
基础理论
策略（strategy）是一种决策选择，指博弈方在每个决策状态下将采取的行动。
正则形式博弈（normal-form game）是博弈的一种表示形式，表示博弈方可能采取的不同策略及其带给博弈方的收益。
占优策略（dominant strategy）是这样一种策略，不管对手如何行动，都能产生最高收益。
原理 运用占优策略，确定是否存在占优策略，如果有的话，运用占优策略。
安全策略（secure strategy）——保证在最坏的情形下获得最高的收益。
安全策略（secure strategy）——保证在最坏的情形下获得最高的收益。
安全策略有两个缺点：首先，这是一种保守策略，只有在极度厌恶风险时才会考虑它；其次，它并没有考虑对手的最优决策，这将影响更高收益的获取。
原理 从对手角度出发
如果没有占优策略，就从对手角度来考察博弈。如果对手有占优策略，就可以预测他会按照占优策略行动。
纳什均衡（Nash equilibrium）的策略是：给定其他博弈方的策略，没有一个博弈方可以通过单方面改变自己的策略而提高自身收益。
它表示在对手行为给定的情况下，每个博弈方所能作出的最佳选择。
一次性博弈的应用
在一次性博弈中，每家厂商的纳什均衡战略是低价策略。在上述一次性博弈中，每家产商的最优策略都是低价策略。
这个利润比两家厂商合谋或者都同意定高价时的利润低。这是经济学中的一个经典结果，称为困境。
在美国合谋是违法的，法律不允许两家厂商密谋定高价。厂商A有动机诱导厂商B定高价，而自己通过欺骗来获取高利润。
当厂商B也有这种动机，这就阻止了合谋协议的初次达成。
广告和质量决策
在大多数寡头垄断市场上，广告之所以能提高产品需求量，是因为它能吸引本行业中其他厂商的消费者。
广告对其利润的提高是以牺牲市场中其他厂商的利润为代价的；厂商间的广告决策同样也会相互影响。
每家厂商做广告仅仅是为了消除其他厂商的广告宣传对自己的负面影响，而不能增加行业需求或厂商需求。这就导致了高额广告支出，并且降低了厂商利润。
协商决策
有两个纳什均衡。一旦两家产商达成一致，它们就不会违背协议。这是一种协商合作的博弈而非利益冲突的博弈。
监督员工
没有纳什均衡。在这个博弈中，管理者和员工都向保密其行为。
这这种情况下，博弈方选择混合（随机）策略（mixed (randomized) strategy）更有利，即博弈方随机选择其策略。
纳什讨价还价
讨价还价博弈中有三个纳什均衡。
一个结果是管理者和工会平分盈余，而另外两个结果是将全部盈余分配给其中一方。
同时行动博弈的结果很难预测，因为存在多个纳什均衡。如果博弈方无法在某个均衡处达成一致，就会导致无效率。
在劳资谈判中，这种僵局现象普遍存在：由于双方索价的总和超过可分配额，导致无法达成协议或协议被推迟。
经验表明，在讨价还价博弈中，博弈方通常认为50-50的分配是公平的。
现实中许多博弈方倾向于选择这种均分策略（即使还存在其他的纳什均衡）。
无限重复博弈
理论
无限重复博弈（infinitely repeated game）是一种永远持续进行的博弈，博弈方从博弈的每次重复中获得收益。
现值分析回顾
企业价值是这个企业所能获得的所有未来利润的现值。
如果利率为i，
Π0是当期利润，Π1是1年以后的利润，Π2是2年以后的利润，
等等，一家将经营T年的企业的价值为：
如果每期企业获得的利润相同（对每期t，Πt=Π），
并且无限期获取（T=∞），
这个公式简化为：
触发策略支持合谋
触发策略（trigger straategy）是指在博弈中根据博弈方过去的行为而制定的策略。
采取触发策略的博弈方会继续同一行为，直到其他博弈方采取了某种行为触发该博弈方改变行为。
如果欺骗成本的现值超过欺骗的当期收益，厂商就不应进行欺骗，所以高价会得以维持。
原理 以触发策略维持合作
假设博弈将无限次重复而且利率为i。
博弈方合作时的收益是Π合作，如果博弈方有欺骗行为，
博弈的最高收益为Π欺骗，博弈纳什均衡的收益为ΠN，并且
如果博弈方在过去没有欺骗行为，合作（合谋）的结果将能在无限重复博弈中实现；
如果任何一方采取欺骗行为，另一方将选择一次性博弈的纳什均衡策略惩罚对方。
上述原理中的实现条件可以写为：
方程左边表示反合谋协议的一次性当期收益。
方程右边表示当期因欺骗所损失的未来各期收益的现值。
如果一次性收益小于因欺骗所损失的收益现值，博弈方遵守协议是有利的。
无限重复博弈的结果是：
当利率较低时，厂商将发现合谋定高价有利可图。
价格博弈中的合谋因素
在以下情形下，厂商通过惩罚策略来维持合谋协议会更容易：（1）厂商知道竞争对手是谁，这样就知道应该惩罚谁（如果有必要的话）；（2）厂商知道竞争对手的顾客是谁，如果有必要采取惩罚行为，就可以通过低价吸引对手的顾客转移；（3）厂商知道竞争对手何时会违背合谋协议，这样才能知道何时开始实施惩罚。它们要能够成功地惩罚违背合谋协议的对手，否则，这种惩罚威胁将不起作用。上述因素与市场结构和行业行为有关。
厂商数量
合谋在只有少数厂商的行业较容易实现。
随着市场中厂商数量增加，监督关系总量迅速增加。
厂商的规模
监督成本占小厂商总成本的比重远大于监督成本占大厂商总成本的比重。
市场的历史
一种方法是厂商公开见面，并口头警告对手不要抢走其顾客，否则将受到惩罚。
另一种方法是厂商根本不需要见面，通过一定时期内的博弈感悟，最终达成默契合谋。
惩罚机制
定价机制会影响到厂商是否有能力惩罚不合作的对手。比如在标准价格市场，厂商对所有顾客采取统一定价，惩罚对手的成本高于采取差异化定价。
无限重复博弈与产品质量
一次性博弈中的厂商有强烈的提供伪劣产品的动机，尤其当消费者在购买前无法确定产品质量的时候。
在一次性博弈中，纳什均衡策略是厂商生产低质量产品且消费者不购买。
产品质量博弈是无限重复博弈，不同于一次性博弈。
启示有两点：首先，如果厂商想持续经营，即长期存活下去，一次性收益被未来损失抵消，欺骗消费者自然不值得。
其次，即使厂商努力提高产品质量，由于疏忽也可能出现一些缺陷产品，这将有损厂商声誉，为此许多厂商提供产品质量保证。
有限重复博弈
不确定终值期的博弈
这是一个有限重复博弈。假设厂商不知道其产品何时过时，博弈的终止期不确定。
假设有限重复博弈在给定回合后即将结束的概率为θ，0＜θ＜1
如果厂商A当期欺骗对方，不论该博弈是1次、2次还是其他任何次数，它的总收益为：
如果厂商A没有欺骗行为，其期望收益是：
式中，θ为给定博弈后博弈终止的概率。尤其需要注意的是，
如果不知道有限重复的次数，合作收益恰好与无限重复博弈时的合作收益一样，即
式中，i为利率。
在一个终止期不确定的重复博弈中，1-θ的作用等同于1/（1+i）；
因为不能确定博弈是否继续，博弈方的未来折现值并非完全归因于利率。
根据上例中的数据，若满足下式，厂商A将不会有欺骗行为：
当θ≤1/5时，该式成立。
由于博弈是对称的，这个结果同样适用于厂商B。
总之，当寡头垄断厂商的博弈是终止期不确定的有限重复博弈时，它们将合谋并所要高价与无限重复博弈一样。
关键在于，博弈持续进行的概率要足够大。
存在终止期的重复博弈：期末问题
合谋难以达成的关键原因在于：最终时刻总会到来，双方肯定不再有明天。前期的任何合作承诺都无意义，因为违背协议后也不会在明天收到惩罚。
实际上博弈方违背协议的动机出现在倒数第2期（因为最后一期无法施加惩罚），所有博弈方都明白这一点，造成倒数第3期的博弈也没有明天。
向后追溯的结果就是在任何时候都不可能施加有效惩罚。所以博弈方每期都定低价直至最后一期。
期末问题的应用
辞退和退出
更好的管理策略是对员工在雇佣期的良好工作表现给予奖励。比如可以强调你的人脉资源，在员工有需要时给他们写推荐信。
这样就向员工传达了一个信号：辞职并不是这场博弈的结束。如果一个员工在合同期末偷懒，你可以告知其潜在雇主来惩罚他们。
”万金油“销售员
在流动商贩处买到假货的教训，究其原因，就是消费者没办法追踪那些卖假货的商贩。这些商贩则利用了期末问题来获利。
多阶段博弈
基本理论
多阶段跨框架允许博弈方作出序贯决策而非同时决策。
一个延展形式博弈（extensive-form game）包括：
博弈方、
在博弈每个阶段博弈方获得的信息、
博弈方的策略、
博弈方的行动顺序即每种策略产生的收益。
与同时行动博弈一样，每个博弈方的收益不仅取决于其自身行动，还取决于其他博弈方的行动。
序贯行动博弈与同时行动博弈存在一个重要区别：
博弈方A必须在博弈方B之前做决策，博弈方A不能根据博弈方B的决策制定自己的决策。博弈方B可以根据博弈方A的行动来选择自己的策略，这就是序贯行动博弈，博弈方B是后行者。
构成子博弈完全均衡（subgame perfect equilibrium）的策略应该满足：
（1）它是纳什均衡；（2）在博弈的每个阶段（决策点）没有博弈方可以通过改变自身策略来增加收益。
子博弈完全均衡是一个包含可置信威胁的纳什均衡。
对阶段博弈的应用
这个博弈存在两个纳什均衡：
第一个均衡是厂商B威胁选择硬策略，厂商A不进入市场。
厂商B的威胁不可信，这个结果是一个纳什均衡而非子博弈完全均衡。
这个博弈的另一个纳什均衡是：
厂商A选择进入，厂商B随之采取软策略。
这是一个子博弈完全均衡，即产商A选择进入，而厂商B选择软策略。
创新
关于进入的博弈分析给管理者的重要启示是：当对手的威胁不可信时，管理者无须理会这个威胁。
序贯讨价还价
这个序贯博弈唯一的子博弈完全均衡是：
管理者分配给工会1美元且工会接受提议。
两阶段序贯博弈的一个显著特征，博弈中的先行者提出”要么接受要么一无所获”，后行者可以接受提议或者拒绝提议（但拒绝将一无所获），
提出“要么接受要么一无所获”提议的博弈方实际上拥有绝对的讨价还价筹码。
现实生活中的一些问题经常导致序贯讨价还价博弈过程复杂化。
首先，博弈方无法知道其他博弈方的真实收益。在讨价还价过程中，投入一些时间去了解对手是值得的。
其次，序贯讨价还价博弈过程的一个重要假设是，一旦博弈方拒绝或者接受了提议。讨价还价就结束了。
如果没有这个假设，博弈方在决定是否接受提议时也许会想：”如果我拒绝提议，对方可能会提出另一个更有吸引力的提议。
“这样就改变了博弈模型并且可能改变博弈方的基本决策。
另外，能够给出可置信的”要么接受要么一无所获“提议的博弈方将在博弈中获益更多。
但如果承诺不可信，当对方还价而拒绝谈判时，提议方可能会吃闭门羹。
第十一章 定价策略
导言
拥有市场势力的企业（如垄断、垄断竞争和寡头垄断企业）的定价策略。
最优定价策略因企业不同而不同，受潜在市场结构和相关工具（比如广告）的影响。
本章以垄断、垄断竞争和寡头垄断中的利润最大化定价作为基本的定价策略，并以此为基础研究使企业实现更高利润的复杂定价策略。
管理者不仅要掌握这些策略，还要掌握每种策略的实施条件。
基本定价策略
利润最大化的基本法则
具有市场势力的企业的基本定价策略：对所有消费者制定同一价格，使边际收益等于边际成本。
产量设定在边际收益（MR）等于边际成本（MC）处。利润最大化价格是消费者在该产量时所支付的每单位最高价格。
垄断和垄断竞争的定价原则
小企业也能从公开信息中
获得一些有关需求和成本的信息。
公式：具有市场势力的企业的边际收益
具有市场势力的企业的边际收益为：
式中，EF是企业产品的需求价格弹性；P是产品定价。
由于利润最大化的产量在边际收益等于边际成本处，
因此
就在利润最大化的产量处。
求解上述方程的P，
得到具有市场势力的企业的利润最大化价格：
利润最大化价格是边际成本的K倍：
式中，K=EF/(1+EF)。
K可视为利润最大化的加成因子。
原理 垄断和垄断竞争的利润最大化的价格加成
利润最大化的价格为：
式中，EF是企业产品的需求价格弹性；MC是企业的边际成本。
括号中的内容为最优加成因子。
管理者在运用该定价原则时须注意两点：
首先，企业产品的需求弹性越大，利润最大化的加成就越低。
其次要注意，边际成本越高，利润最大化的价格就越高。当其他条件不变时，边际成本较高的企业的定价将高于边际成本较低的企业。
务必记住：
当运用加成公式改变某产品或服务的价格时，需求价格弹性也可能改变。
古诺寡头垄断的定价原则
假设一个由N家古诺寡头垄断者组成的行业，
每家的成本结构相同且产品相似。
此时，可利用一个简单公式求出古诺寡头垄断的利润最大化价格。
原理
古诺寡头垄断的利润最大化的价格加成
如果古诺寡头垄断市场有N家相同的企业，
该市场上每家企业的利润最大化价格为：
式中，N是该行业中的企业数量；
EM是市场的需求弹性；MC是边际成本。
微积分表达式
可以将古诺寡头垄断者的需求价格弹性和该市场的需求弹性之间的关系，
代入基于垄断和垄断竞争的加成原则的公式。
对于一个N家企业提供同质产品的古诺寡头垄断市场，
单个企业的产品需求价格弹性是该市场需求价格弹性的N倍：EF=NEM
在垄断和垄断竞争的定价公式中，用NEM代替EF，
结果就是古诺寡头垄断的定价公式。
利用微积分知识分析，如果：
是行业总产量且行业需求为Q=f(P)，市场需求价格弹性为：
单个企业（如企业1）的需求为：
既然一个企业认为其他企业的产量固定，则单个企业的需求价格弹性为：
关于古诺寡头垄断的定价原则必须注意三点：
一是市场需求越富有弹性，利润最大化价格越接近边际成本。
二是当企业数量增加时，利润最大化价格逐渐接近边际成本。
三是边际成本越高，古诺寡头垄断的利润最大化价格也就越高。
获取更多利润的策略
获得消费者剩余
价格歧视
企业对相同的产品或服务索取不同的价格会获得更高利润，
这种策略称为价格歧视（price discrimination）。
价格歧视有三种基本类型——一级、二级和三级。
企业非常愿意选择一级价格歧视——向每个消费者索要其愿意为每单位产品支付的最高价格。通过一级价格歧视，企业将获取全部消费者剩余。
遗憾的是，一级价格歧视（也称完全价格歧视）的实施非常困难，它需要了解每个消费者对可选产品愿意且能够支付的最高价格。
一些与服务相关的行业如汽车经销、机器修理、医疗以及法律事务等，能成功地运用一级价格歧视。
需求曲线和边际成本曲线之间的阴影面积即反映了企业总利润。
这个结果（从企业的视角看）
只有当管理者拥有消费者支付意愿的完全信息时才能实现。
当企业不了解每个消费者为每单位产品所愿意支付的最高价格，
或者针对每单位新增消费量持续降价不可行时，企业很可能采用二级价格歧视策略来获取部分消费者剩余。
二级价格歧视是对不同的购买数量进行离散的递减式定价。
基于一个公开的价格表，根据消费者所选择的产品数量和支付意愿，消费者被自动分类.
二级价格歧视策略带来的利润低于一级价格歧视的利润。
然而较之对所有产品制定相同阿济格的加单策略来说，
二级价格歧视策略的利润仍然要高些。
如果企业能够系统性识别不同消费群体对产品的不同需求，
三级价格歧视策略的应用将更普遍。
在某些情况下，
针对相同产品，企业可以通过向不同的消费群体要价不同而获利。
三级价格歧视之所以能够提高利润，
是因为不同消费者的需求弹性不同。
最后要注意的是，
如果以低价购买产品的消费者能够以稍高价格将产品转售给其他人，价格歧视将失效。
转售使得消费者用低价产品完全替代了企业的高价产品。
这一行为使企业不得不降价而减少利润。
公式：三级价格歧视原则 为实现利润最大化，
具有市场势力的企业在不同消费群体的边际收益
等于边际成本处生产：
两部定价
两部定价（two-part pricing）
是指企业对其产品收取固定费用，
再加上每单位购买量的费用。
通过收取固定费用32美元，
该企业获得了全部消费者剩余。
企业以每单位边际成本2美元销售产品，
所以每单位产品在销售上没有获利。
但是企业获得了固定费用32美元，这就是纯利润。
原理 两部定价通过采用两部定价，企业可以增加利润：
每单位产品定价等于边际成本，加上在这个单位阿济格上的全部消费者剩余的固定费用。
注意：如果边际成本低，每单位的使用费也低。
在边际成本等于零的情况下，一个最优的使利润最大化的两部定价策略是对每次设施使用不收费，
但是固定的入会费等于消费者剩余。两部定价策略下企业利润来源于固定费用。
与价格歧视不同，两部定价不需要消费者对商品具有不同的需求弹性。
整包定价
整包定价（block pricing），
如果你购买过4卷装的卫生纸或者6灌装的苏打水，
就体验了整包定价。
假设企业将8单位产品打包且给定一个打包价格。
消费者不得不在购买8单位商品和什么都不买之间作出决策。
因此只要这8单位商品的大包价格不高于48美元，
消费者将发现购买整包商品是有利可图的。
原理 整包定价通过将一批产品打包且将其作为一个整包进行销售，企业将获得比单一定价更多的利润。
一个整包的利润最大化价格就是消费者从这个整包中获得的总价值。
整包定价迫使消费者作出”要么全部买，要么都不买”的购买决策。
注意：即使消费者对企业产品的需求弹性相同，整包定价也能增加企业利润。
商品捆绑
商品捆绑（commodity bundling）
是指将两种或两种以上的不同产品捆绑在一起
并以单一的捆绑价销售。
当消费者对一家企业的多种商品有不同支付意愿时，商品捆绑能够增加企业利润。
需要强调的是，即使管理者无法区分消费者支付意愿的差异，商品捆绑也能增加利润。
但如果管理者能够准确地知道每个消费者对每个商品的支付意愿，
企业就能获得比价格歧视策略下更高的利润：对那些支付意愿强的消费者索取更高的价格。
特殊成本
—需求架构的定价策略
高峰定价
当高峰期的需求太高
以至于企业无法以相同价格满足所有消费者的需求时，
企业就有条件实施高峰定价策略。
与价格歧视策略一样，
企业会制定两种不同的价格：低峰需求时定价低，高峰需求时定价高。
注意，如果企业在一天中所有时段都定高价PH，
那么在低峰期将无人购买企业产品。
通过在低峰期定低价而在高峰期等高价，企业可以在低峰期向部分消费者销售产品以增加利润。
原理 高峰定价
当需求在一天中的某些时段高于其他时段时，
企业就可以采取高峰定价策略增加利润，
即在高峰期制定比非高峰期更高的价格。
交叉补贴
交叉补贴，该策略通常应用于企业具有成本的互补性，或者消费者对一组产品的需求相互影响之时。
交叉补贴（cross-subsidy）是指利用一种产品产生的利润补贴另一种产品的销售额。
原理 交叉补贴
无论企业的两种产品是成本互补的还是需求相关的，
企业采取交叉补贴策略都可以增加利润：以等于或低于成本的价格销售一种产品，以高于成本的价格销售另一种产品。
转移定价
许多大型企业都有上游和下游生产经理，
他们必须在各自领域对价格和产量作出决策。
此时一个重要的问题是最优转移定价（transfer pricing）——为了使整个企业的利润最大化，
上游部门将以什么内部价格向企业的下游部门销售投入品。
由于上游部门具有垄断势力，
销售给下游部门的投入品价格会超过企业的实际边际成本。
根据这个投入品价格，
下游部门经理将在最终产品市场中边际收益等于其边际成本处进行生产，
使得本部门利润最大化。
这也意味着价格高于边际成本。
简言之，当两个部门的定价都高于边际成本时，
会产生双重边际效应，其结果是利润小于最优的企业总利润。
为了避免双重边际效应问题，设定的转移定价将最大化企业的总利润而非上游部门的利润。
假设下游部门的最终产品的边际成本为MCd，即扣除上游部门的投入品所要求的边际成本之外的部分。
在这种情形下，上游部门的起边际成本（MCu）等于部门净边际收益（NMRd）处生产，企业的总利润能够最大化：
注意，企业生产额外一单位投入品的成本为MCu。
只有在下游部门将投入品转换成最终产品支付额外的MCd时，
该投入品才被转换成另一单位的产量并销售。
这样在最终产品市场上才会产生额外的收益MRd。
因此，企业生产额外一单位的投入品的实际边际收益为NMRd。
令NMRd等于使企业总利润最大化的边际成本。
企业如何形成一种激励机制使得部门经理的行为能够最大化企业总利润。
假设企业的高管决定了最大化企业总利润的最终产量水平Q。
他们令转移价格PT等于上游部门生产投入品的边际成本，其生产投入品的数量取决于下游部门生产Q单位的最终产品。
根据这个内部价格方案，下游部门经理以每单位固定价格PT从上游部门购买所需要的投入品。
上下游的经理被要求使各部门利润最大化，是基于企业高管设定的转移价格情形下的决策。
激烈价格竞争市场的定价策略
价格匹配
价格匹配（price matching）:
采取价格匹配策略的企业需要公布一个价格，
并且承诺将匹配竞争对手制定的任何低价。
如果市场中所有的企业都发出声明，
它们就可以设定较高的垄断价格（P）
并获得更大的利润（而非一般的一次性伯川德寡头垄断中的零利润）。
没有企业有动力为试图从对手那里抢夺消费者而制定较低的价格。
因为如果一个企业降价，对手将跟随其降价并重新抢回市场份额。
降价就触发了价格战，可能导致市场份额提高但利润减少。
因此如果所有企业都采取了价格匹配策略，结果是每个企业都制定垄断价格并获得高利润。
价格匹配策略的一大好处是企业不需要监督对手的定价。
即使一些企业定低价格，
采用价格匹配策略的企业也可以
在发现低价格和没有发现低价格的消费者之间实施价格歧视。
在选择价格匹配策略之前，需要考虑两点：
一是必须设计一种机制排除那些没有发现低价却声称发现了低价的消费者。
避免这种欺骗的一种方法是承诺仅匹配那些在公开报纸上公布的价格，且消费者必须提供这些广告才能获得补偿。
二是如果竞争对手的成本低于本企业，采用价格匹配策略可能陷入困境。
提升品牌忠诚度
通过提高品牌忠诚度，当其他企业降价时，本企业可以减少转移到其他企业的顾客数量。
最普遍的方法是广告战，宣传企业的产品优于其他竞争者。
但是如果消费者认为产品无差异，广告策略将失效。
一些加油站推出”多次加油“活动，这模仿了航空公司的常旅客计划。
随机定价
随机定价（randomized pricing）——企业随时改变其价格。
该策略使企业获利的原因有两点：
一是当企业采取随机定价策略时，消费者难以根据经验了解到市场中哪家企业定价最低。通过增加最低价的不确定性，企业可以降低消费者为了价格信息去逛商店的动力。当消费者对竞争对手的价格信息了解较少时，企业就不容易因竞争对手定低价抢夺顾客而遭受损失。
二是随机定价削弱了竞争对手降价的能力。使得竞争对手无法准确掌握市场上的最低价格，从而不知道如何定价才能低于其他企业。随机定价策略的目的是降低竞争对手开展价格战的动力，从而增加企业利润。
随机定价并非总是有利的。有时，价格匹配等策略可能是增加利润的更有效方法。
促销价格每周都在变化，竞争对手将无法在广告促销中制定低于该企业的价格。
第十二章 信息经济学
导言
首先描述不确定性的含义以及不确定性对消费者行为的影响，其次阐述管理者处理风险的方法，最后将描述特定市场比如拍卖市场中不确定性的重要性。
均值和方差
随机变量x的均值或期望值（mean expected value）被定义为不同的结果出现的概率乘以相应的收益的总和。
如果随机变量可能的结果为X1,X2,X3.....XN，而结果所对应的概率为q1,q2,q3.....qn,则X的期望值为：
式中：q1+q2+q3+....+qn=1
因此随机变量的均值是基于不同结果出现的概率得到的一个统计量。
方差（variance）是最常用的风险测量指标，用来表示各种可能的结果与均值的偏差。
随机变量的方差是不同结果出现的概率乘以结果与均值偏差的平方的总和。
用公式表述，假如随机变量的各种可能结果为X1,X2,X3.....XN，相应的概率为q1,q2,q3.....qn,，x的期望值给定为E[x]，则x的方差为：
标准差（standard deviation）为方差的平方根：
特别是，任何随机变量的结果至少有75%的概率在均值的两个标准差范围内。
不确定和消费者行为
风险规避
风险厌恶（risk averse）指对于一个期望值为M美元的风险结果的偏好胜于确定的M美元。
风险偏好（risk loving）指对于期望值为M美元的风险结果的偏好胜于确定的M美元。
风险中性（risk neutral）指对期望值为M美元的风险结果和确定的M美元的偏好一样。
风险厌恶者的决策
当赌博结果很重要时，
大多数人都是风险厌恶型的。
产品质量风险分析可用于考察消费者在产品质量不确定时的选择。
企业可采用两种策略来吸引风险厌恶者去尝试新产品。
① 降低新产品价格（低于现有品牌价格），弥补消费者尝试新产品的风险。
② 想办法使消费者相信新产品的预期质量高于现有产品的质量。典型的做法如做比较广告。
连锁店 风险厌恶还可以解释为什么企业愿意加盟连锁店，而非独立经营。
保险 若消费者是风险厌恶者，他们愿意为了规避风险而支付一定费用。通过购买保险，放弃小额金钱（相对于潜在损失）而避免巨大损失带来的风险。
一些企业通过”返现“承诺给予消费者一定的保险。
消费者搜索
免费回购指消费者可以在任何时候返回该店以100美元购买手表，不会产生成本。
寻价指其他企业的定价分布不会仅仅因消费者发现一家商店对手表定价100美元而改变。
消费者寻价的期望收益取决于其搜索到的最低价格。
假设已知的最低价格为P，搜索更低价格（低于P）的期望收益（EB）向上倾斜，
即在发现某低价后，搜寻一个更低价格而节省的钱呈递减趋势。
假设每次搜索的成本为水平线c。
如果消费者找到一个低于R的价格，则最好接受该价格（停止寻价，购买该产品），
因为寻到一个更低价格的期望收益小于其成本。
保留价格（reservation price）R指的是一个价格水平，
消费者按该价格购买产品与继续寻找低价是一样的。
用公式表示，如果EB[p]是寻找低于p的价格的期望收益，
c代表每次寻价的成本，保留价格应该满足以下条件：EB[R]=c
原理 消费者搜寻法则
最优搜寻法则指消费者拒绝大于保留价格（R）的价格，
接受低于保留价格的价格。
当企业定价高于消费者的保留价格时，消费者继续搜寻更低的价格；
当企业定价低于消费者的保留价格时，消费者就停止搜寻。
搜索成本的增加使水平线上移到c，从而产生一个更高的保留价格R，
所以搜寻成本上升时，消费者将发现更高的可接受的价格，因此他们将化较少的精力去搜寻。
如果搜寻低价的成本下降，消费者将更乐于搜寻更低的价格。
当消费者对价格拥有不完全信息且面临低搜寻成本时，
管理者设定的最优价格应低于搜寻成本较高时的价格。
必须注意的是，产品定价不能高于消费者的保留价格，否则会促使消费者继续搜寻低价。
如果发现商店里许多消费者犹豫不决，说明产品价格高于他们的保留价格，他们在考虑继续寻求更低的价格。
不确定性和企业
风险厌恶
与消费者一样，管理者对风险结果也有不同的偏好。
多元化——通过多个项目来降低风险。多元化是有利的，但多元化是否为最优选择？
这取决于管理者的风险偏好和管理则规避风险的动机。
股东通常希望管理者是风险中性的。风险中性型管理者更关心风险项目的期望值，而非潜在的风险，其目标是最大化企业的期望利润。
股东可以购买不同企业的股票组合来分散风险，从而消除企业经营中的系统风险。
相比之下，管理者花时间和金钱去分散风险则意义不大。所以，当企业的所有者是风险厌恶者时，他们更希望管理者作出风险中性的决策。
管理者有动力去最大化企业的期望利润。
首次激励，你将以风险中性的方式行事，即使你和企业所有者是风险厌恶者。
生产者搜寻
与消费者搜寻低价商店一样，生产者也搜寻低价的投入品。
当投入品的价格不确定时，企业会采用最优搜寻策略。风险中性的生产者的搜寻策略与风险中性的消费者一样。
利润最大化
风险中性型管理者必须在确定产品需求前决定产量。由于需求不确定，收益也不确定。
为了最大化期望利润，管理者应该在期望边际收益等于边际成本处进行生产：E[MR]=MC
原因在于：如果期望边际收益大于边际成本，管理者可以扩大产量来增加期望利润。
如果期望边际收益小于边际成本，管理者就会减少产量。当产量下降时，企业减少的成本大于其减少的期望收益。
如果管理者是风险中性者，那么需求不确定情况下的利润最大化类似与需求确定情况下的利润最大化。
管理者只需将公式中的边际收益调整为期望边际收益，便可以实现利润最大化。
不确定性和市场
不对称信息
不对称信息（asymmetric information）会导致拥有较少信息的人拒绝参与市场交易。
消费者和企业之间的信息不对称会影响企业利润。
由于信息不对称，消费者可能拒绝购买新产品（即使新产品确实比现有产品好），因为他们不知道新产品是否真的优于现有产品。
不对称信息还将影响其他的管理决策，如雇用员工，向消费者发放贷款。
逆向选择（adverse selecion）是指选择过程导致一些具有经济商不良特征的人出现的情形。
逆向选择通常发生于隐藏特征（hidden characteristics）时——在一笔经济交易中，关于某些特征的信息一方知情而另一方不知情。
相反，道德风险通常出现在一方采取隐藏行为（hidden action）时——该行为是另一方无法察觉的行为。
有时很难区分能力（特征）和努力（行为），所以就难以区分逆向选择和道德风险。
当保险公司提高保费以弥补差司机带来的损失时，只有那些知道自己很有可能发生事故的差司机才愿意付高额保费.
而好司机知道过去的事故都是偶然事件，所以不愿意支付高额保费。
保险公司发现收取低保费且拒绝给有不良驾驶记录者投保，才是有利的选择。否则，那些有不良驾驶记录的差司机会产生逆向选择。
道德风险（moral hazard）如果该合同导致免受损失的一方隐藏行为而损害到对方，就会产生道德风险。
信号传递与信号筛选
激励合同常被用来减少因隐藏行为而导致的道德风险问题。管理者与其他市场参与者利用信号传递和信息筛选来减少逆向选择问题。
信号传递（signaling）发生在拥有信息的一方向不拥有信息的一方发送有关其隐藏特征的信息时。
拥有信息的一方向不拥有信息的一方传递相关信息的信号，必须是可观察的、可置信的，并且难以被模仿。
筛选（screening）指没有信息的一方试图根据相关特征对人们进行分类。
这种分类可以通过自我选择机制（self-selection device）实现：掌握自身特征的人作出一系列选择，这些选择即向没有信息的一方显露的他们的特征。
拍卖
拍卖类型
拍卖过程中，潜在购买者为了得到一种商品、一项服务或者一般意义上任何有价值物品的所有权而竞争。
有些情况下，拍卖者是卖方。有些情况下，拍卖者是买方。在有多个竞拍者的拍卖中，竞拍者之间的竞争有助于拍卖者找到更满意的交易。
竞拍者的风险偏好会影响竞拍策略和拍卖人的期望收益。
假定竞拍者是风险中性者。该假定适用于许多拍卖，因为竞拍者可以通过参加多个拍卖来降低整体风险。
拍卖有四种基本类型：英式拍卖（升价竞拍）；第一价格密封拍卖；第二价格密封拍卖；荷兰式拍卖（降价竞拍）。
这些拍卖的区别在于：（1）竞拍者决策的时间（同时竞拍还是序贯竞拍）；（2）竞拍成功者所需支付的数额。
英式拍卖（English auction）中，单个拍品将被卖给出价最高者。拍卖以公开竞拍方式进行。公开拍卖指拍卖商询问是否有人愿意支付更高价格。
出价在序贯竞拍中不断提高，直到没有参与者希望再加价，最高竞拍者——最后出价者——支付给拍卖商其出价额并拥有该拍品。
注意：英式拍卖中，竞拍者不断地获得另一方的出价信息。
第一价格密封拍卖（first-price，sealed-bid auction），竞拍者在不知道其他参与者出价的情况下，在纸上写下自己的竞价。
拍卖商收集竞价并将拍品交给出价高的竞拍者。该竞拍者支付给拍卖商出价数额。
因此， 第一价格密封拍卖与英式拍卖一样，最高出价者赢得拍品。
然而与英式拍卖不同的是，竞拍者不知道其他参与者的竞价且由拍卖商来收集价格。这个特征会影响竞拍行为。
第二价格密封拍卖（second-price，sealed-bid auction），竞拍者同样在不知道其他人出价的情况下提交竞价。
最高出价的人获胜，但只须支付第二高的出价。
荷兰式拍卖（Dutch auction）中，卖者开始提出一个很高的价格（价格太高以至于没有人愿意支付）。
拍卖商逐步降低价格，直至有买者愿意以此价格买该拍品，于是拍卖结束：竞拍者按最后宣布的价格购买该拍品。
在荷兰式拍卖中，竞拍者所获得的信息与第一价格密封拍卖是一样的。其他参与者的出价信息砸死拍卖结束时才知道。
因此，荷兰式拍卖与第一价格密封拍卖的策略是一样的。
原理 荷兰式拍卖与第一价格密封拍卖的策略等价
荷兰式拍卖与第一价格密封拍卖在策略上是等价的；也就是说，竞拍者的最优出价在这两种拍卖中是一样的。
信息结构
每个竞拍者知道自己对拍品的估价，但不可能知道其他竞拍者的估价，有时甚至无法了解物品的真实价值。这就涉及信息结构问题。
独立的个人估价（independent private values）指每个竞拍者对拍品的估价取决于他们各自的品位，且这种品位只有竞拍者自己知道。
个人估价独立意味着个人不依赖于其他人的估价信息——即使参与者知道其他竞拍者的估价信息，他的估价也不会改变。
但是这些信息可能会改变他在拍卖过程中的竞拍行为。
关联（相关）估价（affiliated （correlated） value estimate）
每个竞拍者基于其对某拍品的估价作出决策，但是竞拍者之间的估价是相关的，也就是存在关联性：某竞拍者的估价高，则其他竞拍者的估价可能更高。
共同价值拍卖（common value auction），拍品的潜在价值对所有竞拍者来说相同。
个人品位对竞拍者的估价毫无影响。不确定性纯粹源自不同竞拍者利用不同的信息对共同价值的不同估价。
风险中性者的最优竞拍策略
独立的个人估价拍卖的策略
原理 英式拍卖的最优竞拍策略
在独立个人估价的英式拍卖中，竞拍者的最优竞拍策略就是拍卖过程中保持活跃，直到竞价超出自己对该拍品的股价。
原理 第二价格密封拍卖的最优竞拍策略
独立个人估价的第二价格密封拍卖中，竞拍者的最优策略就是竞价等于他对拍品的股价。这是占优策略。
原理 第一价格密封拍卖的最优竞拍策略
独立个人估价的第一价格密封拍卖中，
竞拍者的最优策略是出价低于自己对项目的股价。
如果有n个竞拍者且都知道估价平（或一致）
地分布在最低的可能股价L和最高的可能股价H之间。
那么，自身股价为v的竞拍者的最优竞拍价格为：
式中，b代表竞拍者的最优化竞价。
关联（相关）估价拍卖的策略
关联（相关）估价的最优竞拍策略最难描述，原因有二：
一是竞拍者不知道自己对物品的估价，也不知道其他人的估价。二是拍卖过程可能会暴露其他竞拍者的估价信息。
赢家诅咒（winner‘s curse）：获胜向竞拍成功者传递的信息就是，其他所有企业都认为该项目的价值低于获胜者的支付额。
注意：如果某竞拍者能够综合说有竞拍者的信息，那么，它对拍品的评估可能更准确。赢家诅咒代表了一种风险，谨慎的管理者会想办法规避。
原理 避免赢家诅咒在共同价值拍卖中，赢家是对拍品的真实价值最乐观的竞拍者。为了避免赢家诅咒，竞拍者应该根据实际情况下调其个人估价。
在英式拍卖中，关联股价拍卖的最优策略是，只要价格未达你的估价就继续竞价，无论是基于你的个人信息还是拍卖过程中收集的信息。
拍卖中的期望收益
假设拍卖商关注其期望利润最大化。
从拍卖商的视角来看，最好的拍卖取决于竞拍者拥有的信息的特征。
第十三章 企业战略中的高深专题
导言
”如果你不喜欢正在进行的博弈，那就寻找策略来改变博弈。“
区分三种策略，这些策略被管理者用来改变环境。
有两种策略——限制进入定价和掠夺性定价，可用于减少竞争对数的数量。
还有一种降低竞争水平的策略，旨在提高竞争对手的固定成本或边际成本。
这三种策略都涉及经济利益的权衡，选择某种策略之前，管理者必须权衡策略的潜在收益能否弥补相关成本。
策略的合法性：通过消灭对手来提高企业利润的做法，很可能招致一个或多个反垄断机构的诉讼。
先动优势在网络行业（如通信、航空和互联网等行业）如此重要，还将分析渗透定价，这是进入者改变博弈进而突破潜在进入障碍的一种策略。
限制进入定价
限制进入定价的理论基础
限制进入定价（limit pricing）是指一个垄断企业（或具有市场势力的企业）
设定低于垄断水平时的价格以阻止其他企业进入市场。
限制进入定价并没尽善尽美，采用策略时需谨慎。
垄断企业的产品需求曲线DM。当价格为PM时，垄断利润最大为ΠM。
如果一个潜在的进入者拥有与在位企业相同的生产成本，
当它进入市场时，垄断企业的利润将被侵蚀，
该行业将有垄断变为寡头垄断且在位企业的利润减少。
随着进入市场的企业增多，利润将被进一步侵蚀。
假设进入者的成本与在位企业的成本完全一样，
同时进入者拥有在位者的成本即产品需求的完全信息。
在位企业为了限制价格将产量水平定位QL（超过垄断产量QM），因此价格PL低于垄断价格。
若潜在竞争者认为其进入后，在位企业将继续生产QL单位的产量，
则进入者的剩余需求为市场需求DM减去在位企业的产量QL，进入者的需求曲线DM-QL。
进入者的剩余曲线从价格PL开始（此价格水平下DM-QL为零）。
当价格低于PL时，进入者的剩余需求曲线与垄断者需求曲线的水平差距都是QL。
进入者的剩余需求曲线低于平均成本曲线，这意味着进入无利可图。
请注意，如果进入者的产量低于或高于Q，都将面临损失。
如果进入者的产量为Q，则市场总产量增加到Q+QL。
对于进入者来说，价格P=AC，经济利润为零，因此进入该市场无法获得正的经济利润。
一般来说，进入新市场还会发生额外成本，进入者亦无动力进入该市场。
限制进入定价成功阻止了潜在企业的进入，
在位企业得以保持高利润（高于新企业进入的利润，但低于完全垄断时的利润）。
限制进入定价失效
(参考上图）在位企业的低价策略可能无法阻止进入者：进入者选择离开的原因是它认为在位企业的产量至少是QL。
进入者会降低在位企业的边际收益，导致最优产量低于QL。
所以进入者有可能迫使在位企业放弃其产量威胁。认识到这一点，理性进入者会发现即使在位企业的价格在PL处，进入市场仍然有利可图。
为了有效地阻止进入，在位企业有必要采取一些行动来降低进入者进入后的利润。
原理 有效的现值进入定价一个能够有效地阻止理性竞争对手进入的现值进入定价策略，必须将潜在进入者进入前的价格与进入后的利润联系起来。
进入前的价格
与进入后的利润挂钩
承诺机制
在位企业需要以某种”自缚双手“的方式来释放可置信承诺，
使潜在进入者相信，即使其进入，在位企业也不会减少产量，
这种策略能够有效阻止进入。
在位企业的承诺包括建造产能不低于QL的工厂，
表明在位企业的产量会超过QL。
在位企业有先动优势，这个先动优势使它率先决策：
（1）承诺建造产能不低于QL的工厂；
（2）不承诺建造任何产量水平的工厂。
一旦在位企业作出决策，进入者就要决定进入或不进入。
唯一子博弈的完全纳什均衡是不进入市场。
学习曲线效应（learning curve effects）生产历史悠久的企业拥有更多的生产经验，比那些没有经验或者只有少量经验的企业的生产成本低。
学习曲线效应把进入前的价格和进入后的利润挂钩，使得在位企业可以利用限制进入定价来阻止新企业的进入。
由于在位企业第一期生产了更多产品，第一期的产品价格下降。
这个低价格与进入后的利润就挂起钩来（通过学习曲线效应），因此这种限制进入定价能够有效地阻止进入。
不完全信息对企业家和潜在进入者来说，要找到有利的商业机会需要付出成本。
进入前的价格或者利润是一种信号，这种信号使潜在进入者识别获利机会的成本降低，于是在位者可能会伪造进入前的价格和进入后的利润的联系。
限制进入定价就向潜在进入者”隐瞒“了有关的利润信息，从而推迟甚至完全阻止进入（比较少见），其能力取决于潜在进入者从其他渠道获取信息的成本。
还有一种情况就是潜在进入者不清楚在位企业的真实成本。如果潜在进入者得知在位企业的成本高，它会发现进入市场有利可图。
当在位企业定低价时，进入者会推断在位企业的生产成本较低，进而认为进入无利可图。
声誉效应在进入决策中，这种剩余效应可以将进入前的价格与进入后的利润相联系。
动态效应
尽管在位企业可以伪造进入前的价格和进入后的利润之间的关系来阻止进入，
但有时允许进入也可能增加利润。
假设利率为i，ΠM为垄断者定价PM（垄断价格）时的当期利润，
如果该企业能无限期地维持其垄断地位，那么当期与未来利润的现值为：
假设新进入者观察到当期价格PM，决定进入这个市场。
进入改变了第二期及未来各期的竞争，将在位企业的利润从垄断水平（ΠM）降到寡头水平（ΠD）。
寡头利润都低于垄断利润，即ΠD＜ΠM。
如果发生进入，在位企业的当期和未来利润的现值降为：
公式左边是限制进入定价的未来收益现值，右边是限制进入定价的当期收益差额。
根据以上分析，限制进入定价在以下情形下显然更具吸引力：
（a）利率较低；（b）限制进入定价下的利润与垄断价格下的利润相近；（c）垄断利润明显低于限制进入定价下的利润。
缺乏上述条件，在位企业通过限制进入定价策略来阻止进入就不是最优选择。
若企业面对进入时不采用限制进入定价策略来组织，随着时间的推移，将有越来越多的企业进入市场参与竞争，导致市场价格越来越低。
减少竞争的掠夺性定价
掠夺性定价（predatory pricing）是通过消灭现有竞争者来减少竞争。
具体而言，采取掠夺性定价策略时，一个企业把价格设定在其边际成本以下就是为了把竞争对手赶出市场。
一旦”猎物“（竞争对手）离开市场，由于竞争减少，”掠夺者“（发起掠夺性定价的企业）就有能力提高产品价格。
因此，掠夺性定价涉及当前和未来利润的权衡：如果未来利润现值足以弥补当期为了将竞争对手赶出市场而产生的利润损失，掠夺性定价就是有利可图的。
掠夺性定价不仅仅有损”猎物“，也有损”掠夺者“
该策略的成功关键在于：”掠夺者“是否比”猎物”更“健壮”。在市场上建立强势去干竞争对手的声誉，将使其他企业不敢轻易进入该市场，但也可能导致小型竞争对手宁愿以低廉价格卖给大企业，也不愿意被“掠夺者”赶出市场。
当“掠夺者”的售价低于成本时，“猎物”可以减少生产或停止生产（这导致“掠夺者”的损失大于“猎物”的损失），
或者从“掠夺者”处购入产品并囤积。等掠夺性定价结束后再销售。
注意：掠夺性定价对“掠夺者”而言实施成本很高。
一般情况下，掠夺性定价很难将相似的竞争者（相似的规模、成本、财务资源以及产品诉求等）赶走，只能赶走那些小型竞争者（“空口袋”的竞争者）。
垄断性定价容易被起诉违反了《谢尔曼法》，但在法庭上对掠夺性定价往往难以取证。
如果阻止竞争，则会股利那些无效率企业的进入。
如果严格管制反掠夺性定价，还可能导致企业间的合谋，这是因为企业担心被认定为掠夺性定价而遭到起诉，所以通过合谋规避竞争。
此外，厂商为了让新产品进入市场，通过采用低价销售甚至免费赠送方式吸引消费者试用，
等消费者认可后再提高价格，这种行为对企业是有利的。
如果掠夺者具有较大的成本优势，也可能采取类似的策略，此时定低价是为了赶走缺乏效率的企业，
在这种情况下，在位企业无须将价格设定在边际成本之下，仅需要将价格设定在平均成本即可。
但与掠夺性定价相似的是，当无效率企业退出市场后，有效率的企业就可以提高产品价格。
提高竞争对手成本来减少竞争
边际成本策略
通过提高竞争对手的成本
（raising rivals‘ costs），
可以改变竞争对手的决策动机，
最终影响其价格、产量以及进入决策。
企业还可以提高竞争对手的分销成本。
固定成本策略
企业可以通过提高竞争对手的固定成本而获益。
即使企业同时增加了自身的固定成本，仍然可以增加企业利润。
假设在位企业可以成功地游说政府实施管制，
如要求进入市场中的每个企业（包括在位企业）必须从政府处获得经营许可证。
纵向一体化策略
纵向圈定（vertical foreclosure）
是一种提高竞争对手成本的极端策略，通常在企业控制关键的上游投入品以规避下游市场竞争时采用。
企业拒绝向其他下游企业出售必需的投入品，迫使其他下游企业寻找低效替代品，就会增加它们的生产成本。
如果找不到替代品，竞争对手将被赶出下游市场，因为找不到关键投入品。
当通过关键投入品索取高价来驱逐其他下游竞争对手时，纵向一体化企业就放弃了上游投入品的销售利润。
只有当下游市场的利润增加（因为市场势力提升）足以抵消上游投入品的利润损失时，纵向圈定策略才是有利的。
价格 — 成本挤压（price-cost squeeze）
纵向一体化企业可以通过价格 — 成本挤压提高竞争对手的成本。
纵向一体化企业在提高竞争对手的投入品成本同时，
维持最终产品的价格不变甚至降低价格。
企业需权衡短期利润损失和竞争对手退出下游市场后的潜在利润现值。
大型纵向一体化企业可以通过价格—成本挤压策略
“惩罚”下游市场中的非合谋企业（如不参与市场分割或其他合谋协议）。
尽管价格—成本挤压策略导致纵向一体化企业在惩罚对手时面临短期利润损失，
但是这种投资可以形成声誉，从而使纵向一体化企业在重复博弈市场上获取更高的未来利润。
价格歧视
价格歧视作为一种策略性工具，有助于限制进入定价、掠夺性定价以及提高竞争对手成本策略的事实。
决策时机或行动顺序
先动优势
先动优势指企业在其竞争对手作出决策之前先作出决策，从而获得更高的回报。
假设两家企业（企业A和企业B）必须制定产量决策（产量高或低），
它们有两种情形：
情形一，两家企业同时作出产量决策（不存在先动优势的机会）；
情形二，企业A先于企业B作出决策。
描述了同时行动博弈的标准形式。
这个同时行动博弈的唯一纳什均衡是：企业A选择低产量，企业B选择高产量。
假设企业A可以改变决策时机——它能够在企业B之前作出决策。
更重要的是，企业B在作出决策前能够观察到企业A的决策且企业A也知道这一点。
唯一子博弈完全均衡是：企业A选择高产量，企业B（观察到这点）选择低产量。
需要强调的是企业A的先动优势依赖于一下三个重要条件：
（1）企业A的高产量决策不可逆；
（2）企业B观察到企业A的决策之后再制定其决策；
（3）公共信息（企业B知道企业A的决策不可逆，同时企业A知道企业B了解这一点，以此类推）。
即使不存在专利保护，学习曲线效应带来的先动优势仍然存在。
在存在网络效应的行业中，先动优势尤其显著。
活动优势
后来者可以对先行者的投资“搭便车”，这就使得后来者较先行者成本更低，进而可能获得更高的收益。
此外，后来者还可以了解先行者的经验和失误而获得优势，以更低的成本生产出更好的产品。
克服网络效应的渗透定价
网络的含义
许多行业（如航空、电力以及网络拍卖等）存在网络效应，网络效应使在位企业形成潜在进入者很难克服的先动优势。
网络由地理上或经济空间中不同的点（节点）相互连接而成。
最简单的网络是单向网络。在单向网络中，服务单向流动。典型的例子是居民供水系统。
单向网络的显著特征就是：单向网络对于每个消费者的价值与有多少人在使用这个网络无关。
但在双向网络（如电话系统、电子邮件或互联网，包括即时信息）中，网络对于每个客户的价值直接与用户规模（有多少人使用这个网络）相关。
即使不存在规模经济，双向网络的现有提供者仍然拥有显著的先动优势（因为用户数量很大）。
网络的外部性
双向网络显示出正的外部性特征，也称为直接网络外部性（direct network externality）：
随着网络用户数量（网络规模）的扩大，网络所带来的每单位服务的价值也在增加。
一般意义上讲，如果网络中有n个使用者，将出现n(n-)中潜在连接服务。
这意味着网络中每增加一个用户，将增加2n中潜在的连接，这将使所有用户直接受益。
原理 直接网络外部性连接n个用户的双向网络提供了n(n-1)种潜在的连接服务。
当一个新用户加入这个网络，会为整个网络增加2n中潜在连接服务，所有的现有用户会直接从中受益。
双向网络还存在间接网络外部性（indirect network externality），也称网络互补性（network complementarities）。
网络互补性源于特定网络的使用的增加。双向网络和单向网络都会产生网络互补性。
随着越来越多的软件被开发出来并在互联网上广泛使用，互联网对于每个用户的价值不断提高，这就是网络互补性的表现。
注意，网络中也可能出现负的外部性，这取决于网络规模的瓶颈。比如随着网络规模的扩大，最终会达到临界点——现有基础设施无法容纳更多的用户。
超过临界点后，新用户的加入会带来拥堵，从而降低网络对于每个用户的服务价值。
消费者锁定带来的先动优势
由于网络外部性的存在，新网络想取代现有网络或者与现有网络展开竞争都非常困难，
即使新网络在技术上领先于现有网络，也很难战胜现有网络。
因为现有网络拥有更多的用户和互补性服务，
现有网络为每个用户提供的总价值（由于直接或间接的网络外部性）比新网络更高。
网络外部性带来了消费者锁定：消费者陷入一种僵局（均衡）——他们都在使用技术水平低的次级网络。
“改变博弈”的渗透定价
渗透定价（penetration pricing）是后来者可以选择的一种策略，
后来者在初始阶段制定低价、免费赠送，甚至向消费者付钱，
来吸引潜在用户试用自己的产品，从而获得尽可能多的用户。
这使得用户规避风险：用户在继续使用现有网络的同时体验新网络。
通过渗透定价，新进入者能够克服在位企业的先动优势（源于网络外部性）。
第十四章 市场中的政府
导言
政府运用规则与法规在影响着厂商和消费者的每一个决策。
自由市场无法提供社会有效产出的四个原因：（1）市场势力；（2）外部性；（3）公共物品；（4）不完全信息。
市场失灵
市场势力
政府干预市场的原因之一是自由竞争市场有时无法以社会有效价格提供社会有效数量的产品。
拥有市场势力（market power）的企业将使产出低于社会有效产量，
这是因为其定价超过产品的边际成本。
假设垄断企业读市场上所有的消费者制定相同的价格，
则利润最大化产量为QM，产品按垄断价格PM出售。
在此价格处，
消费者为最后一单位产量支付的价格高于生产者的生产成本，
见W区域。
注意：三角形ABC的面积就是垄断带来的无谓损失。
在完全竞争市场中，该面积表现为社会福利的一部分，
但由于垄断企业具有市场势力，该部分福利无法实现。
政府通过反托拉斯政策来阻止垄断的形成和市场势力的提高，
由此避免垄断造成的无谓损失。
但在有些情况下，如由于规模经济的存在，政府又希望形成垄断。
这时政府通常会对垄断企业实施间隔管制，以减少无谓损失。
反托拉斯政策（antitrust policy）
对管理者来说，反托拉斯政策通过规制诸如固定价格协议和其他合谋行为等来减少垄断造成的无谓损失。美国于1890年颁布的《谢尔曼法》第1条和第2条的内容奠定了反托拉斯政策的基石。
反托拉斯政策中的模糊法条及历史案例中的争议内容，很大部分通过法庭作出解释。
合理性原则已经成为法庭裁决反托拉斯案件的重要依据。
合理性原则指出，并非所有的贸易限制皆为非法行为，只有那些不合理的贸易限制才应该被法律禁止。
根据合理性原则，法庭可以认定，单纯根据企业规模大小不足以判定企业有罪。
如果要裁定某企业行为违反了《谢尔曼法》第2条，必须首先证明企业所采取的行为明显削弱了市场竞争。
合理性原则带来的问题是：企业管理者事先往往很难分辨哪些提高利润的定价策略违反了法律规定。
规定只有那些为了减少竞争或促成垄断的价格歧视行为才属于违法行为。
法律上关于价格歧视是否违法仍然描述模糊，而且不同国家的法律不同。
在反托拉斯行动的并购事件中，效率标准只在对竞争影响不大时才会被考虑。
美国司法部反托拉斯局和联邦贸易委员会（FTC）是反托拉斯法的执行机构。
面对“第二次要求”，政府和当事方一般会在起诉之前达成协议
典型的做法是，企业会向第三方出售其在大量重叠的产品或地缘市场中的资产，从而消除政府对反托拉斯问题的顾虑，促使政府允许并购达成。
价格管制
当某产品/服务存在非常大的规模经济时
（一些公用事业公司出现这种情况），
由一家企业向市场提供产品/服务最经济。
此时政府通常会允许一家企业垄断市场，
但会管制垄断企业的定价以减少无谓损失。
如果不实施价格管制，垄断企业的利润最大化定价为PM，产量为PQ，
而完全竞争企业的产量为QC。
法律规定垄断者的定价不能高于PC，所以当产量低于QC时，
尽管消费者的支付意愿高于PC，但企业所能索取的最高价格为PC，
因此垄断者的有效需求曲线为PCBD。
注意：点B左边（产量小于QC）的需求曲线呈水平，类似于完全竞争市场的需求曲线；
当销售量超过QC，产品价格将降至PC之下。
为了实现利润最大化，垄断者将在边际收益（PC）等于边际成本处（即点B）生产，相应的产量为QC。
如上所述，价格管制将导致垄断者在完全竞争的价格水平上提供完全竞争的产量。
价格管制的目的是消除因垄断造成的无谓损失。
政府的管制政策虽然降低了垄断者的利润，但是增加了社会福利。
假设政府将价格设定为P，
此时垄断者的产品需求曲线为PFD，
产量低于Q单位时的边际收益曲线为PF。
被管制的垄断者为了实现利润最大化将在点G
（边际收益等于边际成本处）进行生产，
对应的产量为QR——远低于非管制下垄断者的产量。
存在Q-QR单位的供给短缺。
管制价格下的无谓损失为图中的面积R+W，
也远远大于非管制下的无谓损失（图中的面积W）。
总之，当政府不了解需求和成本的准确信息，
或者由于其他原因限定的管制价格太低时，
就会降低社会福利并造成供给短缺。
垄断企业在点A刚好实现收支平衡，如果不受管制，垄断者的产出为QM且价格为PM。
价格等于产出的平均总成本，在不存在管制的情况下改垄断者的经济利润为零。
加入管制价格为PC，从长期看企业的产量为零。
这是因为企业的平均总成本曲线位于管制价格的上方，
这意味着垄断者只要进行生产就会遭受损失。
所以若价格限定为PC，从长期看垄断者将退出市场，
造成该产品在市场上无处购买，这将有损市场上所有参与者的福利。
为了避免此类现象的发生，政府必须对垄断者的损失进行补贴，
而补贴资金通常来源于税收，为此又得提高税收，导致消费者为低价而付出代价。
此外，由于所有亏损皆由政府来补贴，导致获得政府补贴的垄断企业没有任何动力去降低成本。
垄断和完全竞争进行比较时需要注意：垄断势力的一个关键来源是规模经济。
规模经济使得某些产品在完全竞争市场上无法生产。
完全竞争下产量不可能维持在QC水平，因为此时边际成本曲线与需求曲线的交点定于平均总成本曲线。
外部性
在某些产品的生产过程中，一些没有参与生产或消费过程的人也要付出成本。
这种外部成本称为负外部性（negative externalities）。
典型的负外部性的例子就是污染。
负外部性质污染给社会带来的边际成本。
假设钢铁生产处于完全竞争市场，供给曲线为S，即行业中所有钢铁企业的边际成本之和。
如果允许钢铁企业免费向河流中排放污水，则市场均衡在点B处，市场的均衡产量为QC，钢铁价格为PC。
然而在产量为QC时，社会除了支付给钢铁企业价格PC，还须支付一个边际价格，点A处的价格。
这是污染所带来的额外的社会成本。假设由企业承担污染所带来的社会成本，
则它的边际成本曲线为供给曲线和污染的社会边际成本的加总，加总后的曲线即生产钢铁的社会边际成本。
总和考虑生产钢铁的总成本和总收益，则社会有效产量在点C处。
此时钢铁的社会有效产量为QS，低于完全竞争市场的产量；
钢铁的社会有效价格水平为PS，高于完全竞争的价格水平PC。
换言之，由于存在外部成本，市场均衡的产量高于社会有效产量，市场价格低于社会有效水平。
导致市场失灵的根本原因是产权界定不完善。
政府也可以把自己视为环境所有者来解决污染问题，
政府利用其权力能够使产出和污染达到社会有效水平。
《清洁空气法》
为了解决污染带来的外部性问题，
美国国会于1970年通过了《清洁空气法》，并于1990年度改法进行了修订。
《清洁空气法》所涵盖行业中的企业，只有取得排污许可证才可以排污。
任何价格水平下的产品数量皆有所减少。
因此，《清洁空气法》使市场均衡从Q0降至Q1，市场价格从P0上升到P1。
这正是我们希望看到的解决负外部性问题的所带来的变化——产出减少，价格上升。
这部新法允许许可证在行业内或跨行业的企业间转售。这将从两个方面减少污染：
第一，允许新企业在市场需求增加时进入该行业；
第二，激励在位企业采用新技术及更环保的生产方法。
假如需求增加发生在污染行业且污染许可证不允许在行业内或行业间买卖，那么新企业将无法进入。
所以当某产品需求增加时，排污权的市场化使新企业有条件进入该市场。
排污许可证的出售也给企业提供了一种激励，使企业愿意开发和创新相关技术以减少污染。
公共物品
市场失灵的另一个来源：公共物品（public goods）指那些具有非竞争性和非排他特征的物品，它不仅使购买者收益，而且使其他非购买者收益。
非竞争性消费（nonrival consumption）包括无线信号、灯塔、国防设施以及环境保护等。
公共物品一旦启用，每个人都可以消费它，没有人可以被排除在外。大多数产品或服务的消费都是非排他性消费（nonexclusionary consumption）。
公共物品一旦启用，每个人都可以消费，那么人们就失去了对该物品付钱的动力，而更倾向于让其他人为该物品付费，自己可以在别人努力的基础上搭便车。
政府解决公共物品问题的方法是向每个人强制征税，不管每个人是否希望得到政府的服务。
政府用税收来支持公共项目。如果缺乏政府干预，市场自己不会提供这些商品。
在此要强调一点，政府也可能无法提供社会有效数量的公共物品。政府通常可能会提供给过多的公共物品。
企业在某些市场提供公共物品可能是有利的，这些可以为企业带来声誉，从而形成品牌忠诚度并提高产品需求。
由于公共物品具有非竞争性和非排他性特征，企业不用太昂贵的方式提供公共物品，是大量的消费者从中受益，这在某中情况下是一种有效的广告策略。
企业在公众面前展示了更好的形象。企业很难计算出最优的公共物品提供量。
不完全信息
要保证市场有效运行，市场上的参与者必须获得合理的信息。当参与者拥有不完全信息时，就可能导致要素投入或产出的无效率。
政府往往扮演市场信息提供者的角色。
信息不对称是导致市场势力的重要原因之一。
禁止内幕交易
股票市场上禁止内幕交易，就是政府通过政策来减少不对称信息引致市场失灵的例证。
证书
技能证书及证书鉴定是政府用来传播信息和减少信息不对称的另一项政策。
证书的目的是使收集信息的成本集中。所有由政府颁发许可证的行为都属于证书鉴定的范畴。证书也可以视为一系列最低标准。
《诚实借贷法》
近年来出台的一系列法律是为了是借款人能更容易地收集相关借贷信息。
诚实广告法
一般而言，企业较消费者对产品拥有更多信息，这使得企业有动力利用消费者对产品不了解而对产品的优点进行虚假宣传。
极端情况下，不对称信息可能导致消费者认为所有广告都是虚假的。
强制执行合同
签订合同的目的是防止当事人在博弈最后阶段的机会主义。
政府通过强制执行合同可以解决最后期限导致的市场失灵问题。
寻租
政府的政策通常会牺牲一部分人的利益来保护另一部分人的利益。
因此，有些人（游说者）就会花大量金钱和时间来影响政府政策。这个过程就是寻租（rent seeking）。
假设某政治家有权力管制垄断企业。当前垄断企业制定的产品价格为PM，产量为QM，所获最大利润为图中的阴影区域A。
在垄断价格和垄断产量下，消费者剩余为图中的三角形区域C。
假如消费者能够说服政治家，将垄断产品的价格限定为完全竞争水平的价格PC，则产量将变为QC。
由此垄断企业将失去其所有可获利润，但消费者总剩余将大大增加，即区域A、区域B和区域C之和。
实施管制会导致垄断企业丧失利润，因此垄断企业会极力游说政府以阻止其实施管制措施。
为了规避管制，垄断者愿意花的费用最多为A。这种花费既可能是合法行为，也可能是不合法行为。
注意，消费者也愿意花钱游说政治家对垄断者实施管制。当产品价格限定为PC时，消费者作为整体获得的消费者剩余间增加A+B。
当然，每个消费者的可获利益比消费者整体的利益少得多。
结果是每个消费者都希望搭便车，最终导致消费者整体用于游说的费用非常少。
结果，在游说政府方面，垄断企业通常比消费者花更多的费用，因此垄断企业经常通过寻租来避开法律管制。
在国际市场的交易活动中，寻租通常表现为某种形式的政府干预。
政府政策与国际市场
配额
配额（quota）的目的在于限制外国竞争者所生产的产品进入本国的数量。
本国企业的收益增加以牺牲本国消费者与外国企业的利益为代价。
在实施配额前，本国需求曲线为D，
外国生产上的供给曲线为S外国，
本国生产商的供给曲线为S本国，
市场上的总供给曲线为外国生产商的供给曲线与本国生产商的供给曲线的水平相加，即为S(F+D)。
因此当没有配额限制时，市场的均衡点位于K，K点的均衡价格为P(F+D)，均衡产量为Q(F+D)。
在实施配额后，外国企业的供给曲线为GASF配额，本国企业的供给曲线为S本国。
因此在实施配额后，整个国内市场的总供给曲线为GBC，市场均衡点位于M，本国企业的价格上升至P配额，
本国企业能够获得更高的利润，但造成了无谓损失。
社会总福利减少的原因在于：配额给国内消费者和外国企业带来的总损失要大于本国企业从配额中得到的收益。
所以说，本国企业通常有很强的动力去游说政府对进口产品实行配额限制。
关税
关税也可以保护本国企业的利益，
本国企业的利益同样会有损本国消费者和外国企业的利润。
总量关税（lump-sum tariff）是一笔固定费用，
是外国企业为了合法地在某国市场上销售产品而必须向该国政府交纳的固定费。
从量关税（per-unit (excise) tariff）是根据进口数量交纳的费用，
是本国的进口商为其进口的每单位产品向政府支付的相应费用。
总量关税对企业的边际成本曲线没有影响。
但是总量关税使企业的平均成本曲线由AC1上升到AC2，
除非国内市场的价格高于P2，
否则进口商间不愿意为进入国内市场而支付关税。
在征收关税之前，外国企业的供给曲线为ES(F)，本国企业的供给曲线为ES(D)，
市场总供给曲线为外国企业供给曲线与本国企业供给曲线的水平叠加，即ES(D+F)。
征收总量关税之后，外国企业的供给曲线变为ASF，只有当国内市场的价格高于P²时，
外国企业才愿意进入该市场，
否则进口商将不愿为进入该市场支付总量关税，
所以，征收总量关税之后的国内市场的总的供给曲线为EBCS(D+F)。
如果需求曲线与国内供给曲线相交时的价格低于P²，
外国企业将不进入该国市场，
这意味着实施总量关税的结果是将外国企业从国内市场挤出去。
所以，当国内市场需求较低时，总量关税将增加本国企业的利润，
但如果国内市场需求较高，总量关税对本国的利润不产生影响。
如果对外国企业征收货物关税而非总量关税，
那么，在所有的需求水平上，本国企业都将获益。
征收关税之前，外国企业的供给曲线为S(F)，本国企业的供给曲线为S(D)，
市场总的供给曲线为ABS(D+F)，
所以市场均衡点为H，这是不存在关税时的均衡点。
当对每单位产品征收关税T时，
外国企业的边际成本去下能将向上移动关税量T，导致外国企业的供给曲线变为S(F+D)。
所以市场的总供给曲线变为ABS(D+F+T)，新的均衡点为E。
总之，关税造成国内消费者必须为产品支付更高的价格，
所以在本国企业利润提升的同时损害了国内消费者与外国企业的利益。
第一章 管理经济学基础
导言
管理者
管理者（manager）指为了实现既定目标而引导资源配置的人。
这个定义囊括了有以下行为的所有个体：
（1）指导他人努力工作，包括在组织中分配任务；
（2）购买生产产品/提供服务所需的投入要素；
（3）负责制定决策。
经济学
经济学（economics）是关于如何在稀缺资源条件下制定决策的科学。经济决策涉及稀缺资源的有效配置，而管理者的任务就是有效配置资源以最大限度地实现既定目标。
管理经济学的定义
管理经济学研究的是如何配置稀缺资源以最有效地实现既定的管理目标。
有效决策的关键是掌握决策所需的信息，因此需要收集并处理这些数据信息。最终，管理者需要整合所有信息，分析这些信息并制定决策。
有效管理的经济学
本书所讨论的管理决策将更多地围绕利润最大化
或者一般意义上的公司价值最大化目标。
有效管理的
六个基本原则：
（1）明确目标和约束条件；（2）识别利润的性质和重要性；（3）理解激励；
（4）把握市场；（5）认识货币的时间价值；（6）运用边际分析法。
明确目标和约束条件
不同的目标意味着要作出不同的决策。
约束条件使管理者在实现利润最大化或扩大市场份额等目标时面临难题。
利润最大化目标要求管理者确定产品的最优价格、产量、生产技术、每种投入要素的数量、对竞争者行为的反应等。
识别利润的性质和重要性
经济利润和会计利润
会计利润是销售收入（价格乘以销售量）减去生产该产品或服务的以货币表示的成本。
经济利润（economic profits）是产品或服务所带来的总收益与其机会成本的差额。
一种资源的机会成本
（oportunity cost）包括两部分：
一是该资源的显性（会计）成本
二是该资源放弃其他最佳用途的隐性成本。产品或服务的机会成本通常大于其会计成本
隐性成本难以衡量，导致管理者经常忽视该成本。然而优秀的管理者能通过各种途径找到识别和度量隐性成本的数据。
利润的角色
斯密认为，一个企业通过追求自身利益——利润最大化目标——最终经满足社会的需要。经济利润给稀缺资源的最优配置提供了信号。
通过将稀缺资源向社会最青睐的行业移动，可以提高整个社会的福利。
五力模型和企业盈利能力
潜在进入者在位企业维持长期利润的能力取决于潜在进入者面临的进入壁垒的高低。
供应商能力 当供应商对其产品有较强议价能力时，行业利润往往很低。
购买者力量 与供应商情况类似，当顾客或者购买者在产品/服务的优惠交易条件方面具有讨价还价能力时，该行业的利率率趋于降低。
行业内竞争者 行业利润的可持续性还取决于该行业内厂商的性质和竞争强度。行业利润的影响因素还有产品的差异化程度，以及厂商在价格、数量、产能或者质量/服务品质等方面的博弈。
替代品和互补品 行业利润水平及可持续性还依赖与相关产品/服务的价格和价值。互补品也会影响行业盈利能力。
理解激励
在企业内部，激励影响了资源利用方式及员工工作的努力程度。
管理经济学旨在提供一系列方法，帮助决策者作出正确的决策，并且在组织内部建立有效的激励制度。
本理论建立在“所有你接触的人都是贪婪的”这一假设前提下，即每个人只对自身利益感兴趣，所以必须加以激励。
把握市场
市场运行的最终结果取决于市场中买方和卖方的相对力量。
这种力量即消费者和生产者在市场中讨价还价的地位，
受经济交易中三类竞争的影响：
消费者—生产者竞争，消费者—消费者竞争以及生产者—生产者竞争。
任何一类竞争都是市场运行中的一个约束条件。
管理者实现经营目标的能力亦取决于其产品受到这三类竞争影响的程度。
消费者—生产者竞争 消费者—生产者竞争源于消费者和生产者之间的利益冲突。这两股力量为市场运行提供了天然的制衡力，即便在只有一家厂商（垄断者）的垄断市场中，情况亦是如此。
消费者—消费者竞争 当商品的供给数量有限时，消费者就会竞相购买，愿意为该稀缺商品支付最高价格的消费者将战胜其他消费者得到该商品（消费该商品的权利）。这种竞争显著地存在于垄断市场中，比如拍卖就是消费者—消费者竞争的最佳案例。
生产者—生产者竞争 生产者为获得消费者相互竞争，以最低价格提供最好质量的生产者赢得消费者（为消费者服务的权利）。
政府和市场当市场上的任何一方发现自己在竞争中处于不利地位时，往往会游说政府干预以保护自身利益。
认识货币的时间价值
决策时限涉及项目的成本发生期和获利期之间的时滞。
因为在未来得到1美元的机会成本是今天的1美元所放弃的利息收入。这个机会成本揭示了货币的时间价值。
净现值（net present value，NPV）指项目所带来的收入流的现值（PV）减去项目当前的成本：
运用边际分析法
边际分析通过边际（或增量）收益
和边际（或增量）成本的比较寻求最优决策。
离散决策
控制变量Q不能是分数，只能是整数，
这是离散问题的基本特征。管理者的目标是利润最大化：
边际收益（marginal benefit）指增加一单位管理控制变量带来的增量收益。
边际成本（marginal cost）指增加一单位管理控制变量所产生的增量成本。
边际原理 为使净收益最大，管理者应该增加管理控制变量的使用量直至边际收益等于边际成本，即边际净收益为零
超过此点，继续增加控制变量将不会带来更多收益。
增量决策
管理者有时需要作出接受或放弃的简单选择。
边际分析法在是或否的决策中很有用：如果实施该项目带来的额外收益超过实施该项目的额外成本，则应接受该项目，反之放弃。
从决策中获得的额外收益为增量收益（incremental revenues），
决策产生的额外成本为增量成本（incremental costs）。
如何学习管理经济学
理解经济学术语有两个目的：第一，经济学家使用定义和公式的目的是力求精确。第二，准确的术语有助于实践者更有效的交流和沟通。
第二章
市场力量：
需求与供给
导言
供给和需求，这是美国乃至全球市场经济背后的驱动力。
供求分析法是管理者描述市场情形的常用工具。
供求分析是一种定性的预测分析工具。利用供求分析法可以预测竞争性市场的变动趋势。
需求
需求指在特定时间、特定市场上，
消费者读某种商品愿意并且能够购买的数量。
市场需求曲线（market demandcurve），揭示了消费者在任何水平（包括市场调查未涉及的价格）愿意并且能够购买的商品数量。需求曲线上的每一点都假定价格之外的其他影响因素保持不变。
需求变动
经济学家认识到，除了商品价格，其他变量也会影响需求量。
所有商品价格之外的其他影响需求量变动的因素称为需求变动因素。
沿着需求曲线从A移动到B，称为需求量变动（change in quantity demanded）。
当广告、收入或相关产品价格发生变化时，会导致需求曲线的位置移动而产生需求变动（change in demand）。
需求曲线向右移动称为需求增加，因为同一价格下的需求会更多。需求曲线向左移动称为需求减少。
了解需求曲线移动和沿需求曲线移动的区别，
对于把握五种需求影响因素——消费者收入、相关商品价格、广告和消费者偏好、人口以及消费者预期——的变动如何影响需求变动很重要。
价格
消费者收入
商品需求量随着消费者收入增加而增加
（需求曲线向右移动）的为正常品（normal good）。
当收入增加时，给定价格下消费者的购买数量显著增加。
相反，当消费者收入下降时，正常品的需求会减少（需求曲线向左移动）。
收入增加也会导致某种商品的需求减少，经济学家称这种商品为劣等品（inferior good）。
劣等品并不意味着产品质量差，劣等品这一概念仅仅定义收入与需求量成反比的商品，即那些随消费者收入增加而购买量下降、随收入减少而购买量上升的商品。
相关商品价格
假设A商品的价格上升，导致许多消费者转向购买B商品，
随着越来越多的消费者用B商品替代A商品，在任一价格下B商品需求量将趋于增加。
也就是说，A商品的价格上升导致了B商品的需求增加，用图形表示为B商品的需求曲线向右移动。
以上述方式相互影响的商品称为替代品（substitutes）。
替代商品并非局限于功能相同的产品。
当商品X与商品Y为互补品（complements）时，
商品Y价格下降将增加商品X的需求量（商品X的需求曲线向右移动）：
给定价格下商品X的需求量增加源于其互补品——商品Y的价格下降。
广告和消费者偏好
绘制特定需求曲线时，广告投入通常保持不变。广告增加同样会使需求曲线向右移动。
人口
一般来说，随着人口规模的增加会有更多的人购买某商品，从而推动需求曲线向右移动。
人口结构变化也会影响商品需求。
消费者预期
通常，当消费者预期未来涨价时，他们将以当前购买取代未来购买。这种消费者行为称为备货，多发生于耐用品。现实中经常出现针对便宜耐用品的囤货行为。
其他因素
总之，对某商品而言，任何影响消费者购买意愿或能力的因素都是影响需求变动的潜在因素。
需求函数
将所有影响需求的因素汇总到需求函数（demand function）中。
消费者剩余
消费和从一种商品得到的超出支付费用的那部分价值。此概念对管理者非常重要，它揭示了消费者为得到所购买的一定数量商品而愿意额外付出的钱。
供给
市场供给曲线（market supply curve）反映了当其他影响供给的因素给定不变时，不同价格下所有厂商愿意并能够提供的某种商品的总量。
供给量变动（change in quantity supplied）一种商品的市场供给取决于多种因素，
绘制供给曲线时通常假定商品价格之外的其他因素不变，沿着供给曲线移动
市场供给曲线向上倾斜揭示了供给定律：
当其他因素保持不变是，随着商品价格的上升（下降），该商品的供给量会增加（减少）。价格高的时候，厂商愿意生产更多的产品。
影响供给变动的因素
影响供给曲线位置的变量称为供给变动的影响因素，包括投入品价格、技术水平、市场上厂商数目、税收和价格预期。
当这些变量中的一个或几个发生变化时，整条供给曲线的位置就会发生移动，这种移动称为供给变动（change in supply）。
投入品价格 随着产品成本的变化，在给定的价格下厂商愿意提供产品的意愿会变化。
尤其是当投入品价格上升时，在每给定价格下，厂商会减少产品的生产。这种供给减少表现为供给曲线左移。
技术或政策法规 技术或政策变化若降低了厂商成本，将导致供给增加。反之，则导致供给减少从而推动供给曲线向左移动。
厂商数目 当新厂商进入时，给定价格下该行业的产量会增加，从而导致供给曲线右移。
替代品生产 许多厂商拥有兼容生产多种产品的技术。
税收货物税是针对每单位售出的产品征收的税种，通常向供应商征收。税收使供给曲线向上移动了与税额相同的距离。
由此，在任意给定价格下，厂商愿意提供的数量较税前减少，税收具有减少供给的效应。
从价税将使供给曲线逆时针转动，随着价格的增加，新的供给曲线将更大幅度偏离原曲线。
价格预期 当前供给量与未来供给量之间存在相互替代性。
如果厂商预期未来价格将上涨且该商品为非易腐品，厂商就会“惜售”——减少当前供给量并等待未来以更高的价格卖出。这将推动供给曲线左移。
供给函数
供给函数（supply function）指的是投入品价格以及其他印象因素给定，商品价格取不同值时，该商品的供给量。令Q8X表示商品X的供给量，
PX表示该商品价格，W表示某投入品价格（如劳动工资率），
pr表示相关商品价格，H表示其他影响因素（如当期技术、市场中厂商的数目、税收或价格预期等）。
那么商品X的供给函数可以写为：
假设Q8X与影响因素之间是线性，则线性供给函数（linear supply function）如下：
生产者剩余
供给曲线揭示了给定价格下厂商愿意提供的商品数量，也就是生产者愿意额外提供一单位商品时所要求的价格。
生产者剩余（producer surplus）指生产者获得的超出意愿所得的价值。从几何角度看，生产者剩余就是供给曲线与商品市场价格之间的面积。
市场均衡
竞争性市场上的均衡价格取决于市场上所有买者和卖着的相互作用。
竞争性市场均衡（Market Equilibrium）
竞争性市场均衡由需求曲线和供给曲线的相互作用共同决定。均衡价格是指需求量和供给量相等的价格。假如
Qd（P）和Q8（P）分别代表价格为P时的需求量和供给量，在均衡价格为Pe
时以下公式成立：
价格限制与
市场均衡
最高限价
稀缺经济学的一个基本假设是，当价格为零时没有足够的商品来满足所有消费者的需求，
因此必须采取相应的分配方法来决定谁能获得商品及谁不能获得商品。没有获得商品的人实质上受到了歧视。
市场体制是运用价格来分配商品的，根据价格将商品分配给那些愿意并且能够支付高价的消费者。
最低限价
与最高限价相反，有些竞争性均衡价格对生产者而言太低，于是厂商会游说政府立法为某种商品设置最低价格，这种价格称为最低限价（price floor）。
典型的最低限价就是最低工资，即支付给工人的最低法定工资。
比较静态分析
需求变动
市场的初始均衡位于点A，
即需求曲线D0与供给曲线S的交点。
预测需求将从D0增加到D1，
这将导致均衡点从点A移至点B。
2.6.2 供给变动
市场的初始均衡位于
需求曲线D与供给曲线S0的交点。
投入品价格的增加
使供给从S0移至S1，
新的竞争性均衡移动到B点。
市场价格从P0上升到P1，
均衡数量从Q0下降到Q1。
2.6.3 供给和需求同时变动
市场的初始均衡位于点A，
即需求曲线D0与供给曲线S0的交点。
同时引发供给减少和需求增加，
假设供给曲线从S0下降到S1，
需求曲线从D0增加到D1。
于是市场在点B达到了新的竞争性均衡；
价格从P0上升到P1，
供需数量从Q0增加到Q1。
第三章 定量需求分析
导言
前半部分将介绍需求弹性的概念。 后半部分将介绍回归分析法。
弹性的概念
弹性（elasticity）度量的是一个变量对另一个变量的变化的敏感度。
假如变量G与变量S的弹性用EG,S表示，
那么，变量S的变化百分比（%ΔS)
导致的变量G的变化百分比（%ΔG）
可用公式表示如下：
由于%ΔG=ΔG/G，%ΔS=ΔS/S，弹性亦可记为EG,S=(ΔG/ΔS)(S/G)，
。需要注意的是，ΔG/ΔS就是G与S之间的函数关系的斜率，
反映了给定S的变化所导致G的变化量，乘以S/G之后就将变化量的关系转化为百分比数值，这意味着弹性大小与变量G和S的单位无关。
微积分表达式
假如某因变量G的变化量取决于S的变化，
两者函数关系式为G=f(S),
那么G相对于S的弹性用微积分形式表示如下：
弹性的符号（正或负）
决定了两个变量G和S之间的正相关还是负相关，
如果弹性为正，S的增加将导致G增加；
如果弹性为负，S的增加将导致G减少。
弹性绝对值大于1还是小于1反映出G对S变化的敏感度。
如果弹性绝对值大于1，弹性公式中的分子大于分母，
那么S的一个微小的百分比变化将导致G相对较大的百分比变化。
如果弹性的绝对值小于1，弹性公式的分子小于分母，
此时S的一个给定百分比变化将导致G相对更小的百分比变化。
需求价格弹性
需求价格弹性（own price elasticity of demand）
测度的是需求量对价格变化的敏感程度。
商品X的需求价格弹性用EQx,Px
表示，定义如下：
微积分表达式 假设某商品的需求函数为
其需求价格弹性可用以下微积分形式表示：
如果某商品需求价格弹性的绝对值大于1，则称该商品的需求是富有弹性需求（elastic demand）
如果某商品需求价格弹性的绝对值小于1，则称该商品的需求是缺乏弹性需求（inelastic demand）
如果某商品需求价格弹性的绝对值等于1，则称该商品的需求是单位弹性需求（unitary elastic demand）
如果一种商品的需求富有弹性，表明该商品的需求量相对于商品价格的变动比较敏感。
如果需求缺乏弹性，该商品的需求量相对于商品价格变动不太敏感。
当需求缺乏弹性时，意味着商品价格的上涨造成的需求量下降幅度很小；
但当需求富有弹性时，商品价格的上涨将会导致需求量大幅下降。
弹性与总收益
当沿着需求曲线从点A移动到点I时，需求变得越来越有弹性；
在点E处，需求为单位弹性且总收益达到最大。所有位于点E左上方的点，需求富有弹性且总收益随价格上涨而减少；
所有位于点E右下方的点，需求缺乏弹性且总收益随价格上涨而增加。
分析价格、弹性和总收益之间关系的变化称为总收益检验。
原理 总收益检验 如果产品富有价格弹性，产品价格上升（下降）将导致总收益减少（增加）。
如果产品缺乏价格弹性，产品价格上升（下降）会使总收益增加（减少）。
当产品价格弹性为单位弹性时，总收益最大。
完全弹性需求（perfectly elastic demand）指需求价格弹性绝对值无穷大；
完全无弹性需求（perfectly inelastic demand）指需求价格弹性为零。
影响价格弹性的因素
替代品可获性
决定一种商品的需求弹性大小的关键因素是该商品的近似替代品的数目。
一般来说，大类商品需求比特类商品需求更缺乏弹性。
时间
短期需求相对于长期需求更加缺乏弹性。
支出比重
一般来说，若一种商品的支出占消费预算的比重较小，这种商品较占比较大的商品更缺乏价格弹性。
边际收益与需求价格弹性
相对于线性需求曲线，边际收益曲线恰好位于需求曲线与纵轴的正中间，这意味着每单位产量的边际收益低于没单位产量的销售价格。
原因在于，厂商为了吸引消费者购买更多商品必须降价。当厂商为每一单位商品确定相同价格时，消费者不仅可以用较低的价格购买最后一单位商品，还能以同样低的价格买到在高价位售出的多个单位商品。
一种产品的需求越缺乏弹性，降价虽然带来需求量增加，但总收益下降越多。
上述内容阐明了边际收益和需求弹性之间的一般关系：
P表示商品价格，E表示商品需求价格弹性。
交叉价格弹性
需求的交叉价格弹性（cross-price elasticity）
揭示了一种商品需求对另一种相关商品价格变化的敏感程度。
商品X和商品Y之间的交叉价格弹性可用数学公式表示：
微积分表达式 当商品的需求函数为
商品X和商品Y之间的需求交叉弹性的微积分形式如下：
一般来说，当商品X与商品Y是替代品时，
商品Y的价格上升将导致商品X的需求量增加，
那么
当商品X和商品Y是互补品时，
商品Y的价格上升将导致商品X的需求量下降，
那么
假设某公司的收益来自两种产品
——产品X和产品Y。
该公司的收益为：
其中
指来自产品X的销售收益
是来自产品Y的销售收益。
产品X的价格的微小变化
收入弹性
收入弹性（income elasticity）
度量的是消费者需求
对其收入变化的敏感程度。
收入弹性的数学表达式如下：
微积分表达式 假设商品的需求函数为
其收入弹性的微积分形式如下：
当商品X为正常品时，收入增加使商品的需求量增加，因此
当商品X为劣等品时，收入增加会导致商品X的需求量下降，因此
其他弹性
给定弹性的一般概念，可以将其扩展到其他变量的弹性分析。
利用需求函数计算弹性
线性需求函数的弹性计算
假设线性需求函数如下：
则各种弹性为：
价格弹性：
交叉价格弹性：
收入弹性：
微积分表达式经变形，对数线性函数的反函数
（即原线性需求函数）如下：
对于线性需求函数，弹性大小取决于计算弹性时所对应的特定价格和数量，价格弹性并不等于需求曲线斜率。
一般来说，对于线性需求函数，当产品价格高时，所对应的需求往往富有弹性，
而当产品价格低时，所对应的需求通常缺乏弹性。尤其要注意：当Px=0时，
即当产品价格接近0时，需求缺乏弹性。
也就是说，当产品价格上升时，
随之减少，但弹性的绝对值却随之增大。
非线性需求函数的弹性计算
假设非线性函数如下所示：
式中，c为常数。商品X的需求量与价格和收入是非线性函数。对这个方程取自然对数，将得到用变量的对数来表示的线性表达式：
这个等式称为对数线性需求（log-linear demand）函数。
决定了商品X和商品Y是替代品还是互补品，而
系数lnM的符号决定商品X是正常品还是劣等品。
各种弹性为：
价格弹性：
交叉价格弹性：
收入弹性：
微积分表达式经变形，对数线性函数的反函数
（即原线性需求函数）如下：
式中，c为常数。根据弹性的微积分计算，则有：
同理，可得交叉价格弹性和收入弹性的表达式。
需要注意的是，当需求为对数线性函数时，某给定变量的弹性恰好是所对应对数的系数。
由于所有系数为固定值，因此所有弹性皆为常数。
回归分析
计量经济学是对经济现象进行统计分析的工具。
假设计量经济学家认为Y和X之间存在线性关系，
但是这个关系也存在一些随机偏离。
从数学角度来看，Y和X之间的真实关系可能是：
式中，a和b是未知参数；e是一个均值为零的随机变量（通常称为误差项）。
由于决定Y和X预期关系的参数未知，计量经济学家必须找出参数a和b的具体数值。
回归直线法就是要寻找使估计值与真实值之间偏差平方和最小的那条直线。
a和b的取值通常记为^a^b
，也称参数估计值，相对应的直线通常称为最小二乘回归（least squares regression）。
最小二乘回归法用来寻找一条回归直线，该直线与真实数据值之间的偏差平方和最小。方程
的最小二乘回归直线可由下式给出：
式中，^a和^b为参数估计值，
也就是估计值与真实数据之间偏差平方和最小的a和b的取值。
估计系数的显著性检验
如果ei是相互独立且服从同一正态分布的随机变量，
则系数估计值的标准差可用来构造置信区间并进行显著性检验。
置信区间
假设变量服从上述技术假定，给定一个参数估计值及其标准差，管理者可以通过构造95%的置信区间来估算真实值距离估计系数的上下限。
原理 95%置信区间的经验法则
如果回归方程中参数估计量为^a和^b
，则a和b真实值的95%的置信区间可以分别近似表示为：
t统计量
参数估计的t统计量（t-statistic）
值该参数估计值与其标准差的比率。
假如参数估计值为^a和^b
其
当参数估计值的t统计量的绝对值很大时，可以判定真正的参数值不为零。
因为t统计量的绝对值很大时，相对于参数估计值（注意采用绝对值），参数估计值的标准差很小。
因此可以确认，给定一个真实模型的不同数据样本，新的参数估计值将落在同一取值范围内。
经验法则是：当t统计量的绝对值大于等于2时，相应的参数估计值在统计上显著不为零。
回归软件包给出的P值是准确测量统计显著性的工具。系数估计值的P值越小，估计值越可信。
研究者通常认为P值小于等于0.05，即可判定系数估计值在统计上是显著的。如果P值等于0.05，
即认为系数估计值在5%的水平上统计显著。P值与置信区间一样，
假定回归方程的误差项相互独立且服从同一正态分布。
原理 t统计量的经验法则
当t统计量的绝对值大于2时，可认为回归公式中相关参数的真实值在95%置信度上不等于零。
回归直线的总体拟合程度
R²（也称可决系数）解释被解释变量的总体变动由回归所解释的程度，
R²等于回归误差平方和（SS回归）与总误差平方和（SS总计）之比，
用公式表示如下：
R²的取值范围为0~1，即
时间序列分析中R²通常超过0.9；但在截面数据分析中，R²达到0.5就被认为拟合较好。
R²的一大缺点在于：它只是对拟合优度的主观判断。
有时R²接近于1仅仅是因为观测值的数目少于被估计参数的数目。
调整的R²的计算公式如下：
式中，n为观测值的个数；k为估计系数的个数。
回归分析时待估计参数的个数不能超过观测值的个数。
n-k是回归分析后的残差自由度。
调整的R²是对那些用较小自由度（即利用少量观测值估计多个系数）进行回归的行为的“修订”。
这种“修订”有时很大，调整的R²甚至变为负值。
F统计量
F统计量可以度量总变动幅度中能够被回归方程解释的部分相对于不能解释部分的比例。
F统计量越大，回归直线的拟合越好。
F统计量的有点在于其统计属性已知，
所以能够客观地评价任何F统计量的统计显著性。
同P值一样，F统计量的显著性越小，
回归方程的总体拟合度越好。若F统计量的显著性值小于5%，则通常认为显著。
非线性回归和多元回归
非线性回归函数
为了估计对数线性需求函数，
计量经济学家在采用回归方法之前
先给出价格和数量的对数形式：
上述需求函数变形为：
Q'和P'呈线性关系，
对变形后的函数（转化后为Q'和P'的关系）进行回归求解参数估计值。
对数线性需求的βp估计值就是需求价格弹性。
多元回归
回归技术可以解决多元回归问题——一个因变量对多个自变量。
若需求量与解释变量非线性关系，也可以采用对数线性形式。
注意：观测值数目大于带估计数目。
一个提示
需要重点强调的是，
经济计量学在经济学中是一个非常专业的领域，
需要经过多年的学习及大量的联系才能掌握其分析方法。
一个严谨的管理者通常借助专家（如经济专家或顾问）来开展计量分析并从专家处获取对需求的估计。
除非你精通计量知识，
掌握大量远远超出管理经济学教材和电子表格范畴的计量知识（比如内生性、样本选择、异方差、自相关和不可观测效应），
否则，最好将这些计量经济学知识仅作为与计量专家沟通（解释分析结果）的工具。
第四章 个体行为理论
导言
人的决策思维过程非常复杂，大脑需要处理大量的信息。
正是由于个体行为的复杂性，谁能够更好地理解个体行为，谁就获得了在商业世界中获胜的有效技能。
行为模型必须是对个体的真实决策过程的抽象，所以需要从聚焦与基本属性的简单模型入手，而非一开始就讨论具体行为特征。
消费者行为
一般来说，消费者通常指为了消费目的而购买公司产品/服务的个体。
作为一个公司的管理者，
你不仅仅要关注谁消费了公司产品，还要关注谁购买了这些产品。
消费者机会指消费者有能力购买的产品和服务；
消费者偏好决定了他会购买并消费其中的哪些产品和服务。
在此假定市场上只有两种商品。用X表示一种商品，用Y表示另一种商品。
假设消费者能够对可供选择的商品组合按偏好程度从最高到最低排序。
我们用“＞"表示顺序，
若消费者读组合A的偏好大于组合B，就记为A＞B。
若消费者对组合A和组合B的满意程度相同，
即他认为组合A和组合B是无差别的，用简化符号A~B表示。
假设偏好顺序满足四个基本特性：完备性、越多越好、边际替代率递减、可传递性。
完备性对于任意两种商品组合，比如A和B，消费者偏好可能是A＞B、A＜B或者A~B。
只有假设偏好是完备的，才能判定消费者能够对所有的商品组合进行偏好排序。
越多越好 在一个商品组合A中，如果一种商品的数量和组合B中的至少一样多，而另一些商品比组合B中的更多，那么消费者将更偏好组合A而不是组合B。
越多越好的前提是消费者首先认可商品为”优质品“而非”劣等品“。
无差异曲线（indifference curve）揭示了给消费者带来同等满足程度的商品X和商品Y的不同组合；
对于无差异曲线上的两种商品的任何组合，消费者偏好是无差异的。
无差异曲线的形状与消费者的偏好有关，对于不同的消费者来说，其无差异曲线的形状通常是不同的。
边际替代率（marginal rate of substitution，MRS）是无差异曲线斜率的绝对值。
两种商品的边际替代率就是消费者在满足程度不变的情况下，愿意用一种商品替代另一种商品的比率。
当从组合A移向组合B时，消费者多获得了1单位的商品X，但放弃了2单位的商品Y，所以从点A移到点B时，商品X和商品Y的边际替代率是2。
从点B移动点C，消费者多获得了1单位的商品X，但只愿意放弃1单位商品Y。其原因是无差异曲线有边际替代率递减的特性。
边际替代率递减随着消费者获得更多的商品X，他愿意为多获得1单位商品X而放弃的商品Y的数量呈递减趋势。
这个假设将导致无差异曲线是凸向原点的曲线。
可传递性假设任意三个商品组合A，B和C，如果A＞B，B＞C，那么对于A和C来说，应该有A＞C；
如果A~B，B~C，那么A~C。
消费者的可传递假设与越多越好假设表明无差异曲线不可能相交。这暗示消费者不可能陷入”作不出选择“的境况。
约束
预算约束
预算约束表现为消费者只能选择自己买得起的商品组合。
假如令M表示消费者收入，Px和Py分别表示商品X和商品Y的价格。
预算集可以表示如下：
预算集（budget set）描述了消费者有能力购买的商品X和商品Y的所有组合，
即消费者用于购买商品X和商品Y的支出之和不能超出该消费者的收入。
如果消费者将其全部收入用于购买商品X和商品Y，
则上述公式转化为等式（所表示的关系称为预算线）：
预算线（budget line）指消费者花完全部收入可购买的商品X和商品Y的所有组合。
在预算线两边同时乘以1/Py
，预算线等式变形得到另一种形式：斜率-截距形式，如下所示：
对Y求解，得
注意，Y是X的线性函数，
其中纵轴上的截距为
M/Py，斜率为-Px/Py。
预算线斜率-Px/Py表示商品X与商品Y的市场替代率（market rate of substitution）。
收入的变化
消费者的机会集取决于商品的市场价格和消费者收入。
当两者发生变化时，消费者可选择的商品组合也将发生变化。
接下来我们假设价格保持不变，我们来分析收入变化对机会集的影响。
假设消费者初始收入为M0。若价格不变，当收入从M0增加到M1时，
收入增加并不影响预算线斜率。
但是随着消费者收入的增加，预算线在纵轴的截距和在横轴的截距都将增加，
这是因为收入增加使消费者能买得起更多商品。
价格的变化
假设消费者的收入固定为M，
但商品X的价格下降到P1X
同理，如果商品X的价格上升，
将导致预算线按顺时针方向旋转。
消费者均衡
对于消费者来说，
作出购买决策的目的是
选择能使效用（满足程度）最大化的商品组合。
如果不存在稀缺性，
越多越好意味着消费者希望购买的商品组合无穷多。
由于稀缺性，
消费者只能选择预算集以内的商品组合，
即有能力支付的商品组合。
均衡选择指消费者一旦选择此组合就不再转向有能力支付的其他组合。
消费者均衡（consumer equilibrium）的重要特性是，
当消费者选择均衡消费组合时，
此点处的无差异曲线的斜率与预算线的斜率相等。
无差异曲线斜率的绝对值就是边际替代率，
而预算线斜率为-Px/Py
，在消费者均衡处两者相等，即
比较静态分析
价格变化与消费者行为
消费者最初位于均衡点A，当商品X的价格下降到P1x时，他的机会集扩大了。
在新机会集下，消费者能够得到更大的满足，消费者均衡向新的均衡点B移动。
新均衡点在新预算线上的位置取决于消费者偏好。替代品和互补品的概念在此将起作用。
由于商品X和商品Y是替代品，商品X的价格下降将导致消费者均衡从点A移向点B，消费者在点B比在点A消费更多多的商品X和更少的商品Y。
由于商品X和商品Y是互补品，商品X的价格下降导致消费者均衡从点A移向点B，消费者在点B将比在点A消费更多的商品Y。
对于管理者来说，需要注意的是，
价格变化影响了市场替代率——消费者在不同商品之间进行替代的比率，进而改变了消费者行为。
由于公司改变定价策略，或者竞争对手及其他行业的公司改变价格，导致某商品的价格发生变化。
价格变化将改变消费者对不同商品的购买动机，进而改变其均衡消费组合。
无差异曲线分析法可以帮助管理者理解价格变化对消费者均衡组合的影响，也可以帮助管理者理解收入变化如何影响消费者的均衡商品组合。
收入变化与消费者行为
收入变化会扩大或缩小消费者的预算约束，从而改变消费者的消费方式，
使消费者找到一个新的最优均衡组合。消费者最初的均衡位于点A。
如果消费者收入增加到M1，其预算线向外移动，
消费者能够获得更大的满足，于是消费者会发现选择点B的效用会更大，
注意通过点B的无差异曲线与新的预算线相切。
新均衡点的位置取决于消费者偏好。
由于消费者收入增加，商品X和商品Y的需求量都增加了，
因此商品X和商品Y都是正常品。
替代效应与收入效应
价格变化会产生替代效应（substitution effect）和收入效应（income effect），
两种效应共同决定了价格变化对消费者行为的影响。
假设消费者最初位于均衡点A，点F和G相连形成预算线。
商品X价格上升，预算线沿着顺时针方向选装，形成了新的预算线——点F与点H相连的直线。
关于这个变化需要注意两点：
第一，价格上升导致预算集变小，因而减少消费者福利，
因为价格上升后消费者的实际收入减少了，所有商品组合只能满足更低的无差异曲线。
第二，商品X的价格上升使得预算线斜率更陡峭，这说明两种商品之间的市场替代率变大。
两种因素导致导致消费者从初始的均衡点A移向一个新的均衡点C。
替代效应和收入效应的分离，有助于理解各种效应对消费者选择的影响。
假设价格上升后消费者有足够的收入，连线点J和点H的预算线。
新预算线和原预算线FH斜率相同，但收入水平比FH高。
根据新预算线，消费者将在点B达到均衡，此时商品X的需求量小于初始均衡（点A）时的需求量。
从点A到点B的移动为替代效应，它描述了消费对不同市场替代率做出的反应。替代效应为X0-Xm。
注意从点A到点B消费者仍然在同一条无差异曲线上，
所以这个移动（商品X需求量的减少）只是反映了更高的市场替代率，而不是消费者实际收入的减少。
价格上升时，消费者面对的实际上并不是预算线JI，而是预算线FH。
预算线的移动只是反映了收入的减少，预算线JI和FH的斜率相同。
因此，从点B到点C的移动称为收入效应。收入效应是Xm-X1
，它揭示了价格上升时消费者实际收入减少的事实。
商品X是正常品，因此收入减少将降低商品X的需求量。
价格上升带来的总效应包括替代效应和收入效应两部分。
替代效应表现为沿着同一条无差异曲线的移动，
它反映相对价格变化对消费的影响。
收入效应导致预算的平移，
它反映实际收入下降对消费的影响，如点B向点C的移动。
市场上观察到的价格上升的总效应是从点A到点C的移动，
市场上包括商品X相对价格上升（从点A到点B的移动）的结果，
以及消费者实际收入减少（从点B到点C的移动）的结果。
无差异曲线的应用
消费者的选择
买一赠一
一个消费者最初面对的是连接点A和点B的预算线，消费者在点C处达到均衡。
在买一赠一的促销方式下，消费者预算线从AB变为ADEF。
原因在于如果消费者购买的商品不大于一个，她将得不到优惠，
所以消费者在一个商品左边的那部分预算线保持原形状，即AD。
如果她购买了一个商品，就可以再得到一个免费的商品，
即只要购买一个商品，预算线就变为DEF。
需要注意的是，商品从一个增加到两个时价格为零，
所以在图形上，商品预算线在第一个单位和第二个点位之间的部分应该是水平的
（因为预算线斜率为-Px/Py，而第二个商品的Px为零）。
但是当消费者购买的商品超过两个时，她又要按照正常价格支付。
组合E也成为消费者有能力购买的商品组合，而组合E比组合C更受消费者喜欢
（点E位于更高的无差异曲线上）。
现金礼物、实物礼物即礼券
消费者处于消费均衡状态，商品组合A。
一个价值10美元的礼物商品X。收到这个礼物后，机会集扩大到点B。
根据这个新机会集，无差异曲线移动到通过点B的一条更高的无差异曲线上。
如果得到10美元现金，
预算线相对于原预算线会向外平移，新的预算线将通过点B。
当得到额外的10美元收入时，所有商品的价格并没有变化，
所以其预算线的斜率不变。
面对新的预算线，在点C得到最大化满足（满足程度超过点B），
这意味着得到现金的福利相对于得到礼物的（点B）的福利更大。
由此看来，现金礼物比同等价值的事物礼物更受欢迎，
除非这个实物礼物正是消费者自己特别想买的。
商店为了减少礼物推货量，采取的一种有效的方法是出售礼券。
该礼券可以在X商店（该商店销售商品X）使用，
但不能在Y商店（该店销售商品Y）使用。
直的黑线是消费者得到礼券之前的预算线，
弯折的灰线是消费者得到礼券之后的预算线，
礼券使消费者多得到价值10美元的商品X而不需要花费自己一分钱。
礼券对消费者行为的影响取决于商品X是正常品还是劣等品。
首先假设消费者的初始均衡在点A处。
如果消费者得到一张只能在X商店消费的价值10美元的礼券。
假如商品X和商品Y都是正常品，收入增加后，消费者愿意增加其需求量。
当两种商品都为正常品时，消费者均衡将从点A移到点C。
所以消费者对礼券的反应和对同等价值现金礼物的反应相同。
员工的选择与管理者的选择
收入——休闲选择
收入和休闲在大部分员工眼里就是两种产品，
两者沿着一条无差异曲线以递减的比率相互替代。
员工和管理者的机会集如图直线所示。
员工尝试得到更高的无差异曲线，
知道在点E处，
消费者得到了与机会集相切的无差异曲线。
管理者的决策
管理者的满足程度可以从公司的潜在销售量和公司利润中获得。
高利润和高销售可以扩大公司规模，而大型公司可以提供更多复利。
假设管理者的偏好是将公司的利润和销售量视为商品。
标有”公司利润“的曲线（表示该公司的利润）描述了公司利润和销售量的关系。
该曲线始于原点，依次通过点C、点A和点B，它是销售量的函数，但同时也能反映公司利润。
当公司销售量为零，利润也为零。随着公司的销售量提高，利润随之增加。
企业利润在Qm处达到最大，然后开始下降，到Q0处再次为零。
管理者一直在努力实现越来越高的无差异曲线，直到最终在点A达到均衡。
此时销量为Qu大于利润最大值时的销售量Qm。
也就是说，当管理者把利润和销售量视为商品时，均衡销售量大于利润最大化时的销售量。
如果管理者的偏好仅取决于销售量，无差异曲线是一条垂直的直线。
由此会导致以下情况：
管理者不关心利润，
所以其无差异曲线是垂直的直线且满足程度随着直线向右移动而不断提升。
点B是该管理者的均衡点，此时销售量为Q0，但利润为零。
最后假设管理者只关心公司的利润，此时该管理者的无差异曲线为水平直线。
管理者在点C获得满足最大化，这条无差异曲线是在给定机会集下可能达到的最高的无差异曲线。
与前面两种情形相比，此时利润更高但销售量减少。
无差异曲线与需求曲线的关系
无差异曲线，实际上就是需求函数的基础。
个体需求
消费者起初在点A达到均衡，
此时的收入固定为M，价格为P0x和Py。
当商品X的价格下降到P1x时，机会集扩大，
消费者在点B达到新的均衡。
务必注意：
是消费者从均衡点A移到点B的原因是商品X价格的变化，
而消费者收入和商品Y的价格是固定不变的。
商品X的价格与商品X的需求量之间的关系，这是商品X的个体需求曲线。
市场需求
管理者感兴趣的通常是
所有消费者对公司产品的总需求。
这些信息汇总在市场需求曲线上。
市场需求曲线是个体需求曲线的水平叠加，
描述在任一价格下，
市场上所有消费者可能购买的某种商品的总数量。
第五章 生产过程与成本
导言
公司和非营利组织都要生产产品或提供服务，它们能否成功运营的关键在于管理者在生产过程中能否选择最佳的投入要素数量和类型。
生产函数
技术囊括了将原料投入转化为产出的所有可行方法。
技术是工程专有知识的总称。
管理决策（如研发费用支出等）将影响技术的可获性。
假定生产过程需要投入两种要素：
资本和劳动，
其中，K表示资本数量，L表示劳动数量，
Q表示生产过程中的产量水平。
将资本和劳动转换为产出的技术可用生产函数表示，
所以生产函数（production function）
揭示了给定投入下能够实现的最大产出的技术水平。
生产函数用公式表述如下：
Q=F（K,L）
即投入K单位资本和L单位劳动能够实现的最大产量。
短期决策与长期决策
管理者的重要工作是有效利用生产函数，
确定生产过程中各种投入要素的最佳数量。短期内某些生产要素不变。
短期指存在固定生产要素（fixed factors of production）的生产阶段。
若K表示固定资本值，短期生产函数可以表述为：
长期指所有投入要素都能进行调整的生产阶段。
生产率的测度
管理决策的一个重要内容是确定生产过程中投入要素的生产率。
这一测度对于评估生产过程的有效性
和制定利润最大化的投入决策非常有用。
总产量、平均产量和边际产量是三个最重要的生产率测度指标。
总产量（total product，TP）指给定的投入量下可以实现的最大产出。
平均产出
许多时候决策者往往更关心每单位投入的平均生产率。
一种投入的平均产量（average product，AP）
为总产量除以投入数量。
劳动的平均产量（右公式）。
平均产量度量了每单位投入所带来的产出。
资本的平均产量（APk）为：
边际产量（marginal product，MP）指最后一单位投入所带来的总产量的变化量。
资本的边际产量是总产量的变化量除以资本的变化量：
劳动的边际产量是总产量的变化量除以劳动的变化量：
管理者在生产中的角色
生产过程中管理者的角色包括：
（1）确保公司遵循生产函数（生产技术）运行；
（2）确保公司选择正确的投入水平。
这将保证公司在生产函数的正确决策点上运行，
并且会影响生产效率。
遵循生产函数
生产函数描述了在给定的投入量下可以实现的最大产量。为了确保员工全力以赴，管理者必须建立相应的激励机制来增强员工努力工作的意愿。许多公司制定了利润分红方案来激励员工按生产函数生产。
选择正确的投入水平
边际产量价值（value marginal product）
就是最后一单位投入所生产出来的产品的价值。
假如每单位产品的售价为P，则劳动的边际产量价值是：
资本的边际产量价值是：
原理 利润最大化的投入量
为了实现利润最大化，
投入要素的数量应保证边际利润等于边际成本，
具体来说，若每增加一单位劳动的成本为w，
管理者应该继续投入劳动直至（下方函数）
（该值处于边际产量递减阶段）。
利润最大化时的投入量法则揭示了最求利润最大化的厂商对投入量的追求。
劳动的边际产量价值为劳动量的函数。
当劳动的工资率为W0时，利润组大化的劳动量满足（右方函数）
（该劳动量位于边际报酬递减阶段）。利润最大化时的劳动量为L0单位。
曲线向下倾斜部分揭示了一个利润最大化的厂商对劳动的需求，它解释了企业拟投入的劳动量与价格之间的关系。
由于边际报酬递减规律，对某种投入品的需求呈向下倾斜趋势。
随着某种要素投入量的增加，边际产量减少，边际产量价值也将减少。
由于某种投入品的需求就是边际报酬递减阶段该投入品的边际产量价值，因此对投入要素的需求向下倾斜。
总之，每增加一单位投入品所产生的利润，较之前一单位投入产生的利润要少，
因此追求利润最大化的厂商都希望为额外一单位投入要素支付更少的钱。
生产函数的代数形式
线性生产函数（linear production function）如下：
式中，a和b为常数。
线性函数中，
所有投入要素和总产量之间呈完全线性关系，
投入要素彼此之间可以完全替代。
里昂惕夫生产函数（Leontief production function）：
式中，a和b是常数。
里昂惕夫生产函数也称固定比例生产函数，
其投入品按固定比例投入。
柯布—道格拉斯生产函数（Cobb-Douglas production function）
介于线性生产函数和里昂惕夫生产函数之间，函数形式如下所示：
式中，a和b是常数。
与线性生产函数不同的是，该函数中产出和投入之间为非线性关系。
与里昂惕夫生产函数不同的是，该函数中投入的使用无须遵循固定比例。
柯布道格拉斯生产函数中各投入要素之间存在一定程度的替代性，但不能完全替代。
生产率的代数测度
线性生产函数中，投入要素的边际产量的表达式非常简单，如下所示：
则
微积分表达式 投入要素的边际产量
是生产函数对该要素投入量的导数，
如劳动的边际产量是：
资本的边际产量是：
面对线性生产函数
有
对于线性生产函数，投入要素的边际产量就是该要素投入量的系数，
这意味着投入要素的边际产量与投入量无关，所以线性生产函数不遵循边际产量递减规律。
在柯布—道格拉斯生产函数中，投入要素的边际产量依赖于其投入量的大小。如下列公式所示：
则
微积分表达式
投入要素的边际产量是生产函数对该要素投入量的导数。
对柯布—道格拉斯生产函数求导得出：
要实现利润最大化，
投入要素的最佳用量为该投入要素的边际产出产值等于价格之时。
将此原理应用于生产函数的代数形式
以求解利润最大化时的要素投入量。
等产量线
当资产和劳动都可以改变时，
长期内两种投入要素的最优选择问题。
当生产中存在多种投入要素时，
管理者可以选择不同的投入组合以保证产出的水平相同。
等产量线（isoquant）指能够生产出相同数量产品的各种投入（K和L）的组合。
所以，等产量线上资本和劳动的所有组合能产出相同数量的产品。
投入组合A和B位于同一条等产量线上，所以产出同样数量的产品，即Q0单位。
投入组合A较之投入组合B更趋于资金密集型。
若两种投入都增加，将得到一条更高的等产量线，
在图形中表现为向右上方移动，距离原点越远的等产量线索代表的产量越高。
等产量线呈凸形，这是资本和劳动不可以完全替代。
资本和劳动相互替代的比率称为
边际技术替代率（arginal rate of technical substitution，MRTS）。
资本和劳动的MRTS是等产量线线斜率的绝对值，也是边际产量的比率：
线性生产函数的等产量线也是线性的，
这是因为两种投入要素呈完全替代关系且两种投入要素的替代比率与投入量无关。
尤其对于线性函数Q=aK+bL，当MPL=b和MPK=a时，边际技术替代率是b/a，
与投入要素的使用量毫无关系。
里昂惕夫生产函数所对应的等产量线呈L形，
由于投入要素之间遵循固定比率，
资本和劳动不能相互替代，因此里昂惕夫生产函数不存在边际技术替代率。
对大多数生产过程而言，
等产量线介于完全替代和固定比例之间，
即一种投入要素可以用另一种投入要素替代，但不能完全替代。
沿着等产量线，不同投入要素之间的替代率会发生变化。
柯布-道格拉斯生产函数中等产量线的边际技术替代率呈递减趋势，
一般来说，当等产量线出现边际技术替代率递减时，该等产量线将凸向原点。
边际技术替代率递减是生产函数的一个显著特性，
即保持产量不变时，随着生产者对一种投入要素使用量的减少，
就必须越来越多地增加另一种投入要素的使用量。
等成本线
有时不同投入组合的成本可能是相同的，
这些相同成本的投入组合就构成了一条等成本线（isocost line）。
假设公司所投入要素的成本为C美元，
即劳动成本加上资本成本：
式中，w为工资率（劳动的价格）；r为租金率（资本的价格）。
上述等式就是等成本公式。
等成本线公式可以用等成本线的斜率和截距来表示。
两边分别乘以1/r可得：
成本最小化
等成本线和等产量线可用来确定生产成本最小化（cost minimization）即如何在最低成本处组织生产，时的要素投入量。
点A所表示的投入组合，这个投入组合（K，L）位于等产量线Q0上，因此其产量为Q0；
该组合还位于经过点A的等成本线上，所以管理者如果使用组合A生产Q0单位，需耗费的总成本为C1。
如果使用组合B，将会以更低的成本C2得到同样的产出Q0。
所以对管理者来说，使用组合A是低效率的，组合B能有同等数量的产出且成本更低。
当等产量线的斜率等于等成本线的斜率时，所对应的投入组合就是成本最小化的投入组合。
等产量线斜率的绝对值是边际技术替代率，而等成本线的斜率为-w/r，
成本最小化的投入组合满足：
如果不满足上述条件，
说明L和K之间的技术替代率
不等于两种投入相互替代的市场替代率。
点A的等产量线斜率大于等成本线斜率，
这说明资本相对来说”太贵了”；
生产者会减少资本，
增加劳动投入来生产等量产品，
也就是用劳动替代资本直至边际技术替代率
与投入要素价格比率相等（如点B处）。
成本最小化的投入组合的基本条件
也可以用边际产量表示。
原理 成本最小化投入法则
给定产量水平下，要实现成本最小化，
所有投入要素的单位价格带来的边际产量应该相等：
同样，为了使成本最小化，
投入要素的边际技术替代率等于投入要素价格的比率：
给定产量水平的成本最小化为什么必须满足上述条件呢？假设MPL/w ＞ MPK/r。
对于最后花费的1美元，购买劳动比购买资本划算，因此厂商会减少资本、增加劳动以降低成本。
这是因为，厂商在资本支出上减少1美元，而在劳动支出上增加不到1美元，仍然能够实现相同产量。
所以用劳动来替代资本，厂商可以在保持产量不变的情况下降低成本。
这种替代性将一直持续到在资本上的1美元投入所带来的边际产量
正好等于在劳动上的1美元投入所带来的边际产量。
最优投入替代
投入要素价格变化将导致成本最小化投入组合发生变化。
假设初始等成本线是FG，
生产者在投入组合A上实现成本最小化，产量为Q0单位。
当前工资率提高了，假如厂商的投入总成本固定不变，等成本线将顺时针旋转至FH。
所以，如果厂商的支出成本与工资率上升之前的相同，就无法达到同等产量。
新的等成本线斜率反映出劳动的相对价格提高了。
若保持产量不变，成本最小化的投入组合在点B处，此时新的等成本线IJ与原等成本线相切。
由于劳动价格相对于资本价格提高了，厂商将用资本替代劳动，采用资本密集型的生产模式。
原理 最优的投入替代
为了使给定产量下的成本最小化，当某种投入要素价格上升时，
厂商应该减少该投入要素的使用量而增加其他投入要素的使用量。
成本函数
随着等产量线达到更高水平，
生产成本也随之增加，
因此令C（Q）表示厂商的生产成本
——以成本最小化方式生产时的成本。
该函数C为成本函数。
成本函数为利润最大化下的产出决策
提供了必要信息。
成本函数提炼了生产过程信息。
短期成本
在短期即特定的时期内，一些投入要素的数量固定不变。
短期内管理者可以改变投入的使用量，但不能改变固定投入量。
而固定投入和可变投入都有成本，
所以短期内生产的总成本（totaal cost）
由两方面构成：
（1）固定投入的成本；
（2）可变投入的成本。
固定成本（fixed costs）记为FC，指不随产量变化而变化的成本。
变动成本（variable costs）记为VC（Q），指随产量变化而变化的成本。
固定成本和变动成本之和为产商的短期成本函数（short-run cost function）。
存在固定投入要素时，短期成本函数就是当可变要素以成本最小化方式使用时，
各种产量水平对应的最小可能成本。
固定成本不随产量变化而变化，
它是常数（即使产量为零也必须支付）。
但产量为零时的变动成本为零，
当产量大于零时，变动成本随产量增加而增加。
曲线TC与VC之间的距离就是固定成本。
平均成本和边际成本
资源稀缺性的一个基本内涵就是生产越多，支出越大。
平均固定成本（average fixed cost，AFC）
就是固定成本除以总产量：
固定成本不随产量变化而变化，
所以产量的增加可以使固定成本进一步分摊，
导致平均固定成本随着产量增加而不断降低。
各种成本之间的关系
边际成本曲线与ATC曲线
和AVC曲线的最低点相交。
这是因为当边际成本低于平均成本时，
平均成本是下降的；
当边际成本高于平均成本时，
平均成本将上升。
随着产量的增加，ATC曲线和AVC曲线越来越接近。
这是因为ATC与AVC之差为AFC，而AFC随着产量增加而不断降低。
总成本由变动成本和固定成本组成：
等式两边同时除以产量（Q），可得：
固定成本和沉没成本
固定成本是不随产量变化而变化的成本。
沉没成本（sunk cost）是一旦支付就永远损失的成本。
沉没成本是固定成本中不能撤回的部分。
因为沉没成本一旦支付将无法撤回，所以制定决策时不应考虑。
原理 沉没成本与决策无关
为使利润最大或损失最小，
制定决策时不应该考虑沉没成本。
成本函数的代数形式
三次成本函数（cubic cost function）为：
式中，a，b，c和f均为常数，注意f表示固定成本。
其边际成本函数为：
微积分表达式
边际成本是成本函数对产出求导：
三次成本函数对Q求导：
长期成本
从长期来看，所有成本都是可变的，管理者可以任意调整各种投入要素的使用量。
长期平均成本曲线（long-run average cost curve，LRAC）描述了当可以对所有生产要素
（包括固定要素和可变要素）进行优化选择时，生产不同产量产品的最低平均成本。
长期平均成本曲线是所有短期平均成本曲线的下包络线。
也就是说，长期平均成本曲线位于所有短期平均成本曲线下方，
短期曲线上只有最优利用固定投入要素的组合（短期平均成本最低点）与长期平均成本相切。
规模经济
长期平均成本曲线呈U形，即最初随着产量的增加，
长期平均成本不断降低，从0到Q的这种状态称为规模经济（economies of scale）。
若存在规模经济，扩大生产规模可以降低平均成本。
超过某点，如Q，再增加产量导致平均成本上升，这就是规模不经济（diseconomies of scal）。
有时，某行业中的一项技术可以使厂商在同一最低平均成本上的产量，
称为固定规模收益（constant returns to scale）。
提示：经济成本和会计成本
会计成本指经营活动中实际支出的货币成本，包括为劳动即资本直接支付的所有金额。会计成本体现在厂商的损益表中。
生产活动不仅包括会计成本，而且包括因生产某产品而放弃的机会成本。
多产品成本函数
假设一个多产品厂商的成本函数为C（Q1,Q2），
其中，Q1是产品1的产量，Q2是产品2的产量。
多产品成本函数（multiproduct cost function）
描述了所有投入要素均被有效使用条件下生产
Q1单位产品1和Q2单位产品2的成本。
范围经济
当两种产品联合生产的总成本低于分别生产的总成本时，存在范围经济（economics of scope），即
成本互补
多产品成本函数中，
当一种产品的产出增加而生产另一种产品的边际成本下降时，
则存在成本互补（cost complementarity）。
假设C（Q1,Q2）是一个多产品成本函数，
MC（Q1,Q2）为第一种产品的边际成本。
如果
则存在成本互补。
假设多产品成本函数是二次函数：
当a＜0时，Q2增加导致产品1的边际成本下降。
因此a＜0的成本函数体现了成本互补，
如果a＞0，则不存在成本互补。
针对成本函数，求MC1：
针对成本函数，求MC2：
第六章 企业的组织
导言
成本函数
描述任意产出水平下的最低成本。
点A的成本超出了
给定产量下所能实现的最低成本。
这说明有更好的投入组合，
但如果企业无法实现该组合，
或者员工没有竭尽全力，
企业的成本就会高于最低可能成本。
获取投入要素的三种方法：
现货交易、合同以及纵向一体化。
为了实现生产成本最小化，
企业不仅要有效率地利用所有投入要素，
而且要保证获取投入要素的成本最低。
如果雇员和雇主之间存在利益冲突，
就会出现委托-代理问题。
获取投入品的最佳方式
现货交易获取投入品
现货交易（spot exchange） 买卖双方见面、交易、随后离开，双方之间无任何关系或者合同。
现货交易的买卖双方基本上是“匿名”的，双方甚至在不了解对方名称的情况下就进行交易，而且买卖双方没有正式的（法律意义上的）关系。
现货交易通常用于“标准化”投入品的交易，企业能够从众多供应商中选择一家来购买生产所需的投入品。
通过合同获取投入品
合同（contract）是一份法律文件，它明确了双方在特定时期内进行交易的条款，使交易双方建立起一种长期关系。
如果合同签订过程简单且投入品特征能够在合同中精确描述，那么通过合同获取投入品的方式非常有效。
但合同方式的最大缺点是成本太高。此外，合同很难涵盖未来所有可能的未知事项。复杂环境中合同的不完备性不可避免。
自行生产投入
纵向一体化（vertical integration）避开其他供应商，选择在公司内部生产投入品的方式。
纵向一体化可能有损企业的专业化分工优势，因为企业不仅要从事最终产品的生产，而且要负责投入品的生产，由于造成组织规模扩大及相应的官僚成本。
换个角度来看，由于企业能够自行生产投入品，因此降低了企业对投入品供应商的依赖。
交易成本
交易成本（transaction costs）企业获得一种投入品的成本可能会超过实际支付给供应商的部分。
交易成本在选择最优投入获取方式时具有重要作用。
具体包括：
1.搜寻既定投入品供应商的成本。
2.谈判投入品价格的成本，比如所投入时间的成本、法律费用等。
3.推动交易进行的相关投资或支出等。
专用性投资的类型
专用性投资（specialized investment）是指针对某项特定交易进行的投资，这些投资对其他交易毫无价值。
当需要专用性投资来推动交易时，双方的关系就称为关系专用性交易（relationship-specific exchange）。
关系专用性交易的典型特征是将交易双方“捆绑在一起”，因为双方为开展交易而进行专用性投资。
专用性投资的沉没成本特性通常会引起交易成本。
地点专用性
地点专用性发生于买卖双方开展交易
必须将工厂选址在临近地点之时。
实物资产专用性
实物资产专用性指生产某种投入品的机器设备
不能生产其他投入品。
专用性资产
专用性资产是企业为了同某特定客户进行交易
而作出的一般性投资。
专用性人力资本
在一些雇佣关系中，员工为了在某公司工作必须学习一些专项技能。
这些技能对其他雇主来说无价值或者不能转移，这些技能的学习就是一种专用性投资。
专用性投资的含义
专用性投资之所以提高了交易成本，
是因为它们会引发：
（1）高议价成本；（2）投资不足；（3）机会主义和敲竹杠。
高议价成本
此类投入品通常不存在“市场价格”，
双方当事人需要对投入品进行谈判。谈判的费用很高。
投资不足当需要借助专用性投资推动交易时，
专用性投资水平往往低于最优投资水平。
机会主义和敲竹杠
敲竹杠是指：一旦企业进行了专用性投资，交易方可能凭借专用性投资的沉没特性对其进行“敲诈”。
这使得企业不愿意进行关系专用性投资，除非双方能够通过合同等方式减少敲竹杠的机会。
现实经济活动中，交易双方很多时候需要进行专用性投资，
可能诱发双方的机会主义投机。
最优采购方式
现货交易
企业获取投入品的最直接方式就是现货交易。
当存在专用性投资时，若采取现货交易方式，将无法规避买方的机会主义行为，交易双方最终需要花费大量时间进行价格谈判并承担谈判破裂的成本。
合同
尽管签订合同往往需要预先支付谈判费用和律师费。
但这样做有很多优点：
首先，签订合同可以使双方在进行
专用性投资之前就明确投入品的价格，
这样可以减少机会主义行为。
其次，合同规定双方在较长时期内遵循协议价格，
从而激励双方进行专用性投资。
最优合同期限的确定需要综合权衡合同期内的边际成本和边际收益。
合同期限越长，合同期内的边际成本（MC）就越大，
偶然事件的发生概率越高，签订合同就需要花费越多的时间和金钱。
合同期内的边际成本是向上倾斜的。
对于交易者来说，
续签一年合同的边际收益（MB）
应该等于避免机会主义行为和溢价的机会成本。
理论上讲这些收益会随着合同期限的变化而变化，
但为了简单起见，用一条直线来描述。
所以，合同最优期限L应该位于边际收益与边际成本的交点。
一般来说，当双方之间的专用性投资增加时，合同最优期限也将随之延长。
这是因为，随着专用性投资日益重要，
合同期满后双方将面临更高的交易成本。
签订更长期合同就能避免这些成本，
所以专用性投资越大，签订更长期限合同的边际收益约高。
合同最优期限还受边际成本的影响。
当投入品更加标准化且未来经济环境比较确定时，
签订长期合同的边际成本将从MC0降至MC1，
也就是说，制定合同的环境复杂性降低，从而延长了合同最优期限（从L0到L1）。
合同环境的复杂性会影响合同期限。
由于合同环境的复杂性，交易双方无法通过签订长期合同来规避该成本。
纵向一体化
纵向一体化，即企业的生产流程向上游延伸，新增设备进行投入品的生产。
纵向一体化的优势在于，将原来的专业化厂商整合为综合性公司，从而减少机会主义行为。
存在缺陷，管理者要用内部管理机制取代市场机制。
由于投入品与企业主营业务的运营特征不同，使得企业无法专注于其最擅长的业务。
由于存在这些难题，纵向一体化往往被企业作为最后的选择
——只有当现货交易和合同方式失败时才选用。
经济权衡
薪酬管理与委托-代理问题
大企业的特征是所有权和经营权分离。
所有权与控制权分离会产生委托—代理问题：
所有者不能亲自监督管理者工作，那么如何保证管理者为所有者的利益努力工作？
委托—代理代理问题的本质在于管理者既喜欢赚钱也喜欢享受休闲。
从所有者角度来说，固定工资制是无法激励管理者努力工作的。
在利润共享方案中，管理者的行为依赖于其对休闲和金钱的权衡。
管理者面临的取舍是：如果在工作时间享受休闲，代价是报酬降低。
管理者的约束力量
激励合同
通常情况下，公司CEO会得到股权或者与利润直接挂钩的奖金激励。
如果CEO的收入大部分来源于绩效奖金，那么减少其报酬就是不恰当的。
以绩效为基础的报酬对股东和CEO都有好处，减少CEO报酬将会减少公司利润。
外部激励
声誉
如果管理者能够向其他企业展示其具备提高企业利润的能力，管理者就有更多的工作流动机会。
从长期来看，剩余将提高管理者在经理人市场的溢价。
收购
如果管理者不能很好地经营企业，企业就可能被收购，管理者可能被替代。
管理者与员工之间的委托—代理问题
管理者与员工之间的
委托—代理问题的解决方案
利润共享（profit sharing）是管理者为促使员工努力工作而采取的激励制度，
即将员工报酬与企业盈利能力挂钩。
收益共享（revenue sharing）将员工报酬同企业潜在收益相结合。包括消费和销售佣金等。会产生心理上的不同影响。
当员工的劳动效率和收益相关，但与成本无关时，收益共享方案是非常有效的。
收益共享方案无法激励员工实现成本最小化。
计件工资是按工作量而非固定小时工资计酬的方式。
计件工资的潜在问题是必须努力控制质量，
否则员工可能会因追求数量而牺牲质量。
出勤记录和现场检查
实际上出勤记录对解决委托—代理问题没太大作用。
更有效的监督员工的机制是管理者对工作场所突击检查。
现场检查的缺点是，
管理者出现的频率必须足够多（使员工不敢有侥幸心理），
而且必须对开小差的员工给予相应的惩罚。
现场检查是通过威胁来督促员工作，
而绩效奖金是通过奖励承诺来激励员工，这些做法对员工会产生心理上的不同影响。
第七章 行业的性质
导言
各行业的差异，分析差异的产生原因并研究这些差异如何影响管理决策。
市场结构
不同行业有不同的市场结构，
市场结构因素会影响企业管理者的决策。
企业规模
各行业中最大的企业存在规模差异。
行业集中度
行业集中度的度量
前四家企业集中度比率（four-firm concentration ratio）（CR4），
即行业中规模最大的四家企业的销售额占行业总销售额的比例。
令S1,S2,S3,S4分别代表行业中规模最大的四家企业的销售额，
用ST表示该行业中所有企业的总销售额。则前四家企业集中度比率为：
同理，前四家企业集中度比率是前四家企业市场份额的总和。
前四家企业集中度比率接近零，说明市场中有很多卖方，
为了争夺消费者，生产者之间存在激烈竞争；
前四家企业集中度比率接近1，说明生产商在消费者方面的竞争很小。
赫芬达尔-海希曼指数（Herfindahl-Hirschman index，HHI）指特定行业内各企业市场份额的平方和，为消除小数再乘以10000。
假设企业i在总市场销售额中所占比重为wi=Si/St，
其中，Si是企业i的销售额；St是行业总销售额。则HHI为：
HHI取值在0~10000之间。
如果行业内只有一家企业（市场份额w1=1））时，则HHI为10000；
如果行业中存在大量小企业，HHI接近零。
CR4未考虑第五大企业的市场份额，但HHI考虑了并且计算在内。
其次，HHI基于市场份额的平方，而CR4不是，
所以HHI较CR4给市场份额较大的企业分配了更高的权重。
上述两点导致了CR4和HHI的结论产生差异。
集中度计量法的局限性
全球市场
在计算CR4和HHI时没有考虑已经投资美国市场的外国企业。
在某种程度上扩大了行业真实的集中度。
国家、区域和地方市场
市场对许多行业来说，其相关市场仅仅局限于地方市场（可能只包括少数企业），
若用全国数据就会低估地方市场的集中度水平。
当相关市场是地方市场时，
用全国数据得出的市场结构指数通常会大大低估该地方市场的集中度。
行业定义和产品分类 我们用相互替代性来分类，
相互替代性大（交叉弹性为正切较大）的产品可以归属于同一行业。
技术
技术差异导致不同行业的产品在技术含量方面存在差异。
需求和市场条件
如果行业需求小，市场上通常只有少数企业生存。
如果行业需求很大，则需要大量企业才能满足需求。
消费者信息收集的难易程度因市场不同而不同。
管理者的最优决策受市场中可获信息量的影响非常大。
不同行业中产品的需求弹性也不同。
加入某企业产品在市场上没有近似替代品，
则该企业产品的需求弹性就是市场的需求弹性（因为市场中只有一家企业的产品）；
如果存在多家企业的产品可以替代之，那么，
该企业产品的需求较整个行业的需求更富有弹性。
罗斯查尔德指数（Rothschild index）简称R指数，
指某种产品的行业需求对价格的敏感度与单个企业需求对价格敏感度的比较。
R指数度量了某种产品的整个行业需求价格弹性与单个企业的需求价格弹性的差异。
R指数的计算公式为：
式中，
ET是整个行业市场的需求价格弹性；
EF是单个企业产品的需求价格弹性。
如果行业中有多家企业且每家企业的产品相似，R指数接近零。
进入壁垒
新企业是否进入某行业的市场很大程度上取决于其进入该市场的难易程度。
资本投资需求。专利。 规模经济也会构成进入壁垒。进入壁垒对企业的长期利润有重要影响。
经营行为
定价行为
勒纳指数（Lerner index）的表达式如下：
式中，P是价格；MC是边际成本。
勒纳指数度量了行业内企业的产品定价高于边际成本的幅度，
也就是溢价幅度的大小。
勒纳指数越高，企业的溢价幅度越大。
勒纳指数与企业的溢价幅度有关，将勒纳指数变形得到：
在这个等式中，1/(1-L)称为溢价系数，
边际成本乘以该系数即可得到产品价格。
一体化与并购行为
一体化指的是生产资源的整合。一体化可以通过并购实现；
也可以通过企业自行建厂实现。
企业为减少交易成本、实现规模或范围经济、
增强市场势力或者获取更有利的进入资本市场的机会，可能发生并购行为。
如果并购双方都愿意合并成一家企业，这种并购是善意的；
如果其中有一家公司不愿意因并购而被替代，则可能发生恶意的并购。
许多管理者害怕企业被兼并或收购，
主要是担心会对其职位造成影响。
纵向一体化
纵向一体化指将产品的各个生产阶段囊括于一家企业中。
企业纵向一体化是为了减少交易成本。
横向一体化
横向一体化指把生产相似产品的企业合并成一家企业。
实现横向一体化的原因则是：
（1）获取规模经济或范围经济节约成本；
（2）增强市场势力。横向一体化的成本下降所带来的社会效益，
必须大于行业集中所带来的社会成本，否则这种横向并购可能被限制。
混合并购混合并购指不同类型的产品合并到一家企业内。
并购不相关业务线往往达不到预期目的，混合并购无助于协同效应，甚至会损失专业化优势。
研发行为
企业可以通过研发投入获得技术优势和技术专利。
研发的最优投入量取决于企业所在行业的性质。
广告行为
绩效
绩效指特定行业所实现的利润和社会福利。
利润
社会福利
丹斯比—魏力格绩效指数（Dansby-Willig performance index）（简称DW指数），
可度量当行业内企业以社会有效方式扩大生产时所带来的社会福利
（定位为消费者剩余和生产者剩余之和）的增加量。
当DW指数大于零时，行业提高产出将带来社会福利的改善。
DW指数提供了一种行业分级法。
使我们能够按照行业产出改变时社会福利的增加值对行业进行分级。
机构-行为-绩效范式
因果观
因果观认为，市场结构决定企业行为，企业行为影响资源分配方式，最终导致好或坏的市场绩效。
根据因果观，高集中度的市场会导致高价格和低绩效。
批评意见
批评意见认为结构、行为和绩效三者之间并非单向因果关系，企业行为反过来也能够影响市场结构；
同样，市场绩效也能够影响企业行为和市场结构。
五力模型
五力模型认为以下五种相互关联的作用力影响着行业利润水平、增长性和可持续性：
（1）潜在进入者；（2）供应商力量；（3）购买者力量；（4）行业内竞争者；（5）替代品和互补品。
本书后续内容概述
完全竞争市场
完全竞争市场的特征是：
市场中存在很多企业，每个企业相对于整个市场都很小。
所有企业技术相同且产品同质，不存在更具优势的企业。
完全竞争市场的企业没有市场势力，
单个企业不会对市场价格、产品数量或质量产生显著影响。
在完全竞争市场中，市场集中度比率和R指数接近零。
垄断市场
垄断市场指在相关市场上只有一家企业提供产品。
当市场上只有一家企业提供产品时，卖方通常会利用垄断地位来限制产量、制定高于边际成本的价格。
由于市场上没有其他企业，消费者无法转向其他供给者，
因此要么高价购买，要么放弃购买。垄断企业高度集中，R指数为1。
垄断竞争市场
垄断竞争市场的特征是，市场上有很多企业和消费者。
每家企业的产品与其他企业的产品都略有差异，R指数大于零。
处于垄断竞争市场的企业有一定的定价权。
垄断竞争市场的企业往往会投入大量的广告打造品牌优势，
使消费者相信其品牌”优于“其他品牌，从而当公司提高产品价格时减少转向其他品牌的消费者数量。
寡头垄断市场
寡头垄断市场
通常由几家大企业主导。
在寡头垄断市场，如果一家企业改变其价格或市场战略，不仅会影响自己的利润，还会影响行业中其他企业的利润。
因此当寡头垄断市场中的企业改变其行为时，其他企业也会因此而改变行为。
寡头垄断市场的一个显著特征是行业内的企业存在相互依赖关系。
相互依赖导致企业间在战略上相互影响。
寡头垄断市场上企业的决策
依赖于该企业对行业内其他企业对其行为反应的预期。
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7574 | https://www.youtube.com/watch?v=rkV_OcT0M2k | Proof That the Derivative of e^x is e^x
John's Solution Set
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Posted: 5 Mar 2024
One of the most fascinating things in calculus is how it works so naturally around e^x and ln x. In this video, we prove that the derivative of e^x is again e^x. To do this, we first define e as the number such that when we consider the function e^x and take the derivative at the point x=0, we get 1. Note that all exponential functions will have a slightly different derivative at x=1, but only one will have a derivative of 1. That function is e^x, and that’s how we define the number e. Once we have e defined, we use the limit definition of the derivative to complete the proof.
This video also gives an example of using our new derivative to take the derivative of a function involving e.
ADA-compliant subtitles/captions
28 comments
Transcript:
INSTRUCTOR: Something truly
amazing about calculus is that the derivative of e
to the x is e to the x again. E to the x is the only function
that has this property, and it really shows
how fundamental, I guess how natural, e to the
x is to calculus and really mathematics in general. It's really kind of
along with natural log, e to the x and natural log of
x are kind of the cornerstones or anchors of a lot of calculus,
can be brought back to what happens with e to the x. So let's prove that the
derivative of e to the x is itself e to the x. First, we need a
definition of e. And you'll get a few of
these definitions of e. As you go along with
calculus, there are several, but today we'll
do one as follows. So imagine an exponential
function y equals b to the x, actually imagine
several of these for a bunch of
different bases b. It can be 10 to the x, 2 to the
x, all these different things. Each one of these
exponential functions will have a different
slope of the tangent line at the point x equals 0. So the point of
interest here is 0, 1. So 10 to the x is going
to have a steeper tangent line, 2 to the x will have
a more gradual tangent line, not as steep, but only one of
these exponential functions will have a tangent line with
slope exactly equal to 1. So whatever that
number is, we don't know what it is, we don't
need to know what it is, but let's call it something. Let's call that number
e, so equals e to the x. Any time it has a
slope of 1, there's only one such number
for the base that can give us a slope of 1 for the
tangent line at the point 0, 1. Let's call that number e. And in fact, that will be our
definition of e in this context. So here we go. Here's our definition. Let e be the unique number with
the property that f of x equals e to the x has a
slope 1 at x equals 0. Only one such number
will have that property. Let's call it e. Because how else
could we talk about e? We could say, oh,
2.7 or maybe 2.7, or then do we say
2.718281828459045 blah blah blah. We would have to
talk about it forever to actually say what it is. So we can't say what it
is other than by using these nondirect ways of
talking about it to define it without actually listing out all
the digits because we never can. So we can get even more
detail out of this, which will be useful for the proof. This tells us that this function
f of x equals e to the x, if we take the derivative
at 0, the slope of the tangent line
at 0 must be one. But we have a limit
definition for that. Let's actually go
down here with it. So this is the
limit, if you recall, is h goes to 0 of f of x
plus a or f of a plus h, we use the definition of
the derivative at a point. So f of a plus h, so f of 0 plus
h minus f of a, e to the 0 all over h, all over h. So we know that this
must be equal to 1, but we can simplify
this a little bit. This tells us that the limit
as h goes to 0 of e to the h minus 1 all over h equals 1. OK. So let's stick a pin in that. We'll back to it later and
use that as part of the proof. OK. So now let's prove
the whole thing. So here's our claim. I claim that the derivative
of e to the x is e to the x. OK, proof. OK. And we're still using
our same function here. So here's our proof f
of x equals e to the x. F prime of x then
is the limit as h goes to 0 of f of x plus
h minus f of x all over h, so that's e to the x plus h
minus e to the x all over h. OK. Well, we can factor an e to the
x out of that top bit there. So limit is h goes to
0 of e to the x, what does that leave behind? E to the h minus 1 all over h. Well, this e to the x
has no h in it at all. So that's going to pop out
to the front of the limit since as far as the
limit is concerned, this thing is a constant. Because the limit
is looking at h's not x's, so that comes
out to the front, leaving us with the
limit as h goes to 0. Let's see. Let's bring that e to
the x out to the front. And then we simply have e
to the h minus 1 all over h. But remember that that exact
quantity, that limit as h goes to 0 of e to the
h minus 1 all over h, that's already equal to 1. We already know
that's equal to 1. So this gives us e to the x
times 1, which is e to the x. And that's what we
set out to prove. All right. So we get to do our
little proof box there. And there you have it. Let's do an example of taking
the derivative of functions involving e to the x, or
in this case, e to the w. OK. So here we have the function
in an example g of w equals the quantity e to the
2w plus e to the w all over e to the w. And we're asked to find
g prime of w, which is the derivative, of course. Well, we still don't
have a quotient rule. And honestly, we wouldn't want
one here because we can simplify and you should always
simplify before you go about using some crazy rule. So look, we can
cancel an e to the w. How about this? Let's rewrite this g of w as
let's factor out e to the w How about that? Out of the top there, that
leaves us with e to the w plus 1 left behind. And then we still have e to
the w in the denominator. And look, look at that. They cancel right out. So now we have g of w
equals e to the w plus 1. And we can very quickly
go then to g prime of w-- well, it's simply
e to the w again because remember that the
derivative of e to the anything is e to that thing again. And then the derivative of 1
is 0, since 1 is a constant. So there we have it,
e to the something is pretty much the
easiest derivative we'll have in the entire course
because you just write it again. Now, it will become a
little more challenging when there's a function up here. We'll have to do something
called the chain rule. But even that is not so bad
when you're dealing with e as your base. 6 minutes, 57 seconds |
7575 | https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic | Circuit analysis | Electrical engineering | Science | Khan Academy
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Electrical engineering 9 units · 1 skillsUnit 1 Introduction to electrical engineeringUnit 2 Circuit analysisUnit 3 AmplifiersUnit 4 Semiconductor devicesUnit 5 ElectrostaticsUnit 6 Signals and systemsUnit 7 Home-made robotsUnit 8 Lego roboticsUnit 9 Reverse engineering
Science
Electrical engineering
Unit 2: Circuit analysis
About this unit
Circuit analysis is the process of finding all the currents and voltages in a network of connected components. We look at the basic elements used to build circuits, and find out what happens when elements are connected together into a circuit.
Circuit elements
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Ideal circuit elements (Opens a modal)
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Resistor circuits
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Series resistors (Opens a modal)
Series resistors (Opens a modal)
Parallel resistors (derivation) (Opens a modal)
Parallel resistors (derivation continued) (Opens a modal)
Parallel resistors (Opens a modal)
Parallel resistors (Opens a modal)
Parallel conductance (Opens a modal)
Simplifying resistor networks (Opens a modal)
Simplifying resistor networks (Opens a modal)
Delta-Wye resistor networks (Opens a modal)
Voltage divider (Opens a modal)
Voltage divider (Opens a modal)
Analyzing a resistor circuit with two batteries (Opens a modal)
Practice
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DC circuit analysis
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Circuit analysis overview (Opens a modal)
Kirchhoff's current law (Opens a modal)
Kirchhoff's voltage law (Opens a modal)
Kirchhoff's laws (Opens a modal)
Labeling voltages (Opens a modal)
Application of the fundamental laws (setup) (Opens a modal)
Application of the fundamental laws (solve) (Opens a modal)
Application of the fundamental laws (Opens a modal)
Node voltage method (steps 1 to 4) (Opens a modal)
Node voltage method (step 5) (Opens a modal)
Node voltage method (Opens a modal)
Mesh current method (steps 1 to 3) (Opens a modal)
Mesh current method (step 4) (Opens a modal)
Mesh current method (Opens a modal)
Loop current method (Opens a modal)
Number of required equations (Opens a modal)
Linearity (Opens a modal)
Superposition (Opens a modal)
Natural and forced response
Learn
Capacitor i-v equations (Opens a modal)
A capacitor integrates current (Opens a modal)
Capacitor i-v equation in action (Opens a modal)
Inductor equations (Opens a modal)
Inductor kickback (1 of 2) (Opens a modal)
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RC natural response - intuition (Opens a modal)
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RC natural response - example (Opens a modal)
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RC step response - intuition (Opens a modal)
RC step response setup (1 of 3) (Opens a modal)
RC step response solve (2 of 3) (Opens a modal)
RC step response example (3 of 3) (Opens a modal)
RC step response (Opens a modal)
RL natural response (Opens a modal)
Sketching exponentials (Opens a modal)
Sketching exponentials - examples (Opens a modal)
LC natural response intuition 1 (Opens a modal)
LC natural response intuition 2 (Opens a modal)
LC natural response derivation 1 (Opens a modal)
LC natural response derivation 2 (Opens a modal)
LC natural response derivation 3 (Opens a modal)
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LC natural response example (Opens a modal)
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LC natural response - derivation (Opens a modal)
RLC natural response - intuition (Opens a modal)
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RLC natural response - variations (Opens a modal)
AC circuit analysis
Learn
AC analysis intro 1 (Opens a modal)
AC analysis intro 2 (Opens a modal)
Trigonometry review (Opens a modal)
Sine and cosine come from circles (Opens a modal)
Sine of time (Opens a modal)
Sine and cosine from rotating vector (Opens a modal)
Lead Lag (Opens a modal)
Complex numbers (Opens a modal)
Multiplying by j is rotation (Opens a modal)
Complex rotation (Opens a modal)
Euler's formula (Opens a modal)
Complex exponential magnitude (Opens a modal)
Complex exponentials spin (Opens a modal)
Euler's sine wave (Opens a modal)
Euler's cosine wave (Opens a modal)
Negative frequency (Opens a modal)
AC analysis superposition (Opens a modal)
Impedance (Opens a modal)
Impedance vs frequency (Opens a modal)
ELI the ICE man (Opens a modal)
Impedance of simple networks (Opens a modal)
KVL in the frequency domain (Opens a modal)
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7576 | https://wikidiff.com/peculate/embezzle | Peculate vs Embezzle - What's the difference? | WikiDiff
What's the difference between and
Enter two words to compare and contrast their definitions, origins, and synonyms to better understand how those words are related.
Peculate vs Embezzle - What's the difference?
×
peculate | embezzle |
Peculate is a synonym of embezzle.
As verbs the difference between peculate and embezzle
is that peculate is to embezzle while embezzle is (legal|business) to steal or misappropriate money that one has been trusted with, especially to steal money from the organisation for which one works.
peculate ========
English
Verb
(peculat) To embezzle
Synonyms
defalcate embezzle misappropriate
Related terms
peculation peculator
embezzle ========
English
Verb
(en-verb) (legal, business) To steal or misappropriate money that one has been trusted with, especially to steal money from the organisation for which one works.
1903, , Twelve Stories and a Dream _You waste your education in burglary. You should do one of two things. Either you should forge or you should embezzle _'. For my own part, I '_ embezzle._
1861,_You let Dunsey have it, sir? And how long have you been so thick with Dunsey that you must _collogue_ with him to embezzle my money?_
Synonyms
defalcate misappropriate peculate
Derived terms
embezzler embezzlement
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7577 | https://en.wikipedia.org/wiki/Mass_point_geometry | Jump to content
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Contents
(Top)
1 Definitions
2 Methods
2.1 Concurrent cevians
2.2 Splitting masses
2.3 Other methods
3 Examples
3.1 Problem One
3.2 Problem Two
3.3 Problem Three
4 See also
5 Notes
6 External links
Mass point geometry
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From Wikipedia, the free encyclopedia
Problem-solving technique in geometry
Mass point geometry, colloquially known as mass points, is a problem-solving technique in geometry which applies the physical principle of the center of mass to geometry problems involving triangles and intersecting cevians. All problems that can be solved using mass point geometry can also be solved using either similar triangles, vectors, or area ratios, but many students prefer to use mass points. Though modern mass point geometry was developed in the 1960s by New York high school students, the concept has been found to have been used as early as 1827 by August Ferdinand Möbius in his theory of homogeneous coordinates.
Definitions
[edit]
The theory of mass points is defined according to the following definitions:
Mass Point - A mass point is a pair , also written as , including a mass, , and an ordinary point, on a plane.
Coincidence - We say that two points and coincide if and only if and .
Addition - The sum of two mass points and has mass and point where is the point on such that . In other words, is the fulcrum point that perfectly balances the points and . An example of mass point addition is shown at right. Mass point addition is closed, commutative, and associative.
Scalar Multiplication - Given a mass point and a positive real scalar , we define multiplication to be . Mass point scalar multiplication is distributive over mass point addition.
Methods
[edit]
Concurrent cevians
[edit]
First, a point is assigned with a mass (often a whole number, but it depends on the problem) in the way that other masses are also whole numbers. The principle of calculation is that the foot of a cevian is the addition (defined above) of the two vertices (they are the endpoints of the side where the foot lie). For each cevian, the point of concurrency is the sum of the vertex and the foot. Each length ratio may then be calculated from the masses at the points. See Problem One for an example.
Splitting masses
[edit]
Splitting masses is the slightly more complicated method necessary when a problem contains transversals in addition to cevians. Any vertex that is on both sides the transversal crosses will have a split mass. A point with a split mass may be treated as a normal mass point, except that it has three masses: one used for each of the two sides it is on, and one that is the sum of the other two split masses and is used for any cevians it may have. See Problem Two for an example.
Other methods
[edit]
Routh's theorem - Many problems involving triangles with cevians will ask for areas, and mass points does not provide a method for calculating areas. However, Routh's theorem, which goes hand in hand with mass points, uses ratios of lengths to calculate the ratio of areas between a triangle and a triangle formed by three cevians.
Special cevians - When given cevians with special properties, like an angle bisector or an altitude, other theorems may be used alongside mass point geometry that determine length ratios. One very common theorem used likewise is the angle bisector theorem.
Stewart's theorem - When asked not for the ratios of lengths but for the actual lengths themselves, Stewart's theorem may be used to determine the length of the entire segment, and then mass points may be used to determine the ratios and therefore the necessary lengths of parts of segments.
Higher dimensions - The methods involved in mass point geometry are not limited to two dimensions; the same methods may be used in problems involving tetrahedra, or even higher-dimensional shapes, though it is rare that a problem involving four or more dimensions will require use of mass points.
Examples
[edit]
Problem One
[edit]
Problem. In triangle , is on so that and is on so that . If and intersect at and line intersects at , compute and .
Solution. We may arbitrarily assign the mass of point to be . By ratios of lengths, the masses at and must both be . By summing masses, the masses at and are both . Furthermore, the mass at is , making the mass at have to be Therefore and . See diagram at right.
Problem Two
[edit]
Problem. In triangle , , , and are on , , and , respectively, so that , , and . If and intersect at , compute and .
Solution. As this problem involves a transversal, we must use split masses on point . We may arbitrarily assign the mass of point to be . By ratios of lengths, the mass at must be and the mass at is split towards and towards . By summing masses, we get the masses at , , and to be , , and , respectively. Therefore and .
Problem Three
[edit]
Problem. In triangle , points and are on sides and , respectively, and points and are on side with between and . intersects at point and intersects at point . If , , and , compute .
Solution. This problem involves two central intersection points, and , so we must use multiple systems.
System One. For the first system, we will choose as our central point, and we may therefore ignore segment and points , , and . We may arbitrarily assign the mass at to be , and by ratios of lengths the masses at and are and , respectively. By summing masses, we get the masses at , , and to be 10, 9, and 13, respectively. Therefore, and .
System Two. For the second system, we will choose as our central point, and we may therefore ignore segment and points and . As this system involves a transversal, we must use split masses on point . We may arbitrarily assign the mass at to be , and by ratios of lengths, the mass at is and the mass at is split towards and 2 towards . By summing masses, we get the masses at , , and to be 4, 6, and 10, respectively. Therefore, and .
Original System. We now know all the ratios necessary to put together the ratio we are asked for. The final answer may be found as follows:
See also
[edit]
Cevian
Ceva's theorem
Menelaus's theorem
Stewart's theorem
Angle bisector theorem
Routh's theorem
Barycentric coordinates
Lever
Notes
[edit]
^ Rhoad, R., Milauskas, G., and Whipple, R. Geometry for Enjoyment and Challenge. McDougal, Littell & Company, 1991.
^ "Archived copy". Archived from the original on 2010-07-20. Retrieved 2009-06-13.{{cite web}}: CS1 maint: archived copy as title (link)
^ Rhoad, R., Milauskas, G., and Whipple, R. Geometry for Enjoyment and Challenge. McDougal, Littell & Company, 1991
^ D. Pedoe Notes on the History of Geometrical Ideas I: Homogeneous Coordinates. Math Magazine (1975), 215-217.
^ H. S. M. Coxeter, Introduction to Geometry, pp. 216-221, John Wiley & Sons, Inc. 1969
External links
[edit]
Media related to Mass point geometry at Wikimedia Commons
Retrieved from "
Categories:
Geometric centers
Triangle geometry
Hidden categories:
CS1 maint: archived copy as title
Articles with short description
Short description is different from Wikidata
Commons category link from Wikidata
Mass point geometry
Add topic |
7578 | https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/Biology_(Kimball)/03%3A_The_Cellular_Basis_of_Life/3.23%3A_Diffusion_Active_Transport_and_Membrane_Channels | 3.23: Diffusion, Active Transport and Membrane Channels - Biology LibreTexts
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Unit 3: The Cellular Basis of Life
Biology (Kimball)
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Mon, 17 Mar 2025 11:18:31 GMT
3.23: Diffusion, Active Transport and Membrane Channels
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[ "article:topic", "osmosis", "authorname:kimballj", "concentration gradient", "diffusion", "active transport", "facilitated diffusion", "showtoc:no", "license:ccby", "licenseversion:30", "source@ "cell membranes", "transmembrane proteins", "sodium ions", "potassium ions", "transport" ]
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Contents
1. Home
2. Bookshelves
3. Introductory and General Biology
4. Biology (Kimball)
5. Unit 3: The Cellular Basis of Life
6. 3.23: Diffusion, Active Transport and Membrane Channels
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Biology (Kimball)
Front Matter
Unit 1: The Chemical Basis of Life
Unit 2: The Molecules of Life
Unit 3: The Cellular Basis of Life
Unit 4: Cell Metabolism
Unit 5: DNA
Unit 6: Gene Expression
Unit 7: Cell Division
Unit 8: The Genetic Consequences of Meiosis
Unit 9: Regulation of Gene Expression
Unit 10: Mutation
Unit 11: Genomics
Unit 12: Cancer
Unit 13: Aging
Unit 14: Embryonic Development and its Regulation
Unit 15: The Anatomy and Physiology of Animals
Unit 16: The Anatomy and Physiology of Plants
Unit 17: Ecology
Unit 18: Evolution
Unit 19: The Diversity of Life
Unit 20: General Science
Back Matter
3.23: Diffusion, Active Transport and Membrane Channels
Last updated Mar 17, 2025
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3.22: Chromatophores
3.24: Endocytosis
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Page ID 4070
John W. Kimball
Tufts University & Harvard
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Transport Across Cell Membranes
2. Facilitated Diffusion of Ions
1. Ligand-gated ion channels
1. External ligands
2. Internal ligands
2. Mechanically-gated ion channels/03:_The_Cellular_Basis_of_Life/3.23:_Diffusion_Active_Transport_and_Membrane_Channels#Mechanically-gated_ion_channels)
3. Voltage-gated ion channels/03:_The_Cellular_Basis_of_Life/3.23:_Diffusion_Active_Transport_and_Membrane_Channels#Voltage-gated_ion_channels)
1. The Patch Clamp Technique/03:_The_Cellular_Basis_of_Life/3.23:_Diffusion_Active_Transport_and_Membrane_Channels#The_Patch_Clamp_Technique)
Facilitated Diffusion of Molecules
Active Transport
Direct Active Transport
The Na+/K+ ATPase
The H+/K+ ATPase
The Ca2+ ATPases
ABC Transporters
Indirect Active Transport
Symport Pumps
Antiport Pumps
Some inherited ion-channel diseases
Osmosis
Transport Across Cell Membranes
All cells acquire the molecules and ions they need from their surrounding extracellular fluid (ECF). There is an unceasing traffic of molecules and ions in and out of the cell through its plasma membrane (Examples: glucose, Na+, Ca 2+). In eukaryotic cells, there is also transport in and out of membrane-bounded intracellular compartments such as the nucleus, endoplasmic reticulum, and mitochondria (Examples: proteins, mRNA, Ca 2+, and ATP).
The following problems can occur during transport:
1. Relative concentrations
Molecules and ions move spontaneously down their concentration gradient (i.e., from a region of higher to a region of lower concentration) by diffusion. Molecules and ions can be moved against their concentration gradient, but this process, called active transport, requires the expenditure of energy (usually from ATP).
2. Lipid bilayers are impermeable to most essential molecules and ions.
The lipid bilayer is permeable to water molecules and a few other small, uncharged, molecules like oxygen (O 2) and carbon dioxide (CO 2). These diffuse freely in and out of the cell. The diffusion of water through the plasma membrane is of such importance to the cell that it is given a special name - osmosis. Lipid bilayers are not permeable to ions such as K+, Na+, Ca 2+ (called cations because when subjected to an electric field they migrate toward the cathode [the negatively-charged electrode]) and Cl-, HCO 3- (called anions because they migrate toward the anode [the positively-charged electrode]). They are also not permeable to small hydrophilic molecules like glucose and macromolecules like proteins and RNA. The cells solve the problem of transporting ions and small molecules across their membranes with the help of the following two mechanisms:
Facilitated diffusion:Transmembrane proteins create a water-filled pore through which ions and some small hydrophilic molecules can pass by diffusion. The channels can be opened (or closed) according to the needs of the cell.
Active transport: Transmembrane proteins, called transporters, use the energy of ATP to force ions or small molecules through the membrane against their concentration gradient.
Facilitated Diffusion of Ions
Facilitated diffusion of ions takes place through proteins, or assemblies of proteins, embedded in the plasma membrane. These transmembrane proteins form a water-filled channel through which the ion can pass down its concentration gradient. The transmembrane channels that permit facilitated diffusion can be opened or closed. They are said to be "gated"; some types of gated ion channels:
ligand-gated
mechanically-gated
voltage-gated
light-gated
Ligand-gated ion channels
Many ion channels open or close in response to binding a small signaling molecule or "ligand". Some ion channels are gated by extracellular ligands; some by intracellular ligands. In both cases, the ligand is not the substance that is transported when the channel opens.
External ligands
External ligands (shown here in green) bind to a site on the extracellular side of the channel.
Figure 3.23.1 External Ligands
Examples:
Acetylcholine (ACh). The binding of the neurotransmitter acetylcholine at certain synapses opens channels that admit Na+ and initiate a nerve impulse or muscle contraction.
Gamma amino butyric acid (GABA). Binding of GABA at certain synapses — designated GABA A — in the central nervous system admits Cl- ions into the cell and inhibits the creation of a nerve impulse
Internal ligands
Internal ligands bind to a site on the channel protein exposed to the cytosol. Examples:
"Second messengers", like cyclic AMP (cAMP) and cyclic GMP (cGMP), regulate channels involved in the initiation of impulses in neurons responding to odors and light respectively.
ATP is needed to open the channel that allows chloride (Cl-) and bicarbonate (HCO 3-) ions out of the cell. This channel is defective in patients with cystic fibrosis. Although the energy liberated by the hydrolysis of ATP is needed to open the channel, this is not an example of active transport; the ions diffuse through the open channel following their concentration gradient.
Mechanically-gated ion channels
Sound waves bending the cilia-like projections on the hair cells of the inner ear open up ion channels leading to the creation of nerve impulses that the brain interprets as sound. Mechanical deformation of the cells of stretch receptors opens ion channels leading to the creation of nerve impulses.
Voltage-gated ion channels
In so-called "excitable" cells like neurons and muscle cells, some channels open or close in response to changes in the charge (measured in volts) across the plasma membrane. For example,as an impulse passes down a neuron, the reduction in the voltage opens sodium channels in the adjacent portion of the membrane. This allows the influx of Na+ into the neuron and thus the continuation of the nerve impulse. Some 7000 sodium ions pass through each channel during the brief period (about 1 millisecond) that it remains open. This was learned by use of the patch clamp technique.
The Patch Clamp Technique
The properties of ion channels can be studied by means of the patch clamp technique. A very fine pipette (with an opening of about 0.5 µm) is pressed against the plasma membrane of either an intact cell or the plasma membrane can be pulled away from the cell and the preparation placed in a test solution of desired composition. Current flow through a single ion channel can then be measured.
Figure 3.23.2 The Patch Clamp
Such measurements reveal that each channel is either fully open or fully closed; that is, facilitated diffusion through a single channel is "all-or-none". This technique has provided so much valuable information about ion channels that its inventors, Erwin Neher and Bert Sakmann, were awarded a Nobel Prize in 1991.
Facilitated Diffusion of Molecules
Some small, hydrophilic organic molecules, like sugars, can pass through cell membranes by facilitated diffusion. Once again, the process requires transmembrane proteins. In some cases, these — like ion channels — form water-filled pores that enable the molecule to pass in (or out) of the membrane following its concentration gradient.
Example:
Maltoporin. This homotrimer in the outer membrane of E. coli forms pores that allow the disaccharide maltose and a few related molecules to diffuse into the cell.
Example: The plasma membrane of human red blood cells contain transmembrane proteins that permit the diffusion of glucose from the blood into the cell.
Note that in all cases of facilitated diffusion through channels, the channels are selective; that is, the structure of the protein admits only certain types of molecules through. Whether all cases of facilitated diffusion of small molecules use channels is yet to be proven. Perhaps some molecules are passed through the membrane by a conformational change in the shape of the transmembrane protein when it binds the molecule to be transported.
In either case, the interaction between the molecule being transported and its transporter resembles in many ways the interaction between an enzyme and its substrate.
Active Transport
Active transport is the pumping of molecules or ions through a membrane against their concentration gradient. It requires a transmembrane protein (usually a complex of them) called a transporter and energy. The source of this energy is ATP.
The energy of ATP may be used directly or indirectly.
Direct Active Transport. Some transporters bind ATP directly and use the energy of its hydrolysis to drive active transport.
Indirect Active Transport. Other transporters use the energy already stored in the gradient of a directly-pumped ion. Direct active transport of the ion establishes a concentration gradient. When this is relieved by facilitated diffusion, the energy released can be harnessed to the pumping of some other ion or molecule.
Direct Active Transport
The Na+/K+ ATPase
The cytosol of animal cells contains a concentration of potassium ions (K+) as much as 20 times higher than that in the extracellular fluid. Conversely, the extracellular fluid contains a concentration of sodium ions (Na+) as much as 10 times greater than that within the cell. These concentration gradients are established by the active transport of both ions. And, in fact, the same transporter, called the Na+/K+ ATPase, does both jobs. It uses the energy from the hydrolysis of ATP to
actively transport 3 Na+ ions out of the cell
for each 2 K+ ions pumped into the cell.
This accomplishes several vital functions:
It helps establish a net charge across the plasma membrane with the interior of the cell being negatively charged with respect to the exterior. This resting potential prepares nerve and muscle cells for the propagation of action potentials leading to nerve impulses and muscle contraction.
The accumulation of sodium ions outside of the cell draws water out of the cell and thus enables it to maintain osmotic balance (otherwise it would swell and burst from the inward diffusion of water).
The gradient of sodium ions is harnessed to provide the energy to run several types of indirect pumps.
The crucial roles of the Na+/K+ ATPase are reflected in the fact that almost one-third of all the energy generated by the mitochondria in animal cells is used just to run this pump.
The H+/K+ ATPase
The parietal cells of your stomach use this pump to secrete gastric juice. These cells transport protons (H+) from a concentration of about 4 x 10-8 M within the cell to a concentration of about 0.15 M in the gastric juice (giving it a pH close to 1). Small wonder that parietal cells are stuffed with mitochondria and uses huge amounts of ATP as they carry out this three-million fold concentration of protons.
The Ca 2+ ATPases
A Ca 2+ ATPase is located in the plasma membrane of all eukaryotic cells. It uses the energy provided by one molecule of ATP to pump one Ca 2+ ion out of the cell. The activity of these pumps helps to maintain the ~20,000-fold concentration gradient of Ca 2+ between the cytosol (~ 100 nM) and the ECF (~ 20 mM). In resting skeletal muscle, there is a much higher concentration of calcium ions (Ca 2+) in the sarcoplasmic reticulum than in the cytosol. Activation of the muscle fiber allows some of this Ca 2+ to pass by facilitated diffusion into the cytosol where it triggers contraction.
After contraction, this Ca 2+ is pumped back into the sarcoplasmic reticulum. This is done by another Ca 2+ ATPase that uses the energy from each molecule of ATP to pump 2 Ca 2+ ions.
Pumps 1. - 3. are designated P-type ion transporters because they use the same basic mechanism: a conformational change in the proteins as they are reversibly phosphorylated by ATP. And all three pumps can be made to run backward. That is, if the pumped ions are allowed to diffuse back through the membrane complex, ATP can be synthesized from ADP and inorganic phosphate.
ABC Transporters
ABC ("ATP-Binding Cassette") transporters are transmembrane proteins that
expose a ligand-binding domain at one surface and a
ATP-binding domain at the other surface.
The ligand-binding domain is usually restricted to a single type of molecule.
The ATP bound to its domain provides the energy to pump the ligand across the membrane.
The human genome contains 48 genes for ABC transporters. Some examples:
CFTR — the cystic fibrosis transmembrane conductance regulator
TAP, the transporter associated with antigen processing
The transporter that liver cells use to pump the salts of bile acids out into the bile.
ABC transporters that pump chemotherapeutic drugs out of cancer cells thus reducing their effectiveness.
ABC transporters must have evolved early in the history of life. The ATP-binding domains in archaea, eubacteria, and eukaryotes all share a homologous structure, the ATP-binding "cassette".
Indirect Active Transport
Indirect active transport uses the downhill flow of an ion to pump some other molecule or ion against its gradient. The driving ion is usually sodium (Na+) with its gradient established by the Na+/K+ ATPase.
Symport Pumps
In this type of indirect active transport, the driving ion (Na+) and the pumped molecule pass through the membrane pump in the same direction. Examples:
The Na+/glucose transporter. This transmembrane protein allows sodium ions and glucose to enter the cell together. The sodium ions flow down their concentration gradient while the glucose molecules are pumped up theirs. Later the sodium is pumped back out of the cell by the Na+/K+ ATPase. The Na+/glucose transporter is used to actively transport glucose out of the intestine and also out of the kidney tubules and back into the blood.
All the amino acids can be actively transported, for example out of the kidney tubules and into the blood, by sodium-driven symport pumps.
Sodium-driven symport pumps also return neurotransmitters to the presynaptic neuron.
The Na+/iodide transporter. This symporter pumps iodide ions into the cells of the thyroid gland (for the manufacture of thyroxine) and also into the cells of the mammary gland (to supply the baby's need for iodide).
The permease encoded by the lac operon of E. coli that transports lactose into the cell.
Antiport Pumps
In antiport pumps, the driving ion (again, usually sodium) diffuses through the pump in one direction providing the energy for the active transport of some other molecule or ion in the opposite direction. Example:
Ca 2+ ions are pumped out of cells by sodium-driven antiport pumps. Antiport pumps in the vacuole of some plants harness the outward facilitated diffusion of protons (themselves pumped into the vacuole by a H+ ATPase) to the active inward transport of sodium ions. This sodium/proton antiport pump enables the plant to sequester sodium ions in its vacuole. Transgenic tomato plants that overexpress this sodium/proton antiport pump are able to thrive in saline soils too salty for conventional tomatoes. Antiport pumps to the active inward transport of nitrate ions (NO 3−)
Some inherited ion-channel diseases
A growing number of human diseases have been discovered to be caused by inherited mutations in genes encoding channels.
Examples:
Chloride-channel diseases
cystic fibrosis
inherited tendency to kidney stones (caused by a different kind of chloride channel than the one involved in cystic fibrosis)
Potassium-channel diseases
the majority of cases of long QT syndrome, an inherited disorder of the heartbeat
a rare, inherited tendency to epileptic seizures in the newborn
several types of inherited deafness
Sodium-channel diseases
inherited tendency to certain types of muscle spasms
Liddle's syndrome. Inadequate sodium transport out of the kidneys, because of a mutant sodium channel, leads to elevated osmotic pressure of the blood and resulting hypertension (high blood pressure)
Osmosis
Osmosis is a special term used for the diffusion of water through cell membranes. Although water is a polar molecule, it is able to pass through the lipid bilayer of the plasma membrane. Aquaporins — transmembrane proteins that form hydrophilic channels — greatly accelerate the process, but even without these, water is still able to get through. Water passes by diffusion from a region of higher to a region of lower concentration. Note that this refers to the concentration of water, NOT the concentration of any solutes present in the water. Water is never transported actively; that is, it never moves against its concentration gradient. However, the concentration of water can be altered by the active transport of solutes and in this way the movement of water in and out of the cell can be controlled. Example: the reabsorption of water from the kidney tubules back into the blood depends on the water following behind the active transport of Na+.
Figure 3.23.3: Osmosis effect on blood cells in different solutions. (Public domain; Ladyofhats).
Hypotonic solutions: If the concentration of water in the medium surrounding a cell is greater than that of the cytosol, the medium is said to be hypotonic. Water enters the cell by osmosis. A red blood cell placed in a hypotonic solution (e.g., pure water) bursts immediately ("hemolysis") from the influx of water. Plant cells and bacterial cells avoid bursting in hypotonic surroundings by their strong cell walls. These allow the buildup of turgor within the cell. When the turgor pressure equals the osmotic pressure, osmosis ceases.
Isotonic solutions: When red blood cells are placed in a 0.9% salt solution, they neither gain nor lose water by osmosis. Such a solution is said to be isotonic. The extracellular fluid (ECF) of mammalian cells is isotonic to their cytoplasm. This balance must be actively maintained because of the large number of organic molecules dissolved in the cytosol but not present in the ECF. These organic molecules exert an osmotic effect that, if not compensated for, would cause the cell to take in so much water that it would swell and might even burst. This fate is avoided by pumping sodium ions out of the cell with the Na+/K+ ATPase.
Hypertonic solutions: If red cells are placed in sea water (about 3% salt), they lose water by osmosis and the cells shrivel up. Sea water is hypertonic to their cytosol. Similarly, if a plant tissue is placed in sea water, the cell contents shrink away from the rigid cell wall. This is called plasmolysis. Sea water is also hypertonic to the ECF of most marine vertebrates. To avoid fatal dehydration, these animals (e.g., bony fishes like the cod) must continuously drink sea water and then desalt it by pumping ions out of their gills by active transport.
Marine birds, which may pass long periods of time away from fresh water, and sea turtles use a similar device. They, too, drink salt water to take care of their water needs and use metabolic energy to desalt it. In the herring gull, shown here, the salt is extracted by two glands in the head and released (in a very concentrated solution — it is saltier than the blood) to the outside through the nostrils. Marine snakes use a similar desalting mechanism.
Figure 3.23.4: herring gull salt glands
This page titled 3.23: Diffusion, Active Transport and Membrane Channels is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by John W. Kimball via source content that was edited to the style and standards of the LibreTexts platform.
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EXPLORE MATERIALS SCIENCE TOOLBOX
View the history of American Elements on Wikipedia
Dichloromethane
CAS #: 75-09-2
Linear Formula:
CH2Cl2
MDL Number
MFCD00000881
EC No.:
200-838-9
Chemical Structure
ORDER
| Product | Product Code | SAFETY DATA | Technical data |
--- --- |
| Dichloromethane | METH-CL-01-LIQ | Pricing Add to cart only | SDS > | Data Sheet > |
Question? Ask an American Elements Materials Science Engineer
Dichloromethane Properties (Theoretical)
| | |
--- |
| Compound Formula | CH2Cl2 |
| Molecular Weight | 84.93258 |
| Appearance | Colorless liquid |
| Melting Point | -96.7 °C, |
| Boiling Point | 39.6 °C |
| Density | 1.325 g/mL |
| Solubility in H2O | N/A |
| Exact Mass | 83.953355 |
| Monoisotopic Mass | 83.953355 |
| Charge | 0 |
Dichloromethane Health & Safety Information
| | |
--- |
| Signal Word | Danger |
| Hazard Statements | H351 |
| Hazard Codes | Xn |
| Risk Codes | 40 |
| Safety Statements | 23-24/25-36/37 |
| RTECS Number | PA8050000 |
| Transport Information | UN 1593 6.1/PG 3 |
| WGK Germany | 2 |
About Dichloromethane
Dichloromethane is generally immediately available in most volumes. High purity, submicron and nanopowder forms may be considered. American Elements produces to many standard grades when applicable, including Mil Spec (military grade); ACS, Reagent and Technical Grade; Food, Agricultural and Pharmaceutical Grade; Optical Grade, USP and EP/BP (European Pharmacopoeia/British Pharmacopoeia) and follows applicable ASTM testing standards. Typical and custom packaging is available. Additional technical, research and safety (MSDS) information is available as is a Reference Calculator for converting relevant units of measurement.
Synonyms
Methylene chloride, methylene dichloride, Methylene bichloride, Methane dichloride
Chemical Identifiers
| | |
--- |
| Linear Formula | CH2Cl2 |
| Pubchem CID | 6344 |
| MDL Number | MFCD00000881 |
| EC No. | 200-838-9 |
| IUPAC Name | dichloromethane |
| Beilstein/Reaxys No. | 1730800 |
| SMILES | C(Cl)Cl |
| InchI Identifier | InChI=1S/CH2Cl2/c2-1-3/h1H2 |
| InchI Key | YMWUJEATGCHHMB-UHFFFAOYSA-N |
| Chemical Formula | |
| Molecular Weight | |
| Standard InchI | |
| Appearance | |
| Melting Point | |
| Boiling Point | |
| Density | |
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17 Cl 35.45 Chlorine
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7580 | https://en.wikipedia.org/wiki/KPP%E2%80%93Fisher_equation | Jump to content
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Contents
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1 Details
2 KPP equation
3 See also
4 References
5 External links
KPP–Fisher equation
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From Wikipedia, the free encyclopedia
Partial differential equation in mathematics
Not to be confused with the Fisher equation in financial mathematics.
In mathematics, Fisher-KPP equation (named after Ronald Fisher , Andrey Kolmogorov, Ivan Petrovsky, and Nikolai Piskunov) also known as the Fisher equation, Fisher–KPP equation, or KPP equation is the partial differential equation:
It is a kind of reaction–diffusion system that can be used to model population growth and wave propagation.
Details
[edit]
Fisher-KPP equation belongs to the class of reaction–diffusion equations: in fact, it is one of the simplest semilinear reaction-diffusion equations, the one which has the inhomogeneous term
which can exhibit traveling wave solutions that switch between equilibrium states given by . Such equations occur, e.g., in ecology, physiology, combustion, crystallization, plasma physics, and in general phase transition problems.
Fisher proposed this equation in his 1937 paper The wave of advance of advantageous genes in the context of population dynamics to describe the spatial spread of an advantageous allele and explored its travelling wave solutions. For every wave speed ( in dimensionless form) it admits travelling wave solutions of the form
where is increasing and
That is, the solution switches from the equilibrium state u = 0 to the equilibrium state u = 1. No such solution exists for c < 2. The wave shape for a given wave speed is unique. The travelling-wave solutions are stable against near-field perturbations, but not to far-field perturbations which can thicken the tail. One can prove using the comparison principle and super-solution theory that all solutions with compact initial data converge to waves with the minimum speed.[citation needed]
For the special wave speed , all solutions can be found in a closed form, with
where is arbitrary, and the above limit conditions are satisfied for .
Proof of the existence of travelling wave solutions and analysis of their properties is often done by the phase space method.
KPP equation
[edit]
In the same year (1937) as Fisher, Kolmogorov, Petrovsky and Piskunov introduced the more general reaction-diffusion equation
where is a sufficiently smooth function with the properties that and for all . This too has the travelling wave solutions discussed above. Fisher's equation is obtained upon setting and rescaling the coordinate by a factor of . A more general example is given by with . Kolmogorov, Petrovsky and Piskunov discussed the example with in the context of population genetics.
The minimum speed of a KPP-type traveling wave is given by
which differs from other type of waves, see for example ZFK-type waves.
See also
[edit]
ZFK equation
Allen–Cahn equation
References
[edit]
^ a b c Fisher, R. A. (1937). "The Wave of Advance of Advantageous Genes" (PDF). Annals of Eugenics. 7 (4): 353–369. doi:10.1111/j.1469-1809.1937.tb02153.x. hdl:2440/15125.
^ a b c d A. Kolmogorov, I. Petrovskii, and N. Piskunov. "A study of the diffusion equation with increase in the amount of substance, and its application to a biological problem." In V. M. Tikhomirov, editor, Selected Works of A. N. Kolmogorov I, pages 248–270. Kluwer 1991, ISBN 90-277-2796-1. Translated by V. M. Volosov from Bull. Moscow Univ., Math. Mech. 1, 1–25, 1937
^ Peter Grindrod. The theory and applications of reaction-diffusion equations: Patterns and waves. Oxford Applied Mathematics and Computing Science Series. The Clarendon Press Oxford University Press, New York, second edition, 1996 ISBN 0-19-859676-6; ISBN 0-19-859692-8.
^ Ablowitz, Mark J. and Zeppetella, Anthony, Explicit solutions of Fisher's equation for a special wave speed, Bulletin of Mathematical Biology 41 (1979) 835–840 doi:10.1007/BF02462380
^ Trefethen (August 30, 2001). "Fisher-KPP Equation" (PDF). Fisher 2.
^ Griffiths, Graham W.; Schiesser, William E. (2011). "Fisher–Kolmogorov Equation". Traveling Wave Analysis of Partial Differential Equations. Academy Press. pp. 135–146. ISBN 978-0-12-384652-5.
^ Adomian, G. (1995). "Fisher–Kolmogorov equation". Applied Mathematics Letters. 8 (2): 51–52. doi:10.1016/0893-9659(95)00010-N.
External links
[edit]
Fisher's equation on MathWorld.
Fisher equation on EqWorld.
Retrieved from "
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Ronald Fisher
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7581 | https://www.waterboards.ca.gov/water_issues/programs/outreach/waterlessons/pdf/6part4.pdf | OV E RV I E W After teams have collected data at least twice, students read a one-page information sheet about the living and non-living components of a watershed and water ecosystem, and the factors that affect the survival of those components.
Students take what they learn and relate it to the data they are collecting. Students continue to observe and record data.
Standards: 5c, 5e, 7b A LIVING WATER ECOSYSTEM PART 4 6 T H G R A D E : L I F E SC I E N C E WAT E R Q U A L I T Y U N I T 85 Materials • Information Sheet B – A Living Water Ecosystem – 1 per group Vocabulary Words • Abiotic factor • Biotic factor • Contamination • Decomposition • Dissolved Oxygen • Invertebrate • Microorganism • Photosynthesis • Runoff • Sediment • Watershed Other Resources See Teacher Resources, page 116 for additional activities that relate to watersheds.
Helpful Hints • Provide additional information about the local watershed.
• Watersheds can be as small as a leaking sprinkler head to as large as an entire city basin. Show students the smaller watersheds that can be found around the schoolyard. • Help students investigate where water goes after it leaves their campus.
• Provide students with information about a local body of water. To locate your closest body of water, consult a local map. Go to www.epa.gov/surf to find the name of your watershed.
• If possible, take a field trip to the local waterway to observe how humans may be impacting it.
PA RT 4 – 45 minutes P RO C E D U R E 1.
Have each student read Information Sheet B – A Living Water Ecosystem.
2.
Have student groups discuss what they read and the ways in which it relates to their investigations. Each group can report their main points to the class as part of a group discussion.
3.
Looking at the illustration on Information Sheet B, have students figure out how water moves in their watershed community. Have them investigate further to confirm their ideas. 4.
Have students investigate the pathway water takes from their campus to the nearest body of water.
PA RT 4 : A L I V I N G WAT E R ECOSY ST E M 86 GUIDED QUESTIONS ?
• How do biotic factors differ from abiotic factors?
• What are the biotic and abiotic factors of your watershed?
• Where are there watersheds within our schoolyard?
• When abiotic factors are impacted in a water ecosystem, how does this affect the biotic factors?
• What water services or goods do you depend on?
• What is the closest body of water to our school?
• Where, if at all, does that body of water flow to?
• What impact is what you are observing having on your local water system? How do you know?
Your community, whether it is in a city or rural town, is part of a watershed. A watershed is the land area that directs water to a drainage system or river. It helps supply water to our community by allowing it to seep into the ground or channel it into streams, rivers, and other bodies of water.
Gravity moves water through the watershed from higher to lower areas. A watershed includes living components or biotic factors such as people, wildlife, plants, and insects; as well as non-living components or abiotic factors, including sunlight, oxygen, temperature, and soil. Both components belong to the environment of a watershed community. Look around. What are the biotic and abiotic parts of your watershed?
Your watershed directs water into another system of living and non-living components – a water ecosystem. It is the abiotic factors that make up the environment for the living organisms – water, sunlight, rocks, soil, and air – and allow them to thrive. Without these, living organisms would not survive. Humans depend on the products of a water ecosystem. Water ecosystems provide us with goods and services, including drinking water, recreation, and food. In fact, about 9,000 different species of fish are harvested for food from our water systems. Humans are responsible for protecting these ecosystems. However, pollution can harm these ecosystems and damage their ability not only to provide us with goods, but also maintain the balance of a functioning ecosystem. For example, large rivers in California such as the Sacramento, American, Feather, and lower San Joaquin, provide major fish spawning habitats for salmon, steelhead trout, and striped bass. Young fish depend on small invertebrates – mostly insects and tiny shrimp – for food. When land pollution, field pesticides, and erosion from construction sites run off through a watershed and enter streams and rivers, they kill or seriously harm these food sources and the young fish. This is just one way the condition of the ecosystem is harmed.
Let’s take a closer look at how abiotic factors are affected.
Water: All living things need water to carry out their life processes. Contaminated water from land pollution not only affects the water habitat for fish and other animals but also plants and algae.
A Living Water Ecosystem 6 T H G R A D E : L I F E SC I E N C E WAT E R Q U A L I T Y U N I T INFORMATION SHEET B These organisms use water along with sunlight and carbon dioxide to make food as part of photosynthesis. Other living things eat the plants and algae to obtain energy. Sunlight: Because sunlight is necessary for photosynthesis, it is an important factor for plants, algae, and other living things. If plants or algae do not receive sunlight, they cannot grow. When dirt, sand, and oil that build up on city streets get washed into streams, it decreases the amount of light able to penetrate the water, reducing the amount of light for plants and algae to grow.
Oxygen: Fish, plants, and other water organisms need oxygen to survive. They obtain dissolved oxygen from the water around them. Dissolved oxygen refers to the oxygen stored between water molecules in a river or lake. The amount of oxygen in water is critical to the health of any river system.
Runoff of oxygen-demanding organic matter such as sewage, lawn clippings, and leaves can cause excessive decomposition by microorganisms, using up too much oxygen in the process, and decreasing the amount of dissolved oxygen available for other living organisms.
Temperature: Temperature of the water can also affect oxygen levels. Cold water can hold more dissolved oxygen than warm water. Water temperature can rise when runoff that flows over hot asphalt and concrete pavement drains into the water system. This not only lowers the amount of oxygen available for living organisms, but causes serious problems for organisms adapted to certain water temperatures and already stressed by other contaminants in urban runoff.
Rock and gravel: Rock and gravel provide necessary habitats for living organisms. Fish and amphibians also use them as a spawning ground for laying and hatching their eggs. Runoff of “land pollution” and sediments can cover the available rocks and gravel needed for the fish to lay and cover their eggs.
When affected by “land pollution” and other runoff, all these abiotic factors decrease the availability of resources available to the living organisms within a water ecosystem. Every non-living component is impacted and therefore impacting the living components.
What is happening in your community? Where is the water from your schoolyard going? To a nearby river, stream, lake, or ocean?
The watershed of most cities and school grounds contain up to 90 percent hard surfaces such as rooftops, concrete playgrounds, streets, and parking lots where water collects quickly and runs off into the street. This not only prevents water from seeping into the ground to replenish underground supplies of fresh water, but sends “land pollution” directly into our rivers and the ocean.
Think about the following questions: • What are you observing during your data collection? Do hard surfaces have an impact?
• What about the “land pollution?” What impact on your local water ecosystem do you think it may have? |
7582 | https://www.youtube.com/watch?v=32MnjUdIF0c | Discrete Math - 6.1.1 Counting Rules
Kimberly Brehm
114000 subscribers
1613 likes
Description
162171 views
Posted: 16 Apr 2020
Strategies for finding the number of ways an outcome can occur. This includes the product rule, sum rule, subtraction rule and division rule.
Video Chapters:
Introduction 0:00
Product Rule 0:16
Tree Diagrams 1:13
Sum Rule 3:27
Subtraction Rule (Inclusion-Exclusion) 4:49
Division Rule 8:20
Up Next 11:51
Textbook: Rosen, Discrete Mathematics and Its Applications, 7e
Playlist:
38 comments
Transcript:
Introduction we're going to switch gears now and we're going to start talking soon about probability but before we talk about probability we want to talk about that calculating the number of ways that something can happen and so in this video we're going to take a look at counting rules the product rule occurs Product Rule when we have a procedure that can be broken into a sequence of tasks where the number of outcomes of each task or ways that you can perform each task is expressed as n 1 + 2 + 3 etc then there are N 1 times n 2 times n 3 etc at times and K different ways to perform the procedure so that's just fancy math talk for saying hey if you've got something that can be performed a bunch of different ways then multiply those ways so for instance well let's say I'm getting dressed and I'm going to wear shorts and a t-shirt if I have four different t-shirts and I have three different pair of shorts how many different outfits do I have so I can just use the product rule to say 4 times 3 equals 12 different outfits in the Tree Diagrams last question we were asked to find the number of outfits that you could create with 4 t-shirts and three different pair of shorts if instead you were asked to find what those different outfits were then we might use a strategy called a tree diagram and a tree diagram is just a visual representation it's going to help us see what those outcomes are now tree diagrams are most helpful when you would use the product rule so when you have different outcomes that you would multiply to find the number of options or once we talk about probability the probability of those options but let's get started on this one notice I am going to make four branches because I have four different t-shirts so I'm just going to call this t-shirt one t-shirt two t-shirt three and t-shirt four and then off of each of those I am going to make three branches because there are three options for the next one and so here I would write shorts one shorts two shorts three and I would do that for each of these and again it's not super helpful in this question because I gave us sort of an easy question to look at but the very important last step that a lot of students forget to do is to look at what the total outcomes are so when I'm writing what the outcome is essentially I'm following the branch so I would follow the branch to t1 and then from there follow the branch to s1 so this one is t-shirt one shorts one and then I could do t-shirt one shorts two and t-shirt one shorts three and I could continue that all the way down the list now what's going to be important here is we have already determined that four times three is twelve and there are twelve different options and if I were to list all of these there would also be twelve different outfits that I could wear the sum rule is if I have a task that Sum Rule can be performed in one of n1 ways or one of n2 ways and so or is our keyword here with the sum rule then there are n plus 1 plus sorry n1 plus n2 ways to perform the task so for instance I always always want to take a trip to the beach but let's say I can either travel to one of thirty seven international beaches or one of 14 domestic beaches how many beach vacation choices do I have so all this is saying is if I can take thirty-seven international or I could take 14 domestic then there are a total of 51 beach choice vacations that I have now notice I've written the sum rule out symbolically for you above and that's just saying if I can perform the task and these lines on the side that looks like absolute value that's just the cardinality it's saying how many are inside that Union and so this is saying if I've got some choice of a 1 or a choice of a 2 or a choice in a 3 etc then all I have to do is add the number of each of those choices to find the total number of choices now we have the Subtraction Rule (Inclusion-Exclusion) subtraction rule and quite often you will see this called the principle of inclusion exclusion and we'll talk more in detail in this course and then we'll talk in way more detail if you continue on to combinatorics but the subtraction rule basically says if a task can be done in either one of n1 ways or one of n2 ways then the total number of ways to do the task is n1 plus n2 minus the number of ways that are common to the two different ways so essentially if you've taken any sort of statistics class this is the rule that says hey if there's something here in the middle where those values are not disjoint so there is an overlap of the number of ways in each set then we have to subtract what's in the overlap because essentially it was counted in the first set and in the second set so for instance how many bit strengths bit strings of length 7 either start with one bit or end with three bits 0 0 0 so if I have the number that start with 1 so I'm setting this value to be 1 and then all of these is 1 2 3 4 5 6 different values that can either be 0 or 1 so that should be 2 to the 6th because there's two options here to here to here to here to here and to here remember because a bit string is waiting to have values of 0 or 1 now let's find the number that end with 0 0 0 so I'm setting this to be 0 setting this to be 0 and sending this to be 0 and then again I've got any value that I want of the two 0 or 1 so this could be one of two choices times one of two choices times one of two choices times one of two choices I should have put multiplication up there as well and again that's 2 to the fourth so two to the fourth that end with 0 0 0 and now so I've basically found this value and now I'm going to and I've already found this value I need to find the overlap the ones I counted both here and here so now I'm going to set the first value as 1 the last three values as zeros and these values can be anything I want them to be so that's 2 times 2 times 2 or 2 to the third so my solution would be 2 to the sixth plus 2 to the 4th minus 2 to the 3rd now of course I could go through and determine that answer which is not going to be very difficult because two to the sixth is 64 2 to the 4th is 16 2 to the 3rd is 8 so if I add 64 and 16 I get 80 80 minus 8 is 72 so there's 72 different ways again because there were 8 that I counted in both the numbers that start with 1 and that end with 0 0 0 Division Rule the last rule I would like to go over is the division rule and the division rule essentially says there are n / D ways to do a task if it can be done using a procedure that can be carried out in n ways and there are D corresponding outcomes per group so that seems a little bit sketchy and may be hard to understand so let's take a look at an example and hopefully help us to understand so I have a circular table and I'm going to split this table into six positions and I'm just going to arbitrarily call this position one two three four five and six so it says how many ways can I sit eight people I'm sorry six people around a circular table where two seatings are considered the same when each person has the same left and right neighbor so if I were just going to determine the number and it was not considered the same when the person had the same left and right hand neighbor I could very easily do this by saying there are six options for position one because there are six people once I've sat one person in position one there are five people left for position two and four for position three and three and two and one and so 6 times 4 times 6 times 5 times 4 times 3 times 2 times 1 or 6 factorial would be the total number of ways well that's all well and good but here's the problem I'm going to erase my numbers here oh and my circle apparently so let's say person a set in position 1 and then b c d e and f so Adam and Bob and Kurt and Doug and Eric and Frank look at the idea what the what this is saying is let's say instead Adam set here and Bob set here and I don't remember the names that I made up so we'll just you know stick with my lettering system so I have ABCDEF notice all I did was I took everyone and rotated them clockwise one position and so these two seatings are considered the same and again if I shifted one more and said here's a and here's B and here's C and here's D and here's E and here's F and so we get the idea that what's going to happen is I'm going to end up with six that are all the same because if a started at position one or at two or at three or at four or at five or at six and I kept everyone in the same position all of those are considered the same so this result is six factorial divided by six again six because there are six different positions that are actually the same seating so still the same order of ABCDEF just rotated one position two positions three positions for positions five positions or six positions coming Up Next up next we're going to take a look at permutations and combinations |
7583 | https://archive.org/stream/InorganicChemistry4edHuheeyKeiterKeiter/Inorganic%20chemistry%204ed%20-%20Huheey%2C%20Keiter%20%26%20Keiter_djvu.txt | Full text of "Inorganic Chemistry 4ed Huheey, Keiter & Keiter"
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Full text of "Inorganic Chemistry 4ed Huheey, Keiter & Keiter"
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Inorganic Chemistry
Principles of
Structure and
Reactivity
Fourth Edition
James E. Huheey
University of Maryland
Ellen A. Keiter
Eastern Illinois University
Richard L. Keiter
Eastern Illinois University
m HarperCollin sCollegePublishers
acknowledgment is also given to Ac,a Chemica Scandinavian The American
Association for the Advancement of Science, The American Institute of Physics
Angeweindte Che,me. The Chemical Societv, The International UnS of
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About the Cover
The crystal structure of hoggsile. a recently discovered natural zeolite is comoosed of
a p n , d u,°h x a y n 8 d en i v srfF-
!5ne i Ve 7 i,y , 0fCh ' Ca 8°- Modeling tools used to construct' the cover
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Technolog,e?’lnc ’ C ° mpmer 8raph,c by John M Newsam. BIOSYM
Sponsoring Editor: Jane Piro
Project Coordination: Elm Street Publishing Services Inc
Cover Design: Kay Fulton
Cover Photo: Professor John M, Newsam. BIOSYM Technologies Inc
Compositor: Better Graphics. Inc. oiogies. me.
Printer and Binder: R. R. Donnelley & Sons Company
Cover I rinter: Lehigh Press Lithographers
Inorganic Chemistry: Principles of Structure and Reactivity. Fourth Edition
Copyright © 1993 by HarperCollins College Publishers
AM rights reserved. Printed in the United States of America. No part of this book mav
be used or reproduced in any manner whatsoever without written permission exceot in
the case of brief quotations embodied in critical articles and reviews For information
address HarperCollins College Publishers. 10 East 53rd Street, New York. NY 10022
Library of Congress Cataloging-in-Publication Data
Huheey, James E.
Inorganic chemistry: principles of structure and reactivity /
James E. Huheey, Ellen A. Keiler, Richard L. Keiter
p. cm.
Includes bibliographical references and index
ISBN 0-06-042995-X
I. Chemistry, Inorganic. 1. Keiter. Ellen A. II. Keiter
Richard L. III. Title.
QDI5I.2.H84 1993
546—dc20 92-36083
97 98 99 12 11 10 9
To Catherine. Cathy. Terry. Mercedes. Thorfin. Irene. Alrin, Eric, and Use.
Contents
Preface xi
To the Student xiv
Excerpts from the Preface to the Third Edition xvi
What Is Inorganic Chemistry? 1
Inorganic Chemistry, the Beginnings I
Inorganic Chemistry, an Example 2
Chemical Structure of Zeolites and Other Chemical Systems 3
Chemical Reactivity 5
Conclusion 7
The Structure of the Atom 10
The Hydrogen Atom 10
The Polyelectronic Atom 20
Symmetry and Group Theory 46
Symmetry Elements and Symmetry Operations 46
Point Groups and Molecular Symmetry 53
Irreducible Representations and Character Tables 59
Uses of Point Group Symmetry 63
Crystallography 74
VI
Contents
Chapter 4
Bonding Models in Inorganic Chemistry: 1. Ionic Compounds 92
The Ionic Bond 92
Lattice Energy 99
Size Effects 112
The Predictive Power of Thermochemical Calculations on Ionic Compounds 127
Covalent Character in Predominantly Ionic Bonds 129
Conclusion 134
Chapter 5 Bonding Models in Inorganic Chemistry: 2. The Covalent
Bond loo
Valence Bond Theory 139
Molecular Orbital Theory 153
Electronegativity 182
Chapter 6 The Structure and Reactivity of Molecules 203
The Structure of Molecules 203
Structure and Hybridization 220
Bond Lengths 232
Experimental Determination of Molecular Structure 233
Some Simple Reactions of Covalently Bonded Molecules 237
Contents
Chapter 9 _ Acid-Base Chemistry 318
Acid-Base Concepts 318
Measures of Acid-Base Strength 330
Hard and Soft Acids and Bases 344
Chapter 10 Chemistry in Aqueous and Nonaqueous Solvents 359
Water 360
Nonaqueous Solvents 360
Molten Salts 374
Electrode Potentials and Electromotive Forces 378
Chapter 11 Coordination Chemistry: Bonding, Spectra, and Magnetism 387
Bonding in Coordination Compounds 391
Valence Bond Theory 391
Crystal Field Theory 394
Molecular Orbital Theory 413
Electronic Spectra of Complexes 433
Magnetic Properties of Complexes 459
Chapter 7 The Solid State 252
The Structures of Complex Solids 253
Imperfections in Crystals 263
Conductivity in Ionic Solids 266
Solids Held Together by Covalent Bonding 269
Solid-State Materials with Polar Bonds 276
Chapter 8 Chemical Forces 290
Internuclear Distances and Atomic Radii 290
Types of Chemical Forces 296
Hydrogen Bonding 300
Effects of Chemical Forces 307
Chapter 12 _ Coordination Chemistry: Structure 472
Coordination Number I 472
Coordination Number 2 473
Coordination Number 3 474
Coordination Number 4 474
Coordination Number 5 479
Coordination Number 6 488
Coordination Number 7 503
Coordination Number 8 507
Higher Coordination Numbers 509
Generalizations about Coordination Numbers 511
Linkage Isomerism 513
Other Types of Isomerism 521
The Chelate Effect 522
vii
Contents
viii Contents
l
Chapter 13 Coordination Chemistry: Reactions, Kinetics, and
Mechanisms 537
Substitution Reactions in Square Planar Complexes 538
Thermodynamic and Kinetic Stability 547
Kinetics of Octahedral Substitution 548
Mechanisms of Redox Reactions 557
Chapter 14 Some Descriptive Chemistry of the Metals 577
General Periodic Trends 578
Chemistry of the Various Oxidation States of Transition Metals 580
The Chemistry of Elements Potassium-Zinc: Comparison by Electron
Configuration 582
The Chemistry of the Heavier Transition Metals 587
Oxidation States and EMFs of Groups 1-12 588
The Lanthanide and Actinide Elements 599
Coordination Chemistry 608
The Transactinide Elements 613
Chapter 15 Organometallic Chemistry 623
The 18-Electron Rule 624
Metal Carbonyl Complexes 630
Nitrosyl Complexes 650
Dinitrogen Complexes 653
Metal Alkyls, Carbenes, Carbynes, and Carbides 655
Nonaromatic Alkene and Alkyne Complexes 662
Metallocenes 669
Reactions of Organometallic Complexes 686
Catalysis by Organometallic Compounds 705
Stereochemically Nonrigid Molecules 723
Conclusion 730
Chapter 16 Inorganic Chains, Rings, Cages, and Clusters 738
Chains 738
Rings 765
Cages 785
Boron Cage Compounds 789
Metal Clusters 807
Conclusion 819
Chapter 17 The Chemistry of the Halogens and the Noble Gases 824
Noble Gas Chemistry 825
Halogens in Positive Oxidation States 837
Halides 848
Pseudohalogens 852
Electrochemistry of the Halogens and Pseudohalogens 853
Chapter 18 Periodicity 857
First- and Second-Row Anomalies 858
The Use of p Orbitals in Pi Bonding 861
The Use (or Not) of d Orbitals by Nonmetals 866
Reactivity and d Orbital Participation 875
Periodic Anomalies of the Nonmetals and Posttransition Metals 876
Chapter 19 The Inorganic Chemistry of Biological Systems 889
Energy Sources for Life 889
Metalloporphyrins and Respiration 891
Dioxygen Binding, Transport, and Utilization 895
Electron Transfer. Respiration, and Photosynthesis 911
Enzymes 919
Nitrogen Fixation 933
The Biochemistry of Iron 935
Essential and Trace Elements in Biological Systems 941
Biochemistry of the Nonmetals 953
Medicinal Chemistry 954
Summary 960
Postscript 960
Contents
Appendix A
The Literature of Inorganic Chemistry A-l
Appendix B
Units and Conversion Factors A-3
Appendix C
Atomic States and Term Symbols A-7
Appendix D
Character Tables A-l 3
Appendix E
Bond Energies and Bond Lengths A-21
Appendix F
An Overview of Standard Reduction Potentials of the
Elements A-35
Appendix G
Tanabe-Sugano Diagrams A-38
Appendix H
Models, Stereochemistry, and the Use of Stereopsis A-40
Appendix 1
The Rules of Inorganic Nomenclature A-46
Index A-78
Preface
It has been twenty years since the senior author and Harper & Row, Publishers
produced the first edition of Inorganic Chemistry: Principles of Structure and Reac¬
tivity. In that time: (a) The senior author has become 20 years more senior; (b) two
new authors have joined the project; (c) Harper & Row, Publishers has become
HarperCollins Publishers; and, most important, (d) inorganic chemistry has continued
to grow from its already lusty existence of two decades ago. It is becoming increas¬
ingly impossible for one person to monitor all areas of inorganic chemistry. The new
authors bring to the book their interests in coordination chemistry, organometallics,
and physical methods, as well as fresh viewpoints on a number of other topics.
Nevertheless, the philosophy of the book remains unchanged: To bring to the reader
the essentials of inorganic chemistry in an easily readable format with emphasis on the
fact that inorganic chemistry is an exciting field of research rather than a closed body
of knowledge.
We three authors brought very different undergraduate experiences to the teach¬
ing of inorganic chemistry and the revision of this edition. One of us received a B.S.
degree from a Ph.D. granting institution, one from a private non-Ph.D. liberal arts
college, and one from a public non-Ph.D. liberal arts college. We have taught under¬
graduate and graduate inorganic courses in a variety of settings. When we sat down to
discuss the revision, there were a number of things that we agreed upon: (I) I he book
would be substantially updated. (2) The material presented would continue to be
thoroughly referenced, and the references would continue to appear on the pages of
interest. A relevant reference would not be omitted just because it had appeared in
previous editions. (3) New illustrations, many from the original literature, would be
added. (4) A greater selection of problems, many of them new, would be provided.
Many problems would require library assistance, while others would cover the funda¬
mental aspects of each topic. (5) A chapter on symmetry would be added. (6) Solid
stale chemistry would be given more emphasis. (7) The kinetics chapter would be
more fully developed. (8) The descriptive and organometallic chemistry of the lan¬
thanides and actinides would be included in the corresponding chapters for the
transition metals.
General consensus (among both authors and users) comes more easily than
agreement on specifics. Our discussions of the symmetry chapter are a good example.
All of us agreed that the teaching of symmetry considerations at most institutions had
xi
Preface
Preface
xii
for the most part been delegated to the inorganic chemists. But how much should be
taught, and how much should the remainder of the book depend upon this chapter? At
a minimum we believed that a good introduction to point groups was essential. We
also wanted to include some character table applications but not so much that the
inorganic chemistry in the book couldn't be taught without it. Applications appear
here and there in the text but can be avoided if desired. The chapter, as completed, has
concentrated on familiarizing the student with many applications of symmetry as used
by the inorganic chemist, including spectroscopy and crystallography, without pur¬
porting to be a rigorous exposition of the subject.
We may anticipate an eventual consensus on the amount and place of symmetry in
the chemistry curriculum, but for now we have assumed no prior background in the
subject. We have thus tried to illustrate a wide variety of uses of symmetry without
delving deeply into the background theory. We hope that those new to the topic can
find a useful introduction to the application of symmetry to problems in inorganic
chemistry. On the other hand, those having previous experience with the subject may
wish to use this chapter as a brief review. And, recognizing that things are in a state of
flux, we have attempted to make it possible to study various topics such as orbital
overlap, crystal field theory, and related material, as in the past, with minimal
reference to symmetry if desired.
Students using this book come from exceedingly diverse backgrounds: Some will
have had extensive experience in physical and organic chemistry, perhaps even a
previous course in descriptive inorganic chemistry. For many, however, this will be
the first contact with inorganic chemistry, and some may have had only limited
experience with bonding theory in other courses. For this reason, the early chapters
present the fundamentals of atomic and molecular structure from the inorganic chem¬
ist's perspective. The well-prepared reader may use these chapters as a brief review as
well as mortar to chink between previous blocks of knowledge. The middle chapters of
the book present the "heart of inorganic chemistry," solid-state chemistry beyond
simple salts, acid-base chemistry in a variety of solvents and the gas phase, and
coordination chemistry discussed in terms of bonding, spectra, magnetism, structure,
and reactions.
In line with the philosophy of a topical approach and flexible course content, the
last six chapters of the book are essentially independent of each other, and one or
more may readily be omitted depending on the inclination of the instructor and the
time available.
The fourth edition, in its entirety, works nicely for that unfortunately rare beast,
the two-semester course. But that means that it is balanced and should work equally
well for a one-semester course—the instructor must pick and choose. We firmly
believe that it is more useful to provide a large number of topics, wherein one can
select the topics to be covered, than to dictate a "minimum core.” We hope the book
includes the topics that all instructors find essential, but we hope that it also includes
their favorite topics. It obviously includes ours. A solutions manual that contains
answers to all end-of-chapter problems accompanies the fourth edition.
We would like to thank our colleagues at the University of Maryland at College
Park (UMCP) and Eastern Illinois University (EIU) who have helped in a multitude of
ways. Professor Huheey’s colleagues who helped with previous editions are listed in
the “Excerpts from the Preface to the Third Edition" (page xvi). and their further help
is gratefully acknowledged. In addition, we would like to thank Bryan Eichhorn
(UMCP), William Harwood (UMCP). Mark McGuire (EIU). Robert Pilato (UMCP),
and Rinaldo Poli (UMCP) for special help with this edition. We would also like to
thank colleagues in departments that we have visited on sabbatical leaves: Fred
Hawthorne, Herb Kaesz, Charles Strouse, Joan Selverstone Valentine, and Jeff Zink
(University of California at Los Angeles), and Oren Anderson, Gary Maciel, Jack
Norton. Tony Rappe. and Steve Straus (Colorado State University). We would also
like to thank the Chemistry Departments at UCLA and CSU, the Zoology Department
at Southern Illinois University, as well as our own departments for making possible
sabbatical visits to take advantage of these resources.
We are grateful to Michael W. Anderson. University of Cambridge; Anthony
Arduengo, E. 1. du Pont de Nemours; B. Dubost, Pechiney Institute; Jacek Klinow-
ski. University of Cambridge; John Newsam, BIOSYM Technologies; Joseph J.
Plu’th, University of Chicago; Arnold L. Rheingold, University of Delaware;
P. Sainfort, Pechiney Institute; Charlotte L. Stem, University of Illinois. Urbana-
Champaign; Sir John Meurig Thomas. The Royal Institution of Great Britain; and
Scott Wilson. University of Illinois, Urbana-Champaign, for special help with illustra¬
tions from their work.
The writing of this text has benefitted from the helpful advice of many reviewers.
They include "ivan Bernal, Donald H. Berry, Patricia A. Bianconi, Andrew B.
Bocarsly, P. Michael Boorman, Jeremy Burdett, Ben DeGraff, Russell S. Drago,
Daniel C. Harris. Roald Hoffmann, Joel F. Liebman, John Milne, Terrance Murphy,
Jack Pladziewicz, Philip Power. Arnold L. Rheingold. Richard Thompson. Glenn
Vogel. Marc Walters, James H. Weber, and Jeff Zink.
" We began this preface indicating "changes" that have occurred in the last two
decades. We have dealt with new authorship and new inorganic chemistry above.
Concerning the merger of Harper & Row. Publishers, New York, and Collins, Pub¬
lishers. London, the entropy generated was quite unexpected. When the dust had
settled, there emerged two sterling performers: Jane Piro, Chemistry Editor, and Cate
Rzasa, Project Editor, who helped us in many ways. We are happy to acknowledge
our debt to them.
Finally, there are many, many faculty and students who have helped in the
original writing and further development of this book, often anonymous in the brief
citation of colleagues and reviewers. They know who they are, and we hope they will
accept our sincere thanks for all that they did.
James E. Huheey
Ellen A. Keiler
Richard L. Reiter
To the Student
" ° lder Pr0fesor was hcart «,( he
known hllhl n by h ° f the the0ry and ex P erimenIal spectroscopic methods
knoun by the new chem.stry graduates. A young graduate student was stunned- she
was sure tt would take her years just to learn enough of the chemistry that he already
knew to g he,- degree. Meanwhile, two other professors were arguing heatedly oyer
he relate importance of facts versus theory. One said descriptive chemistry was the
most important because "facts don’t change!"
"Well, some ’facts’ seem to change—I read yesterday that iridium is the densest
element ten years ago when I was a student, I was told that osmium was the densest."
wrnno ' Change aS ^ M theories: Iheories just come and go; besides, what’s
accura,ryalue°" e0ne rePea,mS eXperimenI and doi "S better-getting a more
. ,h lha ‘ S thC P °' nt: neVV L . theories are ^cessary to explain new experimental data,
and theories give us someth,ng to test, a framework around which we can dream "
Dream! We need a little less ’inspiration’ and a lot more perspiration "
oernetu mI 8 ,? 5 ' ' ' h ^ a " d thCSe argUmcnts P^ent a microcosm of
perpetual debates in chemistry and the essence of the great difficulty in writing an
upper level textbook of inorganic chemistry. The field is vast; large numbers of
isol-ulT "T .7 PUbl ' Sh r ed CVery Week ' NCW Synlhc,ic technic l ues allow the
a -I 6 " ^ ° n e 8rCat nUmberS ° f highly reacIive impounds. Thc-
ZZ a de i SCnp,,ons have become increasingly sophisticated, as have spectroscopic
l2 v R ° r 7'| C t m ' Stry mteraCtS Wi ‘ h ° rganic ’ physical ’ and even biological
Cn [" 0leCUlar and SOlid ' State chemis,r y are rapidly disap-
small noln^ otVT “X kn ° W mUny faC,S and theories buI --ealizes it is only a
.mall portion of the whole. The new graduate, with well-developed skills in a few
areas, also has a sense of inadequacy. Perhaps the student faced with his or her first
advanced inorganic course feels this most acutely. The textbook for the course reflects
the instructor s choice of what portion of inorganic chemistry should be taught what
mix of facts and theory, and what relative weight of traditional and new science
tex book T l Cir c hOICeS and th ° Se ^ See " in lhe variet y of available
useful for fi 0 "d T h ° me u fe he3Vily faCtUal ’ USUally bulk y- and specially
elemem Olh 8 ° SOmet ^ abom a " of the Principal compounds of a particular
element. Others present a blend of fact and theory but minimize the book bulk by
To the Student
XV
limiting each topic to a few paragraphs. This has the advantage of including most
topics but the disadvantage of having to look elsewhere for a fuller development. Any
single book, of course, has this problem to a certain degree, thus the need for many
references. Our book, Inorganic Chemistry: Principles of Structure and Reactivity,
fourth edition, is also a blend of fact and theory, but we think it is large enough for a
full meal. There is no reason to expect a book that deals with the chemistry of 109
elements to be smaller than a standard organic chemistry textbook!
We've enjoyed writing this book; we hope that you will enjoy reading it. If you do,
we'd like to hear from you.
James E. Huheey
Ellen A. Keiter
Richard L. Keiter
Chapter
What Is
Inorganic Chemistry?
It is customary for chemistry books to begin with questions of this type questions
that are usually difficult or impossible to answer in simple “twenty-five-words-or-less"
definitions. Simple pictures, whether of words or of art, cannot portray all aspects
of a subject. We most recently had this impressed upon us when our editor asked
us to suggest some aspect of inorganic chemistry for the cover of this book. 'I he very
nature of a cover implies a relatively simple, single item, such as a molecule, a struc¬
ture, a reaction, or a property (or perhaps a simple combination of two or three ol
these). Should we choose the structure of the new high-temperature superconductors
which recentlv gained a Nobel Prize for their discoverers? You probably have read
about them in the popular press and wondered why "high temperature” was colder
than a Siberian winter! Should we choose a metal “cluster compound” that acts, at
the molecular level, like a microscopic fragment of the metal? How about an inorganic
molecule that is optically active (that’s not a subject limited to organic chemistry),
or carboxypeptidase A (that’s an enzyme, but interest in it is certainly not limited to
biochemists)? Maybe a symmetrical crystal of a compound like sodium thiosulfate,
photographer’s "hypo." or a multicolor, polarized micrograph of an inorganic com¬
pound. But no single design can possibly portray the many, varied aspects of in¬
organic chemistry. In the same way, any short and simple definition of a complex
subject is apt to be disappointing and even misleading. So let’s just try to see where
inorganic chemistry came from, what an inorganic chemist does, and, perhaps, where
the subject is going.
$
Inorganic Chemistry,
the Beginnings
The term inorganic chemistry originally meant nonliving chemistry, and it was that
part of chemistry that had arisen from the arts and recipes dealing with minerals
and ores. It began by finding naturally occurring substances that had useful prop¬
erties, such as flint or chert that could be worked into tools (middle Pleistocene, ca.
5 x 10 5 years ago or less). This search continues (see below), but now it is included
in the sciences of mineralogy and geology. Chemistry deals more with the changes
that can be effected in materials. One of the most important early reactions was the
l
l»
80 3 ■ Symmetry and
Table 3.8 _
Symmetry relations between
the four molecules shown
in the unit cell of
Os 3 (CO) 9 (/t 3 -CC 6 F 5 )
depicted in Fig. 3.32
Disorder
roup Theory
Crystallography 81
1
2
3
4
I
identity
inversion
2, screw
c glide
2
inversion
identity
c glide
2, screw
3
2, screw
c glide
identity
inversion
4
c glide
2, screw
inversion
identity
numbered and are related by symmetry as shown in Table 3.8. The identity operation, of
course, leaves the positions of the molecules unchanged. The inversion center lies at the
center of the unit cell, and it interchanges 1 with 2, and 3 with 4. Two screw axes are
found perpendicular to the xz plane and are indicated in the figure. Note that they
do not pass through the center of the unit cell but are found at hair the x distance
and one-fourth the z distance, and at half the x distance and three-fourths the z
distance. The first of these transforms 1 into 3, and at the same time 2 and 4 are
moved to adjacent unit cells. The second screw axis, related to the first by the inver¬
sion center (and therefore rotated opposite that of the first screw axis), takes 2 into 4,
while transforming 1 and 3 into neighboring cells. Glide planes are xz planes and are
at one-fourth and three-fourths the y distance. The first transforms 2 into 3 (with 1
and 4 being reflected into adjacent unit cells), and the latter transforms 1 into 4 (with
2 and 3 being reflected out of the unit cell).
Although P2Jc is one of the simplest space groups, it is also one of the most
common because complicated molecules tend to crystallize in patterns of low sym¬
metry. The above example illustrates the principal difference between point groups
and space groups. The former requires that some point remain unmoved during the
symmetry operation, while the latter does not have that restriction.
In order to keep the section on crystallography relatively short, this discussion has
not included the theoretical basis for X-ray diffraction. However from other courses
you are probably aware of Bragg’s Law and the diffraction of X rays by regularly
spaced atoms comparable to the diffraction of visible light on phonograph records or
compact disks. Thus one necessity for obtaining X-ray diffraction data is the presence
of crystalline material (= regularly spaced atoms). Amorphous materials do not have
the regular spacing necessary for diffraction. However, even seemingly pure crystals
may be subtle mixtures of two related compounds leading to erroneous results. And
since crystallization is a kinetic process, even an otherwise pure compound may not
crystallize in the single thermodynamically most favorable perfect crystal. There may
be a statistical disorder with most of the unit cells having molecules lined up in the
preferred conformation but a fraction of the unit cells with molecules in a different
conformation of slightly higher energy. Since the X-ray diffraction results are summed
over all of the unit cells, the resultant structure will exhibit this disorder.
If a molecule has a high rotational symmetry except for a small symmetry breaking
atom or group (R), the molecule may pack in any of n-fold ways with R having no
effect on the packing, but showing up as 1 /n R at each of the n possible positions.
For example, a model compound for studying the interaction of the dioxygen mole-
37 Obviously, if R is large enough lo affect the packing of the molecule in the crystal then the possibility
of rotational disorder does not exist.
cule with the heme group (FeN 4 C 20 ) in hemoglobin has four-fold rotational disorder
of the distal oxygen atom, and so that atom, labeled O, (Note: In crystallographic
structures the subscript refers to “oxygen number two"; O z does not refer to the
dioxygen molecule as a whole), shows up as four “one-quarter atoms":
In the same way the N-methylimidazole ligand (the group below the iron atom)
shows a two-fold disorder with respect to where the methyl group is located. The
model compound has pseudo-four-fold symmetry and the rotational disorder can
have no effect on the molecular packing. This is especially evident looking at the 0 2
buried in the molecule; see Fig. 19.4 for the complete structure.
In some cases, such as the above, the confusion is minimal, although the presence
of fractional atoms may be startling the first time one encounters them. In other
cases, the problem may be considerably more serious. Perhaps the “classic” case in
inorganic chemistry was the difficulty in the resolution of the structure of Fe 3 (CO), 2 .
Because of disorder in the solid, a "Star of David" of six half atoms was found, as
shown below. 38
The structure of Fe 3 (CO) 12 is an example in which the disordering occurs about a
center of inversion, a symmetry-related disorder. 9 We have seen that the positions
31 Dahl, L. F.; Rundle, R. E. J. Cliem. Phys. 1957, 26, 1751-1752.
39 Whether a molecule is actually dynamically moving in the crystal or merely statistically disordered
among several possible positions is a moot point as far as X-ray crystallography is concerned since the
interaction time of the X ray with the crystal is of the order of 10” 18 s, faster than atomic movements.
In the case of triiron dodecacarbonyl we have other evidence (see Chapter 15) for believing that the
iron atoms are moving. But they need not—the disorder arises from the location of a molecule without
a center of symmetry [Fe 3 (CO)| 2 ] on a symmetry element of the unit cell (the center of inversion).
66
3-Sy
etry and Group Theory
F' 9 ’ 321 Normal mod « of vibration of the water molecule: (a) symmetrical stretching
mode. A,: (b) bending mode, Ap, (c) antisymmetrical stretching mode. B,. and their
transformations under C\,. symmetry operations.
associated with the vibration. Consider the very simplest molecules possible. In X,
molecules there will be only one stretching vibration, there will be no change in dipole
moment (because there is no dipole moment either before or during vibration as long
as the two X atoms are identical), and so that vibration will not be infrared active.
1 he molecules H,. F,, Cl 2 , and N, show no !R absorptions. However, carbon mon¬
oxide. isoelectronic with dinitrogen, has a small dipole moment, and the molecular
vniation is infrared active because the dipole moment changes as the bond length
changes The absorption frequency is 2143 cm"', an important value for coordi¬
nation chemistry.'- The point is that the electromagnetic infrared wave can interact
with the electric dipole moment; in essence, the infrared wave's electric field can “grab”
the vibrating electric dipole moment resulting in a molecular vibration of the same
frequency but increased amplitude.
Consider next the water molecule. As we have seen, it has a dipole moment, so
we expect at least one IR-active mode. We have also seen that it has C 2 „ symmetry,
and wc may use this fact to help sort out the vibrational modes. Each normal mode
o) L ibrarian will Jarm a basis for an irreducible representation of the point group of the
molecule.- A vibration will be infrared active if its normal mode belongs to one of the
irreducible representations corresponding to the x, y and : vectors. The C lo character
table lists four irreducible representations: A u A ,, B„ and B,. If we examine the
three normal vibrational modes for H 2 0, we see that both the symmetrical stretch
and the bending mode are symmetrical not only with respect to the C 2 axis, but also
with respect to the mirror planes (Fig. 3.21). They therefore have A, symmetry and
since r transforms as A u they are IR active. The third mode is not symmetrical with
respect to the C 2 axis, nor is it symmetrical with respect to the aj^xz) plane, so it
has B 2 symmetry. Because y transforms as B,, this mode is also IR active The three
vibrations absorb at 3652 cm-', 1545 cm' 1 , and 3756 cm' 1 , respectively.
22 See Chapters II and 15.
J n° r , a , m ° r L‘ h r Ugh discussion of 8 rou P theory and vibrational spectroscopy, see Harris. D. C.-
Bertolucci, M. D. Symmetry and Spectroscopy, Dover New York, 1989. Chapter 3.
Uses of Point Group Symmetry
67
Table 3.3 _
Derivation of reducible
representation for degrees
of freedom in the water
molecule
E C z
<r„(xx)
<r„(yx)
Unshifted atoms
3 1
1
3
Contribution per atom
3 -1
1
1
r,„,
9 -1
1
3
Table 3.4
Atomic contributions, by
symmetry operation, to the
reducible representation for
the 3N degree!
for a molecule
i of freedom
Contribution
Operation
per atom"
E
3
-1
c>
0
c
1
c 6
2
a
1
i
-3
s>
-2
Ss
-1
St
0
4 C„ = I + 2 cos —
n
S. -
— I + 2 cos
360
n
For a vibration to be Raman active, there must be a change in the polarizability
tensor. We need not go into the details of this here, 24 but merely note that the com¬
ponents of the polarizability tensor transform as the quadratic functions of x, y, and
z. Therefore, in the character tables we are looking for x 2 , y 2 , z 2 , xy, xz, yz, or their
combinations such as x 2 — y 2 . Because the irreducible representation for x 1 is A,
and that for yz is B 2 , all three vibrations of the water molecule are Raman active
as well.
The foregoing analysis for water relied on our ability to describe its fundamental
vibrational motions as a first step. Although this is a relatively simple task for an AB 2
molecule, it becomes formidable for complex structures. An alternative procedure can
be used which requires no knowledge of the forms of the vibrations. It begins with
the derivation of a reducible representation for all 3 N degrees of freedom for the
molecule. This can be accomplished by representing the degrees of freedom for each
atom as a set of Cartesian displacement vectors chosen so that the r vectors are
parallel to the molecule's highest fold rotational axis (Fig. 3.22). The characters of
the reducible representation can then be determined by considering the combined
effect of each symmetry operation on the atomic vectors. A simplification is possible
here because only those atoms that are not shifted by an operation will contribute
to the character for that operation. For water, this includes all three atoms for both
the E and irfyz) operations, but only oxygen for the other two (Table 3.3). The con¬
tribution of each unshifted atom to the total character will be the sum of the effects
of the operation on the atom's displacement vectors. For the C 2u operations, a
vector will either remain stationary (+ 1) or be shifted to minus itself (- I). Thus we
obtain +3 for the identity operation because all (hree vectors are unaffected. In the
case of C 2 , the : vector on the oxygen atom remains stationary while the x and y
vectors are moved to —x and — y, respectively, yielding an overall value of -1.
Because these contributions arc independent of molecular symmetry, they can be
conveniently tabulated for common operations (Table 3.4). A simple multiplication
Fig. 3.22 Cartesian
displacement vectors for
the water molecule.
24 See Ebsworth. E. A. V.; Rankin D. W. H.; Cradock. S. Structural Methods in Inorganic Chemistry, 2nd
ed.; CRC: Boca Raton. EL. 1991; Chapter 5: Drago, R. S. Physical Methods for Chemists, 2nd ed.;
Saunders: Fort Worth. 1992: Chapter 6.
70
3- Symmetry and Group Theory
Fig. 3.24 Normal modes
of vibration of the XeF 4
molecule. Note that both
£„ modes are doubly
degenerate. [Modified from
Harris, D. C.; Bertolucci,
M. D. Symmetry and
Spectroscopy', Dover: New
York, 1989. Used with
permission.]
fact that when the first compounds of noble gases were synthesized there was con¬
siderable uncertainty about their structures. For XeF 4 some chemists of a theoretical
bent opted for a tetrahedral structure, while most inorganic chemists interested in
the problem leaned towards a square planar structure. The matter was resolved by
Claassen, Chernick, and Malm in a paper entitled “Vibrational Spectra and Structure
of Xenon Tetrafluoride.” 27 The opening sentences of their abstract read as follows:
“The infrared spectrum of XeF 4 vapor has strong bands at 123,291, and 586 cm' 1 .
The Raman spectrum of the solid has very intense peaks at 502 and 543 cm -1 and
weaker ones at 235 and 442 cm"'. These data show that the molecule is planar and
of symmetry D Vr "
Claassen and co-workers made the assignments shown in Fig. 3.24. The weak ab¬
sorption at 442 cm' 1 , which does not appear in Fig. 3.24, was ruled out as a funda¬
mental and assigned as an overtone of fl 2u . A glance at the character table for D xh
(for square planar XeFJ shows that the B lu mode is neither TR nor Raman active.
Its frequency was obtained from the overtone (442 cm' 1 ), which is Raman active. 28
The important thing to note is that if XeF 4 is square planar, it will have a center
• 7 Claassen. H. H.; Chernick, C. L.; Malm. J. G. J. Am. Client. Soc. 1963, 85, 1927-1928.
The rcprcscntalion for Ihc overtone may be obtained by squaring the irreducible rcprcscntalion for
x = A,,). The result. transforms as a binary product and therefore is Raman active. For
a discussion of direct products of representations as applied to overtones, see Footnote 24.
Uses of Point Group Symmetry
71
of symmetry, /, and none of the IR and Raman bands can be coincidental; if it
were tetrahedral, there should be IR and Raman bands at the same frequencies. The
presence of six fundamental bands, three in the infrared and three in the Raman,
none coincidental, is very strong evidence of the square planar structure of XeF 4 .
Bonding Covalent bonds can be described with a variety of models, virtually all of which
involve symmetry considerations. As a means of illustrating the role of symmetry in
bonding theory and laying some foundation for discussions to follow, this section will
show the application of symmetry principles in the construction of hybrid orbitals.
Since you will have encountered hybridization before now, but perhaps not in a sym¬
metry context, this provides a facile introduction to the application of symmetry.
You should remember that the basic procedure outlined here (combining appropriate
atomic orbitals to make new orbitals) is applicable also to the derivation of molecular
orbitals and ligand group orbitals, both of which will be encountered in subsequent
chapters.
The atomic orbitals suitable for combination into hybrid orbitals in a given mole¬
cule or ion will be those that meet certain symmetry criteria. The relevant symmetry
properties of orbitals can be extracted from character tables by simple inspection.
We have already pointed out (page 60) that the p x orbital transforms in a particular
point group in the same manner as an x vector. In other words, a p orbital can
serve as a basis function for any irreducible representation that has V listed among
its basis functions in a character table. Likewise, the p y and p. orbitals transform as
y and z vectors. The d orbitals— d xy , d x: , d y ., d x i„ y i, and d.i —transform as the binary
products xy, xz, yz, x 2 - y 2 , and z 2 , respectively. Recall that degenerate groups of
vectors, orbitals, etc., are denoted in character tables by inclusion within parentheses.
An s orbital, because it is spherical, will always be symmetric (i.e., it will remain
unchanged) with respect to all operations of a point group. Thus it will always belong
to a representation for which all characters are equal to 1 (a “totally symmetric"
representation), although this is not explicitly indicated in character tables. The totally
symmetric representation for a point group always appears first in its character table
and has an A designation (/!,, A g , A tg , etc.). When these or any other Mulliken sym¬
bols are used to label orbitals or other one-electron functions, the convention is to
use the lower case: a,, a g , etc.
To see how the s, p. and d atomic orbitals on a central atom are alTcctcd by the
symmetry of the molecule to which they belong, consider the octahedral (0,,), square
pyramidal (C 4 „), and seesaw (C,„) species shown in Fig. 3.25. For the AB 6 case, we
find from the character table for the O h point group (Appendix D) that the p x , p y ,
and p. orbitals belong to the £,„ representation. Since they transform together, they
represent a triply degenerate set. The five d orbitals, on the other hand, form two sets
of degenerate orbitals. The d x ^ yl and d.i orbitals belong to the doubly degenerate
e g representation and the d xy , d x: , and d y . orbitals transform together as a triply
degenerate t 2g set. The s orbital, as always, belongs to the totally symmetric represen¬
tation, a lg .
If we imagine removing one of the B atoms from AB 6 , we are left with square
pyramidal AB S (Fig. 3.25b). By referring to the character table for the C 4v point
group, we can see that the p. orbital now belongs to the a, representation and the p x
and p y orbitals to the e representation. Thus in going from to C 4 „ symmetry, the
triply degenerate p orbitals have been split into two sets, one nondegenerate and one
doubly degenerate. Similarly, the d orbitals are distributed among a larger number
of sets than was the case in the octahedral molecule. The e g level is split into two, a
Excerpts from the Preface
to the Third Edition
It has been my very good fortune to have had contact with exceptional teachers and
researchers when I was an undergraduate (Thomas B. Cameron and Hans H. Jaffe,
University of Cincinnati) and a graduate student (John C. Bailar, Jr., Theodore L.
Brown, and Russell S. Drago, University of Illinois); and to have had stimulating and
helpful colleagues where I have taught (William D. Hobey and Robert C. Plumb.
Worcester Polytechnic Institute; Jon M. Bellama, Alfred C. Boyd, Samuel 0. Grim,
James V. McArdle, Gerald Ray Miller, Carl L. Rollinson, Nancy S. Rowan, and John
A. Tossell, University of Maryland). I have benefitted by having had a variety of
students, undergraduate, graduate, and thesis advisees, who never let me relax with a
false feeling that I “knew it all.” Finally, it has been my distinct privilege to have had
the meaning of research and education exemplified to me by my graduate thesis
advisor, Therald Moeller, and to have had a most patient and understanding friend.
Hobart M. Smith, who gave me the joys of a second profession while infecting me with
the "mihi itch." Professors Moeller and Smith, through their teaching, research, and
writing, planted the seeds that grew into this volume.
Four librarians, George W. Black, Jr., of Southern Illinois University at Carbon-
dale, and Sylvia D. Evans, Elizabeth W. McElroy, and Elizabeth K. Tomlinson, of the
University of Maryland, helped greatly with retrieval and use of the literature. I
should like to give special thanks to Gerald Ray Miller who read the entire manuscript
and proofs at the very beginning, and who has been a ready source of consultation
through all editions. Caroline L. Evans made substantial contributions to the contents
of this book and will always receive my appreciation. Finally, the phrase “best friend
and severest critic” is so hackneyed through casual and unthinking use, paralleled
only by its rarity in the reality, that 1 hesitate to proffer it. The concept of two men
wrangling over manuscripts, impassioned to the point of literally (check Webster’s)
calling each other's ideas “poppycock" may seem incompatible with a friendship
soon to enter its second quarter-century. If you think so, you must choose to ignore
my many trips to Southern Illinois University to work with Ron Brandon, to visit with
him and his family, to return home with both my emotional and intellectual “bat¬
teries” recharged.
My family has contributed much to this book, both tangible and intangible, visible
and (except to me) invisible. My parents have tolerated and provided much over the
xvi
Excerpts from the Preface to the Third Edition
XVII
vears including love, support, and watching their dining room become an impromptu
office : often the same week as holiday dinners. My sister, Cathy Donaldson, and her
husband Terry, themselves university teachers, have both answered and posed
questions ranging from biology to chemical engineering. More important, they ha
b”» .her " when I needed their unipue help. To all of these go my deepest gratitude
and thanks.
James E. Huheey
62 3 Symmetry
Reducible
Representations
and Group Theory
c;,a
Cl
\ I /
Ci-Pt-Cl--}--► C' 2 . a v
/ \
/ | \
/ Cl V
ci\ a.
c;. a.
Fig. 3.19 The PtClj-
ion (D 4 J with x and y
translation vectors.
are aligned along the Pt—Cl bonds, as shown. The C' 2 and C 2 axes are secondary
axes perpendicular to C 4 . The C\ axes are chosen so as to include as many atoms
as possible, and thus they lie along the x and y coordinates. The C' 2 ' axes lie mid¬
way between the x and y axes. The <r„ and a d planes include the C 2 and C 2 axes,
respectively.
Translation of the PtC! 2- ion in the x and y directions can be represented by
the two vectors shown on the platinum atom (Fig. 3.19). In contrast to all of the
cases we have so far considered, certain operations of the D 4)l group lead to new
orientations for both vectors that do not bear a simple +1 or -1 relationship to
the original positions. For example, under a clockwise C 4 operation, the x vector
is rotated to the + y direction, and the y vector is rotated to a -x position. The
character for this operation is zero. (This arises because the diagonal elements of the
matrix for this operation are all zero; other elements in the matrix are nonzero but
do not contribute to the character.) The S 4 and a d operations lead to a similar mixing
of the x and y functions and also have characters of zero. Because of this mixing,
the x and y functions are inseparable within the D ih symmetry group and arc said
to translorm as a doubly degenerate or two-dimensional representation.
The remaining characters of the £„ representation can be generated by con¬
sidering the combined effect of each operation on the x and y vectors. When the
identity operation is applied, both vectors remain unchanged; hence the character
for the operation is two times +1 or +2. Similarly, the cr operation (reflection in
the plane of the molecule) leaves the vectors unmoved and yields a character of +2.
Under the C 2 operation (around the z axis), the x vector is brought to a —x position
and the y vector to a -y position, giving a character of 2(-I) = -2. Inversion
through the center of symmetry leads to the same result. For a C' 2 operation around
the y axis, the y vector is unaltered (+1), while the x vector is rotated to the opposite
direction (- I), yielding a total character of zero. The outcome is identical for reflec¬
tion through the mirror plane that includes the y axis (a a„ operation).
In applying the methods of group theory to problems related to molecular structure
or dynamics, the procedure that is followed usually involves deriving a reducible
representation for the phenomenon of interest, such as molecular vibration, and then
decomposing it into its irreducible components. (A reducible representation will
always be a sum of irreducible ones.) Although the decomposition can sometimes
be accomplished by inspection, for the more general case, the following reduction
Uses of Point Group Symmetry 63
formula can be used:
» = lZz:-xt-» x (3.i)
In this expression, N is the number of times a particular irreducible representation
appears in the representation being reduced, h is the total number of operations in
the group, Xr > s the character for a particular class of operation, x, in the reducible
representation, /f is the character of x in the irreducible representation, n x is the
number of operations in the class, and the summation is taken over all classes. The
derivation of reducible representations will be covered in the next section. For now,
we can illustrate use of the reduction formula by applying it to the following reducible
representation, T„ for the motional degrees of freedom (translation, rotation, and
vibration) in the water molecule:
E C 2 a v (xz) a v (yz)
r r : 9 -1 1 3
To decompose this representation, Eq. 3.1 must be applied for each of the four ir¬
reducible representations in the C 2 „ point group:
A { : N = (i)[(9)(l)(l) + (- 1)(1)(1) + (1)(1)(I) + (3)( I)(1)] = 3
A 2 : N = (i)[(9)(l)(l) + (-1X1)0) + 0X-1X0 + (3)(— DO)] = 1
B,: N = ( 4 ) [(9)0)0) + (-1)(-1)(1) + 0)0)0) + (3)(-1)0)] = 2
B 2 : N = (i) [(9)(l)(l) + (- l)( -1)0) + OX- l)(l) + (3)0)0)] = 3
Thus the reducible representation is resolved into three A { . one A 2 , two J3„ and
three fl 2 species. It can easily be confirmed that the characters for this combination
sum to give the characters of T r .
Uses of Point
Group Symmetry
Optical Activity
The reader will already have encountered chirality extensively in organic chemistry
based upon asymmetric carbon atoms. Although the usual definition of chirality in
organic texts is based upon a nonsuperimposable mirror image 18 and thus allows
chirality in species such as helicene and spiro compounds, few introductory organic
texts discuss chirality other than that based on asymmetric carbon atoms. 19 Inorganic
molecules may be optically active based on asymmetric nitrogen, phosphorus, or
sulfur atoms, 20 but by far the largest number of chiral inorganic compounds do not
have a single asymmetric atom at all, but are chiral because of the overall molecular
symmetry, specifically the absence of an improper axis of rotation. Most of these are
18 As we have seen, the formal definition of optical activity is based upon the absence of an improper axis
of rotation. The two definitions arc equivalent.
18 For discussions of chiral organic molecules that do not contain asymmetric carbon atoms, see: Wade,
L. C., Jr. Organic Chemistry. Prentice-Hall: Englewood Cliffs, NJ. 1987; pp 354-356. Schlogl, K. Topics
Cun. Chem. 1984, 125. 27. Laarhoven, W. H.; Prinscn, W. J. C. Ibid. 1984, 125. 63. Meurcr, K. P.; Vogtle.
F. Ibid. 1985. 127. 1.
20 These are discussed in Chapter 6.
60
3 • Symmetry and Group Theory
A 2 — +1 +1 “1
(a) (b) (c) (d)
Fig. 3.18 Effects of symmetry operations in C 2l . symmetry; rotation about the z axis:
(a) identity, £; (b) rotation about the C 2 axis, (c, d) reflection in a, planes.
Translations (and p orbitals) along the x and z axes in the water molecule con¬
form to different symmetry patterns than the one just developed for the y axis. When
the E , C 2 , cr„(xz), and o v (yz) operations are applied, in that order, to a unit vector
pointing in the +x direction, the labels + 1, — 1, +1, and — 1 are generated. A vector
pointing in the + z direction will be unchanged by any of the symmetry operations
and thus will be described by the set: -I-1, +1, +1, -I-1.
The principles of group theory dictate that the total number of irreducible repre¬
sentations belonging to a point group will be the same as the number of types or
classes of symmetry operations characterizing the group. Hence we expect four ir¬
reducible representations for the C 2 „ point group. We can generate the fourth one by
considering rotation of the water molecule about the z axis. To see this, imagine an
arrow curved clockwise about the z axis (when viewed down this axis; see Fig. 3.18).
Like the linear translations, this motion will be symmetric with respect to any opera¬
tion that causes no change in direction and will be antisymmetric for any operation
that leads to reversal. Both E and C 2 leave the direction unchanged (+1), but reflec¬
tion in either mirror plane causes a reversal (—1). The result is +1, +1, —1, and
— I as the fourth symmetry pattern for the C 2v group.
Many of the symmetry properties of a point group, including its characteristic
operations and irreducible representations, are conveniently displayed in an array
known as a character table. The character table for C 2 „ is 17
c 2 „
E
c 2
ff„(xz)
o'M
l
1
1
1
z
x 2 , y 2 , z 2
2
1
1
-1
-1
R-.
xy
1
-1
1
-1
X, Ry
xz
b 2
1
-1
-1
1
y, R
yz
The column headings are the classes of symmetry operations for the group, and
each row depicts one irreducible representation. The + 1 and — 1 numbers, which
17 Character tables for other point groups can be found in Appendix D.
Irreducible Representations and Character Tables
61
correspond to symmetric and antisymmetric behavior, as we have seen, are called
characters.
In the columns on the right are some of the basis functions which have the
symmetry properties of a given irreducible representation. R x , R y , and R ; stand for
rotations around the specified axes. The binary products on the far right indicate,
for example, how the d atomic orbitals will behave (“transform”) under the operations
of the group.
The symbols in the column on the far left of the character table (Mulliken labels)
are part of the language of symmetry. Each one specifies, in shorthand form, several
features of the representation to which it is attached. One such feature is the dimen¬
sion, which is related to the mathematical origin of the characters. Strictly speaking,
each character is derived from a matrix representing a symmetry operation, and is
in fact equal to the sum of the diagonal elements of the matrix. For the C 2v group,
all of these matrices are of the simplest possible form: They consist ot a single element
(the character) and are thus one-dimensional. However, for groups with rotational
axes of order three or higher, two- and three-dimensional matrices occur, leading to
characters with values as high as 2 and 3, respectively. (This will be illustrated shortly
for the D 4/i point group.) One-dimensional representations, such as those in the C 2v
group, are labeled A if symmetric or B if antisymmetric with respect to the highest
order rotational axis. If two or more representations in a group fit the A or B clas¬
sification, a subscript is added to indicate symmetric (1) or antisymmetric (2) behavior
with respect to a second symmetry element. This second element will be a C 2 axis
perpendicular to the principal axis or, in the absence of such an axis, a vertical mirror
plane. Two-dimensional representations are denoted by E (not to be confused with
the identity operation E), and three-dimensional cases by T. As we have seen before,
the labels g and u may be applied if there is a center of inversion. The superscripts '
and " may be used to signify symmetric and antisymmetric behavior with respect to
a horizontal plane.
The tetrachloroplatinate(ll) anion. PtCI; , was given earlier (Fig. 3.14c) as an
example of a molecule belonging to the D ih point group. The character table for this
group is
2 C'i
i
2S 4
Oh
2 a „
2o d
1
1
1
1
1
1
-1
-1
1
1
1
-1
1
1
-1
1
1 .
-1
x 2 + y 2 ,.
(R„ R f )
Note that two of the irreducible representations in this group are two-dimensionai,
labeled E g and £„. Each has a pair of basis functions listed for it. To see how x and
y translation serve as a basis for the E u representation, refer to Fig. 3.19 which shows
the PtCl^ - ion labeled with a coordinate system assigned according to the usual
conventions. The z axis coincides with the C 4 rotational axis, and the x and y axes
68
3 • Sy mmotry and Group Theory
Table 3.5_
Derivation of r, 0 , for BCI 3
£
2C 3
3C 2
2S 3
3<r„
Unshifted atoms
4
1
2
4
1
2
Contributions
per atom
3
0
-1
1
-2
1
r,„,
12
0
-2
4
-2
2
of the number of unshifted atoms by the contribution for each operation gives the
reducible representation (r, 0 ,) for water shown in Table 3.3.
We determined earlier (page 63) that the irreducible components ot this repre¬
sentation are three A,, one A 2 , two B„ and three B 2 species. To obtain from this
total set the representations for vibration only, it is necessary to subtract the repre¬
sentations for the other two forms of motion: rotation and translation. We can iden¬
tify them by referring to the C 2 „ character table. The three translational modes will
belong to the same representations as the x, y, and z basis functions, and the rota¬
tional modes will transform as R x , R y , and R.. Subtraction gives
r, 0 , = 3A[ + A 2 + 26, + 3B 2
-[r, rnnJ = /t, + b.+ b 2 ]
[r ro , = _ a 2 + b i + b 2 ]
= 2A t + b 2
This is, of course, the same result as obtained above by analyzing the symmetries of
the vibrational modes. . , . . , ID ,
As a second example of the use of character tables in the analysis of IR and
Raman spectra, we turn to BC1 3 with D }h symmetry. Because it has four atoms we
expect six vibrational modes, three of which will be stretching modes (because there
arc three bonds) and three of which will be bending modes. Table 3.5 shows the
derivation of T (0I for the molecule’s twelve degrees of freedom. Application of the
reduction equation and subtraction of the translational and rotational representa¬
tions gives
r,„, = A\ + A'i + 3E' + 2 A 2 + E"
-c r —- , E ' + ]
-rr„,.= A' 2 + + e]
r uifc = A\ + 2 £' + A'i
We see that the six fundamental vibrations of BC1 3 transform as A\, A%, and 2£.
Each £' representation describes two vibrational modes of equal energy. Thus the It
notation refers to four different vibrations, two of one energy and two of another.
The A\ mode is Raman active, the A'± is IR active, and the £' modes are both Raman
and IR active. .... .
In thinking about the actual vibrations associated with these modes, we expect
the A\ mode to be the symmetrical stretch because it remains unchanged under a 1 ol
the symmetry operations (Fig. 3.23a). Another motion is a symmetrical out-of-plane
bending deformation with the boron atom moving in one direction while the three
chlorine atoms move in unison in the opposite direction (Fig. 3.23b) The four re_
maining vibrations (two stretching and two bending) are not as easily categorized
because they are distributed among two doubly degenerate modes, both of which
Uses of Point Group Symmetry
69
I0 BC1,471cm- 1 io BC 1 3 480 cm- 1 I0 BC1 3 995 cnT 1 l0 BClj244cm'
!1 BC1 3 471 cm- 1 "6034600™-' "BC1 3 956 cnT 1 "BCI 3 243 cnT 1
v, v 3 v,
a; av £’ F -
(a) (b) Cc) W)
Fig 3 23 Normal modes of vibralion of (he BC1 3 molecule: (a) symmelrical stretching
mode A',; (b) out-of-plane bending mode. A; (c) unsymmctrical slreiching mode, E'\ and
(d) in-plane bending mode. E'. IModified from Harris. D. C.; Bertolucci. M. D. Symmetry
and Spectroscopy, Dover: New York. 1989. Reproduced with permission.!
belong to £'. Figure 3.23c and d show one component of each of these modes; in each
case, the two components give rise to a single frequency of vibration. Both c and d
are restricted to the xy plane and can thus have only x and y components, and so
both modes can be no more than doubly degenerate. Both are symmetric with regard
to the horizontal mirror plane, so they transform as £', not E".
Another useful "trick” in interpreting spectra is the fact that the characteristic
frequency of a vibrational mode will depend upon the masses of the atoms moving in
that mode. Isotopic substitution can thus be used to assign some of the frequencies.
Note that in a perfectly symmetrical stretch, the boron atom moves not at all, and so
substitution of 10 B for the more abundant "B will leave the absorption unchanged
at 471 cm -1 (see Fig. 3.23a). In contrast, the boron atom moves considerably in b
and c and substitution of the lighter 10 B results in shifts of these absorptions to higher
frequencies. Vibration d is interesting—to a first approximation the boron atom
scarcely moves. But move it does, tending to follow, feebly to be sure, the single Cl
atom in opposition to the pair of Cl atoms. The absorption frequency hardly changes,
from 243 cm -1 to 244 cm -1 , upon isotopic substitution.
We have seen that not all molecules are like water in having all vibrational
modes both IR and Raman active. In fact, there is an extremely useful exclusion rule
for molecules with a center of symmetry, /: If a molecule has a center of symmetry, I R
and Raman active vibrational modes are mutually exclusive; if a vibration is IR active,
it cannot be Raman active, and vice versa. 26
An example to which this rule applies is XeF 4 . In fact, it nicely illustrates the
usefulness of IR and Raman spectroscopy in the assignment of structures. For XeF 4
we expect nine vibrational modes, four stretching and five bending. These are illus¬
trated in Fig. 3.24. Note that like BC1 3 , XeF 4 has two modes that are doubly degen¬
erate (£„). The importance of the vibrational spectroscopy of XeF 4 comes from the
« Other clues may be used of course. For example, il is usually observed that asymmetric stretches occur
at higher frequencies than symmetric stretches, though there arc exceptions.
26 Note that while it is impossible for a molecule with a center of symmetry to have a vibration that is
both IR and Raman active, it is possible for it to have a vibration that is neither. See Keiter, R. L.
J. Chem. Educ. 1983, 60, 625.
Table 3.1
Point groups of chiral and
achiral molecules
six-coordinate complexes with D 3 or closely related symmetry. They will be discussed
at greater length in Chapter 12. Since chiral molecules often possess some elements
of symmetry (e.g., both C 2 and C 3 are found in the D 3 point group), it is not appro¬
priate to refer to them as asymmetric (without symmetry). These molecules have come
to be known as dissymmetric and this is, in fact, the term now used for all chiral
molecules. The absence of an improper axis of rotation defines a molecule as dis¬
symmetric (see Footnote 3). Common point groups are categorized as chiral or achiral
in Table 3.1.
A molecule will have a dipole moment if the summation of all of the individual bond
moment vectors is nonzero. The presence of a center of symmetry, i, requires that
the dipole moment be zero, since any charge on one side of the molecule is canceled
. . . /. . , . -r-1 rr^ . r- 13- /rr:„ i a~\
Dipole Moments
SiF 4 (Fig. 3.4a), PF 5 (Fig. 3.4b), eclipsed ferrocene (Fig. 3.8a), and all D„ molecules
(cf. Fig. 3.14) do not have dipole moments. The presence of a horizontal mirror plane
prevents the possibility of there being a dipole moment, but one or more vertical
mirror planes do not. The dipole moment vector must obviously lie in such planes,
and there may be a C„ axis in the plane along which the dipole lies. Examples of
such molecules are cis-N 2 F 2 (Fig. 3.6) and 0=N—Cl (Fig. 3.11b). Common point
groups are listed as “symmetry allowed" or “symmetry forbidden with respect to
possible dipole moments in Table 3.2. In addition, it is always possible that certain
Symmetry forbidden (symmetry
Symmetry allowed eloment(s) prohibiting dipole)
Point groups for which
dipole moments are
symmetry allowed or
symmetry forbidden
C, (center of symmetry)
S„ (improper axis)
D„ (C„ + nC 2 )
D nh (C„ + nC 2 and <r»)
D n AC:±nC 2 )
T d (4C 3 + 3C 2 )
O h {i, C„ + nC 2 , and of)
I h (i, C, + nC 2 , and ct)
Chiral
Achiral (identifying symmetry element)
C 3 (asymmetric)
C„ (dissymmetric)
D„ (dissymmetric)
C, (plane of symmetry)
C, (center of symmetry)
D„ h (plane of symmetry)
D„ d (plane of symmetry)
S„ (improper axis)
T d (plane of symmetry)
O h (center and plane of symmetry)
l h (center and plane of symmetry)
C„„ (plane of symmetry)
Uses of Point Group Symmetry
65
A Raman experiment
Fiq. 3.20 Schematic representation of infrared and Raman experiments. In infiared
spectroscopy the excitations are detected by absorption of characteristic frequencies. In Raman
spectroscopy the excitations are detected by characteristic shifts in frequencies of the
scattered light. [From Harris, D. C.; Bertolucci, M. D. Symmetry and Spectroscopy, Dover.
New York, 1989. Reproduced with permission.]
heteronuclear bond moments might be zero or two different bonds might have iden¬
tical moments that cancel. Indeed, some molecules have very small but finite dipole
moments: S=C=Te (0.57 x 10" 30 C m; 0.17 D), ds-FN=NF (0.53 x 0 C m
0 16 D), NO (0.50 x 10“ 30 C m; 0.15 D), SbH 3 (0.40 x 10 30 C m; 0.12 D), CO
(0.37 x KT 30 Cm; 0.11 D), FC10 3 (0.077 x I0" 30 C m; 0.023 D).
Infrared and Raman In infrared spectroscopy photons having energies corresponding to the excitation ol
Snecfroscopv certain molecular vibrations arc absorbed and the transmittance of the infrared ltgh
P PY at those frequencies is reduced (Fig. 3.20). In Raman spectroscopy , allowed molecular
excitations result in difTerenccs in the frequency of scattered light. I hat is, incident
light which may be of any wavelength (usually visible), undergoes scattering. Most
of the scattered light has the same frequency as that impinging upon the sample, but
the frequency of a small fraction is shifted by amounts corresponding to the energy
differences of the vibrational states (Fig. 3.20).
The number of vibrational modes or a molecule composed of N atoms is 3 N o
(or 37V — 5 if linear). 21 We may find which of these are infrared and Raman active
by the application of a few simple symmetry arguments. First, infrared energy is
absorbed for certain changes in the vibrational energy levels ol a molecule. For a
vibration to be infrared active, there must be a change in the dipole moment vector
31 The molecule will have a total of 3 N degrees of freedom, of which three will be associated with trans¬
lation and another three (two if linear) with rotation.
Crystallography
77
y
Fig. 3.29 A four-fold
improper axis (4) operation.
The 4 axis is perpendicular
to the plane of the paper,
A, and A 3 ( + ) are above
the plane, A 2 and A 4 (-)
are below. When a point
A, (x, y, 2 coordinates:
a, +b, +c) is rotated
clockwise by 90°, followed
by inversion, it becomes
point A 4 ( —b, +a, — c).
In the same way, A 2
becomes A,, A 3 becomes
A 2 , and A 4 becomes A 3 .
atom. Consideration of these groupings and the overall symmetry which may arise
leads to thirty-two crystallographic point groups. All of the discussion of symmetry
and point groups until now has been in terms of the Schoenjiies system that is pre¬
ferred by spectroscopists and structural chemists who are primarily concerned with
the symmetry properties of isolated molecules. Crystallographers almost always work
with an equivalent but different system, the International system, also known as the
Hermann-Mauguin system. Some of the symmetry elements that can be present in
three-dimensional lattices are the same as we have seen in molecular point groups:
the center of symmetry (center of inversion), mirror planes (given the symbol m), and
simple //-fold rotational axes (designated by the n value, n = 1, 2, 3, 4, and 6). 35 A
mirror plane perpendicular to a principal axis is labeled n/m. In addition, there are
three other symmetry elements: axes of rotatory inversion, glide planes, and screw
axes.
In the Schoenflies system the improper axis is an axis of rotation-reflection (see
page 52). In the International system the axis of rotatory inversion (71) is one of
//-fold rotation followed by inversion (see Fig. 3.29).
A glide plane is a translation followed by a reflection in a plane parallel to the
translation axis. In the simplest case, consider a lattice with unit cell of length a along
the x axis (Fig. 3.30). Movement of a distance a/ 2 along the x axis, followed by re¬
flection, accomplishes the symmetry operation. A glide plane is labeled a, b , or c
depending on the axis along which translation occurs. Additionally, glide operations
may occur along a face diagonal (an n glide) or along a body diagonal (a d glide). If
a glide plane is perpendicular to a principal or screw axis, it is shown as n/c, n/a, etc.
or nfc, nfb , etc., respectively. Note that because of the reflection in the operation,
any chiral molecule will reflect as its enantiomer of opposite chirality. For a glide-
plane to be present in a crystal of a chiral compound, both enantiomers must exist
in the crystal, that is, it must be a racemic mixture.
Another symmetry element that may be present in a crystal is a screw axis
(identified by //,) which combines the rotational symmetry of an axis with translation
along that axis. A simple two-fold (2,) screw axis is shown in Fig. 3.31. In contrast
to the glide plane, only translation and rotation are involved in this operation, and
therefore a chiral molecule retains its particular handedness.
/ /
c /y 3X15
B
7
-A-A-A- x axis
z.
£
7
Fig. 3.30 A glide-plane
operation. The molecule
moves a distance a/2 along
the x axis and then is
reflected by the xy plane.
Note that the chirality of
the molecule changes.
33 It can be shown mathematically that five-fold axes cannot appear in a truly periodic crystal of single
unit cells repeating in space. Nevertheless, some interesting “quasicrystals" have recently been discovered
that have unusual symmetry properties. See Problem 3.39.
74
3 - Symmetry and Gr
p Theory
Reduction of F r (Eq. 3.1) shows that it is composed of e g , a 2u , and b 2u . The D ih
character table reveals that no orbitals transform as b 2u but that p : belongs to a 2u
while cl x . and d y: belong to e g . That these three orbitals on platinum are allowed by
symmetry to participate in out-of-plane n bonding is reasonable since they are all
oriented perpendicular to the plane of the ion (the xy plane). Selection of orbitals on
platinum suitable for in-plane 7r bonds is left as an exercise. (Hint: In choosing vectors
to represent the suitable atomic orbitals, remember that the in-plane and out-of-plane
7t bonds will be perpendicular to each other and that the regions of overlap for the
former will be on each side of a bonding axis. Thus the in-plane vectors should be
positioned perpendicular to the bonding axes.) 30
The symmetry of crystals not only involves the individual point group symmetry of
the molecules composing the crystals, but also the translational symmetry of these
molecules in the crystal. The latter is exemplified by a picket fence or a stationary
row of ducks in a shooting gallery. If we turn on the mechanism so the ducks start
to move and then blink our eyes just right, the ducks appear motionless—the ducks
move the distance between them while we blink, and all ducks are identical. Under
these conditions we could not tell if the ducks were moving or not, because they
would appear identical after the change to the way they appeared before the change.
If we think of the ducks as lattice points, a row of them like this is a one-dimensional
crystal. In a three-dimensional crystal, a stacked array of unit cells, the repeating
units of the system are like the row of ducks and display translational symmetry.
Determining the crystal structure of a compound by X-ray diffraction has be¬
come so important (and so routine) 32 to the inorganic chemist that nearly fifty per¬
cent of the papers currently published in the journal Inorganic Chemistry include at
least one structure. What information is conveyed when we read that the solid state
structure of a substance is monoclinic P2Jdl We can answer this question by starting
with a few basics.
Diffraction patterns can be described in terms of three-dimensional arrays called
lattice points. 33 The simplest array of points from which a crystal can be created is
called a unit cell. In two dimensions, unit cells may be compared to tiles on a fioor.
A unit cell will have one of seven basic shapes (the seven crystal systems), all con¬
structed from parallelepipeds with six sides in parallel pairs. They arc defined ac-
30 Answer: a.y None; h 2a : i e„: (p„ p v ).
31 Ladd, M. F. C. Symmetry in Molecules and Crystals: Wiley: New York, 1989. Hyde, B. G.; Andersson, S.
Inorganic Crystal Structures: Wiley: New York. 1989.
32 There was a time when the collection of data and resolution of a crystallographic structure was a truly
horrendous task, the solution of a single structure often being an accomplishment worthy of a doctoral
dissertation. Today, thanks to automation and computerization, there are research organizations that
will guarantee the resolution of a structure containing a limited number of atoms in one week at a
charge that, in terms of the costs of doing research, is relatively small. For the early days, see WyckofT,
R. W. G,; Posnjak, E. J. Am. Chem. Soc. 1921, 43. 2292-2309. For reprints of this and related papers
together with illuminating commentary, see Classics in Coordination Chemistry, Part 3; Kauffman, G. B.,
Ed.; Dover: New York, 1978. Sec also Pauling, L. In Crystallography in North America; McLachlan, D..
Jr.; Glusker, J. P.. Eds.; American Crystallographic Association: New York. 1983; Chapter I.
33 Brock, P. C.; Lingafelter, C. E. J. Chem. Educ. 1980. 53. 552-554.
Crystallography 31
Crystallography
75
Table 3.6
The seven crystal systems' 1
Relations
between edges
Lengths and
and angles of
angles to be
Characteristic
System
unit cell
specified
symmetry
Triclinic
a^b^c
a, b, c
1-fold (identity or
a P y # 90°
“.Ay
inversion) symmetry
only
Monoclinic
a¥=b # c
a, b, c
2-fold axis (2 or 2)
a = y = 90 ° ±p
P
in one direction only
(y axis)
Orthorhombic
a # b c
a, b, c
2-fold axes in three
o
ers
II
II
II
8
mutually perpendicular
directions
Tetragonal 1 ’
a = b # c
a, c
4-fold axis along
a = P = y = 90°
z axis only
Trigonal 1 ’
and Hexagonal
a = b ys c "I
a = p = 90° l
y = 120° J
a, c
3-fold or 6-fold axis along
z axis only
Cubic
a = b = c
a
Four 3-fold axes each
a = p = y = 90°
inclined at 54°44’ to cell
axes (i.e. parallel to
body-diagonals of unit
cell)
Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Oxford University: Oxford, 1984.
Used with permission.
6 Certain trigonal crystals may also be referred to rhombohcdral axes, the unit cell being a
rhombohedron defined by cell edge a and interaxial angle a (590°)
cording to the symmetry of the crystal, which leads to certain relations between the
unit cell edges and angles for each system (Table 3.6). Although these relations be¬
tween cell dimensions can be said to characterize a particular crystal system, they are
not the criteria by which a crystullographcr assigns a crystal to one of the systems
during a structure determination. Rather, the assignment is made on the basis of the
crystal’s symmetry features. For example, a structure may appear, within experimen¬
tal error, to have all unit cell edges (a, b , and c) of different length and all angles
(a, /?, and y) equal to 90°, suggesting that it is orthorhombic. However, if it is found
to possess only a single two-fold axis, it must be classified as monoclinic.
Only fourteen space lattices, called Bravais lattices, are possible for the seven
crystal systems (Fig. 3.28). Designations are P (primitive), / (body-centered), F (face-
centered), 34 C (face-centered in one set of faces), and R (rhombohedral). Thus our
monoclinic structure P2Jc belongs to the monoclinic crystal system and has a primi¬
tive Bravais lattice.
The internal molecular structure of a unit cell may be complicated because a
lattice point may be occupied by a group of atoms or molecules, rather than a single
34 Further discussion of primitive, face-centered, and body-centered lattices will be found in Chapter 4.
Irreducible Representations and Character Tables
59
fl, = +1 -I -I +1
(a) (b) (c) (d)
Fig. 3.17 Effects of symmetry operations in C 3l , symmetry; translation along the >• axis;
(a) identity, £; (b) rotation about the C. axis; (e.d) reflection in cr v planes.
Irreducible The symmetry operations that belong to a particular point group constitute a mathe-
-;- matical group, which means that as a collection they exhibit certain interrelationships
Representations consistent with a set of formal criteria. An important consequence of these mathe-
and Character matical relationships is that each point group can be decomposed into symmetry
--- patterns known as irreducible representations which aid in analyzing many molecular
Tables _ and e | ectron j c properties. An appreciation for the origin and significance of these
symmetry patterns can be obtained from a qualitative development. 16
Until now we have considered symmetry operations only insofar as they affect
atoms occupying points in molecules, but it is possible to use other references as well.
For example, we might consider how a dynamic property of a molecule, such as
translation along an axis, is transformed by the symmetry operations of the point
group to which the molecule belongs. Recall the symmetry elements and coordinate
system given previously for the water molecule, which belongs to the C 2 „ point group
(Fig. 3.9). The coordinates are assigned according to the convention .that the highest
fold axis of rotation—C\ in this case—is aligned with the z axis, and the x axis is
perpendicular to the plane of the molecule. Now let translation of the molecule in
the +.y direction be represented by unit vectors on the atoms, and imagine how they
will change when undergoing the C 2 „ symmetry operations (Fig. 3.17). At the end of
each symmetry operation, the vectors will point in either the + y or the -y direction,
that is, they will show cither symmetric or antisymmetric behavior with respect to
the operation. If we represent the former with + I and the latter with -1, we can
characterize each operation with one of these labels. Identity (£) does not alter the
position of the arrows ( + 1). Rotation about the C 2 axis causes the +y vectors to
change to —y (- I). Reflection in the <r,,(.xz) plane causes +y to change to —y (- 1),
but reflection in the plane of the molecule, ofyz), results in the vectors remaining
unchanged (+1). The set of four labels ( + 1,-1, - 1, +1) generated in this simple
analysis constitutes one irreducible representation within the C 2 „ point group. It is
irreducible in the sense that it cannot be decomposed into a simpler or more funda¬
mental form. Not only does it describe the effects of C 2v operations on y translation
but on other “y-vector functions" as well, such as the p y orbital. Thus y is said to
serve as a basis function for this irreducible representation within the C 2v point group.
16 For more thorough and mathematical developments in terms of group theory, see the books listed in
Footnote 1.
32
Crystallography
83
3 - Symmetry and Group Theory
of atoms in unit cells are related by symmetry elements. If a molecule lies on a fixed
symmetry element (center of inversion, rotational axis, mirror plane) and does not
itself possess that symmetry element, there will be a superposition of images. Usually
in such a situation, the overall shape of the molecule and distribution of polar bonds
will be similar for the two (or more) disordered fragments. This is the case in triiron
dodecacarbonyl (the complete structure is illustrated in Fig. 15.7). The two possible
orientations of the “iron triangle” are superimposed to give the hexagonal arrange¬
ment observed.
Due to the superposition of two inversionally related half molecules in Fe 3 (CO) 12 ,
the determination of the arrangement of the carbonyl groups proved a difficult task.
In fact, nearly 17 years elapsed before it was successfully solved by a process of
computer simulation and modeling. 40
A current example of a solid that is disordered at room temperature is buck-
minsterfullerene. Until recently, only two allotropes of carbon were known: diamond
and graphite. However, on the basis of ions detected in a mass spectrometer with
m/e = 720, a C 60 molecule was hypothesized. 41 Now this third allotrope, C 60 , has
been isolated from the vaporization products of graphite. One heats carbon rods
in an inert atmosphere of helium or argon, and extracts the soot that forms with
benzene. 42 The proposed structure is analogous to a soccerball 43 with bonds along
the seams and carbon atoms at the junctures of the seams (Fig. 3.33). The name
“buckminsterfullerene” was suggested because of a fancied likeness to a geodesic
dome. This was quickly reduced by the waggish to “buckyball”.
The synthesis of macroscopic amounts of buckyball led to the study of many
interesting properties of this molecule which continue unabated as this book goes to
press. 44 The C ft0 molecule is nearly spherical, and while the molecules themselves
pack nicely in a cubic closest packed structure, each molecule has essentially the free
rotation of a ball bearing, and because of this disorder the structure could not be
determined at room temperature. 45 Being nearly spherical and lacking bond polarities
(a)
(b)
Fig. 3.33 Comparison of (a) a soccer ball and (b) structure of ihc C«, molecule
(Courtesy of D. E. Weeks. II. and W, G. Harter.)
40 Colton. F. A,; Troup, J. M. J. Am. Chem. Soc. 1974. 96. 4155 -4159. See also Wei, C. H.. Dahl, L. F.
J, Am. Chem. Soc. 1969, 91, 1351-1361. For further discussion of this problem together with alternative
approaches to the structural solution, sec Chapter 15.
41 Kroto, H. W.: Heath, J. R.; O’Brien. S. C.; Curl, R. F.; Smalley, R. E. Nature 1985. 313. 162.
42 Kretschmer, W.; Lamb, L. D.; Fostiropoulos, K.; Huffman. D. Nature 1990. 337. 354-358. Sec also
Allcmand. P.-M.; Koch. A.; Wudl, F,; Rubin. Y.; Diedcrich. F.; Alvarez, M. M.; Anz. S. J.; Whctten.
R. L. J. Am. Chem. Soc. 1991. 113. 1050-1051. With regard to the question raised in Chapter I con¬
cerning the earliest chemistry performed by humans, buckminsterfullerene may have been synthesized
quite early and deposited in the soot on cave walls’ In fact, burning processes hold promise for the
large-scale preparation of rullerencs.
45 Or a football in Ihc European and South American sense.
44 Note added in proof. The entire March 1992 issue of Accounts of Chemical Research is devoted to
fullcrcncs.
JJ Pure, sublimed buckminsterfullerene condenses in a regular face-centered cubic (= cubic closest packed)
structure. Complications may occur through the presence of other fullercnc impurities or crystallization
from a variety of solvents. Sec Hawkins, J. M.; Lewis. T. A.; Loren, S. D.; Meyer, A.; Heath, J. R,;
Saykally, R. J.; Hollander, F. J. J. Chem. Soc., Chem. Commun. 1991, 775-776; Guo. Y.; Karasawa, N.:
Goddard, W. A., Ill Nature 1991, 351, 464-467.
that might contribute to lattice energy, the molecules are almost completely rota-
tionally disordered, though their centroids are fixed in space. 46
The problem of rotational disorder was solved by making an osmyl derivative
of buckyball. Osmic acid, OsOj, will add across double bonds:
44 The nearly isotropic rotation is clearly shown by 13 C NMR spectroscopy: Johnson. R, D.; Yannoni,
C. S.; Dorn, H. C.; Salem. J. R.; Bcthune, D. S. Science 1992, 255, 1235-1238.
3 • Symmetry and Group Theory
(a) (b) (c)
Fig. 3.25 Central atom orbital symmetries and degeneracies for 0,, AB 6 , C 4 „ AB 5 , and
C 2 „ AB,i species.
b { (d x j_ y2 ) and an a^d.i), and the triply degenerate t 2g set is converted to e(d x: , d y; )
and b 2 {d xy ). This loss of orbital degeneracy is a characteristic result of reducing the
symmetry of a molecule.
The symmetry will be lowered even further by removing a second B atom to give
seesaw AB 4 , a C 2 „ structure (Fig. 3.25c). 29 The outcome is a complete loss of orbital
degeneracy. The character table shows the following assignments: a^p.), 6,(p x ), b 2 (p y ),
a^d.i), a 2 (d xy ), 6 2 (</„), and b 2 (d ys ). The function x 2 - y 2 is not shown explicitly in the
C 2u character table because when x 2 and y 2 are of the same symmetry, any linear
combination of the two will also have that symmetry. Note that although both the
d s j and d x j_ y2 orbitals transform as a, in this point group, they are not degenerate
because they do not transform together. It would be a worthwhile exercise to confirm
that the s, p, and d orbitals do have the symmetry properties indicated in a C 2 „
molecule. Keep in mind, in attempting such an exercise, that the signs of orbital lobes
are important.
The hybrid orbitals that are utilized by an atom in forming bonds and in accom¬
modating its own outershell nonbonding electrons will have a spatial orientation
consistent with the geometry of the molecule. Thus a tetrahedral molecule or ion,
such as CH 4 , Mn0 4 , or Cr0 4 ~, requires four hybrid orbitals on the central atom
directed toward the vertices of a tetrahedron. The general procedure for determining
what atomic orbitals can be combined to form these hybrid orbitals starts with the
recognition that the hybrid orbitals will constitute a set of basis functions for a repre¬
sentation within the point group of the molecule. This representation, which will be
' Note in Fig. 3.25 that the molecule has not only had a B atom removed in going from AB S to AB 4 but
has also been turned relative to the axis system. This was done to be consistent with the convention of
assigning the principal axis as z.
Fig. 3.26 A tetrahedral
AB 4 species with vectors
representing central atom
hybrid orbitals.
Fig. 3.27 The PtClJ' ion
with vectors representing
orbitals on platinum
suitable for out-of-plane
7r bonding.
Uses of Point Group Symmetry 73
reducible, can be obtained by considering the effect of each symmetry operation of
the point group on the hybrid orbitals. Once generated, the representation can be
factored into its irreducible components (page 62). At that point, we can obtain the
information we are seeking from the character table for the molecule, because atomic
orbitals which transform as these irreducible components will be the ones suitable
for combination into hybrid orbitals.
In carrying out the procedure for a tetrahedral species, it is convenient to let
four vectors on the central atom represent the hybrid orbitals we wish to construct
(Fig. 3.26). Derivation of the reducible representation for these vectors involves per¬
forming on them, in turn, one symmetry operation from each class in the T,, point
group. As in the analysis of vibrational modes presented earlier, only those vectors
that do not move will contribute to the representation. Thus we can determine the
character for each symmetry operation we apply by simply counting the number of
vectors that remain stationary. The result for AB 4 is the reducible representation, T r :
£ 8C 3 3C 2 6S 4 6<r d
T r : 4 1 0 0 2
Application of the reduction formula (Eq. 3.1) yields a, and i 2 as the irreducible
components. Referring to the Tj character table, we see that no orbitals are listed for
the totally symmetric a , representation; however, recall that s orbitals are always in
this class. For the i 2 case, there are two possible sets of degenerate orbitals: p x , p y ,
p. and d xy , d x: , d y; . Thus the four hybrid orbitals of interest can be constructed from
either one s and three p orbitals, to give the familiar sp 3 hybrid orbitals, or from an s
and three d orbitals to yield sd 3 hybrids. Viewed strictly as a symmetry question,
both are equally possible. To decide which mode of hybridization is most likely in a
given molecule or ion, orbital energies must be taken into account. For methane and
other cases involving carbon, the d orbitals lie too high in energy compared to the 2s
orbitals for significant mixing of the two to occur. However, for tetrahedral species
involving transition metals, such as Mn0 4 or Cr0 4 ", there are d and s orbitals
similar enough in energy that the hybrid orbitals involved in bonding may be a mix¬
ture of sp 3 and sd 3 . It is important to understand that a character table tells us only
what orbitals have the right symmetry to interact; only energy considerations can
tell us whether in fact they do.
In deriving hybrid orbitals in the foregoing example, we assumed that these orbit¬
als were directed from the central alom toward the atoms to which it is bound and
that the hybrid orbitals would overlap along the bond axes with appropriate orbitals
of the pendant atoms. In other words, these hybrid orbitals will be involved in sigma
bonds. The same basic approach that was applied to the construction of hybrid or¬
bitals for the a bonds involving a central atom also can be used to select atomic
orbitals that are available for n bonding. As an example consider the square planar
ion, PtCl 2- . Two types of 7r bonds between the platinum atom and each chlorine
atom are possible here: “out-of-plane,” with the two regions of overlap above and
below the plane of the ion, and “in-plane," having both overlap areas in the molecular
plane. Atomic orbitals on platinum that will be capable of participating in out-of-
plane 7i bonding will be perpendicular to the plane of the ion and can be represented
by the vectors shown in Fig. 3.27. As before, a reducible representation may be ob¬
tained by carrying out the operations of the appropriate point group (D 4/l ) and, for
each operation, recording the number of vectors that remain unmoved:
£ 2C 4 C 2 2C 2 2C 2 i 2 S 4 a h 2o„ 3 a d
f,: 4 0 0 -2 0 0 0 -4 2 0
2
1 • What Is Inorganic Chemistry?
What Is Inorganic Chemistry?
3
Inorganic Chemistry,
an Example
reduction of metal oxides, carbonates, and sulfides to the free metals: 1
2Cu,( 0H) 2 C0 3 + 2C -► 4Cu + 4CO, + 2H,0 (I.l)
[Copper/Bronze Age, ca. 4500-7500 years ago]
Fe 3 0 4 + 2C -> 3Fe + 2CO, (1.2)
[Iron Age, from ca. 4500-3500 years ago to present]
This was the first example of applied redox chemistry, but to this day the gain and
loss of electrons is central to inorganic chemistry. The terms oxidation, reduction , and
base (from “basic metal oxide”) are all intimately related to these early metallurgical
processes. [The term acid is derived from vinegar (L. acetum).]
Much of this early work was strictly pragmatic without any theory as we would
understand it. It was necessary to be able to identify the best and richest ores, to be
able to distinguish between superficial resemblances. The familiar properties of fool's
gold, iron pyrites, FeS,, as compared with the element gold is a well-known example.
Some minerals such as the zeolites were poorly understood. The name comes
from the Greek words for boiling (£eiv) and stone (/.tdoa) because, when heated, water
boils away from these minerals in the form of steam. How a solid stone could also
be partly liquid water was, of course, mystifying. The answer seemed to be of no
practical concern, so this question was relegated to "pure” or "basic" chemistry.
Wanting to choose a single chemical system, somehow representative of inorganic
chemistry, for our cover, we have picked a zeolite. The term may not be familiar to
you. However, one or more zeolites are almost certainly to be found in every chemical
research laboratory, in your home, and in many major industrial processes. They,
themselves, are the subject of chemical research from structural determinations to
catalysis to the inorganic chemical aspects of nutrition.
The particular zeolite illustrated on the cover is boggsite, a compound of sodium,
calcium, aluminum, silicon, hydrogen, and oxygen. It had been known for only a few
months when this book went to press. 2 Yet between the time that the earliest obser¬
vations were made on “boiling stones" (1756) and the discovery of boggsite, other
zeolites had achieved major chemical importance. If your home has a water-softening
unit, it contains a zeolite or a related compound. "Hard water” contains metal cations
that interfere with the actions of soaps and synthetic detergents. The material in the
water softener exchanges Na into the water, while removing Mg 2 . Ca 2 . and other
metal ions: 2
Mg 2 + Ca 2 + Na 4 Z -» CaMgZ + 4Na (1.3)
1 The first chemical reactions, such as the discovery of lire, were not consciously applied as “arts and
recipes" that led to chemistry. Perhaps the oldest conscious application of chemistry by humans was
that of the action of yeasts on sugar in baking and brewing, or the somewhat less well delined process
of cooking.
1 It was discovered along the Columbia River, Washington, by a group of amateur mineralogists [MolTat.
A, Science 1990. 247. 1413; Howard, D. G.; Tschernich. R. W,; Smith, J. V.; Klein. G. L. Am. Mineral.
1990, 75. 1200-1204] and the structure determined at the University of Chicago [Plulh. J. J.; Smith.
J. V. Am. Mineral. 1990, 75. 501-507],
1 The symbol Z represents all of the zeolite structure except the exchangeable Na ions.
This discovery was made in the 1850s, and it was the first ion exchange water-softening
process utilized commonly. The ion exchangers used today in home softening units
are closely related in structure and exchange properties, but are more stable for long¬
term use.
More recently, synthetic zeolites have made their appearance in a closely related,
yet quite distinct, application. Not everyone, even in areas of quite hard water, has
a water softener. In an effort to counter the negative effects of hard water, manufac¬
turers early adopted the practice of adding “ builders ” to soaps and synthetic deter¬
gents. At first these were carbonates (“washing soda”) and borates (“borax”). More
recently, these have been polyphosphates, [0 3 P0(P0 3 )„]'" _ (in = n + 3), which com-
plexed the hard water cations, that is, tied them up so that they did not interfere
with the cleaning process. The synthesis of polyphosphates and the study of their
chelating properties with Mg 2 , Ca 2 , and other cations, are other aspects of inor¬
ganic chemistry. However, phosphate is one of the three main ingredients of fertilizer, 4
and too much phosphorus leads to the eutrophication of lakes and streams. In an ef¬
fort to reduce the amount of phosphates used, manufacturers started using a synthetic
zeolite in detergents in the form of microscopic powder to adsorb these unwanted
cations. Today, this is the largest usage of zeolites on a tonnage basis.
Lest you be muttering, “So out with phosphate pollution, in with zeolite pol¬
lution!", zeolites seem to be one of the few things we can add to the ecosystem without
negative consequences. The very structures of zeolites make them thermodynamically
unstable, and they degrade readily to more stable aluminosilicates that are naturally
occurring clays. But that raises other interesting questions: If they are metastable,
why do they form, rather than their more stable decomposition products? How can
we synthesize them?
Another use of zeolites has been as “molecular sieves." This very descriptive, if
slightly misleading, name comes from a remarkable property of those zeolites: their
ability to selectively adsorb molecules on the basis of their size. A mixture of gases
may be separated according to their molecular weights (sizes) just as a coarse mixture
may be separated by a mechanical sieve. Some chemistry labs now have “exhaust-less
hoods” that selectively adsorb larger, noxious molecules, but are inert to smaller,
ubiquitous molecules such as water, dinitrogen (N,), and dioxygen (0 2 ). There are
zeolites that have a special affinity for small molecules (like H ; 0) but exclude larger
molecules. They arc thus excellent drying agents for various laboratory solvents.
Chemical Structure
of Zeolites and Other
Chemical Systems
Before we can understand how these molecular interactions can take place, we must
understand the structures of zeolites. Important for at least a century, the use of struc¬
tural information to understand chemistry is more important now than ever before.
The determination of chemical structures is a combination of careful experimental
technique and of abstract reasoning. Because we have seen pictures of “tinker-toy”
molecules all our lives in TV commercials and company logos, it is almost impossible
for us to realize that it has not been long in terms of human history since arguments
were made that such structures could not be studied (or even could not exist!) because
it was impossible to see atoms (if they existed). The crystallographer's ability to take
a crystal in hand and to determine the arrangement of invisible atoms (Fig. 1.1) is a
When you buy an ordinary ”5-10-5" fertilizer, you arc buying nitrogen (5%, expressed as N), phosphate
(I0"„, expressed as P,O s ), and potassium (5%, expressed as K,0).
56
3 • Symmetry and Group Theory
Point Groups and Molecular Symmetry
57
N \ | ,- N
/j'\
N N
°3
(a)
°\ / H
C
/ C \
H Cl
c 2
(a)
0
/ \
H H
C 2v
(b)
h/ h' h
C 3v
(C)
H 3 N % | 3 ,nh 3
'Co'
✓ | \
h.n L nh.
C 4w
(d)
I —Cl H — C = N
c ~v
(e)
C„ v
(f)
Fig. 3.13 Molecules with C„/„ C„,., and C,„. symmetry: (a) trnns- 1.2-dichloroethene;
(b) water; (c) ammonia; (d) pentaamminechlorocobalt(III) cation; (e) iodine monochloride;
(0 hydrogen cyanide.
molecules such as H 2 0 (Fig. 3.13b), NH 3 (Fig. 3.13c), and pentaammine-
chlorocobalt(III) cation (Fig. 3.13d) 13 possess C nv symmetry.
Question: If the planes of the phenyl rings in triphenylphosphine (Fig.
3.12c) were parallel to the three-fold axis (i.e., if the intersection of their planes
coincided with that axis), what would the point group of triphenylphosphine
be? 14
The point group C xv is a special case for linear molecules such as IC1
and HCN (Fig. 3.13e, f), because it is possible to rotate the molecule about its
principal axis to any desired degree and to draw an infinite number of vertical
planes.
Dihedral groups. Molecules possessing nC 2 axes perpendicular to the prin¬
cipal axis (C„) belong to the dihedral groups. If there are no mirror planes,
as in tris(ethylenediamine)cobalt(III) cation (Fig. 3.14a), the molecule belongs
to the D„ group. Addition of a mirror plane perpendicular to the principal
axis results in the D nh groups which include molecules such as phosphorus
(b)
C '\ / Cl
Pt
/ \
Cl Cl
04
(C)
H 3 N . | - NH 3
'Co
/I \
HjN NH 3
04
<d)
F—Be —F 0 = 0 H —H
0 -
(e)
0-
(f)
0-
(g)
Fig. 3.14 Molecules wi!h D,„ D„,,. and D„,, symmetry: (a) tris(ethylcnediamine)cobalt(III) calion. (b) phosphorus
pentafluoride, (c) tetrachloroplatinate(II) anion, (d) nwix-tetraamminedichlorocobalt(III) cation, (e) beryllium difluoridc.
(f) dioxygen, (g) dihydrogen.
13 The pentaamminechlorocobalt(III) cation has “idealized" C 4c symmetry, that is. the random orienta¬
tion of the hydrogen atoms resulting from free rotation of the ammonia molecules is often ignored for
simplicity.
14 C,„.
m
(a)
(b)
I
I
I
ch 3v | ,CH 3
Sn'
/ I \
(OC)„Fe^ | / Fe(CO) 4
Sn_
(OC) 4 Fe^ j >e(CO) 4
/ S "\
CH, | CH,
(c)
(d>
D u
(=>
Fig. 3.15 Molecules with D,„, symmetry: (a) staggered ethane, (b) octalluorozirconale(lV) anion, (c) bis|dimethyltinbis(/x-
tetracarbonyliron)]tin. (d) octasulfur, (e) dibenzenechromium.
pentafluoride (Fig. 3.14b), the tetrachloroplatinate(II) anion (Fig. 3.14c), the
rrans-tetraamminedichlorocobalt(III) cation (Fig. 3.14d), and eclipsed ferro¬
cene (Fig. 3.8a).
Linear molecules with a center of symmetry, such as BeF ; and all of the
X 2 molecules (Fig. 3.14e-g), possess a horizontal mirror plane and an infinite
number of C 2 axes perpendicular to the principal axis and thus have D^. h
symmetry.
If the mirror planes contain the principal C„ axis and bisect the angle
formed between adjacent C 2 axes, they are termed dihedral planes. Molecules
such as the staggered conformer of ferrocene (Fig. 3.8b), the staggered con-
former of ethane (Fig. 3.15a), the square-antiprismatic octafluorozirconalc(lV)
anion (Fig. 3.15b), bis[dimethyltinbis(/i-tetracarbonyliron)]tin (Fig. 3.15c),
octasulfur (Fig. 3.15d), and the staggered conformer of dibenzenechromium
(Fig. 3.15e) that contain such dihedral planes belong to the D lul groups.
Question: If the triphenylphosphine molecule were planar, what would be
its point group? 15
A flowchart for assigning point symmetry. The symmetry elements, and
the rules and procedures for their use in determining the symmetry of mole¬
cules, can be formalized in a flow chart such as that shown in Fig. 3.16. It
contains all of the point groups discussed above (enclosed in square boxes) as
well as a few others not commonly encountered. In addition, the symmetries
assigned above “by inspection” may be derived in a more systematic way by
the use of this diagram.
4 1 • What Is Ino
Fig. 1.1 The structure of the synthetic zeolite ZSM-5: (a) microscopic crystals; (b) an
electron micrograph of the area marked in (a); (c) the crystal structure of ZSM-5 related to
the electron micrograph. [Courtesy of J. M. Thomas. Royal Institute of Chemistry.]
triumph of abstract reasoning. The determination of the structures of molecules and
extended structures is fundamental to the understanding of inorganic chemistry. It
is not possible to think of modern inorganic chemistry in terms of simple equations
such as Eq. 1.1 to 1.3: A three-dimensional view of the arrangement of atoms is
necessary.
One of the unifying factors in the determination of chemical structures has been
the use of symmetry and group theory. One has only to look at the structure of
boggsite to see that it is highly symmetrical, but symmetry is even more basic to
chemistry than that. Symmetry aids the inorganic chemist in applying a variety of
methods for the determination of structures. Symmetry is even more fundamental:
The very universe seems to hinge upon concepts of symmetry.
The solid-state chemist and solid-state physicist have also developed other tech¬
niques for examining and manipulating solids and surfaces. Of particular interest
recently is a technique known as scanning tunneling microscopy (STM) which allows
us to see and even to move individual atoms. 5 The atoms are imaged and moved by
electrostatic means (Fig. 1.2). 6 Although chemistry is portrayed, correctly, in terms
of single atoms or groups of atoms, it is practiced in terms of moles (6 x 10 23 atoms),
millimoles (6 x 10 20 atoms), or even nanomoles (6 x 10 14 atoms), seldom less. But
perhaps the horizon of atom-by-atom chemistry is not far away.
5 Some people object to the use of the verb "to sec" in this context, correctly arguing that since the wave¬
length of visible light is much greater than the order of magnitude of molecules, the latter cannot be
seen directly, but must be electronically imaged. True, but every year hundreds of millions or people
"see" the Super Bowl on TV! What's the difference?
6 Eigler, D. M.; Schweizer. E. K. Nature 1990, 344, 524-526.
What Is Inorganic Chemistry? 5
Fig. 1.2 Scanning tunneling micrographs of the movement of xenon atoms adsorbed on a
nickel surface. The nickel atoms are not imaged. Each letter is 5 nm from lop to bottom.
[Courtesy of D. M. Eigler. IBM.]
Chemical Reactivity Although it is not possible for the chemist to absolutely control the movement of
individual atoms or molecules in zeolite structures, the nature of the structure itsell
results in channels that direct the molecular motions (Fig. 1.3). Furthermore, the sizes
and shapes of the channels determine which molecules can form most readily, and
which can leave readily. A molecule that cannot leave (Fig. 1.4) is apt to react further.
This may have important consequences: A catalyst (ZSM-5) that is structurally re¬
lated to boggsite is used in the alkylation of toluene by methanol to form puru-xylene.
The methanol can provide methyl groups to make all three (ortho, meta, and para)
Fig. 1.3 Stereoview of the structure of boggsite. Note the channels running in the a
direction. For help in seeing stereoviews, see Appendix H. [From Pluth, J. J.; Smith. J. V.
Am. Mineral. 1990. 75, 501-507. Reproduced with permission.]
54
3-Symmetry and Group Theory
Point Group:
d Molecular Symmetry
55
Fig. 3.10 Point groups and molecules of high symmetry: (a) icosahedron, (b) the ion. (c) octahedron, (d) sulfur
hexafluoride, (e) hexacyanocobaltate(III) anion, (f) tetrahedron, (g) ammonium cation, and (h) tetrafluoroborate anion.
0
II
p
0
II
N
Cl p
/ F
c —C
Fig. 3.11 Molecules wilh
low symmetry: (a) phos-
\
\
/ \"h
phoryl bromide chloride
Cl Br
Cl
F Cl
fluoride. C t ; (b) nitrosyl
C >
c s
c,.
chloride, C v ; (c) the anti
conformer of R.S- 1,2-
(a)
(b)
(c)
dichloro-1,2-difluoroethane, C f .
a. C ( .—Molecules of this symmetry have only the symmetry element E, equi¬
valent to a one-fold rotational axis. Common, simple chiral molecules with
an asymmetric center have only this symmetry (Fig. 3.1 la).
b. C s .—In addition to the symmetry element E, which all molecules possess,
these molecules contain a plane of symmetry. Thus although they have
very low symmetry, they are not chiral (Fig. 3.11b).
c. Cj.—These molecules have only a center of inversion in addition to the iden¬
tity element. The anti conformations of £,S-l,2-dichloro-l,2-difluoroethane
(Fig. 3.11c) and R,S-l,2-dimethyl-l,2-diphenyldiphosphine disulfide (Fig.
3.3) have C, symmetry.
Groups with an n-J'old rotational axis, C„. After the obvious groups with
high or low symmetries have been eliminated by inspection, the remaining
point groups should be assigned by looking for characteristic symmetry ele¬
ments, such as an n-fold axis of rotation. C„. Molecules containing only one
such axis, like gaud ie-H,0, (Fig. 3.12a), tris(2-aminoethoxo)cobalt(I[I) (Fig.
3.12b), or triphenylphosphine in its most stable conformation (Fig. 3.12c), have
C„ symmetry.
If, in addition to the C„ axis, there is a horizontal plane perpendicular
to that axis, the molecule is said to have C„ h symmetry. An example of this
relatively unimportant group is frmis-dichloroethene (Fig. 3.13a). If there are
n mirror planes containing the rotation axis, C„, the planes are designated
vertical planes, and the molecule has C„„ symmetry. Many simple inorganic
planes, a center of symmetry, as well as six S, 0 and ten S 6 axes collinear
with the Cj and C 3 axes.
b. Octahedral, O h . —The octahedron (Fig. 3.10c) is commonly encountered in
both coordination compounds and higher valence nonmetal compounds
(Fig. 3. lOd, e). It has four C 3 axes, three C 4 axes, six C, axes, four axes,
three o h planes, six a d planes, and a center of symmetry. In addition, there
are three C 2 and three S 4 axes that coincide with the C 4 axes.
c. Tetrahedral, T d . —Tetrahedral carbon is fundamental to organic chemistry,
and many simple inorganic molecules and ions have tetrahedral symmetry
as well (Fig. 3.IOg, h). A tetrahedron (Fig. 3.lOf) has four C 3 axes, three C 2
axes, six mirror planes, and three S 4 improper rotational axes.
Groups with low symmetry. There are three groups of low symmetry that
possess only one or two symmetry elements.
Fig. 3.12 Molecules with C„ symmeiry: (a) gauche-H 2 0 2 , (b) tris(2-iiminoclhoxo)cobali(l!i).
(c) triphenylphosphine. The C 2 axis in (a) lies in the plane of the page, the C, axes in (b)
and (c) are perpendicular to the page.
7 8
3 • Symmetry
id Group Theory
Crystallography
79
Fig. 3.31 A screw-axis (2,)
operation. The molecule
moves a distance a/2 along
the x axis while undergoing
a C\ rotation. Note that
the chirality of the molecule
does not change.
C
An equivalence table of the 32 crystallographic point groups in the two systems
is given in Table 3.7. 36 The complete set of symmetry operations for a crystal is
known as its space group. There are 230 possible space groups for three-dimensional
crystals. Note that whereas there is an infinity of possible point groups, the number
of space groups, despite the addition of translational symmetry, is rigorously limited
to 230. For each structure worked out by the crystallographer, an assignment to one
of these possible space groups is essential. Fortunately, this task is made easier by
Toblo 3.7
Comparison of Schoenflios and International notations for the thirty-two crystallographic point groups"
Hormann-Mauguin Schoenflios Hermann-Mauguin Schoenflios
Triclinic 1
c,
Trigonal
3
c 3
I
c,
3
C„
32
D>
Monoclinic 2
C 2
3m
c iv
rn
C,
3m
D u
2/m
Hexagonal
6
c
Orthorhombic 222
D,
6
C ik
mm2
6/m
Cu
mmm
Du
622
0
6ntm
c 6 .
Tetragonal 4
c 4
6m 2
Du
4
s 4
6/mmm
D b „
4/m
CM
422
Cubic
23
T
4 mm
c 4 „
m3
T,
42 m
432
0
4/mmm
d 4 >
43m
T<
m3m
0,
" Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Oxford University: Oxford, 1984; p 48, Used with permission.
36 For a flow chart of the 32 crystallographic point groups in the International system that is analogous
to Fig. 3.16 for the Schoenfiics system, see Brcneman. G. L. J. Chem. Educ. 1987. 64, 216.
the conspicuous evidence left by elements of translational symmetry; All forms of
translational symmetry, including lattice centering, create empty spaces in the diffrac¬
tion pattern called “systematic" absences.
We can now complete our answer to the question, “What information is con¬
veyed when we read that the crystal structure of a substance is monoclinic P2,/c?”
The structure belongs to the monoclinic crystal system and has a primitive Bravais
lattice. It also possesses a two-fold screw axis and a glide plane perpendicular to it.
The existence of these two elements of symmetry requires that there also be a center
of inversion. The latter is not specifically included in the space group notation as it
would be redundant.
Figure 3.32 illustrates the unit cell of 0.s 3 (C0) 9 (/xrCC (l F,), a compound that
crystallizes in the monoclinic space group P2Jc. In addition to the identity element,
it exhibits two two-fold screw axes, two c glide planes, and an inversion center. As an
aid in the identification of these elements, the four molecules of the unit cell are
Fig. 3.32 (a) the molecular structure of Os,(CO).,(p. 1 -CC,,F 5 ), consisting of a triangle of
osmium atoms capped by the CCsFj group to form a triangular pyramid, (b) Stereoview
of unit cell of Os,(CO).,(p.-CC 6 F 5 ). The compound crystalizes in the monoclinic space
group P2Jc with cell parameters a = 1212.7(10) pm, b = 938.6(5) pm, c = 1829.8(15) pm.
P = 98.92(6)’. The center of inversion is indicated by a dot in the center of the unit cell, and
the two two-fold screw axes are perpendicular to the plane of the paper and are marked
with the symbol I . Two glide planes perpendicular to the screw axes in the xy plane (parallel
with the plane of the paper) are not indicated but arc found at distances of one-fourth and
three-fourths unit cell depth. Note that a, b, and c do not correspond exactly to x, y, and z
because one of the three angles of a monoclinic structure is unequal to 90°. The fluorine
atoms have been omitted for clarity. [Modified from Hadj-Bagheri, N.; Strickland, D. S.;
Wilson, S. R.; Shapley, J. R. J. Organomet. Chem. 1991, 410, 231-239; Courtesy of S. R.
Wilson and C. L. Stern.]
Fig. 1.4 meM-Xylene (left) and para-xylene (right) in a channel in the synthetic zeolite
catalyst ZSM-5. [From Thomas, J. M. Angew. Chem. Ini. Ed. Engl. 1988, 27, 1673-1691.
Reproduced with permission.]
xylene isomers. The "linear” para isomer leaves readily (Fig. 1.5), but the angular
ortho and meta isomers do not. They may react further, that is. rearrange, and if
para-xylene forms, it may then leave. 7
In a related process, ZSM-5 may be used to convert methanol into a high-
octane gasoline. Petroleum-poor countries like New Zealand and South Africa are
currently using this process to produce gasoline. If the production of para-xylene
and gasoline sounds too much like “organic chemistry” for the introduction to an
inorganic textbook, it must be pointed out that there is a large branch of chemistry,
Fig. 1.5 Illustration of shape selectivity. [Csicsery, S. M. Chem. Brit. 1985, 21. 473-477.
Reproduced with permission.]
So, in lieu of "chemical tweezers" (STM and related apparatus) we claim to effect particular stereo¬
chemical syntheses by using specially shaped zeolites. But it is stated that these specially shaped zeolites
arc also synthesized—without “chemical tweezers". How? The answer is not as difficult as it may seem.
What Is Inorganic Chemistry?
7
called “organometailic chemistry,” that deals with an area intermediate between
inorganic and organic chemistry and broadly overlapping both. Both organic and
inorganic chemists work in organometailic chemistry, with the broad generalization
often being that the products are “organic” and mostly of interest to the organic
chemist, and the intermediates and catalysts are of more interest to the inorganic
chemist.
Zeolites may be used in purely inorganic catalysis, however. One reaction that
may be used to reduce air pollution from mixed nitrogen oxides, NO, in the indus¬
trial production of nitric acid is catalytic reduction by ammonia over zeolitic catalysts:
6NO + 4.xNH 3 -♦ (3 + 2x)N 2 + 6xH,0 (1.4)
The seriously polluting nitrogen oxides are thus reduced to two harmless molecules.
The strong bond energies of the dinitrogen molecule and the water molecule are the
driving forces; the zeolitic catalyst, in the ideal case, provides the pathway without
being changed in the process.
A related catalytic removal of NO from automobile exhaust may come about
from the reaction:
-jj^q Cull)/Cu(ll)
zeolite
n 2 + o 2
(1.5)
using a Cu(I)/Cu(II) exchanged zeolite as a redox catalyst. 8
To return to the problem of the general invisibility of atoms, how does the chem¬
ist follow the course of a reaction if the molecules cannot be imaged? One way is to
use spectroscopy. Thus the conversion of methanol, first to dimethyl ether, then to
the higher aliphatic and aromatic compounds found in gasoline, can be followed by
nuclear magnetic resonance (NMR) spectroscopy (Fig. 1.6). As the reaction proceeds,
the concentration of the methanol (as measured by the intensity of the NMR peak at
(550 ppm) steadily decreases. The first product, dimethyl ether (<560 ppm), increases at
first and then decreases as the aliphatic and aromatic products eventually predominate.
So why did we pick boggsitc for the cover? Is it “the most important” inorganic com¬
pound known? Certainly not! It is currently known from only one locality and in
the form of extremely small crystal fragments. 9 It is unlikely that it occurs anywhere
on earth in sufficient quantities to be commercially important. Yet its discovery adds
to our knowledge of the structural possibilities of zeolites and the conditions under
which they form. And if we know enough about the structure of a material, we can
usually synthesize it if we try hard enough. The synthesis of zeolites has progressed,
though it must be admitted that there is much yet to be understood in the process.
Boggsite is enough like ZSM-5. yet different, that it has attracted considerable atten¬
tion. There is currently a massive effort in the chemical industry to try to synthesize
this very interesting material. 10 It may become an important industrial catalyst. Then
again, it may not—only time will answer that question.
Iwamolo. M.: Yahiro, H.'.Tanda. K.: Mizuno, N.; Mine. Y.; Kagawa, S.7. Pltvs. Chan. 1991. 95. 3727-
9 Pari of the difficulty in determining the crystal structure was in picking out a suitable crystal fragment
from the matrix in which it was imbedded. Only one was found, 0.07 x 0.08 x 0.16 mm in size. See
Footnote Z
10 Alper, J. Science 1990, 248, 1190-1191.
52
3 -Symmetry and Group Theory
Improper Rotation, S„ A C„ axis is often called a “proper" rotational axis and the rotation about it a
“proper” rotation. An improper rotation may be visualized as occurring in two steps:
rotation by 360% followed by reflection across a plane perpendicular to the rota¬
tional axis. Neither the axis of rotation nor the mirror plane need be true symmetry
elements that can stand alone. For example, we have seen that SiF 4 has C 3 axes but
no C 4 axis. Nevertheless, it has three S 4 axes, one through each pair of opposite faces
of the cube below:
Consider the trans configuration of dinitrogen tetrafluoride. If we perform a C 2
operation followed by a a h operation, we will have a successful S 2 operation. Note,
however, that the same result could have been obtained by an inversion operation:
Point Groups and Molecular Sy
itry
53
h i
Fig. 3.9 Coordinate system
and symmetry elements of
the water molecule.
Thus S 2 ‘ s equivalent to i. Confirm this to your satisfaction with frans-N 2 F 2 , which
contains a center of symmetry and thus must have a two-fold improper axis of rota¬
tion. Note that the SiF 4 molecule, although it possesses true C 2 axes, does not have
a center of symmetry, and thus cannot have an S 2 axis. Furthermore, S, is equivalent
to a because, as we have seen, C L = E and therefore the second step, reflection, yields
Point Groups and
Molecular
Symmetry
wT ;l V
If we analyze the symmetry elements of a molecule such as water (Fig. 3.9), we find
that it has one C 2 axis, two o v planes, and of course E. This set of four symmetry
operations generated by these elements is said to form a symmetry group , or point
group. In the case of the water molecule, this set of four symmetry elements charac¬
terizes the point group C 2 „. The assignment of a point group to a molecule is both
a very simple labeling of a molecule, a shorthand description as it were, 12 and a
useful aid for probing the properties of the molecule.
The assignment of molecules to the appropriate point groups can be done on
a purely formal, mathematical basis. Alternatively, most chemists quickly learn to
classify molecules into the common point groups by inspection. The following ap¬
proach is a combination of the two.
Groups with very high symmetry. These point groups may be defined by the
large number of characteristic symmetry elements, but most readers will rec¬
ognize them immediately as Platonic solids of high symmetry,
a. Icosahedral, I h .—The icosahedron (Fig. 3.10a), typified by the B 12 HJ 2 ion
(Fig. 3.10b), has six C 5 axes, ten C 3 axes, fifteen C 2 axes, fifteen mirror
11 The chief reason for pointing out these relationships is for systematization: All symmetry operations
can be included in C„ and S„. Taken in the order in which they were introduced, tr = S,; / = S 2 ; E = C,.
Thus when we say that chiral molecules arc those without improper axes of rotation, the possibility of
planes of symmetry and inversion centers has been included.
12 For example, the chemist may speak of the T t symmetry of the [FeCIJ ion, the D 4h symmetry of
the [PtCIJ 2 ion, and the C 2o symmetry of the TeCI 4 molecule as alternative ways of describing the
tetrahedral [Fed]" ion, the square planar [PtCI 4 ] 2 ' ion, or the structure of the TeCI 4 molecule which
is sometimes called a “butterfly” molecule.
What Is Inorganic Chemistry?
9
As was pointed out at the beginning of the chapter, many other subjects could
have been chosen for the cover: the new high-temperature superconductors, metal
cluster compounds, an optically active inorganic molecule, a bioinorganic enzyme
(see how far inorganic chemistry has come from the days when it meant “non-living"?),
or a crystal of photographer’s hypo. Indeed, all of these have been used on the covers
of recent inorganic textbooks (one reason why we chose something different), and all
of them are as apprqpriate on the one hand, and as limited in scope on the other,
as boggsite. They will all be discussed in the following chapters. If there is one thought
that you should take away with you after reading this chapter, and eventually this
book, it is the amazing diversity of inorganic chemistry. It deals with 109 elements,
each unique.
It is thus impossible in a single chapter to do more than scratch the surface of
inorganic chemistry: Structure, reactivity, catalysis, thermodynamic stability, sym¬
metry, experimental techniques; gas-phase, solution, and solid-state chemistry; they
are all part of the process. However, it is hoped that some idea of the scope of the
subject may have been formed. The following chapters in this book attempt to pro¬
vide the reader with sufficient basic knowledge of the structure and reactivity of inor¬
ganic systems to ensure a more comprehensive understanding. 11
11 For a recent review of zeolite ealalysis. sec Thomas. J. M. Sci. Amer. 1992. 266 (4), 112-1 IS.
so
3 • Symmetry and Group Theory
Fig. 3.6 The two-fold rotational axis in cis-dinitrogen difluoride.
Figs. 3.2-3.7 that have C 3 axes? 7 ) Note that it also has three C 2 axes, one through
each of the carbonyl groups in the triangular plane that is perpendicular to the C 3
axis.
In contrast, tungsten hexacarbonyl (Fig. 3.7c) has a four-fold axis. In fact, it has
three C 4 axes: (1) one running from top to bottom, (2) one running from left to right,
(3) one running from front to back. (In addition, it has other rotation axes. Can you
find them'? 8 )
A molecule may possess higher order rotational axes. Consider the eclipsed form
of the molecule ferrocene (Fig. 3.8a), which has a C 5 axis through the iron atom and
perpendicular to the cyclopentadienyl rings. Now consider the staggered form of
m y
III V
1
1
c #/
—o
%
, N ,
0 = C —Fc'
o
III
n
1
<-
r-
n
III
0
l/ J \t
1
/ |
/ C V
c'l
■ c
1
1
' III 0
o
1
/> III
/ 0
/
/
/
(a)
(b)
(c)
Fig. 3.7 Additional molecules having n-fold axes: (a) ammonia, (b) pcntacarbonyliron.
(c) hcxacarbonyltungsten.
7 AnsH ’ er: S'F. PF„ [CoF 6 ] j -, NH Jf BjNjH ftl Fc<CO) 3 , and W(CO)„ all have one or more C 3 axes.
8 Answer: W(CO) 6 has three C 4 axes, four C 3 axes (through the octahedral faces), and six C 2 axes (through
the octahedral edges).
Symmetry Elements and Symmetry Operati
51
-Fe- C,
C 5
Fig. 3.8 Molecules con¬
taining live-fold rotational
axes: (a) eclipsed ferrocene,
side and lop view; (b)
staggered ferrocene, side
and top view. Each
molecule has five C ; axes,
only one of which is shown.
Upon rotation about the
C 2 axis, the atoms
interchange: I — 1 ', etc.
(a)
(b)
ferrocene (Fig. 3.8b). Does it have a five-fold rotational axis? 9 Next consider borazine
(Fig. 3.2c). Does it have a C 6 axis? 10
Many molecules have more than one C„ axis. For example, staggered ferrocene
has five C 2 axes, one of which lies in the plane of the paper. Eclipsed ferrocene also
has five C, axes, though they are different from the ones in the staggered conformcr
(Fig. 3.8). In those cases in which more than one rotational axis is present, the one
of highest order is termed the principal axis and is usually the z axis. Planes that
contain the principal axis are termed vertical planes, <r„, and a mirror plane perpen¬
dicular to the principal axis is called a horizontal plane, a h . For example, borazine
(Fig. 3.2c) has three vertical planes (one is shown) and one horizontal plane (the plane
of the molecule).
Identity, E We have seen above that a C, operation (rotation by 360°) results in the same mole¬
cule that we started with. It is therefore an identity operation. The identity operation
is denoted by £. It might appear that such an operation would be unimportant inas¬
much as it would accomplish nothing. Nevertheless, it is included for mathematical
completeness, and some useful relationships can be constructed using it. For example,
we have seen that two consecutive C, operations about the same axis result in
identity. We may therefore write: C 2 x C, = E, and likewise: C 3 x C, x C, = E.
These may also be expressed as C; = E and C, = E.
" Answer: Staggered ferrocene has a five-fold axis, since rotation through 72“ causes both the top and
bottom rings to match their former positions even though they arc staggered with respect to each other.
However, eclipsed ferrocene also has a mirror plane perpendicular to its C 3 axis, that is absent in the
staggered form. For the most stable conformation of ferrocene and related compounds, see Chapter 15,
Answer: No. because rotation of only 60° places a boron atom where formerly there was a nitrogen
atom and vice versa. Borazine has a C 3 axis. However, benzene has a true C 6 axis.
The Structure
of the Atom
A tomic structure is fundamental to inorganic chemistry, perhaps more so even
than organic chemistry because of the variety of elements and their electron config¬
urations that must be dealt with. It will be assumed that readers will have brought
with them from earlier courses some knowledge of quantum mechanical concepts
such as the wave equation, the particle-in-a-box, and atomic spectroscopy.
The Hydrogen When the Schrodingcr equation is solved for the hydrogen atom, it is found that
there are three characteristic quantum numbers n, /, and m, (as expected for a three-
om — dimensional system). The allowed values for these quantum numbers and their rela¬
tion to the physical system will be discussed below, but for now they may be taken
as a set of three integers specifying a particular situation. Each solution found for a
dilTcrcnt set of h, /. and in, is called an eigenfunction and represents an orbital in the
hydrogen atom.
In order to plot the complete wave functions, one would in general require a
four-dimensional graph with coordinates for each of the three spatial dimensions
(x,y, z\ or r, 0, f ) and a fourth value, the wave function.
In order to circumvent this problem and also to make it easier to visualize the
actual distribution of electrons within the atom, it is common to break down the
wave function, l P, into three parts, each of which is a function of but a single vari¬
able. It is most convenient to use polar coordinates, so one obtains
T(r, 0, ) = R(r) ■ 0(0) • <1>((/>) (2.1)
where R(r ) gives the dependence of 'P upon distance from the nucleus and 0 and <1>
give the angular dependence.
10
The Hydrogen Atom
1 1
The Radial Wave The radial functions for the first three orbitals 1 in the hydrogen atom are
Function, R
n= 1,1
= 0, m, = 0
R = 2 (—)
w
3/2
g - Zr/ao
Is orbital
n = 2,1
= 0, in, — 0
\ ( 7 \3/2 / y\
)(_) ( 2 \ e -ZriZn 0
' \ a o) \ a o J
2s orbital
n = 2,1
1, m, - 0
"(zk
(zy i2 Zf e - Zrl2BB
1 \ a oJ a 0
2/j orbital
where Z is the nuclear charge, e is the base of natural logarithms, and a„ is the radius
of the first Bohr orbit. According to the Bohr theory, this was an immutable radius,
but in wave mechanics it is simply the “most probable" radius for the electron to be
located. Its value. 52.9 pm, is determined by a 0 = h 2 /Au 2 me 2 , where li is Planck’s
constant and m and e are the mass and charge of the electron, respectively. In hydro¬
gen, Z = 1. but similar orbitals may be constructed where Z > I for other elements.
For many-electron atoms, exact solutions of the wave equation are impossible to
obtain, and these “hydrogen-like" orbitals are often used as a first approximation. 2
Although the radial functions may appear formidable, the important aspects
may be made apparent by grouping the constants. For a given atom, Z will be
constant and may be combined with the other constants, resulting in considerable
simplification:
n = I, / = 0, m, = 0 R = K U e- Zr, “ n Is orbital
/i = 2./ = 0,m ( = 0 - Ka.^2 - 2s orbital
n = 2, R = K 2p re- Zr ' 2ao 2p orbital
The most apparent feature of the radial wave functions is that they all repre¬
sent an exponential “decay”, and that for n = 2 the decay is slower than for n = I.
This may be generalized for all radial functions: They decay as e‘ Zr/n "". For this rea¬
son. the radius of the various orbitals (actually, the most probable radius) increases
with increasing n. A second feature is the presence of a node in the 2s radial function.
At r = 2aJZ, R = 0 and the value of the radial function changes from positive to
negative. Again, this may be generalized: s orbitals have n - 1 nodes, p orbitals have
n — 2 nodes, etc. The radial functions for the hydrogen Is, 2s, and 2 p orbitals are
shown in Fig. 2.1.
Because we are principally interested in the probability of finding electrons at
various points in space, we shall be more concerned with the squares of the radial
functions than with the functions themselves. It is the square of the wave function
1 The complete wave functions in terms of the quantum numbers n and / arc given by Pauling, L. The
Nature of the Chemical Bond: Cornell University: Ithaca, NY, I960 (n = 1-6) and Porterfield, W. W.
Inorganic Chemistry: A Unified Approach ; Addison-Wesley: Reading, MA. 1984 (n = 1 - 3 ).
3 The use of hydrogen-like orbitals for multielcctron atoms neglects electron-electron repulsion, and this
may often be a serious oversimplification (see pages 20-23).
48
3- Symmetry and Group Theory
(a) (b) (c) <d)
Fig. 3.2 Molecules with and without mirror planes: (a) dichlorophosphine oxide has a single
mirror plane (the sheet of the paper) that reflects the two chlorine atoms into each other;
(b) water has two mirror planes: one which bisects the H—O H angle, the other that lies in
the plane defined by the H —O—H angle; (c) borazine has Four mirror planes: three are of
the vertical type shown, and the fourth is the plane of the ring; (d) methylsulfinyl chloride has
no plane of symmetry. Note that the lone pair is not experimentally observed in the latter
compound, but has been added to rationalize the molecular structure.
tions, the resultant symmetry is referred to as point symmetry. 2 Molecules may have
symmetry axes, a center of symmetry, and mirror planes as symmetry elements. The
reader will already be familiar with the mirror plane used to help determine whether
a molecule is optically active. 3
The Mirror Plane, or Most flowers, cut gems, pairs of gloves and shoes, and simple molecules have a plane
of symmetry. A single hand, a quartz crystal, an optically active molecule, and certain
cats at certain times 4 do not possess such a plane. The symmetry element is a mirror
plane, and the symmetry operation is the reflection of the molecule in the mirror
plane. Some examples of molecules with and without mirror planes are shown in
Fig. 3.2.
Center of Symmetry, i A molecule has a center of symmetry if it is possible to move in a straight line from
every atom in it through a single point to an identical atom at the same distance
on the other side of the center (Fig. 3.3). The center of symmetry is also called an
inversion center. We have encountered inversion about a center with regard to atomic
orbitals, gerade and ungerade , in the previous chapter. Symmetry species of irreducible
representations (see page 59) can also be g or u if a molecule has a center of symmetry.
Of the three most common geometries encountered in inorganic chemistry (Fig. 3.4),
one has a center of symmetry and two do not.
2 Point symmetry of individual molecules is to be contrasted with the translational symmetry of crystals,
to be discussed later in this chapter.
3 One definition of chirality is that the molecule be nonsuperimposablc on its mirror image. An equivalent
criterion is that it not possess an improper axis of rotation (page 52). The absence of a mirror plane does
not insure optical activity because a molecule may have no mirror plane, yet may possess an improper
rotational axis. We can, however, be sure that the molecule with a mirror plane will be optically inactive.
4 Huheey, J. E. J. Chem. Educ. 1986, 63, 598-600.
c " 3 V'
S
-p--ch 3
C fi H,
Fig. 3.3 The center of
symmetry of 1,2-dimethyl-
1 , 2 -diphenyldiphosphine
disulfide.
Symmetry Elements and Symmetry Operations
49
Rotational Axis, C„
Fig. 3.4 Examples of (a)
tetrahedral, (b) trigonal
bipyramidal, and (c)
octahedral geometries. Only
(c) has a center of
symmetry, /.
If rotation of a molecule by 360% results in an indistinguishable configuration, the
molecule is said to have an it-fold rotational axis. Consider irans-dinilrogen difluoride
(Fig. 3.5). If we construct an axis perpendicular to the plane of the paper and midway
between the nitrogen atoms, we can rotate the molecule by 180° and obtain an iden¬
tical configuration. Rotation by 180° is thus a symmetry operation.
The axis about which the rotation takes place is the symmetry element. In this
case fraris-dinitrogen difluoride is said to have a two-fold rotational axis. Note that if
the operation is performed twice, all atoms are back in their initial positions. -
Now consider cis-dinitrogen difluoride (Fig. 3.6). It possesses no axis perpen¬
dicular to the plane of the molecule that allows rotation (other than the trivial 360°
one) and qualifies as a symmetry element. However, it is possible to draw an axis
that lies in the plane of the molecule equidistant between the two nitrogen atoms
and also equidistant between the two fluorine atoms. This is also a two-fold axis.
Rotational axes are denoted by the symbol C„, representing the n-fold axis. Thus
cis- and rrans-dinitrogen difluoride each have a C 2 axis.
Note that SiF 4 (Fig. 3.4a) has-a three-fold axis, C 3 . In fact, it has four of them,
each lying along an Si—F bond. The single three-fold axis in ammonia may be some¬
what less obvious (Fig. 3.7a). It lies on an imaginary line running through the center
of the lone pair and equidistant from the three hydrogen atoms. If all the bond angles
change, as from 107° in NH 3 to 102° in NF 3 , does the symmetry change? 6 Note that
iron pentacarbonyl (Fig. 3.7b) also has a C 3 axis. (Are there any other molecules in
Fig. 3.5 The two-fold rotational axis in trans-dinitrogen difluoride. The two-fold axis is
perpendicular to the plane of the paper and denoted by the symbol ( .
J Some confusion can arise here. We can perform symmetry operations only if the atoms arc indistinguish¬
able from each other. Yet we must attach invisible labels to the atoms in order to recognize when the
atoms are back to their initial positions.
6 Answer: No, both NH 3 and NF 3 have the same symmetry elements. Follow-up question: What if the
bond angles become 120°?
Answer: When the bond angles become 120°, as in BF 3 , the planar molecule has a horizontal plane of
symmetry as well as three new C 2 axes lying in that plane.
2 - The Structure of the Atom
Fig. 2.2 Radial density
functions for n = 2 for the
hydrogen atom. These
functions give the relative
electron density (e pm 3 ) as
a function of distance from
the nucleus. They were »> '
prepared by squaring the «
wave functions given in
Fig. 2.1.
200 400 600
(pm)
too 400 600 800
(pm)
200 400 600
(pm)
Fiq. 2.1 Radial part of the hydrogen eigenfunctions for n - 1.2. 3. [From Herzberg. G.
Atomic Spectra and Atomic Structure . Dover: New York. 1944. Reproduced w,th
permission.)
that provides the electron density or the probability of finding an electron at a point
in space There are two useful ways of doing this. The simplest is merely to square
the functions plotted in Fig. 2.1. We could therefore square the numbers on ‘he ordi-
nates and plot the same curves except that the negative values become positive when
squared (Fig. 2.2). While this seems very simple, it provides us with the relative e e
iron density as a function of the radius. It is important to remember that for s orhituls,
the maximum electron density is at the nucleus ; all other orbitals have zero electron
^"'’'A'more common way of looking at the problem is to consider the atomi to be
composed of “layers" much like an onion and to examine the probability of finding
the electron in the “layer” which extends from rtor + dr, as shown in Fig. -3. T
volume of the thin shell may be considered to be dV. Now the volume of the sphere is
dV = 4nr J dr ^
R 2 dV = 4nr 2 R 2 dr (2-4)
Consider the radial portion of the wave function for the Is orbital as plotted in
Fig. 2.1. When it is squared and multiplied by 47tr 2 , we obtain the probability function
The Hydrogen At
1 3
O Fig. 2.3 Volume of a shell
of thickness dr.
shown in Fig. 2.4. The essential features of this function may be obtained qualitatively
as follows:
At r = 0, 4 nr 2 R 2 = 0; hence the value of the function at the nucleus must be
zero. 3
At large values of r. R approaches zero rapidly and hence 4nr 2 R 2 must ap¬
proach zero.
Fig. 2.4 Radial probability functions for n = 1, 2, 3 for the hydrogen atom. The function gives the probability of finding
the electron in a spherical shell of thickness dr at a distance r from the nucleus. [From Herzberg, Ci. Atomic Spectra and
Atomic Structure; Dover: New York. 1944. Reproduced with permission.]
3 Note that the mathematical function goes to zero because the volume of the incremental shell, iIV, goes
to zero at r = 0. As we have seen, however, there is electron density at the nucleus for s orbitals.
Chap ter
Symmetry Elements
and Symmetry
Operations 1
Symmetry and
Group Theory
Symmetry is a common phenomenon in the world around us. If Nature abhors a
vacuum, it certainly seems to love symmetry! It is difficult to overestimate the im¬
portance of symmetry in many aspects of science, not only chemistry. Just as the
principle known as Occam's razor suggests that the simplest explanation for an ob¬
servation is scientifically the best, so it is true that other things being equal, frequently
the most symmetrical molecular structure is the “preferable" one. More important,
the methods of analysis of symmetry allow simplified treatment of complex problems
related to molecular structure.
Mathematical symmetry is a little more restrictive than is the meaning of the word in
everyday usage. For example, some might say that flowers, diamonds, butterflies, snail
shells, and paisley ties (Fig. 3.1) are all highly symmetrical because of the harmony
and attractiveness of their forms and proportions, but the pattern of a paisley tic
is not “balanced"; in mathematical language, it lacks symmetry elements. A flower,
crystal, or molecule is said to have symmetry if it has two or more orientations in
space that are indistinguishable, and the criteria forjudging these are based on sym¬
metry elements and symmetry operations.
A symmetry operation moves a molecule about an axis, a point, or a plane (the
symmetry element) into a position indistinguishable from the original position. If
there is a point in space that remains unchanged under all of the symmetry opera-
For three very good introductory articles, sec Orchin, M.; Jalfii. H. H. J. Client. Ethic. 1970. 47. 246. 372,
510. Among the best books arc Cotton, F. A. Chemical Applications of Group Theory. 3rd cd.: Wiley:
New York, 1990. Douglas, B. E.; Hollingsworth, C. A. Symmetry in Bantling ami Spectra: An Intraduc-
linn: Academic: Orlando, 1985. Hargitlai, t.; Hargittai, M. Symmetry through the Eyes of a Chemist:
VCH: New York. 1987. Harris, D. C.; Bertolucci, M. D. Symmetry anil Spectroscopy: An Introduction to
Vibrational and Electronic Spectroscopy. Dover: New York. 1989. Orchin, M.; JafTC. H. H. Symmetry.
Orbitals, and Spectra ; Wiley: New York. 1971. See also Wells. A. F. Structural Inorganic Chemistry. 5th
cd.; Clarendon: Oxford, 1984; Chapter 2.
■ag
S
BpillmS
Symmetry Elements and Symmetry Operations
Fig. 3.1 The shapes and patterns of some pleasing designs found in nature or constructed as artifacts: (a) the flowt
black-eyed Susan. Rudbeckia Itirta: (b) the flower, stem, and leaves of the black-eyed Susan; (c) a red eft, Noloplu
viridescens ; (d) a cut diamond: (c) a paisley tie; (f) a snail shell, Cepea nemoralis ; (g) a monarch butterfly,’ Danatts
plexipptts: (h) a suspension bridge. Which are truly symmetrical?
14 2 • The Structure ofthe Atom
In between, r and R both have finite values, so there is a maximum in the piot
of probability (4nr 2 R 2 ) as a function of r. This maximum occurs at r = a 0 , the
value of the Bohr radius.
Similar probability functions (including the factor 4nr 2 ) for the 2s, 2p, 3s, 3p, and
3d orbitals are also shown in Fig. 2.4. Note that although the radial function for the
2s orbital is both positive (r < 2 a Q /Z) and negative (r > 2 a 0 /Z), the probability func¬
tion is everywhere positive (as of course it must be to have any physical meaning) as
a result of the squaring operation.
The presence of a node in the wave function indicates a point in space at which
the probability of finding the electron has gone to zero. This raises the interesting
question, "How does the electron get from one side of the node to the other if it can
never be found exactly at the node?” This is not a valid question as posed, since it
presupposes our macroscopically prejudiced view that the electron is a particle. If
we consider the electron to be a standing wave, no problem arises because it simul¬
taneously exists on both sides of a node. Consider a vibrating string on an instrument
such as a guitar. If the string is stopped at the twelfth fret the note will go up one
octave because the wavelength has been shortened by one-half. Although it is experi¬
mentally difficult (a finger is not an infinitesimally small point!), it is possible to sound
the same note on either half of the octave-stopped string. This vibration can be con¬
tinuous through the node at the fret. In fact, on the open string, overtones occur at
the higher harmonics such that nodes occur at various points along the string. Nodes
are quite common to wave behavior, and conceptual problems arise only when we
try to think of the electron as a “hard" particle with a definite position.
Does the presence of one or more nodes and maxima have any chemical effect?
The answer depends upon the aspect of bonding in which we arc interested. We shall
see later that covalent bonding depends critically upon the overlap of orbitals. Con¬
ceivably, if one atom had a maximum in its radial wave function overlapping with
a region with a node (minimum) in the wave function of a second atom, the overlap
would be poor. 4 However, in every case in which careful calculations have been made,
it has been found that the nodes lie too close to the nucleus to affect the bonding
appreciably.
The presence of nodes and small "subnodal maxima" does have a profound ef¬
fect on the energy of electrons in different orbitals. An electron in an orbital with
these subnodal maxima (particularly s orbitals with higher values of n) are said to
be penetrating, that is. they have considerable electron density in the region of the
nucleus. This is the fundamental reason for the ordering of the energy levels in poly-
electronic atoms: l.v. 2s, 2 p. 3s, 3 p. etc. (see pages 20-22).
Angular Wave The angular part of the wave function determines the shape of the electron cloud
Functions and varies depending upon the type of orbital involved (s, p. d. or /) and its orien¬
tation in space. However, for a given type of orbital, such as s or p., the angular
wave function is independent of the principal quantum number or energy levei. Some
4 Specifically, the overlap integral is the integral, J ‘l' A 'l'„i/r. of the two wave functions (Sec Eq. 5.31). At
the node the product will go to zero, and it will have small values in the region of the node.
The Hydrogen Ato
15
Fig. 2.5 Angular part of
the wave function for
hydrogen-like s orbitals
(left) and p orbitals (right).
Only two dimensions of
the three-dimensional
function have been shown.
Fig. 2.6 Angular
probability function for
hydrogen-like p orbitals.
Only two dimensions of
the three-dimensional
function have been shown.
typical angular functions are
/ = 0, m, = 0 0O = [ 1/( 47t)] 1/2 s orbital
/ = 1, in, = 0 0<D = [3/(47r)] 1/2 cos 9 p. orbital
/ = 2, m, = 0 ©d> = [5/(167£)] 1/2 (3 cos 2 0 - 1) d. 2 orbital
The angular functions for the s and p. orbital are illustrated in Fig. 2.5. For an ,v
orbital, 0 is independent of angle and is of constant value. Hence this graph is
circular or, more properly, in three dimensions—spherical. For the p. orbital we
obtain two tangent spheres. The p x and p y orbitals are identical in shape but are
oriented along the .x and y axes, respectively. We shall defer extensive treatment of
the d orbitals (Chapter 11) and / orbitals (Chapter 14) until bond formation in coor¬
dination compounds is discussed, simply noting here that the basic angular function
for d orbitals is four-lobed and that for / orbitals is six-lobed (see Fig. 2.9).
We are most interested in the probability of finding an electron, and so we shall
wish to examine the function 0 : 2 since it corresponds to the angular part of T 2 .
When the angular functions are squared. dilTcrcni orbitals change in different ways.
For an s orbital squaring causes no change in shape since the function is everywhere
the same: thus another sphere is obtained. For both p and d orbitals, however, the
plot lends to become more elongated (sec Fig. 2.6).
The meaning of Figs. 2.5 and 2.6 is easily misinterpreted. Neither one has any
direct physical meaning. Both arc graphs of mathematical functions, just as Figs.
2.2 and 2.4 are. Both may be used to obtain information about the probable distri¬
bution of electrons, but neither may in any way be regarded as a "picture" of an orbi¬
tal. It is an unfortunate fact that fuzzy drawings of Figs. 2.5 or 2.6 are often presented
as "orbitals". Now one can define an orbital in any way one wishes, corresponding to
y, P 2 , R. R 2 , 0<b. or © 2 2 , but it should be realized that Figs. 2.2, 2.4, 2.5, and 2.6
are mathematical functions and drawing them fuzzily does not represent an atom.
Chemists tend to think in terms of electron clouds, and hence T 2 probably gives the
best intuitive “picture" of an orbital. Methods of showing the total probability of
finding an electron including both radial and angular probabilities arc shown in Figs.
2-7—2.9. Although electron density may be shown either by shading (Fig. 2.7) or by
contours of equal electron density (Figs. 2.8 and 2.9), only the latter method is quan¬
titatively accurate.
44
2 • The Structure of the Atom
What does this drawing represent? Does it suggest that the electron has Protobility
of being found at the nucleus? We know that the probab.l.ty of find.ng a 2 p electron
the nucleus is zero. Is this a paradox? Explain.
2.9 The angular part of the wave function for the d xy orbital is ^^-sin 2 0 sin 20. Show
that this expression corresponds to the d xy orbital. The d : , is actually a simplified way
of representing the d^.^ orbital. Show that this corresponds to the angular function,
3cos 2 0 — 1.
2.10 Consider the following possible electron arrangements for a p i configuration:
Which of these represents the ground state? Which are excited states? Which are impos¬
sible states? In which configuration would exchange energy be maximized: In whic
configuration would coulombic repulsion be maximized?
2.11 The stabilization of a half-filled d shell is even more pronounced than that of the p
subshell. Why?
2.12 Discuss the following question: Does an orbital exist if there is not an electron in it?
2.13 The Pauli exclusion principle forbids certain combinations of m, and m in determining
the term symbols for the states of the nitrogen atom. Consider an excited nitrogen atom
in which the electronic configuration is l S 2 2s 2 2p 2 3p‘. What states now are poss.ble?
2.14 Write out electronic configurations for free atoms of the following elements. Determine
the number of unpaired electrons in the ground state.
B
N Mg
As Lu
2.15 Determine all of the term symbols for the following free atoms. Choose the ground state
term in each case.
B
N Mg V As Lu
2.16 Write out the electronic configurations for the following ions. Determine the number
of unpaired electrons in the ground state.
Ti 3 + Mn !t Cu 2+ Pd 2t Gd 3
2.17 Determine the ground state term symbols for each of the ions in Problem 2.16.
2.18 Clearly distinguish the following aspects of the structure of an atom and sketch the ap¬
propriate function for Is, 2s, 2p, 3s, and 3p orbitals.
a. radial wave function
b. radial probability function
c. angular wave function
d. angular probability function
2.19 Using Slater's rules, calculate Z for the following electrons:
a. a 3p electron in P
b. a 4s electron in Co
c. a 3d electron in Mn
d. a valence electron in Mg
Compare the values of Z thus obtained with those of Clementi and Raimondi.
, C
Problems 4 5
2.20 Which has the higher first ionization energy:
Li or Cs? F or Br? Sc or Cu? Cu or Pt?
2.21 Plot the total ionization energies of Al" + as a function of n from n = 1 to n = 8. Explain
the source of discontinuity in your curve.
2.22 a. Calculate the third ionization energy of lithium. [Hint: This requires no approxima¬
tions or assumptions.]
b. Calculate the first and second ionization energies of lithium using Slater's rules.
c. Calculate the first and second ionization energies of lithium using the rules of Clementi
and Raimondi.
2.23 Which is larger:
K + or Cs? La 3 or Lu 3 ? Cl or Br? Ca 2 or Zn 2+ ? Cs or Fr?
2.24 Which has the highest electron affinity:
Li or Cs? Li or F? Cs or F? F or Cl? Cl or Br?
O or S? S or Se?
2.25 Note that Table 2.6 lists several molecules that have much higher electron affinities
than fluorine (328.0 kJ mol' 1 ) or chlorine (349.0 kJ mol - ')• For example, consider PlF,
(772 kJ mol"'). How can a molecule composed of six fluorine atoms and a metal (to be
sure, not a very electropositive one) have a higher affinity for electrons than a fluorine
atom?
2.26 The electronegativity of a group is determined by many other factors than simply its
electron affinity. Nevertheless, look at the values in Table 2.6 and predict what the most
electronegative groups are.
2.27 Which of the halogens, X 2 , would you expect to be most likely to form a^cation, X ?
Discrete X ions are not known in chemical compounds, but X 2 , X 2 , and XJ are known.
Why should the latter be more stable than X?
1 6
2 -The Structure of the Atom
The Hydrogen Atom
17
Fig. 2.7 (a) Pictorial representation of the electron density in a hydrogen-like 2p orbital
compared with (b) the electron density contours Tor the hydrogen-like 2p ; orbital of carbon.
Contour values are relative to the electron density maximum. The xy plane is a nodal
surface. The signs (+ and -) refer to those of the original wave function. [The contour
diagram is from Ogryzlo. E. A.; Porter, G. B. J. Cliem. Educ. 1963, 40, 258. Reproduced
with permission.]
plane
Fig. 2.8 The electron
density contours for the
hydrogen-like 3p. orbital
of carbon. Contour values
are relative to the electron
density maximum. The xy
plane and a sphere of
radius 52 pm (dashed line)
are nodal surfaces. The
signs (+ and —) refer to
those of the original wave
function. [The contour
diagram is from Ogryzlo.
E. A.; Porter, G. B.
J. Client. Educ. 1963. 40,
256-261. Reproduced with
permission.]
y
(O
Fig. 2.9 Angular wave functions of s, p, d, and / orbitals illustrating gerade and ungerade symmetry: (a) s orbital, gerade;
(b) p orbital, ungerade; (c) pictorial representation of symmetry of p orbital; (d) d x> orbital, gerade; (e) pictorial representation
of symmetry of d orbital; (f) d orbital, gerade; (g) / r , orbital, ungerade.
Symmetry of Orbitals
Since is termed an angular probability function, the question may prop¬
erly be asked what its true meaning is. if not a "picture" of electron distribution. Like
any other graph, it simply plots the value of a function (0 2 <D 2 ) versus the variable
(0 or 9, <p). If one chooses an angle 0, the probability that the electron will be found
in that direction (summed over all distances) is proportional to the magnitude of the
vector connecting the origin with the functional plot at that angle.
In Fig. 2.9 are shown sketches of the angular parts of the wave functions for s , p,
d. and / orbitals. The signs in the lobes represent the sign of the wave function in
those directions. For example, in the p. orbital, for 0 = 90", cos 9 — 0 and for
90° < 9 < 270°, cos 9 is negative. The signs of the wave functions are very important
when considering the overlap of two bonding orbitals. It is customary to speak of
the symmetry of orbitals as gerade or ungerade. These German words meaning even
and uneven refer to the operation shown in the sketches—inversion about the center.
If on moving from any point A to the equivalent point B on the opposite side of the
42
2 • The Stru clui
f the Atom
T able 2.5
Electron affinities of the
elements (kJ mol ')"
z
Element
Value
Z
Element
Value
1
H
72.775
34
Se -♦ Se 1
194.980
2
He
0
Se'- -> Se 2
-410
3
Li
59.63
35
Br
324.6
4
Be
0
36
Kr
0
5
B
26.7
37
Rb
46.887
6
C
153.89
38
Sr
0
7
N -
■ N‘-
7
39
Y
29.6
N'- —
—> N 2
-673
40
Zr
41.1
N 2 - —
-> N 3 -
-1070
41
Nb
86.1
8
O ->
o‘-
140.986
42
Mo
71.9
O'- —
O 2-
-744'’
43
Tc
53
9
F
328.0
44
Ru
101.3
10
Ne
0
45
Rh
109.7
11
Na
52.871
46
Pd
53.7
12
Mg
0
47
Ag
125.6
13
A1
42.5
48
Cd
0
14
Si
133.6
49
In
28.9
15
P-►
pi-
72.02
50
Sn
107.3“
pi- -
-■ P 2 -
-468”
51
Sb
103.2
P 2 ” -
-> P 3 ‘
— 886
52
Te
190.16
16
S ->
S‘-
200.42
53
I
295.18
S'- -
-+ S J -
-456
54
Xe
0
17
Cl
349.0
55
Cs
45.509
18
Ar
0
56
Ba
0
19
K
48.387
57
La
48
20
Ca
0
58-
■7ILn
50
21
Sc
18.1
72
Ilf
0
22
Ti
7.62
73
Ta
31.06
23
V
50.6
74
W
78.63
24
Cr
64.26
75
Re
14.4?
25
Mn
0
76
Os
106.1
26
Fe
15.7
77
Ir
151.0
27
Co
63.7
78
Pt
205.3
28
Ni
111.5
79
Au
222.76
29
Cu
118.4
80
Hg
0
30
Zn
0
81
TI
19.2
31
Ga
28.9
82
Pb
35.1
32
Ge
119.0“
83
Bi
91.2
33
As - 1
. As'-
78
84
Po
183.3
As 1- —
— As 2 ‘
-435
85
At
270.1
As 2 ’ —
— As 3 '
-802
86
Rn
0
° Unless otherwise noted, all values are from Hotop, H.; Lineberger, W. C. J. Phys. Chem. Ref.
Data 1985, 14. 731.
Pearson. R. G. Inorg. Chem. 1991, 30. 2856-2858.
c Miller, T. M.; Miller, A. E. S.; Lineberger. W. C. Phys. Rev. A 1986, 33. 3558-3559.
Problems
43
Table 2.6
Electron affinities of
Experimental
Experimental
molecules'’ Molecule
(kJ mol -1 )
Molecule
(kJ mol -1 )
ch 3
752
OCN
340
C=CH
285
SiH 3
140
C 5 H 5
165
ph 2
150
c 6 h 5
100
PtF 5
630
c 6 h 5 ch 2
85
PtF 6
770
CN
365
SH
223
N 3
266
so 2
107
NH 2
74
S0 3
160
NO
232
SCN
205
NOj
220
SFj
290
no 3
375
SF 6
101
o 2
42
Cl 2
230
o 3
203
Br 2
240
OH
176
TeF 5
430
OCHj
155
TeF 6
320
O-f-QH,
184
h
240
0-neo-C 5 H,,
183
wf 6
330
OC 6 H 3
220
uf 2 o 2
325
o 2 h
104
UF 6
540
“ Lias, S. G.; Bartmcss, J. E.; Licbman, J. F.; Holmes, J. L.;
Levin, R. D.: Mallard, W. G. J. Phys. Chem. Ref. Data 1988,
17. Supplement I, 1-86.
Uncertainty is approximately ±20 except for numbers given
to three significant digits.
Problems _ 2.1 Calculate the r value in pm at which a radial node will appear for the 2s orbital of the
hydrogen atom.
2.2 Which quantum numbers reveal information about the shape, energy, orientation, and
size of orbitals?
2.3 How many orbitals arc possible for tt = 4? Which of these may be described as yeratlel
2.4 How many radial nodes do 3s, 4 p. 3 d and 5/ orbitals exhibit? How many angular nodes?
2.5 Make a photocopy of Fig. 2.8. Draw two lines, one along the r axis, and one at a 45°
angle away from the e axis. Along one of these lines measure the distance from the origin
(nucleus) to each contour line and plot the value of the contour line at that distance (r).
Do this for all contours on both lines. Compare your drawing with Fig. 2.4.
2.6 Determine the maximum number of electrons that can exist in a completely filled n = 5
level. Give four possible quantum numbers for a 5/electron of the hydrogen atom.
2.7 The signs of the unsquared wave functions arc usually shown in plots of the squared
functions. Why do you think this practice exists?
2.8 Sometimes 2p orbitals are drawn as shown below:
18
2 • The Struct'
f the At'
Energies of Orbitals
center the sign of the wave function does not change, the orbital is said to be geracle.
The s orbital is a trivial case in which the sign of the angular wave function is every¬
where the same. The d orbitals (only two of which are shown here) are also gerade.
The p orbitals, however, are unsymmetrical with respect to inversion and the sign
changes on going from A to B; hence the symmetry is ungerade. Likewise, / orbitals
are ungerade. Another way of referring to the symmetry properties of these orbitals
is to say that s and d orbitals have a center of symmetry, and that p and f orbitals
do not. In addition to symmetry with respect to inversion about the center, orbitals
have other symmetry properties with respect to other symmetry operations. These
will be discussed in Chapter 3.
It should be noted that most textbooks, including this one. generally portray
the symmetry of orbitals as in Fig. 2.9a-g with wave functions plotted and the signs
marked. However, an exceedingly common practice in the original literature of both
inorganic and organic chemistry is to indicate the signs of the wave functions by the
shading of stylized orbitals. Fig. 2.9c indicates the symmetry of a p orbital and 2.9e
a d orbital by this convention.
Attention should be called to a rather confusing practice that chemists com¬
monly use. In Figs. 2.7 and 2.8 it will be noted that small plus and minus signs appear.
Although the figure refers to the probability of finding the electron and thus must be
everywhere positive, the signs + and — refer to the sign of the original wave func¬
tion, 4, in these regions of space. In Fig. 2.8, for example, in addition to the inversion
resulting from the ungerade p orbital, there is a second node (actually a spherical nodal
surface) at a distance of 6 aJZ resulting from the radial wave function. Although this
practice may seem confusing, it is useful and hence has been accepted. The 4' 2 plot
is useful in attempting to visualize the physical “picture” of the atom, but the sign
of 4' is important with respect to bonding. 5
The energy levels of the hydrogen atom are found to be determined solely by the
principal quantum number, and their relationship is the same as found for a Bohr
atom:
2k 1 me 4
~li T h r ~
(2.5)
where in is the mass of the electron, e is the electronic charge, n is the principal quan¬
tum number, and U is Planck's constant. Quantization of energy and angular momen¬
tum were introduced as assumptions by Bohr, but they follow naturally from the
wave treatment. The quantum number n may have any positive, integral value from
one to infinity:
n = 1, 2, 3, 4,.... co
The lowest (most negative) energy corresponds to the minimum value of n (n = I)
and the energies increase (become less negative) with increasing n until the contin¬
uum is reached (it = co). Here the electron is no longer bound to the atom and thus
is no longer quantized, but may have any amount of kinetic energy.
The allowed values of / range from zero to n — I:
/ = 0, 1,2, 3,..., it- I
5 See Orchin. M.; Jaflc, H. H.; The Importance of Antihondiny Orbitals', Houghlon Mifflin: Boston. 1967;
pp 5-9, for a good discussion of this point.
The Hydrogen At.
19
The quantum number / is a measure of the orbital angular momentum of the elec¬
tron and determines the “shape" of the orbital. The types of orbitals are designated
by the letters s,p,d,f,g --corresponding to the values of / = 0. 1.2. 3. 4.
The first four letters originate in spectroscopic notation (see page 26) and the re¬
mainder follow alphabetically. In the previous section we have seen the various an¬
gular wave functions and the resulting distribution of electrons. The nature of the
angular wave function is determined by the value of the quantum number /.
The number of equivalent ways that orbitals can be oriented in space is equal to
2 1 + 1. In the absence of an electric or magnetic field these orientations are degenerate ;
that is, they are identical in energy. Consider, for example, the p orbital. It is possible
to have a p orbital in which the maximum electron density lies on the z-axis and the
xv-plane is a nodal plane. Equivalent orientations have the maximum electron density
along the x- or y-axis. Application of a magnetic field splits the degeneracy of the
set of three p orbitals. The magnetic quantum number , m,, is related to the component
of angular momentum along a chosen axis—for example, the z axis—and determines
the orientation of the orbital in space. Values of m, range from — / to +/:
m,= -1,-1+ I,..., -1,0, +1, +2. +1
Thus for / - 1. m, = - 1, 0, +1, and there are three p orbitals possible, p x , p y , and
p.. Similarly, for / = 2 [d orbitals), m, = -2, - I, 0, + 1, +2, and for / = 3 (/’ orbitals),
m,= -3, -2, -1,0, +1, +2, +3. 6
It is an interesting fact that just as the single s orbital is spherically symmetric,
the summation of electron density of a set of three p orbitals, five d orbitals, or seven
/ orbitals is also spherical (Unsold's theorem). Thus, although it might appear as
though an atom such as neon with a filled set of s and p orbitals would have a "lumpy"
electron cloud, the total probability distribution is perfectly spherical.
From the above rules we may obtain the allowed values of n. I, and m,. We have
seen previously (page 10) that a set of particular values for these three quantum num¬
bers determines an eigenfunction or orbital for the hydrogen atom. The possible
orbitals are therefore
n = 1
1 = 0
m, = 0
Is orbital
»! =2
1 = 0
m, = 0
2s orbital
n = 2
/= 1
nt, = - 1, 0, + 1
2 p lx , y , :) orbitals
n = 3
1 = 0
m, = 0
3s orbital
n = 3
1= 1
m, = - 1, 0, + 1
Iptx.y.:) orbitals
n = 3
1=2
m, = — 2, — 1,0, 4-1, +2
2d,.i x ,_ y> xy x . y;) orbitals 7
n = 4
1 = 0
m, = 0
4s orbital
We can
now summarize the relation between the quantum numbers n, /, and
and the physical pictures of electron distribution in orbitals by a few simple rules. It
6 Although the p. and </.: orbitals correspond to m = 0. there is no similar one-to-one correspondence for
the other orbitals and olhcrvalucs of m. The functions arc complex for m, ±1,2 and must be formed into new,
linear combinations for the real p and d orbitals. Sec Moore, W. J. Physical Chemistry'. Prentice-Hall:
Englewood Cliffs. NJ. 1972; p 640; Atkins. P. W. Physical Chemistry, 4th cd.; Freeman: San Francisco, 1990;
p 362; Figgis, B. N. Introduction to Ligand Fields', Wiley: New York. 1966; pp 9-15.
7 These orbitals are sketched and discussed further in Chapter 11.
40
2 • The Structure of the At'
from the trend along the 3 cl series mentioned above. Although this trend is not re¬
sponsible for the effect (as discussed previously), it does give an indication. It thus
appears that as the atomic number goes up, and hence as Z increases, the energy
levels approach more closely to those in a hydrogen atom, namely, all levels having
the same principal quantum number are degenerate and lie below those of the next
quantum number. Now the effective nuclear charge in the ion increases markedly
because of the net ionic charge and the reduced shielding. It is not unreasonable to
suppose, then, that the formation of a dipositive ion accomplishes more than the
gradual changes of the entire transition series were able to—that is, lowering the 3 d
level so far below the 4s that the repulsion energies are overcome and the total energy
is minimized if the 3 d level, rather than the 4s, is occupied. This tendency towards
hydrogen-like orbitals is dramatic with increasing effective nuclear charge. For ex¬
ample, core electrons are scarcely differentiated energetically according to type of
orbital—they closely approach the hydrogenic degeneracy. 29
Electron Affinity Electron affinity is conventionally defined as the energy released when an electron
is added to the valence shell of an atom. Unfortunately, this is in contradiction to
the universal thermodynamic convention that enthalpies of exothermic reactions
shall be assigned negative signs. Since it seems impossible to overthrow the electron
affinity convention at this late date without undue confusion, one can adopt one of
two viewpoints to minimize confusion. One is to let the electron affinities of the most
active nonmetals be positive, even though in thermodynamic calculations the en¬
thalpies are negative:
F + e~ —» F~ EA = +328 kJ mol -1 AW =-328 kJ mol''
A slightly different approach is to consider the electron affinities of the atoms to be
the same as the ionization energies of the anions. Now the positive electron affinity
corresponds to an endothermic reaction:
F~ -» F + e" IE =+328 kJ mol" 1 AH = -328 kJ mol -1
This second approach has the added benefit of calling attention to the very close
relationship between electron affinity and ionization potential. In fact, when the ion¬
ization energies and electron affinities of atoms are plotted, a smooth curve results
and the function may be described rather accurately by the quadratic formula: 30
E = aq + fiq 1 (2.17)
where E is the total energy of the ion (S jE or Z EA ) and q is the ionic charge. See
Fig. 2.13.
It may readily be seen that whereas the acceptance of electrons by active non-
metals is initially exothermic, the atoms become “saturated" relatively quickly, the
energy reaches a minimum, and further addition of electrons is endothermic. In fact.
» For further discussion of the relative energies and reasons, sec Keller. R. N. J. Chem. Educ. 1952. 39. 289;
Hochstrasser. R. M. Ibid. 1965, 42. 154; and Karplus. M.; Porter. R. N. Atoms and Molecules'. Benjamin:
New York. 1970; pp 269-271.
10 Actually, a more accurate expression is a polynomial of the type £ = iq + ftq 2 + yq ! + dq~. The con¬
stants y and 5, however, are small, and Eq. 2.17 is a good approximation. See Iczkowski, R. P.; Margrave,
J. L. J. Am. Chem. Soc. 1961, 83, 3547.
Fig. 2.13 Ionization
energy-electron affinity
curves for oxygen, fluorine,
neon, and chlorine.
Oxidation state
for dinegative ions such as O 2 " and S 2 , the total electron affinity is negative: that is,
their enthalpy of formation is positive. Such ions cannot exist except through stabil¬
ization by their environment, either in a crystal lattice or by solvation in solution.
As might be supposed, electron affinity trends in the periodic chart parallel (hose
of ionization energies (Table 2.5). Elements with large ionization energies lend to
have large electron affinities as weli. There are a few notable exceptions, however.
Fluorine has a lower electron affinity than chlorine, and this apparent anomaly is
even more pronounced for N/P and O/S. It is a result of the smaller size of the first-
row elements and consequent greater electron-electron repulsion in them. Although
they initially have greater tendencies to accept electrons (note the slopes of the lines
as they pass through the origin, or neutral atom, in Fig. 2.13), they quickly become
“saturated” as the electron-electron repulsion rapidly dominates (the flat, bottom
portion of the curve). 31 Fewer data are available for neutral molecules (Table 2.6).
Free radicals made up of electronegative atoms, such as CN, NO,, N0 3 , SF 5 , etc.,
have the expected high electron affinities, and we shall see later that they are among
the most electronegative of groups. As a group the metal hexafluorides have the
highest electron affinities with PtF s having an electron affinity more than double
that of any single atom (see Problem 2.25).
For further discussion of electron affinities, together with useful charts and graphs, see Chen, E. C. M.
Wentworth, W. E. J. Chem. Educ. 1975, 52. 486.
20
2 • The Structure of the Atom
should be emphasized that these rules are no substitute for a thorough understanding
of the previous discussion, but merely serve as handy guides to recall some of the
relations.
Within the hydrogen atom, the lower the value of n, the more stable will be
the orbital. For the hydrogen atom, the energy depends only upon n; for atoms
with more than one electron the quantum number / is important as well.
The type of orbital is determined by the / quantum number:
1 = 0, s orbitals
/ = 1, p orbitals
1 = 2, d orbitals
/ = 3, / orbitals
/ = 4, g orbitals, etc.
There are 2/ + 1 orbitals of each type, that is, one s, three p, five d, and seven
/ orbitals, etc., per set. This is also equal to the number of values that m, may
assume for a given value of /, since m, determines the orientation of orbitals,
and obviously the number of orbitals must be equal to the number of ways
in which they are oriented.
There are n types of orbitals in the nth energy level. Tor example, the third
energy level has s, p, and d orbitals.
There are n - / - 1 nodes in the radial distribution functions of all orbitals,
for example, the 3s orbital has two nodes, the 4 d orbitals each have one.
There are I nodal surfaces in the angular distributional functions of all orbitals,
for example, s orbitals have none, d orbitals have two.
The Polyelectronic
With the exception of Unsold’s theorem, above, everything discussed thus far has dealt
only with the neutral hydrogen atom, the only atom for which the Schrodinger equation
can be solved exactly. This treatment can be extended readily to one-electron ions
isoelectronic with hydrogen, such as He , Li 2+ , and Be 3 f , by using the appropriate
value of the nuclear charge, Z. The next simplest atom, helium, consists of a nucleus
and two electrons. We thus have three interactions: the attraction of electron I Tor
the nucleus, the attraction of electron 2 for the nucleus, and the repulsion between
electrons 1 and 2. This is an example of the classic three-body problem in physics
and cannot be solved exactly. We can, however, approximate a solution to a high
degree of accuracy using successive approximations. For simple atoms such as helium
this is not too difficult, but for heavier atoms the number of interactions which must
be considered rises at an alarming rate and the calculations become extremely la¬
borious. A number of methods of approximation have been used, but we shall not
explore them here beyond describing in conceptual terms one of the more accurate
methods. It is referred to as the Hartree-Fock method, after the men who developed
it, or as the self-consistent field (SCF) method. It consists of (1) assuming a reasonable
wave function for each of the electrons in an atom except one, (2) calculating the
effect which the field of the nucleus and the remainder of the electrons exert on the
chosen electron, and (3) calculating a wave function for the last electron, including
the effects of the field of the other electrons. A different electron is then chosen, and
using the field resulting from the other electrons (including the contribution from the
improved wave function of the formerly chosen electron), an improved wave function
for the second electron is calculated. This process is continued until the wave func¬
tions for all of the electrons have been improved, and the cycle is then started over to
improve further the wave function of the first electron in terms of the field resulting
from the improved wave functions of the other electrons. The cycle is repeated as
many times as necessary until a negligible change takes place in improving the wave
functions. At this point it may be said that the wave functions are self-consistent and
are a reasonably accurate description of the atom.
Such calculations indicate that orbitals in atoms other than hydrogen do not
differ in any radical way from the hydrogen orbitals previously discussed. The prin¬
cipal difference lies in the consequence of the increased nuclear charge—all the or¬
bitals are somewhat contracted. It is common to call such orbitals which have been
adjusted by an appropriate nuclear charge hydrogen-like orbitals. Within a given
major energy level it is found that the energy of these orbitals increases in the order
s < p < d < f. For the higher energy levels these differences are sufficiently pro¬
nounced that a staggering of orbitals may result, such as 6s < 5d = 4/ < 6 p, etc. The
energy of a given orbital depends on the nuclear charge (atomic number) and differ¬
ent types of orbitals are affected to different degrees. Thus there is no single ordering
of energies of orbitals which will be universally correct for all elements. 8 Neverthe¬
less, the order Is < 2s < 2p < 3s < 2p < 4s < 3d < 4 p < 5s < 4d < 5p < 6s < 5 d ^
4/ < 6 p < 7s < 6d < 5/ is found to be extremely useful. This complete order is correct
for no single element; yet, paradoxically, with respect to placement of the outermost or
valence electron, it is remarkably accurate for all elements. For example, the valence
electron in potassium must choose between the 3d and 4s orbitals, and as predicted
by this series it is found in the 4s orbital. The above ordering should be assumed to
be only a rough guide to the filling of energy levels (sec “The aufbau principle", page
23. In many cases the orbitals are very similar in energy and small changes in atomic
structure can invert two levels and change the order of filling. Nevertheless, the above
series is a useful guide to the building up of electronic structure if it is realized that
exceptions may occur. A useful mnemonic diagram was suggested by Moeller 9 (Fig.
2.10). To recall the order of filling, merely follow the arrows and the numbers from
one orbital to the next.
Electron Spin and
the Pauli Principle
As expected from our experience with a particle in a box, three quantum numbers
are necessary to describe the spatial distribution of electrons in atoms. To describe
an electron in an atom completely, a fourth quantum number, m s , called the spin
quantum number must be specified, This is because every electron has associated
with it a magnetic moment which is quantized in one of two possible orientations:
parallel with or opposed to an applied magnetic field. The magnitude of the mag¬
netic moment is given by the expression
Pilar, F. L. J. Chem. Educ. 1978. 55 . 2-6. Scerri, E. R. Ibid. 1989, 66. 481-483. Vanquickenbome, L. G.
. Pierloot, K.; Devoghel, D. Inorg. Chem. 1989, 28, 1805-1813.
0 Moeller. T. Inorganic Chemistry; Wiley: New York, 1952; p 97.
38
2 - The Structure of the At.
Fig. 2.1 2 Relative orbital energies of the elements hydrogen to sodium. Solid lines indicate
one-eleciron orbital energies. Dashed lines represent experimental ionization energies, which
differ as a result of electron-electron interactions.
Tcibl«, 2.4
Ionization energies'
Molecule
(MJ mol -1 )
(eV)
Molecule
(MJ mol -1 )
(eV)
CH,
0.949
9.84
NH,
1.075
11.14
C 2 H 5
0.784
8.13
NO
0.893
9.26
CH,0
0.729
7.56
NO,
0.941
9.75
CN
1.360
14.09
o 2
1.165
12.07
CO
1.352
14.01
OH
1.254
13.00
cf 3
s0.86
S8.9
f 2
1.5146
15.697
N,
1.503
15.58
" Lias, S. 0.; Bartmess, J. E.; Licbman.J. F.; Holmes, J. L.; Levin. R. D.; Mallard, W.G.;
J. Phys. Chem. Ref. Data 1988, 17. Supplement 1.1-861.
Ionization
The ionization energies of a few groups are known (Table 2.4). Although not
generally as useful as atomic values, they can be used in Born-Haber calculations
(see Chapter 4) involving polyatomic cations, such as NO + and O^. They also pro¬
vide a rough estimate of the electron-donating or -withdrawing tendencies of groups.
The electrons that are lost on ionization are those that lie at highest energies and there¬
fore require the least energy to remove. One might expect, therefore, that electrons
would be lost on ionization in the reverse order in which orbitals were filled (see “The
Aufbau Principle”). There is a tendency for this to be true. However, there are some
very important exceptions, notably in the transition elements, which are responsible
The Polyelectronic Ato
39
for the characteristic chemistry of these elements. In general, transition elements react
as follows:
ls 2 2s 2 2p 5 3s 2 3p 6 3t/"4s 2 -♦ ls 2 2s 2 2p 6 3s 2 3p 6 3</"
Atom Dipositive cation
This is true not only for the first transition series but also for the heavier metals: The
ns 2 electrons are lost before the (h - 1 )d or (;i - 2)f electrons. This gives a common
+ 2 oxidation state to transition metals, although in many cases there is a more stable
higher or lower oxidation state.
This phenomenon is puzzling because it appears to contradict simple energetics:
If the 4s level is lower and fills first, then its electrons should be more stable and be
ionized last, shouldn't they? One might ask if there is a possible reversal of energy
levels within the transition series. If the relative energies of the 3 il and 4s levels are
examined, it is found that they lie very close together and that the energy of the 3 tl
level decreases with increasing atomic number. This is often advanced as the explana¬
tion for the electronic configuration of Cu. If the 3 cl level has dropped below the 4s at
atomic number 29, then the ground state must be 3d l0 4s‘. Nevertheless, this can have
no effect on the phenomena that we are investigating, because we are inquiring as to
the difference in configuration between ground state of the neutral atom and the ionic
states of the same element. Because all of the transition metals in the first series (with
exception of Cr and Cu) have a 3</"4s 2 ground state for the neutral atom and a stable
3d"4s° state for the dipositive ion, the source of our problem must be sought in the
difference between atom and ion, not in trends along the scries.
Before the question can be answered adequately, it is necessary to define what is
meant by "orbital energy”. For most purposes, that given by Koopmans' theorem is
adequate: If the nth orbital is the highest orbital used to describe the ground-state
wave function of an atom X. then the energy of orbital n is approximated by the ion¬
ization energy of the atom. These energies can be calculated by the self-consistent-field
method described previously (see page 20). These calculations have been made and
much has been written concerning their interpretation.- x The seeming contradiction
presented by the ground state of the atoms and the ground stale of the ions results
from the fact that it is the total energy of the atom (ion) that is important, not the single
energy of the electron entering the 4s level. We arc reminded again that electrons in¬
teract with each other in different ways depending upon their spatial distribution.
When all of the energies are summed, including all of the electron-electron repulsions ,
it is found that the [Ar] 3t/"4.v 1 configuration is lower in energy than the [Ar] 3 d"~ ‘4s 2
configuration: or for the more common +2 ion, [Ar] 3</"4s° < [Ar] 3</“ -2 4s 2 .
Unfortunately, this does not answer our original question completely, for we
cannot simply say that 4s electrons ionize before the 3 tl electrons because the 4s or¬
bital lies higher in energy—we have seen that we must inspect the total energy. One
could postulate (incorrectly) that because of electron-electron repulsions the correct
configuration for Ti 2 was [Ar] 4s 2 instead of [Ar] 3</ 2 , which runs counter to ex¬
perimental fact. In order to account for the apparent change in stability, depending
on whether the 4s or the 3(/ orbital is occupied, wc must compare the two systems
involved, Ti versus Ti 2 , or more generally, M versus M 2 . A clue may be sought
28 Pilar. F. L. J. Client. Eiittc. 1978. 55. 2-6. Sccrri. E. R. Ibid. 1989, 66. 481-483. Vnnquickcnbornc, L. G.;
Pierloot. K.; Devoghcl. D. Inarg. Cliem. 1989, 28. 1805-1813.
22
2 • The Structure of the At'
/ = 0 1 2 3 4
These "nrhitnlx" ore nonexistent
These orbitals are not filled
in known elements
Fig. 2.10 Mnemonic for
determining the order of
filling of orbitals
(approximate). [Adapted
from Moeller, T. Inorganic
Chemistry, Wiley: New
York, 1952. Reproduced
with permission.]
where the moment (p) is expressed in Bohr magnetons [(e/i)/(4rrm)] 10 and s = |»i,|.
The allowed values of the spin quantum number are ±4. For an atom with two elec¬
trons the spins may be either parallel (S = 1) or opposed and thus cancel (S = 0).
In the latter situation the electrons are referred to as paired . Atoms having only
paired electrons (.S' = ()) arc repelled slightly when placed in a magnetic field and
arc termed diamagnetic. Atoms having one or more unpaired electrons (5 0) are
stongly attracted by a magnetic field and arc termed paramagnetic.
Electrons having the same spin strongly repel each other and tend to occupy
different regions of space. This is a result of a fundamental law of nature known as
the Pauli exclusion principle. It states that total wave functions (including spin) must
change their signs on exchange of any pair of electrons in the system. Briefly, this
means that if two electrons have the same spin they must have different spatial wave
functions (i.e., different orbitals) and if they occupy the same orbital they must have
paired spins. The Pauli principle and the so-called Pauli repulsive forces 11 have far-
10 In SI the Bohr magneton has a value of 9.27 x 10' 24 with units of A m' or J T" 1 .
" The Pauli "force" corresponds to no classical interaction but results from the nature of quantum me¬
chanics. Although it is common in chemistry to speak of "repulsions" and "stabilizing energies" resulting
from the Pauli principle, these do not arise directly from the energetics of spin-spin interactions but
from the electrostatic energy resulting from the spatial distribution due to the requirements of the Pauli
exclusion principle. See Kauzman, W, J. Quantum Chemistry, Academic: New York. 1957; pp 319-320;
Matscn. F. A. J. Am. Chem. Sue. 1970. 92. 3525-3538.
The Polyelectronic Atom
23
reaching consequences in chemistry. For our present discussion the principle may be
stated as follows: In a given atom no two electrons may have all four quantum numbers
identical. This means that in a given orbital specified by n, /, and m h a maximum of
two electrons may exist (m, = +4 ar >d m s ~ — !')■
We can now add Rule 7 to those given on page 20:
Each orbital can contain two electrons, corresponding to the two allowed
values of m s : +4-
The Aufbau Principle The electron configuration, or distribution of electrons among orbitals, may be deter¬
mined by application of the Pauli principle and the ordering of energy levels suggested
above. The method of determining the appropriate electron configuration of mini¬
mum energy (the ground state) makes use of the aufbau principle, or “building up" of
atoms one step at a time. Protons are added to the nucleus and electrons are added
to orbitals to build up the desired atom. It should be emphasized that this is only a
formalism for arriving at the desired electron configuration, but an exceedingly useful
one.
The quantum numbers n, I, and m, in various permutations describe the possible
orbitals of an atom. These may be arranged according to their energies. The ground
state for the hydrogen atom will be the one with the electron in the lowest orbital,
the Is. The spin of the electron may be of either orientation with neither preferred.
We would thus expect a random distribution of spins; indeed, if a stream of hydrogen
atoms were introduced into a magnetic field, half would be deflected in one direction,
the other half in the opposite direction. Thus the four quantum numbers (n, /. m,, mj
for a hydrogen atom are (I, 0, 0, ± 4 ). For the helium atom we can start with a hydro¬
gen atom and add a proton to the nucleus and a second electron, The first three
quantum numbers of this second electron will be identical to those from a hydrogen
atom (i.e., the electron will also seek the lowest possible energy, the Is orbital), but
the spin must be opposed to that of the first electron. So the quantum numbers for
the two electrons in,a helium atom arc (1,0,0, + 4 ) a nd (1,0, 0. - 4 ). The Is orbital
is now filled, and the addition of a third electron to form a lithium atom requires that
the 2.v orbital, the next lowest in energy, be used. The electron configurations of the
first five elements together with the quantum numbers of the last electron are 12
s
H =
Is'
1.0,
±4
He =
Is 2
1,0,
0,
±i
Li =
ls 2 2s'
2, 0,
±i
Be =
ls 2 2s 2
JO
O
0,
±1
B =
ls 2 2s 2 2p‘
2, 1,
1,
±4
This procedure may be continued, one electron at a time, until the entire list of ele¬
ments has been covered. A complete list of electron configurations of the elements
12 The m, values for the unpaired electron in H. Li, and B arc, of course, undefined and may be either
+ J or —It is merely necessary that the values for the second electron entering the s orbital in He and
Be be opposite to the first. Likewise, the last electron in boron may enter the p,. p,„ or p. orbital, all equal
in energy, and so the m, value given above is arbitrary.
z
Element
1
II
ill
IV
V
1
H
1.3120
" 2
He
2.3723
5.2504
3
Li
0.5203
7.2981
11.8149
4
Be
0.8995
1.7571
14.8487
21.0065
5
B
0.8006
2.4270
3.6598
25.0257
32.8266
6
C
1.0864
2.3526
4.6205
6.2226
37.8304
7
N
1.4023
2.8561
4.5781
7.4751
9.4449
8
O
1.3140
3.3882
5.3004
7,4693
10.9895
9
F
1.6810
3.3742
6.0504
8.4077
11.0227
10
Ne
2.0807
3.9523
6.122
9.370
12.178
11
Na
0.4958
4.5624
6.912
9.544
13.353
12
Mg
0.7377
1.4507
7.7328
10.540
13.628
13
Al
0.5776
1.8167
2.7448
11.578
14.831
14
Si
0.7865
1.5771
3.2316
4.3555
16.091
15
P
1.0118
1.9032
2.912
4.957
6.2739
16
S
0.9996
2.251
3.361
4.564
7.013
17
Cl
1.2511
2.297
3.822
5.158
6.54
18
Ar
1.5205
2.6658
3.931
5.771
7.238
19
K
0.4189
3.0514
4.411
5.877
7.976
20
Ca
0.5898
1.1454
4.9120
6.474
8.144
21
Sc
0.631
1.235
2.389
7.089
8.844
22
Ti
0.658
1.310
2.6525
4.1746
9.573
23
V
0.650
1.414
2.8280
4.5066
6.299
24
Cr
0.6528
1.496
2.987
4.74
6.69
. 25
Mn
0.7174
1.5091
3.2484
4.94
6.99
26
Fe
0.7594
1.561
2.9574
5.29
7.24
27
Co
0.758
1.646
3.232
4.95
7.67
28
Ni
0.7367
1.7530
3.393
5.30
7.28
29
Cu
0.7455
1.9579
3.554
5.33
7.71
30
Zn
0.9064
1.7333
3.8327
5.73
7.97
31
Ga
0.5788
1.979
2.963
6.2
32
Ge
0.7622
1.5372
3.302
4.410
9.02
33
As
0.944
1.7978
2.7355
4.837
6.043
34
Se
0.9409
2.045
2.9737
4.1435
6.59
35
Br
1.1399
2.10
3.5
4.56
5.76
36
Kr
1.3507
2.3503
3.565
5.07
6.24
37
Rb
0.4030
2.633
3.9
5.08
6.85
38
Sr
0.5495
1.0643
4.21
5.5
6.91
39
Y
0.616
1.181
1.980
5.96
7.43
40
Zr
0.660
1.267
2.218
3.313
7.86
41
Nb
0.664
1.382
2.416
3.69
4.877
42
Mo
0.6850
1.558
2.621
-4.477
5.91
.43
Tc
0.702
1.472
2.850
' 44
Ru
0.711 ,
1.617
2.747
45
Rh
0.720
1.744
2.997
46
Pd
0.805
1.875
3.177
,
: 47
Ag
0.7310
2.074
3.361
. ,.48
Cd ,.
0.8677
1.6314
3.616
'. 49
In
f 0.5583
1.8206
2.705
'5.2
jQ
Sn
' 0.7086
1.4118
2.9431
3.9303
6.974
5)
Sb '
0.8316
1.595
2.44 :
4.26
5.4 -
52
Te
r 0.8693
1.79
2.698
3.610
' 5.669
12.31
7.883 14.99
To obtain values in
electron volts, multiply
table values by 10.364
9.900 12.1
6.6 12.23
:• -i&■? i■
106.4340
115.3791
28.934
31.643
31.861
33.877
35.867
36.578
38.598
40.760
16.964
18.192
17.370
18.64
19.86
20.19
21.40
22.678
17.959
18.6
19.2
19.6
131.4314
141.3626
35.462
38.457
38.733
40.959
43.138
43.962
46.187
48.576
20.3849
21.741
20.833
22.24
23.58
23.96
25.29
26.6
21.66
22.4
23.0
'Moore, C. E. Ionization Potentials and Ionization Limits Derived from the Analyses of Optical Spectra, NSRDS-NBS 34; National
lureau of Standards: Washington, DC; 1970 and personal communication. Data for the lanthanides and actinides from Martin, W. C.;
lagan, L.; Reader, J.; Sugar, J.J.Phys. Chem. Ref. Data 1974, 3, 771 and Sugar J. J. Opt. Soc. Am. 1975, 65, 1366.
1.0370
0.5094 0.97906
0.49 1.17
0.59 1.11
0.57
1.1704
2.046
0.3757
2.23
0.5029
0.96526
0.5381
1.067
0.528
1.047
.8503
.949
.086
4.82
3.54
3.76
5.552
5.790
5.953
6.046
6.101
6.249
6.413
5.990
6.169
6.282
6.313
6.328
6.445
6.596
6.64
5.40
0.8901
1.0070
0.5893
0.7155
0.7033
0.812
1.98
1.8097
1.9710
1.4504
1.610
24
2- The Structure of the Atom
The Polyelectn
25
Table 2.1
Electron configurations of
the elements"
1
Element
Electron %
configuration
Z
Element
Electron
configuration
1
H
Is 1
31
Ga
[Ar]3d ,0 4s 2 4p'
2
He
Is 2
32
Ge
[Ar]3r/'°4s 2 4p 2
3
Li
[He]2s 1
33
As
[Ar]3d‘°4s 2 4p 3
4
Be
[He]2s 2
34
Se
[Ar]3</ I0 4s 2 4p 4
5
B
[He]2s 2 2p 1
35
Br
[Ar]3c/‘°4s 2 4p 5
6
C
[He]2s 2 2p 2
36
Kr
[Ar]3£y ,0 4s 2 4p 6
7
N
[He]2s 2 2p 3
37
Rb
[Kr]5s 1
8
0
[He]2s 2 2p 4
38
Sr
[Kr]5s 2
9
F
[He]2s 2 2p 3
39
Y
[Kr]4d'5s 2
10
Ne
[He]2s 2 2p 6
40
Zr
[Kr]4c/ 2 5i' 2
11
Na
[Ne]3s 1
41
Nb
[Kr]4t/ J 5s 1
12
Mg
[Ne]3s 2
42
Mo
[Kr]4c/ 5 5s 1
13
Al
[Ne]3s 2 3p 1
43
Tc
[Kr]4c/ 5 5s 2
14
Si
[Ne]3s 2 3p 2
44
Ru
[Kr]4c/ 7 5s '
15
P
[Ne]3s 2 3p 3
45
Rh
[Kr]4d 8 5s 1
16
S
[Ne]3s 2 3p 4
46
Pd
[Kr]4d 10
17
Cl
[Ne]3s 2 3p 5
47
Ag
[Kr]4c/ I0 5j 1
18
Ar
[Ne]3s 2 3p 6
48
Cd
[Kr]4</'°5s 2
19
K
[Ar]4r 1
49
In
[Kr]4</ 10 5sV
20
Ca
[Ar]4.v 2
50
Sn
[Kr)4d ,0 5s 2 5p 2
21
Sc
(Ar]3c/‘4j 2
51
Sb
[Kr]4c/'°5j 2 5p 3
22
Ti
[Ar]3</ 2 4.v 2
52
Te
[Kr]4</'°5s 2 5p 4
23
V
[Ar]3d 3 4j 2
53
I
[Kr]4c/ 10 5s 2 5p 3
24
Cr
[Ar]3c/ 5 4s 1
54
Xe
[Kr]4d l0 5j 2 5p 6
25
Mn
[Ar]3c/ 5 4s 2
55
Cs
[Xe]6s'
26
Fe
[Ar]3</ 6 4j 2
56
Ba
[Xe]6s 2
27
Co
[Ar]3c/ 7 4s 2
57
La
[Xe]5</'6s 2
28
Ni
[Ar]3c/ 8 4s 2
58
Ce
[Xe]4f'5d6s 2
29
Cu
[Ar]3</'"4f 1
59
Pr
[Xe]4/ 3 6s 2
30
Zn
[Ar]3r/'°4s 2
60
Nd
[Xe]4/ 4 6s 2
is given in Table 2.1. It will be seen that there are only a few differences between
these configurations obtained experimentally and a similar table which might be
constructed on the basis of the aufbau principle. In every case in which an exception
occurs the energy levels involved are exceedingly close together and lactors not ac¬
counted for in the above discussion invert the energy levels. For example, the (ti - 1 )d
and ns levels tend to lie very close together when these levels are filling, with the latter
slightly lower in energy. If some special stability arises, such as a filled or half-filled
subshell (see page 27 and Chapter II), the most'stable arrangement may not be
(n - 1 )dns 2 . In Cr and Cu atoms the extra stability associated with half-filled and
filled subshells is apparently sufficient to make the ground-state configuration of the
isolated atoms 3d 5 4s‘ and 3d l0 4s' instead of 3d 4 4s 2 and 3d 9 4s 2 , respectively. Too
much importance should not be placed on this type of deviation, however. Its effect
on the chemistry of these two elements is minimal. It is true that copper has a reason¬
ably stable + 1 oxidation state (corresponding to 3c/ I0 4s°), but the + 2 state is even
more stable in most chemical environments. For chromium the most stable ion in
aqueous solution is Cr 3+ , with the Cr 2+ ion and the Cr(VI) oxidation state (as in
wm
Table 2.1 (Continued)
Electron configurations of
the elements"
ay
o ni c Atom
Z
Element
Electron
configuration
Z
Element
Electron
configuration
61
Pm
[Xe]4/ S 6s 2
83
Bi
[Xe]4/' I4 5c/ I0 6j 2 6p 3
62
Sm
[Xe]4/ fi 6.s 2
84
Po
[Xe]4/ l4 5cf l0 6i 2 6p 4
63
Eu
[Xe]4/ 7 6s 2
85
At
[Xe]4/ u 5d' °6s 2 6p 5
64
Gd
[Xe]4/ 7 5d‘6s 2
86
Rn
[Xe]4/ 14 5</ l0 6s 2 6p 6
65
Tb
[Xe]4/ 9 6s 2
87
Fr
[Rn]7s 1
66
Dy
[Xe]4/ 10 6s 2
88
Ra
[Rn]7s 2
67
Ho
[Xe]4/“6s 2
89
Ac
(Rn]6d'7s 2
68
Er
[Xe]4/ I2 6s 2
90
Th
[Rn]6t/ 2 7.y 2
69
Tm
[Xe]4/ 13 6s 2
91
Pa
[Rn]5/ 2 6c/'7.y 2
70
Yb
[Xe]4/ I4 6.v 2
92
U
[Rn]5/ 3 6t/'7j 2
71
Lu
[Xe]4/ l4 5r/'6s 2
93
Np
[Rn]5/ 4 6c/'7s 2
72
Hf
[Xe]4/ l4 5t/ 2 6s 2
94
Pu
[Rn]5/ 6 7.y 2
73
Ta
[Xe]4/ 14 5<f 3 6.y 2
95
Am
[Rn]5/ 7 7s 2
74
W
[Xe]4/ 14 5t/ 4 6s 2
96
Cm
[Rn]5/ 7 6d's 2
75
Re
[Xe]4/ l4 5c/ 5 6s 2
97
Bk
[Rn]5/ 9 7s 2
76
Os
[Xe]4/ ,4 5r/ 6 6s 2
98
Cf
[Rn]5/'°7s 2
77
Ir
[Xe]4/ I4 5rf 7 6s 2
99
Es
[Rn]5/"7s 2
78
Pt
[Xe]4/ l4 5c/ 9 6s 1
100
Fm
[Rn]5/ I2 7s 2
79
Au
[Xe]4/ I4 5rf ,0 6s l
101
Md fc
[Rn]5/ 13 7s 2
80
Hg
[Xe]4/ I4 5tf l0 6s 2
102
No 6
[Rn]5/ I4 7s 2
81
TI
[Xe]4/ u 5d l0 6s 2 6p 1
103
Lr fc
[Rn]5/ l4 6</7s 2
82
Pb
[Xe]4/ 14 5 c/ I0 6j 2 6p 2
104
Rf b
[Rn]5/ l4 6c/ 2 7s 2
° Moore, C. E. Ionization Potentials and Ionization Limits Derived from the Analyses of
Optical Spectra. NSRDS-NBS 34; National Bureau of Standards: Washington. DC, 1970,
except for the data on the actinides, which are from The Chemistry of the Actinide
Elements; Katz, J. J.; Seaborg, G. T.; Morss, L. R., Eds.; Chapman and Hall: New York.
1986; Vol. 2.
b Predicted configuration.
CrOi") reasonably stable; the Cr(I) oxidation stale is practically unknown. For both
Cu 2 + and Cr 3+ (as well as many other transition metal ions) ligand field effects in
their complexes (see Chapter II) are much more important in determining stable
oxidation states than are electron configurations.
In the case of the lanthanide elements (elements 58-71) and those immediately
following, the 5 d and 4/ levels are exceedingly close. In the lanthanum atom it ap¬
pears that the 57th electron enters the 5 d level rather than the 4 f. Thereafter the 4/
level starts to fill, and some lanthanides appear not to have any 5 d electrons. Here
again, too much attention to details of the electron configuration is not rewarding
from a chemist’s point of view—indeed it may be quite misleading. The difference
in energy between a 5d" l 4f m configuration and a 5d"4f m+ ' configuration is very
small. For mnemonic purposes all lanthanide elements behave as though they had
an electron configuration: 6s 2 5d‘4/"; that is, the most stable oxidation state is always
that corresponding to loss of three electrons (the 6s and 5 d). There are some other
“abnormalities” in the electron configurations of various elements, but they are of
minor importance from a chemical point of view.
34
2 • The Struct
f the At
Toble 2.2 _
Effective nuclear charges
for elements 1 to 36
Element
Is
2s
2p
3s
3 P
4s
3d
4p
H
1.000
He
1.688
Li
2.691
1.279
Be
3.685
1.912
B
4.680
2.576
2.421
C
5.673
3.217
3.136
N
6.665
3.847
3.834
O
7.658
4.492
4.453
F
8.650
5.128
5.100
Ne
9.642
5.758
5.758
Na
10.626
6.571
6.802
2.507
Mg
11.619
7.392
7.826
3.308
A1
12.591
8.214
8.963
4.117
4.066
Si
13.575
9.020
9.945
4.903
4.285
P
14.558
9.825
10.961
5.642
4.886
S
15.541
10.629
11.977
6.367
5.482
Cl
16.524
11.430
12.993
7.068
6.116
Ar
17.508
12.230
14.008
7.757
6.764
K
18.490
13.006
15.027
8.680
7.726
3.495
Ca
19.473
13.776
16.041
9.602
8.658
4.398
Sc
20.457
14.574
17.055
10.340
9.406
4.632
7.120
Ti
21.441
15.377
18.065
11.033
10.104
4.817
8.141
V
22.426
16.181
19.073
11.709
10.785
4.981
8.983
Cr
23.414
16.984
20.075
12.368
11.466
5.133
9.757
Mn
24.396
17.794
21.084
13.018
12.109
5.283
10.528
Fe
25.381
18.599
22.089
13.676
12.778
5.434
11.180
Co
26.367
19.405
23.092
14.322
13.435
5.576
11.855
Ni
27.353
20.213
24.095
14.961
14.085
5.711
12.530
Cu
28.339
21.020
25.097
15.594
14.731
5.858
13.201
Zn
29.325
21.828
26.098
16.219
15.369
5.965
13.878
Ga
30.309
22.599
27.091
16.996
16.204
7.067
15.093
6.222
Ge
31.294
23.365
28.082
17.760
17.014
8.044
16.251
6.780
As
32.278
24.127
29.074
18.596
17.850
8.944
17.378
7.449
Se
33.262
24.888
30.065
19.403
18.705
9.758
18.477
8.287
Br
34.247
25.643
31.056
20.218
19.571
10.553
19.559
9.028
Kr
35.232
26.398
32.047
21.033
20.434
11.316
20.626
9.769
be useful now to discuss trends in atomic sizes without becoming too specific at the
present time about the actual sizes involved.
As we have seen from the radial distribution functions, the most probable radius
tends to increase with increasing n. Counteracting this tendency is the effect of in¬
creasing effective nuclear charge, which tends to contract the orbitals. From these
opposing forces we obtain the following results:
Atoms in a given family tend to increase in size from one period (= horizontal
row of the periodic chart) to the next. Because of shielding, Z increases very
slowly from one period to the next. For example, using Slater’s rules we obtain
The Polyelectronic Atom 35
the following values for Z:
H = 1.0 Li =1.3 Na = 2.2 K = 2.2 Rb = 2.2 Cs = 2.2
The result of the opposing tendencies of n and Z is that atomic size increases
as one progresses down Group IA (1). This is a general property of the periodic
chart with but few minor exceptions, which will be discussed later.
Within a given series, the principal quantum number does not change. [Even
in the “long" series in which the filling may be in the order ns, (ii - l)</, np , the
outermost electrons are always in the nth level.] The effective nuclear charge
increases steadily, however, since electrons added to the valence shell shield
each other very ineffectively. For the second series:
Li =1.3 Be =1.95 B = 2.60 C = 3.25
N = 3.90 O = 4.55 F = 5.20 Ne = 5.85
As a result there is a steady contraction from left to right. The net effect of the
top-to-bottom and the left-to-right trends is a discontinuous variation in
atomic size. There is a steady contraction with increasing atomic number until
there is an increase in the principal quantum number. This causes an abrupt
increase in size followed by a further decrease.
Ionization Energy The energy necessary to remove an electron from an isolated atom in the gas phase
is the ionization energy (often called ionization potential) for that atom. It is the en¬
ergy difference between the highest occupied energy level and that corresponding to
n = oo, that is, complete removal. It is possible to remove more than one electron,
and the succeeding ionization energies are the second, third, fourth, etc. Ionization
energies are always endothermic and thus are always assigned a positive value in
accord with common thermodynamic convention (see Table 2.3). The various ioniza¬
tion energies of an atom arc related to each other by a polynomial equation, which
will be discussed in detail later in this chapter.
For the nontransition elements (alkali and alkaline earth metals and the non-
metals) there arc fairly simple trends with respect to ionization energy and position
in the periodic chart. Within a given family, increasing n tends to cause reduced
ionization energy because of the combined eflects of size and shielding. The transi¬
tion and posttransition elements show some anomalies in this regard, which will be
discussed in Chapters 14 and 18. Within a given series there is a general tendency
for the ionization energy to increase with increase in atomic number. This is a re¬
sult of the tendency for Z to increase progressing from left to right in the periodic
chart. There are two other factors which prevent this increase from being monolonic.
One is the change in type of orbital which occurs as one goes from Group 11A (2)
(,v orbital) to Group MIA (3) (p orbital). The second is the exchange energy between
electrons of like spin. This stabilizes a system of parallel electron spins because
electrons having the same spin tend to avoid each other as a result of the Pauli
exclusion principle. The electrostatic repulsions between electrons are thus reduced.
We have seen previously that this tends to maximize the number of unpaired electrons
(Hund's principle of maximum multiplicity) and also accounts for the “anomalous”
behavior of Cu and Cr. It also tends to make it more difficult to remove the electron
from the nitrogen atom than would otherwise be the case. As a result of this stabiliza¬
tion, the ionization energy of nitrogen is greater than that of oxygen (see Fig. 2.12).
(
26 2 • The Structure
Atomic States,
term Symbols,
and Hund’s Rule
of the Atom
Although the aufbau principle and the ordering of orbitals given previously may
be used reliably to determine electron configurations, it must again be emphasized
that the device is a formalism and may lead to serious error if overextended. For
example, in the atoms of the elements potassium, calcium, and scandium the 4s level
is lower in energy than the 3 d level. This is not true for heavier elements or for charged
ions. The energies of the various orbitals are sensitive to changes in nuclear charge
and to the occupancy of other orbitals by electrons (see ''Shielding”, page 30). and
this prevents the designation of an absolute ordering of orbital energies. It happens
that the ordering suggested by Fig. 2.10 is reasonably accurate when dealing with
orbitals corresponding to the valence shell of an atom; that is, the energies 3d > 4s
and 5 p > 4 d are correct for elements potassium and yttrium, for example, but not
necessarily elsewhere.
It is convenient to be able to specify the energy, angular momentum, and spin multi¬
plicity of an atom by a symbolic representation. For example, for the hydrogen atom
we may define S, P. D, and F states, depending upon whether the single electron oc¬
cupies an s, p. d, or/orbital. The ground state of hydrogen. Is 1 , is an S state: a hydro¬
gen atom excited to a 2 p' configuration is in a P state; etc. For polyelectronic atoms,
an atom in a P state has the same total angular momentum (for ail electrons) as a
hydrogen atom in a P state. Corresponding to states S, P, D, F,... are quantum
numbers L = 0, 1, 2, 3, 4 .which parallel the / values for s, p. d.f,... orbitals. 13
Likewise, there is quantum number S (not to be confused with the S state just men¬
tioned) that is the summation of all the electronic spins. For a closed shell or subshell,
obviously S = 0, since all electrons arc paired. Somewhat less obviously, under these
conditions L = 0, since all of the orbital momenta cancel. This greatly simplifies
working with states and term symbols.
The chemist frequently uses a concept known as multiplicity , originally derived
from the number of lines shown in a spectrum. It is related to the number or un¬
paired electrons and, in general, is given by the expression 2S + I. Thus, if S = 0,
the multiplicity is one and the state is called a singlet; if S = 4, the multiplicity is
two and the state is a doublet; S = I is a triplet; etc. Hund’s rule of maximum multi¬
plicity states that the ground state of an atom will be the one having the greatest
multiplicity (i.c.. the greatest value of S). Consider a carbon atom 1= \s 2 2s 1 2p z ).
We may ignore the closed l.r and 2s 2 . The two 2 p electrons may be paired (5 = 0) or
have parallel spins in different orbitals (5 = I). Hund's rule predicts that the latter will
be the ground stale, that is, a triplet of state. It happens that in this slate L = I. so we
may say that the ground state of carbon is V (pronounced "triplet-P”). The is
said to be the term .symbol.
It is convenient for many purposes to draw "box diagrams" of electron config¬
urations in which boxes represent individual orbitals, and electrons and their spins
are indicated by arrows:
Is 2
Til
2s 2 2p 2
IJ This is (he reverse of the historical process. S, P. D. and F stales were observed spectroscopically and
named after sharp, principal, diffuse, and fundamental characteristics of the spectra. Later the symbols
.s. p, 1 1, and / were applied to orbitals. The methods for ascertaining the various possible values of L and
the determination of term symbols, as well as the general topic of the coupling of orbital angular mo¬
menta and electron angular momenta, are given in Appendix C.
Periodicity of the
Elements
The Polyelectronic Atom 27
Such devices can be very useful for bookkeeping, providing pigeonholes in which
to place electrons. However, the reader is warned that they can be misleading if im¬
properly used, especially with respect to term symbols.
Traditionally, Hund's rule has been explained by assuming that there is less re¬
pulsion between electrons in the high-spin stale, stabilizing it. Yet we have seen that
electrons having the same spin are highly correlated and actually repel each other
more than electrons of opposite spin (page 22). However, because electrons of paral¬
lel spin avoid each other, they shield each other from the nucleus less and the electron-
nucleus attraction is greater and dominates: The overall energy is lowered. 14
The extra stability of parallel-spin configurations is given by the exchange energy:
K
(2.7)
where N is the number of electrons having parallel spins. Because the exchange
energy is a quadratic function of N, it rises rapidly as the number of parallel spins
K K K
increases: 0 — (TV = 1), 2 — (N = 2), 6 — (N = 3). Since the number of parallel spins
is maximized for filled and half-filled subshells, the exchange energy is responsible
for the so-called “special stability" of these configurations. 13
For chemists working with several elements, the periodic chart of the elements is so
indispensable that one is apt to forget that, far from being divinely inspired, it resulted
from the hard work of countless chemists. True, there is a quantum mechanical basis
for the periodicity of the elements, as we shall see shortly. But the inspiration of such
scientists as Mendeleev and the perspiration of a host of nineteenth-century chemists
provided the chemist with the benefits of the periodic table about half a century be¬
fore the existence of the electron was proved! The confidence that Mendeleev had in
his chart, and his predictions based on it, make fascinating reading. 16
The common long form of the periodic chart (Fig. 2.11) may be considered a
graphic portrayal of the rules of atomic structure given previously. The arrangement
of the atoms follows naturally from the aufbau principle. The various groups of the
chart may be classified as follows:
The “s" block elements: Croups IA and IIA (Columns I and 2). the alkali and
alkaline earth metals. These elements arc also sometimes called the “light
metals". They are characterized by an electron configuration of ns 1 or ns 2 over
a core with a noble gas configuration.
The ‘d” block elements: "8" Groups (Columns 3-12). the transition metals.
Characteristically, atoms of these elements in their ground states have electron
configurations that are filling d orbitals. 17 For example, the first transition
series proceeds from Sc(4s 2 3<f 1 ) to Zn(4s 2 3<( 10 ). Each of these ten elements
stands at the head of a family of congeners (e.g., the chromium family, VIB, 6).
IJ Boyd, R. J. Nature 1984. 310. 480-481.
15 Blake. A. B. J. Chem. Educ. 1981. 58. 393-398.
16 See Weeks. M. E.; Leicester, H. M. Discovery of the Elements. 7th cd,; Chemical Education Publishing:
Easton. PA. 1968.
11 There are problems with any simple definition of “transition metal." See the discussion under “Semantics".
32
2- The Structure of the Atom
All of the other electrons in the (ns, np) group, shield the valence electron to an
extent of 0.35 each. 26
All electrons in the n — 1 shell shield to an extent of 0.85 each.
All electrons n — 2 or lower shield completely; that is, their contribution is
1.00 each.
When the electron being shielded is in an nd or nf group, rules 2 and 3 are the
same but rules 4 and 5 become:
All electrons in groups lying to the left of the nd or nf group contribute 1.00.
Examples
Consider the valence electron in the atom 7 N = \s 2 2s 2 2p i . Grouping of the
orbitals gives (Is) 2 (2j, 2 p) 5 . S = (2 x 0.85) + (4 x 0.35) = 3.10. Z =
Z - S ~ 7.0 - 3.1 = 3.9.
Consider the valence (4s) electron in the atom 30 Zn. The grouped electron con¬
figuration is (ls) 2 (2s, 2p) 8 (3s, 3p) 8 (3d) 10 (4s) 2 . S = (10 x 1.00) + (18 x 0.85) +
(1 x 0.35) = 25.65. Z = 4.35.
Consider a 2d electron in Zn. The grouping is as in example 2, but the shield¬
ing is 5 = (18 x 1.00) + (9 x 0.35) = 21.15. Z = 8.85.
It can be seen that the rules are an attempt to generalize and to quantify those
aspects of the radial distributions discussed previously. For example, d and / elec¬
trons are screened more effectively (S = 1.00) than s and p electrons (S = 0.85) by
the electrons lying immediately below them. On the other hand, Slater’s rules as¬
sume that all electrons, s, p, d, or /, shield electrons lying above them equally well
(in computing shielding the nature of the shielding electron is ignored). This is not
quite true, as we have seen above and will lead to some error. For example, in the
Ga atom (= ... 3s 2 3p 6 3d l0 4s 2 4p l ) the rules imply that the 4 p electron is shielded
as effectively by the 3d electrons as by the 3s and 3 p electrons, contrary to Fig. 2.4.
Slater formulated these rules in proposing a set of orbitals for use in quantum
mechanical calculations. Slater orbitals are basically hydrogen-like but differ in two '
important respects:
They contain no nodes. This simplifies them considerably but of course makes
them less accurate.
They make use of Z in place of Z, and for heavier atoms, n is replaced by n,
where for n = 4, n = 3.7; n = 5, n = 4.0; n = 6, n = 4.2. The difference
between n and n is referred to as the quantum defect.
To remove the difficulties and inaccuracies in the simplified Slater treatment of
shielding, Clementi and Raimondi 27 have obtained effective nuclear charges from
26 Except for the Is orbital for which a value of 0.30 seems to work better.
37 Clementi E.; Raimondi, D. L. J. Chem. Phys. 1963, 38, 2686-2689.
The Polyelectronic Atom
33
m
The Sizes of Atoms
self-consistent field wave functions for atoms from hydrogen to krypton and have
generalized these into a set of rules for calculating the shielding of any electron. The
shielding which an electron in the nth energy level and /th orbital (S„,) experiences is
given by:
S ls = 0.3(N ls - 1) + 0.0072 (N 2s + N 2p ) + 0.0158(N 3s . M + N 4s . p ) (2.9)
S 2J = 1.7208 + 0.3601(N 2j - 1 + N 2p ) + 0.2062(N 3s . Pi( , + N 4s . p ) (2.10)
S 2p = 2.5787 + 0.3326(N 2p - 1) - 0.0773N 3s - O.OI61(N 3p + NJ
0.0048N 3d + 0.0085N 4p (2.11)
S 3j = 8.4927 + 0.250 l(N 3s - I + N 3p )
0.0778N 4 , + 0.3382N 3i , + O.I978N 4p (2.12)
S 3p = 9.3345 + 0.3803(N 3p - I) + 0.0526N 4s + 0.3289N 3rf + O.I558N 4p (2.13)
S 4J = 15.505 + 0.0971(N 4J - 1) + 0.8433N 34 + 0.0687N 4p (2.14)
S 3i = 13.5894 + 0.2693(N 3J - 1) - 0.1065N 4p (2.15)
S 4p = 24.7782 + 0.2905(N 4p - 1) (2.16)
where N„, represents the number of electrons in the nl orbital. For the examples given
above, the effective nuclear charges obtained are ZJ = 3.756, Z„ As = 5.965, and
Z„ 3d = 13.987. The shielding rules of Clementi and Raimondi explicitly account for
penetration of outer orbital electrons. They are thus more realistic than Slater's rules,
at the expense, however, of more complex computation with a larger number of
parameters. If accuracy greater than that afforded by Slater’s rules is necessary, it
would appear that direct application of the effective nuclear charges from the SCF
wave functions is not only simple but also accurate. Such values are listed in Table 2.2.
With the accurate values of Table 2.2 available, the chief justification of “rules”,
whether Slater’s or those of Clementi and Raimondi, is the insight they provide into
the phenomenon of shielding.
Atomic size is at best a rather nebulous quantity since an atom can have no well-
defined boundary similar to that of a billiard ball. In order to answer the question,
“How big is an atom?” one must first pose the questions, “How are we going to mea¬
sure the atom?” and “How hard are we going to push?” If we measure the size of a
xenon atom resting in the relatively relaxed situation obtained in solid xenon, we
might expect to get a different value than if the measurement is made through violent
collisions. A sodium ion should be compressed more if it is tightly bound in a crystal
lattice (e.g., NaF) than if it is loosely solvated by molecules of low polarity. The ques¬
tion of how hard we are going to push is particularly important because measuring
atoms is analogous to measuring an overripe grapefruit with a pair of calipers: The
value we get depends on how hard we squeeze. For this reason it is impossible to set
up a single set of values called “atomic radii” applicable under all conditions. It is
necessary to define the conditions under which the atom (or ion) exists and also our
method of measurement. These will be discussed in Chapter 8. Nevertheless, it will
28
2 • The Structure of the Atom
Subshells being completed
Principal
quantum ns (,n-)d
Fig. 2.11 Periodic chart of the elements.
The “f" block elements: Lanthanide and actinide elements. These two series
often appear with a or f in Group NIB (3), but these elements do not belong
to that family. (Note that the transition metals do not belong to group IIA (2),
which they follow.) The most common oxidation state for the lanthanides and
some of the actinides is +3, hence the popularity of the IIIB (3) position. Be¬
cause of their remarkable electronic and chemical properties they should be
set apart, but most periodic tables give no special numerical appellations to
these elements.
The “p” block elements: The nonmetals and posttransition metals, Groups III A
to VIIIA (Columns 13 to 18). This block or elements contains six families
corresponding to the maximum occupancy of six electrons in a set of p orbitals.
The classification between metals and nonmetals is imprecise, principally be¬
cause the distinction between metal and nonmetal is somewhat arbitrary,
though usually associated with a “stair-step” dividing line running from boron
to astatine. All these elements (except He) share the feature of filling p orbitals,
the noble gases (VIIIA and 18) having a completely filled set of p orbitals.
The Polyelectronic Atom
It is possible to trace the aufbau principle simply by following the periodic chart.
Consider the elements from Cs(Z = 55) to Rn(Z = 86). In the elements Cs and Ba
the electrons enter (and fill) the 6s level. The next electron enters the 5 d level and
La(Z = 57) may be considered a transition element. In the elements Ce through Lu
the electrons are added to 4/ levels and these elements are lanthanide or inner transi¬
tion elements. After the 4/ level is filled with Lu, the next electrons continue to fill the
5 d orbitals (the transition elements Hf to Hg). and finally, the bp level is filled in the
elements T1 to Rn, in accord with Table 2.1. The periodic chart may thus be used
to derive the electron configuration of an element as readily as use of the rules given
above. It should be quite apparent, however, that the chart can give us back only
the chemical knowledge that we have used in composing it; it is not a source of knowl¬
edge in itself. It is useful in portraying and correlating the information that has been
obtained with regard to electron configurations and other atomic properties.
Semantics, History,
and Other
Questions
Some chemists would define transition metals strictly as those elements whose
ground state atoms have partially filled d orbitals. This excludes zinc from the first
transition series. One must admit that zinc does have several properties that distin¬
guish it from “typical” transition metal behavior: It forms a single oxidation state,
Zn 2 , which is neither paramagnetic nor colored and which forms rather weak com¬
plexes (for its small size), etc. 18 If we so exclude zinc, do we also exclude Cu + , which
is isoelectronic with Zn 2+ , and metallic copper, which also has a filled 3 d'° configura¬
tion? More importantly, if we exclude zinc and its congeners, cadmium and mercury,
from the transition metals, then to be logically consistent we must exclude the noble
gases from the nonmetals. Some chemists might favor this, but the point is made in
Chapter 17 that the separation of the noble gases from the halogens, which are in
some ways similar, impeded noble gas chemistry. Finally, to be internally logical,
lutetium would have to be removed from the lanthanides and lawrencium from the
actinides. 19
Likewise, the designation of groups as “A” or “B” is purely arbitrary. Assignment
of all of the transition metals to “B’ r groups has an internal consistency as well as
historical precedent. 20 Unfortunately, some periodic charts have used "A” and "B”
in an almost opposite sense. 21 In addition, as a historical carryover from the older
“short form" chart, the iron, cobalt, and nickel families were lumped under the non¬
descript “VIII", This state of confusion led the IUPAC to recommend that the groups
18 Note that these same criticisms apply equally well to scandium (known only as the Sc , 3 d°. ion).
19 That is. there would be only thirteen lanthanides and actinides, just as elimination of zinc, cadmium,
and mercury leaves only nine transition metal groups, and elimination of the noble gases, leaves only/ve
groups to contain the nonmetals and postlransition metals. The main difficulty with any of these sug¬
gestions is that they contradict the simple expectations that (he .r-block should consist of two elements, the
p-block should consist of six elements, the cZ-bloek should consist of ten elements, and the/-block should
consist of fourteen elements.
!n Fcrnclius. W. C.; Powell. W. H. J. Chan. Etlnc. 1982. 59. 501-508.
21 Historically. North American and Russian periodic charts labeled the main group elements as "A" and
the transition metals as “B". Most other European charts labeled the first seven groups on the left (alkali,
alkaline earth, and transition metals in the scandium, titanium, vanadium, chromium, and manganese
families) as “A” and the last seven groups on the right (excluding the noble gases) as the "B" groups
(copper, zinc, boron, carbon, nitrogen, oxygen, and fluorine families). Such ambiguity is anathema to
the indexer and abstracter.
30
2 - The Structi
f the At
be numbered 1 to 18, left to right, from the alkali metals (1) to the noble gases (18).
However this does remove some of the pedagogical value of a simple I—VIII system,
especially for use in introductory courses. On the other hand, a very logical question
can be raised: Why not a 1-32 numbering system with the lanthanides and actinides
incorporated in a “super"-long form of the chart?
To be sure, the form of the periodic chart that is most useful depends upon the
use intended. For the simplest chemistry, perhaps the original short-form may even
be best: It gives the maximum oxidation state for all the elements and reminds us that
it is no accident that perchlorate (C10 4 ) and permanganate (Mn0 4 ) are similar
strong oxidizing agents. On the other hand, for someone whose main interest is in
the symmetry of electron configurations and the periodic chart, perhaps a complete
1-32 chart is the best. 22 The basic conclusions that your authors have used for this
book are: (1) the periodic table is a tool; (2) it should be a useful tool, not a clumsy one:
(3) the usual "long form" of the chart with both "American" A-B labels and 1-18
IUPAC labels is the best compromise at present (Fig. 2.11).
In a similar vein, suggestions have been made that all transactinide elements be
named by a system that translates the atomic number into a latinized name with a
three-letter(!) symbol that is a one-to-one letter equivalent of the atomic number. 22
Within this scheme, rutherfordium (Z = 104) would become unnilquadium, Unq.
according to the formula tin = I, nil = 0, and quad = 4, and hahnium would become
unnilpentium, Unp. One cannot help wondering: If the atomic number, atomic sym¬
bol. and elemental name must needs become completely redundant, why is "element
104" thought to be inadequate for the purpose? 2-1 However, the IUPAC calls these
"temporary names" to be used until a suitable name can be agreed upon, which, it is
hoped, will be chosen in the time-honored manner.
Shielding The energy of an electron in an atom is a function of Z 2 /u 2 . Since the nuclear charge
(= atomic number) increases more rapidly than the principal quantum number, one
might be led to expect that the energy necessary to remove an electron from an atom
would continually increase with increasing atomic number. This is not so, as can be
shown by comparing hydrogen (Z = 1) with lithium (Z = 3). The ionization energies
~ If this is numbered 1-32. immediately the same problem as before arises through simultaneous usage
Of 1-18 and 1-32 charts. Docs column 17 refer to F. Cl. Br. and I. or Lu and Lr? (Sec Jensen. W B.
Chum. Eng. Nun's 1987, (53(33), 2-3.) Strong arguments can be made for extending the A-B system to an
A-B-C system. A logical form is the use of a step-pyramid. (Sec Jensen. W. B. Comp. A Math. Appl. 1986.
I2B. 487-510.) Such a chart makes the logical suffixes: A = "main group elements". B = "transition
elements", C = "inner transition elements" (= lanthanides and actinides). However this means accepting
the North American conventions with regard to A and B. which may not'be politically practical. And
indexes are going to look askance at any future use of A and B. Perhaps internal inclusion of the lan¬
thanides and actinides is not the end of the story: If the "superactinidcs" discussed in Chapter 14 are ever
discovered, a fifty column chart could follow! The step pyramid would accomodate this possibility
with the addition of one more. ”D" layer.
Chau. J. Pt.ru Appl. Cliem. 1979. SI. 381-384.
24 The reason for proposing a change, supposedly only temporary, in the traditional way of naming ele¬
ments (i.c., giving that option and honor to the discoverer) is a result of two important factors: (I) There
should be some provisional way of discussing elements that are as yet undiscovered through, of course,
the atomic number docs this unambiguously. (2) There is an intense national rivalry and chauvinism in
the discovery of these elements, an inability to repeat claimed discoveries, and the fact that the matter
seems to have left the area of science and is now one of politics. Perhaps the current presence of glasnost
and independent European laboratories will resolve this issue. See also Chapter 14.
The Polyelectronic Atom 31
are 1312 kJ mol -1 (H) and 520 kJ mol -1 (Li). The ionization energy of lithium is
lower for two reasons: (1) The average radius of a 2s electron is greater than that
of a Is electron (see Fig. 2.4); (2) the 2s' electron in lithium is repelled by the inner
core Is 2 electrons, so that the former is more easily removed than if the core were not
there. Another way of treating this inner core repulsion is to view it as "shielding" or
“screening” of the nucleus by the inner electrons, so that the valence electron actually
“sees” only part of the total charge. Thus, the ionization energy for lithium corre¬
sponds to an effective nuclear charge of between one and two units. The radial prob¬
ability functions for hydrogen-like orbitals have been discussed previously (Fig. 2.4).
The bulk of the electron density of the Is orbital lies between the nucleus and the bulk
of the 2s density. The laws of electrostatics state that when a lest charge is outside
of a “cage” of charge such as that represented by the Is electrons, the potential is
exactly the same as though the latter were located at the center (nucleus). In this case
the valence electron in the 2s orbital wouid experience a potential equivalent to a
net nuclear charge of one (Z = 1.0). A charge which penetrates the cage will be un¬
shielded and would experience a potential equivalent to the fuil nuclear charge,
Z = 3.0. This is not meant to imply that the energy of the 2s electron varies as it
penetrates the Is orbital, but that the energy is determined by an effective nuclear
charge, Z, which is somewhat less than the actual nuclear charge, Z:
Z = Z - 5 (2.8)
where S is the shielding or screening constant.
As a result of the presence of one or more maxima near the nucleus, s orbitals
are very penetrating and are somewhat less shielded by inner-shell electrons than
are orbitals with larger values of /. In turn, they tend to shield somewhat better than
other orbitals. Orbitals with high / values, such as d and / orbitals, are much less
penetrating and are far poorer at shielding.
In a similar manner the radial distributions of 3s, 3 p, and 3 d orbitals may be
compared (Fig. 2.4). Although the d orbitals are “smaller” in the sense that the most
probable radius decreases in the order 3s >3 p > 3d, the presence of one node and an
intranodal maximum in the 3 p orbital and the presence of two nodes and two intra-
nodal maxima in the 3s orbital cause them to be affected more by the nucleus. Hence
the energies of these orbitals lie 3 d >3 p > 3s as we have seen in filling the various
energy levels previously.
In order to estimate the extent of shielding, a set of empirical rules has been pro¬
posed by Slater. 25 It should be realized that these rules are simplified generalizations
based upon the average behavior of the various electrons. Although the electronic
energies estimated by Slater’s rules are often not very accurate, they permit simple
estimates to be made and will be found useful in understanding related topics such as
atomic size and electronegativity.
To calculate the shielding constant for an electron in an np or ns orbital:
Write out the electronic configuration of the element in the following order and
groupings: (l.v) (2s, 2 p) (3s, 3 p) (3d) (4s, 4p) (4 d) (4 f) (5s, 5 p), etc.
Electrons in any group to the right of the (ns, np) group contribute nothing to
the shielding constant.
23 Staler. J. C. Pltys. Ruv. 1930. 36. 57.
84
3 • Symmetry and Group Theory
Problems
85
Fig. 3.34 ORTEP drawing
of the one-to-one
adduct C6 o(Os 0 4 )
(4-£-Bu-pyridine) 2 showing
the relationship of the
osmyi unit with the carbon
cluster. [From Hawkins,
J. M.; Meyer, A.; Lewis,
T. A.; Loren, S.; Hollander,
F. J. Science 1991, 252,
312-313. Reproduced with
permission.]
The presence of one or more bis(4-r-butylpyridine)osmyl groups provides “handles”
to break the pseudospherical symmetry of the C 60 molecules and anchor them in
the crystal. The X-ray crystal structure of the one-to-one adduct was readily accom¬
plished. This allowed the authors to say:
“The crystal structure [Fig. 3.34] confirms the soccer ball-like arrangement of
carbon atoms in C 60 by clearly showing the 32 faces of the carbon cluster com¬
posed of 20 six-membered rings fused with 12 five-membered rings. No two five-
membered rings are fused together, and each six-membered ring is fused to alternating
six- and five-membered rings. The O—Os—O unit has added across a six-six ring
fusion.. .” 47
A phase change for C 60 occurs at 249 K (face-centered cubic to simple cubic).
The low-temperature phase, unlike that at room temperature, is ordered and thus
crystallographers have been able to obtain a crystal structure of the underivatized
product. 48 In addition, the gas-phase structure has been determined by electron dif¬
fraction. 49 The gas-phase carbon-carbon bond lengths are 140.1 pm for the bond
11 Hawkins, J. M.; Meyer, A.; Lewis, T. A.; Loren, S.; Hollander, F. J. Science 1991, 252, 312-313.
Liu. S.; Lu. Y.-J; Kappcs. M. M.; Ibers, J. A. Science 1991. 254, 408-410. David. W. I. F.; Ibbcrson.
R. M.; Matthcwman, J. C.; Prassides, K.; Dennis. T. J. S.; Hare. J. P.; Kroto, H. W.-.Taylor. R.; Walton.
D. R. M. Nature 1991. 353, 147-149. Burgi, H.-B.; Blanc, E.; Schwarzenbach. D.: Liu. S.; Lu, Y.-J.;
Kappes, M. M.; Ibers. J. A. Angew. Chem. Int. Ed. Engl. 1992. 31, 640-643.
49 Hedberg, K.; Hedberg, L.; Bethune, D. S.; Brown, C. A; Dorn, H. C; Johnson, R. D.; DeVries, M.
Science 1991, 254, 410-412.
‘Problems
fusing six-membered rings and 145.8 pm for the bond fusing five- and six-membered
rings.
Buckminsterfullerene has a high electron affinity. Treatment with up to six moles
of an alkali metal such as potassium or rubidium gives products
nM + C 60 -♦ M„C 60 (3.3)
which show metallic conductivity. If only three moles of potassium or rubidium are
allowed to react, the products consist of K 3 C 60 and Rb 3 C 60 which become super¬
conducting at temperatures below 18 K. and 30 K, respectively. 50 These compounds
have a face-centered cubic cryolite structure with a closest-packed array of
anions with M + ions in the tetrahedral and octahedral holes (see Chapter 4). At
present it does not appear that these superconductors will prove competitive with the
cuprate high-temperature superconductors (Chapter 7) because of their much lower
critical temperature and the fact that they are quite susceptible to oxidation.
Buckminsterfullerene, C 60 , appears to be the first of a large number of allotropic
fullerenes: C 70 is already fairly well known—it probably has the shape of a rugby
ball—and other C„ molecules with n = 76, 84, 90, 94 have been isolated. Even larger
molecules with n equal to 240 and even 540 have been suggested. 51
3.1 Assign the molecules in Figs. 3.2, 3.4, and 3.7 lo their appropriate point groups.
3.2 Assign the molecules in Figs. 3.5, 3.6, and 3.8 to their appropriate point groups.
3.3 Assign the following molecules to their appropriate point groups.
a. cyclopropane
d. B 2 H 6
g- BF 3
j. OSCIj
b. SO,
e. P 4
h. PH 3
H
\ /
k. O —B
\
O —H
c. C0 2
f. C1 2 C=C=CCI 2
i. OjSCI;
I
.1
,P—P‘
rv
3.4 Assign the following to their appropriate point groups.
a. tris(oxalato)chromium(III) ' b. tris(carbonato)cobalt(III)
c. tris(glycinato)cobalt(IH)
3.5 Groups with l h , O h , T d , C„ C„ and C, symmetries were assigned in the text by inspection.
Take the molecules given as illustrations of these symmetries (Figs. 3.10 and 3.11) and
run them through the flowchart (Fig. 3.16) to assign their proper point groups.
3.6 Although most molecules in point groups l h , O h , T d , C„ C s , and C, may be assigned by
inspection, some appear unusual. Consider the cubic symmetry of cubanc, C 8 H a . To
50 Fleming, R. M.; Ramirez, A. P.; Rosseinsky, M. J.; Murphy. D. W.; Haddon, R. C.; Zahurak, S. M.;
Makhija, A. V. Nature 1991, 352, 787.
51 Diederich, F.; Ettl, R.; Rubin, Y.; Whetten, R. L.; Beck, R.; Alvarez, M.; Anz, S.; Sensharma, D.; Wudl,
F.; Kheman, K. C.; Koch, A. Science 1991, 252, 548; Kroto, H. Pure Appl. Chem. 1990, 62, 407-415.
274
7 • The Solid State
Solids Held Together by Covalent Bonding
275
Impurity and Defect
Semiconductors
Instead of silicon or germanium with four valence electrons (to yield a filled band
of 4 + 4 = 8 electrons on band formation), we can form a compound from gallium
(three valence electrons) and arsenic (five valence electrons) to yield gallium arsenide
with a filled valence band. In general, however the A E for the band gap will differ from
those of elemental semiconductors. The band gap will increase as the tendency for
electrons to become more and more localized on atoms increases, and thus it is a
function of the electronegativities of the constituents (Fig. 7.25). Note that conduc¬
tivity is a continuous property ranging from metallic conductance (Sn) through ele¬
mental semiconductors (Ge, Si), compound semiconductors (GaAs, CdS) to insula¬
tors, both elemental (diamond, C) and compounds (NaCl).
Consider a pure crystal of germanium. Like silicon it will have a low intrinsic
conductivity at low temperatures. If we now dope some gallium atoms into this
crystal, we shall have formed holes because each gallium atom contributes only three
electrons rather than the requisite four to fill the band. These holes can conduct
electricity by the process discussed above. By controlling the amount of gallium
impurity, we can control the number of carriers.
Thinking only in terms of electrons or holes that are completely free to move
suggests that there would be no energy gap in a gallium-doped germanium semicon-
Fig. 7.25 Empirical relationship between energy gap and the electronegativities of the
elements present. Note that substances made from a single, fairly electronegative atom
(C, diamond) or from a very low-electronegativity metal and high-electronegativity nonmetal
(NaCl) are good insulators. As the electronegativities approach 1.75, the electronegativity
function rapidly approaches zero. [From Hannay, N. B. Solid-State Chemistry ; Prentice-Hall:
Englewood Cliffs. NJ, 1967. Reproduced with permission.]
ductor. However, note that gallium lies to the left of germanium in the periodic table
and is more electropositive; it thus tends to keep the positive hole. (Alternatively,
germanium is more electronegative, and the electrons tend to stay on the germanium
atoms rather than flow into the hole on the gallium atom.) This electronegativity efFect
creates an energy gap, as shown more graphically in Fig. 7.26. The electronic energy
levels for gallium lie above the corresponding ones for germanium^ and thus above
the germanium valence band. Providing a small ionization energy, A£, generates the
holes for semiconduction. The resulting system is called an acceptor (since gallium
can accept an electron) or p-type (p = positive holes) semiconductor.
In an exactly analogous but opposite manner, doping germanium with arsenic
(five valence electrons) results in an excess of electrons and a donor (the arsenic
donates the fifth electron) or n-type (n = negative electrons) semiconductor. The
conduction can be viewed in terms of an energy diagram in which the electrons can be
removed from the impurity arsenic atoms to the conduction band of the semiconduc¬
tor (Fig. 7.27).
Fig. 7.26 Conduction by
holes in an acceptor or
p-type semiconductor.
Donor levels
Energy gap
Fig. 7.27 Conduction by
electrons in a donor or
n-type semiconductor.
24 The relation between energy levels and electronegativity is presented in Chapter 5.
36
3- Symmetry and Group Theory
what point group does it belong? Consider the pentagonal dodecahedron. (For a model,
see Appendix H.) To what point group does it belong?
3.7 Find the symmetry elements, if any, in the objects shown in Fig. 3.1.
3.8 In the discussion of crystallography, translational symmetry was likened to moving ducks
and blinking eyes. Extend this discussion to the action of a strobe light blinking t times
per second relative to the following.
a. ducks moving with a certain linear velocity
b. a spoked wheel spinning at a certain angular velocity
3.9 Why does SF 4 have C 2 „ symmetry rather than C„? (Sec Fig. 6.4).
3.10 Find all of the symmetry elements in an octahedron.
3.11 Tris(2-aminoethoxo)cobalt(IlI) (Fig. 3.12b) was assigned C 3 symmetry, but the methylene
groups of the ligand were not drawn out explicitly. Does consideration of these groups
change the symmetry? Discuss.
3.12 Gauc/ie-H 2 0 2 has C 2 symmetry. What are the symmetries of the eclipsed (cis) conforma¬
tion and the anti (trans) conformation?
3.13 On page 64 the argument is made that a molecule with a center of symmetry, /, cannot have
a molecular dipole moment. Prove this same rule using a molecule with a center of
symmetry and summing up the individual bond moments.
3.14 Meso molecules such as R,S-l,2-dichloro-I,2-difiuorocthanc are usually cited as achiral
because they possess a mirror plane (when in the perfectly eclipsed conformcr. a stipu¬
lation that is often omitted). What is the symmetry of this conformcr? In other con¬
formations? Discuss why meso molecules show no optical activity,
3.15 Which of the following molecules will have dipole moments?
Problems
87
3.17 Unlike the water molecule, carbon dioxide has no dipole moment. How is it possible for
any of its vibrational modes to be infrared active?
3.18 Sketch the normal vibrational modes for C0 2 and indicate which you expect to be
infrared or Raman active, or both.
3.19 A hydrogen bond consists of a positive hydrogen between two very negative nonmetallic
atoms. One of the strongest hydrogen bonds known, the HF7 ion, will be discussed in
Chapter 8. Possible arrangements of the atoms in [FHF] ~ ion are a) iinear or b) bent,
with either: i)equal, [F—H— F]', orii) unequal, [F—H-F]', bond lengths. The fun¬
damental vibrational absorption frequencies (in cm"') of the hydrogen difluoride anion
and the deutero-substituted anion are as listed below. 52
HF 2 -
df 2
Activity
1550 cm" 1
1140 cm -1
IR
1200 cm' 1
860 cm - 1
IR
675 cm" 1
675 cm' 1
Raman
Suggest the structure of the hydrogen difluoride ion. Explain your reasoning.
3.20 Proceeding through an analysis analogous to that described in this chapter for BCI 3 ,
derive the irreducible representations for the normal vibrations of XeF 4 (Fig. 3.24) and
determine which arc IR active, which are Raman active, and which are neither.
3.21 Determine the number of fundamental vibrations that would be expected for XeF 4 if it
were tetrahedral. How many of these would be infrared or Raman active? Compare these
results as well as those obtained from Problem 3.20 with the experimental data given on
page 70. and give us many reasons as possible lor eliminating the tetrahedral structure.
3.22 Determine the irreducible representation of each of the fundamental vibrations of trans-
[PtCUBr,] 2 " (a square planar structure). Which are IR active?
3.23 Use your answer to Problem 3.22 and Footnote 28 to determine whether the first over¬
tone of any of the fundamental vibrations of rrans-[PtCI 2 Br 2 ] 2 will be IR active.
3.24 How many absorptions would you expect to see in the infrared spectrum of the T-shaped
CIFj molecule?
3.25 Infrared and Raman spectra of crystalline barium rhodizonatc have been reported, 53
From an inspection of the data, the author concluded that the rhodizonate ion, C 6 0$',
(Fig. 16.35) probably has D 6 „ symmetry. Examine the data below (frequency in cm" 1 )
and explain this conclusion.
152 (IR) 275 (IR) 380 (IR) 450 (Raman) 548 (Raman)
1071 (IR) 1278 (IR) 1305 (IR) 1475 (IR) 1551 (Raman)
3.26 Bromine pentafluoride reacts with lithium nitrate to produce BrONO,:
3LiNOj + BrF s -► 3LiF + BrONO, + 2FNO, + O, (3.4)
Draw the structure of the planar BrONO; molecule and determine the irreducible repre¬
sentations for its vibrational modes. Which modes arc IR active and which ones are
Raman active? (Wilson. W. W.; Chrisle. K. O. Inorn. Client. 1987, 26. 1573.)
32 Harris, D. C.; Bertolucci, M. D. Symmetry and Spectroscopy ; Dover New York. 1989; p 160.
33 Bailey, R. T. J. Chem. Soc., Chem. Commun. 1970, 322.
272
7 • The Solid State
Intrinsic and
Photoexcited
Semiconductors
lowered in energy
Fig. 7.22 Effect of an electric field on the energy levels in a metal: (a) no field, no net How
of electrons; (b) field applied, net flow of electrons to the right.
the metal is conducting electricity. If the band is completely filled (Fig. 7.23). there is
no possibility of transfer of electrons and, despite the presence of a potential, equal
numbers of electrons (low either way; therefore, the net current is zero and the
material is an insulator.
All insulators will have a filled valence band plus a number of completely empty bands
at higher energies, which arise from the higher-energy atomic orbitals. For example,
the silicon atom will have core electrons in essentially atomic orbitals U 2 , 2s 2 , and
2p\ and a valence band composed of the 3s and 3 p orbitals. Then there will be empty
orbitals arising out of combinations of 3d, 4s, 4 p. and higher atomic orbitals. If the
temperature is sufficiently high, some electrons will be excited thermally from the
valence band to the lowest-lying empty band, termed the conduction band (Fig. 7.24).
The number excited will be determined by the Boltzmann distribution as a function of
temperature and band gap, A E. Before discussing the source of the magnitude of the
energy gap, let us note typical band-gap and conductivity values for insulators (dia¬
mond, C), semiconductors (Si, Ge), and an "almost metal,” gray tin. 23
C
Si Go
Sn
(gray)
Bandgap, kJ mol 1
580
105 69
7
Conductivity, ft -1 cm' 1
<10-"'
5 x I0- 0.02
10-
n Van Vlack, L. H. Elements of Materials Science and Engineering, 5th ed.; Addison-Wesley:
Reading. MA, 1985; p 303. Gray tin has the same structure as diamond. Metallic tin is called white
tin and has a distorted octahedral environment about each tin atom. It conducts electricity like other
metals.
Solids Held Together by Covalent Bonding
273
Fig. 7.23 Effect of an
electric field on an
insulator. Even with applied
potential, flow is equal in
both directions.
Conduction band
Energy gap
Valence band
Fig. 7.24 Thermal
excitation of electrons in an
intrinsic semiconductor.
The x's represent electrons
and the o's holes.
For every electron excited to the antibonding conduction band, there will remain
behind a hole, or vacancy, in the valence band. The electrons in both the valence band
and the conduction band will be free to move under a potential by the process shown
in Fig. 7.22b, but since the number of electrons (conduction band) and holes (valence
band) is limited, only a limited shift in occupancy from left-bound states to right-bound
states can occur and the conductivity is not high as in a metal. This phenomenon,
known as intrinsic semiconduction, is the basis of thermistors (temperature-sensitive
resistors).
An alternative picture of the conductivity of the electrons and holes in intrinsic
semiconductors is to consider the electrons in the conduction band as migrating, as
expected, toward the positive potential, and to consider the holes as discrete, positive
charges migrating in the opposite direction. Although electrons arc responsible for
conduction in both cases, the hole formalism represents a convenient physical picture.
If, instead of thermal excitation, a photon of light excites an electron from the
valence band to the conduction band, the same situation of electron and hole carriers
obtains, and one observes the phenomenon of photoconductivity, useful in photocells
and similar devices.
88
3- Symmetry and Group Theory
3.27 Fig. 3.25 shows removal of B from octahedral AB 6 to give square pyramidal AB 5 and
loss of a second B to give seesaw AB 4 . Suppose that instead of the geometries shown,
the AB 5 rearranged to give a trigonal bipyramidal structure and AB 4 assumed a square
planar shape. What orbital symmetries and degeneracies would occur for these two cases?
3.28 Consider the following AB„ molecules and determine the symmetries and degeneracies
of the s, p, and d orbitals on A in each.
a. AB 8 (cube) b. AB 4 (square plane)
c. AB 3 (trigonal pyramid) d. AB 3 (trigonal plane)
e. AB 3 (T-shape) f. AB 4 (rectangular plane)
3.29 For each of the following molecules, determine what atomic orbitals on the central atom
are allowed by symmetry to be used in the construction of sigma hybrid orbitals.
a. NH 3 (trigonal pyramid) b. BF 3 (trigonal plane)
c. SF 6 (octahedron) d. PF S (trigonal bipyramid)
3.30 A chemist isolated an unknown transition metal complex with a formula of AB 6 . Five
potential structures were considered, belonging to point groups O h , D }ll , D bh , D 2h , and
D 3 ,/. Spectroscopic studies led to the conclusion that the p orbitals originating on A in the
complex were completely nondegenerate. Sketch a structural formula that is consistent
with each of the five point group assignments and decide which structures can be elimi¬
nated on the basis of the experimental results.
3.31 What atomic orbitals on carbon in the planar C0 3 ~ anion could be used (on the basis
of symmetry) to construct in-plane and out-of-plane n bonds? First answer the question
by thinking about the orientations of the orbitals relative to the geometry of the ion;
then answer it by using reducible representations and the appropriate character table.
3.32 What is the symmetry of buckminsterfullerene? 54 Do you expect it to be chiral? To have
a dipole moment? To be soluble in benzene? Buckminsterfullerene was named after R.
Buckminster Fuller, who became best known for his popularization of the geodesic dome.
Is a geodesic dome the same as a segment of buckminsterfullerene? What is the symmetry
of the bis(4-r-butylpyridine)osmyl derivative of buckminsterfullerene (Fig. 3.34)? Do you
expect it to be chiral? To have a dipole moment? To be soluble in benzene?
3.33 Flow many 13 C NMR signals do you expect to see for C 60 ? How many for C 70 ?
3.34 Look up carbon-carbon bond lengths (single, double, and aromatic) in an organic chem¬
istry textbook and compare with the bond lengths in buckyball. What can you conclude
about the bonding in buckyball?
3.35 Depending upon the conditions, reactant ratios, etc., the products of Eq. 3.2 consist of (1)
a toluene-soluble fraction that gives a single, sharp chromatographic peak for a material
that analyzes C 60 O 4 Os(NCjH 4 C 4 H 9 ) 2 and yields the structure shown in Fig. 3.34, and
(2) a precipitate that analyzes as C 6 o[0 4 Os(NC 5 H 4 C 4 H9)2]2. What is the significance of a
single bis(4-r-butylpyridine)osmyl derivative (1) of buckminsterfullerene? When (2) is
analyzed chromatographically, five peaks are observed. Discuss.
3.36 It has been suggested that if the potassium (or rubidium) atoms in the M 3 C 60 super¬
conductors could somehow be placed inside the buckyballs, they would be protected,
and then these superconductors would not be susceptible to oxidation. Comment.
u In answering this question, you may find it useful to build a model of buckyball. See Vittal. J. J. J.
Chem. Educ. 1989, 66, 282.
Problems
89
Fig. 3.35 Scanning
electron micrograph of a
cluster of quasicrystals of
Al 6 CuLij. [Courtesy of
B. Dubost and P. Sainfort,
Pechiney, France.]
3.37 What are the symmetries of the following?
a. a baseball b. a baseball glove
c. a baseball bat d. a volleyball
e. a hockey puck f. a soccerball
g. a football h. a seamless rubber ball
3.38 Construct models of the tetrahedron, the octahedron (both with and without “propeller
blades” representing chelate rings), and the icosahedron (Appendix H). Find and mark
as many symmetry elements as you can.
3.39 Recently "quasicrystals" having the shape of a triacontahedron have been discovered in
specially prepared alloys of aluminum and other metals. A triacontahedron is a regular
polyhedron with 30 identical, diamond-shaped faces (Fig. 3.35). Quasicrystals seemingly
defy the rules of symmetry that do not allow a periodic structure having unit cells with
five-fold symmetry. 55 What is the symmetry of a triacontahedron? Can you make a model
of it similar to the polyhedra given in Appendix H?
3.40 Figures 3.36 and 3.37 illustrate two woodcuts by artist Maurits Escher. What symmetry
elements can you find? 56
3.41 Among the thirteen possible monoclinic space groups are P2,, P2,/m, and P2Jc. Com¬
pare these space groups by listing the symmetry elements for each.
3.42 Often hydrogen atoms cannot be located crystallographically if there are heavy atoms
present. In a study of H 3 F 2 Sb 2 Fj’|, the hydrogen bonded cation was found to have a
structure of either:
55 Horgan, J. Science 1990. 247, 1020-1022.
56 For a comprehensive volume comparing symmetry in art, music, chemistry, and other human endeavors,
see Symmetry. Unifying Human Understanding; Hargittai, I., Ed.; Pergamon: New York, 1986.
270
7 • The Solid State
germanium, compounds between these elements, such as gallium arsenide, or various
nonstoichiometric or defect structures. In electrical properties they fall between
conductors and nonconductors (insulators).
Band Theory In order to understand the bonding and properties of an infinite array of atoms of a
metallic element in a crystal, we should first examine what happens when a small
number of metal atoms interact. For simplicity we shall examine the lithium atom,
since it has but a single valence electron, 2s', but the principles may be extended to
transition and posttransition metals as well. When two wave functions interact, one of
the resultant wave functions is raised in energy and one is lowered. This is discussed
for the hydrogen molecule in Chapter 5. Similarly, interaction between two Is orbitals
of two lithium atoms would provide the bonding u energy level and the antibonding cr
energy level shown in Fig. 7.19. Interaction of n lithium atoms will result in n energy
levels, some bonding and some antibonding (Fig. 7.20). A mole of lithium metal will
provide an Avogadro's number (A/) of closely spaced energy levels (the aggregate is
termed a band), the more stable of which are bonding and the less stable, antibond¬
ing.- 1 Since each lithium atom has one electron and the number of energy levels is
equal to the number of lithium atoms, half of the energy levels will be filled whether
there are two, a dozen, or N lithium atoms. Thus, in the metal the band will be half
filled (Fig. 7.21), with the most stable half of the energy levels doubly occupied and the
least stable, upper half empty. The preceding statement is true only for absolute zero.
At all real temperatures the Boltzmann distribution 2 together with the closely spaced
energy levels in the band will ensure a large number of half-filled energy levels, and so
Fig. 7.19 Interaction of the 2s orbitals of two lithium atoms to form tr and <r molecular
orbitals. Cf. Figs. 5.1, 5.7.
21 The levels near the center of the band arc essentially nonbonding.
22 In the Boltzmann distribution, the population of higher energy states will be related to the value of
the expression e~ m ‘ where e is the base of natural logarithms, E is the energy of the higher state,
k is Boltzmann's constant, and T is the absolute temperature.
Solids Held Together by Covalent Bonding
271
Fig. 7.20 Interaction of eight 2s orbitals of eight lithium atoms. The spacing of the energy
levels depends upon the geometry of the cluster.
Fig. 7.21 Bonding of a mole of lithium atom 2s orbitals to form a half-filled band. Heavy
shading indicates the filled portion of the band, the top of which is called the Fermi level, c,,.
The real situation is somewhat more complicated because the 2 p orbitals can interact as well.
the sharp cutoff shown in Fig. 7.21 should actually be somewhat fuzzy. The top of the
filled energy levels is termed the Fermi level (e F ).
Each energy state has associated with it a wave momentum either to (he left or to
the right. If there is no potential on the system, the number of states with electrons
moving left is exactly equal to the number with electrons moving right, so that there is
no net flow of current (Fig. 7.22a). However, if an electrostatic potential is applied to
the metal, the potential energy of the states with the electrons moving toward the
positive charge is lower than the states with them moving toward the negative charge;
thus, the occupancy of the states is no longer 50:50 (Fig. 7.22b). The occupancy of
states will change until the energies of the highest left and right states are equal. Thus,
there is a net transfer of electrons into states moving toward the positive charge, and
Fig. 3.36 “Butterfly, Bat,
Bird, Bee," a woodcut
by Maurits Escher.
[Reproduced with
permission; Copyright 1990
M. C. Escher c/o Cordon
Art—Baarn—Holland.]
Fig. 3.37 “Depth," a
woodcut by Maurits
Escher. [Reproduced with
permission; Copyright 1990
M. C. Escher c/o Cordon
Art—Baarn—Holland.]
Problems
91
The authors stated that "[The structure] has space group P I. Owing to the strongly scat¬
tering Sb atoms, the H atoms . . . could not be definitely localized . , . The H 3 F; ion . . ,
is located on a symmetry center of the space group and therefore has [a or b? Choose
one.] conformation.” 57 Discuss how the correct conformation, a or b, can be chosen by
symmetry arguments even if the hydrogen atoms cannot be located.
3.43 Fig. 3.38 is a stereoview of the unit cell of Fe(CO) 4 (r;'-PPh 2 CH 2 CH 2 PPh 2 ) which crys¬
tallizes in the monoclinic space group P2Jc. Find the symmetry elements of the unit
cell. [Hint: Find three easily recognized atoms in the Fe(CO) 4 (// l -PPh,CH 2 CH 2 PPh 2 )
molecule and connect corresponding atoms in the four molecules in the unit cell with tie
lines. Think about the relation of the intersection of these tie lines and the symmetry
elements.]
Fig. 3.38 Stereoview of the unit cell of Fe(CO) 4 (T| l -Ph : PCH 2 CH 2 PPh 2 ). [From Kcilcr,
R. L.; Rheingold. A. L.; Hamerski, J. J.: Castle, C. K. Ori’tinoiiteliillics 1983. 2,
1635-1639. Reproduced with permission.]
3.44 The usual procedure for identifying chiral molecules is to look for a mirror plane iir = 5,).
However, a center of inversion (;' = 5,) also produces an achiral molecule (sec K, 5-1.2-
dimelhyl-l.2-diphenyldiphosphine disulfide; Fig. 3.3). The presence ol an 5., axis will also
result in an achiral molecule. Is the (/?,/?'-2,3-diaminobutane)(S,.S”2.3-diaminobutane)-
zinc(ll) ion (Fig. 3.39) chiral? What about the bis(5.S'-2.3-diuminobulane)zine(II) ion?
H. \
\//
/ H
NH,’
/
n
\
Nil
...-H
-CH,
NH,'
Fig. 3.39 The
</t.W'-2.3-diaminobulanc).
(5.5'-2,3-diaminobutane)-
zinc(ll) ion.
3.45 Look at the drawings accompanying Problem 3.6. Is it possible to superimpose the cube
on the dodecahedron? Castleman and coworkers 58 have recently detected a cation with
m/e = 528, identified as Ti„CjV It is believed that the titanium atoms form a cube with
the addition of twelve carbon atoms to complete a pentagonal dodecahedron. Draw the
proposed structure. What is its point group symmetry?
51 Mootz, D.: Bartmann. K. Anyew. Chem. Int. Ed. Engl. 1988. 27. 391.
58 Guo. B. C: Kerns, K, P.; Castleman. A. W, Jr. Science 1992, 2SS, 1411-1412.
268
7 • The Solid Stote
Solids Held Together by Covalent Bonding
269
Spinel
block
tain composition. It was first thought to be "/3-alumina,’' a polymorph of the common
y-alumina, Al 2 0 3 . Its actual composition is close to the stoichiometric Na 2 AI 22 0 34
(= Na 2 0-1 IA1 2 0 3 ), but there is always an excess of sodium, as, for example,
Na 2 J8 AI,, 8 0 34 . The structure is closely related to spinel, with 50 of the 58 atoms in
the unit cell arranged in exactly the same position as in the spinel structure. 19 In fact,
sodium beta alumina may be thought of as infinite sandwiches composed of slices of
spinel structure with a filling of sodium ions. It is the presence of the sodium between
the spinel-like layers that provides the high conductivity of sodium beta alumina. The
Al—0—A1 linkages between layers act like pillars in a parking garage (Fig. 7.17) and
keep the layers far enough apart that the sodium ions can move readily, yielding
conductivities as high as 0.030 fT'cm -1 . There is a related structure called sodium /3"
alumina with the layers held farther apart and with even higher conductivities of up to
0.18 fl _l cm -1
There are many potential uses for solid electrolytes, but perhaps the most attrac¬
tive is in batteries. Recall that a battery consists of two very reactive substances (the
more so, the better), one a reducing agent and one an oxidizing agent (see Chapter 10
for a discussion of inorganic electrochemistry). To prevent them from reacting di¬
rectly, these reactants must be separated by a substance that is unreactive towards
both, and which is an electrolytic conductor but an electronic insulator. Generally (as
in the lead storage battery, the dry cell, and the nickel alkaline battery), solutions of
electrolytes in water serve the last purpose, but in most common batteries this reduces
the weight efficiency of the battery at the expense of reactants. The attractiveness of
solid electrolytes is that they might provide more efficient batteries.
Fig. 7.17 Relation of the spinel structure (left) to the structure of sodium beta alumina
(right). The sodium ions arc free to move in the open spaces between spinel blocks, held
apart by Al—O—Al pillars in the "parking garage" structure. [In part from Wells, A. F.
Structural Inorganic Chemistry. 5th ed.; Oxford University: Oxford. 1984. Reproduced with
permission.)
19 Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Clarendon: Oxford. 1984; pp 598-599.
Solids Held Together
by Covalent Bonding
Types of Solids
2Na + 3S “_ 2Na + S 3 "
Fig. 7.18 Sodium/sulfur battery with a sodium beta alumina solid electrolyte.
Consider the battery in Fig. 7.18. The sodium beta alumina barrier allows sodium
ions formed at the anode to flow across to the sulfur compartment, where, together
with the reduction products of the sulfur, it forms a solution of sodium trisulfidc in (he
sulfur. The latter is held at 300 °C to keep it molten. The sodium beta alumina also acts
like an electronic insulator to prevent short circuits, and it is inert toward both sodium
and sulfur. The reaction is reversible. At the present state of development, when
compared with lead storage cells, batteries of this sort develop twice the power on a
volume basis or four times the power on a weight basis.
Because some of the properties of solids that contain no ionic bonds may be conve¬
niently compared with those of ionic solids, it is useful to include them here despite
the fact that (his chapler deals primarily with ionic compounds.
We may classify solids broadly into three types based on their electrical conductivity.
Metals conduct electricity very well. In contrast, insulators do not. Insulators may
consist of discrete small molecules, such as phosphorus triiodide, in which the energy
necessary to ionize an electron from one molecule and transfer it to a second is too
great to be effected under ordinary potentials. 20 We have seen that most ionic solids
are nonconductors. Finally, solids that contain infinite covalent bonding such as
diamond and quartz are usually good insulators (but see Problem 7.5).
The third type of solid comprises the group known as semiconductors. These are
either elements on the borderline between metals and nonmetals, such as silicon and
30 Given sufficient energy, of course, any insulator can be made to break down.
C h a p t e r
Bonding Models in
Inorganic Chemistry:
1 . Ionic Compounds
Structure and bonding lie at the heart of modern inorganic chemistry. It is not too
much to say that the renaissance of inorganic chemistry following World War II was
concurrent with the development of a myriad of spectroscopic methods of structure
determination. Methods of rationalizing and predicting structures soon followed. In
this and following chapters we shall encounter methods of explaining and predicting
the bonding in a variety of compounds.
The Ionic Bond
Properties of Ionic Several properties distinguish ionic compounds from covalent compounds. These
Substances may be related rather simply to the crystal structure of ionic compounds, namely,
a lattice composed of positive and negative ions in such a way that the attractive
forces between oppositely charged ions are maximized and the repulsive forces be¬
tween ions of the same charge are minimized. Before discussing some of the possible
geometries, a few simple properties of ionic compounds may be mentioned: 1
Ionic compounds tend to have very low electrical conductivities as solids but
conduct electricity quite well when molten. This conductivity is attributed to
the presence of ions, atoms charged either positively or negatively, which are
free to move under the influence of an electric field. In the solid, the ions are
1 Some very interesting ionic compounds prove to be exceptions to these rules. They are discussed in
Chapter 7.
Although there is no sharp boundary between ionic bonding and covalent bonding,
it is convenient to consider each of these as a separate entity before attempting to
discuss molecules and lattices, in which both are important. Furthermore, because
the purely ionic bond may be described with a simple electrostatic model, it is ad¬
vantageous to discuss it first. The simplicity of the electrostatic model has caused
chemists to think of many solids as systems of ions. We shall see that this view needs
some modification, and there are, of course, many solids, ranging from diamond to
metals, which require alternative theories of bonding.
92
The Ionic Bond
93
bound tightly in the lattice and are not free to migrate and carry electrical
current. It should be noted that we have no absolute proof of the existence
of ions in solid sodium chloride, for example, though our best evidence will
be discussed later in this chapter (pages 111-113). The fact that ions are found
when sodium chloride is melted or dissolved in water does not prove that they
existed in the solid crystal. However, their existence in the solid is usually
assumed, since the properties of these materials may readily be interpreted in
terms of electrostatic attractions.
Ionic compounds tend to have high melting points. Ionic bonds usually are
quite strong and they are omnidirectional. The second point is quite important,
since ignoring it could lead one to conclude that ionic bonding was much
stronger than covalent bonding—which is not the case. We shall see that sub¬
stances containing strong, multidirectional covalent bonds, such as diamond,
also have very high melting points. The high melting point of sodium chloride,
for example, results from the strong electrostatic attractions between the so¬
dium cations and the chloride anions, and from the lattice structure, in which
each sodium ion attracts six chloride ions, each of which in turn attracts six
sodium ions, etc., throughout the crystal. The relation between bonding, struc¬
ture, and the physical properties of substances will be discussed at greater
length in Chapter 8.
Ionic compounds usually are very hard but brittle substances. The hardness of
ionic substances follows naturally from the argument presented above, except
in this case we are relating the multivalent attractions between the ions with
mechanical separation rather than separation through thermal energy. The
tendency toward brittleness results from the nature of ionic bonding. If one
can apply sufficient force to displace the ions slightly (e.g., the length of one-
half of the unit cell in NaCl), the formerly attractive forces become repulsive
as anion-anion and cation-cation contacts occur; hence the crystal flies apart.
This accounts for the well-known cleavage properties of many minerals.
Ionic compounds are often soluble in polar solvents with high permittivities
(dielectric constants). The energy of interaction of two charged particles is
given by
(4.1)
where q and q~ are the charges, r is the distance of separation, and e
is the permittivity of the medium. The permittivity of a vacuum, e 0 , is
8.85 x 10~ 12 C 2 m" 1 J' 1 . For common polar solvents, however, the per¬
mittivity values are considerably higher. For example, the permittivity is
7.25 x 10' 10 C 2 m' 1 J _1 for water, 2.9 x IO~ 10 C 2 m' 1 J' 1 for acetonitrile,
and 2.2 x 10' 10 C 2 m" 1 J' 1 for ammonia, giving relative permittivities of
82 £q (H 2 0), 33 £q (CH 3 CN), and 25 £ 0 (NH 3 ). Since the permittivity of am¬
monia is 25 times that of a vacuum, the attraction between ions dissolved in
ammonia, for example, is only 4% as great as in the absence of solvent. For
solvents with higher permittivities the effect is even more pronounced.
Another way of looking at this phenomenon is to consider the interaction
between the dipole moments of the polar solvent and the ions. Such solvation
will provide considerable energy to offset the otherwise unfavorable energetics
of breaking up the crystal lattice (see Chapter 8).
266
7 • The Solid Slate
267
Conductivity
in Ionic Solids
Conductivity by
Ion Migration 17
Normally, ionic solids have very low conductivities. An ordinary crystal like sodium
chloride must conduct by ion conduction since it does not have partially filled bands
(metals) or accessible bands (semiconductors) for electronic conduction. The conduc¬
tivities that do obtain usually relate to the defects discussed in the previous section.
The migration of ions may be classified into three types.
Vacancy mechanism. If there is a vacancy in a lattice, it may be possible for
an adjacent ion of the type that is missing, normally a cation, to migrate into it,
the difficulty of migration being related to the sizes of the migrating ion and the
ions that surround it and tend to impede it.
Interstitial mechanism. As we have seen with regard to Frenkel defects, if an
ion is small enough (again, usualy a cation), it can occupy an interstitial site,
such as a tetrahedral hole in an octahedral lattice. It may then move to other
interstitial sites.
Interstitialcy mechanism. This mechanism is a combination of the two
above. It is a concerted mechanism, with one ion moving into an interstitial site
and another ion moving into the vacancy thus created. These three mecha¬
nisms are shown in Fig. 7.15.
In purely ionic compounds, the conductivity from these mechanisms is intrinsic
and relates only to the entropy-driven Boltzmann distribution; the conductivity will
thus increase with increase in temperature. Because the number of defects is quite
limited, the conductivities are low, of the order of 10" 6 !l _l cm '. In addition,
extrinsic vacancies will be induced by ions of different charge (sec page 264). There
exist, however, a few ionic compounds that as solids have conductivities several
orders of magnitude higher. One of the first to be studied and the one with the highest
room-temperature conductivity, 0.27 FT 1 cm -1 , is rubidium silver iodide, RbAg 4 I 5 . lx
-Vacancy -- Interstitial -»- Interstitialcy
Fig. 7.15 Mechanisms of ionic conduction in crystals with defect structures: (a) vacancy
(Schottky defect) mechanism, (b) interstitial (Frenkel defect) mechanism, (c) interstitialcy
(concerted Schotlky-Frenkel) mechanism.
17 Farrington, G. C.; Briant, J. L. Science 1979. 204. 1371. West, A. R. Basic Solid State Chemistry;
Wiley: New York. 1988: pp 300-330.
18 Geller, S. Acc. Chem. Res. 1978. II. 87. See also Footnote 17.
Conductivity in Ionic Solids
The conductivity may be compared with that of a 35% aqueous solution of sulfuric
acid. 0.8 fi -1 cm -1 . The structure consists of a complex (not a simple closest-packed)
arrangement of iodide ions with Rb + ions in octahedral holes and Ag + ions in
tetrahedral holes. Of the 56 tetrahedral sites available to the Ag + ions, only 16 are
occupied, leaving many vacancies. The relatively small size of the silver ion (114 pm)
compared with the rubidium (166 pm) and iodide (206 pm) ions give the silver ion more
mobility in the relatively rigid latice of the latter ions. Furthermore, the vacant sites
are arranged in channels, down which the Ag can readily move (Fig. 7.16).
Another solid electrolyte that may lead to important practical applications is
sodium beta alumina. Its unusual name comes from a misidentification and an uncer-
Fig. 7.16 Structure of RbAgJ, crystal. Iodide ions are represented by large spheres,
rubidium ions by small white spheres. Tetrahedral sites suitable for silver ions are marked
w.th short sleeves on horizontal arms. (The easiest to see is perhaps the one formed by the
triangle of iodide ions front left with the fourth iodide ion behind and to the right.)
Conduction is by movement of Ag ions from one tetrahedral site to the next, down channels
in the crystal. One channel may be seen curving downward from upper center to lower left
(including the site mentioned above). [From Geller. S. Science 1967, 157. 310. Reproduced
with permission.)
94 4 ■ Bonding Model
Occurrence of Ionic
Bonding
Structures of Crystal
Lattices
s in Inorganic Chemistry: 1. Ionic Compounds
Simple ionic compounds form only between very active metallic elements and very
active nonmetals. 1 2 Two important requisites are that the ionization energy to form
the cation, and the electron affinity to form the anion, must be energetically favorable.
This does not mean that these two reactions must be exothermic (an impossibility—
see Problem 4.13), but means, rather, that they must not cost too much energy. Thus
the requirements for ionic bonding are (1) the atoms of one element must be able to
lose one or two (rarely three) electrons without undue energy input and (2) the atoms
of the other element must be able to accept one or two electrons (almost never three)
without undue energy input. This restricts ionic bonding to compounds between the
most active metals: Groups IA(1), IIA(2), part of IIIA(3) and some lower oxidation
states of the transition metals (forming cations), and the most active nonmetals:
Groups VIIA(17), VIA(16), 3 and nitrogen (forming anions). -1 All ionization energies
are endothermic, but for the metals named above they are not prohibitively so. For
these elements, electron affinities are exothermic only for the halogens, but they are
not excessively endothermic for the chalcogens and nitrogen.
Before discussing the energetics of lattice formation, it will be instructive to examine
some of the most common arrangements of ions in crystals. Although only a few of
the many possible arrangements are discussed, they indicate some of the possibilities
available for the formation of lattices. We shall return to the subject of structure after
some basic principles have been developed.
The first four structures described below contain equal numbers of cations and
anions, that is, the I: I and 2:2 salts. Most simple ionic compounds with such for¬
mulations crystallize in one of these four structures. They differ principally in the
coordination number, that is, the number of counterions grouped about a given ion,
in these examples four, six, and eight.
The sodium chloride structure. Sodium chloride crystallizes in a face-centered
cubic structure (Fig. 4.1a). To visualize the face-centered arrangement, consider
only the sodium ions or the chloride ions (this will require extensions of the sketch
of the lattice). Eight sodium ions form the corners of a cube and six more are
centered on the faces of the cube. The chloride ions are similarly arranged, so that
the sodium chloride lattice consists of two interpenetrating face-centered cubic
lattices. The coordination number (C.N.) of both ions in the sodium chloride
lattice is 6, that is, there are six chloride ions about each sodium ion and six
sodium ions about each chloride ion.
Sodium chloride crystallizes in the cubic space group Fmlm (see Table 3.7).
that is, it is face-centered, has a three-fold axis, and has two mirror planes of
different class. If there is one C 3 axis, however, three others must exist, and the
1 It is true that ionic compounds such as [NH il ]'[B(C 6 Hj) 1 ]~ arc known in which there are no extremely
aclive metals or nonmetals. Nevertheless, the above statement is for all practical purposes correct, and
we can consider compounds such as ammonium tetraphenylboratc to result from the particular covalent
bonding properties of nitrogen and boron.
3 Recall from the discussion in Chapter 2: Roman numerals arc from the "American System" and Arabic
numerals arc from the “1-18 System" of labeling the periodic table.
Since the transition between ionic bonding and covalent bonding is not a sharp one. it is impossible
to define precisely the conditions under which it will occur. However, the generalization is helpful and
does not rule out the possibility of unusual ionic bonds, for example, between two metals: Cs'Au~. Sec
Chapter 12.
The Ionic Bond 95
(a)
(b)
Fig. 4.1 Crystal structures of two 1:1 ionic compounds: (a) unit cell of sodium chloride,
cubic, space group Fm3m; (b) unit cell of cesium chloride, cubic, space group Pm3m. [From
Ladd. M. F. C. Structure and Bonding in Solid State Chemistry; Wiley: New York, 1979.
Reproduced with permission.]
presence of two different mirror planes requires seven others. In fact, this com¬
pact symmetry label is enough to tell us that all elements of symmetry found in
an octahedron are present. Thus, the Schoenfiies equivalent of Fm3m is O h .
The sodium chloride structure is adopted by most of the alkali metal halides:
All of the lithium, sodium, potassium, and rubidium halides plus cesium fluoride.
It is also found in the oxides of magnesium, calcium, strontium, barium, and
cadmium.
The cesium chloride structure. Cesium chloride crystallizes in the cubic ar¬
rangement shown in Fig. 4.1b. The cesium or chloride ions occupy the eight
corners of the cube and the counterion occupies the center of the cube. 5 Again,
5 The structure of CsCl should not be referred to, incorrectly, as "body-ccntcrcd cubic”. True body-
centered cubic lattices have the same species on the corners and the center of the unit cell, as in the
alkali metals, for example.
264
7-The Solid State
© © © ©
© © © ©
© © ©
Fig. 7.11 Schottky defect
(cation vacancy) induced
and balanced by the
presence of a higher
valence cation. [Hannay,
N. B. Solid-State
Chemistry-, Prentice-Hall:
Englewood Cliffs, NJ. 1967.
Reproduced with
permission.]
amount of cadmium chloride, the Cd 2+ ion fits easily into the silver chloride lattice (cf.
ionic radii. Table 4.4). The dipositive charge necessitates a vacancy to balance the
change in charge (Fig. 7.11). Closely related is the concept of "controlled valency," in
which a differently charged, stable cation is introduced into a compound of a transition
metal. Because the latter has a variable oxidation state, balance is achieved by gain or
loss of electrons by the transition metal. For example, consider Fig. 7.12.
Stoichiometric nickel(ll) oxide, like aqueous solutions containing the Ni 2+ ion, is pale
green. Doping it with a little LUO causes a few of the cation sites to be occupied by + I
lithium in place of +2 nickel’ This induces a few Ni2 + ions to lose electrons and
become Ni 3+ ions, thus preserving the electrical neutrality of the crystal. The proper¬
ties of the NiO change drastically: The color changes to gray-black, and the former
insulator (to be expected of an ionic crystal, see Chapter 4), is now a semiconductor. 1 - 1
A rather similar effect can occur with the formation of nonstoichiometric com¬
pounds. For example, coppcr(I) sulfide may not have the exact ratio of 2:1 expected
from the formula, Cu 2 S. Some of the Cu ions may be absent if they are compensated
by an equivalent number of Cu 2+ ions. Since both Cu^ and Cu - "' ions are stable, it
Fig. 7.12 Controlled
valency (Ni 2 — Nf 1 ) by
addition of Li ions to
NiO. [From Hannay. N. B.
Solid-State Chemistry,
Prentice-Hall: Englewood
Cliffs, NJ. 1967.
Reproduced with
permission.]
© © © © ©
13 Semiconductors are discussed on pp 274-276 of this chapter.
© © © © ©
© © © © ©
© © © © ©
Imperfections in Crystals 265
is possible to obtain stoichiometries ranging from the ideal to Cu, 77 . If the “vacancy"
is not a true vacancy but contains a trapped electron at that site, the imperfection is
called an F center. For example, if a small amount of sodium metal is doped into a
sodium chloride crystal, the crystal energy causes the sodium to ionize to Na + + e -
and the electron occupies a site that would otherwise be filled by a chloride ion (Fig.
7.13) . The resulting trapped electron can absorb light in the visible region and the
compound is colored (F = Ger. Farbe, color). The material may be considered a
nonstoichiometric compound, Na l+s Cl, 14 or as a dilute solution of "sodium
electride." 15
If the missing ion has not been completely removed as in a Schottky defect, but
only dislocated to a nearby interstitial site, the result is called a Frenkel defect (Fig.
7.14) . The vacancy and corresponding interstitial ion may be caused by a cation or an
anion, but because the cation is generally smaller than the anion, it will usually be
easier to fit a cation into an interstitial hole other than the one in which it belongs. For
the same reason, although it is theoretically possible to have both interstitial cations
and anions at the same time, at least one will ordinarily be energetically unfavorable
because of size. 16
Fig. 7.13 An F center: an
electron occupying an
anionic site.
o» f
Cl' Na
site Fig. 7.14 A Frenkel defect:
a cation displaced from its
normal site.
Vacant cation site
Oo
Anion Cation
14 Where S is small with respect to I.
' 5 See Chapter 10 for solutions of sodium electride in liquid ammonia.
Fine, M. E. In Treatise on Solid State Chemistry, Vol. I: The Chemical Structure of Solids', Hannay,
N. B.. Ed.; Plenum: New York. 1973; pp 287-290.
96
4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compound
The Ionic Bond
97
we must consider a lattice composed either of the cesium ions or of the chloride
ions, both of which have simple cubic symmetry. The coordination number of
both ions in cesium chloride is 8; that is, there are eight anions about each cation
and eight cations about each anion. The space group is Pm3m: The lattice is
primitive, but otherwise the symmetry elements are the same as in NaCl.
Among the alkali halides, the cesium chloride structure is found only in
CsCl, CsBr, and Csl at ordinary pressures, but all of the alkali halides except
the salts of lithium can be forced into the CsCl structure at higher pressures.
It is also adopted by the ammonium halides (except NH 4 F), T1C1, TIBr, T1CN,
CsCN, CsSH, CsSeH, and CsNH 2 .
The zinc blende and wurtzite structures. Zinc sulfide crystallizes in two dis¬
tinct lattices: hexagonal wurtzite (Fig. 4.2a) and cubic zinc blende (Fig. 4.2b). We
shall not ekiborate upon them now (see page 121), but simply note that in both
the coordination number is 4 for both cations and anions. The space groups are
P6j/nc and F43m. Can you tell which is which?
(a)
Fig. 4.2 Unit cells of two zinc sulfiide (2:2) structures; circles in order of decreasing size
are S and Zn: (a) wurtzite. hexagonal, space group P6 3 mc; (b) zinc blende, cubic, space
group F43m. [From Ladd. M. F. C. Structure and Bonding in Solid State Chemistry-. Wiley:
New York, 1979. Reproduced with permission.]
Fig. 4.3 Unit cell of the
fluorite structure; smaller
circle is Ca (not drawn to
scale): cubic, space group
Fm3m. [From Ladd,
M. F. C. Structure and
Bonding in Solid State
Chemistry-. Wiley: New
York, 1979. Reproduced
with permission.]
Many divalent metal oxides and sulfides such as BeO, ZnO, BeS, MnS, ZnS,
CdS, and HgS adopt the zinc blende or wurtzite structures, or occasionally both.
Other compounds with these structures include Agl, NH 4 F, and SiC.
All the following structures have twice as many anions as cations (1:2 struc¬
tures); thus the coordination number of the cation must be twice that of the
anion: 8:4, 6:3, 4:2, etc. The inverse structures are also known where the cations
outnumber the anions by two to one.
TheJluorite structure. Calcium fluoride crystallizes in the fluorite structure,
cubic Fm3m (Fig. 4.3). The coordination numbers are 8 for the cation (eight
fluoride ions form a cube about each calcium ion) and 4 for the anion (four
Ca 2+ ions tetrahedrally arranged about each F" ion).
Many difiuorides and dioxides are found with the fluorite structure. Exam¬
ples are the fluorides of Ca, Sr, Ba, Cd, Hg, and Pb, and the dioxides of Zr, Hf,
and some lanthanides and actinides. If the numbers and positions of the cations
and anions are reversed, one obtains the anlijluorite structure which is adopted
by the oxides and the sulfides of Li, Na, K, and Rb.
The rutile structure. Titanium dioxide crystallizes in three crystal forms
at atmospheric pressure: anatase, brookitc, and rutile (Fig. 4.4a). Only the last
(tetragonal P4 2 lmnm) will be considered here. The coordination numbers are 6
for the cation (six oxide anions arranged approximately octahedraliy about the
titanium ions) and 3 for the anion (three titanium ions trigonally about the oxide
ions). The rutile structure is also found in the dioxides of Cr, Mn, Ge, Ru, Rh,
Sn, Os, Ir, Pt, and Pb.
The [i-cristobalite structure. Silicon dioxide crystallizes in several forms
(some of which are stabilized by foreign atoms). One is /Tcristobaiite (Fig. 4.4b),
which is related to zinc blende (Fig. 4.2b) having a silicon atom where every zinc
and sulfur atom is in zinc blende, and with oxygen atoms between the silicon
atoms. 6 Other compounds adopting the /1-cristobalite structure are BcF 2 , ZnCI 2 ,
SiS 2 at high pressures, and Be(OH) 2 and Zn(OH) 2 , although the latter are dis¬
torted by hydrogen bonding. Another form of Si0 2 , tridymite, is related to the
<• The structure of /3-cristobalite has been determined several times over the past 60 years, but crystal
disorder has led to uncertainty in the space group assignment (Hyde, B. G.; Andersson, S. Inorganic
Crystal Structures; Wiley: New York. 1989; pp 393-395.
262
Imperfections in Crystal:
263
7• The Solid State
There is no simple explanation for the precise arrangements of all of the layered
structures. But certainly the forces involved are complex, often subtle, and a hard-
sphere ionic model will not come close to accounting for them. As Adams 10 has
pointed out:
“. . . ionic theory is a good starting point forgetting some general guidance on the
relative importance of factors such as size and coordination arrangement and is very
important in energetics, but for anything beyond this we must use the concepts and
language of modern valence theory and talk in terms of orbital overlap and band
structure. Ionic theory has had a good run . . . [now about three-fourths of a cen¬
tury] ... it is still heavily overemphasized; so far as detailed considerations of crystal
structure are considered it is time it was interred.”
These are the words of someone devoted to the details of solid state chemistry,
and they may be a bit overstated for the student merely wishing a general knowledge
of inorganic solids, but they are well taken, and certainly all of the interesting
subtleties of structure will be found to arise out of forces other than the electrostatics
of hard spheres. We shall examine these more closely later in this chapter.
Another Look at It was noted in Chapter 4 that the Madeiung constant of a structure may be expressed
Modelling Constants in various ways. The way that is conceptually simplest in terms of the Born-Lande
equation is the simple geometric factor, A, such that when combined with the true
ionic charges, Z + and Z~, the correct electrostatic energy is formulated. It was noted
that some workers have favored using another constant, A, combined with the highest
common factor of Z + and Z~, Z.
Further insight into the stability of predominantly ionic compounds can be gained
by inspection of the reduced Madeiung constant. A'. 11 The reduced Madeiung con¬
stant is closely related to the derivation of the Kapustinskii equation given earlier
(Chapter 4). Templeton 1 - showed that if the lattice energy of a compound M„,X t is
formulated as:
,, 1.389 x I0 5 kJ mol -1 pm 2 Z Z _ A'(m + .r)
u o ~ -) r (7.2)
ir o
then all Madeiung constants reduce to a value of about 1.7 (Table 7.1). The usefulness
of this viewpoint is that it indicates that despite the wide variety of ionic sizes,
compound formulations, and structures, there is basically an upper limit to lattice
energy set by the constraints of geometry; inefficient structures may be somewhat
below it (though not far: There will always be an alternative structure near the limit)
and even efficient structures may never rise above it.
10 Adams, D. M. Inorganic Solids', Wiley: New York. 1974; p 105. This is one of those rare books lhal
arc interesting to read for pleasure as well as for information. A second quote from Adams is:
" Plagiarise, plagiarise, let no one's work escape your eyes.' {Tom Lehrer}" We have obviously
felt that both of Adams' statements arc worth noting and following!
11 In place of the symbols A. A, A', some texts (see Jolly, W. L. Modern Inorganic Chemistry:
McGraw-Hill: New York. 1984, and Porterfield, W. M. Inorganic Chemistry: A Unified Approach:
Addison-Wcsley: Reading, MA, 1984) use M for the geometric factor, M Tor the '‘conventional"
factor, and M' for the reduced factor. The symbol used by Templeton (see Footnote 12), who
introduced the concept, was a,
' 2 Templeton. D. H. J. Chem. Phys. 1955, 23. 1826-1829.
Table 7,1 _
A comparison of Madeiung Compound, Geometric Conventional Reduced
constants M m X, factor, A factor, A factor, A
ai 2 o,
4.172
25.031
1.68
CaF 2
2.519
5.039
1.68
CdCl,
2.244
4.489
1.50
Cdl,
2.192
4.383
1.46
CsCI
1.763
1.763
1.76
NaCI
1.748
1.748
1.75
SiCV
2.298
4.597
1.47
Ti<V
2.408
4.816
1.60
ZnS'
1.641
1.641
1.64
ZnS7
1.638
1.638
1.64
° Johnson, Q.
C.; Templelon,
D. H. J, Chem.
Phys. 1961, 34.
Exact values depend upon details of structure.
c Cristobalite.
J Rutile.
<■ Wurtzite.
7 2inc blende.
Imperfections
in Crystals
To this point the discussion of crystals has implicitly assumed that the crystals were
perfect. Obviously, a perfect crystal will maximize the cation-anion interactions and
minimize the cation-cation and anion-anion repulsions, and this is the source of the
very strong driving force that causes gaseous sodium chloride, for example, to
condense to the solid phase. In undergoing this condensation, however, it suffers a
loss of entropy from the random gas to the highly ordered solid. This en¬
thalpy-entropy antagonism is largely resolved in favor of the enthalpy because of the
tremendous crystal energies involved, but the entropy factor will always result in
equilibrium defects at all temperatures above absolute zero.
The simplest type of defect is called the Schottky or Schotlky-Wagner defect. It is
simply the absence of an atom or ion from a lattice site. In an ionic crystal, electrical
neutrality requires that the missing charge be balanced in some way. The simplest way
is for the missing cation, for example, to be balanced by another Schottky defect, a
missing anion, elsewhere (Fig. 7.10).
Alternatively, the missing ion can be balanced by the presence of an impurity ion
of higher charge. For example, if a crystal of silver chloride is “doped” with a small
Fig. 7.10 Two Schottky
defects balancing each
other for no net charge.
93 4 •
(b)
Fig. 4.4 Crystal structures of two more 1:2 compounds; oxygen is the larger circle in both:
(a) unit cell of rutile, TiO,, tetragonal, space group P4 2 /mnrn; (b) unit cell of /J-cristobalite.
SiO[From Ladd, M. F. C. Structure anil Bonding in Solid State Chemistry, Wiley:
New York, 1979. Reproduced with permission.]
Lattice Energy 99
Fig. 4.5 Crystal structures of two forms of calcium carbonate: (a) unit cell of calcitc,
rhombohedral. space group R3c; (b) unit cell of aragonite, orthorhombic, space group Pcmn.
Circles in decreasing order of size are oxygen, calcium, and carbon. [From Ladd. M. F. C.
Structure and Bonding in Solid State Chemistry, Wiley: New York, 1979. Reproduced with
permission.]
wurtzite structure in the same way that /i-cristobalite is related to zinc blende.
The coordination numbers in (1-cristobaiite and tridymite are 4 for silicon and 2
for oxygen.
The calcite and aragonite structures. Almost all of the discussion in this
chapter is of compounds containing simple cations and anions. Nevertheless,
most of the principles developed here are applicable to crystals containing poly¬
atomic cations or anions, though often the situation is more complicated. Exam¬
ples of two structures containing the carbonate ion, CO]", are calcite (Fig. 4.5a)
and aragonite (Fig. 4.5b). Both are calcium carbonate. In addition MgC0 3 ,
FeC0 3 , LiN0 3 , NaN0 3 , InB0 3 , and YB0 3 have the calcite structure (rhombo¬
hedral R3c). The coordination number of the metal ion is 6. Larger metal ions
adopt the aragonite structure (orthorhombic Pcmn) with nine oxygen atoms
about the metal ion. Examples are, in addition to calcium carbonate, SrC0 3 ,
KN0 3 , and LaB0 3 .
Lattice Energy The energy of the crystal lattice of an ionic compound is the energy released when
ions come together from infinite separation to form a crystal:
M(g> + Xig) - MXisi (4.2)
It may be treated adequately by a simple electrostatic model. Although we shall
include nonelectrostatic energies, such as the repulsions of closed shells, and more
sophisticated treatments include such factors as dispersion forces and zero-point
energy, simple electrostatics accounts for about 90% of the bonding energies. The
theoretical treatment of the ionic lattice energy was initiated by Born and Lande,
and a simple equation for predicting lattice energies bears their names. The deriva¬
tion follows.
260
7 • The Solid State
The Structures of Complex Solids
261
Layered Structures
should expect for a complex such as [RhF 6 ]~. Bridging halide ions are well known in
coordination compounds. Furthermore, according to Fajans’ rules, we should be
suspicious of an ionic structure containing a cation with a +5 charge.
Sanderson 8 went so far as to say that even crystals such as alkali halides should be
considered as infinite coordination polymers with each cation surrounded by an
octahedral coordination sphere of six halide ions, which in turn bridge to five more
alkali metal atoms. Although this point of view is probably of considerable use when
discussing transition metal compounds, most chemists would not extend it to all ionic
lattices.
This brings us to a class of compounds too often overlooked in the discussion of
simple ionic compounds: the transition metal halides. In general, these compounds
(except fluorides) crystallize in structures that are hard to reconcile with the structures
of simple ionic compounds seen previously (Figs. 4.1-4.3). For example, consider the
cadmium iodide structure (Fig. 7.8). It is true that the cadmium atoms occupy
octahedral holes in a hexagonal closest packed structure of iodine atoms, but in a
definite layered structure that can be described accurately only in terms of covalent
bonding and infinite layer molecules.
Layered structures form, in some ways, an awkward bridge between simple
compounds with a high degree of ionicity (for which NaCl seems always to be the
prototype), less ionic compounds with considerable covalency but similar structure
(both AgCl and AgBr have the NaCl structure), and solids such as HgCI, and AI,Br 6
wherein the presence of discrete molecules seems apparent. A schematic illustrating
the relationships among some of these structures in terms of size and electronic
structure might look like this:
Increasing polarization due to
electron configuration
and cation
electronegativity
Increasing
polarization
due to
small cation
and large
anions
CaF,
Ti0 2
CdCI,
SiO,
Cdl,
HgCI,
Layered structures are extremely prevalent among transition metal halides. Examples
of compounds adopting the cadmium iodide structure or the related cadmium chloride
structure (Fig. 7.9) are:
Cdl, MCI 2 (M = Ti, V)
MBr 2 (M = Mg, Fe, Co, Cd)
MI 2 (M = Mg, Ca, Ti. V, Mn, Fe. Co. Cd, Ge, Pb. Th)
CdCI 2 MCI, (M = Mg, Mn, Fe, Co, Ni, Zn, Cd)
MBr 2 (M = Ni, Zn)
ML, (M = Ni, Zn)
8 Sanderson. R. T. J. Chem. Educ. 1967. 44, 516; Polar Covalence ; Academic: New York. 1983; pp
165-173.
Fig. 7.8 Stereoview of the
unit cell of the cadmium
iodide, Cdl,, structure type;
hexagonal, space group
/’3ml. Large circles, I;
small circles, Cd. [From
Ladd, M. F. C. Structure
and Bonding in Solid State
Chemistry, Ellis Horwood:
Chicester, 1979.
Reproduced with
permission.!
B
mm
dSS&M
mfk$.
Fig. 7.9 Layered structure
of cadmium chloride.
CdCI,. Note relationship to
extended NaCl structure
(missing atoms are dashed).
[From Wells, A. F.
Structural Inorganic
Chemistry, 5th ed.; Oxford
University: Oxford, 1984.
Reproduced with
permission.!
In the Cdl, structure, octahedral sites between every other pair of hep layers of
iodine atoms are occupied by cadmium atoms; CdCI, has cadmium in octahedral sites
between ccp layers. Actually, matters can gel much more complicated than lhat:
There are three different Cdl, structures with varied layers. To indicate the potential
complexity without going into the details of the several possibilities ,' 9 the structure of
one form of CdBrI will be mentioned: It has a twelve-layer repeating unit of halide
ions, ABCBCABABCAC, surely long-range order of a fairly high degree.
We have seen (Problem 4.1) that the change in stoichiometry from CsCI to CaF 2
can be accommodated readily in the simple cubic system by merely “omitting" every
other metal ion; the resulting fluorite structure is highly symmetrical and quite stable.
This is not the case when covalency becomes important. Fig. 7.9 illustrates this well:
Although the CdCI, structure can be readily related to the NaCl structure, the
omission of alternate cations is not such as to leave a structure with the same high
symmetry: The 3-D cubic symmetry has been transformed into a 2-D layer symmetry.
9 For more nearly complelc discussions sec Adams, D. M. Inorganic Solids', Wiley: New York. 1974;
pp 229-232; Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Oxford University: Oxford. 1984;
pp 258-265.
100
4 ■ Bonding Models in Inorganic Ch
istry: 1. Ionic Compounds
A
Fig. 4.6 Energy curves for
an ion pair.
Consider the energy of an ion pair, M, X , separated by a distance r. The
electrostatic energy of attraction is obtained from Coulomb’s law. 7
(4.3)
Since one of the charges is negative, the energy is negative (with respect to the energy
at infinite separation) and becomes increasingly so as the interionic distance de¬
creases. Figure 4.6 shows the coulombic energy of an ion pair (dotted line). Be¬
cause it is common to express Z + and Z“ as multiples of the electronic charge,
e = 1.6 x 10" 19 coulomb, we may write:
Z + Z~e 2
4ns 0 r
(4.4)
Now in the crystal lattice there will be more interactions than the simple one
in an ion pair. In the sodium chloride lattice, for example, there are attractions to
the six nearest neighbors of opposite charge, repulsions by the twelve next nearest
neighbors of like charge, etc. The summation of all of these geometrical interactions
is known as the Modelling constant, A. The energy of a pair of ions in the crystal is
then:
AZ+Z~e 2
Ait e 0 r
(4.5)
The evaluation of the Madelung constant for a particular lattice is straightforward.
Consider the sodium ion (®) at the center of the cube in Fig. 4.7. Its nearest neighbors
are the six face-centered chloride ions (•), each at a characteristic distance deter¬
mined by the size of the ions involved. The next nearest neighbors are the twelve
sodium ions (O) centered on the edges of that unit cell (cf. Fig. 4.1a inverted). The
distance of these repelling ions can be related to the first distance by simple geometry,
as can the distance of eight chloride ions in the next shell (those at the corners of
the cube). If this process is followed until every ion in the crystal is included, the
7 Note that these are ionic charges and not nuclear charges for which Z is also used.
Fig. 4.7 An extended lattice of sodium chloride. Starting with the sodium ion marked ®,
there are six nearest neighbors (•), twelve next nearest neighbors (O). eight next, next
nearest neighbors (darkly shaded), and so on.
Madelung constant, A, may be obtained from the summation of all interactions. The
first three terms for the interactions described above are
Fortunately, the Madelung constant may be obtained mathematically from a
converging series, and there are computer programs that converge rapidly. However,
we need not delve into these procedures, but may simply employ the values obtained
by other workers (Table 4.1). The value of the Madelung constant is determined
Table 4.1
Coordination
number
Geometrical
factor, A
Conventional
factor, A°
Madelung constants
of some common
crystal lattices
Structure
Sodium chloride
6:6
1.74756
1.74756
Cesium chloride
8:8
1.76267
1.76267
Zinc blende
4:4
1.63806
1.63806
Wurtzite
4:4
1.64132
1.64132
Fluorite
8:4
2.51939
5.03878
Rutile
6:3
2.408
4.816
/J-Cristobalite
4:2
2.298
4.597
Corundum
6:4
■4.1719
25.0312
“ Use Z ± = highest common factor.
r l %: m
Exact values depend upon details of structure.
■---» Sr — . iv
.•AS. -iU ii-
x i. a _ . 1
258
7 • The Solid State
A Second Look
at the Transition
from Ionic to
Covalent Solid
In the usual discussion of Fajans' rules as given in Chapter 4, emphasis is placed on
physical properties such as melting points, solubility, etc. The possible effects of
covalency on structure were also mentioned with regard to the fact that sp 1 bonding in
HgS could favor the tetrahedral zinc sulfide structure (Chapter 4). Therefore in
moving from an empirical structure field map to a semiempirical model (or semi-
theoretical, depending upon the viewpoint), we may take other factors into account.
For example, we have previously seen that covalent character might be expected to
cause a switch from coordination number 6 to coordination number 4. Therefore, we
might attempt to bring in covalent corrections to improve a purely mechanical or
radius ratio approach. There are two factors that increase covalency: (I) Small
differences in electronegativity produce highly covalent bonds; (2) other things being
equal, smaller atoms form stronger covalent bonds than larger atoms (see Chapter 9).
To incorporate these two variables, Pearson- 1 has plotted the principal quantum
number, n (a rough indicator of size), versus a function of electronegativity difference
(Ax) and radius ratio. r+/r_. and has shown that compounds with coordination
number 4 segregate quite well from those with coordination number 6.
More recently, Shankar and Parr 4 have designed structure stability diagrams in
which the compounds are plotted according to electronegativity and hardness (Fig.
7.6). 5 When the data are presented in this way, the compounds segregate quite well by
structure and coordination number. The slope of any line passing through the origin is
given by:
u l h
b A + b \
(7.1)
This is the simplest expression for charge (ionicity) in the Mulliken-Jaff6 system with
electronegativity equalization (see Eq. 5.85). The boundary lines in Fig. 7.6 radiate
more or less from the origin with the slope of m from Eq. 7.1. We may thus infer that
each represents a line of constant ionicity that is responsible for the changeover from
one structure type to the next.
Other workers have developed the ideas presented here to increased levels of
understanding.'’ However, often increased precision is purchased at the expense of
simplicity.
Because coordination compounds are usually considered to be covalently
bonded, to a first approximation (sec Chapter 11), extended complex structures in the
solid can readily be related to them. Consider, for example, rhodium pentafluoride.
Obviously, it could be considered as an ionic structure, Rh s 5F - , and indeed the
crystal structure 7 consists, in part, of /tep-arranged fluoride ions with rhodium in
octahedral holes. However, closer inspection of the structure reveals that it consists
of tetrameric units, Rh 4 F 2ll , that are distinct from one another (Fig. 7.7). The environ¬
ment about each rhodium atom, an octahedron of six fluorine atoms, is what we
1 Pearson. W. B. J. Phys. Chan. Soluls 1962.27, 103,
4 Shankar. S.: Parr, R. G. Proc. Noll. Acad. Sci. U.S.A. 1985. 82. 264-266.
5 Hardness is a property of an atom or ion approximately inversely proportional to its polarizability. It
is useful in acid-base discussions. Sec Chapter 9.
h For discussions of the prediction of the structures of solids, starting with the material presented here
and going far beyond, see Burden. J. K. In Structure and Bonding in Crystals', O'KecfFc. M. K.;
Navrotsky, A.. Eds.; Academic: New York. 1981; Vol. I. Chapter It; Adv. Cltem. Phys. 1982. 49.
47.
7 Morrell. B. K.; Zalkin. A.; Tressaud, A.; Bartlett. N. Inorg. Client. 1973. 12, 2640-2644.
Fig. 7.6 Crystal structures correlated with atomic charge (slope of lines) for IA-VIIA (1-17),
IIA-VIA (2-16). and IIIA-VA (13-15) compounds. The abscissa is the difference in
electronegativity. Ax («,\ “ "uh and (he ordinate is the sum of charge coefficients (>j = }/>),
Note that the charge. 6 ■= (« A - «„)/(/> A + /; H ), is approximately twice the reciprocal of the
slopes as plotted here. [Modified from Shankar. S.; Parr. R. G. Proc. Nall. Acad. Sci. U. S. A.
1985, $2. 264-266. Reproduced with permission.]
Fig. 7.7 Stereoview of the tetrameric unit of Rh.,F : „. The rhodium atoms are at the centers
of the octahedra of fluorine atoms. Note bridging fluorine atoms. [Front Morrell, B. K.;
Zalkin. A.; Tressaud. A.; Bartlett. N. Inorg. Cltem. 1973. 12, 2640-2644. Reproduced with
permission.)
259
1 02
4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
only by the geometry of the lattice and is independent of ionic radius and charge.
Unfortunately, previous workers have often incorporated ionic charge into the value
which they used for the Madelung constant. The practice appears to have arisen
from a desire to consider the energy of a “molecule” such as MX,:
E -
— A Zje 2
4ns 0 r
(4.7)
where A = 2.4 and Z\ is the highest common factor of Z + and Z~ (1 for NaCI,
CaF,, and AL0 3 ; 2 for MgO, TiO,, and Re0 3 ; etc.). We could ignore this confusing
practice and use the geometric Madelung constant, A, only, except that values re¬
ported in the literature are almost invariably given in terms of Eq. 4.7. Values for
both A and A are given in Table 4.1, and the reader may readily confirm that use
of either Eq. 4.5 or 4.7 yields identical results. 8
Returning to Eq. 4.5 we see that unless there is a repulsion energy to balance
the attractive coulombic energy, no stable lattice can result. The attractive energy be¬
comes infinite at infinitesimally small distances. Ions are, of course, not point charges
but consist of electron clouds which repel each other at very close distances. This
repulsion is shown by the dashed line in Fig. 4.6. It is negligible at large distances
but increases very rapidly as the ions approach each other closely.
Born suggested that this repulsive energy could be expressed by
(4.8)
where B is a constant. Experimentally, information on the Born exponent, n, may be
obtained from compressibility data, because the latter measure the resistance which
the ions exhibit when forced to approach each other more closeiy. The total energy
for a mole of the crystal lattice containing an Avogadro's number, N. of units is
U = E c + E r =
ANZZ~e 2
The total lattice energy is shown by the solid line in Fig. 4.6. The minimum in the
curve, corresponding to the equilibrium situation, may be found readily:
dU ANZ + Z'e 2 nNB
dr ~ 4jT£ 0 r r" 77
(4.10)
Physically this corresponds to equating the force of electrostatic attraction with the
repulsive forces between the ions. It is now possible to evaluate the constant B and
remove it from Eq. 4.9. Since we have fixed the energy at the minimum, we shall use
For further discussion of the problem of defining Madelung constants, sec Quanc, D. J. Chum. EJuc.
47 . 396.
_
Lattice Energy
103
Table 4.2
Values of the Born
exponent, n
Ion configuration n
He 5
Ne 7
Ar, Cu 9
Kr, Ag- 10
Xe. Au 12
U 0 and r 0 to represent this energy and the equilibrium distance. From Eq. 4.10:
B -AZ + Z~e 2 r-'
47tE 0 n
AZZ~Ne 2 ANZ Z~ e 2
0 4n£ 0 r 0 4ne 0 r 0 n
47t£ 0 r 0 V n )
(4.11)
(4.12)
(4.13)
This is the Born-Lande equation for the lattice energy of an ionic compound. As we
shall see, it is quite successful in predicting accurate values, although it omits certain
energy factors to be discussed below. It requires only a knowledge of the crystal
structure (in order to choose the correct value for A) and the interionic distance, r 0 ,
both of which are readily available from X-ray diffraction studies.
The Bom exponent depends upon the type of ion involved, with larger ions
having relatively higher electron densities and hence larger values of n. For most
calculations the generalized values suggested by Pauling (see Table 4.2) are suffi¬
ciently accurate for ions with the electron configurations shown.
The use of Eq. 4.13 to predict the lattice energy of an ionic compound may be
illustrated as follows. For sodium chloride the various factors are
A = 1.74756 (Table 4.1)
(V = 6.022 x I0 23 ion pairs mol’ 1 , Avogadro's number
Z = +1. the charge of the Na + ion
Z~ = — 1, the charge of the Cl ion
e = 1.60218 x 10 19 C. the charge on the electron (Appendix B)
?r = 3.14159
Co = 8.854188 x 10” 12 C 3 J 1 m~‘ (Appendix B)
r 0 = 2.814 x 10" 10 m, the experimental value. If this is not available, it may
be estimated as 2.83 x 10" 10 m, the sum of radii of Na + and Cl"
(Table 4.4).
n = 8, the average of the values for Na and Cl (Table 4.2).
Performing the arithmetic, we obtain U 0 = — 755 k J mol l , which may be compared
with the best experimental value (Table 4.3) of — 770 kJ mol" 1 . We may feel con¬
fident using values predicted by the Born-Lande equation where we have no experi¬
mental values.
As long as we do not neglect to understand each of the factors in the Born-Lande
equation (4.13), we can simplify the calculations. It should be realized that the only
variables in the Born-Lande equation are the charges on the ions, the internuclear <
distance, the Madelung constant, and the value of n. Equation 4.13 may thus be
simplified with no loss of accuracy by grouping the constants to give:
256
7 • The Solid State
257
)
Key:
O
O
o
Mg
lop layer
O, lower layer
Si above 0
Si beneath 0
Fig. 7.3 Structure of an olivine. Mg 2 Si0 4 . portrayed three ways. Left: Discrete SiOlT
tetrahedra and Mg 2 ions. Center: Network of Mg—O illustrating the extended structure.
The Si atoms have been omitted for clarity. Right: Mg~ + ions in an hep array of oxide ions.
The Si atoms have been omitted. Can you find the tetrahedral holes that they occupy?
[Modified from Wells. A. F. Structural Inorganic Chemistry, 5th ed.; Clarendon: Oxford.
1984. Reproduced with permission.]
Fig. 7.4 Structure field map for A 2 B0 4 compounds as a function of cation size. Note that
only the more common structures are plotted. Each point on this plot represents at least one
compound having the indicated structure and size of cations A(r A ) and B(r 8 ). [From Muller.
O.; Roy. R. The Major Ternary Structural Families’, Springer-Verlag: New York, 1974.
Reproduced with permission.]
The Structures of Complex Solids
but three spinels—sodium molybdate, sodium tungstate, and silver molybdate—fall
well outside their field. 1
Once we have established the fields shown in Fig. 7.4, we can use the map as
follows: If we discover a new mineral with r A = 90 pm and r a = 30 pm. we should
expect it to have the same structure as the mineral olivine, (Mg,Fe) 2 SiO d , but we
should not be too surprised if it turned out to be isomorphous with thenardite,
Na 2 S0 4 .
A second structure field map is shown in Fig. 7.5. This is a much more ambitious
and generalized undertaking, since the oxidation states for species A range from + I to
+ 4 and for B from +2 to +6, with X = O or F. By combining such a large number of
compounds and somewhat oversimplifying the resultant diagram, we lose some ac¬
curacy in predictability, but we gain in the knowledge that a large and diverse set of
compounds can be understood in terms of such simple parameters as relative sizes.
It may be noted in passing that although by far the largest amount of work on
crystal structures has been done in terms of the ordering of cations in a closest packed
structure of anions, this is not the only viewpoint. As is often the case in inorganic
chemistry, it is usually possible and often profitable to turn the model around 180°, so
to speak. Thus interesting insights can be gained by considering the alternative:
placing the emphasis on the arrangements of cations. 2
Ionic radius (r„) in pm-•-
Fig. 7.5 Composite
structure field map for
ABX.| structures, X = F or
0. [From Muller, O.: Roy,
R. The Major Ternary
Structural Families-.
Springer-Verlag: New
York, 1974. Reproduced
with permission.)
1 Muller, O.; Roy. R. The Major Ternary Structural Families-, Springer-Verlag: New York. 1974; pp
76-78.
2 O'Keeffe. M.; Hyde. B. G. In Structure and Bonding in Crystals, O'KccITc. M.; Navrotsky. A.,
Eds.; Academic: New York. 1981; Vol. I. pp 227-254; Struct. Bonding (Berlin) 1985, 61, 77.
104
4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compi
Lattice Energy
105
The Born—Haber
Cycle
o unds
U 0 = 1.39 x 10 5 kJ mol 1 pm^ Z f ~ < 4 ' 14)
Note that the internuclear distance should have the units of picometers, as given in
Table 4.4. If working with angstrom units and kcal mol -1 , the value of the grouped
constants is 332 kcal mol -1 A.
Equation 4.13 accounts for about 98% of the total energy of the lattice. For
more precise work several other functions have been suggested to replace the one
given above for the repulsion energy. In addition, there are three other energy terms
which affect the result by a dozen or so kJ mol -1 : van der Waals or London forces
(see Chapter 8), zero-point energy, and correction for heat capacity. The latter arises
because we are usually interested in applying the results to calculations at tempera¬
tures higher than absolute zero, in which case we must add a quantity:
= J Q (C U(MX) — C„ (M ., — C vlx -))dT (4.15)
where the C v terms are the heat capacities of the species involved. 9
The best calculated values, taking into account these factors, increase the ac¬
curacy somewhat: U 0 = -778, overestimating the experimental value by slightly less
than 1%. Unless one is interested in extreme accuracy, Eq. 4.13 is quite adequate.
Hess’s law states that the enthalpy of a reaction is the same whether the reaction
takes place in one or several steps; it is a necessary consequence of the first law of
thermodynamics concerning the conservation of energy. If this were not true, one
could “manufacture” energy by an appropriate cyclic process. Born and Haber 10
applied Hess’s law to the enthalpy of formation of an ionic solid. For the formation
of an ionic crystal from the elements, the Born—Haber cycle may most simply be
depicted as
Mw
AH a m
M w
m,:,
i X 2( 9 )
A II,
X,;,
|u°
MX,
It is necessary that
A H f = A H Am + A// Ax + A// ie + A // EA + U 0 (4.16)
The terms A// Am and Atf Ax are the enthalpies of atomization of the metal and the
nonmetal, respectively. For gaseous diatomic nonmetals, A H A is the enthalpy of dis¬
sociation (bond energy plus RT) of the diatomic molecule. For metals which vaporize
to form monatomic gases, A H A is identical to the enthalpy of sublimation. If sub¬
limation occurs to a diatomic molecule, M 2 , then the dissociation enthalpy of the
reaction must also be included:
9 It is commonly assumed that the independent cations and anions will behave as ideal monatomic gases
with heat capacities (at constant volume) of f R.
10 Born. M. Verhandl. Deut. Physik. Ges. 1919. 21. 13; Haber, F.; Ibid. 1919. 21. 750.
Uses of Born-Haber-
Type Calculations
M 2 -> 2M (4.17)
Values for the ionization energy, IE, and the electron affinity, EA, may be ob¬
tained from Tables 2.3 and 2.5. Bond dissociation energies for many molecules are
given in Appendix E. A useful source of many data of use to the inorganic chemist
has been written by Ball and Norbury. 11
The enthalpy of formation of an ionic compound can be calculated with an accuracy
of a few percent by means of the Born-Lande equation (Eq. 4.13) and the Born-
Haber cycle. Consider NaCI. for example. We have seen that by using the pre¬
dicted internuclear distance of 283 pm (or the experimental value of 281.4 pm), the
Madelung constant of 1.748, the Born exponent, n, and various constants, a value
of -755 kJ mol -1 could be calculated for the lattice energy. The heat capacity
correction is 2,1 kJ mol -1 , which yields U^ 98 = —757 kJ mol -1 . The Born-Haber
summation
is
then
U 2 0 98
—
-757
kJ
mol -1
A H lE
=
496
kJ
mol -1
W Ea
=
-349
kJ
mol -1
AHac.
=
121
kJ
mol -1
ah A n .
=
108
kJ
mol -1
£ = -381 kJ mol -1
This can be compared with an experimental value for the enthalpy of formation,
A H} 98 = -411 kJ mol -1 .
Separation of the energy terms in the Born-Haber cycle gives us some insight
into their relative importance in chemical bonding. For example, the Af/ A terms are
always positive, but are usually of relatively small size compared with the other terms
and do not vary greatly from compound to compound. 12 The ionization energies
are always greatly endothermic. Electron affinities for the halogens are exothermic,
but for the chalcogens they are endothermic as a result of forcing the second elec¬
tron into the negatively charged X - ion. In either case, the summation of ionization
energy and electron affinity is always endothermic, and it is only the overwhelming
exothermicity of the attraction of the ions for each other that makes ionic compounds
stable with respect to dissociation into the elements. At room temperature this energy
appears as the lattice energy. It should not be supposed, however, that at tempera¬
tures above the boiling point of the compound (1413 °C for NaCI, for example) no
reaction would occur between an active metal and nonmetal. Even in the gas phase
there will be electrostatic stabilization of the ions through the formation of ion pairs,
M + X - . The latter should be added to the Born-Haber cycle, and to clarify the
nature of the energy relationships, it is best to draw it in more explicit form as in
Fig. 4.8. In such a diagram the individual enthalpies can be portrayed and related
to the original enthalpy of the starting materials. 13
11 Ball, M. C.; Norbury, A. H. Physical Data for Inorganic Chemists'. Longman: London, 1974.
12 This statement is strictly true only for the halogens. The dissociation energies of 0 2 and Nj are con¬
siderably larger.
13 For a discussion of this point as well as several others concerning Born-Haber-type cycles, see Haight,
G. P., Jr. J. Chem. Educ. 1968. 45. 420.
Fig. 7.2 Geometric relationships and interconversions (1-16) among various molecules and
lattices (a-r): A “real" sulfur hexafluoride molecule (a) is transformed (I) into a space-filling
model (b), which is transformed (2) into a "stick-and-ball" model (c) of SF ft . The SF h
molecule is symmetrically identical (3) to the hypothetical [NaC!,,I 5 ~ ion (d), which is a
portion (4) of the NaCI lattice (e), which may be depicted (5) by the unit cell of NaCI shown
as a "see through" lattice (f). or depicted (6) as a space-filling model (g). The unit cell (f.g)
may also be depicted (6') with fractional atoms to show the actual number of atoms per unit
cell (h). The unit cell is (7) part of the extended lattice (i). Removal of the chloride ion
nearest the viewer (8) reveals an underlying triangular set of three sodium ions lying on top
of a triangular set of six chloride ions (j). Removal of these three sodium ions and six
chloride ions (9) reveals a triangle of ten sodium ions lying on top of a triangle of fifteen
Fig. 7.2 (Continued)
chloride ions (k). The fifteen chloride ions form (10) a closest packed array of ions (I).
Compare (I) with Figs. 4.12 and 4.14. Taking (I I) a portion (m) of the previous array provides
(12) a tetrahedral hole (n) or (13) an octahedral hole (o) depending upon the covering atoms.
Adding a second closest packed layer of three ions forms (13) an octahedral hole. There is a
Cj axis perpendicular to the plane of the paper (marked A). Addition (14) of geometric lines
shows the octahedral symmetry (p). and the structure may be converted (15) into an
octahedron (q) as in SF fi which is identical to the stick-and-ball octahedral model (c) seen
previously. Likewise the formation of a tetrahedral hole can be shown (12) by addition of one
atom on top to give a tetrahedral space-filling model (n) which, in turn may be converted (16)
to a “real" molecule of electron clouds (r).
254
255
106
4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
800
720
640
560
480
400
320
E?|"o240
“ -g E
“ w 3 160
80
o!
Nu’,„+ e - + Cl (gJ
N “i»' +Cl (l>
Na (|) -t-j Cl;„, A// Ac , = 120.9 -450.2
A// Ava = 108.4
Na (il + 5 CI JU)
-80
-160
-240
Ml, =-410.9
Na'cr,,,
" 336.8
-320
-400
NaCI,,,
Fig. 4.8 Born-Haber diagram showing relative magnitudes of various terms for sodium
chloride. [Adapted from Haight. G. P„ Jr. J. Cliem. Ethic. 1968. 45, 420-422. Reproduced
with permission.]
Lattice Energy 1 07
Table 4.3
Experimental and calculated
lattice energies ( —U 0 ) of
alkali halides (kJ mol -1 )
Salt
Experimental Simple model
(Born-Haber cycle) (Eq. 4.13) “Best values
Kapustinskii
approximation 6
LiF
1034
1008
1033
952.7
LiCI
840.1
811.3
845.2
803.7
LiBr
781.2
766.1
797.9
792.9
Lil
718.4
708.4
739.7
713.0
NaF
914.2
902.0
915.0
884.9
NaCI
770.3
755.2
777.8
752.9
NaBr
728.4
718.8
739.3
713.4
Nal
680.7
663.2
692.0
673.6
KF
812.1
797.5
813.4
788.7
KCI
701.2
676.5
708.8
680.7
KBr
671.1
659.8
679.5
674.9
KJ
632.2
623.0
640.2
613.8
RbF
780.3
761.1
777.8
760.2
RbCI
682.4
661.5
686.2
661.9
RbBr
654.0
636.4
659.0
626.3
Rbl
616.7
602.5
622.2
589.9
CsF
743.9
723.0
747.7
713.0
CsCl
629.7
622.6
652.3
625.1
CsBr
612.5
599.6
632.2
602.1
Csl
584.5
568.2
601.2
563.6
■ Calculated using a modified Born equation with corrections for polarization effects, repulsion
between nearest and next nearest neighbors, and zero-point energy (Cubicciotti, D. J. Cliem
Phys. 1959, 3J. 1646-1651; ibid., 1961, 34, 2189).
See Eq. 4.20.
Most of the enthalpies associated with steps in the cycle can be estimated, to a
greater or less accuracy, by experimental methods. The lattice energy, however, is
almost always obtained theoretically rather than from experimental measurement. It
might be supposed that the “enthalpy of dissociation" of a lattice could be measured
in the same way as the enthalpy of atomization of the metal and nonmctal, that is,
by heating the crystal and determining how much energy is necessary to dissociate it
into ions. Unfortunately, this is experimentally very difficult. When a crystal sublimes
(A H s ), the result is not isolated gaseous ions but ion pairs and other clusters. For
this reason it is necessary to use Eq. 4.13 or some more accurate version of it. We
can then use the Born-Haber cycle to check the accuracy of our predictions if we can
obtain accurate data on every other step in the cycle. Values computed from the
Born-Haber cycle are compared with those predicted by Eq. 4.13 and its modifica¬
tions in Table 4.3.
Once we have convinced ourselves that we are justified in using theoretical values
for U 0 , we can use the cycle to help obtain information on any other step in the cycle
which is experimentally difficult to measure. For many years electron affinities were
obtained almost exclusively by this method since accurate estimates were difficult to
obtain by direct experiment.
Finally, it is possible to predict the heat of formation of a new and previously
unknown compound. Reasonably good estimates of enthalpies of atomization, ion¬
ization energies, and electron affinities are now available for most elements. It is
then necessary to make some good guesses as to the most probable lattice structure,
including internuclear distances and geometry. The internuclcar distance can be esti¬
mated with the aid of tables of ionic radii. Sometimes it is also possible to predict
the geometry (in order to know the correct Madelung constant) from a knowledge
of these radii (sec next section). In such a case it is possible to predict the lattice
energy and the enthalpy of formation (the latter almost as accurately as it could be
measured if the compound were available). Examples of calculations on hypothetical
compounds are given below, and a final example utilizing several methods associated
with ionic compounds is given on page 127.
Consideration of the terms in a Born-Haber cycle helps rationalize the existence
of certain compounds and the nonexistence of others. For example, consider the hy¬
pothetical sodium dichloride, Na 2+ , 2CP. Because of the +2 charge on the sodium
ion. we might expect the lattice energy to be considerably larger than that of NaCI,
adding to the stability of the compound. But if all the terms are evaluated, it is found
that the increased energy necessary to ionize sodium to Na 2+ is more than that which
is returned by the increased lattice energy. We can make a very rough calculation
assuming that the internuclear distance in NaCl 2 is the same as in NaCI' 4 and that
Wc shall see that this ovcrestimalcs the distance, but for the present approximation it should be
adequate.
Chapter
7
The Solid State
In previous chapters we have seen how simple bonding models (the electrostatic one
for ionic compounds, various theories of covalent bonding, partial ionic and covalent
character, etc.) can be applied to the chemical and physical properties of compounds
of interest to the inorganic chemist. Of course, there are other important factors such
as dipole moments and van der Waals forces that influence these properties, and we
shall encounter them later. In this chapter we shall examine examples of the solids
held together by ionic or covalent bonds or mixtures of the two. Crystals held together
by predominantly ionic forces (e.g., magnesium oxide, which has the NaCI structure,
see Fig. 4.1) and those held together by purely covalent forces (e.g., diamond, see Fig.
7.1) are surprisingly similar in their physical properties. Both types of crystals are
mechanically strong and hard, are insulators, and have very high melting points (MgO
= 2852 °C, diamond = 3550 °C). Neither type is soluble in most solvents. The
conspicuous difference between the two types of crystals is that there are a few
solvents of high permittivity that will dissolve some ionic compounds (water is most
notable, but see Chapter 10). The second difference is that these solutions as well as
252
Fig. 7.1 Unit cell of the structure of diamond (carbon). Note the tetrahedral ( sp 3 )
configuration about each atom. Cf. Fig. 4.2b. [From Ladd, M. F. C. Structure and Bonding
in Solid State Chemistry, Ellis Horwood: Chicester. 1979. Reproduced with permission.]
The Structures of Complex Solids
253
the molten ionic compounds conduct electricity, but that is not a property of the solid
itself.
The Structures of
Complex Solids
Chapter 4 considered the topic of simple ionic compounds such as NaCI, CsCl, CaF 2 ,
etc., as well as the concepts of tetrahedral and octahedral holes in closest packed
lattices, the idea of efficiency in packing, and the radius ratio rule. Chapters 5 and 6
discussed covalent bonding and the structure of molecules. These ideas are summa¬
rized and illustrated in Fig. 7.2 (carefully correlate the parts and processes in the figure
and legend). In addition, the drawings in this figure should be of help in visualizing
various structures by showing the different methods used by chemists to depict atoms
and ions.
Given the difficulties and exceptions that we have seen with the radius ratio rule,
we might despair that any predictive power was available to the inorganic chemist
studying complex crystal structures. If simple M + X - compounds violate the radius
ratio rules as often as they do (see Fig. 4.18), how is the geochemist to deal in a
rational way with the structures of minerals like olivine (Mg 2 Si0 4 ), spinel (MgAI 2 0 4 ),
and other silicates and aluminosilicates containing a variety of metal ions? These
minerals form most of the earth's crust and mantle (see Chapter 16). In addition to
being important minerals, some of these compounds are important in the laboratory as
well. The class of compounds called spinels played an important early role in the
development of crystal field theory (see Chapter 11). The current intensity of interest
in high-temperature superconductors centers on mixed metal oxides with structures
similar to the mineral perovskite (see page 285).
There are various ways of looking at structures of this sort. One may formulate
them as silicate anions with isolated tetrahedra or linked into rings, chains, sheets, etc.
This viewpoint will be pursued further in Chapter 16. Alternatively, one can view
them as closest packed structures, in the case of olivine as Si 4+ ions occupying
tetrahedral holes and Mg 2+ ions (or Fe 2+ ions by isomorphous replacement) in
octahedral holes in a hexagonal closest packed array of oxide ions (Fig. 7.3). But
considering the difficulties encountered with the radius ratio approach, one may well
ask: "Is it possible to make accurate predictions?" Fortunately, the answer is "Yes.”
The chief difficulty with the radius ratio approach is that it is based purely on
geometric considerations, not chemical ones. If we include chemical factors, such as
partial covalency, our predictive power is considerably enhanced. There are several
approaches to the problem, but only two will be mentioned here. The simplest is a
purely empirical approach, and like so many methods in inorganic chemistry its
strength lies in its experimental basis: Possible unknown errors and hidden factors are
built into and accommodated by it. One takes a list of known structures of a given
general formula, say A,B0 4 (where A is a metal and B is a higher valent metal or
nonmetal). The radii of A and B are plotted against each other along two coordinate
axes. This is the graphical equivalent of looking at arithmetic radius ratios. In the
resultant structure field map (Fig. 7.4) it is found that similar structures cluster
together. The olivines cluster around 90 pm for A (r Mg > = 86 pm, r Fc 2 = 92 pm) and
40 pm for B (r si >+ = 40 pnt). In contrast, the spinels cluster around r A =» r B » 60-90 pm
(r A| j. = 68 pm).
A structure field map is remarkably accurate. That exceptions do occur, usually
on the borders of the fields, should not be surprising. Serious errors are relatively rare,
108
4-Bonding Models in Inorganic Chemistry: 1. Ionic Compo
d s
it would crystallize in the fluorite structure with a Madelung constant of A - 2.52.
The lattice energy is then U 0 = -2180 kJ mol" 1 . The summation of Born-Haber
terms is
('o =
-2180
ah Ano =
108
a// ie , =
496
Atf.E, -
4562
2Atf EA =
-698
A# Ac. =
242
ll
2530 kJ mol" 1
Although the estimation of U 0 by our crude approximation may be off by 10-20%,
it cannot be in error by over 100%, or 2500 kJ mol" 1 . Hence we can see why NaCl 2
does not exist: The extra stabilization of the lattice is insufficient to compensate for
the very large second ionization energy.
A slightly different problem arises when we consider the lower oxidation states
of metals. We know that CaF, is stable. Why not CaF as well? Assuming that
CaF would crystallize in the same geometry as KF and that the internuclear dis¬
tance would be about the same, we can calculate a lattice energy for CaF, L 0 -
-795 kJ mol -1 . The terms in the Born-Haber cycle are
U o =
-795
Atf Ac „ =
178
A//,u =
590
A H ea =
-328
A H Af =
79
AH f =
-276 kJ mol -1
An enthalpy of formation of -276 kJ mol" 1 , though not large, is perfectly ac¬
ceptable because it is about the same as that of Lil, for example. Why then does
CaF not exist? Because if one were able to prepare it, it would spontaneously dis¬
proportionate into CaF 2 and Ca exothermically. 15
2CaF -♦ CaF, + Ca ( 4 - 18 )
2Atf / =-550 AHf = —1220 AH f = 0 AH r = -670 kJ mol" 1
An examination of the ionic compounds of the main group elements would show
that all of the ions present have electronic configurations that are isoelectronic with
noble gases; hence the supposed “stability of noble gas configurations". But what
type of stability? It is true that the halogens are from 295 to 350 kJ mol ^ lower
in energy as halide ions than as free atoms. But the formation of the O ", S ,
N 3 ", Li + , Na + , Mg 2 + , and Ca 2+ ions is endothermic by 250 to 2200 kJ mol . Even
though these ions possess noble gas configurations, they represent higher energy states
than the free atoms. The “stability” of noble gas configurations is meaningless unless
one considers the stabilization of the ionic lattice. For the main group elements the
13 The direction of chemical reaction will be determined by the free energy, AG, not the enthalpy, AH.
However, in the present reaction the entropy term. AS, is apt to be comparatively small and once
AG = AH - TAS, the free energy will be dominated by the enthalpy at moderate temperatures.
noble gas configuration is that which maximizes the gain from high charges (and
large lattice energies) while holding the cost (in terms of ionization potential-electron
affinity energies) as low as possible. This is shown graphically in Fig. 4.9. Although
the second ionization energy for a metal is always larger than the first, and the third
larger than the second, the increase is moderate except when a noble gas configura¬
tion is broken. Then the ionization energy increases markedly because the electron
is being removed from the n — 1 shell. Below this limit the lattice energy increases
faster with oxidation state than does the ionization energy, so that the most stable
oxidation state is the one that maximizes the charge without breaking the noble gas
configuration. This is why aluminum always exists as Al 3+ when in ionic crystals
despite the fact that it costs 5140 kJ mol -1 to remove three electrons from the atom!
For transition metals, all electrons lost on ionization are either ns or (n — 1 )d
electrons which, as we have seen, are very similar in energy. Hence there are no
abrupt increases in ionization energy, only the more gradual change accumulating
from loss of electrons to form higher Z n , and these will be compensated by higher
lattice energies. Consider, for example, CuCl and CuCl,. We may calculate (cf. Prob-
Fig. 4.9 Energies of free
cations and of ionic
compounds as a function
of the oxidation state of
the cation. Top: Lines
represent the ionization
energy necessary to form
ihe +1, +2, +3, and
+ 4 cations of sodium,
magnesium, and aluminum.
Note that although the
ionization energy increases
most sharply when a noble
gas configuration is
“broken,” isolated cations
are always less stable in
higher oxidation states.
Bottom: Lines represent the
sum of ionization energy
and ionic bonding energy
for hypothetical molecules
MX, MX 2 , MXj, and MX 4
in which the interatomic
distance, r 0 , has been
arbitrarily set at 200 pm.
Note that the most stable
compounds (identified by
arrows) arc NaX, MgX 2 ,
and AIXj. (All of these
molecules will be stabilized
additionally to a small
extent by the electron
affinity of X.)
Oxidation slate
Oxidation state
2S0
6-The Structure and Reactivity of Molecules
angle decreases. Bond angles For NO;’', N0 : . and NO : are 180°. 134°. and 115°. respec¬
tively. Account For this trend.
6.29 How many sets of symmetry equivalent atoms are Found in the following molecules?
a. CIFj b. SF 4 c. Fe(CO)j (D } „)
d. Mn07 e. B^Hj. f. naphthalene. C| 0 H g
g. COf- h. AB 6 (D Vi )
6.30 Figures 6.4b and 6.7a illustrate unobserved but possible structures for SF 4 and ICI7-
Assign point group symmetries to these hypothetical structures. Would the difference in
symmetry affect the spectroscopic properties?
6.31 On page 235 the statement is made that there are other symmetry operations which exist
that illustrate the equivalence of the bridging bromo groups. Give an example.
6.32 Atoms of molecules which are chemically nonequivalent are also magnetically non-
equivalent and will, in general, give rise to different NMR chemical shifts. Sometimes
atoms may be chemically equivalent, but at the same time be magnetically noncquivalent.
Atoms are magnetically equivalent if they couple equally to all other atoms in the molecule.
For example, in methane each of the four hydrogen atoms couples to l5 C and to each other
in exactly the same way and they are therefore magnetically equivalent (as well as
chemically equivalent). In rro/ts-CjHjF;, however, a fluorine atom trans to hydrogen
couples to it differently than a fluorine atom which is cis. As a result, even though the two
fluorine atoms are chemically (symmetry) equivalent, they are magnetically nonequivalenl.
as are the two hydrogen atoms. In NMR terms, we would say that the irons- CjHjF; has an
AA'XX' rather than an A : X : spin system, and that its NMR spectrum would reflect that
complexity. Refer to Problem 6.29 and determine which atoms of each molecule or ion arc
magnetically equivalent.
6.33 The structures of all of the mixed (/i = 1-5) X , -chlorofluorophosphancs MI (PCI„Fj_„) and
' chlorofluoroarsanes (AsCI„Fj_„) have recently been determined.'' 1 Before reading the
experimental results, predict these structures.
6.34 Methyl- and trifluoromethylsulfur chloride were fluorinated with silver difluoride. w
AsT.
CH ' SCI cold O-ci, win ’ <»
CF ' SC1 < 2 > < 6 - 29 >
Elemental analysis of the products gave:' 1 '
(1) C = 11.5%. H = 2.8%. S = 30.8%. F = 54.8%;
(2) C = 7.6%. S = 20.2%, F = 72.2%.
Predict the molecular structures of the products as completely as possible. Name the
products according to IUPAC nomenclature.
“In IUPAC nomenclature, a "phosphanc" is normally a derivative of what we ordinarily call
“phosphine." PHj. Pentavalent derivatives of the hypothetical PfU (phosphorane) are labeled X\
Similarly, derivatives of H;S would be sulfancs. and the tetravalent and hcxavalcnt derivatives
would be labeled X 4 and respectively.
61 Minkwitz. R.; Prcnzcl, H.; Schardcy. A,; Obcrhammer. H. Inorg. Cliem. 1987. 26. 2730-2732.
Macho. C.; Minkwitz, R.; Rohmann, J.; Stcgcr. B.; Wdlfel, V,: Obcrhammer. H. I bill. 1986. 25.
2828-2835.
62 Downs. A. J.: McGrudy, G. S.; Barnfield, E. A.; Rankin. D. W. H.; Robertson, H. E.; Boggs. J. E.;
Dobbs. K. D. Inorg. Client. 1989. 28. 3286-3292.
The authors did not report elemental analyses, but they are given here (calculated on the basis of the
empirical formulas) to aid in the solution of this problem.
^ [ [| | | ^ ,f. i| i ill _ ...‘.L. .d!_t..
Problems
251
6.35 In the opening paragraph of Chapter 3 the statement was made that “frequently the most
symmetrical molecular structure is the ‘preferable - one." But in that same chapter we saw
that XeF 4 is square planar, not tetrahedral, which some theoreticians had argued for
because of the higher symmetry of the latter. Other exceptions have been discussed in this
chapter. Rationalize these apparent paradoxes.
6.36 The original investigators of the diazirine compound shown in Fig. 6.22 claim that the
following features can be found in that figure:
(a) bent bonds in the C—N—N triangle;
(b) the N=N double bond;
(c) the greater electronegativity of nitrogen compared with carbon;
(d) the presence of lone pairs on the nitrogen atoms:
(e) the absence of C 2 ,. symmetry in the electron density of the diazirine ring, nor should
such symmetry be expected. Note the difference of an order of magnitude between the
contours in (a) and (b).
Confirm each of these observations.
.
_J
1 10 4 • Bonding Mod'
Some Simplifications
and “Rules of
Thumb”
is in Inorganic Chomistry: 1. Ionic Compounds
lem 4.25) the enthalpies of formation as follows (kJ mol -1 ):
Term
CuCl
CuCl 2
^Ac
338
338
A^IEi
746
746
M lE2
1958
A ^A C1!
121
242
ah ea
-349
-698
U 0
-973
-2772
AH,
-117
-186
The enthalpy of atomization of copper does not differ at all for the two compounds,
and the atomization of chlorine adds only a small difference for the second mole of
chlorine. The major energy cost for CuCl 2 is the second ionization energy of copper
which is compensated by the electron affinity to form the second chloride ion and
especially the lattice energy. Since the electron ionized to form Cu 2+ is a d electron
and does not break a noble gas structure, IE 2 is not excessive, and both CuCl and
CuCl 2 are stable compounds.
In the same way that Fig. 4.9 was sketched with “average” values to illustrate the
stability of compounds with noble gas configurations, we can simplify Eq. 4.14 further
by inserting some "average" values. It must be clearly understood that this is merely
clearing away some of the numerical shrubbery to lay out the picture of the chemical
forest in clearer detail. Let us assume that we are studying compounds M + X“ with
an internuclear distance of about 200 pm. Of course, Z = -Z - = I. To be as gen¬
eral as possible, let’s use an average value of A = 2, which is not too inaccurate
for present purposes (about 20% error) for NaCI, CsCl, CaF 2 , TiO : , and both ZnS
structures. Equation 4.14 reduces to
t/ 0 « — 1400 kJ mol“ 1 = —330 kcal mol -1 ss -14eV (4.19)
This approximation is somewhat high for most compounds chiefly because an inter¬
nuclear distance of 200 pm is too small for most compounds. But it has the useful
asset of requiring that only the coefficients of Eq. 4.14 be remembered. Furthermore,
it allows some simple predictions to be made without involving the detailed calcula¬
tion of the above examples. For example, can we make a "rule of thumb" to predict
when a compound M + X - will be readily oxidized to M 2 2X - ? Using Eq. 4.14,
we predict that the lattice energy will double, or increase by one to one-and-a-half
MJ mol -1 , upon conversion to MX 2 . By far the major energy that has to be paid
to accomplish this change is IE, of the metal. While a thorough examination of all
of the energy terms is necessary for a careful analysis of the situation, we are led to
believe that if the additional cost of ionization is less than about 1.3—1.5 MJ mol -1
(13-15 eV) for the higher oxidation state, it may well be stable, too. In the case of
copper, given above, we have
IE, =0.75 MJ mol -1 IE 2 = 2.0 MJ mol -1 IE 3 = 3.5 MJ mol -1
Our rule of thumb follows the more careful calculations above and predicts that both
Cu(I) and Cu(II) compounds will be stable and, furthermore, it also works where
data are not available for a more careful analysis: Cu(III) compounds are predicted
to be unstable or marginally stable (Chapter 14).
Lattice Energy 111
On the other hand, if the succeeding ionization energies are too near each other,
as was the case for IE, and IE, of calcium above:
IE, =0.6 MJ moi -1 IE 2 = 1.1 MJ mol -1 IE 3 = 4.9 MJ moi -1
then the lower oxidation state (Ca + ) is unstable because it is too readily oxidized to
Ca 2- . Of course, Ca 3+ is unavailable because it is too prohibitively expensive.
Ahrens, 16 who was the first to point out this rule of thumb, contrasted the be¬
havior of titanium:
IE, = 0.66 MJ mol -1 IE, = 1.3 MJ mol -1
IE 3 = 2.6 MJ mol -1 IE 4 = 4.2 MJ mol -1
with that of zirconium:
IE, = 0.66 MJ mol -1 IE 2 = 1.3 MJ mol -1
IE 3 = 2.2 MJ mol -1 IE 4 = 3.3 MJ mol -1
The differences between the successive oxidation states for titanium are just sufficient
to allow marginally stable Ti(II) and Ti(III) oxidation states in addition to Ti(IV). The
corresponding lower oxidation states are uncommon for zirconium whose chemistry is
dominated by Zr(IV).
Of intermediate accuracy between the rough rule of thumb given above and the
precise Born-Lande equation is a suggestion made by Kapustinskii. 17 He noted that
the Madelung constant, the internuclear distance, and the empirical formula of a
compound are all interrelated. 18 He has suggested that in the absence of knowledge
of crystal structure (and hence of the appropriate Madelung constant) a reasonable
estimation of the lattice energy can be obtained from the equation:
U o =
120,200t>Z+Z -
r o
(kJ mol ‘)
(4.20)
where v is the number of ions per "molecule” of the compound and r 0 is estimated as
the sum of the ionic radii (Table 4.4), r,. + /•_ (pm). For the sodium chloride example
given previously, v = 2 and r 0 = 281 pm, yielding a lattice energy of - 750 kJ mol -1 ,
or about 98% of the experimental value, comparing favorably with that obtained
from Eq. 4.13. Of course, the usefulness of Eq. 4.20 lies not in its prediction of the
16 Ahrens, L. H. Geochim. Cosmochim. Acta 1953, 3, 1. Ahrens values, 8-10 eV, seem low in the light of
subsequent experience. A careful analysis has suggested that differences of 13-15 eV (1.3-1.5 MJ mol 1 )
between successive ionization energies will lead to multiple, stable oxidation states (Porterfield, W. W.
Inorganic Chemistry: A Unified Approach ; Addison-Wesley: Reading, MA, 1984; pp 416-420).
17 Kapustinskii, A. F. Z. Phys. Chem. (Leipzig ) 1933, 022, 257; Zh. Fiz. Kilim. 1943, 5, 59; Quart. Rev.
Chem. Soc. 1956, 10, 283.
18 This follows from the fact that, given a certain number of ions of certain sizes, the number of ways of
packing them efficiently is severely limited. Simple cases or this arc discussed in the sections entitled
"Efficiency of Packing and Crystal Lattices" and “Radius Ratio". For more thorough discussions of
Kapustinskii's work, see Waddington, T. C. Adv. Inorg. Chem. Radiochem. 1959, /, 157; or Dasenl, W. E.
Inorganic Energetics, 2nd cd.; Cambridge University: Cambridge, 1982; pp 76-79.
248
6-The Structure and Reactivity of Molecules
Problems
249
Fig. 6.31 Square pyramidal dioxo-A’-phosphanc with five- and four-membered rings. [From
Howard. J. A.; Russell. D.;Trippett. S. Chem. Commun. 1973. 856-857. Reproduced with
permission.)
6.18 Suggest the most likely stereochemistry of the phosphinatc ester resulting from ethanolysis
of the following compound:
CH j, .O
p-^ blOH
CHjCh/ 'Cl
6.19 Predict the geometries of (CH,) 2 P(CF 3 ) 3 and (CH,) 3 P(CF 3 ) 2 . Do you expect these mole¬
cules to undergo pseudorotalion? Explain. (See The. K. I.; Cavcll. R. G. Chem. Commun.
1975. 716.)
6.20 In an sp 3 d hybridized phosphorus atom in a TBP molecule, will the atom have a greater
electronegativity when bonding through equatorial or axial orbitals? Explain.
6.21 Earlier (page 212) it was staled that the repulsive effects of a lone pair and a doubly bonded
oxygen atom in VSEPR theory were very similar. Discuss qualitative and quantitative
differences that you feel should exist. (See Christe K. O.; Oberhammer. H. Inorg. Chem.
1981. 20. 296.)
6.22 Predict and say as much as you can about the probable structure of solid InCI. Be careful!
(Him: Why do you think this problem was included in this chapter rather than in Chap¬
ter 4?)
6.23 Consider Fig. 6.32 which is an electron density contour map of the sodium cyanide crystal.
Interpret this diagram in terms of everything that you know about the structure of solid
sodium cyanide.
Fig. 6.32 Differential electron density map of crystalline sodium cyanide. NaCN. Solid
contours indicate increased electron density upon compound formation from the aloms,
dashed contours represent decreased electron density. [From Coppcns, P. Angew. Chem.
Ini. EJ. Engl. 1977. 16. 32.)
6.24 Identify Ihe symmetry elements and operations in the molecules and ions shown in the
figures listed below. Determine the appropriate point group for each molecule and ion.
a. 6.4 b. 6.5 c. 6.6
d. 6.7 e. 6.8 f. 6.10
g. 6.11 h. 6.18
6.25 Calculate the hybridization of the carbon and nitrogen atoms in Fig. 6.22.
6.26 Considering the molecular orbital diagram of carbon monoxide (Fig. 5.20) and the discus¬
sion concerning hybridization and energy (pages 225-227), predict which end of the
carbon monoxide molecule will be the more basic (i.e., will donate electrons more readily
and form the stronger, direct covalent bond).
6.27 Consult the molecular orbital diagram in Fig. 6.16 and predict whether BH 2 will be linear or
bent. What would you predict for the excited stale configuration, 2a\b\b\l
6.28 Refer to the molecular orbital diagram for N0 2 in Fig. 5.31. Walsh diagrams, similar to
those in Fig. 6.16, predict that the HOMO la,) becomes more stable as the O—N—O bond
112
4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
lattice energy or sodium chloride, which is well known and provides a check on its
accuracy, but in giving reasonably accurate estimates for compounds that are not
well known (see Problem 4.24). . .
In summary, in addition to allowing simple calculations of the energetics ol
ionic compounds, the Bom-Haber cycle provides insight into the energetic factors
operating. Furthermore, it is an excellent example of the application of thermo¬
dynamic methods to inorganic chemistry and serves as a model for other, similar
calculations not only for solids, but also for reactions in solution and m the gas phase.
The determination of the sizes of ions has been a fundamental problem in inorganic
chemistry for many years. Many indirect methods have been suggested for appor¬
tioning the internuclear distance between two ions, relatively easy to obtain, into
cationic and anionic radii. Although these have been ingenious and provide insight
into atomic properties, they are no longer necessary.
When an X-ray crystallographer determines the structure of a compound such as
NaCl (Fig. 4.1a), usually only the spacing of ions is determined, because the repeated
spacings of the atoms diffract the X rays as the grooves on a phonograph record
diffract visible light. However, if very careful measurements are made, accurate maps
of electron density can be constructed since, after all, it is the electrons of the in-
Fig. 4.10 Electron density contours in sodium chloride. Numbers indicate the electron ^
density (electrons A’ 3 = 10~ 6 electrons pm -3 ) along each contour line. The "boundary" of
each ion is defined as the minimum in electron density between the ions. The internuclear
distance is 281 pm (= 2.81 A). [Modified from Schoknecht, G. Z. Naturforsch. 1957, 12A,
983. Reproduced with permission.]
Size Effects
Ionic Radii
Size Effects
113
dividual atoms that scatter the X rays. The result is Fig. 4.10. One may now appor¬
tion the interatomic distance in NaCl, 281 pm, using the minimum in electron density
as the operational definition of “where one ion stops and the other starts .
Although not many simple ionic compounds have been studied with the requisite
accuracy to provide data on ionic radii, there are enough to provide a basis for
a complete set of ionic radii. Such a set has been provided in the crystal radii of
Shannon and Prewitt . 19 Values of these radii are given in Table 4.4.
Factors Affecting A comparison of the values given in Table 4.4 allows one to make some conclusions
the Radii of Ions regarding the various factors that affect ionic size. We have already seen that pro¬
gressing to the right in a periodic series should cause a decrease in size. II the ionic
charge remains constant, as in the +3 lanthanide cations, the decrease is smooth
and moderate. Progressing across the main group metals, however, the ionic charge
is increasing as well, which causes a precipitous drop in cationic radii: Na + (116 pm),
Mg 2+ (86 pm), Al 3+ (67.5 pm). In the same way, for a given metal, increasing oxida¬
tion state causes a shrinkage in size, not only because the ion becomes smaller as it
loses electron density, but also because the increasing cationic charge pulls the anions
in closer. This change can be illustrated by comparing the bond lengths in the com¬
plex anions FeCl^ and FeCl;. The Fe(lII)—Cl bond length is 11 pm shorter than
the Fe(II)—Cl bond length . 20
For transition metals the multiplicity of the spin state affects the way in which
the anions can approach the cation; this alters the effective radius. Although this is
an important factor in determining cationic radii, it is beyond the scope of the present
chapter and will be deferred to Chapter 11.
For both cations and anions the crystal radius increases with the increase in co¬
ordination number. As the coordination number increases, the repulsions among the
coordinating counterions become greater and cause them to “back off” a bit. Alter¬
natively, one can view a lower coordination number as allowing the counter-ions to
compress the central ion and reduce its crystal radius.
As we shall see over and over again, the simple picture of billiard-ball-like ions
of invariant radius is easy to describe but generally unrealistic. The fluorides and
oxides come closest to this picture, and so the values in Table 4.4 work best with
them. Larger, softer anions in general will present more problems. Little work has
been done in this area, but Shannon 21 has presented a table, analogous to Table 4.4,
for sulfides.
Radii of Polyatomic The sizes of polyatomic ions such as NH and SO 4 are of interest for the under-
j ons standing of the properties of ionic compounds such as (NH.J 2 SO.i, but the experimen¬
tal difficulties attending their determination exceed those of simple ions. In addition,
the problem of constancy of size from one compound to the next—always a problem
19 Shannon, R.; Prewilt, C. T. Ada Crystalloyr. 1969. B25, 925; Shannon, R. D. ibid. 1976. A32. 751. Most
inorganic books in ihc past, including the first edition of the present one, have given some set of
"traditional" ionic radii based on indirect estimates. The Shannon and Prewitt crystal radii given in
Table 4.4 are about 14 pm larger for cations and 14 pm smaller for anions than the best set of traditional
radii.
20 Lauhcr, J. W.; Ibcrs, J. A. Inorg. Chem. 1975, 14. 348.
Shannon. R. D. In Structure and Bonding in Crystals', O'Keefe, M.; Navrotsky. A., Eds.; Academic:
New York. 1981. Vol. II. Chapter 16.
246
6'The Structure and Reactivity of Molecules
The process responsible for the "ozone hole" over Antarctica is thought to be
similar, though it may be heterogeneous, taking place on ice particles . 58
2fCI' + 0 3 -• CIO' + 0 2 ] (6.23)
CIO’ + CIO' + M -♦ (CIO), 4- M (6.24)
(CIO) 2 + hv - CP + ClOO- (6.25)
ClOO- + M -♦ CP + 0 2 + M (6.26)
Net: 20 3 -► 30, (6.27)
"T—r; 6.1 Draw Lewis structures for the following molecules and predict the molecular geometry:
Problems
a. BC1 3 b. BeH,
c. SnBr 4 d. TeF fi
e. AsF s f. Xe0 4
6.2 Draw Lewis structures for the following molecules and predict the molecular geometry
including expected distortions:
a. TeCI 4 b. ICIf
c. CIFj d. SO,
e. XcF, f. XeF 4
g. XeOj h. CIO,F,
6.3 What is the simplest reason for believing that molecular structure is more often governed
by BP-BP/BP-LP/LP-LP repulsions than by the van der Waals repulsions of the nonbond¬
ing electrons on the substituent atoms?
6.4 Use Eq. 6.1 to derive the bond angles in sp, sp-. and sp hybrid orbitals.
6.5 Assuming that the orbitals are directed along the intemuclear axes (i.e., the bonds are not
bent) use Eq. 6 .1 to calculate thep character in the bonds of NHj. The bond angle in NH j is
107.5°. What is the p character of the lone pair?
6.6 The bond angles in the nuoromethanes are:
Moloculo
H—C-H
F—C-F
CH,F
110 - 112 °
_
ch,f 2
II1.9 ± 0.4°
108.3 ± 0.1°
CHF,
—
108.8 ± 0.75°
a. Calculate the s character used by the carbon atom in the orbitals directed to the
hydrogen and fluorine atoms.
b. Discuss the results in terms of Bent's rule.
6.7 Show in a qualitative way why the energy levels of AH 2 in Fig. 6.16 increase or decrease in
the way they do upon bending the molecule. Attempt to account for small and large
changes.
5" Zurer. P. S. Chem. Eng. News 1990. 68(1), 15-16.
Problems
247
6.8 Consider the free radicals CH 3 and CFj . One is planar, the other pyramidal. Which is
which? Why?
6.9 Group VIA (16) tetrafluorides act as Lewis acids and form anions : 59
Cs + F _ + SF 4 —
— Cs[SF 5 ] _
(6.28)
Cs-F- + SeF 4 —
— Cs+[SeF 5 J _
(6.29)
Cs-F- + TeF 4 —
-■ Cs-[TeF 5 r
(6.30)
Predict the structures of these anions.
6.10 From Fig. 6.23, derive an equation for tungsten analogous to Eq. 6.8 for carbon.
6.11 a. Predict the carbon-carbon bond length(s) in benzene.
b. Predict the carbon-carbon bond length(s) in buckminsterfullerene (Fig. 3.33).
6.12 Consider the molecule CH 5 C=CH. Applying Bent's rule in its classical form, predict
whether the bond angles, H—C—H, are greater or less than 1094°. Considering the
arguments on overlap on page 228, predict again. (The experimental result is given by
Costain, C. C. J. Cliem. Phys. 1958, 29, 864.)
6.13 Consider the molecule CIF 3 0 2 (with chlorine the central atom). How many isomers are
possible? Which is the most stable? Assign point group designations to each of the isomers.
6.14 The structure for AI,Br ft (Fig. 6 .1 h) is assumed by both AI,Br 6 and AI,CI 6 in the gas phase.
In the solid, however, the structures can best be described as closest packed arrays of
halogen atoms (or ions) with aluminum atoms (or ions) in tetrahedral or octahedral holes. In
solid aluminum bromide the aluminum atoms are found in pairs in adjacent tetrahedral
holes. In solid aluminum chloride, atoms are found in one-third of the octahedral holes.
a. Discuss these two structures in terms of an ionic model for the solid. What factors favor
or disfavor this interpretation?
b. Discuss these two structures in terms of covalent bonding in the solid. What factors
favor or disfavor this interpretation?
6.15 Obtain the covalent and van dcr Waals radii of phosphorus and the halogens from Table
8 . 1 .
a. Show that for an assumed bond angle of 109J° in the phosphorus trihalidcs there must be
van dcr Waals contacts among the halogen atoms.
b. Show that because of the concomitant increase in both covalent and van dcr Waals
radii, the repulsion between the halogens docs not become worse as one progresses
from PFj to PI j.
6.16 One of the few phosphorus compounds that exhibit square pyramidal geometry is shown in
Fig. 6.31. Rationalize the preferred geometry of SP over TBP in terms of the presence of
the four- and five-membered rings. (Holmes, R. R. J. Am. Chem. Soc. 1975, 97, 5379.)
6.17 Consider the cyclic compounds I and II. In I the rapid exchange of the fluorine atoms is
inhibited just as it is in (CH 3 ),PF 3 . However, exchange in II is very rapid. Suggest a reason.
F F
I II
59 Christe. K. 0.; Curtis. E. C.; Schack, C. J.; Pilipovich, D. Inorg. Chem. 1972, //, 1679-1682.
114
4 Bonding Models in Inorganic Chemistry: 1. Ionic Comp
Table 4.4
Effective ionic radii of
Coordination
Coordination
Coordination
the elements" Ion
number 4
pm
Ion
number 4
pm
Ion
number 4
pm
Ac 3+
6
126
Bk 4
6
97
Co 4
4
54
Ag' +
2
81
8
107
6 HS
67
4
114
Br"
6
182
Cr 2
6 LS
87
4 SQ
116
Br 3
4 SQ
73
HS
94
5
123
Br 3
3 PY
45
Cr 3
6
75.5
6
129
Br 7
4
39
Cr 4
4
55
7
136
6
53
6
69
8
142
C J
3
6
Cr 3
4
48.5
Ag 2+
4 SQ
93
4
29
6
63
6
108
6
30
8
71
Ag 3+
4 SQ
81
Ca 2
6
114
Cr 6
4
40
ai 3
6
89
7
120
6
58
4
53
8
126
Cs 1
6
181
5
62
9
132
8
188
Am 2
6
67.5
10
137
9
192
7
135
12
148
10
195
8
140
Cd 2
4
92
11
199
Am 3
9
145
5
101
12
202
6
111.5
6
109
Cs"
10
348-
8
123
7
117
Cu"
2
60
Am
6
99
8
124
4
74
8
109
12
145
6
91
As 3-
6
210 “'
Ce 3
6
115
Cu 2
4
71
As 3
6
72
7
121
4 SQ
71
As 3
4
47.5
8
128.3
5
79
At 7
6
60
9
133.6
6
87
6
76
10
139
Cu 3
6 LS
68
Alt'
6
151
12
148
Dy'
2
4
Au 3
4 SQ
82
Ce“
6
101
Dy 2
6
121
6
99
8
III
7
127
Au 3
6
71
10
121
8
133
B 3
3
15
Cf 3
12
128
Dy 3
6
105.2
4
25
6
109
7
III
Ba 2
6
41
Cf J
6
96.1
8
116.7
6
149
8
106
9
122.3
7
152
Cl"
6
167
Er 3
6
103
8
156
Cl 3
3 PY
26
7
108.5
9
161
Cl 7
4
22
8
114.4
10
166
6
41
9
120.2
II
171
Cm 3
6
111
Eu 2
6
131
Be 2
12
175
Cm 4
6
99
7
134
3
30
8
109
8
139
4
41
Co 2
4 HS
72
9
144
6
59
5
81
10
149
Bi 3
5
no
6 LS r
79
Eu 3
6
108.7
6
117
HS
88.5
7
115
Bi 3
8
131
8
104
8
120.6
6
90
Co 3
6 LS
68.5
9
126
Bk 3
6
110
HS
75
F"
2
114.5
Continued
Size Effect'
1 1 5
Table 4.4 ( Continued )
Effective ionic radii of
the elements"
Coordination Coordination Coordination
Ion
number 4
pm
Ion
number 4
pm
Ion
number 4
pm
3
116
In 3
4
76
6
75
4
117
6
94
Mo 6
4
55
6
119
8
106
5
64
F 7
6
22
Ir 3
6
82
6
73
Fe 2
4 HS
77
Ir 4
6
76.5
7
87
4 SQ HS
78
lr s
6
71
N"
4
132
6 LS
75
K"
—
313'
N 3
6
30
HS
92
K'
4
151
N 5
3
4.4
8 HS
106
6
152
6
27
Fe 3
4 HS
63
7
160
Na"
_
276'
5
72
8
165
Na 1
4
113
6 LS
69
9
169
5
114
HS
78.5
10
173
6
116
8 HS
92
12
178
7
126
Fe 4
6
72.5
La 3
6
117.2
8
132
Fe 6
4
39
7
124
9
138
Fr'
6
194
8
130
12
153
Ga 3
4
61
9
135.6
Nb 3
6
86
5
69
10
141
Mb 4
6
82
6
76
12
150
8
93
Gd 3
6
107.8
Li'
4
73
Nb 3
4
62
7
114
6
90
6
78
8
119.3
8
106
7
83
9
124.7
Lu 3
6
100.1
8
88
Gc 2
6
87
8
111.7
Nd 2
8
143
Ge 4
4
53
9
117.2
9
149
6
67
Mg 2
4
71
Nd 3
6
112.3
H 1
1
-24
5
80
8
124.9
2
-4
6
86
9
130.3
Hr"
4
72
8
103
12
141
6
85
Mn 2
4 HS
80
Ni 2
4
69
7
90
5 HS
89
4 SQ
63
8
97
6 LS
81
5
77
Hg'
3
111
HS
97
6
83
6
133
7 HS
104
Ni 3
6 LS
70
Hg 2
2
83
8
no
HS
74
4
110
Mn 3
5
72
Ni 4
6 LS
62
6
116
6 LS
72
No 2
6
124
8
128
HS
78.5
Np 2
6
124
Ho 3
6
104.1
Mn 4
4
53
Np 3
6
115
8
115.5
6
67
Np 4
6
101
9
121.2
Mn 3
4
47
8
112
10
126
Mn 6
4
39.5
Np 3
6
89
l"
6
206
Mn 7
4
39
Np 6
6
86
I 3
3 PY
58
6
60
Np 7
6
85
6
109
Mo 3
6
83
O 2
2
121
I 7
4
56
Mo 4
6
79
3
122
6
67
Mo 3
4
60
4
124
Continued
1
6-The Structure and Reactivity of Molecules
ee Radical
'echanisms
inversion or the less common retention, there is a contrast with the loss of stereo¬
chemistry associated with a carbocatanion mechanism.
The stability of five-coordinate intermediates also makes possible the ready
racemization of optically active silanes by catalytic amounts of base. The base can add
readily to form a five-coordinate intermediate. The latter can undergo Berry pseudoro¬
tation with complete scrambling of substituents followed by loss of the base to yield
the racemized silane.
Most of the reactions the inorganic chemist encounters in the laboratory involve ionic
species such as the reactants and products in the reactions just discussed or those of
coordination compounds (Chapter 13). However, in the atmosphere there are many
free radical reactions initiated by sunlight. One of the most important and controver¬
sial sets of atmospheric reactions at present is that concerning stratospheric ozone.
The importance of ozone and the effect of ultraviolet (UV) radiation on life has been
much discussed. Here we note briefly that only a small portion of the sun's spectrum
reaches the surface of the earth and that parts of the UV portion that are largely
screened can cause various ill effects to living systems.
The earth is screened from far-UV (extremely high energy) radiation by oxygen in
the atmosphere. The UV radiation cleaves the oxygen molecule to form two free
radicals (oxygen atoms):
0 2 + hv (below 242 nm) - O" + O" (6.11)
The oxygen atoms can then attack oxygen molecules to form ozone:
0" + 0 2 + M -► 0 3 + M (6.12)
The neutral body M carries off some of the kinetic energy of the oxygen atoms. This
reduces the energy of the system and allows the bond to form to make ozone. The net
reaction is therefore:
30 2 + hv - 20 3 (6.13)
This process protects the earth from the very energetic, short-wavelength UV radia¬
tion and at the same time produces ozone, which absorbs somewhat longer
wavelength radiation (moderately high energy) by a similar process:
0 3 + hv (220-320 nm) -► 0 2 + 0" (6.14)
The products of this reaction can recombine as in Eq. 6.12, in which case the ozone
has been regenerated and the energy of the ultraviolet radiation has been degraded to
thermal energy. Alternatively, the oxygen atoms can recombine to form oxygen
molecules by the reverse of Eq. 6.11, thereby reducing the concentration of ozone. An
equilibrium is set up between this destruction of ozone and its generation via Eq. 6.13
and so under normal conditions the concentration of ozone remains constant.
The controversy over supersonic transports (SSTs) of the Concorde type revolves
around the production of nitrogen oxides whenever air containing oxygen and nitrogen
passes through the very high temperatures of a jet engine. One of these products,
nitric oxide, reacts directly with ozone, thereby reducing its concentration in the
stratosphere:
NO + O
N0 2 + O-
(6.15)
v. ..
Some Simple Reactions of Covalently Bonded Molecules
245
Furthermore, nitrogen dioxide formed in Eq. 6.15 or directly in the combustion
process can react to scavenge oxygen free radicals and prevent their possible recom¬
bination with molecular oxygen to regenerate ozone (Eq. 6.12):
N0 2 + O" -♦ NO + 0 2 (6.16)
Note that a combination of reactions (Eq. 6.15 and 6.16) results in the net conversion
of ozone to oxygen:
and that the nitrogen oxides, either NO or N0 2 , continuously recycle and thus act as
catalysts for the decomposition of ozone:
The current controversy revolves around the extent to which nitrogen oxides,
NO,, would be formed by SSTs and how much the ozone concentration would be
affected.«
The ozone question is complicated by the fact that other chemicals are implicated
in its destruction. Chlorofluorocarbons were formerly widely used as propellants in
spray cans, and they continue to be used as refrigerants.56 They are extremely stable
and long-lived in the environment. However, they too can undergo photolysis in the
upper atmosphere:
FjCCI + hv (190-220 nm)
F,C- + CP
The chlorine free radical can then interact with ozone in several different ways
analogous to the NO t . At mid-latitudes the reactions are
Cl + 0 3
CIO- + O"
CIO- + 0 2
Cl- + 0,
for a net reaction of:
0“ + 0 3 -► 20 2 (6.22)
with regeneration of the monoatomic chlorine. The chlorine thus acts as a catalyst,
and present evidence indicates that the CIO, cycle may be three times more efficient in
the destruction of ozone than is the NO cycle.57
53 Johnston. H. S.; Kinnison, D. E.; Wuebbles, D. J. J. Geophys. Res. 1989, 94, 16351 —16363.
Manahan, S. E. Environmental Chemistry. 5th cd.; Lewis: Chelsea, Ml, 1991.
56 The U.S. will cease production of ozone-depleting chlorofluorocarbons by the end of 1995. Chern.
Eng. News 1992. 70(6). 7-13.
37 A recent report suggests that N0 2 actually protects ozone by scavenging the much more destructive
CIO: N0 2 + CIO -» C10N0 2 . Chem. Eng. News 1992, 70(2). 4-5.
116 4'Bonding Mo-
Table 4.4 ( C ontinued )
Effective ionic radii of
the elements"
in 1n orgo
mic Chemistry
r; 1 . Ion
ic Compounds
Ion
Coordination
number 6
pm
Coordination
Ion number 6
pm 1
6
126
9
i3i.9 ;
8
128
Pt 4
6
99
OH 1 "
2
118
8
110 I
3
120
Pt 2
4 SQ
74
4
121
6
94 :
6
123
Pt 4
6
76.5
Os 4
6
77
Pt 3
6
71
Os 5
6
71.5
Pu 3
6
114
Os 6
5
63
Pu 4
6
100
6
68.5
8
110
Os 7
‘6
66.5
Pu 5
6
88
Os 8
4
53
Pit 6
6
85
P 3 "
6
200 J
Ra 2
8
162
P 3
6
58
12
184
P 5
4
31
Rb 1 "
—
317 c
5
43
Rb 1
6
166
6
52
7
170
Pa 3
6
118
8
175
Pa 4
6
104
9
177
8
115
10
180
Pa 5
6
92
11
183
8
105
12
186
9
109
14
197
Pb 2+
4 PY
112
Re 4
6
77
6
133
Re 5
6
72
7
137
Re 6
6
69
8
143
Re 7
4
52
9
149
6
67
10
154
Rh 3
6
80.5
11
159
Rh 4
6
74
12
163
Rh 5
6
69
Pb 4
4
79
Ru 3
6
82
5
87
Ru 4
6
76
6
91.5
Ru 5
6
70.5
8
108
Ru 7
4
52
Pd'
2
73
Ru 8
4
50
Pd 2
4 SQ
78
S 2 "
6
170
6
100
S 4
6
51
Pd 3
6
90
s 6
4
26
Pd 4
6
75.5
6
43
Pm 5
6
111
Sb 3
4 PY
90
8
123.3
5
94
9
128.4
6
90
Po 4
6
108
Sb 5
6
74
8
122
Sc 3
6
88.5
Po 6
6 •
81
8
101
Pr 3
6
113
Se 2 "
6
184
8
126.6 Se 4
6 •
64
Table 4.4 ( Continued )
Effective ionic radii of
the elements"
Size Effects 117
Coordination Coordination Coordination
Ion
number 6
pm Ion
number 6 pm
Ion
number 6
pm
11
132
9
119
6
62
12
135
12
131
Y 3
6
104
Ti 2
6
100 U 5
6
90
7
110
Ti 3
6
81
7
98
8
115.9
Ti 4
4
56 . U 6
2
59
9
121.5
5
65
4
66
Yb 2
6
116
6
74.5
6
87
7
122
8
88
7
95
8
128
TI'
6
164
8
100
Yb 3
6
100.8
8
173 V 2
6
93
7
106.5
12
184 V 3
6
78
8
112.5
TI 3
4
89 V 4
5
67
9
118.2
6
102.5
6
72
Zn 2
4
74
8
112
8
86
5
82
Tm 2
6
117 V 5
4
49.5
6
88
7
123
5
60
8
104
Tm 3
6
102
6
68
Zr 4
4
73
8
113.4 W 4
6
80
5
80
9
119.2 W 5
6
76
6
86
U 3
6
116.5 W 6
4
56
7
92
U 4
6
103
5
65
8
98
7
109
6
74
9
103
8
114 Xe 8
4
54
“ Values of crystal radii from Shannon, R. D.
Acta Crystalloyr. 1976.
A32. 751-767.
6 SQ
= square planer, PY = pyramidal; HS
= high spin; LS
= low spin.
c Huang. R. H.;
Ward. D. L.; Dye, J.
L. J. .
4m. Chem. Soc.
1989, III. 5707-5708.
J Modified from Pauling, L. Nature of the Chemical Bond, 3rd ed.: Cornell University:
Ithaca, NY, 1960. These values are only approximate._
even in simple ions—often becomes much worse. For example, one set of data in¬
dicates that the radius of the ammonium ion is consistently 175 pm, but a dillercnt
set indicates that it is the same size as Rb, 166 ppm. 2 - This is not a serious dis¬
crepancy, but it is a disturbing one since its source is not obvious.
Yatsimirskii 23 has provided an ingenious method for estimating the radii ol poly¬
atomic ions. A Born-Haber calculation utilizing the enthalpy ol formation and re¬
lated data can provide an estimate of the lattice energy. It is then possible to find
what value of the radius of the ion in question is consistent with this lattice energy.
These values are thus termed thermochemical radii. The most recent set of such values
is given in Table 4.5. In many cases the Tact that the ions (such as CO 2 ,", CNS",
CHjCOO - ) are markedly nonspherical limits the use of these radii. Obviously they
22 Shannon, R. D, Acta Crystalloyr. 1976, A32, 751.
23 Yatsimirskii K. B. Izv. Akad. Nauk SSSR. Otdel, Khim. Nauk 1947, 453; 1948. 398. See also Mingos.
D. M. P.; Rolf, A. L. Inary. Cheat. 1991. 30. 3769-3771. where the shape of the ion is taken into con¬
sideration as well as its size (see Problem 4.42).
242
6 The Structure and Reactivity of Molecules
Fig. 6.30 l9 F NMR spectra of a solution of PCLFj in isopentane at various temperatures. All
of the fluorine resonances are doublets from 3, P- ,9 F coupling, (a) At -22 °C only a single
doublet is observed, indicating that all of the fluorine atoms are equivalent, (b) At - 109 °C
this resonance disappears, (c) At - 127 °C two new absorptions appear, (d) At - 143 °C two
types of fluorine atoms are seen: a doublet of doublets at low field (two axial Fs) and a
doublet of triplets at higher field (one equatorial F). [From Holmes, R. R.; Carter. R. P., Jr.;
Peterson, G. E. Inorg. Chem. 1964. 3. 1748-1754. Reproduced with permission.]
Substitution of alkyl groups on the phosphorus atom provides some interesting
effects. If a single methyl group replaces a fluorine atom, it occupies one of the
equatorial positions as expected and rapid exchange of the two axial and the two
equatorial fluorine atoms is observed, as in PF 5 . If two methyl groups arc present,
(CHj),PF 3 , the molecule becomes rigid and there is no observable exchange among
the three remaining fluorine atoms. This dramatic change in behavior appears to be
attributable to the intermediate which is formed, shown in Fig. 6.29. In the
pseudorotalion of CH 3 PF 4 the methyl group can remain at position E , and thus remain
in an equatorial position both before and after the pseudorotation. In contrast, in
(CH 3 ) 2 PF 3 one of the methyl groups is forced to occupy either £, or £ 3 ; therefore,
after one pseudorotation it is forced to occupy the energetically unfavorable (for a
substituent of low electronegativity) axial position. Apparently the difference in en¬
ergy between equatorial and axial substitution of the methyi group is enough to inhibit
the pseudorotation.
This difference in energy can be shown dramatically in the sulfurane
Some Simple Reactions of Covalently Bonded Molecules
243
Nucleophilic
Displacement
This molecule has an approximately TBP electronic structure and is chiral. However,
potentially it could racemize via a series of Berry pseudorotations. 52 That it does not
do so readily, and is therefore the first optically active sulfurane to have been isolated,
has been attributed to the fact that all racemization pathways must proceed through a
TBP with an apical lone pair. 53 As we have seen in the preceding chapter, there is a
very strong tendency for the lone pair to seek an equatorial site. The reluctance of the
lone pair to occupy an apical site appears to be a sufficient barrier to allow the
enantiomers to be isolated.
The question might be asked: Are there similar mechanisms for changing the
configuration of molecules without breaking bonds in molecules with coordination
numbers other than 3 and 5? The answer is “yes." One of the most important series of
inorganic compounds consists of six-coordinate chelate compounds exemplified by the
tris(ethylenediamine)cobalt(III) ion. Because of the presence of the three chelate
rings, the ion is chiral and racemization can take place by a mechanism that is closely
related to atomic inversion or Berry pseudorotation (the mechanism for six-coordina¬
tion is termed the “Bailar twist"; see Chapter 13).
The crux of organic mechanistic stereochemistry may be the Walden inversion, the
inversion of stereochemistry about a four-coordinate carbon atom by nucleophilic
attack of, for example, a hydroxide ion on an alkyl halide. Many reactions of inorganic
molecules follow the same mechanism. In contrast, the dissociative mechanism of
tertiary halides to form tertiary carbocatanion intermediates is essentially unknown
among the nonmetallic elements silicon, germanium, phosphorus, etc. The reason for
this is the generally lower stability of species with coordination numbers of less than 4,
together with an increased stability of five-coordinate intermediates. This difference is
attributable to the presence of (/orbitals in the heavier elements (Chapter 18).
The simplest reaction path for nucleophilic displacement may be illustrated by the
solvolysis of a chlorodialkylphosphine oxide:
We would expect the reaction to proceed with inversion of configuration of the
phosphorus atom. This is generally observed, especially when the entering and leaving
groups are highly electronegative and are thus favorably disposed at the axial posi¬
tions, and when the leaving group is one that is easily displaced. In contrast, in some
cases when the leaving group is a poor one it appears as though front side attack takes
place because there is a retention of configuration. 54 In either case, the common
52 Of course, there are other potential mechanisms for racemization. If there were a trace of free Cl -
ion. an S N 2 displacement might be possible, or merely a simple S N I dissociation of the Cl atom and
racemization. Neither of these reactions appears to take place either.
53 Martin. J. C.; Balthazor, T. M. J. Am. Chem. Soc. 1977. 99. 152-162.
54 For a discussion of the various possibilities, see Tobe, M. L. Inorganic Reaction Mechanisms',
Thomas Nelson: London, 1972; pp 25—37; Katakis, D.; Gordon, G. Mechanisms of Inorganic
Reactions: Wiley: New York, 1987; pp 190-191.
— ' __I
118 4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
Table 4.5
Thermochemical radii of Ion
pm
Ion
pm
Ion
pm
1 Ion
pm
Cations
Anions
Anions
Anions
nh;
151
CoFi-
230
MnCl 2 "
308
PtF|-
282
Me 4 N +
215
CrF|“
238
MnF 2 "
242
pur
328
ph;
17!
CrO 2 '
242
MnOT
215
SbCl
337
Anions
CuClj
307
NJ
181
SeO 2
225
AICI4
NCO
189
SeOj
235
bci;
296
GaCU"
275
nh 2 ch 2 coj
176
SiFg-
245
bf;
218
GeCI 2 -
314
no 2
178
SnBrr
349
bh;
179
GeF 2 '
252
NOj-
165
SnCir
335
BrOj
140
HCIj
187
o 2 -
144
Snir
382
CH3COO-
148
HCOj
155
or
159
sor
244
C 103 -
157
HCO3
142
OH’
119
TiBr 2 "
338
cio 4
hf 2 -
158
PbCl 2 -
334
Ticir
317
CN-
177
HS-
193
PdCI 2 -
305
TiF|-
275
CNS~
199
HSe
191
PtBr|-
328
VO3-
168
CO 2 -
164
IO3-
108
PtCI 2 '
279
vor
246
C0CI4 -
305
io 2 f 2
163
PtCI 2 '
299
ZnBrr
285
IrCir
221
ZnCI 2 -
272
Znir
309
“ Data from Jenkins. H. D. B.; Thakur, K. P. J. Clmm. Ednc. 1979. 56. 576-577. adjusted to be
compatible with Shannon-Prewitt crystal radii. Used with permission.
can be reinserted into further thermochemical calculations and thus provide such
data as the anticipated lattice energy of a new (sometimes hypothetical) compound.
In the case of tetrahedral and especially octahedral ions, the symmetry is suf¬
ficiently high that the ions may be considered pseudospherical. and so the values
more closely represent the physical picture that we have of ionic radii.
Efficiency of Packing If we consider atoms and ions to be hard spheres, we find that there are certain
and Crystal Lattices geometric arrangements for packing them which are more efficient than others. This
can be confirmed readily in two dimensions with a handful of coins. For example, if
a set of coins of the same size (dimes, for example) is arranged, it will be found that
six of them fit perfectly around another (i.e., touching each other and the central
dime), giving a coordination number of 6 . However, only five quarters or four silver
dollars will fit around a dime , 2 illustrating the importance of size in determining
the optimum coordination number. The effect of charge can also be illustrated. If all
of the atoms are the same, the most efficient two-dimensional lattice is the closest
packed, six-coordinate arrangement. If they are of the same size but opposite charge,
the six-coordinate structure is not stable since it will have too many repulsions or
like-charge ions. This can also be readily shown with coins (using heads and tails to
24 1 he fit is not exact in the latter two cases.
Size Effects
119
■
S3
Fig. 4.11 Two-dimensional lattices: (a) stable, six-coordinate, closest packed lattice
of uncharged atoms; (b) unstable, six-coordinate lattice of charged ions; (c) stable,
four-coordinate lattice of charged ions.
Fig. 4.12 (a) Sites created by layer 1 and available to accept atoms in layer 2. (b) Covering
all f sites by atoms in the second layer, making the t' sites (relabeled o) unavailable for
occupancy by close-packed atoms.
represent charge), and it can be seen that the most stable arrangement is a square
lattice of alternating charge (Fig. 4.11c).
The same principles hold for three-dimensional lattices. Consider first a lattice
composed only of uncharged atoms as in a metal or a crystal of noble gas atoms.
The first layer will consist of a two-dimensional, closest packed layer (Fig. 4.1 la). The
second layer will be of the same type but centered over the “depressions" that exist
where three atoms in the first layer come in contact (Fig. 4.12a ). 25 A layer containing
n atoms will have 2 n such sites capable of accepting atoms (marked l and r'), but
once an atom has been placed in either of the two equivalent sets ft and r’) the
remainder of that layer must continue to utilize that type of site (Fig. 4.12b), and
the remaining n sites (labeled o) are not utilized by the packing atoms.
The third layer again has a choice of n sites out of a possible 2 n available (f and
r' types again). One alternative places the atoms of the third layer over those of the
first; the other places the atoms of the third layer over the o sites of the first layer. In
25 The reader is strongly urged to build these structures using Styrofoam spheres and to consult texts
on structural chemistry such as Wells, A. F. Structural Inorganic Chemistry, 5th ed.; 1984; The Third
Dimension in Chemistry, Clarendon: Oxford, 1956. The present discussion merely presents the more
salient features of the subject.
0
6 The Structure and Reactivity of Molecules
material can take place via the mechanism shown in Eq. 6.9. It is of interest that the
energy barrier to inversion is strongly dependent on the nature of the central atom and
that of the substituents. For example, the barrier to inversion of ethylpropylphenyl-
phosphine (Fig. 6.28b) is about 120 kJ mol -1 . This is sufficient to allow the separation
of optical isomers, and their racemization may be followed by classical techniques. In
contrast, the barrier to inversion in most amines is low (~- 40 kJ mol -1 in methyl-
propylphenylamine; about 25 kJ mol -1 in ammonia). With such low barriers to
inversion, optical isomers cannot be separated because racemization takes place faster
than the resolution can be effected. Since traditional chemical separations cannot
effect the resolution of the racemic mixture, the chemist must turn to spectroscopy to
study the rate of interconversion of the enantiomers. The techniques involved are
similar to those employed in the study of fluxional organometallic molecules (Chapter
15), and for now we may simply note that for inversion barriers of 20-100 kJ mol -1 ,
nuclear magnetic resonance is the tool of choice.
Because the transition state in the atomic inversion process of Eq. 6.9 involves a
planar, sp 2 hybridized central atom, the barrier to inversion will be related to the ease
with which the molecule can be converted from its pyramidal ground state. We should
therefore expect that highly strained rings such as that shown in Fig. 6.28c would
inhibit inversion, and this is found (145 kJ mol -1 ). Furthermore, all of the effects we
have seen previously affecting the bond angles in amines and phosphines should be
parallel in the inversion phenomenon. For example, the smaller bond angles in
phosphines require more energy to open up to the planar transition state than those of
the corresponding amines; hence the optical stability of phosphines in contrast to the
usual instability of most amines. In addition, the presence of electron-withdrawing
substituents tends to increase the height of the barrier, but electron-donating groups
can lower it. Just as in the case of the stereochemistry of pyramidal molecules, the
results can be rationally accommodated by a variety of interpretations.
Berry We have seen previously that in PF S the fluorine atoms are indistinguishable by means
seudorotation of l9 F NMR (page 236). This means that they are exchanging with each other faster
than the NMR instrument can distinguish them. The mechanism for this exchange is
closely related to the inversion reaction we have seen for amines and phosphines. The
exchange is believed to take place through conversion of the ground state trigonal
bipyramid (TBP) into a square pyramidal (SP) transition state and back to a new TBP
structure (Fig. 6.29). This process results in complete scrambling of the fluorine atoms
at the equatorial and axial positions in phosphorus pentafluoride, and if it occurs faster
than the time scale of the NMR experiment (as it does), then all of the fluorine atoms
Fig. 6.28 Chiral amines and phosphines.
Some Simple Reactions of Covalently Bonded Molecules
241
appear to be identical. Because it was first suggested by Berry, 51 and because, if all of
the substituents are the same as in PF S , the two TBP arrangements (Fig. 6.29) are
related to each other by simple rotation, the entire process is called a Berry
pseudorotation. Note that the process can take place very readily because of the
similarity in energy between TBP and SP structures (page 223). In fact, the series of
five-coordinate structures collected by Muetterties and Guggenberger, which are
intermediate between TBP and SP geometries (Table 6.3), effectively provides a
reaction coordinate between the extreme structures in the Berry pseudorotation.
The exchange of fluorine atoms in PF S is too rapid to monitor with NMR spec¬
troscopy. The atoms in some other molecules exchange more slowly, especially at
lower temperatures. For example. PC1 2 F 3 is expected to be a trigonal bipyramid with
two apicophilic fluorine atoms in the axial positions, and two chlorine atoms and the
third fluorine in equatorial positions. At temperatures of -22 °C and above, the
resonance of fluorine is observed as a single doublet (Fig. 6.30a). However, if the
temperature is lowered to - 143 °C, the two axial fluorine atoms can be distinguished
from the single equatorial fluorine (Fig. 6.30d).
All three l9 F nuclei (/ = 4) are split by the 3I P nucleus (/ = i) with a coupling
constant of J P _ P = 1048 Hz. At - 143 °C this produces a downfield doublet (8 -67.4)
for the axial atoms and an upfield doublet (8 41.5) for the equatorial fluorine atom. The
single equatorial fluorine atom splits each component of the doublet of the two axial
fluorine atoms into another doublet (2 nl + I; n = I) with J F _ F = 124 Hz. In the same
way the doublet pattern of the single equatorial atom is split into triplets (In! + I, n =
2) by the two axial fluorine atoms. The overall intensity of the axial fluorine resonance
is twice that of the single equatorial fluorine. Also, the weighted average of the
chemical shifts at - 143 °C [2 x (— 67.4 ppm) + I x (+41.5 ppm)] is the same as that
at — 22 °C [3 x ( — 31.1 ppm)] indicating that the structure does not change on
warming, even though the fluorine exchange accelerates. The fact that phos¬
phorus-fluorine coupling is preserved on warming indicates that there is no inter-
molecular exchange, but that the coalescence of the spectrum at higher temperatures
is the result of an m/ramolecular rearrangement.
D,k c,. D s
Fig. 6.29 Berry pseudorotation in a pentavalcnt phosphorus compound.
51 Berry, R. S. J. Chem. Pliys. I960, 32. 933-938. For recent reviews of phosphorus compounds, see
Corbridge. D. E. C. Phosphorus: An Outline of Its Chemistry. Biochemistry and Technology, 4th
ed.: Elsevier: Amsterdam, 1990; pp 994-1003; Cavell, R. G. In Phosplwrus-31 NMR Spectroscopy
in Stereochemical Analysis', Verkade, J. G.: Quin, L. D.. Eds; VCH: Deerfield Beach. FL, 1987;
Chapter 7.
1 20
4 - B<
ding Models in Inorganic Chemistry: 1. Ionic Comp
Size Effects
121
the first type the layers alternate ABABAB and the lattice is known as the hexagonal
closest packed (hep) system. Alternatively, the cubic closest packed (cep) system has
three different layers, ABCABC. Both lattices provide a coordination number of 12
and are equally efficient at packing atoms into a volume.
It is easy to see the unit cell and the origin of the term hexagonal closest packed.
In Fig. 4.13a the unit cell can be constructed by drawing a hexagon through the
nuclei of the six outer atoms in layer A and a parallel hexagon in the next A layer
above, and then connecting the corresponding vertices of the hexagons with perpen¬
dicular lines to form a hexagonal prism (Fig. 4.13a).
One could follow a similar practice and construct a similar hexagonal ‘‘sand¬
wich" with two layers (B, C) of “filler,” but a cubic cell of higher symmetry can be
constructed; the second system is thus characterized as cubic closest packed. The re¬
lation between the cubic unit cell (which is identical to the face-centered cubic cell we
have already seen) is not easy to visualize unless one is quite familiar with this system.
The easiest way is to take a face-centered cubic array (Fig. 4.14c), and by removing
(a) (b) (c) (d)
Fig. 4.14 Unit cells in the cubic closest packed systems, (a) A face-centcrcd array of atoms. Note that the exposed layer
consists of a closest packed array of fifteen atoms. Consider this the "A layer", (b) A closest packed layer of six atoms
placed on (a). Consider this the “B layer”, (c) The final atom, a member of the “C layer," is added to complete the
cube. The fee unit cell is redrawn in (d). Note that the single atom that composes the “C layer" does not lie above any atom
in the "A layer" (as it would if this were hep).
Fig. 4.13 Arrangement of
layers in hexagonal closest
packed (a) and cubic closest
packed (b) structures. These
are “side views” compared
with the “top views” shown
in the preceding figures.
an atom (Fig. 4.14b), then a few more (Fig. 4.14a), reveal the closest packed layers
corresponding to A. B, and C in Fig. 4.13b.
The noble gases and most metals crystallize in either the hep or the ccp structure
as would be expected for neutral atoms. The alkali metals, barium, and a few transi¬
tion metals crystallize in the body-centered cubic system, though the reasons for this
choice are unknown.
If all the packing atoms are no longer neutral (e.g., half are cations and half are
anions), the closest packed structures are no longer the most stable, as can be seen
from the similar two-dimensional case (see above). However, these structures may still
be useful when considered as limiting cases for certain ionic crystals. Consider lithium
iodide, in which the iodide anions are so much larger than the lithium cations that
they may be assumed to touch or nearly touch. They can be considered to provide
the framework for the crystal. The much smaller lithium ions can then fit into the
small interstices between the anions. If they expand the lattice slightly to remove the
anion-anion contact, the anionic repulsion will be reduced and the crystal stabilized,
but the simple model based on a closest packed system of anions may still be taken
as the limiting case and a useful approximation.
Where the lithium ions fit best will be determined by their size relative to the
iodide ions. Note from above that there are two types of interstices in a closest packed
structure. These represent tetrahedral (t) and octahedral (o) holes because the co¬
ordination of a small ion fitted into them is either tetrahedral or octahedral (see
Fig. 4.12). The octahedral holes are considerably larger than the tetrahedral holes
and can accommodate larger cations without severe distortion of the structure. In
lithium iodide the lithium ions fit into the octahedral holes in a cubic closest packed
lattice of iodide ions. The resulting structure is the same as found in sodium chloride
and is face-centered (note that face-centered cubic and cubic closest packed describe
the same lattice).
Consider a closest packed lattice of sulfide ions. Zinc ions tend to occupy tetra¬
hedral holes in such a framework since they are quite small (74 pm) compared with
the larger sulfide ions (170 pm). If the sulfide ions form a ccp array, the resulting
structure is zinc blende; if they form an hep array, the resulting structure is wurtzite.
See Fig. 4.15.
Although in the present discussion size is the only parameter considered in deter¬
mining the choice of octahedral versus tetrahedral sites, the presence of covalent
bonding (d 2 sp 3 versus sp 3 hybridization, see Chapter 5) and/or ligand field stabiliza¬
tion (see Chapter 11) can affect the stability of ions in particular sites. Size will usually
be the determining factor when these additional factors are of small importance—for
example, when considering alkali and alkaline earth ions. The concept of closest
packing of anions is also very useful in considering polar covalent macromolecules
such as the silicates and iso- and heteropolyanions. 26
If the cations and anions are of approximately the same size, the limiting case of
the framework being determined by the larger ion is inappropriate, and we simply
determine the most efficient lattice for oppositely charged ions of equal size. This
turns out to be the CsCI lattice, which maximizes cation-anion interaction (C.N. = 8)
and is the most stable structure when the sizes of the cation and anion are comparable.
26 Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Clarendon: Oxford, 1984, For a comprehensive
and detailed discussion of the broad usefulness of classifying structures in terms of closest-packed
structures, see Douglas, B. E.; McDaniel, D. H.; Alexander, J. J. Concepts and Models of Inorganic
Chemistry, 2nd ed.; Wiley: New York, 1983; pp 198-208.
Toble 6.5 _
Comparison of some physical techniques for structural studies 0
Technique Nature of the effect Information Interaction time Sensitivity Comments
X-ray diffraction Scattering, mainly by Electron density map 10 -l8 sbut crystal Location of light atoms or
electrons, followed by of crystal averaged over ~10 3 cm 3 distinction between
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122
Radius
©5 OZn
( 3 S OZn
o
(a)
(b)
Fig. 4.15 (a) The structure of wurtzite. The sulfide ions form an hep array with A (gray)
and B (black) alternating layers (Cf. Fig. 4.13a). (b) The structure of zinc blende. The sulfide
ions form a irp array with A (white), B (black), and C (gray) layers. (Cf. Figs. 4.13b and
4.14.) Note that in both structures the zinc atoms (small white circles) occupy tetrahedral
holes.
Ratio
It is not difficult to calculate the size of the octahedral hole in a lattice of closest
packed anions. Figure 4.16 illustrates the geometric arrangement resulting from six
anions in contact with each other and with a cation in the octahedral hole. Simple
geometry allows us to (ix the diagonal of the square as 2r_ + 2/>. The angle formed
by the diagonal in the corner must be 45°, so we can say:
2 r_
2 r_ + 2 1 \
= cos 45° = 0.707
(4.21)
r_ = 0.707r_ + 0.707r +
(4.22)
Fig. 4.16 (a) Small cation
(dashed line) in octahedral
hole formed by six anions,
(b) Dissection of octahedron
to illustrate geometric
relationships shown in (c).
Size Effects 123
Table 4.6
Radius ratio and
coordination number
Coordination
number
Geometry
4
Tetrahedral
6
Octahedral
8
Cubic
12
Cuboctohedral
Limiting radius
ratio"
0.414; 2.42
0.732; 1.37
1.000
Possible lattice
structures
Wurtzite. zinc blende
NaCl, rutile
CsCl, fluorite
° The second ratio is merely the reciprocal of the first. It is often convenient to
have both values.
h The atoms in the top three layers of Fig. 4.13b form a cuboctohedron.
<• Coordination number 12 is not found in simple ionic crystals. It occurs in
complex metal oxides and in closest packed lattices of atoms.
0.293r_ = 0.707r +
(4.23)
r_
0.293
0.707
= 0.414
(4.24)
This will be the limiting ratio since a cation will be stable in an octahedral
hole only if it is at least large enough to keep the anions from touching, that is,
r + /r_ > 0.414. Smaller cations will preferentially fit into tetrahedral holes in the
lattice. By a similar geometric calculation it is possible to determine that the lower
limit for tetrahedral coordination is r+/r_ = 0.225. For radius ratios ranging from
0.225 to 0.414, tetrahedral sites will be preferred. Above 0.414, octahedral coordi¬
nation is favored. By similar calculations it is possible to find the ratio when one
cation can accommodate eight anions (0.732) or twelve anions (1.000). A partial list
of limiting radius ratio values is given in Table 4.6.
The use of radius ratios to rationalize structures and to predict coordination
numbers may be illustrated as follows. 27 Consider beryllium sulfide, in which
r l)cl ./r s2 - = 59 pm/170 pm = 0.35. We should thus expect a coordination number
of 4 as the Bc 2+ ion fits most readily into the tetrahedral holes of the closest packed
lattice, and indeed this is found experimentally: BcS adopts a wurtzite structure.
In the same way we can predict that sodium ions will prefer octahedral holes in
a closest packed lattice of chloride ions (r Nn ./r cl - = 116 pm/167 pm = 0.69), forming
the well-known sodium chloride lattice with a coordination number of 6 (Fig. 4.1a).
With larger cations, such as cesium, the radius ratio (r C5 /r cl - = 181 pm/
167 pm = 1.08) increases beyond the acceptable limit for a coordination number of
6 ; the coordination number of the cations (and anions) increases to 8, and the cesium
chloride lattice (Fig. 4.1b) results. As we have seen, although this is an efficient struc¬
ture for cations and anions of about the same size, it cannot be directly related to a
closest packed structure of anions.
Table 4.6 indicates that a coordination number of 12 should be possible when
the radius ratio is 1.00. Geometrically it is possible to fit 12 atoms about a central
11 Since crystal radii vary slightly with coordination number, values from Table 4.4 were taken for
C.N. = 6 as "average” values.
236
6'The Structure and Reactivity of Molecules
atoms to be identical spectroscopically. The l9 F NMR spectrum of PF 5 , however,
consists of a single doublet, indicating that all fluorine atoms are equivalent. This
means that: (I) all of the P—F bonds are identical or (2) the fluorine atoms are
exchanging positions faster than the NMR technique can follow. We have seen that it
is structurally impossible to have a three-dimensional arrangement of five equivalent
points in space. We are therefore led to accept the second alternative. This process of
interconversion will be discussed later in this chapter.
Other spectroscopic techniques also may be applied to the resolution of molecular
structure. For now. one general method may be illustrated by the Mossbauer elucida¬
tion of the structure of I 2 CI 4 Br,. By analogy with I,CI 6 (Fig. 6.26a), we might expect a
bridged structure with either chloro or bromo bridges (Fig. 6.26b-d). The experimen-
Cl \
| /C N
, /CI
Cl \,
c,/
\ C1 /
^Cl
Cl^
S^
Si
D:„
0 ,,
(a)
(b)
c, \,
, /C, ^l
/ Br
Cl \,
/ Cl \,
,/ Br
Br'''
\ C |/
' S 'CI
\ c ,/
^Br
C„
c 2 „
(c)
(d>
Fig. 6.26 Some possible
structures of iodine
trihalides. Note that
structure (d) has a different
environment for each iodine
atom.
Velocity (cm s' 1 )
Fig. 6.27 Partial l2 »I
Mossbauer spectra of I 2 CI 6
(top) and l 2 Br 2 Cl.i (bottom).
Note splitting of peak A in
NCIfc into A and A' and B
into B' in I 2 Br 2 Cl a .
Presumably C' is hidden
under a shoulder of B,
[From Pasternak. M.;
Sonnino, T. J. Chem. Pliys.
1968. 48, 1997. Reproduced
with permission.!
1.5 1.0 0.5 0
Velocity (cm s' 1 )
Some Simple Reactions of Covalently Bonded Molecules
237
tal result 4 that the two iodine atoms are in different environments (Fig. 6.27) rules out
the symmetrical structures shown in Fig. 6.26b, c and strongly suggests that the
correct structure is Fig. 6.26d.
Summary of This has been a brief survey of some of the methods available to the inorganic chemist
Structural for the determination of structure. Further examples will be encountered later in the
Methods text illustrating methods. A useful summary of some of the methods of structure
determination has been provided by Beattie. 49 listing some of the characteristics that
have been discussed above as well as some other features of their use (Table 6.5).
Some Simple One of the major differences between organic and inorganic chemistry is the relative
Reactions of
emphasis placed on structure and reactivity. Structural organic chemistry is relatively
simple, as it is based on digonal, trigonal, or tetrahedral carbon. Thus organic chem-
Covalently
Bonded Molecules
istry has turned to the various mechanisms of reaction as one of the more exciting
aspects of the subject. In contrast, inorganic chemistry has a wide variety of structural
types to consider, and even for a given element there are many factors to consider.
Inorganic chemistry has been, and to a large extent still is, more concerned with the
"static” structures of reactants or products than with the way in which they intercon¬
vert. This has also been largely a result of the paucity of unambiguous data on reaction
mechanisms. However, this situation is changing. Interest is increasingly centering on
how inorganic molecules change and react. Most of this work has been done on
coordination chemistry, and much of it will be considered in Chapter 13, but a few
simple reactions of covalent molecules will be discussed here.
Atomic Inversion The simplest reaction a molecule such as ammonia can undergo is the inversion of the
hydrogen atoms about the nitrogen atom, analogous to the inversion of an umbrella in
a high wind:
/
H
N.
V
H
H
H { H
(6.9)
One might argue that Eq. 6.9 does not represent a reaction because the “product” is
identical to the "reactant” and no bonds were formed or broken in the process. 50
Semantics aside, the process illustrated in Eq. 6.9 is of chemical interest and worthy of
chemical study. For example, consider the trisubstituted amines and phosphines
shown in Fig. 6.28. Because these molecules are nonsuperimposable upon their mirror
images (i.e., they are chiral), they are potentially optically active, and separation of
the enantiomers is at least theoretically possible. Racemization of the optically active
4 " Pasternak. M.; Sonnino. T. J. Chem. Phys. 1968, 48. 1997-2003.
49 Beattie, I. R. Chem. Soc. Rev. 1975, 4, 107-153. See also Footnote 45.
50 Obviously, the same result can be obtained by dissociating a hydrogen atom from Ihe nitrogen atom
and allowing it to recombine to form the opposite configuration. For a discussion of the various
competing mechanisms that must be distinguished in studying Eq. 6.9, as well as values for barrier
energies and methods for obtaining them, see Lambert. J. B. Topics Slereochem. 1971, 6, 19-105.
4 • Bonding Models in Inorganic Chemislry: 1. Ionic Compounds
C.N. = 8
C.N. = 6
C.N. = 4
Fig. 4.17 The total energy of a cubic lattice of rigid anions and cations as a function of r,.
with r. fixed, for different coordination configurations. When the anions come into mutual
contact as a result of decreasing r + their repulsion determines the lattice constant and the
cohesive energy becomes constant when expressed in terms of r_. Thus near the values of
rjr. at which anion-anion contact takes place, the radius ratio model predicts phase
transitions to structures of successively lower coordination numbers. Note that the breaks
in the curves correspond to the values listed in Table 4.6. [From Treatise on Solid State
Chemistry; Hannay, N. B., Ed.; Plenum: New York, 1973.]
atom (see the discussion of closest packing in metals, page i 19), but it is impossible
to obtain mutual twelve-coordination of cations and anions because of the limitations
of geometry. Twelve-coordination does occur in complex crystal structures of mixed
metal oxides in which one metal acts as one of the closest packing atoms and others
fit into octahedral holes, but a complete discussion of such structures is more appro¬
priate in a book devoted to the structures of solids. 28
The change in coordination number as a result of the ratio of ionic radii is
shown graphically in Fig. 4.17. In general, as the cation decreases in size the lattice
is stabilized (lattice energy becomes more negative) until anion-anion contact occurs.
Further shrinkage of the lattice is impossible without a reduction in coordination
number; therefore, zinc sulfide adopts the wurtzitc or the zinc blende structure,
gaining additional energy over what would be possible in a structure with a higher
coordination number. Note that although there is a significant difference in energy
between structures having coordination numbers 4 and 6, there is little difference
between 6 and 8 (the two lines almost coincide in Fig. 4.17 on the left). The difference
in energy between six- and eight-coordinate structures is less than 1% based on
electrostatics.
In a 1:1 or 2:2 salt, the appropriate radius ratio is obviously the ratio of the
smaller ion (usually the cation) to the larger to determine how many of the latter will
fit around the smaller ion. In compounds containing different numbers of cations
and anions (e.g., SrF 2 , Ti0 2 , Li 2 0, Rb 2 S) it may not be immediately obvious how
to apply the ratio. In such cases it is usually best to perform two calculations. For
28 See Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Clarendon: Oxford, 1984; pp 480-589.
Sixe Effect:
125
example, consider SrF 2 :
= — =1.11 maximum C.N. of Sr 2+ = 8
r F - 119
rp — - 11^ = 0.90 maximum C.N. of F" =8
/•sr- 132
Now there must be twice as many fluoride ions as strontium ions, so the coordination
number of the strontium ion must be twice as large as that of fluoride. Coordination
numbers of 8 (Sr 2 + ) and 4 (F ') are compatible with the maximum allowable co¬
ordination numbers and with the stoichiometry of the crystal. Strontium fluoride
crystallizes in the fluorite lattice (Fig. 4.3).
A second example is Sn0 2 :
= -H- — o,66 maximum C.N. of Sn 4+ = 6
r 0 2- 126
= 1.52 maximum C.N. of O 2 =6
r Sn .. 83
Considering the stoichiometry of the salt, the only feasible arrangement is with
C.N . 0 2 - = 3,C.N. Sn . = 6; tin dioxide assumes the TiO z or rutile structure of Fig. 4.4.
Note that the radius ratio would allow three more tin(IV) ions in the coordination
sphere of the oxide ion, but the stoichiometry forbids it.
One final example is K 2 0:
k
r Q i-
r o‘-
'•x
Considering the stoichiometry of the salt, the structure must be antifluoritc (Fig. 4.3,
reversed) with C.N. 0 i- = 8, C.N. K . =4.
The radius ratio quite often predicts the correct coordination numbers ol ions in
crystal lattices. It must be used with caution, however, when covalent bonding be¬
comes important. The reader may have been puzzled as to why beryllium sulfide was
chosen to illustrate the radius ratio rule for coordination number 4 (page 123) instead
of zinc sulfide, which was used repeatedly earlier in this chapter to illustrate foui-
coordinate structures such as wurtzite and zinc blende. The reason is simple. If ZnS
had been used, it would have caused more confusion than enlightenment: It violates
the radius ratio rule! Proceeding as above, we have r + /r_ = 88 pm/170 pm = 0.52,
indicating a coordination number of 6, yet both forms of ZnS, wurtzite and zinc
blende, have a C.N. of 4, for both cations and anions. If one argues that 0.52 does
not differ greatly from 0.41, the point is well taken, but there exist more vexing cases.
The radius ratio for mercury(II) sulfide. HgS, is 0.68, yet it crystallizes in the zinc
blende structure. In both of these examples the .vp 3 hybridized covalent bonding
seems to be the dominant factor. Both ZnS and especially HgS are better regarded
as infinite covalent lattices (see Chapter 7) than as ionic lattices.
= —- = 1.21 maximum C.N. of K + =8
126
= y^- = 0.83 maximum C.N. of O 2 " = 8
234
Experimental Determination of Molecular Structure
235
6-The Structure and Reactivity of Molecules
In order to solve a structure by X-ray diffraction, one generally needs a single
crystal. Although powder data can provide ••fingerprint” information and, in simple
cases, considerable data, it is generally necessary to be able to grow crystals for more
extensive analysis. X rays are diffracted by electrons; therefore what are located are
the centers of electron clouds, mainly the core electrons. This has two important
consequences. First, if there is a great disparity in atomic number between the heavy
and light atoms in a molecule, it may not be possible to locate the light atom
(especially if it is hydrogen), or to locate it as accurately as the heavier atom. Second,
there is a small but systematic tendency for the hydrogen atom to appear to be shifted
10-20 pm toward the atom to which it is bonded. 43 This is because hydrogen is unique
in not having a core centered on the nucleus (which is what we are seeking) and the
bonding electrons are concentrated toward the binding atom.
Because the location of an atom in a molecule as obtained by X rays is the time
average of all positions it occupied while the structure was being determined, the
resultant structure is often presented in terms of thermal ellipsoids, which are proba¬
bility indicators of where the atoms are most likely to be found (see Fig. 6.25).
Occasionally, from the size and orientation of an ellipsoid, something may be ventured
on the bonding in a molecule. If the ellipsoid is prolate (ovoid. American football
shaped), the motion of the atom is mostly back and forth along the bond axis, and if
oblate (curling-stone shaped), the motion is mostly wobbling about the bond axis.
Obviously, the less the atom moves in the molecule, the smaller its thermal ellipsoid.
In the molecule shown in Fig. 6.25, the carbon atoms in trimethylphosphine and in the
phenyl rings "'waggle" a good deal; the atoms "locked" in the central five-membered
ring move relatively little. This is especially true of the platinum atoms; they are heavy
and so they have less thermal motion. For someone used to "ball-and-stick" models,
this is perhaps the most confusing aspect of an ORTEP diagram 4 '- viewed for the first
Fig. 6.25 ORTEP diagram of complex containing trimethylphosphine and nitrosobenzene
ligands, and two platinum atoms. Note the differences in the sizes of the thermal ellipsoids
for C. N. and Pt. For a discussion of the types of bonding in this complex, see Chapter 12.
(From Packet!. D. L.; Troglcr. W. C.; Rheingold. A. L. Inorg. Chem. 1987. 26, 4309.
Reproduced with permission.!
44 ORTEP is an acronym for "Oak Ridge Thermal Ellipsoid Program," a computer program frequently
used in structural analysis. The acronym is often used as a short label to indicate a drawing in which
ellipsoids indicate the extent of thermal motions of the atoms.
time. We become accustomed to the size of an atom being reflected in the size of its
model—ORTEP drawings are often quite the opposite.
Neutron diffraction is very similar in principle to X-ray diffraction. However, it
differs in two important characteristics: (1) Since neutrons are diffracted by the nuclei
(rather than the electrons), one indeed locates the nuclei directly. (2) Furthermore, the
hydrogen nucleus is a good scatterer; thus the hydrogen atoms can be located easily
and precisely. The chief drawback of neutron diffraction is that one must have a
source of neutrons, and so the method is expensive and not readily available. X-ray
diffraction and neutron diffraction may be used to complement each other to obtain
extremely useful results (cf. Fig. 12.24).
Methods Based on Since a molecule with a center of symmetry, such as one belonging to point groups
Molecular Symmetry D nh (n even), C nl , (n even). D nJ (n odd), O h , and /,„ cannot have a dipole moment, no
matter how polar the individual bonds (Chapter 3), dipole moments have proved to be
useful in distinguishing between two structures. Much of the classic chemistry of
square planar coordination compounds of the type MA,B 2 was elucidated on the basis
of cis isomers having dipole moments and trans isomers having none (see Chan¬
ter 12). 4 ?
Both infrared (IR) and Raman spectroscopy have selection rules based on the
symmetry of the molecule. Any molecular vibration that results in a change of dipole
moment is infrared active. Fora vibration to be Raman active, there must be a change
of polarizability of the molecule as the transition occurs. It is thus possible to
determine which modes will be IR active, Raman active, both, or neither from the
symmetry of the molecule (see Chapter 3). In general, these two modes of spec¬
troscopy are complementary; specifically, if a molecule has a center of symmetry, no
IR active vibration is also Raman active.
There are many methods that give spectroscopic shifts, sometimes called chem¬
ical shifts, depending on the electronic environment of the atoms involved. For many
of these methods, it is necessary to use symmetry considerations to decide whether
atoms are chemically equivalent (symmetry equivalent) in interpreting the spec¬
troscopic results. Two atoms will be symmetry equivalent if there is at least one
symmetry operation that will exchange them. For example, the chloro groups of
PtCI- (Dji,) are all equivalent since any one chloro group can be moved into the
position occupied by another one by a C A rotation or by a rellcction [o t . or a ). On the
other hand, in AI : Br h (D, v Fig. 6.1 h) the bridging bromo groups are not equivalent to
the terminal bromo groups since no symmetry operation within D 2h allows them to
interchange positions. Operations do exist, however, that interchange the two bridg¬
ing bromo groups with each other and they therefore are equivalent. Likewise, the
four terminal groups are equivalent because they can be interchanged by a symmetry
operation.
Perhaps more subtle is a molecule such as PF 5 (Fig. 6.le), which has D Vl
symmetry. The three equatorial fluorine atoms can be interchanged by reflection or by
rotation about the C 3 axis. Similarly, the two axial atoms can be reflected or rotated
into each other. However, no operation allows interchange of an axial and an equa¬
torial fluorine atom. Thus we have two sets of symmetry (and chemically) equivalent
fluorine atoms. As a consequence, we would not expect P— F„, bond lengths to be the
same as P— F cq bond lengths (and they are not), nor would we expect the five fluorine
47 For further examples of the use of dipole moments in structure analysis, see Moody, G. J.; Thomas,
J. D. R. Dipole Moments in Inorganic Chemistry: Edward Arnold: London 1971
126 4-
Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
It should be kept clearly in mind that the radius ratio rules apply strictly only to
the packing of hard spheres of known size. As this is seldom the case, it is surprising
that the rules work as well as they do. Anions are not “hard" like billiard balls, but
polarizable under the influence of cations. To whatever extent such polarization or
covalency occurs, errors are apt to result from application of the radius ratio rules.
Covalent bonds are directed in space unlike electrostatic attractions, and so certain
orientations are preferred.
There are, however, other exceptions that are difficult to attribute to directional
covalent bonds. The heavier lithium halides only marginally obey the rule, and per¬
haps a case could be made for C.N. =4 for Lil (Fig. 4.18). Much more serious,
however, is the problem of coordination number 6 versus 8. The relative lack of
eight-coordinate structures—CsCl, CsBr, and Csl being the only known alkaii metal
examples is commonly found, if hard to explain. There are no eight-coordinate
Fig. 4.1 8 Actual crystal structures of the alkali halides (as shown by the symbols) contrasted
with the predictions of the radius ratio rule. The figure is divided into three regions by the
lines r + /r_ = 0.414 and r t /r_ = 0.732, predicting coordination number 4 (wurtzite or zinc
blende, upper left), coordination number 6 (rock salt, NaCl, middle), and coordination
number 8 (CsCl, lower right). The crystal radius of lithium, and to a lesser extent that of
sodium, changes with coordination number, so both the radii with C.N. = 4 (left) and
C.N. = 6 (right) have been plotted.
The Predictive Power of Then
I Calculations on Ionic Compounds 127
oxides, MO. even though the larger divalent metal ions, such as Sr' 1 '. Ba 2+ , and
Pb 2+ , are large enough that the radius ratio rule would predict the CsCl struc¬
ture. There is no simple explanation for these observations. We have seen that the
Madelung constant for C.N. = 8 is only marginally larger than that for C.N. = 6.
Thus small energies coming from other sources can tip the balance.
The radius ratio is a useful, though imperfect, tool in our arsenal for predicting
and understanding the behavior of ionic compounds. 29 From a theoretical point of
view it rationalizes the choice of lattice for various ionic or partially ionic compounds.
Its failings call our attention to forces in solids other than purely electrostatic ones
acting on billiard-ball-like ions. We shall encounter modifications and improvements
of the model in Chapter 7.
The Predictive
Power of
Thermochemical
Calculations on
Ionic Compounds
The following example will illustrate the way in which the previously discussed
parameters, such as ionic radii and ionization energies, can be used advantageously
to explore the possible existence of an unknown compound. Suppose one were in¬
terested in dioxygenyl tetrafluoroborate. [0,] + [BF 4 ]'. At first thought it might
seem an unlikely candidate for existence since oxygen tends to gain electrons rather
than lose them. However, the ionization energy of molecular oxygen is not exces¬
sively high (1165 kJ mol' 1 ; cf. Hg, 1009 kJ mol' 1 ), so some trial calculations might
be made as follows.
The first values necessary are some estimates of the ionic radii of 07 and BF 4 .
For the latter we may use the value obtained ihermochcmically by Yalsimirskii,
218 pm. An educated guess has to be made for 0 2 + , since if we arc attempting to
make it for the first lime (as was assumed above), we will not have any experimental
data available for this species. However, we note that the CN" ion, a diatomic ion
which should be similar in size, has a thcrmochemical radius of 177 ppm. Further¬
more, an estimate based on covalent and van der Waals radii (see Chapter 8) gives
a similar value. Because 07 has lost one electron and is positively charged, it will
probably be somewhat smaller than this. We can thus take 177 pm as a conservative
estimate; if the cation is smaller than this, the compound will be more stable than
our prediction and even more likely to exist. Adding the radii we obtain an estimate
of 395 pm for the intcrionic distance.
Next the lattice energy can be calculated. One method would be to assume
that wc know nothing about the probable structure and use the Kapuslinskii equa¬
tion (Eq. 4.20) and r 0 = 395 pm. The resulting lattice energy is calculated to be
— 555 kJ mol~
Alternatively, we might examine the radius ratio of 07BF 4 and get a crude esti-
mate of ^ = 0.8. The accuracy of our values does not permit us to choose between
coordination number 6 and 8, but since the value of the Madelung constant does
not differ appreciably between the sodium chloride and cesium chloride structures, a
value of 1.75 may be taken which will suffice for our present rough calculations. We
may then use the Born-Lande equation (Eq. 4.13), which provides an estimate of
-616 kJ mol" 1 for the attractive energy, which will be decreased by about 10% (if
3 " An analysis of 227 compounds indicated that the radius ratio rule worked about two-thirds of the
time. Particularly troublesome were Group IB (I I) and IIB (12) chalcogenidcs like HgS. Nathan, L. C.
J. Chem. Etiuc. 1985, 62, 215-218.
232
6-The Structure and Reactivity of Molecules
R aV Bond lengths and multiple bonding were discussed in Chapter 5. and a comparison of
Bond Length s- various lypes of atomic radii will be discussed in Chapter 8 , but a short d.scuss.on ot
factors that affect the distance between two bonded atoms will be given here to
complement the previous discussion of stenc factors.
Rond Multiplicity One of the most obvious factors affecting the distance between two atoms is the bond
Bond Multiplicity multj , idty> single bonds are longer than double bonds which are longer than tnpte
bonds: C-C = 154 pm, C=C = 134 pm, C=C = 120 pm; N-N - 145 pm, N-N
= ,25 pm N=N = 110 pm; 0-0 = 148 pm, 0=0 = 121 pm. etc. For carbon.
Pauling 44 has derived the following empirical relationship between bond length (£> in
pm) and bond order («):
D n =D, - 71 log n 6-8)
This relationship holds not only for integral bond orders but also for fractional ones (in
molecules with resonance, etc.). One can thus assign variable bond orders depending
upon the length of the bond. In view of the many factors affecting bond lengths, to be
discussed below, it does not seem wise to attempt to quantify the bond order-bond
length relationship accurately. Nevertheless, bonds formed by elements other than
carbon show similar trends (Fig. 6.23), and the general concept certainly is a valid one
We have seen in Chapter 5 that the strength of a bond depends to a certain extern
upon the hybridizations of the atoms forming the bond. We should therefore expect
bond length to vary with hybridization. Bent has shown that this vanat,on is quite
regular: C-C bond lengths are proportional to p character (Fig. 6.24) or, to say
Bond order
Fig. 6.23 Tungsten-tungsten
bond lengths as a function
of bond order. Each of the
points represents the mean
and range of several values.
See Chapter 16 for
discussion of multiple
bonds in metal clusters.
Bond orders are plotted on
a logarithmic scale in
accord with Eq. 6.8.
<4 Pauling. L. The Nature of,he Chemical Bond. 3rd ed.: Cornell University: Ithaca. NY, 1960; p 239.
Experimental Determination of Molecular Structure
233
Fig. 6.24 Carbon-carbon
bond lengths as a function
of the hybridization of the
bonding atoms. [Data from
Bent. H. A. Chem. Rev.
1961. 61. 275-311.]
another way, increasing s character increases overlap and bond strength and thus
shortens bonds.
Another factor that affects bond length is electronegativity. Bonds lend to be
shortened, relative to the expectations for nonpolar bonds, in proportion to the
electronegativity difference of the component atoms. I hus the experimental bond
length in HF is 91.8 pm versus an expected value of 108 pm. The quantitative
shortening of bonds because of electronegativity differences and multiple bonding in
elements other than carbon will be discussed in Chapter 8 .
Experimental
Determination of
It is impossible to present the theory and practice of the various methods of determin¬
ing molecular structure completely. No attempt will be made here to go into these
methods in depth, but a general feeling of the importance of the different techniques
Molecular Structure can be gathered together with their strengths and shortcomings. For more material on
these subjects, the reader is referred to texts on the application of physical methods to
inorganic chemistry . 45
X-ray Diffraction X-ray diffraction (Chapter 3) has provided more structural information for the in¬
organic chemist than any other technique. It allows the precise measurement of bond
angles and bond lengths. Unfortunately, in the past it was a time-consuming and
difficult process, and molecular structures were solved only when there was reason to
believe they would be worth the considerable effort involved. The advent of more
efficient methods of gathering data and doing the computations has made it relatively
easy to solve most structures.
45 Drago. R. S. Physical Methods for Chemists. 2nd ed.; Saunders: Philadelphia. 1992. Ebsworth. E.
A. V.; Rankin, D. W. H.; Cradock. S. Structural Methods in Inorganic Chemistry. 2nd ed.;
Blackwell: Oxford, 1991.
128
4 • Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
n = 10) to 20% (if n = 5). The two calculations thus agree that the lattice energy will
probably be in the range —480 to —560 kJ mol 1 ( — 115 to — 134 kcal mol '). This
is a quite stable lattice and might be sufficient to stabilize the compound.
Next we might investigate the possible ways of producing the desired compound.
Because the oxidation of oxygen is expected to be difficult to accomplish we might
choose vigorous oxidizing conditions, such as the use of elemental fluorine:
0 2 + F 2 + BF 3 -► [0 2 ] + [BF 4 y (4.25)
It is possible to evaluate each term in a Born-Haber cycle based on Eq. 4.25.
The usual terms we have encountered in previous Born-Haber cycles may be
evaluated readily:
Ionization energy of 0 2 = 1165 kJ mol -1
Dissociation of £F 2 = 79 kJ mol' 1
Electron affinity of F = —328 kJ mol 1
One additional term occurs in this Born-Haber cycle: the formation of the tetra-
fluoroborate ion in the gas phase:
BF 3 ,b) + F<g) - BFii B ) (4-26)
Fortunately, the enthalpy of this reaction has been experimentally measured 30 to be
-423 kJ mol" 1 . Adding in the value of -500 ± 20 kJ mol -1 for the lattice energy
provides an estimate of the heat of the reaction in Eq. 4.25 that is essentially zero.
This is somewhat discouraging, since if Eq. 4.25 is not exothermic, entropy will drive
the reaction to the left because all of those species are gases, and dioxygenyl tetra-
fluoroborate would not be expected to be stable. Recall, however, that our estimates
were on the conservative side. We would therefore expect that dioxygenyl tetra-
fluoroborate is either energetically unfavorable or may form with a relatively low
stability. It certainly is worth an attempt at synthesis.
In fact, dioxygenyl tetrafluoroborate has been synthesized by a reaction similar
to Eq. 4.25, although in two steps: the formation of intermediate oxygen fluorides and
then combination with boron trifluoride. 31 It is a white crystalline solid that slowly
decomposes at room temperature. Energy calculations of this type arc exceedingly
useful in guiding research on the synthesis of new compounds. Usually it is not neces¬
sary to start with the complete absence of knowledge assumed in the present example.
Often one or more factors can be evaluated from similar compounds. It was the
observation of the formation of dioxygenyl hexafluoroplatinate(V) and similar cal¬
culations that led Bartlett to perform his first experiment in an attempt to synthesize
compounds of xenon. This successful synthesis overturned prior chemical dogma (see
Chapter 17).
Now that we have seen that dioxygenyl compounds can be prepared, we might
be interested in preparing the exotic and intriguing compound dioxygenyl super¬
oxide, 0 2 + 0 2 . Using methods similar to those discussed above, we can set up a
30 Srivastava, R. D.; Uy, O. M.; Farbcr, M. J. Chem. Soc., Faraday Trans. I 1974, 70, 1033.
31 Keith, J. N.; Solomon, I. J.; Shell, I.; Hyman, H. H. Inorg. Chem. 1968, 7, 230-234. Goetschel, C.T.;
Campanile, V. A.; Wagner, C. D.; Wilson. J. N. J. Am. Chem. Soc. 1969. 91. 4702-4707
Covalent Character in Predominantly h
lie Bonds
129
(a) (b) < c >
Fig. 4.19 Polarization effects: (a) idealized ion pair with no polarization, (b) mutually
polarized ion pair, (c) polarization sufficient to form covalent bond. Dashed lines represent
hypothetical unpolarized ions.
Born-Haber cycle and evaluate the following terms.
O, -♦ 0 2 + + e" A H = 1165 kJ mol' 1
0 2 + e" - 1 - OJ A H = -42 kJ mol' 1
Lattice energy AH = —500 kJ mol
A H f = +623 kJ mol -1
The calculations support our intuitive feelings about this compound. If it were
somehow possible to make an ionic compound 07 0 2 , it would decompose with
the release of a large amount of energy:
0 + 07 - 20 2 A H » -623 kJ mol' 1
Covalent Char acter
in Predominantly^
■Ionic Bonds _
Dioxygenyl superoxide is not a likely candidate for successful synthesis.
It is probable that every heteronuclear bond the chemist has to deal with contains a
mixture of covalent and ionic character. Ordinarily we speak glibly of an ionic com¬
pound or a covalent compound as long as the compound in question is predominantly
one or the other. In many cases, however, it is convenient to be able to say some¬
thing about intermediate situations. In general, there are two ways of treating iomc-
covalcnt bonding. The method that has proved most successful is to consider the bond
to be covalent and then consider the effect of increasing charge displacement from
one atom toward another. This method will be discussed in the next chapter. Another
method is to consider the bond to be ionic and then allow for a certain amount of
covalency to occur. The second method was championed by Kasinur Fajans in
his quanticule theory. The latter theory has found no place in the repertoire of the
theoretical chemist largely because it has not proved amenable to the quantitative
calculations which other theories have developed. Nevertheless, the qualitative ideas
embodied in “Fajans’ rules” offer simple if inexact approaches to the problem ol
partial covalent character in ionic compounds.
Fajans considered the effect which a small, highly charged cation would have on
an anion. If the anion were large and “soft" enough, the cation should be capable of
polarizing it, and the extreme of this situation would be the cation actually pene¬
trating the anionic electron cloud giving a covalent (shared electron) bond (Fig. 4.19).
.'2 Fajans. K. Nattirwissensclwflen 1923. //. 165. For a more recent discussion or Ihc same subject, see
Fajans. K. Struct. Bonding Berlin 1967, 3. 88-105. For an interesting short sketch on the theory and the
man. sec Hurwic. J. J. Cltem. Educ. 1987, 64. 122.
230
6-The Structure and Reactivity of Molecules
Bent Bonds
and methylene halides, HCX, and H 2 CX 2 , substitution of fluorine causes the bond
angles to decrease (108.6°, 108.3°), but the corresponding chlorine compounds show
larger bond angles (III .3°, 111.8°).
An interesting example of this effect is isobutylene, CH,=C(CH 3 ) 2 . Naively we
might assume substituents on an ethylene to be bonded at 120°. We have seen that
VSEPR predicts that the double bond will tend to close down the CH 3 —C—CH 3
angle, but how much? Bartell has pointed out that if you assume that the two methyl
groups and the methylene group can be portrayed by spheres representing the van der
Waals radius of the carbon plus hydrogen substituents, then the short C=C double
bond naturally causes repulsions at 120° that can be relaxed if the CH 3 —C—CH 3
angle closes. Furthermore, it should close exactly enough to make all repulsions
equal; thus the three substituent carbon atoms should lie very nearly on the corners of
an equilateral triangle, which they do (Fig. 6.20). Furthermore, it is possible to
quantify these qualitative arguments and reproduce the bond angles in some represen¬
tative hydrocarbons quite well (Fig. 6.21).
The overlap of two atomic orbitals is maximized if they bond "head on,” that is, with
the maximum electron density directly between the nuclei. However, small rings
exhibit "ring strain" in the form of poorer overlap. Although rehybridization occurs
as ring size decreases, which places more p character within the ring and more s
character in exocyclic bonds, the minimum interorbital angle possible with only s and
p orbitals is 90° (pure p). In three-mcmbered rings the orbitals cannot follow the
internuclear axis; therefore the so-called a bonds are not symmetrical about that axis
but are distinctly bent. The bending of these bonds can actually be observed experi¬
mentally with a buildup of electron density outside the ring (Fig. 6.22). For example,
in 3-[(p-nitrophenoxy)methyl]-3-chlorodiazirine
the increase in electron density is clearly seen both in the cyclic bonds (outside the
C—N—N triangle) and in the N = N double bond. 4 - 1
Fig. 6.20 Relaxation ot nonbonded repulsions in a perfectly trigonal isobutylene molecule to
torm 114° bond angle observed experimentally. [From Bartell. L. S. J. Chem. Educ. 1968. 45.
754. Used with permission.!
43 Coppens. P. Angew. Chem. Ini. Ed. Engl.. 1977. 16. 32-40. Cameron. T. S.; Bakshi, P. K.;
Borecka. B.; Liu. M. T. H. J. Am. Chem. Soc. 1992. 114. 1889-1890.
Structure and Hybridization
231
Calculated CCC angle (degrees)
Fig. 6.21 Comparison of calculated C—C—C bond angles with experimental values.
Calculations are based on nonbonded interactions. The solid bar represents a range of values
found in n-alkancs. [From Bartell. L. S. J. Chem. Educ. 1968, 45, 754. Used with
permission.)
Fig. 6.22 I he three-membered C—N—N ring in a diazirine molecule: (a) Electron density
through the plane of the diazirine ring. Contours are at 5 x I0-? c pm" 1 , (b) Differential
electron density map through the plane of the diazirine ring showing increases (solid lines)
and decreases (broken lines) of electron density upon bond formation. Contours are at
4 x IO-» e pm -3 . Note the build-up of electron density om.iidc the C—N—N triangle.
The effects can best be seen by superimposing a transparency (as made for overhead
projectors, for example) of part (b) over part (a), (c) Interpretation of (a) and (b) in terms of
hybnd orbitals. [From Cameron. T. S.; Bakshi. P. K..; Borccka. B.; Liu, M. T. H. J. Am. '
Chem. Soc. 1992. 114. 1889-1890. Reproduced with permission.)
1 30
4 ■ Bonding Models in Inorganic Chemistry: 1. Ionic Compounds
Fajans suggested the following rules to estimate the extent to which a cation could
polarize an anion and thus induce covalent character. Polarization will be increased
by:
High charge and small size of the cation. Small, highly charged cations will
exert a greater effect in polarizing anions than large and/or singly charged
cations. This is often expressed by the ionic potential 33 of the cation: = Z + /r.
For some simple ions, ionic potentials are as follows (r in nm):
Li + = 14 v Be 2 + = 48,^ B 3+ = 120
Na + = 9 ''"-Mg 2 + = 28 '"^AI 3 + = 56
K + = 7 Ca 2+ = 18 Ga 3+ = 49
Obviously there is no compelling reason for choosing Z/r instead of Z/r 2
or several other functions that could be suggested, and the values above are
meant merely to be suggestive. Nevertheless, polarization does follow some
charge-to-size relationship, and those cations with large ionic potentials are
those which have a tendency to combine with polarizable anions to yield
partially covalent compounds. The ionic potentials listed also rationalize an
interesting empirical observation indicated by the dashed arrows: The first
element in any given family of the periodic chart tends to resemble the second
element in the family to the right. Thus lithium and magnesium have much
in common (the best known examples are the organometallic compounds of
these elements) and the chemistry of beryllium and aluminum is surprisingly
similar despite the difference in preferred oxidation state. 34 This relationship
extends across the periodic chart; for example, phosphorus and carbon re¬
semble each other in their electronegativities (see Chapter 18).
A word should be said here concerning unusually high ionic charges often
found in charts of ionic radii. Ionic radii are often listed for Si 4 , P 5 , and
even Cl . Although at one time it was popular, especially among geochemists,
to discuss silicates, phosphates, and chlorates as though they contained these
highly charged ions, no one today believes that such highly charged ions have
any physical reality. The only possible meaning such radii can have is to in¬
dicate that if an ion such as P 5 or Cl 7 could exist, its high charge combined
with small size would cause it immediately to polarize some adjacent anion
and form a covalent bond.
High charge and large size of the anion. The polarizability of the anion will
be related to its “softness,” that is, to the deformability of its electron cloud.
Both increasing charge and increasing size will cause this cloud to be less
under the influence of the nuclear charge of the anion and more easily in¬
fluenced by the charge on the cation. Thus large anions such as I“, Se 2 ~, and
31 Cartlcdge. G. H. J. Am. Cl,cm. Soc. 1928, 50. 2855. 2863; ibid. 1930. 52. 3076.
34 It is true that the value of the ionic potential or Li r is closer to that or Ca 3 than to that of Mg 3 , and
a strong argument has been made that Li resembles Ca 3 more than Mg 3 [Hanusa, T. P. J. Chem.
Educ. 1987, 64. 686.] The strength of the Fajans approach and the related idea of diagonal resemblance
rests on its qualitative success. The diagonal rule and the ionic potential should be used as guides rather
than as substitutes for close inspection of each individual situation.
Results of
Polarization
Covalent Character in Predominantly Ionic Bonds 131
Te 2 and highly charged ones such as As 3- and P 3 ' are especially prone to
polarization and covalent character.
A question naturally occurs: What about the polarization of a large cation
by a small anion? Although this occurs, the results are not apt to be so spec¬
tacular as in the reverse situation. Even though large, a cation is not likely
to be particularly “soft" because the cationic charge will tend to hold on to
the electrons. Likewise, a small anion can tend to polarize a cation, that is,
repel the outside electrons and thus make it possible to “see” the nuclear
charge better, but this is not going to lead to covaleni bond formation. No
convincing examples of reverse polarization have been suggested.
Electron configuration of the cation. The simple form of the ionic potenlial
considers only the net ionic charge of the ion with respccl to its size. Actually
an anion or polarizable molecule will feel a potential resulting from the total
positive charge minus whatever shielding the electrons provide. To use the
ionic charge is to assume implicitly that the shielding of the remaining elec¬
trons is perfect, that is, 100% effective. The most serious problems with this
assumption occur with the transition metal ions since they have one or more
d electrons which shield the nucleus poorly. Thus for two ions of the same
size and charge, one with an {n - 1 )d x ns° electronic configuration (typical of
the transition elements) will be more polarizing than a cation with a noble
gas configuration (n - l)s 2 (a - l)p 6 tis° (alkali and alkaline earth metals, for
example). As an example, Hg 2+ has an ionic radius (C.N. = 6) of 116 pm, yet
it is considerably more polarizing and its compounds are considerably more
covalent than those ofCa 2 with almost identical size (114 pm) and the same
charge.
One of the most common examples of covalency resulting from polarization can be
seen in the melting and boiling points of compounds of various metals. 35 Comparing
the melting points of compounds having the same anion, but cations of dilTcrcnt size,
we have BcCK = 405°C. CaCI : = 782 °C; for cations of different charge, we have
NaBr - 747°C, MgBr, = 700 °C, AlBrj = 97.5°C; for a constant cation, but anions
of different sizes, wc have LiF = 845 °C, LiCl = 605 °C, LiBr = 550 C, Li I = 449 ’C;
and for ions having the same size and charge; the effect of electron configuration can
be seen from CaCl 2 = 782 ’C, HgCI 2 = 276 °C. Care must be taken not to interpret
melting points and boiling points too literally as indicators of the degree of covalent
bonding; there arc many effects operative in addition to covalency and these will be
discussed at some length in Chapter 8.
A second area in which polarization effects show up is the solubility of salts in
polar solvents such as water. For example, consider the silver halides, in which we
have a polarizing cation and increasingly polarizable anions. Silver fluoride, which is
quite ionic, is soluble in water, but the less ionic silver chloride is soluble only with
the inducement of complexing ammonia. Silver bromide is only slightly soluble and
silver iodide is insoluble even with the addition of ammonia. Increasing covalency
from fluoride to iodide is expected and decreased solubility in water is observed.
35 One learns in general chemistry courses that ionic compounds have high melting points and covalent
ones have low melting points. Although this oversimplification can be misleading, it may be applied to
the present discussion. A more thorough discussion of the factors involved in melting and boiling points
will be found in Chapter 8.
228
6-The Structure and Reactivity of Molecules
prepared by Martin and coworkers.” These, as well as related phosphoranes, provide
interesting insight into certain molecular rearrangements (see page 240).
Bent’s rule is also consistent with, and may provide alternative rationalization tor.
Gillespie's VSEPR model. Thus the Bent’s rule predication that highly electronegative
substituents will ■’attract” p character and reduce bond angles is compatible with the
reduction in angular volume of the bonding pair when held tightly by an elec¬
tronegative substituent. Strong, 5-rich covalent bonds require a larger volume in
which to bond. Thus, doubly bonded oxygen, despite the high electronegativity of
oxygen, seeks 5 -rich orbitals because of the shortness and better overlap of the double
bond Again, the explanation, whether in purely 5-character terms (Bent s rule) or in
larger angular volume for a double bond (VSEPR), predicts the correct structure.
It is sometimes philosophically unsettling to have multiple “explanations when
we are trying to understand what makes molecules behave as they do; it is only human
“to want to know for sure" why things happen. On the other hand, alternative,
nonconflicting hypotheses give us additional ways of remembenng and predicting
facts, and if we seem to find them in conflict, we have made either: (1) a mistake
(which is good to catch) or (2) a discovery!
The mechanism operating behind Bent’s rule is not completely clear. One factor
favoring increased p character in electronegative substituents is the decreased bond
angles of p orbitals and the decreased steric requirements of electronegative substit¬
uents. There may also be an optimum "strategy” of bonding for a molecule >n which j
character (and hence improved overlap) is concentrated in those bonds in which the
electronegativity difference is small and covalent bonding is important. The p char¬
acter. if any. is then directed toward bonds to electronegative groups. The latter will
result in greater ionic bonding in a situation in which covalent bonding would be low
anyway (because of electronegativity differences). 38
Some light may be shed on the workings of Bent’s rule by observations ot
apparent exceptions to it. 39 The rare exceptions to broadly useful rules arc unfortu¬
nate with respect to the universal application of those rules. They also have the
annoying tendency to be confusing to someone who is encountering the rule for the
first time. On the other hand, any such exception or apparent exception is a boon to
the research scientist since it almost always provides insight into the mechanism
operating behind the rule. Consider the cyclic bromophosphate ester:
PH Rr
Br
The phosphorus atom is in an approximately tetrahedral environment using four
u bonds of approximately 5 P 3 character. We should expect the more electronegative
oxygen atoms to bond to 5 -poor orbitals on the phosphorus and the two oxygen atoms
in the ring do attract hybridizations of about 20% 5.-" The most electropositive substit¬
uent on the phosphorus is the bromine atom and Bent’s rule would predict an 5-nch
37 For a review, see Martin. J. C. Science 1983. 221. 509-514.
38 For a further discussion of ionic versus covalent bonding and the total bond energy resulting from
the sum of the two. see Chapter 9, “Electronegativity and Hardness and Softness.
31 In addition to the example given here, see Problem 6.12.
40 Grim. S. O.; Plastas. H. J.; Huheey, C. L.; Huheey. J. E.. Phosphorus 1971. /. 61-66.
Structure and Hybridization
229
orbital, but instead it draws another 5-poor orbital, slightly less than 20% 5. The only 5-
rich orbital on the phosphorus atom is that involved in the cr bond to the exocyclic
oxygen. This orbital has nearly 40% 5 character! This oxygen ought to be about as
electronegative as the other two. so why the difference? The answer probably lies in
the overlap aspect. (1) The large bromine atom has diffuse orbitals that overlap poorly
with the relatively small phosphorus atom; thus, even though the bromine is less
electronegative than the oxygen, it probably does not form as strong a covalent bond.
(2) The presence of a rr bond shortens the exocyclic double bond and increases the
overlap of the <j orbitals. If molecules respond to increases in overlap by rehybridiza¬
tion in order to profit from it, the increased 5 character then becomes reasonable.
From this point of view. Bent's rule.might be reworded: The p character lends to
concentrate in orbitals with weak covalency [from either electronegativity or overlap
considerations), and s character tends to concentrate in orbitals with strong
covalency (matched electronegativities and good overlap).
Some quantitative support for the above qualitative arguments comes from aver¬
age bond energies of phosphorus, bromine, and oxygen (Appendix E):
P— Br 264 kJ mol -1
P—O 335 kJ mol -1
P=0 544 kJ mol-'
Bent’s rule is a useful tool in inorganic and organic chemistry. For example, it has
been used to supplement the VSEPR interpretation of the structures of various
nonmetal fluorides, 4 ' and should be applicable to a wide range of questions on
molecular structure.
Nonbonded For anyone who has encountered steric hindrance in organic chemistry, the emphasis
Repulsions thus far placed on electronic effects as almost the only determinant must seem
j . puzzling. However, the preeminence of electronic over steric effects can be ra-
° n rUC UrS tionalized in terms of the following points: (1) Even the largest "inorganic atomic
substituent," the iodine atom, is no larger than a methyl group (see Table 8.1 for van
der Waals radii), to say nothing of /-butyl or di-orthosubstituted phenyl groups. (2)
The wider range of hybridizations, the larger variety of varying electronegativities,
and the greater importance of smaller molecules all combine to enhance electronic
effects, so much so. in fact, that it is easy to forget nonbonded interactions in the
discussions of electron-pair repulsions, overlap, etc. in the preceding sections. Bartel!
has called attention to the importance of nonbonded repulsions and to the situations in
which they may be expected to be important. 42 In general, the latter are apt to be
molecules in which a small central atom is surrounded by large substituent atoms.
Consider the water molecule, for example, with a bond angle of 104.5°. Replacing the
hydrogen atoms with more electronegative halogen atoms should reduce the bond
angle in terms of either Bent’s rule or VSEPR-electronegativity models (rule 4, page
217). Indeed. OF, has a slightly smaller bond angle, 103.2°. On the other hand, the
bond angle in Cl,0 is larger than tetrahedral; it is 110.8°. Similarly, in the haloforms
41 Huheey, J. E. Inorg. Chem. 1981, 20. 4033-4035.
42 Bartell. L. S. J. Chem. Educ. 1968. 45 . 754-767. This is another example of Ihe principle that if
overlap is included, antibonding orbitals arc more destabilizing than the corresponding bonding
orbitals are stabilizing. Nonbonded repulsions are merely another name for describing the forced
overlap of two orbitals already filled with electrons and the destabilization that occurs as the
antibonding orbital rises in energy faster than the bonding one lowers.
4-Bonding Model. In Inorganic Chemistry: 1. Ionic Compounds
n 3 o I" 3
Fig. 4.20 Hydration
energies as a function of
size and charge of cations.
[From Philips, C. S. G.;
Williams, R. J. P. Inorganic
Chemistry. ; Clarendon:
Oxford, 1965. Reproduced
with permission.]
ou’V TV H
-500 - Zn J VmV" Co2
c d J \^Mg 2
-400 -
Pb 2 -!_^V \r 3 Mn J
-300 - r » j -—
/ ^ Ba 2
100 -
Kjg£— Tl
° ^ 1 2 3 4 5 6 7
Silver halide
K s„
Silver fluoride
Soluble
Silver chloride
2 x 10" 10
Silver bromide
5 x 1(T 13
Silver iodide
8 x KT 17
As in the case of melting points, solubility is a complex process, and there are many
factors involved in addition to covalency. _ ,
Closely related to solubility are the hydration enthalpies of ions. It has been
found 36 that it is possible to correlate the hydration enthalpies of cations with their
“effective ionic radii” by the expression (see Fig. 4.20)
A H = - 69,500(Z 2 /r eff ) kJ mol" 1 (r eff in pm) (<U7)
36 Latimer, W. M.; Pitzer, K. S.; Slansky, C. M. J. Chem. Phys. 1939. 7, 108-111.
Covalent Character i
Predomina
ly Ionic Bonds
133
In this case the reason lor the correlation is fairly obvious. The parameter r eff is
equal to the ionic radius plus a constant, 85 pm, the radius of the oxygen atom in
water. Therefore, r c(r is effectively the interatomic distance in the hydrate, and the
Born-Lande equation (Eq. 4.13) can be applied.
A third and perhaps the most fundamental, aspect of polarization can be seen
in the bond lengths of silver halides. If we predict these distances using the ionic
radii of Table 4.4, our accuracy decreases markedly in the direction AgF > AgCl >
AgBr > Agl:
Compound r + r~ cxp ^
AgF 248 246 -2
AgCl 296 277 -19
AgBr 311 289 —22
Agl 320 281 -39
The Shannon-Prewitt ionic radii (r + + r") are based on the most ionic compounds,
the fluorides and oxides for the radii of the metal cations, and the alkali halides for
the radii of the anions of the remaining halides. The shortening of silver halide bond
lengths is attributable to polarization and covalency.
The basis for other correlations between size, charge, and chemical properties is
not so clearcut. Chemical reactions can often be rationalized in terms of the polarizing
power of a particular cation. In the alkaline earth carbonates, for example, there is
a tendency toward decomposition with the evolution of carbon dioxide:
MC0 3 -♦ MO + C0 2 (4 ' 28)
The ease with which this reaction proceeds (as indicated by the temperature necessary
to induce it) decreases with increasing cation size: BeC0 3 , unstable: MgCOj, 350 °C;
CaC0 3 , 900 °C; SrC0 3 , 1290°C; BaCOj, 1360"C. The effect of d electrons is also
clear: Both CdC0 3 and PbC0 3 decompose at approximately 350 °C despite the fact
that Cd 2+ and Pb 2+ are approximately the same size as Ca 2 + . The decomposition
of these carbonates occurs as the cation polarizes the carbonate ion, splitting it into
an O 2- ion and C0 2 . . . ,
Stern 37 has extended the qualitative argument on decomposition by showing
that the enthalpies of decomposition of carbonates, sulfates, nitrates, and phosphates
are linearly related to a charge/sizc function, in this case r u /Z (see Fig. 4-1).
Although the exact theoretical basis of this correlation is not clear, it provides an¬
other interesting example of the general principle that size and charge are the im¬
portant factors that govern the polarizing power of ions and, consequently, many
of their chemical properties.
From the preceding, it might be supposed that covalent character in predom¬
inantly ionic compounds always destabilizes the compound. This is not so. Instability
results from polarization of the anion causing it to split into a more stable compound
(in the above cases the oxides) with the release of gaseous acidic anhydrides. As will
be seen in Chapter 16, many very stable, very hard minerals have covalent-iomc
bonding.
37 Stern, K. H. J. Chem. Educ. 1969, 46. 645.
226
6-The Structure and Reactivity of Molecules
In Ihe phosphorus atom there is little initial promotion energy: The ground state is
trivalent, as is the valence state. Note that any hybridization will cost energy as a filled
3i orbital is raised in energy and half-filled 3 p orbitals are lowered in energy:
This energy of hybridization is of the order of magnitude of bond energies and can
thus be important in determining the structure of molecules. It is responsible for the
tendency of some lone pairs to occupy spherical, nonstereochemically active y orbitals
rather than stereochemically active hybrid orbitals (see page 215). For example, the
hydrides of the Group VA (15) and VIA (16) elements are found to have bond angles
considerably reduced as one progresses from the first element in each group to those
that follow (Table 6.4). An energy factor that favors reduction in bond angle in these
compounds is the hybridization discussed above. It costs about 600 kJ moM to
hybridize the central phosphorus atom. From the standpoint of this energy factor
alone, the most stable arrangement would be utilizing pure p orbitals in bonding and
letting the lone pair “sink" into a pure j orbital. Opposing this tendency is the
repulsion of electrons, both bonding and nonbonding (VSEPR). This favors an ap¬
proximately tetrahedral arrangement. In the case of the elements N and O the steric
effects are most pronounced because of the small size of atoms of these elements. In
the larger atoms, such as those or P. As, Sb, S, Se. and Te. these effects are somewhat
relaxed, allowing the reduced hybridization energy of more p character in the bonding
orbitals to come into play.
Another factor which affects the most stable arrangement of the atom in a
molecule is the variation of bond energy with hybridization. The directed lobes of s-p
hybrid orbitals overlap more effectively than the undirected j orbitals, the two-lobed p
orbitals, or the diffuse d orbitals. The increased overlap results in stronger bonds
Tablo 6.4
Bond angles in tho hydridos NH 3 = 107.2° PH, = 93.8° AsH, = 91.8° SbH, = 91 3°
VIACOMo VA (15) a " d 0H 2 = 04.5° SH 2 = 92° SeHj = 91° TeH 2 = 89.5°
" P ala f rom Wells. A. F. Structural Inorganic Chemistry, 5th ed.; Oxford
University: Oxford, 1984. Reproduced with permission.
227
Structure and Hybridization
(Chapter 5). The molecule is thus forced to choose” between higher promotion
energies and better overlap for an i-rich3-> hybrid, or lower promotion energies and
poorer overlap for an j-poor hybrid.
Good examples of the effect of the differences in hybrid bond strengths are shown
by the bond lengths in MX„ molecules with both equatorial and axial substituents.35
r eq (Pm)
r ax <Pm)
PF 5
153.4
157.7
PC1 5
202
214
SbCIj
231
243
SF„
154
164
cif 3
159.8
169.8
BrF 3
172.1
181.0
An spd hybrid orbital set may be considered to be a combination of p.d.i hybrids
and sp x p y hybrids. The former make two linear hybrid orbitals bonding axially and the
latter form the trigonal, equatorial bonds. The sp hybrid orbitals are capable of
forming stronger bonds, and they are shorter than the weaker axial bonds. When the
electronegativities of the substituents on the phosphorus atom differ, as in the mixed
chlorofiuorides, PCI t F s _ r , and the alkylphosphorus fluorides, R t PF s _ r , it is experi¬
mentally observed that Ihe more electronegative substituent occupies the axial posi¬
tion and the less electronegative substituent is equatorially situated. This is an
example ot Bent s rule , 3A which states: More electronegative substituents "prefer"
hybrid orbitals having less s character, and more electropositive substituents “pre¬
fer' hybrid orbitals having more s character. Although proposed as an empirical rule,
in the chlorofiuorides of phosphorus, it is substantiated by molecular orbital
calculations.
A second example of Bent's rule is provided by the fluoromethanes. In CH,F, the
F ~ C ~ F bond an 8le is less than 1091°, indicating less than 25% j character, but the
H—C—H bond angle is larger and the C—H bond has more s character. The bond
angles in the other fluoromethanes yield similar results (Problem 6.6).
The tendency of more electronegative substituents to seek out the low elec¬
tronegativity p.d.: apical orbital in TBP structures is often termed “apicophilicity." It
is well illustrated in a scries of oxysulfuranes of the type
I ho molecule, of course, does not "choose." bul in order lo understand its behavior, Ihe chemist
must choose the relative importance.
,J Since most orbitals range from 0% to 50% s character, .r-rich orbitals arc those with more than 25°/ j
character, and j-poor arc those with less than 25%.
33 Dala from Wclls - A - F - Structural Inorganic Chemistry. 5th ed.; Oxford University: Oxford. 1984.
36 Bent, H. A. J. Chem. Educ. I960. 37, 616-624: Chem. Rev. 1961, 61, 275-311.
134 4 B
Conclusion
Problems
onding Models in Inorganic Chemistry: 1. Ionic Compounds
Fig. 4.21 Enthalpy of
decomposition of sulfates
and carbonates as a
function of size and charge
of the metal cation. [From
Stern, K. H. J. Chem. Educ.
1969, 46, 645-649.
Reproduced with
permission.]
Ionic crystals may be viewed quite simply in terms of an electrostatic model of lattices
of hard-spherc ions of opposing charges. Although conceptually simple, this model is
not completely adequate, and we have seen that modifications must be made in it.
First, the bonding is not completely ionic with compounds ranging from the alkali
halides, for which complete ionicity is a very good approximation, to compounds for
which the assumption of the presence of ions is rather poor. Secondly, the assumption
ol a perfect, infinite mathematical lattice with no defects is an oversimplification.
As with all models, the use of the ionic model does not necessarily imply that it is
true , merely that it is convenient and useful, and if proper caution is taken and
adjustments are made, it proves to be a fruitful approach.
4.1 Both CsCl and CaF 2 exhibit a coordination number of 8 for the cations. What is the
structural relationship between these two lattices?
4.2 The contents of the unit cell of any compound must contain an integral number of
formula units. (Why?) Note that unit cell boundaries "slice” atoms into fragments: An
atom on a face will be split in half between two cells; one on an edge will be split into
quarters among four cells, etc. Identify the number of Na and Cl' ions in the unit cell
of sodium chloride illustrated in Fig. 4.1a and state how many formula units of NaCI the
unit cell contains. Give a complete analysis.
4J The measured density of sodium chloride is 2.167 gem' 3 . From your answer to Prob¬
lem 4.2 and your knowledge of the relationships among density, volume, Avogadro's
Problems 135
number, and formula weight, calculate the volume of the unit cell and thence the length
of the edge of the cell. Calculate the length r f + r_. Check your answer, r + + r_, against
values from Table 4.4.
4.4 Study Figs. 4.1-4.3 and convince yourself of the structural relatcdness of all of the cubic
structures and of all of the hexagonal structures.
4.5 The structure of diamond, a covalent crystal, is shown in Fig. 7.1. How is it related to
some of the structures of ionic compounds discussed in this chapter?
4.6 What simple mathematical relationship exists between the empirical formula, numbers
of cations and anions in the unit cell, and the coordination numbers of the cations and
anions in a binary metal halide, M 0 X,,?
4.7 If you did not do Problem 2.21 when you read Chapter 2, do so now.
4.8 One generalization of the descriptive chemistry of the transition metals is that the heavier
congeners (c.g., Mo, W) more readily show the highest oxidation state than does the
lightest congener (e.g„ Cr). Discuss this in terms of ionization energies.
4.9 Show your understanding of the Born-Haber cycle by calculating the heal of formation
of potassium fluoride analogous to the one in the text for sodium chloride.
4.10 Using any necessary data from appropriate sources, predict the enthalpy of formation
of KCI by means of a Born-Haber cycle. You can check your lattice energy against
Table 4.3.
4.11 Using any necessary data from appropriate sources, predict the enthalpy of formation
of CaS by means of a Born-Haber cycle.
4.12 Show your understanding of the meaning of the Madelung constant by calculating A
for the isolated F'Bc 3 F' fragment considered as a purely ionic species.
4.13 The ionic bond is often described as "the metal wants to lose an electron and the non-
metal wants to accept an electron, so the two react with each other." Criticize this state¬
ment quantitatively using appropriate thermodynamic quantities.
4.14 Why is the thermite reaction:
2AI + M.O] = 2M + AI.Oj (M = Fe. Cr. etc.) (4.29)
so violently exothermic? (The ingredients start at room temperature and the metallic
product, iron, etc., is molten at the end of the reaction.)
4.15 We have seen, in Chapter 2. that platinum hexafluoride has an electron allinity more
than twice as great as fluorine. Yet when lithium metal reacts with platinum hexafluoride,
the crystalline product is LiF', not LiPtF s '. Explain.
4.16 To ionize Mg to Mg 3 costs two times as much energy as to form Mg\ The formation
°f O’ is endothermic rather than exothermic as for O'. Nevertheless, magnesium oxide
is always formulated as Mg 3 '0 3 ' rather than as Mg0\
a. What theoretical reason can be given for the Mg 2 0 3 ' formulation?
b. What simple experiment could be performed to prove that magnesium oxide was not
Mg0'?
4.17 Some experimental values of the Born exponent are: LiF, 5.9; LiCl, 8.0; LiBr. 8.7; NaCI.
9.1; NaBr, 9.5. What is the percent error incurred in the calculation of lattice energies
by Eq. 4.13 when Pauling's generalization (He = 5, Ne = 7, etc.) is used instead of the
experimental value of n?
4.18 Using Fig. 4.7 generate the first five terms of the series for the Madelung constant for
NaCI. How close is the summation of these terms to the limiting value given in Table 4.1?
4.19 The enthalpy of formation of sodium fluoride is -571 kJ mol' 1 . Estimate the electron
affinity of fluorine. Compare your value with that given in Table 2.5.
6-The Structure and Reactivity of Molecules
Dihedral
angle
Dihedral
.deal and observed angles
'og) in ML 5 complexes-
Compound
Ideal trigonal bipyramid 53.1.53.1 53.1
CdCl 3- 53.8, 53.8 53.8
Ni'P^C H 5)3 r 54.2, 57.3 50.8
CoV 6 H 7 NO)l + 54.5, 58.5 37.8
NKCN) 3- 62.7, ^ 8 -6 32-2
Nb(NC 5 H l0 ) 5 (Nb2) 65.4, 67.0 23.2
Nb(NC 5 H, 0 ) 5 (Nbl) 68.6, 70.6 5.8
Nb(NMe 2 ) 5 70.2, 70.2 5.6
(C 6 H 5 ) 5 Sb 68.5, 69.2 14.4
NifCN), - 75.0, 79.4 0.3
Ideal tetragonal pyra mid 75.7, 75.7 _0d)
Muetterties, E. L.; Guggenberger, L. J. J. Am. Chem.
Sac. 1974, 96, 1748-1756.
fc These two structures occur in the unit cell of the Crfe
salt.
Ni(CN).
fy the extent of distortion from either ideal structure
g the dihedral angles in a particular five-coordinate
describe the intermediate structure as 60% TBP and
aracter is more difficult if either d orbitals participate
the nrhitals is equivalent to another to form a subset.
Seel. B 1980. 36. 2316-2323.
J. M. J. Coord.'Chem. 1981
Structure and Hybridization 225
Bent's Rule and
the Energetics
of Hybridization
Fig. 6.19 Directional
properties of hybrid orbitals
from s, p, and d atomic
orbitals. [From Kasha, M.;
adapted from Kimball, G.
Ann. Rev. Pliys. Chem.
1951, 2, 177. Reproduced
with permission.]
In the case of d orbitals, the relation between hybridization and bond angles is given in
Fig. 6.19, although the accuracy is somewhat less than can be obtained from computa¬
tion. For completely nonequivalent hybrid orbitals, simultaneous equations involving
all of the bond angles and hybridizations may be solved. 27
When a set of hybrid orbitals is constructed by a linear combination of atomic orbitals,
the energy of the resulting hybrids is a weighted average of the energies of the
participating atomic orbitals. For example, when carbon forms four covalent bonds,
although there is a promotion energy from 1 s 2 2s 2 2p : - li 2 2i'2p 3 , this is independent
of the hybridization to the valence state: 32
32 The levels shown are one-electron levels and do not show correlation effects. See discussion of the
energetics of the valence state. Chapter 5.
1 36
4 ' Bonding Model in Inorganic Chemi.try: 1. Ionic Compound;
4.20 Calculate the proton affinities of the halide ions. The enthalpies in question are those of
the type:
Compare your values with those given in Table 9.5.
4.21 Perform radius ratio calculations to show which alkali halides violate the radius ratio 1
rule.
4 22 Even if there are exceptions to the radius ratio rule, or if exact data are hard to come by, <
it is still a valid guiding principle. Cite three independent examples of pairs of compounds I
illustrating structural differences resulting from differences in ionic radii.
4 23 Berkelium is currently available in microgram quantities—sufficient to determine struc- j
tural parameters but not enough for thermochemical measurements.
a. Using the tabulated ionic radii and the radius ratio rule, estimate the lattice energy j
of berkelium dioxide, Bk0 2 .
b. Assume that the radius ratio rule is violated (it is!). How much difference does this j
make in your answer?
4.24 The crystal structure of LaF 3 is different from those discussed. Assume it is unknown.
Using the equation of Kapustinskii, estimate the lattice energy.
4.25 Copper(I) halides crystallize in a zinc blende structure. Copper(Il) fluoride crystallizes in ;
a distorted rutile structure (for the purposes of this problem assume there ^ no d.sto ion |
Calculate the enthalpies of formation of CuF and CuF 2 . Discuss. (All of the necessary ,
data should be readily available, but if you have difficulty finding a quantity, see ho
much of an argument you can make without it.)
4.26 Thallium has two stable oxidation states, +1 and +3. Use ‘he Kapustinskh equation
to predict the lattice energies of T1F and T1F,. Predict the enthalpies of formation of
these compounds. Discuss.
4.27 Plot the radii of the lanthanide(lll) (Ln 3+ ) ions from Table 4.4 versus atomic number.
Discuss.
4.28 All of the alkaline earth oxides, MO, except one crystallize in the rock salt (NaCI) struc¬
ture. What is the exception and what is the likely structure for it? (Wells, A. F. Structural
Inorganic Chemistry, 5th ed.; Oxford University: Oxford, 1984.)
4.29 It is not difficult to show mathematically that with the hard sphere model, anion-anion
contact occurs at r/r = 0.414 for C.N. = 6. Yet Wells ( Structural Inorganic Chemistry,
5th ed.; Oxford University: Oxford, 1984) states that even with the hard sphere model, we
should not expect the change to take place until rjr . a 0.35. Rationalize this apparent
contradiction. (Hint: Cf. Fig. 4.17.)
4.30 There exists the possibility that a certain circularity may develop in the radius ratio
arguments on page 125. By assuming a coordination number of 6 were the calculations
biased? Discuss.
4.31 Perform a calculation similar to that on page 127 for the formation of dioxygenyl
hexafluoroplatinate(V):
0 2 + PtF 6 -> 0 2 PtF (4J0)
All data (or approximations, if necessary) may be obtained from Chapters 2 and 4. Predict
the enthalpy of reaction Tor this equation. Carefully note any assumptions you must
make.
4.32 Repeat the calculation in Problem 4.31, but for the reaction:
Xe + PtF 6 -> Xe + PtFft (4J,)
Should xenon react with platinum hexafluoride?
I
Problems
137
4.33 Suppose that someone argues with you that your answer to Problem 4.32 is invalid, and
that any prediction that Neil Bartlett might have made on the basis of similar reasoning
(see Chapter 17) is equally invalid—he was just lucky—the reaction product of Eq. 4.31
is not a simple ionic compound, Xe + PtFJ , but a mixture of compounds, and apparently
the xenon is covalently bound. What is your reply.’
4.34 Calculate the enthalpy of the reaction CuI 2 , s) -» Cul (s , + 5l 2l „. Carefully list any assumptions.
4.35 Which of the following will exhibit the greater polarizing power?
a. K or Ag b. K + or Li + c. Li or Be 2
d. Cu 2+ orCa 2 + e. Ti 2+ or Ti 4
4.36 As one progresses across a transition series (e.g., Sc to Zn) the polarizing power of Ni¬
tons increases perceptibly. In contrast, in the lanthanides, the change in polarizing power
of M 3 changes much more slowly. Suggest two reasons for this difference.
4.37 Some general chemistry textbooks say that if a fluorine atom, Z = 9, gains an electron, it
will become a fluoride ion with len electrons that cannot be bound as tightly (because of
electron-electron repulsion) as the nine of the neutral atom, so the radius of the fluoride
ion (119 pm) is much greater than the radius of the neutral fluorine atom (71 pm). Discuss
and criticize.
4.38 If the addition of an electron F + e" - F" causes a great increase in size, why does not
the addition of two electrons to form the oxide ion (r_ = 126 pm) cause it to be much
larger than the fluoride ion (r_ = 119 pm)?
4.39 A single crystal of sodium chloride for an X-ray structure determination is a cube 0.3 mm
on a side,
a. Using data from Table 4.4, calculate how many unit cells are contained in this crystal.
b. Compute the density of NaCI. Compare your value with that in a handbook.
4.40 There has been a recent flurry of interest in the possibility ol “cold fusion of hydrogen
atoms (the deuterium isotope) in metallic palladium. 38 The original idea came from the
enormous solubility-of hydrogen gas in palladium. Palladium metal has an fee lattice.
Hydrogen atoms occupy the octahedral holes. If 70% of the octahedral holes are filled
by hydrogen atoms and the lattice does not expand upon hydrogenation, how many
grams of hydrogen will be contained in one cubic centimeter of the palladium hydride?
Compare this to the density of liquid hydrogen in gem -3 . Comment. (Rieck, D. F. J.
Client. Educ. 1989, 66, 1034.)
4.41 Mingos and Rolf 39 have discussed the packing of molecular ions in terms of their shape
as well as size. Three indices, each ranging in value from 0.00 to 1.00, are used to describe
the shape of an ion: the spherical index, F r ; the cylindrical index, F c \ and the discoidal
index, Fj. Consider the following index values and try to correlate them with what you
know of the shapes of the ions. If you arc uncertain as to the shapes, refer to Chapters 6
and 12.
a. NH;. NMei, BF 4 '. C10 4 ' (T rf ). PF. and OsClJ" (0) all have values F, = 100.
F c = 0.00, F,, = 0.00.
b. Au(CN)j and If (DJ have values F , = 0.00, F c = 1.00, F d = 1.00.
c. AuBrJ, PtClJ (D 4 „) both have values F, = 0.00, F c = 0.50, F d = 1.00, and Ni(CN);
has values F, = 0.00, F e = 0.54, F d = 1.00.
d. When it is trigonal bipyramidal (D lh ), Ni(CN)j" has values /•, = 0.75, F, = 0.25,
F d = 0.14 but when it is square pyramidal (C 4 „), the values are F, = 0.68, F e = 0.16,
F d = 0.32.
38 Flcischmann, M.; Pons, S. J. Electroanal. Chem. 1989, 261, 301-308.
39 Mingos, D. M. P.; Rolf, A. L. Inorg. Chem. 1991, 30. 3769-3771; J. Chem. Soc. Dalton 1991,
3419-3425.
222
6-The Structure and Reactivity of Molecules
orbita.3 are not equivalent. We shall see that the trigonal bipyramidal hybridization
results in three strong equatorial bonds (sp trigonal orbitals) and two weaker axial
bonds (dp linear orbitals). The square pyramidal hybridization is approximately a
square planar Ap set plus th ep z orbital. As in the trigonal bipyramidal hybridization
the bond lengths and strengths are different.
The relation between hybridization and bond angle is simple for s-p hybrids. For
two or more equivalent orbitals, the percent s character (S) or percent p character (P)
is given by the relationship:’’
where 9 is the angle between the equivalent orbitals (°) and the s and p characters are
expressed as decimal fractions. In methane, for example.
cos 9 =
= -0.333; 9 = 109.5°
In hybridizations involving nonequivalent hybrid orbitals, such as spd, it is
usually possible to resolve the set of hybrid orbitals into subsets of orbitals that are
equivalent within the subset, as the sp 2 subset and the dp subset. We have seen
(Chapter 5) that the nonequivalent hybrids may contain fractional s- and p character,
e.g.. the water molecule which uses bonding orbitals midway between pure p and sp
hybrids. For molecules such as this, we can divide the four orbitals into the bonding
subset (the bond angle is I04f) and the nonbonding subset (angle unknown). We can
then apply Eq. 6.1 to each subset of equivalent orbitals. In water, for example the
bond angle is I04£°, so
cos 9 = -0.250 =
p ~ 0 80 = 80% p character and 20% s character ( 6 . 4 )
Now ot course the total p character summed over all four orbitals on oxygen must be
3.00 ( = Px + + p .) and the total s character must be 1.00. If the bonding orbitals
contain proportionately more p character, then the nonbonding orbitals (the two lone
pairs) must contain proportionately less p character, 70%: [0.80 + 0.80 + 0 70 +
0.70 = 3.00 (p); 0.20 + 0.20 + 0.30 + 0.30 = 1.00 ()]. The opening of some bond
angles and closing of others in nominally 1 ‘tetrahedral” molecules is a common
phenomenon. Usually the distortion is only a few degrees, but it should remind us that
the terms "trigonal." "tetrahedral," etc. usually are only approximations. Exactly
trigonal and tetrahedral hybridizations are probably restricted to molecules such as
BCI 3 and CH 4 , in which all the substituents on the central atom are identical. We see
Equation 6.1 is restricted to molecules such as water in which the angle is known between Iwo
equivalent orbitals (e.g., the two orbitals binding the hydrogen atoms). Equivalent and non-
cquivalcnl hybrids are discussed further on page 227, See Bingcl. W. A.; Luttkc W Aneew Chen,
.98,, 20 899-9.1; Boisen. M. B„ Jr, Gibbs. G. V. Phys. ChL Minerats
S O Plastas, H. J.; Huhecy, C. L, Huhcey. J. E. Phosphorus 1971. /. 61-66. See also McWeeny
R. Coulson s Valence-, Oxford University: Oxford. 1979; pp 195-198. The validity of this method
has been questioned: Magnusson. E. J. Am.Chem.Soc. 1984, 106, 1177-1185. 1185-1191. The lack
of agreement revolves around the question of the orthogonality of hybrid orbitals.
I
Structure and Hybridization
223
this distortion epitomized in AL 3 molecules. Unlike coordination numbers, 2, 3, 4,28
and 6, there is no unique, highly symmetrical set of equivalent orbitals that can be
constructed for five-coordination. Of the two hybridizations shown in Fig. 6.17, most
compounds of nonmetals favor the trigonal bipyramidal (TBP) structure.29 Many
coordination compounds are known, however, with square pyramidal (SP) structures
(see Chapter 12). More important for the present discussion, however, is the fact that
there are many compounds that cannot be classified readily into TBP or SP geome¬
tries. Muetterties and Guggenberger™ have shown that there is a continuous spectrum
of compounds ranging from TBP to SP.
f
~~ <6 ' 5)
I
TBP SP
A list of compounds showing the gradual change from trigonal bipyramidal to
square pyramidal geometry is given in Table 6.3 (see Fig. 6.18). The gradual change
can be quantified in terms of the dihedral angles between the faces of the polyhedra.
For example, in the conversions shown above, when TBP—► SP, the dihedral angle of
the edge furthest from the viewer opens up gradually until it reaches 180°, i.e., the
back face of the square pyramid becomes a plane. The angles of the edges (two of
which are labeled e, and e 2 in the figures accompanying the table) are opened up (16°),
approaching right angles in the pyramid. Now as the reverse change takes place. SP-»
TBP, edge e 3 reappears as a "real” edge and <?,, e 2 , and e 3 all close until they reach
identical values in the idealized trigonal bipyramid. The gradual change in these angles
as one progresses through the list of compounds in Table 6.3 indicates that just about
every possible intermediate between the two limiting geometries is known. When the
various substituents arc different (and occasionally even when they are not—see
Chapter 12), these intermediate structures are the rule rather than the exception, and
they should warn us to avoid overgeneralizing a structure to make it fit a preconceived
pigeonhole. As Muetterties and Guggenberger™ have said, describing a molecule as
“a distorted trigonal bipyramid'' conveys little information as to the extent of the
distortion and the shape of the molecule. Methods are available for calculating the
location of an intermediate on the TBP-SP spectrum, using the dihedral angles shown
The apparent exceplion of two slable slruclurcs. of T,, and D tl , symmetry, for coordinalion number
4 can be misleading. The square planar structure is known only where there arc special stabilizing
energies resulting from the d electron configuration in transition metal compounds.
M Three exceptions arc pentaphcnylantimony. pcnlacyclopropylanlimony, and a bicyclic phosphor-
anc (sec Problem 6.16). Howard. J. A, Russell. D. R, Tripped. S. Chem. Common. 1973. 856-857
and references therein.
30 Muetterties. E. L, Guggenberger. L. J. J. Am. Chem. Soc. 1974, 96, 1748-1756.
I
Chap ter
5
Bonding Models in
Inorganic Chemistry:
2. The Covalent Bond
This chapter and the one following will be devoted to a preliminary analysis of
covalent bonding. Most of the ideas presented here may be found elsewhere and with
greater rigor, and many will have been encountered in previous courses. However,
since they form the basis for subsequent chapters, a brief presentation is in order here.
Covalent bonding will also be discussed in Chapters 6 and II.
The Lewis Structure This method of thinking about bonding, learned in high school and too often forgotten
in graduate school or before, is a most useful first step in thinking about molecules.
Before delving into quantum mechanical ideas or even deciding whether molecular
orbital or valence bond theory is likely to be more helpful, a Lewis structure should be
sketched. The following is a brief review of the rules for Lewis structures:
Normally two electrons pair up to form each bond. This is a consequence of
the Pauli exclusion principle—two electrons must have paired spins if they are
both to occupy the same region of space between the nuclei and thereby attract
both nuclei. The definition of a bond as a shared pair of electrons, however, is
overly restrictive, and we shall see that the early emphasis on electron pairing
in bond formation is unnecessary and even misleading.
lor most atoms there will be a maximum of eight electrons in the valence shell
( = Lewis octet structure). This is absolutely necessary for atoms of the ele¬
ments lithium through fluorine since they have only four orbitals (an s and
three p orbitals) in the valence shell. It is quite common, as well, for atoms of
other elements to utilize only their j and p orbitals. Under these conditions the
sum of shared pairs (bonds) and unshared pairs (lone pairs) must equal the
number of orbitals—four. This is the maximum, and for elements having fewer
than four valence electrons, the octet will usually not be filled. The following
compounds illustrate these possibilities:
.. : 9. H
: C |: C:CI : :P:H :0:H :C|:B:C1: H :Be :H Li:CH 3
=C1: H H :CI:”
138
Valence Bond Theory
139
For elements with available d orbitals, the valence shell can be expanded
beyond an octet. Because (/orbitals first appear in the third energy level, they
are low enough in energy to be available for bonding in elements of Period 3
and beyond. These elements are nonmetals in the higher valence compounds
and transition metals in complexes. In the nonmetals, where the number of
valence electrons is usually the limiting factor, we have maximum covalencies
of 3, 6, 7, and 8 in Groups VA (15), VIA (16), VIIA (17), and VIIIA (18),
respectively. Note that covalency (the number of covalent bonds to an atom)
and coordination number (the number of atoms bound to another atom) are not
always the same. Factors determining covalencies and coordination numbers
in complexes are of several kinds, and discussion of them will be deferred.
Examples ot molecules and ions containing more than eight electrons in the
valence shell of the central atom are:
NH,
f/ F
l/
\l/
, 3
I/™,
T\
P S F
f'K F
HjN-Co-NH 3
F F
F F
F F
H.N^ 1
1
X
z
1
/?\
O O o
Bonding Theory
It has been assumed implicitly in all of these rules that the molecule will seek
the lowest overall energy. This means that, in general, the maximum number of
bonds will form, that the strongest possible bonds will form, and that the
arrangement of the atoms in the molecule will be such as to minimize adverse
repulsion energies.
In modern times there have been two "contenders for the throne” of bonding theory:
valence bond theory (VI3 T) and molecular orbital theory (MOT). The allusion is an apt
one since ,t seems that much of the history of these two theories consisted of
contention between their respective proponents as to which was best. Sometimes
ovcrzealous supporters of one theory have given the impression that the other is
wrong Granted that any theory can be used unwisely, it remains nonetheless a fact
that neither theory should be regarded as true to the exclusion of the other Given a
specific question one theory may prove distinctly superior in insight, ease of calcula-
tton, or simplicity and clarity of results, but a different question may reverse the
picture completely Surely the inorganic chemist who does not become thoroughly
familiar with both theories is like the carpenter who refuses to carry a saw because he
already has a hammer! Both are severely limiting their skills by limiting their tools.
Valence Bond
Theory
The valence bond (VB) theory grew directly out of the ideas of electron pairing by
Lewis and others. In 1927 W. Heitler and F. London proposed a quantum-mechanical
treatment of the hydrogen molecule. Their method has come to be known as the
valence bond approach and was developed extensively by men such as Linus Pauling
220 6-The Structure and Reactivity of Molecules
MO configuration will be 2cr; kr 2 „ It^Itt 2 , (or 2a]\b\ia\b]). Because the for¬
merly nonbonding It r,„ orbital is greatly stabilized (3a,) on bendmg. the water mole¬
cule is bent rather than linear. ......
The Walsh diagram shown in Fig. 6.16 is accurate only for molecules in which
there is a large separation between the ns and np energy levels of the central atom. If
the ns-np separation is small (as in SrF 2 and BaF 2 ), the 1 b 2 level (ofTig 6.16) does
not rise as rapidly as 2a, falls, and the molecule may be stabilized on bending.- Note
that in MF, molecules of this type, the 3 a, and I b t levels are unoccupied.
This brief discussion cannot do justice to the MO approach to stereochemistry,
but it does illustrate the reduced importance of electron-electron repulsions (usually
omitted in simple approximations) and the increased importance of overlap in this
approach. Although the VSEPR approach and the LCAO-MO approach to stereo¬
chemistry appear on the surface to be very different, all valid theories of bonding,
when carried sufficiently far, are in agreement that the most stable molecule will have
the best compromise of (1) maximizing electron-nucleus attractions and (2) mini¬
mizing electron-electron repulsions.
As we have seen in Chapter 5, it is not correct to say that a particular structure is
"caused” by a particular hybridization, though such factors as overlap and eneigy are
related to hybridization. We have also seen the usefulness of viewing structures in
terms of VSEPR. We shall encounter yet further factors later in this chapter, ev-
ertheless it is appropriate and useful to note here that certain structures and hybridiza¬
tions arc associated with each other. Some of the most common geometries and their
corresponding hybridizations are shown in Fig. 6.17. In addition, there are many
hybridizations possible for higher coordination numbers, but they are less frequent .
encountered and will be introduced as needed in later discussions.
The possible structures may be classified in terms of the coordination number of
the central atom and the symmetry of the resulting molecule (Fig. 6.17). Two groups
about a central atom will form angular (p 2 orbitals, C 2 „ symmetry) or linear (sp hybrid
n , symmetry) molecules; three will form pyramidal (p 3 . C 3 „) or trigonal planar (sp ,
o') molecules; four will usually form tetrahedral (.vp 3 . 7,,) or square planar (dsp ,
D '); five usually form a trigonal bipyramidal (D 3 „), more rarely a square pyramidal
(C,’.) molecule (both dsp 3 hybrids, but using different orbitals, see I able 6-). and six
groups will usually form an octahedral molecule (d-sp 3 , O h ). .
If in addition to the bonding pairs there are stereochemically active lone pairs, the
symmetry will be lowered (BF 3 is D 3 „; NF 3 is C 3v ). Furthermore, the hybridization of
the lone pair(s) will be different from that of the bonding pairs (see belowT
Most hybridizations result in equivalent hybrid orbitals, i.e., all the hybrid or¬
bitals are identical in composition (% 5 and % p character) and in spatial orientation
with respect to each other. They have very high symmetries, culminating in
tetrahedral and octahedral symmetry. In the case of dsp> hybrid orbitals, the resulting
26 See Burdelt. J. K. Molecular Shapes'. Wiley: New York. .980;<
J. K..; Whangbo, M. H. Orbital Interactions in Chemistry: W.ley: New York. 1985. Chapter .
Structure and
Hybridization
Structure and Hybridization
221
/ 90° I 1/7
/; \
// \
T" 90°
A -7 ''
\V 1 S
\ > /
Fiq. 6.17 Geometries of some common hybrid and nonhybrid bonding arrangements. Solid
lines represent bonds formed from the orbitals on the central atom ©. The dashed lines are
geometric lines added for perspective.
Component atomic orbitals
involved in hybrid orbital
formation
Hybridization Atomic orbitals _
sp, sp 2 , p 3 s + arbitrary p"
dsp 2 d x i _ y 2 + J + P x + Py
dsp 3 (TBP) d.2 + s + p x + P y + P z
dsp 3 (SP) d x t_ y 2 + .v + p, + P y + P ;
cPsp 3 d x 2_ y 2 + dj + p x + p y + p.
140
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
and J. C. Slater. The following discussion is adapted from the works of Pauling and
^Suppose we have two isolated hydrogen atoms. We may describe them by the
wave functions A and B , each having the form given in Chapter - for a \s orbital. If
the atoms are sufficiently isolated so that they do not interact, the wave function for
the system of two atoms is
, , , (5-D
<l< = i PA(I>Vb<2>
where A and B designate the atoms and the numbers 1 and 2 designate electrons
number I and 2. Now, we know that when the two atoms are brought together to form
a molecule they will affect each other and that the individual wave functions t/i A and ^
will change, but we may assume that Eq. 5.1 is a good starting place as a trial function
for the hydrogen molecule and then try to improve it. When we solve for energy as
function of distance, we find that the energy curve for Eq. 5.1 does '"deed have a
minimum (curve fl . Fig. 5.1) of about -24 kJ mol" at a distance of about 90 pm. The
actual observed bond distance is 74 pm, which is not too different from our first
approximation, but the experimental bond energy of H, is -458 kJ mol , almost _0
times greater than our first approximation. .
If we examine Eq. 5.1, we must decide that we have been overly restrictive in
using it to describe a hydrogen molecule. First, we are not justified in labeling
electrons since all electrons are indistinguishable from each other. Moreover even if
we could we would not be sure that electron I will always be on atom A and electron 2
on atom B. We must alter Eq. 5.1 in such a way that the artificial restrictions are
removed. We can do this by adding a second term in which the electrons have changed
positions:
'P = </'a(I)'Ab< 2) + l /'A(2) l /'B(i) (
This improvement was suggested by Heitler and London. If we solve for the
energy associated with Eq. 5.2, we obtain curve b in Fig. 5.1. The energy has
improved greatly (-303 kJ mol'') and also the distance has improved slightly. Since
Fig. 5.1 Theoretical energy
curves ( a-d. f ) for the
hydrogen molecule, H 2 ,
compared with the
experimental curve (e).
Curves a-d show
successive approximations
in the wave function as
discussed in the text. Curve
/is the repulsive interaction
of two electrons of like
spin.
50 100 150 200
Inlcmuclear distance, r (pm)
i Pauling. L. The Nature of the Chemical Bond, 3rd ed; Cornell University: Ithaca NY. I960.
Coulson. C.A. Valence. 2nd ed.; Oxford University: London. 1961. McWeeny. R. Coulson
Valence-, Oxford University: London, 1979.
Valence Bond Theory 141
the improvement is a result of our "allowing" the electrons to exchange places, the
increase in bonding energy is often termed the exchange energy. One should not be
too literal in ascribing this large part of the bonding energy to “exchange,” however,
since the lack of exchange in Eq. 5.1 was merely a result of our inaccuracies in
approximating a correct molecular wave function. If a physical picture is desired to
account for the exchange energy, it is probably best to ascribe the lowering of energy
of the molecule to the fact that the electrons now have a larger volume in which to
move. Recall that the energy of a particle in a box is inversely related to the size of the
box; that is, as the box increases in size, the energy of the particle is lowered. By
providing two nuclei at a short distance from each other, we have "enlarged the box"
in which the electrons are confined.
A further improvement can be made if we recall that electrons shield each other
(Chapter 2) and that the effective atomic number Z will be somewhat less than Z. If
we adjust our wave functions, i// A and <// B , to account for the shielding from the second
electron, we obtain energy curve c—a further improvement.
Lastly, we must again correct our molecular wave function for an overrestriction
which we have placed upon it. Although we have allowed the electrons to exchange in
Eq. 5.2, we have demanded that they must exchange simultaneously, that is, that only
one electron can be associated with a given nucleus at a given time. Obviously this is
too restrictive. Although we might suppose that the electrons would tend to avoid
each other because of mutual repulsion and thus tend to stay one on each atom, we
cannot go so far as to say that they will always be in such an arrangement. It is
common to call the arrangement given by Eq. 5.2 the "covalent structure ’ and to
consider the influence of "ionic structures" on the overall wave function:
H-H «— H + H- «—♦ H-H +
Covalent Ionic
We then write
'll = </'A(l) l /'B(2) + ■/ , A(2 >Ab(I) + Ai /'a<|) , /'a< 2> + ^BO^Btt) (5-3)
where the first two terms represent the covalent structure and the second two terms
represent ionic structures in which both electrons are on atom A or B. Because the
electrons tend to repel each other somewhat, there is a smaller probability of finding
them both on the same atom than on different atoms, so the second two terms are
weighted somewhat less (A < 1). Equation 5.3 can be expressed more succinctly as
'll = •Z'cov + A| Z'HH- + a /'h-h +
When we investigate the energetics of the wave function in Eq. 5.3, we find further
improvement in energy and distance (curve d , Fig. 5.1).
This is the first example we have had of the phenomenon of resonance, which we
shall discuss at some length in the next section. It should be pointed out now,
however, that the hydrogen molecule has one structure which is described by one
wave function, li as a combination of two or more wave functions, each of which only partially
describes the hydrogen molecule. Table 5.1 lists values for the energy and equilibrium
distance for the various stages of our approximation, together with the experimental
values.
Now, if one wishes, additional “corrections" can be included in our wave
function, to make it more nearly descriptive of the actual situation obtained in the
218
219
6-The Structure and Reactivity o( Molecules
To the above Drago 21 suggested the following empirical rule which rationalizes the
very small angles (—90°) in phosphine, arsine, hydrogen sulfide, etc., and which is
compatible with the energetics of hydridization (page 225):
If the central atom is in the third row or below in the periodic table, the lone
pair will occupy a stereochemically inactive s orbital and the bonding will be
through p orbitals and near 90° bond angles if the substituent electronegativity
is < —2.5.
Because of its intuitive appeal and its high degree of accuracy, the VSEPR model
has been well received by inorganic chemists, but the theoretical basis has been a
matter of some dispute. 22 More recently, there have been strong theoretical arguments
for localized, stereochemically active orbitals. 22
Because VB methods deal with easily visualized, localized orbitals, stereochemical
arguments (such as the VSEPR model) have tended to be couched in VB terminology.
Several workers 2 - have attempted to modify simple LCAO-MO methods to improve
their predictive power with respect to geometry. The basis for these methods consists
of Walsh diagrams 25 that correlate changes in the energies of molecular orbitals
between a reference geometry, usually of high symmetry, and a deformed structure of
lower symmetry. Consider the BeH, molecule discussed previously. In the preceding
discussion only the filled, bonding orbitals were emphasized although other types of
orbitals were mentioned. Figure 6.16 illustrates what happens to the energies of all the
orbitals in BeH,—bonding, nonbonding, and antibonding—as the molecule is bent.
Consider first the 2 tr K orbital. It is constructed from atomic wave functions that are
everywhere positive, and hence on bending there is an increase in overlap since the
two hydrogen wave functions will overlap to a slightly greater extent (recall that the
wave function of an atom never goes completely to zero despite our diagrammatic
representation as a finite circle). The energy of the 2 <r g orbital (relabeled 2a,) is
lowered somewhat. In contrast, the energy of the I<r„ orbital increases on bending.
This is because the wave function changes sign (as shown by the shading), and overlap
of the terminal hydrogen wave functions will be negative. In addition, the overlap of
the hydrogen atoms with the linear p orbital must be poorer in the bent molecule, and
so the energy of the lcr„ orbital (relabeled 16,) will increase more than 2«, will
decrease. BeH, has a molecular orbital electron configuration 2<r].\ of,,or 2a] I h\, and
since I/;, loses more energy than 2a, gains, BeH, is linear, not bent.
Similar arguments can be applied to the nonbonding and antibonding orbitals (Fig.
6.16; see Problem 6.7). In the water molecule. H,0, with eight valence electrons, the
Molecular Orbitals
and Molecular
Structure
21 Drago, R. S. J. Chem. Educ. 1973, 50, 244-245.
22 Edmiston. C.; Bartlcson, J.; Jarvie, J. J. Am. Chem. Soc. 198ft. 108. 3593-35%; Palke, W, E.;
Kirtman, B. J. Am. Chem. Soc. 1978. 100. 5717-5721. See also Footnote 3.
23 Bader, R. F. W.; Gillespie, R. J.; MacDougall, P. J. J. Am. Chem. Soc. 1988. 110. 7329-7336; In
From Atoms to Polymers: Isoeiectronic Analogies', Licbman, J. F.; Greenberg, A., Eds; VCH: New
York. 1989; pp 1-51. Bader, R. F. W. Atoms in Molecules-. Oxford University: Oxford, 1990.
Bnrtcll. L. S.: Barshad, Y. Z. J. Am. Chem. Soc. 1984, 106. 7700-7703.
24 See. for example, Gimarc. B. M. Acc. Chem. Res. 1974, 7, 384-392; Molecular Structure and
Bonding-. Academic: New York, 1980, and Footnote 25.
25 Walsh, A. D. J. Chem. Soc. 1953, 2260-2331.
The Structure of Molecules
Fig. 6.16 Molecular orbital
pictures and qualitative
energies of linear and bent
AH 2 molecules. Open and
shaded areas represent
differences in sign (+ or
—) of the wave functions.
Changes in shape which
increase in-phase overlap
lower the molecular orbital
energy. [From Gimarc,
B. M. J. Am. Chem. Soc.
1971, 93. 593. Reproduced
with permission.)
142
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Table 5.1
Energies and equilibrium
Type of wave function
Energy (kJ mol -1 )
Distance (pm)
distances (or VB wave
functions 0
Uncorrected, ijt = i/< a i/< b
24
90
Heitler-London
303
86.9
Addition of shielding
365
74.3
Addition of ionic contributions
388
74.9
Observed values
458.0
74.1
° McWeeny, R. Coulson's Valence; Oxford Unversity: London, 1979; p 120.
Used with permission.
hydrogen molecule. 2 3 - 2 However, the present simplified treatment has included the
three important contributions to bonding: delocalization of electrons over two or more
nuclei, mutual screening, and partial ionic character.
There is an implicit assumption contained in all of the above: The two bonding
electrons are of opposite spin. If two electrons are of parallel spin, no bonding occurs,
but repulsion instead (curve /, Fig. 5.1). This is a result of the Pauli exclusion
principle. Because of the necessity for pairing in each bond formed, the valence bond
theory is often referred to as the electron pair theory, and it forms a logical quantum-
mechanical extension of Lewis's theory of electron pair formation.
Resonance When using valence bond theory it is often found that more than one acceptable
structure can be drawn for a molecule or, more precisely, more than one wave
tunclion can be written. We have already seen in the case of the hydrogen molecule
that we could formulate it either as H—H or as H + H~. Both are acceptable struc¬
tures, but the second or ionic form would be considerably higher in energy than the
"covalent” structure (because of the high ionization energy and low electron affinity
of hydrogen). However, we may write the wave function for the hydrogen molecule as
a linear combination of the ionic and covalent functions:
•/' = (!- AW cov + A icn (5.5)
where A determines the contribution of the two wave functions. When this is done, it is
found that the new wave function is lower in energy than either of the contributing
structures. This is a case of covalent-ionic resonance which will be discussed at
greater length in the section on electronegativity.
Another type of resonance arises in the case of the carbonate ion. A simple Lewis
structure suggests that the ion should have three tr bonds and one n bond. However,
when it comes to the placement of the tt bond, it becomes obvious that there is no
unique way to draw the n bond. There is no a priori reason for choosing one oxygen
atom over the other two to receive the it bond. We also find experimentally that it is
impossible to distinguish one oxygen atom as being in any way different from the other
two.
2 A 100-icrm function (see Footnote 3) has reproduced the experimental value to within 0.01 kJ moM.
See McWeeny, R. Coulson's Valence; Oxford University: London. 1979: pp 120-121. and Pilar,
F. L. Elementary Quantum Chemistry; McGraw-Hill: New York. 1968; pp 234-249.
3 Kolos. W.; Wolniewicz, L. J. Chem. Phys. 1968, 49. 404. See Hcrzberg, G. J. Mol. Spectrosc. 1970,
33. 147, for the experimental value.
Valence Bond Theory
143
Each of these structures may be described by a wave function, t/.,, .//„, or iThe
actual structure of the carbonate ion is none of the above, but a resonance hybrid
formed by a linear combination of the three canonical structures;
“ fl t + btp u + c„| ( S. 6)
There is no simple Lewis structure that can be drawn to picture the resonance hybrid,
but the following gives a qualitative idea of the correct structure:
It is found that the energy of IV is lower than that of I, II, or III. It is common to
speak of the difference in energy between I and IV as the resonance energy of the
carbonate ,on. One should realize, however, that the resonance energy arises only
because our wave functions m are rather poor descriptions of the actual
structure of the ion. In a sense, then, the resonance energy is simply a measure or our
ignorance of the true wave function. More accurately, the resonance energy and the
entire phenomenon of resonance are merely a result of the overly restrictive approach
we have adopted in valence bond theory in insisting that a "bond" be a localized pair
of electrons between two nuclei. When we encounter a molecule or ion in which one
or more pairs of electrons are delocalized we must then remedy the situation by
invoking resonance. We should not conclude, however, that valence bond theory is
wrong—it merely gets cumbersome sometimes when we have many delocalized
electrons to consider. In contrast, in molecules in which the electrons are localized,
the valence bond theory often proves to be especially useful.
In the carbonate ion. the energies of the three contributing structures are identi¬
cal, and so all three contribute equally (a = b = c) and the hybrid is exactly
intermediate between the three, In many cases, however, the energies of the contrib¬
uting structures differ (the hydrogen molecule was an example), and in these cases we
find that the contribution of a canonical structure is inversely proportional to its
energy, that is, high energy, unstable structures contribute very little, and for reso¬
nance to be appreciable, the energies of the contributing structures must be compara¬
ble. Using the energy of the contributing structures as a basis, we can draw up a set of
general rules for determining the possibility of contribution of a canonical structure.
216
The Structure of Molecules
217
6 • The Structure
and Reactivity of Molecules
Fig. 6.14 (a) Molecular structure of /ranj-tetrachlorobis(tetramethylthiourea)tellurium(IV) in
the monoclinic form. Note that the line Cl(4)—Te—Cl(2) approximates an axis for a
pentagonal bipyramid if it is assumed that a lone pair occupies a position between SO) and
Cl(3) (b). [In part from Esper&s. S.; George. J. W.; Husebye. S.; Mikalsen. 0. Ada Chem.
Scand. 1975. 29A, !4I-I48.|
Finally, it should be noted that the XeF fi molecule exhibits a definite tendency to
donate a fluoride ion and form the XeF^ cation, which is isoelectronic and isostruc-
tural with IF S as expected from the VSEPR model. The structure of solid XeF 6 is
complex,'# with 144 molecules of XeF 6 per unit cell; however there are no^discrete
XeF fi molecules. The simplest way to view the solid is as pyramidal XeFj cations
extensively bridged by "free” fluoride ions. Obviously, these bridges must contain
considerably covalent character. They cause the xenon-containing fragments to clus¬
ter into tetrahedral and octahedral units (Fig. 6.15a,b). There are 24 tetrahedra and
eight octahedra per unit cell, packed very efficiently as pseudospheres into a Cu 3 Au
structure (Fig. 6.15c)." 19 The structure thus provides us with no information about
molecular XeFfi, but it does reinforce the idea that the VSEPR-correct, square
pyramidal tXeF, is structurally stable.
m There arc actually four phases known of solid xenon hexafluoride. All have several structural
features in common, and the phase described here, phase IV or the cubic phase, is the easiest to
describe.
Burbank. R. D.; Jones. G. R. J. Am. Chem. Soc. 1974. 96. 43-48.
Fig. 6.15 The structure of solid XeF 6 . Each Xe atom sits at the base of a square pyramid of
five fluorine atoms. The bridging fluorine atoms are shown as larger circles, (a) The
tetrameric unit with the Xe atoms forming a tetrahedron, (b) the hexameric unit with the Xe
atoms forming an octahedron, (c) the Cu 3 Au structure: The shaded circles represent the
octahedral clusters and the open circles the tetrahedral clusters. [In part from Burbank R. D.;
Jones, G. R. J. Am. Chem. Soc. 1974, 96. 43-48. Used with permission.]
Another problem arises with alkaline halide molecules, MX 2 . These molecules
exist only in the gas phase—the solids are ionic lattices (cf. CaF 2 , Fig. 4.3). Most MX,
molecules are linear, but some, such as SrF, and BaF 2 , are bent. 211 If if is argued that
the bonding in these molecules is principally ionic and therefore not covered by the
VSEPR model, the problem remains. Electrostatic repulsion of the negative anions
should also favor a 180° bond angle. At present, there is no simple explanation of these
difficulties, but the phenomenon has been treated by means of Walsh diagrams (see
page 218).
The preceding can be surfimed up in a few rules:
Electron pairs tend to minimize repulsions. Ideal geometries are:
a. for two electron pairs, linear.
b. for three electron pairs, trigonal.
c. for four electron pairs, tetrahedral.
d. for five electron pairs, trigonal bipyramidal.
e. for six electron pairs, octahedral.
Repulsions are of the order LP-LP > LP-BP > BP-BP.
a. When lone pairs are present, the bond angles are smaller than predicted by
rule 1.
b. Lone pairs choose the largest site, e.g., equatorial in trigonal bipyramid.
c. If all sites are equal, lone pairs will be trans to each other.
Double bonds occupy more space than single bonds.
Bonding pairs to electronegative substituents occupy less space than those to
more electropositive substituents.
20 Buchler, A.: Stauffer, J. L.; Klemperer. W. J. Am. Cliem. Soc. 1964. 56, 4544-4550. Calcier, V.;
Mann. D. E.; Seshadri. K. S.; Allavena. M.; White, D. J. Chem. Phys. 1969. 51. 2093-2099.
144
5 • Bonding Models in Inorgonic Chemistry: 2. The Covolen. Bond
example, the structure
=^5rsSSSrHia==
structures Chaste following state,ures for phosphorous acid represent an
equilibrium between two distinct chemical spectes, not resonance.
H:0:P:0:H
H:0:P:0:H
: 9. :
H
hypothetical resonance for nitrous oxide.
:N::N::0:
“ +
Aside from the fact that II is a strained structure and therefore less stable than
U wi 1 not contribute to the resonance of N 2 0 because the bond angle .180
L md 60“ in I! For any intermediate hybrid, the contnbut.on of e.ther 1 or
would be unfavorable because of the high energy cost when I .s bent or when
^ ° Tfetwords should be said about the difference between resonance and
molecJ^ vibrations. Although vibrations take place, they are osc.llat.on
Tout an equilibrium position determined by the structure of the resonan
ySd and' sholld not be confused^ f/^VTom
nr—
££L n of resonance be,ween coupled pendulum,
or other mechanical systems.
be ignored; ,b=, will be discussed in Rule 3 ■> .el, .s in .be neb.
Valence Bond Theory 145
Distribution of formal charges in a contributing structure must be reasonable.
Formal charge, which will be more fully explained in the next section, may be
defined as the charge an atom in a molecule would have if all of the atoms had
the same electronegativity. Canonical forms in which adjacent like charges
appear will probably be unstable as a result of the electrostatic repulsion. A
structure such as A - —B + —C + —D~ is therefore unlikely to play a major role
in hybrid formation.
In the case of adjacent charges which are not of the same sign, one must
use some chemical discretion in estimating the contribution of a particular
structure. This is best accomplished by examining the respective elec¬
tronegativities of the atoms involved. A structure in which a positive charge
resides on an electropositive element and a negative charge resides on an
electronegative element may be quite stable, but the reverse will represent an
unstable structure. For example, in the following two molecules
X F F +
X — P=0 «- X — P — 0~ F —B -- F —B.
XX F F
I II I 'I
canonical form II contributes very much to the actual structure of phosphoryl
compounds, but contributes much less to BF 3 and, indeed, the actual contribu¬
tion in compounds of this sort is still a matter of some dispute.
Furthermore, placement of adjacent charges of opposite sign will be more
favorable than when these charges are separated. When adjacent, charges of
opposite sign contribute electrostatic energy toward stabilizing a molecule
(similar to that found in ionic compounds), but this is reduced when the
charges are far apart.
Contributing forms must have the same number of unpaired electrons. For
molecules of the type discussed previously, structures having unpaired elec¬
trons should not be considered since they usually involve loss of a bond
A = B -A—B-
I II
and higher energy for structure II. We shall see when considering coordination
compounds, however, that complexes of the type ML,, (where M = metal,
L = ligand) can exist with varying numbers of unpaired electrons but com¬
parable energies. Nevertheless, resonance between such structures is still
forbidden because the spin of electrons is quantized and a molecule either has
its electrons paired or unpaired (an intermediate or “hybrid" situation is
impossible).
These rules may be applied to nitrous oxide, N 2 0. Two structures which
are important are
:N::N::6: «— :N:::N:o:
•-•
i
ii
214
6-The Structure and Reactivity of Molecules
little help in studying XeF,, because the pentagonal bipyramidal structure was the first
to be experimentally eliminated as a possibility. A number of other lluoro complexes
with coordination number 7 are known: [ZrF 7 ] 3_ , known in both pentagonal bi¬
pyramidal and capped trigonal prismatic forms; the [NbF 7 ] 2_ anion is a capped
octahedron. 9
Determining the exact structure of the gaseous XeF 6 molecule proved to be
unexpectedly difficult. It is known to be a slightly distorted octahedron. In contrast to
the molecules discussed previously, however, the lone pair appears lo occupy less
space than the bonding pairs. The best model for the molecule (Fig. 6.13) appears to
be a distorted octahedron in which the lone pair extends either through a face (C 3l ,
symmetry) or through an edge (C 2v symmetry). 10 -"
The conformation of lowest energy appears to be that of C 3l , symmetry. Part of the
experimental difficulties stems from the fact that the molecule is highly dynamic and
probably passes through several conformations. In either of the two models shown in
Fig. 6.13, the Xe—F bonds near the lone pair appear to be somewhat lengthened and
distorted away from the lone pair; however, the distortion is less than would have
been expected on the basis of the VSEPR model. That the latter model correctly
predicted a distortion at all at a time when others were predicting a highly symmetrical
octahedral molecule (all other hexafluoride such as SF and UF ft are perfectly oc¬
tahedral) is a signal success, however.
The powerful technique of X-ray diffraction cannot be applied to the resolution of
this question since solid XeF,, polymerizes with a completely different structure (see
below). However, the isoelectronic compound Xe(OTeF 5 ) h crystallizes as a simple
molecular solid, so that it may be studied by X-ray diffraction. Each molecule has C 3l ,
symmetry (for the oxygen coordination shell about the xenon) indicating a stereo-
chemically active lone pair (Fig. 6.13a). suggesting the same structure in XeF,, as
well."
Even more puzzling are the structures of anions isoelectronic with XeF,,. Raman
spectroscopy indicates that the IF7 anion, like XeF. has lower symmetry than
Fig. 6.13 Possible molecular structures of xenon hexafluoride: (a) lone pair emerging through
face of the octahedron. C,,. symmetry; (b) lone pair emerging through edge of octahedron.
C’,. symmetry. [From Gavin, R. M.. Jr,; Bartell. L. S. J. Client. Pliys. 1968. 48. 2466.
Reproduced with permission.)
9 Wc shall see that the number of d electrons in a complex can aflccl its stability and geometry
(Chapters 11 and 12). The examples given here were chosen to have a d" configuration.
10 Gavin. R. M.. Jr.: Bartell. L. S. J. Client. Plivs. 1968. 48. 2466.
11 Seppelt, K.; Lentz. D. Progr. Inorg. Client. 1982. 29. 172-180.
' - ■ ' - '' - • '-'if -
The Structure of Molecules
215
octahedral, but that BrF is octahedral on the spectroscopic time scale. 1 - Both are
fluxional on the NMR time scale. 15 Anions such as SbX“, TeX;;~ (X = Cl. Br. or I),
and BrF have been assigned perfectly octahedral structures on the basis of X-ray
crystallography. 14 For these structures in which the lone pair is stereochemically
inactive, it is thought that the pair resides in an s orbital. This could result from steric
crowding of the ligands or the stability of the lower energy s orbital, or both. Raman
and infrared spectroscopy have indicated, however, that these ions may be either
nonoctahedral or extremely susceptible to deformation. 15
Complexes of arsenictIII), antimony(III), [electron configuration = (/; — |)
rf,0 « 2 l. lead(II), and bismuth(III) [(« - 2)/' 4 (n - )d "»/ ( .y2] with polydentate ligands
occupying six coordination sites have been found to have a stereochemically active
lone pair. However, the dichotomy of behavior of the heavier elements that have a
lone pair is reflected in the crystal chemistry of Bi 3+ , When forced into sites of high
symmetry, the Bi'' ion responds by assuming a spherical shape; in crystals of lower
symmetry the lone pair asserts itself and becomes stereochemically active.'' 1
There appears to be no simple “best" interpretation of the stereochemistry of
species with 14 valence electrons. Rather, it should be noted that there seem to be
several structures of comparable stability and small forces may tip the balance in favor
of one or the other. An example of the balance of these forces is the trans isomer of
tetrachIorobis(tetramethylthiourea)tellurium(IV) which provides a very interesting
story. I his compound was synthesized and separated as orthorhombic crystals which
contained centrosymmetric molecules consisting of approximately octahedral arrange¬
ments of four chlorine and two sulfur atoms about each tellurium atom. However,
when these crystals were examined several years later, it was found that most had
changed to a monoclinic form. When the X-ray structure was determined, it was found
that the monoclinic crystals contained severely distorted molecules (Fig. 6.14a) with
most angles decreased to about 80°, but one opened to 106° (compared to approx¬
imately 90° in the orthorhombic form).' 7 Apparently the orthorhombic form contains
molecules in which the lone pair has been forced into a stereochemically inert s orbital
(as is TeX ), while the monoclinic form has a stereochemically active lone pair
presumably protruding within the 106° bond angle (Fig. 6.14b). Consistent with this
interpretation is the lengthening of the Te—S and Te—Cl bonds adjacent to the lone
pair. Such lengthening is commonly found and may be interpreted in terms of the in¬
creased LP-BP repulsions. It is tempting to suggest that crystal symmetry forces the
lone pair to be stereochemically inactive in the orthorhombic form and that the pro¬
trusion of the lone pair forces the crystal symmetry to change. However, (his tends to
oversimplify what must be a delicate balance of crystal packing forces and electronic
effects.
'- Christc. K. O.: Wilson. W. W. Inorg. Client. 1989. 28. 3276-3277.
13 The lim c scale of spectroscopic and oilier techniques is discussed in Table 6.5, and fluxional or
rearranging molecules arc discussed on pp 237-243 and in Chapter 15.
14 Lawton. S. L.; Jacobson. R. A. Inorg. Client. 1966. 5. 743. Mahjoub, A. R.: l-loscr. A.; Fuchs J •
Seppelt, K. Angew. Client, hit. Ed. Engl. 1989. 28. 1526-1527.
15 Adams. C. ].; Downs. A, J. Client. Commun. 1970, 1699.
Shannon. R. D. Acta Cryslallogr. 1976. A32. 751. Kepcrt. D. L. Progr. Inorg. Client. 1977. 23. I.
Sec also Abriel. W. Acta Cryslallogr.. Sect. B 1986. B42. 449.
17 Experts. S.; George. J. W.; Huscbye, S.: Mikalsen, 0. Acta. Client. Stand.. Sect. A 1975. 29.
141-148.
146
5 • Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Both of these structures have four bonds and the charges are reasonably
placed. A third structure
:N:N:::0:
-2 + +
III
is unfavorable because it places a positive charge on the electronegative
oxygen atom and also has adjacent positive charges.
Other possibilities are
:N:N::0: :N::N:0:
iv v
and the cyclic structure discussed under Rule 2. This last structure has been
shown above to be unfavorable. Likewise IV and V should be bent and are
energetically unfavorable when forced to be linear to resonate with I and II. In
addition, both have only three bonds instead of four and are therefore less
stable. Furthermore, V has widely separated charges, but they are exactly
opposite to those expected from electronegativity considerations.
It is almost impossible to overemphasize the fact that the resonance hybrid is the
only structure which is actually observed and that the canonical forms are merely
constructs which enable us to describe accurately the experimentally observed mole¬
cule. The analogy is often made that the resonance hybrid is like a mule, which is a
genetic hybrid between a horse and a donkey. The mule is a mule and does not
“resonate" back and forth between being a horse and a donkey. It is as though one
were trying to describe a mule to someone who had never seen one before and had
available only photographs of a jackass and a mare. One could then explain that their
offspring, intermediate between them, was a mule . 5 There is perhaps a better analogy,
though one that will be unfamiliar to those not versed in ancient mythology: Consider
a falcon (a real animal) described as a hybrid of Re (the falcon-headed Egyptian sun
god) and a harpy (a creature with a woman's head and the body of a raptor), although
neither of the latter has an independent existence.
Formal Charge In the discussion of several preceding molecules, including nitrous oxide, formal
charges were assigned without explanation. As we said earlier, formal charge may be
regarded as the charge that an atom in a molecule would have if all of the atoms had
the same electronegativity. It may approximate the real ionic charge as in the phos-
phonium ion, PH 4 . Phosphorus and hydrogen have approximately the same elec¬
tronegativity. The formal charge on each hydrogen atom is zero and the phosphorus
atom carries a single positive formal charge corresponding to the ionic charge. On the
other hand, some molecules, such as N 2 0. exhibit formal charges in otherwise neutral
molecules.
:N=N=0: «—» :N=N— 6 :
"
1 11
5 This analogy, like any other, can be pushed too far. The contributing structures should not be
considered as “parents" of the hybrid.
Valence Bond Theory
147
In the case of N 2 0, the electronegativities of nitrogen and oxygen are different. In
both cases, the calculated formal charges indicate the presence of real electrical
charges on the atoms in question, though not necessarily exactly +1 or - I. Spec¬
ifically, the charge density about the two nitrogen atoms is not the same.
To obtain the formal charge on an atom, it is assumed that all electrons are shared
equally and that each atom "owns” one-half of the electrons it shares with neighbor¬
ing atoms. The formal charge, Q F , is then:
Qf = - N m = ,V A - yV LP - i N ap ( 5 . 7 )
where /V A is the number of electrons in the valence shell in the free atom and N M is the
number of electrons "belonging to the atom in the molecule"; N u , and N 0P are the
numbers of electrons in unshared pairs and bonding pairs, respectively.
Applied to the Lewis structure of the phosphonium ion.
H
H:P:H
H
H —P —H
we obtain the following formal charges:
Gp = 5 - 4 = 5 - 0 - 1(8) = +1
Qh = = | - 0 - 1 ( 2 ) = 0
This is in contrast to the phosphine molecule.
H: P: H or H —P—H
for which the formal charges arc
Gp = 5- 5 = 5 — 2 - 1(6) = 0 (5.10)
Qh — I — I = I — 0 — 1 ( 2 ) = 0 (5.11)
To return now to nitrous oxide, N 2 0, specifically structure I, we have a Lewis
structure:
:N,::N C "6: or N, = N c = O
where / = terminal (or left) and c = central, merely as identifying labels. Hence:
Gn, = 5- 6 = 5- 4- }(4) = - I (5.12)
Gn,. = 5- 4 = 5- 0 — 1(8) = +1 (5.13)
Go = 6 — 6 = 6- 4— 1(4) = 0 (5.14)
Likewise, structure II gives a + I charge on N ( . and - I on the oxygen.
Recently, the concept of formal charge has been made more quantifiable by
combining it with the idea of electronegativity to estimate the relative effects of each in
212
id Reactivity of Molecules
The Structure of Molecules
213
6 - The Structure an
120 °
-120° C - — Q
^ - 120 °
F
(a)
126°
108°
126°
(b)
Fig. 6.10 (a) Possible
structure of OCF 2 .
Arrows indicate small
distortions resulting from
electronegativity and size
effects, (b) Actual
molecular structure of
OCF 2 . Note small FCF
bond angle.
bond angles (Fig. 6.10a). The molecule is in “
severely from a symmetrical trigonal arrangement (Fig. 6.10b). It is apparent
he oxygen atom requires considerably more room than the fluonne . oms^The
are at least two steric reasons for this. First, the oxygen atom .s doubl>^bonded to
the carbon and the C=0 bond length (120 pm) ts somewhat less than that of
C _F ( i35 pm)- thus, the van der Waals repulsion of the oxygen atom will be
greater More important in the present case is the fact that the double bon
contains two pairs of electrons, and whether viewed as a < 7-77 pair or twin bent
bonds, it is reasonable to assume that they will requ.re more space than single
b ° n< Thfs assumption is strengthened by other compounds with double bonds. In
the OSF, molecule the doubly bonded oxygen atom seeks the more spac.ous
equatorial position, and the fluorine atoms are bent away—a fom
other two equatorial and the two axial positions (Fig. 6.1 lb). Further examples
are listed in Table 6.1. Note that the behavior of the doubly bonded oxygen ato
ways similar .0 iira, of a ioae pair. Both
single bonding pair, both seek the equatorial position, andbothrepd adjac
fable 6.1
Bond angles in molecules
containing doubly bonded
oxygen and electron lone
pairs
oj<
F
1/
0= N
F
\ l- F
C = S
S P'
Axial FSF angle
(idealized =
,80°, 1790
178.5°
170°
Equatorial FSF ■
(idealized =
an8lc 103°
120°) l0J
110°
97°
(a)
(b)
(c)
fluorine atoms.
,, , ■ no 1 in 7 % Note that the bond angles accepted by
4 Christe Md Oberhammer'were soSa. d.fferen, than .‘hose given here and were somewhat more
favorable for their argument.
Molecule
X
1
-<
1
Molecule
X—Y—X‘
O
II
n
71
108°
:GeF 2
94° ± 4°
o 2 sf 2
96°
:SF,
98°
0=PC1 3
103.3°
:PCI,
100.3°
o=sf 4
110,178.5°
:SF 4
103,179°
o=if 5
82,89°
:IFj
82°
“ Y = central atom. C, S, P. I. Ge; X = halogen atom.
Cl, F.
than another. However, the plane of the ir bond can only be inferred from the
bond angles, leading to a possible circularity in reasoning. More straightforward is
the CH 2 = SF 4 molecule (Fig. 6.1 lc). Because the hydrogen atoms lie in the CSF-,
axial plane; we know that the n bond involving a p orbital on the carbon atom
must lie in the equatorial plane of the molecule. And the resulting repulsion
between the v electrons and the electron pairs bonding the equatorial fluorine
atoms is dramatic: The F^-F^ angle has been reduced to 97V
No discussion of the VSEPR model of molecular structure would be complete
without a brief discussion of some problems remaining. One interesting problem is the
molecular structure of XeF 6 . The simplest MO treatment of this molecule predicted
that the molecule would be perfectly octahedral. 6 In contrast, the VSEPR model
considers the fact that there will be seven pairs of electrons in the valence shell (six
bonding pairs and one lone pair) and predicts a structure based on seven-coordina¬
tion. 7 Unfortunately, we have little to guide us in choosing the preferred arrangement.
Gillespie suggested three possibilities for XeF 6 : a distorted pentagonal bipyramid, a
distorted octahedron, or a distorted trigonal prism. The lone pair should occupy a
definite geometric position and a volume as great as or greater than a bonding pair.
Unfortunately, only three neutral fluoride molecules with seven bonding pairs are
known: IF 7 , ReF 7 , and OsF 7 . The structures are known with varying degrees of
certainty, but all three appear to have approximate D ih symmetry, a distorted pen¬
tagonal bipyramid (Fig. 6.I2). 8 Unfortunately, knowing the MF 7 structures was of
Fig. 6.12 Molecular
structure of iodine
heptafluoride.
\N / ?/
5 Huheey, J. E. Inorg. Chem. 1981, 20, 4033.
6 A discussion of molecular orbital theory applied to noble gas compounds will be found in Chapter 17.
7 Gillespie, R. J. In Noble Gas Compounds', Hyman, H. H., Ed.; University of Chicago: Chicago,
1963; p 333.
8 Drew, M. G. B. Progr. Inorg. Chem. 1977, 23, 67.
148
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
I
Hybridization
determining the total charge Q r .« The equation suggested for determining the charge
on A in the molecule AB„ is
The sum over B represents the cumulative electronegativity effects built up over n
bonds in the molecule AB„. When the electronegativities of the bonding atoms are
equal, Eq. 5.15 reduces to the simple equation for formal charge, Eq. 5.7. When the
electronegativity difference is large the ionic character of the bond is large^and the
electrons are shifted towards one atom or the other. The limit of 2S as x A « b 1S
and the electrons have been completely transferred from atom A to atom(s) B or,
conversely, when A » „• 22 is 2 - and lhe eleclrons have been , completely trans¬
ferred from atom(s) B to atom A. Applied to the N 2 0 molecule, Eq. 5.15 gives
estimated charges of 6 7
-0.33 +1.10 -0.77
Note that this allocation confirms the + 1 charge on N r and splits the negative charge
between the terminal nitrogen atom and the oxygen atom, with the latter getting the
larger fraction because of its greater electronegativity.
Although formal charges do not represent real, ionic charges, they do represent a
tendency for buildup of positive and negative charges. For example, consider carbon
monoxide. The only reasonable Lewis structure that maximizes the bonding is the
normal triple-bonded one:
Note, however, that this places a formal positive charge on the oxygen and a formal
negative charge on the carbon. If the electronegativities of carbon and oxygen were
the same, carbon monoxide would have a sizable dipole moment in the direction
but since the electronegativity difference draws electron density back from the carbon
atom to the oxygen atom, the effect is canceled and carbon monoxide has a very low
dipole moment, 0.4 x 10 30 C m (0.12 D).
In the valence bond theory, hybridization of orbitals is an integral part of bond
formation. As we shall see, the concept need not be explicitly considered in molecular
orbital theory but may be helpful in visualizing the process of bond formation.
Consider the methane molecule, CH 4 . The ground state of a carbon atom is V
corresponding to the electron configuration of \s 2 2s 2 2p' t 2p' y . Carbon in this state would
be divalent because only two unpaired electrons are available for bonding in the and
p orbitals. Although divalent carbon is well known in methylene and carbene inter¬
mediates in organic chemistry, stable carbon compounds are tetravalent. In order for
four bonds to form, the carbon atom must be raised to its valence stale. This requires
6 Allen. L. C. J. Am. Chem. Soc. 1989. Ill, 9115-9116.
7 The charges shown are averages obtained by weighting resonance structure II twice structure I. See
reference in Footnote 6.
Valence Bond Theory 149
the promotion of one of the electrons from the 2s orbital to the formerly empty 2 p
orbital. This excited S state has an electron configuration \s 2 2s'2p\2p' y 2pl This
promotion costs 406 kJ mol' 1 . Because the valence state, V 4 , is defined as the state of
an atom in a molecule, but without the addition of bonded atoms, it is necessary to
supply a further amount of energy to randomize the spins of the 5 5 state, that is, to
supply enough energy to overcome the normal tendency toward parallel spins. 8
Despite all of the energy necessary to reach the valence state, the formation of two
additional bonds makes CH 4 895 kJ mol - ' more stable than CH, + 2H.
Hybridization consists of mixing or linear combination of the “pure" atomic
orbitals in such a way as to form new hybrid orbitals. Thus we say that the single 2.v
orbital plus the three 2 p orbitals of the carbon atom have combined to form a set of
four spatially and energetically equivalent sp hybrid orbitals. This is illustrated in
Fig. 5.2 for the conceptually simpler case of the sp hybrid formed from an s orbital and
a single p orbital. Combination of the s and p orbitals causes a reinforcement in the
region in which the signs of the wave function are the same, cancellation where the
signs are opposite.
If we let and represent the wave functions of an j and a p orbital, then we
combine them to make two equivalent orbitals as follows:
•Pdi, = 'A (, + <!>„) (5 - 16)
tp dh = >p p ) (5 - 17)
where V is the normalizing coefficient and ifij,, and l/ Jh are the new digonal ( di) or sp
orbitals.
Mathematically, the formation of sp or tetrahedral orbitals for methane is more
complicated but not basically different. The results are four equivalent hybrid orbitals,
each containing one part s to three parts p in each wave function, directed to the
corners of a tetrahedron. As in the case of sp hybrids, the hybridization of a and p has
Fig. 5.2 Formation of sp
hybrid orbitals by the
addition and subtraction of
angular wave functions.
» The existence of this extra valence state excitation energy may be clearer if the reverse process is
considered, ir (in a thought experiment) four hydrogen atoms are removed from methane but the
carbon is not allowed to change in any way, the resulting spins would be perfectly randomized.
Energy would then be released if the spins were allowed to become parallel. See McWeeny. R.
Coulson's Valence: Oxford University: London. 1979; pp 150, 201-203, 208-209. It should be noted
that unlike P. 'S. etc,, V 4 is not an observable spectroscopic state but is calculated by adding
promotion energies related to the electron spins.
211
6 • The Structure and Reactivity of Molecules
Fig. 6.6 (a) The pentafluorotellurate(IV) anion. Approximately octahedral arrangement of
bonding and nonbonding electrons, (b) Experimentally determined structure. The tellurium
atom is below the plane of the fluorine atoms. [From Mastin. S. H.; Ryan. R. R.; Asprey.
L. B. Inorg. Client. 1970. 9. 2100-2103. Reproduced with permission.!
Fig. 6.7 The tetrachloroiodatc(lll) ion. (a) Octahedral arrangement of bonding and
nonbonding electrons with lone pairs cis to each other, (b) Octahedral arrangement of
bonding and nonbonding electrons with lone pairs trans to each other, (c) Experimentally
determined structure.
m
Nitrogen dioxide (C 3l ,). nitrite ion (O,,). and nitryl ion (D x/l ). The three
species. NO,, NO,. and NOV". show the effect of steric repulsion of bonding and
nonbonding electrons. The Lewis structures are
[:0::n::0:] + :0::n:0: [:0::N:0:] -
The nitryl ion. NOT. is isoelectronic with carbon dioxide and will, like it. adopt a
linear structure with two tt bonds (Fig. 6.8a). The nitrite ion. NOT. will have one
77 bond (stereochemieally inactive), two a bonds, and one lone pair. The resulting
structure is therefore expected to be trigonal, with 120° sp- bonds to a first
approximation. The lone pair should be expected to expand at the expense of the
bonding pairs, however, and the bond angle is found to be 115° (Fig. 6.8b).
The nitrogen dioxide molecule is a free radical, i.e., it contains an unpaired
electron. It may be considered to be a nitrite ion from which one electron has been
removed from the least electronegative atom, nitrogen. Instead of having a lone
pair on the nitrogen, it has a single electron in an approximately trigonal orbital.
Since a single electron would be expected to repel less than two. the bonding
i
The Structure of Molecules
Fig. 6.8 (a) The linear nitryl ion, NOT. (b) The effect of the lone pair in the nitrite ion, NOT.
Resonance has been omitted to simplify the discussion, (c) The effect of the unpaired
electron, half of a lone pair, in nitrogen dioxide.
electrons can move so as to open up the bond angle and reduce the repulsion
between them (Fig. 6.8c).
Phosphorus trihalides (C 3 ,.). The importance of electron repulsions near the
nucleus of the central atom is nicely shown by the bond angles in phosphorus
trihalide molecules: PF 3 = 97.7°, PC1 3 = 100.3°, PBr 3 = 101.0°, PI 3 = 102°. The
immediate inclination to ascribe the opening of the bond angles to van der Waals
repulsions between the halogens must be rejected. Although the van der Waals
radii increase F < Cl < Br < I, the covalent radii and hence the P—X bond
lengths also increase in the same order. The two effects cancel each other (see
Problem 6. 15). The important factor appears to be the ionicity of the P—X bond.
The more electronegative fluorine atom attracts the bonding electron pairs away
from the phosphorus nucleus and allows the lone pair to expand while the
p p p an Ele closes. Reduced bond angles in nonmetal fluorides are commonly
observed. For the small atoms nitrogen and oxygen, where the VSEPR interac¬
tions seem to be especially important, the fluorides have smaller bond angles than
the hydrides (NF 3 = 102.3°, NH 3 = 107.2°, OF, = 103.1°, OH 2 = 104.5°).
Gillespie' has discussed the effect of substituent electronegativity and pointed out
that the expansion of lone pairs relative to bonding pairs may be viewed simply as
an example of the extreme effect when the nonexistent “substituent” on the lone
pair has no electronegativity at all (sec Fig. 6.9).
Carbonyl fluoride (C 2l ,). Fluorine and oxygen atoms arc about the same size
and similar in electronegativity; therefore we might expect OCF, to have a rather
symmetrical structure. There are no lone pairs on the carbon atom, so to a first
approximation we might expect the molecule to be planar with approximately 120°
Fig. 6.9 Effect of decreasing electronegativity of X on the size of a bonding pair of
electrons: (a) electronegativity X > A. (b) electronegativity X = A, (c) electronegativity
X < A. (d) X = lone pair of electrons; effective electronegativity of X is zero. [From
Gillespie. R. J. J. Chem. Educ. 1970. 47. 18. Reproduced with permission.)
150
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Valence Bond Theory
151
Nodal surface
Fig. 5.3 Electron density
contours for an sp i hybrid
orbital. Note that the nodal
surface does not pass
through the nucleus.
resulted in one lobe of the hybrid orbital being much larger than the other (see
Fig. 5.3). Hybrid orbitals may be pictured in many ways: by several contour surfaces
(Fig. 5.3); a single, outer contour surface (Fig. 5.4a); cloud pictures (Fig. 5.4b); or by
simpler, diagrammatic sketches which ignore the small lobe of the orbital and picture
the larger lobe (Fig. 5.4c). The latter, though badly distorted, are commonly used in
drawing molecules containing several hybrid orbitals.
It is possible to form a third type of s-p hybrid containing one s orbital and two p
orbitals. This is called an sp- or trigonal (tr) hybrid. It consists of three identical
orbitals, each of which does not differ appreciably in shape from Fig. 5.3 and is
directed toward the corner of an equilateral triangle. The angles between the axes of
the orbitals in a trigonal hybrid are thus all 120°.
Although promotion and hybridization are connected in the formation of methane
from carbon and hydrogen, care should be taken to distinguish between them. Promo¬
tion involves the addition of energy to raise an electron to a higher energy level in
order that the two additional bonds may form. It is conceivable that after promotion
the carbon atom could have formed three bonds with the three p orbitals and the
fourth with the s orbital. That carbon forms tetrahedral bonds instead is a consc-
Fig. 5.4 Other ways of representing hybrid orbitals: (a) orbital shape shown by a single
contour, (b) cloud representation, (c) simplified representation. The small back lobes have
been omitted and the shape streamlined to make it easier to draw molecules containing
several hybrid orbitals.
quence of the greater stability of the latter, not a necessary result of promotion. Thus,
although promotion and hybridization often occur together, either could occur without
the other.
A second point to be made with regard to hybrids is the source of the driving force
resulting in hybridization. Statements are often made to the effect that “methane is
tetrahedral because the carbon is hybridized sp 3 .” This is very loose usage and gets
the cart before the horse. The methane molecule is tetrahedral because the energy of
the molecule is lowest in that configuration, principally because of increased bond
energies and decreased repulsion energies. For this molecule to be tetrahedral, VB
theory demands that sp 3 hybridization take place. Thus it is incorrect to attribute the
shape of a molecule to hybridization—the latter prohibits certain configurations and
allows others but does not indicate a preferred one. For example, consider the
following possibilities for the methane molecule:
I sp 3 II sp 2 + p III sp + p- IV T + p 3 V
The first three geometries involve the tetrahedral, trigonal, and digonal hybrids
discussed above and the fourth involves the use of pure ■ and p orbitals as discussed
on page 149. The last structure contains three equivalent bonds at mutual angles of 60°
and a fourth bond at an angle of approximately 145° to the others. It is impossible to
construct s-p hybrid orbitals with angles less than 90°, and so structure V is ruled out.
In this sense it may be said that hybridization does not •‘allow’’ structure V, but it may
not be said that it ’’chooses” one of the others. Carbon hybridizes sp, sp 2 , and .sp 3 in
various compounds, and the choice of sp in methane is a result of the fact that the
tetrahedral structure is the most stable possible.
Although we shall not make explicit use of them, the reader may be interested in
the form of the s-p hybrids we have seen. 9
Vj>, +
(5.18)
vT, - + Vj> Pv
(5.19)
je-
l
P
l
(5.20)
Ms + M p , + M Py + M p .
(5.21)
Ms ~ M P , - M Py + M P:
(5.22)
Ms + M p , - 4 li Py - J 4> Pt
(5.23)
i
i
(5.24)
9 The percent s and p character is proportional to the squares of the coefficients. Taken from Hsu.
C. Y.; Orchin, M. J. Chem. Educ. 1973. 50, 114-118.
_08
6 -The Structure and Reactivity af Molecules
5
greater reduction of the bond angle by two lone pairs.
J!M - ■»« - ihc approxima,e mo -
lecular shape but also distortions which will take place.
will be a trigonal bipyramid, as in PF 5 . a 4 -,i or axially (Fig. 6.4b). The
„„e of .wo possible ways, ->h=r e^.o™! Y J,g 6.4^0 r ax.a » F,g ^
, , (b) (c)
(a)
structure of sulfur tetrafluonde.
The Structure of Molecules
209
tions (with the equatorial bonding pairs). Presumably the 120° interactions are
sufficiently relaxed that they play no important role in determining the most stable
arrangement. This is consistent with the fact that repulsive forces are important
only at very small distances. In any event, lone pairs always adopt positions
which minimize 90° interactions.
Bromine trifluoride (C 2 „). The :BrF 3 molecule also has ten electrons in the
valence shell of the central atom, in this case three bonding pairs and two lone
pairs. Again, the approximate structure is trigonal bipyramidal with the lone pairs
occupying equatorial positions. The distortion from lone pair repulsion causes the
axial fluorine atoms to be bent away from a linear arrangement so that the
molecule is a slightly “bent T” with bond angles of 86° (Fig. 6.5a).
Dichloroiodate(l) anion (£>„,,). The : ICIT anion has a linear structure as
might have been supposed naively. However, note that three lone pairs are
presumably still stereochemically active, but by adopting the three equatorial
positions they cause no distortion (Fig. 6.5b). [A note on bookkeeping for ions:
Add 7 electrons (I) + 2 electrons (2C1) + 1 electron (ionic charge) = 10 =
5 pairs.]
Pentafluorotellurate(IV) anion (C 4l ,). In the :TeFJ ion the tellurium atom
has twelve electrons in its valence shell, five bonding pairs and one nonbonding.
The most stable arrangement for six pairs of electrons is the octahedron which we
should expect for a first approximation. Repulsion from the single lone pair should
cause the adjacent fluorine atoms to move upward somewhat (Fig. 6.6a). The
resulting structure is a square pyramid with the tellurium atom 40 pm below the
plane of the four fluorine atoms (Fig. 6.6b).
Tetrachloroiodate(IIl) anion (D ih ). The : IC1 4 anion is isoelectronic with
the TeFf ion with respect to the central atom. In this case, however, there are
four bonding pairs and two lone pairs. In an undistorted octahedron, all six points
are equivalent, and the lone pairs could be adjacent, or cis (Fig. 6.7a); or trans
(Fig 6.7b), opposite to one another. In the cis arrangement the lone pairs will
compete with each other for volume into which to expand, a less desirable
arrangement than trans, in which they can expand at the expense of the bonding
pairs. Since the lone pairs are not seen in a normal structural determination, the
resulting arrangement of atoms is square planar (Fig. 6.7c).
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Toble 5.2 _
Bond angles of hybrid
Hybrid
Geometry
Bond angle(s)
sp ( di )
Linear (digonal)
180°
sp 2 (tr)
Trigonal
120 °
sp 2 (te)
Tetrahedral
109$°
dsp 2
Trigonal bipyramidal or
Square pyramidal
90°, 120°
90°, <90°
d 2 sp 2
Octahedral
90°
It is not necessary to limit hybridization to s and p orbitals. The criteria are that
the wave functions of the orbitals being hybridized must be of appropriate symmetry
(Chapter 3) and be similar in energy. If the orbitals are not close in energy, the wave
function of the hybrid will be unsuited for bonding because the electron density would
be spread too thinly. In practice this means that hybrids are formed among orbitals
lying in the same principal energy level or, occasionally, in adjacent energy levels^
Some hybrid orbitals containing s, p , and d orbitals are listed in Table 5.2. I he
structural aspects of various hybrid orbitals will be discussed in Chapter 6, but the
bond angles between orbitals of a given hybridization are also listed in Table 5.2 for
reference.
Most sets of hybrid orbitals are equivalent and symmetric, that is, four sp 2 orbitals
directed to the corners of a regular tetrahedron, six tPsp 2 orbitals to the corners ot an
octahedron, etc. In the case of sp 2 d hybrids the resulting orbitals are not equivalent. In
the trigonal bipyramidal arrangement three orbitals directed tngonally form one set of
equivalent orbitals (these may be considered sp 2 hybrids) and two orbitals directed
linearly (and perpendicular to the plane of the first three) form a second set of two
(these may be considered dp hybrids). The former set is known as the equatorial
orbitals and the latter as the axial orbitals. Because of the nature of the different
orbitals involved, bonds formed from the two are intrinsically different and will have
different properties even when bonded to identical atoms. For example, in molecules
like PF, bond lengths differ for axial and equatorial bonds (see Chapter 6)-
Even in the case of s-p orbitals it is not necessary that all the orbitals be
equivalent. Consider the water molecule, in which the H—O—H angle is 104 .
which does not correspond to any of the hybrids described above, but lies between the
109 $° angle for sp 2 and 90° for pure p orbitals. Presumably the two bonding orbitals in
water are approximately tetrahedral orbitals but contain a little more p c J ar ^‘ er '
which correlates with the tendency of the bond angle to diminish toward the 90 ot
pure p orbitals. The driving forces for this effect will be discussed in Chapter 6.
The relationship between p or s character and bond angle will also be discussed in
Chapter 6. For now we need only consider the possibility of s-p hybridization other
than sp sp 2 . and sp 2 . If we take the ratio of the 5 contribution to the total orbital
complement in these hybrids, we obtain 50%, 33%. and 25% character, respectively,
for these hybrids. A pure 5 orbital would be 100% s, and a p orbital would have 0/ o s
character. Since hybrid orbitals are constructed as linear combinations of and p
orbital wave functions,
= aip s + bili p <5 ‘ 25)
there is no constraint that a and b must have values such that the 4 character is exactly
25%, 33%, or 50%. A value of 20% 4 character is quite acceptable, for example, and
Molecular Orbital Theory
153
Ta ble 5.3 _
Effect of hybridization on
overlap and bond
properties"
Molecule
Hybridization
C — H bond energy
(kJ mol” 1 )
C — H bond
length (pm)
H—C=C—H
h 2 c=ch 2
ch 4
CH radical
sp
sp 2
sp 2
~p
500
400
410
335
106.1
108.6
109.3
112.0
» McWeeny, R. Coulson's Valence-. Oxford University: London, 1979; p 204.
Used with permission.
Hybridization
and Overlap
indeed this happens to be the value in water. When the hybridization is defined as
above, the % p character is always the complement of % s , in the case of water, 80%.
We may make the generalization that the strength of a bond will be roughly propor¬
tional to the extent of overlap of the atomic orbitals. Both pure 4- and pure p orbitals
provide relatively inefficient overlap compared with that of hybrid orbitals. The
relative overlap of hybrid orbitals decreases in the order sp > sp 2 > sp 2 » p. The
differences in bonding resulting from hybridization effects on overlap can be seen m
Table 5.3. The C—H bond in acetylene is shorter and stronger than in hydrocarbons
having less s character in the bonding orbital. The hybridization in the hydrocarbons
listed in Table 5.3 is dictated by the stoichiometry and stereochemistry. In molecules
where variable hybridization is possible, various possible hybridizations, overlaps,
and bond strengths are possible. Other things being equal, we should expect molecules
to maximize bond energies through the use of appropriate hybridizations.
Molecular Orbital
Theory
A second approach to bonding in molecules is known as the molecular orbital (MO)
theory. The assumption here is that if two nuclei are positioned at an equilibrium
distance, and electrons are added, they will go into molecular orbitals that are in many
ways analogous to the atomic orbitals discussed in Chapter 2. In the atom there are s ,
p d, f. . . . orbitals determined by various sets of quantum numbers and in the
molecule we have tr, ir, 8 _orbitals determined by quantum numbers. We should
expect to find the Pauli exclusion principle and Hund's principle of maximum multi¬
plicity obeyed in these molecular orbitals as well as in the atomic orbitals.
When we attempt to solve the Schrddinger equation to obtain the various mo¬
lecular orbitals, we run into the same problem found earlier for atoms heavier than
hydrogen. We are unable to solve the Schrddinger equation exactly and therefore
must make some approximations concerning the form of the wave functions for the
molecular orbitals.
Of the various methods of approximating the correct molecular orbitals, we shall
discuss only one: the linear combination of atomic orbitals (LCAO) method. We
assume that we can approximate the correct molecular orbitals by combining the
atomic orbitals of the atoms that form the molecule. The rationale is that most of the
time the electrons will be nearer and hence “controlled” by one or the other of the two
nuclei and when this is so, the molecular orbital should be very nearly the same as the
atomic orbital for that atom. The basic process is the same as the one we employed in
constructing hybrid atomic orbitals except that now we are combining orbitals on
different atoms to form new orbitals that are associated with the entire molecule. We
206 6-The
Structures of
Molecules
Containing
Lone Pairs
of Electrons
Structure and Reactivity of Molecules
because the aluminum atom can accept an additional pair of electrons (Lewis
acid, see Chapter 9) in its unused p orbital and rehybridize from sp 2 to sp 3 . We
should expect the bond angles about the aluminum to be approximately
tetrahedral except for the strain involved in the A1—Br—A1—Br four-membered
ring. Since the average bond angle within the ring must be 90°, we might expect
both the aluminum and bromine atoms to use orbitals which are essentially purely
p in character for the ring in order to reduce the strain. The structure of the Al 2 Br 6
molecule is shown in Fig. 6.1h.
Although the discussions of the preceding molecules have been couched in valence
bond terms (Lewis structures, hybridization, etc.), recall that the criterion for molec¬
ular shape (rule 2 above) was that the a bonds of the central atom should be allowed to
get as far from each other as possible: 2 at 180°, 3 at 120°, 4 at 109.5°, etc. This is the
heart of the VSEPR method of predicting molecular structures, and is, indeed,
independent of valence bond hybridization schemes, although it is most readily
applied in a VB context.
The source of the repulsions that maximize bond angles is not completely clear.
For molecules such as C0 2 , B(CH 3 ) 3 , or 0=PF 3 we might suppose that van der
Waals repulsions (analogous to the Born repulsions in ionic crystals. Chapter 4)
among, for example, the three methyl groups might open the bond angles to the
maximum possible value of 120°. In the next section we shall see that nonbonding
pairs of electrons (lone pairs) are at least as effective as bonding pairs (or bonded
groups) in repulsion, and so attention focuses on the electron pairs themselves.
Although a number of theories have been advanced, the consensus seems to be that
the physical force behind VSEPR is the Pauli force: Two electrons of the same spin
cannot occupy the same space. However, it should be noted that there has been some
disagreement over the matter. Nevertheless, as we shall see, the VSEPR model is an
extremely powerful one for predicting molecular structures.
When we investigate molecules containing lone (unshared) electron pairs, we must
take into account the differences between the bonding electrons and the nonbonding
electrons. First, before considering hybridization and the energies implicit in the
bonding rules (Chapter 5) let us consider the simplest possible viewpoint. Consider the
water molecule in which the oxygen atom has a ground state electron configuration of
\s 2 2s 2 2p\2p\2p[. The unpaired electrons in the p x and p v orbitals may now be paired
with electrons on two hydrogen atoms to give H 2 0. Since the p x and p y orbitals lie at
right angles to one another, maximum overlap is obtained with an H—O—H bond
angle of 90°. The experimentally observed bond angle in water is, however, about
104j°, much closer to a tetrahedral angle. Inclusion of repulsion of positive charges on
the adjacent hydrogen atoms (resulting from the fact that the oxygen does not share
the electrons equally with the hydrogens) might cause the bond angle to open up
somewhat, but cannot account for the large deviation from 90°. Not only must the
H-H repulsions be taken into consideration, but also every other energetic interaction
in the molecule: all repulsions and all changes in bond energies as a function of angle
and hybridization. It is impossible to treat this problem in a rigorous way, mainly as a
result of our ignorance of the magnitude of the various energies involved; however,
certain empirical rules have been formulated. 1
First, as we have seen in examples on the previous pages, bond angles in
molecules tend to open up as much as possible as a result of the repulsions between
the electrons bonding the substituents to the central atoms. Repulsions between
The Structure of Molecules 207
unshared electrons on the central atom and other unshared electrons or bonding
electrons will affect the geometry. In fact, it is found that the repulsions between lone
pair electrons are greater than those between the bonding electrons. The order of
repulsive energies is lone pair-lone pair > lone pair-bonding pair > bonding
pair-bonding pair.-' This results from the absence of a second nucleus at the distal end
of the lone pair which would tend to localize the electron cloud in the region between
the nuclei. Because the lone pair does not have this second nucleus, it is attracted only
by its own nucleus and tends to occupy a greater angular volume (Fig. 6.2).
The difference in spatial requirements between lone pairs and bonding pairs may
perhaps be seen most clearly from the following example. Consider an atom or ion
with a noble gas configuration such as C~, N 3- , O 2- , F - , or Ne (Ii 2 2r2p 5 ). Assume
that the eight electrons in the outer shell occupy four equivalent tetrahedral orbitals.
Now let a proton interact with one pair of electrons to form an X—H bond (HC 3- ,
NH 2 “. OH". HF. NeH + ). The proton will polarize the pair of electrons to which it
attaches in the same way that a proton or other small, positive ion polarizes an anion
(Fajans' rules. Chapter 4). Electron density will be removed from the vicinity of the
nucleus of the first atom and attracted toward the hydrogen nucleus. The remaining,
nonbonding pairs may thus expand at the expense of the bonding pair. Addition of a
second proton produces two polarized, bonding pairs and two expanded lone pairs
(H 2 C 2 ~. H 2 N~, H 2 0, H,F + ). A third proton forms H 3 C~, NH 3 , and H 3 0 + with one
expanded lone pair. A fourth proton produces CH 4 and NH„ + in which all four pairs of
electrons have been polarized toward the hydrogen nuclei, are once more equivalent,
and hence directed at tetrahedral angles.
From this point of view, the water molecule can be considered to be hybridized
tetrahcdrally to a first approximation. Since the two lone pairs will occupy a greater
angular volume than the two bonding pairs, the angle between the latter two is reduced
somewhat (from 1091° to 1041°). allowing the angle between the lone pairs to open up
slightly. The series methane, CH 4 (no lone pairs, bond angle = 1091°); ammonia, NH ,
(one lone pair, bond angle = 107°); and water, H,0 (two lone pairs, bond angle =
1041°) illustrates an isoelectronic series in which the increasing requirements of the
nonbonding pairs reduce the bond angle (Fig. 6.3).
Fig. 6.2 (u) Four
equivalent bonding electron
pairs, (b) Three bonding
electron pairs repelled by
a nonbonding pair of
electrons.
(a)
lb)
3 Although the electron pair repulsion ranking has been widely used to rationalize geometries, some
theoretical studies suggest that bonding pair-bonding pair repulsion is important in keeping them
apart; the tendency for the nonbonding pairs to assume .v character (see p 225) can be used rather
than lone pair-lone pair repulsion to explain reductions in bond angles. See Hall. M. B. J. Am.
Chem. Sac. 1978. 100. 6333-6338; Inarg. Clicm. 1978, 17. 2261-2269: Shustorovich. E.; Dobosh.
P. A. J. Am. Chem. Soc. 1979. 101. 4090-4095.
154
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
therefore combine the atomic orbitals i// A and i/> B on atoms A and B to obtain two
molecular orbitals: 10
h = «Aa + A b (5.26)
= '/'A - 'Ab (5.27)
The one-electron molecular orbitals thus formed consist of a bonding molecular
orbital ( ip h ) and an antibonding molecular orbital (iA (J ). If we allow a single electron to
occupy the bonding molecular orbital (as in HT. for example), the approximate wave
function for the molecule is
^ = fan = t aid + fan (5.28)
For a two-electron system such as H 2 , the total wave function is the product of the
wave functions for each electron:
= fan+tm = [fan + fanUfai) + faij (5.29)
^ = fan fan + fat) fait + l /'A(l)'/'B(2) + 'I 1 Am fa I) (5.30)
The results for the MO treatment are similar to those obtained by VB theory. Equation
5.30 is the same (when rearranged) as Eq. 5.3 except that the ionic terms (</' A( i ) </' A(2)
and tfi B( i) l /'B(2 ) ) are weighted as heavily as the covalent ones (^ A(I) iAb(2 ) and 'A Amfaij-
This is not surprising, since we did not take into account the repulsion of electrons in
obtaining Eq. 5.29. This is a general result: Simple molecular orbitals obtained in this
way from the linear combination of atomic orbitals (LCAO-MO theory) tend to
exaggerate the ionicity of molecules, and the chief problem in adjusting this simple
method to make the results more realistic consists of taking into account electron
correlation. As in the case of VB theory it is possible to optimize the wave function by
the addition of correcting terms. Some typical results for the hydrogen molecule are
listed in Table 5.4.
The two orbitals ip h and differ from each other as follows. In the bonding
molecular orbital the wave functions for the component atoms reinforce each other in
the region between the nuclei (Fig. 5.5a, b), but in the antibonding molecular orbital
they cancel, forming a node between the nuclei (Fig. 5.5d). We are, of course.
Table 5.4
Energies and equilibrium
Type of wave function
Energy (kJ moM)
Distance (pm)
distances for MO wave
functions in H 2
Uncorrected, «A = «A a + <A B
260
85
Addition of shielding
337
73
MO, SCF limit
349
74
Observed values
458.0
74.1
" McWeeny, R. Coulson's Valence-. Oxford University: London, 1979;
p 120. Used wilh permission.
10 The combination fa - does not represent a third MO, but is another form of fa.
Molecular Orbital Theory
155
(d) (e)
Fig. 5.5 (a) iA a and fa for individual hydrogen atoms (cf. Fig. 2.1). (b) fa = i/> A + fa.
(c) Probability function for the bonding orbital, (d) ip„ = t/, A - fa, (e) Probability
function for the antibonding orbital, fa. Note that the bonding orbital increases the electron
density between the nuclei (c), but that the antibonding orbital decreases electron density
between the nuclei (e).
interested in learning of the electron distribution in the hydrogen molecule, and will
therefore be interested in the square of the wave functions:
’/'/.= 'A a + 2iA a iA b + <Ab (5.31)
'/'a = (Aa “ 2 <Aa<Ab + <An (5.32)
The difference between the two probability functions lies in the cross term 2 iA a iA b . The
integral / s known as the overlap integral, S. and is very important in bonding
theory. In the bonding orbital the overlap is positive and the electron density between
the nuclei is increased, whereas in the antibonding orbital the electron density be¬
tween the nuclei is decreased. (See Fig. 5.5c, e.) In the former case the nuclei are
shielded from each other and the attraction of both nuclei for the electrons is en¬
hanced. This results in a lowering of the energy of the molecule and is therefore a
bonding situation. In the second case the nuclei are partially bared toward each other
and the electrons tend to be in those regions of space in which mutual attraction by
_i
6-The Structure and Reactivity of Molecules
and no nonbonding electrons on the carbon atom, ana
tion is for the <7 bonds to form on opposite sides of f
require hybridization of the carbon 2s and 2 p, orbitals
with a bond angle of 180°.
Once the structure of the (/-bonded molecule has b«
may be added as necessary to complete the molecule,
and p v orbitals on the carbon atom were unused t
available for the formation of ir bonds. A complete sti
would thus be as shown in Fig. 6.1a.
iese simplified VSEPR rules may seem a far cry from
f symmetry and molecular orbitals to the berylliuml
ion (Chapter 5), or the BH, molecule (Problem 6.-7
I approach can rationalize these structures, the direct
is bv far the easier way to approach a new structure.
The Structure of Molecules
205
Trimethylborane (D 3/l ). 2 We may assume that the methyl groups will have
their usual configuration found in organic compounds. The Lewis structure of
(CH 3 ) 3 B will place six electrons in the valence shell of the boron atom, and in
order that the electron pairs be as fair apart as possible, the methyl groups should
be located at the comers of an equilateral triangle. This results in sp 2 , or trigonal
(tr), hybridization for the boron atom (Fig. 6.1b).
Phosgene (C 2v ). A Lewis structure for OCCl 2 has eight electrons about the
carbon, but one pair forms the v bond of the double bond, so again an sp 2 , or
trigonal, hybridization will be the most stable (Fig. 6.1c).
Phosphorus oxyfluoride (C 3v ). Two Lewis structures can be drawn for the
OPF 3 molecule.
;F: _
:0"P:F: — :0:P:F:
'F• ; F=
(a) (b)
To a first approximation, the three fluorine atoms and the single oxygen atom will
be bonded to the phosphorus atom with a bonds from sp 3 tetrahedral orbitals. One
of the five 3 d orbitals on the phosphorus atom also can overlap with a 2 p orbital on
the oxygen atom (Fig. 6. Id) and form a fifth bond, d n -p n , further stabilizing the
molecule.
Phosphorus pentafluoride (D v ,). A Lewis structure for the PF 3 molecule
requires ten electrons in the valence shell of the phosphorus atom and the use of
3.r, 3 p, and 3 d orbitals and five a bonds. It is impossible to form five bonds in three
dimensions such that they are all equidistant from one another, but the trigonal
bipyramidal (Fig. 6.1e) and square pyramidal arrangements tend to minimize
repulsions. Almost every five-coordinate molecule (coordination compounds ex¬
cepted) which has been carefully investigated has been found to have a trigonal
bipyramidal structure. The structure of the PF 5 molecule is shown in Fig. 6.1e
(sp 2 d hybrid). The bonds are of two types: axial, the linear F—P—F system; and
equatorial, the three P—F bonds forming a trigonal plane.
Sulfur hexafluoride ( O h ). Six sulfur-fluorine a bonds require 12 electrons in
the valence shell. Six equivalent bonds require an octahedron and so sulfur will be
hybridized sp 2 d 2 as shown in Fig. 6. If.
Ammonium tetrafluoroborate (T (/ ). Both the ammonium (NHj) and
tetrafluoroborate (BF4) ions are isoelectronic with the methane molecule and we
might therefore reasonably expect them to have similar structures. Indeed, all
four bonds are equivalent, and since the electrons avoid each other as much as
possible, the most stable arrangement is a tetrahedron (Fig. 6. Ig).
Aluminum bromide (D lh ). For the molecule AIBr 3 , a structure similar to
that of trimethylborane would be expected with 120° bond angles. Experimen¬
tally, however, it is found that aluminum bromide is a dimer, AI 2 Br 6 . This is
readily explainable as a result of the tendency to maximize the number of bonds
formed since AI 2 Br 6 contains four bonds per aluminum atom. This is possible
2 The point group symmetry of each molecule is given in parentheses. See Chapter 3.
5-Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
both nuclei is severely reduced. This is a repulsive, or antibonding, situation. An
electron density map for the hydrogen molecule ion. H 2 “, IS shown in Fig. 5.6 illustrat¬
ing the differences in electron densities between the bonding and antibonding
conditions. 11
We have postponed normalization of the molecular orbitals until now. Because
j> 2t = 1 for the probability of finding an electron somewhere in space, the integral of
Eq. 5.31 becomes
jN 2 h ip 2 h dT = N 2 h jil> 2 A dj + ~ 1
where N„ is the normalizing constant. If we let 5 be the overlap integral, /(Mb. we
have
fields = f'l'ids + f ip 2 t,dr + 26']
Now since the atomic wave functions i Ji A and i/i b were previously normalized, Jip\dt
and each equal one. Hence
Nj> 2 + 26
N >- = V2 + 2
(5.35) 1
For most simple calculations the value of the overlap integral. 6, is numerically rather
small and may thus be neglected without incurring too great an error. This simplifies
Fig. 5.6 Electron density contours for the Ht ion: bonding (a) and antibonding (b) orbitals.
The electron density map is easier to obtain for tit than for H 2 because it is not necessary to correct
for electron interactions. The differences are not great.
Molecular Orbital Theory 157
the algebra considerably and is sufficiently accurate for most purposes. With complete
neglect of overlap, our molecular wave functions become
i l> h = VICiAa + ^
li„ - ~ M ^ ^
The idea of “complete neglect of overlap" refers only to the omission of the
mathematical value of the overlap integral in the normalization calculation. Note,
however that “good overlap" in the qualitative sense is necessary lor good bonding
because the covalent energy, A£,, is proportional to the extent that the atomic orbitals
overlap. If overlap is neglected in the calculations, the stabilization and destabilization
of bonding and anlibonding orbitals are equal (Fig. 5.7), and the value tor both
normalization constants is (Eqs. 5.36 and 5.37) N„ = N h = 0.71. If the overlap is
explicitly included in the calculations, the normalization coefficients are N, ■
and N h = 0.56. Other molecules have smaller overlap integrals than H, and so the
effect is less.
c „nrl As we have seen from Eqs. 5.31 and 5.32 the only difference between the electron
^ m T 7 2 distribution in the bonding and antibonding molecular orbitals and the atomic orbitals
° verlap is in those regions of space for which both A and .// B have appreciable values, so that
their product (6 = J> A 0, and for antibonding, 5 < 0. The condition 6 = 0 is termed nonbond,ng
and corresponds to no interaction between the orbitals. That 5 serves as a criterion for
bonding, antibonding, and nonbonding conditions is consistent with our earlier asser¬
tion that bond strength depends on the degree of overlap of atomic orbitals. In general,
we should expect that bonds will form in such a way as to maximize overlap.
In j orbitals the sign of the wave function is everywhere the same (with the
exception of small, in.ranodal regions for « > D. and so there is no problem with
matching the sign of the wave functions to achieve positive overlap. With p and d
orbitals.'however, there are several possible ways of arranging the orbitals, some
resulting in positive overlap, some in negative overlap, and some in which the overlap
is exactly zero (Fig. 5.8). Bonding can take place only when the overlap is positive.
Fiq. 5.7 Energy levels for the H 2 molecule with neglect of overlap. The quantity t£,.
represents the difference in energy between the energy levels of the separated atoms and the
bonding molecular orbital. It is equal to 458 kJ mol -1 .
2 Only a minimum of symmelry is included in this and the following section, and to understand
bonding in these terms does not require prior reading of Chapter 3. I he use of symmeti y and1 group
theory in bonding will be explored more explicitly later in this chapter and in following chapters.
202
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
electrons would each of these three species have? Would you expect the nonbonding
electron pairs on nitrogen or those on oxygen to be more reactive? Explain.
5.26 Construct a qualitative molecular orbital diagram for CI0 2 and compare it to the one
presented in Figure 5.31 for NOT.
5.27 Return to Problem 2.25. Answer it now in terms of group electronegativity.
5.28 The methyl group is usually considered to be electron-donating with regard to hydrogen,
yet its electronegativity is not lower than hydrogen but slightly higher. Explain. (Hint:
Think specifically about situations in which the methyl group is a good donor.)
5.29 You may have learned in organic chemistry that the acidity of acids R—C(0)0H depends
upon R . Discuss in terms of R = H, CHj, and CC1 3 .
5.30 The FjSeO group is extremely electronegative. On the basis of the 'H NMR chemical shifts
(Fig. 5.36) of methyl compounds, CH 3 X, Lentz and Seppelt have suggested that this group
may be even more electronegative than fluorine. Discuss.
Fig. 5.36 (Translated from the original paper:| Correlation of the 'H chemical shift of methyl
compounds CH,X with the electronegativity (Allrcd-Rochow) of the group X. Extrapolation
to the OSeF, group gives an electronegativity slightly greater than that of fluorine. (From
Huppmann, P.; Lentz, D.; Seppelt, K. Z. Anorg. All. Cltern. 1981. 472. 26-32. Reproduced
with permission.]
Chapter
6
I The Structure and
Reactivity of Molecules
The Structure In this chapter a few simple rules for predicting molecular structures will be investi¬
gated. We shall examine first the valence shell electron pair repulsion (VSEPR)
model, and then a purely molecular orbital treatment.
Valence Shell We begin by considering the simplest molecules—those in which Ihe electrons on the
Electron Pair central atom are all involved in bonds. It should be kept in mind that each molecule is
Repulsion Theory 1 a unique structure resulting from the interplay of several energy factors and that the
following rules can only be a crude attempt to average the various forces.
of Molecules
First, from the electronic configuration of the elements, determine a reason¬
able Lewis structure. For example, in the carbon dioxide molecule, there will
be a total of 16 valence electrons to distribute among three atoms:
:6-"C "6: or :'6"C
•O'
(a) (b)
Note that a Lewis structure says nothing about the bond angles in Ihe molecule
since both (a) and (b) meet all the criteria for a valid Lewis structure.
A structure should now be considered which lets all the electron pairs in the
valence shell of the central atom(s) get as far away from each other as possible.
In the usual cr-7r treatment this usually means ignoring the n bonds temporarily
since they will follow the cr bonds. In carbon dioxide there will be two a bonds
1 Gillespie. R. J. Chem. Soc. Rev. 1992 .21. 59-69. Gillespie. R. ].. Hargittai. I. The VSEPR Model of
Molecular Geometry-. Allyn and Bacon: Boston, 1991.
203
!8
Molecular Orbital Theory
159
It may occur to the reader that it is always possible to bring the orbitals together in
such a way that the overlap is positive. For example, in Fig. 5.8g, h if negative overlap
is obtained, one need only invert one of the atoms to achieve positive overlap. This is
true for diatomic molecules or even for polyatomic linear molecules. However, when
we come to cyclic compounds, we no longer have the freedom arbitrarily to invert
atoms to obtain proper overlap matches. One example will suffice to illustrate this.
There is a large class of compounds of formula (PNX 2 )„ (X = F, Cl, Br),
containing the phosphazene ring system (see Chapter 16). The trimer, P 3 N 3 X 6 , is
illustrated in Fig. 5.9. Note the resemblance to benzene in the alternating single and
double bonds. Like benzene, the phosphazene ring is aromatic, that is, the n electrons
are delocalized over a conjugated system with resonance stabilization. The description
of the 7r bonding in the phosphazene ring, which involves p orbitals on the nitrogen
atoms and d orbitals on the phosphorus atoms, has been a matter of considerable
debate. One view is illustrated in Fig. 5.10, in which the phosphazene ring has been
split open and arranged linearly for clarity. We start on nitrogen atom number one (N,)
and assume an arbitrary assignment of the positive and negative lobes of the p orbital.
The phosphorus atom tt bonds through its d orbitals, and so for the P, atom we draw a
H
H -cAc- H
H /C ^C /C ^H
C S/ CI
Fig. 5.9 Comparison of
bonding in the ring systems
of (a) benzene and (b)
hexachlorotriphospliazene.
(b)
Fig. 5.10 Overlap of the orbitals in the p-p -n system in benzene (a) and the p-d it system in
the phosphazene ring (b). Note the mismatch of orbital symmetry in the latter.
200
5 - Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Problems
5 l Draw Lewis structures for CS,_, PF 3 , SnH 4 , HONH
5.2 Draw Lewis structures for H 2 CO„ HN0 3 , NO. Be(CH 3 h.
5 3 Draw Lewis structures for BF 3 , SF 6 , XeF 2 , PF 5 , IF 7 - flPN rn
„hnf,he sien of the wave function in the rr system of (PNC1 2 ) 4
5.4 Show that there is no mismatch of the sign
in contrast to (PNC1,) 3 .
5.5 Write the MO electron configuration for the NO" ion.
a. What is the bond order?
I, Will the bond length be shorter or longer than in NO?
rr_rrr: r.r_.
orders, bond lengths, and unpaired electrons. . ...
5 - ?
Suggest a macule for which charges completely rnle out resonance.
5 8 Write resonance structures, including formal charges, for 0 3 . S0 3 , N0 2 .
, The assumption was made that the carbon-carbon rr^jond IS Same
in CH,CH 3 . In reality, it is probably somewhat stronger. Discuss.
5.. 0 The NNO -to w„ - «« ^
Would you expect it to be more stable or less stable than NNU. wny
iheOCO arrangement rather than COO ? rND” arc
5U ss" s s,iBh,ly diffcrcm
5.. 2 Calculate' the electronegativity of hydrogen from the ionization potential and the electron
(X = hydrogen or halogen) arc stable, that is. whether the reaction
N 2 + 3X 2 2NX 3
(Appendix E). but you must predict the bond ene gy
5.14 Which do you expect to be more acidic:
CHj —P—OH or
Explain (See Cook, A. G.; Mason. G. W. J. Org. Chem. 1972. J7. 3342-3345.)
MS Table 5.6 ,he electa,Wife of .he nob.e «. - » ^ ta " '
Problems
201
being higher even than those of the halogens. Yet we all know that the noble gases do not
accept electrons from elements of low electronegativity:
Na + A - NaA“
Discuss the meaning of the electronegativities of the noble gases.
5.16 In discussing ionic resonance, AB - AB~, where:
Ip = Ol/W + b>l>i onic
Pauling assumed that ip iomc made a negligible contribution if A = x B - The bond energy of
Cl 2 is 240 kJ mol -1 and the bond length is 199 pm. In the Cl 2 molecule. x A = B - Show by
means of a Bom-Haber-type calculation that the canonical structure. Cf CP, cannot
contribute appreciably to the stability of the molecule. (You may check your answer with
Pauling, L. The Nature of the Chemical Bond, 3rd cd.; Cornell University: Ithaca, NY,
I960; p 73.)
5.17 The energy necessary to break a bond is not always constant from molecule to molecule.
For example:
NCI, - NCI, + Cl A H = -375 kJ mol" 1
ONCI -♦ NO + Cl A H = 158 kJ moM
Suggest a reason for the difference of -200 kJ moM between these two enthalpies.
5.18 From what you know of the relationship between ionization energies, electron affinities,
and electronegativities, would you expect the addition of some d character to a hybrid to
raise or lower the electronegativity; for example, will sulfur be more electronegative when
hybridized sp i or spd~1
5.19 The dipole moment of H—C=C—Cl is in the direction —. Explain, carefully.
5.20 The legend to Fig. 5.20 says “The Itrand 3o MOs are essentially nonbonding." Describe
these nonbonding orbitals more explicitly, perhaps in VB terms.
5.21 Look at Figs. 5.14 and 5.15 carefully. Identify:
a. the nodal planes responsible for the symmetry of the MOs (i.e.. sigma, pi, etc.).
b. any changes in electron density that you can ascribe to bonding versus antibonding
situations.
5.22 Oxygen is more electronegative than carbon and Fig. 5.18a indicates that there is more
electron density on oxygen than on carbon in carbon monoxide. Yet the dipole moment of
CO is quite small (0.373 x lO" 50 C m; 0.112 D) and it is known that the oxygen atom is the
positive end of the dipole'. Explain. Hint: Does a comparison with the isoelectronic
dinitrogen molecule (Fig. 5.18b) help?
5.23 Using the MO treatment of BeH, (page 175) as a starting point, write linear combinations
for the molecular orbitals in CO,.
5.24 Construct a molecular orbital diagram for water using the proper character table and
deriving reducible representations. The Lewis structure for water suggests two equivalent
nonbonding and two equivalent bonding pairs of electrons, but your molecular orbital
diagram should show four nonequivalent molecular orbitals. How can you rationalize this
difference? Which molecular orbitals do you think best represent the two nonbonding pairs
and the bonding pairs? Compare your result with that found in Shustorovich, £.; Dobosh,
P. A. J. Am. Chem. Soc. 1979, 101, 4090.
5.25 The HOMO (a,) of NOJ is somewhat antibonding. On this basis, what predictions would
you make about the N—O bond lengths in NOj. N0 2 , and N0 2 ? How many unpaired
160
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Molecular Orbital Theory
161
Symmetry of
Molecular Orbitals
Molecular Orbitals
in Homonuclear
Diatomic Molecules
d orbital with appropriate symmetry such that the overlap between N, and P, is
greater than zero. We continue with N 2 , P 2 , N„ and P 3 , each time matching the orbital
symmetries to achieve positive overlap. However, when we come to the overlap
between P, and N, (to close the ring) we find that we would like to have the N, orbital
lie as shown on the right, but we have previously assigned it the arrangement shown
on the left. It is impossible to draw the six orbitals in such a way as to avoid a
mismatch or node in the system.
Some of the possible combinations of atomic orbitals are shown in Fig. 5.11. Those
orbitals which are cylindrical^ symmetrical about the internuclear axis are called
(j orbitals, analogous to an j orbital, the atomic orbital of highest symmetry. If the
internuclear axis lies in a nodal plane, a rr bond results. In 5 bonds (Chapter 16) the
internuclear axis lies in two mutually perpendicular nodal planes. All antibonding
orbitals (identified with an ) possess an additional nodal plane perpendicular to the
internuclear axis and lying between the nuclei. In addition, the molecular orbitals may
or may not have a center of symmetry. Of particular interest in this regard are 7r p _ p
orbitals, which are ungerade , and tt_ p orbitals, which are gerade.
Molecules containing two atoms of the same element are the simplest molecules to
discuss. We have already seen the results for the hydrogen molecule (page 157;
Fig 5 7) and for the linear combination of 5 and p orbitals (Fig. 5.8). We shall now
investigate the general case for molecular orbitals formed from two atoms having
atomic orbitals 1$, 2s, 2p, 35, etc.
There are two criteria that must be met for the formation of bonding molecular
orbitals, that is, orbitals that are more stable (lower in energy) than the contributing
atomic orbitals. One is that the overlap between the atomic orbitals must be positive.
Furthermore, in order that there be effective interaction between orbitals on different
atoms, the energies of the atomic orbitals must be approximately the same. For now
we will assume that molecular orbitals will form from corresponding orbitals on the
two atoms (i.e.. Is + Is, 2s + 2s, etc.). We shall soon see that under some
circumstances this assumption will have to be modified. When we combine the atomic
orbitals in this way, the energy levels shown in Fig. 5.12 are obtained. The appropriate
combinations are: 13
ou = Ia + ,5 b
< = Is A “ 1 -Vb
v - 1 , = 2s a + 2s b
° z» = 2 s a - 2s b
0? p = 2 PzA + 2 P 7 .b
02 p = 2 P;A - 2 Pz D
™lp, ~ 2 Py A + 2 Py\i
7T 2pr = 2 P, A + 2 PxB
= Lp > ~ 2P vB
2p., = 2p, A “ 2P,B
The a-,, and of, orbitals correspond to the molecular orbitals seen previously for the
hydrogen molecule. The atomic 2s orbitals form a similar set of <r and <r orbitals. The
13 Symbols A and B represent atoms; . y. and z represent orientation of the p orbitals. Note that the
“atomic orbital labels” (e.g.. Is) can apply only lo molecular orbitals constructed by m.xrng two
orbitals of the same type. We shall see later (page 164) that this is an oversimplification.
(a)
oo
00
p + p
oo
oo
p - p
Fig. 5.11 Symmetry of molecular orbitals formed from atomic orbitals illustrating a (a-d)
and rr (e, f) orbitals, and bonding (a, c, e) and anlibonding (b, d. f) orbitals. 1 lie orbitals are
depicted by electron density sketches with the sign of superimposed.
atomic p orbitals can form <r bonds from direct (“head on”) overlap of the p z orbitals
and two tt bonds from parallel overlap of the p y and p x orbitals. Because the overlap is
greater in the former case, we should expect the covalent energy to be greater also
(page 153), and a bonds are generally stronger than it bonds. Hence the c r 2p orbital is
stabilized (lowered in energy) more than the v 2p orbitals, and conversely the corres¬
ponding antibonding orbitals are raised accordingly. By analogy with atomic electron
configurations, we can write molecular electron configurations. For H 2 we have
H 2 = o u
198 5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Methods of
Estimating Charges:
Electronegativity
Equalization
For many reasons, chemists would like to be able to estimate the charges on the
constituent atoms in a molecule. There have been many attempts to do this, but none
has proved to be completely successful. The ideal way would be to solve the wave
equation for a molecule without the use of any simplifying assumptions, and then to
calculate the electron distribution. Such ab initio calculations are possible for small
molecules 52 but become increasingly difficult as the number of atoms increases. Even
when the calculations are possible, there is not complete agreement among chemists
as to the best way of apportioning the charge density among the atoms in the
molecule. 53
Several workers have suggested semiempirical methods based on elec¬
tronegativity for the estimation of charge. Only one method will be discussed here.
Sanderson 54 has proposed that when a bond forms between two atoms electron
density will shift from one atom to the other until the electronegativities have become
equalized. Initially the more electronegative element will have a greater attraction for
electrons (Fig. 5.34), but as the electron density shifts toward that atom it will become
negative and lend to attract electrons less. Conversely, the atom which is losing
electrons becomes somewhat positive and attracts electrons better than it did when
neutral. This process will continue until the two atoms attract the electrons equally, at
which point the electronegativities will have been equalized and charge transfer will
cease (Fig. 5.35):
Ta = «a + Ma = Xb - “b ~ Ma
«A
“n ~ “a
/; A + l> a
(5.84)
(5.85)
Fig. 5.34 Relation between ionization energy-electron affinity curve (solid line) and inherent
electronegativity (dashed line) for a less electronegative element (A) and a more
electronegative element (B).
52 Figures 5.14-5.18 were obtained from such calculations.
53 Politzcr, P.; Reggio. P. H. J. Am. Chem. Soc. 1972. 94. 8308-8311. Evans. R. S.; Huheey. J. E.
Chem. Phys. Lett. 1973, 19. 114-116.
54 Sanderson. R. T. J. Chem. Etluc. 1954, 31, 2-7. See also Footnote 48.
Electronegativity 199
Fig. 5.35 Superposition of
ionization energy-electron
affinity curves for a more
electronegative (B) and less
electronegative (A) element.
The common tangent ( =
equalized electronegativity)
is given by the dashed line.
The partial charges in the HC1 molecule may be estimated with Eq. 5.85 by using
the appropriate a and b values from Table 5.6: a H = 7.17, = 12.84, a ri ,.,, =
12-15, and 11.55. CKI4 ' 1 ' 1
= 12.15 - 7.17
H 11.55+12.84
0.20
(5.86)
The charge estimated by this method is often different from a similar estimate based on
dipole moments. If the total ionization energy (including the electron affinity) were the
only energy involved in the charge distribution, Eq. 5.85 would be rigorously correct.
In a molecule, however, other energy terms are important. The exchange energy
associated with the overlap of orbitals will be reduced if the charge transfer is too
great. The Madelung energy (so named because of resemblance to that found in ionic
crystals) resulting from the electrostatic attraction of A + for B _ (within the molecule)
tends to increase ionicity. These energies tend to cancel each other in effect because
they work in opposite directions, so Eq. 5.85 can be considered a useful, qualitative
approximation.
Although there is no universal agreement on the “real" charges in molecules,
various attempts have been made to improve upon simple electronegativity calcula¬
tions. One method is to estimate the exchange and Madelung energies by simple
bonding models, and then to use them to adjust the values obtained by the elec¬
tronegativity equalization method. This modification has been found to correlate well
with some ab initio calculations for some simple molecules. 53
There is a maxim that when there are many treatments for a disease, none of them
is completely adequate. This same idea could be applied to electronegativity in view of
the many attempts to define and quantify it. Nevertheless, bond energies, polarities,
and the inductive effect are fundamental to much of inorganic, organic, and physical
chemistry, hence the efforts applied to electronegativity theory. While there is as yet
no complete agreement on all aspects of electronegativity, defining it some way in
terms of the energetics of the valence electrons is generally accepted as the best
approach, although the last word has undoubtedly not yet been said on the matter.
162
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Fig. 5.12 Simplified
molecular orbital energy
levels for diatomic
molecules of elements in
the second period,
assuming no mixing of s
and p orbitals. The three 2 p
orbitals are degenerate, that
is, they all have the same
energy and might also be
shown as-.
The molecule shown is
dioxygen.
Using Fig. 5.12 as a guide, we can proceed to build up various diatomic molecules
in much the same way as the aufbau principle was used to build up atoms.
I. Molecules containing one to four electrons. We have already seen the H,
molecule in which there are two electrons in the a lt orbital. Two bonding
electrons constitute a chemical bond. The molecular orbital theory does not
restrict itself to even numbers of bonding electrons, and so the bond order is
given as one-half the difference between the number of bonding electrons and
the number of antibonding electrons:
Bond order = J (N h - N„) (5.40)
The molecule He 2 is unknown since the number of antibonding electrons
(2) is equal to the number of bonding electrons (2) and the net bond order is
zero. With no bond energy 14 to overcome the dispersive tendencies of entropy,
two helium atoms in a “molecule" will not remain together but fly apart. If it
existed, molecular helium would have the electron configuration:
He 2 = cr] s <r
14 Actually, if the calculation is made carefully, the bond energy of He 2 is positive. If overlap is not
neglected in the calculation, the antibonding orbital is more destabilizing than the bonding orbital is
stabilizing, and so He 2 has a repulsive energy forcing it apart. This is another aspect of the Pauli
principle—two electrons of the same spin cannot occupy the same region of space.
Molecular Orbital Theory
163
II helium is ionized, it is possible to form diatomic helium molecule-ions,
He 2 . Such a molecule will contain three electrons, two bonding and one
antibonding, for a net bond order of one-half. Such a species, although held
together with only about one-half the bonding energy of the hydrogen mole¬
cule, should be expected to exist. In fact it does, and it has been observed
spectroscopically in highly energetic situations sufficient to ionize the helium.
That it is not found under more familiar chemical situations as, for example, in
salts, He-TX - . is not a result of any unusual weakness in the He—He bond,
but because contact with just about any substance will supply the missing
fourth electron with resultant conversion into helium atoms.
Isoelectronic in a formal sense, but quite different in the energies involved
is the Xe 2 ion believed to exist in certain very acidic solvents (see Chapter 17).
The energetics of the situation are not completely understood, but presumably
the much lower ionization energy of xenon can more readily be compensated
by the solvation energy of the polar solvent, thus stabilizing the Xe 2 cation.
Lithium and beryllium. Two lithium atoms contain six electrons. Four will fill
the cr u and o s orbitals with no bonding. The last two electrons will enter the
a 2 s orbital, giving a net bond order of one in the Li 2 molecule. The electron
configuration will be
Li, = KKo-i
where K stands for the K (1 j:) shell.u
Eight electrons from two beryllium atoms fill the four lowest energy levels,
a \s' o’!' °Zf yielding a net bond order of zero, as in He,, with an electron
configuration of:
Be, = KKoj, irJ
Likc the dihelium molecule. Be, is not expected to exist. The experimental
facts are that lithium is diatomic in the gas phase but beryllium is monatomic.
Oxygen, fluorine, and neon. These three molecules can be treated with the
same energy diagram that we have been using for other diatomic molecules of
the second-row elements. As we shall see shortly, the intervening molecules,
B,, C 2 , and N,. require additional considerations, which lead to an alteration in
the relative energies of the molecular orbitals.
I he oxygen molecule was one ol the first applications of molecular orbital
theory in which it proved more successful than valence bond theory. The
molecule contains sixteen electrons. Four of these lie in the v u and cr (
orbitals, which cancel each other and may thus be ignored. The next four
electrons occupy er,, and it?, orbitals and also contribute nothing to the net
bonding. The remaining eight electrons occupy the os,,, ir 2p and n 2p levels
giving as the electron configuration:
O, = KKoi^jal p nt„n 2 2 p
However, examination of the energy level diagram in Fig. 5.12 indicates that
the 7 r 2p level is doubly degenerate from the two equivalent n orbitals, v 2p and
15 Thc inner shells of core electrons are often abbreviated since no net bonding takes place in them.
The symbols used. K. L. M, etc., refer to the older system of designating the principal energy
levels, n = I (K), n = 2 (L). etc. Thus Na 2 = KK LL <r 3 ,2.
196
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Group
Electronegativity
phorus, and sulfur do not fit well into such simple schemes. Tte a , because
hybrids are often some nonintegral mix of 5 and p character. Methods have been
proposed for determining hybridizations from bond angles, but th ^ are approve
at best (see Chapter 6). Recently. Bratseh^ has suggested a purely numerical rule
based orfan extension of the hybnd properties of the
each row. Elements with group numbers. N - 1 GA. D. 2 (IIA _), ■ '
(IV A 14) form hybrids of the type sp »-t. For the nonmetals to the ngh. of he
periodic table. Bratseh suggests working hybridizations of nttrogeu Graup VA U5> -
methane, and the hybridization must be sp\
I, is often convenient to have an estimate of the inductive ability
erouD As we have seen previously, we cannot use a single value of carbon ( 2.5) to
represent 3 the electronegativity of carbon in both CH ancI CF„ The ele cW negahvtnr
of these two groups will be the electronegativity of carbon as it is adjusted by
presence of three hydrogen or three fluorine atoms. Estimation of group elec, ™ neg ^
tivities has been approached from a variety of ways. Organic chemists have developed
S of substituent constants from kinetic data- and these have proven use adm
certain inorganic systems as well. Other values have been obtained from physical
measurements of electronic effects, and calculated directly from atomic electronega-
“ Some comparative values are listed in Table
details of the various methods, but note that there is general agreement, and that wo
simple rules-of-thumb hold for group electronegativities: (1) The inherent group elec¬
tronegativity flG , is approximated by the simple average of the inherent atomic
dectron gativitie . (2) The charge coefficient, is an average as well, but aversely
tro^Z, to the number of atoms in the group. Thus the electronegativity of a
group is given by:
ATc = fl o + /; c 5 g
l- %
It is intuitively reasonable that these values should be an average over the n values of
!he T. ■ ■ nth atom. In the same way. the charge coefficient should be inverse l
proportional (i.e., the charge capacity should be directly proportional) to the number^
n, of atoms over which the charge may be spread. This is the most < m P° r ‘° n 'P r ° p ’ y
of group electronegativities: Groups are "superatoms capable °f a ^' n tt £™
amount of positive or negative charge. This means that groups can donate or accept
49 Bratseh. S. G. J. Chem. Educ. 1988, 65. 34-41.
5 0 For a review of various sets of substituent constants see Lowry. T. H, ^R.chanlson ^ •
Mechanism and Theory in Organic Chemistry. 3rd ed., Harper & Row. New York.
3i For methods of calculating group electronegativities from ^
of the constituent atoms, see Bratseh, S. G. J. Chem. Educ. 1985, 62. 101-103.
Electronegativity
197
Table 5.7 _
Estimates of the
electronegativities of
some common groups
Experimental
Calculated values 0 values 1
Group
Xc
Pauling
scale
a G
(eV)
b G
(eV)
XG
Pauling
scale
ch 3
2.31
7.44
3.11
2.34
2.30
7.45
4.64
ch 2 ch 3
2.32
7.48
1.77
—
2.32
7.52
3.58
CFj
3.47
12.85
3.90
3.35
3.32
10.50
5.32
CCI 3
2.95
10.24
2.88
3.03
3.19
10.12
4.33
COOH
3.04
10.68
3.41
2.8-3.5
3.36
10.62
5.43
CN
3.32
12.09
6.23
3.3
3.76
11.83
6.47
nh 2
2.47
8.10
4.31
1.7-3.4^
2.78
8.92
5.92
nf 2
3.53
13.16
5.31
—
3.78
11.89
6.77
OH
2.82
9.61
7.02
2.3-3.9'
3.42
10.80
8.86
OCHj
2.52
8.30
2.59
—
3.12
9.90
5.74
Calculated by methods of electronegativity equalization. Two
values arc given to indicate variation. Roman values are from
Bratseh, S. G. J. Chem. Educ. 1988. 65. 223-227. Italic values
are from Watts. J. C., Ph.D. Dissertation, University of
Maryland. College Park, 1971. See also Footnote b.
t> Various experimental methods, mostly infrared spectroscopy.
For details, see Huheey, J. E. J. Phys. Chem. 1965, 69.
3284-3291.
«• xhc wide range of values results from the possibility of
conjugation of lone pair of it electrons with the remainder of
the molecule.
charge better than would be indicated by their inherent electronegativities (a values)
alone. For example, the methyl group is slightly, though not significantly, more
electronegative than hydrogen. Yet the methyl group is generally considered a better
donor than is the hydrogen atom. It is Ihe greater charge capacity, which results from
the ability to spread the charge around that allows Ihe methyl group to donate more
electron density than the smaller hydrogen atom:
H 8
H 8 — C 8 —
I
H 8+
164
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
ir p . Hund's rule of maximum multiplicity predicts that the two electrons
entering the tt level will occupy two different orbitals, so the electronic
configuration can be written more explicitly as
0 2 = KKo-fj tTs °2p 17 2/> ‘ n 2p x 7r 2p r
This has no effect on the bond order, which is still two [|(6 - 2)], as
anticipated by valence bond theory. The difference lies in the paramagnetism
of molecular oxygen resulting from the two unpaired electrons. (In this regard
0 2 is analogous to atomic carbon in which the last two electrons remained
unpaired by entering different, degenerate orbitals.) The simple valence bond
theory predicts that all electrons in oxygen wiil be paired; in fact, the formation
of two bonds demands that the maximum number of electrons be paired. This
is the first case of several we shall encounter in which the stress placed on
paired bonding electrons is exaggerated by the valence bond theory. The
molecular orbital theory does not require such pairing as it merely counts the
number of bonding versus antibonding electrons. The experimentally mea¬
sured paramagnetism of 0 2 confirms the accuracy of the MO treatment.
For the fluorine molecule, there will be a total of 18 electrons distributed:
F 2 = KK <4s<rt 2 <ri p v 4 2 P ^J
The net bond order is one, corresponding to the cr bond, and agreeing with the
valence bond picture.
The addition of two more electrons to form the Ne 2 molecule will result in
filling the last antibonding orbital, the o-J p orbital. This will reduce the bond
order to zero and Ne 2 , like He 2 , will not exist.
Boron, carbon , and nitrogen. According to Fig. 5.12, the B ; molecule would
be predicted to have a single a bond and be diamagnetic. Experimentally the
B-, molecule is found to have two unpaired electrons. The C 2 molecule would
be predicted to have an electron configuration KKer; 5 o-?;<7;,,7r 2( ,7r 2/ , and be
paramagnetic. The experimental evidence indicates that the ground state of C 2
is diamagnetic.
The problem here is that in constructing Fig. 5.12 mixing was allowed only
between orbitals on atoms A and B that were identical in energy. Actually,
mixing will take place between all orbitals of proper symmetry, inhibited only
by the fact that if the energy mismatch between orbitals is large, mixing will be
reduced. We are therefore justified in dismissing mixing between the Is and 2s
orbitals. The energy difference between the 2s and 2 p orbitals is less and varies
with the effective nuclear charge. With a larger Z, as in fluorine, the energy
difference is greater and mixing may again be neglected. The difference in
energy between the 2s and 2 p levels dramatically increases from about 200 kJ
mol -1 in the lithium atom to about 2500 kJ mol -1 in fluorine. In the case of the
elements to the left of the series, the lower effective nuclear charge allows the
2s and 2 p orbitals to come sufficiently close to mix. This phenomenon is the
equivalent of hybridization in the valence bond theory.
Another way to view this phenomenon is to ignore s-p mixing in the initial
construction of molecular orbitals but then recognize that molecular orbitals of
the same symmetry will interact if they are close enough in energy. Thus the
cr (2s) and <r R (2 p) molecular orbitals in a molecule such as B 2 will interact. As a
result, the lower-energy orbital [cr^(2s)] will be stabilized while the higher-
energy one [a (2p)] will become less stable. This leads to a reversal in the
fc -Si
Molecular Orbital Theory 165
B B, B
/-\6o„
/ \
/ s
/ \
/ -- 2n \
/ ' N \
/ -- x v
Fig. 5.13 Correct molecular orbital energy levels for early elements of the first long row.
Some mixing (hybridization) has occurred between the 2s and 2 p orbitals. Note that it is
somewhat more difficult "to keep books" and determine the bond order here than in
Fig. 5.12: 3cr t . and lw„ are clearly bonding (they lie below the atomic orbitals contributing to
them); 4<r„ and 5 <r„ are essentially nonbonding since they lie between the atomic orbitals
contributing to them and roughly symmetrically spaced about the “center of gravity." The
maximum net bond order is therefore one cr bond plus two tt bonds. The electronic
configuration shown is for the B; molecule. Note the unpaired pi electrons.
energy ordering of the ir u (2p) and cr K (2p) molecular orbitals (Fig. 5.13) com¬
pared to the case for molecules such as F 2 , where essentially no mixing occurs.
There will also be some interaction between the rr(2.r) and the <r{2p) orbitals,
with the lower-energy orbital becoming stabilized and the higher-energy orbital
being destabilized. However, because these orbitals do not approach each
other closely in energy, the interaction will be negligible. Note in Fig. 5.13 that
it is no longer appropriate to use labels such as 2 a- and 2p to identify the origins
of molecular orbitals, so we merely label them according to their symmetry and
number them in order from the most to the least stable. Thus (r k ,(2s) becomes
3cr g , o- g (2p) becomes 5 tr K , etc.
The magnetic properties of B 2 and C 2 provide strong experimental verifica¬
tion that their electron configurations are based on Fig. 5.13 rather than on Fig.
5.12. For N 2 (fourteen electrons), either diagram would predict a triple bond
(one cr and two tt) and diamagnetism, consistent with physical measurements.
Experimental evidence supporting one configuration over the other for N, has
been sought in photoelectron spectroscopy. The method involves ionizing
electrons in a molecule or atom by subjecting them to radiation of appropriate
energy. When ionizing photons in the ultraviolet range are used, valence-level
electrons are ejected, whereas X rays can be used to ionize inner, core
electrons. The energy of the impinging photons is known from their frequency
(E = hv), and the kinetic energy (£) of the ejected electrons can be measured.
194
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Other Methods of
Estimating
Electronegativity
monofluoride. CIF. is aboul 255 kJ moP 1 , greater than that of either CI-, (242 kJ mol -1 )
or F 2 (158 kJ mol" 1 ). Pauling suggested that molecules formed from atoms of different
electronegativity would be stabilized by ionic resonance energy resulting from reso¬
nance of the sort:
Aab = + H a + b- + ci /'a-b + (5.79)
For molecules in which atoms A and B are identical, b = c « a (see page 141 for the
H : molecule), and the contribution of the ionic structures is small. If B is more
electronegative than A. then the energy of the contributing structure A + B~ ap¬
proaches more nearly that of the purely covalent structure A—B and resonance is
enhanced. On the other hand, the energy of B + A" is so prohibitively high that this
structure may be dismissed from further consideration. For a predominantly covalent,
but polar, bond, a > b » c. The greater the contribution of the ionic structure (i.e.,
the closer it comes to being equivalent in energy to the covalent structure) the greater
the resonance between the contributing structures and the greater the stabilizing
resonance energy. Pauling suggested that electronegativity could be estimated from
calculations involving this ionic resonance energy. The interested reader is referred to
Pauling's discussions of the subject for the details of the methods he used. 47 but an
outline follows.
Pauling assumed that if the CIF bond were completely covalent, its bond energy
would be simply the average of the Cl 2 and F, bond energies:
242 + 158
2
200 kJ mol -1
(5.80)
Hie ionic resonance energy is the difference between the experimental bond energy of
CIF, 255 kJ mol -1 , and the calculated value, 200 kJ mol -1 , or 55 kJ mol -1 . Pauling
defined the difference in electronegativity between chlorine and fluorine as the square
root of the ionic resonance energy.-3
(5.81)
This may be compared with the tabular value for the difference in electronegativities
of fluorine and chlorine. 3.98 - 3.16 = 0.82. which is based on many experimental
data, not just the single calculation illustrated here. Once again, the details of the
calculation are not particularly important since Pauling's method of obtaining elec¬
tronegativity data is probably mainly of historical interest.-- The concept of
covalent-ionic resonance is still quite useful, however. Unfortunately, as alternative
methods or treating electronegativity have developed, the fact that a bond with partial
ionic character can be stronger than either a purely covalent or purely ionic bond has
often been overlooked. Energies associated with electronegativity differences can be
useful in accounting for the total bonding energies of molecules.
Many other methods have been suggested for determining the electronegativity values
of the elements. Only one general method will be discussed here. It is to consider
electronegativity to be some function of size and charge. These methods differ among
42 Pauling. L. The Nature of the Chemical Bond. 3rd ed.: Cornell University: Ithaca. NY. I960;
Chapter 3.
" The conversion factor96.5 kJ mof'cV 1 is included because Pauling set up his scale based on bond
energies measured in electron volts.
44 Not all chemists would agree with this statement.
Electronegativity
195
themselves only in the choice of function (energy, force, etc.) and the method of
estimating the effective charge. Allred and Rochow- defined electronegativity as the
electrostatic force exerted by the nucleus on the valence electrons. They used effec¬
tive nuclear charges obtained from Slater’s rules and obtained the formula:
Aar = (3590 Z/r) + 0.744 (5>82)
where r is the covalent radius (pm). The Allred-Rochow scale has been widely
accepted as an alternative to Pauling's thermochemical method for determining elec¬
tronegativities. Allred-Rochow values are listed in Table 5.6.
Another definition that is based on size and charge, but in a unique way, is the
definition of Sanderson,-’ which is based on relative electron density. This method has
never been accepted widely, although Sanderson has applied it successfully to a
variety of problems. 4 and his values were the first to illustrate the interesting elec¬
tronegativity properties of the posttransition elements (see Chapter 18).
Values for each of the electronegativity systems discussed here are listed in Table 5.6.
With more than one valid system available, the choice of the "best" one is not always
easy. We can arbitrarily divide the various methods into two groups. One consists of
the methods that depend on orbital energies: the Mulliken-Jafte theory, density
tunctional theory, and the spectroscopic theory. They may be termed “theoretical” or
“absolute" scales because they are based only on the fundamental orbital energies of
isolated atoms. The other scales are “empirical" and "relative” because they utilize
experimentally obtained data such as enthalpies of formation, covalent radii, etc. Both
gioups of systems have advantages. In general, the energy scales are more satisfying
because they are. in a sense, more fundamental and basic. The empirical methods also
have an advantage, resulting directly from their methods of derivation. In other
words, variables such as hybridization, etc., are often “built in" as long as the atom
under consideration is in a fairly typical environment. Each of the empirical methods
has advantages and disadvantages, adherents and detractors, and they do not really
differ greatly among themselves. If the situation is sufficiently nonspecific to make it
necessary to use an empirical system, it probably will not make a great deal of differ-
ence which is chosen. However, one must be consistent and avoid picking the value
for one element from Pauling, another from Allen, and a third from Allred-Rochow
By judicious mixing or systems like this, one could probably “prove" anything!
Choice of Hybrids Choosing the appropriate hybridization for use with Mulliken-Jaffe electronegativities
for Nonmetals sometimes presents problems. Only the elements to the left in the periodic table have
unambiguous hybridization assignable by structure. Thus few would argue with an
assumption of sp 7 for boron in its tricovalent compounds, and organic chemistry is
based on the successful assumption of digonal (sp), trigonal (jp 7 ), and tetrahedral (sp)
hybridizations for carbon. However, the hybridizations of nitrogen, oxygen phos-
45 Allrcd - A - L -; Rochow, E. G. J. I norg. Nucl. Chem. 1958, 5. 26-1-268.
Allrcd and Rochow counted all of the electrons in a particular atom as shielding the electron coming
from another atom, so their 7- values arc 0.35 higher than those obtained by the usual application of
Slater's rales. Such dili'crences are unimportant as long as one is consistent in the application: They
will be absorbed into the appropriate parameters for fitting Eq. 5.82.
47 Sanderson. R. T. J. Chem. Educ. 1952. 29, 539-544; 1954. 31, 2-7.
■“ Sanderson. R. T. Polar Covalence-, Academic: New York. 1983: Simple Inorganic Substances-.
Kneger: Malabar. FL, 1989.
Choice of
Electronegativity
System
166
5-Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
The difference between these two quantities (IE) is the amount of energy that
must be provided to overcome the attraction of the nuclei for the ionized
electron:
IE = E iw ~ E k (5.41)
The technique thus can provide valuable information regarding energies of
occupied molecule orbitals in a molecule since, by Koopmans’ theorem, IE =
-E„ where E n is the energy of an atomic or molecular orbital. The theorem
assumes that orbital energies are the same in the ion (N 2 in this case) produced
in the photoelectron experiment as in the original molecule. The photoelectron
spectrum of N, shows that the IE values for the Stx K and I -n u electrons are
about 15.6 and 16.7 eV, respectively, giving - 15.6 and - 16.7 eV as the orbital
energies and suggesting that sufficient s-p mixing (or molecular orbital interac¬
tion) occurs in this molecule to make the 5<z ff level higher in energy than the
Itt-,,. 16 However, ab initio calculations reveal that these two levels are quite
close in energy and may undergo a reversal in their respective orders during
the photoionization process. 17 In other words, the Koopmans approximation
cannot be assumed to hold for N,.
Some molecular orbital results for first- and second-row diatomic molecules, as
well as relevant experimental data, are summarized in Table 5.5.
Bond Lengths and Further support for the MO descriptions presented in the preceding section comes
Ionization Energies from investigation of the bond lengths in some diatomic molecules and ions. For
example, consider the oxygen molecule. As we have seen previously, it has a double
bond resulting from two cr-bonding electrons, four 7r-bonding electrons, and two
Table 5.5
Molecular orbital predictions __E xperim en tal
Unpaired Bond energy Dia- or Bond length
Molecule Electro ns J4ot bonds electrons (Id moM)° paramagnetic (pm)
H ? 2 I 0 432.00 D 74 2
He 2 4 () o _ _ _
e '2 6 I 0 105 D 267 2
Be, 8 0 0 _ _
B 2 10 ] 2 293 P 158,9
C 2 12 2 0 602 D 134
N 2 14 3 0 941.69 D 109.8
°2 16 2 2 493.59 P 120.7
E 2 18 I 0 155 D 141 8
Ne, 20 0 0 _ _ _
“ Sec discussion of bond energies in Appendix E.
Ift Gardner, J. L.; Samson, J. A. R. J. Chem. Phys. 1975, 62. 1447-1452.
17 Ermlcr. W. C.; McLean. A. D. J. Chem. Phys. 1980. 73. 2297-2303. DeKock. R. L.; Gray. H. B,
238 242 <,/ S ' r " C "” e anti Bont/ "W ; Benjamin/Cummings: Menlo Park, CA, 1980: pp 212-217 and
Molecular orbital results for
selected diatomic molecules
Molecular Orbital Theory
167
7 r-antibonding electrons. The bond length is 121 pm. Addition of two electrons to the
oxygen molecule results in the well-known peroxide ion, 0 2 _ :
O, + 2e“ -♦ (5.42)
According to Fig. 5.12 these two electrons will enter the rr orbitals, decreasing the
bond order to one. Since the compressive forces (bond energy) are reduced and the
repulsive forces (nonbonding electron repulsions) remain the same, the bond length is
increased to 149 pm. If only one electron is added to an oxygen molecule, the
superoxide ion. O,, results. Because there is one less antibonding electron than in
0 ;~, the bond order is 1$ and the bond length is 126 pm.
Furthermore, ionization of O, to a cation:
°2 0, + + e- (5 43)
causes a decrease in bond length to 112 pm. The electron ionized is a rr antibonding
electron and the bond order in 0 2 is 24.
The nitric oxide molecule, NO, has a bond length of 115 pm and a bond order of
21. Ionization to the nitrosyl ion, NO + , removes an antibonding t r electron and
results in a bond order of three (isoelectronic with N,) and a shortening of the bond
length to 106 pm. In contrast, addition of an electron (to a it orbital) causes a
decrease in bond order and an increase in bond length.
The fact that the formation of the nitrosyl ion results from the removal of an
imlibonding electron makes the ionization energy (IE) for the reaction
NO -► NO + + e - IE = 894 kJ mol -1
(5.44)
lower than it is for the unbound atoms of nitrogen (IE = 1402 kJ mol - ') and oxygen
(IE = 1314 kJ mol '). The nitrosyl ion is thus stabilized and exists in several
compounds, such as NO + HSO^ and NO f BFj .
A comparison of the ionization energies of molecular oxygen and nitrogen illus¬
trates the same point. The ionization energy of molecular nitrogen is 1503 kJ mol ',
greater than that of atomic nitrogen, in agreement with Fig. 5.13 that a bonding (and
therefore more stable) electron is removed. In contrast, the ionization energy of
molecular oxygen is 1164 kJ mol ', less than that of atomic oxygen. In this case the
ionized electron is removed from an antibonding orbital, requiring less energy.
Electron Density in The approximate shapes of molecular orbitals have been given previously (Fig. 5 . 11 ).
Molecules Li 2 These give a general idea of the electron distribution in diatomic molecules. Wahl" 1
through F 2 has computed electron density contours for the molecular orbitals of diatomic mole¬
cules for H, to Ne 2 . Some examples are shown in Figs. 5.14 and 5.15. Note particu¬
larly that: (I) bonding orbitals cause an increase in electron density between the
nuclei; (2) antibonding orbitals have nodes and reduced electron density between
nuclei; and (3) inner shells (1 j in Li. for example) are so contracted from the higher
effective nuclear charge that they are nearly spherical with almost no overlap and thus
contribute little to the overall bonding. We are thus justified in ignoring these core
electrons in determining the molecular electron configuration (page 163).
Molecular Orbitals In developing a molecular orbital description for heteronuclear diatomics, we need to
in Heteronuclear lake into account the fact that different types of atoms have different capacities to
Diatomic Molecules attract electrons. The ionization potential of fluorine is considerably greater than that
,s Wahl, A. C. Science 1966, 151. 961.
5 - Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
192
=ss;=“=r==s
^ and «h« hydrogen atom J ^aSene has hydrogen
electronegativity of carbon with 33/c Ca2+c = C 2 - form rather easily. In
SiTS ZXy bSS’cibon atom (50% , character) has about the same
-TSSSS of Ltes"“arcuon of ,he
aliphatic amrne, pyridine.
and nitriles, exhibits this property:
-25% 5
Me 3 N: + H 2 0 Me 3 NH + OH-
P t = 4 -2
(5.72)
33% 5
tQNt.H.O^QNH-fOH-
pK h = 8.8
(5.73)
50% s
M eC = N:+ HjO No reaction
(5.74)
The electronegativity of the nitrogen atom increases as the , character of the hybridi-
been fonnd ,ha, strained ring systems of the “ 7^.
S,“™n% n r;oV,r„l P gIn C a,o m and reduced basicity co.pared to the
unconstrained analogues. 41 . rti . c u arKC
induced ethyl iodide hy^y^es as
expected for alkyl halides, but trifluoromethyl iodide gives unusual products.
CH 3 I + OH" - CHjOH + I" (
CFjI + OH" -♦ CF 3 H + 10"
properties.
39 Sec Footnote 38. „ , , „
Streitwieser. A.. Jr,. Ziegler. G. R, Mowery. P. C, Lewis. A,. Lawler. R. G. J. Am. Chem. Soc.
90, 1357-1358.
•" Markgraf. J. H.; Rail. R. J. J. Org. Chem. 1972. 37, 717-718.
Electronegativity
193
Fig. 5.33 Biphenylene. Shaded orbitals have increased p character; hence unshaded orbital
has increased 5 character, increased electronegativity. [From Streitwieser. A., Jr, Ziegler.
G. R.; Mowry, P. C.; Lewis, A.; Lawler, R. G. J. Ain. Chem. Soc. 1968, 90, 1357 1359.
Pauling's
Electronegativity
Although the products differ considerably in these two reactions, presumably the
mechanisms are not drastically different. The negative hydroxide ion attacks the most
positive atom in the organic iodide. In methyl iodide this is the carbon atom (i > Afc’
and the iodide ion is displaced. In the trifluoromethyl iodide the fluorine atoms induce
a positive charge on the carbon which increases its electronegativity until it is greater
than that of iodine and thus induces a positive charge on the iodine. The latter is thus
attacked by the hydroxide ion with the formation of hypoiodous acid, which then loses
an H + in the alkaline medium to form 10".
It may seem paradoxical that the carbon atom can induce a greater positive charge
on the iodine than that which the carbon itself bears but a simple calculation based on
electronegativity equalization (see pages 198-199) indicates that the charges are S, =
0.21, S c = +0.15, and 5 F = -0.12. Although it is exceedingly unlikely that the real
charges have these exact values, they arc probably qualitatively accurate. 1 his is an
example of the importance of the ability of an atom to donate or accept charge. Iodine
is the most polarizable atom in this molecule. The large, soft, polarizable nature ol the
iodine atom allows it to accept the larger charge.
A similar reaction of more interest to inorganic chemists is the reaction between
carbonylate anions and alkyl iodides:
CH 3 I + Na + [Mn(CO) s ]~ - Nal + CH,Mn(CO), (5-77)
2CFjI + Na + [Mn(CO) 5 ]" -» Nal + C 2 F„ + Mn(CO) 5 l (5-78)
In this reaction also, the polarity of the C—1 bond depends upon the substituents on
the carbon atom. .
It is an interesting paradox that most of the examples of variable electronegativity
come from organic chemistry, although it is probable that electronegativity variation is
much more important in inorganic chemistry. For example, there must be a large
difference in electronegativity between Cr(IIl) in [Cr(NH 3 ) h ]‘ and ip- 1 Cr(Vl) in
CrOj + . The fact that it is not so well documented as yet speaks to the difficulties of
treating the electronegativities of transition metals. Some examples that will be
discussed include the basicity of NH 3 versus NF 3 , the oxidation state of oxyacids, the
tendency of metals to hydrolyze, and the effect of ring strain on basicity (Chapter 9).
Pauling observed that bonds between dissimilar atoms were almost always stronger
than might have been expected from the strength of bonds of the same elements when
bonded in homonuclear (nonpolar) bonds. For example, the bond energy of chlorine
168
5 - Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
of lithium Likewise, the electron affinity of fluorine is strongly exothermic out inai o.
Swum "much less so. and some metals have endothermic electron affimt.es. A bond
between lithium and fluorine is predominantly ionic, consisting (to a firsl
lion) of transfer of an electron from the lithium atom to the fluonne atom Hydrogen is
nature in contrast to the bonds discussed prev.ously (page 167). Charge y
Molecular Orbital Theory
169
<0
Fig. 5.15 Electron density contours for various orbitals in the 0 2 molecule, (a) <7 2 .,i <b) ozt'.
(c) tr,; (d) 7 r 2p ; (e) (f) total electron density. [From Wahl, A. C. Science 1966, 151, 961.
Reproduced with permission.!
distributions for these molecules are shown in Fig. 5.16, which may be compared with
the nonpolar, homonuclear bonds in Figs. 5.14 and 5.15. Cross-sectional density
profiles of several homonuclear and heteronuclear molecules are shown in Fig. 5.17.
Although LiF gives an appearance of being (again, to a first approximation) an ion
pair, in HF the hydrogen atom is deeply embedded in the electron cloud of the fluoride
ion as predicted by Fajans' rules (Chapter 4).
190
5 Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Table 5.6
Electronegativities of the elements (Continued)
Mulliken-
-Jaffe, 0 °
Allred-
Orbital
a
b
Pauling
Sanderson
Rochow
Allen
or
Pauling
Volts /
Element
-jj.
f
AR d
A'spec
hybrid
scaled
Volts
electro
Ta
1.5
1.33
W(II)
2.36
0.73
W(III)
0.98
1.40
W(VI)
1.67
Re
1.9
1.46
Os
2.2
1.52
Ir
2.20
1.55
Pt
2.28
1.44
Au
2.54
1.41
S
1.87
5.77
6.92
Hg
2.00
2.20
1.44
sp
1.81
5.55
5.81
TI(I)
1.62
0.99
p
0.76
2.50
5.92
Tl(III)
2.04
2.25
1.44
sp 2
1.96
6.08
6.40
Pb(II)
1.87
1.92
1.5
p
1.16
3.52
7.47
Pb(IV)
2.33
2.29 h
sp 3
2.41
7.82
5.32
Bi
2.02
2.34
1.67
20% j
2.15
6.81
8.09
Po
2.0
1.76
17%r
2.48
8.14
8.81
At
2.2
1.90
14%$
2.85
9.76
10.03
P
2.55
Rn
2.06u
P
2.12
6.92
_
Fr
0.7
0.86
s
0.68
2.30
3.40
Ra
0.9
0.97
sp
0.92
2.88
3.69
Ac
1.1
1.00
Th
1.3
1.11
Pa
1.5
1.14
U
1.38
1.22
Np
1.36
1.22
Pu
1.28
1.22
Am
1.3
(1.2)
Cm
1.3
(1.2)
Bk
1.3
(1.2)
Cf
1.3
(1.2)
Es
1.3
(1.2)
Fm
1.3
(1.2)
Md
1.3
(1-2)
No
1.3
(1.2)
" Bratsch, S. G. J.
Chem. Educ. 1988, 65, 34-41. See text for definitions of a and b.
fc Values io two decimal places are by Allred, A. L. J. lnorg. Nucl. Chem. 1961. 17, 215, unless otherwise noted
Values to one decimal place are from Pauling, L. The Nature of the Chemical Bond, 3rd ed.; Cornell University
Ithaca, NY, I960; p 93.
r Sanderson, R. T. Simple Inorganic Substances ; Kricgcr: Malabar, FL. 1989; p 23, unless otherwise noted.
J Allrcd - A - L -i Rochow. E. G. J. lnorg. Nucl. Chem. 1958, 5. 264, except for italicized values from Little, E. J„
Jr.; Jones, M. M. J. Chem. Educ. 1960, 37, 231, or otherwise noted. Values in parentheses arc estimates.
' Allen, L. C.J. Am. Chem. Soc. 1989, III, 9003-9014.
'V« T.35(T M j) i/2 - 1.37 (from ref. a).
Allen, L. C.; Huhcey, J. E. J. lnorg. Nucl. Chem. 1980, 42, 1523-1524.
Sanderson, R. T. Polar Covalence-, Academic: New York, 1983; p 41.
Electronegativity
191
Variation of
Electronegativity
Two recent approaches should be mentioned. In one, Parr and others have
followed Mulliken by defining electronegativity in terms of ionization energy and
electron affinity. They have also advanced the idea of electronegativity in terms of
density functional theory. This is a close parallel to the Mulliken-Jaffe system in its
emphasis upon the first and second derivatives of the ionization potential-electron
affinity energy curves. In addition, it provides quantum mechanical support for the
intuitively appealing idea of electronegativity equalization (see page 198). It differs
only to the extent that those using this method have tended to use ground stale values
instead of valence state values.
More recently Allen” has proposed that electronegativity be defined in terms of
the average one-electron energy of valence shell electrons in ground-state free atoms
which may be obtained spectroscopically. This quantity is termed the configuration
energy:
CE
m e p + ne s
m + n
(5.71)
where e p and e s , m and n are the energies and numbers of electrons in the p and s
orbitals of the valence shell, respectively. The result is another strong argument for
defining electronegativity ( spec ) in terms of orbital energy. This system of elec¬
tronegativity has been successfully applied to periodic properties of the elements such
as covalent, metallic, and ionic bonding; atomic radii; multiple bonding; oxidation
states; and the unique properties of carbon and hydrogen. Note, however, that these
energies are not valence state energies, nor does the calculation include electron
affinities. 36 This prevents application lo the effect of charge capacity (hardness and
softness) as related to electronegativity or to the variability of electronegativity with
hybridization (see below). On the other hand it presents an unambiguous measure of
an atom's average electronegativity. Further efforts to apply the method lo transition
metals and to changes in oxidation states will probably be forthcoming.
The articles by Allen cited above arc also recommended as (he most recent review
of various aspects of electronegativity theories and for the idea that electronegativity
is "the third dimension of the periodic chart.”
Although electronegativity is often treated as though it were an invariant property of
an atom, we have seen that it depends on two properties: valence state (hybridization)
and atomic charge. Hybridization affects electronegativity because of the lower en¬
ergy and hence greater electron-attracting power of .v orbitals. We might expect the
electronegativity of an atom to vary slightly with hybridization, with those orbitals
having greater j character being more electronegative. Some results of the variation in
M Parr, R. G. Ann. Rev. Fltys. Chem. 1983, 34. 631-656. Polilzer. P.; Weinstein, H. J. Chem. Pltys.
1979, 71, 4218-4220. Van Genechtcn. K. A.; Mortier, W. J. Zeolites 1988, 8, 273-283. Pearson
R. G. Acc. Chem. Res. 1990, 23. 1-2.
35 Allen, L. C. J. Am. Chem. Soc. 1989, III, 9003-9014; Acc. Chem. Res. 1990. 23, 175-176; J. Am.
Chem. Soc. 1992, 114, 1510-1511; Can.J. Chem. 1992, 70. 631-635. Note that Allen's spectroscopic
electronegativity, like Mulliken's, will normally be expressed in terms of electron volts, but can be
converted to the Pauling scale if desired.
It should be noted that inasmuch as the ionization energy of most atoms is an order of magnitude
larger than the electron affinity, electronegativity methods which are fundamentally related only to
ionization energies are still successful.
Molecular Orbital Theory
171
The treatment of heteronuclear bonds revolves around the concept of elec¬
tronegativity. This is simultaneously one of the most important and one of the most
difficult problems in chemistry. In the previous discussion of molecular orbitals it was
assumed that the atomic orbitals of the bonding atoms were at the same energy. In
general, this will be true only for homonuclear bonds. Heteronuclear bonds will be
formed between atoms with orbitals at different energies. When this occurs, the
bonding electrons will be more stable in the presence of the nucleus of the atom having
the greater attraction (greater electronegativity), that is, the atom having the lower
atomic energy levels. They will thus spend more time nearer that nucleus. The
electron cloud will be distorted toward that nucleus (see Fig. 5.16) and the bonding
MO will resemble that AO more than the AO on the less electronegative atom.
Consider the carbon monoxide molecule, CO, isoelectronic with the N-, molecule.
Oxygen is more electronegative than carbon, so the bonding electrons are more stable
if they can spend a larger proportion of their time in the region of the oxygen nucleus.
The electron density on the oxygen atom is greater than that on the carbon atom in
contrast to the symmetrical distribution in the N 2 molecule (Fig. 5.18). For homo-
nuclear diatomic molecules we have seen that the’molecular orbitals are
‘Ph = + 'Pb (5.4S)
'Pc, = <Pa - •Pb (5.46)
Both orbitals contribute equally. Now if one atomic orbital is lower in energy than the
other, it will contribute more to the bonding orbital:
•Ph = ‘"Pa + P'Pb (5.47)
where b > a if atom B is more electronegative than atom A. Conversely, the more
stable orbital contributes less to the antibonding orbital:
‘P„ = H a - H b (5.48,
Fig. 5.18 (a) Total electron density contours for the carbon monoxide molecule. The carbon atom is on the left, (b) Total
electron density contours for the dinitrogen molecule. [From Bader, R. F. W.; Bandrauk. A. D. J. Chem. Phvs 1968 49
1653. Reproduced with permission.) ’ ’
jndinq Models in Inorganic Chemistry: 2. The Covalent Bond
Table 5.6 __
Electronegativities of the elements (Continued)
Mulliken-Jaffe, Tmj 0
Element
Pauling
_
Sc(lll)
1.36
Ti(II)
1.54
Ti(lll)
Ti(IV)
VdD
1.63
V(III)
V(V)
Cr(Il)
1.66
Cr(Ill)
Cr(Vl)
Mn(ll)
1.55
Mn(III)
Mn(lV)
Mn(V)
Mn(VI)
Mn(VlI)
Fe(II)
Fe(III)
Co(II)
Co(IlI)
Ni(II)
Cu(l)
Cu(II)
Zn
Ga(I)
Ga(lll)
Ge(ll)
Sanderson
Jj_
1.02
0.73
1.09
1.50
0.69
1.39
2.51
VJCVI V t
As(lll)
2.18
2.82
2.20
2.21
sp>
20% i
P
Se
2.55
3.01
2.48
2.42
sp 3
17%.?
P
Br
2.%
3.22
2.74
2.68
14%
P
Kr
3.00
2.91
2.94
' 2.97
12%
37 . Rb 0.82 0.31
Sr 0.95 0.72 0.97
Y(IID 1-22 ^ ;' ;
ZrdD 1-33 0.52
Zr(lV) 0- 90
Nb(II) 0- 77
Nb(V) 1.6 1-42
/.2J
Electronegativity
189
Table 5.6 ___
Electronegativities of the elements (Continued)
Mulliken-Jaffe, Tmj"
Element
Pauling
Sanderson
1
Allred-
Rochow
y 6
AR
Allen
A'sp^c
Orbital
or
hybrid
Modi)
2.16
0.90
Mo(IlI)
2.19
1.15
Mo(IV)
2.24
1.40
1.30
Mo(V)
2.27
1.73
Mo(VI)
2.35
2.20
Tc
1.9
1.36
Ru
2.2
1.42
Rh
2.28
1.45
Pd
2.20
1.35
Ag(I)
1.93
1.83
1.42
sp
Cd(II)
1.69
1.98
1.46
sp
Ind)
P
In(III)
1.78
2.14
1.49
1.66
P 2
Sn(II)
1.80
1.49
30%
Sn(IV)
1.96
1.72
1.82
p 3
Sb
2.05
2.46'• _
1.82
1.98
p 3
-in"/ ..
2.66 2.78 2.21 2.36
2.60' 2.34 2.40 2.58
Cs
0.79
0.22
0.86-
Ba
0.89
0.68
0.97
La
1.10
1.08
Ce
1.12
1.08
Pr
1.13
1.07
Nd
1.14
1.07
Pm
■
Sm
1.17
1.07
Eu
1.01
Gd
1.20
■ y
Tb
1.10
Dy
1.22
1.10
Ho
Er
1.23
1.24
ljL
v
1.10
■
Tm
Yb
1.25
£ 14 '
£
■A /S:-4
4 3m
m !: ■
Lu
Hf
1.27
S jp.
w§.
1.14
lilsilM
1.47
4.44
6.27
1.53
4.62
5.91
1.12
3.40
6.11
1.76
5.39
6.63
1.85
6.09
6.97
2.21
7.05
5.04
2.22
7.09
8.16
2.12
6.68
8.02
1.67
5.08
7.45
2.57
8.52
.8.74
2.41
7.83
8.64
2.06
6.46
8.44
2.95
10.26
9.38
2.74
9.29
9.33
2.45
8.00
9.25
3.01
10.52
9.52
2.73
9.24
9.36
2.44"
7.96"
9.21
2.40
7.76
—
0.62
2.18
3.42
0.88
2.79
3.93
172
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
In carbon monoxide the bonding molecular orbitals will resemble the atomic orbitals
of oxygen more than they resemble those of carbon. The antibonding orbitals resem¬
ble the least electronegative element more, in this case the carbon (see Fig. 5.19). This
results from what might be termed the conservation of orbitals. The number ot
molecular orbitals obtained is equal to the total number of atomic orbitals combined,
and each orbital must be used to the same extent. Thus, if the carbon atomic orbital
contributes less to the bonding molecular orbital, it must contribute more to the
antibonding molecular orbital. The energy level diagram for CO is shown in Fig. 5.20.
Fig. 5.19 Diagrammatic sketches of the molecular orbitals in carbon monoxide: (a) one
7 T bonding orbital, (b) one it antibonding orbital.
V "
■+K
Fig. 5.20 Energy level diagram for the molecular orbitals of carbon monoxide. Note that
upon bond formation electrons occupy orbitals that are more oxygen-like than carbon-like.
Note carefully the bond order: The Itr and 3cr MOs are essentially nonbonding. The bond
order, as in the N 2 molecule is three.
Molecular Orbital Theory
173
A second feature of heteronuclear molecular orbitals which has been mentioned
previously is the diminished covalent energy of bonds formed from atomic orbitals
of different energies. This may be shown qualitatively by comparing Fig. 5.21 with
Fig. 5.22. This can be seen even more readily in Fig. 5.23, in which the elec¬
tronegativity difference between atoms A and B is so great as to preclude covalent
bonding. In this case the bonding MO does not differ significantly from the atomic
orbital of B, and so transfer of the two bonding electrons to the bonding MO is
indistinguishable from the simple picture of an ionic bond; the electron on A has been
transferred completely to B.
This extreme situation in which the energy level of B is so much lower than that ot
A that the latter cannot contribute to the bonding may be visualized as tollows: II the
energy of atomic orbital B is very much lower than A, the electron will spend essen¬
tially all of its time in the vicinity of nucleus B. Although this may be a very stable
situation, it hardly qualifies as a covalent bond or sharing of electrons. In this case, the
sharing of electrons has been drastically reduced, and the covalent energy is negligi¬
ble. All chemical bonds lie somewhere on the spectrum defined by Figs. 5.21-5.23.
There has been some confusion in the literature concerning the strength of bonds
in situations such as shown in Figs. 5.22 and 5.23. Because good energy match is
necessary for a large A E c and good covalent bonding, some workers have concluded
that Figs. 5.22 and 5.23 represent increasingly weak bonding. This is not true, for the
Fig. 5.21 Homonuclear
diatomic molecule, A 3 .
The covalent energy is
maximized.
Fig. 5.22 Heteronuclear molecule. A+B-. with relatively small electronegativity difference
between A and B. Covalency reduced with respect to A : (Fig. 5.21). but still more important
than the ionic contribution.
Fig. 5.23 Heteronuclear molecule, AB~. with large electronegativity difference. Covalency
is insignificant; the bond is essentially ionic.
186 5 • Bonding
Recent Advances
Electronegativity
Theory
Models in Inorganic Chemistry: 2. The Covalent Bond
charge. Eq. 5.68 allows us to quantify that effect: An iodine atom with a partial charge
of about +0.4 is almost as electronegative as a neutral fluorine atom. The significance
of the parameters a and b is clear. The inherent or neutral atom electronegativity is
given by a. This is the electronegativity of an atom in a particular valence state as
estimated by the Mulliken method and corresponds to similar estimates of elec¬
tronegativity by Pauling ( P ), Allred and Rochow ( AR ). and others (see Table 5.6). It
may be used alone ixm = t0 sa y> for example, that in LiH the hydrogen atom (a =
2.25) is much more electronegative than is lithium (a = 0.97), and should be written
Li 5+ H s ~ or even Li + H _ .
The parameter b is the charge coefficient. It measures the rate of change of
electronegativity with charge. Mathematically, b is the second derivative of energy
(first derivative of electronegativity) with respect to charge:
b = d 2 E/d8 2 = dx/d8 = IE V - EA V (5.69)
It thus defines the curvature of the energy-charge parabola. Chemically, it is the
inverse of the charge capacity («•) or polarizability: 31
b = I/ (5.70)
Large, soft, polarizable atoms have low values of b. and small, hard, nonpolarizable
atoms tend to have higher values. An atom with a large charge coefficient will change
electronegativity much more rapidly than one with a lower value of b. Thus a small
atom (low «, large b) has only a limited ability to donate or absorb electron density
before its electronegativity changes too much for further electron transfer to take
place. One of the most important examples is the very electronegative but very small
fluorine atom. Although initially very electronegative when neutral (note the steep
slope at the origin of Fig. 5.32), it rapidly becomes “saturated" as it accepts electron
density (note how quickly the slope flattens out between -0.4 and -0.6), and beyond
-0.7, it is necessary "to push" to get more electron density onto a fluorine atom. This
is closely related to the comparatively low electron affinity of fluorine (Chapter 2).
The charge capacity effect is responsible for the well known inductive effect of
alkyl groups (see page 196). It is also important in hard and soft acid-base theory (see
Chapter 9), and causes several other unexpected effects. 33 It is basically a polarization
effect in which larger atoms and groups can acquire or donate large amounts of
electron density without unfavorable energy changes. 33
in The advances in recent years have been more evolutionary than revolutionary. In¬
creasingly, Mulliken’s original idea of expressing electronegativity in terms of the
energy of valence electrons has come into favor, and the other definitions in terms of
resonance energy or algebraic relationships of size and charge have been viewed as
useful approximations when orbital energies are not available. In addition, the rela¬
tionship between electronegativity and acidity and basicity, always intimate, has been
extended further (see Chapter 9).
31 Politzcr. P. J. Client. Phys. 1987. 86. 1072-1073.
32 Politzcr, P.; Huhecy, J. E.; Murray, J. S.; Grodzicki. M. J. Mol. Structure ( THEOCHEM). 1992,
259. 99-120.
33 Huhecy, J. E. J. Org. Chem. 1971. 36. 204-205. Politzcr. P.; Murray. J. S.; Grices. J. E. In
Chemical Hardness'. Sen, K. D.. Ed.: Springer-Verlag: Berlin, in press.
Electronegativity
Table 5.6
Electronegativities of the elements
Pauling Sanderson
Element
Allred-
Rochow
v d
Allen
A'spoc
Orbital
or
hybrid
Mulliken'
Pauling
scaled
-Jaffe,
a
Volts
b
Volts/
electron
H
2.20
2.59
2.20
2.30
J
2.25
7.17
12.84
He
5.50
S
3.49“
12.98“
23.22
4.86
15.08
Li
0.98
0.89
0.97
0.91
s
0.97
3.00
4.77
Be
1.57
1.81
1.47
1.58
sp
1.54
4.65
6.58
B
2.04
2.28
2.01
2.05
sp 2
2.04
6.37
8.74
sp 2
1.90
5.86
8.64
C
2.55
2.75
2.50
2.54
sp
2.99
10.42
11.70
sp 2
2.66
8.91
11.50
sp 3
2.48
8.15
11.39
N
3.04
3.19
3.07
3.07
sp
3.68
14.00
13.32
sp 2
3.26
11.78
13.22
sp 3
3.04
10.66
13.16
20%
2.90
10.00
13.13
P
2.28
7.32
13.00
O
3.44
3.65
3.50
3.61
sp 2
3.94
15.48
15.62
sp 3
3.68
14.02
15.55
17%
3.41
12.55
15.47
P
2.82
9.63
15.33
F
3.98
4.00
4.10
4.19
sp 3
4.30
17.63
17.99
14 %
3.91
15.30
17.81
P
3.35
12.20
17.57
Ne
4.50
4.84
4.79
sp 3
4.49
18.86
18.92
12%
3.98
15.71
18.50
P
3.41“
12.56“
18.08
4.26
13.29
_
Na
0.93
0.56
1.01
0.87
s
0.91
2.84
4.59
Mg
1.31
1.32
1.23
1.29
sp
1.37
4.11
5.27
Al(l)
0.84
p
0.91
2.86
6.23
Al(lll)
1.61
1.71
1.47
1.61
sp 2
1.83
5.61
6.12
sp 3
1.71
5.21
5.92
Si
1.90
2.14
1.74
1.92
sp 3
2.28
7.30
7.13
P
2.19
2.52
2.06
2.25
sp 3
2.41
7.84
9.53
20%
2.30
7.41
9.39
P
1.84
5.67
8.83
S
2.58
2.96
2.44
2.59
sp 3
2.86
9.84
10.36
I7%
2.69
9.04
10.20
P
2.31
7.44
10.12
Cl
3.16
3.48
2.83
2.87
14%.?
3.10
12.15
11.55
P
2.76
10.95
11.40
Ar
3.31
3.20
3.24
sp 3
3.49
12.98
12.38
12%
3.19
11.41
12.18
P
2.86“
9.83“
11.98
• «;
3.11
9.87
■-j
K
0.82
0.45
0.91
0.73
s
0.73
2.42
3.84
Ca
1.00
0.95
1.04
1.03
sp
1.08
3.29
1.08
187
174
5-Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
loss of covalent bonding may be compensated by an increase in ionic bonding,
which, as we have seen previously, can be quite strong. In fact, the total of ionic and
covalent bonding may make a very strong bond in the intermediate situation shown in
Fig. 5.22. (Note that the ionic contribution to the bonding does not appear in these
figures.) In fact, the strengthening of a polar bond over a corresponding pureiv
covalent one is an important phenomenon.
A second example of a heteronuclear diatomic molecule is hydrogen chloride. In
this molecule the attraction of the chlorine nucleus for electrons is greater than that of
the hydrogen nucleus. The energies of the 3 s and 3 p orbitals on the chlorine atom are
less than that ot the Is orbital on hydrogen as a result of the imperfect shielding of the
much larger nuclear charge of chlorine. The molecular orbitals for the hydrogen
chloride molecule are illustrated in Fig. 5.24. There is one cr bond holding the atoms
together. The remaining six electrons from chlorine occupy nonbonding orbitals,
which are almost unchanged atomic orbitals of chlorine. These nonbonding molecular
orbitals correspond to the lone pair electrons of valence bond theory. They represent
the two p orbitals on the chlorine atom that lie perpendicular to the bond axis. They
are therefore orthogonal to the hydrogen \s orbital (Fig. 5.8) and have a net overlap of
zero with it. As such, they cannot mix with the hydrogen orbital to form bonding and
antibonding molecular orbitals. The third nonbonding MO is the second of the
hybridized atomic orbitals resulting from some s-p mixing. If the mixing were com¬
plete (50% s character, which is far from the case in HCI), it would be the second di
orbital directed away from the bond. Since little mixing of the s orbital of the chlorine
into the bonding MO occurs, the third lone pair is a largely s orbital, a distorted sphere
of electron density with the major portion behind the chlorine atom.
Although mixing of 5 and p orbitals is represented in Fig. 5.24 as a separate step
preceding the formation ol molecular orbitals, the entire process can be combined into
a single step. For example, the bonding molecular orbital in hydrogen chloride may be
considered to be formed as
'h, = a ) P + cifi ls (5.49)
where ip 2s , tp 3f> , ip ls are the atomic orbitals on the chlorine (3r and 3 p) and hydrogen
(Ij) atoms. Now a and b can be varied relative to each other in such a way that any
amount of p character can be involved in the molecular orbital. For example, if a = 0,
Fig. 5.24 Energy level diagram for the hydrogen chloride molecule. HCI. The mixing of the
s and p orbitals has been emphasized.
19 Ferreira, R. Chem. Phys. Lett. 1968, 2, 233.
Molecular Orbital Theory
175
the chlorine atom uses a pure p orbital, but if a 2 = \b 2 , the p character will be 75% (an
“sp hybrid" in VB terminology).^ And, of course, the relative weighting of a and b
versus c indicates the relative contributions of chlorine versus hydrogen wave func¬
tions to the bonding molecular orbital.
Molecular Orbitals The linear molecule BeH 2 will serve as our first example of a triatomic species. The
in Triatomic molecular orbitals for this molecule are constructed from the \s orbitals on the
Molecules and Ions hydrogen atoms (labeled H and H') and the 2s and one of the 2 p orbitals of beryllium
(the one directed along the H—Be—H bond axis). The remaining two Ip orbitals of
beryllium cannot enter into the bonding because they are perpendicular to the mo¬
lecular axis and thus have zero net overlap with the hydrogen orbitals.
Because four atomic orbitals enter into the bonding, we anticipate the formation
of four molecular orbitals. As always, the bonding molecular orbitals are formed by
linear combination of the atomic orbitals to give maximum overlap. Prior to forming
molecular orbitals, we can combine the orbitals of the two hydrogen atoms into group
orbitals that are consistent with the linear geometry of the molecule and with the
symmetries of the atomic orbitals on beryllium. The group orbitals are formed by
simply taking linear combinations of the \s orbitals on H and H'. There are only two
possibilities, so the group orbitals correspond to <b H + <// H , and ib H - ip H .. The first
one is appropriate for overlap with the beryllium 2s orbital, which is everywhere
positive. The second one will form a bonding MO by overlapping with the 2 p orbital of
beryllium, which has one positive and one negative lobe. The antibonding orbitals will
be formed by opposite combinations, which give nodes between the bonded atoms.
The molecular orbitals can be represented as
^ + bit H + <!>»■) = <r g - 2t k (5. 50)
= CI P 2 P + diib H ~ i/'h’) = cr u = lcr„ ( 5 . 51 )
</£ = bib - a(ib H + i/, H .) = o£ = 3cr A , ( 5 . 52 )
ilC, = dib 2p - c(ib H - ib H .) = = 2 <r„ (5.53)
1 he parameters «. c, and d are weighting coefficients, which are necessary because
of differences in electronegativity between Be and H.
The energies of the BeH, molecular orbitals are shown in Fig. 5.25 and their
electron density boundary surfaces are sketched in Fig. 5.26. Both of I lie bonding
molecular orbitals are delocalized over all three atoms. This is a general result of the
MO treatment of polyatomic molecules. Note that the lowest-energy orbital, the lcr„
is not shown in either Fig. 5.25 or 5.26. It would be formed from the \s orbital on
beryllium, which interacts very little with the hydrogen orbitals because of the large
energy difference between them. This molecular orbital is therefore nonbonding and is
essentially indistinguishable from the beryllium Ir atomic orbital.
The nitrite ion, NOT. is an example of a nonlinear triatomic species with rras.well
as o- bonds. In the valence bond description of the ion, resonance structures are used
to allow for the distribution of the 7r electrons over all three atoms. The molecular
“ The percent s and p character is proportional to the square of the coefficients. In the case of
chlorine, the difference in energy between the 3 s and 3 p orbitals is so great as to preclude very much
mixing of s and p character, but it is always possible; the closer the energy levels are, the more likely
it is to occur.
Models in Inorganic Chemistry: 2. The Covalent Bond
formulation provides clear intuitive perspective: It indicates tnat an
fv electronegative if it releases much energy (because its energy curve is steep) as
quireselectron density; another atom will be less electronegative became
’ Is less Sleep and, when combined with ,he more electronego ««<»». “
lost os much energy climbing US own energy curve. A mole °“ ,e cm moTecule
F»-, and it will be more stable than a hypothetical nonpolar CIF
The relationship between the Mulliken definition and that of Jaffe can be sho n
'simply Taking Eq. 5.62 and substituting , = + 1. we know that the energy, £ .of
fhat of the + 1 cation, or the first ionization energy. Likewise tor
-EA V = £_| = a(-D + ^( _1 ) 2
Subtracting Eq. 5.65 from Eq. 5.64 gives;
IE V + EA V = 2 a
which yields a as the Mulliken electronegativity (Eq. 5.59).
Fig. 5.32 Ionization
energy-electron affinity
curves for fluorine and
chlorine. The electro¬
negativities are given by the
slopes of these curves. This
figure is an enlarged portion
of Fig. 2.13.
Charge
of electron affinity does nol follow the usual thermodynamic convention in
affinity is exothermic. See Chapter 2.
Electronegativity
185
A word must be said here concerning the subscript v's in the above equations.
They refer to the valence state. Just as Pauling’s definition of electronegativity is tor
an atom in a molecule, the Mulliken-Jaffd definition of electronegativity tor various
hybridizations involves the computation of valence state ionization energies and
valence state electron affinities by adjusting for the promotion energy from the ground
state. The valence state ionization energy and electron affinity are not the experimen¬
tally observed values but those calculated for the atom in its valence state as it exists
in a molecule. Two short examples will clarify the nature ot these quantities.
Divalent beryllium bonds through two equivalent sp, or digonal, hybrids. The
appropriate ionization energy therefore is not that of ground state beryllium, l.v 2 2i-,
but an average of those energies necessary to remove electrons Irom the promoted,
valence state:
lj 2 2s'2pl -♦ \s 2 2s'2p° (IE p ) and I s 2 2s'2p' -♦ \s 2 2s°2p' (IE t ) (5.67)
It is thus possible to calculate the hypothetical energy necessary to remove an electron
from an sp hybrid orbital. This VSIE (and the corresponding valence state electron
affinity) can be used to calculate the electronegativity of an sp (di) orbital.
A chlorine atom may well be assumed to use a pure p orbital with hybridization
neglected, but still the valence state ionization energy does not correspond to the
experimentally observed quantity. We may consider that the use of ionization energies
and electron affinities relates to the occurrence of covalent-ionic resonance as shown
previously in Eq. 5.57. Now one of the requisites for resonance to occur is that all
contributing forms have the same number of unpaired electrons (page 145), so the
wave function and energy for any contributions from Cl' must be lor zero spin (all
simple molecules containing chlorine are diamagnetic). The ground state ionization
energy corresponds to the process Cl(V)-> Cl + m but the VSIE is for ionization to
a singlet state for Cl + , a suitably weighted average of >S and 1 D. We need not concern
ourselves with the mechanics™ of calculating the necessary promotion energies for
either beryllium or chlorine, but we should remember that it is not possible to
calculate accurate electronegativities simply from ground state ionization energies and
electron affinities alone.
One of the strengths of the Mulliken-Jaffd approach is that it is capable of treating
the electronegativity of partially charged atoms. We should not expect an atom that
has lost electron density to have the same electronegativity as a similar atom that has
not had such a loss; the former should be expected to hold on to its remaining electron
density more tightly. Conversely, as an atom acquires a partial negative charge, its
attraction for more electron density will decrease. This can be shown readily by
rewriting Eq. 5.63 as
= dEfdS = a + h5 (5-68)
in which the partial charge, 5, replaces the ionic charge, q, and the constants have
been changed for convenience (a = a; h = 2/3).
The importance of this equation lies in illustrating the large effect that charge can
have on the electronegativity of an atom. Intuitively, one would expect an atom with a
positive charge to be more electronegative than that same atom with a negative
» See Molfitt. W Proc. Roy. Soc. (London) 1950. A202, 548. For very readable accounts of the
valence state sec McWeeny. R. Coulsons Valence, Oxford University: London. 1979; pp 150,
201-203. 208-209; Johnson. D. A. Some Thermodynamic Aspects of Inorganic Chemistry, 2nd ed„
Cambridge University: Cambridge. 1982; pp 176-177. 200-206.
5 • Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Molecular Orbital Theory 177
orbital treatment provides an alternate approach that does not require resonance
structures because electrons are automatically delocalized as the method is applied.
We can consider the cr system in N07 to consist of two N—O single bonds
formed by overlap of sp 2 hybrid orbitals on nitrogen and oxygen. 21 Also part of the
cr network will be four pairs of electrons in nonbonding orbitals that are essentially
oxygen sp 2 hybrids. (These are the oxygen lone pairs in the valence bond description.)
Remaining on the nitrogen and oxygen atoms are parallel p orbitals (Fig. 5.27). These
orbitals will interact to form bonding and antibonding combinations:
h = Po, + Po 3 + Pn ^
•I 1 ,, = Po, + Poz ~ Pn
(5.54)
(5.55)
As in the cases we have seen before, the bonding orbital yields a concentration ol
electron density between the atoms and the antibonding orbital has nodes between the
atoms (Fig. 5.28). There is a third combination possible:
<Pn = Po, ~ Poz ± Pn (5,56)
Fig. 5.27 Sigma bonds and
lone pairs in the nitrite ion.
N0 2 '.
(a)
Fig. 5.28 Atomic orbitals
(left) and resulting
molecular orbitals (right) in
the nitrite ion: (a) bonding
and (b) antibonding.
21 Hybridization is. of course, completely unnecessary when using molecular orbital theory, but is a
convenience here since we are primarily concerned with the 7r system. The VB and MO treatments
of the <r system are not significantly different in their results.
182
5- Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
nonbonding electron pairs. It should be emphasized that the diagram in Fig. 5.31 is
oversimplified because it shows only the major interactions between orbitals. There
will actually be some interaction between all orbitals of the same symmetry. In other
words, the diagram suggests more localization of electrons than is actually the case.
The orbitals shown in Figs. 5.28 and 5.29 correspond to the h, n molecular orbitals
(bonding and antibonding) and the a 2 nonbonding orbital in the MO diagram.
Electronegativity Linus Pauling first defined electronegativity and suggested methods for its estimation.
Pauling's definition 24 has not been improved upon: The power of an atom in a
molecule to attract electrons to itself. It is evident from this definition that elec¬
tronegativity is not a property of the isolated atom (although it may be related to such
properties) but rather a property of an atom in a molecule, in the environment and
under the influence of surrounding atoms. One must also note that the "power to
attract" is merely another way of describing the “reluctance to release" electrons
from itself to a more electronegative element.
Pauling based his scale on thermochemical data. We shall examine his methods
shortly, but we may note that his scale is an arbitrary one chosen so that hydrogen is
given a value of about 2 and the most electronegative element, fluorine, has a value of
about 4:
H = 2.2
Li = 1.0 Be = 1.6 B = 2.0 C = 2.5 N = 3.0 0 = 3.4 F = 4.0
Na = 0.9 Mg = 1.3 Al = 1.6 Si = 1.9 P = 2.2 S = 2.6 Cl = 3.2
There are other scales that have absolute units, and whereas it might seem at first
glance that an absolute scale would be preferable, the Pauling scale has a familiarity
and attendant literature that no absolute scale can come close to matching. Familiarity
as a virtue should not be discounted in unthinking attempts to "standardize" things.
Several times workers have reported erroneous electronegativity values in electron
volts or in MJ mol -1 that they would have instantaneously noticed and rejected if they
had converted their values to the Pauling scale. A value of 3.3 for fluorine stands oul
like the proverbial sore thumb! 25
Consider Pauling's approach to the treatment of a molecule of hydrogen chloride
which is usually represented today as
H + CI«-
The use of 5± to represent partial charges in a polar molecule is relatively recent.
Pauling would have pictured it as
H—Cl <—♦ H + CP <-► H - CI + (5.57)
l II III
24 Pauling, L. The Nature of the Chemical Bond. 3rd cd.: Cornell University: Ithaca. NY, I960: p 88.
25 Liebman, J. F.; Huheey, J. E. Phys. Rev. D 1987. 36. 1559-1561.
Electronegativity
183
where the wave function of the resonance hybrid would be
-^cov + y + - + (5.58)
where .r, y. and z are weighting coefficients and cov, + - , and - + are labels for
contributing canonical forms I, II, and III. If chlorine is more electronegative than
hydrogen, contributing form II wiil be important as well as form I (HCI is predomi¬
nantly covalent), but form III, which places a positive charge on the chlorine atom,
makes a negligible contribution. Pauling assumed that resonance would stabilize the
molecule of HCI and that the greater the contribution of II, the more polar the
molecule and the greater its stability. Soon after Pauling published his first paper on
electronegativity, Mulliken 2 '- suggested a method for estimating how much each of the
forms I, II, and III would contribute to the hybrid and used it to establish the
electronegativity scale that bears his name.
Mulliken-Jaffe
Electronegativities
The method of treating electronegativities that has the firmest theoretical basis is the
Mulliken-Jaffe system. Recall that canonical forms that are low in energy and stable
contribute the most to the resonance hybrid, and that high-energy forms contribute
little. Mulliken suggested that two energies associated with an atom should reflect a
measure of its electronegativity: (1) the ionization energy , as a measure of the
difficulty of removing an electron (or, more generally, electron density) to form a
positive species; (2) the electron affinity , as a measure of the tendency of an atom to
form a negative species. Structure II is stable because chlorine has a high electron
affinity and hydrogen has a relatively low ionization energy fora nonmetal. Structure
III is unstable because chlorine has a high ionization energy and hydrogen has a low
electron affinity. Mulliken's definition of electronegativity is given simply as
1E V + EA V
(5.59)
or, when the energies are in electron volts (the most common unit used in the past),
putting the values on the Pauling scale:
or. when the energies are in MJ mol - ': 27
Xm = 3 - 48 [ ' Ev 2 EAv " °- 0602 ]
(5.6!)
Now it should be recalled that the first ionization energy and the electron affinity
are merely two of the multiple ionization potential-electron affinity energies that fit a
polynomial equation (see Chapter 2) that is quite close to being quadratic (the coef¬
ficients of the higher order terms are small). Jaffe and coworkers 21 have pointed out
26 Mulliken, R. S. J. Client. Phys. 1934, 2. 782-793; 1935, J. 573-585.
27 This simple, linear relationship is the most frequently used. However. Bratsch (sec Footnote 28 and
Table 5.6) has presented evidence for a somewhat belter-fitting quadratic relationship.
28 Hinze. J.; Jaffe. H. H. J. Am. Chem. Soc. 1962. 84. 540-546; J. Phys. Chem. 1963. 67. 1501-1506;
Hinze. J.; Whitehead. M. A.; Jaffe. H. H. J. Am. Chem. Soc. 1963. 85. 148-154. See also Bratsch
S. G. J. Chem. Educ. 1988, 65. 34-41. 223-227.
178
179
5 Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
Fig. 5.29 Two schematic representations of the nonbonding orbital in the nitrite ion.
In this case, regardless of the sign given to the nitrogen orbital, there will be an
inevitable mismatch—to whatever extent there is positive overlap with one p 0 orbital
there will be negative overlap with the other—producing a nonbonding situation
(Fig. 5.29).
The molecular orbital result for NOT is similar to that obtained by the valence
bond picture with resonance. There is a bonding pair of n electrons spread over the
nitrogen and two oxygen atoms. The second pair of tt electrons is nonbonding in the
MO description and effectively localized on the two oxygen atoms (the node cuts
through the nitrogen atom), but as in the VB picture, they too are smeared over both
oxygen atoms rather than occupying discrete atomic orbitals.
The molecular orbital description for the nitrite ion just presented was developed
without the aid of symmetry considerations and as a starting point, it assumed that
cr bonds were formed from sp 2 hybrid orbitals on the nitrogen and oxygen atoms. Let
us now see how we could have obtained a similar end result by using a method that
involves a more formal application of symmetry and does not invoke hybridization.
(For a review of symmetry in bonding, see Chapter 3.)
As we have seen pictorially in this chapter, in order for a bond to form, the
overlapping orbitals must meet symmetry requirements determined by the type of
bond (o-, 7 r, etc.) and the spatial positions of the bonded atoms. This means that the
molecular orbitals for N0 2 and the atomic orbitals from which they are constructed
must conform to the C 2l . symmetry of the ion. Consider first the sigma MOs. which
will form a basis for a representation within the C 2v point group. By determining this
representation and its irreducible components, we will establish the symmetry criteria
that the contributing atomic orbitals on nitrogen and oxygen must meet, and then can
identify those orbitals. The sigma MOs can be represented as vectors along the N—O
bond axes:
z
N
/ \
O O
Carrying out the C 2v symmetry operations on these two vectors and recording the
number that are invariant for each operation generates the reducible representation.
Reduction of V r (Eq. 3.1) yields an u, and a b 2 as the irreducible components. In order
for atomic orbitals on nitrogen and oxygen to be suitable for linear combination into
Molecular Orbital Theory
sigma MOs. they must belong to one of these representations. Our next task is to
choose these orbitals.
The atomic orbitals on nitrogen contributing to the bonding will be the outer shell
2s and 2 p. By again referring to the C 2v character table, we see that the s and p.
orbitals belong to the a, representation while the p x and p v belong to />, and b-'
respectively. Thus the nitrogen orbitals qualifying for participation in sigma MOs are
the p y and the s and p.. Note that these aiso are the three orbitals we would have
chosen for construction of .sp 2 hybrid orbitals.
By a similar analysis, the nitrogen orbitals capable of participating in n bonds can
be identified:
0 -t
U ^
E C 2 tr r . <7 V .
F r = 2 0" o' -2 = a 2 + b t
The C 2v character table shows that none of the nitrogen valence orbitals transforms as
a 2 . However, p x transforms as b { and therefore can participate in v bonding.
Having determined which nitrogen orbitals can participate in cr and tt MOs, we
must now identify the oxygen orbitals with which they can combine. More precisely,
we must derive the combinations of oxygen orbitals, or the oxygen group orbitals, that
will meet the symmetry criteria for MO formation. 22 As shown in the treatment of
BeH 2 (page 175). group orbitals can be derived by taking linear combinations (additive
and subtractive) of atomic orbitals. The orbitals of interest here are the outer shell (2a-
and 2 p) orbitals of oxygen. The 2s orbitals on the two atoms can be added (s + ,r) or
subtracted (i - s) to give two group orbitals (Fig. 5.30). By applying the C 2v symme¬
try operations to these group orbitals, we find that the (,v + v) combination is sym¬
metric with respect to all operations, and thus belongs to a,, while the (j - s) combin¬
ation belongs to b 2 . Continuing in a similar fashion with the p orbitals, we see that the
Pj) combination transforms as c;, while its subtractive counterpart (p. - p.)
belongs to b 2 . For (p v + p v ) and (p v - p y ) we obtain b 2 and respectively. Finally,
(p t + p x ) transforms as b, and (pj - p f ) as a 2 .
We now have all the symmetry information needed to construct a molecular
orbital diagram for the nitrite ion. Of course, our symmetry analysis provides no
indication of the relative energies of the orbitals involved. To obtain such information
would require detailed calculations or experimentation. 22 We can, however, make
some reasonable assumptions that will allow us to construct a qualitative diagram.
Because oxygen is more electronegative than nitrogen, the 2s and 2 p orbitals on
oxygen will lie lower in energy than the same orbitals on nitrogen. It is on this basis
that the atomic and group orbitals are arranged on the left and right sides of Fig. 5.31.
The nitrogen and oxygen orbitals that can combine to form sigma MOs are all of those
Whereas the pictorial approach used here for deriving the oxygen group orbitals and determining
their symmetries is suitable for N0 2 . more complicated molecules and ions (especially those
involving degeneracies) generally require more sophisticated methods. For coverage of these
methods, the interested reader should refer to the group theory texts in Footnote I or Chapter 3.
23 Harris. L. E. J. Chem. Phys. 1973, 58, 5615-5626.
180
5 ■ Bonding Models in Inorganic Chemistry: 2. The Covalent Bond
(s + s)
(i -s)
(p. + p.)
(P.-P;)
(P v -P.)
£ C 2 o„ a r:
I I I 1 = "|
1 - 1-1 1 = 6 2
1 1 1 1 = “i
1 -1 -1 1 = b l
1-1-1 I = h
1 1 1 1 = fl|
that have a, and b 2 symmetries. Taking the «, category first, we find a total of five
orbitals (two from nitrogen and three oxygen group orbitals), which will result in five
molecular orbitals. In estimating the relative energies of these molecular orbitals, we
should bear in mind that two factors will promote stabilization of a bonding molecular
orbital formed from two interacting orbitals: favorable overlap and similarity in
energy. The degree to which these two factors are present or absent will determine
whether a molecular orbital is bonding (with an antibonding counterpart) or essentially
nonbonding. Thus we would predict (on the basis of a rather large energy gap) that the
(j + j) oxygen group orbital and the nitrogen a, orbitals will lead to an u, molecular
orbital that is only slightly bonding. Overlap and energy match are more favorable for
Molecular Orbital Theory 181
2<Py)
b |(P,)
<i|(P.)
a,(i)
a ’ <P.,-P,)
b 2
a | (Py-Py)
b \ < P.. + Pl )
b 2
a | (p. p.)
b 2 (s - s)
a, (s + i)
Atomic orbitals Molecular orbitals Group orbitals
of nitrogen of NOf of oxygen
Fig. 5.31 Molecular orbital diagram for the nitrite ion.
the (/?. + p.) group orbital and the nitrogen </, orbitals, so they would be expected to
combfne to form a strongly bonding and strongly antibonding pair of MOs. On the
basis of poor overlap between Ihe (p y - p y ) group orbital and the n, orbitals on
nitrogen, we can assign this group orbital as an essentially nonbonding n, MO.
By applying similar reasoning to the b 2 orbitals (four in all), we conclude that the
nitrogen orbital of this symmetry will overlap to yield a slightly bonding MO with the
(s - s) group orbital and a distinctly bonding MO with the (p. - p.) group orbital.
The (p v + p v ) group orbital will be essentially nonbonding and there will be one
antibonding MO. Finally, the nitrogen />, orbital will combine with the (p, + p x ) group
orbital to form bonding and antibonding 7r MOs.
Of the twelve molecular orbitals constructed, the nine that are lowest in energy
will be occupied by the valence electrons in NOJ. The electrons in the lowest level u,
and b 2 orbitals are only slightly bonding and may be thought of as lone pairs on the
oxygen atoms. The Lewis structure for N07 includes three additional lone pairs on
the two oxygen atoms. In the MO picture, they arc ihe electrons in the </, (p v - p y )
and b 2 (p v + p y ) orbitals—which we have already identified as essentially nonbond-
ing—and those in the a 2 orbital, which is strictly nonbonding because the a 2 group
orbital has no symmetry match on nitrogen. The nonbonding electron pair that is
shown in the Lewis structure as residing on nitrogen corresponds to the pair in the
highest occupied molecular orbital (the HOMO) in the MO diagram. The final result,
then, for our molecular orbital description is two a bonds, one rr bond, and six
276 7-The Solid State
Solid-State
Materials with
Polar Bonds
Various imperfections can lead to semiconductivity in analogous ways. For
example, nickel(II) oxide may be doped by lithium oxide (see Fig. 7.12). The Ni 3 "' ions
now behave as holes as they are reduced and produce new Ni 3 ions at adjacent sites.
These holes can migrate under a potential (indicated by the signs on the extremes ot
the series of nickel ions):
e
( + )Ni 2+ . .
. . . Ni 3+ _
-0
. Ni 2+ _
e~
. Ni 2+ . . . .
. Ni 2+ (—)
( + )Ni 2+ . .
. . . Ni 2+ -
. Ni-_
"Ni-....
^ e~
• Ni 2+ (-)
( + )Ni 2+ . .
. . . Ni 2+ _
. Ni 2+ _
. Ni 3t _
^Ni 2+ (~)
The range of possibilities for semiconduction is very great, and the applications to the
operation of transistors and related devices have revolutionized the electronics indus¬
try, but an extensive discussion of these topics is beyond the scope of this text.- 5
Note, however, that inorganic compounds are receiving intensive attention as the
source of semiconductors, superconductors (page 285), and one-dimensional conduc¬
tors (Chapter 16).
We have seen basically two models for bonding in several types of solids. The ionic
model (Chapter 4) has complete localization of electrons on the ions, patently untrue
but a useful approximation in crystals containing very electropositive metals and very
electronegative nonmetals. A completely covalent insulating solid such as diamond is
basically the same: All of the electrons are localized in C—C bonds. The band model
for conductors (above) has the valence electrons completely and evenly delocalized
over the whole crystal. Semiconductors fall betweeen insulators and conductors in
that the electrons are localized but with a small energy gap. Most solid compounds will
not fall neatly into any of these simple pictures, just as most molecules are neither
completely covalent nor ionic. The current picture for bonding in many interesting
materials is a composite of these two extremes. It will be illustrated by an examination
of compounds belonging to the ThCr 2 Si 2 structure type.
2.1 See Jolly. W. L. Modern Inorganic Chemistry, McGraw-Hill: New York. 1984; West. A. R. Basic
Solid State Chemistry, Wiley: New York. 1988: pp 294-300.
26 See HofTmann. R. Solids and Surfaces-, VCH: New York. 1988. for a clear discussion of solid stale
chemistry and physics in chemist's language for this structure as well as many others. Th.s
discussion may also be found in HofTmann, R. Angew. Chem. 1987. 26, 846-878; Rev. Mod. Phys.
1988, 60, 601-628. The specific example for this section is described in an article with the same title
as the heading above; see Footnote 34.
Solid-State Materials with Polar Bonds 277
The ThCr 2 Si 2 More than 400 compounds of AB 2 X 2 stoichiometry adopt the ThCr 2 Si 2 type struc-
Structure Type 26 ‘ore.” In these A is typically an alkali, alkaline earth, or rare earth metal. B may be a
transition metal or a main-group metal. X is a group VA (15), IVA (14), or occasionally
III A (13) nonmetal. The compounds in which we shall be most interested are com¬
posed of an alkaline earth metal (A = Ca, Sr. Ba), a transition metal (B = Mn, Fe. Co.
Ni, Cu). and phosphorus (see Table 7.2). These compounds are isostructural and
crystallize in the ThCr 2 Si 2 structure with space group 14/mmm. The unit cell (Fig.
7.28) consists of eight A 11 ions at the corners of a rectangular parallelepiped plus one
Table 7.2
Some interatomic distances in AB 2 Pj compounds with the
Compound r Co -p Pb-p r P-P Compound
ThCr 2 Si 2
fSr-P
structure.
ffc-p
r P-P
Compound
rBo-p
r B-P
f>_p
BuMn,P,
341
245
373
CaFe,P,
304
224
271
SrFc 2 P 2
320
225
343
BaFe 2 P 2
332
226
384
CaCo 2 P,
299
226
245
SrCo,P 2
318
224
342
CaNi 2 P,
300
-229
230
CaCu. ,,P?
305
238
225
316
243
230
302
229‘
243
318
23 l r
305
336
236'
378
±CT
±3
±6
±21
±2
±11
±67
±6
±13
±8
« Data from Mewis, A. Z. Naturforsch. 1980, 35B, 141-145. All distances in pm.
This compound contains both Cu" and Cu" 1 .
«• Statistically meaningless since r 0 varies down the series.
O a <™
O B(Cr)
• P(Si)
Fig. 7.28 Unit cell of an
alkaline earth (A)/
transition metal (B)/
phosphide (P) of the
ThCr 2 Si 2 -type structure.
The distances listed in
Table 7.2 arc indicated.
| Modified from Hoffmann,
R.; Zheng. C. J. Phys.
Chem. 1985, 89, 4175-4181.
Reproduced with
permission.!
n In crystallography, as in systematic botany and zoology, the type is merely a “name-bearer"—
something to which a name may be attached unambiguously. It is not necessarily the typical species
in the everyday meaning of that word (= representative, usual). Although this can cause some
initial confusion, note that the overall stoichiometry and the total number of s and p electrons (as
given by the Roman numeral group numbers) arc the same in both Th Cr 2 Si 2 and A B 2 P 2 .
450
11 ■ Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.21 _
Configurations for which
Jahn-Teller distortions are
expected in ML^ complexes
Configuration Ground-state term
Jahn-Teller distortion?
d'
2 t 2k
Yes
d 2
X
Yes
d 3
No
d 4
5 E k (high spin)
Yes
3 7j A , (low spin)
Yes
d 5
6 A Is (high spin)
No
2 T 1r (low spin)
Yes
d 6
s T 2i , (high spin)
Yes
'A lx (low spin)
No
d 1
4 7j j . (high spin)
Yes
d 8
2 E (low spin)
Yes
3 a 2 ,
No
d"
%
Yes
will be subject to Jahn-Teller distortions by considering ground state degeneracies.
The Tanabe-Sugano diagrams in Appendix G reveal that the only configurations
having nondegenerate ground states are d } ( 4 A 2); ), high spin d 5 ( 6 A, K ), low spin d k
and d H ( 3 A 2 J. Thus spontaneous Jahn-Teller distortions are expected for all
other configurations: d l , d 2 , d 4 , low spin d s , high spin t/ 6 . d'. and d' (Table 11.21).
Basic insight into the nature of the Jahn-Teller effect can be obtained by returning
to an orbital picture. Consider Fig. 11,47a in which the two ligands on the z axis of an
Fig. 11.47 Alteration of
octahedral orbital energies
(center) under tetragonal
distortion: (a) z ligands
out: (b) z ligands in.
Drawing is not to scale:
A„ >> 5, > S : .
Electronic Spectra of Complexes
451
ML fi complex have moved away from the central metal. In so doing, they have
reduced their interaction with the metal d orbitals that have a z component, i.e., the
d.z, d xv and d yz . As a result, these orbitals are stabilized. Because of the "center-of-
gravity" rule, the orbitals without a z component, the and d ry , will be raised a
corresponding amount. It is not possible, a priori, to predict the magnitude of these
splittings because the extent of distortion cannot be predicted. However, we can say
that the splitting of the strongly a antibonding e orbitals (5,) will be significantly
larger than that of the r 2x orbitals (S 2 ) because the latter are either nonbonding or are
involved in weaker tt interactions with the ligands. Also, both 5, and 5, will be
relatively small with respect to A„, so we are justified in regarding the distortion as a
perturbation of an octahedral geometry.
The Jahn-Teller theorem per se does not predict which type of distortion will take
place other than that the center of symmetry will remain. The z ligands can move out
as in the example discussed above or they can move in. For a “z-in" distortion, the
splitting pattern is similar to that observed for a “z-out," but the energy ordering
within the e g and t 2g levels is inverted (see Fig. 11.47b).
Consider a complex that is subject to Jahn-Teller distortion, [TiClJ 3- . The Ti 3+
ion is a d 1 species and the ~D ground term which arises from it in an octahedral field is
split into % and 2 E x terms. The % K term is the ground state and, because it is triply
\ degenerate, the Jahn-Teller theorem would predict a distortion. 48 The structure of
' [TiCl 6 ] 3- does show a slight compression of the axial ligands at low temperature, but
this is thought to be due to packing forces rather than the Jahn-Teller effect. Evidence
for a Jahn-Teller distortion is seen, however, in the electronic spectrum of the
complex. There are two absorption peaks separated by approximately 1400 cm -1 , one
resulting from excitation of an electron from the ground state ( 2 B 2k ) to the excited 2 B l/!
state and the other to the excited 2 A„ state (Fig. 11.48). 4y This assignment is sup¬
ported by EPR results, which are consistent with a tetragonal compression. For some
d' complexes, the Jahn-Teller splitting is not of sufficient magnitude to produce well-
Fig. 11.48 Electronic
spectra of Rb 2 Na|TiCl h l (I).
Cs,K|TiCU (II). and
Rb,|TiCIJ (III). [From
Ameis. R.; Kremer, S.;
Reincn, D. Innrg. Client.
2751-2754. Used
with permission.!
8 10 12 14 16 18 20- 10 W
48 In thinking about degeneracy, it is important to distinguish between terms and orbitals. For
example, a cl 3 configuration (Cr 3 ) gives rise to three electrons occupying the i l, y . and </„.
orbitals in an octahedral complex. Although the three orbitals arc degenerate, the ground state term
for this configuration is J A lc which is nondcgencratc. Thus, despite the degeneracy of the orbitals,
no Jahn-Teller distortion would be predicted for this configuration.
49 Ameis, R.; Kremer. S.; Reinen, D. Inorg. Client. 1985. 24. 2751-2754.
278
279
7 • The Solid Slate
body-centered A 11 ion. The transition metal atoms (B 11 ) and the phosphorus atoms
occur in [B^P,]" - layers, each in a square array such that each metal atom is
surrounded by a tetrahedron of phosphorus "ligands":
•=p 0=8
Isolated
tetrahedron
Note the capping phosphorus atom atop the square pyramid: It is coordinated to four
metal atoms, all on one side, highly unusual for an ion. However, if we ask whether
this is an extraordinary covalent structure for a nonmetal, we note that it is not at all
unusual for sulfur (cf. SF 4 , Fig. 6.4). 28 Although currently unknown for phosphorus in
a simple molecule, a similar structure would be expected for the isoelectronic :PF7
anion if it existed.- 2
If we examine the distances listed in Table 7.2 some interesting facts emerge. For
a given metal A, the A—P distance is constant as we might expect for an ionic alkaline
earth metal-phosphide bond. Furthermore, these distances increase calcium < stron¬
tium < barium in increments of about 15 pm as do the ionic radii of Ca 2 , Sr + , and
Ba 2 + (Table 4.4). However, the B—P distances vary somewhat more with no periodic
trends (Mn. Cu larger; Ni, Fe, Co smaller). Most interesting, however, is the larger
variability in the P—P distance from about 380 pm (Mn, Fe) to 225 pm (Cu). As it
turns out, the lower limit of 225 pm (Cu) is a typical value fora P—P bond (Table E.l,
Appendix E) and 380 pm is approximately twice the van der Waals radius of phos¬
phorus. 30 Furthermore, there is a steady reduction of this distance as one progresses
across the transition series. All of this is consistent with the hypothesis that this is an
electronic (covalent) effect.
Before examining these electronic effects, we should delve a little more deeply
into the theory and terminology of solid state chemistry than was done on pages
269-272. Specifically, if there is an infinite array of identical orbitals (say H l.v)
represented by </>„, <A|, </>,, . . . related by translational symmetry and spaced at
distance a, then we can have linear combinations, ^ A : 31
28 It might be objected that the four bonds in "trigonal bipyramidal" SF 4 consist of two short
equatorial bonds and two longer axial bonds. Recall however the very slight difference in energy
between the TBP structure and the square pyramidal structure. In a rectangular lattice, the latter
will be favoured.
w This would be the conjugate anion of l-IPF 4 acting as a BrOnsted acid. The chemistry of the
fiuorophosphorancs is not well known, but the corresponding alkyl analogues (c.g.. CH 5 PF 4 ) arc
well known and have structures similar to the one suggested (sec Chapter 6).
30 Covalent radii and van der Waals radii have not yet been discussed in any detail (see Chapter 8), but
the reader should be familiar with the general concepts from previous courses.
31 See the discussion of LCAO-MO theory. Chapter 5 (Eqs. 5.26. 5.27). where for diatomic molecules,
a = 4>n. iba = t. <!'h = ^ n. and = -At-
Solid-State Materials with Polar Bonds
a = 0 1 2 3
• • • •
4>o z <£i
-A = 2c' k ""<b„
4 . . .
(7.3)
where <A„ is the basis function of the nth orbital, t = VTTJ, an d e is the base of natural
logarithms. The linear combinations. t// A , are called Bloch functions and k is an index
that indicates which combination (irreducible representation) is involved. To return to
the previous discussion of the analogy between band structure and a mole of mo¬
lecular orbitals, consider two values for k.
If k = 0: -A„ = = 2„
<Afl = <A 0 + 'At + 2 + <A.r + 4
(7.4)
(7.5)
From simple MO theory, we expect that this nodeless function will be the most
bonding state. 32 It thus represents the bottom of the band.
,f k = Tv ‘hi,, = 2 e= £ „(~ D"
II /I
Kt„ = <A 0 - <6, + <A ; - <A, + <A 4
(7.6)
(7.7)
This is the most antibonding state. These two define the bottom and the top of the
band. The situation is the same as we have seen previously for lithium, whether there
are two, eight, or a mole of hydrogen atoms in metallic polyhydrogen. 33 Now what we
12 It should not be assumed that k = 0 is always the lowest stale. If the basis functions arc p. orbitals
(where the translational symmetry operates along the j axis), iAn will have the most nodes and be
highest in energy; iwill have no nodes and be lowest in energy. Sec Footnote 26,
33 Another array of orbitals with the same sort of linear combinations is that of the it system of
cyclopolyencs (benzene, naphthalene) with which you are probably already familiar. The molecular
orbitals constructed from sets of parallel p orbitals arc both bonding and antibonding at various
energies. The system is particularly stable if Huckel's rule (2n + 2 electrons; see also Chapter 15) is
obeyed. In the case of an infinite array of hydrogen atoms (l.v 1 ). the situation is unstable (as you
should have questioned immediately): it reverts to an array of H 2 molecules:
H H H H H H - H —H H —H H —H
This is why lithium was chosen as an example in the earlier discussion: Polyhydrogen is unstable at
ordinary pressures. Now what the chemist takes as intuitive, the solid state physicist calls a
“Peierl's distortion." In the case of lithium, a solid results because of the mixing of s and p orbitals.
There are many other interesting and important (and certain to become more so) processes hiding
under different names and different viewpoints in solid state chemistry and physics, but they are
beyond the scope of this book. The interested reader is referred to Footnote 26. Whether hydrogen
becomes metallic at attainable, high pressures is still a matter of some uncertainty. Mao, H. K.;
Heniiey, R. J. Science 1989, 244, 1462; 1990, 247. 863-864. Silvera, I. F. Ibid. 1990, 247. 863.
448
11 • Coordination Chemistry: Bonding, Spectra, and Magnetisi
Tetragonal
Distortions from
Octahedral
Symmetry
We have seen that the electronic spectra of octahedral ML 6 and tetrahedral ML 4
complexes may be analyzed with the aid of appropriate correlation, Orgel. orTanabe-
Sugano diagrams. When we move away from these highly ordered cubic structures to
complexes having lower symmetries, spectra generally become more complex. A
general consequence of reducing the symmetry is that energy levels that were degen¬
erate in the more symmetric geometry are split. With more energy levels, the number
of possible transitions increases and so does the number of spectral bands. In this
section we will examine departures from octahedral symmetry in six-coordinate
complexes.
There are a number of circumstances that can lead to a symmetry that is less than
octahedral in a six-coordinate complex. One is simple replacement of some of the
ligands of an ML 6 molecule or ion with ligands of another type. For example, if we
replace two L groups to give either cis- or /ran.(-MX 3 L 4 , the symmetry becomes C 2l , or
D 4Ii , respectively. More subtle alterations in symmetry frequently occur in complexes
having bidentate or chelating ligands. For instance, the chelated complexes.
[Cr(en) 3 ] 3+ and [Cr(ox) 3 ] 3- are not perfectly octahedral. Because of the rings associ¬
ated with the bidentate ligands, these complexes belong to the lower symmetry point
group D y Nevertheless, the perturbation is slight enough that we were able to
successfully analyze the spectra of these complexes (Fig. 11.13) as though they were
purely octahedral, and the expected absorptions for a symmetrical [CrL 6 ] 3+ species
were observed. However, if one of the ethylenediamine ligands of [Cr(en) 3 ] 3+ is
replaced with two F _ ligands to give /ran$-[Cr(en) 2 F 2 ] + , the change in symmetry is
drastic enough that to treat the new complex as pseudooctahedral is no longer valid;
rather, it must be analyzed as a tetragonal (D 4/l ) species. The alteration in energy
levels that accompanies this progression from octahedral to tetragonal symmetry is
shown in Fig. 11.45. Each of the triply degenerate T terms is split into two new terms
(an E and an A or B), with the result that six transitions are now expected instead of
three. The four lowest energy absorptions for /ra/i5-[Cr(en) 2 F 2 ] + are shown in Fig.
Fig. 11.45 Alteration of
energy levels for a d y ion as
the symmetry of its
environment changes from
octahedral (O h ) to
tetragonal (£>.,;,).
Electronic Spectra of Complexes
449
Frequency (cm" 1 x 10~ 3 )
Fig. 11.46 Electronic spectrum of//wi.s-[Cr(en) 2 F 2 ] + ; transition frequencies are given in
Table 11.20. [Modified from Dubicki, L.; Hitchman. M. A.; Day, P. Inorg. Cltem. 1970, 9.
188-290. Used with permission.]
11.46. A fifth band appears as a shoulder in the charge transfer region and the sixth has
been calculated (Table 11.20).
Tetragonal distortion from octahedral symmetry often occurs even when all six
ligands of a complex are the same: Two L groups that are trans to each other are found
to be either closer to or farther from the metal ion than are the other four ligands. A
distortion of this type actually is favored by certain conditions described by the
Jahn-Teller theorem. The theorem states that for a nonlinear molecule in an elec¬
tronically degenerate state, distortion must occur to lower the symmetry, remove the
degeneracy, and lower the energy. 47 We can determine which octahedral complexes
Tabl e 11.20
Spectral data for trans-
[Cr(en) 2 F2]CI0 4 at 4 K“
Observed
frequency (cm"')
Assignment
18.500
x-
21,700
X -
X
25,300
X -
29,300
X -
-x
41,000 (shoulder)
X -
4 A,JP)
43,655 (calculated)
x-
— 4 E I/ (P)
“ Sec Fig. 11.46.
47 Jahn, H. A.: Teller. E. Proc. R. Soc. Lomt. 1937, A/6/, 220-225. Jahn. H. A. Proc. R. Soc. Lond.
1938. AIM. 117-131.
280
7 • The Solid State
have drawn previously as a block to represent the aggregation of a very large number
of orbitals (cf. Figs. 7.19-7.21 with the simplified diagrams here):
H, H„ Hj
the physicist plots as an energy function:
o —
This graph conveys the same important information given by the energy level diagram:
The number of states (molecular orbitals) generated by the linear combination of
atomic orbitals in Eq. 7.3 is not evenly distributed over the energy range, but is
densest at the bottom and top. The number of states in the interval E + dE is known as
the density of states (DOS):
0 t _ it/a 0 - DOS —-
As can be readily seen, the density of states at any given energy is inversely propor¬
tional to the slope of the energy function at that energy.
The density of states for this type of system may be worked out qualitatively
rather easily. 34 We shall make some simplifications. First, we shall assume that in the
compound BaMn 2 P 2 the barium occurs as simple cations having no covalent interac-
34 For the complete discussion, see Hoffmann. R.: Zheng. C. J. Phys. Chem. 1985. 89. 41.5-4181.
Solid-State Materials with Polar Bonds
281
lions with the remaining atoms. 33 Then we shall let the phosphorus atoms interact with
the metal atoms as though we were dealing with a discrete, tetrahedral, molecular
complex of the sort [Mn(PR 3 ) 4 r- + . 36 In such a situation, molecular orbital theory gives
a set of four bonding MOs (a, + t 2 ) which come from the atomic 4s and 4 p orbitals ol
the metal and the phosphorus lone pair orbitals ot the same symmetry:
m |M(PR 3 ) 4 ] “PR 3
The orbitals which are principally nonbonding metal 3 d in character are split into an e
set and a /, set. The latter splitting will be discussed at some length in Chapter 11 and
need not concern us too much at present. Finally, there are the a, and t 2 anlibonding
° fb Correspondingly, we can calculate the band structure and density of states for the
extended Mn 2 P;‘ layer (Fig. 7.29). We have seen previously (Chapter 5) that if the
interacting AOs’are distinctly separate in energy, we can treat the resulting MOs as
though they came essentially from only the AOs of a given energy. We thus can look at
the DOS for the extended Mn,P; layer and find the origin of the bands. I he lowest
(- 19 eV) corresponds to the a, orbital of the isolated complex and comes from the
manganese 4s orbital and the phosphorus 3.v orbitals. The next (- 15 eV) corresponds
to manganese 4 p and phosphorus 3/;. It is possible to decompose these bands into the
relative contribution of manganese and phosphorus (Fig. 7.30) and. as we should
expect from the lower electronegativity of manganese, these bands are dominated by
the phosphorus. In contrast to these two bands which are mostly phosphorus but
partly manganese, at higher energies (between -13 and -8 eV) we find that tie
electron density is almost entirely on the manganese. In isolated metal complexes
these are the approximately nonbonding metal d orbitals.
jj X his turns out to be an oversimplification; we have seen that there is no such thing as a perfectly
ionic bond, but the simplification docs not cause serious errors (sec Footnote 34).
3 Discrete tetrahedral lMn(PR 3 ) 4 F + complexes have apparently not been prepared, but MnIPRjfeb
consists of distorted tetrahedral molecules. As we shall sec (Chapter 11 ) phosphine complexes with
large positive charges on the transition metal will be less stable than when there is electro
density on the metal, as in the [B,P : ]?- layer. The present MO discussion and .he MO diagram
above anticipate the discussion of molecular orbital theory in complexes in Chapter 11 and may
most profitably be read again after reading that chapter.
446
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
the electronic repulsion parameter 6. The apparent value of B in complexes (referred
to as B') is always smaller than that of the free ion. This results from a phenomenon
known as the nephelauxetic effect and is attributed to delocalization of the metal
electrons over molecular orbitals that encompass both the metal and the ligands. As a
consequence of this delocalization or “cloud expanding," the average interelectronic
repulsion is reduced, making B' smaller than B. The nephelauxetic ratio, /3, is given
by:
P = B'/B
(11.17)
The value of /3 is always less than one and it decreases with increasing delocalization.
There are several approaches to extracting information from a Tanabe-Sugano
diagram. One is to fit observed transition energy ratios to the diagram and thereby
obtain a value for MB. The accuracy with which this can be done by hand depends to
some extent on the precision of the Tanabe-Sugano diagrams that are used. With
small versions such as those presented in Fig. 11.42 and Appendix G, accuracy is
limited, but nevertheless they can be used to illustrate the principles of fitting. The low
spin complex [Co(en) 3 ] 3+ will serve as an example. Its spectrum (Fig. 11.43) shows
two bands, at 21,550 cm -1 and 29,600 cm -1 , which are assigned as follows:
X- 21,550 cm -1
X- 'T 2ii 29,600 cm" 1
The ratio of these two energies, given by
(11.18)
can be fitted to the diagram (Fig. 11.42) by sliding a ruler along the abscissa until a
point is found at which the measured energy level separations have this same ratio.
This is achieved at MB = 40. At this point, the value of EIB (actually E/B') for the
lowest energy transition can be read from the diagram as 38. or:
21,550 cm" 1
zr
= 38
(11.19)
Solving for B' yields 570 cm -1 , which is much smaller than B for the free Co 3+ ion
(1100 cm ') given in Table 11.18. Finally, A can be calculated from MB’ (40) and
B‘ (570 cm -1 ). The result, 23,000 cm -1 , is in very good agreement with the more
precise value (23,160 cm -1 ) found in Table 11.6.
The spectra of the three Cr 3 complexes shown in Fig. 11.13 may be analyzed
similarly. The splitting of 4 Fand 4 P terms, including mixing of the two states, is shown
in Fig. 11.44. The value of A is obtained directly from the spectrum as the energy of
the lowest energy transition (v,). When all three transitions are observed, it is a simple
matter to assign a value for 5’ since the following equation must hold:
1 5B ' = v 3 + k, - 3v, (11.20)
where the absorption frequencies increase in the order < v 2 < For example, the
value of B' for the Duoro complex is:
B'= [34,400 + 22,700 - 3(14,900)] = 827 cm -1 (11.21)
If only two transitions are observed (as, for example, in Fig. 11.13b where v 3 is
obscured by a charge transfer band), it is still possible to evaluate B‘ by other
Electronic Spectra of Complexes
447
T oble 11,19
Calculated and
experimental spectral
transitions (cm - ) for
chromium(lll) complexes
Mixing
-(/>>
Fig. 11.44 Splitting ofV
and 'P terms in an
octahedral field. Note
mixing of 4 T, K terms.
methods. 45 Once values for A and B 1 have been determined, it is possible to estimate
all of the transition frequencies for a complex. The appropriate relationships for high
spin octahedral d 3 and d 8 and tetrahedral d 2 and d 1 species are:
", = -- T 2k = A (11.22)
I/, = A 2k - T^[F) = 7.55' + I.5A - i[225 B a + (A) 2 - 18.05' A] 1 (11.23)
v 3 = A 2„ - r,„(5) = 7.56' + I.5A + J[2256' 2 + (A) 2 - 18.06' A| (11.24)
These equations, which can be solved by iterative processes. 4,1 lead to accurate
estimates of transition frequencies (Table 11.19) and quite satisfactory fitting of
spectra. Parallel equations for octahedral c/ 6 complexes are:
j/, = A - 45’ + 865 ,2 / A (11.25)
i> 2 = A + 125' + 25 ,2 / A (11.26)
[CrF 4 ] 3-
[Cr(o:
Os] 3 "
[Cr(on) 3 ] 3+
diagram
Calc.
Exp.
Calc.
Exp.
Calc.
Exp.
4 T.JP)
34,800
34.400
38,100
_«
46,500
_«
a T,„(F)
22,400
22,700
24,000
23,900
28,700
28,500
4 7V.
14,900
14.900
17,500
17,500
21,850
21,800
x.
"l V 1 V 3
■' This transition is not experimentally observed because it is masked by the charge transfer
spectrum.
45 Lever. A. B. P. J. Client. Educ. 1968. 45. 711-712.
Brown. D. R.; Pavlis. R. R. J. Client. Educ. 1985. 62. 807-808.
282
7 • The Solid State
Mn 4i-
Mn 3d =
P3 P^EEE
P 3s -
Separate atoms
Fig. 7.29 Left: Energy levels of separated Mn and P atoms. Mn-P MO's from adjacent atoms, and extended bonding.
Right: Band structure of a single [MnjPjIS- layer. [Modified from Hoffmann, R.; Zheng. C. J. Phys. Cliem. 1985, 89,
4175-4181. Reproduced with permission.)
Fig. 7.30 Total DOS of the
extended [Mn : P:;J;~ layer.
The relative contributions
of the manganese (dark
area) and the phosphorus
(light area) are indicated.
Note that the bonding
states at - 19 and - 15 cV
are dominated by the
phosphorus, that is. there is
more electron density on
the phosphorus than on the
manganese. [From
Hoffmann, R.; Zheng. C. J.
Phys. Chem. 1985. 89,
4175-4181. Reproduced
with permission.!
Now, what can we say about the phosphorus-phosphorus interaction between
layers? Comparing the layer structure of Mn 2 P^~ with the unit cell in Fig. 7.28, we see
that the 3-D structure of BaMn 2 P 2 consists of alternating Mn 2 P?“ and Ba 2+ layers.
This brings the apical phosphorus atoms of one layer close to those of the next layer,
and they interact along the z axis. If we look at the DOS for a single Mn-,P; - layer (Fig.
7.30), but now inquire as to the contribution of the phosphorus 3 p. orbital, we see that
Solid-State Materials with Polar Bond:
283
Fig. 7.31 Phosphorus 3 p.,
orbital contribution (dark
area) to the total DOS
(dashed line; cf. Figs. 7.29
and 7.30) of the [Mn 2 P 2 ]J~
layer. [Modified from
Hoffmann. R.; Zheng, C. J.
Phys. Chem. 1985, 89,
4175-4181. Reproduced
with permission.)
DOS —-
most (70%) of it is in a rather narrow band at - 15 eV (Fig. 7.31). The narrowness of a
band is an indication of its localization; these are the lone pairs that were postulated on
the basis of the :PF7 analogue (page 278). If these orbitals are completely filled, (he
lone pairs on adjacent layers will repel each other. If half filled, (hey could form inter¬
layer covalent bonds. When the layers come together, we expect the P t/ , orbitals to
interact strongly with shifts to higher (antibonding) energies and lower (bonding)
energies. In fact, all orbitals of the system with z components will interact, but only
the P ip orbitals will be sufficiently close to have much overlap.
The three-dimensional (total) DOS is illustrated in Fig. 7.32. We see a low-lying
band at - 16.5 eV corresponding essentially to a P-P bonding interaction and
Fig. 7.32 Phosphorus 3 p.
orbital contribution (dark
area) to the total DOS
(dashed line) of the three-
dimensional (total)
[Mn 2 P 2 ) 2 “ lattice. The P—P
interactions are labeled ir
and cr. The square bracket
encloses the bands arising
principally from the
manganese 3 d orbitals.
[Modified from Hoffmann,
R.; Zheng, C. J. Phys.
Chem. 1985. 89, 4175-4181.
Reproduced with
permission.)
11 - Coordination Chemistry: Bonding, Spectra, and Magnetism
The spectrum of any octahedral d b complex can be assigned with the help of
Fig. 11.42. For high spin species such as [CoFJ 3- , the only spin allowed transition is
5 T 2t! —» ’E k and only one absorption should be observed. Indeed the blue color of this
complex results from an absorption centered at 13,100 cm -1 . 44 For low spin Co 3+
complexes there are two spin allowed transitions at relatively low energies: 'A, A , -
'T Ia , and 1 A lA , — 'T 2u . There are additional spin allowed transitions at higher energies,
44 Actually the E g state is split by a Jahn-Teller effect (page 453) resulting in two peaks.
Electronic Spectra of Complexes
445
Table 11.18 _
Free ion values of
parameters B and C
(in cm -1 ) for gaseous
transition metal ions°
Configuration Ion
B
C
3d 2
Ti 2+
718
2629
V 3+
861
4165
Cr 4+
1039
4238
3d 2
Sc +
480
V 2+
766
2855
Cr 3+
918
3850
Mn 4f
1064
3 d A
Cr 2+
830
3430
Mn 3+
1140
3675
3d 5
Mn 2+
960
3325
3 d b
Fe 2+
1058
3901
Co 3+
1100
3d 7
Co 2+
971
4366
3 d s
Ni 2+
1041
4831
4 d 7
Mo 3+
610
4 d b
Rh }+
720
4 d 7
Rh 2 +
620
4002
4 d»
Pd 2+
683
2620
5 d 2
Os 6+
780
5 d 2
Re 4+
650
Ir 6+
810
5d A
Os 4+
700
5 d"
Ir 3+
660
Pt 4+
720
5 d
Pt 2+
600
" Lever, A. B. P. Inorganic Electronic
Spectroscopy, 2nd ed.; Elsevier: New York,
1986; p 115.
but they generally are masked by totally allowed transitions and hence are not
observed. Because the slope of changes more rapidly than that of the two ob¬
served peaks will be further apart in energy at larger values of A. The spectra of yellow
[Co(cn)j] 3+ and green [Co(ox) 3 ] 3 " (Fig. 11.43) confirm these expectations.
Fitting an observed spectrum to its corresponding Tanabe-Sugano diagram en¬
ables one to obtain the value of A for a complex. In addition, it is possible to evaluate
Fig. 11.43 Spectra of
[Co(cn)j|' (—) and
[Cotoxb] 3- (-). [From
Mead, A. Trans. Faraday
Soc. 1934, 30, 1052-1058.
Reproduced with
permission.]
284
7 • The Solid State
another at -6 eV that is essentially a P-P antibonding interaction. If only the lower
band is filled, we shall have P—P bonds between layers; if both are filled there will be
nonbonding (van der Waals) contacts.
We must now compare how these bands lie with respect to the energies of the
electrons in the bands arising from the metal 3 d orbitals. The bottom and top of the 3 d
band and the Fermi level change as one progresses across the transition series: 37
Mn Fe Co Ni Cu
There are two factors involved. The fraction of the band filled with electrons increases
with each increase in atomic number and addition of a valence electron. At the same
time, the level and width of the band decrease as a result of the increase in effective
atomic number. (Recall that d electrons shield poorly.) The overall result is a slow
lowering of the Fermi level from Mn to Cu. Now if we superimpose the calculated
levels of the <r P _ P and the o P _ P interactions (Fig. 7.32) upon the Fermi level diagram,
we note an interesting difference between early and late transition metals:
Mn Fc Co Ni Cu
37 Mackintosh. A. R.; Andersen. 0. K. In Electrons at the Fermi Surface : Springford. M.. Ed.;
Cambridge University: Cambridge, 1980. Andersen. O. K. In The Electronic Structure of Complex
Systems ; Phariscau. P.; Temmcrman. W. M.. Eds.; Plenum: New York. 1984. Andersen. O. K. In
Highlights of Condensed Matter Theory, Bassani. F.; Fumi. F.; Tossi. M. P.. Eds.; North-Holland:
New York. 1985. Varma. C. M.; Wilson, A. J. Phys. Rev. B: Condens. Matter 1980. B22. 3795.
Solid-State Materials with Polar Bonds
285
The P-P band is always filled, corresponding to a P—P bond (225 pm) in the coppet
compound. At the other extreme, the P-P band is also filled, giving an ant.bond.ng
interaction in addition. Thus, overall, there is a nonbonded interaction between the
two phosphorus atoms and so we should not be surprised that the P-P distance is
approximately twice the van der Waals radius of phosphorus (2 x 185 pm « 384 pm)_
We can view the progression from Mn to Cu as a redox tuning of the occupancy o
these energy levels: 38
We see that our intuition concerning the oddly coordinated phosphorus atoms, that
they seemed to resemble phosphorus atoms in discrete molecules, has borne fruit. 33
Hiah-Temperature Superconductivity was discovered in mercury metal in 1911. Below 4.2 K the re-
SuDerconductors 40 sistance of mercury drops to zero. Currently much interest is focused on h.gh-
Superconductors temperature superc onductors such as YBa 2 Cu 3 0 7 _ B . In this case, high-tem¬
perature” is about 100 ± 20 K, greater than the boiling point of nitrogen (77 K), but
much lower than climatic temperatures on Earth. Earlier superconductors need ® d t0
be cooled by the more expensive and difficultly handled liquid helium (bp - 4.. ).
Superconductivity has generated much excitement in the popular press because of the
Meissner effect illustrated by the now familiar picture of a magnet floating over the
superconductor.-" _, „
The first breakthrough superconductors were formulated as La 2 _ t ba r cuu 4 _ ft
{x < 0.2, (5 unspecified but small) and have the tetragonal, layered K : NiF 4 perovskite
structure. They had a critical temperature of about 35 K. 47 -
Observation that the critical temperature increased with pressure suggested that it
depended upon lattice distances. Therefore strontium (r + = 132 pm) was substituted
for barium (r + = 149) with some increase in T,. but dramatic improvement occurred
when Y (r. = 104 pm) was substituted for La (r + = 117 pm), and a new type ol
compound, YBa,Cu 3 0 7 _ s , was formed. 43 This is the so-called 1-2-3 superconductor
3" Here again, both ihc diphosphoranc-lypc system on the left and Ihe anionic structures on the right
arc unknown in simple phosphorus molecules, but S 2 F,„ is known, so these are reasonable
structures.
.'■» For details of the calculations and their interpretation, see Footnote 34.
« Whangbo. M-H, Torardi. C. C. Aec. Chem. Res. .991,24. 127-133 Williams. J. M : Benm A,.
Carlson, K. D.. Geiser. U.: Kao. H. C. I.; Kini. A.M.; Porter. L. C.; Schultz. A J.. Thorn R. J..
Wang. H. H.; Whangbo. M-H.;Evain. M. Acc. Chem. Res. 1988.2/. 1-7. Holland. G. F.. Stacy. A.
M Ibid. 1988. 21, 8-15. Ellis. A. B. J. Chem. Educ. 1987, 64. 836-841.
4i See Jacob. A. T.; Pechmann. C. I.; Ellis. A. B. J. Chem. Educ. 1988. 65. 1094-1095; Gunn. M.;
Porter. J. New Scient. 1988, 118 (1618), 58-63.
43 Bcdnorz J. G.; Mfiller, K. A. Z. Phys. B.: Condens. Matter 1986. 64. 189. By today' numbers, this
was not a large increase over previous values (23 K for a niobium alloy), but at that time it was
relatively large, and it opened a completely new class of materials for the study ol superconduc¬
tivity. Bcdnorz and Muller were awarded the 1987 Nobel Prize in Physics. Their acceptance lecture
is Bcdnorz. J. G.; Muller. K. A. Rev. Mod. Phys. 1988. 60. 585-599.
43 Wu. M. K.; Ashburn. J. R.; Torng. C. J.; Hor. P. H.; Mens. R. L.tGao. L.:lHuang. Z. J.. Wang. Y.
O • Chu C W Phys. Rev. Lett. 1987. 58. 908-910. The values of tonic radii arc from Table 4.4 and
arc' for C.N. = 6. In the perovskite structure the C.N. =8 for Y and 10 for Ba. so we can expect the
ions to be about 10% larger.
442 11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Fig. 11.40 Orgel diagram for d 2 , d 3 , d 1 , and d ions in octahedral and tetrahedral fields.
[From Lever, A. B. P. Inorganic Electronic Spectroscopy, 2nd ed.; Elsevier: New York.
1986; p 85. Used with permission.)
[Ti(H 2 0) 6 ] 3+ (Fig. 11.8). The single absorption band is quite broad, extending over
several thousand wave numbers. The breadth of the absorption can be attributed
mainly to the fact that the complex is not a rigid, static structure. Rather, the
metal-ligand bonds are constantly vibrating, with the result that an absorption peak is
integrated over a collection of molecules with slightly different molecular structures
and A„ values. Such ligand motions will be exaggerated through molecular collisions in
solution. In the solid state, however, it is sometimes possible to resolve spectral bands
into their vibrational components.
Sharp peaks also occur in solution spectra when the transitions involve ground
and excited states that are either insensitive to changes in A„ or are affected identically
by the changes. These terms will appear on energy level diagrams as parallel lines,
which will be horizontal in the event that energies arc independent of A„. This is the
rationale offered, for example, for the relatively sharp peaks (a few hundred cm -1 in
width) observed in the spectra of Mn' f complexes (Fig. 11.41).
Two additional factors that can contribute to line breadth and shape are spin-orbit
coupling, which is particularly prevalent in complexes of the heavier transition metals,
and departures from cubic symmetry, such as through the Jahn-Teller effect. This
latter effect, which will be discussed later in this chapter, is believed to be responsi¬
ble for the low-frequency shoulder observed on the absorption line for [Ti(H,0) 6 ] :+
(Fig. 11.8).
Tanabe—Sugano In order to treat fully the problem of interpretation of spectra, it is common to use
Diagrams diagrams provided by Tanabe and Sugano, 41 which provide an alternative means of
depicting the variation of term energies with field strength. Tanabe-Sugano diagrams
41 Tanabe. Y.: Sugano. S. J. Phys. Soc. Jpn. 1954, 9. 753-766. 766-779.
Electronic Spectra of Complexes 443
Fig. 11.41 Absorption spectrum for octahedral MnF 2 . Note the narrow lines. [From Lever,
A. B. P. Inorganic Electronic Spectroscopy, 2nd ed.; Elsevier: New York. 1986; p 451. Used
with permission.]
include both weak and strong fields and hence are more comprehensive than Orgel
diagrams. They are similar to correlation diagrams but are more useful for extracting
quantitative information.
A simplified version of the Tanabe-Sugano diagram for d b octahedral complexes
is shown in Fig. 11.42. 4 - The ground slate is always taken as the abscissa in these
diagrams with the energies of the other states being plotted relative to it. Inter-
electronic repulsion is expressed in terms of the Racah parameters B and C, which are
linear combinations of certain coulomb and exchange integrals pertaining to the
uncomplexed ion. 43 Accurate evaluation of these integrals is in general not feasible
and so these factors are instead treated as empirical parameters and are obtained from
the spectra of free ions. The parameter B is usually sufficient to evaluate the difference
in energy between states of the same spin multiplicity; however, both parameters are
necessary for terms of different multiplicity. A relationship that will prove to be quite
useful in analyzing spectra is that the difference in energy between a free ion ground
state F term and an excited P term of the same spin multiplicity (as found for d 1 , d 3 ,
d 1 , and d x configurations) is I5ZL Energy (E) and field strength are expressed on a
Tanabe-Sugano diagram in terms of the parameter B as ElB and A IB, respectively. In
order to represent the energy levels with any accuracy, it is necessary to make some
assumptions about the relative values of C and B. The ratio CIB for the diagram in
Fig. 11.42 is 4.8. For most transition metal ions B can be estimated as approximately
1000 cm -1 and C = ~4 B. More precise values are given in Table 11.18.
At weak octahedral fields, the ground state for a d b complex is 5 T 2k , which
originates from the free ion S D term (Fig. 11.42). Among the excited terms at the weak
field limit is a ljf (from the free ion '/), which falls precipitously in energy with
increasing A, eventually displacing 5 T 2x as the ground term at MB = 20. At this point
spin pairing takes place, resulting in a discontinuity in the diagram, marked by the
vertical line. From this boundary on, the low spin 'A,,, term remains the ground stale.
42 A complete set of Tanabe-Sugano diagrams is given in Appendix G.
43 Occasionally the Slater-Condon-Shortley parameters F 2 and F A arc used instead. Their relation to
the Racah parameters is B = F 2 - SF and C = 35F 4 .
286
7 • The Solid State
(from the ratio ol Y-Ba-Cu). and perhaps is the best studied. It may be prepared by
various methods, but the pH-adjusted precipitation and high-temperature decomposi¬
tion of the carbonates is typical:
2Y 3+ + 3HC0 3 - Y 2 (C0 3 ) 3 l (7. 9)
Ba 2+ + HCOj —BaCOj j, (7.10)
Cu 2+ + HCOj -=£+ CuC0 3 1 (71I)
Y 2 (C° 3 ) 3 + 4BaC0 3 + 6CuC0 3 2YBa 2 Cu 3 0 7 _ fi + !3C0 2 t (7.12)
Other procedures start with the oxides, or mixtures of oxides and carbonates. The rate
and conditions of cooling are also important. 44
The 1-2-3 superconductor has a perovskite-like structure (7.33a.c). There are
systematic oxygen atom vacancies in the unit cell compared to a stack of simple
perovskite unit cells (Fig. 7.33b). These occur between adjacent copper atoms in the
chains along the c axis. The vacancies are in the yttrium atom plane. There are also
vacancies between copper atoms along the a axis in the copper-and-oxygen planes
O m"
• M' v
O Y • Cu
^ Bn O O
w <»> (O
F.g. 7.33 (a) Unit cell of the 1-2-3 superconductor, orthorhombic, space group Pmmm. One-dimensional CuO, chains run
along the b ax.s. and two-d.mens.onal CuO, layers lie in the ah plane, (b) The cubic structure of perovskite. SrTiO, Three
urn. cells are shown stacked vertically, (c) The unit cell of the 1-2-3 superconductor in the context of the surrounding
crystal Copper atoms are surrounded e.ther by five oxygen atoms in a square pyramid or four oxygen atoms in a square
plane. |From Holland. G. F.; Stacy, A. M. Acc. Chem. Res. 1988. 21. 8-15. Reproduced with permission.]
The preparation of these superconductors is still much of an art with grinding, heating, annealing or
slow cooling, etc., and each lab has its own recipe. Mixtures are often formed with different phases
present. Procedures arc given in Footnotes 40, 41. and in Porter. L. C.; Thorn R J ■ Gciscr U •
Umczawa. A.; Wang. H. H.: Kwok. W. K.; Kao. H-C. I.: Monaghan. M. R.; Crabtree. G.V!
Carlson. K. D.; Williams. J. M. Inorg. Chem. 1987, 26. 1645-1646: Engler. E M • Lee V Y •
Nazzal, A. I.; Beyers. R. B.; Urn. G.; Grant. P. M.; Parkin. S. S. P.; Ramirez. M. L.: Vazquez.’
J. E.; Savoy. R. J. J. Am. Chem. Soc. 1987, 109. 2848-2849; Garbauskas. M. F.; Green. R. W.;
Arcndl. R. H.: Kasper. J. S. Inorg. Chem. 1988, 27. 871-873.
Solid-State Materials with Polar Bonds
287
that Se between the planes of barium atoms. The structural unit that is thought to be
responsible for the superconductivity is the Ba 2 Cu 3 0 7 ~ slab. The odd stoichiometry.
YBa 2 Cu 3 0 7 _ 5 , results from additional oxygen vacancies (defect structure) at the 01
and 02 positions such that 0.0 < 5 < 0.4; usually 5 = 0.19.
More recently, other metals such as thallium, bismuth, and lead have been
included in superconductor formulation. In one interesting series, the critical tem¬
perature has been found to increase with increasing n in susperconductors of the type
TIBa 2 Ca n _ l Cu„0 : „ +2 to a maximum of 122 K for n = 4 (Fig. 7.34).-»5 The current
maximum critical temperature is 125 K for a closely related TI 2 Ba,Cu 3 0 H) .
The following generalizations can be made about all of the high-temperature
superconductors examined to date: (I) The structures can be derived by stacking
different amounts and sequences of rock salt and perovskite-like layers of metal and
oxygen; (2) superconductivity occurs in the Cu0 2 layers; (3) the-similarity in energy
between the copper 3d and oxygen 2 p levels causes them to mix extensively in the
electronic band at the Fermi level; (4) the non-Cu0 2 layers (part of the Cu0 3 chains in
the 1-2-3 compounds, the TI-0 and Bi-0 layers in others) furnish electron density
that tunes the electronic state of the Cu0 2 layers. 46 Detailed discussion of supercon¬
ductivity theory or of band theory applied to these crystals is beyond the scope of this
Fig. 7.34 Unit cells (with idealized atomic positions) of the first four members of the homologous series
TIBa,Ca„-|Cu„0 2n > 2 . [From Haidar. P.; Chen. K.; Maheshwaran, B.; Roig-Janicki, A.; Jaggi, N. K.; Markiewicz, R. S.;
Giessen. B. C. Science 1988, 2-11. 1198-1200. Reproduced with permission.]
45 Haidar. P.;Chcn. K.; Maheswaran. B.; Roig-Janicki, A.; Jaggi. N. K.; Markiewicz. R. S.; Giessen.
B. C. Science 1988. 241, 1198-1200. For a discussion with many drawings of the various supercon¬
ductor structures, see Muller-Buschbaum. H. Angew. Chem. Ini. Engl. Ed. 1989. 28. 1472-1493.
4fi Cava. R. J. Science 1990, 247. 656-662.
440
11 - Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.16
Molar absorptivities (f) for
various types of electronic
transitions observed in
complexes 0
Type of
transition
€
(L mol -1 cm -1 )
Typical
complexes
Spin forbidden
IO -3 — 1
Many octahedral complexes
Laporte forbidden
of d 5 ions, e.g.,
[Mn(H 2 0) 6 ] 2
Spin allowed
1-10
Many octahedral complexes.
Laporte forbidden
e.g., [Ni(H 2 0)J 2+
10 -IO 2
Some square planar
complexes, e.g., [PdCI 4 ] 2-
IO 2 -IO 3
Six-coordinate complexes of
low symmetry; many square
planar complexes,
particularly with organic
ligands
Spin allowed
IO 2 -IO 3
Some metal-to-ligand charge
Laporte allowed
transfer bands in molecules
with unsaturated ligands
10 2 -10 J
Acentric complexes with
ligands such as acac or
those having P, As, etc. as
donor atoms
IO 3 — IO 6
Many charge transfer bands;
transitions in organic
species
" Lever, A. B. P. Inorganic Electronic Spectroscopy, 2nd cd.; Elsevier: New
York. 1986; Chapter 4.
Table 11.17 summarizes this information for weak field octahedral and tetrahedral
complexes. Octahedral complexes having d', d 4 , d 6 , and d 9 configurations and weak
field ligands should each give one absorption corresponding to A„. Configurations d 2 ,
d 3 , d 1 , and d g in weak octahedral fields each have three spin allowed transitions. (In
each case A„ is the energy difference between adjacent A 2ll and terms.) As we have
already seen, there are no spin allowed transitions for d 5 octahedral complexes having
weak field ligands. We shall see later that other factors, such as spin-orbit coupling
and Jahn-Teller distortions, often lead to more complex spectra than predicted with
the spin selection rule.
Another popular way of representing ground and excited states of the same
multiplicity for a particular configuration is with Orgel diagrams. Like correlation
diagrams, they portray the energies of states as a function of field strength; however,
Orgel diagrams are much simpler because excited states of multiplicities different from
that of the ground state are omitted and only weak field cases are included. An Orgel
diagram forCo 2+ (d 1 ) in tetrahedral and octahedral ligand fields is shown in Fig. 11.39.
Once again, we see the inverse relationship between the two symmetries, which arises
because a tetrahedral field is, in effect, a negative octahedral field. The diagram also
illustrates the effects of mixing of terms. As a general rule, terms having indentical
symmetry will mix, with the extent of mixing being inversely proportional to the
energy difference between them. For Co 2+ the terms involved are the two 4 T t
(tetrahedral) and 4 T lg (octahedral) levels. Mixing of terms exactly parallels the mixing
of molecular orbitals we encountered earlier (Chapter 5) and it leads to an identical
Electronic Spectra of Complexes
441
Toble 11.17 _
Ground and excited terms
having the same spin
multiplicities for weak field
octahedral (oct) and
tetrahedral (tet) complexes
Configuration
Ground term 0
Excited terms with the same spin
multiplicity as the ground term 0
d' oct, d 9
tet
% g ,
2 ^2(«)
d 1 oct, d 8
tet
3t «JV
3 ^2<>> 3/l 2(s)> 3y t
d 3 oct, d'
tet
%,. XJP)
d 4 oct, d b
tet
5 f
C 2< g )
5j
1 2(.«?)
d 3 oct, d 5
tet
6 J
None
d b oct. d 4
tet
5 t
J 2Ut)
d' oct, d 3
tet
Xjf,
d g oct, d 2
tet
% g » %„,(/>>
d 9 oct, d 1
tet
%0O
%„>
° The g subscripts are appropriate only for octahedral stereochemistry.
The ( F) and (/) notations designate the free ion term from which the
listed term originates.
result: The upper level is raised in energy while the lower level falls. This is repre¬
sented in the Co 2+ diagram as diverging lines for the pairs of 4 T^ K and J 7j levels; the
condition of no mixing is shown as dashed lines. Note that for the tetrahedral case in
the absence of mixing, the two 4 T t terms gradually approach each other in energy as
the field strength increases while just the opposite is true for octahedral complexes. As
a result, the extent of mixing is greater for tetrahedral complexes.
Orgel diagrams provide a convenient means of predicting the number of spin
allowed absorption bands to expect in a UV/visible spectrum for a complex. From
Fig. 11.39, it is clear that a complex of Co 2+ (or any other d 7 ion) should produce a
spectrum with three absorptions. A more general Orgel diagram pertaining to high spin
octahedral or tetrahedral complexes of metals with two to eight d electrons is shown in
Fig. 11.40.
Up to this point we have considered two central issues involved in interpreting
electronic spectra of transition metal complexes—the number and intensities of spec¬
tral lines. There is a third important spectral feature, the widths of observed bands,
which we have not yet discussed. Consider again the visible spectrum for
Fig. 11.39 Orgel diagram
for the Co :+ ion in
tetrahedral (left) and
octahedral (right) fields.
The dashed lines represent
the i T, terms before
mixing. (From Orgel, L. E.
J. Chan. Phys. 1955, 23.
1004-1014. Reproduced
with permission.!
288
7 • The Solid State
text (but see Problem 7.13), but it may be noted that these compounds are testing both
experimental technique and basic theory. 47
Problems 7.1 Find the spinel exceptions to the structure field map in Fig. 7.4.
7.2 Predict the structures of the following (i.e., to what mineral classes do they belong?),
a. MgCr 2 Oj b. K 2 MgF 4
7.3 Rationalize the fact that the fluorite field lies above and to the right of the rutile field (Fig.
7.5) from what you know about these structures. Does this insight enable you to predict
anything about the silicon dioxide structure?
7.4 With regard to each of the following, does it make any difference whether one uses correct
radii, such as empirically derived Shannon-Prewitt radii, or whether one uses theoretically
reasonable but somewhat misassigned traditional radii?
a. prediction of the interionic distance in a new compound. MX.
b. calculation of the radius ratio in M 3 X.
c. calculation of the enthalpy of formation of a hypothetical compound. MXj.
d. construction of a structure field map as shown in Figs. 7.4 and 7.5.
7.5 Why is graphite a good conductor whereas diamond is not? (Both contain infinite lattices of
covalently bound carbon atoms.)
7.6 It was stated casually (page 275) that the energy levels of gallium are above those of
germanium and, later, that those of arsenic lie below those of germanium. Can you provide
any arguments, data, etc., to substantiate this?
7.7 Cadmium sulfide is often used in the photometers of cameras to measure the available
visible light. Suppose you were interested in infrared photography. Using Fig. 7.25. suggest
some compounds that might be suitable for an infrared photocell.
7.8 Using Fig. 7.25. calculate the wavelength of light at which photoconduction will begin for a
CdS light meter. If you are interested in black and white photography, can you tell why this
wavelength is particularly appropriate?
7.9 A very important photographic reaction is the photolytic decomposition of silver bromide
described approximately by the following equation:
AgBr(s) --» Ag(s) + lBr,(l) ,7 - l3>
Assuming that the enthalpy of the reaction described in the equation can be equated with
the energy of the photon, use a Bom-Haber-lypc cycle to calculate the wavelength of light
that is sufficiently energetic to effect the decomposition of silver bromide. What are some
sources of error in your estimate?
7.10 There are two structures illustrated by figures in this chapter that are not identified as being
the same, although they depict the same crystal structure. Examine all the crystal struc¬
tures in this chapter and identify the two figures that are the same structure.
7.11 Convince yourself that if there were no defect vacancies in the I -2-3 superconducting slab,
its empirical formula would be Ba 3 Cu 3 C$ .
7.12 If you are certain that the true formula of the 1-2-3 superconductor is YBa 3 Cu 3 07 _« with
0.0 < 5 < 0.4, what does that imply concerning the copper atoms?
47 Whangbo. M.-H.; Evain. M.; Bono. M. A.; Williams. J. M. Inorg. Chem. 1987.26. 1829. 1831. 1832.
Matsen. F. A. J. Chem. Educ. 1987 . 64. 842. Burdett, J. K. In Perspectives in Coordination
Chemistry-. Williams, A. F.; Floriani. C.: Merbach. A. E.. Eds.; VCH: New York. 1992.
Problems
289
7.13 To follow up on Problem 7.12. the band structure arising from the copper 3d orbitals has
been calculated to be:
s a
| r - y
e V
-12
J .r - r
The x- - y 2 band lies in the Cu0 3 layers (ab plane between the Ba and Y atoms), and the
z 2 - yZ band lies along the CuO, [Cu0 3 —Cu0 3 — | chains (b axis between adjacent Ba
atoms). What can you say about the electron density on the different Cu atoms? (See
Footnotes 40. 47.)
7.14 Slishovite is a dense, metastable polymorph of SiO ; with aC.N. = 6 for silicon. It forms at
pressures above 8.5 GPa. In the meteoritic impact vs. vulcanism controversy over the
nonconformity at the Cretaceous-Tertiary boundary ("What killed the dinosaurs?"), the
presence of slishovite at the K/T boundary has been used as an argument in favor of
meteoritic impact rather than volcanic activity (See McHone, J. F.; Nieman, R. A.; Lewis.
C. F.; Yates. A. M. Science 1989. 243. 1182-1184). Discuss the possible changes involved
in the quartz-to-stishovitc 4 ’ 1 phase transitions in terms of heat and pressure, and how they
relate to meteorites vs. volcanoes. (See also Sigurdsson, H.; D'Hondt, S.; Arthur, M. A.;
Bralower, T. J.: Zachos. J. C.; van Fosscn. M.; Channell, J. E. T. Nature, 1991. 349,
482-487.)
“ Quartz has C.N. = 4 for silicon, much like p-cristobalite.
438
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Electronic Spectra of Complexes
439
Selection rules reflect the restrictions on state changes available to an atom or
molecule. Any transition in violation of a selection rule is said to be ■'forbidden,'' but
as we shall see, some transitions are "more forbidden than others” (to paraphrase
George Orwell 40 ). We shall not pursue the theoretical bases of the rules in any detail
but merely outline simple tests for their application.
The first selection rule, known as the Laporte rule, states that the only allowed
transitions are those with a change of parity: gerade to ungerade (g — u) and ungerade
to gerade (// — g) are allowed, but not g — g and u —> u. Since all d orbitals have
gerade symmetry in centrosymmetric molecules, this means that all d-d transitions in
octahedral complexes are formally forbidden. This being true, it may seem strange
that UV/visible spectroscopy for such complexes is even possible. In fact, optical
spectroscopy is not only possible but has been an important source of experimental
support for current bonding theories for complexes. The key element here is that there
are various mechanisms by which selection rules can be relaxed so that transitions can
occur, even if only at low intensities. For example, unsymmetrical vibrations of an
octahedral complex can temporarily destroy its center of symmetry and allow transi¬
tions that would otherwise be Laporte forbidden. Such vibronic (vfbrational-
electron/c) transitions will be observable, though weak (the number of molecules in an
unsymmetrical conformation at any instant will be a small fraction of the total).
Typically, molar absorptivities for octahedral complexes are in the range of I to
l() 2 L mol -1 cm -1 . In practical terms, this means that if you made upaO. 10 M solution
of a typical ML,, complex and obtained its UV/visible spectrum, the d-d absorptions
probably would be observable. On (he other hand, a 0.10 M solution of a substance
such as benzene, which has fully allowed transitions, would yield absorption peaks
that would be grossly off scale.
in tetrahedral complexes, there is no center of symmetry and thus orbitals have
no g or ti designation. However, the atomic orbitals from which the e and /, orbitals
are derived do have parity properties that have a bearing on the molecular orbitals.
The nonbonding e orbitals are purely metal d atomic orbitals (Fig. 11.21) and hence
retain their g character even in the complex. The t 2 molecular orbitals, on the other
hand, are formed from atomic d (gerade ) and p ( ungerade ) orbitals. Through this
d-p mixing, which imparts some it character to the t 2 level in the complex, the La¬
porte selection rule is relaxed. As a result, extinction coefficients for tetrahedral com¬
plexes are about It) 2 greater than those for octahedral complexes, ranging from I0‘ to
10 3 L mol -1 cm -1 .
A second selection rule states that any transition for which AS A 0 is forbidden ;
i.e., in order to be allowed, a transition must involve no change in spin stale. Looking
at the correlation diagram for a d 2 configuration in an octahedral field (Fig. 11.35). we
note that the ground state has a multiplicity of 3 (5 = 1) and that there are three
excited states with this same multiplicity: i T 2ll , 3 A 2k , and 3 7', Jf (from the 3 P). Thus we
can envision three transitions that are spin allowed:
40 "All animals arc equal, but some arc more equal than others." Orwell. G. Animal Farm: Harcourt.
Brace, and World: New York. 1946.
Transitions from 3 T lg to any of the singlet excited states are spin forbidden. A dr
octahedral complex should, therefore, give rise to an electronic spectrum consisting of
three absorptions. This will be true whether the field is weak or strong. However, it
should be observed that as the field strength increases, the separation between the
triplet ground and excited states becomes larger. Thus with increasing field strength,
transition energies become higher and spectral bands are shifted toward the UV
region. For blue [V(H 2 0) 6 ] 3+ , two of the three expected absorptions are observed in
the visible region (Fig. 11.38a): The transition to the 3 T 2a , state occurs at 17,200 cm -1
and the transition to the 3 T ig {P) state is found at 25.700 cm" 1 . The transition to ’/l-,,, is
calculated to be at 36.000 cm -1 , but because it is of low intensity and is in the high
energy portion of the spectrum where it is masked by many totally allowed transitions,
it is not observed. In the solid state (V 3+ /Al 2 0 3 ), this transition is seen at 38,000 cm" 1 .
The d 5 correlation diagram (Fig. Il.37d) is particularly interesting. The ground
state ( 6 A U .) is the only state on the diagram with a multiplicity of 6. This means that for
a d 5 octahedral complex, all transitions are not only Laporte forbidden but also spin
forbidden. Absorptions associated with doubly forbidden transitions are extremely
weak, with extinction coefficients several hundred times smaller than those for singly
forbidden transitions. It is understandable, then, that dilute solutions of Mn(II) are
colorless and only with a substantial increase in concentration is the characteristic
faint pink color of [Mn(H 2 0) 6 ] 2t observable (Fig. 11.38b).
The spin selection rule breaks down somewhat in complexes that exhibit spin-
orbit coupling. This behavior is particularly common for complexes of the heavier
transition elements with the result that bands associated with formally spin forbidden
transitions (generally limited to A .S' = ± I) gain enough intensity to be observed. Table
11.16 summarizes band intensities for various types of electronic transitions, including
fully allowed charge transfer absorptions, which will be discussed later in the chapter.
If one's goal is merely to predict the number of spin-allowed transitions expected
for a complex, a complete correlation diagram is not needed. It is only necessary to
know the number of excited states having the same multiplicity as the ground state.
Fig. 11.38 Electronic absorption spectra for (a) [V(H 2 0) 6 | 3 and (b) [Mn(H 2 0),,| 2+ . |From
Figgis. B. N. Introduction to Ligand Fields: John Wiley: New York. 1966; pp 221 and 224.
Used with permission.]
Internuclear Distances and Atomic Radii
291
Chap ter
8
Chemical Forces
In the preceding chapters attention has been called to the importance of the forces
between atoms and ions in determining chemical properties. In this chapter these
forces will be examined more closely and comparisons made among them. The
important aspects of each type of force are its relative strength, how rapidly it
decreases with increasing distance, and whether it is directional or not. The last
property is extremely important when considering the effects of a force in determining
molecular and crystal structures. Because distance is an important factor in all
interaction energies, a brief discussion of interatomic distances should preface any
discussion of energies and forces.
Internuclear
Distances and
Atomic Radii
It is valuable to be able to predict the internuclear distance of atoms within and
between molecules, and so there has been much work done in attempting to set up
tables of atomic radii such that the sum of two will reproduce the internuclear
distances. Unfortunately there has been a proliferation of these tables and a bewilder¬
ing array of terms including bonded, nonbonded, ionic, covalent, metallic, and van dcr
Waals radii, as well as the vague term atomic radii. This plethora of radii is a reflection
of the necessity of specifying what is being measured by an atomic radius. Nev¬
ertheless, it is possible to simplify the treatment of atomic radii without causing
unwarranted errors.
Van der Waals Radii If two noble gas atoms are brought together with no kinetic energy tending to disrupt
them, they will "stick” together. The forces holding them together are the weak
London dispersion forces discussed in a later section (pages 299-300). The inter¬
nuclear distance will be such that the weak attractive forces are exactly balanced by
the Pauli repulsive forces of the dosed shells. If the two noble gas atoms are identical,
one-half of the internuclear distance may be assigned to each atom as its nonbonded or
van der Waals radius. Solid argon (Fig. 8.1), for example, consists of argon atoms
spaced at a distance of 380 pm yielding a van der Waals radius of 190 pm for argon.
Although the van der Waals radius of an atom might thus seem to be a simple,
invariant quantity, such is not the case. The size of an atom depends upon how much it
290 ls compressed by external forces and upon substituent effects. For example, in XeF 4
Ionic Radii
Covalent Radii
Fig. 8.1 Unit cell of argon. Note that the connecting lines are for geometric perspective only
and do not represent bonds. [From Ladd. M. F. C. Structure and Boudin}; in Solid State
Chemistry. Ellis Horwood: Chichester. 1979. Reproduced with permission.]
the van der Waals radius of xenon appears to be closer to 170 pm than the accepted
value of 220 pm obtained from solid xenon. The explanation is that the xenon is
reduced in size because electron density is shifted to the more electronegative fluorine
atom. In addition, the partial charges induced (Xe 8+ , F 5- ) may cause the xenon and
fluorine atoms to attract each other and approach more closely. 1
Although we must therefore expect van dcr Waals radii to vary somewhat depend¬
ing upon the environment of the atom, we can use them to estimate nonbonded
distances with reasonable success. Table 8.1 lists the van der Waals radii of some
atoms.
Ionic radii are discussed thoroughly in Chapters 4 and 7. For the present discussion it
is only necessary to point out that the principal difference between ionic and van der
Waals radii lies in the difference in the attractive force, not the difference in repulsion.
The interionic distance in LiF, for example, represents the distance at which the
repulsion of a He core (Li + ) and a Ne core (F _ ) counterbalances the strong elec¬
trostatic or Madelung force. The attractive energy for Li + F“is considerably over
50(1 kJ mol -1 and the London energy of He-Ne is of the order of 4 kJ mol -1 . The
forces in the LiF crystal are therefore considerably greater and the interionic distance
(201 pm) is less than expected for the addition of He and Ne van der Waals radii
(340 pm).
The internuclear distance in the fluorine molecule is 142 pm, which is shorter than the
sum of two van der Waals radii. The difference obviously comes from the fact that the
electron clouds of the fluorine atoms overlap extensively in the formation of the F—F
bond whereas little overlap of the van der Waals radii occurs between the molecules
1 Hamilton, W. C.: Ibers.J. A. In Noble Gas Compounds-, Hyman, H. H., Ed.; University of Chicago:
Chicago. 1963; pp 195-202. Templeton, D. H,; Zalkin, A.; Forrester J. D.; Williamson. S. M. Ibid.
pp 203-210. Bums. J. H.; Agron. P. A.; Levy, H. Ibid, pp 211-220. In XcF 4 the xenon atoms do not
touch each other. The estimate of the van der Waals radius must be made by subtracting the van der
Waals radius of fluorine from the shortest nonbonded (i.e.. between molecules) xenon-fluorine
distance (320-330 pm).
436 11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.15 _
Splitting of d" terms in an
octahedral field
Term Components in an octahedral field
5 > A lK
P T„
D —> E g + T 2g
F ~
A 2f
7j s 4
E +
T ht +
T 2«
' b K
f 7 I +
7 1 +
7 2 K
' +
T \ K
tions are shown in Fig. 11.36. A d 9 metal ion has an electron vacancy or “hole" in its d
level and thus can be regarded as the inverse of a d' arrangement. These two
configurations also have identical ground state free ion terms, 2 D which will be split by
an octahedral field into the same two levels ( 2 7 2s and 2 E g ). However, the energy order
for the two levels in a d 9 metal will be just the inverse of what is shown in
Fig. 11.36 for the d' case. This hole formalism applies to all other d" configurations as
well: d 2 and c/ R , d 3 and d 7 , etc. have identical ground state terms but octahedral field
splittings that are the inverse of each other.
An inverse relationship also exists between fields of octahedral and tetrahedral
symmetries. We saw earlier in this chapter that crystal fields of these two symmetries
produce inverse splitting patterns for one-electron d orbitals. This relationship also
holds when electron-electron repulsions are added to the picture; any free-ion term
will be split into the same new terms (except for g and u designations, which are
inappropriate for tetrahedral complexes) by tetrahedral and octahedral fields, but the
energy ordering will be opposite for the two symmetries.
Correlation diagrams for d 2 to d 8 octahedral and tetrahedral complexes are shown
in Fig. 11.37. By taking advantage of the hole formalism and the octahedral-
tetrahedral inversion, all seven configurations in both geometries can be represented
with just four diagrams. In each diagram (except the one for d 5 ), free ion terms are
shown in the center, with the d" octahedral and d ,0 ~" tetrahedral splittings on the
right and the d l0- " octahedral and d" tetrahedral splittings on the left. Field strength
increases in both directions outward from the center. Only lower energy terms are
D
/ f
2
/
I
r 1
A„
1
L r
0.2 A 2
T
r
0.6 A u
\ j
_ T
'1
l.2A 0
(a) (b)
Fig. 11.36 Splitting by an octahedral field of the ground-state terms arising from (a) d' and
(b) d 2 electron configurations.
Fig. 11.37 Correlation diagrams for d" and d w ~" ions in octahedral and tetrahedral fields: (a) d 2 and </. (b) d' and d‘,
(c) d 4 and d 6 , (d) d $ . For (a-c), free ion terms are in the center, with field strength increasing in both directions. At the
two extremes arc strong field configurations for octahedral d'°~" and tetrahedral d" complexes (left) and octahedral d"
and tetrahedral d m ~" complexes (right). [From Figgis. B. N. In Comprehensive Coordination Chemistry, Wilkinson, G.:
Gillard, R. D.; McCleverty, J. A.. Eds.; Pergamon: Oxford. 1987; Vol. I, Chapter 6. pp 236-237. Used with permission.)
included. A feature that is common to all of the diagrams is that as the magnitude of
the ligand field increases, a number of energy level crossovers occur. A general rule
that governs all such crossovers is that they always involve slates of different symme¬
try and spin multiplicity: Levels of identical designation never cross. For some
configurations (c/ 4 , d 5 , d b , and d 7 ) the crossovers lead to a change in the ground state.
For d 5 , for example, the lowest energy term for the free ion is a 6 S, which splits in a
weak octahedral field to give 6 A, K as the ground state. At an intermediate field,
however, the ~T 2g state drops to lower energy than b A ^ and becomes the ground state.
The change in spin multiplicity from six to two here corresponds to a decrease in the
number of unpaired electrons from five to one; there is a transition from high to low
spin. In contrast, the ground states of octahedral d 2 , d 3 , d H , and d 111 complexes remain
the same under all field strengths. For example, the ground term of a (TiL f ,] 2+ (d 2 )
complex is X at all values of A„, which means that all such complexes will have two
unpaired electrons regardless of the nature of the ligand L. All of this is consistent, of
course, with the bonding models that have been previously discussed in this chapter.
In order to use the correlation diagrams shown in Fig. 11.37 or simplifications of
them, it is necessary to know the selection rules that govern electronic transitions.
Toble 8.1 __
Atomic radii and multiple bonding parameters (pm)
Element
H
He
Li
Be
B
C
N
8 . O
F
Ne
Na
Mg
A1
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
'Vow 0
120"-I45'
180-
180
^-HO"
155
150
150-160
160''
230
170
C (Eq. 8.4)
37
(32)
90(+1) 134
59( 4 2) 125
41( + 3) 90
77
75
126( — 2) 73
119( — 1) 71
(69)
116(+ 1) 154
86( 4 2) 145
68( + 3) 130
118
110
170( -2) 102
167 (— 1) 99
(97)
152(4-1) 196
114(4-2)
88( + 3)
74( + 4)
Element 'Vow"
Y
C (Eq. 8.4)
160
170
108(+ 1)
160
109( +2)
190
94( + 3)
220
140
20
143
31
210
207( — 2)
135
27
195-212
206(- 1)
133
26
220'
130
139^
Os
125"
Ir
126
Pt
170-180
160
l21(Td) / ’
Au
170
151(41)
116(Sq)
Hg
150
116(4 2)
140
91(4-1)
Tl
200
102(4 3)
140
88( + 2)
120
Pb
200
190
76( + 3)
120
Bi
122
22
Po
122
35
At
190
184(— 2)
117
28
Rn
190
180-200
182( - 1)
114
27
U
200'
110
Organic groups
166(4-1)
CH-,
200'
132(4-2)
C 6 H 5
170M
. Values of van dcr Waals radii from Bondi. A. J. Phys. Chem. 1964. 68. 441. unless othenvise noted.
b ionic radii (C.N. = 6) arc from Table 4.4 and are listed for comparative purposes only. For additional values, see that table,
r Covalent radii estimated from homonuclear bond lengths where available and from selected heteronuclear bonds otherw.se.
Bond lengths from Tables of Interactomic Distances and Configuration in Molecules and lons\ Sutton. L„ Ed Spec Publ. •
II and 18; The Chemical Society: London. 1958, 1965. except where noted. Values in parentheses are for ~»le ga»es not kno
to form compounds and are extrapolated from the values of neighbonng nonmetals; Allen. L. C.. Huheey. J. E. J. Inorg. Nucl.
Client. 1980. 42. 1523.
J N. L. Allinger; Hirsch. J. A.; Miller. M. A.; Tyminski, I. J.; Van-Catledge. F. A. J. Am. Chem. Soc. 1968. 90. 1199.
Cook. G. A. Argon. Helium and the Rare Gases'. Wiley (lnterscicnce): New York. 1961; Vol. I. p 13.
/ Cotton. F. A.; Richardson. D. C. Inorg. Chem. 1966, 5, 1851.
Dahl. L. F.; Rodulfo de Gil. E.; Feltham. R. D. J. Am. Chem. Soc. 1969. 91, 1655.
Kilboum. B. T.; Powell. H. M. J. Chem. Soc.. A. 1970, 1688.
' Pauling. L. The Nature of the Chemical Bond. 3rd ed.; Cornell University: Ithaca. NY, 1960.
-J In direction perpendicular to ring.
Internuclear Distances and Atomic Radii
293
(Fig. 8.2) because of the rapidity with which repulsive energies increase with decreas¬
ing distance. Now it might be supposed that the equilibrium distance in the F 2
molecule is that at which the maximum overlap of the bonding orbitals occurs.
However, if this were the sole criterion, the F 2 molecule would "collapse" until the
two nuclei were superimposed. This would cause the orbital wave functions to have
identical spatial distributions and the maximum possible overlap. Obviously this does
not occur because of repulsions between the two positive nuclei, and repulsions
between the inner electron core and the electrons of the other atom. We can estimate
the radius of the He core (Z = 9) by using Pauling's estimate for the isoelectromc F
ion 7 pm. 2 To this we add the radius of the overlapping orbital from the second F
atom. For the latter we can use the van der Waals (VDW) radius of fluorine (150 pm.
Fig 8 2, almost certainly too large) or the ionic radius of fluoride (119 pm. Fig. 8.3a,
probably too small). The F + [He]-F- (or [He]~-F vow ) distance will be about 130
Fig. 8.3 (a) Hypothetical F+F" ion-pair molecule illustrating repulsion between the inner He
core and the "lone pair" of the “F- ion.” (b) More realistic representation of repulsions
between inner core and valence shell electrons. (The He core is not drawn to scale in either
sketch.)
2 Pauling L The Nature of the Chemical Bond. 3rd ed.; Cornell University: Ithaca. NY I960; p 514.
This assumes that the size of the He core will be unaffected by penetration of (he Is and 2 p electrons,
which is not quite true.
434
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Thus far in this chapter we have seen electronic spectra of four complexes:
[Ti(H 2 0) 6 ] 3+ (Fig. 11.8), and [Cr(en) 3 ] 3+ , [Cr(ox) 3 l 3 ~, and [CrFJ 3 " (Fig. 11.13).
Casual inspection of these examples reveals that the number of absorptions varies. At
the heart of the interpretation of electronic spectra is the question of how many
absorptions are expected for a given complex. Answering this question requires an
accurate energy level diagram for the complex of interest as well as familiarity with the
selection rules governing electronic transitions.
The energy level diagrams that have been presented thus far for transition metal
complexes are based on the so-called one-electron model even if the central metal ion
has more than one d electron. In other words, the effects of electron-electron repul¬
sions have been ignored. Because these repulsions will make a significant contribution
to electron energies in any complex that has more than one d electron or more than
one J-level vacancy, they must be taken into account in interpreting spectra. The
approach that is usually followed in developing an energy level diagram for a complex
is to begin with the d" configuration of the free ion and then to add, in turn, the effects
of interelectronic repulsions and the effects of surrounding ligands. Our discussion will
be mainly qualitative and will use the concept of fields produced by ligands, as
introduced in the treatment of crystal field theory (page 394). However, from here on
we will use ligand field terminology to emphasize that the discussion applies quite
broadly to bonding models that range from the pure crystal field theory through the
qualitative molecular orbital approach described in the foregoing section of this
chapter (page 413).
As we saw in Chapter 2, electron-electron repulsions cause a given electron
configuration to be split into terms. However, for the simplest case, d', there will be
no such splitting of the free ion levels because there is only one electron. Thus we
have only one term, the ground state : D, because the five d orbitals are degenerate and
the electron has an equal probability of being in any one of them. As we have also seen
previously, these five d orbitals will, under the influence of an octahedral field (either
weak or strong), be split into / 2j , and e e orbitals. The ~D term likewise will be split into
2 T 2t , and 2 E f! terms in an octahedral complex.
For the d 2 configuration, electron-electron interactions come into play, giving rise
to not only a ground state free-ion term ( 3 F) but a number of excited state terms ( /’,
'G. 'D, and 'S) as well. Now we must be concerned with how each of these terms is
affected by the ligand field. If the separation between terms is large compared to the
perturbation produced by the ligands, we have the weak field case. It. on the other
hand, the ligand field splitting is large in comparison to the energy difference between
terms, we have the strong field condition. Figure 11.35 shows the free ion terms of a d
configuration and how they are split in the presence of a weak octahedral field (leit
side of the diagram). The right side of the diagram shows the effects of a strong
octahedral field. The lines connecting the weak and strong field extremes allow one to
estimate the relative energies of states resulting from intermediate fields. Construction
of the strong field side of a correlation diagram such as this one for d 2 is beyond the
scope of this text, but development of the weak field portion is more easily
accomplished.
The wave functions for S. P. D, F. etc. terms have the same symmetry as the
wave functions for the corresponding sets of s. p. d.f, etc. orbitals. This means that a
D term is split by an octahedral field in exactly the same manner as a set of d orbitals
39 The excited state terms may be obtained by methods described in Appendix C.
:Ay‘A'.T;-TT"
Electronic Spectra of Complexes
435
/ \
'/ '
// --. \ \
^ \ \
JjL. ' \ ' \
(9) \ \ \ V \
'£ W SN \
—r— M \ \
(2) \ y 3 V
—1- ' T vv'
(3) \ ' - \ \
' \ W
\ \\
w \ w
\ ' \ \
\ ' \ \
Hh _A\j__
(9)
\ \ \
\ \ \
12 Dq
2Dq
9 Dq
(a) (b) tc) (d) (c)
Fig. 11.35 Correlation diagram for a d~ ion in an octahedral field, (a) Free ion terms;
(bl weak field terms; (c) strong field ground and excited configurations; (d) strong field terms;
(e) intermediate field region. The numbers in parentheses indicate how many microstales are
associated with each term or configuration. |From Lever. A. B. P. Inorganic Electronic
Spectroscopy, 2nd ed.; Elsevier: New York. 1986; p 82. Used with permission.!
and that the splitting for an F term is the same as that for a set of/orbitals, and so on.
Transformations for terms S through / in an octahedral field are given in Table 11.15.
The orbital degeneracies associated with the terms A, E, T, D, and Fare 1,2, 3, 5, and
7, respectively. Note that the sum of the degeneracies of the individual components in
an octahedral field is equal to that of the original term; in other words, overall
degeneracy is conserved. The spin multiplicity of each component will be the same as
that of its parent term because the spin state of an electron is unaffected by the
symmetry of an external field. Thus the ground state 3 F term for a d 2 configuration will
be split into three terms in an octahedral field: a 3 r,,,a 3 r 2 ,. and a 3 A 2k , consistent
with the three levels shown in Fig. 11.35.
The energies of terms in a weak octahedral field will be such that the average
energy, or the barycenter. is equal to that of the originating free-ion term in a spherical
field. Energy diagrams for the ground-state terms associated with d' and d 2 configura-
294
8 • Chemical Forces
Internuclear Distances and Atomic Radii
Table 8.2 _
Comparison of additive
and experimental bond
distances (pm)
(160) pm. The experimental bond distance in F 2 is 142 pm. about halfway between the
two admittedly crude estimates. Corresponding values for the other halogens are 190
(210) versus 199 pm for Cl, 220 (230) versus 228 pm for Br. and 250 (260) versus 267 pm
for I. This is not meant to imply that the covalent bond in F, is either an ionic F + F~ or
a van der Waals [He] — F; it isn’t (see Problems 8.30 and 8.31). The point here is not
the crude estimation of values easily obtained experimentally, but the physical model
that explains why the covalent radii of the halogens are 71, 99. 114, and 133 pm,
respectively. The chief factor in determining the covalent radii of atoms is the size of
the core electron cloud beneath the valence shell. This might be loosely termed the
“van der Waals radius of the core.”
Table 8.1 lists covalent radii obtained by dividing homonuclear bond distances by
two. In many cases the appropriate homonuclear single bond has not been measured
and the assigned covalent radius is obtained indirectly by subtracting the covalent
radius of element B in a heteronuclear bond AB to obtain the radius of atom A.
The values in Table 8.1 are reasonably additive, that is, the covalent bond
distance in a molecule AB„ can be estimated reasonably well from r A + r B . Some
typical values are listed in Table 8.2. The agreement is fairly good. In the case of
molecules with several large substituent atoms around a small central atom such as
CBr 4 and CCI 4 , the crowding apparently causes some lengthening of the bond. There
are other cases in which the additivity of the radii is rather poor. For example, the
^ an d F F bond distances are 74 and 142, respectively, yielding covalent radii
of 37 and 71 pm. However, the bond length in the HF molecule is not 108 pm. but 92
pm. If we assume that the size of the fluorine atom is constant, then the radius of
hydrogen in HF is 21. Alternatively, we could assume that the fluorine atom is
somewhat smaller in the HF molecule than in the F, molecule, an extremely unlikely
situation. Or more realistically, we can admit that the hydrogen atom is unique, that it
has no inner repulsive core to determine its covalent radius but that in bonding the
proton often partially penetrates the electron cloud of the other atom and that the bond
distance is determined by a delicate balance of electron-nucleus attractions and
nucleus-nucleus repulsions. However, this does not really solve our problem, for a
widespread deviation from additivity results from the effect of differences in elec¬
tronegativity between the bonding atoms. It is usually observed that the bond length
between an electropositive atom and an electronegative atom is somewhat shorter
Moloculo
Bond
fA + r b
r., P
HF
HF
108
92
HCI
HCI
136
128
HBr
HBr
151
142
HI
HI
170
161
CIF
CIF
170
163
BrF
BrF
185
176
BrCI
BrCI
213
214
ICI
ICI
232
232
CH„
CH
114
109
cf 4
CF
148
136
cci„
CCI
176
176
CBr 4
CBr
191
194
ci<
Cl
210
215
295
than expected on the basis of their assigned covalent radii. Over fifty years ago
Schomaker and Stevenson 5 suggested the relation
r AB - r A + r B - 9A (8.1)
where r is in pm, and is the difference in electronegativity between atoms A and B
in Pauling units. Several workers have suggested modifications to improve the ac-
j curacy, but only one will be mentioned here. Porterfield has found that Eq. 8.2 is
somewhat more accurate and has a better theoretical justification: 4
Cab = r A + ''b ~ 7(A) 2 (8.2)
The significance of the bond shortening in highly polar molecules is reasonably clear.
Heteropolar bonds are almost always stronger than expected on the basis of the
corresponding homopolar bonds (see discussion of ionic resonance energy, Chapter
5). The atoms in the molecule AB are therefore held together more tightly and
compressed somewhat relative to their situation in the molecules AA and BB, which
are the basis of the covalent radii. It is helpful to analyze the source of this stabiliza¬
tion somewhat more closely than merely labeling it "ionic resonance engery.” To a
first approximation, it is caused by the extra bonding energy (“ionic” or Madelung
energy) resulting from the partial charges on the atoms:
H + — F‘~
(8.3)
The difference in electronegativity between fluorine and hydrogen is about 1.8 Pauling
units, predicting a shortening of about 16 pm (Eq. 8.1). The exact fit with the
experimental data (108 - 92 = 16 pm) is fortuitous (Eq. 8.2 yields A r = 23 pm), and
the importance of these equations lies in the predicted shortening and strengthening of
heteropolar bonds. This is an important aspect of covalent bonding.
For a polyvalent atom the partial charge builds up every time another highly
electronegative substituent is added. Thus the partial charge on the carbon atom in
carbon tetrafluoride is considerably larger than it is in the methyl fluoride molecule,
and so all of the C—F bonds shrink, though the effect is not as great for the last
fluorine as for the first:
C—F (pm)
CH 3 F
139.1
ch 2 f 2
135.8
CHFj
133.2
cf 4
132.3
Peter 5 has combined the Schomaker-Stevenson equation with Eq. 6.8, which
relates bond length to bond order, and obtained:
r AB = r A + r a ~ I0| A x I ~ (Ca + C B - I7| A |) log n (8.4)
3 Schomaker. V.; Stevenson. D. P. J. Am. Chem. Soc. 1941, 63, 37-40.
4 Porterfield. W. W. Inorganic Chemistry; Addison-Wesley: Reading. PA, 1984; p 167, and personal
communication.
5 Peter. L. J. Chem. Educ. 1986. 63. 123.
432
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
A B, B,B,B 4 B s B 6 By C, C 2
8 10 12 14 16 18 20
Ionization energy (eV)
Fig. 11.32 He(l) ultraviolet pholoelectron spectrum of Cr(CO) ft . Peak positions correspond
to relative energies of molecular orbitals in the complex. [From Higginson. B. R.; Lloyd. D.
R.; Burroughs. P.; Gibson. D. M.; Orchard, A. F. J. Cliem. Soc.. Faraday Trans. 2 1973. 69.
1659-1668. Used with permission.]
band in the photoelectron spectrum is examined under very high resolution and its first
derivative is taken (Fig. 11.33), one observes a vibrational progression that has
frequency spacings corresponding to M—C stretching. The vibrational fine structure
results because the [Cr(CO) 6 ] + ions that form may be in a vibrational ground state or
in one of several vibrational excited states (Fig. 11.34). The value of r MC for
[Cr(CO) 6 ] + obtained from the fine structure is 325 cm' 1 , compared to 379 cm -1 for the
neutral molecule, the reduction being consistent with involvement of the l 2t! electron
Ionization energy (eV)
12.8 13.0 13.2
Electron kinetic energy (eV)
Fig. 11.33 Expanded view
of the peak labeled B| in
the PES spectrum shown in
Fig, 11.32 along with its
first derivative. [From
Hubbard. J. L.;
Lichtenbergcr, D. L. J.
Am. Chem. Soc. 1982. 104.
2132-2138. Reproduced
with permission.]
Electronic Spectra of Complexes
433
Fig. 11.34 Potential well
representations for M(CO),,
and M(CO), showing origin
of vibrational fine structure
in PES spectrum. [Modified
from Hubbard. J. L.;
Lichtenberger. D. L. J.
Am. Chem. Soc. 1982, 104.
2132-2138. Used with
permission.]
lntemuclcar distance
in n bonding. It is estimated that this frequency shift corresponds to a I4-pm increase
in the M—C bond length upon ionization. Further support that such a structural
change occurs is provided by intensity data in the spectrum. If the most intense
transition that is observed is to the ground vibrational state, it is an indication that
there is little alteration of structure when ionization occurs (by the Franck-Condon
principle). However, if the most intense transition involves a vibrational excited stale,
it can be concluded that a substantial perturbation in geometry has taken place in
going from the neutral molecule to its positive ion. In the case of Cr(CO),., the latter
was observed, substantiating the conclusion drawn from the frequency shift, namely,
that the t 2lt orbital is a rr-bonding orbital.
Electronic Spectra
of Complexes 38
The variety of colors among transition metal complexes has long fascinated ihe
observer. For example, aqueous solutions of octahedral [Co(H 2 0) fi ] 2+ are pink but
those of tetrahedral [CoCIJ 2- are blue. The green color of aqueous [Ni(H 2 0) 6 ] 2+
turns blue when ammonia is added to the solution to give [Ni(NHj) 6 ] 2+ . The reduction
of violet [Cr(H,0) 6 ] 3+ gives bright blue [Cr(H 2 0)J 2 . As with all colors, these arise
from electronic transitions between levels whose spacings correspond to the wave¬
lengths available in visible light. (Of course, when a photon of visible light is absorbed,
it is its complementary color that we actually see.) In complexes, these transitions are
frequently referred to as d-d transitions because they involve the molecular orbitals
that are mainly metal d in character (the e K and i 2k or e and i 2 orbitals in octahedral and
tetrahedral complexes, respectively). Obviously, the colors produced are intimately
related to the magnitude of the spacing between these levels. Since this spacing
depends on factors such as the geometry of the complex, the nature of the ligands
present, and the oxidation state of the central metal atom, electronic spectra of
complexes can provide valuable information related to bonding and structure.
58 Lever, A. B. P. Inorganic Electronic Spectroscopy. 2nd ed.; Elsevier: New York. 1986. Figgis.
B. N. In Comprehensive Coordination Chemistry: Wilkinson. G.; Gillard, R. D.; McCleverty ,J. A.,
Eds.: Pergamon: Oxford, 1987; Vol. 2. Chapter 6. Figgis. B. N. Introduction to Ligand Fields: John
Wiley: New York. 1966.
296
8 • Chemical Forces
where r A and r R are single bond covalent radii, and C A and C„ are unitless multiple-
bond parameters for each element (Table 8.1). With a few notable exceptions. Eq. 8.4
gives reasonable estimates of bond lengths over a wide range of bond order and
electronegativity differences. 6
Types of
Chemical Forces
Covalent Bonding
This topic has been discussed extensively in Chapters 5 and 6, so only those aspects
pertinent to comparison with other forces will be reviewed here. In general, the
covalent bond is strongly directional as a result of the overlap criterion for maximum
bond strength. We have seen previously the implications that this has for determining
molecular structures. In addition, the covalent bond is very strong. Some typical
values 7 for purely covalent bonds are P—P, '-200 kJ mol 1 ; C C, 346 kJ mol ; and
H — h, 432 kJ mol -1 . The smaller atoms can effect better overlap and hence have
stronger bonds. Bond polarity can increase bond strength (cl. Pauling's electronega¬
tivity calculations. Chapter 5), and so we find a few much stronger bonds such as
Si—F (which probably includes some n bonding as well), 565 kJ mol Homopolar
bonds between small atoms with repulsive lone pairs tend to be somewhat weaker than
average, for example, N—N, 167 kJ mol" 1 , and F— F, 155 kJ mol - '. Nevertheless, a
good rule of thumb is that a typical covalent bond will have a strength of about
250-400 kJ mol -1 . As we shall see. this is stronger than all other chemical interactions
with the exception of ionic bonds.
Because of the complexity of the forces operating in the covalent bond, it is not
possible to write a simple potential energy function as for the electrostatic forces such
as ion-ion and dipole-dipole. Nevertheless, it is possible to describe the covalent
energy qualitatively as a fairly short-range force (as the atoms are forced apart, the
overlap decreases).
Ionic Bonding
The strength of a purely ionic bond between two ions can be obtained quite accurately
by means of the Born-Land<5 equation (Chapter 4). Neglecting repulsive forces, van
der Waals forces, and other small contributions, we can estimate the energy of an ion
pair simply as
E
Z+Z~e
4-nre,,
(8.5)
For a pair of very small ions, such as Li + and F . we can estimate a bond energy of
about 665 kJ mol"'. The experimental values are 573 kJ mol 1 (Appendix E) for
dissociation to atoms and 765 kJ mol 1 [573 + A//| U (Li) + A// ea (F)| tor dissociation
to ions. For a pair of larger ions, such as Cs and l", the energy is correspondingly
We shall see in Chapter 16 lhal some bonds, such as the Cr—Cr bond, arc particularly sensitive lo
the nature of the substituents.
7 Tables of bond energies can be found in Appendix E.
» The "bond strength" thus obtained refers, of course, to the dissociation of the ion pair to the
separated ions. M"X' -» M + X". It is somewhat easier to dissociate an ion pair into the
uncharged constituent atoms, M X" - M + X. because the ionization energy of the metal is greater
than the electron affinity of the nonmetal.
Types of Chemical Forces
297
smaller or about half as much. It is evident that the strength of ionic bonds is of the
same order of magnitude as covalent bonds. The common notion that ionic bonds are
considerably stronger than covalent bonds probably results from mistaken interpreta¬
tions of melting-point and boiling-point phenomena, which will be discussed later.
Ionic bonding is nondirectional insofar as it is purely electrostatic. The attraction
of one ion for another is completely independent of direction, but the sizes and
numbers of ions determine crystal structures. Compared with the forces to follow,
ionic bonding is relatively insensitive to distance. It is true that the force between two
ions is inversely proportional to the square of the distance between them and hence
decreases fairly rapidly with distance, but much less so than most other chemical
forces.
Ion-Dipole Forces The various factors affecting the magnitude of the dipole moment in a polar molecule
were discussed in previous chapters. For the present discussion it is sufficient to
picture a molecular dipole as two equal and opposite charges U/~) separated by a
distance r'. The dipole moment, /a. is given by
M = qr' < 8 - 6 >
When placed in an electric field, a dipole will attempt to orient and become aligned
with the field. If the field results from an ion. the dipole will orient itself so that the
attractive end (the end with charge opposite to that of the ion) will be directed toward
the ion and the other, repulsive end directed away. In this sense, ion-dipole forces
may be thought of as "directional," in lhal they result in preferred orientations of
molecules even though electrostatic forces are nondirectional.
The potential energy of an ion-dipole interaction is given as
= 1 (8.7)
4 nr 2 e„
where Z- is the charge on the ion and r is the distance between the ion and the
molecular dipole:
lon-dipole interactions are similar to ion-ion interactions, except that they are more
sensitive to distance (1//- instead of \/r ) and tend lo be somewhat weaker since the
charges (q. q~) comprising the dipole are usually considerably less than a full
electronic charge.
Ion-dipole forces are important in solutions of ionic compounds in polar solvents
where solvated species such as Na(OH 2 )J" and F(H 2 0) v (for solutions of NaF in H 2 0)
exist. In the case of some metal ions these solvated species can be sufficiently stable to
be considered as discrete species, such as [Co(NH 3 ) A ] 3+ . Complex ions such as the
latter may thus be considered as electrostatic ion-dipole interactions, but this over¬
simplification (Crystal Field Theory; see Chapter 11) is less accurate than are alter¬
native viewpoints.
430
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.14 _
Infrared carbonyl stretching
frequencies (cm' 1 ) for some
W(CO)sL complexes"
h 2 s
1916
1935
2076
C
Ph 2 S
1930
1943
2074
e
Me 2 S
1932
1937
2071
c
CyNHj''
1894
1929
2071
e
pyridine
1895
1933
2076
r
aniline
1916
1929
2071
f
CH 3 CN
1931
1948
2083
b
H 3 P
1921
1953
2083
c
Me 3 P
1947
1937
2070
/
Me,PhP
1947
1938
2071
/
MePh 2 P
1947
1940
2072
I
Ph 3 P
1942
1939
2072
/
(MeS) 3 P
1946
1940
2062
X
(EtO) 3 P
1959
1945
2078
(PhO) 3 P
1965
1958
2083
g
V
1981
1979
2087
h
Br 3 P
1991
1982
2093
h
C1 3 P
1990
1984
2095
1
F 3 P
2007
1983
2103
X
c 2 h 2
1952
1967
2095
J
C(OEt)Me
1958
1945
2072
k
c 2 h 4
1973
1953
2088
J
" A variety of solvents were used in these studies and because
frequency values are generally solvent dependent, small
differences should not be taken as significant. The forbidden B
mode, frequently present in these spectra as a weak absorption,
has been omitted.
Stolz. I. W.; Dobson. G. R.; Shcline, R. K. Inorg. Chem.
2, 323-326.
«' Hcrbcrhold, M.; Suss. G. J. Chem. Res. (M) 1977, 2720.
d Cy = cyclohcxyl.
' Kraihanzcl. C. S.; Cotton. F. A. Inorg. Chem. 1963, 2.
533-540.
f Bancroft. M.; Dignard-Bailey. L.; Puddephatt. R. J. Inorg.
Chem. 1986, 25. 3675-3680.
Kciter, R. L.; Verkade, J. G. Inorg. Chem. 1969 . 8.
2115-2120.
• Fischer, E. O.; Knauss. L. Chem. Ber. 1969. 102. 223-229.
L
A n>
£
Af 1
Reference
Me,NCHO
1847
1917
2067
b
Me,CO
1847
1920
2067
b
Et 2 0
1908
1931
2074
b
1 Poilblanc, R.; Bigorgnc, M. Bull. Chim. Soc. Fr. 1962, 1301.
J Stoltz. 1. W.; Dobson. G. R.; Shcline, R. K. Inorg. Chem.
1963, 2. 1264-1267. j .
4 Darcnsbourg, M. Y.; Darensbourg. D. J. Inorg. Chem. 1970,
9. 32-39.
Molecular Orbital Theory
431
L L L L
Fig. 11.31 Vibrational
stretching modes and their
symmetries for M(CO) s L
complexes.
NO > CO. RNC, PF 3 > PCI 3 , C,H 4 , P(OPh) 3 > P(OEt) 3 >
C(OR)R > C 2 H 2 > P(SR) 3 > PPh 3 > PR 3 > R 2 S > Ph 3 >
RCN > aniline > alkyl amines > ethers > alcohols
This series shows many of the trends that might have been expected on the basis
of electronegativity, especially for the phosphorus-bearing ligands: PF 3 > PC1 3 >
P(OR) 3 > PR 3 . The similarity of phosphites and phosphines is more than might have
been predicted from electronegativity arguments indicating that there may be signi¬
ficant O—P 7r bonding in the phosphites and competition for the phosphorus cl orb¬
itals. Alkyl amines, ethers, and alcohols have no empty low-lying orbitals and hence
form the weak end of the 7r-acceptor scale.
Care must be taken in applying a 7r-acidity scale such as the one just presented. In
using it to interpret 1R data for carbonyl complexes, one should keep in mind that
factors other than inherent ligand 7r acidity can also influence y co values. Given that
a bonding is much more important than it bonding in these systems, differences in
cr-donating capacity among ligands may outweigh trends in 7r-accepting ability. For
example, replacing the phenyl groups on phosphorus with methyl groups has little
effect upon the carbonyl stretching frequencies of W(PR 3 )(CO) s complexes even
though the 7r-acidity series would predict that this substitution should cause an
increase in v co . It is possible that the expected increase is not observed because, in
addition to being a better 7r acid. Ph 3 P is a better <j donor compared to Me 3 P, 33 making
the total amount of electron density that is available on the metal greater in the Ph 3 P
complex. It is also possible that the change in ligand substituents introduces a steric
perturbation that is substantial enough to have a greater effect than electronic factors
(see Chapter 15).
Plioloelectron spectroscopy. Photoelectron spectroscopy (see Chapter 5) has
been used to obtain metal-carbon stretching frequencies for Group VIB (6) carbonyl
complexes. 16 - 37 The full spectrum for Cr(CO) 6 is shown in Fig. 11.32. The signal
labeled corresponds to the ionization of an electron from a l 2ll orbital, which of
course is the orbital having appropriate symmetry to interact with the 7r orbital of
carbon monoxide. If in fact this interaction exists, removal of the electron should
weaken the metal-carbon bond and decrease its stretching frequency. If, on the other
hand, there is no interaction, one would expect to see little change in v MC upon
ionization since the electron would be coming from a nonbonding orbital. When the B,
55 Photoclcctron spectra have been interpreted in terms of PhjP being a better <r donor than Me 3 P; see
Bancroft. M.; Dignard-Bailey. L.; Puddephatt. R. J. Inorg. Chem. 1986, 25, 3675-3680. For a
discussion of phosphine basicity, see Angelici, R, J.: Bush. R. C. Inorg. Chem. 1988, 27. 681-686;
Sowa. Jr.. J. R.; Angelici. R. J. Inorg. Chem. 1991. 30. 3534-3537. For recent attempts at separating
a and w effects, see Rahman. M. M.; Hong. Y. L.; Prock. A.; Gicring, W. P. Organomelalllcs 1987,
6. 650-658; Wang, S. P.; Richmond. M. G.; Schwartz. M. J. Am. Chem. Soc. 1992, 114. 7595-7596.
36 Hubbard, J. L.; Lichtcnbergcr, D, L. J. Am. Chem. Soc. 1982, 104, 2132-2138.
37 For another view, see Hu, Y.-F.; Bancroft. G. M.; Bozck. J. R.; Liu, Z.; Sutherland, D. G. J.;Tan.
K. H. J. Chem. Soc. Chem. Common. 1992. 1276-1278.
298
8 Chemical Forces
Dipole-Dipole
Interactions
Induced Dipole
Interactions
The energy of interaction ot two dipoles 9 may be expressed as
r_ -W 2
47rr 3 e 0 (8.8)
This energy corresponds to the "head-to-tail" arrangement shown in Fig. 8.4a. An
alternative arrangement is the antiparallel arrangement in Fig. 8.4b. The second
arrangement will be the more stable if the molecules are not too ’‘fat.'' It can be shown
that the energies of the two arrangements are equal if the long axis is 1.12 times as lone
as the short axis. Both arrangements can exist only in situations in which the attractive
energy is larger than thermal energies (RT = 2.5 kJ mol -1 at room temperature). In
the solids and liquids in which we shall be interested, this will generally be true. At
higher temperatures and in the gas phase there will be a tendency for thermal motion
to randomize the orientation ol the dipoles and the energy of interaction will be
considerably reduced.
Dipole-dipole interactions tend to be even weaker than ion-dipole interactions
and to fall off more rapidly with distance (I/r>). Like ion-dipole forces, they are
directional in the sense that there are certain preferred orientations and they are
responsible for the association and structure of polar liquids.
If a charged particle, such as an ion, is introduced into the neighborhood of an
uncharged, nonpolar molecule (e.g., an atom of a noble gas such as xenon), it will
distort the electron cloud of the atom or molecule in much the same way that a charged
cation can distort the electron cloud of a large, soft anion (Fajans' rules. Chapter 4).
The polarization ol the neutral species will depend upon its inherent polarizability
("softness"), a. and on the polarizing field afforded by the charged ion, Z±. The
energy of such an interaction is
.. I Z'-ae 2
E ~ ~2~H~ (M)
In a similar manner, a dipole can induce another dipole in an otherwise uncharged,
nonpolar species. The energy of such an interaction is
where p. is the moment of the inducing dipole.
Both of these interactions tend to be very weak since the polarizabilities of most
species are not large. Because the energies vary inversely with high powers of r, they
are effective only at very short distances. Their importance in chemistry is limited to
situations such as solutions of ionic or polar compounds in nonpolar solvents.
OO
Fig. 8.4 (a) Head-to-tail
arrangement of dipoles;
(b) antiparallel arrangement
of dipoles.
v Multiplying Eqs. 8.5, 8.7. 8.9. and 8.10 by Avogadro's number yields the correct energy for one mole
of each species interacting. Since Eq. 8.8 involves two molecules of the polar species, multiplying by
N yields the energy of two moles of dipoles.
Types of Chemical Forces
299
Instantaneous
Dipole-Induced
Dipole Interactions 10
Even in atoms in molecules which have no permanent dipole, instantaneous dipoles
will arise as a result of momentary imbalances in electron distribution. Consider the
helium atom, for example. It is extremely improbable that the two electrons in the \s
orbital of helium will be diametrically opposite each other at all times. Hence there
will be instantaneous dipoles capable of inducing dipoles in adjacent atoms or mole¬
cules. Another way of looking at this phenomenon is to consider the electrons in two
or more "nonpolar” molecules as synchronizing their movements (at least partially) to
minimize electron-electron repulsion and maximize electron-nucleus attraction.
Such attractions are extremely short ranged and weak, as are dipole-induced dipole
forces. The energy of such interactions may be expressed as
( 8 . 11 )
where p. is the mean instantaneous dipole, or more conveniently as
( 8 . 12 )
Repulsive Forces
where a is the polarizability and / is the ionization energy of the species.
London forces are extremely short range in action (depending upon I//•<>) and the
weakest of all attractive forces of interest to the chemist. As a result of the cf- term,
London forces increase rapidly with molecular weight, or more properly, with the
molecular volume and the number of polarizable electrons.
It can readily be seen that molecular weight per se is not important in determining
the magnitude or London forces as reflected by the boiling points of IL, MW = 2,
bp = 20 K; D ; , MW = 4 (a factor of two different), bp = 23 K; T 2 , MW = 6, bp =
25 K—as well as similar compounds, such as hydrocarbons containing different
isotopes of hydrogen. Fluorocarbons have unusually low boiling points because
tightly held electrons in the fluorine atoms have a small polarizability.
All of the interactions discussed thus far are inherently attractive and would become
infinitely large at r = 0. Countering these attractive forces are repulsive forces
resulting from nucleus-nucleus repulsion (important in the H, molecule) and, more
important, the repulsion of inner or core electrons. At extremely short interatomic
distances the inner electron clouds of the interacting atoms begin to overlap and Pauli
repulsion becomes extremely large. The repulsive energy is given by
F - +k
(8-13)
where k is a constant and n may have various values, comparatively large. For ionic
compounds, values of n ranging from 5 to 12 prove useful (Chapter 4), and the
Lennard-Jones function, often used to describe the behavior of molecules, is some¬
times referred to as the 6-12 function because it employs r 6 for the attractive energies
111 These arc also sometimes referred to as London dispersion forces or van der Wauls forces The
former name, although widely used (including in this text), is unfortunate inasmuch as it seems to
imply that the forces tend to “disperse" the molecules, whereas they arc always attractive. Usage
of the term "van der Waals forces" varies: Some authors use it synonymously with London forces;
others use it to mean all of the forces which cause deviation from ideal behavior by real gases The
latter would include not only London forces but also dipole interactions etc
428
11- Coordination Chemistry: Bonding, Spectra, and Magnetism
Molecular Orbital Theory 429
Table 11.13
Infrared absorptions of some
metal carbonyl complexes
Compound
Frequency
(cm" 1 )
[Mn(C0) 6 ] +
2090
[CrfCOIJ
2000
[V(CO) h ]-
I860
[Ti(CO) A J 2
1748
[Ni(CO) 4 ]
2060
[Co(CO) 4 ]-
1890
[Fe(CO) 4 ] 2-
1790
metal-carbonyl bonds from structural data because other factors can influence bond
lengths.
Infrared spectroscopy. The most widely used experimental method for analyz¬
ing metal carbonyl complexes is infrared (IR) spectroscopy. The frequency of the IR
absorption (or more properly, the force constant, k) associated with a C—0 stretching
vibration is a measure of the resistance of the bond to displacement of its atoms.
Hence the stretching frequency provides a qualitative measure of bond strength, with
stronger bonds in general giving rise to IR absorptions at higher frequencies. Consider
Table 11.13, which lists IR data for two isoelectronic series of metal carbonyls. On the
basis of the absorption maxima, we can say that the C—O bond strengths in these two
series decrease in the order [Mn(CO) 6 ] + > [CrfCOJJ > [V(CO) 6 ] _ > [Ti(CO) 6 ] 2- and
[Ni(CO) 4 J > [Co(CO) 4 J~ > [Fe(CO) 4 ] 2_ . These qualitative results are consistent with
the Tr-bonding model described earlier: As M—C 7r bonding increases, the C—O bond
becomes weaker. The greater the positive charge on the central metal atom, the less
readily the metal can donate electron density into the 7r orbitals of the carbon
monoxide ligands to weaken the C—O bond. In contrast, in the carbonylate anions
the metal has a greater electron density to be dispersed, with the result that M—C
■n bonding is enhanced and the C—0 bond is diminished in strength.
The usefulness of the C—0 stretching frequency as a measure of C—O bond
strength (and hence of the extent of metal-carbonyl n bonding) derives from the
sensitivity of this absorption to the electron population of the CO antibonding orbitals.
In the isolated carbon monoxide ligand, the lone electron pair on carbon resides in the
3 ct orbital, which is the HOMO (highest occupied molecular orbital) for this molecule
(Fig. 5.20). Promotion of one of these electrons in gaseous CO to a 7r level to give the
3<r'27T 1 excited state causes the C—0 stretching frequency to drop from 2143 to 1489
cm -1 . This dramatic change is a strong indication that even a small amount of electron
shift from a central metal into the 7r orbital of a bound CO can be easily detected via
IR measurements. 31
As has already been stated, the lone pair on carbon in the carbon monoxide ligand
resides in a molecular orbital that is slightly antibonding. Support for this assertion is
also provided by IR data. When one of the lone pair (HOMO) electrons is removed
from CO to form CO + , the C—O stretching frequency increases from 2143 to 2184
cm -1 , showing that the C—O bond order also increases. Protonation of the molecule,
which can be considered as a coordination of CO to H + , also leads to an increase in
the stretching frequency. It would be expected that, upon donation of the CO lone pair
to a metal atom, a similar increase in v co should occur, provided there is no other
concomitant electron shift. What is actually observed, however, is that v co almost
always decreases upon complex formation, an indication that n electron density
flowing from the metal into the n orbital of the ligand more than compensates for the
increase in C—O bond order that would accrue from the ligand-to-metal cr donation. 32
In a preceding section we saw crystallographic evidence that substitution of
phosphorus ligands for carbon monoxide in Cr(CO) 6 leads to a strengthening of Cr—C
bonds, particularly those trans to the phosphorus groups, and this was interpreted in
31 Johnson. J. B.; Klemperer, W. G. J. Am. Chem. Soc. 1977, 99, 7132-7137.
52 Exceptions are ClAuCO. for which v co = 2152cm -1 (Dell'Amico, D. B.: Calderazzo, F.: Dcll'Am-
ico. G. Gazz. Cltim. I nil. 1977, 107. 101-105) and AgCO(B(OTcF 5 ) 4 |. for which v C o = 2204 cm"'
(Hurlburt, P. K.; Anderson, O. P.: Strauss. S. H. J. Am. Chem. Soc. 1991. 113. 6277-6278). An
example involving a main-group metal is MejAICO. for which v co = 2185 cm (Sanchez, R.;
Arrington. C.; Arrington. C. A.. Jr. J. Am. Chem. Soc. 1989, ///. 9110-9111).
terms of competition of the ligands for available v electron density. Changes in the CO
infrared absorptions also occur and can be evaluated in the same vein. In general,
substitution of CO with a ligand L will alter the CO stretching frequencies of the
remaining carbonyl ligands in a manner that reflects the net electron density transmit¬
ted by L to the central metal atom. This in turn will depend both on the o-donating
capacity and the n acidity of L.
It is instructive to look at a set of W(CO) 5 L complexes to see how a variety of
ligands (L) perturb the C—O stretching frequencies. These complexes all have C 4v
symmetry, at least ideally, and give rise to three allowed IR absorptions (two non¬
degenerate and one doubly degenerate), having the symmetry labels A 1 ,' 1 , A l2> , and E.
In Chapter 15 the procedure for obtaining symmetry assignments for vibrational
modes from the appropriate character table will be illustrated, but for now we will
simply use the results (Table 11.14). The particular vibrational stretching modes
involved are shown in Fig. 11.31. The important one to focus on is A 1 , 1 , which
corresponds to the symmetrical stretching motion of the CO group lying opposite the
ligand L. It is this CO that competes most directly with L for available 7r electron
density and therefore is in a position to best reflect the it acidity of L.
For ligands in Table 11.14 having little or no 77 acidity (e.g., those in which oxygen
or nitrogen is the donor atom), the CO in a trans position can absorb significant
electron density into its antibonding orbital, and relatively low C—O stretching
frequencies are observed. In the case of the phosphorus ligands, the t r acidity in¬
creases as the electronegativity of any substitutuent on P increases. As these ligands
become more and more compelitive for rt electrons, CO receives less and less v
electron density and the C—O stretching frequency increases accordingly. The very
high C—O stretching frequency of the PF 3 complex indicates that this ligand is
comparable in its v acidity to carbon monoxide itself. Groups in Table 11.4 having
carbon as the ligating atom, which will be discussed in Chapter 15, are quite effective
it acceptors, as shown by the relatively high A 1 ,' 1 stretching frequencies of their
complexes.
Although IR frequencies provide a useful measure of the extent of it bonding in
carbonyl complexes, a better quantitative picture can be obtained from C—O force
constants. These values are commonly derived from IR data by means of the Colton-
Kraihanzel force-field technique. 33 This procedure makes certain simplifying assump¬
tions in order to provide a practical solution to a problem that would be extremely
difficult to solve rigorously. Among the important assumptions are that the C—O
vibrations are not coupled to any other vibrational modes of the molecule and that the
observed frequencies can be used without correction for anharmonic effects. The
results of force constant calculations of this type provide a means of setting up a tr-
acceptor series: 34
33 Cotton, F. A.; Kraihanzel, C. S. J. Am. Chem. Soc. 1962, 84, 4432-4438. See also Cotton, F. A.;
Wilkinson, G. Advanced Inorganic Chemistry, 5th ed.; Wiley: New York, 1988; pp 1038-1040;
Timney, J. A. Inorg. Chem. 1979, 18, 2502-2506.
34 There is not universal agreement on the ordering of this series because many of the ligands have rr-
acccpting tendencies which are virtually indistinguishable. Arsenic and antimony compounds fall
into the scries alongside the corresponding phosphorus compounds. Although there is some uncer¬
tainty as to whether these compounds are slightly better or slightly poorer acceptors than their
phosphorus analogues, the accepting ability is generally believed to be in the order P > As > Sb. In
the above series R can represent phenyl or alkyl groups with the phenyl ligands usually exhibiting
better acceptor ability.
300
8 • Chemical Forces
Table 8.3
Summary of chemical forces and interactions
Type of interaction
Strength
Energy-distance function
Covalent bond
Very strong
Complex, but comparatively long range
Ionic bond
Very strong
1/r, comparatively long range
Ion-dipole
Strong
1/r 2 , short range
Dipole-dipole
Moderately strong l/r 2 , short range
Ion-induced dipole
Weak
I fr, very short range
Dipole-induced dipole
Very weak
1/r 6 , extremely short range
London dispersion forces
Very weak"
I/r 6 , extremely short range
« Since London forces increase with increasing size
and there is no limit to the size of molecules, these
forces can become rather large. In general, however
, they arc very weak.
Summary
(cf. Eq. 8.12) and r 12 for repulsions. In any event, repulsive energies come into play
only at extremely short distances.
Various forces acting on chemical species are summarized in Table 8.3. The forces are
listed in order of decreasing strength from the ionic and covalent bonds to the very
weak London forces. The application of a knowledge of these forces to interpretation
of chemical phenomena requires a certain amount of practice and chemical intuition.
In general, the importance of a particular force in affecting chemical and physical
properties is related to its position in Table 8.3. For example, the boiling points of the
noble gases arc determined by London forces because no other forces are in opera¬
tion. In a crystal of an ionic compound, however, although the London forces are still
present they are dwarfed in comparison to the very strong ionic interactions and may
be neglected to a first approximation (as was done in Chapter 4).
Hydrogen
Bonding
Although some would contend that hydrogen bonding is merely an extreme manifesta¬
tion of dipole-dipole interactions, it appears to be sufficiently different to warrant a
short, separate discussion. In addition, there is no universal agreement on the best
description of the nature of the forces in the hydrogen bond.
We shall adopt an operational definition or the hydrogen bond: A hydrogen bond
exists when a hydrogen atom is bonded to two or more other atoms." This definition
implies that the hydrogen bond cannot be an ordinary covalent bond since the
hydrogen atom has only one orbital (Is) at sufficiently low energy to engage in
covalent bonding.
Macroscopically the effects of hydrogen bonding are seen indirectly in the greatly
increased melting and boiling points of such species as NH 3 , H 2 0, and HF. This
11 Hamilton. W. C.; Ibcrs. J. A. Hydrogen Bonding in Solids'. W. A. Benjamin: New York. I%8; p 13.
Similar definitions arc offered by Pimentel, G. C. ; McClellan. A. L. The Hydrogen Bond'. Freeman:
San Francisco, I960; Joestcn. M. D.; Schaad, L. J. Hydrogen Bonding', Marcel Dckkcr: New York.
1974; The Hydrogen Bond', Schuster. P.; Zundel. G.; Sandorfy. C.; Eds.; North-Holland: Amster¬
dam. 1976; Vols. Mil; Jeffrey. G. A.; Saenger. W. Hydrogen Bonding in Biological Structures',
Springer-Vcrtag: New York. 1991. See also Allen. L. C. J. Am. Chem. Soc. 1975, 97. 6921-6940;
Emsley. J. Chem. Soc. Rev. 1980, 57, 91-124; Legon. A. C.; Millen. D. J. Chem. Rev. 1986, 86,
635-657; Joesten. M. D. J. Chem. Educ. 1982, 59. 362.
Hydrogen Bonding
301
phenomenon is well documented in introductory texts and need not be discussed
further here. On the molecular level we can observe hydrogen bonding in the greatly
reduced distances between atoms, distances that fall below that expected from van der
Waals radii. Indeed this is a practical method of distinguishing between a true bonding
situation and one in which a hydrogen atom is close to two atoms but bonded to only
one. Table 8.4 lists some distances in hydrogen bonded A—H ■ • • B systems com¬
pared with the sum of the van der Waals radii for the species involved. In many
hydrogen bonds, the atoms A and B are closer together than the sum of the van der
Waals radii. Even more characteristic is that the hydrogen atom is considerably closer
to atom B than predicted from the sum of the van der Waals radii, indicating
penetration (or compression) of atom B's electron cloud by the hydrogen.
In the typical hydrogen bonding situation the hydrogen atom is attached to two
very electronegative atoms. The system is usually nearly linear and the hydrogen atom
is nearer one nucleus than the other. Thus, for most of the systems in Table 8.4, the
hydrogen atom is assumed to be attached to atom A by a short, normal covalent bond
and attached to atom B by a longer, weaker hydrogen bond of about 50 kJ mol -1 or
less. This situation usually obtains even if both A and B are the same element as in the
hydrogen bonding between oxygen atoms in water. There are important exceptions,
however. These include salts of the type M + HA 2 , where A" may be the fluoride ion
(less frequently another halide) or the anions of certain monoprotic organic acids such
as acetic or benzoic acid. Alternatively, HA AH may be a diprotic acid such as maleic
or phthalic acid:
Tablo 8.4
Van dor Waals distances
and obsorved distances (pm)
Bond typo
A • ■ • B 6
(calc)
A • • B
(obs)
H • • • B
(calc)
H • • • B
(obs)
for somo common hydrogen
F—H — F
270
240
260
120
bonds"
O— H • •
O
280
270
260
170
O— H ■ •
F
280
270
260
170
O— H • •
N
290
280
270
190
O— H • •
Cl
320
310
300
220
N—H • •
O
290
290
260
200
N—H • •
F
290
280
260
190
N—H • •
Cl
330
330
300
240
N—H • •
N
300
310
270
220
N—H • •
S
340
340
310
240
C—H • •
O
300
320
260
230
• Hamilton W. C.; Ibers, J. A. Hydrogen Bonding in Solids;
W. A. Benjamin: New York, 1968; p 16. Used with permission.
The values in column 2 arc not those to be obtained by the
use of Table 8.1 because Hamilton and Ibers used van der
Waals radii from Pauling.
426
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
The metal-carbon bond length in carbonyl complexes provides a better measure
of double bond character because these lengths are more sensitive to changes in bond
order. If the covalent radii of an sp carbon atom and of the metal atom to which it is
bound are known, summation of the two should give the length expected for a metal-
carbonyl single bond, that is, one with no tt bonding. This value could then be
compared with the measured bond length to determine the extent of 7r bonding
present. A factor that can cause problems with this strategy is that the covalent radius
of the metal atom may not be known with certainty. However, this difficulty can be
circumvented by choosing a complex that contains both carbonyl and alkyl ligands. In
such a complex the covalent radius of the metal atom will be the same for both ligands
in the absence of n bonding, and can be derived from the measured M—C (alkyl) bond
length and the known covalent radius for an sp 3 carbon. This value can in turn be used
to calculate an expected metal-carbonyl single bond length. The methyl derivative of a
rhenium carbonyl complex, Re(CH 3 )(CO) 5 , may be used to illustrate the procedure: 2 ' 3
Re—C single bond length in ReCH 3 = 231 pm
— (Covalent radius for sp 3 C) = - 77 pm
Covalent radius for Re =154 pm
Covalent radius for Re =154 pm
(Covalent radius for sp C) = + 70 pm
Re—C single bond length for ReCO = 224 pm
The experimentally determined Re—CO bond distance for this complex is 200.4 ±
0.4 pm, about 24 pm shorter than that predicted for a <r-only bond. When this type of
analysis is applied to other complexes, similar decreases in metal-carbon lengths are
observed, substantiating the view that the M—CO bond has considerable double bond
character.
Further crystallographic evidence for metal-carbonyl tt bonding is found in phos¬
phine and phosphite derivatives of hexacarbonylchromium. Substitution of R 3 P for
CO in Cr(CO) h creates a complex of C 4l . symmetry in which one CO group lies trans to
the phosphorus ligand (Fig. 11.29). The two trans ligands will compete for the same 7r
orbital, but carbon monoxide is a better tx acid {n acceptor) than the phosphine
(Fig. 11.30). As a result, the Cr—CO, n bond should be shorter relative to Cr—CO,. I(
and to Cr—CO in Cr(CO) h . The data in Table 11.12 show that these predictions are
borne out. in keeping with the substantial -jt character in the metal-carbonyl bond.
In general, the 7r-accepting ability of a phosphine increases as the elec¬
tronegativities of its substituents increase. Thus we would expect the 7r acidity of
Fig. 11.29 Structure of the
phosphine and phosphite
derivatives of
hexacarbonylchromium.
:v Rankin. D. W. H.: Robertson. A. J. Organomet. Client. 1976. 105. 331-340.
Toble 11.12 _
Bond lengths (pm) in
chromium carbonyl
complexes, Cr(CO) 5 L°
Molecular Orbital Theory 427
Both bonds very
strong, equal
Stronger
bond
Weaker
bond
(a) (b)
Fig. 11.30 Competition by ligands for the it bonding d orbital of a central metal atom.
Relative overlap is symbolized by the shaded areas, (a) Equal and strong tt bonds resulting
from equal and good overlap of the two carbon monoxide 7r orbitals with the metal d orbital,
(b) Superior overlap of carbon monoxide n' orbital with polarized metal d orbital compared
to poorer overlap between ligand d and metal d orbitals. Polarization (mixing of higher energy
wave functions) occurs so as to maximize total overlap. Recall that the overlap integral
includes both spatial and intensive properties; the representation above is a graphic
simplification.
P(OPh) 3 to be greater than that of PPh 3 . Consistent with this view, it is found that the
Cr—P bond length in the phosphite complex is shorter lhan in the phosphine complex.
The carbonyl ligand trans to the phosphorus ligand would be expected to receive more
7r electron density in the phosphine complex than in the phosphite complex, which
would lead to a shorter metal-carbon bond for the phosphine derivative, and that is
observed as well.
The P(CH-,CHXN) 3 ligand appears to be inconsistent with the model; however,
its Cr—P bond distance suggests that it is a poorer tt bonding ligand than P(OPh) 3 ,
whereas the Cr—C bond lengths suggest that it is a better it acceptor. Some of the
ambiguity may arise because of the different steric requirements of the phosphorus
ligands. Steric interaction with equatorial carbonyl ligands could lead to a Cr—P bond
lengthening which obscures the intrinsic electronic effect.- 10 it should be apparent from
this discussion that it is often not a simple matter to sort out crand it contributions to
L
Cr—P
Cr-C
(trans to P)
Cr—C
(trans to CO)
C—O
(trans to P)
C—O
(trans to CO)
CO
—
—
191.5(2) av
_
114.0(2) av
P(OPh) 3
230.9(1)
186.1(4)
189.6(4) av
113.6(6)
113.1(6) av
P(CH,CH 2 CN) 3
236.4(1)
187.6(4)
189.1(4) av
113.6(4)
113.8(4) av
PPh 3
242.2(1)
184.4(4)
188.0(4) av
115.4(5)
114.7(6) av
11 Rees, B.; Mitschler, A. J. Am. Chem. Sac. 1976, 98, 7918-7924 (Chromium hcxacarbonyl).
Plastas, H. J.; Stewart, J. M.; Grim, S. 0. J. Am. Chem. Sac. 1969, 91, 4326-4327 (triphenyl-
phosphinc and triphcnylphosphitc complexes). Cotton, F. A.; Darensbourg, D. J.; Ilslcy, W. H.
Inorg. Chem. 1981, 20, 578-583 (tris(2-cyanocthyl)phospliinc complex].
30 Cone angles of phosphines are discussed in Chapter 15. The cone angles of P(OPh) 3 ,
P(CH 2 CH 2 CN) 3 , and PPhj are 128°, 132°, and 145°, respectively.
I
302
Hydrogen Bonding
303
8-Chemical Forces
This type of hydrogen bonding is termed symmetric in contrast to the more common
unsymmetric form. Symmetrical hydrogen bonds form in only the strongest bonded
systems. These are frequently anionic like FHF - and the carboxylates mentioned
above. An example of a strong, symmetric hydrogen bond in a cation is the
bis(A/-nitrosopyrrolidine)hydrogen cation (Fig. 8.5). 12
Although the subject of symmetric versus unsymmetric hydrogen bonding has
received considerable attention, there is yet little understanding of the factors in¬
volved. Certainly, for the long, weak hydrogen bond we can approximate the situation
by assuming the hydrogen atom to be covalently bonded to one atom and to be
attracting the other. Obviously, it will be closer to the covalently bound atom than to
the dipole-attracted atom. It is not so easy to see when or why the bond will become
symmetrical, although if a resonance or delocalized molecular orbital model is in¬
voked, an analogy with the equivalent bond lengths in benzene can be appealed to.
The situation is more complicated than that, however, for unsymmetrical FHF" ions
are known in some crystals. 13 In the same way the hydrogen maleate and hydrogen
phthalate anions, found to be symmetrical in the crystal, appear to be unsymmetrical
in the less ordered aqueous solution. 14 Whether symmetrical hydrogen bonds are
forced by a symmetrical environment or whether unsymmetrical bonding is induced
by crystals and solutions of lower symmetry is, perhaps, a moot point.
Since hydrogen bonding generally occurs only when the hydrogen atom is bound
to a highly electronegative atom, 15 the first suggestion concerning the nature of the
hydrogen bond was that it consists of a dipole-ion or dipole-dipole interaction of the
sort A' s —H A+ ■ • • 13 or A - — H fi+ • • • B' s ~ — R 1 ' 1 ', where R is simply the remainder
of a molecule containing the electronegative atom B. Support for this viewpoint comes
from the fact that the strongest hydrogen bonds are formed in systems in which the
hydrogen is bonded to the most electronegative elements:
F" + HF - FHF -
(CH,),CO + HF - (CH,) 2 CO • • • HF
H,0 + HOH H,0 • • • HOH (ice)
HCN + HCN » HCN • ■ • HCN
A// = - 161 ± 8 kJ mol-' (8.14)'
A H = -46 kJ mol-1 (8.15)
A// = -25 kJ mol-' (8.16)
AH = -12 kJ mol-' (8.17)
V
3f
12 Keefer, L. K.; Hrabic, J. A.; Ohanncsian, L.: Flippen-Anderson. J. L.; George. C. J. Am. Client.
Soc. 1988, 110. 3701-3702.
13 Williams. J. M.: Schnecmeycr, L. F. J. Am. Client. Sue. 1973. 95. 5780.
14 Perrin, C. L.; Thoburn. J. D. J. Am. Client. Soc. 1989. III. 8010-8012.
15 Even rather electropositive elements such as carbon can cause attached hydrogen atoms to form
hydrogen bonds. In HCN. for example, the effective electronegativity of the carbon is still com¬
paratively high, the hydrogen atom is positive, and it would be expected to hydrogen bond to the
negative nitrogen atom of an adjacent molecule. It has a boiling point of 26 °C compared to 20 °C for
HF and 100 °C for H : 0. Even in the methyl group of CHjCN some hydrogen bonding can
apparently lake place since the boiling point of acetonitrile is higher (82 °C) than expected on the
basis of its molecular weight (London forces) alone: The boiling point of n-propane = -42 ”C.
Acetonitrile is also completely miscible with water. Sec Green, R. D. Hydrogen Boitdinp by C—H
Groups-, Wiley: New York. 1974; Mueller-Westcrholf, U. T.; Nazzal, A.; Prdssdorf. W. J. Am.
Client. Soc. 1981, 103, 7678-7682; Desiraju, G. R. Ate. Client. Res. 1991, 24, 290-296.
16 Lias, S. G.; Bartmess, J. E.; Liebman, J. F.; Holmes, J. L.; Levin, R. D. Mallard. W. G. J. Phys.
Client. Ref. Data 1988, 17. Suppl. I.
Fig. 8.5 Short, symmetrical hydrogen bond in the cation of bis(A'-nitrosopyrrolidine)-
hydrogen hexalluorophosphate which crystallizes in the monoclinic space group P2,/c. The
hydrogen bonded proton lies on a center of symmetry, with only one of the nilrosopyrrolidine
molecules being crystallographically unique. Note that the O—0 distance is only 247 pm
whereas two times the van der Waals radius of oxygen (Table 8.1) is 300 pm. [From Keefer,
L. K.; Hrabie. J. A.; Ohannesian. L.; Flippen-Anderson. J. L.; George. C. J. Am. Client.
Soc. 1988, 110, 3701-3702. Reproduced with permission.)
The simplistic electrostatic model qualitatively accounts for relative bond ener¬
gies and the geometry (a linear arrangement maximizes the attractive forces and
minimizes the repulsions). Nevertheless, there are reasons to believe that more is
involved in hydrogen bonding than simply an exaggerated dipole-dipole or ion-dipole
interaction. First, the shortness of hydrogen bonds indicates considerable overlap of
van der Waals radii, and this should lead to considerable repulsive forces unless
otherwise compensated. Secondly, symmetrical hydrogen bonds of the type
F—H — F would not be expected if the hydrogen atom were covalently bound to one
fluorine atom but weakly attracted by an ion-dipole force to the other. Of course, one
can invoke resonance in this situation to account for the observed properties:
F—H • • • F - --- F - • • • H—F
This implies delocalization of the covalent bond over both sides of the hydrogen atom.
One might then ask whether a simpler molecular orbital treatment of the delocalization
would be more straightforward. The answer is yes. The mechanics will not be given
here (see Chapter 17, the three-center four-electron bond), but the results are that the
covalent bond is “smeared” over all three atoms. In symmetric hydrogen bonds it is
equal on both sides; in unsymmetric hydrogen bonds more electron density is concen¬
trated in the shorter link. Several workers have calculated and analyzed hydrogen
bond energies. 17 The interpretations are not identical, but all indicate strong contribu¬
tions from both electrostatic (ion-dipole, dipole-dipole) and covalent (delocalization,
resonance) interactions.
Systematic analyses of crystallographic data for hydrogen bonds have revealed a
range of geometries and have led to proposals for rules to rationalize or predict
hydrogen bonding patterns." 1 An energetic preference for linear or near-linear
17 For example, see Basch, H.; Stevens, W. J. J. Am. Client. Soc. 1991, 113, 95-101; Dykslra, C. E.
Acc. Client. Res. 1988, 21. 355-361; Legon, A. C. Client. Soc. Rev. 1990, 19, 197-237; Curtiss.
L. A.; Blander. M. Client. Rev. 1988. 88, 827-841.
18 Taylor. R.; Kcnnard. O. Acc. Chem. Res. 1984, 17, 320-326. Etter. M. C. Ate. Chem. Res. 1990,
23. 120-126. Gorbitz, C. H.; Etter, M. C. J. Am. Chem. Soc. 1992. 114. 627-631.
424
'11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
including tt interactions, for an octahedral M(CO),, complex is shown in Fig. 11.28.
The increase in A ( , caused by -n bonding is substantial enough in many cases that the
absorption maximum for the t 2f ,-to-e electronic transition is blue-shifted out of the
visible region into the ultraviolet portion of the electromagnetic spectrum, with
the result that the complexes are colorless. This is the case for the metal carbonyls, for
example.
Halide ions such as Cl - , Br - , and I - present a different situation. Like the
fluoride ion, they have filled p orbitals, but unlike fluoride, their empty d orbitals may
participate in 7r bonding. It is difficult to predict which set of r 2jf LGOs for these ions
(those constructed from filled p or from empty d orbitals) will interact more strongly
with the t 2 orbitals of the metal. Empirically, we observe that all of the halide ions lie
at the weak-field end of the spectrochemical series, indicating that the p-orbital
interaction is more important than that of the d orbitals.
The potential for u orbitals to serve as tt acceptors has become apparent in
recent times. Phosphines, instead of using empty pure d orbitals as tt acceptors, may
accept 7r donation into low-lying a orbitals or into hybrids involving a and 3 d
M M(CO) 6 6 co
Fig. 11.28 MO diagram for an octahedral M(CO),, complex: both a and 7r interactions are
included. Correlation lines are drawn only to those molecular orbitals to which the metal d
electrons contribute.
Molecular Orbital Theory
425
Experimental
Evidence for
Pi Bonding
■
orbitals. 27 In addition, the coordination of dihydrogen (see Chapter 15) is thought to
involve cr electron donation to the metal from the I-J—H a bond and back donation
from the metal into the H 2 cr orbitals.
Few topics in coordination chemistry have received more attention than tt bonding.
We have seen in the preceding section that it provides a reasonable rationale for much
of the spectrochemical series. In Chapter 13 we shall find that tt bonding is important
in determining patterns of ligand substitution reactions. It is also central to under¬
standing reactivity and stability in organometallic complexes (Chapter 15). In this
section various experimental methods of evaluating tt bonding in metal carbonyl
complexes and their derivatives will be examined.
There is strong agreement among inorganic chemists, theoreticians and experi¬
mentalists alike, that the stability of metal carbonyl complexes depends on the ability
of carbon monoxide to accept metal electron density into its tt orbitals. 28 Although
carbon monoxide is a weak base toward hydrogen ion or BHj, it has a significant
affinity for electron-rich metals (see Chapter 15). For example, it reacts with metallic
nickel at modest temperatures to form gaseous Ni(CO) 4 . This is especially impressive
when one considers that the metal-metal bonds in nickel, which must be broken for
the complex to form, are quite strong. Observations such as these cannot be easily
explained by cr bonding alone. The currently accepted bonding model views carbon
monoxide as a cr donor (OC—>M) and a tt acceptor (OC<— M), with the two interac¬
tions synergistically enhancing each other to yield a strong bond:
M + C=0 -► M - —C=0 + -—► M=C=0 (11.16)
Crystallography. The M—CO bonding model described above suggests that the
greater the extent of it bonding, the more the C—O bond will be lengthened and the
M—C bond shortened. The cr interaction, on the other hand, should have the opposite
effect on the C—0 bond length because the lone pair on carbon that is utilized in
forming the cr bond is in a slightly antibonding MO of the carbon monoxide ligand (the
3cr orbital in Fig. 5.20). Donation of this pair of electrons to a Lewis acid would be
expected to make the C—O bond stronger and shorter relative to that of carbon
monoxide. It would seem from this analysis that one could merely compare the C—0
bond distance in carbon monoxide (112.8 pm) with that in a carbonyl complex and. if
the latter is found to be longer, this could be taken as evidence for tt bonding. A
problem arises with this approach because the C—O bond length (and that of multiple
bonds in general) is relatively insensitive to bond order: The difference in length
between the triple bond in CO (113 pm) and typical C=0 double bonds in organic
molecules (~ 123 pm) is small. Moreover, observed C—0 bond lengths among metal
carbonyls fall within a very short range—about 114 to 115 pm. Unless the measure¬
ment is made with exceptional accuracy, any bond length difference within this range
cannot be regarded as statistically significant, let alone as a meaningful indication of
bond order.
27 McAulilTc, C. A. In Comprehensive Coordination Chemistry, Wilkinson. G.; Gillard. R. D.:
McCIcvcrty, J. A.. Eds.; Pergamon: Oxford, 1987; Vol. 2. Marynick, D. S. J. Am. Cheat. Soe.
1984, 106, 4064-4065. Green. J. C.: Kallsoyannis. N.; Sze, K. H.; MacDonald. M. A. J. Cheat. Sue.
Dalton Trans. 1991, 2371-2375. Also see Chapter 18.
28 Sherwood. D. E„ Jr.: Hall, M. B. Inorg. Chem. 1980, 19. 1805-1809. Burster. B. E.; Freicr, D. G.;
Fenske, R. F. Inorg. Chem. 1980. 19, 1810-1811. Bauschlicher, C. W.. Jr.; Bagus. P. S. J. Chem.
Phys. 1984, 81, 5889-5898.
304
8 Chemical Forces
A—h • • • B configurations, at least in the crystalline state, is confirmed by the
experimental data. The stereochemical requirements of hydrogen bonds determine the
structure of ice and lead to the well-known fact that solid water is less dense than
liquid water at the melting point. This is because the structure of ice is rather open as a
result of an extensive network of hydrogen bonds (Fig. 8.6). Hydrogen bond energet¬
ics and stereochemsitry have wide-ranging implications in the areas of catalysis,
molecular recognition, and design of new materials.
Finally, there are other systems such as W—H —W and B—H—B (Chapter 16)
which formally meet the operational definition of hydrogen bonding given above. They
differ, however, in having electropositive atoms bonded to the hydrogen atom. To
distinguish them from the electronegative hydrogen bonded systems, they are olten
termed hydrogen-bridged systems.
Hvdrates and The hydration of ions upon solution in water has been mentioned previously and its
rinthrnt importance to solution chemistry discussed. In the solid crystalline hydrates, hydro¬
gen bonding becomes important in addition to the ion-dipole attractions. 1 '’ Often the
water molecules serve to fill in the interstices and bind together a structure which
would otherwise be unstable because of disproportionate sizes of the cation and anion.
For example, both FeSiF 6 -6H 2 0 and Na 4 Xe0 6 -8H 2 0 are well-defined, crystalline
solids. The anhydrous materials are unknown. The large, highly charged anions
presumably repel each other too much to form a stable lattice unless there are water
molecules present. In general, some water molecules will be found coordinated
Fig. 8.6 The open structure of normal ice that results from the directionality of the hydrogen
bonding. [From Dickerson. R. E.; Geis. I. Chemistry. Matter, and the Universe : W. A.
Benjamin: Menlo Park, 1976.1
i» Hamilton; W. C.; lbcrs. J. A. Hydrogen Bonding in Solids-, W. A. Benjamin: New York. 1968; pp
204-221.
Hydrogen Bonding 305
directly to the cation and some will not. All the water molecules will be hydrogen
bonded, either to the anion or to another water molecule.
A specific example of these types of hydrates is CuS0 4 -5H 2 0. Although there
are five molecules of water for every Cu 2+ ion, only four are coordinated to the cation
its six-coordination being completed by coordination from SOj (Fig. 8.7a). The filth
water molecule is held in place by hydrogen bonds, O-H • 0. between it and two
coordinated water molecules and the coordinated sulfate anion (Fig. 8.7b). Dehydra¬
tion to CuS0 4 -3H,0, CuS 0 4 -H,0. and eventually anhydrous CuS0 4 results in the
water molecules coordinated to The copper being gradually replaced by oxygen atoms
from the sulfate.- 0
An interesting hydrate is that of the hydronium ion in the gas phase. It consists of
a dodecahedral cage of water molecules enclosing the hydronium ion: H ,0 (H 2 O) 20 .
Each water molecule is bonded to three others in the dodecahedron (Fig. 8.8a). Of the
various possible hydrates of HjO + in the gas phase, H 3 O h (H 2 O) 20 is by far the most
The dodecahedral structure may carry over into the solid phase. Note that hall of
the oxygen atoms in Fig. 8.8a have their fourth coordination position occupied by a
hydrogen atom that can bond to adjacent polyhedra (Fig. 8.8b), and the other hall have
a lone" pair at the fourth position which can donate a pair of electrons to form an
external hydrogen bond (Fig. 8.8c). Thus in the solid, these dodecahedra can pack
together to form larger structures with relatively large voids in the centers of the
dodecahedra. Guest molecules such as Ar, Kr, Xe. CH 4 , etc., may occupy these
spaces. These gas hydrates in which the guest molecules are no! bound chemically but
arc retained by the structure of the host are called clathrates. Since the structure can
exis. with incomplete filling of holes, the formulas of these clathrates are variable.
OSO,
• = S of SO vjg) “ Oxygen of SOJ
Q = Cu O = Oxygen of fifth H,0
Q = Oxygen of
coordinated H 2 O
(a) (b)
Fiq. 8.7 Structure of copper(ll) sulfate pentahydrate. (a) Coordination sphere of Cu^. four
water molecules and two sulfate ions; (b) Position of fifth water molecule (oxygen shown by
heavy circle). Normal covalent bonds depicted by solid lines; O-H • • • 0 hydrogen bonds
depicted by dashed lines.
20 Wells, A. G. Structural Inorganic Chemistry, 5th cd.; Clarendon: Oxford. 1984. pp 678-680.
21 Wei. S.; Shi. Z.; Castleman, A. W., Jr. J. Chem. Phys. 1991, 94. 3268-3270.
422
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Molecular Orbital Theory
423
Fig. 11.25 Pi overlap of t 2ll
LGOs with a metal t 2l! (d tl .)
orbital. There are two
additional sets
perpendicular to the
one shown.
the fluorine 2 p orbitals lie at a lower energy than the corresponding metal 3d orbitals.
Under these circumstances, the bonding tt MOs will resemble the fluorine orbitals
more than the metal orbitals, and conversely the tt MOs will more closely resemble
the metal orbitals (see Chapter 5). The molecular orbital energy diagram for the n
system in [CoF fi ] 3- is shown in Fig. 11.26. Since the 2 p orbitals on the fluoride ligands
are filled, these electrons will fill the resultant molecular t 2g tt orbitals. The electrons
from the 3d (t 2fl ) orbitals of the cobalt are therefore in tt antibonding orbitals (it) at a
higher energy than they would be if tt bonding had not taken place. Since the level of
the e orbitals is unaffected by the tt interaction, A„ is reduced as a result of the 7r
bonding. It is felt that this is the source of the position of fluoride (and other halides) at
the weak field extreme in the spectrochemical series (weaker than most cr-only
ligands). In the same way, the weaker field of OH - compared to H,0. so puzzling in
terms of a purely electrostatic model, can be rationalized in terms of the hydroxide ion
being a better tt donor. Note that the overall gain in bond energy as a result of
tt bonding is slight: The filled t 2l! orbitals are lowered in energy somewhat, but the
nearly filled t 2g orbitals are raised an equal amount. Thus the only net stabilizing
energy is that derived from the slightly different populations of the two sets of orbitals.
Finally, it should be mentioned that oxyanions of transition metals in high oxidation
states such as CrO 2 , Mn0 4 , and Fe0 4 ~ probably contain appreciable tt bonding. In
principle this tt bonding between metal and 0 2 “ ligands can be treated as above, but
because the complexes are tetrahedral, the problem of tt bonding is somewhat more
complex and will not be discussed here.
Ligands such as R,P may also participate in 7r bonding. In these molecules, as in
NHj, the ligating atom can er-bond to the metal through an approximately sp } hybrid
orbital. Unlike nitrogen, however, phosphorus has empty 3d and cr orbitals lying low
enough in energy that they can receive electron density from the metal. These orbitals
Fig. 11.26 MO diagram for
the tt system of |CoF ft )' _ .
Left: MOs for the u system
of the complex: right:
LGOs of t 2g symmetry:
center: MOs after n
interaction. Note that A„ is
diminished by the tt
interaction.
o Comple
Ligand ir orbitals
have fairly low electronegativities (compared to the metal orbitals), and so the /,.
LGOs formed from them will lie at a higher energy than the corresponding metal
orbitals. The resulting energy level diagram is shown in Fig. 11.27. Although the
orbital of the complex is lowered and the t_ g raised in a manner almost identical to that
of the previous case, the fact that the ligand t 2g orbitals are empty allows the r.
orbitals to rise with no cost of energy while the bonding t 2g orbitals are stabilized. Pi
bonding of this type thus can stabilize a complex by increasing the bond energy. In
addition, the resulting t 2g tt orbital is delocalized over both metal and ligand as
opposed to being a nonbonding t 2g orbital localized on the metal, which would have
been the case in the absence of tt bonding. Electron density is thus removed from the
metal as a result of tt bonding. This will not be particularly desirable in a complex
containing a metai in a high formal oxidation state since the metal will already carry a
partial positive charge.-'’ In low oxidation states, on the other hand, electron density
that tends to be built up via the cr system can be dispersed through the n system; that
is, a synergistic effect can cause the two systems to help each other. The more
electron density that the tt system can transfer from the metal to the ligand, the more
the metal is able to accept via the a system. In turn, the more electron density the a
system removes from the ligand, the more readily the ligand can accept electron
density through the tt system. Up to a certain point, then, each system can augment
the bonding possibilities of the other.
Pi bonding between metal and ligands provides a simple raison d'etre for strong
field ligands, an issue that crystal field theory could not resolve. If we examine the
strong field end of the spectrochemical series (page 405), we find ligands such as nitrite
ion, cyanide ion. carbon monoxide, phosphites, and phosphines. The latter two owe
their positions in the series to their ability to serve as tt acceptors, as described above,
which increases the value of A„ relative to what it would be in a cr-only system (Fig.
11.27). The other three ligands tt bond in a very similar fashion except that the
acceptor orbital is a 7r orbital as shown in Fig. 11,23c. The net result is the same as for
ligands in which either d or cr orbitals or both serve as tt acceptors: The bonding t 2 ,
level is lowered so that the quantity A„ is increased. A molecular orbital diagram.
a Complex
Ligand ir orbitals
Fig. 11.27 MO diagram for
I he tt system of an
octahedral complex with
acceptor ligands such as
CO, PR,, or SRj. Note that
the 7r interaction in this
case increases A„.
26 ,n recent years many complexes of early transition metals in relatively high oxidation states have
been prepared which contain neutral phosphorus donor ligands. Although the metal atom may pre¬
fer a harder Lewis base, in the absence of one, bonding to a phosphine may occur. Fryzuk. M. D.;
Haddad. T. S.; Berg, D. J. Coord. Chan. Rev. 1990, 99. 137-212.
306
8 • Chemical Forces
Fig. 8.8 (a) Pentagonal dodecahedron composed of twenty H 2 0 molecules connected by hydrogen bonds, (b) Apex of
dodecahedron with external hydrogen atom capable of accepting a lone pair from an adjacent polyhedron to form linking
hydrogen bond, (c) Apex of dodecahedron with external lone pair capable of hydrogen bonding to a hydrogen in an adjacent
polyhedron.
Hydrate clathrates of organic compounds are thought to be responsible for the
behavior of "ice” in the heads of comets and in wet methane under pressured Unless
methane is carefully dried, high-pressure lines may become clogged with the ice-like
gas hydrate. There may be large deposits of methane hydrates, “the ice that burns,”
beneath the ocean floor.
Not all clathrates are hydrates. Other well-known examples have host lattices
formed from hydrogen bonded aggregates of hydroquinone, phenol, and similar
organic compounds. Non-hydrogen bonded host structures are also known. One
example is a cyclotriphosphazene. (C 6 H 4 0 2 PN) 3 , that traps molecules such as ben¬
zene in tunnels in the crystal.--' In addition, coordination polymers are formed by
ambidenlate ligands, such as CN and SCN~, which coordinate to metal ions at both
ends (Chapter 12). Perhaps the best known of this type of compound is the series of
Ni(CN) 2 NH 3 -M compounds, where M may be benzene, thiophene, furan. pyrrole,
aniline, or phenol.
Current interest in clathratc structures focuses on molecular recognition, a broad
topic that includes resolution of enantiomers (Chapter 12), macrocyclic chelates
(Chapter 12), and key-and-lock enzyme activity (Chapter 19). In terms of clathrates,
the challenge is to structure the vacancy in such a way that particular molecules will
be incorporated as guests. 24
Blake, D.; Allamandola, L.; Sandford, S.; Hudgins, D.; Freund, F. Science 1991 254 548-551
Appcnzeller, T. Science 1991, 252, 1790-1792.
J Allcock. H. R.; Allen, R. W.; Bisscll, E. C.; Smcltz, L. A.; Teeter, M. J. Am. Chem. Soc 1976 98
5120-5125.
24 Molecular Inclusions and Molecular Recognition—Clathrates I; Gerdil, R.; Weber. E., Eds.;
Topics in Current Chemistry 140: Springer-Verlag: Berlin. 1987. Molecular Inclusions and Mo¬
lecular Recognition—Clathrates II; Weber, E.; Bishop, R.. Eds.; Topics in Current Chemistry 149;
Springer-Verlag; Berlin, 1988.
Effects of Chemical Forces
307
Fusion and vaporization result from supplying sufficient thermal energy to a crystal to
overcome the potential energy holding it together. It should be noted that in most
cases the melting and vaporization of a crystal does not result in atomization, that is,
the complete breaking of all chemical forces. In order to understand the relationship
between chemical forces and physical properties such as melting and boiling points, it
is necessary to compare the binding energies of the species in the vapor with those in
the crystal. Only the difference between these two energies must be supplied in order
to vaporize the solid. The following discussion will emphasize energy differences with
respect to variation in melting and boiling points, but it should be realized that entropy
effects can also be very important.
Crystals held together solely by London dispersion forces melt at comparatively
low temperatures and the resulting liquids vaporize easily. Examples of this type are
the noble gases which boil at temperatures ranging from -269 °C (He) to -62 °C (Rn).
Many organic and inorganic molecules with zero dipole moments such as CH„ (bp =
- 162 °C), BF 3 (bp = - 10! °C). and SF„ (sublimes at -64 °C) fall into this category.
Because London forces increase greatly with polarizability, many larger molecules
form liquids or even solids at room temperature despite having only this type of
attraction between molecules. Examples are Ni(CO) 4 (bp = 43 °C), CC1, (bp =
^4 °C) k° raZ ' ne ' = 53 °C), and trimeric phosphazene, P 3 N 3 CI 6 (mp =
It should be noted that these compounds are a trivial illustration of the principle
stated in the first paragraph. Although all of the molecules contain very strong
covalent bonds, none is broken on melting or vaporization, and hence they play no
part in determining the melting and boiling points.
The melting point of a compound is another property to which symmetry is an
important contributor. Symmetrical molecules tend to have higher melting points
than their less symmetrical isomers. For example, the melting point of neopentane is
-17 °C, that of n-pentane -130 °C. If the molecule has very high symmetry, the
melting point may be raised until the substance sublimes rather than melts. If you
think about some substance that you know sublimes (in addition to SF 6 mentioned
above), you will note that it has high symmetry. Common examples are I,, CO-,, and
camphor. The extreme example is perhaps dodecahcdrane. C 20 H 20 :
It has a mp of 430 ± 10 °C, difficult to obtain because of sublimation. This is about
100 °C higher than the bp of the straight chain hydrocarbon of similar molecular
weight, /i-eicosane (bp = 343 °C), and almost 400 °C higher than its mp (37 °C). The
dilference lies in the high symmetry (/,,) of dodecahedrane. As the temperature rises,
the molecules can pick up energy in the form of rotations, and even diffusion (transla¬
tion). without disrupting the lattice and forcing melting to occur.
Molecules in polar liquids such as water, liquid ammonia, sulfiiric acid, and
chloroform are held together by dipole-dipole and hydrogen bonding interactions. For
molecules of comparable size, these are stronger than London forces resulting in the
familiar trends in boiling points of nonmetal hydrides. For the heavier molecules, such
as H,S, H 2 Se, PH 3 , and HI. dipole effects are not particularly important (the elec-
Effects of
Chemical
Forces
Melting and Boiling
Points
420
H • Coordination Chemistry: Bonding, Spectra, and Magnetism
of square planar d 8 complexes: All of the bonding molecular orbitals are filled and all
antibonding orbitals remain unoccupied. Adding additional electrons would de¬
stabilize a complex because the electrons would occupy antibonding levels. Fewer
than sixteen electrons would also lead to lower stability, all other things being equal,
because the bonding interaction would be diminished.
It should be apparent from the molecular orbital diagrams in Figs. 11.20. 11.21.
and 11.22 that there are strong resemblances between the molecular orbital and crystal
field descriptions for transition metal complexes. The energy levels that appear in the
central portion of each MO diagram match the splitting pattern derived for a crystal
field environment of the same symmetry. In the molecular orbital description some of
these levels are antibonding, a concept that of course has no place in the crystal field
model. From our qualitative development here, it may appear that the two models are
quite similar. However, it is important to recognize that the common ground between
them is limited to the symmetry aspects of the bonding description. The two theories
differ fundamentally in how they describe the metal-ligand bond, with the MO view
being more realistic and leading to far better quantitative predictions of properties.
Pi Bonding and In addition to forming a bonds, many ligands are capable of a n bonding interaction
Molecular Orbital with 3 metal. There are no disputes over which ligand orbitals have the correct
Theory symmetry to participate in tt bonding, but as we shall see in later sections, the extent
to which this actually occurs for some ligands is vigorously debated. Even when ligand
and metal orbitals have the proper symmetry for 7r bond formation, an energy or size
mismatch may lead to insignificant interaction.
Recall (Chapter 5) that a tt bond has a nodal surface that includes the bond axis
and that a it bonding orbital will have lobes of opposite sign on each side of this nodal
surface. From the standpoint of orbital symmetry, an octahedral complex could have
up to twelve such bonds—two between the metal and each of the six ligands, although
this number is never realized in an actual complex. Metal and ligand orbitals par¬
ticipating in tt bonds will lie perpendicular to the intemuclear axes. Consider four
potential metal-ligand tt interactions: (I) d v -p n , (2) d^-d v , (3) d^-tr, and (4) d n -<r
(Fig. 11.23). Examples of ligands capable of each type are shown in Table 11.11. In
principle, either the ligand or the metal can function as the electron donor. Filled metal
d orbitals can donate electron density to an empty orbital on the ligand, or an empty d
orbital on the metal can receive electron density from a filled orbital of the ligand.
The ligand group orbitals capable of 7r interactions in an octahedral complex fall
into four symmetry categories (Fig. 11.24): t 2e , r Ul , and r l(f . Of these, a transition
metal will possess orbitals of only two of the types: l 2y (d xv , r/ ( _, and d yi ) and t lu (p r ,
p r and pj. Conceivably, the metal could use all of these orbitals for tt bonds.
However, the members of the t lu set are directed towards the ligands and therefore
participate in strong <r bonds. Formation of tt bonds using these orbitals would tend to
ftft ftQft Oft o Oft
Oft) ftSft Oft'"o Oft
Meial Ligand
(a)
Metal Ligand
Metal Ligand
(c)
Metal Ligand
(d)
Fig. 11.23 Pi overlap of a metal d orbital with various types of ligand orbitals: (a) p, (b) d, (c) tt , and (d) a.
: (> or :; -v feiWpKp• •
Molecular Orbital Theory 421
Table 11.11 _
Pi bonding in coordination
compounds 0
Type
Description
Ligand examples'
P„~ d „
Donation of electrons from
filled p orbitals of ligand to
empty d orbitals of metal
RO“, RS", O 2- , F~, cr,
Br~, I', R 2 N~
Donation of electrons from
filled d orbitals of metal to
empty d orbitals of ligand
R 3 P, RjAs, r 2 s
d n -TT
Donation of electrons from
filled d orbitals of metal to
empty tt antibonding
orbitals of ligand
CO, RNC, pyridine, CN ,
N,, N0 2 , ethylene
d^-ff
Donation of electrons from
filled d orbitals of metal to
empty cr orbitals of ligand
H 2 , RjP, alkanes
° Nugent. W. A.; Mayer, J. M. Metal-Ligand Multiple Bonds', Wiley: New
York, 1988.
Some of these ligands fit into more than one category. For example, |- not
only has filled p orbitals to donate electrons but also has low-lying empty d
orbitals to accept electrons. R 3 P, as shown, may accept tt electron donation into
its empty d or P—R cr orbitals; the a’ contribution is generally regarded as
more significant.
weaken the a system and hence will not be favored. The t 2ll orbitals, on the other
hand, are directed between the ligands which, as we saw earlier, restricts them to a
nonbonding status in a tr-only system (Fig. 11.17b). They can, however, readily
tr-bond to LGOs of matching symmetry (Fig. 11.25). The and r lK ligand group
orbitals must remain nonbonding for the simple reason that there are no orbitals of
matching symmetry on the metal. Pi bonding in an octahedral complex is thus limited
to the orbitals of t 2g symmetry.
One of the simplest cases of tt bonding in octahedral complexes is found in
[CoF 6 ] 3- . Its <t system will be similar to that in Fig. 11.20. The t 2g orbitals of the metal
can interact with t 2g LGOs constructed from the fluorine 2 p orbitals to form 7r-bonding
and antibonding molecular orbitals. Since fluorine is more electronegative than cobalt.
° h £ 8 C 3 6 C; 6 ^ 3C 2 ( = C) i 4S, 85, 30, 6 ^
r„ 12 0 0 0 -4 0 0 0 0 0
r n = 'l, + % + 'l» + '2-
Fig. 11.24 Identification of the symmetries of ligand group orbitals and metal orbitals
capable of participating in tt bonds (represented as vectors) in an octahedral ML„ complex.
The characters and irreducible components of the reducible representation. r„, were derived
by application of the same methods used for the cr-only system (Fig. 11.18).
308
8 ■ Chemical Forces
tronegativities of the nonmetals are very similar to that of hydrogen) and the boiling
points are low and increase with increasing molecular weight. The first member ot
each series (H 2 0. NH 3 , HF) is strongly hydrogen bonded in the liquid state and has a
higher boiling point.
Ionic compounds are characterized by very strong electrostatic forces holding the
ions together. Vaporization results in ion pairs and other small clusters in the vapor
phase. Although the stabilizing energies of these species are large, they are consider¬
ably less than those of the crystals. Assuming a hard-sphere model as a first approx¬
imation, the difference in electrostatic energies of an ion pair in the gas and the solid
lattice would lie in their Madelung constants. For NaF, A = 1.00 tor an ion pair. 1.75
for the lattice. We should thus expect that if crystalline sodium fluoride vaporized to
form ion pairs, the bond energy would be slightly more than halt (1.00/1.75 = 0.57) ot
the lattice energy. There are several factors that help stabilize the species in the gas
phase and make their formation somewhat less costly. Polarization can occur more
readily in a single ion pair than in the lattice. This results in a somewhat greater
covalent contribution and shorter bond distances in the gas phase. Secondly, in
addition to ion pairs there are small clusters of ions with a greater number ot
interactions and more attractive energy. It is not surprising to learn, therefore, that
vaporization costs only about one-fourth of the lattice energy, not almost one-half
(Table 8.5). Nevertheless, since lattice energies are large, the energy necessary to
vaporize an ionic compound is large and responsible for the high boiling points of ionic
compounds.
Table 8.5
Dissociation energies of the
alkali halides for the solid
and gas phases (kJ mol )°
Compound_M X(g)-M"(g ) + X (g) M X(s)-M"(g) + X (g) _ F. u bl
LiF 766 1033 268
LiCl 636 845 209
LiBr 615 799 84
Gas-phase data arc from the bond energies in Appendix E corrected to the ionic
ase by addition of the ionization energy and electron affinity. Lattice energies are
rom the best values in Table 4.3. Energies of sublimation (assuming ion pairing) are
he difference between the energy of the lattice and that of the ion pairs. The ratio is
hat of E, ubl /U which yields the fraction of the energy 'Tost" on sublimation.
Effects of Chemical Forces
309
Increasing the ionic charges will certainly increase the lattice energy of a crystal.
For compounds which are predominantly ionic, increased ionic charges will result in
increased melting and boiling points. Examples are NaF, mp = 997 °C, and MgO,
mp = 2800 °C.
The situation is not always so simple as in the comparison of sodium fluoride and
magnesium oxide. According to Fajans' rules, increasing charge results in increasing
covalency especially for small cations and large anions. Covalency per se does not
necessarily favor either high or low melting and boiling points. For species which are
strongly covalently bonded in the solid, but have weaker or fewer covalent bonds in
the gas phase, melting and boiling points can be extremely high. Examples are carbon
in the diamond and graphite forms (sublimes about 3700 °C) and silicon dioxide (melts
at 1710 °C, boils above 2200 °C). For example, in the latter compound the transition
consists of changing four strong tetrahedral a bonds in the solid polymer to two «r and
two relatively weak -n bonds in the isolated gas molecules.
,Si
"o
O = Si = 0
On the other hand, if the covalent bonds are almost as stable and as numerous in the
gas-phase molecules as in the solid, vaporization takes place readily. Examples are the
depolymerization reactions that take place at a few hundred degrees. For example, red
phosphorus sublimes and recondenscs as white phosphorus.- 5
-p p —p p-
\ D /
2 „ pH-p
Thus increased covalent bonding resulting from Fajans-type phenomena can lower _the
transition temperatures. For example, the alkali halides (except CsCl CsBr, and Csl)
and the silver halides (except Agl) crystallize in the NaCI structure. 1he sizes of he
cations are comparable: Na = 116 pm, Ag‘ - 129 pm. K - pm. y e L
melting points of the halides are considerably different (Table 8.6). The greater
covalent character of the silver halide bond (resulting from the </'<> electron configura¬
tion) compared with those in the alkali halides helps stabilize discrete AgX molecules
in ,he liquid and thus makes the melting points of the silver compounds lower than
those of the potassium compounds. A similar comparison can be made between the
Melting points of potassium
and silver halides
KF = 858 °C
KC1 = 770 °C
KBr = 734 °C
AgF = 435 °C
AgCI = 455 °C
AgBr = 432 °C
NaF = 993 °C
NaCI = 801 °C
NaBr = 747 °C
23 The exact structure of red phosphorus is unknown, but this structure has been suggested. The
argument here is not dependent on knowledge of the exact structure.
418
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Tetrahedral and
Square Planar
Complexes
a Jn antl l iu molecular orbitals are the lowest in energy and the corresponding a, and
antibonding orbitals the highest in energy. The e K and e orbitals arising from the 3d
oibitals are displaced less from their barycenter because of poorer overlap. The /,
orbitals, being nonbonding in a cr-only system, are not displaced at all from their
original energy.
Electrons may now be added to the molecular orbitals of the complex in order of
increasing energy. For [Co(NH 3 ) 6 ] 3+ there will be a total of eighteen electrons to
assign, twelve^from lone pairs on the ammonia ligands and six from the 3d 6 configura¬
tion of the Co- + ion. The electron configuration of the complex can be represented as
or in a more abbreviated form as simply t\ K . Note that [Co(NH 3 ) 6 ] 3+ is
diamagnetic because electrons pair in the t 2g level rather than enter the higher energy
e level. If the energy difference between the t 2g and e levels is small, as in [CoFJ 3- ,
the electrons will be distributed t 4 2f! e 2 . Thus both molecular orbital theory and crystal
field theory account for magnetic and spectral properties of octahedral complexes on
the basis of two sets of orbitals separated by an energy gap A„. If this energy gap is
greater than the pairing energy, low spin complexes will be formed, but if the energy
necessary to pair electrons is greater than A n , high spin complexes will result.
The procedures used in the preceding section may also be applied to the generation of
MO diagrams for complexes of other geometries. The metal atom or ion in each case
will have the same nine valence orbitals available for bonding, but their symmetry
properties will vary from one geometry to another. For a tetrahedral ML, complex (7 (/
symmetry), the metal s and p orbitals have and t 2 symmetries, respectively (see the
I lt character table in Appendix D). The five d orbitals are split into two sets: e (d.-. and
d.r-y-) an d >2 W.r.y djfg, and d y: ). The four LGOs constructed from ligand lone-pair
orbitals will consist of a i 2 set and one orbital of «, symmetry. The /, LGOs can
interact with both sets of metal t 2 orbitals (p and d) to give three sets of o- MOs—one
bonding, one slightly antibonding, and one clearly antibonding.
A it MO diagram for a tetrahedral complex is shown in Fig. 11.21. Note that in
contrast to the octahedral case, the metal e orbitals are now nonbonding. The separa¬
tion between the e and the next highest /, orbitals is labeled A,, as in crystal field
theory. For a complex such as [CoCIJ" . the ligands provide two electrons each for a
total of eight, and the d 1 Co~ ion furnishes seven, giving an overall total of fifteen.
1 welve electrons will fill the six lowest energy molecular orbitals (through the e set)
with the final three electrons remaining unpaired and occupying the slightly anlibond-
ing r 2 molecular orbitals.
A number of four-coordinate complexes adopt a square planar geometry, which
for lour identical ligands, leads to D ih symmetry. In this environment, the metal d
level is split into (</.:), e„ (</,.. d yz ). b 2g (d xv ), and b, K (</,-•_,..•) orbitals. The p level
also loses its degeneracy, appearing as a 2ll ( p .) and e„ (p x . p y ). the four ligands, which
will be oriented along the .r and y axes, will give rise to ligand group orbitals of
b \K' and e„ symmetry. They will interact with metal orbitals of the same symmetry
leading to the rr MO diagram shown in Fig. 11.22. Note that the a l/f LGO overlaps with
both a tll metal orbitals, producing three MOs of this symmetry. Several metal orbitals,
the a 2ll , e g , and the b 2k „ remain nonbonding because they engage in no net overlap
with the ligand orbitals.
The square planar geometry is particularly common for complexes of d 8 metal
ions. For such a complex, there will be sixteen electrons, eight from the metal and
eight more from the four ligands, to be assigned to the molecular orbitals in Fig. 11.22.
These electrons will occupy the eight lowest energy MOs and the complex will be
diamagnetic. The MO description provides a clear rationale for the observed stabilities
(« + Dp
M 4L (LGOs)
Fig. 11.21 A it MO diagram for a complex of tetrahedral symmetry.
M ML 4 4L (LGOs)
Fig. 11.22 A cr MO diagram for a square planar ML, complex (£> 4 „ symmetry).
310
8 ■ Chemical Forces
Melting and boiling points
of some alkali and alkaline
earth halides
Melting point (°C)
KBr = 734 CaBr, = 730
CsF = 682 BaF,‘ = 1355
Boiling point (°C)
KBr = 1435 CaBr, = 812
CsF = 1251 BaFl = 2137
Solubility
predominanlly ionic species CsF and BaF, and the more covalent species KBr and
CaBr, (Table 8.7). The change from 1:1 to 1:2 composition in the highly ionic fluorides
produces the expected increase in lattice energy and corresponding increase in the
transition temperatures. For the more covalent bromides, however, the molecular
species CaBr, (in the gas phase and possibly to some extent in the liquid) has sufficient
stability via its covalency so that the melting point is about the same as that of KBr.
and the boiling point is actually lower.
In extreme cases of Fajans' effects, as in Bel, and transition metal bromides and
iodides, the stabilization resulting from covalency is very large. Distortion of the
lattice occurs and direct comparison with ionic halides is difficult. For metal halides
the boiling points of these compounds are comparatively low as expected: Bel, =
590 °C. Znl, = 624 °C, FeCI 3 = 315 °C. The extreme of this trend is for the covalent
forces to become so strong as to define discrete molecules even in the solid (e.g.,
AI,Br fi , mp = 97 °C, bp = 263 °C). At this point we have come full circle and are back
at the SF 6 and CCI., situation.
Solubility and the behavior of solutes is a complicated subject, 26 and only a brief
outline will be given here. A further discussion of solutions will be found in
Chapter 10.
Solutions of nonpolar solutes in nonpolar solvents represent the simplest type.
The forces involved in solute-solvent and solvent-solvent interactions are all London
dispersion forces and relatively weak. The presence of these forces resulting in a
condensed phase is the only difference from the mixing of ideal gases. As in the latter
case, the only driving force is the entropy (randomness) of mixing. In an ideal solution
^^mixing = aI constant temperature the free energy change will be composed
solely of the entropy term:
40 ' AH - TAS ( 8.201
AG = -TAS (for Ml = 0) (8 21)
The change in entropy for the formation of a solution of this type is 27
AS = ~R(n A In .r A + n B In .t B ) (8 22)
where .r A and r B are the mole fractions of solute and solvent. For an equimolar
mixture of "solute" and "solvent" the change in free energy upon solution at room
temperature is rather small, only - 1.7 kJ mol -1 .
76 For- more detailed discussions of solute behavior, sec Hildebrand. J. H ; Scott, R. L. The Solubility
of Non-electrolytes-, Van Nostrand-Reinhold: New York. 1950; Robinson: R. A.; Stokes. R. H
blectrolyte Solutions, 2nd cd.; Bulterworths: London. 1959.
7 For the origin of this expression and an excellent discussion of the thermodynamics of solution
formation, see Barrow. G. Physical Chemistry, 4th ed.; McGraw-Hill: New York. 1979.
Effects of Chemical Forces
311
At the other extreme from the ideal solutions of nonpolar substances are solutions
of ionic compounds in a very polar solvent such as water. The entropy change for such
a process may be positive or negative unlike in the ideal solute-solvent interaction
described above. In addition to the increased disorder expected as ions go from solid
to solution, there will also be an ordering of solvent molecules as these ions become
solvated. The positive term will be dominant for large ions of low charge, but for ions
that interact strongly with water (small size and high charge), the negative term
becomes more important. For many salts the entropy contribution to the free energy
change for dissolution is comparable in magnitude to the enthalpy change and both
terms must be considered. 2
In order lor an ionic compound to dissolve, the Madelung energy or electrostatic
attraction between the ions in the lattice must be overcome. In a solution in which
the ions are separated by molecules of a solvent with a high dielectric constant
<e H : o = 8 1 • 7e 0 ) lhe attractive force will be considerably less. The process of solution
of an ionic compound in water may be considered by a Born-Haber type of cycle. The
overall enthalpy of the process is the sum of two terms, the enthalpy of dissociating
the ions from the lattice (the lattice energy) and the enthalpy of introducing the
dissociated ions into the solvent (the solvation energy):
I wo factors will contribute to the magnitude of the enthalpy of solvation. One is the
inherent ability of the solvent to coordinate strongly to the ions involved. Polar
solvents are able to coordinate well through the attraction of the solvent dipole to the
solute ions. The second factor is the type of ion involved, particularly its size. The
strength and number of interactions between solvent molecules and an ion will depend
upon how large the latter is. The lattice energy of the solute also depends upon ionic
size. The forces in the lattice are inherently stronger (ion-ion) than those holding the
solvent molecules to the ion (ion-dipole), but there are several of the latter interac¬
tions for each ion. As a result, the enthalpy of solvation is roughly of the same order of
magnitude as the lattice enthalpy, and so the total enthalpy of solution can be either
positive or negative depending upon the particular compound. When the enthalpy of
solution is negative and the entropy of solution is positive, the free energy of solution
is especially favorable since then the enthalpy and entropy of solution reinforce each
other.
In many cases the enthalpy of solution for ionic compounds in water is positive. In
these cases we find the solution cooling as the solute dissolves. The mixing tendency
of entropy is forcing the solution to do work to pull the ions apart, and since in an
adiabatic process such work can be done only at the expense of internal energy, the
solution cools. If the enthalpy of solution is sufficiently positive, favorable entropy
may not be able to overcome it and the compound will be insoluble. Thus some ionic
compounds, such as KCI0 4 , are essentially insoluble in water at room temperature.
The fact that the solubility of a salt depends critically upon the enthalpy of
solution raises an interesting question concerning the magnitude of this quantity.
28 Cox. B. G.; Parker. A. J. J. Chem. Soc. 1973, 95. 6879-6884.
416
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
L
L
O h E 8 C 3 6 C, 6 C 4 3C, ( = Cj) i 6 S 4 SS 6 3 <s h 6a d
r a = 6 T IT T 2 00042
r a = a U + e s + '■»
Fig. 11.18 Identificalion of the symmetries of ligand group orbitals and metal orbitals
involved in the a bonds (represented as vectors) of an octahedral ML complex. The
characters of the reducible representation.,r, r , are derived by counting the number of vectors
that remain unmoved under each symmetry operation of the O,, point group. The irreducible
components of T,, are obtained by application of Eq. 3.1.
Fig. 11.19 Ligand group orbitals (LGOs) and symmetry-matched metal atomic orbitals
appropriate for a bonding in an octahedral ML complex.
same, which is taken to be positive. The six ligands can interact equally with this
orbital, and each contributing orbital must also have a positive sign. Thus the a lg LGO
can be constructed from an additive combination of the six ligand orbitals:
2„,„ = v|(ff r + <y~ x + cr y + <r_ y + <r z + <r _ z ) (11.13)
where 2 and cr represent the wave functions for the ligand group orbital and the
contributing ligand orbitals, respectively, and I/V6 is a normalization constant. The
Molecular Orbital Theory
417
LGO that can interact with the orbital will have components only along the .r
and y axes:
= i(<r, + <J- X ~ °V “ °--v) (11,14)
The LGO that matches the d.z metal orbital is
20-^ - tr x - cr_ x - tr, - <r_ y ) (H.15)
The three LGOs of symmetry that will overlap with the metal p orbitals are
constructed in a similar manner, as shown in Fig. 11.19. Since the metal l 2l! orbitals
cannot participate in cr overlap, they are considered nonbonding molecular orbitals in
complexes where there is no possibility for ir bonding. In cases where there are ligand
orbitals of appropriate symmetry available, the t 2K orbitals will be involved in -n bonds.
A molecular orbital energy diagram for the cr bonds in an octahedral complex such
as ICo(NH 3 ) 6 ] 3+ is shown in Fig. 11.20. There are several approximations involved
and the diagram shown is only qualitatively accurate; even the ordering of the energy
levels is somewhat uncertain. However, this does not detract from the usefulness of
the diagram. It is certain that the overlap of the metal 4 s and 4 p orbitals with ligand
group orbitals is considerably better than that of the 3 d orbitals.25 Consequently, the
M ML 6 6 L (LGOs)
Fig. 11.20 A o--bond molecular orbiial diagram for a complex of octahedral symmetry.
w In general, d orbitals lend lo be large and diffuse and, as a result, overlap of d orbitals with others
may be quantitatively poor even when qualitatively favorable. This problem is discussed in Chap¬
ter 18 .
312
8 • Chemical Forces
Obviously, a large solvation enthalpy contributes toward a favorable enthalpy of
solution. However, we find that the solvation enthalpy alone provides us with
little predictive usefulness. Water soluble salts are known with both large (Cal,,
-2180 kJ mol -1 ) and small (KI, -611 kJ mol - ') hydration energies; insoluble salts
are also known with large (CaF,, -6782 kJ mol"') or small (LiF. - 1004 kJ mol )
hydration energies. It is apparent that the hydration energies alone do not determine
the solubility. Countering the hydration energies in these cases is the lattice energy.
Both lattice energy and hydration energy (Fig. 4.20) are favored by large charge (Z)
and small size (r). The difference lies in the nature of the dependence upon distance.
The Born-Land6 equation for the lattice energy (Eq. 4.13) may be written as a
function of distance:
(8.24)
The simplest equation for the enthalpies of hydration of the cation and anion (Eq. 4.27)
may be rewritten as:
Now the lattice energy is inversely proportional to the sum of the radii, whereas the
hydration enthalpy is the sum of two quantities inversely proportional to the individual
radii. Clearly the two functions will respond differently to variation in r + and r_.
Without delving into the details of the calculations, we may note that Eq. 8.24 is
favored relative to Eq. 8.25 when r + = r_ and the reverse is true for r_ « r + or
r _ » r+ . To express it in terms of a physical picture, the lattice energy is favored
when the ions are similar in size—the presence of either a much larger cation or a
much larger anion can effectively reduce it. In contrast, the hydration enthalpy is the
sum of the two individual ion enthalpies, and if just one of these is very large (from a
single, small ion), the total may still be sizable even if the counterion is unfavorable
(because it is large). The effects of this principle may be seen from the solubility of the
alkali halides in water. Lithium fluoride is simultaneously the least soluble lithium
halide and the least soluble alkali fluoride. Cesium iodide is the least soluble cesium
halide and the least soluble alkali iodide. The most soluble salts in the series are those
with the most disparate sizes, cesium fluoride and lithium iodide. 29
The enthalpy of solution has been discussed somewhat more quantitatively by
Morris.30 He has pointed out the relation between the enthalpy of solution and the
difference between the hydration enthalpy of the cation and that of the anion. This
difference will be largest when the cation and anion differ most in size (Fig. 8.9). In
these cases the enthalpy of solution tends to be large and negative and favors solution.
When the hydration enthalpies (and the sizes) are more nearly alike, the crystal is
favored. When entropy effects are added, a very nice correlation with the solubility
and the free energy solution is found (Fig. 8.10).
There is a very practical consequence of the relation of solubility to size. It is
often possible to make a large, complex ion from a metal and several ligands that is
stable in solution but difficult to isolate without decomposition. Isolation of such
2 » For a very thorough discussion of enthalpy, entropy, and the solubility of ionic compounds, see
Johnson. D. A. Some Thermodynamic Aspects of Inorganic Chemistry; Cambridge University:
London, 1968, Chapter 5.
30 Morris, D. F. C. Struct. Bonding (Berlin) 1968, 4, 63; 1969, 6, 157.
Effects of Chemical Forces
313
Fig. 8.9 Relation between
the heat of solution of a salt
and the individual heats ot
hydration of the component
ions. [From Morris, D. F. C.
Struct. Bonding (Berlin )
1969, 6. 157. Reproduced
with permission.]
Difference in All between anion
and calion (kJ mol’ 1 )
-200 -100 0 +100 +200
Difference in AC„ yd between an.o.i
and cation (kJ mol' 1 )
Fig. 8.10 Relation between
the free energy of solution
of a salt and the individual
free energies of hydration
of the component ions.
[From Morris, D. F. C.
Struct. Bonding (Berlin)
1969, 6, 157. Reproduced
with permission.]
large complex ions is facilitated by attempting to precipitate them as salts ot equa ly
large counterions. This favors the stability of the crystalline stale relative to solu¬
tion and makes it easier to obtain crystals or the desired complex. For example,
the [Ni(CN),] 3- ion was known to exist in solution but when solutions were
evaporated, even in the presence of saturated KCN, only K,Ni(CN) 4 H,0 could
be isolated. However, addition of large complex ions of chromium such as hex-
aamminechromium(III), [Cr(NH 3 )J 3+ , and tnsfethylen^iam.n^chrommmf III),
[Cr(NH 2 CH 2 CH 2 NH 2 ) 3 ] 3+ , allows the separation of [Cr(NH 3 ) h ][Ni(CN) 5 ]-2H 2 0 and
414
Molecular Orbital Theory
415
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Additional evidence for covalency in metal-ligand bonds is provided by electron
paramagnetic resonance (EPR). As a result of their spins, unpaired electrons behave
like magnets and align themselves either parallel or antiparallel to an applied magnetic
field. These two alignments will have slightly different energies, and transitions from
one level to the other can be induced and detected by applying resonant energy in the
form of electromagnetic radiation. An unpaired electron that is not subject to interac¬
tions with other unpaired electrons or with magnetic nuclei will show a single absorp¬
tion for this transition. The EPR spectra of many complexes, however, show
hyperfine splitting patterns that arise from the interaction of the unpaired metal
electron with magnetic nuclei on the ligands (Fig. 11.16). This clearly indicates that the
electron is at least partially delocalized over the ligands.
Octahedral The construction of molecular orbitals for an octahedral complex involves the same
Complexes general approach that was used in Chapter 5 for simpler molecules and ions. In the
case of the complex, there will merely be more overlapping orbitals and electrons to
consider. For a complex ion such as [Co(NH 3 ) 6 ] 3+ , the valence orbitals available on
the central metal will be the 3d, 4 j, and 4 p. The ligand orbitals involved in <j bonds to
the metal will be the six approximately jp 3 hybrid lone pair orbitals on the ammonia
molecules. (For 7r-bonding ligands, additional orbitals would have to be considered.)
Although it may at first appear that finding the proper linear combinations of nine
metal orbitals and six ligand orbitals would be a formidable task, we can draw on our
previous experience, which has shown that using the symmetry properties of the
orbitals greatly expedites the procedure.
Since the molecular orbitals we are seeking will be linear combinations of metal
and ligand atomic orbitals having the same symmetry, it is appropriate to begin by
constructing linear combinations of the ligand orbitals, or ligand group orbitals
( LGOs ), that will overlap with metal orbitals along the octahedral bonding axes.
(Recall that this approach was used previously for BeH 2 and NOf in Chapter 5.)
These LGOs must match the symmetries of the metal orbitals available for bonding.
As has already been stated, the metal valence orbitals of interest are the ns, np, and
{n - I )d. Their symmetry properties in an octahedral complex can be determined by
reference to an O h character table (Appendix D), which reveals that the s orbital
transforms as a lR and the set of p orbitals as r,„. whereas the five d orbitals lose their
degeneracy to form e R (d.j and t/ r :_ y2 ) and (t/ rv ,and d y .) sets. (Note that these
two groups of d orbitals are the same as we saw previously in crystal field theory.)
In order to actually participate in a cr bond within the complex, a metal orbital
must be capable of positive overlap with a ligand group orbital directed along the
bonding axes. For the moment, let us merely consider the directional requirement and
— ~75G—
Fig. 11.16 The electron
paramagnetic resonance
spectrum of K 2 lrCl h in
K;PtCI„. [From Owen. J.
Faraday Disc. Client. Soc.
1955. 19. 127-134.
Reproduced with
permission.]
ignore the fact that for positive overlap, the metal and ligand orbitals must also have
the same sign. The a lf! metal orbital will be spherical, therefore being capable of
overlapping with LGOs on all axes. The and e M sets all have lobes concentrated
along the bond directions and thus also are capable of bond participation. This is
shown in Fig. 11.17a for the d x :_ y : orbital. The t 2R set, however, will have lobes
directed between the bonding axes and thus will yield no net overlap with ligand
orbitals (Fig. 11.17b).
The foregoing analysis has shown that the LGOs to be constructed for the cr bonds
of an octahedral complex must be of a iR , and e R symmetries. The same conclusion
could have been reached in a more formal manner by considering the six cr bonds of an
octahedral complex as vectors (Fig. 11.18) and determining how they transform under
the symmetry operations of the O h point group. (For a review of the procedure
involved, see Chapter 3.) The reducible representation, T, r , that results from counting
the number of vectors that remain unchanged by each symmetry operation is shown in
Fig. 11.18. Reduction of this representation (Eq. 3.1) yields a if ,, e K , and t Ul as its
irreducible components, indicating that these are the symmetries of the metal and
ligand group orbitals that will be suitable for forming a molecular orbitals in the
complex—the same conclusion reached earlier by considering the capability of the
metal valence orbitals for overlap.
Determining the symmetries of the orbitals that will participate in the bonding
does not tell us what specific combinations of ligand lone-pair orbitals should be used
in constructing the six LGOs.- 11 Some insight into this process can be obtained from
Fig. 11.19. The sign of the wave function for the metal orbital is everywhere the
Fig. 11.17 Overlap of ligand orbitals in the .ty plane with metal (a) and d, y (b) orbitals.
Note that an appropriate choice of sign for the ligand orbitals provides positive overlap with
the c/,!_ V ! orbital; however, no single sign choice for the ligand wave function produces
positive overlap with the d, y orbital.
:4 Within a group theoretical analysis, this is normally accomplished with projection operators. For a
discussion of this method, see the group theory texts listed in Footnote I of Chapter 3.
314
8 - Chemical Forces
315
Problems
[Ci(NH,CH : CH : NH;,) 3 ][Ni(CN) 5 J'I.5H-.0, both of which are stable at room tem¬
perature.-• ,|
The insolubility of ionic compounds in nonpolar solvents is a similar phe¬
nomenon. The solvation energies are limited to those from ion-induced dipole forces,
which are considerably weaker than ion-dipole forces and not large enough to over¬
come the very strong ion-ion forces of the lattice.
The reason tor the insolubility of nonpolar solutes in some polar solvents such as
water is less apparent. The forces holding the solute molecules to each other (i.e., the
forces tending to keep the crystal from dissolving) are very weak London forces. The
interactions between water and the solute (dipole-induced dipole) are also weak but
expected to be somewhat stronger than London forces. It might be supposed that this
small solvation energy plus the entropy of mixing would be sufficient to cause a
nonpolar solute to dissolve. In fact, it does not because any entropy resulting from the
disordering of the hydrogen bonded structure of the solvent water is more than offset
by the loss ol energy from the breaking of hydrogen bonds. Anthropomorphically we
might say that the solute would willingly dissolve but that the water would rather
associate with itself.
We can summarize the energetics of solution as follows. There will usually be an
entropy driving force favoring solution. In cases where the enthalpy is negative, zero,
or slightly positive, solution will take place. If the enthalpy change accompanying
solution is too positive, solution will not occur. In qualitatively estimating the en¬
thalpy effect, solute-solute, solvent-solvent, and solute-solvent interactions must be
considered:
^^solutlon ^^solute—solvent ~ ^^solute-solute — ^^solvent-solvent (8.26)
where the various energies result from ion-ion. ion-dipole, ion-induced dipole,
dipole-dipole, and London forces.
8.1 From Fig. 8.1 describe the crystal structure of solid argon.
8.2 Confirm the statement made on page 290 that the van dcr Waals radius of argon is 190 pm.
using Fig. 8.1 and the knowledge that the unit cell is 535 pm on each edge.
8.3 Predict the internuclear distances in the following molecules and lattices by use of the
appropriate van der Wauls, ionic, covalent, and atomic radii. In those cases where two or
more sets of values are applicable, determine which yield the results closest to the
experimental values.
System
Distance
'•(pm)
LiF molecule
Li—F
155
LiF crystal
Li—F
201
Csl molecule
Cs—1
332
Csl crystal
Cs—1
395
Lil molecule
Li—1
239
Lil crystal
Li—1
302
XeF, molecule
Xe—F
194
XeF., crystal
F—F (different molecules)
313
H,0 molecule
H—O
96
SnCl d molecule
Sn—Cl
233
" For discussion of this subject, see Basoto. F. Coord. Cliem. Rev. 1963. J. 213: Mingos. D. M. P.;
Rolf. A. L. Inorg. Chan. 1991. JO. 3769-3771.
Problems
8.4 At one time the melting points of the fiuorides of the third-row elements were taken to
indicate a discontinuity between ionic bonding (AIF,) and covalent bonding (SiF.,). Explain
the observed trend assuming that the bond polarity decreases uniformly from NaF to SF,,.
NaF = 993 °C AIF 3 = 1291 °C PF, = -83 °C
MgF, = 1261 °C SiF d = -90 °C SF„ = -51 °C
8.5 List the following in order of increasing boiling point:
FLO Xe LiF Lil H, BaO SiCl„ SiO,
8.6 The majority of clathrate compounds involve hydrogen bonding in the host cages. Discuss
how the intermediate nature of the hydrogen bond (i.e., stronger than van der Waals
forces, weaker than ionic forces) is related to the prevalence or hydrogen bonded
clathrates.
8.7 Two forms of boron nitride are known. The ordinary form is a slippery while materiul. The
second, formed artificially at high pressures, is the second hardest substance known. Both
remain as solids at temperatures approaching 3000 °C. Suggest structures.
8.8 Predict bond lengths in the following: FLO. HCI, NF 3 . CF„, H,S, SF,.
8.9 The Schomaker-Stevenson relationship states that heteropolar bonds are always stronger
and shorter than hypothetical, purely covalent bonds between the same atoms. In an ionic
crystal, would you expect some covalency to shorten or lengthen the bond? Explain.
(Shannon. R. D.; Vincent. H. Struct. Bondinn (Berlin) 1974, 19. I.)
8.10 Find the melting points and boiling points of the elements or compounds listed. For each
series, tabulate the data and explain the trends you observe in terms of the forces involved:
a. He. Ne. Ar. Kr, Xe
b. H 2 0. H,S. H,Se. H,Te
c. CH 4 , CHjCI. CH,CL, CHClj, CO,
d. Carbon, nitrogen, oxygen, fluorine, neon
8.11 Consider the sizes of the isoelectronic species N , . 0 : . F~, and Ne. Discuss the forces
operating. Caveat'. Be careful in choosing which numbers to use in your discussion.
8.12 The stability of noble gas configurations was discussed in Chapter 4, where it was pointed
out that many ions are not stable, that is. they arc endothermic with respect to the
corresponding atoms, but they are stabilized by the ionic lattice. However, some chemists
argue that these ions are stable because they exist in solution as well as in lattices. Discuss.
8.13 Consider the ions (Ph,B— C=N — BPh,] - and |Ph,P=N—PPh,] + .
a. Work out the electronic structures of these ions in detail including assigning formal
charges.
b. Compare the geometries and other similarities or differences.
c. How should these ions prove useful in inorganic synthesis?
8.14 Water is well known to have an unusually high heat capacity. Not so well known is that
liquid XeF ft also has a high heat capacity compared to “normal" liquids such as argon,
carbon tetrachloride, or sulfur dioxide. From your knowledge of the structures of the solids
and the gaseous molecules or these materials (most of them are sketched in this text),
explain the "anomalous" heat capacity of XcF„,
8.15 Find the solubilities in water of the alkali halides. Calculate the molarity or molality (as
convenient) of a saturated solution for each and plot them in matrix form with columns
headed F. Cl. Br. I. and horizontal rows labeled Li, Na, K. Rb. and Cs. Discuss any trends
you notice.
8.16 In these first eight chapters you have encountered many tables of atomic and molecular
412
11 - Coordination Chemistry: Bonding, Spectra, and Magnetism
Several factors undoubtedly play a role in determining whether an oxide adopts
the normal or inverse spinel structure, one of which is d orbital splitting energy.
Although the CFSE contribution to the total bonding energy of a system is only about
5-10%, it may be the deciding factor when the other contributions are reasonably
constant. The crystal field contribution for spinels can be assessed by considering the
difference in crystal field stabilization energies for octahedral compared to tetrahedral
coordination for the metal ions involved. For purposes of estimating this difference, it
can be assumed that the oxide ions will provide a moderately weak crystal field similar
to that for water, for which a number of A„ values have been measured. Values of A,
for the four-coordinate sites can be approximated by the relationship A, = JA„.
Octahedral site preference energies determined in this manner for di- and trivalent ions
of the first transition series are given in Table 11.9.
Table 11.8 reveals that most spinels involving Fe 3+ (AFe 2 0 4 ) have the inverse
structure. The d 5 Fe 3+ ion will have a CFSE of zero for both tetrahedral and
octahedral coordination, so if there is to be a site preference it will be due to the A(II)
ion. This is clearly the case for NiFe 2 0 4 , for example, the Ni 2 ' ion having an
octahedral site preference energy of 86 kJ mol -1 . In magnetite, Fe 3 0 4 , both A and B
ions are iron, with some in the +2 oxidation state and others in +3: Fe ll Fe lll 0 4 . For
the d 6 Fe 2+ ion, octahedral coordination is more favorable than tetrahedral by about
15 kJ mol -1 , whieh, although only a modest amount, is apparently sufficient to invert
the structure. In contrast, the similar oxide Mn 3 0 4 has the normal structure. In this
instance, the d 5 Mn 2 ' has no CFSE in either octahedral or tetrahedral fields, but d A
Mn 3 + shows a preference of 106 kJ mol -1 for octahedral sites. For Co 3 0 4 , another
mixed-valence oxide, there is an additional factor to take into account—Co 3 ‘ is low
Table 11.9
Crystal field data for aqua complexes of metal ions in the first transition series
No. of d
electrons
Ion
Free
ion
ground
state
Octahedral Tetrahedral
field field Ao®
configuration configuration (cm )
A» b ,
(cm ')
CFSE
(kJ mol - ')
Octahedral
site preference
energy
(Id mar’)
Oct.
Tetr.
1
Ti 3+
2 d
e 1
20.100
8930
96.2
64.1
32.1
2
V 3+
3 f
e 2
19,950
8870
190.9
127
64
3
V 2
F
•In
e 2 t{
12,100
5380
174
51.5
122
Cr 3+
V
•In
4
17,400
7730
250
74.0
176
4
Cr 2+
f D
'2n e 'n
f2, 2
14,000
6220
101
29.8
71
Mn ,+
S D
fi e'
2n c x
e 2 , 2
21,000
9330
151
44.6
106
5
Mn 2+
b S
,3 „2
‘2ii e n
e 2 t\
7500
3330
0
0
0
Fe 3+
b S
t\A
e 2 t\
14,000
6220
0
0
0
6
Fe 2+
5 d
e',\
9350
4160
44.7
29.9
14.8
Co 3+
’d
•%
e>,\
20,760
—
—
—
— r
7
Co 2+
4 f
,'f!
e 4 t\
9200
4090
88.0
58.7
29.3
8
Ni 2+
3 f
8500
3780
122
36.2
86
9
Cu 2+
2 d
V2
12,000
5330
86.1
25.5
60.6
" Experimental values for (M(H 2 0) h ] : ' + . Lever, A. B. P. Inorganic Electronic Spectroscopy. 2nd ed.; Elsevier: New York, 1986;
Chapter 6. >
h Calculated as IA„.
c Octahedral site preference energy not calculated because [CotHiOIJ 3 is low spin.
Molecular Orbital Theory 413
spin in the field produced by six oxide ions. This makes for complications in estimating
an octahedral site preference energy for Co 3+ because in a tetrahedral site it would be
high spin. However, the low spin d b configuration imparts additional stabilization to
Co 3+ in an octahedral hole. Thus the octahedral preference for Co 3+ will clearly
outweigh that for Co 2+ (29.3 kJ mol -1 ), favoring the normal arrangement.
Although crystal field theory quite successfully rationalizes observed structures
of the spinels of the first transition series, it must be applied with care to other
examples. In comparing structures in which other factors (ionic radii, covalency, etc.)
are more dissimilar, d orbital splittings alone generally do not explain the observa¬
tions. In these cases, a broader analysis is required.
Molecular Orbital Although the crystal field theory adequately accounts for a surprisingly large quantity
-- of data on transition metal complexes, the theory has serious limitations. There are
Theory several experimental and semitheoretical arguments that can be presented against the
assumption that the splitting of metal d orbitals is a result solely of electrostatic effects
and that the bonding between metal and ligand is ionic with no covalent character.
Indirect evidence that electrons are shared between the ligands and the central
metal ion comes from the nephelauxetic effect. It is found that the electron-electron
repulsion in complexes is somewhat less than that in the free ion. From data derived
from electronic spectra of complexes, separate nephelauxetic series may be set up for
metal ions and for ligands, indicating the order of decreasing electron-electron repul¬
sion (or increasing nephelauxetic effect) (Table 11.10). The observed decrease in
electronic repulsion that occurs upon bond formation may be attributed to an effective
increase in the distance between electrons that results when metal and ligand orbitals
combine to form larger molecular orbitals. (Nephelauxetic means "cloud expand¬
ing.”) The ligands that are most effective in delocalizing metal electrons display the
largest values of the nephelauxetic parameter, h.
Tobl e 11.10 _
The nephelauxetic series of
ligands and metal ions°
Ligand
h
Metal
k
F -
0.8
Mn(II)
0.07
h 2 o
1.0
V(II)
0.1
urea
1.2
Ni(II)
0.12
NH 3
1.4
Mo(Ill)
0.15
en
1.5
Cr(III)
0.20
ox
1.5
Fe(IIl)
0.24
cr
2.0
Rh(III)
0.28
CN -
2.1
Ir(III)
0.28
Br -
2.3
Co(lII)
0.33
Nr
2.4
Pt(IV)
0.6
r
2.7
Ptl(IV)
0.7
« The total nephelauxetic effect in a
complex MX„ is
proportional to the
product hg-k M .
For ligand
abbreviations, see Table 11.7.
Jprgensen C. K. Oxidation Numbers
and Oxidation States-, Springer: New
York; 1969; p 106. Used with
permission.
Problems
317
316 8 • Chemical Forces
properties. They may be classified into two groups: (I) radial wave functions, ionization
energies, electron affinities, etc.; (2) ionic radii, covalent radii, electronegativities, etc.
a. What distinguishes and separates these two groups?
b. Lee Allen (pers. comm.) has suggested that it is the problems associated with the second
group that make chemistry a distinct and interesting science, not just a sub-branch of
chemical physics. Discuss.
8.17 Predict which of the following bonding interactions will be the stronger:
a. 0=0 or 0—0 b. C—C or Si—Si
c. Ne—NeorXe— Xe <1. Li'F” or Mg : 0 : (ion pair)
e. LiF - or Ba 3+ Te J " (ion pair) f. LiF' or C—C (in diamond)
8.18 Predict the following bond lengths:
C—O (ketones) C—O (carbon monoxide)
C—N (cyanide) P—O [single vs. double in (H0)jP=0)
N—O (nitric oxide) N—O (nitrogen dioxide)
Compare your answers with the values given in Table E.l. Appendix E.
8.19 Reconsider Problem 4.15. Extend your explanation: Suggest a means of stabilizing hexa-
fluoroplatinale(V) salts.
8.20 How will the IR and Raman spectra of FHF" and CIHF differ?
8.21 On page 302 it was stated that although the FHF" anion was usually symmetrical, occa¬
sionally it was found to be unsymmetrical in the solid. What physical methods could you
use to detect unsymmetrical FHF ions in a solid?
8.22 If you did not do Problems 4.37 and 4.38 when you read Chapter 4. do them now.
8.23 In the preparation of the bis(N-nitrosopyrrolidine)hydrogen cation with the short hydrogen
bond (Fig. 8.5). the complex was made by evaporating a solution of N-nitrosopyrrohdinc
and hexafluorophosphoric acid in ether to obtain crystals. Why hexafluorophosphonc acid?
Why not hydrofluoric acid?
8.24 If you did not do Problem 6.15 when you read Chapter 6, do it now.
8.25 How would you characterize the hydrogen bond described in Problem 3.42?
8.26 The ammonium ion is about the same size (r, = 151 pm) as the potassium ion (r, =
152 pm) and this is a useful fact to remember when explaining the resemblance in properties
between these two ions. For example, the solubilities of ammonium salts are similar to
those of potassium salts. Explain the relation between ionic radius and solubility. On the
other hand, all of the potassium halides crystallize in the NaCI structure with C.N. = 6 (see
Chapter 4). but none of the ammonium halides does so. The coordination numbers of the
ammonium halides are either four or eight. Suggest an explanation.
8.27 Find as many data as you can (distances, energies, etc.) on hydrogen bonds and hydrogen
bridges. Use this chapter. Chapters 15 and 16. and any other sources. Arrange the data and
argue (however you conclude) that hydrogen bonds such as F—H—F and hydrogen
bridges such as W—H—W (are. are not) merely the ends of a continuous spectrum of
bonding. (See Bau. R.; Teller. R. G.; Kirtley, S. W.; Koetzlc. T. F. Ate. Chem. Res. 1979.
12. 176-183.)
8.28 If you did not do Problem 3.34 when you read Chapter 3. do it now.
8.29 The classical qualitative analysis scheme is based on solubility rules: Acetates, nitrates,
and chlorides (except for Ag\ Hg;. and Pb 2+ ) are always soluble. There are specific
solubility patterns for sulfides, carbonates, and phosphates. Find a qual scheme and
explain it in terms of your understanding of solution processes.
8.30 The point is made early in this chapter that comparisons of atomic radii should be restricted
lo the same type of radii: van der Waals vs. van der Waals, covalent vs. covalent, etc. In
class one day a student asked. "Which is larger, the van der Waals radius of neon or the
covalent radius of sodium?" (!) Was that a nonsensical question? Discuss.
8.31 Superimpose Figs. 8.2 and 8.3 (a or b. but not both at the same time). Discuss the physical
picture of van der Waals. ionic, and covalent radii and the problems associated with the
calculations on page 293.
410
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Fig. 11.15 Radii of the
divalent ions Ca 2 to Zn 2
(above) and the trivalent
ions Sc 3 to Ga 3 (below)
as a function of the number
of d electrons. Low spin
ions are indicated by solid
circles. [Data from
Shannon, R. D.; Prewitt.
C. T. Acta Cryst. 1970.
B26, 1076.1
90
I 80
“ 70
0123456789 10
Number of id electrons
The crystal field model can also be used to account for the stability of particular
oxidation states. In aqueous solution Co(III) is unstable with respect to reduction by
water to form Co(ll). Although there are several energy terms involved, this may be
viewed as a reflection of the high third ionization energy of cobalt. If various
moderate-to-strong field ligands are present in the solution, however, the Co(III) ion is
perfectly stable. In fact, in some cases it is difficult or impossible to prevent the
oxidation of Co(ll) to Co(III).
For example, the appropriate emfs (in volts) are
[Co(H 2 0)J 2 -♦ [Co(H,0)„] 3+ + e“ E° = -1.83 (11.6)
[Co(ox)j] 4_ - [Co(ox),p- + e" E° = -0.57 (11.7)
[Co(phen)j] 2+ -♦ [Co(phen) 3 ] ,+ + e“ E° = -0.42 (11.8)
(Co(edta)] 2- -♦ [Co(edta)]“ + e" E° = -0.37 (11.9)
[Co(en),] 2 -♦ [Co(en) 3 ] 3 + e" E" = -0.18 (11.10)
[Co(NH 3 )„] 2 -► [Co(NHj) a ) 3+ + e" E° = -0.11 (11.11)
[Co(CN),H 2 0] 3- + CN~ -► [Co(CN)J 3- + H 2 0 + e _ £° = +0.83 (11.12)
Note that the order of ligands in Eqs. 11.6-11.12 is approximately that of the spec-
trochemical series and hence that of increasing crystal field stabilization energies. The
Number of id electrons
..... ; .. j -
Crystal Field Theory 411
oxidation of Co(II) to Co(III) results in a change from high to low spin. We can think of
the oxidation as taking place in two steps, the first being the rearrangement of
electrons to the low spin state and the second the removal of the e g electron to
produce Co(lII):
C° 2+ (4> ~ Co 2 + (f^) — Co 3+ (4$
This is not to imply that the process actually occurs in this manner, but we may
consider the thermodynamics as the sum of this hypothetical sequence. The first step
involves pairing of electrons, and the energy required for this will be in part compen¬
sated by the additional CFSE of the low spin configuration (l.8A„ versus 0.8A„). The
stronger the field, the larger will be the magnitude of A„. The second step, removal of
an electron from the e g level, is endothermic because of the high ionization energy
(Co 2+ to Co 3+ ), but the increase in CFSE (1.8A„ to 2.4A„) will favor ionization. It
should be pointed out lhat CFSE is only one of a number of factors affecting the emf.
In particular, entropy effects associated with chelate rings can be important and are
largely responsible for the fact that the order of ligands in Eqs. 11.6-11.12 is different
from that in the spectrochemical series. In any event, the emf of a couple can be
"tuned” by varing the nature of the ligands, a phenomenon that becomes exquisitely
important in biological systems (see Chapter 19).
Crystal field factors may also be used to help account for observed site prefer¬
ences in certain crystalline materials such as the spinels. Spinels have the formula
AB 2 0 4 , where A can be a Group IIA (2) metal ora transition metal in the +2 oxidation
state and B is a Group 1IIA (3) metal or a transition metal in the +3 oxidation state.
The oxide ions form a close-packed cubic lattice with eight tetrahedral holes and four
octahedral holes per AB 2 0,, unit. In a so-called normal spinel such as MgAI-,0.,, the
Mg 2 ions occupy one-eighth of the tetrahedral holes and the Al 3+ ions occupy one-
half of the available octahedral holes. This is the arrangement that would be predicted
to be most stable inasmuch as it yields a coordination number of 4 for the divalent ion
and 6 for the trivalent ion (cf. [Be(H 2 0) 4 ] 2 and [AI(H 2 0)J 3 , page 393). Very
interesting, therefore, arc spinels having the inverse structure in which the A(I1) ions
and one-half of the B(III) ions have exchanged places; i.e., the A(II) ions occupy
octahedral holes along with one-half of the B(III) ions while the other one-half of the
B(III) ions are in tetrahedral holes. Also observed are cases that are intermediate
between the normal and inverse distributions. It is common to describe the structure
of a spinel by the parameter A, defined as the fraction of B ions in tetrahedral holes.
The value of A ranges from zero for normal spinels to 0.50 for those having the inverse
composition. Cation distributions (as A values) in a number of common spinels are
given in Table 11.8.
Table 11.8
Values of A for spinels,
A"Byo 4 »
\ A 2
B 3 \
Mg 2+
Mn 2
Fe 2+
Co 2+
Ni 2+
Cu 2+
Zn 2+
ai 3+
0
0
0
0
0.38
_
0
Cr 3+
0
0
. 0
0
0
0
0
\
Fe 3
0.45
0.1
0.5
0.5
0.5
0.5
0
Mn 3
—
0
—
—
—
—
0
Co 3
—
J —
—
0
—
■ —
0
“ Wells, A. F. Structural Inorganic Chemistry, 5th ed.; Clarendon: Oxford, 1986; p 595. Used
with permission.
Chap ter
Acid-Base Chemistry
Acids and bases are fundamental to inorganic chemistry. Together with the closely
related subjects of redox and coordination chemistry, they form the basis of descrip¬
tive inorganic chemistry. Because they are so fundamental, there has been much work
(and sometimes much disagreement) attempting to find the -best” way of treating the
subject.
T he first point to be made concerning acids and bases is that so-called acid-base
theories” are in reality definitions of what an acid or base is; they are not theories in
the sense ot valence bond theory or molecular orbital theory. In a very real sense, we
can make an acid be anything we wish; the dift'erences between the various acid-base
concepts are not concerned with which is "right” but which is most convenient to use
in a particular situation. All of the current definitions of acid-base behavior are
compatible with each other. In fact, one of the objects in the following presentation of
many different definitions is to emphasize their basic parallelism and hence to direct
the students toward a cosmopolitan attitude toward acids and bases which will stand
them in good stead in dealing with various chemical situations, whether they be in
aqueous solutions of ions, organic reactions, nonaqueous titrations, or other
situations.
Bronsted-Lowry In 1923 J. N. Brpnsted and T. M. Lowry independently' suggested that acids be
Definition defined as proton 2 donors and bases as proton acceptors. For aqueous solutions the
1 Br0nsled, J. N. Reel. Trav. Cliim. Pays-Bos 1923, 42. 718-728; Lowry. T. M. Chem. Ind. ( London )
1923 , 42. 43.
2 As (his book was going io press the International Union of Pure and Applied Chemistry recom¬
mended that the word proton be used only when the 'H isotope was intended, and that the more
general hydron be used everywhere else, as in hydron donor. Sec Appendix I. Section 8. We have not
attempted at the last minute to change all of these "protons" to "hydrons." Like the St system of
units, this change, if accepted by the world's chemists, will take some time, and the term "proton
donor" will not soon disappear.
Acid-Base
Concepts
318
Acid—Bose Concepts
319
Lux-Flood Definition
Br0nsted-Lowry definition does not differ appreciably from the Arrhenius definition
of hydrogen ions (acids) and hydroxide ions (bases);
2H 2 0 ;=± H 3 0 + + OH" (9.1)
Pure solvent Acid Base
The usefulness of the Br0nsted-Lowry definition lies in its ability to handle any
protonic solvent such as liquid ammonia or sulfuric acid:
NH 4 + NH 2 " - 2NH 3 (9.2)
Acid Base Neutralization
product
H 3 S0 4 + + HS0 4 " - 2H 2 S0 4 (9.3)
Acid Base Neutralization
product
In addition, other proton-transfer reactions that would not normally be called neu¬
tralization reactions but which are obviously acid-base in character may be treated as
readily:
NH 4 + + S 2 ~ -► NH 3 + H S~ (9.4)
Acid Base Base Acid
Chemical species that differ from each other only lo the extent of the transferred
proton are termed conjugates (connected by brackets in Eq. 9.4). Reactions such as
the above proceed in the direction of forming weaker species. The stronger acid and
the stronger base of each conjugate pair react to form the weaker acid and base. The
emphasis which the Br 0 nsted-Lowry definition places on competition for protons is
one of the assets of working in this context, but il also limits the flexibility of Ihe
concept. However, as long as one is dealing with a protonic solvent system, the
Br0nsted-Lowry definition is as useful as any. The acid-base definitions given below
were formulated in an attempt lo extend acid-base concepts to systems not containing
protons.
In contrast to the Br0nsted-Lowry theory, which emphasizes Ihe proton as (he
principal species in acid-base reactions, the definition proposed by Lux and extended
by Flood 3 describes acid-base behavior in terms of the oxide ion. This acid-base
concept was advanced to treat nonprotonic systems which were not amenable to the
Brpnsted-Lowry definition. For example, in high-temperature inorganic melts, reac¬
tions such as the following take place:
CaO + Si0 2 -> CaSi0 3 (9.5)
Base Acid
3 Lux. H. Z. Eleklrochem. 1939, 45. 303. Flood. H.; Forland, T. Acta Chem. Scnnd. 1947. /. 592-604,
781. Flood, H.; Forland. T.: Roald. B. Ibid. 1947. /. 790-798.
408
11 - Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.7
Selected values of /and g Ligand
f factor
Metal ion
g factor
factors" g-
0.72
Mn(Il)
8.0
SCN -
0.73
Ni(II)
8.7
cr
0.78
Co(Ii)
9
N,~
0.83
V(II)
12.0
F -
0.9
Fe(III)
14.0
ox = C 2 Oj -
0.99
Cr(IIl)
17.4
H,0
1.00
Co(III)
18.2
NCS -
1.02
Ru(II)
20
gly“ = NH 2 CH,C0 2
1.18
Mn(IV)
23
py = C,H 5 N
1.23
Mo(III)
24.6
NHj
1.25
Rh(III)
27.0
en = NH 2 CH 2 CH 2 NHj
1.28
Tc(IV)
30
bpy = 2,2'-bipyridine
1.33
will)
32
CN“
1.7
Pt(IV)
36
" Jprgensen, C. K. Modern Aspects of Ligand Field Theory ;
Elsevier: New York, 1971; Chapter 26.
In units of kK (= 1000 cm -1 ). Ligating atoms are shown in
boldface type.
central metal ions and led eventually to the development of bonding descriptions that
include covalent interactions between ligands and metal. Despite its imperfections, the
basic crystal field theory can be used to interpret a number of effects in coordination
chemistry, several of which are discussed in the next section.
Applications of Among the early successes of crystal field theory was its ability to account for
Crystal Field Theory magnetic and spectral properties of complexes. In addition, it provided a basis for
understanding and predicting a number of their structural and thermodynamic proper¬
ties. Several such properties are described in this section from the crystal field point of
view. Certainly other bonding models, such as molecular orbital theory, can also be
used to interpret these observations. Even when they are, however, concepts from
crystal field theory, such as crystal (or ligand) field stabilization energy, are often
invoked within the discussion.
One of the first indications that crystal field stabilization energy might be impor¬
tant in transition metal compounds arose from the computation of lattice energies. We
have seen in Chapter 4 that, although the prediction of lattice energies is not highly
accurate, such predictions are much better for ions such as Na + , K + , Ca 2+ , Mn~ + ,
and Zn 2+ than they are for Cr + , Fe 2 , Co 2+ , Ni 2+ , and Cu 2+ . Wherever a serious
discrepancy for a six-coordinate metal ion is found, it may be attributed to the CFSE.
The ions that do not show such discrepancies are those with d°, d 5 (high spin), and d w
configurations, which all have in common that CFSE = 0. Consider the lattice
energies of the halides from CaX 2 to ZnX 2 , in which the metal ions occupy octahedral
holes. Inasmuch as we might expect a gradual decrease in ionic radius from Ca 2 ‘ to
Zn 2+ (Chapter 2), we should also expect a gradual and smooth increase in lattice
energy based on the Born-Lande equation (Chapter 4). However, as shown by
Fig, 11.14, the expected smooth curve is not observed. The ions Ca 2 + , Mn -+ , and
Zn 2+ lie on a curve that is nearly a straight line. Moreover, deviations from this
approximate line are maximized in two places: in the region of V 2+ and the region of
Ni 2+ . Table 11.3 indicates that for a weak octahedral field (recall that the halide ions
Crystal Field Theory 409
Fig. 11.14 Lattice energies
of the divalent metal halides
of Ihe first transition series.
Vertical bars indicate
uncertainties in
experimental values.
[Modified from George. P.;
McClure, D. S. Prog.
Inorg. Client. 19S9, /,
381-463. Reproduced with
permission.]
are on the weak end of the spectrochemical series), V 2 (z/ 3 ) and Ni 2+ (t/ 8 ) have the
greatest CFSE values (1,2A„). The d 2 , d 4 , d 1 , and d ions have somewhat less (0.6 and
0.8A„) and the d°, d s , and d'° cases have zero CFSE, qualitatively confirming the
shape of the curve within the unfortunately large experimental errors.
Somewhat better data are available for the enthalpies of hydration of transition
metal ions. Although this enthalpy is measured at (or more properly, extrapolated to)
infinite dilution, only six water molecules enter the coordination sphere of the metal
ion to form an octahedral aqua complex. The enthalpy of hydration is thus closely
related to the enthalpy of formation of the hexaaqua complex. If the values of AH,, yd
for the +2 and +3 ions of the first transition elements (except Sc 2+ , which is unstable)
are plotted as a function of atomic number, curves much like those in Fig. 11.14 are
obtained. If one subtracts the predicted CFSE from the experimental enthalpies, the
resulting points lie very nearly on a straight line from Ca 2+ to Zn“ + and from Sc^ to
Fe 3+ (the +3 oxidation state is unstable in water for the remainder of the first
transition series). Many thermodynamic data for coordination compounds follow this
pattern of a double-humped curve, which can be accounted for by variations in CFSE
with d orbital configuration.
A slightly different form of the typical two-humped curve is shown by the ionic
radii of the 3 d divalent metals. These are plotted in Fig. 11.15 (from Table 4\4). For
both dipositive and tripositive ions there is a steady decrease in radius for the strong
field case until the r 2jf configuration is reached. At this point the next electron enters
the e level, into an orbital directed at the ligands, repelling them and causing an
increase in the effective radius of the metal. In the case of high spin ions the increase in
radius occurs with the if- e' x configuration for the same reason.
320
9- Acid-Base Chemistry
Solvent System
Definition
The base (CaO) is an oxide donor and the acid (Si0 2 ) is an oxide acceptor. The
usefulness of the Lux-Flood definition is mostly limited to systems such as molten
oxides.
This approach emphasizes the acid- and basic-anhydride aspects of acid-base
chemistry, certainly useful though often neglected. The Lux-Flood base is a basic
anhydride:
Ca 2 + + O 2- + H 2 0 -♦ Ca 2+ + 20H" (9.6)
and the Lux-Flood acid is an acid anhydride:
SiO, + H,0 -► H 2 Si0 3 (9.7)
(This latter reaction is very slow as written and is of more importance in the reverse,
dehydration reaction.) The characterization of these metal and nonmetal oxides as
acids and bases is of help in rationalizing the workings, for example, of a basic
Bessemer converter in steelmaking. The identification of these acidic and basic
species will also prove useful in developing a general definition of acid-base behavior.
An acidity scale has been proposed in which the difference in the acidity param¬
eters, (a B - a A ), of a metal oxide and a nonmetal oxide is the square root of the
enthalpy of reaction of the acid and base. 4 Thus for reaction 9.5, the enthalpy of
reaction is -86 kJ mol -1 and so the a values of CaO and Si0 2 differ by about 9 units.
Selected values are listed in Table 9.1. Although based on the Lux-Flood concept, the
values are obviously of more general interest. The most basic oxide, as expected, is
cesium oxide, amphoteric oxides have values near zero (water was chosen to calibrate
the scale at a value of 0.0), and the most acidic oxide is C1 2 0 7 , the anhydride of
perchloric acid.
The decomposition of carbonates and sulfates discussed previously (Chapter 4)
can be viewed as the tendency of metal cations to behave as oxide ion acceptors
(Lux-Flood acids), and the ordering shown in Fig. 4.21 can be related to measures of
acid-base behavior of these metals (e.g., cf. Table 9.3). The nonexistence of iron(lll)
carbonate, for example, in Fig. 4.21 indicates that the Fe 3+ ion is too strong an acid to
allow it to coexist with carbonate: or in terms of the a parameter above, the enthalpy
of reaction to form iron(III) carbonate is expected to be only about -(—1.7 - 5.5) 2 or
about -52 kJ mol -1 , not enough to overcome the TA5 term of the free energy change,
arising largely because of gaseous carbon dioxide:
Fe 2 0 3 (s) + 3C0 2 (g) -♦ Fe 2 (C0 3 ) 3 (s) (9.8)
Many solvents autoionize with the formation of a cationic and an anionic species as
does water:
2H 2 0 ,-= H 3 0 + + OH' (9.9)
2NH 3 NH 4 + + NH 2 - (9.10)
2H 2 S0 4 t—^ H 3 S0 4 + HS0 4 (9-11)
20PCI 3 OPCl 2 + + OPCi; (9.12)
For the treatment of acid-base reactions, especially neutralizations, it is often conve¬
nient to define an acid as a species that increases the concentration of the
4 Smith, D. W. J. Chem. Educ. 1987. 64. 480-481.
Acid-Base Concepts
321
Table 9.1 _
Selected acidity parameters,
a, for acidic, basic, and
amphoteric oxides"
Oxide
a
Oxide
a
h 2 o
0.0
FeO
-3.4
Li-,0
-9.2
Fe 2 0 3
-1.7
Na-,0
-12.5
CoO
-3.8
K.,0
-14.6
NiO
-2.4
Rb,0
-15.0
Cu,0
-1.0
Cs 2 0
-15.2
CuO
-2.5
BeO
-2.2
Ag-,0
-5.0
MgO
-4.5
ZnO
-3.2
CaO
-7.5
CdO
-4.4
SrO
-9.4
HgO
-3.5
BaO
-10.8
b 2 o 3
1.5
RaO
-11.5
ai 2 o 3
-2.0
Y 2 0 3
-6.5
co 2
5.5
La-,0,
-6.1
Si0 2
0.9
Lu-jOj
-3.3
n 2 o 3
6.6
Ti0 2
0.7
n 2 o 5
9.3
Zr0 2
0.1
PAo
7.5
ThO,
-3.8
As 2 0 5
5.4
V 2 0,
3.0
so.
7.1
Cr0 3
6.6
so 3
10.5
Mo0 3
5.2
Se0 2
5.2
WO,
4.7
Se0 3
9.8
MnO
-4.8
ci 2 o 7
11.5
Mn,0 7
9.6
l 2 O s
7.1
9.6
Re 2 0 7
9.0
“ Values from Smith, D. W. J. Chem.
Educ. 1987, 64. 480.
characteristic cation of the solvent, and a base as a species that increases the
concentration of the characteristic anion. The advantages of this approach are prin¬
cipally those of convenience. One may treat nonaqueous solvents by analogy with
water. For example:
K w = [H 3 0 + ][OH-] = 1(T 14 (9-13)
K ab = [A + ][B-] (9-14)
where [A + ] and [B“] are the concentrations of the cationic and anionic species
characteristic of a particular solvent. Similarly, scales analogous to the pH scale of
water may be constructed with the neutral point equal to -i log K AB , although, in
practice, little work of this type has actually been done. Some examples of data of this
type for nonaqueous solvents are listed in Table 9.2. The "leveling" effect follows
quite naturally from this viewpoint. All acids and bases stronger than the charac¬
teristic cation and anion of the solvent will be "leveled" to the latter. Acids and bases
weaker than those of the solvent system will remain in equilibrium with them. For
example:
h 2 o + HC10 4
h 3 o + + CIO;
(9.15)
406
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.6 _
Some values of A for
transition metal complexes"
Complex
Oxidation state
of metal
Symmetry
A (cm -1 )
[VCIJ 2 -
4
0„
15,400
VCI 4
4
T„
7900
[CrFJ 2 "
4
o„
22,000
[CrFJ 3 "
3
o„
15,060
[Cr(H 2 0) 6 ] 3+
3
o h
17,400
[Cr(en) 3 J 3+
3
o„
22,300
[Cr(CN)J 3 -
3
o„
26,600
[Mo(H 2 0) 6 ] 3+
3
o h
26,000
IMnFJ 2 -
4
o h
21,800
[TcF/-
4
o„
28,400
IReFJ-
4
Op
32.800
[Fe(H 2 0) h ] 3+
3
0„
14,000
[Fe(ox),] 3-
3
0„
14,140
[Fe(CN) 6 ] 3 -
3
O h
35,000
[Fe(CN) h | 4 -
2
o„
32,200
[Ru(H 2 0) 6 ] 3+
3
0„
28.600
(Ru(ox) 3 ] 3 “
3
On
28.700
[Ru(H 2 0)J 2+
2
O h
19,800
[Ru(CN)J‘-
2
O h
33.800
[CoF h ) 2 -
4
O h
20,300
[CoFJ 3 -
3
On
13,100
[Co(H,0) 6 ] 3+
3
On
20,760
[Co(NH,)J 3+
3
O h
22,870
[Co(en)j] 3+
3
On
23,160
[Co(H 2 0),J 2+
2
On
9200
[Co(NH 3 )J 2+
2
On
10,200
[Co(NH 3 )„] 2+
2
Tj
5900
(RhFJ 2 -
4
On
20.500
[Rh(H,0)J ,+
3
On
27.200
(Rh(NH 1 ) 6 J 3+
3
0„
34,100
[IrFJ 2 -
4
On
27,000
[lr(NH 3 )J 3+
3
0„
41.200
" Lever, A. B. P. Inorganic Electronic Spectroscopy. 2nd ed.;
Elsevier: New York, 1986: Chapters 6 and 9.
Abbreviations: ox = oxalate: en = ethylcnediamine.
Jytrgensen has developed a means of estimating the value of A for an octahedral
complex by treating it as the product of two independent factors: 2 -’
A „=/•,' (11.5)
The quantity / describes the field strength of a ligand relative to water, which is
assigned a value of 1.00. Values range from about 0.7 for weak field bromide ions to
Jorgensen. C. K. Modern Aspects of Ligand Field Theory. Elsevier: New York. 1971: Chapter 26.
Crystal Field Theory
407
Frequency (cm -1 )
JOO J00 400 500
kJ mol' 1
Fig. 11.13 Spectra of three
chromiinn(Ill) complexes:
(a) [Cr(en)j] 3+ ; (b)
[Crfox),] 3 -; (c) [CrFJ 3- ;
i’, corresponds to A„;
CT = charge transfer band
(see page 455). The
coordinating atoms for the
bidentate ethylenediamine
(en) and oxalate (ox)
ligands are N and O.
respectively. (Spectra a and
b from Schlafer, H. L.
Z. Phys. Chem. (Frankfurt
am Main) 1957, II. 65.
Reproduced with
permission. Spectrum c
sketched from data given
by Jprgensen, C. K.
Absorption Spectra and
Chemical Ponding in
Complexes-. Pergamon:
Elmsford, NY. 1962.]
about 1.7 for the very strong field cyanide ion. The g factor is characteristic of the
metal ion and varies from 8000 to 36.000 cm -1 (Table 11.7). Equation 11.5 is useful for
approximating A and. in combination with pairing energies (Table 11.4), for predicting
whether a new octahedral complex will be high or low spin.
Although the spectrochemical series and other trends described in this section
allow one to rationalize differences in spectra and permit some predictability, they
present serious difficulties in interpretation for crystal field theory. If the splitting of
the d orbitals resulted simply from the effect of point charges (ions or dipoles), one
should expect that anionic ligands would exert the greatest effect. To the contrary,
most anionic ligands lie at the low end of the spectrochemical series. Furthermore,
OH - lies below the neutral H-,0 molecule and NH 3 produces a greater splitting than
H,0, although the dipole moments are in the reverse order (jt NHi = 4.90 x IO -30
C m; ji H;0 = 6.17 x I0 _3 ° C m). The model is also unable to account for the fact that
with certain strong field ligands (such as CN~), A„ varies only slightly for analogous
complexes within a group. These apparent weaknesses in the theory called into
question the assumption of purely electrostatic interactions between ligands and
322
9 Acid — Base Chemistry
Table 9.2
Ion products, pH ranges,
_ j ,_r • , ,
Solvent
Ion product
pH range
Neutral point
and neutral points of some
solvents 0
h 2 so„
itr 4
0-4
2
CH 3 COOH
itr 13
0-13
6.5
H-,0
itr 14
0-14
7
C 2 H 5 OH
itr 20
0-20
10
nh 3
1 (T 29
0-29
14.5
“ Data from Jander, ].; Lafrenze, C. Ionizing Solvents', Verlag
Chemie Gmbh: Weinheim, 1970.
but
/OH O-
H,0 + CHjC H 3 0 + CH } C'
o
Similarly,
NH 3 + HC10 4 - NH 4 + + C10 4
and
nh 3 + hc 2 h 3 o 2 —♦ nh; + c 2 h 3 o 2 -
but
(9.16)
(9.17)
(9.18)
NH 3 + NH 2 CONH 2 NH 4 + + NHjCONH" (9.19)
The solvent system concept has been used extensively as a method of classifying
solvolysis reactions. For example, one can compare the hydrolysis of nonmetal
halides with their solvolysis by nonaqueous solvents:
3H 2 0 + OPCl 3 -- OP(OH) 3 + 3HC1T
3ROH + OPCI 3 - OP(OR) 3 + 3HCIf
6 NH 3 + OPCI 3 - OP(NH 2 ) 3 + 3NH 4 C1
(9.20)
(9.21)
(9.22P
Considerable use has been made of these analogies, especially with reference to
nitrogen compounds and their relation to liquid ammonia as a solvent. 6
One criticism of the solvent system concept is that it concentrates too heavily
on ionic reactions in solution and on the chemical properties of the solvent to the ne¬
glect of the physical properties. For example, reactions in phosphorus oxychloride ( =
phosphoryl chloride) have been systematized in terms of the hypothetical
autoionization:
OPCl 3 OPCl 2 +CT
or
(9.23)
20PCI 3 OPCl 2 + OPCl 4 -
Although this reaction appears to be different from the others in stoichiometry and products, the
difference lies merely in the relative basicity of H 2 0. ROH, and NH 3 and the stability of their
conjugate acids toward dissociation: BH + + Cl - B + HCI.
6 Audrieth, L. F.; Kleinberg, J. Non-aqueous Solvents; Wiley: New York, 1953.
Acid—Bose Concepts
323
Substances which increase the chloride ion concentration may be considered bases
and substances which strip chloride ion away from the solvent with the formation of
the dichlorophosphoryl ion may be considered acids:
OPCI 3 + PC1 5 OPCl 2 + + PC1 6 - (9.25)
Extensive studies of reactions between chloride ion donors (bases) and chloride ion
acceptors (acids) have been conducted by Gutmann , 7 who interpreted them in terms
of the above equilibria. An example is the reaction between tetramethylammonium
chloride and iron(lll) chloride, which may be carried out as a titration and followed
conductometrically:
(CH 3 ) 4 N + Cr + FeCl 3 — (CH 3 ) 4 N + FeCl 4 - (9.26)
which was interpreted by Gutmann in terms of
(CH 3 ) 4 N + cr ( ch 3 ) 4 n + + cr (9.27)
FeCl 3 + OPCI 3 OPCI 2 + + FeCl 4 (9.28)
OPCI 2 + + cr - OPCl 3 (9.29)
Meek and Drago 8 showed that the reaction between tetramethylammonium chloride
and iron(III) chloride can take place just as readily in triethyl phosphate, OP(OEt) 3 , as
in phosphorus oxychloride, OPCl 3 . They suggested that the similarities in physical
properties of the two solvents, principally the dielectric constant, were more impor¬
tant in this reaction than the difference in chemical properties, namely, the presence or
absence of autoionization to form chloride ions . 9
One of the chief difficulties with the solvent system concept is that in the absence
of data, one is tempted to push it further than can be justified. For example, the
reaction of thionyl halides with sulfites in liquid sulfur dioxide might be supposed to
occur as follows, assuming that autoionization occurs:
2SO z S0 2 + + SO^ - (9.30)
Accordingly, sulfite salts may be considered bases because they increase the sulfite
ion concentration. It might then be supposed that thionyl halides behave as acids
because of dissociation to form thionyl and halide ions:
SOCl 2 S0 2 + +2CP (9.31)
The reaction between cesium sulfite and thionyl chloride might now be considered to
be a neutralization reaction in which the thionyl and sulfite ions combine to form
solvent molecules:
S0 2+ +SO§" -♦ 2S0 2 (9.32)
Indeed, solutions of cesium sulfite and thionyl chloride in liquid sulfur dioxide yield
the expected products:
Cs 2 S0 3 + SOCI 2 -► 2CsCl + 2S0 2 (9.33)
7 Gutmann, V. J. Phys. Chem. 1959. 63. 378-383. Baaz. M.; Gutmann, V.; Htibner, L. J. Inorg. Nucl.
Client. 1961. 18, 276-285.
8 Meek. D. W.; Drago. R. S. J. Am. Chem. Soc. 1961.83. 4322-4325. For a complete discussion of this
point of view and critique of the solvent system approach, see Drago. R. S.; Purcell, K. F. Prog.
Inorg. Chem. 1964. 6, 271-322.
9 See Chapter 10 for further discussion of this point.
404
n - Coordination Chemistry: Bonding, Spectra, and Magnetism
Fig. 11.12 An octahedral
complex (a) undergoing z
axis elongation such that it
becomes tetragonally
distorted (b) and finally
reaches the square planar
limit (c). The a,,, Id-i)
orbital may lie below the e
(d,. and d zy ) orbitals in the
square planar complex.
spacings for square planar complexes (and those of any other noncubic symmetry)
requires additional parameters, commonly labeled Dt and Ds. 2 '
The square planar geometry is favored by metal ions having a d configuration in
the presence of a strong field. This combination gives low spin complexes with the
eight d electrons occupying the low-energy d xv d yz , ds. and d orbitals, while the
high-energy ds v = orbital remains unoccupied. The stronger the surrounding field,
the higher the d x ,. y : orbital will be raised. As long as this level ,s unoccupied,
however, the overall effect on the complex will be stabilization because the lower,
occupied orbitals will drop in energy by a corresponding amount. 1 yp.ca low spin
square planar'complexes are [Ni(CN) 4 ] 2 -, [PdCI 4 ] 2 . [Pt(NH 3 ) 4 ] , [PtCl 4 ] , and
[AuCI„T, all d 8 species.
Orbital Splittings in
Fields of Other
Symmetries
Although a large number of complexes can be accommodated under the symmetries
that have been discussed, there are also many that exist in other geometric configura¬
tions. These configurations will not be treated in detail, but their d orbital energies are
included in Table 11.5 along with energies for the more common geometries. From the
values in the table, energy level diagrams similar to those given previously for
octahedral, tetrahedral, and square planar complexes can be constructed for any of
the symmetries listed.
Factors Affectinq the There are a number of factors that affect the extent to which metal d orbitals are split
Factors Attectmg surrounding ligan ds. Representative values of A for a variety of complexes are
Magnitude of A jn Jable ,, 6 From [he data seV eral important variables and trends can be
identified. ...
Oxidation slate of the metal ion. The magnitude of A increases with increasing
ionic charge on the central metal ion. Several complexes in Table 116 involving
different oxidation states for a particular metal ion with the same ligand illustrate this
trend. Note, for example, [Ru(H 2 0) 5 ] 2+ <A„ = 19,800 cm" 1 ) and [Ru(H,0) 6 ]‘ (A„ - -
28,600 cm -1 ). . .
Nature of the metal ion. Significant differences in A also occur for analogous
complexes within a given group, the trend being 3 d <4d<5d.\o progressing from Cr
2' Schlafer, H. L.; Gliemann, G. Basic Principles of Ligand Field Theory. Wiley-Interscience: New
York. 1969; pp 63-65 and 343-347.
Crystal Field Theory 405
Table 13.S _
The energy levels of d
orbitals in crystal fields of
different symmetries 0 ' 1 ’
C.N.
Structure
ds
ds-S
d, y
d„
dr
1
Linear
5.14
-3.14
-3.14
0.57
0.57
2
Linear'
10.28
-6.28
-6.28
1.14
1.14
3
Trigonal' 1
-3.21
5.46
5.46
-3.86
-3.86
4
Tetrahedral
-2.67
-2.67
1.78
1.78
1.78
4
Square planar 11
-4.28
12.28
2.28
-5.14
-5.14
5
Trigonal bipyramidal' -
7.07
-0.82
-0.82
-2.72
-2.72
5
Square pyramidaL
0.86
9.14
-0.86
-4.57
-4.57
6
Octahedral
6.00
6.00
-4.00
-4.00
-4.00
6
Trigonal prismatic
0.96
-5.84
-5.84
5.36
5.36
7
Pentagonal bipyramidal
4.93
2.82
2.82
-5.28
-5.28
8
Cubic
-5.34
-5.34
3.56
3.56
3.56
8
Square antiprismatic
-5.34
-0.89
-0.89
3.56
3.56
9
[ReH 9 ] 2- structure (see Fig. 12.40)
-2.25
-0.38
-0.38
1.51
1.51
12
Icosahedral
0.00
0.00
0.00
0.00
0.00
“ Zuckerman. J. J. J. Client. Educ. 1965, 42. 315. Krishnamurthy. R.; Schaap, W. B. J. Client.
Educ. 1969. 46. 799. Used wilh permission.
b All energies are in Dq units; 10 Dq = A„.
c Ligands lie along i axis.
d Ligands lie in xy plane.
e Pyramid base in xy plane.
to Mo or Co to Rh, the value of A„ increases by as much as 50%. Likewise, the values
for Ir complexes are some 25% greater than for Rh complexes. An important result of
this trend is that complexes of the second and third transition series have a much
greater tendency to be low spin than do complexes of the first transition series.
Number and geometry of the ligands. As we have seen, the point-charge model
predicts that A for a tetrahedral complex will be only about 50% as large as for an
octahedral complex, all other factors being equal. This approximate relationship is
observed for VC1 4 and [VCI 6 ] 2- , as well as for [Co(NH 3 ) 4 ] 2+ and [Co(NH 3 ) ft ] 2 %
Nature of the ligands. The effect of different ligands on the degree of splitting is
illustrated in Fig. 11.13, which shows the absorption spectra of three CrL ( , complexes.
Three d-d transitions are predicted for each one, based on an analysis that includes
the effects of electron-electron repulsion (see pages 433-437). The transitions labeled
v, correspond to A„ or \0Dq. Note that there is a steady increase in the frequency of
this absorption as the ligating atom changes from F to O to N, corresponding to a
progressive increase in the field strength. Based on similar data for a wide variety of
complexes, it is possible to list ligands in order of increasing field strength in a
spectrochemical series. Although it is not possible to form a complete series of all
ligands with a single metal ion, it is possible to construct one from overlapping
sequences, each constituting a portion of the series: 22
I - < Br~ < S 2- < SCN - < Cl" < Nj, F - < urea, OH - < ox,
O 2- < H 2 0 < NCS - < py, NH 3 < en < bpy, phen < NOJ < CHj, '
c„h 5 -< CN - < CO
22 Abbreviations are ox = oxalate, py = pyridine, cn = cthylcnediamine, bpy = 2,2'-bipyridine,
phen = 1,10-phenanthroline. For SCN” and NCS - , the bound atom is given first. Lever. A. B. P.
Inorganic Electronic Spectroscopy. 2nd ed.; Elsevier: New York, 1986. Jorgensen, C. K. Modern
Aspects of Ligand Field Theory, Elsevier: New York, 1971; Chapter 26.
324
9-Acid-Base Chemistry
Furthermore, the amphoteric behavior of the aluminum ion can be shown in sulfur
dioxide as readily as in water. Just as AI(OH), is insoluble in water but dissolves
readily in either a strong acid or basic solution. A1 2 (S0 3 ) 3 is insoluble in liquid sulfur
dioxide. Addition of either base (SO$~) or acid (SO 2 ) causes the aluminum sulfite to
dissolve, and it may be reprecipitated upon neutralization.
The application of the solvent system concept to liquid sulfur dioxide chemistry
stimulated the elucidation of reactions such as those of aluminum sulfite. However,
there is no direct evidence at all for the formation of SO 2 in solutions of thionyl
halides. In fact, there is evidence to the contrary. When solutions of thionyl bromide
or thionyl chloride are prepared in «S-labeled (S) sulfur dioxide, almost no exchange
takes place. The half-life for the exchange is about two years or more. If ionization
took place:
2S 0, S0 2 + S0 3 _ (9.34)
SOCl 2 S0 2+ +2C1" (9.35)
one would expect rapid scrambling of the tagged and untagged sulfur in the two
compounds. The lack of such a rapid exchange indicates that either Eq. 9.34 or 9.35
(or both) is incorrect.
The fact that labeled thionyl bromide exchanges with thionyl chloride indicates
that perhaps the ionization shown in Eq. 9.35 actually occurs as 10
soci 2 socr + cr ( 9 - 36 >
In a solvent with a permittivity as low as sulfur dioxide (e = I5.6e 0 at 0 °C) the
formation of highly charged ions such as SO 2 is energetically unfavorable.
When the ionic species formed in solution are known, the solvent system ap¬
proach can be useful. In solvents that are not conducive to ion formation and for
which little or nothing is known of the nature or even the existence of ions, one must
be cautious. Our familiarity with aqueous solutions of high permittivity (e H , 0 =
81.7e 0 ) characterized by ionic reactions tends to prejudice us toward parallels in other
solvents and thus tempts us to overextend the solvent system concept.
Lewis Definition In 1923 G. N. Lewis" proposed a definition of acid-base behavior in terms of
electron-pair donation and acceptance. The Lewis definition is perhaps the most
widely used of all because of its simplicity and wide applicability, especially in the
field of organic reactions. Lewis defined a base as an electron-pair donor and an acid
as an electron-pair acceptor. In addition to all of the reactions discussed above, the
Lewis definition includes reactions in which no ions are formed and no hydrogen ions
or other ions are transferred: 12
R 3 N + BF 3 - R 3 NBF 3 (9-37)
4CO + Ni -» Ni(CO) 4 (9-38)
'« Norris, T. H. J. Phys. Chem. 1959, 63. 383.
11 Lewis, G. N. Valence and the Structure of Atoms and Molecules ; Chemical Catalogue: New York.
1923. See also Luder. W, F.; Zuffanti. S. The Electronic Theory of Acids and Bases. 2nd rev. ed.;
Dover: New York. 1961. Drago. R. S.; Matwiyoff. N. A. Acids and Bases-. Heath: Lexington, MA,
1968.
12 L = electron pair donating ligand such as acetone, various amines, or halide ion.
Acid-Base Concepts
325
2L + SnCi 4 -♦ SnCI 4 L 2 (9.39)
2NH 3 + Ag - Ag(NH 3 ) 2 (9.40)
The Lewis definition thus encompasses all reactions entailing hydrogen ion. oxide ion,
or solvent interactions, as well as the formation of acid-base adducts such as R,NBF 3
and all coordination compounds. Usage of the Lewis concept is extensive in both
inorganic and organic chemistry, and so no further examples will be given here, but
many will be encountered throughout the remainder of Ihe book."
Usanovich Definition The Usanovich definition" of acids and bases has not been widely used, probably
because of (1) the relative inaccessibility of the original to non-Russian-reading chem¬
ists and (2) the awkwardness and circularity of Usanovieh's original definition. The
Usanovich definition includes all reactions of Lewis acids and bases and extends the
latter concept by removing the restriction that the donation or acceptance of electrons
be as shared pairs. The complete definition is as follows: An acid is any chemical
species which reads with bases, gives up cations, or accepts anions or electrons, and.
conversely, a base is any chemical species which reacts with acids, gives up anions or
electrons, or combines with cations. Although perhaps unnecessarily complicated,
this definition simply includes all Lewis acid-base reactions plus redox reactions,
which may consist of complete transfer of one or more electrons. Usanovich also
stressed unsaturation involved in certain acid-base reactions:
OH" + 0=0=0 - HOCOJ (9- 4 »)
Unfortunately the Usanovich definition of acids and bases is often dismissed
casually with the statement that it includes "almost all of chemistry and the term
‘acid-base reaction - is no longer necessary; the term 'reaction' is sufficient." If some
chemical reactions were called acid-base reactions simply to distinguish them from
other, non-acid-base reactions, this might be a valid criticism. However, most work¬
ers who like to talk in terms of one or more acid-base definitions do so because of
the great systematizing power which they provide. As an example, Pearson has
shown that the inclusion of many species, even organic compounds not normally
considered acidic or basic, in his principle of hard and soft acids and bases helps the
understanding of the nature of chemical reactions (pages 344-355). It is unfortunate
that a good deai of faddism and provincialism has been shown by chemists in this area.
As each new concept came along, it was opposed by those who felt ill at ease with the
new definitions. For example, when the solvent system was first proposed, some
chemists refused to call the species involved acids and bases, but insisted that they
were "acid analogues" and "base analogues"! This is semantics, not chemistry. A
similar controversy took place when the Lewis definition became widely used and
later when the Usanovich concept was popularized. Because the latter included redox
reactions, the criticism that it included too much was especially vehement. That the
dividing line between electron-pair donation-acceptance (Lewis definition) and oxida¬
tion-reduction (Usanovich definition) is not a sharp one may be seen from the
following example. The compound C 5 H 5 NO, pyridine oxide, can be formed by the
13 A very useful book discussing many aspects of acid-base chemistry in terms of the Lewis definition
is Jensen. W. B. The Lewis Acid-Base Concepts: An Overview. Wiley: New York. 1980.
M Usanovich. M. Zhur. Obschei Khim. 1939. 9. 182. Finston. H. L.; Rychtman, A. C. A New View of
Current Acid-Base Theories'. Wiley: New York. 1982.
402
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Table 11.4 _
Pairing energies lor some
3 d metal ions°
Ion
^coul
Pc
Pr
d
Cr 2+
Mn 3+
71.2 (5950)
87.9 (7350)
173.1 (14,475)
213.7 (17,865)
244.3 (20,425)
301.6 (25,215)
d 5
Cr-
Mn !t
Fe 3+
67.3 (5625)
91.0 (7610)
120.2 (10,050)
144.3 (12,062)
194.0 (16,215)
237.1 (19,825)
211.6 (17,687)
285.0 (23,825)
357.4 (29,875)
d b
Mn +
Fe 2
Co 3+
73.5 (6145)
89.2 (7460)
113.0 (9450)
100.6 (8418)
139.8 (11,690)
169.6 (14,175)
174.2 (14,563)
229.1 (19,150)
282.6 (23,625)
d 1
Fe +
Co 2+
87.9 (7350)
100 (8400)
123.6 (10,330)
150 (12,400)
211.5 (17,680)
250 (20,800)
" Pairing energies in kJ mol -1 (and cm -1 ) calculated from formulas and
data given by Orgel, L. E. J. Cliem. Phys. 1955, 23, 1819, and Griffith. J.
S. J. Inorg. Nucl. Client. 1956, 2, I, 229. The values pertain to the free
ion and may be expected to be from 15% to 30% smaller for the
complexed ion as a result of the nephelauxetic effect. Pcoui. Pc.- and P T
refer to the coulombic, exchange, and total energy opposing pairing of
electrons.
however, the ligands do not directly approach any of the metal d orbitals, but they
come closer to the orbitals directed to the edges of the cube (the d iv , d x .. and d y .) than
to those directed to the centers of the cube faces (the d r :_ y z and the d.z). Hence the t ■,
orbitals are raised in energy while the e K orbitals are stabilized relative to the bary-
center. Furthermore, since the center-of-gravity rule holds, the upper levels arc raised
by 0.4A and the lower ones stabilized by 0.6A from the barycenter, giving an energy
level scheme that is exactly the inverse of that for octahedral symmetry.
If four ligands are removed from alternate corners of the cube in Fig. 11.10. the
remaining ligands form a tetrahedron about the metal. The energy level scheme for
tetrahedral symmetry (Fig. II.II) is qualitatively the same as that for cubic, but the
splitting (A,) is only half as large because there are half as many ligands. The labels we
apply to the two sets of degenerate orbitals are consistent with their symmetry
Fig. 11.10 Complete set of
d orbitals in a cubic field.
Either set of tetrahedral
ligands (0 or O) produces a
field one-half as strong as
the cubic field.
Crystal Field Theory 403
properties in a tetrahedral environment (see the character table for the 7 (/ point group
in Appendix D): /, for the d xy , d x: , and d y . orbitals, and e for the d x z_ y z and d.z orbitals.
The g subscripts which were used for the octahedral and cubic fields are'no longer
appropriate because the tetrahedron lacks a center of inversion.
The crystal field splitting in a tetrahedral field is intrinsically smaller than that in
an octahedral field because there are only two-thirds as many ligands and they have a
less direct effect on the d orbitals. The point-charge model predicts that for the same
metal ion, ligands, and metal-ligand distances. A, = j|A„. As a result, orbital splitting
energies in tetrahedral complexes generally are not large enough to force electrons to
pair, and low spin configurations are rarely observed.™ Rather, under conditions
favoring strong crystal fields, other geometries are preferred over tetrahedral
structures.
Tetragonal If two trans ligands in an octahedral ML h complex (for example those along the z axis)
Symmetry: Square are moved either towards or away from the metal ion, the resulting complex is said to
Planar Complexes b e tetragonally distorted. Ordinarily such distortions arc not favored since they result
in a net loss of bonding energy. In certain situations, however, such a distortion is
favored because of a Jahn-Teller effect (page 449). A complex of general formula
trims- MA : B 4 also will have tetragonal (Z> 4/l ) symmetry. For now. we will merely
consider the limiting case of tetragonal elongation, a square planar ML., complex, for
the purpose of deriving its d orbital splitting pattern.
Figure 11.12 illustrates the effect of z-axis stretching on the e and /- orbitals in
an octahedral complex. Orbitals having a z component (the d.z. d x „ and d y .) will
experience a decrease in electrostatic repulsions from the ligands and will therefore be
stabilized. At the same time, the "non-z" orbitals will be raised in energy, with the
barycenter remaining constant. The overall result is that the e K level is split into two
levels, an upper b, K (d x z_ y z) and a lower a tK (d.z) and the t 2u set is split into a , (d )
and a doubly degenerate e K (d x . d yz ). Assignment of these symmetry labels can be
confirmed by referring to the D ih character table in Appendix D. The energy spacing
between the b 2u (d xy ) and />, A . (d x z_ y z) levels is defined as A. As in the octahedral case,
this splitting is equal to 10 Dq. However, the full crystal field description of the orbital
I he first example of a low spin tetrahedral complex of a first-row transition metal, letrakisd-
norbornyl)cobalt(IV). was recently confirmed. See Byrne, E. K.: Richeson. D. S.;Thcopold. K. II.
Cliem. Commun. 1986. 1491-1492.
326
9-Acid—Base Chemistry
A Generalized
Acid-Base Concept
oxidation of pyridine. Now this may be considered to be a Lewis adduct of pyridine
and atomic oxygen:
Yet no one would deny that this is a redox reaction, even though no electron transfer
has occurred between ionic species.
An example of the different points of view and different tastes in the matter of
acid-base definitions was provided to one of the authors in graduate school while
attending lectures on acid-base chemistry from two professors. One felt that the
solvent system was very useful, but that the Lewis concept went too far because it
included coordination chemistry. The second used Lewis concepts in all of his work,
but felt uncomfortable with the Usanovich definition because it included redox chem¬
istry! To the latter’s credit, however, he realized that the separation was an artificial
one, and he suggested the pyridine oxide example given above.
In the presence of such a plethora of definitions, one can well ask which is the
“best” one. Each concept, properly used, has its strong points and its weaknesses.
One can do no better than to quote the concluding remarks of one of the best
discussions of acid-base concepts. 15 “Actually each approach is correct as far as it
goes, and knowledge of the fundamentals of all is essential.”
One justification for discussing a large number of acid-base definitions, including a
few that are little used today, is to illustrate their fundamental similarity.All define
the acid in terms of donating a positive species (a hydrogen ion or solvent cation) or
accepting a negative species (an oxide ion. a pair of electrons, etc.). A base is defined
as donating a negative species (a pair of electrons, an oxide ion. a solvent anion) or
accepting a positive species (hydrogen ion). We can generalize all these definitions by
defining acidity as a positive character of a chemical species which is decreased hy
reaction with a base ; similarly basicity is a negative character of a chemical species
which is decreased by reaction with an acid. The advantages of such a generalization
are twofold: (I) It incorporates the information content of the various other acid-base
definitions; (2) it provides a useful criterion for correlating acid-base strength with
electron density and molecular structure. Some examples may be useful in illustrating
this approach. It should be kept in mind that acid-base concepts do not explain the
observed properties; these lie in the principles of structure and bonding. Acid-base
concepts help correlate empirical observations.
I. Basicity of metal oxides. In a given periodic group, basicity of oxides tends to
increase as one progresses down the periodic chart (see Table 9.1). For
example, in group 11A (2) BeO is amphoteric, but the heavier oxides (MgO,
Moeller, T. Inorganic Chemistry-. Wiley: New York. 1952; p 330. Sec also Moeller. T. Inorganic
Chemistry. A Modern Introduction-. Wiley: New York. 1982: pp 585-603. Two other excellent
discussions giving different insights are: Douglas. B. E.: McDaniel. D. H.; Alexander. J. J.
Concepts and Models of Inorganic Chemistry. 2nd ed.; Wiley: New York. 1983; pp 511-553;
Porterfield, W. W. Inorganic Chemistry: A Unified Approach-. Addison-Wesley: Reading, MA
1984; pp 292-324.
16 J. E. Inorganic Chemistry: Principles of Structure and Reactivity-. Harper & Row: New
York. 1972; pp 213-216. See also Finston. H. L.: Rychtman. A. C. A New View of Current
Acid-Base Theories-. Wiley: New York. 1982: Chapter 5.
Acid-Base Concept:
327
CaO. SrO. BaO) are basic. In this case the charge on the metal ion is the same
in each species, but in the Be 2 ion it is packed into a much smaller volume,
hence its effect is more pronounced. As a result. BeO is more acidic and less
basic than the oxides of the larger metals. In this case, “positiveness” is a
matter of the size and charge of the cation. This is closely related, of course, to
the Fajans polarizing ability (Chapter 4).
Acidity of nonmetal oxides. With increasing covalency oxides become less
basic and more acidic. Nonmetal oxides are acid anhydrides. This effect is seen
in several metal and nonmetal oxides (see Table 9.1). It can be shown that
these acidities and basicities are directly related to the electronegativities of the
metals and nonmetals involved. 17
Hydration and hydrolysis reactions. We have seen (Chapters 4 and 8) that
large charge-to-size ratios for cations result in an increase in hydration ener¬
gies. Closely related to hydration and, in fact, inseparable from it except in
degree is the phenomenon of hydrolysis. In general, we speak of hydration if
no reaction beyond simple coordination of water molecules to the cation
occurs:
Na + + nH 2 0 -♦ [Na(H 2 0)„] + (9.43)
In the case of hydrolysis reactions, the acidity (charge-to-size ratio) of the
cation is so great as to cause rupture of H—0 bonds with ionization of the
hydrate to yield hydronium ions:
Al' + + 6H,0 -♦ [AI(H,0)J 3+ —» H,0 + + [AI(H 2 0) 5 0H] 2+ (9.44)
Cations that hydrolyze extensively are those which are either small (e.g..
Be") or are highly charged (e.g., Fe , + , Sn 4+ ) or both, and have a high charge-
to-size density. Values of pK h (negative log of the hydrolysis constant) are
compared to the (charge)-/(size) ratio 1 in Table 9.3. The correlation is good
for the main group elements and La 3+ but less so for the transition metals,
especially the heavier ones. The reason for the apparently anomalous behavior
of metal ions such as Hg _ + , Sn 2 '. and Pb 2+ is not completely clear, but it may
be related to their "softness" (see page 345).
The concept of hydrolysis may also be extended to the closely related
phenomenon of the reaction of nonmetal halides with water:
PCI 3 + 6H 2 0 -► H 3 P0 3 + 3H 3 0 + + 3C1" (9.45)
In this case the water attacks and hydrolyzes not a cation but a small, highly
charged center (the trivalent phosphorus atom) resulting from the inductive
effect of the chlorine atoms.
Acidity of oxyacids. The strength of an oxyacid is dependent upon several
factors that relate to the inductive effect of the central atom on the hydroxyl
group: (a) The inherent electronegativity of the centra! atom. Perchloric acid.
HCI0 4 , and nitric acid, HNO,, are among the strongest acids known; sulfuric
acid, H 2 S0 4 , is only slightly weaker. In contrast, phosphoric acid. H 3 P0 4 , and
carbonic acid, H 2 C0 3 , are considerably weaker and boric acid, H 3 B0 3 , is
extremely weak, (b) The inductive effect of substituents. Although acetic acid,
CH 3 COOH, is rather weak, successive substitution of chlorine atoms on the
17 Bratsch, S. G. J. Chem. Educ. 1988. 65. 877-878.
18 Z~/r has been used here, but any of the Z n !r m functions would give similar results. See Chapter 4.
400
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
This quantity is termed the crystal field stabilization enemy (CFSE For d and d
metal ions, we would expect the electrons to obey Hund s rule and thus to occupy
different degenerate t 2 . orbitals and to remain unpaired. The resulting configurat.ons,
,2 and ,3 wm have CSFEs of 0.8A„ and I -2A 0 , respectively. When one more electron
is added'to form the d 4 case, two possibilities arise: either the eiectron may enter th
higher energy level or it may pair with another electron in one of the orbitals
The actual configuration adopted will, of course, be the lowest energy one and w 1
depend on the relative magnitudes of A„ and P. the energy necessary to caie^^
pairing in a single orbital. For A„ < P (the weak field or high spin cond, , t,0n >- J
electron will enter one of the e, orbitals rather than ‘pay the ■>"«L°‘
one in a t 2y orbital. The configuration will be ^e 1 ,. and the net CFSE will be
CFSE = (3 X + 0.4A,,) - (I x +0.6A,,) = 0.6A„ ( n - 4 )
The addition of a fifth electron to a weak field complex gives a configuration 4 e J and
a CFSE of zero. The two electrons in the unfavorable e g level exactly balance the
stabilization associated with three in the t 2g level (Fig. 11.9a). p .
If the splitting of the d orbitals is large with respect to the pairing energy (A„ > P),
it is more favorable for electrons to pair in the t 2g level than to enter the strong y
unfavorable <?„ level. In these strong field or low spin complexes. the ^'^ elr ^" S
unoccupied for d' through ions (Fig. 11.9b). As a result, the crystal A^ld stab l.za-
tion energies of complexes having four to seven d electrons will be greater for sirang
field than for weak field cases. For d\ for example, the low spin configuration wi I be
/•' giving a CFSE of l.6A„, compared to 0.6A o for the high spin arrangement,
summary of configurations, crystal field stabilization energies, and numbers of un¬
paired electrons for d' through d'° in both strong and weak field situations is given in
1 ab A comparison of total energies for strong and weak field cases, including electron¬
pairing energies (/>). may be computed: The CFSE for a low spin
be 2 4A - 3 P The corresponding high spin configuration would have a CFSE ot
0. 4 a', -P for a difference between the two of 2.0A„ - IP. Since the two configura¬
te < b >
Fig. 11.9 Electron configurations of (a) a d 5 ion in a weak octahedral field and (b) a d 6 ion in
a strong octahedral field.
Crystal Field Theory
401
Table 11.3 _ _
Crystal field effects for
weak and strong octahedral
fields 0
d n
Weak field
Strong field
Configuration
Unpaired
electrons
CFSE
Configuration
Unpaired
electrons
CFSE
d'
4
I
0.4A„
! 2k
1
0.4A„
d 2
4
2
0.8A„
2
0.8A„
d>
4
3
1.2A C
4
3
l.2A„
d 4
'IA
4
0.6A„
4
2
l.6A„
d 5
AA
5
0.0A„
4
1
2.0A„
d 6
AA
4
0.4A„
4
0
2.4A„
d 1
,5 p 2
3
0.8A„
4i
1
1.8A„
d 8
,<< p2
2
1.2A„
6 p 2
2
1.2A„
d #
1
0.6A„
$A
1
0.6A„
d'°
AA
0
0.0A„
tA
0
0.0A„
" This table is somewhat simplified because pairing energies and electron-electron effects have
been neglected.
tions differ in the spins of two electrons, this amounts to an energy factor of (1.0A„ -
P) per electron spin.
The electron-pairing energy is composed of two terms. One is the inherent
coulombic repulsion that must be overcome when forcing two electrons to occupy the
same orbital. A gradual decrease in the magnitude of this contribution is observed as
one proceeds from top to bottom within a given group of the periodic table. The larger,
more diffuse 5 d orbitals in the heavier transition metals more readily accommodate
two negative charges than the smaller 3 d orbitals. The second factor of importance is
the loss of exchange energy (customarily taken as the basis of Hund’s rule. Chapter 2)
that occurs as electrons with parallel spins are forced to have antiparallel spins. The
exchange energy for a given configuration is proportional to the number of pairs 19 of
electrons having parallel spins. Within a d subshell the greatest loss of exchange
energy is expected when the d 5 configuration is forced to pair. Hence r/ 5 complexes
(e.g.. Mn : and Fe J+ ) frequently are high spin. Some typical values of pairing energies
for free gaseous transition metal ions are listed in Table 11,4.
Tetrahedral The two most common geometries for four-coordinate complexes are the tetrahedral
Symmetry and square planar arrangements. The square planar geometry, discussed in the next
section, is a special case of the more general D 4h symmetry, which also includes
tetragonal distortion of octahedral complexes (page 448). Tetrahedral coordination is
closely related to cubic coordination. Although the latter is not common in coordina¬
tion chemistry, it provides a convenient starting point for deriving the crystal field
splitting pattern for a tetrahedral ML 4 complex.
Consider eight ligands aligned on the corners of a cube approaching a metal atom
located in the center as shown in Fig. 11.10. A complex such as this would belong tc^
the same point group (O h ) as an octahedral one so the d orbitals will be split into two
degenerate sets, t 2l! and e g , as for the octahedral case. In the cubic arrangement.
19 The word "pair” as used here simply means a set of two electrons, not two electrons with their
spins paired.
328
9-Acid-Base Chemistry
» Values of pK h from Yatsimirksii. K. B.; VasiFev, V. P. Instability Constants of Complex
Compounds-. Pergamon; Elmsford. NY. I960, except for Bi, Hf. Lu. Pu. Sc. and Tl. which are
from Stability Constants of Metal-Ion Complexes: Part II. Inorgan,c Ltgands-. Bjerntm. J.;
Schwarzcnbach. G.: Sillen. L. G.. Eds.; The Chemical Society; London. 1958. For many
elements there is considerable uncertainty in the hydrolysis constants not only as a result of
experimental errors but also because some have not been corrected to infin.te dilution. Z /r
values were calculated from ionic radii in Table 4.4.
C : m"‘ x I0 2H
Increasing tendency to hydrolyze because of charge-size function
Acid-Base Concepts
329
methyl group increases the dissociation of the proton until trichloroacetic acid
is considerably stronger than phosphoric acid, for example.
More important for inorganic oxyacids is the number of oxygen atoms
surrounding the central atom. Thus in the series of chlorine oxyacids, acid
strength increases in the order HOCI < HOCIO < H0C10 2 < HOCIO v I he
trends in acidity of oxyacids. and even reasonably accurate predictions of their
p/C, values, can be obtained from; 19
p/C, = 10.5 - 5.0/? - x.x < 9 - 46)
for acids of the formula X(0H)„,0„, and where x, is the electronegativity of the
central atom. Both effects (a) and (b) are included in Eq. 9.46.
Basicity of substituted amines. In water, ammonia is a weak base, but nitro¬
gen trifiuoride shows no basicity whatsoever. In the NH-, molecule, the nitro¬
gen atom is partially charged negatively from the inductive effects of the
hydrogen atoms, but the reverse is true in the NF 3 molecule. Replacement of a
hydrogen atom in the ammonia molecule with an electron-withdrawing group
such as —OH or —NH 2 also results in decreased basicity. Because alkyl
groups are normally electron donating (more so than hydrogen) toward elec¬
tronegative elements, we might expect that replacement of a hydrogen atom by
a methyl group would increase the basicity of the nitrogen atom. This effect is
readily seen in the familiar equilibrium constants for weak bases in water
(Table 9.4).
As expected, substitution of an alkyl group for a hydrogen atom in the
ammonia molecule results in increased electron density on the nitrogen atom
and increased basicity. Substitution of a second alkyl group also increases the
basicity, although less than might have been expected from the previous
substitutional effect. The trialkylamines do not continue this trend and surpris¬
ingly are as weak as or weaker than the monoalkylamines. Although the
explanation of this apparent anomaly is fairly simple, it does not depend upon
electron density and so will be postponed to the next section.
"Ultimate acids and bases." Familiarity with the idea that acidity and
basicity are related to electron density at reacting sites and charge-to-size ratio
Table 9.4
Basicity ot ammonia and amines
Pb
NH 3 = 4.74
Electron-withdrawing substitution
Less basic
Electron-donating substitution
More basic
NH,OH = 7.97 NH 2 NH 2 = 5.77
MeNH 2 = 3.36 Me 2 NH = 3.29 Me 3 N = 4.28
EtNH 2 = 3.25 Et 2 NH = 2.90 Et 3 N = 3.25
/-PrNH, = 3.28 ?-Pr 2 NH = 2.95
«-BuNH 2 = 3.51 i-Bu 2 NH = 3.32 f-BUjN = 3.58
19 The first such relationships were by Ricci. J. E. J. Am. Chem. Soc.. 1948. 70, 109-113; Pauling, L.
General Chemistry. 3rd ed.; Freeman; San Francisco, 1970; pp 499-502. The present formulation is
by Somasekharan. K. N.; Kalpagam. V. Chem. Educ. 1986. 2(4). 43-46.
398
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
three. Division into these two groups arises from the symmetry properties of
the d orbitals within an octahedral environment, which we can confirm by referring to
the character table for the O h point group (Appendix D). We see that the d and d r z_ v z
orbitals transform as the E K representation and the d xv , d xv and d yz orbitals transform
as the T 2f! representation. Our earlier conclusion that the orbitals are split into two
sets, one doubly degenerate and the other triply degenerate, is thus substantiated. The
labels customarily given to these two groups also denote these symmetry properties:
for the triply degenerate set and e K for the doubly degenerate pair, the lower case
being appropriate for orbitals. Of course a symmetry analysis does not tell us which of
the two sets of orbitals is higher in energy. To determine relative orbital energies, the
nature of the interactions giving rise to the splitting must be considered.
The extent to which the e K and t 2y orbitals are separated in an octahedral complex
is denoted by A„ (the a subscript signifying octahedral) or 10 Dq (Fig. 11.7). 15 More
insight into the nature of this splitting can be obtained by viewing formation of a
complex as a two-step process. In the first step, the ligands approach the central
metal, producing a hypothetical spherical field which repels all of the d orbitals to the
same extent. In the second, the ligands exert an octahedral field, which splits the
orbital degeneracy. In going from the first to the second step, we find that the bary-
center, or “center of gravity," of the orbitals remains constant. That is to say, the
energy of all of the orbitals will be raised by the repulsion of the advancing ligands in
step one, but merely rearranging the ligands from a hypothetical spherical field to an
octahedral field does not alter the average energy of the five d orbitals. To maintain
the constant barycenter, it is necessary for the two e g orbitals to be further repelled
by 0.6A„ while the three t 2v orbitals are stabilized to an extent of 0.4A„ as shown in
Fig. 11.7. This constancy of the barycenter of the d orbitals holds for all complexes,
regardless of geometry. 16
To gain some appreciation for the magnitude of A„ and how it may be measured,
let us consider the d' complex. [Ti(H-,0)J 3 + . This ion exists in aqueous solutions of
Ti 3+ and gives rise to a purple color. The single d electron in the complex will occupy
Fig. 11.7 Splitting of the
five d orbitals by an
octahedral field. The
condition represented by
the degenerate levels on the
left is a hypothetical
spherical field.
The terms D and </ are quantities inherent in the formal mathematical derivation of the electrostatic
model. They depend on the charge on the metal ion. the radial distribution of the valence d
electrons, and the metal-ligand distance. The factor of 10 in 10 Dq arises specifically for a single
electron in an electrostatic potential of octahedral geometry.
"• Because the electrons of all of the d orbitals arc repelled, it has been suggested that this be
represented on orbital splitting diagrams (such as Fig. 11.7) by tic lines showing the increased orbital
energy upon approach of the ligands. On the other hand, attraction between the metal ion and the
ligand ions or dipoles will result in an overall lowering of energy (as it must if the complex is to be
stable) and it seems unnecessary to include all of these effects in simple splitting diagrams as long as
one is aware of their existence.
Crystal Field Theory 399
the lowest energy orbital available to it. i.e., one of the three degenera
The purple color is the result of absorption of light and promotion of the
the e g level. The transition can be represented as
The absorption spectrum of [Ti(H-,0) 6 ] 3+ (Fig. 11.8) reveals that this transition
with a maximum at 20,300 cm" 1 , which corresponds to 243 kJ mol" 1 of ene
A„- 17 By comparison, the absorption maximum 18 of ReF 6 (also a d' species) is 32,500
cm ', or 338 kJ mol" 1 . These are typical values for A„ and are of the same order of
magnitude as the energy of a chemical bond.
The d' case is the simplest possible because the observed spectral transition
reflects the actual energy difference between the e g and t 2/ . levels. For the more
general d" situation, electron-electron interactions must be taken into account and the
calculation becomes somewhat more involved. The appropriate methods are dis¬
cussed on pages 433 and following.
In the d 1 case discussed above, the electron occupies a l 2f! orbital, which has an energy
of —0.4A„ relative to the barycenter of the e/ orbitals. The complex can thus be said to
be stabilized to the extent of 0.4A„ compared to the hypothetical spherical-field case.
Crystal Field
Stabilization Energy
Wavelength (nm)
-0.5 I-
10.000
30,000
Frequency (cm" 1 )
Fig. 11.8 Electronic spectrum of a 0.1 M aqueous solution of [Ti(H : 0),,| 3+ . The
top indicate the colors associated with portions of the visible spectrum. (From H;
Schliifcr. H. L.; Hansen. K. H. Z. Anorg. Allg. Client. 1956, 284, 153-161. Reprc
permission.]
It is suggested that the reader verify this equivalence by means of appropriate conversion factors.
Moflitt, W.; Goodman. G. L.; Fred. M.: Weinstock. B. Mol. Phy.x. 1959, 2, 109-122.
330
9-Acid-Base Chemistry
might lead one to ask if there exists a single strongest acid or base species. A
little reflection would suggest the bare proton as having the highest positive
charge-to-size ratio. Of course the proton never occurs uncoordinated or
unsolvated in chemical systems, but attaches itself to any chemical species
containing electrons. It is too strong an acid to coexist with any base without
reacting. Even a noble gas atom, not normally considered to be a base, will
combine with the exceedingly acidic proton (see Table 9.5). The choice of the
proton as the “characteristic” exchanged species in the Br0nsted-Lowry
concept was not fortuitous.
Concerning the "ultimate" base, one might choose various small, highly
charged ions such as H , F , or O - , all of which are indeed quite basic.
However, the electron appears to be the complement of the proton. It might be
objected that the isolated electron has even less justification as a chemical
entity than the proton, but solutions (and even solids) are known in which
electrons are the anionic species! And interestingly, solutions containing
"free” electrons are very basic. This topic will be discussed in further detail in
the section on liquid ammonia chemistry in the next chapter.
Measures of Historically, acid-base chemistry has been strongly tied to solution chemistry, not
Acid-Base Strenqth ° nly ' n WalCr bUt nonac l ueous solvents as well (Chapter 10). Chemists knew that
-9- there were strong solvation effects that might be altering or obscuring inherent
acid-base properties, and they tried various means of estimating these effects or
eliminating them through the use of nonpolar solvents. Nevertheless, for many years
the solution thermodynamics of acid-base behavior was only poorly understood. In
the past 10-15 years a remarkable amount of data on solutionless, that is. gas-phase.
acid-base chemistry has been collected. Since it is easiest to see inherent acid-base
effects in the absence of competing solvent effects, the following discussion will
proceed: gas-phase —» nonpolar solvents -> polar solvents, though this treats the
subject in reverse chronological order.
Gas-phase
Basicities:
Proton Affinities
The most fundamental measure of the inherent basicity of a species is the proton
affinity. It is defined as the energy released for the reaction:
B(g) [or 8“(g)] + H + (g) -► BH + (g) [or BH(g)] (9.47)
Note that the proton affinity (PA) has the opposite sign from the enthalpy of reaction
of Eq. 9.47: Proton affinities are always listed as positive numbers despite referring to
exothermic reactions (recall the same convention with electron affinities. Chapter 2).
Proton affinities may be obtained in a number of ways. The simplest, and most
fundamental for defining an absolute scale of proton affinities, is to use a Born-Haber
cycle of the sort:
8(g) + H(g)
BH(g)
8(g) + H(g)
SHa -PA
BH(g)
Acid-Base Concepts
331
The molecule BH must be sufficiently stable that its bond energy (enthalpy of
atomization, A H A J and ionization potential (IE BH ) can be measured. Once several
proton affinities have been established in this way, many more may be obtained by a
technique known as ion cyclotron resonance spectroscopy and related methods, 20
which measure the equilibrium concentrations of the species involved in the
competition:
B(g) + B'H + (g) — BH + (g) + B'(g) (9 49)
Gas-phase proton affinities (Table 9.5) confirm many of our intuitive ideas about
the basicities of ions and molecules, though some of the first to be obtained contra¬
dicted our prejudices based on solution data (see page 344). The greatest proton
affinity estimated to date is that of the trinegative nitride ion, N 3- , because of the large
electrostatic attraction of the -3 ion. 2 ' The dinegative imide ion, NH 2 ", has a very
large but somewhat lower value, followed by amide, NH 2 “, and ammonia, NH,.
Inductive effects are readily observed with values ranging from nitrogen trifluoride,
NF, = 604 kJ mol , through ammonia, NH 3 = 872 kJ mol" 1 , to trimethylamine,’
(CHjjjN - 974 kJ mol Similar effects can be seen for toluene vs. benzene,
acetonitrile vs. hydrogen cyanide, ethers vs. water, and several other comparisons.
Since the proton affinity of a cation indicates its tendency to attract and hold a proton,
its value will also be the enthalpy of dissociation of its conjugate acid in the gas phase'
Consider HF (PA F - = 1554 kJ mol" 1 ):
HF -♦ H + + F“ A// = +1554 kJ mol" 1 ( 9 . 50 )
The more endothermic Eq. 9.50 is. the weaker the acid will be. Therefore Table 9.5
may readily be used to compare gas-phase acid strengths, and HF is a weaker acid in
the gas phase than are the other HX acids, as it is also in aqueous solution. In the same
way. acetic acid (PA CHiCOO = +1459 kJ mol“') is a weaker acid than triffuoroacetic
acid (PA CFiCOO = +1351 kJ mol '). Which is the stronger acid, methane or toluene?
Does Table 9.5 confirm or contradict your memory from organic chemistry?
Gas-phase Acidities: The Bronsted gas-phase acidity will be related to the proton affinity of the conjugate
Electron Affinities base. However, this gives us no estimate of the relative acidity of nonprotonic (Lewis)
acids. If the electron is the basic analogue of the acidic proton, then electron affinities
should provide an inherent gas-phase measure of acidity that parallels proton affinities
for bases. 22 That they have not been more frequently used in this connection is
20 Auc - D Bowers, M. T.. In Gas Phase ton Chemistry. Bowers. M. T.. Ed.; Academic: New
York. 1979: Vol. 2, Chapter 9; Lias. S. G.; Liebman. J. F.; Levin. R. D. J. Phys. Chem Ref. Data
1984. 13, 695-808.
21 ll should be noted lhal the proton affinities of all of the trinegalive and dinegative anions are
calculated by means of a Born-Haber cycle. They are not experimentally accessible since these ions
have no existence outside of a stabilizing crystal environment—they would exothermically expel an
electron (sec Chapter 2).
22 There is a complicating factor here in that acidity can refer to the acceptance of a single electron or
an electron pair. Thus a free radical might have a high electron affinity but not have an empty, low-
lying orbital to accept an clcctronpair. Thus, comparison of SO, (EA = 160 kJ moP') as a stronger
acid than SO,^EA = 107 kJ mol ') is valid, but a similar comparison with the free radical OH (EA
- 176 kJ mol ). which docs not have a completely empty low-lying orbital, is not.
Gas-phase Acidities:
Proton Loss
3'
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Fig. 11.4 Electron density contours for (a) d : i and (b) d, : orbitals. The triangles represent
points of maximum electron density. [From Perlmutter-Hayman. B. J. Chem. Educ. 1969. 46,
428-430. Reproduced with permission.]
Crystal Field Theory 397
Fig. 11.5 Representation of the d.'. orbital as a linear combination of d.:- x : and
orbitals. The d.: label is actually shorthand notation for
the negative field and the negative electrons in the orbitals. If the field results from the
influence of real ligands (either anions or the negative ends of dipolar ligands such as
NH, or H,0). the symmetry of the field will be less than spherical and the degeneracy
of the d orbitals will be removed. It is this splitting of d orbital energies and its
consequences that are at the heart of crystal field theory.
Crystal Field Effects:
Octahedral
Symmetry
Let us consider first the case of six ligands forming an octahedral complex. For
convenience, we may regard the ligands as being symmetrically positioned along the
axes of a Cartesian coordinate system with the metal ion at the origin, as shown in Fig.
11.6. As in the case of a spherical field, all of the d orbitals will be raised in energy
relative to the free ion because of negative charge repulsions. However, it should be
pictorially obvious that not all of the orbitals will be alfcctcd to the same extent. The
orbitals lying along the axes (the d.: and the d x 2 v :) will be more strongly repelled than
the orbitals with lobes directed between the axes (the d xy . d x: , and d V7 ). The d orbitals
are thus split into two sets with the d.: and d x ;_ y : at a higher energy than the other
Fig. 11.6 Complete set of
d orbitals in an octahedral
field produced by six
ligands. The e„ orbitals arc
shaded and the orbitals
arc unshaded. The torus of
the d .2 orbital has been
omitted for clarity.
332
9-Acid-Base Chemistry
probably because of the fact that formerly there were few electron affinity values
known for molecules (Table 2.6). Those Lewis acids having large electron affinities are
apt to be strong acids. This idea is especially powerful when applied to metal cations.
Recall that the electron affinity of a monopositive cation is the same as the ionization
energy of the metal atom. From this point of view it is readily apparent why the alkali
and alkaline earth metals are such weak Lewis acids when compared to the transition
metals:
Measures of Acid-Base Strength 333
Table 9.5 ( Continued )
Gas-phase proton affinities (kJ mol ')°
Trinegative ions Dinegative ions b
Uninegative ions Neutral molecules
CIO - = 1502
NO-T = 1421
NOj- = 1358
CF 3 C(0)0~ =1351
FSOj" = 1285'
CF 3 SO 7 = 1280'
CF 3 C(0)OH = 707
■23
8 8
m rr
II II
1 1
CHjS - = 1493
SH" = 1469
SeH~ = 1466
(CH 3 ),SO = 884
CO, = 548
O’ = 422
(C1-I 3 ),S = 839
CHjSH = 784
H,S =712
H,Se = 717
F" = 1554
cr = 1395
Br" = 1354
T = 1315
HF = 489.5
HC1 = 564
HBr = 569
HI = 628
He = 178
Ne = 210
Ar = 371
Kr = 425
Xe = 496
1315 < Mn(CO)j < 1340'
1380 < Re(CO)J < 1395'
Co(PF 3 ) 4 - < 1280'
(CH 3 ),Hg - 778
« Unless otherwise noted, all data for neutral molecules are from Lias, S. G.; Liebman. J. P.:
Levin. R. D. J. Phys. Chem. Ref. Dam 1984, 13. 695-808, and for anions are from Lias, S. G.:
Bartmcss, J. E.; Liebman, J. F.; Holmes. J. L.; Levin. R. D.; Mallard, W. G. J. Phys. Chem.
Ref. Data 1988, 17. 1-861.
b Dinegative and trinegativc ions have no existence outside of a system, such as a lattice, that
stabili7.es them. These values arc rather crude estimates from Born-Habcr cycles and indicate
the relative difficulty of pulling a proton away from their conjugate acids.
c Mcot-Ner, M.; Sicck, L. W. J. Am. Chem. Soc. 1991, 113, 4448-4460.
d Waddington, T. C. Adv. Inorg. Chem. Radiochem. 1959, /, 157-221.
' Miller, A. E. S.; Kawamura. A. R.; Miller, T. M. J. Am. Chem. Soc. 1990. 112. 457-458.
Viggino, A. A.; Henchman, M. J.; Dale. F.; Deakyne, C.; Paulson, J. F. J. Am. Chem. Soc.
1992. 114. 4299-4306. HCo(PF 3 ) 4 is the strongest gas-phase acid reported to date, donating
protons to all anions listed above it.
K + +e” —
—» K
EA = 419 kJ mol -1
(9.51)
Ca 2+ +e" —
— Ca +
EA = 1145 kJ mol -1
(9.52)
Mn 2+ + e~ —
—> Mn +
EA = 1509 kJ mol -1
(9.53)
Pt 2+ + e“ —
—> Pt +
EA = 1791 kJ mol -1
(9.54)
Co 3+ + e" —
— Co 2+
EA = 3232 kJ mol -1
(9.55)
394
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Fig. 11.2 Effect of metal-ligand 7r bonding: the Cr—C bond order is increased and the
C—0 bond order is decreased, (a) VB viewpoint: w bond between d orbital on Cr atom and
p orbital on C atom, (b) MO viewpoint: tt bond between d orbital on Cr atom and
antibonding orbital (tt) on the CO ligand.
In contrast. Al' h can adequately accommodate six water molecules; however, the
nitrogen donor of the ammonia ligands is not sufficiently electronegative to prevent
the buildup of excess electron density on aluminum in [AI(NH 3 )J 3+ , with the result
that the complex is unstable.
In apparent contradiction to the electroneutrality principle, there are many com¬
plexes in which the metal exists in a low oxidation state and yet is bonded to an
element of fairly low electronegativity. Among the most prominent examples are the
transition metal carbonyls, a large class of compounds in which the ligand (CO) is
bound to the central metal through carbon. The source of stability in these complexes
is the capacity of the carbon monoxide ligand to accept a "back donation" of electron
density from the metal atom. Within valence bond theory, this process can be
described in terms of resonance:
Cr—C= 6 : ~ Cr=C=0
(I) (ID
To whatever extent canonical form II contributes to the resonance hybrid, electron
density will be shifted from chromium to oxygen. A more precise examination of this
process indicates that the delocalization of the electron density occurs via tt overlap of
a cl orbital on the metal with an orbital of appropriate symmetry on the carbonyl
ligand. In valence bond theory, the ligand orbital would be one of the p orbitals of the
carbon atom (thus making it unavailable for a n bond to the oxygen), whereas the
molecular orbital theory would speak in terms of overlap with the n antibonding
orbital of carbon monoxide (Fig. 11.2). In either representation a tt bond forms
between metal and ligand and provides a mechanism for a shift in electron density
away from the metal toward the ligand. The formation of tt bonds of this type will be
discussed at greater length later in this chapter.
Crystal Field
Theory
The model that largely replaced valence bond theory for interpreting the chemistry of
coordination compounds was the crystal field theory, first proposed in 1929 by Hans
Dethe . 11 As originally conceived, it was a model based on a purely electrostatic
" Bethe. H. Ann. Pliys. 1929. 3. 135-206.
Crystal Field Theory 395
interaction between the ligands and the metal ion. Subsequent modifications, which
began as early as 1935 with papers by J. H. Van Vleck,'2 allow some covalency in the
interaction. These adjusted versions of the original theory generally are called ligand
field theory.'3 It is an interesting feature of scientific history that, although' the
development of crystal and ligand field theories was contemporary with that of
valence bond theory, they remained largely within the province of solid state physics
for about 20 years. Only in the 1950s did chemists begin to apply crystal field theory to
transition metal complexes . 14
Pure crystal field theory assumes that the only interaction between the metal ion
and the ligands is an electrostatic or ionic one with the ligands being regarded as
negative point charges. Despite this rather unrealistic premise, the theory is quite
successful in interpreting many important properties of complexes. Moreover, the
symmetry considerations involved in the crystal field approach are identical to those
of the molecular orbital method. The electrostatic model thus serves as a good
introduction to modern theories of coordination chemistry.
In order to understand clearly the interactions that are responsible for crystal or
ligand field effects in transition metal complexes, it is necessary to have a firm grasp of
the geometrical relationships of the d orbitals. There is no unique way of representing
the five d orbitals, but the most convenient representations are shown in Figs. 11.3 and
11.4. In fact, there are six wave functions that can be written for orbitals having the
typical four-lobed form (d xy and d x i _ y :, for example). Inasmuch as there can be only
five d orbitals having any physical reality, one of them (the d.:) is conventionally
regarded as a linear combination of two others, the d.i _ v 2 and d z C x j. Thus these latler
two orbitals have no independent existence, but the dj can be thought of as having the
average properties of the two (Fig. 11.5). Therefore, since both have high electron
density along the z axis, the t/..- orbital has a large fraction of its electron density
concentrated along the same axis. Also, since one of the component wave functions
(d.t-p) has lobes along the x axis and the other (</.:_ v :) along the y axis, the resultant
d .2 orbital has a torus of electron density in the ary plane. This .ry component, which is
often referred to as a "doughnut" or a "collar." is frequently neglected in pictorial
representations, especially when an attempt is being made to portray all five d orbitals
simultaneously. Nevertheless, it is important to remember this xy segment of the d.z
orbital.
The five d orbitals in an isolated, gaseous metal ion are degenerate. If a spherically
symmetric field of negative charges is placed around the metal, the orbitals will remain
degenerate, but all of them will be raised in energy as a result of the repulsion between
Van Vleck. J. H. J. Cltem. Plivs. 1935. 3. 803-806. 807-813.
'3 There is some inconsistency in the use of the label "ligand field theory" among textbooks and other
sources. In some instances it is taken as essentially a substitute for the label crystal field theory on
the premise that the latter is misleading (see. for example. Schlafcr, H. L.; Gliemann. G. Basic
Principles of Lipand Field Theory. Wiley-lntersciencc: New York. 1969; p 17). At the other extreme
arc those who use the term to mean a molecular orbital description of bonding in complexes. It is
significant that Van Vleck in his 1935 paper was seeking to reconcile Bethc's theory with Mulliken's
molecular orbital approach. This efTort has been described by Ballhausen (see Footnote 14) as
"incorporating the best features of both the pure crystal field theory and molecular orbital theory."
It seems appropriate, then, to view ligand field theory as a model that owes its origins to crystal field
theory and shares with it a central emphasis on the perturbation of metal valence orbitals by ligands,
while at the same time serving as a bridge to the full molecular orbital treatment of complexes.
14 lhc evolu,ion of bonding models for inorganic complexes from the 1920s to the mid-1950s is
described in Ballhausen. C. J. J. Cltem. Educ. 1979. 56. 194-197. 215-218, 357-361.
334 9-Acid-Base
Lewis Interactions
in Nonpolar
Solvents
Chemistry
This gets us back to the fundamental property of the ionization energy of a metal that
determines not only its redox chemistry but also its tendency to bond to anions and
other Lewis bases. 23
The evaluation and correlation of strengths of Lewis acids and bases have attracted
the interest of many inorganic chemists. Recently gas-phase data have become avail¬
able. but before that many systems were studied in aprotic, nonpolar solvents. Even
today, such solvents allow the collection of large amounts of data by various methods.
Solvation effects will be small and. it is hoped, approximately equal for reactants and
products such that their neglect will not cause serious errors.
It is common to equate the strength of interaction of an acid and a base with the
enthalpy of reaction. In some cases this enthalpy may be measured by direct calorime¬
try: A H = q for an adiabatic process at constant pressure.
Often the enthalpy of reaction is obtained by measuring the equilibrium constant
of an acid-base reaction over a range of temperatures. If In K is plotted versus I IT. the
slope will be equal to — A H/R. Thus various experimental methods have been devised
to measure the equilibrium constant by spectrophotometric methods. Any absorption
which differs between one of the reactants (either acid or base) and the acid-base
adduct is a potential source of information on the magnitude of the equilibrium
constant since it gives the concentration of two of the three species involved in the
equilibrium directly and the third indirectly from a knowledge of the stoichiometry of
the reaction. For example, consider the extensively studied reaction between organic
carbonyl compounds and iodine. The infrared carbonyl absorption is shifted in fre¬
quency in the adduct with respect to the free carbonyl compound. Thus the equi¬
librium mixture exhibits two absorption bands in the carbonyl region of the spectrum
(Fig. 9.1) and the relative concentrations of the free carbonyl and the adduct can be
obtained. 24 Alternatively, one can observe the absorption of the iodine molecule, I 2 , in
Frequency (cm' 1 )
Fig. 9.1 Infrared absorption spectra of dimethylacetamidc-iodine system: (I) dimethylaceta-
mide only; (2-6) increasing concentrations of iodine. Peak at 1662 cm -1 is from free di-
methylacetamide, that at 1619 cm' 1 is from the DMA-L adduct. |From Schmulbach. C. D.:
Drago, R. S. J. Am. Client. Soc. I960, 82. 4484. Reproduced with permission.]
25 For a very useful discussion of the chemical properties of metals, as related to their ionization
energies, see Ahrens. L. H. Ionization Potentials'. Pcrgamon: Oxford, 1983.
24 Schmulbach. C. D.; Drago. R. S. J. Am. Cliem. Soc. 1960. 82. 4484-4487.
Measures of Acid-Base Strength 335
the 300-600 nm portion of the visible spectrum. Again, the adduct absorbs at a
different frequency than the free iodine and the two absorption maxima provide
information on the relative concentrations of the species present. 23
Two complications can prevent a simple determination of the concentration of
each species from a measurement of absorbance at a chosen frequency. Although
most of the acid-base reactions of interest result in a one-to-one stoichiometry, one
cannot assume this a priori, and two-to-one and three-to-one adducts might also be
present. Fortunately, this is usually an easy point to resolve. The presence of an
isosbestic point or point of constant absorbance (see Fig. 9.1) is usually a reliable
criterion that only two absorbing species (the free acid or base and a single adduct) are
present. 26
The second problem is somewhat more troublesome. The separation between the
absorption maximum of the adduct and that of the free acid (or base) is seldom large
and so there is considerable overlapping of bands (see Fig. 9.1). If the absorptivities of
each of the species at each frequency were known, it would be a simple matter to
ascribe a proportion of the total absorbance at a given frequency to each species. It is
usually a relatively simple matter to measure the absorptivity, E , of the free acid (or
base) over the entire working range. Since it is often impossible to prepare the pure
adduct (in the absence of equilibrium concentrations of free acid and base), its
absorptivity cannot be measured. However, if the equilibrium is studied at two
different concentrations of acid (or base), it is possible to set up two simultaneous
equations in terms of the two unknowns K and e and solve for both. 22
Alternative methods of measuring the enthalpy of acid-base reactions involve
measuring some physical property which depends upon the strength of the interaction.
In general, such methods must be calibrated against one of the previous types of
measurement, but once this is done they may often be extended to reactions that
prove difficult to measure by other means. One example is the study of phenol as a
Lewis acid. 28 Phenol forms strong hydrogen bonds to Lewis bases, especially those
that have a donor atom with a large negative charge. The formation of the hydrogen
bond alters the electron density in the O—H group of the phenol and the O—H
stretching frequency in the infrared. Once the frequencies of a scries of known
phenol-base adducts have been used for calibration (Fig. 9.2). it is possible to estimate
the enthalpy of adduct formation of bases with similar functional groups directly from
IR spectra.
A second method involves the relation between .v character and NMR coupling
constants. Drago and coworkers 29 have shown that there is a good correlation be¬
tween the ll9 Sn-H coupling constant in chlorolrimethylstannane-ba.se adducts and
the strength of the base-tin bond. In the free stannane the strongly bonding methyl
groups can maximize their bond strength, s character, and thus 7n« Sn _, H . It has been
suggested that the stronger bases force the tin to rehybridize to a greater extent {-~sp }
25 Drago. R. S.: Carlson. R. L.: Rose. N. J.; Wcnz. D. A. J. Am. Cliem. Soc. 1961, 8J. 3572-3575.
26 The present discussion of isosbestic points is oversimplified: the reader is warned that one can get
into difficulties with such oversimplifications. For further discussion of the use of isosbestic points,
see Cohen, M. D.; Fischer. E. J. Cliem. Soc. 1962, 3044-3052; Mayer, R. G.: Drago, R, S, horn■
Cliem. 1976, 15, 2010-2011; Styncs. D. V. horn. Cliem. 1975, 14. 453-454.
27 Rose. N. J.; Drago, R. S. J. Am. Cliem. Soc. 1959, 81. 6138-6141.
28 Eplcy. T. D.; Drago. R. S. J. Am. Cliem. Soc. 1967. 89. 5770-5773; 1969. 91. 2883-2890. One must
be careful in the choice of bases. Sec Vogel. G. C.; Drago. R. S. Ibid. 1970, 92. 5347-5351.
29 Bolles, T. F.; Drago. R. S. J. Am. Cliem. Soc. 1966, 88. 5730-5734.
392
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
include pairing of electrons as well as ligand-metal-ligand bond angles of 90°. Pauling
suggested this occurs via hybridization of one (n - I)c/, the ns. and two up orbitals to
form four equivalent dsp 2 hybrids directed toward the corners of a square. These
orbitals then participate in covalent a bonds with the ligands, the bonding electron
pairs being furnished by the ligands. The eight electrons that were distributed among
the five d orbitals in the free ion are assigned as pairs to the four unhybridized metal d
orbitals (Fig. II.la).
With some ligands, such as Cl - , Ni(II) forms four-coordinate complexes that are
paramagnetic and tetrahedral. For these cases, VB theory assumes the d orbital
occupation of the complex to be the same as that of the free ion, which eliminates the
possibility that valence-level d orbitals can accept electron pairs from the ligands.
Hybrid orbitals of either the sp or sd } type (the latter involving rt-level d orbitals) or a
combination of the two provide the proper symmetry for the cr bonds as well as
«n Co,™,,;;- JUiiLil_JiiLiI_y.il.
2d -
d~sp •' hybrid orbitals
Electrons from NH , ligands
Octahedral geometry
Fig. 11.1 Orbital diagrams depicting the valence bond description of the metal-ligand bonds
in (a) [PtCIj]' - , (b) [NiCU 2- . (c) (CoFj' - . and (d) [Co(NH,)„]’.
Valence Bond Theory
393
allowing for the magnetic properties (Fig. 11.1b). The examples presented here illus¬
trate a useful rule, originally called “the magnetic criterion of bond type," which
allows prediction of the geometry of a four-coordinate d 8 complex if its magnetic
properties are known: diamagnetic = square planar; paramagnetic = tetrahedral. 9
The valence bond picture for six-coordinate octahedral complexes involves d 2 sp i
hybridization of the metal (Fig. 11.1c, d). The specific d orbitals that meet the sym¬
metry requirements for the metal-ligand a bonds are the d j and dp _ i (page 396). As
with the four-coordinate c/ 8 complexes discussed above, the presence of unpaired
electrons in some octahedral compounds renders the valence level (n - I )d orbitals
unavailable for bonding. This is true, for instance, for paramagnetic [CoFJ 3- (Fig.
Il.lc). In these cases, the VB model invokes participation of u-level d orbitals in the
hybridization.
The Electroneutrality One difficulty with the VB assumption of electron donation from ligands to metal ions
Principle and Back is the buildup of formal negative charge on the metal. Since this is a problem that
Bonding arises, in one form or another, in all complete treatments of coordination compounds,
the following discussion is appropriate to all current bonding models.
Consider a complex of Co(II) such as [CoL 6 ] 2+ . The six ligands share twelve
electrons with the metal atom, thereby contributing to the formal charge on the metal a
total of -6, which is only partially canceled by the metal's ionic charge of +2. From a
formal charge point of view, the cobalt acquires a net -4 charge. However, Pauling
pointed out why metals would not in fact exist with such unfavorable negative
charges. Because donor atoms on ligands are in general highly electronegative (e.g.,
N, O, and the halogens), the bonding electrons will not be shared equally between the
metal and ligands. Pauling suggested that complexes would be most stable when the
electronegativity of the ligand was such that the metal achieved a condition of
essentially zero net electrical charge. This tendency for zero or low electrical charges
on atoms is a rule-of-thumb known as the electroneutrality principle, and it is used to
make predictions regarding electronic structure in many types of compounds, not only
complexes. Pauling has made semiquantitative calculations correlating the stability of
complexes with the charges on the central metal atom. 10 Some typical results are:
[Be(H,0) 4 ) 2f [Be(H 2 0)J 2+ [AI(H,0)J ,+ |A1(NH 3 )J 3+
Be = -0.08 Be = —1.12 Al = -0.12 AI--I.08
40 = -0.24 60 = -0.36 60 = -0.36 6N = 1.20
8H = 2.32 I2H = 3.48 I2H = 3.48 I8H = 2.88
Total = +2.00 Total = +2.00 Total = +3.00 Total = +3.00
Although the above values involve very rough approximations, they do indicate
qualitatively how buildup of excessive negative charge on a metal can destabilize a
complex. Within the group of complexes shown. |Be(H 2 0) 4 l 2+ and [AI(H 2 0) 6 ] 3+ are /
stable, but the other two are not. Four water molecules effectively neutralize the +2
ionic charge of beryllium, but six water molecules donate too much electron density.
9 An exception to this rule is the paramagnetic complex, |R : P(0)NR'];Ni (R = !3u'; R' = Pr'), in
which Ihe central Ni(ll) is bound to two oxygen and two nitrogen atoms in a planar arrangement. Sec
Frommcl. T.; Peters, W.; Wunderlich, H.; Kuchen. W. Angcw. Chem. Int. Ed. Engl. 1992. 31,
612-613.
10 Pauling. L. The Nature of the Chemical Bond. 3rd ed.; Cornell University: Ithaca. NY. I960;
pp 172-174.
I
336
9-Acid-Base Chemistry
A" 0 _„ <«»">
Fig. 9.2 Relation between the enthalpy of formation of base-phenol adducts and the
stretching Frequencies of the O—H bond in the phenol. Bases: (a) acetonitrile, (b) ethyl
acetate, (c) acetone, (d) tetrahydrofuran. (e) dimethylacetamide. (f) pyridine, (g) triethyl-
amine. [From Epley. T. D.; Drago. R. S. J. Am. Chem. Soc. 1967. 89. 5770. Reproduced
with permission.]
in free Me 3 SnCI; ~sp i d in the limit of strong base adducts) than weaker ones, and thus
the change in s character of the Sn—C bonds.
Drago and coworkers have proposed a number of ways of expressing enthalpies of
reactions in terms of contributing parameters of acids and bases. The first was
-A tf = E A £ B + C A C B (956)
where A H is the enthalpy of formation of a Lewis acid-base adduct. E A and C A are
parameters characteristic of the acid, and £ B and C B are parameters characteristic of
the base.3° The E parameters are interpreted as the susceptibility of the species
to undergo electrostatic (‘•ionic” or dipole-dipole) interactions and the C parameters
are the susceptibility to form covalent bonds. From this we expect those acids which
bond well electrostatically (E A is large) to form the most stable adducts with bases that
bond well electrostatically (since the product E A £ B will then be large). Conversely,
acids that bond well covalently will tend to form their most stable adducts with bases
that bond well covalently. Some typical values of £ A , E B , C A , and C B are listed in
Table 9.6. The application of Eq. 9.56 may be illustrated with the reaction between
pyridine (E = 1.78. C = 3.54) and iodine (E = 0.50, C = 2.00; by definition, the origin
of the scale; see Footnotes 30. 36).
A// culc = E A £ B + C A C B = (0.50 x 1.78) 4- (2.00 x 3.54)
= (7.97 kcal mol -1 ) = 33.3 kJ mol" 1 (9.57)31
-AW cxp = (7.8 kcal mol -1 ) = 32.6 kJ mol -1
The importance of the E-C parameters is manifold. First, they enable predictions
to be made of the enthalpies of reactions that have not been studied. Thus the
parameters in Table 9.6 and comparable values were obtained from a few hundred
jo Drago. R. S.: Wayland. B. B. J. Am. Chem. Soc. 1965, 87. 3571-3577. Drago. R. S.; Vogel. G. C.;
Needham, T. E. J. Am. Cliem. Soc. 1971. 93. 6014-6028.
Ji The original E-C scale and all subsequent modifications have been in kcal mol '. One can either
multiply the result by 4.184 to obtain enthalpies in kJ mol "\ or create a new scale in wh.ch all the
parameters in Table 9.6 are multiplied by 2.05 (kJ mol ') ‘/(kcal mol ) '.
Systematics of
Lewis Acid—Base
Interactions
■ __
Table 9.6__
Acid and base parameters 0
Acid
£a
C A
«A
Acid
Ea
Ca
Ra
0.50
2.00
_
H +
(X)
13.03
130.21
h 2 o
1.54
0.13
0.20
CHt
19.70
12.61
55.09
so 2
0.56
1.52
0.85
Li +
11.72
1.45
24.21
HF b
2.03
0.30
0.47
K +
3.78
O.IO"
20.79
HCN
1.77
0.50
0.54
NO +
0.I
6.86
45.99
CH 1 OH
1.25
0.75
0.39
nh; 6
4.31
4.31
18.52
H 2 S fc
0.77
1.46
0.56
(CHjKNH^
3.21
0.70
20.72
hq
3.69
0.74
0.55
(CH 3 ) 4 N + "
1.96
2.36
8.33
c 6 h,oh
2.27
1.07
0.39
c 5 h,nh +/ '
1.81
1.33
21.72
(CH 3 ) 3 COH
1.36
0.51
0.48
(C,H,) 1 NH +fc
2.43
2.05
11.81
HCCIj
1.49
0.46
0.45
(CH,) 3 NH’ l ' ft
2.60
1.33
15.95
CHjCOiH
1.72
0.86
0.63
H 3 cr
13.27
7.89
20.01
CFjCHjOH
2.07
1.06
0.38
(I-LO) 2 H +
11.39
6.03
7.36
C,H 5 OH
1.34
0.69
0.41
(H 2 0) 3 H +
11.21
4.66
2.34
i-C 3 H 7 OH
1.14
0.90
0.46
(H,0).,H + '’
10.68
4.11
3.25
| -
PF
0.61
0.36
0.87
(CH 3 ) 3 Sn +
7.05
3.15
26.93
B(OCH 3 )
0.54
1.22
0.84
(C 5 H 5 )Ni +
11.88
3.49
32,64
AsF
1.48
1.14
0.78
(CH 3 )NH 3 ''
2.18
2.38
20.68
Fe(CO)j
0.10
0.27
1.00
CHFj
1.32
0.91
0.27
B(C 2 H s )?
1.70
2.71
0.61
Base"
f B
C B
7b
Base c
Jb
C B
r B
nh 3
2.31
2.04
0.56
C,H,NO
2.29
2.33
0.67
ch 3 nh 2
2.16
3.12
0.59
(CH 3 ) 3 P
1.46
3.44
0.90
(CH 3 ) 2 NH
1.80
4.21
0.64
(CH 3 ) 2 0
1.68
1.50
0.73
(CH 3 )jN
1.21
5.61
0.75
(CH 3 ) 2 S
0.25
3.75
1.07
c 2 h 5 nh 2
2.35
3.30
0.54
ch 3 oh
1.80
0.65
0.70
(C 2 h 5 ) 3 n
1.32
5.73
0.76
c 2 h,oh
1.85
1.09
0.70
HC(C,H 4 ) 3 N
0.80
6.72
0.83^
c„h 6
0.70
0.45
0.81
C 5 HjN
1.78
3.54
0.73
h 2 s
0.04
1.56
1.13
4-CH,C 5 H 4 N
1.74
3.93
0.73"
hcn
1.19
0.10
0.90
3-CH 3 C 5 H 4 N
1.76
3.72
0.74"
HXO"
1.56
0.10
0.76
3-CIC 5 H 4 N
1.78
2.81
0.75"
CHjCI 6
2.54
0.10
0.23
CHjCN
1.64
0.71
0.83
CH 3 CHO h
1.76
0.81
0.74
CH 3 C(0)CHj
1.74
1.26
0.80
H,0'’
2.28
0.10
0.43
CH 1 C(0)OCH 3
1.63
0.95
0.86
(CH 3 ) 3 COH'’
1.92
1.22
0.71
CH,C(0)OC,H s
1.62
0.98
0.89
C^HjCN
1.75
0.62
0.85
HC(0)N(CH 3 ) 2
2.19
1.31
0.74'/
F -
9.73
4.28
37.40
(C 2 H s ) 2 0
1.80
1.63
0.76
a~
7.50
3.76
12.30
0(CHXH,)jO
1.86
1.29
0.71
Br“
6.74
3.21
5.86
(Ch 2 ) 4 o
1.64
2.18
0.75
r
5.48
2.97
6.26
(CH 2 ) 5 0
1.70
2.02
0.74-/
CN -
7.23
6.52
9.20
(C 2 h 5 ) 2 s
0.24
3.92
uo/
OH - "
10.43
4.60
50.73
(CHj) 2 SO
2.40
1.47
0.65
CH,0 -fc
10.03
4.42
33.77
° Drago. R. S.; Ferris, D. C.; Wong, N, J. Am. Chem. Soc. 1990, 112. 8953-8961, Drago, R. S.;
Wong, N.; Ferris, D. C. Ibid. 1991. 113. 1970-1977. Reproduced with permission.
h Tentative parameters from limited enthalpy data.
c If not indicated otherwise, the bases in this table have E u and C B determined from the fit of
neutral acid-neutral base adducts (Footnote 30).
d The Eg and C» for these bases arc well determined. The T„ values are tentative for they have
been determined from limited data. Sec the original papers for methods and accuracy of the
values.
337
390 11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
A second important contribution that Werner made to the study of coordination
chemistry was the postulate that the bonds to the ligands were fixed in space and
therefore could be treated by application of structural principles. By means of the
numbers and properties of the isomers obtained. Werner was able to assign the correct
geometric structures to many coordination compounds long before any direct experi¬
mental method was available for structure determination. Werner's method was that
used previously by organic chemists to elucidate the structures of substituted ben¬
zenes, namely isomer counting. Werner postulated that the six ligands in a complex
such as [Co(NH-,) 6 P+ were situated in some symmetrical fashion with each NH 3
group equidistant from the central cobalt atom. Three such arrangements come to
mind: a planar hexagon—similar to the benzene ring—and two polyhedra. the trigonal
prism and the octahedron. The trigonal prism is closely related to the octahedron,
being formed by a 60° rotation of one of the trigonal faces of the octahedron (in fact,
the octahedron can be considered a trigonal ant/prism). For a "disubstituted” com¬
plex. M A 4 B 2 , the planar arrangement gives three isomers, the familiar ortho, meta.
and para arrangements of organic chemistry. The trigonal prism yields three isomers
also, but there are only two octahedral arrangements for this formulation. The total
number of isomers expected for each geometrical arrangement together with the
experimental results for various compositions are listed in Table 11.2.
In every case Werner investigated, the number of isomers found was equal to that
expected for an octahedral complex. For [Co(NH 3 ) 4 CI 2 JCI. for example, two isomers
(one violet and one green) were observed. Although the correlation here was perfect,
it must be borne in mind that the observation of two instead of three isomers for this
compound and others constitutes negative evidence concerning the structure of these
complexes. Even, though Werner worked carefully and examined many systems, there
was always the possibility, admittedly small, that the third isomer had escaped his
detection. The failure to synthesize a compound, to observe a particular property, or
to effect a particular reaction can never be positive proof of the nonexistence of that
compound, property, or reaction. It may simply reflect some failure in technique on
the part of the chemist. One well-known example of the fallacy of negative evidence
involves the overthrow of the dogmatic belief in the chemical inertness of the noble
gases (see Chapter 17).
Werner was correct, however, in his conclusions concerning the octahedral
geometry of coordination number 6 for cobalt(III) and platinum(IV). He was also
correct, and on a firmer logical footing, in his assignment of square planar geometries
Table 11.2 _
Numbers of isomers
expected and found for
C.N. = 6
Valence Bond Theory
391
to the four-coordinate complexes of palladium and platinum from the fact that two
isomers had been isolated for compounds of formula MA 2 B,. The most likely alter¬
native structure, the tetrahedron, would produce only one isomer for this composi¬
tion. 6 The ability of Werner and others to assign the correct structures from indirect
data and logic was hailed by Henry Eyring: 7
"The ingenuity and effective logic that enabled chemists to determine complex
molecular structures from the number of isomers, the reactivity of the molecule and of
its fragments, the freezing point, the empirical formula, the molecular weight, etc., is
one of the outstanding triumphs of the human mind.”
There has been much work done in attempting to formulate theories to describe the
bonding in coordination compounds and to rationalize and predict their properties.
The first success along these lines was the valence bond(VB) theory applied by Linus
Pauling and others in the 1930s and following years. In the 1950s and 1960s the crystal
field {CF) theory and its modifications, generally known under the label ligand field
(LF) theory, gained preeminence and in turn gradually gave way to the molecular
orbital (MO ) theory. Although both the valence bond and crystal field theories have
been largely displaced as working models for the practicing inorganic chemist, they
continue to contribute to current discussions of coordination compounds. Because
they shaped the thinking about these compounds in the very recent past, the earlier
models still serve as the background against which newer ones are evaluated. More¬
over, certain of their features remain part of the conceptual framework and vocabu¬
lary used by current chemists. Hence they must be appreciated in order to have a full
understanding of modern constructs.
From the valence bond point of view, formation of a complex involves reaction
between Lewis bases (ligands) and a Lewis acid (metal or metal ion) with the forma¬
tion of coordinate covalent (or dative) bonds between them. The model utilizes
hybridization of metal s, p. and d valence orbitals to account for the observed
structures and magnetic properties of complexes. For example, complexes of Pd(II)
and Pt(II) are usually four-coordinate, square planar, and diamagnetic, and this
arrangement is often found lor Ni(II) complexes as well. Inasmuch as the free ion in
the ground state in each case is paramagnetic (t/\ i F), the bonding picture has to
The first crystallographic confirmation of Werner's assignment of octahedral geometry to Pt(IV)
complexes was not published until 1921. some twenty years after his theories were completed
(Wyckoff. R. W. G.; Posnjak. E. J. Am. Chem. Soc. 1921 .43, 2292-2309; [NHJjlPtCU). The square
planar structure of Pt(II) complexes was confirmed the next year (Dickinson. R. G. J. Am. Chem.
Soc. 1922. -14. 2404-2411; K 2 (PtCI 4 |). Interestingly, neither paper mentions Werner and in fact, the
second one stales: "It would probably be anticipated that in the chloroplulinates, four chlorine atoms
would be grouped about each platinum atom; but the form and dimensions of this group as well as its
situation in the structure could scarcely be predicted with safety."
7 Eyring. H. Chem. Eng. News 1963. 4/(1). 5.
K The classical account of valence bond theory as applied to coordination compounds is probably
Pauling's book. The Nature of the Chemical Bond. 3rd cd.; Cornell University: Ithaca. NY. I960.
For a more recent discussion pertaining to coordination compounds and organomctallic compounds,
see Mingos. D. M. P.: Zhenyang. L. Struct. Bonding (Berlin) 1990. 72 . 73-1II.
Valence Bond
Theory 8
Bonding in
Coordination
Compounds
338
9-Acid-Base Chemistry
Measures of Acid-Base Strength
339
reactions of acids and bases, but they can be used to predict the enthalpies of
thousands of reactions. For example, accurate values can be obtained for reactions
such as:
(CH,) 3 N + S0 2 - (CH 3 ) 3 NS0 2 (9.58)
A// calc = 38.5 kJ mol -1 (9.2 kcal mol -1 )
A// exp = 40.2 kJ moF 1 (9.6 kcal mol -1 )
^CH,CH a
O
n ch,ch 2
p + HOC 6 H 5
/C H,CH,
0
n ch,ch,
o—HOC 6 Hj
(9.59)
A// ca | C = 23.4 kJ mol 1 (5.6 kcal mol ‘)
A// exp = 23.4 kJ mol -1 (5.6 kcal mol -1 )
The second item of importance with respect to parameters of this sort is that they
enable us to obtain some insight into the nature of the bonding in various systems.
Thus, if we compare the C A and £ A parameters of iodine and phenol, we find that l 2 is
twice 32 as good a “covalent-bonder” as phenol, but that the latter is about five times
as effective through electrostatic attractions. This is not unexpected inasmuch as
phenol, C 6 H s OH, is a very strong hydrogen bonding species. In contrast, iodine has
no dipole moment but must react with a Lewis base by expanding its octet and
accepting electrons to form a covalent bond.
A similar effect can be observed in the bases. The E a value of dimethylsulfoxide,
(CH 3 ) 2 SO, is much larger than that of diethylsulfide, (C 2 H 5 ) 2 S, corresponding to the
large dipole moment of (CH 3 ) 2 SO (/x = 13.2 x 1() -30 C m: 3.96 D) compared with that
of the latter (fx - 5.14 x 10 -3<> C m; 1.54 D). On the other hand, the C B values are
reversed, corresponding to the enhanced ability of the sulfur atom to bond covalently
to the acid.
Drago and coworkers have modified Eq. 9.56 by adding a constant specified by
the acid (acceptor): 33
-A H = E A E n + C A C n -W (9.60)
Eq. 9.56 deals with a simplified situation: the approach of A and B to the bond
distance with a resultant E A E B electrostatic energy based on the inherent electrostatic
bonding capabilities (dipole-dipole interactions, etc.) and a resultant C A C B term based
on the inherent covalent bonding capabilities (related to overlap, etc.). This approx¬
imation is quite good for neutral species, and small discrepancies (such as the increase
in covalency through electrostatic polarization) could be (and have been) accommo¬
dated by incorporating them into the E and C parameters. 34
32 One must be careful in making comparisons using these numbers. A comparison of £|, to Ephon is
valid, but one cannot compare the £ and C parameters directly against each other for a single
species because of the necessary, arbitrary assignments: £|, = 0.50; C| : = 2.00.
53 Drago. R. S,; Wong, N.; Bilgrien, C.; Vogel, G. C. Inorg. Client. 1987. 26. 9-14. The constant Wis
for an energy always associated with a particular reactant, such as the enthalpy of dissociation of a
dimer allowing it to react as a monomer.
34 By adjusting £ and C to give the best fit of the experimental data, some of the neglect of transfer
energy can be alleviated.
If the amount of charge transfer from base to acid is large, however, as it must be
if either species, or both, is an ion, this energy must be explicitly accounted for.
Fundamentally, it is the energy change as an atom (ion) runs along the energy-charge
curve of Fig. 5.32. Thus, this transfer energy term parallels ionization energies and. to
a lesser extent, also involves electron affinities. 33 It can be treated by adding a term
composed of two additional parameters: R A (receptance, acid) and T B (transmittance.
base): 36
-A H = E a E b + C a C b +R a T b (9.61)
Two processes illustrate the application of Eq. 9.61:
H 2 0 + H + -<• H 3 0' -Atf cxp = 695 kJ mol -1 (166 kcal mol -1 ) (9.62)
-A// calc = E a E b (102.6) + C A C B (1.3) + R A T B (56.0) = (9.63)
669 kJ mol -1 (159.9 kcal moF 1 )
(CH 3 ) 2 0 + H + -► (CH 3 ) 2 OH + (9.64)
~A// t . xp = 803 kJ mol -1 (192 kcal mol -1 )
-A// calc = E a E b (75.6) + C A C B (19.5) + R A T B (95.0) = (9.65)
795 kJ moF 1 (190.1 kcal mol -1 )
The last term represents the energy accompanying the transfer of electron density
from an electron-rich base to an electron-poor acid. The details need not concern us,
but the results certainly are of interest:
To obtain a complete picture of bonding in acid-base interactions, three
separate factors must be taken into account: a) the electrostatic energy of the
acid-base interaction; b) the covalent energy of the acid-base interaction; c)
the energy involved when electron transfer takes place. These results were
anticipated in principle on the basis of Mulliken-Jalfe electronegativity. 32
In the gas phase the proton is a tremendous acceptor of electron density, and
the transfer energy is very large. This transfer energy has already been largely
"spent" for the solvated proton (e.g., hydronium ion) in solution where the
reactions are of a displacement type:
H 3 0 + NH 3 -► H 2 0 + NH; ,9.66)
35 The neglect of the electron affinity in electronegativity can often be justified because in xm =
!(IE V + EA V ) the value of IE V may be an order of magnitude larger than that of EA V . Thus a two-
parameter system may be approximated by a one-parameter equation. However, the recent devel¬
opments in acid-base theory reported from here to the end of the chapter reflect the realization that
acid-base interactions arc more subtle than one-parameter equations predict. For a direct com¬
parison. note how well a one-parameter electronegativity system (Pauling) works, in general, even
though electronegativity is obviously a two-parameter function (Mulliken-JalTd). Sec also discus¬
sion on page 341.
36 Drago. R. S.; Ferris. D. C.; Wong. N. J. Am. Client. Soc. 1990 , 112. 8953-8961. Drago, R. S.;
Wong, N.; Ferris. D. C. Ibid. 1991, 113. 1970-1977.
37 Evans. R. S.; Huhecy. J. E. J. lnor K . Nucl. Client. 1970, 32. 777-793. The approach here involved
the reverse process: First electronegativity equalization was assumed (equals optimizing the elec¬
tron transfer energy) and then modifications for ionic and covalent contributions were added. This
worked best on ion-ion interactions (Li + F ► LiF ) and more poorly on neutral mole¬
cules (Huheey and Evans, unpublished). See page 351.
388
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
ond to none, and had he not been prejudiced by some of the theories of valence
current in his day. he might well have achieved the same results and fame as Werner. 2
Werner, in formulating his ideas about the structure of coordination compounds,
had before him facts such as the following. Four complexes of cobalt(III) chloride with
ammonia had been discovered and named according to their colors:
Complex
Color
Early name
CoCI 3 -6NH 3
Yellow
Luteo complex
CoCI 3 -5NH 3
Purple
Purpttreo complex
CoCI 3 -4NH 3
Green
Praseo complex
CoCI 3 -4NH 3
Violet
Violeo complex
One of the more interesting facts about this series is that two compounds have
identical empirical formulas, CoCI 3 -4NH 3 , but distinct properties, the most noticeable
being a difference in color. Furthermore, Werner noted that the reactivities of the
chloride ions in these four compounds differed considerably. Addition of silver nitrate
resulted in different amounts of precipitated silver chloride:
CoC 1 3 -6NH 3 + excess Ag + -» 3AgCl(s)
CoCl 3 -5NH 3 + excess Ag + - 2AgCl(s)
CoCl 3 -4NH 3 + excess Ag" -► 1 AgCl(s)
Reaction 11.3 occurs for both the praseo and violeo complexes.
The correlation between the number of ammonia molecules present and the
number of equivalents of silver chloride precipitated led Werner to the following
conclusion: 3
“We can thus make the general statement: Compounds M(NH 3 ) 5 X 3 [M = Cr. Co;
X = Cl, Br. etc.| are derived from compounds M(NH 3 ) h X 3 by loss of one ammonia
molecule.
With this loss of an ammonia molecule, however, a simultaneous change in
function of one acid residue X [= chloride ion] occurs .... [In] Co(NH 3 ) 5 CI 3 . . .
two chlorine atoms behave as ions and are precipitated by silver nitrate at room
temperature, while the third behaves completely analogously to chlorine in chlo-
roethane, that is, it no longer acts as an ion."
From this conclusion Werner postulated perhaps the most important part of his
theory: that in this series of compounds cobalt exhibits a constant coordination
number of 6, and as ammonia molecules are removed, they are replaced by chloride
ions which then act as though they are covalently bound to the cobalt rather than as
free chloride ions. To describe the complex chemistry of cobalt, one must therefore
consider not only the oxidation state of the metal but also its coordination number.
2 For discussion or the earliest work in coordination chemistry, sec Kauffman. G. B. J. Cliem. Educ.
1959. 36. 521-527; Kauffman. G. B. Classics in Coordination Chemistry-. Dover: New York. 1968
(Part I: The Selected Papers of Alfred Werner): 1976 ( Part 2: Selected Papers. 1798-1899): 1978
(Part 3: Twentieth-Century Papers. 1904-1935).
J Werner. A. Z. Anorg. Cliem. 1893. 3. 267-342. For a translation see the second reference in Footnote
2. All bracketed material and ellipses are ours.
Coordination Chemistry: Bonding, Spectra, and Magnetism
389
Werner thus formulated these four salts as (Co(NH 3 ),JCI 3 , ICofNHj^CllCU. i 'nd
[Co(NH 3 ) 4 C1 2 ]C1. j
Realizing that these formulations implied a precise statement of the number of
ions formed in solution. Werner chose as one of his first experimental studies mea¬
surement of the conductivities of a large number of coordination compounds. 5 Some
of the results of this work are listed in Table 11.1 together with values for simple ionic
compounds for comparison.
Table 11.1 _
Conductivities of
coordination compounds
Empirical
formula
Conductivity 0
Werner formulation
Nonelectrolytes
PtCI 4 -2NH 3
3.52
[Pt(N IT 3 ) 2 CI 4 ] (trans)
PtCI 4 -2NH 3
6.99
1 :1 Electrolytes
[Pi(NH 3 ) 2 C1 4 ] (cis)
NaCI
123.7
—
PtCl 4 -3NH }
96.8
[Pt(NH 3 ) 3 Cl 3 ]Cl
PtCl 4 NH 3 KCI
106.8
K[Pt(NH 3 )Cl 5 ]
1 : 2 and 2 : 1 Electrolytes
CaCI,
260.8
—
CoCI 3 -5NH 3
261.3
[Co(NH 3 ),CI]CU
CoBr,-5NH 3
Cr€l,'5NH,
CrBty5NH 3
257.6
|Co(NH,),Br|Br z
260.2
[Cr(NH 3 ),CI]CU
280.1
[Cr(NH,),BrlBr z
PtCI 4 -4NH 3
228.9
[Pt(NH,) 4 CU]CI 2
PtCI 4 -2KCI
256.8
K 2 |PtCi h ]
1 : 3 and 3 : 1 Electrolytes
LaCl 3
393.5
—
CoCI 3 -6NH 3
431.6
[Co(NH 3 )JCI 3
CoBr 3 -6NH 3
426.9
[Co(NH 3 )„)Br 3
CrCI 3 -6NH 3
PtCI 4 -5NH 3
441.7
[Cr(NH 3 )JCl 3
404
1 :4 Eloctrolytcs
[Pt(NH 3 ) 5 CI|CI 3
PtCI 4 -6NH 3
522.9
[Pt(NH,)„]CI 4
■' This is the molar conductivity measured at a concentration of
0.001 M. Values arc from Werner and Miolati, except for
[Pt(NHj) s CI]Clj. which is from Vladimirov, N.; Chugacv. L. A.
Compt. Rend. 1915, 160. 840.
The theoretical value is. of course, zero, but impurities or a
reaction with the solvent water could produce a small
concentration of ions.
4 Werner's terminology and symbolism differed in small, relatively unimportant ways Irom that used
today. For example. Werner referred to oxidation stale as “primary valence" (Hauptvulenz) and
coordination number as "secondary valence" (Nehenvalenz). Also, he wrote formulas as
{Co'^'lCU. instead of 1 Co(NH 3 KC1]CI 3 .
5 Werner. A.: Miolati. A. Z. Phys. Cliem. (Leipzig) 1893, 12. 35-55. Ibid. 1894. 14. 506-521.
340
9-Acid-Base Chemistry
Measures of Acid—Base Strength 341
Bond Energies
The nearest anionic analogue of H + is the F _ ion. Some of the calculations for
it are at first surprising, but parallel those of the proton and are acceptable
under closer scrutiny: (a) F“ forms a stronger covalent bond than Cl (> Br >
I )• 38 (b) F~ is a very strong base with a large transfer of electron density to
the acid. This is a result of the low charge capacity (low electron affinity) of
fluorine.
As with any successful model, exceptions call attention to themselves and
signal the existence of unusual effects: repulsions, 7r-bonding, or adduct geom¬
etry variation.
There are two ways to approach the formation of a polar bond —Y A . We have
already encountered both. One is to consider the formation of a nonpolar molecule.
X—Y, followed by an electronegativity-controlled shift of electron density from X
towards Y. Alternatively, one can form the ions X' and Y , followed by their
interaction. We can view the latter as the Fajans polarization of Y by X , or basic
attack of :Y~ upon X + . Whatever model is used, there occur, in one form or another,
three contributing energies:
The covalent energy , £,., arising from electron sharing. It is a maximum in a
homopolar bond and decreases with ionicity.
The Modelling energy , £ M , arising from the coulombic attraction of the partial
charges:
X s+ —Y s "
_ S + S~e 2
~ 4t rre 0
(9.67)
This energy is termed the Madelung energy since it represents a "lattice
energy” internal to the molecule with a Madelung constant, of course, equal to
1.00. It is a maximum in a purely ionic bond (6 = Z + ) and decreases to
whatever extent the charges on X and Y decrease.
The electronegativity energy, £ x , or IE-EA energy arising from ionization
energy-electron affinity terms in the total energy sum. It is a more complex
function than £ M and E c but it will be clarified by some examples below«
Consider Pauling's ionic resonance energy of a principally covalent bond with a
little ionic character from the difference in electronegativity. To a first approximation,
the ionic resonance energy may be equated with the sum of the Madelung energy, E M ,
and the electronegativity energy, £ x , which stabilizes the XY molecule more than the
small loss of covalent energy destabilizes it. The simplest example of these three terms
has already been encountered in the form of the Bom-Haber cycle and the
Bom-Land6 Equation (taken only to the point of isolated gas-phase molecules with a
Madelung constant of 1.00). The equation in that form shows that the gas-phase ion
pairs (ionic molecules) are stabilized by the Madelung energy holding them together
and are destabilized by the ionization energy-electron affinity energy that had to be
paid to form the ions. Note that if we brought two ions together (acid-base reaction)
38 This is merely the consequence of smaller atoms having better overlap. The usual prejudice that
fluorine cannot bond covalently arises from the unfortunate tendency to overemphasize the differ¬
ences between covalent and ionic bonding. Bonding is too often characterized as covalent or ionic,
rather than possibly covalent and ionic.
59 See also Evans, R. S.: Huheey. J. E. Chem. Phys. Lett. 1973. 19. 114-116.
Steric Effects
and they were not completely ionic (no compounds are), electron density would flow
from the anion (base) to the cation (acid); there would be some loss of Madelung
energy (the charges decrease) but a stabilization of the formerly unfavorable £, E _ EA
(£.) as the metal flows down the energetically steep part of curve Fig. 5.34a (A + ) and
the nonmetal flows up (in the case of F _ . initially down) the relatively mild slope of
Fig. 5.34b (B~). To say it another way Fig. 5.35 will be approximated, no matter
whether we approach equilibrium from A- + B or A T + :B . Finally, inherent in the
bond A—B is the covalent energy term arising from the overlap of orbitals, whether it
be from the covalent bond picture or polarization of ionic species.
We can now examine in further detail the £, C, R, and 7 parameters of Drago's
system. We have seen that £ A £„ and C A C B terms indicate tendencies to form
electrostatic (£ M ) or covalent (£,.) acid-base interactions. Finally, the R A T B term
provides a measure of E x which can be clearly shown to be related to an IE-EA energy
(see page 339). Note, however, that in contrast to the ion-pair example given above,
where £, E _ EA (£ ) was destabilizing, in covalent acid-base reactions of ionic species,
the £ term will be strongly stabilizing. This is especially true of species such as H f
and F", in accordance with the electronegativity argument given above, and the R A T B
term will be a major contributor to the stabilization of the donor-acceptor bond.
An example of the importance of the energy associated with this transfer of
electron density is discussed by Drago. Because of the very large value of electron
transference energy (resulting from the high ionization energy of hydrogen), the
energy associated with the gas-phase attachment of H + to bases (proton affinity) is
unique; other acids, including the aqueous hydronium ion, are different. This is
because the bare proton has an extremely high charge/size ratio and releases an
extremely large amount of £ x upon adduct formation {R\j• = 130.21) compared to all
other acids that have several atoms over which to delocalize the cationic charge. I his
is analogous to the b parameter of Mulliken-Jafte electronegativity (see Chapter 5),
which measures the capacity of a group to "soak up" charge and stabilize it. Thus
methyl groups can stabilize charge on a cation (have a low electronegativity b term)
for exactly the same reason that as ions they have lower R A values than the proton
(/W = 55.09).
In reactions between Lewis acids and bases such as amines and boranes or boron
halides, bulky substituents on one or both species can affect the stability of the
acid-base adduct. Perhaps the most straightforward type of effect is simple steric
hindrance between substituents on the nitrogen atom and similar large substituents on
the boron atom. Figure 9.3 is a diagrammatic sketch of the adduct between molecules
of tripropylamine and triethylborane. This phenomenon is known as front or "F-
Fig. 9.3 Tripropylamine-
triclhylborane adduct
illustrating steric hindrance
between the bulky sub¬
stituents on the nitrogen
and the boron.
386
10- Chemistry in Aqueous and Nonaqueous Solvents
c. when solutions (a) or (b) are evaporated carefully in vacuo.
d. when (a) is treated with Fe 2 Oj.
How can (d) be considered a leveling reaction?
10.10 Consider each of the following solvents individually: (I) ammonia, (2) acetic acid, (3)
sulfuric acid.
a. Give equations for autoionization of the pure solvent.
b. Discuss what will happen if CHjCOOH is dissolved in each of the solvents, that is.
what ions will form. Give appropriate equations. Will the solution be acidic or basic
with respect to the pure solvent? Will the solute act as a weak or a strong acid (base)?
c. Give an example of a strong base, a strong acid, and a neutralization reaction.
10.11 As a working hypothesis, assume that you accept the solvent system picture of OPCIj and
a value of 5 x I(T 14 mol 2 L~ 2 for the ion product. Set up a pCI scale for OPCIj. draw the
equivalent of Fig. 10.1 for it, and discuss how you would go about obtaining data for
compounds to complete your diagram.
10.12 The stability constant, K, for Au(CN)? is defined as f AUl 5’,‘l >: i, .
[Au ][CN ]
a. From the £° of +0.60 V for Eq. 10.131 estimate K.
b. Qualitatively describe why this complex is so stable.
10.13 Correlate the behavior of various solutes in "superacids" with their gas-phase proton
affinities. What factors besides proton affinities affect their solution chemistry? Predict
what species will be present when XH, (Group VA. 15). H : X (Group VIA, 16) and HX
(Group VUA, 17) are dissolved in "superacids.“ 3J
10.14 Single-crystal “cesium electridc" is almost entirely diamagnetic. Reconcile this with the
formulation [Cs(ligand)f e~. Is there a paradox here?
10.15 On page 372, HgClj is mentioned as an exception to the obviously intuitive rule that
“good acceptors should not be good donors, and vice versa." Can you suggest a reason
why Hg(ll) might be paradoxical?
10.16 Suggest equilibria for the redox chemistry at an aluminum electrode and show how the
potential can be related to the (Cl - ] (page 374).
10.17 On page 376 it was stated that one of the difficulties with the cell described there was
diffusion of CuCN. Explain.
10.18 Is the diffusion discussed on page 376 and in Problem 10. 17 a fatal flaw? (Hint: Recall what
you know from general chemistry about a simple aqueous cell: Zn | Zn 2 || Cu 2 | Cu.)
34 Olah. G. A.; Shen, J. J. Am. Client. Soc. 1973, 95. 3582-3584.
Coordination Chemistry:
Bonding, Spectra,
and Magnetism
Coordination compounds have been a challenge to the inorganic chemist since they
were identified in the nineteenth century. In the early days they seemed unusual
because they appeared to defy the usual rules of valence (hence the label "complex”
compounds). Today they comprise a large body of current inorganic research. Al¬
though the usjtal bonding theories can be extended to accommodate these compounds,
they still present stimulating theoretical problems and in the laboratory they continue
to provide synthetic challenges. One class of coordination compounds, those involv¬
ing metal-carbon bonds, is the focus of an entire subdiscipline known as
organometallic chemistry (Chapter 15), and the field of bioinorganic chemistry (Chap¬
ter 19) is centered on coordination compounds present in living systems.
The modern study of coordination compounds began with two men, Alfred
Werner and Sophus Mads Jprgensen. Both were astute chemists, not only in labora¬
tory aspects but also in the areas of interpretation and theory. As it turned out, they
differed fundamentally in their interpretation of the phenomena they observed and
thus served as protagonists, each spurring the other to perform further experiments to
augment the evidence for his point of view. From our vantage point nearly a century
later, we can conclude that Werner was "right" and Jprgensen was "wrong" in the
interpretation of the experimental evidence they had. Indeed, Werner was the first
inorganic chemist to be awarded the Nobel Prize in chemistry (1913). 1 Nevertheless,
Jdrgcnsen's contributions should not be slighted—as an experimentalist he was sec-
1 After Alfred Werner won the prize, it long appeared that he might be the only inorganic chemist to
receive it. Then sixty years later in 1973, Geoffrey Wilkinson and E. O. Fischer shared the Nobel
Prize in chemistry for their work on ferrocene (see Chapter 15). (For a personal account from the
American side of the Atlantic, see Wilkinson, G. J. Or/fanomet. Client. 1975, 100, 273-278.)
However, in the past couple of decades, a number of prizes have been awarded for inorganic
chemistry or closely related work: William Lipscomb (1967) and H. C. Brown (1979) for their
theoretical and synthetic work in boranc chemistry (sec Chapter 16): Roald Hoffmann (1981) for his
theoretical work in organometallic chemistry (see Chapter 15); and Henry Taubc (1983) for his
research in inorganic kinetics (sec Chapter 13).
387
342
9-Acid—Base Chemistry
Strain" anti can have a considerable influence on the stability of the adduct since the
alkyl groups tend to sweep out large volumes as they rotate randomly.
A second, similar effect is known as back or "B-strain." It results from the
structural necessity for the nitrogen atom in amines to be approximately tetrahedral
(tfp 3 ) in order to bond effectively through its lone pair. If the alkyl groups on the
nitrogen atom are sufficiently bulky, presumably they can force the bond angles of the
amine to open up, causing more s character to be used in these bonds and more p
character to be left in the lone pair. The extreme result of this wouid be the formation
of a planar, trigonal molecule with a lone pair in a pure p orbital, poorly suited for
donation to an acid (Figure 9.4).
Related to B-strain, but less well understood, is “I-strain” (for internal strain). In
cyclic amines and ethers, such as (CH,)„0, the basicity varies with ring size. In such
compounds the hybridization (and hence the overlapping ability and electronegativity)
of not only the basic center (N, O, etc.), but also of the carbon atoms in the ring will
vary with ring size, and there are no simple rules for predicting the results.
When the basic center is exocyclic, as in lactams, lactones, etc., the results can be
interpreted in a straightforward way analogous to the argument given previously
(Chapter 5) for biphenylene. Consider the series of lactams:
O
As the ring size is reduced, the internal bond angles must reduce, and the hybridiza¬
tion of the cyclic atoms must have less s character and lower electronegativity.
Toward the exocyclic oxygen atom, the basic center,-' 0 the cyclic carbon atom must in
turn exhibit greater s character and a higher electronegativity. The carbonyl groups in
small ring compounds are therefore less basic.-"
"Proton Sponges" Although steric effects and strain always work against basicity in simple molecules
and monocyclic compounds, there are a few compounds in which the steric and strain
effects stabilize the H f adduct, increasing the basicity. An example is the 1,8-
bis(dimethylamino)naphthalene molecule in which steric hindrance of the methyl
groups and repulsions of lone pairs on the nitrogen destabilize the free base:
(») (b) to
Fig. 9.4 B-strain in substituted amines: (a) small substituents, no strain, good base;
(b) moderate strain from intermediate-sized substituents, some rehybridization: (c) extreme
bulkiness of substituents, nitrogen atom forced into planar, sp- + p hybridization, weak
base.
411 Sec Problem 9.16.
41 Filgueiras, C. A. L.; Huheey. J. E. J. Org. Client. 1976, 41. 49-53.
Measures of Acid-Base Strength
343
Solvation Effects
and Acid-Base
"Anomalies" 43
Such compounds are very basic (pA b = 1.9; cf. NH 3 , pA b = 4.74; l-dimethyl-
aminonapthalene. pAj, = 4.57). and have been nicknamed “proton sponges” from
their avidity for hydrogen ions. 42 The strong, symmetric N —H —N hydrogen bond
(see Chapter 8) stabilizes (he conjugate acid. Note, however, that a second proton
cannot be added without incurring the original steric problem: Diprotonation is only
half complete in 86% sulfuric acid!
Paradoxically, data on acid-base behavior in aqueous solution have been collected
over a longer period than for any other system, yet until recently have been under¬
stood less. This is partially due to the fact that theories had been constructed to
account for the aqueous data without knowledge of acid-base behavior in the absence
or near absence of solvation effects. For example, F-strain and B-strain have been
invoked to account for the anomaly of the pA b of trimethylamine when compared to
the less substituted amines and ammonia (page 329). We now know that when the
basicities of methylamines are measured in the gas phase, they increase regularly NH,
< CHjNH, < (CHj) 2 NH < (CH ,),N (see Table 9.5). Therefore the “anomaly” of the
basicity of trimethylamines must lie in some solution effect. Solvation through hydro¬
gen bonding will tend to increase the apparent strength of all amines because the
positively charged ammonium ions will be more extensively solvated (A// ten to one
hundred times larger) than the uncharged amine.
R
I /H
RjN + 2H,0 = OH- + R —N —H--0 (9.69)
I H
R
H H
\ /
O
H
I /H
RNH 2 + 4H,0 = OH- + R — N + — H — O (9.70)
I ""H
H
A
H H
42 For a review of steric effects and basicity, including cyclic compounds and proton sponges, see
Alder. R. W. Cltem. Rev. 1989. 89. 1215-1223. Note that many proton sponges have low solubilities
in pure water.
45 Arnett. E. M . J. Client. Educ. 1985. 62. 385-391. For a longer discussion than given here and more
examples, see Lowry, T. H.; Richardson, K. S. Mechanism anti Theory in Organic Chemistry. 3rd
ed.; Harper & Row: New York. 1987; pp 296-316.
Table 10.6
A comparison of Pyrometallurgy Hydrometallurgy _
pyrometallurgy and Energy Because high temperatures Because low temperatures are
hydrometa urgy a consumption (about 1500 °C) are involved, involved in dissolution pro¬
reaction rates are high but cesses, they require little
much energy is consumed. energy, although reaction
Heat recovery systems are rates are slow. However, a
needed to make the process requirement for electrowin-
economical. Heat can be re- ning or for cleaning effluents
covered readily from hot and recovering reagents may
gases (although the equip- more than offset this energy
ment needed is bulky and advantage,
expensive), but is rarely re¬
covered from molten slag or
metal, so that a great deal of
energy is lost.
Dust Most processes emit large No problem, because materials
amounts of dust, which must handled usually are wet.
be recovered to abate pollu¬
tion or because the dust itself
contains valuable metals;
equipment for dust recovery
is bulky and expensive.
Toxic gases Many processes generate toxic Many processes do not generate
gases, so that reactors must gases, and if they do, reactors
be gas-tight and the gases re- can be made gas-tight easily,
moved by scrubbers or other
systems; this is expensive, es¬
pecially when the gases are
hot and corrosive.
Solid residues Many residues, such as slags. Most residues are finely divided
are coarse and harmless, so solids that, when dry, create
that they can be stored in ex- dust problems and, when wet,
posed piles without danger of gradually release metal ions
dissolution, although the piles in solution that may contami-
may be esthetically nate the environment,
unacceptable.
Treatment of Sulfur dioxide is generated. Ores can be treated without
sulfide ores which in high concentrations generating sulfur dioxide,
must be converted to sulfuric eliminating the need to make
acid (for which a market must and market sulfuric acid;
be found) and in low concen- sulfide sulfur can be re-
trations must be disposed in covered in elemental form,
other ways (available but
expensive).
Treatment of Unsuitable because separation Suitable,
complex ores is difficult.
Treatment of Unsuitable because large Suitable if a selective leaching
low-grade ores amounts of energy are re- agent can be used,
quired to melt gangue
materials.
Economics Best suited for large-scale oper- Can be used for small-scale op-
ations requiring a large capital erations requiring a low
investment. capital investment.
" From Habashi. F. Chem. Eng. News. 1982.60(6). 46. Used with permission.
384
Problems
385
Problems 10.1 Suggest the specific chemical and physical interactions responsible for the reversal of
-Eqs. 10.3 and 10.4 in water and ammonia solutions.
10.2 Using a Bom-Haber cycle employing the various energies contributing to the formation of
M + . e(NH 3 ).7 species in ammonia solutions, explain why such solutions form only with
the most active metals.
10.3 When I mole of N 2 0, is dissolved in sulfuric acid, 3 equivalents of base are produced.
Conductivity studies indicate that v = 6 for N 2 0 5 . Propose an equation representing the
solvolysis of N 2 0 5 by sulfuric acid.
10.4 What is the strongest acid listed in Fig. 10.1? The strongest base?
10.5 From Fig. 10.1 determine how the following solutes will react with the solvents, and how
the equilibria will lie, that is. will the solute be completely leveled or in equilibrium? State
whether the solution formed in each case will be more acidic or more basic than the pure
solvent.
Solute
Solvent
H,SO,
Acetic acid
h,so 4
Water
h,so 4
Ammonia
CH 3 C(0)CH 3
Ammonia
CH 3 C(0)CH 3
Water
PhNH -
Ammonia
PhNH"
Water
PhNH”
Acetic Acid
PhNH'
Sulfuric acid
10.6 Construct the Latimer diagram for manganese in basic solution (from values in Table F.l).
and predict which oxidation states will be stable. Explain the source of instability for each
unstable species.
10.7 Calculate the potential for the oxidation of U0 2 to U0 2 in acid solution trom the
following information.
U0 2 + e - U0 2 E° = 0.66 V
u0 :+ +. c - -► U0 2 E" = 0.16 V
10.8 Use the Latimer diagram for plutonium in acid solution below to answer the following
questions.
1.25
a. Would you expect plutonium metal to react with water?
b. Pu 4 is stable in concentrated acid but disproporlionates to PuO; and Pu" at low
acidities. Explain. .
c. Pu0 2 tends to disproportionate to Pu' 1 and PuCK'. Under what pH conditions would
this reaction be least likely to occur?
10.9 With equations and words describe what happens
a. when metallic potassium is dissolved in ammonia to form a dilute solution.
b. when more potassium is added to form concentrated solutions.
344
9-Acid-Base Chemistry
Hence Ihe basicity of the amines is enhanced in proportion to the extent of solvation of
the conjugate ammonium ion, and the energies of solvation are RNH^ > R 2 NH 2 >
R,NH + . This is the reverse order of increase in basicity that results from electronic
(inductive) effects. Two opposing, nonlinear trends will give a maximum or a mini¬
mum. Therefore it is not surprising to find a maximum in basicity (as measured in
aqueous solutions) for the dialkylamines. When these reactions are analyzed by a
Born-Haber-type cycle (see Problem 9.28) the effect of solvation can readily be seen.
When each hydrogen bonding positive atom on an ammonium ion is replaced by a non¬
hydrogen bonding alkyl group, the ion loses about 30 kJ mol 1 of hydration energy. 43
Many of the “anomalies" are historical artifacts: Accurate experimental data
for species in solution had been accumulated for decades, and corresponding theories
had been proposed long before the first gas-phase data were collected. For exam¬
ple. it has been found that the acidity of water and alcohols goes in the order H 2 0 >
R(l°)OH > R(2°)OH > R(3°)OH with the “explanation" being that the electron¬
releasing alkyl groups force electron density onto the oxygen of the conjugate base
making it more basic. But note that the electronegativities of branched and un¬
branched alkyl groups are practically identical, and if there is any trend, those groups
having more carbon atoms are slightly more electronegative: Me = 2.30, Et = 2.32,
/-Pr = 2.34, /-Bu = 2.36 (see Table 5.7). However, these groups differ significantly in
their charge capacity (Chapter 5). Thus highly branched groups are both better donors
(when attached to electronegative centers) and better acceptors (when attached to
electropositive centers). Seemingly paradoxically (but refer again to Figs. 2.13 and
5.34) O - is electropositive: The oxygen atom will be stabilized if the anionic charge is
delocalized. This can best be accomplished by groups with larger charge capacities.
Relative to the hydrogen atom (1.0). the charge capacities of groups are Me = 2.8,
Et = 3.4, i-Pr = 3.9, and /-Bu = 4.2. The net result is that in gas-phase reactions with
no complicating solvation energies, the order of basicity is OH > R(l°)0 >
R(2°)0“ > R(3°)0 - .^
So why the reversal of basicity as one proceeds from the gas phase to solution?
Once again, solvation effects overcome inherent electronic effects. As in the case of
amines, hydrogen bonding is the predominant factor, and as the organic portion of the
ion grows, it becomes increasingly like a ball of wax. The anion loses the special
solvation stability normally enjoyed over neutral molecules and thus more readily
accepts a proton. The enhanced basicity of the /-butoxide ion arises not because the
electron density on the oxygen is higher (it is lower). 45 but because the anion lacks
stabilizing solvation.
Hard and Soft
Acids and Bases
44 Brauman, J. I.; Blair, L. K. J. Am. Chem. Soc. 1968. 90. 6561-6562; 1970. 92, 5986-5992.
« Baird. N. C. Can. J. Chem. 1969. 47. 2306-2307: Lewis. T. P. Tetrahedron 1969. 25. 4117-4126;
Huheey. J. E. J. Org. Chem. 1971, 36, 204-205.
46 Irving. H.: Williams, R. J. P. J. Chem. Soc. 1953. 3192-3210.
For some time coordination chemists were aware of certain trends in the stability of
metal complexes. One of the earliest correlations was the Irving-Williams series of
stability. 46 For a given ligand, the stability of complexes with dipositive metal ions
follows the order: Ba 24 < Sr 24 < Ca 2+ < Mg 24 < Mn 2+ < Fe 2+ < Co 24 < Ni 24 <
Cu 2+ > Zn 24 . This order arises in part from a decrease in size across the series and in
Hard and Soft Acids and Bases
345
Classification of
Acids and Bases as
Hard or Soft
part from ligand field effects (Chapter 11). A second observation is that certain ligands
form their most stable complexes with metal ions such as Ag 4 , Hg 2+ , and Pt 2+ . but
other ligands seem to prefer ions such as Al' 1 , Ti 41 , and Co 1 . 47 Ligands and metal
ions were classified 4 as belonging to type (a) or {b) w according to their preferential
bonding. Class (a) metal ions include those of alkali metals, alkaline earth metals, and
lighter transition metals in higher oxidation states such asTi 4+ , Cr ‘, Fe 34 , Co 3 ‘ and
the hydrogen ion. H 4 . Class ( b) metal ions include those of the heavier transition
metals and those in lower oxidation states such as Cu 4 , Ag 4 , Hg 4 , Hg- 4 , Pd~ + , and
p t 2+ .5o According to their preferences toward either class (a) or class ( b) metal ions,
ligands may be classified as type (a) or (/>), respectively. Stability of these complexes
may be summarized as follows:
Tendency to complex with Tendency to complex with
class (a) metal ions class (b) metal ions
N » P > As > Sb N « P > As > Sb
O » S > Se > Te O « S < Se ~ Te
F > Cl > Br > I F < Cl < Br < 1
For example, phosphines (R,P) and thioethers (R 2 S) have a much greater tendency to
coordinate with Hg 2+ , Pd 24 , and Pt 24 , but ammonia, amines (R 3 N), water, and
fluoride ions prefer Be 24 , Ti 44 , and Co 34 . Such a classification has proved very useful
in accounting for and predicting the stability of coordination compounds.
Pearson 51 suggested the terms “hard” and "soft" to describe the members of
class (a) and (b). Thus a hard acid is a type (a) metal ion and a hard base is a ligand
such as ammonia or the fluoride ion. Conversely, a soft acid is a type (b) metal ion and
a soft base is a ligand such as a phosphine or the iodide ion. A thorough discussion ol
the factors operating in hard and soft interactions will be postponed temporarily, but it
may be noted now that the hard species, both acids and bases, tend to be small,
slightly polarizable species and that soft acids and bases tend to be larger and more
polarizable. Pearson suggested a simple rule (sometimes called Pearson's principle)
for predicting the stability of complexes formed between acids and bases: Hard acids
prefer to bind to hard bases and soft acids prefer to bind to soft bases. It should be
noted that this statement is not an explanation or a theory, but a simple rule of thumb
which enables the user to predict qualitatively the relative stability of acid-base
adducts.
In addition to the (a) and ( b) species discussed above that provide Hie nucleus for a set
of hard and soft acids and bases, il is possible to classify any given acid or base as hard
or soft by its apparent preference for hard or soft reactants. For example, a given
47 The existence of isolated ions of high charge such as Ti 44 in chemical systems is energetically
unfavorable. Nevertheless, complexes exist with these elements in high formal oxidation states.
4 " Ahrland, S.; Chatt, J.; Davies. N. R. Quart. Rev. Chem. Soc. 1958, 12, 265-276. See also
Schwarzcnbach, G. Experientia Suppl. 1956, 5, 162.
4tf The la) and (6) symbolism is arbitrary. Sometimes the symbols A and B arc used. Neither should be
confused with the A and B subgroups of the periodic table or A and B as generic representations of
acids and bases.
so Only a limited number of examples of class (u) and (6) metal ions is given here. A more complete
listing is provided in Table 9.7.
5t Pearson. R. G. J. Am. Chem. Soc. 1963. 85, 3533-3539,
382
10- Chemistry in Aqueous and Nonaqueous Solvents
Hydrometallurgy
A single example of the application of electrode potentials to chemistry in am¬
monia will suffice. The Latimer diagram for mercury in acidic solution is
and for the insoluble mercury(I) iodide the diagram is
Hg
It may readily be seen that the mercurous ion (whether free or in Hg-,1-,) is ther¬
modynamically unstable with respect to disproportionation in ammonia, in contrast to
its stability in water.
Electrochemistry in nonaqueous solvents is not merely a laboratory curiosity. We
have already seen batteries made with solid electrolytes (sodium beta alumina, see
Chapter 7) that are certainly “nonaqueous." In looking for high-efficiency cells one
desires the cathode and anode to be highly reactive (large positive emf) and to have a
low equivalent weight. In these terms, lithium appears to be highly desirable. Its very
reactivity, however, precludes its use in aqueous systems or even liquid ammonia.
One successful battery utilizing lithium has been developed using sulfur dioxide or
thionyl chloride (OSCL) as solvent and oxidant. Others involve weight-efficient
lithium metal with other oxidants and solvents.32 Highly efficient batteries of this sort
are widely used in specialized applications where light weight and long life are
important.
Traditionally the winning of metals from their ores has been achieved by
pyrometallurgy: the reduction of relatively concentrated metallic ores at high tem-
P®^ tures - The reactions of the blast furnace form a typical example (see also page
Fc 2 0 3 + 3CO -► 2Fe + 3C0 2 (10.127)
C0 2 + C -► 2CO (10.128)
Carbon monoxide for the reduction of the iron is formed not only from the recycling of
carbon dioxide (Eq. 10.128) but also from the direct oxidation of the coke in the charge
by hot air:
2C + 0 2 -► 2CO (10.129)
The energy released by the combustion is sufficient to raise the temperature well
above the melting point of iron, 1535 °C. One of the incentives for development of
alternative methods of producing metals is the hope of finding less energy-intensive
processes.
Hydrometallurgy is not new; it has been used for almost a century in the separa¬
tion ol gold from low-grade ores. This process is typical of the methods used. Gold is
normally a very unreactive metal:
Au -♦ Au + + e" E° = -1.69 V (10.130)
52 Jones. K. J.; Hatch, E. S.. Jr. lnd. Re. r. Dev. Feb. 1982. 24. 182; Mar. 1982. 24. 89.
Problems
383
With such a negative oxidation emf, it is too noble to react with either 0-, ( E° =
+ 1.185 V) or CL (£° = +1.36 V). By complexation of the Au(I) ion, however, the
emf can be shifted until it is much more favorable:
Au + 2CN~ - Au(CN),- + e“ £° = +0.60 V (10.131)
Oxygen in the air is now a sufficiently strong (and cheap!) oxidizing agent to effect the
solution of the gold. It may then be reduced and precipitated by an active metal such
as zinc powder (£° = -0.763 V). Such hydrometallurgical processes offer definite
advantages:
Low-grade ores may be leached, with complexing agents if necessary, and
profitably exploited.
Complex ores may be successfully treated and multiple metals separated under
more carefully controlled processes.
Since the reactions are carried out at room temperature, energy savings are
possible.
Because no stack gases are involved, air pollution does not present the prob¬
lem faced by pyrometallurgy.
These aspects do not form an unmixed blessing, however. If the metal must be re¬
duced by electrolysis, the process may become energy intensive. Thus attractive
solutions to this problem are reduction of more valuable gold by less expensive zinc
and of more valuable copper by scrap iron. Finally, in view of the large amounts of
waste water formed as by-product, one may be trading an air pollution problem for a
water pollution problem. A comparison of the two types of processes is given in Table
10 . 6 .
Related hydrometallurgic methods may allow the use of bacteria to release copper
from low-grade ores, or the use of aigae to concentrate precious metals such as gold
(see Chapter 19 ). 33
33 Bricrlcy. C. L. Sci. Amer. 1982. 247(2). 44-53. Darnall, D. W.; Greene. B.; Henzl. M. T.; Hosea, J.
M.; McPherson, R. A.; Sneddon, J.; Alexander, M. D. Environ. Sci. Technol. 1986, 20. 206-208.
346
Hard and Soft Acids and Bases
347
9> Acid-Base Chemistry
base, B, may be classified as hard or soft by the behavior of the following
equilibrium : 52
BH + + CH 3 Hg + CH,HgB + H + (9.71)
In this competition between a hard acid (H) and a soft acid (CH 3 Hg), a hard base
will cause the reaction to go to the left, but a soft base will cause the reaction to
proceed to the right . 55 The methylmercury cation is convenient to use because it is a
typical soft acid and, being monovalent like the proton, simplifies the treatment of the
equilibria.
An important point to remember in considering the information in Table 9.7 is that
the terms hard and soft are relative, with no sharp dividing line between them. This is
illustrated in part by the third category, ‘•borderline,’' for both acids and bases. But
even within a group of hard or soft, not all will have equivalent hardness or softness.
Thus, although all alkali metal ions are hard, the larger, more polarizable cesium ion
will be somewhat softer than the lithium ion. Similarly, although nitrogen is usually
hard because of its small size, the presence of polarizable substituents can affect its
behavior. Pyridine, for example, is sufficiently softer than ammonia to be considered
borderline.
Acid-Base Strength Hardness and softness refer to special stability of hard-hard and soft-soft interactions
and Hardness and and should be carefully distinguished from inherent acid or base strength. Forexam-
Softness pl e - OH and F are hard bases; yet the basicity of the hydroxide ion is about
•O ' 3 times that of the fluoride ion. Similarly, both SO 2 " and Et 3 P may be considered
soft bases; however, the latter is 10 7 times as strong (toward CH 3 Hg + ). It is possible
for a strong acid or base to displace a weaker one, even though this appears to violate
the principle of hard and soft acids and bases. For example, the stronger, softer base,
the sulfite ion, can displace the weak, hard base, fluoride ion. from the hard acid, the
proton, H:
SO 7 " + HF — HSOJ + F" K eq = I0 4 (9.72)
Likewise the very strong, hard base, hydroxide ion, can displace the weaker soft
base, sulfite ion, from the soft acid, methylmercury cation:
OH" + CH 3 HgSOJ =: CH,HgOH + SOj" K cq = 10 (9.73)
In these cases the strengths of the bases (SO}" > F". Eq. 9.72: OH" > SO 5 ", Eq.
9.73), are sufficient to force these reactions to the right in spite of hard-soft considera¬
tions. Nevertheless, if a competitive situation is set up in which both strength and
hardness-softness are considered, the hard-soft rule works:
CH,HgF +
HSOf
CH,HgSO, + HF K cq ~ I0 3
(9.74)
Soft-hard
Hard-soft
Soft-soft Hard-hard
CH,HgOH +
HSO," ?=;
CH,HgSOJ + HOH K cq > I0 7
(9.75)
52 If this equilibrium is studied in aqueous solution, as is usually the case, the acids will occur as
CH,Hg(H,0) and H,0 with additional waters of hydration. For data on equilibria of this type,
sec Schwarzenbach, G.; Schcllenberg, M. Helv. Chim. Acta 1965, 48. 28.
53 An interesting historical sidelight on this type of soft-soft interaction is the origin of the name
"mercaptan." a mercury capturer: Hg 2 + 2RSH -♦ Hg(SR), + 2H.
Table 9.7 _
Hard and soft acids and
bases
Acids
Hard acids
H + , Li. Na + . K(Rb, Cs)
Be 2+ . Be(CH,),, Mg 2 . Ca 2 , Sr(Ba 2 )
Sc 3 , La 3 , Ce 4 , Gd 3 , Lu 3 , Th 4+ . U 4 , UO 2 ,, Pu 4
Ti 4 , Zr 4 . Hf 4 , VO 2 , Cr 3 , Cr 6 , MoO 3 , WO, Mn 2+ , Mn 7+ , Fe 3 ,
Co 3
BF 3 , BCI 3 , B(OR)„ Al 3 . AI(CH,),. AICI,. AIH,, Ga 3 , In 3
CO,, RCO, NC, Si 4 . Sn 4 , CH,Sn 3 .' (CH,j,Sn 2
N 3 . RPO, ROPOr, As 3
SO„ RSO?, ROSO^
Cl 3 , Cl 7 , I 5 , l 7 ‘
HX (hydrogen bonding molecules)
Borderline acids
Fe 2 . Co 2 , Ni 2 , Cu 2 , Zn 2+
Rh 3 , Ir 3 , Ru 3 . Os 2
B(CH 3 )„ GaH 3
RjC. C.HJ, Sn 2 , Pb 2
NO. Sb 4 '. Bi 3 ^
SO,
Soft acids
Co(CN)j~. Pd 2 . Pt 2 . Pt 4
Cu. Ag. Au. Cd 2 . Hg 2 . Hg 2 . CH,Hg
BHj, Ga(CH,)„ GaCI 3 , GaBr„ Gal,. Tl! TI(CH,),
CH 2 , carbencs
7 r-acceptors: trinitrobenzene, chloroanil, quinoncs. tetracyanoethylene, etc.
HO. RO. RS. RSe, Te 4 . RTe
Br„ Br. I,. I, ICN. etc.
O.'CI, Br, f. N, RO-, RO,-
M° (metal atoms) and bulk metals
Bases
Hard bases
NHj. RNH„ N,H 4
H,0. OH", O 2 ", ROH. RO". R,0
CH,COO", CO 5 ", NO". PO 3 ', SO 2 ", cio;
F"(CI") _
Borderline bases
C 6 HjNH 2 , CjHjN, NJ. N 2
NO,", SO 2 ,"
Br"'
Soft bases
H"
R", C 2 H 4 , C 6 H 6 , CN", RNC, CO
SCN", R,P, (RO),P, R,As
R,S, RSH, RS", S,0 2 "
I"
380
10- Chemistry in Aqueous and Nonaqueous Solvents
The Nernst equation applies to the potentials of both half-reactions and total
redox reactions:
E = E?-^\nQ (10-107)
nF
where £° represents the overall potential with all species at unit activity and Q
represents the reaction quotient.
Reactions resulting in a decrease in free energy (AG < 0) are spontaneous.
This is a requirement of the second law of thermodynamics. Concomitantly,
redox reactions in which E > 0 are therefore spontaneous.
In aqueous solutions two half-reactions are of special importance: (a) the
reduction of hydrogen in water or hydronium ions:
I M acid H 3 0 + + e" -► H 2 0 + JH 2 £° = 0.00 V (10.108)
Neutral solution H 2 0 + e _ - OH~ + 1H 2 E = —0.414 V (10.109)
1 M base H,0 + e“ - OH' + H, £° = -0.828 V (10.110)
and (b) the oxidation of oxygen in water or hydroxide ions:
I M acid H 2 0 -> -i0 2 + 2H+ + 2e“ £° = -1.229 V (10.111)
Neutral solution H 2 0 -♦ 10 2 + 2H + + 2e E = —0.815 V (10.112)
I M base 20H" -♦ J0 2 + H 2 0 + 2e' E° = -0.401 V (10.113)
These reactions limit the thermodynamic stability 28 of species in aqueous
solution.
6 . In calculating the “ skip-step emf" for a multivalent species it is necessary to
take into account the total change in free energy. Suppose we know the emfs
for the oxidation of Fe to Fe 2+ and Fe 2+ to Fe 3+ and wish to calculate the skip-
step emf for Fe to Fe 3+ :
Fe - Fc 2+
£° = 0.44
A0° = -2 x 0.44 x F
(10.114)
Fc 2+ - Fe 3+
£° = -0.77
AG° = -1 x -0.77 x F
(10.115)
Fe - Fe 3+
A C° = -0.11 F
(10.116)
£° = — = 0.1
11/3 = 0.037 V
(10.117)
Although the emfs are not additive, the free energies are. allowing simple
calculation of the overall emf for the three-electron change.
Standard potential or ‘‘Latimer” diagrams are useful for summarizing a con¬
siderable amount of thermodynamic information about the oxidation states of
an element in a convenient way. For example, the following half-reactions may
be taken from Table F.l. Appendix F:
28 As is always the case when dealing with thermodynamic stabilities, it must be bome in mind that a
species may possibly be thermodynamically unstable yet kinetically inert, that is. no mechanism of
low activation energy may exist for its decay.
Electrode Potentials and Electromotive Forces
381
Mn 2+ + 2 e _
- Mn
E°
=
-1.18 V
(10.118)
Mn 3+ + e"
- Mn 2+
E°
=
1.56 V
(10.119)
Mn0 2 + 4H + + e
-> Mn 3+ + 2H 2 0
£°
=
0.90 V
(10.120)
HMnOa + 3H + + 2e“
- Mn0 2 + 2H 2 0
£°
=
2.09 V
(10.121)
H + + MnO; + e“
-> HM 11 O 4
£°
=
0.90 V
(10.122)
MnO, + 4H + + 2e
-♦ Mn 2+ + 2H 2 0
£°
=
1.23 V
(10.123)
MnOJ + 4H + + 3e~
- Mn0 2 + 2H 2 0
E°
=
1.70 V
(10.124)
MnC >4 + 8 H + + 5e~
-♦ Mn :+ + 4H 2 0
£°
=
1.51 V
(10.125)
By omitting species such as H,0, H + , and OH , all of the above information
can be summarized as:
^•1.51
♦ 1.70 +1 -23
The highest oxidation state is listed on the left and the reduction emfs arc listed
between each species and the next reduced form, with the lowest oxidation
stale appearing on the right . 29
Flpctrochemistrv Although the entire discussion of electrochemistry thus far has been in terms ot
• M nnueniK aqueous solutions, the same principles apply equally well to nonaqueous solvents. As
m Nonaqueous ^ resu|t of differe nces in solvation energies, electrode potentials may vary cons.der-
Solul.ons ably [hose fQund jn aqueous so i u ti on . In addition the oxidation and reduction
potentials characteristic of the solvent vary with the chemical behavior of the solvent.
As a result of these two effects, it is often possible to carry out reactions in a
nonaqueous solvent that would be impossible in water. For example, both sodium and
beryllium are loo reactive to be electroplated from aqueous solution, but beryllium
can be electroplated from liquid ammonia and sodium from solutions in pyridine .-' 0
Unfortunately, the thermodynamic data necessary to construct complete tables of
standard potential values are lacking for most solvents other than water. Jolly - 11 has
compiled such a table for liquid ammonia. The hydrogen electrode is used as the
reference point to establish the scale as in water.
NH ; + e - -» }H 2 + NH 3 £° = 0.000 V (10.126)
» This convenlion originated with Latimer and is widespread in the inorganic chemical lileralurc.
Unfortunately. Latimer used oxidation emfs. and so his diagram is a mirror image of the one drawn
on the basis of reduction polcntials. This has resulted in a wide variety of modified ‘‘Latimer
diagrams.” often with no indication of ihe convention employed concerning the spontaneity of the
half-reaction. To avoid confusion, arrows (not present in the original Latimer diagram) arc recom¬
mended. Further discussion of Latimer diagrams and related topics may be found in Chapter 14.
jo parry. R. W.; Lyons. E. H., Jr. In The Chemistry of Coordination Compounds: Bailar. J. C.. Jr.,
Ed.; Van Nostrand-Rcinhold: New York. 1956; pp 669-671.
51 Jolly. W. L. J. Chem. Educ. 1956. 33. 512-517.
348
9-Acid-Base Chemistry
Table 9.8 lists the strengths of various bases toward the proton (H + ) and the methyl-
mercury cation (CH 3 Hg + ). Bases such as the sulfide ion (S 2 ~) and triethylphosphine
(Et 3 P) are very strong toward both the methylmercury ion and the proton, but about a
million times better toward the former; hence they are considered soft. The hydroxide
ion is a strong base toward both acids, but in this case about a million times better
toward the proton; hence it is hard. The aqueous fluoride ion, F~, is not a particularly
good base toward either acid but slightly better toward the proton as expected from its
hard character.
The importance of both inherent acidity and a hard-soft factor is well shown by
the Irving-Williams series and some oxygen, nitrogen, and sulfur chelates (Fig. 9.5).
The Irving-Williams series of increasing stability from Ba 2+ to Cu'' is a measure of
increasing inherent acidity of the metal (largely due to decreasing size). Superimposed
upon this is a hardness-softness factor in which the softer species coming later in the
series (greater number of d electrons, see page 350) favor ligands S > N > O. The
harder alkaline earth and early transition metal ions (few or no d electrons) preferen¬
tially bind in the order O > N > S.
As noted above, the hardness or softness of an acidic or basic site is not an inherent
property of the particular atom at that site, but can be influenced by the substituent
atoms. The addition of soft, polarizable substituents can soften an otherwise hard
center and the presence of electron-withdrawing substituents can reduce the softness
of a site. The acidic boron atom is borderline between hard and soft. Addition of three
hard, electronegative fluorine atoms hardens the boron and makes it a hard Lewis
acid. Conversely, addition of three soft, electropositive hydrogens 4 softens the boron
and makes it a soft Lewis acid. Examples of the difference in hardness of these two
boron acids are
R 2 SBF 3 + R 2 0 -► R 2 OBF 3 + R 2 S (9.76)
R 2 OBH 3 + R 2 S -► R 2 SBH 3 + R 2 0 (9.77)
In a similar manner, the hard BF 3 molecule will prefer to bond to another fluoride ion,
but the soft BH 3 acid will prefer a softer hydride ion:
BF 3 + F" -► BF 4 - (9-78)
B 2 H 6 + 2H- -► 2BH; (9.79)
Therefore, the following competitive reaction will proceed to the right:
bf 3 h _ + bh 3 f~-► bf 4 ~ + bh; < 9 - 8 °)
54 In a manner analogous to the usual treatment of balancing redox equations, it is necessary here to do
some careful "bookkeeping." Although this is merely a formalism, it is necessary to make certain
that the proper comparison is being made. In the present example, the formation of BFj is formally
considered to be B + 3F~. The three F~ ions harden the B 3 + . The analogous comparison is
B 3 + 3H" = BHj. In this case the soft hydride ions soften the B 3 . One must be careful to
distinguish between the small, hard proton <H) and the large (r_ = 208 pm), soft hydride ion (H ).
55 The simple BH 3 molecule does not exist in appreciable quantities, but always dimerizes to B 2 H h .
See Chapter 16.
Hard and Soft Acids and Bases
349
Table 9.8
Basicity toward the proton
and methylmercury cation
Base
Linking atom pK, a (CH 3 Hg + )
P Kh b (H "")
F~
F
1.50
2.85
cr
Cl
5.25
-7.0
Br"
Br
6.62
-9.0
r
I
8.60
-9.5
OH"
0
9.37
15.7
HPO 2 "
0
5.03
6.79
S 2 "
S
21.2
14.2
HOC,H 4 S"
S
16.12
9.52
SCN"
s
6.05
-4
SO 2 "
s
8.11
6.79
s 2 o 2 -
s
10.90
negative
NH 3
N
7.60
9.42
p-NH 2 C 6 H 4 S0 3
N
2.60
3.06
Ph 2 PC 6 H 4 S0 3 "
P
9.15
-0
Et 2 PC 2 H 4 OH
P
14.6
8.1
Et,P
P
15.0
8.8
CN"
C
14.1
9.14
“ p K, = log[CH 3 HgB]/[CH,Hg+]lBl
p K h = !og[HB]/[H+]lB]
Fig. 9.5 The Irving-
Williams effect: The
stability increases in the
series Ba-Cu. decreases
with Zn. (From Sigel, H.
McCormick. D. B. Acc\
Client. Res. 1970, 3, 201.
Reproduced with
permission.!
378
10 Chemistry in Aqueous and Nonaqueous Solvents
Electrode Potentials and Electromotive Forces
379
Solid Acid and Base
Catalysts
Electrode Potentials
and Electromotive
Forces
the concentration of the chloride ion in molten lithium chloride is about 35 M.
Furthermore, there are no other competing ligands (such as H 2 0) present to interfere.
As a result, it is possible to form not only complex ions that are well known in aqueous
solution:
CoCl 2 + 2C1" - CoCir (10.98)
but also those that cannot exist in aqueous solution because of their susceptibility to
hydrolysis:
FeC ! 2 + 2C1 - —
FeCI 2-
(10.99)
CrCl 3 + 3C1" —
— CrCir
( 10 . 100 )
TiCIj + 3C1" —
— TiCir
( 10 . 101 )
Some of these complexes are discussed further in Chapter 11.
While they are not solvents and solutions in the usual sense of the word, it is con¬
venient here to introduce the concept of solid acids and bases. For example, consider
the class of compounds known as zeolites. These are aluminosilicate structures with
variable amounts of AI(III), Si(IV), metal cations, and water (see Chapter 16).
Zeolites may behave as Lewis acids at Al 3+ sites, or as Br0nsted-Lowry acids by
means of absorbed H ions. Because they have relatively open structures, a variety of
small molecules may be accommodated within the —O—Al—0—Si— framework.
These molecules may then be catalyzed to react by the acidic centers. Coordinatively,
unsaturaled oxide ions can act as basic sites, and in some catalytic reactions both
types of centers are believed to be important. Catalysis by zeolites is discussed further
in Chapter 15.
Solid superacids may be made by treating ordinary solid acid catalysts with strong
Br0nsted or Lewis acids. For example, if freshly precipitated titanium hydroxide or
zirconium hydroxide is treated with sulfuric acid and calcined in air at 500 °C, a very
active solid acid catalyst results. The solids consist mainly of the metal dioxides with
sulfate ions coordinated to the metal ions on the surface. Likewise, a superacid solid
catalyst can be made by treating these metal oxides with antimony pentafluoride. Both
catalysts contain both Brpnsted and Lewis acid sites, and they are sufficiently active
to catalyze the isomerization of n-butane at room temperature . 26
As we have seen, acidity and basicity are intimately connected with electron transfer.
When the electron transfer involves an integral number of electrons it is customary to
refer to the process as a redox reaction. This is not the place for a thorough discussion
of the thermodynamics of electrochemistry; that may be found in any good textbook
of physical chemistry. Rather, we shall investigate the applications of electromotive
force (emf) of interest to the inorganic chemist. Nevertheless, a very brief review of
the conventions and thermodynamics of electrode potentials and half-reactions will be
presented.
26 Olah, G. A.; Prakash, G. K. S.; Sommer, J. Superacids: Wiley: New York, 1985; pp 53—61. See also
Tanabe, K.; Misono, M.: Ono, Y.; Hattori, H. New Solids. Acids and Bases ; Elsevier: Amslerdam.
1989.
The standard hydrogen electrode ( a H ■ = 1.00; / H = 1.00) is arbitrarily
assigned an electrode potential of 0.00 V.
If we construct a cell composed of a hydrogen electrode and a second electrode
(M" + /M) of metal M immersed in a solution of M' ,+ of unit activity, we can
measure the potential between the electrodes of the cell. Since the hydrogen
electrode was assigned a potential of 0.00 V, the potential of the electrode,
M n+ /M, is by definition the same as the measured potential of the cell. If the
metal electrode is positively charged with respect to the hydrogen electrode
(e.g., Cu 2+ /Cu), the electrode potential of the metal is assigned a positive sign
(^cirVcu = +0-337 V). If the metal tends to lose electrons more readily than
hydrogen and thus becomes negatively charged (e.g., Zn 2+ /Zn), the electrode
is assigned a negative sign (E zn -.- /Zn = -0.763 V). This convention is conve¬
nient in that it results in a single, invariant quantity for the electrode potential
for each electrode (the zinc electrode is always electrostatically negative
whether the reaction under consideration occurs in a galvanic cell or an
electrolytic cell). Most physical and inorganic chemists are more interested in
the thermodynamics of half-reactions rather than the electrostatic potential
that obtains in conjunction with the standard hydrogen electrode. The conven¬
tion related to thermodynamics may be termed the thermodynamic conven¬
tion. This convention assigns to the electromotive force (£) a sign such that
AG = -nFE (10.102)
where AG is the change in Gibbs free energy, n is the number of equivalents
reacting, and F is Faraday's constant, 96,485 coulombs equivalent -1 . It is
necessary to specify the direction in which a reaction is proceeding. Thus if we
consider the reaction
Zn + 2H + - Zn 2+ + H-, (10.103)
and find that for the reaction, as written, AG < 0, then (since H + /H-, is defined
as 0.00 V):
Zn -► Zn 2+ + 2e - E> 0 (10.104)
For the nonspontaneous reaction:
H 2 + Zn 2+ -» Zn + 2H - (10.105)
AG > 0, and so for
Zn 2+ + 2e - -► Zn E < 0 (10.106)
Accordingly, the sign of the emf of either a half-reaction (“electrode”) or the
overall redox reaction depends upon the direction in which the equation for the
reaction is written (as is true for any thermodynamic quantity such as enthalpy,
entropy, or free energy). The sign of the reduction electrode is always al¬
gebraically the same as that of the electrostatic potential. 27
27 In Ihc past the electrostatic convention has often been called the "European convention" and the
thermodynamic convention popularized by Latimer (The Oxidation Potentials of the Elements and
Their Values in Aqueous Solution ; Prentice-Hall: Englewood Cliffs, NJ, 1952) the “American
convention." In an effort to reduce confusion, the IUPAC adopted the "Stockholm convention" in
which electrode potentials refer to the electrostatic potential and emfs refer to the thermodynamic
quantity. Furthermore, the recommendation is that standard reduction potentials be listed as
"electrode potentials" to avoid the possibility of confusion over signs.
350
351
9"Acid-Base Chemistry
The isoelectronic fluorinated methanes behave in a similar manner:
CF 3 H + CH 3 F-► CF 4 + CH 4 (9.81)
J0rgensen 57 has referred to this tendency of fluoride ions to favor further coordination
by a fourth fluoride (the same is true for hydrides) as "symbiosis.” Although other
factors can work to oppose the symbiotic tendency, it has widespread effect in
inorganic chemistry and helps to explain the tendency for compounds to be sym¬
metrically substituted rather than to have mixed substituents. We have seen (Chap¬
ter 5) that the electrostatic stabilization of C—F bonds (ionic resonance energy) will
be maximized in CF 4 , and similar arguments can be made for maximizing hard-hard or
soft-soft interactions.
Theoretical Basis of Although the hard-soft rule is basically a pragmatic one allowing the prediction of
Hardness and chemical properties, it is of interest to investigate the theoretical basis of the effect. In
Softness this regard there is no complete unanimity among chemists concerning the relative
importance of the various possible factors that might affect the strength of hard-hard
and soft-soft interactions. Indeed, it is probable that the various factors may have
differing importance depending upon the particular situation.
A simple explanation for hard-hard interactions would be to consider them to be
primarily electrostatic or ionic interactions. Most of the typical hard acids and bases
are those that we might suppose to form ionic bonds such as Li + , Na + . K + . F", and
OH . Because the electrostatic or Madelung energy of an ion pair is inversely
proportional to the interatomic distance, the smaller the ions involved, the greater is
the attraction between the hard acid and base. Since an electrostatic explanation
cannot account for the apparent stability of soft-soft interactions (the Madelung
energy of a pair of large ions should be relatively small), it has been suggested that the
predominant factor here is a covalent one. This would correlate well for transition
metals, Ag, Hg, etc., since it is usually assumed that bonds such as Ag—Cl are
considerably more covalent than the corresponding ones of the alkali metals. In this
regard the polarizing power and the polarizability of d electrons becomes important. It
has been pointed out that all really soft acids are transition metals with six or more d
electrons, with the d 10 configuration (Ag 1- , Hg-) being extremely good . 58 From this
point of view the polarization effects in soft-soft interactions resemble in some ways
the ideas of Fajans (Chapter 4). although there are notable differences.
In general, species having relatively high electronegativities arc hard and those having
low electronegativities are soft. In this regard it should be recalled that we are
considering ions and that although Li. for example, has a low electronegativity, the
Li + ion has a relatively high electronegativity resulting from the extremely high
second ionization potential. In contrast, transition metals in low oxidation states
(Cu , Ag + . etc.) have relatively low ionization energies and low electronegativities.
The same may be said of hard and soft bases. This relation between hardness and
Electronegativity
and Hardness and
Softness
Equation 9,81 is not meant to imply that a mixture of trifluoromethane anti fluoromcthanc would
react to lorm letrafiuoromelhanc and methane, although the reaction would be exothermic if it
occurred. In this case, as in many others in chemistry, kinetic considerations (lack of a suitable
mechanism) override favorable thermodynamics.
-' 7 Jorgensen, C. K. Inorg. Chem. 1964. 3. 1201-1202.
' Ahrland. S. Struct. Bonding (Berlin) 1966. /. 207.
Hard and Soft Acids and Bases
electronegativity helps explain the fact that the trifluoromethyl group is considerably
harder than the methyl group and boron trifluoride is harder than borane.
Recall that the Mulliken-Jaffd definition of electronegativity involves two param¬
eters, a. the first derivative of the ionization energy-electron affinity curve, and h. the
second derivative (see page 186). The a term is identical with the original Mulliken
electronegativity, and the h term is the inverse of the charge capacity of an atom or
group. It appears that the association between electronegativity and hardness actually
refers to the b parameter, but the values of a and b for elements tend to parallel each
other, hence the similarity. It was early suggested that since the b parameter is the
inverse of the charge capacity, hard atoms will have high values of b. and soft atoms
will have smaller values. 57 ' 5y Thus fluorine not only forms a hard anion (note the very
high value of b in Table 5.6) but it likewise hardens the trifluoromethyl group by
contributing to a higher b value for it than for methyl.
Recently Parr and Pearson 6 1 have used the b parameter to investigate the hard and
soft properties of metal ions and ligands. They have termed this the absolute hardness
in comparison to the Mulliken-Jaffe a parameter which they call absolute elec¬
tronegativity. They provide strong arguments for the use of the absolute hardness
parameter in treating hard-soft acid-base (HSAB) interactions.
Almost from the beginning of HSAB theory, attention has been directed to
frontier orbitals.'' These are the highest occupied molecular orbital (HOMO) and
the lowest unoccupied molecular orbital (LUMO). According to Koopmans - theorem,
the energy of the HOMO should represent the ionization energy and the LUMO the
electron affinity for a closed-shell species. These orbitals are thus involved in the
electronegativity and HSAB relationships just discussed: Hard species have a large
HOMO-LUMO gap whereas soft species have a small gap. The presence of low-lying
unoccupied MOs capable of mixing with the ground state accounts for the polar¬
izability of soft atoms and, indeed. Politzer has shown a close correlation between
atomic polarizabilities and the b parameter . 62 Such mutual polarizability allows distor¬
tion of electron clouds to reduce repulsion. In addition, for polarizable species
synergistically coupled a donation and 7 r backbonding will be enhanced . 65
The idea of equating hard-hard interactions with electrostatics has probably been
overemphasized. It is natural, since a typical hard-hard interaction is Li + F“. But the
isoelectronic scries Li—F, Be—O. B—N, C—C all form strong bonds. The Li—F
bond is the strongest since it is a resonance hybrid of Li'F~ Li—F. Some
calculations based on electronegativity theory and a simple bonding model suggested
that in LiF about one-fourth of the bond energy comes from covalent bonding, one-
half from ionic bonding, and one-fourth from the transfer of electron density from the
less electronegative lithium atom to the more electronegative fluorine atom . 64 The
latter corresponds roughly to Pauling's ionic resonance energy.
» Huheey. J. E.; Evans, R. S. J. Inorg. Nitcl. Cliem. 1970. 32. 383-390.
60 Parr. R. G.; Pearson, R. G. J. Am. Cliem. Sot-. 1983. 105. 7512-7516: Pearson, R. G. J. Chem.
Editc. 1987. 64. 561-567. Initially the Parr-Pearson parameter was arbitrarily defined as p =
(I - A)/2. This results in >j = 6/2. It should also be noted that the Parr-Pearson numbers arc not, in
general, for atoms in their valence states, so caution must be used in their application.
61 Klopman, G. J. Am. Chem. Soc. 1964. 86. 1463-1469; 1968, 90. 223-234.
62 Politzer. P. J. Chem. Phys. 1987. 86. 1072-1073.
63 Backbonding in metal complexes is discussed in Chapter II. For a clear discussion of HOMOs,
LUMOs, and HSAB. see Pearson, R. G. J. Am. Chem. Soc. 1985, 107. 6801-6806: 1986. 108.
6109-6114; and Footnote 60.
M See Footnote 38 and the associated discussion of strong covalent bonding by fluoride (fluorine).
376
10- Chemistry in Aqueous and Nonaqueous Solvents
[CuCl 2 ]" + CuCI
[cu 2 ci 3 r
(10.85)
2[CuCl,]~
[Cu 2 Cl 3 ]~ + Cl"
(10.86)
[cuci 2 y +cr
;=± [CuCl 3 ] 2 "
(10.87)
Evidence for these equilibria comes from the Raman spectra, which show an
absorption peak (or unresolved peaks), probably attributable to Cu,Cl 3 . Addition ot
CuCI or Cl - causes this peak to increase or decrease as expected by the above
equilibria. The system thus probably contains at least four anionic species, and the
"impurities” account for the depression of the melting point. In accordance with this
interpretation is the fact that the material is oily and never forms a crystalline solid
with a true freezing point, but congeals to a glass at about 0 °C.
Of interest is the use of this system as both solvent and reactant in a voltaic cell. If
two platinum gauze electrodes are immersed in liquid chlorocuprates and a potential is
applied, the cell begins charging. At less than 1% of full charge, the potential stabilizes
at 0.85 V and remains at that value until the cell is fully charged. The half-reactions for
charging are
CuCIT + e“ -♦ Cu + 2CI" (10-88)
CuCI," -♦ CuCI, + e" (10.89)
Allowing the reactions to proceed spontaneously (reverse of Eqs. 10.88 and 10.89)
produces 0.85 V with low current (low. The chief difficulty with the cell is the fact that
CuCI, is soluble in the melt. It thus diffuses and allows the cell to decay through direct
reaction of the electrode materials:
CuCI, + Cu + 2C1" -► 2CuCl 2 " (10.90)
The fact that the solvent can be both oxidized and reduced is an asset in the above
reactions, but it is a handicap when the system is used merely as a solvent. For
example, the chlorocuprate solvent must be handled in the absence of air to prevent
oxidation. Some solutes cannot be studied. Even so gentle an oxidizing agent as FeCI,
oxidizes the solvent:
FeCl 3 + Cl" + CuCI 2 " -► FeCir + CuCI 2 (10.91)
Unreactivity of
Molten Salts
KHF, clcclr ‘" MS > }F, + JH, + KF 00.92)
NaC , JC ^ + Na (10.93)
The latter reaction is also important in the commercial production of sodium, which,
like the halogens, is too reactive to coexist with water.
The reactions in Eqs. 10.92 and 10.93 are typical of the many important industrial
processes involving high-temperature molten salts. Other examples are the production
Many reactions that cannot take place in aqueous solutions because of the reactivity
of water may be performed readily in molten salts. Both chlorine and fluorine react
with water (the latter vigorously), and so the use of these oxidizing agents in aqueous
solution produces hydrogen halides, etc., in addition to the desired oxidation prod¬
ucts. The use of the appropriate molten halide obviates this difficulty. Even more
important is the use of molten halides in the preparation of these halogens:
Molten Salts
377
of magnesium and aluminum and the removal of silica impurities (in a blast furnace,
for example) by a high-temperature acid-base reaction:
SiO z + CaO -♦ CaSi0 3 < 10 94 >
Gangue Flux Slag
Solutions of Metals One of the most interesting aspects of molten salt chemistry is the readiness with
which metals dissolve. For example, the alkali halides dissolve large amounts of the
corresponding alkali metal, and some systems (e.g., cesium in cesium halides) are
completely miscible at all temperatures above the melting point. On the other hand,
the halides of zinc, lead, and tin dissolve such small amounts of the corresponding free
metal that special analytical techniques must be devised in order to estimate the
concentration accurately.
At one time solutions of metals in their molten salts were thought to be colloidal in
nature, but this has been shown not to be true. However, no completely satisfactory
theory has been advanced to account for all the properties of these solutions. One
hypothesis involves reduction of the cation of the molten salt to a lower oxidation
state. For example, the solution of mercury in mercuric chloride undoubtedly involves
reduction:
Hg + HgCl, - Hg 2 CI,
and mercury(I) chloride remains when the melt is allowed to solidify. For most
transition and posttransition metals the evidence for the formation of "subhalides” is
considerably weaker. The Cd? + ion is believed to exist in solutions of cadmium in
molten cadmium chloride but can be isolated only through the addition of aluminum
chloride:
Cd + CdCl, - [Cd,CI,] —i Cd,[AICI 4 ], (10.96)
In many cases, although the presence of reduced species is suspected, it is impossible
to isolate them. On solidification the melts disproportionate to solid metal and solid
cadmium(II) salt.
In solutions of alkali metals in alkali halides, reduction of the cation, at least in the
sense of forming discrete species such as M,', is untenable. It is probable that in these
salts ionization takes place upon solution:
M -► M + e" 00.97)
The presence of free electrons thus bears a certain similarity to solutions of these same
metals in liquid ammonia. If the electrons are thought to be trapped in anion vacancies
in the melt, an analogy to F-centers (see Chapter 7) may be made. Undoubtedly the
situation is considerably more complex with the possibility of the electron being
delocalized in energy levels or bands characteristic of several atoms, but a thorough
discussion of this problem is beyond the scope of this book. 25
Complex Formation Molten salts provide a medium in which the concentration of anionicAigands can be
much higher than is possible in aqueous solutions. The concentration of the chloride
ion in concentrated aqueous hydrochloric acid is about 12 M, for example. In contrast,
25 Corbett. J. D. In Fused Sails', Sundhcim. B. R., Ed.; McGraw-Hill: New York, 1964: p 341.
352
Hard and Sofj Acids and Bases
353
9-Acid-Base Chemistry
Larger atoms do not have the advantage of good overlap for strong covalency, rtor
short interatomic distances for strong electrostatic interactions. In fact, if one investi¬
gates a typical HSAB reaction in the gas phase when all of the energies are known, we
find
HgF 2 + Bel 2 - BeF 2 + Hgl 2 AH = —397 kJ mol -1 (9.82)
Soft-hard Hard-soft Hard-hard Soft-soft
The HSAB rule works and the reaction is exothermic as written. If we look at the
individual heats of atomization of the species (from bond energies. Appendix E) we
find BeF, = +1264; HgF, = +536; Bel, = +578; Hgl, = +291 kJ mol -1 . The
driving force in Eq. 9.82 is almost entirely the strong bonding in the hard-hard
interaction.
One advantage that softer acids and bases generally do have is the ability to
•tr bond. This is made possible by large numbers of electrons in d orbitals on the metals
and empty, low-lying acceptor orbitals on the ligands. Once again, “bigger" is not
necessarily “better”: Once an atom is large enough, that is. has n - 3. it will have
available d orbitals. Sulfur and phosphorus are exemplars of soft-atom behavior. They
are much better at binding soft metals than are their larger congeners such as se¬
lenium, arsenic, tellurium, and antimony. This is for the same reason we have seen
before: Long 7r bonds are not very strong just as long a bonds are not very strong, but
the effect is even more pronounced because of the sideways overlap to form n bonds.
The premier soft ligand and n bonder is carbon monoxide which has low-lying tr
acceptor orbitals and which has the advantage of small size to obtain good overlap.
Both the HSAB principle and the £ A £ B -C A C B system were proposed and devel¬
oped in the 1960s. Insofar as the HSAB principle employs ideas of electrostatic and
covalent bonding to account for hardness and softness, it was natural to attempt a
correlation with the £ and C parameters. The early 1970s showed repeated attempts to
correlate the two ideas, prove one superior to the other, or to improve their theoretical
bases. 65 For example, both Drago and Pearson have discussed the possible quantifica¬
tion of the HSAB principle along the lines of the E-C system, but have come to
diametrically opposed conclusions. Drago and coworkers have even suggested that
the HSAB model is no longer tenable.
Part of the difficulties encountered in comparing these two approaches results
from the different ways in which they are used. The E-C approach treats the interac¬
tion of only two species at a time; to the extent that the nonpolar solvents used in these
studies minimize solvation effects, the results are comparable to gas-phase proton
affinities. In contrast, the HSAB principle is usually applied to exchange or competi¬
tion reactions of the sort:
A,B, + A,B, ;= A,B, + A,B, (9.83)
We have already seen that in the gas phase the stability of all metal halides follows the
order F _ > Cl - > Br _ > I", contrary to the simplest possible interpretation of the
HSAB rule. Perhaps the rule should be restated as follows: Soft acids prefer to bond to
soft bases when hard acids are preferentially bonding to hard bases. Although the
HSAB rule works in the gas phase, by far its greatest usefulness lies in the interpreta-
« Drago. R. S.; Kabler. R. A. Inorg. Client. 1972. II. 3144-3145. Drago. R. S. Ibid. 1973. 12.
2211-2212. Pearson. R. G. tnorg. Client. 1972. II. 3146. Drago, R. S. J. Client. Edttc. 1974. 51.
300-307.
tion of complexes in aqueous solution. These ions will always be hydrated though this
may not be explicitly stated. Under these circumstances, it is somewhat surprising
that the HSAB rule works as well as it does.
McDaniel and coworkers w > have presented a graphical means ot portraying some
of the ideas discussed in this chapter. For the reaction of hard and soft acids and bases
A b B s + A s B h -+ A„B h + A S B S (9,84)
it can be shown that the enthalpy change for this reaction, \H r . can be related to the
affinities of the bases for the two acids as shown in Fig. 9.6. If the affinities for a hard
acid (e.g.. H + ) and a softer acid (e.g.. CH 3 + ) are plotted and lines of unit slope are
drawn through them. AH r for the reaction can be measured by the distance between
the lines in either the .v or y direction. Furthermore, if two bases were to tall on the
same line in Fig. 9.6. they would be equally soft. If the line for a given base lies above
and to the left of that for another, the first base is softer and the second is harder.
Finally, since strength is related to the magnitude ol acid-base interactions, the
further a given base lies from the origin, the stronger it is.
Some typical anionic bases are plotted with their proton affinities and their methyl
cation affinities in Fig. 9.7. The solid line was drawn by the original investigators as a
least squares fit of all of their data. The dashed line was added for this discussion, and
it is arbitrarily drawn through F _ and OH" (the archetypical hardest bases) with unit
slope (see Fig 9.6). Hard bases lie close to the dashed line, soft bases lie further away
from it. The reader is urged to find analogous pairs, such as 1“ and F~, SH and OH ,
CN” and NH, _ , and interpret their positions on this graph in terms of inherent
strength, hardness, and softness.
Staley and coworkers'’ 7 Ipve provided direct measurement of HSAB efiects in
gas-phase dissociation energies between transition metals (where the principle has
Fig. 9.6 McDaniel diagram
illustrating HSAB param¬
eters. A|, is the difference in
affinity of two bases, B„
and B|„ for the hard acid
A,,. The reaction enthalpy,
AH,, of Eq. 9.84 is given by
the horizontal distance (or
vertical distance) between
the two lines of unit slope.
[Courtesy of D. H.
McDaniel.)
m. McDaniel. D. H.. The University of Cincinnati, personal communication. 1975.
7 Kappes. M. M.; Staley. R. H. J. Am. Client. Soc. 1982. 104. 1813, 1819. Jones, R. W.; Staley. R. H.
Ibid. 1982, 104. 2296. J. Pltys. Client. 1982. 86. 1387.
374
10 • Chemistry in Aqueous and Nonaqueous Solvents
Molten Salts The chemistry of molten salts as nonaqueous solvent systems is one that has devel-
oped extensively from the 1960s till the present, and only a brief survey can be given
here. The most obvious difference when compared with the chemistry of aqueous
solutions are the strongly bonded and stable nature of the solvent, a concomitant
resistance to destruction of the solvent by vigorous reactions, and higher concentra¬
tions of various species, particularly coordinating anions, than can be obtained in
saturated solutions in water.
Solvent Properties On the basis of the structure of the liquid, molten salts can be conveniently classified
into two groups although there is no distinct boundary between the two. The first
consists of compounds such as the alkali halides that are bonded chiefly by ionic
forces. On melting, very little change takes place in these materials. The coordination
of the ions tends to drop from six in the crystal to about four in the melt and the long-
range order found in the crystal is destroyed; but a local order, each cation surrounded
by anions, etc., is still present. These fused salts are all very good electrolytes because
of the presence of large numbers of ions. They behave normally with respect to
cryoscopy and this is a useful means of study. The number of ions, v, may be
determined in these systems just as in the sulfuric acid system (page 365). For
example, if sodium chloride is the solvent, t> KF = 2, »» BaF , = 3, etc. One interesting
point is that a salt with a common ion behaves somewhat anomalously in that the
common ion does not behave as a "foreign particle” and v is correspondingly lower.
In sodium chloride solutions, v NuF = I.
The second group consists of compounds in which covalent bonding is important.
These compounds tend to melt with the formation of discrete molecules although
autoionization may occur. For example, the mercury(II) halides ionize as follows:
2HgX, =; HgX + + HgXj (10.73)
This is analogous to the aprotic halide solvents discussed in the previous section.
Acidic solutions may be prepared by increasing the concentration of HgX^ and basic
solutions by increasing the concentration of HgXJ:
Hg(C10J 2 + HgX 2 -► 2HgX + + 2CIO; (10.74)
KX + HgXj -► K + + HgXJ (10.75)
and neutralization reactions occur on mixing the two:
HgX + + CIO; + K + + HgX 3 - -♦ 2HgX 2 + K + CIO; (10.76)
If aluminum chloride is added to an alkali metal chloride melt, an alkali metal
tetrachloroaluminate forms:
2[M + Cr] + AI 2 Cl fi - 2M + + 2AICIJ (10.77)
The tetrachloroaluminate ion undergoes autoionization
2AICI; == AUCly + Cl" K eq = 1.06 x I0“ 7 (10.78)
and one can clearly relate basicity to the concentration of the chloride ion. At 175 °C
the neutral melt has [A1 2 CIJ] = [Cl"] = 3.26 x I0" 4 M and a pCI scale can be set up
with a neutral point of 3.5. Basic solutions have lower values of pCI (a saturated
solution of NaCI has pCI = 1.1) and acidic solutions (made by adding excess AI 2 CI 6 )
Molten Salts
375
Room-Temperature
Molten Salts
have higher values. The pCI can be monitored electrochemically with an aluminum
electrode. 21
Although the term “molten salts" conjures up images of very high-temperature fused
systems, some salts are liquid at or near room temperature. For example, if alkyI-
pyridinium chlorides are added to aluminum chloride, the resultant compounds are
very similar to the alkali metal tetrachloroaluminates, but they are often liquids: 22
The chemistry in these melts is very similar to that in MAIC1 4 except that it can be
carried on at about 25 °C instead of 175 °C!
One problem with chloroaluminate melts is that aluminum chloride and most
transition metal chlorides (cf. Eqs. 10.99 to 10.101) are hygroscopic, and even if very
carefully handled will hydrolyze from any moisture in the atmosphere:
[AICIJ" + H,0 -> [CI 2 A10]" + 2HC1 (10.80)
[C1 2 A10]" + [TiCI 6 ] 2 ~ ?=: [TiOCl 4 ] 2 " + [AICI„]" (10.81)
Such impurities are, of course, a problem whenever careful measurements are at¬
tempted. It has been found 22 that phosgene quantitatively removes the oxide
impurities
[TiOCIJ 2- + OCCI 2 -♦ [TiCI 6 J 2 " + C0 2 (10.82)
[NbOClJ" + OCCI, -♦ [NbCy- + C0 2 (10.83)
and this has proved a useful way to keep the systems anhydrous.
Although the chloroaluminates are the best known room-temperature molten
salts, there are several other interesting systems. For example, if one mixes the
crystalline solids triethylammonium chloride and copper(I) chloride, an endothermic
reaction takes place to form a light green oil. The most reasonable reaction is the
coordination of a second chloride ion to the copper(I) ion 24
[EtjNHJCI + CuCI - [Et,NH][CuCl 2 ] (10.84)
to form the dichlorocuprate(I) ion. The source of the low melting point seems to be the
following equilibria:
21 Chum. H. L.; Osteryoung. R. A. Ill Ionic Liquids', Inman. D.; Lovering, D. G.. Eds.; Plenum: New
York. 1981: pp 407-423.
— Sec Hussey, C. L. Adv. Molten Salt Client. 1983, 5. 185. Gale, R. I.; Osteryoung, R. A. In Molten
Salt Techniques'. Lovering, D. G.; Gale, R. J., Eds.; Plenum: New York, 1983: Vol. I, pp 55-78.
22 Sun. I-W.; Ward. E. H.; Hussey. C. L. Inorp. Client. 1987, 26, 4309-4311.
2,1 Porterfield. W. W.; Yoke, J. T. In Inor/ianlc Compounds with Unusual Properties', King, R. 8.,
Ed.; Advances in Chemistry 150; American Chemical Society: Washington. DC. 1976; Chapter 10.
Hussey, C. H. Adv. Molten Salt Chem. 1983, 5, 185.
355
354
Fig. 9.7 Methyl cation affinity vs. proton affinity of a series of anionic bases of varying
hardness. The solid line is a least-squares fit of the data drawn by the original investigators.
The dashed line has been arbitrarily drawn through F _ and OH - with unit slope (see Fig.
9.6). Hard bases lie close to the dashed line, soft bases lie further away from it. [From
Brauman, J. I.; Han, C.-C. J. Am. Chern. Soc. 1988, I/O. 5612. Reproduced with
permission.]
Fig. 9.8 Comparison of relative ligand dissociation energies for Mn and Co. Zero points
for the scales have been arbitrarily chosen. Note that for the soft ligands. McSH. HCN.
EtSH. and Me 2 S, the data points lie above and to the left of those for the oxygen bases. This
indicates relatively stronger hard-hard bonding (0—Mn) or soft-soft bonding (S—Co), or
both. [Modified from Jones. R. W.; Staley. R. H. J. Phys. Chem. 1982. 86. 1387. Reproduced
with permission.]
Problems
Problems
always proved most useful) and various ligands, both hard and soft. As expected. Cu
(J ') is significantly softer than Co (J H ), which in turn is softer than Mn (</ 6 ). A
comparison of the results for Co and Mn is given in Fig. 9.8. There is the expected
correlation of dissociation energies for a large series of oxygen bases, the variation
along the line resulting from differences in substituents, hybridization, elec¬
tronegativity, etc. However, as soon as the hard oxygen bases are replaced by softer
bases such as MeSH. EtSH. Me,S. and HCN, a new line is generated with the softer
Co ion showing ca. 30 kJ mol -1 greater dissociation energies. The reader is referred
to the original articles that contain far more data and interesting figures than can be
presented here. In summary, all of these data are consistent with the HSAB effect
acting in the absence of complicating solvent effects to stabilize either hard-hard
interactions, or soft-soft interactions, or both.
In summary, acid-base chemistry is conceptually rather simple, but the multi¬
plicity of factors involved makes its treatment somewhat involved. Until more unify¬
ing concepts are developed, as they undoubtedly will be, it will be necessary to apply
to each problem that is encountered the ideas, rules, and (when available) the param¬
eters applicable to it.
9.1 Use the Lewis definition of acids and bases to explain the examples given for the
Brpnsted-Lowry. Lux-Flood, and solvent system definitions (Eqs 9 1-9 4 9 5-9 8
9.9-9.36). .
9.2 Use the generalized definition of acids and bases to explain the examples given for the
Bronsted-Lowry. Lux-Flood, solvent system, and Lewis definitions (Eqs 9 1-9 4
9.5-9.8. 9.9-9.36; 9.37-9.40).
9.3 Which would you expect to be a better Lewis acid. BCI, or B(CH,),? Explain.
9.4 The order of acidity of boron halides is BF, < BG, < BBr,. Is there anything unexpected
in this order? Suggest possible explanations.
9.5 Plot the acidity parameters, a. from Table 9.1 vs. the pA.' h values in Table 9.3 for those
metals that occur in both tables. Interpret your plot.
9.6 B-strain can occur in amines to lower their basicity. Will B-strain inhibit or enhance the
acidic behavior of boranes?
9.7 Predict which way the following reactions will go (left or right) in the gas phase:
HI + NaF = HF + Nal
Allj + 3NaF = AIFj + 3NaI
Cul 2 + 2CuF = CuF 2 + 2CuI
TiF 4 + 2TiI 2 = Til 4 + 2TiF 2
CoF 2 + HgBr, = CoBr 2 + HgF 2
9.8 Calculate the values for the proton affinities of the halide anions shown in Table 9.5 from a
Born-Habcr thermochcmical cycle and values for ionization energies, electron affinities,
and bond energies.
9.9 a. Using Eq. 9.61. calculate the proton affinities of the following bases: NH, CH,NH->
<CH,) : NH. (CH,),N. py. H,0. (CH,) : 0. H 2 S, (CH,) 2 S. and (CH,) 2 SO
b. Compare your answers with experimental values as given in Table 9.5. Which com¬
pounds show the greatest differences between AW tllk . and A/7,. >p ? Discuss possible
reasons for the differences.
372
10 - Chemistry in Aqueous and Nonaqueous Solvents
Most molecular chlorides behave as acids:
FeCI 3 OPC '-> OPCI 2 + + FeCi; (10-66>
SbCI 5 0PC '' -> OPCtf + SbCl (10.67)
As might be expected, basic solutions may be titrated with acidic solutions and the
neutralization followed by conductometric, potentiometric, photometric, and similar
methods. Some metal and nonmetal chlorides are amphoteric in phosphorus
oxychloride:
K + +• C!" + AlClj ■ 0PC '' -> K + + AICI^ (10-68)
SbCl 5 + AlClj AlClj + SbCI (10.69)
A table of relative chloride ion donor and acceptor abilities can be established 17 from
equilibrium and displacement reactions (Table 10.5). As expected, good donors are
generally poor acceptors and vice versa with but few exceptions (e.g., HgCI,).
There has been some controversy in the literature over the proper interpretation
of reactions in solvents such as phosphorus oxychloride. Drago and coworkers 18 have
suggested the "coordination model" as an alternative to the solvent system approach.
They have stressed the errors incurred when the solvent system concept has been
pushed further than warranted by the facts. In addition, they have pointed out that
iron!Ill) chloride dissolves in triethyl phosphate with the formation of tetrachloro-
Table 10.5
Relative chloride ion donor
Chloride ion donors
Chloride ion acceptors
and acceptor abilities
[R,N]C1
KCI
AlClj
AlClj
TiCI 4
ZnCl 2
PCI,
PC1 5 ‘
,Tt
ZnCl 2
TiCl 4
1
HgCI 2
1
BCIj
BCIj
|
BF 3
|
|
InCI 3
|
SnCI 4
SnCI 4
aici 2 +
HgCI,
SbClj
SbCl 3
FeCl 3
1 7 The ordering of (his list is not invariant: Some of the compounds listed have very similar donor and
acceptor abilities and exchange places depending upon the nature of the other ions in solution. This
is to be expected in a solvent of relatively low permittivity (e/« 0 = 13.9) where ion pair formation
will be important and the nature of the counterion can affect the stability of a chloride adduct,
ts Meek. D. W.; Drago. R. S. J. Am. Chem.Soc. 1961. 83. 4322-4325. Drago. R. S.; Purcell, K. F. In
Non-Aqueous Solvent Systems: Waddington. T. C.. Ed.: Academic: New York, 1965; pp 211-251.
Nonaqueous Solvents
373
ferrate(lll) ions. FeClj - . just as in phosphorus oxychloride. In triethyl phosphate,
however, the solvent cannot behave as a chloride ion donor and so a reaction such as
Eq. 10.66 is not applicable. In triethyl phosphate the chloride ion transfer must take
place from one FeCI 3 molecule to another with the formation of a cationic iron(Ill)
species:
2FeCI 3 [FeCl 2 {OP(OEt) 3 }„] + + FeC!^ (10.70)
Drago and coworkers argue that in view of the similarity in physical and chemi¬
cal properties between phosphorus oxychloride, OPCI 3 , and triethyl phosphate,
OP(OEt) 3 , it is probable that the formation of FeClj" in phosphorus oxychloride
proceeds by a reaction similar to Eq. 10.70.
2FeCl 3 0PC '' > [FeCI 2 (OPCl 3 )„] + + FeClJ (10.71)
They argue that the similar coordinating ability of these phosphoryl (— P=0) sol¬
vents (and to a lesser extent their dielectric constants) is more important than their
chemical differences (supposed autoionization and chloride ion transfer in phosphorus
oxychloride).
Gutmann 19 has rejoined that the dichloroiron(III) ion. [FeCI, (solvent)] + , is not
found in dilute solutions in phosphorus oxychloride but only in concentrated solutions
or those to which a strong acid such as SbCI 5 has been added. In such cases the
chloride donor ability of the solvent has been exceeded and chloride ions are ab¬
stracted from the iron(UI) chloride. This point was made earlier- 11 by the observation
that the controversy is at least partly a semantic one. The only "characteristic
property" of the solvo-cations and solvo-anions in the solvent system autoionization
is that they tire the strongest acids and bases that can exist in that particular solvent
without being leveled. In triethyl phosphate (a nonleveling solvent) the dichloro-
iron(III) ion is perfectly stable. In phosphorous oxychloride a mechanism for leveling
exists, namely:
FeCtf + OPCl 3 ;==± OPC1+ + FeCl 3 (10.72)
This equilibrium will lie to the right if the dichloroiron(III) ion is a stronger acid than
the dichlorophosphoryl ion and to the left if the acid strengths are reversed. The
important point is that neither the solvent system approach nor the coordination
model can. a priori, predict the nature of the equilibrium in Eq. 10.72. To make this
prediction, one must turn to the generalized acid-base definition given above together
with some knowledge of the relative electron densities on the central atoms in FeCI 2
and OPCI 2 . The essence of the acidity of iron(III) chloride lies in its tripositive ion of
rather small radius and high charge, which is compensated only in part by three
coordinated chloride ions and which seeks elsewhere for electron density to reduce its
positive character. It is thus an acid irrespective of the solvent chosen and will accept
the strongest base available to it. If the basicity of the phosphoryl group is sufficient
(as it must of necessity be in triethyl phosphate or in phosphorus oxychloride if the
chloride ion concentration is too low), then the iron(III) chloride is less acidic than if it
can abstract a chloride ion (possible only in phosphorus oxychloride).
iv Gutmann. V. Coordination Chemistry in Non-Aqueous Solutions: Springer: New York, 1968. and
references therein.
20 Huheey, J. E. J . Inorg. Nucl. Chent . 1962. 24, 1011-1012.
356
9-Acid-Base Chemistry
9.10 a. Do you expect dimethylsulfoxide. (CH,) : SO. to bind to H + through the sulfur or "the
oxygen atom? Support your prediction with numbers,
b. Calculate the affinities of the following bases for trimethylstannyl cation. (CH,),Sn:
H 2 0, NHj, CH 3 NH 2 , (CH 3 ) 2 NH, (CH 3 ),N
9.11 In general, the best data for correlating acid-base phenomena are obtained in gas-phase
experiments rather than in solution. Discuss factors present in solution, especially in polar
solvents, that make solution data suspect.
9.12 In contrast to the generalization made in Problem 9.11 there is reason to believe the
solution data for CH 3 0(CH 2 )„NH 2 may be more indicative of inherent basicity than the
gas-phase work. Can you suggest a reason? (Hint: Consider the possibilities for hydrogen
bonding.) 68
9.13 a. Estimate the approximate pof phosphoric and arsenic acids.
b. Refine your answer by deciding whether H 3 PO.i or H,As0 4 is stronger.
9.14 Phosphorous acid can exist as either of two tautomers,
/ 0H 11
P —OH or H — P—OH
X OH 1
0H OH
From the p K a of phosphorous acid (1.8) assign a structure to the form of phosphorous acid
in aqueous solution. The pof hypophosphorous acid. H 3 P0 2 , is 2.00. Assign a reason¬
able structure. (See Chapter 18.)
9.15 In Fig. 9.1. could you have assigned the peaks if the legend had not?
9.16 The discussions of basicity of amides on pages 334 and 342 are based upon the carbonyl
oxygen being the basic site. It is also possible that the amide nitrogen atom could act as a
base. What experimental work could you suggest to determine which atom is the most
basic site?
9.17 If you did not answer Problem 5.14 when you read Chapter 5, do so now.
9.18 Using Fig. 9.6. explain why H 2 S is considered softer than H : 0 even though it binds more
tightly to the hard acid H.
9.19 Complete and balance the following equations, identifying the acids and the bases.
a. S0 3 I- KjO —
b. MgO + Al : 0. -
c. Al : 0, + Na 2 0 —
d. CaO + PjOki —
e. Si0 2 + K : 0 + Al,0, + MgO —
9.20 If you did not answer Problem 6.26 when you read Chapter 6. do so now.
9.21 Potassium metal reacts with graphite to form an intercalation compound approximating
C„K. Will this material act as an acidic or basic catalyst? 69
9.22 Predict the order of proton affinities for the following bases: NR 3 , S 2 '. NF 3 . 0‘ . NH,,
OH', NCI 3 , N 5 '. Pick any pair of bases from this scries and explain why you decided that
one was stronger than the other.
68 Love. P.; Cohen. R. B.: Taft. R. W. J. Am. Cltem. Sac. 1968. 90. 2455.
69 Bergbreiter. D. E.: Killough. J. M. J. Chem. Soc., Chem. Commun. 1976. 913-914.
Problems
357
9.23 Calculate the enthalpies for all of the possible 1:1 reactions between the acids H. CH{,
PF 3 . and CF 3 CH 2 OH. and the bases OH'. NH,. (CH,),P. and Cl'. You may check some
of your answers against Table 9.5. How accurate are your calculations?
9.24 Examine Fig. 9.2 and provide a rationale for the relationship therein.
9.25 Most frequently the change in frequency of the carbonyl group upon coordination to a
Lewis acid is stated in terms of bond order. Develop such an argument.
9.26 a. We learn in organic chemistry that C,.H 5 NH 2 and C 5 H 5 N are weaker bases than NH,,
but Table 9.5 indicates otherwise. Discuss, including the important molecular attributes
of each molecule. (Hint: See RO , page 344).
I). Water is a weak acid, but most hydrocarbons are usually considered to have virtually no
acidity whatsoever. However, in the gas phase Q.HjCH, is 10stronger as an acid than
H : 0. Discuss the particular molecular properties that cause the gas-phase values to be
different from solution data and to differ so much between these two species.
9.27 Reconcile the values of the proton affinities of pyridine (953 kJ mol' 1 ) and ammonia (872 k.l
mol' 1 ) with the argument on page 343 concerning the relationship between p/C h and
electronegativity. The latter argument seems to go with the '•conventional wisdom" rather
than the discussion in this chapter. Criticize.
9.28 Using a Bom-Haber cycle, clearly show all of the terms that one should evaluate in
considering (he energetics involved in transferring the competition (as given by the en¬
thalpy of reaction):
BH'(g) + B’(g) B(g) + B'H-fg) (9-85)
into solution:
BH (aq) + B'(aq) tyaq) + B'H + (aq) (9.86)
Once you have your Born-Habcr diagram drawn, return to Problem 9.26 and see if it helps
to clarify your answers. 70
9.29 Dioxygen. 0 2 . is not a very good ligand, but it is fairly soft. Hemoglobin contains Fe"
which is only of borderline softness.
a. Look at the structure of the heme group and suggest how (he iron in heme is softened.
b. Carbon monoxide is poisonous because it bonds more tightly to the hemoglobin in red
blood cells than does dioxygen. Why docs carbon monoxide out-compete dioxygen as a
ligand?
9.30 What is the significance of the two lines shown in Fig. 9.7? How do they differ? What arc
the softest bases shown in this figure?
9.31 How is it possible for the noble gases to have exothermic proton affinities, indicating the
formation of X—H chemical hands' ?
9.32 Presumably, in the gas phase IF,0 2 has the same structure as C1F,0 2 . but the structure of
the solid is a bit of a surprise. Suggest possibilities. (Hint: Remember that the current
chapter deals with Lewis acid-base interactions.)
9.33 a. Acid rain is defined as any precipitation with pH < 5.6. Why 5.6; why not 7.0?
b. Some ill effects of acid rain come not from the low pH. per se, but from the toxicity of
metals ions. Explain.
70 Arnett. E. M. J. Chem. Etluc. 1985. 62. 385.
370
10-Chemistry in Aqueous and Nonaqueous Solvents
Table 10.4 _
Donor number (DN),
acceptor number (AN), and
relative permittivity
(dielectric constant, e/c 0 ) of
selected solvents
Solvent
DN
AN
e/eo
Acetic acid
_
52.9
6.2
Acetone
17.0
12.5
20.7
Acetonitrile
14.1
19.3
36
Antimony pentachloride
—
100.0
_
Benzene
0.1
8.2
2.3
Carbon tetrachloride
_
8.6
2.2
Chloroform
—
23.1
4.8
Dichloromethane
_
20.4
Diethyl ether
19.2
3.9
4.3
Dimethylacetamide
27.3
13.6
37.8
Dimethylformamide (dmf)
24.0
16.0
36.7
Dimethylsulfoxide (dmso)
29.8
19.3
45
Dioxane
14.8
10.8
2.2
Hexamethylphosphoric triamide (hmpa)
38.8
10.6
_
Nitromethane
2.7
20.5
38.6
Phosphorus oxychloride
11.7
—
_
Propylene carbonate
15.1
_
65.1
Pyridine (py)
33.1
14.2
12.3
Tetrahydrofuran
20.0
8.0
7.3
Trifluoroacetic acid
_
105.3
Trifluorosulfonic acid
_
129.1
Water
18
54.8
81.7
" The ratio c/c„ is more convenient to use than the value of the permittivity in
absolute units.
h From Gutmann, V. Chimin, 1977, 31, I; The Donor-Acceptor Approach to
Molecular Interactions', Plenum: New York. 1978. Used with permission.
to include an acceptor number (AN), that measures the electrophilic behavior of a
solvent (Table 10.4). 13 Drago 14 has criticized the donor-number concept because it
does not go far enough in accounting for differences in hardness and softness (or
electrostatic and covalent differences). By limiting the evaluation of donor numbers to
a single acid (SbCI 5 ), the donor-number system in effect presents only half of the
information available from the £ A £ U + C A C B four-parameter equation.
The third group of solvents consists of those that are highly polar and autoioniz-
ing. They are usually highly reactive and are difficult to keep pure because they react
with traces of moisture and other contaminants. Some even react slowly with silica
containers or dissolve electrodes of gold and platinum. An example of one of the more
reactive of these solvents is bromine trifluoride. Nonfluoride salts, such as oxides,
carbonates, nitrates, iodates, and other halides, are fluorinated:
Sb 2 0 5
BrF 2 + SbF~
(10.50)
Ge0 2
2BrF 2 + + GeF~
(10.51)
PBr 5
BrF 2 + + PF^
(10.52)
NOCI
NO + + BrF 4 "
(10.53)
Gutmann, V. Electrochim. Acta 1976. 21, 661-670; Chimia 1977. 31, 1-7.
14 Drago, R. S. Inorp. Cltem. 1990. 29. 1379-1382.
Nonaqueous Solvents
371
Fluoride salts dissolve unchanged except for fluoride ion transfer to form conducting
solutions:
KF — K + BrF 4
BrF,
AgF- Ag + BrF 4
SbF 5 ~ BrF 2 + + SbF^
SnF 4 2BrF 2 + SnF^"
(10.54)
(10.55)
(10.56)
(10.57)
These solutions can be considered acids or bases by analogy to the presumed auto¬
ionization of BrF 3 : 15
2BrF 3 BrF 2 + BrF 4 (10.58)
Reactions 10.50 to 10.52, 10.56, and 10.57 above may be considered to form acid
solutions (BrF 2 ion formed) and reactions 10.53 to 10.55 may be considered to form
basic solutions (BrF 4 ions formed). Acid solutions may be readily titrated by bases:
(BrF 2 )SbF 6 + AgBrF 4 -♦ AgSbF 6 + 2BrF 3 (10.59)
Such reactions may be followed conveniently by measuring the conductivity of the
solution: A minimum occurs at the 1:1 endpoint. Solutions of SnF 4 behave as dibasic
acids:
(BrF 2 ),SnF 6 + 2KBrF 4 -► K 2 SnF 6 + 4BrF 3 (10.60)
with minimum conductivities corresponding to 1:2 mole ratios.
A similar, although less reactive, aprotic solvent is phosphorus oxychloride
(phosphoryl chloride). A tremendous amount of work on the properties of this solvent
has been done by Gutmann and coworkers. 16 They have interpreted their results in
solvent system terms based on the supposed autoionization:
opci 3 opci 2 + cr (10.61)
or the more general solvated forms:
(in + n)OPCI 3 ^[OPCI 2 (OPCI 3 )„_,] + + [CI(OPCI 3 ),„]“ (10.62)
It is extremely difficult to measure this autoionization because contamination with
traces of water yields conducting solutions which may be described approximately as:
3H 2 0 + 20PC1 3 - 2(H 3 0)CI + CI 2 P(0)0P(0)CI 2 (10.63)
If autoionization does occur, the ion product, [OPCI 2 ][Cr], is equal to or less than
5 x KT 14 .
Salts which dissolve in phosphorus oxychloride to yield solutions with high
chloride ion concentrations are considered bases:
ora,
KCI - K + Cl strong base
ora,
Et 3 N == [Et 3 NP(0)CI 2 ] + + CP weak base
(10.64)
(10.65)
13 This expression for the autoionization of BrF 3 is based on the conductivity of pure BrF 3 and the
characterization of the BrF7 salts such as KBrF 4 . The evidence for BrF 2 is weaker. Further
support for this formulation is obtained from the ICI 3 system where X-ray evidence for ICI 2 and
ICI 4 has been obtained.
16 Gutmann, V. Coordination Chemistry in Non-Aqueous Solutions', Springer: New York, 1968.
358
9-Acid-Base Chemistry
9.34 Explain what effect acid rain would have on the condition of each of the following and why:
a. The Taj Mahal, at Agra. India
b. A limestone barn near Antietam Battlefield. Maryland, dating from the Civil War
c. The (Caryatides, the Acropolis, Athens, Greece
d. The ability of an aquatic snail to form its shell in a lake in the Adirondack Mountains
e. The asbestos-shingled roof on the house of one of the authors in College Park, Maryland
f. The integrity of the copper eaves-troughs and downspouting on that house
g. The integrity of the brick siding of that house
h. The growth of the azaleas planted along the foundation of that house
i. The integrity of the aluminum siding on a neighbor’s house
j. The slate roof on another neighbor's house
k. The longevity of galvanized steel fencing in the neighborhood
9.35 Throughout this chapter, the fluoride ion is referred to as a strong base, yet good,
illustrative examples are seemingly not common. Suggest a few such examples. Why are
there not more simple examples of F" acting as a strong base or nucleophile? Suggest ways
of making strong F~ bases. 71
71 Schwesinger. R.; Link. R.; Thiele. G.; Rotter. H.; Honert. D.: Limbach, H-H.; Miinnlc, F. Angew.
Client. Inr. Eel. Ellul. 1991.30, 1372-1375.
Chapter
Chemistry in Aqueous
and Nonaqueous Solvents
Almost all of the reactions that the practicing inorganic chemist observes in the
laboratory take place in solution. Although water is the best-known solvent, it is not
the only one of importance to the chemist. The organic chemist often uses nonpolar
solvents such as carbon tetrachloride and benzene to dissolve nonpolar compounds.
These are also of interest to the inorganic chemist and, in addition, polar solvents such
as liquid ammonia, sulfuric acid, glacial acetic acid, sulfur dioxide, and various
nonmetal halides have been studied extensively. The study of solution chemistry is
intimately connected with acid-base theory, and the separation of this material into a
separate chapter is merely a matter of convenience. For example, nonaqueous sol¬
vents are often interpreted in terms of the solvent system concept, the formation of
solvates involve acid-base interactions, and even redox reactions may be included
within the Usanovich definition of acid-base reactions.
There are several physical properties of a solvent that are of importance in
determining its behavior. Two of the most important from a pragmatic point of view
are the melting and boiling points. These determine the liquid range and hence the
potential range of chemical operations. More fundamental is the permittivity (di¬
electric constant). A high permittivity is necessary if solutions of ionic substances are
to form readily. Coulombic attractions between ions are inversely proportional to the
permittivity of the medium:
F -
E ~ 4nre M-D
where e is the permittivity. In water, for example, the attraction between two ions is
only slightly greater than 1% of the attraction between the same two ions in the
absence of the solvent:
e H : o = 8I.7e„ (10.2)
where e 0 is the permittivity of a vacuum. Solvents with high permittivities will tend to
be water-like in their ability to dissolve salts.
359
368
10-Chemistry in Aqueous and Nonaqueous Solvents
Fig. 10.1 Relative acidity and basicity of solvents. Solvents and solutes are listed from lop
to bottom in order of decreasing basicity and increasing acidity. Solutes are listed in order or
decreasing p K u as determined in water. Some values of p K„ are estimated. In ideal aqueous
solutions, equimolar mixtures of an acid and its conjugate base will have a pH equal to the
p K a . The range of acidity and basicity over which a particular solvent is differentiating is
shown at the right. All acids lying below and all bases lying above the enclosed box will be
leveled to the characteristic cation and anion of the solvent. The arrows involving CH,COOH
and NH, illustrate the fact that an acid will readily donate a proton to a base above it and to
the right.
Nonaqueous Solvents
369
Aprotic Solvents
equilibrium (the “weak" case, differentiation) or alternatively to go essentially to
completion (the "strong” case, leveled). Finally, it must be recalled that only solvents
of high dielectric constants can support electrolytic solutions. Solvents of low di¬
electric constants will result in weak electrolytes irrespective of acidity or basicity
arguments.
Thus far. the solvents discussed have had one feature in common with water, namely,
the presence of a transferable hydrogen and the formation of onium ions. In this
section we shall look briefly at solvents which do not ionize in this way. These may be
conveniently classified into three groups. The first group consists of solvents such as
carbon tetrachloride and cyclohexane which are nonpolar, essentially nonsolvating,
and do not undergo auloionization. These arc useful when it is desired that the solvent
play a minimum role in the chemistry being studied, for example, in the determination
of E and C parameters discussed in the previous chapter.
The second group consists of those solvents that are polar, yet do not ionize.
Some examples of solvents of this type are acetonitrile, CH 3 C=N; dimethyl-
acetamide. CH,C(0)N(CH 3 ) : ; dimethyl sulfoxide (dmso), (CH ? )-,S=0; and sulfur
dioxide. S0 2 . Although these solvents do not ionize to a significant extent, they are
good coordinating solvents because of their polarity. The polarity ranges from low
(SO.) to extremely high (dmso). Most are basic solvents tending to coordinate strongly
with cations and other acidic centers:
CoBr, + 6dmso -♦ [Co(dmso)J 2 ' + 2Br (10.46)
SbCI, + CH 3 C=N -♦ CH 3 C=NSbCI, (10.47)
A few. the nonmetal oxides and halides, can behave as acceptor solvents, reacting
with anions and other basic centers:
Ph,CCI + SO. - Ph,C + + SO.Cr (10.48)
This group of solvents ranges from the limiting case of a nonpolar solvent
(Group I) to an autoionizing solvent (Group III, see below). Within this range a wide
variety of reactivity is obtained. Gutmann 1 - has defined the donor number (DN) as a
measure of the basicity or donor ability of a solvent. It is defined as the negative
enthalpy or reaction of a base with the Lewis acid antimony pentachloride, SbCI 5 :
B + SbCI 5 - BSbCI, DN ShCI ,= -A/Y (10.49)
These donor numbers provide an interesting comparison of the relative donor abili¬
ties of the various solvents (Table 10.4). ranging from the practically nonpolar 1,2-
dichloroethane to the highly polar hcxamethylphosphoramide. [(CH 3 ),N|,PO. Note,
however, that there is no exact correlation between donor number and permittivity.
Some solvents with relatively high permittivities such as nitromethane and propylene
carbonate (e/e,, = 38.6 and 65.1) may be very poor donors (DN = 2.7 and 15.1).
Conversely, the best donors do not always have high permittivities: pyridine (DN =
33.1. e/e 0 = 12.3) and diethyl ether (DN = 19.2, e/e„ = 4.3). This should serve to
remind us that solubility is not merely an electrostatic interaction but that solvation
also involves the ability to form covalent donor bonds. Note that pyridine may be
considered to be a relatively soft base (Chapter 9). Gutmann has extended the concept
/
12 Gutmann. V. The Donor-Acceptor Approach to Molecular Interactions', Plenum: New York. 1978.
360
10-Chemistry in Aqueous and Nonaqueous Solvents
Physical properties of water
Boiling point
Freezing point
Density
Permittivity (dielectric constant)
Specific conductance
Viscosity
Ion product constant
100 °C
0 °C
1.00 g cm -3 (4 °C)
81.7e (18 °C)
4 x 10" 8 IT 1 cm" 1 (18 °C)
1.01 gem" 1 s" 1 (20 °C)
1.008 x 10" 14 mol 2 L" 2 (25 °C)
Water will be discussed only briefly here but a summary of its physical properties is
given in Table 10.1 for comparison with the nonaqueous solvents to follow. One
notable property is the very high permittivity which makes it a good solvent for ionic
and polar compounds. The solvating properties of water and some of the related
effects have been discussed in Chapter 8. Electrochemical reactions in water are
discussed on pages 378-381.
Nonaqueous
Solvents
Ammonia
Although many nonaqueous solvent systems have been studied, the discussion here
will be limited to a few representative solvents: ammonia, a basic solvent; sulfuric
acid, an acidic solvent; and bromine trifluoride, an aprotic solvent. In addition a short
discussion of the chemistry taking place in solutions of molten salts is included.
Ammonia has probably been studied more extensively than any other nonaqueous
solvent. Its physical properties resemble those of water except that the permittivity is
considerably smaller (Table 10.2). The lower dielectric constant results in a generally
decreased ability to dissolve ionic compounds, especially those containing highly
charged ions (e.g., carbonates, sulfates, and phosphates are practically insoluble). In
some cases the solubility is higher than might be expected on the basis of the
permittivity alone. In these cases there is a stabilizing interaction between the solute
and the ammonia. One type of interaction is between certain metal ions such as Nr \
Cu 2 \ and Zn 2+ and the ammonia molecule, which acts as a ligand to form stable
ammine complexes. A second type is between the polarizing and polarizable ammonia
molecule and polarizable solute molecules or ions. Ammonia may thus be a better
solvent than water toward nonpolar molecules. Ionic compounds containing large,
polarizable ions such as iodide and thiocyanate also are quite soluble.
Physical properties
of ammonia
Boiling point
Freezing point
Density
Permittivity (dielectric constant)
Specific conductance
Viscosity
Ion product constant
-33.38 °C
-77.7 °C
0.725 gem 3 (-70 °C)
26.7« 0 (-60 °C)
1 x 10" 11 A" 1 cm" 1
0.254 g cm'
(-33 °C)
5.1 x 10" 27 mol 2 L" 2
Nonaqueous Solvents
361
Precipitation reactions take place in ammonia just as they do in water. Because of
the differences in solubility between the two solvents, the results may be considerably
different. As an example, consider the precipitation of silver chloride in aqueous
solution:
KC1 + AgN0 3 -♦ AgClj + KN0 3 (10 - 3)
In ammonia solution the direction of the reaction is reversed so that:
AgCl + KN0 3 -♦ KCli + AgN0 3 (10-4)
Ammonia undergoes autoionization with the formation of ammonium and amide
2Nh 3 nh; + NH 2 "
Neutralization reactions can be run that parallel those in water:
KNH, + NH 4 I - KI + 2NH 3 (10.6)
Furthermore, amphoteric behavior resulting from complex formation with excess
amide also parallels that in water:
Zn 2+ + 20H" -♦ Zn(OH), ! -♦ Zn(OH)J (10.7)
Zn 2+ + 2NH; -» Zn(NH 2 ),l -♦ Zn(NH 2 ) 2 " (10.8)
All acids that behave as strong acids in water react completely with ammonia (are
leveled) to form ammonium ions:
hcio 4 + NH 3
HN0 3 + NH 3
nh; + cio;
nh; + no 3 -
In addition, some acids which behave as weak acids in water (with pA'„ up to about 12)
react completely with ammonia and hence are strong acids in this solvent:
HC 2 H 3 0 2 + NH 3 -♦ NH 4 + + C 2 H 3 0 2 " (10.11)
Furthermore, molecules that show no acidic behavior at all in water may behave as
weak acids in ammonia:
NH 2 C(0)NH 2 + nh 3 nh; + nh 2 c(o»nh- < 10.12)
The basic solvent ammonia levels all species showing significant acidic tendencies and
enhances the acidity of very weakly acidic species.
Most species that would be considered bases in water are either insoluble or
behave as weak bases in ammonia. Extremely strong bases, however, may be leveled
to the amide ion and behave as strong bases:
H~ + NH 3 -
O 2 " +NH 3
NH," + H 2 T
► NH 2 " + OH'
Solvolysis reactions are well known in ammonia, and again many reactions
parallel those in water. For example, the solvolysis and disproportionation of halogens
may be illustrated by
366
10 Chemistry in Aqueous and Nonaqueous Solvents
Perchloric acid is one of the strongest acids known, but in sulfuric acid it is
practically a nonelectrolyte, behaving as a very weak acid:
HC10 4 + H 2 S0 4 H 3 S0 4 + + CIO; (10.38)
One of the few substances found to behave as an acid in sulfuric acid is disulfuric
(pyrosulfuric) acid. It is formed from sulfur trioxide and sulfuric acid:
S0 3 + H 2 S0 4 -♦ H 2 S 2 0 7 (10.39)
H 2 S 2 0 7 + H 2 S0 4 ,_=i H 3 S0 4 + + HS 2 0 7 - (10.40)
An exceptionally strong acid in sulfuric acid is hydrogen tetrakis(hydrogensulfato)-
borate, HB(HS0 4 ) 4 . The compound has not been prepared and isolated in pure form,
but solutions of it may be prepared in sulfuric acid:
H 3 B0 3 + 6H 2 S0 4 -► BIHSOj; + 3H 3 0 + + 2HS0 4 - V = 6 (10.41)
Addition of S0 3 removes the H 3 0 + and HS0 4 ions:
B(HS0 4 ) 4 - + 3H 3 0 + + 2HS0 4 + 3S0 3 -♦
H 3 SO; + B(HS0 4 ) 4 + 4H 2 S0 4 ( 10.42)
Some very strong acids have been termed “superacids." 9 They consist of simple
very strong Brpnsted acids such as disulfuric acid, very strong Lewis acids such as
antimony pentafluoride, or combinations of the two. One of the most interesting is
“magic acid," a solution of antimony pentafluoride in lluorosulfonic acid. It acquired
its name when a postdoctoral fellow happened to drop a small piece of a Christmas
candle (following a lab party) into such a solution: The paraffin candle dissolved! The
wax, composed of long-chain alkanes, would not be expected to be soluble in such a
very polar solvent. Furthermore, a 'H NMR spectrum of the sample showed a sharp
singlet characteristic of the /-butyl cation, indicating much cleavage and rearrange¬
ment. This type of reaction is most simply shown by the reaction with neopentane:
SbF 3 + 2HS0 3 F - FS0 3 SbFJ + H 2 S0 3 F (superacid) (10.43)
ch 3
H 3 C — C — CH 3 + superacid
CH t !_ CHj J (10.44)
The strongest known superacid is a solution of antimony pentafluoride in hydrogen
fluoride: 10
SbFj + 2HF - H 2 F + + SbF 6 - (10.45)
9 This term originated in the title of a paper by Hall. N. F.; Conant, J, B. J. Am. Client. Soc. 1927.49 ,
3047-3061, but modern work on superacids dates from the 1960s and 1970s: Gillespie, R. J.: Pcel.T.
E. Adv. Phys. Org. Client. 1972, 9. 1-24; Olah, G. A. Ant few. Client, hit. Ed. Engl. 1973, 12. 173-212.
Olah, G. A.; Prakash, G. K. S.; Sommer, J. Superacids'. Wiley: New York. 1985.
10 Gillespie, R. J.: Liang. I.J. Am. Cltem. Soc. 1988. 110. 6053-6057. Eq. 10.45 is simplified for dilute
solutions. As the concentration of SbF 5 increases, polyfluoroantimonate species are formed with
complex equilibria.
Nonaqueous Solvents
367
Even such unlikely bases as Xe, H 2 , Cl 2 , Br 2 , and C0 2 have been shown to accept H +
ions from superacids, though perhaps only to a small extent. There is no evidence that
Ar, 0 2 , or N 2 become protonated.
Summary of Despite certain differences, the three protonic solvents discussed above (water, am-
Protonic Solvents monia, and sulfuric acid) share a similarity in their acid-base behavior. All are
autoionizing, with the ionization taking place through the transfer of a proton from one
molecule of solvent to another with the formation of a solvated proton (Br0nsted acid,
solvent system acid) and a deprotonated anion (Br0nsted and Lewis base, solvent
system base). The inherent acidities and basicities of these three solvents differ,
however, and so their tendencies to protonate or deprotonate solutes differ. It is
possible to list solvents in order of their inherent acidity or basicity. Water is ob¬
viously less acidic than sulfuric acid but more so than ammonia. Glacial acetic acid lies
between water and sulfuric acid in acidity. Figure 10.1 graphically illustrates the
relative acidities and basicities of four solvents, together with various acid-base
conjugate pairs. They are listed in order of the p K t , in water. In an ideal aqueous
solution the pH of an equimolar mixture of conjugates is given by the p K a . and similar
acidity scales may be used in other solvents. The p K a is thus a rough estimate of
acidity in solvents other than water. Any given acid is stronger than the acids listed
above it and, conversely, any base is stronger than the bases below it. All species that
lie within the extremes of a particular solvent behave as weak electrolytes in that
solvent and form weakly acidic or weakly basic solutions. All species that lie beyond
the enclosed range are leveled by the solvent.
An example may serve to illustrate the information that may be obtained from
Fig. 10.1. Consider acetic acid. In water, acetic acid behaves as an acid or, to be more
precise, an equimolar mixture of acetic acid and an acetate salt will have a pH of 4.74.
If acetic acid is added to sulfuric acid, it will behave as a base and be leveled to
CH 3 C(OH) 2 , the acetic acidium ion, and HSOJ (cf. Eq. 10.36; note the equilibrium
lying at about -9 on the scale in Fig. 10.1).
If dissolved in ammonia, acetic acid will behave as a strong acid and be leveled to
NH 4 and CH 3 COO (cf. Eq. 10.11; note equilibrium lying at about 4.7 on the scale in
Fig. 10.1). The different behavior of acetic acid as a base (sulfuric acid), a strong acid
(ammonia), or a weak acid (water) depends upon the acidity or basicity of the solvent.
The “equilibrium boxes" for the solvents (Fig. 10.1) indicate the range over
which differentiation occurs; outside the range of a particular solvent, all species are
leveled. For example, water can differentiate species (i.e., they are weak acids and
bases) with p/Cj/s from about 0 to 14 (such as acetic acid). Ammonia, on the other
hand, behaves the same toward acetic acid and sulfuric acid because both lie below
the differentiating limit of ~12. The extent of these ranges is determined by the
autoionization constant of the solvent (e.g., ~I4 units for water). The acid-base
behavior of several species discussed previously may be seen to correlate with
Fig. 10.1."
A complete discussion of relative acidities and basicities would be too extensive
to be covered here. Nevertheless it is possible to summarize the behavior of acids and
bases as involving (I) the inherent acidity-basicity of the solvent, (2) the inherent
acidity—basicity of the solute, and (3) the interaction of solute and solvent to form an
11 For a more extensive discussion of the use of conjugate acid-base charts like Fig. 10.1, see
Treptow, R. S. J. Chem. Educ. 1986, 63. 938-941.
362
10 Chemistry in Aqueous and Nonaqueous Solvents
Cl 2 + 2H 2 0-► HOC1 + H 3 0 + + CP (10.15)
Cl 2 + 2NH 3 - NH 2 C1 + NH 4 + + Cl" (10.16)
Since it is more basic than water, ammonia can cause the disproportionation of sulfur:
5S 8 + 16NH 3 - 4S 4 N~ + 4S|~ + 12NH 4 + (10.17)
The hexasulfide ion is in dissociative equilibrium: 1 2
Sl~ < -r 2S3; (10.18)
The S 3 ion is responsible for the deep blue color of these solutions (A max = 610 nm).
This ion is also responsible for the color of sulfur dissolved in chloride melts (see
below) and in the aluminosilicate known as ultramarine (see Chapter 16). Many
nonmetal halides behave as acid halides in solvolysis reactions:
OPCl 3 + 6H 2 0 -► OP(OH) 3 + 3H 3 0 + + 3CP (10.19)
opci 3 + 6nh 3 —► op(nh 2 ) 3 + 3nh; + 3cr (10.20)
The resemblence of these two reactions and the structural resemblance between
phosphoric acid [OP(OH) 3 ] and phosphoramide [OP(NH,) 3 ] has led some people to
use the term "ammono acid" to describe the latter.
In a manner analogous to that used for water, a pH scale can be set up for
ammonia: pH = 0(1 M NH 4 + ); pH = 13 ([NH 4 + ] = [NHTD. neutrality: pH = 26(1 M
NH 2 ). Likewise oxidation-reduction potentials may be obtained, based on the hydro¬
gen electrode (see page 379):
NH 4 + e" = NH, + |H 2 &' = 0 (10.21)
In summary, the chemistry of ammonia solutions is remarkably parallel to that of
aqueous solutions. The principal differences are in the increased basicity of ammonia
and its reduced dielectric constant. The latter not only reduces the solubility of ionic
materials, it promotes the formation of ion pairs and ion clusters. Hence even strong
acids, bases, and salts are highly associated.
Solutions of Metals If a small piece of an alkali metal is dropped into a Dewar flask containing liquefied
in Ammonia ammonia, the solution immediately assumes an intense deep blue color. If more alkali
metal is dissolved in the ammonia, eventually a point is reached where a bronze-
colored phase separates and floats on the blue solution.- Further addition of alkali
metal results in the gradual conversion of blue solution to bronze solution until the
former disappears. Evaporation of the ammonia from the bronze solution allows
one to recover the alkali metal unchanged. 3 This unusual behavior has fascinated
chemists since its discovery in 1864. Complete agreement on the theoretical interpre¬
tation of experimental observations made on these solutions has not been achieved.
1 The solution is complex with further equilibria and disproportionation reactions. Sec Dubois. P.:
Lclicur, J. P.: Lepoutre, G. Inorg. Client. 1988, 27, 1883-1890; 1989, 28. 195-200.
2 Cesium appears to be an exception. Although the solution changes from blue to bronze with
increasing concentration, a two-phase system is never obtained.
3 One must be very careful to exclude water and other materials which might react with the alkali
metal and thus prevent the reversibility of the solution.
Nonaqueous Solvents
363
but the following somewhat simplified discussion will indicate the most popular
interpretations. 4
The blue solution is characterized by (1) its color, which is independent of the
metal involved: (2) its density, which is very similar to that of pure ammonia; (3) its
conductivity, which is in the range of electrolytes dissolved in ammonia; and (4) its
paramagnetism, indicating unpaired electrons, and its electron paramagnetic reso¬
nance g-factor, which is very close to that of the free electron. This has been
interpreted as indicating that in dilute solution, alkali metals dissociate to form alkali
metal cations and solvated electrons:
M M+ + [e(NH3) J- (10 . 22)
Hie dissociation into cation and anion accounts for the electrolytic conductivity. The
solution contains a very large number of unpaired electrons, hence the paramag¬
netism, and the g value indicates that the interaction between solvent and electrons is
rather weak. It is common to talk of the electron existing in a cavity in the ammonia,
loosely solvated by the surrounding molecules. The blue color is a result of a broad
absorption peak that has a maximum at about 1500 nm. This peak results from an
absorption of photons by the electron as it is excited to a higher energy level, but not
all workers are in agreement as to the nature of the excited state.
The very dilute solutions of alkali metals in ammonia thus come close to present¬
ing the chemist with the hypothetical "ultimate" base, the free electron (Chapter 9).
As might be expected, such solutions are metastable, and when catalyzed, the elec¬
tron is "leveled" to the amide ion:
[e(NH 3 )J“ NHJ + H 2 + (x - 1)NH 3 (10.23)
The bronze solutions have the following characteristics: (I) a bronze color with a
definite metallic luster; (2) very low densities; (3) conductivities in the range of metals;
and (4) magnetic susceptibilities similar to those of pure metals. All of these properties
are consistent with a model describing the solution as a "dilute metal” or an "alloy"
in which the electrons behave essentially as in a metal, but the metal atoms have been
moved apart (compared with the pure metal) by interspersed molecules of ammonia.
The nature of these two phases helps to throw light on the metal-nonmelal
transition. For example there has been much speculation that hydrogen molecules at
sufficiently high pressure, such as those occurring on the planet Jupiter, might undergo
a transition to an "alkali metal." The fundamental transition is one of a dramatic
change of the van der Waals interactions of H, molecules into metallic cohesion. 5
Solutions of alkali metals in ammonia have been the best studied, but other metals
and other solvents give similar results. The alkaline earth metals except beryllium
form similar solutions readily, but upon evaporation a solid "ammoniate," M(NH 3 ) r ,
is formed. Lanthanide elements with stable + 2 oxidation states (europium, ytterbium)
also form solutions. Cathodic reduction of solutions of aluminum iodide, beryllium
chloride, and tetraalkylammonium halides yields blue solutions, presumably contain¬
ing Al ,+ , 3e ; Be 2+ , 2e"; and R 4 N ', e" respectively. Other solvents such as various
amines, ethers, and hexamethylphosphoramide have been investigated and show
some propensity to form this type of solution. Although none does so as readily as
ammonia, stabilization of the cation by complexation results in typical blue solutions
4 Edwards. P. P. Adv. Inorg. Chem. Radiochem. 1982. 25, 135-185.
3 Edwards. P. P.; Sienko. M. J. J. Am.-Chem. Soc. 1981, 103. 2967. See also Footnote 33 in Chapter 7.
364
365
1 0 ■ Chemistry ii
Sulfuric Acid
Toblo 10.3
Physicol properties of
sulfuric acid
Aqueous and Nonaqueous Solvents
in ethers/’ The solvated electron is known even in aqueous solution, but it has a very
short (~I0 -3 s) lifetime.
These solutions of electrons are not mere laboratory curiosities. In addition to
being strong bases, they are also good one-electron reducing agents. For example,
pure samples of alkali metal superoxides may be readily prepared in these solutions:
M + + e" + O, -> M + + 07 ( 10 - 24 >
The superoxide ion is further reducible to peroxide:
M + + e - + 07 - M + + (10 ' 25)
Some metal complexes may also be forced into unusual oxidation states:
[Pt(NH 3 )J 2+ + 2M + + 2e" - [Pt(NH 3 ) 4 ] + 2M" (10.26)
Mo(CO) ( , + 6Na + + 6e" -♦ Na 4 [Mo(CO) 4 ] + Na,C,0, (10.27)
Au + M + + e - - M + + Au“ (10.28)
The chemistry of metal electrides has been extensively studied and although the
formulation M + e - is undoubtedly the best, most chemists have the all-too-human
emotion of feeling more secure in their science if they have something more tangible
than solutions and equations on paper. Therefore the isolation and structural charac¬
terization of cesium electride, [Cs(ligand)] ^e . as single crystals was welcome, in¬
deed. 7 The crystals are dark blue with a single absorption maximum at 1500 nm, have
no likely anions present (the empirical formula is 1:1, Cs:ligand. with a trace of lithium
impurity, an artifact of the synthetic technique), and are most readily formulated as a
complex of cesium electride.
The physical properties of sulfuric acid are listed in Table 10.3. The dielectric constant
is even higher than that of water, making it a good solvent for ionic substances and
leading to extensive autoionization. The high viscosity, some 25 times that ol water,
introduces experimental difficulties: Solutes are slow to dissolve and slow to
crystallize. It is also difficult to remove adhering solvent from crystallized materials.
Furthermore, solvent that has not drained from prepared crystals is not readily
removed by evaporation because ot the very low vapor pressure of sulluric acid.
Boiling point
Freezing point
Density
Permittivity (dielectric constant)
Specific conductance
Viscosity
Ion product constant
300 °C (with decomposition)
10.371 °C
1.83 g cm -3 (25 °C)
110e o (20 °C)
1.04 x I0 -2 n _l cm 1 (25 °C)
24.54 g cm -1 s -1 (20 °C)
2.7 x I0 -4 mol 2 L -2 (25 °C)
o Dye. J. L.; DcBackcr, M. G.; Nicely. V. A. J. Am. Chem. Soc. 1970. 92. 5226-5228.
7 Issa. D. ; Dye. J. L. J. Am. Chem. Soc. 1982. 104. 3781. For the crystal structure of this compound,
see Chapter 12. Fig. 12.50b.
Nonaqueous Solvents
Autoionization of sulfuric acid results in the formation of the hydrogen sulfate
(bisulfate) ion and a solvated proton:
2H,S0 4 H 3 S0 4 + HS07 (10-29)
As expected, a solution of potassium hydrogen sulfate is a strong base and may be
titrated with a solution containing H 3 S0 4 ions. Such a titration may readily be
followed conductometrically with a minimum in conductivity at the neutralization
point. 8
Another method that has proved extremely useful in obtaining information about
the nature of solutes in sulfuric acid solution is the measurement of freezing point
depressions. The freezing point constant ( k ) for sulfuric acid is 6.12 kg °C mol -1 . For
ideal solutions, the depression of the freezing point is given by
AT = kmv (10.30)
where m is the stoichiometric molality and v is the number of particles formed when
one molecule of solute dissolves in sulfuric acid. For example, ethanol reacts with
sulfuric acid as follows:
C,H 5 OH + 2H 2 S0 4 -♦ C 2 H 5 HS0 4 + HS07 + H 3 0 + v = 3 (10.31)
It is found that all species that are basic in water are also basic in sulfuric acid:
OH" + 2H 2 S0 4 -♦ 2HSO; + H 3 0 + v = 3 (10.32)
NH 3 + H 2 S0 4 -♦ HSO; + NH; v = 2 (10.33)
Likewise, water behaves as a base in sulfuric acid:
H 2 0 + H 2 S0 4 -♦ HSO; + HjO + v = 2 (10.34)
Amides, such as urea, which are nonelectrolytes in water and acids in ammonia accept
protons from sulfuric acid: I
NH 2 C(0)NH 2 + h 2 so 4 -♦ HSO; + NH 2 C(0)NH3 + V = 2 (10.35)
Acetic acid is a weak acid in aqueous solution and nitric acid a strong acid, but both
behave as bases in sulfuric acid!
^.OH
CHjC + H,S0 4 -► HSO; + CH 3 C' v = 2 (10.36)
^OH ‘ ^OH
HN0 3 + 2H 2 S0 4 -- 2HSO; + NO; + H 3 0 v = 4 (10.37)
Sulfuric acid is a very acidic medium, and so almost all chemical species which react
upon solution do so with the formation of hydrogen sulfate ions and are bases.
Because of the extreme tendency of the H 2 S0 4 molecule to donate protons, molecules
exhibiting basic tendencies will be leveled to HSOJ.
8 This statement is not quite true. The concentration of ions is at a minimum at the neutralization
point, but since the conductivity depends on viscosity as well (which changes with composition), the
absolute minimum conductivity docs not occur exactly when (H 3 S0 4 ] = [HS0 4 ]. The slight
difference is not important in practice, however.
452
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Electronic Spectra of Complexes
453
separated bands in the electronic spectra. A case in point is [Ti(H 2 0) 6 ] 3+ , for which
the two absorptions appear as one broad peak with a low-energy shoulder (Fig. 11.8).
As we have seen (Table 11.21), all of the ML 6 complexes that are susceptible to
Jahn-Teller distortion have octahedral configurations that involve asymmetric elec¬
tron occupation of either the e or t 2g orbitals. Generally speaking, the former leads to
considerably more pronounced distortions than does the latter. This occurs because
the e level is much more involved in the <r bonding than is the r 2g . Hence complexes
with e 1 or e } configurations (from high spin d 4 and d b , low spin d 1 , or d 9 ) often
exhibit substantial distortions. It is not uncommon for these complexes to have bond
length differences (two longer and four shorter or vice versa) that can be detected
crystallographically at room temperature. In fact, some of the strongest evidence for
Jahn-Teller effects in transition metal compounds comes from structural studies of
solids containing the d 9 Cu 2+ ion. Distortion by either elongation or compression will
lead to stabilization of a copper(II) complex. However, experimental measurements
show that the distortion is generally elongation along the z axis (Fig. 11.49). Table
11.22 lists some bond distances found in crystals containing hexacoordinate Cu(Il)
ions. Each compound has both shorter and longer bonds. It is of interest that the
“short” bonds represent a nearly constant radius for the Cu _+ ion, whereas the
“long” bonds show no such constancy. This suggests that the short bonds represent a
lower limit or starting point from which various degrees of distortion in the form of
bond lengthening can occur.
We have fewer data to support Jahn-Teller distortion in high spin d 4 or low spin
d 1 complexes. Chromium(II) and manganese(III) are d 4 ions, and both have been
found to be distorted in some compounds (see Table 11.22). Furthermore, extensive
studies of six-coordinate manganese(lll) compounds have shown that their spectra
can be readily interpreted in terms of elongation along the z axis. The d 1 configuration
of low spin Co 2+ is less straightforward. With ligands that are sufficiently strong field
to induce pairing, the ion tends to form five- or four- rather than six-coordinate com¬
plexes. For example, the expected hexacyano complex, [Co(CN) 6 ] 4 , is not found, but
instead the principal species in solution has five cyano groups per cobalt and is probably
[Co(CN) 5 H 2 0] 3_ . This might be viewed as an extreme form of Jahn-Teller distortion,
namely complete dissociation of one cyanide from the hypothetical [Co(CN)J 4 ~ ion.
Six-coordinate Co 2+ complexes are observed for the ligands bis(salicylidene)ethyl-
enedianiine (H 2 salen) and nitrite. The spectra for these species are quite complicated,
but they are consistent with axially elongated structures.
Fig. 11.49 Orbital energy
level diagrams for d 9
configuration in octahedral
(O h ) and c-out tetragonal
(D ih ) fields.
Table 11.22 _
Some typical metal-ligand
distances in Cu(ll), Cr(ll),
and Mn(lll) compounds 0
Compound
Short distances
rut"
Long distances
r M b
CuF 2
4F at 193
122
2F at 227
156
CuFy2H 2 0
2F at 190
119
2F at 247
176
20 at 194
121
NaXuF 4
4F at 191
120
2F at 237
166
K,CuF 4
4F at 192
121
2F at 222
151
NaCuF 3
2F at 188
117
2 F at 226
155
2F at 197
126
KCuF 3
2F at 189
118
2F at 225
154
2F at 196
125
CuCI,
4C1 at 230
131
2CI at 295
196
CuC1,-2H 2 0
2C1 at 229
130
2CI at 294
195
20 at 196
123
CuCI,-2C,H,N
2N at 202
127
2CI at 305
206
2CI at 228
129
Cu(NH 3 ) 2+
4N at 207‘‘
132
2N at 262<
187
CrF 2
4F at 200
119
2F at 243
172
KCrFj
2F at 200
119
4F at 214
143
MnFj
2F at 179
108
2 F at 209
138
2F at 191
120
K 2 MnF,H,0
4F at 183
112
2 F at 207
136
“ Data, unless otherwise noted, front Wells, A. F. Structural Inorganic
Chemistry, 5th cd.; Oxford University Press: London, 1986. Wells lists many
additional data.
'' All distances arc in picomcters. The radius of the metal was obtained by
subtracting the covalent radius of the ligating atom (Table 8.1) from the
M—X distance.
c Tistler, T.; Vaughan, P. A. Inorg. Chem. 1967, 6, 126.
Complexes having measurable bond-length differences, such as those reported in
Table 11.22, are examples of static Jahn-Teller behavior. In some other complexes,
no distortion can be detected in the room temperature crystal structure, but additional
evidence shows that the Jahn-Teller effect is nonetheless operative. The supplemen¬
tary evidence may consist of a low-temperature crystal structure showing distortion or
spectroscopic data consistent with tetragonal geometry. These complexes are display¬
ing dynamic Jahn-Teller behavior. In its simplest form, this can be thought of as a
process in which a complex oscillates among three equivalent tetragonal structures. At
any instant, the complex is distorted, but if the oscillation between forms is rapid
enough, the structure observed by a particular physical method may be time-averaged
and therefore appear undistorted. Sometimes cooling a sample will slow the oscilla¬
tions enough that a single distorted structure is “frozen out." In some instances,
however, a distorted structure produced upon cooling does not represent a true static
condition but rather a different form of dynamic behavior.
There is an interesting series of compounds, all of them containing the hexa-
nitrocuprate(II) ion ([Cu(N0 2 ) 6 ] 4- ), which exhibit the full range of static and dy¬
namic Jahn-Teller effects described above. 50 In some members of the series, such as
o.
50 Hathaway B. J. Struct. Bonding ( Berlin ) 1984, 57, 54-118.
582
14 • Some Descriptive Chemistry of the Metal:
substitution of Cr 3+ for Al 3 + . In this case the two cations have the same charge and
similar radii (r AI 3+ = 67.5 pm; r Cr 3> = 75.5 pm). There are also examples of re¬
semblance between Mg 2 + (r = 71 pm), Mn 2 + (r = 80 pm), and Zn 2+ (r = 74 pm).<'
despite the fact that one has a noble gas configuration ( s 2 p 6 ) and the others do not ( d 5
and d' : )J The remarkable resemblances among the lanthanides bear witness to the
overwhelming influence of identical charge and similar size in these species. Like¬
nesses that depend more on charge than on electron configuration might be termed
physical. They relate to crystal structure and hence to solubilities and tendencies to
precipitate. Coprecipitation is often more closely related to oxidation state than to
family relationships. Thus carriers for radioactive tracers need not be of the same
chemical family as the radioisotope. Technetium(VII) may be carried not only by
pcrrhenate but also by perchlorate, periodate, and tetrafluoroborate. Lead(II) has the
same solubility characteristics as the heavier alkaline earth metals. Thallium(I) (r =
164 pm) often resembles potassium ion (r = 151 pm), especially in association with
oxygen and other highly electronegative elements. Thus, like K + , Tl + forms a soluble
nitrate, carbonate, phosphate, sulfate, and fluoride. Thallium(I) can also be incorpo¬
rated into many potassium enzymes and is exceedingly poisonous. Of course, some
properties of cations (especially polarization of anions) are affected by electronic
structure (see Chapter 4). Thus we should not be surprised that with respect to the
heavier halogens, Tl + resembles Ag + more closely than it does K + .
Finally, another chemical property that depends on the cationic charge is the
coordination number. Although it is greatly influenced by size (Chapter 4), there is a
tendency for cations with larger charges to have larger coordination numbers, e.g.,
Co 2 (C.N. = 4 and 6) versus Co 3 + (C.N. = 6), Mn 2+ (C.N. = 4 in IMnCIJ 2- )
versus Mn‘‘ (C.N. = 6 in [MnF fi ] 2 ). This is a consequence of the electroneutrality
principle (see Chapter II). On the other hand, metals in extremely high oxidation
states [Cr(VI), Mn(VII), Os(VIII)) have a tendency to form metal-oxygen double
bonds (considerable u bonding from the oxygen to the metal), and four-coordinate
tetrahedral species ([CrOJ 2 , [MnOJ - , 0s0 4 ) result.
The Chemistry
of Elements
Potassium-Zinc:
Comparison
by Electron
Although there are resemblances that depend only on the charge or oxidation state,
transition metal chemistry is more often governed by the electron configurations of the
metal ions. Thus, despite a natural tendency for lower oxidation states to be reducing
in character and higher ones to be oxidizing, the electron configuration may well make
the divalent, trivalent, or even higher oxidation states the most stable for a particular
metal. In this section the properties of the metals of the first transition scries are
briefly examined in terms of electron configuration. 8
Configuration
A The similarity of the names of magnesium and manganese results from the confusion of these two
elements by early chemists, an error which has persisted among neophyte chemists to this day. Both
names derive ultimately from Magnesia, an ancient city in Asia Minor.
7 Note that Mg' most closely resembles Zn 3 and high spin Mn 3 . The resemblance of Mg 3 to the
other +2 transition metal cations is less because of LFSEs in complexes of the latter.
8 It is often helpful to view the descriptive chemistry of the transition metals from different perspec¬
tives in a comparative study. For a thorough review of transition metal chemistry in an clemcnt-by-
clcment approach, sec Cotton. F. A.: Wilkinson, G. Advanced Inorganic Chemistry, 5th cd.: Wiley:
New York. 1988. or Greenwood, N. N.; Eamshaw, A. Chemistry of the Elements: Pergamon:
Oxford. 1984.
The Chemistry of Elements Potassium-Zinc
583
The d 1
Configuration
I
The d 2
Configuration
This configuration occurs for simple ions such as K + , Ca 2 + , and Sc 3+ and for the
formal oxidation states equal to the group numbers for many of the transition metals.
This holds true as far as Mn(VII) (Fe(VIII) is unknown].
All metal ions with d° configurations are hard acids and prefer to interact with
hard bases such as oxide, hydroxide, or fluoride.'' Complexation chemistry is less
extensive than for other configurations, but work in this area continues to expand.
There is no tendency for metals with this configuration to behave as reducing agents
(there are no electrons to lose) and little tendency for them to behave as oxidizing
agents until species such as [Cr0 4 ] 2 ~, [Cr0 3 Cl]-, CrCLCL. and [Mn0 4 r are
reached. In general, therefore, their aqueous chemistry may be described simply: The
lower charged species (K , Ca 2 + , Sc 3+ ) behave as simple, uncomplexed (other than
by water) free ions in aqueous solution. 10 Complex ions of scandium, such as [ScF 6 ] 3 ~
and (Sc(OH)J 3- . are known and result when excess F" or OH - is added to insoluble
ScFj or Sc(OH),. The "crown" complexes of the alkali metals have been discussed
previously (Chapter 12). and Ca 2 may be complexed by polydentate ligands such as
edta." The higher oxidation states [Cr(VI). Mn(VII)] tend to form oxyanions. which
are good oxidizing agents, especially in acidic solution: the oxides of the intermediate
species are insoluble (TiO,) or amphoteric (V ; 0„ [VOJ 3 ).
A quick survey of the metals with the d° configuration yields the following
descriptive information. For potassium, calcium, and scandium it is the only stable
electron configuration. It is by far the most stable for titanium (e.g., TiO,, TiCI 4 ,
[TiFJ 2 ”). Vanadium! V) occurs in the vanadate ion. [V0 4 ] 3- , and a variety'of poly-
vanadates. It is a mild oxidizing agent giving way to vanadium(lV) as the most stable
oxidation state,': The strongly oxidizing species |HCrOJ (£° to Cr 3 is 1.37 V) and
]Mn0 4 ] (£° to Mn' ' is 1.51 V) are unstable relative to lower oxidation states.
This does not tend to be a stable configuration. It is completely unknown for scandium
and strongly reducing in Tit III). The later members of the series tend to disproportion¬
ate to more stable configurations:
3[Cr0 4 ] 3 + 10H’ -► 2 [HCrO,] + Cr 3 + 4H,0 (14.5)
3[Mn0 4 ] 2 + 4H -► 2 [MnOJ - + Mn0 3 + 2H ; 0 (14.6)
The only d' species of importance is the vanadyl ion. VO 2 \ which is the most stable
form of vanadium in aqueous solution.
This configuration ranges from Ti(II). very strongly reducing, to Fc(VI). very strongly
oxidizing. It is not a particularly stable configuration. Both Ti(II) and V(IJI) are
reducing agents and Cr(IV) and Mn(V) are relatively unimportant. The ferrate(Vl) ion.
Sofl hases wiI1 lorm complexes in many cases if hard ligands are absent (Fryzuk. M. D.; Haddad,
T. S.; Berg, D. J. Coord. Client. Rev. 199(1. 99. 137-212).
1,1 Hydrolysis of Sc is extensive in aqueous solution tscc Brown, P. I..; Ellis J.: Sylva, R, N. J.
Client. Soc. Dulton Trans. 1983. 35-43).
11 Scc Sohmidbaur. H,: Classen. H. G.; Helbig, J. Angew. Client. Ini. Ed. Engl. 1990, 29. 1090-1103
for biologically important complexes of alkaline earth and alkali metal ions. Coordination chemistry
of alkali metal and alkaline earth ions is also discussed in Poonia. N. S.: Bajaj, A V. Client Rev
1979. 79. 389-445.
12 See Chapter 16 and Fig. 16.9 for a full discussion of the oxyanions of vanadium.
454
1 1 • Coordination Chemistry: Bonding, Spectra, and Magnetism
K 2 Ba[Cu(N0 2 ) 6 J. K 2 Ca[Cu(N0 2 ) fi ], and K 2 Sr[Cu(N0 2 ) h ], the Cu(ll) anions are elon¬
gated at room temperature (298 K), while in others, such as K 2 Pb[Cu(N0 2 ) 6 ], they are
undistorted. In yet a third category are Cs 2 Ba[Cu(N0 2 ) 6 ] and Rb 2 Ba[Cu(N0 2 ) 6 ], for
which the room temperature structures appear to be axially compressed octahedra but
actually are the dynamic averages of two tetragonally elongated structures. Upon
being cooled to 276 K. K 2 Pb[Cu(N0 2 ) 6 ] also assumes this "pseudo compressed"
geometry. The subtle structural variations in these and other complexes exhibiting
similar behavior have been elucidated with a combination of physical methods, most
often crystallography in conjunction with EPR and electronic spectroscopies. 51
No discussion of the Jahn-Teller effect in coordination compounds would be
complete without including the special features of chelated compounds (see Chapter 12
for a more thorough discussion of chelated complexes). The very nature of the
chelated ring tends to restrict the distortion of a complex from a perfect octahedron
because the ligand will have a preferred "bite" or distance between the coordinating
atoms:
/CH, —CH,.
H,N ' NH,
2_2 '
bite
An example of the conflict between stabilization from the Jahn-Teller effect and
chelate geometrical requirements is found in the ethylenediamine complexes of Cu 2 .
Most divalent transition metal ions form complexes with ethylenediamine (en) by
stepwise replacement of water:
[M(H 2 0) 6 ] 2 + + en - [M(H 2 0) 4 en] 2 + 2H,0 (11.27)
[M(H 2 0) 4 en] 2 + + en - [M(H 2 0) 2 (en) 2 ] 2 + 2H 2 0 (11.28)
[M(H 2 0) 2 (en) 2 ] 2+ + en -+ [M(en) 3 ] 2+ + 2H,0 (11.29)
Each step has associated with it an equilibrium or stability constant, Kj, A' 2 . or K y
which measures the tendency for formation of a mono-, bis-, or tris(cthylenediamine)
complex, respectively. The values of these constants for the ions Mn’ to Zn 2+ show
a rather uniform trend of gradually increasing stability from left to right across the
series (the Irving-Williams order) (Fig. 11.50). The Cu 2+ ion provides a striking
exception, however, with the tris(ethylenediamine)copper(!I) complex. [Cu(en)-,] 2+ ,
being remarkably unstable. In fact, at one time the very existence of this species was
questioned. Although it was subsequently prepared. 52 the value of K i (a measure of
the tendency to add the third ethylenediamine ligand) is the lowest of the ions in the
scries even though the K , and K 2 values are the highest. This lack of stability for the
Iris complex can be traced directly to the tendency for a six-coordinate d 9 ion to
undergo distortion. The bis(ethylenediamine) complex. [Cu(en) 2 (H 2 0) 2 ) 2 . can distort
readily by letting the two trans water molecules move out from the copper, leaving the
51 Ammeter. J. H.: Burgi, H, B.; Gamp. E.; Mcycr-Sandrin. V.: Jensen, W. P. inarg. Cliem. 1979. IS.
733-750.
52 Gordon. G.; Birdwhistelt. R. K. J. Am. Chem. Soc. 1959. SI. 3567-3569.
Electronic Spectra of Complexes
455
Charge Transfer
Spectra 54
Fig. 11.50 Stepwise
stability constants. K h K 2 .
and K } . for ethylenediamine
complexes of several first-
row transition metals in
aqueous solution.
two ethylenediamine rings relatively unchanged. In contrast, the tris(ethylenediamine)
complex cannot distort tetragonally without straining at least two of the chelate rings:
(11.30)
Alternatively, it is possible that the constraint of a chelate ring system can prevent
tetragonal distortion altogether, but the resulting perfectly octahedral complex will
lack the stabilization inherent in Jahn-Teller distortion. Despite the restraining in¬
fluence of a bidentate ligand, a number of distorted chelated structures are known. For
example, [Cu(bpy)(hfa) 2 ] is known to be tetragonally distorted. 53 Both bipyridine and
hexafluoroacetylacetonate are chelating ligands that form bonds through nitrogen and
oxygen atoms, respectively. The structure of this molecule is shown in Fig. 11.51. The
two nitrogen atoms of bipyridine bind to the copper at a distance of 200 pm, consistent
with the short bonds from other nitrogen ligands to copper(II) (Table 11.22). One
oxygen from each hexafluoroacetylacetonate ligand binds at a distance of 197 pm,
again consistent with short Cu—O bonds. The two remaining oxygen atoms form
Cu—O bonds which are some 33 pm longer, indicating severe Jahn-Teller distortion.
Although most of the visible spectra studied by inorganic chemists in evaluating
coordination compounds have been of the d-d (or ligand field) type, perhaps more
53 bpy = 2.2'-bipyridinc =
o~o
\=N N=/
hfa = hexafluoroacetylacetonate. [CFjC(O)CHC(0)CFjl” (= 1,1,1,5.5.5-hcxalluoropentanc-
2.4-dionate).
54 Lever. A. B. P. Inorganic Electronic Spectroscopy. 2nd ed,; Elsevier: New York. 1986; Chapter 5.
Lever. A. B. P. J. Chem. Educ. 1974. 51. 612-616.
580
14'Some Descriptive Chemistry of the Metals
Table 14,2 _
Some oxidation states of the metals of the first transition series
Group
Oxidation
state
II IB
(3)
IVB
(4)
VB
(5)
VIB
(6)
VIIB
(7)
7
[MnOJ -
6
[CrOJ 2 -
[MnOJ 2-
5
[vo 4 ] 3 -
[Cr0 4 ] 3 "
[Mn0 4 ] 3_
4
TiCl 4
[VOFJ 2 -
[Cr0 4 ] 4-
Mn0 2
3
[Sc(H 2 0) 6 ] 3+
[Ti(H 2 0) 6 ] 3+
[V(H 2 0) 6 ] 3+
[Cr(H 2 0) 6 ] 3+
[Mn(CN) 6 ] 3-
2
TiCI 2
[V(H 2 0) 6 ] j+
[CrfHjOJJ 24
[Mn(H 2 0) 6 ] 2+
1
[V(bpy) 3 J +
[Cr(bpy) 3 ] +
[Mn(CN) 6 ] 5_
0
Ti(bpy) 3
V(bpy) 3
Cr(bpy) 3
Mn(bpy) 3
-1
[Ti(bpy) 3 r
[V(bpy) 3 r
[Cr(bpy) 3 r
[Mn(bpy) 3 r
-2
[THCOy 2 -
[Cr(CO) 5 ] 2-
[Mn(H 2 pc] 2 ~
-3
[V(CO) 5 ] 3-
[Cr(bpy) 3 ] 3 -
[Mn(CO) 4 ] 3 -
-4
[Cr(CO) 4 J 4-
The +8 oxidation state is found in Ru0 4 and 0s0 4 and the +7 oxidation state is found in their anions, [Ru0 4 ] and [0s0 4 ] -
Chemistry of the
Various Oxidation
States of
Transition Metals
Low and Negative
Oxidation States
The entire question of oxidation state 5 is an arbitrary one and the assignment of
appropriate oxidation states is often merely a matter of convenience (or inconven¬
ience!). The concept of oxidation state is best defined in compounds between
elements of considerably different electronegativity in which the resulting molecular
orbitals are clearly more closely related to the atomic orbitals of one atom than
another. In those cases in which the differences in electronegativity are small and
especially those in which there are extensive delocalized molecular orbitals that are
nonbonding, weakly bonding, or antibonding, the situation becomes difficult. The
former situation is found with complexes containing halogen, oxygen, or nitrogen
<7-bonding ligands. The latter condition is common among organometallic complexes
for which often no attempt is made to assign oxidation states. Ligands stabilizing low
(if imprecise) oxidation states are cyanide and phosphorus trifluoride, both excellent
7 r-bonding ligands. Hence it is possible to prepare zero-valent nickel complexes with
these ligands:
Ni(CO) 4 + 4PFj
Nj2 + jccn^ Kz [- Ni(CN)4 j
Ni(PF 3 ) 4 + 4CO
.]— K 4 [Ni(CN) 4 ]
Ligands with extensively delocalized molecular orbitals that are essentially non¬
bonding can make the assignment of precise oxidation states difficult or impossible.
For example, we have already seen this in the thiolene-thiolate ligands (see Chap¬
ter 12). A similar ligand is bipyridine, which forms complexes that may formally be
classified as containing + 1,0, or even — 1 oxidation states for the metal. A substantial
portion of the electron density on the metal in these low oxidation states is delocalized
over the ligand tr system. Other instances of extensive delocalization stabilizing
5 For a book devoted entirely to the problems related to defining oxidation states, see Jprgensen, C. K.
Oxidation Numbers and Oxidation States; Springer: New York, 1969.
Chemistry of the Various Oxidation States of Transition Metals
581
Group
VIIIB
IB
IIB
(8)
(9)
(10)
(ID
(12)
[Fe0 4 ] 2 ~
(FeOJ 3 -
[FeOJ 4 -
[Fe(H,0) 6 ] 3+
[FetHjOJ 2
[Fe(N0)(H 2 0) 5 ] 2+
Fe(bpy) 3
[Fe(bpy) 3 ]
[Fe(CO) 4 ] 2 -
[Co( I -norborny 1) 4 ] +
[CoFJ 2 -
[Co(CN) 5 ] 3 ~
[Co(H 2 0) 6 ] 2+
[Co(bpy) 3 ] +
Co(bpy) 3
[Co(CO) 4 r
[NiFJ 2 -
[NiFJ 3 -
[Ni(H 2 0)J 2+
Ni(PPh 3 ) 3 Br
Ni(bpy) 2
[Ni^COlJ 2 -
[CuF 6 ] 2-
[CuF 6 J 3 -
[Cu(H 2 0) 6 ] 2+
[Cu(CN),]~
[ZnfHjOy 24
Zn 2 Cl 2
Zn(bpy) 2
metals in low oxidation states are often encountered in biological systems (see Chap¬
ter 19).
Range of Although the definition of low oxidation states is somewhat subjective, it is possible to
Oxidation States discuss the range of oxidation states exhibited by various metals. When the metals of
the first transition series are examined, the results shown in Table 14.2 are found.
There is a general trend between a minimum number of oxidation states (one or two) at
each end of the series (Sc 3 + and Zn 2 + ) to a maximum number in the middle (man¬
ganese, -3 to +7). The paucity of oxidation states at the extremes stems from either
too few electrons to lose or share (Sc, Ti) or too many d electrons (and hence fewer
open orbitals through which to share electrons with ligands) for high valency (Cu, Zn).
A second factor tending to reduce the stability of high oxidation states toward the end
of the transition series is the steady increase in effective atomic number. This acts to
decrease the energies of the d orbitals and draw them into the core of electrons not
readily available for bonding. Thus, early in the series it is difficult to form species that
do not utilize the d electrons: Scandium(U) is virtually unknown and Ti(IV) is more
stable than Ti(III), which is much more stable than Ti(ll). At the other extreme, the
only oxidation state for zinc is +2 (no d electrons are involved) and for nickel, Ni(U)
is much more stable than Ni(III). As a result, maximum oxidation states of reasonable
stability occur equal in value to the sum of the s and d electrons through manganese
(Ti lv 0 2 , [V v 0 2 ] + , [Cr vl 0 4 ] 2- , (Mn vll 0 4 ] _ ), followed by a rather abrupt decrease in
stability for higher oxidation states, so that the typical species to follow are Fe(ll, III),
Co(II, III), Ni(II), Cu(I, II), Zn(II).
Comparison of There are certain resemblances among metal ions that can be discussed in terms of
Properties by oxidation state but which are relatively independent of electron configuration. They
Oxidation State relate principally to size and charge phenomena. For example, the ordinary alums,
KAI(S0 4 ) 2 - 12H 2 0, are isomorphous with the chrome alums, KCr(S0 4 ) 2 - 12H 2 0, and
mixed crystals of any composition between the two extremes may be prepared by
456
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
Fig. 11.51 Structure of
[Cu(bpy)(hfa) 2 ] showing
Cu—O and Cu—N bond
lengths (in picometers).
[From Veidis, M. V.;
Schreiber, G. H.; Gough,
T. E.: Palenik. G. J. J. Am.
Chem. Soc. 1969, 91. 1859.
Reproduced with
permission.)
important from an applied standpoint have been those involving charge transfer
transitions. As the term implies, these transitions involve electron transfer from one
part of a complex to another. More specifically, an electron moves from an orbital that
is mainly ligand in character to one that is mainly metal in character (ligand-to-metal
charge transfer, LMCT) or vice versa (metal-to-ligand charge transfer, MLCT). Un¬
like d-d transitions, those involving charge transfer are fully allowed and hence give
rise to much more intense absorptions (see Table 11.16). When these absorptions fall
within the visible region, they often produce rich colors and therein lies the source of
practical interest in these types of transitions. 55
In a primitive sense, a charge transfer transition may be regarded as an internal
redox process. This makes it possible to use familiar ideas, such as ionization energies
and electron affinities, to predict the conditions that will favor such a transition.
Consider a crystal of sodium chloride. Imagine ionizing an electron from a chloride ion
(A H = electron affinity) and transferring it to the sodium ion (A H = negative of
ionization energy). It could be imagined that the overall energy (including U^)
required to effect this process might be supplied by a photon. Indeed such photons
exist, but their energy is so high that they belong to the ultraviolet portion of the
spectrum. Hence, sodium chloride does not absorb visible light: It is colorless.
Now consider how we might modify the metal-ligand combination to make the
electron transfer from ligand to metal more favorable. We would want a metal with a
relatively high ionization energy so that it would have empty orbitals at fairly low
energies. Good candidates would be transition or posttransition metals, especially in
higher oxidation states. An ideal ligand would be a nonmetal with a relatively low
electron affinity, which would mean that it would have filled orbitals of fairly high
energy and would be readily oxidizable. Chalcogenides or heavier halides would be
examples of good choices. The net result of such a metal-ligand combination would be
that the orbitals involved in an LMCT process would be close enough in energy that
the transition could be induced by a photon in the visible or near-ultraviolet region.
The permanganate ion, MnO^, meets the criteria set forth in the preceding
paragraph: Manganese is in a formal oxidation state of +7 and combined with four
oxide ions. The molecular orbital diagram for tetrahedral complexes in Fig. 11.52
allows us to identify possible LMCT transitions. In any tetrahedral complex, the four
55 For a discussion of color arising from charge transfer transitions, see Nassau. K. The Physics and
Chemistry of Color. John Wiley: New York. 1983; Chapter 7.
Electronic Spectra of Complexes
457
M ML 4 4L
Fig. 11.52 Molecular orbital diagram for a tetrahedral ML 4 complex, showing possible
ligand-to-metal charge transfer (LMCT) transitions.
lowest energy a-bonding orbitals will be filled and will be primarily ligand in character.
Next there are two sets of cr-nonbonding MO’s, one ligand-centered and one metal-
centered. In permanganate, these orbitals would correspond to filled oxygen n p
orbitals and empty manganese 3 d orbitals, respectively. All of the higher energy
antibonding molecular orbitals would be unoccupied for a manganese!VII) complex.
Hence there are four possible ligand-to-metal transitions:
Ur,) -» M(e)
L(t,) -> M(/ 2 )
Uh) - M(e)
L(£ 2 ) - M(/J)
For MnOTall four of these transitions have been observed: 17,700 cm" 1 (/, —> <?);
29,500 cm -1 (r, —► /£); 30,300 cm" 1 ( l 2 -» e)\ and 44,400 cm -1 (/ 2 - /)- 56 Only the
absorption at 17,700 cm -1 falls within the visible range (14,000-28.000 cm -1 ), and it is
responsible for the familiar deep purple color of MnO^.
56 Lever. A. B. P. Inorganic Electronic Spectroscopy. 2nd ed.; Elsevier: New York, 1986; p 324.
578 14- Some Descriptive Chemistry of the Metals
Even so, it sometimes seems as though no facts are retained. Each of us has a
“silver chloride is a pale green gas" story. Our local one concerns the organic
graduate student who reported that he couldn’t get bromine to dissolve in carbon
tetrachloride (!). He even tried grinding it with a mortar and pestle (!!). When his
professor investigated, it turned out that the bromine was still in an unopened con¬
tainer in the bottom of the packing can, and the material in question, being subjected
to grinding and solubilization tests, was the vermiculite packing material (!!!). Truly,
A little learning is a dangerous thing;
Drink deep, or taste not the Pierian spring.2
In this chapter the theories developed previously will be used to help correlate the
important facts of the chemistry of groups 1-12. Much of the chemistry of these
elements, in particular the transition metals, has already been included in the chapters
on coordination chemistry (Chapters II, 12, and 13). More will be discussed in the
chapters on organometallic chemistry (Chapter 15), clusters (Chapter 16), and the
descriptive biological chemistry of the transition metals (Chapter 19). The present
chapter will concentrate on the trends within the series (Sc to Zn, Y to Cd, Lu to Hg,
La to Lu, and Ac to Lr), the differences between groups (Ti —> Zr -» Hf; Cu —► Ag —
Au), and the stable oxidation states of the various metals.
As the effective atomic number increases across a series of transition metals, the size
decreases from poor shielding by the d electrons. For transition metal ions the results
of ligand held effects override a smooth decrease and so minima in the ionic radii
curves arc found for d b low spin ions. etc. (Fig. 11.15). The decrease in ionic radii
favors the formation of stable complexes, and this, together with ligand field stabiliza¬
tion energies (LFSEs) arising from incompletely filled z/ orbitals, is responsible for the
general order of stability of complexes. The increasing availability of d electrons for
back bonding via v orbitals (especially in low oxidation states) increases the softness
ol the metaj ions in going from left to right in a series. However, the d orbitals in ions
such as Cu- ’ and Zn 2t have become so stabilized that tt bonding for them becomes
relatively unimportant.
The dilferences between one series and another are discussed later in this chapter,
but we may note some seeming paradoxes: The heavier metals tend to be somewhat
less reactive as elements, and yet they arc more easily oxidized to higher oxidation
stales; the first and second series appear to be more closely related to each other than
to the third on the basis of ionization energies, and yet it is common to group the
second and third series together on the basis of chemical properties and to differentiate
them from the first series.
The first ionization energies of the main group and transition metals are listed in
Table 14.1. The first two series do not differ significantly from each other—sometimes
an element of one series is higher, sometimes the other. Beginning with cesium the
third series has a noticeably lower ionization energy as we might expect on the basis of
General Periodic
Trends
2 Pope. A. An Essay on Criticism. Pari II. I. 15. See Bartlett. J. Familiar Quotations : Beck. E. M..
Ed.; Little. Brown: Boston. 1968: p 402b. Evans. B. Dictionary of Quotations: Bonanza Books - New
York. 1968: p 381:10.
General Periodic Trends
579
Table 14.1
Ground state ionization energies of the main group and transition metals in kJ mol -1 (eV mor , ) a
Group
number
IA
(1)
IIA
(2)
IIIB
(3)
IVB
(4)
VB
(5)
VIB
(6)
VIIB
(7)
(8)
VIIIB
(9)
(10)
IB
(11)
IIB
(12)
4th period
419
590
631
658
650
653
717
759
758
737
746
906
(K-Zn)
(4.34)
(6.11)
(6.54)
(6.82)
(6.74)
(6.77)
(7.44)
(7.87)
(7.86)
(7.64)
(7.73)
(9.39)
5th period
403
550
616
660
664
685
702
711
720
805
731
868
(Rb-Cd)
(4.18)
(5.70)
(6.38)
(6.84)
(6.88)
(7.10)
(7.28)
(7.37)
(7.46)
(8.34)
(7.58)
(9.00)
6 th period
376
503
538
654
761
770
760
840
880
870
890
1007
(Cs-Hg)
(3.89)
(5.21)
(5.58)
(6.79)
(7.89)
(7.98)
(7.88)
(8.7)
(9.1)
(9.0)
(9.22)
(10.44)
" Moore, C. E. Ionization Potentials and Ionization Limits Derived from the Analyses of Optical Spectra. NSRDS-NBS 34;
National Bureau of Standards: Washington, DC, 1970.
larger size (Cs > Rb, Ba > Sr, La > Y). A reversal of the ionization trend, beginning
with Hf, occurs after the lanthanides are added. The addition of 14 poorly shielding 4 f
electrons and the enhanced importance of relativistic effects results in increased
ionization energies and decreased size. The “lanthanide contraction" and other
phenomena related to it are discussed in the chapter on periodicity (Chapter 18), but
for now we may note that atoms of postlanthanide metals are (I) smaller than would
otherwise have been expected—so much so that they are often essentially the same
size as their lighter congeners and (2) more difficult to oxidize. The latter phenomenon
proceeds with ever increasing ionization energies until the most noble metals are
reached: iridium, platinum, and gold. As a result of the relatively small size, heavy
nuclei, and tightly held electrons of their atoms, the elements in this series are also the
densest elements (</ 0 , = 22.61 g cm -3 ; d u = 22.65 g cm -3 ).
Although in many respects relativistic effects and the lanthanide contraction serve
to make the postlanthanide elements less reactive than would otherwise be the case, in
other respects their bonding ability is increased. For example, bis(phosphine)plati-
num(0) complexes react with molecular hydrogen, but the analogous palladium com¬
plexes do not: 3
R,P
Pt + H,-
R,P
R,P H R.P
\ / I
M <—X— H, + Pd
/ \ 'I
R,P H R.P
(14.2)
Comparisons of gold and silver further illustrate the point. The Au—H bond is much
stronger than the Ag—H bond and AulT is much more stable than Agl-T. The
relatively high effective nuclear charge leads to relativistic stabilization (contraction)
of 6.v orbitals and destabilization (expansion) of 5 d orbitals. The net result is greater
participation in M — L bonding by the 5 d orbitals and thus stronger bonds. 4
3 Low. J. L.; Goddard, W, A.. Ill Orgnnometallics 1986, 5, 609-622.
4 Schwerdtfcgcr, P.; Boyd, P. D. W.: Burrell, A. K.; Robinson, W. T.; Taylor, M. J. Inorg. Client.
1990. 29. 3593-3607.
458
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
In a similar way, the charge transfer spectrum of orange CrO;;~ ion can be
analyzed, the LMCT process being facilitated by the high oxidation state of chro-
mium(VI). Many iodide salts also are colored because of charge transfer transitions of
this type. Examples are Hgl, (red), Bil, (orange-red), and Pbl 2 (yellow). The metal
ions in these substances certainly are not outstanding oxidizing agents, but the
transitions occur because the iodide ion is easily oxidized.
If the energy difference between the lowest unoccupied molecular orbital
(LUMO) centered on the metal ion and the highest occupied molecular orbital
(HOMO) centered on the ligand is very small (less than 10,000 cm -1 ), total electron
transfer between the two may occur. This will be the case if the metal ion is a
sufficiently good oxidizing agent and the ligand a good enough reducing agent to cause
a spontaneous redox process. The result is breakdown of the complex. Examples of
complexes in which this occurs are [Co(H 2 0) 6 ] 3+ and Fel 3 : Water is oxidized by Co''
and the iodide ion is oxidized by Fe 3 + . Despite the thermodynamic instability of these
complexes, however, it has been possible to isolate both of them by procedures that
take advantage of their kinetic inertness. 57
Some examples of pigments that owe the nature and intensity of their colors to
ligand-to-metal charge transfer transitions are listed in Table 11.23. Some of these
have been long known and used by people in their efforts to beautify their immediate
environments. The use of ochres as pigments, dating from prehistoric times, followed
from their natural abundance [the reds and yellows of the deserts and some other soils
are caused by iron(IIl) oxide]. Pigments used in antiquity included orpiment found in
Tutankhamen's tomb near Thebes and lead antimonate (later dubbed “Naples
yellow"), which was important in Babylonian glazes. 5H Later, vermillion became an
important pigment in the Venetian school of painting, and chrome yellow continues to
be an important coloring agent in several contexts because of its brightness. 59
Pigment Primary orbitals involved b
Cadmium yellow (CdS) Ligand ir p -» metal 5s
Vermilion (HgS) Ligand tr p - metal 6 s
Naples yellow [Pb 3 (Sb0 4 ) 2 ] Ligand ir p -► metal 5s or 5p
Massicot (PbO) Ligand tr p -► metal (is
Chrome yellow (PbCr0 4 ) Ligand ir p -► metal 3 d
Red and yellow ochres (iron oxides) Ligand ir p - metal 3 d
" Brill. T. B. Light, Its Interaction with Art and Antiquities-, Plenum: New
York. 1980. Used with permission.
Simplified notation. See text and Fig. 11.52.
57 The synthesis of Cs|Co(H 2 0),,)|S0. 1 b'6H ; 0 is reported in Johnson. D. A.; Sharpe. A. G.
J. Client. Stic. (A). 1966. 798-801. The preparation of Fell is described in Yoon. K. B.; Kochi. J. K.
Inorg. Client. 1990. 29, 869-874. This report dispelled a long-held assumption that the compound
could not exist.
•'"A History of Technology. Vol. I: From Early Times to Fall of Ancient Empires ; Singer. C. J.;
Holmyard. E. J.: Hall. A. R.. Eds.: Oxford: London. 1954.
59 For an interesting discussion of inorganic pigments and their relation to art. see Brill. T. B. Light,
Its Interaction with Art and Antiquities- Plenum: New York. 1980.
Table 11.23
Pigments in which color is
produced by ligand-to-metal
charge transfer transitions"
Magnetic Properties of Complexes
459
Magnetic
Properties
of Complexes 60
M ML 6 6L
Fig. 11.53 Simplified molecular orbital diagram for an octahedral ML h complex showing
possible mclul-to-ligand charge transfer (MLCT) transitions when both the N A . and <• orbitals
arc occupied and the ligands have empty it' orbitals.
Charge transfer processes in the opposite direction, from metal to ligand, are
favored in complexes that have occupied metal-centered orbitals and vacant low lying
ligand-centered orbitals. Prime examples are complexes in which the ligands have
empty n antibonding orbitals. Ligands falling into this category include carbon mon¬
oxide, pyridine, bipyridine, pyrazine, and o-phenanthroline. Figure 11.53 shows possi¬
ble MLCT transitions for an octahedral complex in which both the l 2t! and e orbitals
are occupied.
It should be quite clear from the foregoing discussion that electronic spectroscopy is a
powerful method for investigating transition metal complexes. Additional and comple¬
mentary information can be provided by magnetic measurements. Because complexes
generally have partially filled metal d or/orbitals, a range of magnetic properties can
be expected, depending on the oxidation state, electron configuration, and coordina¬
tion number of the central metal.
60 Figgis, B. N. In Comprehensive Coordination Chemistry-, Wilkinson. G.; Gillard. R. D.; Mc-
Clcvcrty, J. A., Eds.; Pcrgamon: New York. 1987; Vol. 1, Chapter 6. Gcrloch, M. Magnetism and
Ligand-Field Analysis', Cambridge University: New York. 1983. Carlin, R. L. Magnetochemistry-,
Springer-Verlag: New York. 1986. O'Connor. C. J. Prog. Inorg. Client. 1982, 29. 203-283.
13 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
13.31 If reaction 13.76 proceeds by an inner sphere mechanism with the formation of
[(NC) 5 Fe—CsN—Co(CN) s l A ~ as an intermediate, what can you say about the rate of
Fe—C bond breaking relative to the rate of Co—N bond breaking?
13.32 The reaction of Cr 2+ with
CONH,] ;
in the presence of acid initially gives
Can we conclude that this reaction proceeds by remote attack? 70
13.33 Define the following terms: fluorescence, phosphorescence, luminescence, photolumines¬
cence, and chemiluminescence.
13.34 If I mol of 452-nm photons are absorbed by [Ru(bpy)jJ J+ , how many kJ are absorbed?
How many eV does this correspond to? If [Ru(bpy) 3 ] 2 lies 2.1 eV higher in energy than
[Ru(bpy) 3 ] 2+ , what wavelength of light will be emitted when the excited state relaxes to
the ground state?
13.35 What wavelength of light would be needed to provide the minimum energy for the reaction
N 2 (g) + 3H 2 0(I) -♦ 2NH 3 (g) + 30 2 (g)
for which AG° = 678 kJ mol -1 ?
13.36 The intcrvalence spectrum of
[{Ru(NH 3 ),} 2 (N
I)] 5 ' 1 ' consists of an absorption at 1050 nm. In what
part of the electromagnetic spectrum is it found? Explain how this absorption arises.
13.37 A cyclic voltammogram of |Ru(bpy) 3 ) 2 in acetonitrile is shown below. 71
How do you account for the observation of three reduction potentials?
13.38 If the potentials for Eqs. 13.62 and 13.63 had not been given, could you have calculated
them (Chapter 10)?
70 Taube, H.; Gould. E. S. Acc. Client. Res. 1969, 2. 321-329.
71 Juris. A.; Balzani. V.; Barigelletti. F.; Campagna, S.; Belser. P.; von Zelewsky, A. Coord. Chem.
Rev. 1988. 84. 85-277.
Chapter
Some Descriptive
Chemistry of the Metals
In the preceding chapters principles guiding the structure and reactions of transition
metal complexes have been considered. The present chapter will concentrate on the
properties of individual metals in various oxidation states. The stabilities of these
oxidation states will be examined and the similarities and differences compared. The
material in this chapter and much of the next may be characterized as the “descriptive
chemistry" of the alkali, alkaline earth, transition, lanthanide, and actinide metals.
Unfortunately, descriptive chemistry has not always been especially popular with
students and teachers. Admittedly, complete mastery of all the properties and all the
reactions of the compounds of one element would be an impossible lask, to say
nothing of attempting it with 109. Furthermore, new reactions and properties are
constantly being discovered that require the continual revision of one’s knowledge.
Nevertheless, it is impossible to ignore descriptive chemistry and try only for mastery
of the theoretical side of chemistry. Theory can only be built upon and checked
against facts. Actually, in reading this book you will encounter a vast body of
descriptive chemistry, perhaps without consciously being aware of it. As each theory
or model has been presented, an appeal has been made to the real world to support or
modify that concept. Much of this descriptive chemistry may go unnoticed, but
consider that almost all metal carbonyls are diamagnetic (Chapter 15), that magnetite
has an inverse spinel structure (Chapter II), and that potassium permanganate is
purple (Chapter II) or that it is a strong oxidizing agent (Chapter 10), Furthermore
(Chapter 13),
[PtClJ 2- + 2NH 3 -♦ c/i-PtCI 2 (NH 3 ) 2 + 2CI" (14.1)
When it comes to what is "important" descriptive chemistry, chacun a son goihl'
1 This is why any single volume, even the so-called descriptive ones, must pick and choose which facts
are to be presented. The reader should become familiar with three sets of volumes on inorganic
chemistry: Comprehensive Coordination Chemistry, Wilkinson. G.; Gillard, R. D.; McClevcrty,
J. A., Eds.; Pergamon: Oxford. 1987. Comprehensive Inorganic Chemistry, Bailar, J. C., Jr.;
Emeldus, H. J.; Nyholnt. R.; Trotman-Dickenson. A. F., Eds.; Pergamon: Oxford. 1973. MTP
International Review of Science: Inorganic Chemistry, Emel6us. H. J.. Ed.; Butterworths: London,
1972. 1975; Series One and Two.
577
460
11 - Coordination Chemistry: Bonding, Spectra, and Magnetism
Magnetic Properties of Complexes
461
Substances were first classified as diamagnetic or paramagnetic by Michael Fara¬
day in 1845, but it was not until many years later that these phenomena came to be
understood in terms of electronic structure. When any substance is placed in an
external magnetic field, there is an induced circulation of electrons producing a net
magnetic moment aligned in opposition to the applied field. This is the diamagnetic
effect and it arises from paired electrons within a sample. Since all compounds contain
some paired electrons, diamagnetism is a universal property of matter. If a substance
has only paired electrons, this effect will dominate, the material will be classified as
diamagnetic, and it will be slightly repelled by a magnetic field.
Paramagnetism is produced by unpaired electrons in a sample. The spins and
orbital motions of these electrons give rise to permanent molecular magnetic moments
that tend to align themselves with an applied field. Because it is much larger than the
diamagnetic effect, the paramagnetic effect cancels any repulsions between an applied
field and paired electrons in a sample. Thus even substances having only one unpaired
electron per molecule will show a net attraction into a magnetic field. The paramag¬
netic effect is observed only in the presence of an external field: When the field is
removed, individual molecular moments are randomized by thermal motion and the
bulk sample has no overall moment. When a field is present, there is competition
between the thermal tendency toward randomness and the field's capacity to force
alignment. Consequently, paramagnetic effects decrease in magnitude as the tem¬
perature is increased.
When any substance is placed in a magnetic field, the field produced within the
sample will either be greater than or less than the applied field, depending on whether
the material is paramagnetic or diamagnetic. The difference between the two (AH) can
be expressed as
A H = B-H 0 (11 Jl)
where B is the induced field inside the sample and H 0 is the free-fieid value. AH will be
negative (B < H 0 ) for a diamagnetic substance and positive (B > //„) for one that is
paramagnetic. More commonly the difference between the applied field and that
induced in the sample is expressed in terms of /, the intensity of magnetization, which
is the magnetic moment per unit volume:
4t rl = B - H 0 (11.32)
Because both B and / will tend to be proportional to an external field, dividing
Eq. 11.32 by //„ yields ratios (////„ and BIH n ) that will be essentially constant for a
given substance. The term BIH 0 , called the magnetic permeability, is a ratio of the
density of the magnetic force lines in the presence of the sample to the same density
with no sample; for a vacuum this ratio will be equal to 1. The term ////„ is the
magnetic susceptibility per unit volume (<c), which expresses the degree to which a
material is subject to magnetization:
4ttk = B/H a — I (11.33)
The value of k will be negative for a diamagnetic substance and positive for one that is
paramagnetic.
The quantity that is most frequently obtained from experimental measurements of
magnetism is the specific (or mass) susceptibility, x- It is related to the volume
susceptibility through the density, d:
By multiplying the specific susceptibility of a compound by its molecular weight, we
can obtain the molar susceptibility, Xm'-
Xm = X- MW (11.35)
A number of methods exist for laboratory measurement of magnetic suscep¬
tibilities. 61 Two that are very common and quite similar to each other are the Gouy
and Faraday methods. Both techniques are based on the determination of the force
exerted on a sample by an inhomogeneous magnetic field and both of them involve
measuring the weight of a substance in the presence and absence of the field. The
Faraday method has two distinct advantages over the Gouy method. The first pertains
to sample size: The Faraday method requires several milligrams of material whereas
the Gouy technique requires approximately I gram. A second advantage of the
Faraday method is that it gives specific susceptibility directly. The Gouy experiment
yields volume susceptibility, which must in turn be converted to specific suscep¬
tibility. This conversion can be problematic because it requires an accurate value for
density, which can be difficult to obtain for solids because the value varies according
to how the material is packed.
The setup for a Faraday experiment is represented schematically in Fig. 11.54. If a
sample of mass m and specific susceptibility x is placed in a nonuniform field H that
has a gradient in the x direction ( SHISx ) the sample will experience a force (/) along .v
due to the gradient:
f=(mxH 0 ){^j (11.36)
This force can be measured by weighing the sample both in the field and out of the
field, the difference between the two weights being equal to/. Commonly the experi¬
ment is simplified by determining the force exerted on a standard of known suscep-
Fig. 11.54 Schematic
diagram of apparatus used
for the Faraday
determination of magnetic
susceptibility. The sample
is suspended between
magnet poles that have
been carefully shaped so
that the value of H„(SH/8x)
is constant over the region
occupied by the sample.
61 For discussion of various torque and induction methods, including those employing the vibrating
sample magnetometer and the superconducting quantum interference device (SQUID), see Gerloch.
M. Magnetism and Ligand-Field Analysis 4 , Cambridge University: New York, 1983; O'Connor,
C. J. Prog. Inorg. Client. 1982. 29. 203-283.
X = x/d
(11.34)
574
575
1 3 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
/\
Ph,P PPh,
'I I
Br—Pt—Pt —Br + CF
/\
Ph,P PPh,
'I I '
Br—Pt— Pt — Cl + Br
Ph,P PPh
2
Comment on the trans effect exhibited by the Pt—Pt bond.
13.14 The platinum-carbon NMR coupling constants for /ran.s-[Pt(PEt 3 ) 2 (CO)Hr and
r/wij-[Pt(PEt 3 ) 2 (CO)D] + are 996.2 ± 0.6 and 986.3 ± 0.6 Hz. respectively. Discuss the
relative trans influences of H and D. 6J
13.15 When (Pt(NHj) 4 ]~^ is allowed to stand in 0.1 M HCI for many days at 30 °C. no reaction is
observed. Only under forcing conditions is the cation converted to /ra/u-Pt(NHj) 2 Ck
However, when [PtCIJ 1- is treated with NH,. substitution to give cw-Pt(NH 3 ) 2 CI 2 is
rapid. w ' Account for the difference in the rates of these two reactions.
13.16 In the synthesis of c/j-[Pt(Me 2 S) 2 (NH 3 ) 2 ] 2+ from m-Pt(NH 3 ) 2 CI 2 , the reactant is first
treated with AgCI0 4 and then with Me 2 S. What role does silver perchlorate play in this
reaction?
13.17 For the reaction:
[Pd(L)CI] + + Y~ -► [Pd(L)Y] + CF
the following data were collected: 67
Y"
L
A,
) 5V (cm 3 mor’)
A 2 (dm 3 mol 1 s 1
) 5V\ (cm 3 mol' 1 )
OH”
l,4.7-Me,dien
24.9
-12.2
223
21.1
OH'
l,l,7,7-Me„dien
0.90
-15.5
4.53
25.2
I
l,4,7-Me 3 dien
21.9
-9.2
4318
-18.9
r
l,l,7,7-Me 4 dien
0.99
-13.4
0.28
Account for the following: (a) Differences in A, and A 2 values for the attack of substrate by
Y . (b) Differences in volumes of activation for I" and OH - .
13.18 Sketch energy/rcaction coordinate diagrams for ligand-substitution reactions in which
products are more stable than reactants, and
a. no intermediate is formed.
b. an intermediate is formed and bond breaking and bond making arc equally important.
c. an intermediate is formed and bond breaking is more important than bond making.
d. an intermediate is formed and bond breaking is less important than bond making.
13.19 Arrange the following in order of increasing rate of water exchange:
[V(H 2 0) a ] 3 \ [Cr(H 2 0)„] 3> , [Mg(H 2 0) A ] 2 \ [AI(H 2 0) 6 ] 3
13.20 a. The rate of water exchange for [Mo(H 2 0) 6 1 j is very slow. Why?
h. The rate constant for the formation of IMolH.OMNCS)! 2 at 298 K is 0.317 M" 1 s' 1
and lor [Mo(H : 0) 5 CI|' is 4.6 x I0~ 3 M' 1 s' 1 . Seven-coordinate molybdcnum(III)
65 Crabtree. R. H.: Habib. A. toff, Chem. 1986. 25. 3698-3699.
66 Annibale. G.: Canovese. L.; Cattalini. L.: Marangoni. G.; Michclon. G.; Tobe. M. L. Inorg Chen,
1984. 23. 2705-2708.
67 Breet. E. L. J.; van Eldik, R. Inorg. Chem. 1984. 23. 1865-1869.
Problems
complexes are known. The volume of activation for the NCS~ reaction is - 11.4 cm J
mol -1 . Explain how each of the factors suggests that these anation reactions proceed
by an /„ mechanism. (See Richens, D. T.; Ducommun, Y.; Merbach, A. E. J. Am.
Chem. Soc. 1987. 109. 603-604.)
13.21 The rate constant for the aquation of [Co(NH 2 Me) 5 Cl] 2+ is 22 times larger than for the
aquation of [Co(NH 3 ) 5 CI] 2 ‘'’. Provide an explanation. For analogous chromium complexes
the rate constant is larger for the ammine complex than for the methylamine complex. The
explanation for the reversal is controversial. Read Lay, P. A. Inorg. Chem. 1987, 26.
2144-2149. and summarize arguments for and against an l„ mechanism for the chromium
complexes.
13.22 The hydrolysis of chelated carbonato complexes of cobalt(Ill) is much faster in acid than
in neutral solution. Explain.
13.23 The reactions of [M(H : 0)„| : ^ (M = Zn and Cd) with 2,2'-bipyridine have been studied.
For which metal is the reaction most likely to proceed by an associative interchange
mechanism? (See Ducommun. Y.: Laurenczy, G.; Merbach. A. E. Inorg. Chem. 1988. 27.
1148-1152.)
13.24 Water exchange by dissociation becomes increasingly important for both IM(H 2 0)<J' +
and [M(H 2 0 )a] 3 as one moves from left to right in the periodic table. Explain.
13.25 The outer sphere electron exchange reaction:
[Ru(NH 3 ) 5 ] 2 + [Ru(NH 3 ) 6 ] 3+ - [Ru(NH 3 ) 6 ] 3+ + [Ru(NH 3 ) 6 ] 2 +
is 200 times faster than the reaction:
[Ru(H 2 0) 6 ] 2 + [Ru(H 2 0) 6 ] 3 -♦ [Ru(H 2 0) 6 ] 3 + + [Ru{H 2 0) 6 ] j+
Account for this difference by giving an evaluation of the relative importance of the
factors contributing to the energies of activation. 1
13.26 Ligand exchange in |Fe(phen) 3 )' and in |Fe(bpy),| 2 ‘ is much slower than the transfer of
an electron from the bipyridine complex to the phenanthroline complex. Why does this
rule out an inner sphere mechanism for the electron transfer?
13.27 Calculate k iz for the reaction:
Fc’‘(aq) + |Ru(NH,)J-” -♦ Fe 3+ (aq) + [Ru(NH,)„r
Self-exchange rates lor the oxidant and rcductnnt are 4.2 M' 1 s 1 and 4,0 x It)' M' 1 s' 1 ,
respectively. The equilibrium constant for the reaction is 2,1 x K) 11 .
13.28 The lCo(H 2 0)h| : " “ electron exchange reaction proceeds It) times faster than predicted
hy the Marcus equation. What does this suggest about the mechanism of electron
transfer?
13.29 The reaction:
[Co(NH 3 ) 5 OH] 2 + [Cr(OHj) h ] 2 -♦ [Co(NH 3 ) 5 OH] + + [Cr(OH,) 6 ] 3
proceeds rapidly (A = 1.5 x 10’) by an inner sphere mechanism. When OH' is replaced
by H 2 0 in the cobalt reactant, the reaction slows considerably (A = 0.1). It is observed,
however, that the reaction rate is inversely dependent upon the concentration of hydrogen
ion. Provide an explanation that is consistent with these facts.
13.30 When [Co(NH,),SCN] 2 reacts with [Cr(OH.)„] 2 \ both [S=C=N— Cr(OH 2 ) 5 | 2 and
|N=C=S—Cr(OH 2 ) 5 ]-' are produced. It has been postulated that both remote and
adjacent attacks arc involved in the formation of these products. Draw bridging intermedi¬
ates consistent with this view for the formation of both products. 61 '
6,1 Bernhard. P.; Btirgi, H.-B.; Hauser. J.; Lehmann. H.; Ludi. A. Inorg. Chem. 1982, 21. 3936-3941.
69 Haim. A. Prog. Inorg. Chem. 1983. 30. 273-357.
462
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
tibility, such as Hg[Co(SCN)J ( = 16.44 x I0 -6 cm 3 mol -1 ). If the same magnetic
field and gradient are used for both the standard (s) and the unknown (//), it is not
necessary to know the precise value of either. Hence
Solving this equation for the susceptibility of the unknown gives
(I, -»
The molar susceptibility of the sample, Xm' can obtained from x„ by applying
Eq. 11.35.
Once an experimental value of Xm has been obtained for a paramagnetic sub¬
stance, it can be used to determine how many unpaired electrons there are per
molecule or ion. In order to translate the experimental result into the number of
unpaired spins, it must first be recognized that a measured susceptibility will include
contributions from both paramagnetism and diamagnetism in the sample. Even though
the latter will be small, it is not always valid to consider it negligible. The most
common procedure is to correct a measured susceptibility for the diamagnetic contri¬
bution. Compilations of data from susceptibility measurements on a number of dia¬
magnetic materials make it possible to estimate the appropriate correction factors. The
diamagnetic susceptibility for a particular substance can be obtained as a sum of
contributions from its constituent units: atoms, ions, bonds, etc. (Table 11.24). The
basic assumption underlying such a procedure, namely, that the diamagnetism associ¬
ated with an individual atom or other unit is independent of environment, has been
shown to be valid.
The next step is to connect the macroscopic susceptibility to individual molecular
moments and finally to the number of unpaired electrons. From classical theory, the
corrected or paramagnetic molar susceptibility is related to the permanent paramag¬
netic moment of a molecule, /x, by:
where N is Avogadro's number. R is the ideal gas constant. T is the absolute
temperature, and g. is expressed in Bohr magnetons (BM) (I BM = ehlAirm). Solving
this expression for the magnetic moment gives
M = = 2.84< a ,D ,/2 (11-40)
As we know, this paramagnetic moment originates in the spins and orbital motions of
the unpaired electrons in the substance. There are three possible modes of coupling
between these components: spin-spin, orbital-orbital, and spin-orbital. For some
complexes, particularly those of the lanthanides, we must consider all three types of
coupling. The theoretical paramagnetic moment for such a complex is given by
H = g[J{J + l)] l/2 (11.41)
where J is the total angular momentum quantum number and g is the Land6 splitting
factor for the electron, defined as
8 = 1 +
■/(■/ + I) + 5(5 + I) - UL + 1)
2 J{J + I)
(11-42)
Magnetic Properties of Complexes
463
Table IT24 _
Diamagnetic susceptibilities
[x 10 & (cgs units) mol -, ]°
Cations^
Anions
Li +
-1.0
F-
-9.1
Na +
-6.8
cr
-23.4
K +
-14.9
Br -
-34.6
Rb +
-22.5
I -
-50.6
Cs +
-35.0
NOj -
-18.9
Tl +
-35.7
CI0 3 -
-30.2
nh;
•13.3
cio;
-32.0
Hg 2+
40.0
CN -
-13.0
Mg 2+
-5.0
NCS -
-31.0
Zn 2+
15.0
OH -
-12.0
Pb 2+
32.0
so 2-
-40.1
Ca 2+
10.4
o 2-
-12.0
Neutral Atoms
H
-2.93
As(III)
-20.9
C
-6.00
Sb(lII)
-74.0
N (ring)
-4.61
F
-6.3
N (open chain)
-5.57
Cl
-20.1
N (imide)
-2.11
Br
-30.6
0 (ether or alcohol)
-4.61
I
-44.6
0 (aldehyde or ketone)
-1.73
S
-15.0
P
-26.3
Se
-23.0
As(V)
-43.0
Some Common Ligands
H,0
-13
C,0 2-
-25
nh 3
-18
acetylacetonate
-52
C 2 H 4
-15
pyridine
-49
CHjCOO"
-30
bipyridyl
-105
H,NCH 2 CH 2 NH 2
-46
o-phenanthroline
-128
Constitutive Corrections
c=c
5.5
N = N
1.8
c=c—c=c
10.6
C=N — R
8.2
c=c
0.8
C—Cl
3.1
C in benzene ring
0.24
C—Br
4.1
° Carlin, R. L. Magnelochemisiry, Springcr-Vcrlag: New York, 1986; p 3.
b The inner core diamagnetism of the first-row transition metals can be taken as
approximately -13 x IO -6 (cgs units) mol -1 .
The value of J depends on the total orbital angular momentum quantum number, L,
and the total spin angular momentum quantum number, 5 (Appendix C). Some
calculated and experimental magnetic moments for lanthanide complexes are shown in
Table 11.25.
For complexes in which spin-orbit coupling is nonexistent or negligible but spin
and orbital contributions are both significant, the predicted expression for /x is
H = (4S(J + 1) + L(L + I)] ,/z (11.43)
Equation 11.43 describes a condition that is never fully realized in complexes because
the actual orbital contribution is always somewhat less than the ideal value. This
572
13 Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Problems
573
Problems
Fig. 13.16 Alternative
electron tunneling pathways
from ruthenated His62 to
the heme in a mutant yeast
cytochrome c. [From
Bowler, B. E.; Meade.
T. J.: Mayo. S. L.;
Richard, J. H.; Gray, H. B.
J. Am. Chem. Soc. 1989,
III. 8757-8759.
Reproduced with
permission.]
13.1 Metal-halogen bonds are more labile than metal-nitrogen bonds. Use this information and
the trans effect to devise syntheses for the following geometric isomers from [PtCI 4 ] 2 ~.
a. Cl Br b. Cl Br
X x
Py nh 3 h 3 n Py
13.2 Predict the geometries of the complexes which result from the following reactions:
a. [Pt(NOj)CI 3 ] 2 - + NHj - [Pt(NO,)(NHj)Cl 2 ]" +CI"
b. cis-[Pt(RNH 2 ) 2 (NH 3 )(N0 2 )] + Cl' -♦ Pt(RNH 2 )(NH 3 )(N0 2 )Cl + RNH 2
13.3 Predict the products of the following reactions (I mol of each reactant):
a. [Pt(CO)Cl 3 ]- + NH 3 -►
b. [Pt(NH 3 )Br 3 ]- + NHj -►
13.4 Trialkyl phosphines are rather good trans directors and, as expected, the reaction of Bu 3 P
with [PtCI 4 ] 2- gives the trans isomer as a major product. However, when one uses Ph 3 P in
this reaction, only the insoluble cis product is obtained. Offer an explanation for this
apparent violation of the trans-effect prediction. (Problem 13.5 may be helpful.)
13.5 When pure Pwi.s-PtCI 2 (Bu 3 P) 2 is placed in solution with a trace of Bu,P. isomerization
occurs to give a mixture of cis and trans isomers. Provide a plausible mechanism.
13.6 Nickel complexes are observed to undergo substitution much faster than platinum com¬
plexes. Offer an explanation.
13.7 The following data were collected for the reaction [dien = diethylenetriamine,
HN(CH 2 CH 2 NH 2 ) 2 ]:
[Pd(dien)SCN] + + py -♦ [Pd(dien)py] 2+ + SCN“
fcob.
[py]
6.6 x
I0 -3
1.24
X
I0~ 3
8.2 x
10 -3
2.48
X
I0~ 3
2.5 x
I0~ 2
1.24
X
10 -2
Use the data to calculate k , and k 2 for substitution in this square planar complex.
13.8 Sketch plots of it obs versus [Y] for two cases of substitution of a square planar complex:
(a) one in which the solvent pathway is insignificant and (b) one in which the solvent
pathway is exclusive.
13.9 The rate of substitution in a square planar complex often depends on the identity of the
leaving group, X. For the reaction:
[Pt(dien)X]'' + py - [PUdienlpy] 2 + X"
the following data were collected:
Ligand, X
104. ("')
H,0
1900
cr
35
Br“
23
r
10
N07
0.05
CN"
0.017
Of these ligands, CN~ has the least effect and H>0 has the greatest effect on the rate of the
reaction. Yet as trans directors, just the opposite order is observed for these two ligands.
Explain.
13.10 The hydroxide ion is a stronger base than ammonia, and yet it reacts more slowly than
ammonia with a square planar complex. Explain.
13.11 Rationalize the order of the following A ob , values for the reaction of «s-Pt(PEl 3 ) 2 LCl with
py-
L
^obt
phenyl
0.08
o-tolyl
0.0002
mesityl
0.000001
13.12 Sketch a reaction profile for substitution in a square planar complex in which (a) a five-
coordinate intermediate exists, but bond breaking is more important than bond making:
(b) a five-coordinate intermediate exists, but bond making is more important than bond
breaking.
13.13 Substitution reactions of dinuclear platinum(I) complexes have been investigated. 4 The
rate constant for the reaction below is 93 ± 20 M 1 s 1 at 10 °C in dichloromcthane. Rate
constants for halide substitution in irans-( R 3 P) 2 PlX 2 complexes are typically 10 4 M
64 Shimura, M.; Espenson, J. H. Inorg. Chem. 1984, 23. 4069-4071.
464
11 - Coordination Chemistry: Bonding, Spectra, and Magnetism
Magnetic properties
Central
metal
No. of f
electrons
Ground
state
Compound
p(expt)
BM
/i(calc) b
BM
compounds of the
Ce 3+
1
Ce,Mg 3 (N0 3 ) 6 -24H 2 0
2.28
2.54
lanthanide metals 0
Pr 3+
2
3 h 4
Pr,(S0 4 ) 3 -8H,0
3.40
3.58
Nd 3+
3
Nd,(S0 4 ) 3 -8H,0
3.50
3.62
Sm 3+
5
6u"
Sm,(S0 4 ) 3 ■ 8H 2 0
1.58
1.6 C
Eu 3+
6
7 f
Eu-.fSO^ • 8H,0
3.42
3.61 f
Sm 2
6
7 F
SmBr 2
3.57
3.6k '
Gd 3+
7
Gd 2 (S0 4 ) 3 -8H 2 0
7.91
7.94
Eu 2+
7
%/2
EuCl,
7.91
7.94
Tb 3+
8
7 F 6
Tb,(S0 4 ) 3 -8H 2 0
9.50
9.72
Dy 3+
9
%sn
Dy 2 (S0 4 ) 3 -8H 2 0
10.4
10.63
Ho 3+
10
%
10.4
10.60
Er }+
11
4 /
ErifSO^j ■ 8H 2 0
9.4
9.57
Tm 3+
12
TmjfSOJj-SHjO
7.1
7.63
Yb 3+
13
2r
112
Yb-,(S0 4 ) 3 -8H 2 0
4.86
4.50
a Figgis, B. N. In Comprehensive Coordination Chemistry: Wilkinson, G.; Gillard, R. D.;
McCleverty, J. A.. Eds.; Pergamon: New York, 1987; Vol. t, p 261.
g[J(J + l)]' n , except where noted.
C Calculation includes effects of mixing of ground and higher energy terms.
occurs because the orbital angular momentum is reduced from what it would be in the
free metal ion by the presence of ligands. In the extreme case, where L is effectively
zero, the orbital contribution to the magnetic moment is said to be quenched. This is
the general situation in complexes having A or E ground states, which would include
octahedral d i , d 4 (high spin), d 5 (high spin), c/ 6 (low spin), cf (low spin), and d cases.
Furthermore, when a complex involves a first-row transition element, even if the
ground state is T. the orbital contribution generally may be ignored. For the L = 0
condition, Eq. 11.43 reduces to
fi = [45(5 + 1)] l/2 = 2[5(5 + l)J ,/2 < 1L44)
which is known as the spin-only formula for magnetic moment. By recognizing that 5
will be related to the number of unpaired electrons («) by 5 = nil, the expression may
be further simplified to
p. = [«(/» + 2)] l/2 (1L45)
A number of calculated and experimental magnetic moments for first-row transition
metal complexes are given in Table 11.26, showing that the spin-only formula gives
results that are in reasonably good agreement.
As we know, a number of transition metal ions form both high and low spin
complexes, and we have now seen that magnetic susceptibility measurements allow us
to experimentally distinguish one from the other. Within ligand field theory, these two
spin configurations in octahedral complexes are explained in terms of relative magni¬
tudes of A„ and pairing energy IP): We associate high spin complexes with the
condition A„ < P and low spin complexes with A„ > P. For complexes in which the
energy difference between A„ and P is relatively small, an intermediate field situation,
it is possible for the two spin states to coexist in equilibrium with each other. Consider
the Fe 2+ ion. At the two extremes, it forms high spin paramagnetic [Fe(H 2 0)] . (5 =
2) and low spin diamagnetic [Fe(CN) 6 ] 4- (5 = 0). The Tanabe-Sugano diagram
Magnetic Properties of Complexes
465
Magnetic properties of some complexes of the first-row transition metals'
High spin complexes
Low spin complexes
unpaired
electrons
No. of d
electrons
Central
metal
electrons
Coordination Chemistry: Experimental Methods ; Butlcrworth: London, 1973.
» Burger, K.
Spin-only value.
pertaining to these d b complexes (Appendix G) shows that near the crossover point
between weak and strong fields the difference in energy between the spin-free ( 5 7^)
and spin-paired ('A lff ) ground states becomes very small (Fig. 11.55). Within Ihis
region, it is reasonable to expect that both spin states may be present simultaneously
and that the degree to which each is represented will depend on the temperature (A„ -
P = kT). A complex illustrating these effects is [Fe(phen) 2 (NCS) 2 ]. A plot of its
magnetic moment against temperature appears in Fig. 11.56. At high temperatures a
moment consistent with four unpaired electrons is observed, but as the temperature is
Fig. 11.55 Variation in
energies of the 5 T 3w and
'/ti„ terms with increasing
A„ for d" octahedral
complexes. At weak fields
(high spin complexes) the
ground term is 5 T :ji „ while
at strong fields (low spin
complexes) il is 'A lu . Note
that in the region
immediately on each side of
the spin crossover point,
the energy difference
between the two terms is
small; thus high and low
spin complexes may
coexist.
570 13 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Table 13.10 _
Calculated rate constants
for electron transfer in
[Ru(bpy) 2 CI] 2 l-L
complexes and distances (r)
separating the metal
centers"
L—L
r, A It, s"'
6.8 3 x 10 9
7.1 1 x 10 8
6.0 6 x IO 10
11.3 1 x 10 8
nQ^-C = C-^Qn 13.8 2 x IO 7
H
" Creutz. C. Prog. Inorg. Chem. 1983, 30. 1-73.
braies). It is an electron carrier for oxidative phosphorylation, transferring electrons
to O-,. The energy released in this process is used to synthesize ATP. The heme group
of cytochrome c lies near the surface of the protein. The iron atom is six-coordinate,
with bonds to five nitrogen atoms (four from the porphyrin, one from a histidine
nitrogen) and one sulfur atom from a cysteine. Since all six iron coordination sites arc
occupied, direct electron transfer to iron is not possible, and the electron must pass
through the surrounding protein bridgework (see Fig. 19.3). Electron transfer rates
between horse heart cytochrome c (very stable and commercially available) and many
transition metal complexes have been studied. For example, the observed rate con¬
stant tor reduction by [Ru(NHj) ft ]‘' + is 3.8 x IO 11 M -1 s -1 , which compares well with
the value ot 7.8 x 10 1 M 1 s 1 calculated from the Marcus cross-reaction equation.
Calculating rate constants for electron transfers between two metalloproteins is a
much more complicated affair. Distances between metal sites of two such proteins are
often large and uncertain. A clever approach to gaining information about the distance
dependence of electron transfer in these systems is to bind a second metal center to
the surface of a single metalloprotein. A number of different electron transfer proteins
(including horse heart cytochrome c, cytochrome c 55I . azurin. plastocyanin, and an
iron-sulfur protein) as well as sperm whale myoglobin have been modified by attach¬
ing [Ru(NH,) 5 ] 2+ to a surface histidine nitrogen atom. The distance between metal
centers is thereby fixed and can be determined. In cytochrome c, for example, both
the Ru ol the surface [Ru(NH 3 ) 5 HisI : ' and the heme Fe begin in the +3 oxidation
state. The ruthenium moiety is reduced to [Ru(NH 3 ) 5 Hisl 2 ' t ' chemically or by pulse
radiolysis. The transfer of an electron from Ru 2+ to Fe 3+ can then be followed by
monitoring the decreasing Ru 21 ' absorption. Since, in general, several surface histidine
nitrogen atoms are available for binding, it is possible, within the same protein, to
place |Ru(NH 3 ) 5 ] 2+ units at various distances from the heme center. In fact four
different [Ru(NH 3 ) 3 ] 2+ derivatives of sperm whaie myoglobin have been characterized
(Fig. 13.15). It turns out that there is generally an exponential dependence of electron
Mechanisms of Redox Reactions 571
His 12
Fig. 13.15 Computer¬
generated view of the heme
group and four ruthenium
surface histidines in sperm-
whale myoglobin. Closest
heme-a s -Ru(His) edge-to-
edge distances are 14.6
(His48), 19.1 (His81). 20.1
(Hisl 16), and 22.1 A
(Hisl2). [From Mayo,
S. L.; Ellis, W. R., Jr.;
Crutchley. R. J.; Gray,
H. B. Science 1986, 233.
948-952. Used with
permission.]
transfer rates upon distance in these sorts of systems. 60 However, not all pathways are
identical so it is possible for a particular silualion to be an exception to this rule. A
theory has been developed that suggests that electron transfer in proteins is regulated
by pathways that are optimal combinations of through-bond, hydrogen bond, and
through-space links. 61 According to this model, the electron transfer will follow the
best overall linkage. 62 Hydrogen bond bridges will be less efficient than through-bond
linkages, but much better than through-space electron transport. Thus the optimum
pathway will be the one with the best compromise of minimizing distance (covalent
links) and avoiding, if possible, hydrogen bond linkages and. especially, through-
space gaps. These optimum pathways can be computed making use of assumptions
about the relative effectiveness of through-bond, hydrogen bond, and through-space
links. Figure 13.16 illustrates such a pathway.
The driving force of a reaction, as noted earlier and as predicted by Marcus
theory, also affects the rate of electron transfer in and between proteins. Replacing the
heme in a ruthenium-modified myoglobin with another metal system gives one a
chance to evaluate these effects. When the heme is replaced with a photoactive
palladium porphyrin, which is a good reducing agent in its electronic excited state, the
electron transfer driving force increases. 63 Studies such as these aliow reorganization
energies, which strongly influence reaction rates, to be evaluated.
60 Axup. A. W.; Albin. M.; Mayo. S. L.; Crutchley, R. J.; Gray. H. B. J. Am. Chem. Soc. 1988. I/O.
435-439. Cowan. J. A.; Gray. H. B. Inorg. Chem. 1989, 28. 2074-2078.
61 Bcratan, D. N.; Onuchic, J. N. Photosynthesis Research 1989. 22. 173-186. Bcratan, D. N.;
Onuchic. J. N.; Betts. J. N.; Bowler. B. E.; Gray, H. B. J. Am. Chem. Soc. 1990, U2. 7915-7921.
62 In many ways, the electronic tunneling pathway is analogous to various electric circuits with
different resistors: The longer the “circuit" in these systems, the greater the resistance. Hydrogen
bonds tend to increase resistance and open space has the greatest of all.
63 Karas. J. L.; Lieber, C. M.; Gray, H. B. J. Am. Chem. Soc. 1988, 110. 599-600.
470
471
11 • Coordination Chemistry: Bonding, Spectra, and Magnetism
11.19 Why is a solution of copper(II) sulfate blue?
11.20 II one [CuL„J- solution is blue and another is green, which would be expected to have the
higher value of A„?
11.21 Consider the following electronic transition frequencies (in cm - ') for a series of nickcl(ll)
complexes (dma = dimethylacetamide):
Complex
"i
12
"3
INi(H 2 0) f ,]- +
8500
15.400
26.000
[NitNH,)] 2 "
10.750
17.500
28,200
[Ni(OSMe : )„] 2+
7728
12,970
24.038
[Ni(dma) 6 ] 2+
7576
12.738
23.809
Determine appropriate values of A„ and /?' for these complexes.
11.22 The following absorption bands are found in the spectrum of [Cr(CN) A | 3_ : 264 nm (charge
transfer), 310 nm. and 378 nm. Determine the values of A„ and B'.
11.23 When visible light passes through a solution of nickel(ll) sulfate, a green solution results.
What are the spin allowed transitions responsible for this color? Would you expect a
Jahn-Teller distortion for this complex?
11.24 Chromium!II) (luoride and manganese(ll) fluoride both have a central metal ion sur¬
rounded by six fluoride ligands. The Mn—F bond lengths are equidistant, but four of the
Cr—F distances are long and two arc short. Provide an explanation.
11.25 If a molecule having a center of symmetry undergoes a Jahn-Teller distortion, the center
ol symmetry must be maintained, and this is the case when octahedra undergo tetragonal
distortions. Can you think of any other distortion of an octahedral complex that would be
consistent with this principle?
11.26 Both FeF, and K,[CoF„| contain six-coordinate high spin metal ions. The electronic
spectrum of the former shows absorptions at 6990 and 10.660 cm -1 , while the latter has
absorptions at 10.200 and 14.5(H) cm '. For which complex is A„ largest? Why? How
many multiplicity allowed electronic transitions would you expect for these complexes?
How can you account for the presence of two bands in each spectrum?
11.27 1 he ligand-to-mctal charge transfer bands increase in energy in the scries: ICol.l - <
|CoBr 4 | < |CoClj| . Explain.
11.28 Explain the following (from Lever. A. B. P. J. Chon. EJuc. 1974. 51. 612-616):
a. I he anhydrous solids CuCI; and CuBr. arc green and black, respectively, while CuL is
not stable.
b. The NCS" ion is a good colorimetric reagent for Fe(III).
c. Complexes in which two metals of different oxidation state arc close together arc
frequently highly colored.
<1. Many complexes exhibiting charge transfer bands in the visible region are unstable in
sunlight.
11.29 From the reaction of NiBr, and Ph : EtP. it is possible to isolate green crystals of
|Ni(Ph.EtP) ; Br : |. which have a magnetic moment of 3.20 Bohr magnetons, and red
crystals of |Ni(Ph.EtP) : Br : |, which have a magnetic moment of zero. When either of
these is dissolved in dichloromcthanc at 40 °C. the resulting solution has a magnetic
moment of 2.69 BM. Suggest structures for the green and red crystals and offer an
explanation for the solution magnetic moment. (Sec LaMar. G. N.: Sherman. E. O. J.
Am. Client. Soc. 1970. 92. 2691.
11.30 Calculate the magnetic moment of Dy 2 (S0 4 )r8H 2 0.
Problems
11.31 Show that the ground state term of Er’ is J /,. v; . What magnetic moment would you
expect for Er : (S0 4 ),-8H : 0?
11.32 The complexes [Mn(H : 0)„] : % [Fe(H,0) 6 ] 3 . lMnCI 4 ] 2- , and (FeCl.,) - all have mag¬
netic moments of nearly 5.92 BM. What does this tell you about the geometric and
electronic structures of these complexes? Why is the spin-only formula so precise in these
cases?
11.33 One means of determining the magnetic susceptibility of a transition metal complex, the
Evans method, utilizes NMR. As an illustration of the procedure, consider the following
experiment utilizing a 60-MHz spectrometer. A capillary tube is filled with an aqueous
solution consisting of 8.0 mg mL 1 of CuS0 4 in 2% /-butyl alcohol. The capillary tube is
sealed and placed in an NMR tube that also contains a 2% /-butyl alcohol solution with no
CuS0 4 . The volume susceptibilities of the two solutions differ and as a result the /-butyl
group shows a different proton chemical shift in each. For this particular experiment, the
chemical shift of the /-butyl group in the copper solution is 8.6 Hz upfield from that in the
noncopper solution at 310 K. Use these data and help from Loligcr. J.: Scheffold. R. J.
Cheat. Edni. 1972. 49. 646-647. to determine the magnetic moment of CuS0 4 . How does
the value you obtain compare to that expected from the spin-only formula?
11.34 Explain the following experimental results: At ambient temperature a 0.192-g sample of
Fe|HB(l-pyrazolyl),) : weighs the same in a magnetic field as it does out of the field (to
three significant figures). However, in an identical magnetic field at 449 K, a 0.192-g
sample of the same material gains 0.015 g over its out-of-field weight. (See Hutchinson.
B.; Hancc. R. L.: Hardegree. E. L.; Russell S. A. J. Client. Ecltic. 1980. 57. 830-831.)
11.35 When ethylencdiumine is added to a solution of cobalt(ll) chloride hexahydrate in concen¬
trated hydrochloric acid, a blue crystalline solid is obtained in 80% yield. Analysis of this
compound shows that it contains 14.16% N. 12.13% C. 5.09% H, and 53.70% Cl. The
effective magnetic moment is measured as 4.6 BM. The blue complex dissolves in water to
give a pink solution, the conductivity of which is 852 ohm -1 cm 2 mol -1 at 25 °C. The
visible spectrum of a dmso solution of the complex has bands centered at 3217. 5610. and
15.150 cm 1 (molar absorptivity = 590 mol -1 L cm - '), but for a water solution, the ab¬
sorptions occur at 8000. -~ 16.000 and 19.400cm' 1 (molar absorptivity = 5mol -1 Lem -1 ).
In a titration with sodium hydroxide, each mole of the complex neutralizes four moles of
base. Determine the formula and structure of the complex. Account for all reactions and
observations.
11.36 Addition of TiCI, to an aqueous solution of urea followed by addition of Kl gave deep blue
crystals of a complex containing titanium, urea, and iodine. The visible spectrum of the
material showed one absorption at 18,DIM) cm 1 and its magnetic moment was determined
lobe 1.76 BM, When 1.000 g of the compound was decomposed at high temperatures in an
oxygen atmosphere, all ligands volatilized and 0.101 g of TiO : formed. Deduce the
formula and structure of the complex. Do you think urea or water lies higher in the
speclrochemical series? How might you determine whether urea is bound to titanium
through oxygen or through nitrogen?
11.37 In the solid state. Co(py) ; CI : is violet and has a magnetic moment of 5.5 BM, butaCH.CL
solution of this compound is blue and has a magnetic moment of 4.42 BM. In contrast,
Co(py) 2 Br 2 is blue in both the solid state and in a CH 2 CL solution and has a magnetic
moment of 4.6 BM in both forms. Explain these observations. Predict the colors and
magnetic moments of Co(2-Mepy),CI : and Co(py) : L.
11.38 It has been known at least since the 15th century that blue glass could be obtained by
including cobalt! II) in the formulation. Aside from its beauty, this type of glass is useful
for absorbing sodium emission thereby allowing one to observe the characteristic flame
test for potassium in the presence of sodium. Explain. Low alkali borate glasses contain¬
ing cobalt(ll) do not give good cobalt blue—they tend to be pink. Addition of some NaCI
to the giass deepens the blue color. Why? (Paul. A. Chemistry of Glasses. 2nd ed.;
Chapman and Hall: London. 1990; pp 323ff.)
568
13 "Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Mechanisms of Redox Reactions
569
Mixed Valence
Complexes
Theoretical treatments of electron transfer between two transition metal ions in
solution are complicated by contributions arising from solvent reorganization and by
transfer pathway uncertainties. If, however, the reducing and oxidizing agents are
separated by a bridge within a single bimetallic complex, there will be no solvent
molecules between the metal ions and the pathway will be defined. Furthermore,
electron transfer over various distances can be studied by varying the length of the
bridge and this can provide some insight into important biological processes. Com¬
plexes that contain a metal atom in more than one oxidation state are referred to as
mixed valence complexes. One could envision some systems in which the two metal
ions are so far removed from one another that electron transfer does not take place nor
can it be induced:
M 2+ -M 3+
In other systems the two metal ions may be so strongly coupled that properties of
the separate +2 and +3 ions are lost and the entire unit is best represented as two
+ 2i ions.
M 2 ' 4+ —M 2 ' 4+
Of greater interest are systems in which modest coupling exists between metal
centers, for in these it is possible to photolytically induce electron transfer. The
potential energy diagram in Fig. 13.13 shows by means of a vertical arrow the
photochemical energy necessary for an electron to pass from the potential energy
surface of M 2+ ~M 3+ to the potential energy surface for M 3+ ~M 2+ . This means that
intervalence transitions are observed in the electronic spectra (often in the near-
infrared region) of these complexes but are not found in the spectra for monometallic
complexes of either M 2+ or M ,+ . 55 Of course the electron could also pass thermally
from one surface to the other, as for outer sphere electron transfer, through equaliza¬
tion of orbital energies by vibrational elongation and contraction of metal-ligand
bonds. A comparison of optical and thermal electron transfer is shown in Fig. 13.14. In
the optical process depicted in the top part of the figure, we see that transfer occurs
prior to bond length adjustment, but in the thermal process (bottom of diagram), bond
length changes take place (as required by the Franck-Condon principle) prior to
electron transfer.
Fig. 13.13 Potential
energy diagram for a
photochemically induced
homonuclear electron
transfer in a mixed valence
complex (R = reactants;
P = products).
55 A tight ion pair may also give rise to a mixed valence transition.
Applications to
Bioinorganic
Chemistry
Fig. 13.14 A comparison of photochemical and thermal electron-transfer processes in mixed
valence systems. The photochemical pathway (top) allows electron transfer prior to bond-
length adjustment, while the thermal route (bottom) requires adjustment prior to electron
transfer. [Creutz, C. Prog, tnorg. Client. 1983. 30, 1-73. Used with permission.]
One photoactive system which has been extensively studied is a bimetallic com¬
plex of Ru 2+ /Ru 3+ in which 4,4'-bipyridine functions as the bridge.
[(H,N) 5 Ru 2+ — N
N — Ru 3 ^(NH 3 ) 5 ]
[(H,N) 5 Ru 3+ — N
(13.78)
N— Ru 2 (NH 3 ) 5
An absorption (1050 mm) found in the near-infrared spectrum of this complex arises
from a mixed valence transition. Light-induced metal-to-metal charge transfer was
predicted by Hush 56 for systems of this type before it was observed experimentally.
Further, his theory relates the energy of absorption to that required for thermal
electron transfer (hv = 4 x E a ) and from this it is possible to calculate the thermal
electron transfer rate constant (5 x 10 8 s -1 ). 57
The expectation that the rate of electron transfer will slow with increasing dis¬
tance between the two ions is realized, as shown for [Ru(bpy) 2 CI) 2 L—L complexes in
Table 13.10. Distance alone, of course, does not determine how quickly an electron
can pass through a bridge. The nature of the bridge itself is important with some
bridges being more resistant to electron transport than others. A comparison of the
diphosphine and pyrazine bridges, which are nearly the same length, shows the rate
constant of the latter to be 30 times that of the former. 58
Cytochrome c. discussed more extensively in Chapter 19, is an important biological
intermediate in electron transfer. 59 This metalloprotein, found in all cells, has a
molecular weight of approximately 12,800 and contains 104 amino acids (in verte-
56 Hush. N. S. Prog. Inorg. Chem. 1967. 8, 391-444.
57 Brown. G. M.; Krentzien. H. J.; Abe. M.; Taube H. Inorg. Client. 1979, 18, 3374-3379.
58 Creutz, C. Prog. Inorg. Client. 1983. 30, 1-73.
59 Struct. Bonding (Berlin) 1991. 75 (entire volume is devoted to "Long-Range Electron Transfer in
Biology"). Electron Transfer in Inorganic and Biological Systems : Bolton, i. R.; Malaga, N.;
McLendon, G., Eds.; Advances in Chemistry 228; American Chemical Society: Washington. DC,
1991. McLendon, G. Acc. Chem. Res. 1988 . 21. 160-167. Marcus, R. A.; Sutin, N. Biochim.
Biophys. Acta 1985, 811, 265-322. Bowler, B. E.; Raphael, A. L.; Gray, H. 13. Prog. Inorg. Chem.
1990, 38, 259-322. Mayo. S. L.; Ellis, W, R., Jr.; Crulchley, R. J.; Gray, H. B. Science 1986, 233,
948-952. Scott. R. A.;Mauk. A. G.;Gray, H. B.J. Chem. Educ. 1985.62.932-938. McLendon, G.;
Guarr. T.; McGuire, M.; Simolo, K.; Strauch, S.; Taylor, K. Coord. Chem. Rev. 1985,64, 113-124.
Sykes, A. G. Chem. Br. 24. 1988, 551-554.
C h a p t e r
Coordination Chemistry:
Structure
The previous chapter described the bonding principles responsible for the energetics
and structure of coordination compounds. In this chapter the resulting structures will
be examined in more detail with particular regard to the existence of various coordina¬
tion numbers and molecular structures, and the effect of these structures on their
chemical and physical properties.
The coordination numbers of metal ions range from I, as in ion pairs such as
Na + CI _ in the vapor phase, to 12 in some mixed metal oxides. The tower limit, I. is
barely within the realm of coordination chemistry, since the Na"Cl ion pair would
not normally be considered a coordination compound, and there are few other exam¬
ples. Likewise, the upper limit of 12 is not particularly important since it is rarely
encountered in discrete molecules, and the treatment of solid crystal lattices such as
hexagonal BaTiO, and perovskite 1 as coordination compounds is not done frequently.
The lowest and highest coordination numbers found in '•typical” coordination com¬
pounds are 2 and 9 with the intermediate number 6 being the most important.
We have seen in Chapter 4 that the coordination number of ions in lattices is
related to the ratio of the radii of the ions. The same general principles apply to
coordination compounds, especially when a single coordination number, such as 4
has two common geometries-tetrahedral and square planar. An extended list of
radius ratios is given in Table 12.1.
Coordination
Number 1
As mentioned above, ion pairs in the gas phase may be considered as examples of
coordination number 1. There are a few other examples known. For instance, the aryl
radical derived from the highly sterically hindered 1 , 3 , 5 -triphenylbenzene. forms one-
, it if' ti > „„,l 14 IP 14 1
IdUlWl uv.u -- j -J .
to-one 2 organometallic compounds of the type CuC 6 H 2 (C A H 5 ) 3 and AgC 6 H 2 (L A il 5 ) 3 ,
as shown in the margin.
1 Note, however, that some of the properties of metal ions in these systems can be described in terms
of coordination chemistry. Sec Chapter 7.
2 Lingnau. R.: Strahle. J. Angew. Chem. In,. Ed. Engl. 1988. 27.436. Note that what we often naively
write as "LiCHj," "LiC 6 H„" “KCHj.” and "CuCHj" are often considerably more complex:
(LiCH,) 46 (hexameric molecules in hydrocarbon solvents, tetramcric molecules in THF and the
solid). K'CHj (ionic solid), and (CuCH,), (bridged polymeric solid). See Elschenbroich, C.; Salzer.
A. Organomelallics. 2nd ed.; VCH: Wcinheim. Germany. 1992; pp 20-21.
Coordination Number 2
473
Minimum
C.N. radius ratio Coordinati on polyhedron _
4 0.225 Tetrahedron
6 0.414 Octahedron/square plane
6 0.528 Trigonal prism
7 0.592 Capped octahedron
8 0.645 Square antiprism
8 0.668 Dodecahedron (bisdisphenoid)
8 0.732 Cube
9 0.732 Tricapped trigonal prism
12 0.902 Icosahedron
12 1.000 Cuboctahedron
fnnrrlinotion Few complex ions are known with a coordination number of 2. They are generally
loorainaTlQ- limited t0 lhe +1 ions of the Group IB (II) metals and the closely related Hg(U)
Number 2 species W 10 ). Examples are [Cu(NH 3 ) 2 ] + , [Ag(NH 3 ) 2 ] + , [CuCI 2 ] , [AgCl 2 ] ,
[AuCL]", [Ag(CN) 2 ]“, [Au(CN) 2 ] - , and [Hg(CN) 2 ], Even these may react with
additional ligands to" form higher-coordinate complexes such as:
[Ag(NH 3 ) 2 ] + + 2NH 3 -♦ [Ag(NH 3 ) 4 ] + (12 ' 1)
[Hg(CN) 2 ] + 2CN~ -♦ [Hg(CN) 4 ] 2_ (12l2)
The low stability of two-coordinate complexes with respect to other possible struc¬
tures is well illustrated by the cyano complexes. Although silver(I) and gold(l) form
discrete bis(cyano) complexes, solid KCu(CN) 2 possesses a chain structure in which
the coordination number of the copper(l) is 3.
If the ligand is sterically hindered sufficiently, such as [N(SiMe 3 ) 2 l ,
|N(SiMePh,),r, [NPhBMes-,] - , and [NPhBXyLP, two-coordinate complexes may
also be formed by ions such'as Mn 2+ , Fe 2+ , Co 2+ , and Ni 2 V The last two ligands
have the advantage that the boryl group draws off one of the lone pairs on the nitrogen
through N->B dative tt bonding and reduces the tendency of the nitrogen to bridge
and form dimeric complexes.
The geometry of coordination number 2 would be expected to be linear, eithei
from the point of view of simple electrostatics or from the use of sp hybrids by the
metal (but there are exceptions; see Chapter 6). If the (« - I )d orbitals of the metal are
sufficiently close in energy to the ns and np orbitals, the d .2 orbital can enter into this
hybridization to remove electron density from the region of the ligands. The tendency
for this to occur will be in the order Hg = Au > Ag > Cu because of relativistic
effects 4 (see Chapter 18). This, in turn, may be partially responsible for the increased
softness of Au(I) and Hg(II) (see Chapter 9).
J Mes = Cfcl-UMcj, Xyl = C 6 H 3 Me 2 . See Chen. H.: Barllclt. R. A.; Olmstead. M M.: Power, fcftj
Shoner. S. C. J. Am. Chem. Soc. 1990. 112. 1048-1055. Power. P. P. Comments Inorg. Chem. 1987,
8. 177-202; and references cited iherein.
4 Schwerdtfeger. P.; Boyd. P. D. W.; Burrell. A. K.; Robinson, W. T.; Taylor. M. J. Inorg. Chem.
1990. 29. 3593-3607.
Table 12.1
Radius ratios
[Ruf Hciltu)(H,0)]~ |Ru(Hc<lia)(NJ|
2NH_ f ' \ r
566
Mechanisms of Redox Reactions 567
In the above instance the kinetically favored nitrito complex isomerizes to the ther¬
modynamically favored nitro complex in seconds.
13 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
two metals is intimately involved in the electron transfer. The classic example of this
type of mechanism was provided by Taube and coworkers. 52 Their system involved
the reduction of cobalt(III) (in [Co(NH 3 ) 5 Cl] 2+ ) by chromium(II) (in [Cr(H 2 0) 6 ] 2+ )
and was specifically chosen because (1) both Co(III) and Cr(III) form inert complexes
and (2) the complexes of Co(II) and Cr(II) are labile (see page 549). Under these
circumstances the chlorine atom, while remaining firmly attached to the inert Co(lll)
ion, can displace a water molecule from the labile Cr(II) complex to form a bridged
intermediate:
Fig. 13.12 A system for the photochemical conversion of N 2 to NH,. Electrons (e - ) are
promoted to the conduction band (CB) leaving holes (h) in the valence band (VB). (Modified
from Khan. M. M. T.; Bhardwaj. R. C.; Bhardwaj, C. Angew. Chern. hit. Ed. Engl. 1988. 27.
923-925. Reproduced with permission.)
55 The rate-determining step in most inner sphere reactions is the electron transfer step, not the
formation of the bridged complex. If dissociation of a reactant complex were rate determining, first-
order kinetics would be expected.
54 Haim. A. Prog. Inorg. Client. 1983. 30. 273-357.
The importance of the nature of the bridging ligand in an inner sphere reaction is
shown in Table 13.9. The reduction of [Co(NH 3 ) 5 CI] 2+ is about lO 10 faster than the
reduction of [ColNH-^J 3 . The bound ammonia ligand has no nonbonding pairs of
electrons to donate to a second metal. Thus the reduction of the hexaammine complex
cannot proceed by an inner sphere mechanism. If ligands are not available which can
bridge two metals, an inner sphere mechanism can always be ruled out. A second
important feature of an inner sphere reaction is that its rate can be no faster than the
rate of exchange of the ligand in the absence of a redox reaction, since exchange of the
ligand is an intimate part of the process. As was noted earlier, electron transfer
reactions must be outer sphere if they proceed faster than ligand exchange.
It is often difficult to distinguish between outer and inner sphere mechanisms. The
rate law is of little help since both kinds of electron transfer reactions usually are
second order (first order with respect to each reactant): 53
rate = A'[oxidant][reductant] (13.75)
Furthermore, although the chloro ligand in Eqs. 13.73 and 13.74 is transferred from
oxidant to reductant. it is not always the case that the bridging ligand is transferred in
an inner sphere reaction. After electron transfer takes place in the dinuclear complex,
the subsequent dissociation may leave the ligand that functioned as a bridge attached
to the metal with which it began. 54 If the bridging ligand stabilized its original complex
more than the newly formed complex, failure of its transfer would be no surprise. For
example:
[Fe(CN)„] 3 - + [Co(CN)j) 3- -♦ [Fe(CN) ( ,] J - + [Co(CN)j] 2 “ (13.76)
Presumably the C-bound cyano group stabilizes the d 6 (Fe 2+ ) configuration of
[Fe(CN),,] 4 " more than the N-bound cyano group would stabilize a d b (Co 3+ ) con¬
figuration in [CotCN)j(NC)] 3- .
If the bridging ligand contains only one atom (e.g.. CD , both metal atoms of the
complex must be bound to it. However, if the bridging ligand contains more than one
atom (e.g.. SCN“), the two metal atoms may or may not be bound to the same
bridging-ligand atom (see Problem 13.30). The two conditions are called adjacent and
remote attack, respectively. A remote attack may lead to both linkage isomers:
(H 3 N) 3 Co — N +[Co(CN) 5 | 3 -
O
O—Co(CN) s
O —N + [Co(NH 3 ) 5 ]> (13.77)
(NC),Co — N.
(Co(NH 3 ) 5 CI] 2+ + [Cr(H 2 0) 6 ] 2+ -► [(H 3 N) 5 Co~CI~Cr(OH 2 ) 5 ] 4+ + H,0
(13.73)
The redox reaction now takes place within this dinuclear complex with formation of
reduced Co(II) and oxidized Cr(III). The latter species forms an inert chloroaqua
complex, but the cobalt(II) is labile, so the intermediate dissociates with the chlorine
atom remaining with the chromium:
[(HjN)jCo—C l—Cr(OH 2 ) 3 ] 4+ - [(H 3 N) 5 Co] 2+ + [ClCr(OH 2 ) 5 ] 2+ (13.74)
The five-coordinate cobalt(II) species presumably immediately picks up a water
molecule to fill its sixth coordination position and then hydrolyzes rapidly to
[Co(H 2 0) 6 ] 2 ' . Formally, such an inner sphere reaction consists of the transfer of a
chlorine atom from cobalt to chromium, decreasing the oxidation state of the former
but increasing that of the latter. In addition to the self-consistency of the above model
(inert and labile species) and the observed formation of a chlorochromium complex,
further evidence for this mechanism has been obtained by running the reaction in the
presence of free radioisotopes of chloride ion in the solution. Very little of this labeled
chloride is ever found in the product, indicating that the chloride transfer has indeed
been through the bridge rather than indirectly through free chloride.
Table 13.9
52 Taube. H.; Myers. H.; Rich. R. L. J. Am. Cltem. Soc. 1953. 75. 4118.
Rate constants for the
reaction of [CofNHjlsXD
with Cr 2+ °
X
k. M _1
s' 1
NH 3
8.9 x
IO -5
H,0
0.5
OH"
1.5 x
10 6
F~
2.5 x
IO 5
cn
6 x
10 5
Br~
1.4 x
IO 6
r
3 x
10 6
NT
3 x
IO 5
" Basolo, F.; Pearson, R.
G. Mechanisms of
Inorganic Reactions. 2nd
ed.; Wiley: New York,
1967.
474
12- Coordination Chemistry: Structure
Coordination
Number 3 s
This is a rare coordination number. Many compounds which might appear to be three-
coordinate as judged from their stoichiometry are found upon examination to have
higher coordination numbers. Examples are CsCuCI 3 (infinite single chains.
Cl CuCU Cl , C.N. = 4 at 228 and 236 pm; two more Cl - from adjacent
segments at 278 pm; note the operation of the Jahn-Teller effect), KCuCI-, (infinite
double chains, CI 4 (Cu,CI-,)—Cl 4 , C.N. = 6, distorted octahedron), and NH 4 CdCI 3
(infinite double chains, C.N. = 6, undistorted).
The KCu(CN), chain described above (—CN— Cu(CN)—(CN)—Cu(CN)—) is
an example of true three-coordination. Some other examples of three-coordination
that have been verified by X-ray studies 6 are tris(trimethylphosphine sulfide)copper(I)
perchlorate, [Cu(SPMe 3 ) 3 J[CI0 4 ] (Fig. 12. la), cyc/o-tris(chloro-p.-trimethylphosphine
sulfidc)copper(I)(Fig. 12.1b), the tris(/-butylthiolato)mercurate(II) anion 7 (Fig. 12. Ic),
the iriiodomercurate(II) anion, [Hgl 3 j“, and tris(triphenylphosphine)platinum(0),
f(Ph 3 P) 3 Pt], In all examples the geometry approximates an equilateral triangle with the
metal atom at the center of the plane as expected for sp 2 hybridization. Some d orbital
participation can be expected as in the case of linear hybrids, since a trigonal sd 2
hybrid is also possible. C.N. = 3 is also favored by steric considerations over the
more common C.N. = 4, and because electronic factors do not favor it. the former
must be dominant.
A few complexes are known in which the geometry is planar but not equilateral.
One angle may be much greater than 120° (T-shaped. Fig. 12. Id) or much less than
120 (Y-shaped, Fig. 12.le). Just as the bond angles are no longer equal, the bond
lengths are no longer equal (see Problem 12.29).”
Coordination
Number 4 9
This is the first coordination number to be discussed that has an important place in
coordination chemistry. It is also the first for which isomerism is to be expected. The
structures formed with coordination number 4 can be conveniently divided into
tetrahedral and square planar forms although intermediate and distorted structures
are common.
Tetrahedral Tetrahedral complexes are favored by steric requirements, either simple electrostatic
Complexes repulsions of charged ligands or van der Waals repulsions of large ones. A valence
bond (VB) point of view ascribes tetrahedral structures to sp } hybridization. From a
crystal field (CF) or molecular orbital (MO) viewpoint we have seen that, in general,
tetrahedral structures are not stabilized by large LFSE. Tetrahedral complexes arc
thus favored by large ligands like Cl - , Br .and I - and small metal ions of three types:
' Eller. P. G.; Bradley, D. C.: Hurslhousc, M. B.; Meek. D. W. Coord. Client. Re 1977. 2d. 1-95.
6 Tiethof. J. A.; Stalick. J. K.: Meek. D. W. Inorg. Chen,. 197 3. 12. 1170-1174. Tielhof, J, A ; Hclcy
A. T.; Meek. D. W. Inorg. Chen,. 1974. 13. 2505-2509. Eller. P. G.: Corfield. P. W. R. Chem.
Conumm. 1971. 105-106. Fenn. R. H.; Oldham. J. W. H.; Phillips. D. C. Nature 1963. 118. 381-382.
Albano, V.; Bellon. P. L.; Scalturin. V. Chem. Commim. 1966. 507-509.
7 This anion has been proposed as a model compound for the receptor site in a mercuric biosensor
(Walton. S. P.; Wright. J. G.: MacDonncll, F. M.; Bryson. 1. W.; Sabat, M.; O'Haltoran T V
J. Am. Chem. Sue. 1990, 112. 2824-2826). Sec Chapter 19.
For example, sec Munakata. M.; Maekawa. M.: Kitagawa. S.; Matsuyama. S.; Masuda. H. Inorg.
Chem. 1989. 28. 4300-4302, and references therein.
9 Favas. M. C.; Kepcrt. D. L. Prog. Inorg. Chem. 1980. 27. 325-463.
Coordination Number 4
475
?
M> (c)
Fig- 12.1 Three-coordinate complexes: (a) Cu[SP(Mc,),J: (b) |Cu(SP(CH,) 3 )CI|,; (c) [HglSBu'),] - ; (d)
|C,,H>CH 2 N(CH 2 CH 2 C,H.|N) 2 Cu) ; (c) [(CH 3 CN)CuN 2 C| 2 H 4 (CH 3 ) 2 )PF ( ,. Carbon atoms are unmarked and hydrogen atoms
have been omitted. (From Eller. P. G.: Corfield. P. W. R. Cliem. Commim. 1971, 105-106. Tiethof. J. A.; Stalick, J. K.;
Meek. D. W. Inorg. Chem. 1973, 12. 1170-1174. Tiethof. J. A.; Hetey. A. T.; Meek, D. W. Inorg. Chem. 1974, 13.
2505-2509. Walton. S. P.; Wright. J. G.; MacDonnel. F. M.; Bryson. J. W.; Sabat. M.; O’Halloran. T. V. J. Am. Chem.
Soc. 1990. 112. 2824-2826. Blackburn, N. J.: Karlin. K. D.: Concannon. M.; Hayes. J. C.; Gultneh, Y.; Zubieta, J. Chem.
Common. 1984. 939-940. Munakata. M.; Maekawa. M.; Kitagawa. S.: Matsuyama, S.; Masuda. H. Inorg. Chem. 1989. 28.
4300-4302. Reproduced with permission.)
(I) those with a noble gas configuration such as Be 2+ (n.v°); (2) those with a pseudo¬
noble gas configuration (/t - I )d'"ns' , np ( \ such as Zn 2+ and Ga 3+ ; and (3) those
transition metal ions which do not strongly favor other structures by virtue of the
LFSE, such as Co 2+ , d 1 .
Ietrahedral complexes do not exhibit geometrical isomerism. However, they are
potentially chiral just as is tetrahedral carbon. The simple form of optical isomerism
exhibited by most organic enantiomers, namely four different substituents, is rarely
observed because substituents in tetrahedral complexes are usually too labile"' for the
complex to be resolved, i.e., they racemize rapidly. However, an interesting series of
cyclopentadienyliron phosphine carbonyl compounds (see Chapter 15 for further
10 Labile refers not to thermodynamic stability per sc but. rather, to the ease of substitution by other
ligands. In addition to bond strength, the accessibility of a suitable mechanism also contributes to
the inertness or lability of a complex (sec Chapter 13). Labile is one of those words the pronuncia¬
tion of which American chemists seem unable to agree upon. Some follow the dictionary and rhyme
it more or less with its antonym, stable. Others rhyme it with Mobile, and a third group shows an
English or Australian bent and rhymes it with hay-stile.
564
13 Coordination Chemistry: Reactions, Kinetics, and Mechanisms
The excited state cation has the potential to reduce water.
2[Ru(bpy) 3 ] 2+ -<■ 2[Ru(bpy) 3 ] 3+ + 2e _ 0.86 V (13.65)
2e~ + 2H 2 0 -■ 20H~ + H 2 -0.41 V (13.66)
2[Ru(bpy) 3 ] 2+ + 2H 2 0- 2[Ru(bpy) 3 l 3+ + 20H~ + H 2 0.45 V (13.67)
The [Ru(bpy) 3 f + generated in reaction 13.67 could then oxidize water.
1.26 V (13.68)
0.82 V (13.69)
0.44 V (13.70)
If we sum Eqs. 13.64-13.70 we obtain Eq. 13.61. The absence of a ruthenium complex
in the overall equation reveals its catalytic nature.
In practice, however, the scheme fails for several reasons. First, reaction 13.67 is
much slower than decay of [Ru(bpy) 3 ] 2+ to its ground state. Second, the production
of one mole of hydrogen requires two electrons and the production of one mole of
oxygen involves four electrons, but the ruthenium complexes provide or accept only
one electron at a time. This means that various intermediates arise as the reactants
proceed to products and the catalyst would be required somehow to stabilize these
intermediates while sufficient electrons are being provided. A variety of quenching
agents, both oxidizing and reducing, have been used in attempts to circumvent these
problems. The cycle presented in Fig. 13.11 is an example of reductive quenching. An
electron is passed from [Ru(bpy) 3 ] 2+ to methylviologen, MV 2+ .
2[Ru(bpy) 3 ] 3+ + 2e" - 2[Ru(bpy) 3 ] 2+
H 2 0 - 40, + 2H + + 2e"
2[Ru(bpy) 3 ] 3+ + H 2 0-» 2[Ru(bpy) 3 ] 2+ + 40, + 2H
Triethanolamine (TEOA) is added to the initial reaction mixture to reduce
[Ru(bpy) 3 ] 3 + as it is generated; thus the back reaction (producing [Ru(bpy) 3 ] 2+ and
MV 2+ ) is retarded. The reduced methylviologen MV + reduces water to hydrogen in
the presence of collodial platinum and is oxidized back to MV 2+ , thereby completing
the cycle. Many other cycles have been devised, some of which succeed in splitting
water into hydrogen and oxygen with visible light, but none of which are yet of
practical importance. 49
The cleavage of water to produce hydrogen and oxygen is only one of several
photochemical redox reactions with major economic potential to be studied in recent
49 Energy Sources through Photochemistry and Catalysis-, Gratzel. M., Ed.; Academic; New York.
1983. See O'Regan, B.; Gratzel, M. Nature 1991, 353, Til -740 for the description of a new solar cell
based on TiO, with a monolayer of RuL,[(/t-CN)Ru(CN)Lj], (L = 2,2'-bipyridinc-4,4'-dicarboxylic
acid; L' = 2,2'-bipyridine). This inexpensive cell is more efficient than the natural solar cells used in
photosynthesis and shows great commercial promise.
Mechanisms of Redox Reactions
565
Inner Sphere
Mechanisms
hv
Fig. 13.11 A system for the photochemical reduction of H,0 to H,. (From Kutal, C. J.
Client. Educ. 1983, 60. 882-887. Used with permission.]
times. Another is the reported reaction of nitrogen with water to produce ammonia 50
(see Chapters 15 and 19 for further discussions of nitrogen fixation):
N, + 3H,0 -Z- 2NH 3 + §0, (13.72)
This conversion is catalyzed by (Ru(Hedta)(H,0)] - (Hedta = trianion of eth-
ylenediaminetetraacetic acid) at 30 °C and 10 s Pa in the presence of a solid semi¬
conductor mixture (CdS/Pt/RuO,). The photocatalytic production of ammonia is
initiated by absorption of visible light (505 nm) by the CdS semiconductor (Fig. 13.12).
Presumably, the incoming photons promote electrons from the valence band (VB) of
CdS to its conducting band (CB), a process that leaves holes in the valence band.
Water is photooxidized by RuO,, releasing electrons which are trapped by holes in the
valence band of CdS. The electrons in the conducting band are transferred to the
ruthenium complex via platinum metal. Protons from the water oxidation are attracted
to the reduced ruthenium complex, interact with coordinated N, in some unknown
fashion, and are expelled as NH 3 . The cycle is complete when the coordination site
left by NH, becomes occupied once again by H 2 0. It remains to be seen whether
proposed cycles such as this one measure up to their promise.
Inner sphere reactions are more complicated than outer sphere reactions because, in
addition to electron transfer, bonds are broken and made. 51 A ligand which bridges
50 Khan. M. M. T.; Bhardwaj. R. C.; Bhardwaj, C. Angew. Chem. Int. Ed. Engl. 1988, 27, 923-925.
51 Haim. A. Prog. Inorg. Chem. 1983, 30, 273. Endicott, J. F.; Kumar, K.; Ramasami, T.; Rotzinger,
F. P. Prog. Inorg. Chem. 1983. 30, 141-187.
476
12 Coordination Chemistry: Structure
0
(b)
Fig. 12.2 (a) Line drawing of an acetyl(carbonyl)(cyclopentadienyl)(phosphine)iron complex,
(b) Stereoview of the same molecule. [From Korp, J. D.; Bernal, I. J. Organomet. Chem.
1981, 220, 355-364. Reproduced with permission.)
discussion of organometallic compounds) has been synthesized and characterized. 11 A
line drawing and stereoview of one of these is shown in Fig. 12.2. Note that the large
C 5 H 5 ring forces the other ligands back until the bond angles are essentially 90° rather
than I09i°. Indeed an argument could be made for considering the complex to be eight-
coordinate, though little is gained by such a view. The chirality of the molecule is the
important feature to be noted.
A second form of optical isomerism analogous to that shown by organic
spirocyclic compounds has been demonstrated. Any molecule will be optically active
if it is not superimposable on its mirror image. The two enantiomers of bis(ben-
zoy!acetonato)beryllium are illustrated in Fig. 12.3. In order for the complex to be
chiral, the chelating ligand must be unsymmetric (not necessarily asymmetric or
chiral, itself); [Be(acac) 2 ] is not chiral.
11 Brunner, H. Adv. Organomet. Cltem. 1980, 18, 151-206. Korp, J. D.; Bernal. I. J. Organomet.
Chem. 1981 ,220. 355-364. J. Organomet. Chem. 1981, 370 (entire issue devoted to "Organometallic
Compounds and Optical Activity," edited by H. Brunner). Saura-Llamas, I.; Dalton. D. M.; Arif,
A. M.; Gladysz, J. A. Organometallics 1992. //. 683-693.
Coordination Number 4
477
Square Planar
Complexes
Fig. 12.3 Enantiomers of bis(benzoylacetonato)beryllium.
Square planar complexes are less favored sterically than tetrahedral complexes (see
Table 12.1) and so are prohibitively crowded by large ligands. On the other hand, if the
ligands are small enough to form a square planar complex, an octahedral complex with
two additional cr bonds can usually form with little or no additional steric repulsion.
Square planar complexes are thus formed by only a few metal ions. The best known
are the </ 8 species such as Ni 2+ , Pd 2+ , Pt 2+ , and Au 3+ (Chapter II). There are also
complexes of Cu 2+ (d 9 ), Co 2+ (d 7 ), Cr + (d 4 ), and even Co 3+ (d 6 ) that are square
planar, but such complexes are not common. 12 The prerequisite for stability of these
square planar complexes is the presence of nonbulky, strong field ligands which tt
bond sufficiently well to compensate for the energy “lost” through four- rather than
six-coordination. For Ni 2+ , for example, the cyanide ion forms a square planar
complex, whereas ammonia and water form six-coordinate octahedral species, and
chloride, bromide, and iodide form tetrahedral complexes. For the heavier metals the
steric requirements are relaxed and the effective field strength of all ligands is in¬
creased. Under these conditions, even the tetrachloropalladate(II), tetrachloroplati-
nate(II), and tetrachloroaurate(III) anions are low spin square planar.
One unexpected square planar complex [Cd(OAr) 2 (thf) 2 ] (Fig. 12.4) has recently
been reported. 12 It is the first example in this geometry for d w Cd 2+ . Inasmuch as a
closely related complex of the smaller zinc ion, [Zn(OAr') 2 (thf) 2 ], has distorted
tetrahedral geometry, 14 simple steric factors cannot account for the pseudo-D 2/ , sym¬
metry of the cadmium complex, though to be sure, the steric relaxation of the larger
metal atom and the perpendicular planes of the thf and phenoxide rings are nicely
accommodated by it. Furthermore, octahedral geometry would not be unusual in ad 10
12 Hermes, A. R.; Morris, R. J.; Girolami, G. S. Organometallics 1988, 7, 2372-2379. Brewer, J. C.;
Collins, T. J.; Smith, M. R.; Santarsicro, B. D. J. Am. Chem. Soc. 1988, 110. 423-428,
■J Gocl, S. C.; Chiang. M. Y.; Buliro. W. E. J. Am. Chem. Soc. 1990, 112, 6724-6725. "OAr = 2.6-
di-f-butylphcnoxidc. thf = Ictrahydrofuran. Another complex in which this geometry occurs unex¬
pectedly is (RjP(0)NR'] 2 Ni (R = Bu'; R' = Pi 4 ), which has been shown to be square planar and
paramagnetic, a combination not predicted for a d metal ion. (FrcSmmel, T.; Peters, W.; Wun¬
derlich. H.; Kuchen. W, Angew. Chem. Int. Ed. Engl. 1992, 31, 612-613).
14 C 2l . microsymmetry at the zinc with an Ar'O-Zn-OAr' angle of 122° and a thf-Zn-thf angle of 95°.
“OAr‘ = 2.4,6-tri-t-butylphenoxide.
562
13 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Toblo 13.8
Calculated and observed
log
^12ob»d
^12calcd
rate constants for outer
Reaction
ffl2
(M _1 s' 1 )
(M- 1 s'" 1 )
sphere cross reactions 0
Ru(NH 3 ) 2+ + Ru(NH 3 ) 5 py 3+
1.4 x 10 6
4.40
4 x 10 6
Ru(NH 3 ) 5 py 2+ + Ru(NH 3 ) 4 (bpy) 3+
3.39
1.1 x 10 8
4 x 10 7
Ru(NH 3 ) 2+ + Co(phen) 3+
5.42
1.5 x 10 4
1 x I0 5
Ru(NH 3 ) 5 py 2+ + Co(phen) 3 +
1.01
2.0 x 10 3
1 x !0 4
V 2 ; + Co(en) 3+
0.25
5.8 X 10 -4
7 x 10“ 4
V 2 ; + Ru(NH 3 ) 3+
5.19
1.3 x 10 3
1 x I0 3
V£ + Fe 3 q +
16.90
1.8 x 10 4
2 x I0 6
Fe 2 q + Os(bpy) 3+
1.53
1.4 x 10 3
5 x 10 5
1
p ejq + Fe(bpy) 3+
3.90
2.7 x 10 4
6 x 10 6
!
Ru(NH 3 ) 2+ + Fe 3 ;
11.23
J.4 x 10 5
2 x 10 6
r
Ru(en) 2+ + Fe 34
9.40
8.4 x 10 4
4 x 10 5
i
Mo(CN) 4 " + IrCl 2 ~
2.18
1.9 x 10 6
8 x 10 s
j
Mo(CN) 4- + MnOJ
-4.07
2.7 x 10 2
6 x 10 1
Mo(CN) 4- + HMn0 4
8.48
1.9 x 10 7
2 x 10 7
[
Fe(CN) 4 " + IrCIf
4.08
3.8 x 10 5
1 x 10 6
Fe(CN) 4 " + Mo(CN)j} -
2.00
3.0 x 10 4
4 x I0 4
I
Fe(CN) 4 " + MnO;
3.40
1.7 x 10 s
6 x 10 4
Marcus, R. A.; Sutin, N. Biochim. Biophys. Acia 1985, SI I, 265.
[Ru(bpy) 3 ] 2+ . 46 When this cation absorbs light at 452 nm, the excited state species that
initially forms, [Ru(bpy) 3 ] 2+ , relaxes to a relatively long-lived one. [Ru(bpy) 3 ] 2 + . 47
The electronic transition involved in the absorption is an example of a mctal-to-ligand
charge transfer in which a d electron of ruthenium is promoted to a tr antibonding
orbital of one of the bipyridine ligands. Thus the excited state complex,
[Ru"(bpy) 3 ] 2+ , may be formulated as [Ru"‘(bpy),(bpy~)] 2+ . The availability of an
electron in a ligand antibonding orbital makes this excited state cation a much better
reducing agent than the ground state cation. Furthermore, the "hole'' created at the
ruthenium center enhances its electron-seeking power and as a result, the excited
cation is also a much better oxidizing agent than it was in its ground state. A
comparison of the redox properties of the ground state with those of the excited state
is shown in Fig. 13.10. Here we see that [Ru(bpy) 3 J 2+ is a better oxidizing agent than
[Ru(bpy) 3 ] 2+ by 2.12 volts (+0.84 V + 1.28V) and a better reducing agent by 2.12
volts ( + 0.86 V + 1.26 V). These are large voltages from which one can easily see the
potential for a wide range of redox chemistry.
Whether a complex in an excited state can manifest its enhanced redox properties
will depend on whether it can undergo electron transfer faster than it undergoes
something else, such as relaxation to the ground state (luminescence). The emission
lifetime of [Ru(bpy) 3 J 2+ in aqueous solution at 25 °C is 0.6 /xs and it increases
4,1 Juris, A.; Balzani, V.; Barigelletti. F.; Campagna. S.; Belser. P.; von Zelewsky, A. Coord. Chem.
Rev. 1988. S4. 85-277. Meyer, T. J. Prop. Inorg. Chem. 1983. JO. 389-140. Walls R. J. J. Chem.
Educ. 1983, 60. 834-842. Kutal. C. J. Chem. Educ. 1983. 60. 882-887.
47 Earlier we used M to indicate a radioactive isotope. Here we use M to indicate an excited state.
Mechanisms of Redox Reactions
563
Radiationless
decay
Luminescence\»o,H4 v
Fig. 13.10 Absorption of blue light by [Ru(bpy) 3 ] 2 gives [••Rulbpy).,] 2 which relaxes to
[Ru(bpy) 3 ] 2+ without light emission. [Ru(bpy) 3 ] 2 may emit orange-red light (luminesce) or
undergo oxidation or reduction. Standard reduction potentials associated with individual
processes are shown in the diagram. [Modified from Juris, A.; Balzani. V.: Barigelletti, F.;
Campagna. S.; Belser, P.: von Zelewsky. A. Coord. Chem. Rev. 1988, 84. 85-277. Used with
permission.]
substantially at low temperatures. This can be compared with the rates for electron
transfer by self-exchange.:
[Ru(bpy) 3 ] 2+ + [Ru(bpy) 3 ] 3+ - 4 [Rulbpy),] 2 + [Ru(bpy) 3 ] 3+ (13.59)
|Ru“(bpy) 3 ) 2+ + [Ru(bpy) 3 r - [Ru(bpy) 3 j 2+ + [Ru(bpy) 3 ]^ (13.60)
for which rate constants have been estimated to be 10 s M“ 1 s -1 . It is quite clear that
[Ru(bpy),J 2 ’' may exist in solution long enough to participate in electron transfer
reactions.
Interest in [Ru(bpy) 3 | 2 + skyrocketed after Creutz and Sutin suggested that it had
possibilities for the photochemical cleavage of water: 4
H,0(l> H,(g) + !0 : (g) (13.61)
AG"= 238 kJ mol -1
The suggestion led to the speculation that solar energy could be used to make
hydrogen gas which could then be used as fuel. At pH 7 and 10 s Pa. potentials for the
reduction and oxidation of water are as follows:
2e~ + 2H.0
20H" + H,
2H + + 40, + 2e“
Water does not absorb visible light, but one could envision a sequence of reactions
that utilize [Ru(bpy) 3 ] 24 ' as a photosensilizer for the decomposition. The first step
would be absorption of solar energy by [Ru(bpy) 3 ] 21 ’:
[Ru(bpy) 3 ] 2
[Ru(bpy) 3 .] 2 +
4 Creutz. C.: Sutin. N. Proc. Nall. Acad. Sci. USA 1975. 72 , 2858. Sutin. N.; Creutz, C. Pure Appl.
Chem. 1980. 52. 2727. Sykora, J.; Sima. J. Coord. Chem. Rev. 1990, 107. 1-212.
478
12- Coordination Chemistry: Structure
Fig. 12.4 ORTEP of
[Cd(OAr),(thf) 2 ]. Hydrogen
atoms have been omitted
for clarity. The Cd atom
resides on a crystallo¬
graphic inversion center.
[From Goel, S. C.; Chiang,
M. Y.; Buhro, W. E. J.
Am. Chem. Soc. 1990, 112,
6724-6725. Reproduced
with permission.]
species, but the 2,6-di-t-butyl groups may prevent coordination of a fifth and sixth
ligand. From an electronic viewpoint, it may be that the stronger cadmium-phenoxide
bonds dominate the bonding picture, leaving the weakly basic thf molecules to bind
as best they can. 15 If so, this would be an example of Bent's rule (the maximization
of s character towards the strongest bonding ligand) acting in a complex ion. See Prob¬
lem 12.38.
Square planar complexes of the formula [Ma 2 B 2 ] may exhibit cis-trans isomerism:
CI- < .- > .NH 3
ci nh 3
i .Pt i /t/0
Cl. -nh 3
i/ Pt O
HjN- " ” C1
M = 0
cis-diammincdichloroplatinum (II) frans-diamminedichloroplatinum(II)
If such complexes are neutral molecules as in the above example, they may be readily
distinguished (and often separated as well) by the presence of a dipole moment (p.) in
the cis isomer but none in the trans isomer. Only in the unlikely event that the M — A
and M—B bond moments were identical could the cis isomer have a zero dipole
moment.
Square planar complexes rarely show optical isomerism. The plane formed by the
four ligating atoms and the metal ion will ordinarily be a mirror plane and prevent the
possibility of chirality. An unusual exception to this general rule was used in an
ingenious experiment to prove that platinum(II) and palladium(II) complexes were not
tetrahedral. 16 Carefully designed complexes (Fig. 12.5a) with square planar structures
have no improper axes of rotation and hence are chiral. If these complexes were
tetrahedral (Fig. 12.5b), there would be a mirror plane (defined by the metal and two
nitrogen atoms from isobutylenediamine) reflecting the phenyl groups, methyl groups,
etc. Inasmuch as optical activity is found experimentally, these complexes cannot be
tetrahedral and, barring some unusual geometry, must be square planar.
IJ Haaland, A. Angew. Chem. Int. Ed. Engl. 1989. 28, 992-1007. See also Footnote 13.
,A Mills. W. H.; Quibell, T. H. H. J. Chem. Soc. 1935 . 839-846. Lidstonc. A. G.: Mills. W. H.
J. Chem. Soc. 1939, 1754-1759.
Coordination
Number 5 17
Coordination Number 5 479
Fig. 12.5 Possible
structures of ( meso -
stilbenediamine)
(isobutylenediamine)
palladium(II) and
platinum(II) complexes: (a)
planar structure, optically
active; (b) tetrahedral
structure, optically inactive.
In the past a coordination number of 5 was considered almost as rare as a coordination
number of 3. Again, many of the compounds which might appear to be five-coordinate
on the basis of stoichiometry are found upon close examination to have other coordi¬
nation numbers. Thus Cs 3 CoCI 5 and (NH 4 ) 3 ZnCl 5 contain discrete tetrahedral
MCI 7- anions and free chloride ions. Thallium fluoroaluminate, TI 2 AIF 5 , is com¬
posed of infinite chains, —F—AIF 4 —F, in which the coordination number of the
aluminum is 6. The complex of cobalt(II) chloride and diethylenetriamine,
H 2 NCH 2 CH 2 NHCH 2 CH 2 NH 2 , of empirical formula [CoCUdien] is not a five-coordi¬
nate molecule but a salt, [Co(dien) 2 ][CoCI 4 ], containing octahedral cations and
tetrahedral anions.
If electrostatic forces were the only forces operating in bonding, five-coordinate
compounds would always disproportionate into four- and six-coordinate species (as
does the “Co(dien)CI 2 " complex above). Since covalent bonding is obviously of great
importance in coordination compounds, it is possible to have stable five-coordinate
complexes, but it is true that there is a delicate balance of forces in these complexes,
and their stability with respect to other possible structures is not great. For example,
the compound [Ni(PNP)X 2 ] (where PNP = (C 6 H 5 ) 2 PCH 2 CH 2 NRCH 2 CH 2 P(C 6 H 5 ) 2 )
is a true five-coordinate species, but on warming slightly it converts to
[Ni(PNP)X] 2 [NiX 4 ], which contains both square planar and tetrahedral species. An¬
other example is the pair of compounds of empirical formula MX 2 (Et 4 dien) where
M = Co or Ni, and Et 4 dien = Et 2 NCH 2 CH 2 NHCH 2 CH 2 NEt 2 . The cobalt complex
is five-coordinate, but the corresponding nickel compound is four-coordinate,
[NiX(Et 4 dien))X.
17 For reviews of coordination number 5, sec Holmes, R. R. Prog. Inorg. Chem. 1984. 32, 119-235;
Kepert, D. L. In Comprehensive Coordination Chemistry; Wilkinson. G.; Gillard, R. D.; Mc-
Clcvcrty, J. A.. Eds.; Pergamon: Oxford. 1987; Vol. I, pp 39-48.
560 13-Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Fig. 13.9 (a) Potential energy diagram for a homonuclear electron transfer reaction such as
[Fe(H 2 0) fi ] J+ + [Fe(H 2 OU + -♦ [Fe(H 2 0) 6 ] 3+ + [Fe(H 2 0) A ] 2+
(b) Potential energy diagram for a heteronuclear electron transfer reaction. Each diagram represents a cross section of the
total potential energy of reactants (R) and products (P) in configurational space. Electron transfer may occur at I, the
intersection of the two potential energy surfaces, where the energies of the participating orbitals are equal.
Co(II)—N bond is necessary before electron transfer can occur. In contrast, the self¬
exchange rate constant for the |Ru(NH,) 6 ] 2+ /[Ri‘(NH 3 ) 6 ] 3+ couple is 8.2 x I0 2 M~‘
s -1 and the Ru—N bond length difference is 0.04(6) A. 4l > This much faster rate for the
ruthenium exchange is consistent with a small bond length adjustment prior to elec¬
tron transfer. The cobalt and ruthenium systems are not entirely analogous, however,
since cobalt goes from a low spin d h complex to a high spin d 1 complex while
ruthenium remains low spin in both the oxidized and reduced forms. It has been
argued that the cobalt reaction is anomalously slow because it is spin forbidden;
however, recent work does not support this hypothesis. 41
It should be noted that not all self-exchange reactions between Co(III) and Co(ll)
are slow. The nature of the bound ligand has a significant influence on the reaction
rate. In particular, ligands with tr systems provide easy passage of electrons. For
[Co(phen) 3 ] 3+ /!Co(phen) 3 ] 2+ exchange, for example, k is 40 M -1 s _l , many orders of
magnitude faster than for the cobalt ammine system. 42
Marcus has derived a relationship from first principles that enables one to calcu¬
late rate constants for outer sphere reactions: 43
k n = (k u k 22 K n f l2 )"2 (13.55)
To illustrate the application of this equation and the definitions of all of its terms, we
will calculate the rate constant, k l2 , for the reaction of Eq. 13.47 (called a cross
40 Slyncs. H. C.; Ibcrs. J. A. Inorg. Clwm. 1971. 10. 2304-2308.
41 Lars son, S.; St4hl, K..; Zcrner, M. C. Inorg. Client. 1986. 25. 3033-3037.
42 Farina. R; Wilkins. R. G. Inorg. Chem. 1968. 7, 514-518.
43 Marcus, R. A. Anno. Rev. Phvs. Chem. 1964, 15. 155-196. Newton. T. W. J. Chem. Educ. 1968. 45.
571-575.
Excited State Outer
Sphere Electron
Transfer Reactions
Mechanisms of Redox Reactions 561
reaction as compared to a self-exchange reaction). For this reaction, k tl is (he rate
constant for the self-exchange process involving the hexacyanoferrate complexes:
[Fe(CN)J 4 - + [Fe(CN)J 3 - -► [Fe(CN) ft l J " + [Fe(CN)J 4 " (13.56)
k u = 7.4 x I0 2 M - ' s _1
k 22 is the rate constant for the similar reaction involving the octacyanomolybdenum
complexes:
(Mo(CN)J 3 - + [Mo(CN)j,] 4 - - [Mo(CN) 8 | 4 - + |Mo(CN) K ] 3 ~ (13.57)
k 22 = 3 x I0 4 M -1 s -1
K i2 is the equilibrium constant for the overall reaction (Eq. 13.47):
[Fe(CN) h ] 4- + [Mo(CNy 3 ~ ^ [Fe(CN)J 3 ~ + [Mo(CN) H ] 4 - (13.58)
K a = 1.0 x I0 2
and log/,, = (log K ]2 ) 2 /4 \og(k tl k 22 /Z 2 ). This last term contains Z. which is the
collision frequency of two uncharged particles in solution and is taken as I0 11 M _l
s“ 1 . The factor/, 2 has been described as a correction for the difference in free energies
of the two reactants and is often close to unity (in this case it is 0.85). 44 When all of the
appropriate values are substituted into Eq. 13.55, k l2 is calculated to be 4 x 10 4 M _l
S"‘. which compares quite well with the experimental value of 3.0 x I0 4 M -1 s“'. 45
Table 13.8 summarizes results for a number of other outer sphere cross reactions.
Confidence in the Marcus equation is high enough that, if it leads to a calculated rate
constant that is in strong disagreement with an experimental value, a mechanism other
than outer sphere should be considered.
The Marcus equation (13.55) connects thermodynamics and kinetics, as shown by
the dependence of k l2 on K l2 : As K l2 increases, the reaction rate increases. Thus
outer sphere reactions which are thermodynamically more favorable tend to proceed
faster than those which are less favorable. These observations may be surprising to
you since most elementary treatments of reaction dynamics keep thermodynamics
(how far) and kinetics (how fast) separate. Here we see that how fast a reaction occurs
can depend to some degree on how far it goes, or the driving force, AG. The simplified
Marcus equation we have presented here, however, breaks down when K t2 becomes
large. The complete theory reveals that rate increases rapidly with increasing spon¬
taneity, reaching a maximum when the change in free energy is equal to the sum of
reorganization energies, and then decreases as the driving force increases further.
The redox properties of a transition metal complex may change dramatically if it
has absorbed energy and exists in an excited state. One of the most famous and
widely studied complexes in this area is tris(2,2'-bipyridine)ruthenium(II) cation,
4-1 Pennington. D. E. In Coordination Chemistry. Martell. A. E.. Ed.: ACS Monograph 174; American
Chemical Society: Washington. DC. 1978: Vol. 2. Chapter 3.
4 'Campion. R. J.; Purdie. N.: Sutin. N. Inorg. Chem. 1964. 3. 1091-1094. Chou. M.: Creutz. C.;
Sutin, N. J. Am. Chem. Soe. 1977. 99. 5615-5623. Sutin. N.; Creutz. C. J. Chem. Edttc. 1983. 60.
809-814.
480
12 • Coordination Chemistry: Structure
Coordination Number 5
481
Limiting Geometries:
Trigonal Bipyramidal
and Square
Pyramidal
Although five-coordinate compounds are still less common than those of either coordi¬
nation number 4 or 6, recently there has been considerable interest in them, and the
number of compounds with known structures has increased rapidly. The complexes
can be described as ‘Tegular" or "distorted" trigonal bipyramidal (TBP; Fig. 12.6),
“regular” or "distorted” square pyramidal (SP; Fig. 12.7), or as "highly distorted
structures," i.e., something between TBP and SP. As we have seen, however, every
intermediate structure between “perfectly TBP” and "perfectly SP” is possible
(Chapter 6), and it serves little purpose to try to fit them into neat "pigeonholes.” The
differences between the various structures are often slight and the energy barriers
tending to prevent interconversion are also small.
Of particular interest in regard to the delicate balance between the forces
favoring TBP versus SP structures are two pentacyanonickelate(ll) salts with differ¬
ent but very similar cations. Tris(l,3-diaminopropane)chromium(Ill) pentacyano-
nickelate(II), [Cr(tn) 3 ][Ni(CN) 5 ], contains square pyramidal anions. In contrast,
crystalline tris(ethy!enediamine)chromium(III) pentacyanonickelate(II) sesquihy-
drate, [Cr(en) 3 ][Ni(CN) 5 ]- 1.5H 2 0, contains both square pyramidal anions (Fig. 12.7)
and slightly distorted trigonal bipyramidal anions. The IR and Raman spectra of this
solid exhibit two sets of bands, one of which (the TBP set) disappears when the
sesquihydrate is dehydrated. In aqueous solution the structure is apparently also
square pyramidal. It would appear that the SP structure is inherently more stable but
Fig. 12.6 Trigonal
bipyramidal structure of the
pentachlorocuprate(II)
anion in the compound
[CitNHjWICuCI,]. Note
difference in bond lengths.
[From Raymond, K. N.;
Meek. D. W.; Ibers. J. A.
Inorg. Client. 1968 , 7.
1111-1118. Reproduced
with permission.)
Fig. 12.7 Square pyramidal
structure of the penta-
cyanonickelate(II) anion in
[Crten)jJ[Ni(CN) s H.5H 2 0.
Note the difference in bond
lengths. [From Raymond,
K. N.; Corfield, P. W. R.;
Ibers, J. A. Inorg. Client.
1968. 7. 1362-1372.
Reproduced with
permission.)
by such a slight margin that forces arising in the hydrated crystal can stabilize a TBP
structure. The forces favoring each of the limiting structures are not completely
understood, but the following generalizations can be made. On the basis of ligand
repulsions alone, whether they be considered naively as purely electrostatic or as
Pauli repulsions from the bonding pairs, the trigonal bipyramid is favored (see Chapter
6). For this reason almost every five-coordinate compound with a nonmetallic central
element (such as PF 5 ) has the TBP structure (unless there are lone pairs), since effects
arising from incompletely filled d orbitals are not present. Likewise, we should expect
d° and d'° to favor the TBP structure. Comparison of the relative energies of the
orbitals in TBP ( D Jh ) versus SP (C 4 „) geometry (Fig. 12.8) shows that d\ d 2 , d \ and
d 4 configurations should also favor TBP versus SP as much or more since the e"
orbitals of D v , are more stable than the e of C 4v . In contrast, low spin d 6 should favor
the SP configuration since the e orbitals of the latter are lower in energy than are the e'
orbitals of a TBP complex. 18 For d R the order of stability again switches back to favor
TBP ( e’ is lower in energy than a,) and this continues through d 9 and d"\ Unfortu¬
nately, there are few data available to test these predictions. The low spin d 7 complex
[Co(dpe) 2 Cl] + [dpe = l,2-bis(diphenylphosphino)ethane] crystallizes in two forms: a
red solid that contains SP ions and a green form that contains TBP ions (Fig. 12.9). 19
Apparently, the slight ligand field stabilization energy favoring the SP arrangement
Cu -
Fig. 12.8 Wave function and energy changes along a Berry pseudorotation coordinate. Note
relative energies of the e' {Du,) and a, and e (C 4l .) levels. [From Rossi, A. R.; Hoffmann, R.
Inorg. Cltern. 1975. 14, 365-374. Reproduced with permission.)
18 Note, however, that octahedral low spin d b and square planar low spin d" are favored even more.
'» Stalick. J. K.; Corfield. P. W. R.; Meek. D. W. Inorg. Chem. 1973, 12, 1668-1675.
558
559
13 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Outer Sphere
Mechanisms
involving transition metal complexes have been divided into two broad mechanistic
classes called outer sphere and inner sphere. In this section we will compare these
mechanisms and examine factors which influence reaction rates for each.
In this type of reaction bonds are neither made nor broken. Consider the reaction:
[Fe(CN) 6 ] 4 " + [Mo(CN) 8 ] 3_ -♦ [Fe(CN) 6 ] 3 - + [Mo(CN) 8 ] 4 - (13.47)
Such a reaction may be considered to approximate a simple collision model. The rate
of the reaction is faster than cyanide exchange for either reactant so we consider the
process to consist of electron transfer from one stable complex to another with no
breaking of Fe—CN or Mo—CN bonds.
An outer sphere electron transfer may be represented as follows:
0 + R -► [0-R]
(13.48)
[0-R] -► [0—R]
(13.49)
[O—R] - [CT-R]
(13.50)
--r+] -► o' + R +
(13.51)
First the oxidant (O) and reductant (R) come together to form a precursor complex.
Activation of the precursor complex, which includes reorganization of solvent mole¬
cules and changes in metal-ligand bond lengths, must occur before electron transfer
can take place. The final step is the dissociation of the ion pair into product ions.
A specific example further clarifies the activation and electron transfer steps. The
exchange reaction between solvated Fe(III) and Fe(II) has been studied with radioac¬
tive isotopes (Fe) of iron. 36
[Fe(H 2 0),J 3+ + [Fe(H 2 0) 6 ] 2> -♦ [Fe(H 2 0)J 2+ + [Fe(H 2 0) 6 ] 3+ (13.52)
The energy of activation, AC, for this reaction is 33 kJ mol~ 1 . One might ask why it is
not zero since the reactants and products are the same. In order for electron transfer
to occur, the energies of the participating electronic orbitals must be the same, as
required by the Franck-Condon principle. 37 In this reaction an electron is transferred
from a t 2k . orbital of Fe(ll) to a t 2ll orbital of Fe(III). The bond lengths in Fe 2 ‘ and Fe 3+
complexes are unequal (see Table 4.4), which tells us that the energies of the orbitals
are not equivalent. If the electron transfer could take place without an input of energy,
we would obtain as products the Fe(II) complex with bond lengths typical of Fe 3+ and
the Fe(III) complex with bond lengths typical of Fe 2+ ; both could then relax with the
release of energy. This would clearly violate the first law of thermodynamics. In fact,
there must be an input of energy in order for electron transfer to take place. The actual
process occurs with shortening of the bonds in the Fe(II) complex and lengthening of
the bonds in the Fe(III) complex until the participating orbitals are of the same energy
(Fig. 13.8). Vibrational stretching and compression along the metal-ligand bonds
allow the required configuration to be achieved.
The energy of activation may be expressed as the sum of three terms:
AG = A Gf + AG} + AG}, (13.53)
36 Sykes, A. G. Kinetics of Inorganic Reactions', Pcrgamon: Oxford, 1966; Chaplcr 2.
37 Lewis. N. A. J. Chem. Educ. 1980. 57, 478-483.
£1
Mechanisms of Redox Reactions
h ,°\I ;° h , H '-°\ | / 0H ’
Fe
/IV
Fe
/I V
3+
OH,
H.O | =OH,
x •
H.O
H.O
• \
/
H j0
3+
OH,
i ; 0H :
I'' ’
OH,™’
Fig. 13.8 Extension and compression of the Fe—O bonds in [FelHoO)] 5 and
[Fe(H : 0) 6 ] 2 ‘’'. respectively, to form an activated complex in which all metal-ligand distances
are identical, a prerequisite for electron transfer between the two complexes. [From Lewis,
N. A. J. Client. Educ. 1980.37, 478-483. Used with permission.]
in which AG} is the energy required to bring the oxidant and reductant into a
configuration in which they are separated by the required distance (for charged
reactants this includes work to overcome coulombic repulsion), A Gf is the energy
required for bond compression and stretching to achieve orbitals of equal energy, and
AG is the energy needed for solvent reorganization outside of the first coordination
sphere.
Potential energy diagrams further clarify the connection between molecular mo¬
tion and electron transfer. The potential energy of all reactant and associated solvent
nuclei before electron transfer can be approximated as a harmonic potential well
(Fig. 13.9a. R). The potential energy of all product and solvent nuclei after electron
transfer can be described similarly (Fig. 13.9a, P). The reactants and products of Eq.
13.52 have the same energy, as shown in Fig. 13.9a. At the intersection of potential
energy surfaces, 1, the requirement of equal orbital energies is met. For electron
transfer to occur at I, however, coupling of vibrational and electronic motion must
take place. The extent of this interaction is related to A E shown in the figure. If the
coupling interaction is strong, which is the condition when bond distortions are small,
electron transfer is favorable. If the interaction is very weak, associated with large
bond distortions, AG will be large and the reaction will be slow. These considerations
are equally applicable to heteronuclear reactions as depicted in Fig. 13.9b.
The importance of bond distortion magnitudes is revealed in the self-exchange
reaction of hexaamminecobalt complexes:
[Co(NH,)J 3+ + [Co [Co(NH,)„] 2+ + ICo(NH 3 )J 3+ (13.54)
The second-order rate constant for this slow reaction is I0 -6 M _l s -1 . 38 The Co—N
bond length in Co(lII) is 1.936(15) A while in Co(II) it is 2.114(9) A, a difference of
0.178 A. 3y Considerable elongation of (he Co(III)—N bond and compression of the
5,1 Hammershi. A.; Geselowitz. D.; Taubc. H. Inorg. Chem. 1984, 23, 979-982.
39 Kime N. E.; Ibers. J. A. Acta Cryslallogr., Sect. B: Struct. Sci. 1969, 25, 168.
482
12’ Coordination Chemistry: Structure
Coordination Number 5
483
Cl
Cl
Fig. 12.9 The structures of
the red, square pyramidal
isomer (a) and the green
trigonal bipyramidal isomer
(b) of the chlorobis[l,2-
bis(diphenylphosphino)-
ethane]cobalt(II) cation.
Phenyl groups and other
substituents have been
removed for clarity. [From
Stalick. J. K.; Corfieid. P.
W. R.; Meek, D. W. Inorg.
Chem. 1973, 12, 1668-1675.
Reproduced with
permission.)
balances the inherent superiority of the TBP arrangement and allows the isolation of
both isomers. In solution, the two forms interconvert readily, either by a Berry
pseudorotation or through dissociation and recombination (see Chapter 6).
Finally, polydentate ligands can affect the geometry of a complex merely as a
result of their own steric requirements. For example, we find some tetradentate
ligands such as tris(2-dimethylaminoethyl)amine, [Me 6 tren = ((CH 3 ) 2 NCH 2 CH 2 ) 3 N),
form only five-coordinate complexes (Fig. 12.10), apparently because the polydentate
ligand cannot span a four-coordinate tetrahedral or square planar complex and cannot
conform (“fold") to fit a portion of an octahedral coordination sphere.
Fig. 12.10 Molecular
structure of bromotris(2-
dimethylaminoethy!)-
aminecobalt(II) cation in
[CoBr(Me 6 tren)JBr, [From
Di Vaira, M.; and Orioli,
P. L. Inorg. Chem. 1967, 6,
955. Reproduced with
permission.)
Site Preference in
Trigonal
Bipyramidal
Complexes
We have seen that with nonmetallic central atoms such as phosphorus (d°), more
electronegative elements prefer the axial positions of a TBP structure. A molecular
orbital analysis of metal complexes 20 indicates that most d" configurations follow this
same pattern. A notable exception is d 5 , which favors electropositive substituents at
apical sites and electronegative substituents at equatorial sites. In the same way, the
normally weak bonding of axial substituents is reversed with the d 8 configuration.
Thus we find the methyl group in the axial position in the d 8 iridium(I) complex shown
in Fig. 12.11a in contrast to its universal equatorial position in phosphoranes. Also, in
contrast to the phosphoranes, the axial bonds are shorter in Fe(CO) 5 than are the
equatorial bonds (Fig. 12.1 lb); however, it must be stressed that there are exceptions
to this behavior (Table 12.2). The same type of analysis 21 predicts that in d s com¬
plexes good ir-accepting ligands wiil prefer the equatorial position. The compounds
shown in Figs. 12.11 and 12.12 allow us to test this. Note that most of the ligands occur
in both axial and equatorial positions depending upon what other ligands are present.
Fig. 12.11 Complexes showing apparent exceptions to the rules for trigonal bipyramidal
bonding, (a) The methyl ligand seeks the axial position and allows the strong ir acceptors to
occupy equatorial positions. Substituent groups on the phosphine ligands omitted for clarity.
[From Rossi. A. R.; Hoffmann, R. Inorg. Chem. 1975, 14, 365-374. Reproduced with
permission.) (b) The equatorial bonds in Fe(CO) 5 are slightly longer than the axial bonds.
Table 12.2
Bond lengths in some d B TBP
complexes 0
M
—L (pm)
Complex
Axial
Equatorial
Fe(CO) 5
181
183
[Co(CNCH,) 5 ) +
[Pt(SnCl 3 ) 5 r
184
188
254
254
[Mn(CO),)-
182
180
° Rossi, A. R.; Hoffmann, R. Inorg.
Chem. 1975. 14, 365-374.
20 Rossi. A. R.; Hoffmann. R. Inorg. Chem. 1975, 14. 365-374.
21 There is insufficient space here to go through the complete derivation, but it may be noted that the
method is not unlike that given previously for octahedral complexes (Chapter 11). For the complete
method, see Footnote 20.
556
557
13 Coordination Chemistry: Reactions, Kinetics, and Mechanisms
The dissymmetry would thus be lost, and when the chelate ring reforms, it would have
a 50-50 chance of producing either the A or A isomer (Chapter 12). Since the rate¬
determining step in this mechanism is the dissociation of the ligand, the rate of
racemization ( k r ) would have to equal the rate of dissociation (k d ). For example,
tris(phenanthroline)nickel(II) racemizes at the same rate ( k r = 1.5 x 10~ 4 s~') as it
dissociates (k d = 1.6 x 10~ 4 s _I ), which implies dissociation is part of the racemiza¬
tion mechanism. If racemization takes place faster than dissociation [as it does, e.g.,
for tris(phenanthroiine)iron(II): k r = 6.7 x 10 -4 s _l and k d = 0.70 x 10 -4 s _1 ], this
mechanism can be eliminated.
For cases that involve dissociation, it is probable that only one end of the chelate
detaches with formation of a five-coordinate complex. This complex could have either
trigonal bipyramidal or square pyramidal geometry and either of these could undergo
Berry pseudorotation (see Chapter 6) with scrambling of ligand sites. Reattachment of
the dangling end of the bidentate ligand to reform the chelate ring would give a racemic
mixture of A and A isomers.
For many complexes, k r > k d , which means that racemization occurs without
bond rupture; for these, an intramolecular pathway must be operative. Four symmetry
allowed intramolecular pathways have been identified (Fig. 13.7). 29 The ‘‘push
through” (six coplanar ligands) and the “crossover" (four coplanar ligands) mecha¬
nisms both require large metal-ligand bond stretches to relieve steric hindrance and
are energetically unfavorable. "Twist" mechanisms require more modest bond stretches
and are believed to account for the racemization. The earliest twist mechanism,
proposed by R&y and Dutt, 30 is known as the rhombic twist. Some years later, Bailar
suggested a trigonal twist mechanism. 31 The RAy-Dutt twist involves rotating a
trigonal face that is not associated with a threefold axis of the complex through a C 2v
transition state into its mirror image. The Bailar twist can be seen as twisting a
complex about a threefold axis through a trigonal prismatic transition state into its
mirror image. It is a commentary on the experimental difficulties encountered that so
many years have passed with little firm experimental evidence in support of one or the
other of the mechanisms. It appears that rigid chelates and those with small bite angles
favor a trigonal twist. 32 Small bite angles are known to stabilize trigonal prismatic
geometries (Chapter 12) and thus might be expected to reduce energy barriers to such
a twist. Calculations suggest that the trigonal twist is favored when the bite b , defined
as the distance between donor atoms in the same chelate ligand, is substantially
smaller than /, the distance between donor atoms on neighboring chelate ligands (see
Fig. 13.7). 33 On the other hand, for a rhombic twist to be favored, b must be much
greater than /. It would appear that both mechanisms are rather common because
many complexes belong to each geometrical class. In cases where b and / are not
significantly different, both twist mechanisms may operate simultaneously. Intra¬
molecular isomerization of cis and trans octahedral complexes, M(CO) 4 LL\ is also
well established for some complexes and probably proceeds through a trigonal twist
mechanism. 34
29 Rodger. A.: Johnson. B. F. G. Inorg. Client. 1988, 27. 3061-3062.
30 RSy. P. C.; Dull. N. K. J. Indian Chem. Soc. 1943. 20. 81.
31 Bailar, J. C., Jr. J. Inorg. Nuci. Client. 1958. 8. 165-175.
32 Kepcrt. D. L. Prog. Inorg. Chem. 1977. 23. 1-65.
33 Rodger. A.; Johnson. B. F. G. Inorg. Chem. 1988. 27. 3061-3062.
34 Darensbourg. D. J.; Gray. R. L. Inorg. Chem. 1984. 23. 2993-2996.
Mechanisms of Redox Reactions
Fig. 13.7 Four intramolecular mechanisms for racemization of a tris(chelate) octahedral
complex (R = reactant; T = iransition stale; P = product). Only the Bailar and R&y-Dutt
twists are energetically acceptable. [Modified from Rodger, A.; Johnson, B. F. G. Inorg.
Chem. 1988, 27. 3061-3062. Used with permission.)
Mechanisms of
Redox Reactions 35
It might be assumed that there would be little to study in the mechanism of electron
transfer—that the reducing agent and the oxidizing agent would simply bump into
each other and electron transfer would take place. Reactions in solution are compli¬
cated. however, by the fact that the oxidized and reduced species are often metal ions
surrounded by shields of ligands and solvating molecules. Electron transfer reactions
35 Taubc, H. Electron Transfer Reactions of Complex Ions in Solution ; Academic: New York. 1970;
Chapter 2. Haim A. Acc. Chem. Res. 1975 . 8. 264-272. Pennington. D. E. In Coordination
Chemistry-. Martcll. A. E., Ed.; ACS Monograph 174; American Chemical Society: Washington,
DC, 1978; Vol. 2, pp 476-590. Taube. H. Science 1984. 226. 1028-1036 (Professor Taube's Nobel
Prize address). “An Appreciation of Henry Taube," Prog. Inorg. Chem. 1983 , 30. Rudoiph A.
Marcus Commemorative Issue, J. Phys. Chem. 1986, 90, 3453-3862.
484
12 Coordination Chemistry: Structure
Fig. 12.12 A series of
trigonal bipyramidal
complexes which ailow a
ir-acceptor series to be
arranged by noting
equatorial vs. axial site
preference. [From Rossi.
A. R.; Hoffmann. R. Inorg.
Chem. 1975, 14, 365-374.
Reproduced with
permission.)
If we assume that the best ir acceptors will always choose an equatorial position, we
can arrange them in the following order: 22
NO + > CO > CN~ > SnCl^> CP > PR 3 > C 2 H 4 > CH^
This series may be compared with that given in Chapter 11 derived from completely
different assumptions. The general concurrence is reassuring.
Site Preference in
Square Pyramidal
Complexes
In SP geometry the central atom may be in the plane of the basal ligands or above it to
varying degrees. The following discussion assumes that the metal atom is lying
somewhat above the basal plane, as is commonly found. Under these conditions the
“normal” situation ( d°-d 6 , and d'°) is for the apical bond to be the strongest with
weaker basal bonds. As in the TBP case, the d s configuration is reversed with stronger
basal bonds and a weak apical bond. Likewise, good donors usually (d°-d b , d'°) seek
the apical position, but in t/ 8 complexes electronegative ligands should prefer the
apical position. The bond lengths shown in Table 12.3 generally support these conclu¬
sions although, as before, there are some puzzling exceptions.
If the five-coordinate complex is a result of the addition of a fifth, weakly bound
ligand to a strongly ir bonded, square planar complex:
[Pd(diars) 2 ] 2+ + X" -> [Pd(diars) 2 Xr (12.3)
Toblo 12.3 _
Bond lengths in some SP
complexes"
Complex
M
~ L (pH
Apical
Basal
d "
[Nb(NMe 2 ) 5 ]"
198
204
d i
[MnClj] 2-
258
230
d
[Ni(CN) 3 ] 3 ~
217
186
d
[InCl 5 ] 2-
242
246
d >°
Sb(C 6 H 3 ) 5
212
222
d “>
" Rossi, A. R.; Hoffmann. R. Inorg. Chem.
1975, 14, 365-374.
22 Since CN - and SnCIJ do not occur in the same complex in this series, the inequality shown for
them is uncertain.
Coordination Number 5
485
Magnetic and
Spectroscopic
Properties
then the tr-bonding requirements of the former ligands [diars = o-phenylenebis-
(dimethylarsine)] require that they remain coplanar or nearly so with the metal atom;
hence the SP arrangement is strongly favored.
Five-coordinate complexes with d 5 , d b , d 1 , and d s may be either high or low spin. The
magnetic susceptibility of the low spin complexes is that expected if one of the d
orbitals is unavailable for occupancy by the metal d electrons. Thus S equals 0 {d s ),
(d 7 ), 1 (d b ), and | (d 5 ). The magnetic susceptibilities of the low spin five-coordinate
complexes thus differ significantly from corresponding low spin octahedral com¬
plexes. The unavailability of the fifth d orbital can be rationalized in terms of dsp 3
hybrid bonding. The TBP structure results from djsp } hybridization and SP from
d x 2 _ y ^sp i hybridization. 23 In this sense valence bond theory is in qualitative agreement
with simple crystal field theory or more elaborate molecular orbital schemes. The
latter two methods, however, also account for the energy levels of the other d orbitals
as well as assign the difference between high and low spin complexes to the relative
energies of the </.:and dp_ y i orbitals (see Fig. 12.8). The placement of these energy
levels can be reasonably interpreted by means of spectral measurements. In some low
spin complexes the interelectron repulsions can be neglected to a first approximation
and the spectrum interpreted solely on the basis of the simple one-electron energy
level diagrams (Fig. 12.13). In the more general case, however, these elec¬
tron-electron effects must be treated in a manner analogous to that given in Chapter 11
for octahedral complexes.
'■=fi=«= =t#= =ff#=
. . I || ,, || ,, I
' =fP= = fl = = =fP=
Fig. 12.13 Spectra of (a) bromotris(methylmercapto-o-phenyl)phosphinenickel(II) and (b) bromotris(dimethylarsino-o-
phenyl)arsinenickel(ll) cations. [From Furlani, C. Coord. Chem. Rev. 1968, 3, 141. Reproduced with permission.)
23 These arc the simplest hybridizations to visualize. It is possible to substitute more d character for p
character and arrive at the same symmetry. For example. d l sp also forms TBP. and ds and d p
also form SP. The actual percentages of s, p, and d character will depend upon energetic factors
such as promotion energy and quality of overlap of the resulting hybrids.
1 3 • Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Kinetics of Octahedral Substitution
When this reaction was carried out in 0.10 M hydroxide and 1,0 M azide, both the
hydroxo (90.3%) and azido (9.7%) complexes were formed. Keep in mind that without
base, azide substitution would not be observed within the experimental time period.
Furthermore, the rate of hydrolysis is not much different when azide is replaced with
perchlorate or acetate. It becomes clear that these anions are spectators of the five-
coordinate activated complex and are not involved in the loss of X~.
The preceding discussion of substitution mechanisms barely scratches the surface
of a field that has occupied the attention of many of the world’s best coordination
chemists. It is an area which seems to have an infinity of problems as well as methods
of attack. It is unfortunate that it is not possible here to present a more comprehensive
theory of substitution mechanisms. The discussion presented errs on the side of
omission of fine points and controversial interpretations. For every experiment de¬
signed to confirm a mechanism, an alternative explanation can usually be found. As
one noted researcher once said: “[The members of the other school of thought] are
extremely ingenious at coming up with alternative explanations for all of the con¬
clusive experiments that we seem to do.” 28 This should serve to remind us of the
truism that it is not possible to prove that a particular mechanism is the correct one; it
is only possible sometimes to prove that an alternative mechanism is not correct. To
this might be added a corollary: Often it is extremely difficult to prove that the
alternative is impossible.
ammonia, which possesses no free lone pairs when it is bound to a metal, is the leaving
group, no acceleration is observed. 2
Hydroxide ion also may have an appreciable effect on the rate of hydrolysis of
octahedral complexes. The rate constant for hydrolysis of [Co(NH 3 ) 5 CI] 2+ in basic
solution is a million times that found for acidic solutions. Furthermore, the reaction is
found to be second order and dependent on the hydroxide ion concentration:
rate = [Co(NH 3 ) 5 C1 2+ ][OH - ] (13.38)
Although an associative mechanism is consistent with these results, the prevailing
opinion is that the reaction takes place via proton abstraction:
[Co(NH 3 ) 5 CI] 2+ + OH" [Co(NH 3 ) 4 (NH 2 )CI] + + H ; 0
[Co(NH 3 ) 4 (NH 2 )Cl] + - [Co(NH 3 ) 4 (NH 2 )] 2+ + cr
[Co(NH 3 ) 4 (NH 2 )] 2 + + H 2 0 [Co(NH 3 ) 5 (OH)] 2+
According to this viewpoint, the hydroxide ion rapidly sets up an equilibrium with the
amidocobalt complex. The rate-determining step is the dissociation of this complex
(Eq. 13.40), but since its concentration depends on the hydroxide ion concentra¬
tion through equilibrium, the reaction rate is proportional to the hydroxide ion
concentration.
This mechanism, assigned the symbolism S N \ CB for a first-order reaction involv¬
ing the conjugate base of the complex, is supported by a number of observations. It
rationalizes the fact that the hydroxide ion is unique in its millionfold increase in rate
over acid hydrolysis; other anions which are incapable of abstracting protons from the
complex, but which would otherwise be expected to be good nucleophiles, do not
show this increase. Furthermore, the mechanism can apply only to complexes
in which one or more ligands have ionizable hydrogen atoms. Thus complexes such as
[Co(py) 4 Cl 2 ] + and [Co(CN) 5 CI] 3- would not be expected to exhibit typical base
hydrolysis and indeed they do not. In these cases, the hydrolysis proceeds slowly and
without dependence on hydroxide ion.
If the hydroxide ion accelerates reactions by proton abstraction rather than by
direct attack, it might be supposed that it would be possible to trap the five-coordinate
intermediate by addition of large amounts of anion other than hydroxide. One system
for which this is possible is the base hydrolysis of [CofNH-jRKX] 2 f (X = OSO-CF 3 )
with NJ as the trapping agent: 27
Racemization and
Isomerization
Another set of reactions that has received considerable attention is that in which
optically active complexes, especially tris(chelale) compounds racemize:
There are several mechanisms that are possible for such an inversion, some of which
can be eliminated by appropriate experiments. For example, one mechanism would
involve complete dissociation of one chelating ligand with formation of a square planar
or a trans diaqua complex as a first step:
[Co(NH,R) 5 X] 2t + OH-
[Co(NH 2 R) 4 (NHR)Xr + H,0
[Co(NH 2 R) 4 (NHR)X]-
[Co(NH 2 R) 4 (NHR)P + X
[Co(NH 2 R) 5 OH] 2 '
[Co(NH 2 R) 5 N 3 ] :
26 Wilkins. R. G. Kinetics and Mechanisms of Reactions of Transition Metal Complexes, 2nd ed.
VCH: New York. 1991.
27 Curtis. N. J.; Lawrence, G. A.; Lay. P. A.; Sargeson, A. M. Inorg. Chem. 1986, 25. 484-488.
28 Pearson. R. G. In Mechanisms of Inorganic Reactions: Klcinberg, J.. Ed.; Advances in Chemistry
49; American Chemical Society: Washington, DC, 1965; p 25.
486 12 • Coordination Chemistry: Structure
Isomerism in Although it has long been known that various geometric and optical isomers are
Five-Coordinate possible for coordination number 5, examples have been few. We have already seen
Complexes examples of TBP-SP isomerism in an Ni(II) complex (page 480) and a Co(II) complex
(page 481). Another example is [(C 6 H 5 ) 3 P] 2 Ru(CO)[(CF 3 ) 2 C 2 S 2 J. It forms two iso¬
mers, one orange and one violet, which coexist in solution. Recrystallization from
most solvents (e.g., acetonitrile) yields only the orange isomer, but dichloromethane/
hexane as solvent yields a mixture of orange crystals and violet crystals. Both isomers
are square pyramidal, but the orange isomer has the carbon monoxide ligand in the
apical position, and the violet isomer has a basal carbon monoxide with one of the
phosphine ligands at the apex of the pyramid (Fig. 12.14). 24 A related type of geo¬
metric isomerism is found in the organometailic complex dibromodicarbonylcy-
clopentadienylrhenium(lll). Both isomers have the cyclopentadienyl ring at the apex
of a square pyramid with the basal ligands in either a cis or a trans arrangement: 25
Finally, optical isomerism is even more rare. The first example of the determination of
the absolute configuration of such a chiral complex is shown in Fig. 12.15. 26 This
..Re.
oc / v'-x
X CO
.Re.
X / v ' co
X CO
Fig. 12.14 Stereoviews of
the inner coordination
spheres about the central
ruthenium atom in the
orange (top) and violet
(bottom) isomers of
1(C 6 H,),P) 2 Ru((CFj) : C 2 S 2 |-
(CO). [From Bernal. I.;
Clearfield, A.; Ricci. J. S.,
Jr. J. Cryst. Mol. Struct.
1974. 4. 43-54. Reproduced
with permission.)
2J Bernal, I.; Clearfield. A.; Ricci. J. S., Jr. J. Cryst. Mol. Struct. 1974. 4. 43-54. Clearfield. A.;
Epstein, E. F.; Bernal. I. J. Coord. Client. 1977. 6, 227-240.
25 King. R. B.: Reimann, R. H. Inorg. Client. 1976. 15. 179-183,
26 LaPlaca. S. J.; Bernal, I.; Brunner, H.; Herrmann, W. A. Angew. Chem. Int. Ed. Engl. 1975. 14,
353-354. Bernal. I.; LaPlaca, S. J.: Korp, J.; Brunner, H.: Herrmann, W. A. Inorg. Chem. 1978, 17,
382-388. See also Brunner, H. Adv. Organomet. Chem. 1980. 18. 151-206.
Coordination Number 5 487
Fig. 12.15 A Schiffbase complex of dicarbonylcyclopentadienylmolybdcnum(II) cation.
(From Bernal, I.; LaPlaca, S. J.; Korp, J.; Brunner, H.; Herrmann, W. A. Inorg. Chem.
1978, 17, 382-388. Reproduced with permission.]
isomer forms along with a diastereomer in the reaction of tricarbonyI(cyclopenta-
dienyl)chloromolybdenum with an asymmetric Schiff base:
A B
Note that these compounds are not enantiomers, but true diastereomers with different
properties, and they may be separated by fractional crystallization. The asymmetric
carbon atom has an S configuration in both diastereomers, but the chirality about the
molybdenum atom is different. Thus the asymmetric carbon aids in the resolution of
the molybdenum center, but its presence is not necessary for the complex to be chiral.
It is merely necessary for the Schiffbase to be unsymmetric, i.e., have one pyridine
nitrogen and one imino nitrogen. If the bidentate ligand had been ethylenediamine,
bipyridine, or the oxalate ion, there would have been a mirror plane and no chirality at
the molybdenum.
5$2
13-Coordination Chemistry: Reactions, Kinetics, and Mechanisms
Fig. 13.6 Dependence of
k 0 bs on concentration of
entering group (SCN") in
H 2 0 substitution involving
a Co(IH) hematoporphyrin
complex. [From Fleischer,
E. B.; Jacobs, S.;
Mestichelli, L. J. Am.
Chem. Soc. 1968. 90.
2527-2531. Used with
permission.!
Table 13.4
Rato constants for the
roaction of
[Co(NH 3 ) 5 (H 2 0)] 3 with x-
in watur at 45
°C“
x «-
k (M- 1 "’)
NCS"
1.3 x 1(T 6
h 2 po 4 -
2.0 x I0"‘
cr
2.1 x I0 -6
no 3 _
2.3 x I0" 6
SOj"
1.5 x IO" 3
" Basolo, F.;
Pearson,
R. G. Mechanisms of
Inorganic Reactions, 2nd
ed.; Wiley: New York,
Table 13.5
Rate constants for the
reaction of [Co(NH 3 ) 5 X] m+
with H 20 a
X"~
k (s" 1 )
NCS"
5.0 x IO" 10
h,po;
2.6 x IO" 7
cr
1.7 x IO" 6
no 3 -
1
©
X
<N
SO 2 '
1.2 x I0" 6
« Basolo, F. -
Pearson,
R. G. Mechanisms of
Inorganic Reactions. 2nd
ed.; Wiley: New York.
This expression simplifies to Eq. 13.30 with k obs - kK\J[ under pseudo first-order
conditions. Furthermore, the form of the rate law does not change if bond making
becomes more important than bond breaking (/„). Since rate laws for D. I d . and l a
cannot be distinguished with certainty (knowledge of rate and equilibrium constants
for individual reaction steps may provide clarification), it is not surprising that there
have been considerable debate and controversy over the mechanistic details of many
octahedral substitution reactions. Few reactions appear to fit into the limiting D and A
categories; thus most discussion centers around the I d and /„ mechanisms.
Because of the inertness of Co(III) and Cr(III) complexes, their substitution
reactions were the first among those of octahedral complexes to be extensively
studied. Most evidence supports the I d mechanism for substitution in Co(l11) com¬
plexes. First, there is little dependence of reaction rates on the nature of the incoming
ligand. If bond making were of significant importance, the opposite would be ex¬
pected. Data are presented in Table 13.4 for the anation reaction of penta-
ammineaquacobalt(III):
[Co(NH 3 ) 5 (H 2 0)] 3+ + X"- -♦ [Co(NH 3 ) 5 xr + + H 2 0 (13.35)
We see only a small variation in rate constants for a variety of anionic X' 1 ligands. It
is instructive to also consider the reverse reaction of Eq. 13.35, aquation ot the Cot III)
complex. If this is an / (/ reaction, M—X bond strength should correlate with reaction
rate since most of the activation energy would be associated with bond breaking.
Table 13.5 satisfies our expectation that the reaction rate depends on the kind of
M—X bond being broken. The entering group and leaving group data provide con¬
vincing evidence for a dissociative mechanism and this view is further supported by
steric arguments. The reaction in which water replaces Cl in [Co(NMeH 2 ) 5 CI)“
takes place 22 times faster than the same reaction for [Co(NH 3 ) 5 CI] . ,8 The greater
steric requirements of methylamine encourage dissociation of the Cl ligand. If the
reaction proceeded by an /„ or A pathway, the order of rates would be the opposite
because increased steric repulsion of the incoming ligand would be expected to slow
the reaction. Finally, it should be noted that the absence of a trans effect (so important
in square planar substitution) for Co(III) complexes is consistent with a dissociative
mechanism.
There is growing evidence that substitution reactions in Co(III) complexes may
not be typical of octahedral transition metal complexes. Early studies of substitution
reactions for Cr(IlI) complexes revealed a rather strong dependence of reaction rate
is Buckingham, D. A.; Foxman. B. M.: Sargeson, A. M. Inorg. Chem. 1970, 9, 1790-1795.
Kinetics of Octahedral Substitution
553
Table 13.6 _
Volumes of activation for
solvent exchange in
[M(NH 3 ) 5 (OH 2 )] 3+ °
M AY, cm 3 mol 1
5 +1.2 (300 K)
Cr 3+ -5.8 (298 K)
Rh ,+ -4.1 (308 K)
Ir 3 -3.2 (344 K)
a Ducommun, Y.;
Merbach. A. E. In
Inorganic High Pressure
Chemistry ; van Eldik, R.,
Ed.; Elsevier: Amsterdam,
1986.
Reaction Rates
Influenced by Acid
and Base
Tab l e 13.7
Volumes of activation for
water exchange in
hexaaqua complexes of
transition metal ions of the
first row”
M AY, cm 3 mol -1
Ti 3+ -12.1
V 3+ -8.9
Cr 3 -9.6
Fe 3 -5.4
“ Hugi, A. D.; Helm, L.;
Merbach, A. E. Inorg.
Chem. 1987. 26. 1763-1768.
Xu, F.-C.; Krousc, H. R.;
Swaddle, T. W. Inorg.
Chem. 1985, 24. 267-270.
Swaddle, T. W.; Merbach.
A. E. Inorg. Chem. 1981,
20. 4212-4216.
on the nature of the entering group, 19 which supported the /„ mechanism. More
recently, high pressure oxygen-17 NMR spectroscopy has come into widespread use
for obtaining mechanistic details about fast reactions and, as a result, many water-
stable transition metal complex ions have been investigated. The parameter of interest
that is yielded by these experiments is volume of activation. AV, which is a measure
of the change in compressibility that occurs as the reaction proceeds from the ground
state to the transition state (see page 542). The data in Table 13.6, obtained for solvent
exchange with [M(NH 3 ) 5 (H,0)] 3+ complexes, show a positive AV for Co + but
negative values for Cr 3 , Rh 3+ , and Ir 3 ^, suggesting an I d mechanism for the cobalt
ion but /„ for chromium, rhodium, and iridium ions. 20 Data for water exchange
reactions of first-row hexaaqua tripositive ions are shown in Table 13.7. We see a
general increase in volumes of activation as we move across the periodic table Irom
Ti(III) to Fe(III). In fact the value for Ti(III) approaches that predicted lor an A
mechanism. 21 The trend may be viewed as a gradual change from strongly associative
to moderately associative. Similar NMR studies ot solvent exchange reactions also
have been carried out for divalent transition metal ions, [M(H 2 0) 6 ] \ In these
experiments, volumes of activation indicate a change from / u to l d across the first row;
i.e., the dissociative mechanism is more important for Ni(II) than for Fe(II). 22 Since
volumes of activation also include volume changes in solvents and reactants, inter¬
pretation is not always straightforward, and some believe that the power of the
method has been overstated. For example, Langford 23 and Swaddle 24 have presented
opposing views on this matter. Undoubtedly, the /„ pathway is more common for
octahedral substitution than once thought.
Substitution reactions taking place in water solution can often be accelerated bv the
presence of an acid or base. If the coordinated leaving group (X) has lone pairs which
can interact with H + or metal ions such as Ag‘ or Hg 21 , the M—X bond may
be weakened and loss of X facilitated. 23 This effect is seen in the aquation of
[Cr(H 2 0) 5 F] 2+ :
[Cr(H 2 0) 5 F] 2+ + H + ;= [Cr(H 2 0)j~F-H] 3+ (>3-36)
[Cr(H 2 0) 5 —F—H] 3+ + H 2 0 -♦ [Cr(H 2 0)„] 3+ + HF (,3l37)
Available lone pairs of the bound fluorine group are attracted to the hydrogen ion,
leading to the formation of a weak acid. The rate constant for the overall reaction is
6.2 x 10 -I ° s -1 in neutral solution, but 1.4 x 10 s s 1 in acid solution. When
19 Espcnson, J. H. Inorg. Chem. 1969,8, 1554-1556.
M Docommun. Y.; Merbach. A. E. In Inorganic High Pressure Chemistry; van E ‘ dlk> E ' ;
Elsevier: Amsterdam. 1986. van Eldik. R.; Asano. T.; Le Noble, W. J. Chem. Rev. 1989, 89.
549-688.
31 Hugi. A. D.; Helm. L.; Merbach. A. E. Inorg. Chem. 1987, 26. 1763-1768.
33 Lincoln. S. F.; Hounslow. A. M.; BolTa, A. N. Inorg. Chem. 1986. 25. 1038-1041.
33 Langford, C. H. Inorg. Chem. 1979. 18. 3288-3289.
34 Swaddle. T. W. Inorg. Chem. 1980. 19. 3203-3205. ^
33 Hard ligands such as F are most effectively removed by hard metal ions such as Be' and Al ,
but soft ligands such as Cl", Br", and 1" are better removed by soft metal ions such as Ag and
Hg 2+ -
488
12 ■ Coordination Chemistry: Structure
Another interesting example that combines both geometric isomerism and chi¬
rality consists of complexes of the type: 27
V-.I '' . - MO-, \
C^- \ • O
RPhjP^'.' CO
cis (chiral)
irons (achiral)
Note that the cis isomer lacks an improper axis of rotation and is therefore chiral, but
that the trans isomer has a plane of symmetry and will be achiral in the absence of an
asymmetric carbon in the phosphine ligand. 28 As in the case of the previously encoun¬
tered cyclopentadienyl complex (page 476), it can be argued whether the coordination
number is 5 or 9. In either semantic interpretation these compounds are of consider¬
able interest since isomerism in nine-coordinate complexes is even less well docu¬
mented than in those with coordination number 5.
Coordination This is by far the most common coordination number. With certain ions six-coordinate
— - complexes are predominant. For example. chromium(III) and cobalt(III) are almost
N umb er 6 29 exclusively octahedral in their complexes. 20 It was this large series of octahedral
Cr(lII) and Co(III) complexes which led Werner to formulate his theories of coordina¬
tion chemistry and which, with square planar platinum(Il) complexes, formed the
basis for almost all of the classic work on complex compounds. Before discussing the
various isomeric possibilities for octahedral complexes, it is convenient to dispose of
the few nonoctahedral geometries.
Distortions from
Perfect Octahedral
Symmetry
Two forms of distortion of octahedral complexes are of some importance. The first is
tetragonal distortion, either elongation or compression along one of the fourfold
rotational axes of the octahedron (Fig. 12.16a). This type of distortion has been
discussed previously in connection with the Jahn-Teller effect. Another possibility is
elongation or compression along one of the four threefold rotational axes of the
octahedron that pass through the centers of the faces (Fig. 12.16b), resulting in a
trigonal antiprism. Another configuration that is not really a distortion but involves a
reduction of symmetry may be mentioned here. It consists of the replacement of six
unidentate ligands in a complex such as [Co(NH 3 ) 6 ] 3+ with chelate rings such as
n Reisner. G. M.; Bernal. I.; Brunner. H.; Muschiol. M.; Siebrecht. B. Chem. Commun. 1978,
691-692.
28 In fact, chiral ligands are often used in synthesizing molecules of this type, both for observing stcric
effects and for aid in solving the X-ray crystal structures. See the discussion of absolute configura¬
tion of the previous complex with C.N. = 5 and octahedral complexes with C.N. = 6 (page 495).
29 Kepert. D. L. Prog, tnorg. Chem. 1977, 23. 1-65. Kepert. D. L. In Comprehensive Coorilinalion
Chemistry-. Wilkinson. G.;GilIard, R. D.; McCleverty, J. A., Eds.; Pcrgamon: Oxford. 1987; Vol. I,
pp 49-68.
30 Co(III) is tetrahedral in garnets, occasionally square planar (page 477), or five-coordinate (TBP or
SP). Likewise, CrfHI) is rarely trigonal, tetrahedral, or TBP.
Coordination Number 6
489
Trigonal Prism
(a) Tetragonal distortion (b) Trigonal distortion
Fig. 12.16 (a) Tetragonal and (b) trigonal distortion of an octahedral complex. Either may
occur via elongation or compression.
ethylenediamine to form [Co(en) 3 ] 3+ . The latter complex has no mirror plane and the
symmetry of the complex has been reduced from O h to £> 3 . For most purposes this
reduction in symmetry has little effect (the visible spectrum of the ethylenediamine
complex is very similar to that of the hexaammine complex) except to make it possible
to resolve optically active isomers (see page 492).
Although by far the greatest number of six-coordinate complexes may be derived from
the octahedron, a few interesting complexes have the geometry of the trigonal prism.
For many years the only examples of trigonal prismatic coordination were in crystal
lattices such as the sulfides of heavy metals (MoS, and WS 2 , for example). 31 The first
example of this geometry in a discrete molecular complex was tris(m-l,2-di-
phcnylethene-I,2-dithiolato)rhenium. Re[S 2 C 2 (C 6 H 5 ) 2 ] 3 (Fig. 12.17). 32 Following that,
a significant series of trigonal prismatic complexes of ligands of the type R 2 C 2 S 2 was
fully characterized with rhenium, molybdenum, tungsten, vanadium, zirconium, and
niobium, and suggested for other metals. 33
There is considerable ambiguity concerning the charge on this type of ligand. This
is because it may be formulated either as a neutral dithioketone or the dianion of an
unsaturated dithiol. The difference of two electrons can be represented formally as:
S W S
’ S \ / S ~
„X D
R R
HS SH
,C=C
Since the electrons involved are delocalized over molecular orbitals not only of the
ligand but of the metal as well, it is impossible to assign a formal charge to the ligand or
the metal. Nevertheless, the actual distribution of electron density may be quite
important in determining how a reaction proceeds, as suggested for molybdenum
enzymes (see also Chapter 19). 34
31 Wells. A. F. Structural Inorganic Chemistry. 5th ed.; Oxford University: London, 1984; p 757.
32 Eisenberg, R.; lbers. J. A. J. Am. Chem. Soc. 1965.57. 3776-3778; Inorg. Chem. 1966,5.411-416.
33 Bennett. M. J.; Cowie, M.; Martin, J. L.; Takats, J. J. Am. Chem. Soc. 1973. 95, 7504-7505.
34 Barnard. K.'R.; Wedd. A. G.; Tiekink. E. R. T. Inorg. Chem. 1990. 29, 891-892.
13 Coordination Chemistry: Reactions, Kinetics, and Mechani
As mentioned above, the k t term of the rate law shown in Eq. 13.10 could also
arise from dissociation ( D ) of X to give a three-coordinate complex which then reacts
with Y.
ML 2 TX — ML,T ML 2 TY (,3.1 5)
In other words, the form of the rate law does not help one distinguish between an A (or
4) and D (or I d ) mechanism for the k, pathway. The ambiguity in the interpretation of
the k, term has caused much discussion and experimentation. It is found that reactions
take place faster in more nucleophilic solvents, suggesting that solvent attack plays an
important role.
Also, dissociative reactions should be accelerated by the presence of sterically
demanding ligands; just the opposite is observed, in keeping with an A or / mech¬
anism. "
Further insight into the question of an associative versus a dissociative mecha¬
nism can be provided by thermodynamic data such as that shown in Table 13.1 for the
substitution of bromide by iodide or thiourea in fra/ij-[Pt(PEt 3 ) 2 (R)Br]:
R—Pt —Br+ X
R— Pt — X + Br-
X = r or SC(NH 2 ) 2 ; R = 2,4,6-Me 3 C 6 H 2
The reaction rate is primarily determined by the enthalpy of activation (A//), which is
usually the case in square planar nucleophilic substitution reactions. Of greater
importance, so far as a dissociative versus an associative mechanism is concerned, are
the entropies and volumes of activation, AS and AY, respectively. Note that the
values are negative for both the k, and the k 2 steps. The observed decrease in entropy
is what we would expect for a mechanism in which two particles come together to give
an activated complex. The volume of activation is determined by doing the reaction
under high pressure;
_ RT\n (,/,)
" {P> _ p 2) (13.17)
An activated complex with a smaller volume than the reacting species will give rise to
a negative AY and is characteristic of association (see page 553 for further discus-
Toble 13.1
Activation parameters (or
the reaction of trans-
[Pt(PEt3) 2 (2,4,6-Me 3 C 4 Hj)Br]
with I and with SC(NH 2 ) 2
in methanol 0
A//, Id mof
AS. J K.-' mof
AY. cm 1 mol -1
SC(NH 2 ) 2
k] k 7
-115 -80
-17 -II
" van Eld 'k. R : : Palmer. D. A.; Kelm, H. Inorg. Chem. 1979,
Substitution Reactions in Square Planar Complexes
543
sion). 10 For all of the above reasons it is believed that square planar nucleophilic
substitution reactions proceed by association rather than by dissociation. 1 '
Many experiments have been carried out to gain a clearer understanding of the
details of the associative mechanism. There are two key questions to be answered;
What effect does the nature of the entering group have on the rate of reaction and how
does this effect alter our view of the intimate mechanism? The same questions have
been asked with regard to the leaving group, the ligand trans to the leaving group, the
ligand cis to the leaving group, and the nature of the central metal itself. In all
observed reactions, the entering group occupies the site vacated by the leaving group,
and any reasonable mechanism must account for this experimental fact.
The Trans Effect None of the above factors has been studied more exhaustively than the effect of the
ligand trans to the leaving group. By varying the nature of this ligand, it is possible to
cause rate changes of many orders of magnitude. Furthermore, the effect can be used
to advantage in designing syntheses.
The presence of large deposits of platinum ores in Russia led to an intensive study
of the coordination compounds of platinum early in the development of coordination
chemistry. As a result of these studies by the Russian school, the first stereospecific
displacement reaction (and first example of the trans effect) was discovered. Consider
two means of forming diamminedichloroplatinum(II): (1) displacement of CF ions
from [PtCIJ 2- by NH,; (2) displacement of NH 3 from [Pt(NH 3 ) 4 ] 2+ by Cl - ions. It is
found that two different isomers are formed:
H 3 N Cl
X
Cl NH,
irrms-diamminedichloro-
platinum(II) (Nol found
in ihis reaction) (jq jg)
C, X NH ’
Cl NH,
ri.v-diamminedichloro-
pluiinum(ll)
(Exclusive product)
10 Negalive values for A5 and AY do not prove that a reaction is associative. Solvent reorganization
can lead to unexpected entropy changes and contribute to overall volume changes. However, large
negative values, such as those in Table 13.1. are generally accepted as indicating an associative
mechanism. See Inorganic High Pressure Chemistry, Kinetics and Mechanisms', van Eldik. R.,
Ed.; Elsevier: Amsterdam. 1986; van Eldik, R.; Asano, T.; LcNoble. W. J. Chem. Rev. 1989. 89,
549-688.
11 Evidence for a dissociative mechanism has been reported: Lanza, S.; Minnili, D.; Moore. P.;
Sachinidis. J.; Romeo, R.: Tobc, M. L. Inorg. Chem. 1984 , 23, 4428—1433. The reaction, which
involves substitution of dmso in PtR,(dmso) 2 , proceeds by loss of one dmso ligand. However,
because it is possible that the dmso ligand that remains coordinated can function in a chelating
capacity, it can be debated, as these workers acknowledge, whether this reaction can be called a
true dissociation.
490
12 Coordination Chemistry: Structure
In addition to the neutral complexes, it is possible to add one, two, or three
electrons to form reduced species of the type [M(S 2 C 2 R 2 ) 3 ]”~. Present evidence is that
the reduced species tend to retain the trigonal prismatic coordination with some
distortion towards a regular octahedron with increasing addition of electrons (see
Table 12.4).
One of the most interesting features of the 1,2-ethenedithiolate or 1,2-dithiolene
complexes is the short distance between the two sulfur atoms within a chelate ring.
This distance is remarkably constant at about 305 pm, some 60 pm less than the sum of
the van der Waais radii (Table 8.1), indicating the strong possibility of some S—S
bonding that may stabilize the trigonal structure. 35 One way in which this might come
about is by pulling the sulfur atoms towards each other, reducing the bite angle (it is
about 81° in the complex in Fig. 12.17). Perfect octahedral coordination requires 90°.
Another way to look at it is to imagine the molecule in Fig. 12.17 undergoing a 60°
twist of one of the S 3 triangles to form an octahedron. If the other dimensions remain
the same, the sulfur atoms would have to move away from each other, and this would
be inhibited if there is any S—S bonding.
There are few trigonal prismatic complexes with unidentate ligands, but both
[Zr(CH 3 ) 6 ] 2- and W(CH 3 ) 6 have £> 3/ , symmetry, as shown in the margin. 36 The factors
favoring O h symmetry and inhibiting trigonal distortion are not difficult to see: steric
effects of bulky ligands, large partial charges on the six ligands, and relatively small
metal size. To be sure, these are absent in these hexamethyl compounds and in the
hypothetical CrH fi studied by ab initio calculations. The factors favoring D 3/ , symme¬
try are more subtle: for example, the relative stabilization of formally nonbonding
orbitals similar to that in the MO analysis of BeH 2 and H 2 0 in Chapter 5. 37
Map of twist angles ( 0 ) in tris(dithiolato)metal complexes, ML 3
Group IVB (4) Group VB (5) Group VIB (6)
_ VL 3 ( 6 = 0°)
ZrL 3 - (0 = 19.6°) NbLj (6 = 0.6°) MoL,(0 = 0°)
TaL," (0 ~ 16°)
Group VIIB (7) Group VIIIB (8)
" Trigonal prism, 6 = 0°; octahedron. 0 = 60°
55 For reviews of complexes conlaining these ligands, sec references in Footnote 29.
36 Morse. P. M.; Girolami. G. S. J. Am. Client. Soc. 1988. ///. 4114-4116. Haaland. A.; Hammel. A.;
Rypdal. K.; Volden, H. V. J. Am. Chem. Soc. 1990, 112. 4547-4549.
37 Kang. S. K.; Albright. T. A.: Eiscnstein. O. Inorg. Chem. 1989, 28. 1611-1613.
Coordination Number 6
491
Geometric
Isomerism in
Octahedral
Complexes 38
Fig. 12.17 Structure of
[Re(S 2 C 2 Ph 2 ) 3 ], The phenyl
rings have been omitted for
clarity. [From Eisenberg,
R.; Ibers, J. A. Inorg.
Chem. 1966,5, 411-446.
Reproduced with
permission.)
There are two simple types of geometric isomerism possible for octahedral complexes.
The first exists for complexes of the type M A 2 B 4 in which the A ligands may be either
next to each other (Fig. 12.18a) or on opposite apexes of the octahedron (Fig. 12.18b).
Complexes of this type were studied by Werner, who showed that the praseo and
violeo complexes of tetraamminedichlorocobalt(III) were of this type (see Chapter 11).
A very large number of these complexes is known, and classically they provided a
fertile area for the study of structural effects. More recently there has been renewed
interest in them as indicators of the effects of lowered symmetry on electronic
transition spectra.
Two geometric isomers are also possible for complexes of the type MA 3 B 3 : (1)
The ligands of one type may form an equilateral triangle on one of the faces (the facial
Fig. 12.18 Examples of cis and trans octahedral isomers: (a) cis-
tetraamminedichlorocobalt(III). Werner's violeo complex, and (b) trans-
tetraamminedichlorocobalt(III), Werner's praseo complex.
_ I
38 For a full discussion of geometric isomerism, see Saito, Y. Inorganic Molecular Dissymmetry,
Springer-Verlag: Berlin, 1979; pp 6-7, 73-88.
540
13-Coordination Chemistry: Reactions, Kinetics, and Mechanisms
The Rate Law
for Nucleophilic
Substitution in a
Square Planar
Complex
L L
in which Y is the entering nucleophilic ligand, X is the leaving ligand, and T is the
ligand trans to X. Kineticists try to simplify their experiments as much as possible,
and one way to do that in this case is to run the reaction under pseudo first-order
conditions. 9 Practically, this means that the concentration of Y is made large com¬
pared to that of the starting complex so that changes in [Y] will be insignificant during
the course of the reaction ([Y] = constant). For reactions in which reverse processes
are insignificant, the observed pseudo first-order rate law for square planar substitu¬
tion is:
rate = - d[ML 2 TX]/<// = A,[ML 2 TX] + A 2 [ML 2 TX][Y] (13.10)
This expression may be rearranged to give:
rate = (A, + A 2 [Y])[ML 2 TX] = A^JMLVTX] (13.11)
and
obs = i + 2 m (i3.i2)
From Eq. 13.12 we can see that by repeating the reaction at various concentrations of
Y, we can obtain both A, and A, because a plot of A obs against [Y] will give a straight
line with A, as the intercept and k 2 as the slope (Fig. 13.1).
What does the rate law tell us about the nature of the reaction? Substitution
reactions in inorganic chemistry have been divided into four classes based on the
relative importance of bond making and bond breaking in the rate-determining step:
Associative, A. The M—Y bond is fully formed before M—X begins to break.
Interchange associative, /„. The M—X bond begins to break before the
M—Y bond is fully formed, but bond making is more important than bond
breaking.
Dissociative, D. The M—X bond is fully broken before the M — Y bond
begins to form.
Interchange dissociative, l d . The M—Y bond begins to form before the
M—-X bond is fully broken, but bond breaking is more important than bond
making.
A first step in elucidating a mechanism for a reaction is to determine the rate law
experimentally. The reaction of interest here may be represented as
T —M —X+ Y
T —M —Y + X
Nonzero values for both A, and k 2 in Eqs. 13.10-13.12 indicate that ML 2 TX is
reacting by two different pathways. The k 2 term, first order with respect to both
complex and Y, indicates an associative pathway, A, similar to the S N 2 reaction of
organic chemistry. The term arises from the nucleophilic attack of ML 2 TX by Y. As
9 To leam how a kineticist thinks, see Espenson, J. H, Chemical Kinetics and Reaction Mechanisms-,
McGraw-Hill: New York, 1981.
Substitution Reactions in Square Planar Complexes
541
2 4 6 8 10
I0' 2 [Y]
Fig. 13.1 Rate constants
(A obs , s' 1 ) as a function of
nucleophile concentration
([Y]) for reaction of
«wis-[Pt(Py) 2 Cl 2 ] with
various nucleophiles in
methanol at 30 °C. [From
Belluco, U.; Cattalini, L.;
Basolo, F.; Pearson, R. G.;
Turco, A. J. Am. Client.
Soc. 1965, 87. 241-246.
Used with permission.]
would be expected for a reaction in which bond making is important, rates of reaction
depend markedly upon concentration of Y. Furthermore, the rates are significantly
dependent on the nature of Y.
At first glance the A, term, first order with respect to complex and independent of
Y, would suggest a dissociative pathway. Strong evidence, however, supports the
view that this pathway also is associative. It must be recognized that, in general,
solvent (S) molecules will be nucleophiles and will therefore compete with Y for
ML^TX to form ML 2 TS (sometimes called the solvento complex). Thus the two-term
rate law could be written as:
rate = - dlML 2 TX]/d/ = A'[ML 2 TX][S] + A 2 [ML 2 TX][Y] (13.13)
However, because the solvent is present in large excess, its concentration is essen¬
tially constant and therefore A'[S] = A,. As a result, Eq. 13.13 simplifies to Eq. 13.10.
The two associative pathways are summarized in the following reaction triangle:
ML 2 TX --- ML 2 TY
ML,TS
(13.14)
492
12-Coordination Chemistry: Structure
HjO H,0
facial meridional
(a) (b)
Fig. 12.19 Examples of facial and meridional octahedral isomers: (a ) fac-
triaqiiatrichlororuthenium(lll) and (b) mer-triaquatrichlororuthenium(lll).
isomer, Fig. 12.19a, abbreviated fac in the name) or (2) they may span three positions
such that two are opposite, or trans, to each other (the meridional isomer. Fig. 12.19b,
abbreviated mer in the name). 39 In contrast to cis-trans isomers of MA : B 4 , which
number in the hundreds, far fewer facial-meridional isomers have been characterized;
examples are [Ru(H, 0) 3 C1 3 ], [Pt(NH 3 ) 3 Br 3 ] + , [Pt(NH 3 ) 3 I 3 ]", [Ir(H 2 0) 3 CI 3 ],
[Rh(CH 3 CN) 3 Cl 3 ], [Co(NH 3 ) 3 (N0 2 ) 3 ], and [M(CO) 3 (PR 3 ) 3 ] (M = Cr, Mo. W).
It was mentioned above that tris(chelate) complexes of the type [Co(en) 3 ] 3 lack an
improper axis of rotation. As a result, such complexes can exist in either of two
enantiomeric forms (or a racemic mixture of the two). Figure 12.20 illustrates the
complex ions [Co(en) 3 ] 3+ and [Cr(ox) 3 ) 3_ , each of which is chiral with D } symmetry.
It is not necessary to have three chelate rings present. The cation dichloro-
bis(ethylenediamine)cobalt(III) exists as two geometric isomers, cis and trans. The
trans isomer has approximate D 2/i symmetry (Fig. 12.21b). Because it has three
internal mirror planes, it is achiral. 40 The cis isomer has C 2 symmetry and is chiral
(Fig. 12.21a). Since the two chloride ions replace two nitrogen atoms from an eth-
Optical Isomerism
in Octahedral
Complexes
Fig. 12.20 Structures of the optically active complexes [Co(en) 3 ] 3 and (Cr(ox) 3 ] 3 “ (one
enantiomer of each) and a stylized drawing of the two enantiomers of any tris(chelate)
complex.
39 These are also sometimes called cis-trans isomers, but the fac-mer nomenclature is unambiguous.
However, the 1UPAC recommendations arc moving away from the use of terms like facial and
meridional.
4U This assumes that the ethylencdiamine-metal rings are strictly planar, which they are not. We shall
see that these rings are skewed. However, since the two skew forms interconvert rapidly, for
purposes of optical activity, the compound behaves as though it were achiral.
Coordination Number 6
493
Mirror plane Mirror plane
(a) (b)
Fig. 12.21 (a) Optical isomers of cu-dichlorobis(ethylenediamine)cobali(UI) ion (violeo salt),
(b) /ron.v-Dichlorobis(ethylenediamine)coball(Hl) ion [ftraseo salt) showing one internal plane
of symmetry (there are two others perpendicular to the one shown).
ylenediamine ring without disturbing the remaining geometry, the optical activity is
preserved. 41 For purposes of dissymmetry, the change of two nitrogen atoms to
two chlorides is insignificant. Further replacement of chelate rings by paired uniden-
tate ligands can take place as in c7s'-c/^-c/'j-[Co(NH 3 ) 2 (H 2 0) 2 (CN) 2 ] + , which is chiral, 42
but in ordinary practice the resolution of enantiomers is limited to complexes con¬
taining chelate rings. The chelate rings provide the complex with additional stability
(see “The Chelate Effect," page 522). In addition, the nonchelate complexes are diffi¬
cult to synthesize, and it is also difficult to separate the large number of geometric
isomers.
Werner synthesized and resolved optical isomers as strong corroborative evi¬
dence for his theory of octahedral coordination. Although he did not discuss the
subject in his first paper, the idea appears to have come to him about four or five years
later. 43 and he was able to resolve amminechlorobis(ethylenediamine)cobalt(lII) cat¬
ion (cf. Fig. 12.21 with one chloro group replaced by ammonia) about fourteen years
after he first became interested in this aspect. 44 Werner realized that the presence of
enantiomers in the cis but not the trans compound was incompatible with alternative
formulations of these complexes. He succeeded in resolving a large series of com¬
plexes including Co(III), Cr(IlI), Fe(II), and Rh(III) as metal ions, and ethylene-
diamine, oxalate, and bipyridine as chelating ligands. Nevertheless, a few of his critics
pointed out that all his ligands contained carbon. By associating optical activity
somehow with “organic” versus “inorganic" compounds and ignoring all symmetry
arguments, they proceeded to discount his results. In order to silence these specious
arguments, Werner synthesized and resolved a polynuclear complex containing no
carbon atoms, tris[tetraammine-(x-dihydroxocobalt(III)]cobalt(IIl) (Fig. 12.22). 45 This
41 The “replacement" of elhylenediamine by chloride ions, cited here is a paper-and-pencil reaction
and refers to the formal change. It does not imply that in solution two chloride ions could attack a
[Co(en)j) 3> cation with retention of configuration.
42 Ito. T.; Shibata, M. Inorg. Client. 1977, 16. 108-116.
43 Werner mentioned the possibility in a letter to his coworker Arturo Miolati in 1897. See Classics in
Coordination Chemistry. Kauffman, G. B., Ed.; Dover: New York, 1968; pp 155-158.
44 Werner, A. Cliem. Ber. 1911, 44. 1887. A translation may be found in Classics in Coordination
Chemistry-, Kauffman. G. B., Ed.; Dover: New York, 1968; pp 159-173.
45 Werner, A. Cltem. Ber. 1914. 47. 3087. A translation may be found in Classics in Coordination
Chemistry. Kauffman. G. B.. Ed.; Dover: New York, 1968; pp 177-184.
13 Coordination Chemistry: Reactions, Kinetics, and Mechanisms
[M(NH,),ai 2+ [M(NH 3 ) 5 (S0 2 CFj)J 2+ — [M(NH 3 ) 6 ] 3+ (13.3)
This reaction has fairly general utility because it can be used to synthesize a wide
variety of [M(NH 3 ) 5 L]" + complexes.
Ideally, chemists hope to understand a number of reaction mechanisms well
enough that predictions about a diverse assortment of complexes involving different
metals, ligands, and reaction conditions can be made. A good example of a type of
reaction for which this level of understanding has been achieved is substitution in four-
coordinate square planar complexes.
Substitution
Reactions in
Square Planar
Complexes
Complexes with J 8 electronic configurations usually are four-coordinate and have
square planar geometries (see Chapter 12). These include complexes of Pt(Il), Pd(II),
Ni(H) (also sometimes tetrahedral, often octahedral), Ir(I), Rh(I), Co(I), and Au(lll).
Among the </ 8 ions. Pt(II) was a particular favorite of early kineticists. Complexes of
Pt(II) have been attractive for rate studies because they are stable, relatively easy to
synthesize, and undergo ligand exchange reactions at rates that are slow enough to
allow easy monitoring. Reaction rate ratios for Pt(II):Pd(II):Ni(II) are approximately
I: IO 5 :10 7 . Furthermore, because isomerization of less stable Pt(II) isomers to ther¬
modynamically more stable ones is a slow process, scrambling of ligands is not
generally a problem.
I here are several pathways by which one ligand may replace another in a square
planar complex, including nucleophilic substitution, electrophilic substitution, and
oxidative addition followed by reductive elimination. The first two of these are
probably familiar from courses in organic chemistry. Oxidative addition and reductive
elimination reactions will be covered in detail in Chapter 15. All three of these classes
have been effectively illustrated by Cross for reactions of PtMeCKPMe^Ph)-,.-
PtMeCl(PMe,Ph),
PtMe(N 3 )(PMe 2 Ph) 2 + CF
NucleophiUc
substitution
PtCl,(PMe 2 Ph), + MeHgCI
Electrophilic
substitution
PtMeCl 3 (PMe 2 Ph) 2
PtCI 2 (PMe,Ph) 2 + MeCl
Reductive
elimination
The reaction in Eq. 13.5 can be thought of as an electrophilic attack by Hg(II) on the
platinum-carbon bond. The oxidative addition reaction shows oxidation of Pt(II) to
Pt(IV) with simultaneous expansion of the coordination number of Pt from 4 to 6.
2 Cross. R. J. Chem. Soc. Rev. 1985, 14. 197-223.
Substitution Reactions in Square Planar Complexes
539
Elimination of methyl chloride returns the oxidation state to +2 and the coordination
number to 4 with a net substitution of chloride for methanide.
Much of what is currently known about substitution reactions of square planar
complexes came from a large number of careful studies executed in the 1960s and
1970s. 3 You should not conclude, however, that details of the mechanisms of these
reactions are of historical interest only. 4 Work in this area continues unabated as
studies focus on chelation, steric effects, biological reactions, and homogeneous
catalysts. For example, the mechanism for the Wacker process (Chapter 15), which
utilizes square planar [PdCl 4 ] 2 “ as a homogeneous catalyst for the industrial conver¬
sion of ethylene to acetaldehyde, is still a subject of investigation. 5 The overall
reaction for the process is:
PdCI,
CH 2 =CH 2 + 10 2 CH 3 CHO (13.7)
Knowledge of the mechanism may suggest changes in reaction conditions (solvent,
temperature, pressure, etc.) that could improve the efficiency of the overall process.
As another example, early studies 6 of the hydrolysis of c«-Pt(NH 3 ) 2 CI 2 are still of
interest because of the ability of this complex to inhibit the growth of malignant
tumors (Chapter 19). The biological activity of this compound is believed to involve
coordination of DNA to the Pt, and the details of this interaction are under intense
investigation. 7 However, it is generally agreed that prior to DNA complexation,
chloro groups of Pt(NH 3 ) 2 Cl 2 are reversibly replaced by water, thereby assisting in the
transfer of the drug from the blood to the tumor cells, where the water or chloride
ligands can be displaced by donor groups of DNA. 8
C k / Cl
^Pt^ + H 2 0
H.N NH,
V
H 3 N NH 3
For the remainder of this section on square planar substitution reactions, we will
confine our attention to those proceeding by a nucleophilic path. We turn now to
consideration of the mechanistic details of these reactions.
’ Basolo. F.: Pearson, R. G. Mechanisms of Inorganic Reactions. 2nd ed.: Wiley: New York. I9(>7.
Wilkins. K. G. The Study of Kinetics and Mechanisms of Reactions of Transition Metal Complexes'.
Allyn and Bacon: Boslon. 1974. Langford, C. H.; Gray, H. B. Ligand Substitution Processes',
Benjamin: New York. 1965. Tobc. M. L. Inorganic Reaction Mechanisms', Nelson: London. 1972.
4 Among recent books arc: Jordan. R. B. Reaction Mechanisms of Inorganic and Organometallic
Systems'. Oxford University: New York. 1991. Wilkins. R. G. Kinetics and Mechanisms of Reac¬
tions of Transition Metal Complexes. 2nd ed.; VCH: New York. 1991. Katakis, D.; Gordon, G.
Mechanisms of Inorganic Reactions'. Wiley: New York. 1987. Atwood. J. D. Inorganic and
Organometallic Reaction Mechanisms". Brooks/Cole: Monterey. 1985.
5 Akermark. B.; Sodcrberg. B. C.; Hull. S. S. Organometallics 1987, 6. 2608. Bdckvall. J. E.;
Akermark. B.; Ljunggren. S. 0. J. Am. Chem. Soc. 1979. 101. 2411-2416.
6 Reishus, J. W.; Martin. D. S., Jr. J. Am. Client. Soc. 1961. SJ. 2457-2462.
7 Inagaki. K.; Dijt. F. J.; Lempers. E. L. M.; Rccdijk, J. Inorg. Chem. 1988. 27, 382-387. Mukundan,
S., Jr.: Xu. Y.; Zon. G.; Marzilli, L. G. J. Am. Chem. Soc. 1991, 113. 3021-3027.
" Bruhn. S. L.; Toney. J. H.: Lippard, S. J. Prog. Inorg. Chem. 1990. 38. 477-516. Rccdijk. J.;
Fichlinger-Schepman. A. M. J.; van Oostcrom. A. T.: van de Puttc. P. Struct. Handing IBerlin) 1988,
67, 53-89. Caradonna, J. P.; Lippard, S. J. In Platinum Coordination Complexes in Cancer Chemo¬
therapy, Hacker, M. P.; Douple, E. B.: Krakoff, 1. H.. Eds.: Martinus Nijhoff: Boston, 1984.
494
12 ■ Coordination Chemistry: Structure
Resolution of
Optically Active
Complexes
Fig. 12.22 One enantiomer
of the tris[tetraammine-£i-
dihydroxocobalt(! II)]
cobalt(lll) cation.
laid to rest the distinction between carbon and the other elements that should have
died in 1828 with Wohler's work. It is interesting to note that in all of the work since
Werner only one other completely carbon-free complex has been resolved, sodium
c/.y-diaquabis(sulfamido)rhodate(IIl) (Fig. 12.23), 46 chiefly because of the difficulty of
preparing noncarbon chelating agents.
Few inorganic chemists have been as lucky as Pasteur in having their optically active
compounds crystallize as recognizable, hemihedral crystals of the two enantiomers
which may be separated by visual inspection. 47 Various chemical methods have been
devised to effect the resolution of coordination compounds. They all involve interac¬
tion of the enantiomeric pair of the racemic mixture with some other chiral species.
For optically active cations, interaction with one enantiomer of a chiral tartrate,
a-bromocamphor-iT-sulfonate, or one enantiomer of a metal complex anion will result
in differentially soluble diastereomers, the shifting of equilibria in solution, or various
other changes in physical properties that allow the separation to be effected. For the
resolution of an anion, chiral bases such as strychnine or brucine (protonated to form
cations) may be used. Neutral complexes present difficulties since it is generally
impossible to form salts. Differential physical properties toward dissymmetric sub¬
strates may be employed such as preferential extraction into a dissymmetric solvent or
preferential adsorption chromatographically upon quartz, sugar (both consisting of a
single enantiomer), or other chiral substrate.
Fig. 12.23 One enantiomer
of the c/i-diaquabis-
(Sulfamido)rhodatc(III)
anion.
46 Mann. F. G. J. Chem. Soc. 1933. 412.
47 However, some relatively simple cobalt complexes such as c7j-bis(cthylcncdiaminc)dinitro-
cobalt(lll) salts exhibit this behavior. Some of these were prepared by Werner and were, unknown
to him. resolving themselves spontaneously as he recrystallized them. This occurred at the very
time that he was spending 14 years pursuing the resolution of enantiomeric complexes! See Bernal.
Kauffman. G. B. J. Chem. Educ. 1987. 64. 604-610.
Coordination Number 6
495
Absolute
Configuration
of Complexes
A or D isomer A or l isomer
The determination of the absolute spatial relationship (the chirality or “handedness”)
of the atoms in a dissymmetric coordination compound is a problem that has intrigued
inorganic chemists from the days of Werner. The latter had none of the physical
methods now available for such determinations. Note that it is not possible to assign
the absolute configuration simply on the basis of the direction of rotation of the plane
of polarized light, 48 although we shall see that, through analysis of the rotatory
properties of enantiomers, strong clues can be provided as to the configuration.
Before discussing the methods of experimentally determining absolute configura¬
tions, let us briefly discuss means of denoting such configurations. As was the case in
organic chemistry, these rules grew up before much was known about the absolute
configurations, so they leave much to be desired in terms of logical interrelationships.
The simplest method, already mentioned, is notation of the experimental direction of
rotation of polarized light, ( + ) or d, or (-) or /. Thus one may speak of ( + ) 589 -
[Co(en) 3 ] 3+ , which is dextrorotatory with respect to light of wavelength equal to
589 nm. 49 This label identifies enantiomers with respect to each other, but serves little
purpose otherwise, although we shall see that by using certain techniques and assump¬
tions, strong clues can be provided with regard to configuration.
Next, as in Emil Fischer’s system for o-glyceraldehyde and sugars, we can
arbitrarily assign the d configuration to (+)-[Co(en) 3 ] 3+ and compare all known
configurations with it. This now immediately tells us the configuration of a D-isomer
(by comparing it with a figure of D-[Co(en) 3 ] 3+ ), but the d symbol is an arbitrary one
that could have as readily been zzz versus www. Furthermore, it can even seduce the
unwary into thinking that d has something to do with being “dextrorotatory” or
having "right-handedness."
A systematic basis can be gained by viewing a tris(chelate) complex, the most
common species of chiral coordination compounds, down the threefold rotation axis.
If the helix thus viewed is right-handed, the isomer is the A-isomer, and its mirror
image is the A-isomer. 50 The D-, l-, A-, and A-isomers may thus be portrayed as
shown in the margin. Note that it is a result of these systems that A A l and A A d,
furthering possible confusion.
In ordinary X-ray diffraction work both enantiomers give exactly the same diffrac¬
tion pattern, and thus this method gives no information on the absolute configuration
about the metal atom. However, absolute configurations of coordination compounds
can be directly determined by means of the anomalous dispersion of X rays, called
Bijvaet analysis.' The method has not been widely applied, but as in the related
problem in organic chemistry of the absolute configuration of D-glyceraldehyde, once
one absolute configuration is known, there are methods to correlate others. The abso¬
lute configuration of D-( + )-[Co(en) 3 ] 3+ has been determined as the chloride and bro-
• ul For example, it is known that the enantiomers of [Cotenlj]' 1 ' and (Rh(cn)j) 3+ . which rotate sodium
D light in the same direction, have the opposite absolute configuration; i.e., they arc of opposite
handedness—mirror images of each other (ignoring the difference between Co and Rh).
49 This is the "sodium D line." For it. too. the "D" has nothing to do with "dextrorotatory." Since
the sodium D line is so frequently used, reference to it is often omitted, but other wavelengths
should ulways be explicitly stated.
50 This suggestion was made by Piper. T. S. J. Am. Chem. Soc. 1961. 83. 3908-3909. The IUPAC
recommendations have extended this Ufa considerably more generalized system (sec Appendix I).
For tris(chelates) both systems have the same designation, as given above.
51 Bijvoet, J. M. Endeavor. 1955, 14. 71-77. Sec also Saito, Y. Inorganic Molecular Dissymmetry-,
Springer-Verlag: Berlin, 1979; Chapter 2.
536
12-Coordination Chemistry: Structure
Fig. 12.54 ORTEP drawings of (a) dichloro-/V.iV'-(2-pyridylmethylurea)copper(II); (b)
dichloro-iV,0-(2-pyridylmethylurea)zinc(II). [From Maslak. P.; Sczepanski. J. J.. Parvez. M.
J. Am. Chem. Soc. 1991. 113. 1062-1063. Reproduced with permission.)
12.37 On page 528 the statement is made that "crystalline [Na(C222)) + Na consists approx¬
imately of closest packed, large, complex cations with sodide anions in the octahedral
holes (Fig. 12.50a)." Review the discussion of closest packing in Chapter 4 and label the
atoms in Fig. 12.50 as belonging to layers A and B and indicate the octahedral holes.
12.38 Assume that the geometry of [Cd(OAr) 2 (thf) 2 ] (Fig. 12.4) is the result of the maximization
of s character toward the strongly bonding phenoxide ions (Bent's rule), allowed (90°
interactions) by the stericaily relaxed cadmium atom. If zinc is too small to accommodate
square planar geometry and must be pseudotetrahedral. how can it follow Bent's rule?
Can you cite evidence to support your case?
12.39 In the discussion of [Cd(OAr) 2 (thf) 2 l. it was implied that it does not have true D 2I ,
symmetry. Look closely at Fig. 12.4 and assign a point group symmetry to it.
12.40 A chemist performs the following reactions:
(1) K 2 [PtCI 4 ] + 2NHj - "A" + 2KCI
(2) [Pt(NH,) 4 )(NOj) 2 + 2KCI -♦ "B" + 2NH, + 2KNO,
She finds that both A and B are white, diamagnetic, crystalline compounds that give
elemental analyses for empirical formula PtCl 2 (NH 3 ) 2 . However. A is most soluble in
polar solvents, such as ethanol, while B is soluble in petroleum ether (a mixture of
hydrocarbons) and carbon tetrachloride. Draw the structures of A and B.
12.41 Another chemist reads a report of the experiment described in Problem 12.40 and
immediately identifies A and B. Since nickel is in the same group of the periodic table as
platinum, he decides to perform the same experiment with nickel(ll) but is unsuccessful.
He is unable to perform Reaction 2 because the starting material. [Ni(NHj) 4 )X 2 . is not
listed in any chemical catalogs (he checks for X = halide, nitrate, etc.), and none of his
colleagues has ever heard of it. He obtains some KjNiCU, but when he attempts to run
Reaction I, the only products he is able to isolate are Ni(NH 3 )<,CI 2 and KCI. When
triphenylphosphine, Ph 3 P. is used in place of ammonia as one ligand (chloride ion is still a
second ligand), compound C is isolated. This compound analyzes for empirical formula
NiCl 2 (Ph 3 P) 2 , and it is greenish, paramagnetic, and soluble in organic polar solvents.
Regardless of the reaction conditions or concentrations chosen, only C is found; no other
isomers are observed. Draw the structure of C. Will it have a dipole moment?
Coordination Chemistry:
Reactions, Kinetics,
and Mechanisms
Despite extensive study, inorganic chemistry has yet to achieve the understanding of
reaction mechanisms enjoyed by organic chemistry. This situation, to which we
alluded previously in Chapter 6, can be attributed to the inherent difficulties involved
in trying to systematize the reactions of more than one hundred elements. Even
attempts to predict from one element to another in the same group are not always
successful. The classical synthetic schemes developed for the hexaamminecobalt(III)
and hexaamminerhodium(III) cations illustrate this lack of generality. The preparation
of the cobalt complex involves a combination of ligand displacement and redox
chemistry. To the starting material, a stable and common cobalt(ll) salt (such as the
nitrate or chloride), the desired ligand (ammonia) is added in high concentrations to
replace those present; then an oxidizing agent (air or hydrogen peroxide with a
charcoal catalyst) is used to effect the change in oxidation state:
[CoCIJ 2 - — [Co(NH 3 ) h r- + — [Co(NH 3 ) 6 ] 3+ (13.1)
Simple rhodium(IIl) salts, unlike their cobalt counterparts, are stable in water. Thus
the hexaammincrhodium(III) cation can be prepared from reactants that are already in
the +3 oxidation state:
RhCl 3 H 2 0 [Rh(NH 3 ) 6 ] 3 + (13.2)
Because forcing conditions are required in order to remove the last chloro ligand Irom
rhodium and displace it with NH 3 , the reaction is often carried out in a sealed tube at
elevated temperature and pressure.
The hexaammine preparations described above were both devised in the 19th
century, long before reaction mechanisms were investigated. Mechanistic insight
acquired in the latter half of the 20th century has led to the development of more
systematic syntheses for many complexes of cobalt and rhodium. For example, the
hexaammine complexes may be easily prepared by simple substitution of NH 3 for
the CF 3 SO^ (trifluoromethanesulfonato or triflate) ligand in [M(NH 3 ) 5 (0S0 2 CF 3 )|'''
(M = Co. Rh). a starting material that can easily be obtained from the penlaam-
minechloro complex. 1
1 Dixon. N. E.; Lawrance, G. A.; Lay. P. A.; Sargeson. A. M. /norg. Chem. 1983. 22. 846-847.
537
498 12 Coordination Chemistry: Structure
\
Fig. 12.25 The Cotton effect: (a) positive Cotton effect; (b) negative Cotton effect. The
absorption band is not shown; it would be a positive Gaussian curve centered on \ mu . but
off scale. The dashed line represents the ORD curve (and relates to the refractive index scale
on left). The solid line represents the CD curve («, - e r , scale on right). The maximum
absorption, zero values of ORD. and maxima and minima of CD values occur at A ml „. The
two figures, (a) and (b). represent two enantiomers. These are ideal curves for an absorption
peak well separated from other absorptions. [Modified from Gillard, R. D. Prog. Irtorg.
Chem. 1966. 7, 215-276. Reproduced with permission.)
Although ORD was used extensively at one time because of simpler instrumenta¬
tion, circular dichroism is currently much more useful. The CD effect arises because
there is differential absorption of left and right circularly polarized light associated
with transitions such as 'A, - 'E and 'A, —► 'A,. The circular dichroism is the
difference between the molar absorptivities of the left and right polarized light, e, - e r
(solid curves in Fig. I2.25). 5fi Complexes having the same sign of CD for a given
absorption band will have the same absolute configuration. Some typical values are
listed in Table 12.5.
Stereoselectivity and In addition to the dissymmetry generated by the tris(chelate) structure of octahedral
he Conformation of complexes, it is possible to have dissymmetry in the ligand as well. For example, the
Chelate Rings gauche conformation of ethylenediamine is dissymetric (Fig. 12.27) and could be
resolved were it not for the almost complete absence of an energy barrier preventing
56 Sometimes the value of - « r is given as At. If At is positive (e, > e r ). it is called a positive Cotton
effect, etc.
Coordination Number 6
499
Fig. 12.26 The absolute configurations and ORD spectra of (a) A- [Co(en),] 51 ; (b) A-|Co(S-
ala),| (.S-ala = the anion of S-(L)-alanine); (c) A-|Co(en);(5-glu))' H (5-glu = the dianion of
5-(i.)-glutamic acid). All of these complexes have the A or d configuration.
Table 12.5
Circular dichroism data
for some A-(o)- ond A-(l)-
Formula of complex 6
u (cm ')
<1 “ <r
Absolute
configuration
tris(chelate) complexes of
cobalt(lll)°
( + )j« 9 '[Co(en) 3 ] 3 +
20,280
23,310
2.18
0.20
A
( — )jtw'lCo(en)j) 3 +
20,280
23.310
2.18
0.20
A 1 '
( + ) w -[Co(S-pn),) 3 +
20,280
22,780
1.95
0.58
A
( + WtCo(R- P n),] 3 +
21,000
2.47
A
(+) 389 -[Co(S-ala) 3 ]
18,500
21,000
1.3
0.2
A
(+ ) 495 -[Co(A'-glu)(en) 2 ] 2 +
19,600
2.5
A
" Saito. Y. Inorganic Molecular Dissymmetry : Springer-Verlag; Berlin,
1979; p 136. Note that the tris(ala) complex has a different chromophore
(CoNjOj) than the tris(diaminc) complexes (CoN 6 ). and so the
resemblance is only approximate.
en = ethylenediamine; pn = propylcnediamine; ala = alaninatc;
glu = glutamate.
c Since in each case the A isomer will have the same value with opposite
sign for its specific rotation, the same value for the absorption maximum,
and the same value(s) with opposite sign for the circular dichroism
compared to the A isomer, these values are not all listed here.
534
12 • Coordination Chemistry: Structure
Problems
535
12.25 It is till well and good to say that macrocyclic polyethers "stabilize metals, such as
sodium, that might not otherwise react ... to form (Na(macrocycle)]’Au - salts." but
specifically, in terms of a Born-Haber cycle, what part does the macrocyclic ligand play in
the reaction described on page 526?
12.26 Assuming that Piutli was at least a reasonably careful worker, how was it possible for him
to get "identical" spectra from solutions of the nitro and nitrito isomers discussed on page
513?
12.27 Read the section on point groups in Chapter 3 again, and identify the symmetry elements
and operations in the molecules and ions shown in the figures listed below. Determine the
appropriate point group for each molecule and ion.
a. 12.1c
r.
12.10
k. 12.21a. b
p. 12.39
b. 12.2a
B-
12.17
12.22
q. 12.40
C. 12.4
h.
12.18a. b
m. 12.23
r. 12.41a
d. 12.5a. b
i.
12.19a. b
n. 12.33
s. 12.53
e. 12.7
j-
12.20
o. 12.35b
12.2S Occasionally, in the preparation of the artwork for a research article or a textbook, the
photographic negative taken from the original line drawing of the artist is inserted
"upside-down" (or reverse, front-to-back) and the resulting image is reversed. Does this
make any difference? Discuss. Are there any exceptions to the general rule? Illustrate
your argument with sketches.
12.29 Consider the shapes (i.e.. bond angles. 0, and bond lengths, d) of T-shaped (0 3 > 0 , = 0 ; )
and Y-shaped (0., < 0, = 0 2 ) molecules (Fig. 12. Id. e).
a. What causes these molecules to be T-shaped or Y-shaped rather than equilateral?
b. For one of these shapes. </, > d, = </ ; ; for the other, dy < d, = </ 2 . Which is which?
Why?
c. What is the point group symmetry of these molecules?
12.30 The circular dichroism spectral data for the dextrorotatory enantiomer of tris(/?./?-
Irans- 1.2-diaminocyclohexane)cobalt(III) cation. ( +) w ,-(Co(/?./?-chxn),| w . if listed in
Table 12.5. would appear as follows:
v (cm -1 ) £| - £,
( + ),,-[Co(R.R-chxn) 3 ] 3t 20,000 - 2.28
22,500 +0.69
Assign the correct (A or A) configuration to the metal. Do you get the same answer as in
Problem 12.31? Discuss.
12.31 The X-ray crystal structure of the dextrorotatory enantiomer of lris(/?./?-fr<mj-l.2-di-
aminocyclohexane)cobalt(lll) cation. ( + ). w -[Co(/?./?-chxn)jj', may be solved to give
Fig. 12.53 as one of the two possible enantiomeric solutions.
a. Is Fig. 12.53 a A or a A enantiomer?
b. Does (+1.™, (Co(/?,/?-chxn)j|' have a A or a A configuration about the metal? Is your
answer the same as in Problem 12.30? Explain.
c. Is the isomer lei or obi
d. Are the chelate rings ft or A?
12.32 If either pure A or pure B from Equation 12.4 is heated at 75 °C in dimethvlformamide. the
isomers interconvert until there is a mixture of 40% A and 60% B. Why is this not a 50-50
mixture? Is this a racemizalion?
Fig. 12.53 One of two
possible enantiomeric
solutions to the X-ray
crystal structure of the
dextrorotatory enantiomer
of lris(/?./?-/r«/r.v-1,2-di-
aminocyclohexane)
cobalt(III) cation.
( + U M -[Co{R.R-
//wu-chxnlil' 1 '. See
Problem 12.31. (Modified
from Saito. Y. Inorganic
Molecular Dissymmetry ;
Springer-Verlng: Berlin,
1979; p 62. Reproduced
with permission.]
12.33 Which of the following is the most likely structure for pentacyanocoball(III)-/x-cyanopen-
laammineocobalt(lll)? |(NH,).sCo—CN—Co(CN).,] or |(NH,)<Co—NC—Co(CN) 5 ]
Why?"'
12.34 Metallothionein is found in humans and other animals. It appears to remove toxic
cadmium ion from the kidneys. This protein has 27 sulfhydryl groups and can function as a
protective chelating agent. Each metal ion is thought to be bound to three sulfhydryl
groups. Show the probable geometry around the cadmium ion when bound to this
chelating agent.
12.35 Urea ligands usually bond through the oxygen atom, although there are two potential
nitrogen donor atoms available as well. Recently ambidentatc behavior of a substituted
urea has been demonstrated. 120 (2-Pyridylmethyl)ureu forms N.N'-bonded square planar
complexes with Ni : ' and Cu :> (Fig. 12.54a). but N.O-bonded complexes with Zir" (Fig,
12.54b).
a. Suggest a reason for this behavior.
b. If you did not have the X-ray crystal structure, what other experimental evidence
might you use to differentiate between O-bonding and N-bonding in ureas? {Hint:
Compare the analogous problem with proton-binding sites. Chapter 9.)
12.36 Twelve-coordination is certainly rare, as discussed on pages 510-511, yet there is another
discrete complex discussed in this chapter with C.N. = 12. though not so identified. Can
you find it?
"9 Wang. B-C.: Schaefer. W. P.; Marsh. R. E. Iiwrg. Chem. 1971. 10. 1492-1497,
120 Maslak. P.; Sczepanski. J. J.; Parvez. M. J. Am. Chem. Soc. 1991. 113. 1062-1063.
500
12 Coordination Chemisfry: Structure
NH;
H
5
NH,
H
\
Fig. 12.27 Enantiomeric
conformations of gauche
ethylenediamine (1,2-
diaminoethane). Note that 8
represents a right-handed -
helix and A a left-handed
helix.
racemization. Attachment of the chelate ligand to a metal retains the chirality of the
gauche form, but the two enantiomers can still interconvert through a planar confor¬
mation at a very low energy, similar to the interconversions of organic ring systems.
Thus, although it is possible in principle to describe two enantiomers of a complex
such as [Co(NH 3 ) 4 (en)] 3+ , in practice it proves to be impossible to isolate them
because of the rapid interconversion of the ring conformers. 57
If two or more rings are present in one complex, they can interact with each other
and certain conformations might be expected to be stabilized as a result of possible
reductions in interatomic repulsions. For example, consider a square planar complex
containing two chelated rings of ethylenediamine. From a purely statistical point of
view we might expect to find three structures, which may be formulated MSS, MAA,
and MSA (which is identical to MAS). The first two molecules lack a plane of symme¬
try, but MSA is a meso form. Corey and Bailar 58 were the first to show that the MSS
and MAA should predominate over the meso form since the latter has unfavorable
H—H interactions of the axial-axial and equatorial-equatorial type between the two
rings (Fig. 12.28). The enantiomeric MSS and MAA forms are expected to be about 4 kJ
mol -1 more stable than the meso isomer, other factors being equal.
More important consequences result for octahedral tris(chelate) complexes.
Again, from purely statistical arguments, we might expect to find MSSS, MSSA, M5AA,
and MAAA forms. In addition, these will all be optically active from the tris(chelate)
structure as well, so there are expected to be eight distinct isomers formed. In general,
a much smaller number is found, usually only two. This stereoselectivity is most easily
followed by using a chiral ligand such as propylenediamine, CH 3 CH(NH 2 )CH 2 NH 2 .
The five-membered chelate ring will give rise to two types of substituent positions,
those that are essentially axial and those that are essentially equatorial. All substi¬
tuents larger than hydrogen will cause the ring to adopt a conformation in which the
substituent is in an equatorial position. As a result of this strong conformational
propensity, /?-(-)-propyIenediamine bonds preferentially as a A chelate and
S-( + )propylenediamine bonds as a 5 chelate. This reduces the number of expected
isomers to four: A.-M888 (= d-MSSS), A-MAAA (= d-MAAA), A-M55S (= l-MSSS), and
37 Note that in this case the chirality is not due to the arrangement about the metal atom, but results
from the S or A chirality of the ethylencdiamine-mctal ring. Sec Shimura. Y. Bull. Chem. Soc.
(Japan) 1958. 31, 311.
38 Corey, E. J.; Bailar, J. C., Jr. J. Am. Chem. Soc. 1959, 81 , 2620-2629. This is the classic paper in
the field upon which ail of the subsequent work has been based. See also Saito, Y. Inorganic
Molecular Dissymmetry, Springer-Veriag: Berlin, 1979; Chapter 3.
Coordination Number 6
501
(a) (b)
Fig. 12.28 Conformational interactions in bis(chelate) square planar complexes: (a) AA form;
(b) AS form. All hydrogen atoms except four have been omitted for greater clarity. Dashed
lines represent inter-ring H-H repulsions. (Modified from Corey. E. J.; Bailar, J. C., Jr.
J. Am. Chem. Soc. 1959, 81. 2620-2629. Reproduced with permission.]
A-MAAA (= l-MAAA) where A, A, d, and l refer to the absolute configuration about
the metal related to A-(-f) 589 -[Co(en) 3 ] 3+ (= the d enantiomer; see Fig. 12.26). In a
typical reaction such as the oxidation of cobalt(II) chloride in the presence of racemic
ft.S-propylenediamine, only two isomers were isolated:
[Co(H,0) 6 ] 2+ + R- pn + A’-pn -^A-[Co<A-pn) 3 ] 3+ + A-[Co(J?-pn) 3 J 3+ (12.7)
The difference in stability between the various isomers has been related to
preferred packing arrangements of chelate rings about the central metal atom. Thus,
for 5-propylenediamine forming a 8 chelate ring, the most efficient method of fitting
around a metal will be in the form of a left-handed helix. This arrangement minimizes
the various repulsions. It has been termed the lei isomer since the C—C bonds are
parallel to the threefold axis of the complex (Fig. 12.29). The alternative isomer, in
which the ligands form a right-handed helix about the metal, is known as the ob isomer
since the C—C bonds are oblique to the threefold axis (Fig. 12.30). 59 The interactions
Fig. 12.29 The lei
conformer of the A or d
enantiomer of tris(diamine)
metal complexes. The
hatched circles represent
the positions of the methyl
groups in the propylene¬
diamine complex. For
propylenediamine, this
represents the A888 or d8S8
isomer. [Modified from
Corey, E. J.; Bailar, J. C.,
Jr. J. Am. Chem. Soc.
1959, 81. 2620-2629.
Reproduced with
permission.]
39 Specifically since there are three chelate rings, each of which can potentially be either lei or ob,
these two isomers should be labeled as /e/j and ofij since it is possible to have mixed ob-lel
complexes.
5J2
12-Coordination Chemistry: Structure
12.10 Draw the most likely structure of pentaamminecobalt(III)-i-thiocyanatopentacyano-
cobalt(lll)." J
12.11 Which of the two isomers, Co(Hdmg) 2 (SCN)py or Co(Hdmg) : tNCS)py. would you pre¬
dict to be thermodynamically the most stable? Hdmg represents the monoanion di-
methylglyoximate, HON=C(CHj)C(CHj)=NO . ll5
12.12 Draw the molecular structure of the following complexes:
a. ci's-dichlorotetracyanochromate(III)
b. /nt'Mriamminelrichlorocobalt(lll)
c. f/wi.v-dichlorobis(trimelhylphosphinc)palladium(Il)
d. /flc-triaquatrinitrocobalt(lll)
12.13 Why does iron(ll) hexacyanofcrrate(lll) spontaneously isomerize to Prussian blue?
12.14 With the aid of the table of ligand abbreviations given in Appendix I find the names of each
of the ligands listed below. Sketch the structure of each ligand and classify it as uniden-
tate, bidentale. tridentate, etc. Sketch the mode of attachment of the ligand to a metal ion.
a. acac g. dtp m. phen
b. bpy h. edta n. pn
c. C222 i. fod o. py
d. chxn j. Hedta p. tap
e. dien k. ox q. tn
f. dmf I. pc r. trien
12.15 Arrange the following six ligands in order of increasing ability to form stable complexes
and account for your order.
12.16 A recent review of steric effects in coordination compounds includes the following
statements concerning stability:" 6
a. Tetrahedral geometry should be more stable than square planar for C.N. = 4.
b. Octahedral geometry should be more stable than trigonal prismatic for C.N. = 6.
c. Square anliprismatic geometry should be slightly more stable than dodecahedral,
which is considerably more stable than cubic for C.N. = 8.
Discuss these statements in terms of Table 12.1.
"• Fronczck. F. R.; Schaefer, W. P. Inorg. Chcm. 1975. 14, 2066.
" 5 Raghunathan. S. J. Inorg. Nucl. Chem. 1975, 37, 2133.
" 6 Kepert. D. L. In Comprehensive Coordination Chemistry; Wilkinson. G.. Gillard, R. D., Mc-
Cleverty, J. A.. Eds: Pergamon: Oxford. 1987; Vol. I. Chapter 2.
B>'K - '•' . <-s. -— • • •; - • • •
Problems
533
12.17 The macrocyclic ligand entcrobactin (Fig. 19.27c) has an extraordinarily high affinity for
Fe J with a stability constant of 10 s " (the largest known stability constant for Fe , ' + with a
naturally occurring substance)." 7
a. Suggest a structure for the Fe(III)-enterobactin complex that explains its high stability.
b. If the concentration of the FeGID-enterobactin complex within the microorganism is
10~ 7 mol L~\ how many liters of bacteria would have to be searched to find a single
free Fe ,+ ion?
12.18 In Chapter 6 it was pointed out that X rays are diffracted by electrons. Yet on page 528 it
is stated that the anionic electrons of complexed cesium electride "do not show" in the
structure determination. Discuss this apparent paradox. (Hint: Why is it hard to locate
hydrogen atoms in an X-ray crystallographic determination?)
12.19 Alkali metal and other cationic cryptates have been known for a number of years. More
recently, anionic cryptates have been characterized. Suggest a structure for
[C1N(CH : CH ; NHCH,CH 3 NHCH : CH : )jN] _ . (See Footnote 112.)
12.20 There is a complex (C.N. = 6) with a chiral metal center illustrated in this chapter that is
not so labeled. Find it and determine if the isomer shown is A or A.
12.21 Figure 12.52 illustrates two forms of the pentanitrocuprate(II) ion. [Cu(N0 3 ) 5 ] 3_ . Discuss
all of the types of isomerism exhibited in these ions." 1
12.22 The molecules shown in Fig. 12.1a, c appear to be "left-handed." Do these molecules
have chirality? Explain.
12.23 If you learned the Cahn-lngold-Prelog rules in organic chemistry, test your recall by
assigning the appropriate R.S notation to the molecule shown in Fig. 12.2. Assume that
the C S H 5 ligand is a "single atom" of mass 65 (5 x 13).
12.24 Commercial mayonnaise, salad dressings, kidney beans, house-plant food, and liquid
dishwashing detergents usually contain edta (see Problem 12.14h) or some derivative of it.
Why? See Hart. J. R. J. Chem. Ednc. 1985. 62, 75-76.
Fig. 12.52 Two forms of the [Cu(NO;)j] J_ ion. All bond distances in angstroms. (From
Kiandcrman. K. A.; Hamilton, W. C.: Bernal. I. Inorg. Chem. Acta 1977. 23, 117-129.
Reproduced with permission.]
117 Harris, W. R.; Carrano. C. J.; Cooper. S. R.; Soften. S. R.: Aydeef, A. E.; McArdle, J. V.;
Raymond. K. N. J. Am. Chem. Soc. 1979. 101, 6097-6104.
" Klanderman.K. A.; Hamilton. W. C.; Bernal. I. Inorg. Chim. Ada 1977, 23, 117-129.
502
12 Coordination Chemistry: Structure
503
Asymmetric
Syntheses Catalyzed
by Coordination
Compounds
Fig. 12.30 The ob
conformer of the A or l
isomer of tris(diamine)metal
complexes. The hatched
circles represent the
positions of the methyl
groups in the propylene-
dianiine complex. For
propylenediamine this
represents the A558 or l 85S
isomer. [Modified from
Corey, E. J.; Bailar. J. C..
Jr. J. Am. Client. Soc.
1959, 81. 2620-2629.
Reproduced with
permission.)
between the hydrogen atoms of the various rings stabilize the lei isomer by a few
kilojoules per mole. Note that this greater stability of the one configuration over
another is the source of the stereospecific synthesis of A-tris(/?-propylenediamine)-
cobalt(IH) seen previously (page 496).
There has been considerable interest in the stereospecific synthesis of organic com¬
pounds using optically active coordination compounds. Chiral catalysts are sought for
the production of drugs, pesticides, pheromones, and fragrances. Quite recently a
significant advance was made with the discovery that asymmetric cpoxidation can be
achieved with manganese(III) complexes containing chiral chelating agents. One of
the catalysts used is generated from manganese(II) acetate, ( R. R')- or {S. S')-1,2-
diamino-1,2-diphenylethanc, a substituted salicylaldehyde. and lithium chloride. In
the presence of this catalyst and sodium hypochlorite, alkcnes are converted to
epoxides:
R 1 , R 2 . R 3 = H. alkyl, aryl
R = Ph
X = H, OMe, Me, Cl, N0 2
Some cis substituted alkenes can be converted to chiral epoxides in greater than
90% enantiomeric excess by this process.'’ 0
Coordination Number 7
Id) (e) (f)
Fig. 12.31 The chiral diphosphine ligands (a) R.R- (-)-diop, (bi R-{ + )-binup. and
(c) S-(-)-binap. The free ligands are shown at the top and coordinated to a rhodium)I)
or ruthenium(ll) atom, below: (d) (/?,/?-(— )-diop)Rh(I) moiety, (e) (5-1 -)-binap)Ru(II)
moiety. Note the chirality faced by incoming ligands. The view is down the C : axis.
(0 (S-(-)-binap)bis(carboxylnto)ruthcnium(II) looking down the pseudo-D.i axis.
Hydrogenation of double bonds using a rhodium(l) catalyst (see Chapter 15) may
be carried out stereospecifically using a chiral diphosphinc such as "diop” or "binap”
(Fig. 12.31). The chirality of the diphosphine makes the two possible transition states
(leading to the two enantiomeric products) diastereomers and therefore subject to
differences in equilibrium concentrations, energies of activations, and reaction rates—
one enantiomeric product may thus form to the exclusion of the other. The
(diop)Rh(D-catalyzed enantioselective hydrogenation is used commercially to make
S-(l)-DOPA. a drug used to treat Parkinson's disease, and aspartame [S-(i..)-phen-
ylalanyl-.V-(i.)-aspartic acid), an artificial sweetener. The tbinap)Ru(ll) catalysis may
become even more useful. 6 '
Coordination
Number 7 62
Coordination number 7 cannot be considered at all common. The relative instability of
these species can be attributed to the fact that the additional energy of the seventh
bond is offset by (I) increased ligand-ligand repulsion. (2) weaker bonds, and (3)
generally reduced ligand field stabilization energy as a result of nonoctahedral geome¬
try. There are three geometries known: (I) a pentagonal bipyramid (Fig. 12.32), which
is also found in the main-group compound IF, (Fig. 6.12): (2) a capped octahedron in
which a seventh ligand has been added to a triangular face (Fig. 12.33): and (3) a
60 Jacobsen, E. N.; Zhang, E.; Gillcr, M, L. J. Am. Chem. Soc. 1991, 113. 6703-6704. For additional
discussions of asymmetric syntheses with coordination compounds, see Bosnich. B.; Fryzuk. M. D.
Top. Slereochem. 1981, 12. 119-154: Kagan. H. B. In Comprehensive OrganometuUic Chemistry:
Wilkinson, G.; Stone. F. G. A.; Abel. E. W., Eds.; Pergamon: Oxford. 1982; Vol. 8. pp 463-498:
Scott, J. W. Top. Slereochem. 1989. 19. 209-226.
61 Noyori. R. Science 1990 . 248. 1194-1199. Noyori, R.; Takaya. H. Acc. Chem. Res. 1990 . 23.
345-350. Brunner. H. Top. Slereochem. 1988. 18. 129-247.
62 Kcpert, D. L. In Comprehensive Coordination Chemistry: Wilkinson. G.; Gillard. R. D.: Me-
Cleverty, J. A., Eds.: Pergamon; Oxford. 1987; Vol. I. pp 69-83.
530
12 Coordination Chemistry: Structure
called sodium pump necessary for the proper Na + /K + ionic balance responsible for
electrical gradients and potentials in muscle action. 111
The ultimate in encirclement of metal ions by the ligand is shown by encapsulation
reactions in which the ligand forms a three-dimensional cage about the metal. The
resulting case (Fig. 12.51) is called a clathro-chelate or a cryptale. One class of
cryptate-forming ligands of the type NCCH^CHiOCH^CH^OCHjCHoJjN has been
called “football ligands” because the polyether bridges between the two nitrogen
atoms resemble the seams of a football. Ligands of this type form exceptionally stable
complexes with alkali metals and show high selectivity when the size of the “football”
is adjusted to fit the desired cation." 2
Closely related to the football ligands are the so-called sepulchrate ligands. One
can be formed by the condensation of formaldehyde and ammonia onto the nitrogen
atoms of tris(ethylenediamine)cobalt(IlI). This results in tris(methylene)amino caps on
opposite faces of the coordination octahedron. If the synthesis utilizes one of the
(A, A)-enantiomers, the chirality of the complex is retained. Furthermore, the com¬
plex may be reduced to the corresponding cobalt(ll) cation and reoxiriizea to co-
balt(lll) without loss of chirality. This is particularly unusual in that, as we shall see in
the following chapter, cobalt(II) complexes are quite labile in contrast to the stability
of cobalt(III) complexes. Once again the extra stability of polydentate complexes is
demonstrated.
In contrast, Bernal 113 has isolated a self-resolving complex whose chirality de¬
pends only on the conformation of an acyl bridge:
Single crystals consist of only a single enantiomer, and so in the solid state the
enantiomers do not racemize despite the seeming lack of barriers to rotation of the
rings (see Chapter 13). Presumably crystal-packing forces “lock in" the chirality.
Immediately upon solution, the complex racemizes, indicating the fragility of the
forces stabilizing it.
111 Rawn. J. D. Biochemistry, Neil Patterson: Burlington, NC, 1989; pp 1024-1034. Strycr, L.
Biochemistry, Freeman: New York, 1988; pp 948-954.
1,2 Lchn. J. M. Angew. Client. Int. Ed. Engl. 1988 , 27. 89-112.
113 Bernal. I., personal communication.
Problems
531
(a) (b)
Fig. 12.51 Cryptate derived from dimcthylglyoxime, boron trifluoride, and cobalt(Ill): (a)
formula; (b) geometry of cryptate; boron atoms form apexes on lower left and upper right;
coordinate positions on the octahedron are occupied by nitrogen atoms. The heavier lines
represent edges of the polyhedron spanned by chelate rings. [From Boston, D, R.; Rose, N
i. J. Am. Chem. Soc. 1968, 90, 6860. Reproduced with permission.]
Problems _ 12.1 A complex of nickel(Il), [NiCI 2 (PPhj) 2 ], is paramagnetic. The analogous complex of
palladium(U) is diamagnetic. Predict the number of isomers that will exist for each of
these formulations.
12.2 a. Discuss the identification of cis-trans isomers of compounds MA 2 B 2 by dipole
moments.
b. Discuss the possibilities of identifying the cis-trans isomers of compounds MA 2 Bj by
dipole moments.
c. Discuss possibilities of identifying facial-meridional isomers of MAjBj by dipole
moments.
d. Are there any problems arising in octahedral complexes that make this application less
certain than in square planar complexes?
12.3 The anions of the five-coordinate complexes [CuCI 5 ]' - , [ZnCI,] 3 , and [CdCI,] 3- were
isolated as salts of [Co(NHi),,]' and [CrtNHdcJ' 1 ' • Explain why these counterions were
chosen.
12.4 The cations shown in Fig. 12.1 were isolated as perchlorate, tetrafluoroborate. or liexa-
lluorophosphatc salts. Discuss.
12.5 Which diastcrcomer, A or B, from Eq. 12.4 is shown in Fig. 12.15?
12.6 Use the data in Table 42.6 to calculate equilibrium constants for the formation of
[Ni(NH 3 )<J‘ and [Ni(en) 3 ]' from aqueous Ni‘. Also calculate the equilibrium constant
for the reaction:
iNi(NHj) fc ]“' + 3 en :=: [Ni(en) } J : + 6NH, (12.44)
12.7 On the basis of the structures of [Cs(C222)]Cs~ and |Cs( I6C6),J f Cs", the ccside anion
has been suggested as the largest known monoatomic ion. What other candidates for this
distinction are possible? Do you think that any of them will prove to be larger than Cs"?
12.8 Why is the product of Eq. 12.42 [K(I8-crown-6fNa~ instead of |Na(l8-crown-6)] H K~?
Use a Bom-Haber cycle to analyze the effect of the various factors responsible.
12.9 Draw out all the isomers, geometric and optical, of the following: (Co(cn).CI-.],
[Co(en),(NHj)CI] 2+ . [Co(en)(NH,) 2 CI 2 ].
504
12 • Coordination Chemistry: Structure
Fig. 12.32 Molecular
structure of the seven-
coordinate complex of iron
involving 2,13-
dimethyl-3.6,9,12,18-
pentaazabicyclol-[12.3.1]-
octadeca-I(l8),2.12,14,16-
pentaene and two axial
SCN~ ligands. Hydrogen
atoms omitted. [From
Fleischer, E.; Hawkinson,
S. J. Am. Chem. Soc. 1967,
89. 720-721. Reproduced
with permission.]
Fig. 12.33 The molecular
unit of tris(diphenylpro-
panedionatolaquahol-
mium(III) projected down
the threefold axis. The
water molecule is directly
above the holmium atom
but has been displaced
slightly in this drawing to
show the structure better.
[From Zalkin, A.;
Templeton, D. H.;
Karraker. D. G. Inorg.
Chem. 1969. 8. 2680-2684.
Reproduced with
permission.]
capped trigonal prism in which a seventh ligand has been added to a rectangular face
(Fig. 12.34). They are of comparable stability and easily interconvertible. Therefore
there are also many intermediate cases, and the situation is reminiscent of five-
coordinate geometries.
In many of these complexes the requirements of polydentate ligands favor coordi¬
nation number 7. Thus it is not difficult to see the effects of five macrocyclic and
coplanar nitrogen atoms in Fig. 12.32 on the resulting pentagonal bipyramidal struc¬
ture. In some cases, even unfavorable interactions may be forced by the ligand
geometry. For example, in one type of seven-coordinate complex, it appears as
though the seventh coordination, forced by the geometry of the other six coordinating
atoms, might in some cases better be considered as an "antibond” rather than a bond.
This effect is seen in the series of [M(py 3 tren)] 2+ complexes (M = Mn, Fe, Co, Ni,
Cu, Zn; py 3 tren = [C 5 H 4 NCH=NCH 2 CH 2 ] 3 N) in which the three imine nitrogen
Coordination Number 7
505
Fig. 12.34 Structure of the
heplafluoroniobate( V)
anion. [From Hoard, J. L.
J. Am. Chem. Soc. 1939,
61. 1252. Reproduced with
permission.]
atoms are at the vertices of one equilateral triangle and the three pyridine nitrogen
atoms are at the vertices of another. 63 One could refer to this arrangement as an
approximate octahedron, but since the metal ion is closer to the imine nitrogens than
to the pyridine nitrogens, it is best to refer to it as a trigonal antiprism (Fig. 12.35). 64
Even that is not a perfect description because there is some trigonal distortion
(rotation of opposing triangular faces tending toward a trigonal prism) in each com-
(a) tb • (c)
Fig. 12.35 [M(pyjtren)] 2 complexes: (a) The py_,tren ligand: N(CH 2 CH : N=CHC J H,|N) 3 ; the nitrogen atoms are labeled
N py . the pyridine nitrogen. Nj. the imine nitrogen, and N 7 , the seventh or unique nitrogen atom; (b) a diagrammatic
representation of the [M(pyjtren)]‘ + complex, viewed down the threefold axis; (c) the molecular structure of the zinc(ll)
complex, viewed perpendicular to the threefold axis. [Courtesy of E. C. Lingafelter.J
63 Kirchner, R. M.; Mealli, C.; Bailey, M.; Howe. N.; Torree, L. P.; Wilson, L. J.; Andrews, L. C.;
Rose, N. J.; Lingafelter. E. C. Coord. Chem. Rev. 1987, 77. 89-163.
64 Note that an octahedron is a special case of trigonal antiprism in which all edges have the same
length. However, in contrast to the usual choice of coordinates (x, y. and z lie on the fourfold axes
for an octahedron), the highest fold axis in a trigonal antiprism is, in general, the threefold axis
which is assigned to the z axis. Note that (except for the choice of coordinates and corresponding
labels) the trigonal splitting closely resembles a pscudooctahcdral splitting (Chapter 11): three lower
orbitals, a, and \e (r 2 „ in Oh) and two higher orbitals, 2e {e„ in O/,). The effect of trigonal
antiprismatic distortion on the relative energy of the a t orbital with respect to the \e and 2c orbitals
will depend upon the interaction of the </ c : orbital with the N 7 lone pair orbital. It is shown above as
lower than le, but it has been suggested that in some of these complexes it may lie between te and
2e. As in the octahedral case, jt bonding can be added to the cr-only system.
528
12 Coordination Chemistry: Structure
differing stability constants of the [M(ligand)] + complexes, it is possible to form mixed
metal compounds:
Na + K + l8-crown-6- [K(l8-crown-6)] + + Na~ (12.42)
As expected, all of these systems are strongly reducing and tend to decompose on
exposure to air and moisture, so very careful work is necessary to study them. In a
few cases, single crystals have been grown and structures determined
crystallographically. 108 Thus crystalline [Na(macrocycle)] + Na~ consists approx¬
imately of closest packed, large, complex cations with sodide anions in the octahedral
holes (Fig. 12.50a). The alkalide ions can be found and measured; as expected they are
considerably larger than the alkali metal cations. In fact, the Cs - anion is the largest
known monoalomic ion with a radius of about 310-350 pm. In one salt there is
anion-anion contact giving an effective radius of 319 pm. In addition to the usual
difficulties of assigning an exact radius to an ion, the ceside anion has the added
property that it is not only the largest ion, but also the most polarizable.
The structure of [Cs(18-crown-6),]' f e _ has been determined, i" 9 Because the
electride anions are extremely poor scatterers compared to the large cesium cation
(and to a lesser extent the C and 0 atoms of the crown ether), the structure has the odd
appearance of complexed metal cations with no corresponding anions (Fig. 12.50b).
However, the most likely position of the electrons can be inferred from the presence
of cavities of 240-pm radius; presumably the electrons are located in these cavities.
One final example of macrocyclic complexation will be given: From the argu¬
ments presented in Chapter 9, the fluoride ion, F - , should be a strong base and
(a) <b)
Fig. 12.50 (a) Packing of [NaC222f f (large circles) and Na - (small solid circles) ions in
(cryptand)sodium sodide. (b) ORTEP stereo packing diagram of bis(18-crown-6)cesium
electride. The anionic hole centers are indicated by the symbol O. [From Tehan, F. J.;
Barnett, B. L.; Dye, J. L. J. Am. Chem. Sac. 1974, 96, 7203; Dawes. S. B.; Ward, D. L.;
Huang, R. H.; Dye, J. L. J. Am. Chem. Soc. 1986, 108. 3534-3535. Reproduced with
permission.]
">s For example, ]K(C222)]e": Ward, D. L.; Huang, R. H.; Dye, J. L. Ada Crystallgr Sed. C:
Cryst. Struct. Commun. 1988, C44, 1374-1376. [Cs(C222)f Cs" and [Cs(l8C6) 2 fCs - : Huang,
R. H.; Ward. D.L.; Kuchenmeister, M. E.; Dye, J. L. J. Am. Chem. Soc. 1987, 109. 5561-5563.
»» Dawes. S. B.; Ward. D. L.; Huang. R. H.; Dye. J. L. J. Am. Chem. Soc. 1986. 108. 3534-3535.
The Chelate Effect
529
Tabl e 12.7 _
Thermodynamic
contributions to the
macrocyclic effect in
complexes of 18-crown-6
and pentaglyme,
CH 3 (OCH 2 CH 2 ) s OCH 3 , in
methanol"
nucleophile. Normally, however, it does not show these expected properties because
its very basicity attracts it to its countercation so strongly that it is ion-paired in
solution and not free to react. However, addition of, for example, crown-6 to a
solution of potassium fluoride in benzene increases the solubility tenfold and also
increases the nucleophilicity of the fluoride ion. We shall encounter this phenomenon
again in Chapter 15, which covers organometallic chemistry.
Earlier we saw that chelating ligands form complexes of greater stability than
those of unidentate ligands. This greater stability was attributed primarily to entropy
effects but enthalpy effects are of some importance. Macrocyclic ligands are even
more stable than open-chain chelating ligands. A thermodynamic comparison of
!8-crown-6 complexes of Na + , K + , and Ba 2+ with those of pentaglyme,
CH 3 (OCH 2 CH-,) 5 OCH 3 , is shown in Table 12.7. The additional stability is primarily an
enthalpy effect due to preorganization of the macrocycle. This is not to say that the
conformation of l8-crown-6 is the same as that found in the complex:
Nevertheless, the energy required to rearrange the macrocyclic ligands for complex
formation is less than the energy required to rearrange pentaglyme into a suitable
conformation."" In addition to their direct structural relationship to biological mole¬
cules, macrocycles such as the polyethers may provide clues to the discrimination
shown by biological tissues toward various ions. This selectivity provides the so-
Na"
K-
Sa 2+
log a:,
18-crown-6
4.36
6.06
7.04
pentaglyme
1.44
2.1
2.3
log K difference
2.92
3.96
4.74
A H
18-crown-6
-35.1
-56.0
-43.5
pentaglyme
-16.7
-36.4
-23.8
AH difference
-18.4
-19.6
-19.7
AS
18-crown-6
-33
-71
' -13
pentaglyme
-29
-84
-33
AS difference
-4
13
20
a Modified from Hancock, R. D.; Martell, A. E. Comments Inorg. Chem. 1988, 6, 237-284. Free
energy and enthalpy changes are expressed in kJ mol -1 . Entropy changes arc expressed in J
mol -1 K - '.
110 Cram, D. J. Angew. Chem. Int. Ed. Engl. 1988. 27. 1009-1020.
506
12 Coordination Chemistry: Structure
plex. The tr-only ligand field splitting pattern for a trigonal antiprism (D 3 ) consists of
three levels, the a x (cb ), \e (d x z_ v 2 , d xy ). and the 2e (d xz , d yz ):
The lone pair of the tertiary amine nitrogen atom (N 7 ) capping the trigonal antiprism is 'JgS
directed at the center of a trigonal antiprismatic face and thus interacts directly with 3
the a, orbital.
Figure 12.36 illustrates the metal-nitrogen distances in this series of complexes.
The upper trace represents the M — N 7 distance. An increase in electrons in orbitals
directed at any of the seven nitrogen atoms causes an increase in bond length. J
Specifically, in the case of the Fe 2 h complex, the only low spin complex in the series,
there is a dramatic decrease in one distance and concomitant increase in the other.^
The low spin («,) 2 (le) 4 configuration maximizes electron density toward the axial j l8|
nitrogen while minimizing it in the direction of the trigonal antiprismatic nitrogen aBa
atoms.
The M—N 7 distances for the nickel, copper, and zinc complexes (323, 311, and jjffll
301 pm, respectively) are about 10 pm longer than the sum of the van der Waals radii
,r Mvi.w + r N v „ w )’ w h' c h are 315, 295, and 295 pm, respectively (see Table 8.1). The .
Fig. 12.36 Effect of orbital occupancy on metal-nitrogen distances. Upper part of diagram
gives the M — N 7 distances (shown as ■). The lone pair of the tertiary amine atom (N 7 ) is
directed at the a, orbital on the metal ion. The lower part of the diagram gives the M—N,
and M—N 2 weighted mean bond distances. The inline and pyridine nitrogen atoms are
directed at the 2e orbitals on the metal ion. The distances (shown as A) for inline nitrogen
atoms, N|, are slightly shorter and thus are located just below the values (shown as O) for
pyridine nitrogen atoms, N : . Dashed lines (—) represent a constant electron configuration,
while the dotted line (••■•) represents an increasing number of electrons in a given type of
orbital on the metal ion. The orbital occupancy, (<i,)'( lt’) , (2e) : . is given under each metal ion
as .v, y, c. [Modified from Kirchncr. R. M.; Mealli, C.; Bailey. M.; Howe. N.; Torrce. L. P.;
Wilson. L. J.; Andrew, L. C.: Rose, N. J.: Lingafcller. E. C. Coord. Chem. Rev. 1987. 77,
89-163. Reproduced with permission.!
orbital
occupancy
Coordination Number 8
507
other six M—N bond distances (201-223 pm) are considerably shorter than the van
der Waals radii (an X—Y distance less than the van der Waals sum is often the
accepted criterion for bonding). In spite of constraints that would be expected to
restrict the movement of N 7 away from the metal, it appears that for most of these
complexes, the M—N 7 distance is too long to be a true bond. For Mn(II) and Co(II),
the M—N 7 distances are shorter than the van der Waals interactions, implying a weak
bond. This is not an unexpected result for high spin Mn(II) because the orbital is
only half-filled. In the case of Co(II), it has been argued that the a , and le levels have
been interchanged, resulting in just one electron in the a, orbital in this complex as
well.
At the same time the bond angles (C—N—C) at the N 7 position vary from 112°
(~sp 3 as expected for an amine ligand) in the manganese complex (where repulsion is
least) up to a maximum value of 120° in the iron complex with maximum repulsions.
The tertiary amine nitrogen atom (N 7 ) corresponds to a three-ribbed umbrella that has
been inverted by the wind (the handle is the lone pair directed at the metal). As the
and \e levels fill, the repulsions increase, the metal-nitrogen distance increases, and
the umbrella begins to flatten:
112 °
Although both geometrical and optical isomerism are in principle possible in
seven-coordinate complexes, no examples are known. Note, for example, that the
py 3 tren complexes must be optically active (see Fig. 12.35b), subject of course to
kinetic stability with respect to racemization.
Although coordination number 8 cannot be regarded as common, the number of
known compounds has increased rapidly in recent years, so that it is now exceeded
only by four- and six-coordination. The factors important in this increase can be
traced largely to improved three-dimensional X-ray techniques and to increased inter¬
est in the coordination chemistry of lanthanide and actinide elements (see Chapter 14).
Two factors are important in favoring eight-coordination. One is the size of the
metal cation. It must be sufficiently large to accommodate eight ligands without undue
crowding. Relatively few eight-coordinate complexes are known for the first transition
series. The largest numbers of this type of complex are found for the lanthanides and
actinides, and it is fairly common for zirconium, hafnium, niobium, tantalum, molyb¬
denum, and tungsten. A corollary is the requirement that the ligands be relatively
small and electronegative. The commonest ligating atoms are carbon, nitrogen, oxy¬
gen, and fluorine. The second factor is the oxidation state of the metal, a high formal
65 Kepert. D. L. In Comprehensive Coordination Chemistry-. Wilkinson. G.: Gillard. R. D.. Mc-
Cleverty, J. A., Eds.; Pergamon: Oxford, 1987; Vol. I, pp 83-95.
Coordination
Number 8 65
• Coordination Chemistry: Structure
Fig. 12.49 Structure of the macrocyclic complex of potassium with l8-crown-6. The planes
bear a 50-pm grid and the lighting source is at infiinity so that shadow size is meaningful.
This drawing was made with the kanvas computer graphics program. This program is based
on the program schakal of E. Keller (Kristallographisches Institut der Universitat Frieburg,
Germany), which was modified by A. J. Arduengo. Ill (E. I. du Pont de Nemours & Co..
Wilmington, DE) to produce the back and shadowed planes.
compound Cs ' Au". Its existence might at first appear to be improbable since it is an
ionic compound between two metals rather than between a metal and a nonmetal. Yet
inspection of the ionization energy of cesium and the electron affinity of gold (or
merely comparing the Pauling electronegativity of gold, 2.54. with that of iodine. 2.66)
sparks curiosity: It should be investigated. Indeed, based on the suggestion of the late
Sir Ronald Nyholm, the first edition of this book (1972), as well as this edition
(Footnote 4, Chapter 4), both mention the possibility of such a bond. The experimen¬
tal stumbling block is obvious, however: If you mix two metals, how do you determine
that you do indeed have an ionic compound and not an alloy? When cesium was
allowed to react with gold, the conductivity in the melt was characteristic of an ionic
compound. Further evidence was desirable, however. Two approaches were used.
One was to take advantage of the solubility of alkali metals in liquid ammonia, the
strong reducing power of the free electron, and the stability of reduced species in this
medium. (Note that Au - should be a powerful reductant.) The second was to stabilize
metals, such as sodium, that might not otherwise react (recall that the ionization of
any metal, even sodium or cesium, is endothermic), with macrocyclic polyethers to
form [Na(macrocycle)]Au _ salts. A combination of these techniques with physical
The Chelate Effect 527
methods such as nonaqueous electrochemistry and those showing the anomalously
low ESCA binding energy and the Mossbauer chemical shift were used to characterize
the - 1 gold.ms Unfortunately Sir Ronald did not live to see his prediction verified: he
died in an automobile accident in 1971.
In a similar manner the so-called Zintl salts composed of alkali metal cations and
clusters of metals as anions (see Chapter 16) were known in liquid ammonia solution
but proved to be impossible to isolate: Upon removal of the solvent they reverted to
alloys. Stabilization of the cations by complexation with macrocyclic ligands allowed
the isolation and determination of the structures of these compounds.
This general trait of crown ethers and cryptands (to be discussed later) to stabilize
alkali metal salts has been extended to even more improbable compounds, the al-
kalides and electrides, which exist as complexed alkali metal cations and alkalide or
electride anions. For example, we saw in Chapter 10 that alkali metals dissolve in
liquid ammonia (and some amines and ethers) to give solutions of alkali electrides: l0( >
M
M + + e'
(12.38)
However, the situation is somewhat more complicated than Eq. 12.38 would indicate,
because the electrons can react further with the metal to form alkalide ions:
M + e -» M (12.39)
Thus, in general, there are alkali metal cations, alkalide anions, and electride anions
(and perhaps other minor species) in an equilibrium mixture dictated by various
energetic factors. Note, for example, that because of the excellent solvating of the
electride ion by liquid ammonia, the alkalide ions are not favored in these solutions,
and they are somewhat atypical. By using methylamine and ether solvents, even
adding solvents of low polarity such as /i-penlane, the equilibria can be shifted and
crystals sometimes grown. 10 7 The affinity of alkali metal ions for crown ethers and
cryptands causes cationic complexes to form readily in solution. (See Eq. 12.37
above.) This, too will affect the equilibrium, with Eq. 12.38 being shifted to the right
and, therefore. Eq. 12.39 being shifted to the left, by the addition of these ligands. A
rough generalization can be made that excess ligand will favor the formation of the
electride inasmuch as all of the metal may be complexed as the cation.
Cs + excess l8-crown-6 - [Cs(l8-crown-6) 2 ] + + e" (12.40)
On the other hand, a mole ratio of 2:1, mctakligund, tends to favor the alkalide:
2Na + l8-crown-6 - (Na(l8-crown-6)] + + Na" (12.41)
Aside from the effects of ligand stoichiometry and the nature of the solvent, there are
also differences in stability of the alkalide ions. The sodide anion is the most stable,
and the ceside ion the least. Because of differential stabilities of the alkalide ions and
105 Batchelor. R. J.: Birchall. T.: Burns, R. C. Inorg. Chem. 1986, 25, 2009-2015, Jaganalhan, R.;
Wallace. D. W.; Lagowski. J, J. Inorg. Chan. 1985. 24. 113. Knccht. J.; Fischer. R.; Overhof. H.;
Hcnsel. F. Chem. Commun. 1978, 905-906. Peer, W. J.: Lagowski, J. J. ./. Am. Chan. Soc. 1978,
100. 6260-6261. For further discussion of the oxidation states of gold, see Chapter 14.
"• In Eq. 10.22 the electron was written as |c(NH}),r to emphasize the solvation effects. However,
just as in water Na(H 2 0), may be written as Nil or more simply as Na. so also in liquid
ammonia the solvation may be indicated by the subscript "am,'' e,7 m , or omitted altogether.
107 For an extensive review of this chemistry, see Dye, J. L. Prog. Inorg. Chem. 1984, 32. 327-441.
For a shorter, more recent review, see Dye. J. L. Sci. Am. 1987, 257(3), 66-75.
508
12 Coordination Chemistry: Structure
oxidation state favoring eight-coordination. Jhis requirement arises out of the elec
troneutrality principle. The formation of eight dative a bonds to a metal in a low
oxidation state would result in excess electron density on the metal. The common
oxidation states are thus 4-3 or greater, resulting in electron configurations with few
remaining d electrons such as d°, dd 2 . 66
There are several coordination polyhedra available for eight-coordination. The
most regular, the cube, is almost never found in discrete complexes but occurs in
lattices such as CsCl. 67 The two common structures are the trigonal dodecahedron
(Fig. 12.37a) and the square antiprism (Fig. 12.37b). Both may be considered to be
distortions of the simple cube resulting in reduced ligand-ligand repulsions. The
square antiprism has slightly less ligand-ligand repulsion than the dodecahedron
Other geometries are known, but these two are the most important.
From a valence bond point of view, the formation of both the dodecahedron and
the square antiprism can arise from sp^d hybridization. The necessity of using four d
Fig. 12.37 Distortions of
the cube to form: (a) the
trigonal dodecahedron: (b)
the square antiprism. Note
the similarity of the
dodecahedron, viewed
down the puckered CDHG
face ( C 2 axis) and the
antiprism, viewed down the
ABCD face (5-axis). [From
Lippard, S. J. Prog. Inorg.
Chem. 1968.5. 109-193.
Reproduced with
permission.)
i i
c, s.
'■' The picture of the formation of. for example. [Mo(CN)l' 1- . as a combination of Mo‘(d') + SCN
has little physical meaning and is merely a bookkeeping device, but hardly more so than Co +■
6NHj discussed extensively earlier.
47 There appears to be at least one exception to the general rule: (Et J N)4[U(NCS) s l has cubic anions in
the solid, but that structure is not retained in solution (Countryman. R.; McDonald, R. S. J. Inorg.
Nucl. Chem. 1971 .33. 2213-2220). Note that the cube is a special case of the square prism, and like
the trigonal prism with respect to the octahedron, the ligand-ligand interactions are greater com¬
pared to the square 3ntiprism.
orbitals in the hybridization for the ligand bonding rationalizes the common occur¬
rence of d°.d', and d 2 configurations from a valence bond model. Crystal field theory
and molecular orbital theory give a similar picture in that both the square antiprism
and dodecahedron give a nondegenerate lower level (to accept the one or two d
electrons) and the remaining four metal d orbitals (or molecular orbitals derived
principally from metal d orbitals) lie at higher levels (Fig. 12.38). The LFSEs of both
structures are comparable, and the choice between the two is a delicate balance ot
forces, as was the case for five-coordination. Although there is no completely satisfac¬
tory treatment of the problem of proper selection of geometry for eight-coordination,
both VB and MO calculations can correctly predict adjustments of a few degrees
depending upon the electron configuration of Ihe metal.
There are extensive possibilities for the formation of geometric and optical iso¬
mers in eight-coordinate complexes. Thus far. apparently only one pair has been
completely characterized. The diglyme [= di(2-methoxyethyl)ether) adduct of sa¬
marium iodide, SmI : [0(CH : CH ; OCH 3 ) : j : . has been isolated in both cis and trans
forms. The trans complex (Fig. 12.39a) has a center of symmetry. Thus, the 1—Sm—1
angle is exactly 180°, and the molecule is a bicapped trigonal antiprism. The cis isomer
(Fig. 12.39b) has the lower symmetry of a distorted dodecahedron with I—Sm—I
angles of 92°. 0 »
There are few structures known with coordination numbers larger than 8. The exis¬
tence of coordination number 12 in some crystal lattices was mentioned above. Dis¬
crete nine-coordinate structures are known for complexes such as [Ln(H,0) 9 ] 3 '‘‘ and
for the hydride complexes [MH V ]“ (where M = Tc or Re). These structures are
formed by adding a ligand to each of the rectangular faces of a trigonal prism (Fig.
12.40).
Burden. J. K.: Hoffmann. R.; Fay. R. C. Inorg. Chem. 1978. 17. 2353-2568. Mingos. D. M. P.;
Zhcnyang. L. Slrucl. Bonding ( Berlin ) 1990 , 72. 94-98.
69 Sen. A.: Chcbolu, V.: Rhcingold, A. L. Inorg. Chem. 1987. 26. 1821-1823.
70 Kepert. D. L. In Comprehensive Coordination Chemistry. Wilkinson. G.. Gillard. R. D.. Mc-
Clcverty. J. A.. Eds.: Pergamon: Oxford. 1987: Vol. I. pp 95-101.
Higher Coordination
Numbers 70
524
12-Coordination Chemistry: Structure
The Chelate Effect
525
With trivalent metals, acetylacetone thus forms neutral tris complexes such as
[Al(acac) 3 ], [Ti(acac) 3 ], [Cr(acac) 3 ], and [Co(acac) 3 ]. As a result of resonance, the two
M—O bonds in each of these complexes are equal in length, as are the two C—O and
the two ring C—C bonds, giving a symmetric structure (only one ring shown):
H 3 % /C V CH3
H,C
CH CH,
O
(12.36)
Ligand-metal -tr bonding enhances the delocalization of electrons compared to that in
the free enolate, producing some resonance stabilization.
An interesting example of at least partial destruction of resonance from
Jahn-Teller distortion is given by bipyridinebis(hexafluoroacetylacetonato)copper(II)
(see Fig. 11.51 and accompanying discussion of the Jahn-Teller effect). As a result of
this distortion, the two Cu—0 distances are not equivalent (197 vs. 230 pm) and
presumably the ir bonding is not equivalent. Therefore one resonance form is favored
over the other, and as a result there is an alternation of bond lengths throughout the
ring (Fig. 12.48).
The chelate effect is amplified in the case of polydentate ligands that form sev¬
eral rings with a single metal atom. The extreme of this form of stabilization is
found with hexadentate ligands such as ethylenediaminetetraacetate (edta),
COOCCH 2 ) 2 NCH 2 CH 2 N(CH 2 COCT) 2 , which has six ligating atoms.
We have seen that chelate rings obey much the same type of steric requirements
with respect to conformations as do organic rings. Unlike organic ring systems,
maximum stability in chelate rings usually arises from five-membered rings because
the metal atom is larger than a carbon atom and the bond angles at the metal
(L—M—L) will be 90° in square planar and octahedral complexes in contrast to an
optimum angle of I09i° for tetrahedral carbon. For rings exhibiting significant reso¬
nance effects, such as acetylacetonates, six-membered rings are quite stable. Larger
and smaller chelate rings are known, but they are not nearly so stable as the five- and
six-membered species.
Macrocycles 101
An area of particular research interest in recent years has been the construction of
planar, macrocyclic ligands. These are special types of polydentate ligands in which 1
the ligating atoms are constrained in a large ring encircling the metal atom. Examples
are polyethers in which the ether oxygen atoms, separated by two methylene groups
each, lie in a nearly planar arrangement about the central metal atom (Fig. 12.49) and
the remainder of the molecule lies in a “crown" arrangement, hence the name "crown
ethers." 102 All of the oxygen atoms “point" inward toward the metal atom, and these
macrocycles have the unusual property of forming stable complexes with alkali
metals. This exceptional stability has been attributed to the close fitting of the alkali
metal ion into the hole in the center of the ligand. However, some data seem to
contradict this simple model. Although calculations indicate that the Li' 1 ' should
preferentially fit crown-4, in solution crown-4, crown-5, crown-6, and crown-7 all
prefer K + . Hancock 103 has suggested that this may arise because the five-membered
ring formed when any of the above crown ethers binds to an alkali metal cation best
fits the size of K + .
Gas-phase studies 104 show that crown-5 prefers Li + more than the other alkali
metal cations:
crown-5: Li + » Na + > K + > Cs +
On the other hand crown-6 shows more affinity for Na + and crown-7 prefers K + :
crown-6: Na + £: K + > Li + > Rb + > Cs +
crown-7: K + > Na + > Rb + > Li + > Cs +
The difference in affinities shown in gas and solution phases suggests that solvent
effects are quite important. These ligands have the unusual ability to promote the
solubility of alkali salts in organic solvents as a result of the large hydrophobic organic
ring. For example, alkali metals do not normally dissolve in ethers as they do in
ammonia (see Chapter 10). but they will do so if crown ligands are present:
K + crown-6 [K(crown-6)] + + [e(solvent)J“ (12.37)
The ability to complex and stabilize alkali metal ions has been exploited several
times to effect syntheses that might otherwise be difficult or impossible. Consider the
Fig. 12.48 Bond lengths in
the acetylacetonate ring of
[Cu(py)(hfa) 2 ]. [From
Veidis, M. V.; Schreiber,
G. H.; Gough, T. E.;
Palenik, G. J. J. Am.
Chem. Soc. 1969, 91,
1859-1860. Reproduced
with permission.)
101 This is an area in which three chemists recently shared the Nobel Prize in chemistry (1987): C. J.
Pedersen (Footnote 104). J. M. Lehn (Footnote 112), and D. J. Cram (Footnote 110). The materials
cited in these references contain their Nobel laureate addresses. See also Izatt, R. M.; Pawlak, K.;
Bradshaw. J. S.; Bruening, R. L. Chem. Rev. 1991. 91, 1721-2085.
i °2 Pedersen, C. J. Angew. Chem. Ini. Ed. Engl. 1988. 27. 1021-1027. The nomenclature of the organic
ring systems is complex (the ligand in Fig. 12.49 is 1,4,7,10,13,16-hexaoxacyclooctadecane), and
Pedersen condensed this to " !8-crown-6," in which the number 18 refers to the macrocyclic ring
size, "crown" is a trivial generic name for the class, and 6 refers to the number of oxygen atoms.
This name is sometimes further abbreviated to I8C6 in formulas. For purposes of the present
discussion, these compounds may be referred to as “crown-4," "crown-5," and "crown-6," with
the implication that the ring size is equal to three times the number of oxygen atoms. The
cryptands, to be discussed shortly, may be abbreviated to the point of C222, standing for "cryp-
tand with three pairs of oxygen atoms on the ‘scams of the football"' (see page 530).
103 Hancock. R. D. J. Chem. Educ. 1992. 69, 615-621; Perspectives in Coordination Chemistry-,
Williams, A. F.; Floriani, C.; Merbach, A. E., Eds.; VCH: Weinheim, 1992; pp 129-151.
104 Maleknia. S.; Brodbelt, J. J. Am. Chem. Soc. 1992. 114, 4295-4298.
12 • Coordination Chemistry: Structure
(b)
Fig. 12.39 Stcreoviews of the structure of (a) fra/is-SmI 2 (0(CH;CH : 0CHj);]2 and (b) cis-
SrnljI0(CH 2 CH20CHj)2] 2 . [From Sen, A.; Chcbolu, V.; Rheingold, A. L. Inorg. Chem.
1987. 26, 1821-1823. Reproduced with permission.]
Fig. 12.40 Molecular
structure of the [ReH 9 ] 2_
anion. [From Abrahams, S.
C.; Ginsberg, A. P.. Knox.
K. Inorg. Chem. 1964. J.
558-567. Reproduced with
permission.]
H
No compounds are known with ten or more distinct ligands (i.e., nonchelate
slructures); however, a few ten-coordinate chelate structures have been described.
One possible structure is a '"double trigonal bipyramid" with bidentate nitrate or
carbonate ions at each of the TBP sites (Fig. 12.41a). Coordination numbers as high as
12 are also known with six bidentate nitrate ligands along the edges of a
dodecahedron, but these complexes are rare (Fig. 12.41b).
Generalizations about Coordination Numbers
511
Fig. 12.41 (a) Structure of the pentanitratocerate(III) anion. Each nitrate ion may be thought of as occupying an apex of a
trigonal bipyramid. The resulting coordination number is 10 since each nitrate ion is bidentate. The view is down the
principal axis of the trigonal bipyramid and the axial nitrogen atoms as well as the central cerium atom are partially
obscured, (b) Structure of the hexanitratocerate(lll) anion. Each nitrate ligand is bidentate to give a coordination number of
12. [From Al-Karaghouli, A. R.; Wood, J. S. Chem. Commtm. 1970. 135-136. Beineke. T. A.; Delgaudio, J. Inorg. Chem.
1968, 7, 715-721. Reproduced with permission.)
Generalizaiions What generalizations can be made concerning high and low coordination numbers? All
—--—-—-;- generalizations have exceptions, but we can list the following trends. The factors
ab out Coordination favoring low coordination number are:
Numbers
-I. Soft ligands and metals in low oxidation states. Electronically, these will
favor low coordination numbers because extensive rr bonding will compensate
in part for Ihc absence of additional (potential) a bonding. Metals in low
oxidation stales are electron rich and do noi seek additional contributions of
electron density from additional ligands.
Large, balky ligands. If the complex is coordinatively unsaturated (in an
electronic sense), then steric hindrance may prevent additional ligands from
coordinating to the metal.
Counterions of low basicity. Any cationic complex with a low coordination
number is a Lewis acid potentially susceptible to attack and coordination by its
anionic counterion. For this reason, anions of low basicity and coordinating
ability are chosen for counterions. Among the oxyanions. nitrate and perchlo¬
rate have a long history of use. Both show some coordinating ability (cf. the
nitrate complexes discussed in the preceding section), and as strong oxidizers
their presence with organic ligands can be potentially hazardous. The trifiate
anion, CFjOSOf obviates the explosive hazard and also reduces the tendency
toward rotational disorder in crystals. Fluoride adducts of strong Lewis fluoro
acids such as BF.^, PFJ~, and SbF^are frequently chosen as counterions be-
522
12> Coordination Chemistry: Structure
minecobalt(III) hexacyanochromate(III), [Co(NH 3 ) 6 ][Cr(CN) 6 ], and its coordination
isomer, [Cr(NH 3 ) 6 ][Co(CN) 6 ], are known. Another example is [Cu(NH 3 ) 4 ][PtCI 4 ] and
[Pt(NH 3 ) 4 ][CuCI 4 J in which the isomers differ in color (as a result of the d 9 Cu 2+
chromophore), being violet and green, respectively. There are many cases of this type
of isomerism.
A special case of coordination isomerism has sometimes been given the name
“polymerization isomerism" since the various isomers differ in formula weight from
one another. However, the term is unfortunate since polymerization is normally used
to refer to the reaction in which a monomeric unit builds a larger structure consisting
of repeating units. The isomers in question are represented by compounds such as
[Co(NH 3 ) 4 (N0 2 ) 2 ][Co(NH 3 ) 2 (N0 2 ) 4 ], [Co(NH 3 ) 6 ][Co(N0 2 ) 6 ], [Co(NH 3 ) 5 NO,]
[Co(NH 3 ) 2 (N0 2 ) 4 J 2 , [Co(NH 3 ) 6 ][Co(NH 3 ) 2 (N0 2 ) 4 ] 3 , [Co(NH 3 ) 4 (N0 2 ) 2 ] 3 (Co(N 0 2 ) 6 ],
and [Co(NH 3 ) 5 N0 2 ] 3 (Co(N 0 2 ) 6 ] 2 . These all have the empirical formula
Co(NH 3 ) 3 (N0 2 ) 3 , but they have formula weights that are 2, 2, 3, 4, 4, and 5 times this,
respectively.
Reference has been made previously to the enhanced stability of complexes contain¬
ing chelate rings. This extra stability is termed the chelate effect. It is chiefly an
entropy effect common to all chelate systems, but often additional stabilization results
from enthalpy changes. Entropy changes associated with chelation are complex. 98
With regard to translational entropy there are two points of view which are essentially
equivalent in that they are both statistical and probabilistic in nature. They therefore
relate to the entropy of the system but they look at the problem from somewhat
different aspects. One is simply to consider the difference in dissociation between
ethylenediamine complexes and ammonia complexes, for example, in terms of the
effect of the ethylenediamine ring (the electronic effects of the nitrogen atoms in
ethylenediamine and ammonia are similar). If a molecule of ammonia dissociates from
the complex, it is quickly swept off into the solution, and the probability of its ever
returning to its former site is remote. On the other hand, if one of the amine groups of
ethylenediamine dissociates from a complex, the ligand is retained by the end still
attached to the metal. The nitrogen atom can move only a few hundred picometers
away and can swing back and attach to the metal again. The complex has a smaller
probability of dissociating and is therefore experimentally found to be more stable
toward dissociation.
A more sophisticated approach would be to consider the reaction:
[Ni(NH 3 ) 6 ] 2+ + 3en [Ni(en) 3 ] 2+ + 6NH 3 (12.34)
in terms of the enthalpy and entropy. Since the bonding characteristics of ammonia
and ethylenediamine are very similar 99 we expect A// for this reaction to be small. To
a first approximation the change in entropy would be expected to be proportional to
1,7 Hancock, R. D.; Martcll. A. E. Comments Inorg. Client. 1988, 6. 237-284.
98 Chung, C.-S. J. Chem. EJttc. 1984, 61. 1062-1064. Meyers, R. T. Inorg. Client. 1978, 17. 952-958.
99 Ethylenediamine and ammonia arc almost identical in field strengths (their/ factors differ by less
than 3%, Chapter It), and so the LFSE for complexes with the same metal ion will be almost
identical. However, ethylenediamine is also a stronger base than ammonia from the inductive effect
of the methylene groups. In addition there may be enthalpy differences in the form of ring strain or
other steric effects. Sec Table 12.6 and the following discussion.
The Chelafe
Effect 97
The Chelate Effect
523
Table 12.6
Thermodynamic contributions to the chelate effect in complexes of nickel(ll) and coppor(ll) a
Ammonia Ethylenediamine _ Chelate effect
complexes
AG
AH
AS
complexes
AG
AH
AS
AG
AH
AS b
33.4n
[Ni(NH 3 ) 2 (H 2 0)J 2+
-29.0
-33
-12
[Nien(H,0) 4 ] 2+
-41.9
-38
17
-12.9
-5
29
33
[Ni(NH 3 ) 4 (H 2 0)->] 2+
-46.3
-65
-63
[Ni(en) 2 (H,0) 2 ] 2+
-77.2
-77
12
-30.9
-11
74
67
[Ni(NH 3 ) 6 ] 2+
-51.8
-100
-163
[Ni(en) 3 ] 2+
-101.8
-117
-42
-50.0
-17
121
100
[Cu(NH 3 ) 2 (H,0) 4 ] 2+
-44.7
-46
-4
[Cuen(H,0) 4 ] 2+
-60.1
-55
25
-15.5
-8
29
33
[Cu(NH 3 ) 4 (H 2 0),] 2+
-74.2
-92
-58
[Cu(en) 2 (H 2 0) 2 ] 2+
-111.8
-107
29
-37.6
-15
88
67
“ Modified from Hancock, R. D.; Martell, A. E. Comments Inorg. Chem. 1988, 6, 237-284. Free energy and enthalpy changes arc ex¬
pressed in kj mol -1 . Entropy changes are expressed in J mol -1 K _ 1
h Entropies of chelation should be compared with 33.4n (n = number of chelate rings) based on AS = nil In 55.5.
the difference in the number of particles present at the beginning and end of Ihe
reaction. The reaction proceeds to the right with an increase in number of particles,
and hence translational entropy favors the production of the chelate system instead of
the hexaammine complex. 100 In the replacement of water molecules by chelates, the
increase in number of molecules in solution causes an increase of entropy given by
AS = nR In 55.5 = 33.4/1 J mol -1 K~' where n is the number of chelate rings,
contributing 10.0 kJ mol -1 to the free-energy stabilization of the complex at 300 K
for each chelate ring formed. As seen in Table 12.6, the calculated entropy values
(33.4« J moP 1 K _l ) arc in reasonable agreement with Ihe observed values. Given the
complexity of the thermodynamics involved in chelation, it is somewhat fortuitous
that this simple approach is as successful as it is.
There are also decided enthalpy effects present. These may be most simply
viewed in terms of the chelate being "preformed." In other words, certain energy
costs that have to be paid to form complexes, such as steric interference between two
adjacent unidentate ligands, repulsions between the dipoles of iwo adjacent ligands,
etc., may have been paid, in part, when the potentially bidentate ligand was originally
formed-and need not be expended again upon complex formation. To whatever extern
that this is true, the enthalpy of formation of the chelate will benefil with regard to lhat
of the unidentate complex.
Finally, chelating ligands such as acelylacctone enjoy resonance stabilization as a
result of forming six-membered rings having some aromatic character. Acetylacetone
(2,4-pentanedione) coordinates as an anionic enolate ligand:
CH 3 —c—ch 2 —c— ch 3
O OH
CH 3 —C—CH=C—CH 3
100 Entropy changes associated with solvation and rotational differences are also important. The
driving force for this reaction (AG = -50 kJ moP 1 ) comes predominantly from the T&S term.
512
12-Coordination Chemistry: Structure
Linkage Isomerism
513
cause they have a low tendency to transfer fluoride ions to the acidic catio'
though such transfer can occur if the cation is sufficiently acidic. For example
often a successful way of abstracting a. chloride ion from the coordinatio
sphere of a metal is to allow the complex to react with silver tetrafluoroboratl
Silver chloride is precipitated, and the low-basicity and (one hopes!) noncobi
dinating BF^ is introduced as a coumterion. Thus, potentially, trisf
butylpyrazolyl)hydroboratochlorocobalt(II) could react with silver tetrafluo
borate to give the very stable silver chloriide and the tetrafluoroborate salt of
three-coordinate cation. However, when ihis reaction was attempted, fluoride
abstraction (from BF^) was found instead (Fig. 12.42). 71
Similar three-coordinate “tripod” ligands are known of the type"
[C(R 2 PY) 3 ]“ (where Y can be 0, S, or a pair of electrons on the phosphorus
atoms). 72 These tripod ligands strongly determine three coordination position
in a pyramidal arrangement, leading to a fourth coordination by a halide.
Bridging by fluorides may also occur. For example SbF^ may coordinate to
the very strong Lewis acid SbF 5 to fornn [F 5 Sb—F—SbF 5 ] _ (see Proble
3.42), but it is very difficult to abstract a fluoride ion from SbF^. It has been
suggested that [SbFJ - should be the anion of choice in the synthesis of
reactive yet potentially isolable cationic Lewis acids. 73
Factors favoring high coordination numbers are:
I. High oxidation states and hard ligands. These will serve to maximize the
electrostatic contribution to stabilizing the complexes. Because of their high
electronegativity, fluoride and oxygen-containing ligands can stabilize higl
oxidation states.
Ag'BF~
AgCU + BFj
Fig. 12.42 Reaction of tris(3-/-butylpyrazolyl)hydroboratochlorocobalt(Il) with silver tetrafluoroborate. A fluoride ion has
been abstracted from the BFf anion to give tris(3-t-butylpyrazolyl)hydroboratofluorocobalt(II). All of the unlabeled atoms
are carbons and the hydrogen atoms are not shown. [Modified from Gorrell, I. B.; Parkin, G. Inorg. Chem. 1990, 29,
2452-2456. Reproduced with permission.)
71 Gorcll, I. B.; Parkin, G. Inorg. Chem. 1990. 29, 2452-1456.
77 Grim, S. O.: Smith. P. H.; Nittolo, S.; Ammon, H. L.: Satek, L. C.; Sangokoya, S. A.; Khanna
R. K.; Colquhoun, I. J.; McFarlanc. W.; Holden, J. R. Inorg. Chem. 1985, 24, 2889-2895. Grim.-S
O.; Sangokoya, S. A.; Colquhoun. I. J.; McFarlanc. W.; Khanna. R. K. Inorg. Chem. 1986. 25
2699-2704.
73 Honeychuck, R. V.; Hersh, W. H. Inorg. Chem. 1989. 28, 2869-2886.
Small steric requirements of the ligands. Again, fluorine and oxygen serve
well. —
Large, nonacidic cations. Even though the metals are usually in high oxida¬
tion states, the large number of negative fluoride ions, oxide ions, alkoxide
ions, etc., tend to make these complexes anionic overall. Since high coordina¬
tion numbers will tend to make large ions, even with small ligands, large
cationic counterions will tend to stabilize the crystal lattice (see Chapter 8).
Also, small, polarizing cations such as Li + are to be avoided since they will
have a tendency to polarize the anion and abstract a fluoride or oxide ion (see
Chapter 4).
In addition to the geometric and optical isomerism discussed previously, there is
another type that is important in inorganic chemistry. It deals with ligands having two
potentially ligating atoms that are capable of bonding through one type of donor atom
in one situation but a different donor atom in another complex. The first example of
this type of isomerism was provided by Jprgensen, Werner’s contemporary. His
method of preparation was as follows: 74
[Co(NH3) 5 C1]C1 2
NaNO, , .
■ -► Solution A
(12.10)
. ... Let sinnd in cold
‘Solution A -
[(NH 3 ) 5 CoONO]CI 2 red
(12.11)
"Solution A” —[(NH 3 ) 5 CoN 0 2 ]C1 2 yellow
(12.12)
J0rgensen and Werner agreed that the difference between the two isomers resides in
the linkage of the N0 2 group to the cobalt. The N-bonded (or “nitro”) structure was
assigned to the yellow isomer and the O-bonded (or "nitrito") structure to the red
isomer on the basis of the color of similar compounds. 75 For example, both the
hexaammine and tris(ethylenediamine) complexes of cobalt (assuredly N-bonded) are
yellow, and the aquapentaammine and nitratopentaammine complexes, containing one
oxygen atom and five nitrogen atoms in the coordination sphere, are red. Thus long
before the electronic explanation of spectra had evolved, the correct assignment of
structure was made on the basis of color.
In the following years, these compounds were the subject of considerable con¬
troversy. A brief history of the disputing claims is given here both because it indi¬
cates some of the methods applicable in such studies, and also because it indicates
the errors that can be perpetuated if reports in the literature are accepted uncritically.
The red isomer is less stable than the yellow isomer and is slowly converted to the lat¬
ter on standing or more rapidly by heating or addition of hydrochloric acid to a
solution. Piutti 76 claimed that the absorption spectra of the two forms were identical.
74 Jorgensen, S. M. Z. Anorg. Chem. 1894, 5, 168. Actually the two isomers date back much further:
Gibbs, W.;Genth, F. A. Am. J. Sci. 1857, 24, 86. Sec Kauffman, G. B. Coord. Chem. Rev. 1973, II,
161-188.
75 From what you know, now, almost one hundred years later, was Ihis a good or a bad assumption?
76 Piutti. A. Ber. Deut. Chem. Ges. 1912. 45, 1832.
520
12-Coordination Chemistry: Structure
Fig. 12.47 Portion of the crystal structure of Prussian blue showing the bridging by
ambidentate cyanide ions. Circles represent iron(Il) (O), iron(IIl) (O), and oxygen in water
(•). The remaining interstitial or "zeolitic" water in the cubic sites has been omitted for
clarity, as have most of the cyanide ions. In addition, some of the cyanide ions are replaced
by water molecules coordinated to iron(IU), and there are also vacancies in the structure.
(Modified from Buser. H. J.; Schwarzenbach, D.; Petter, W.; Ludi. A. Inorg. Chem. 1977,
16, 2704-2710. Reproduced with permission.]
microenvironment about the iron atoms, this confirms the identity. Ferric ferricyanide
(Berlin green) consists of iron(III) at all iron sites, and white Everitt's salt (actually
KofFe'^CNJ^Fe 11 ]) consists of iron(II) at all iron sites and potassium ions in all of the
interstices.
If one pyrolyzes Prussian blue gently in a vacuum or, better, precipitates ferrous
ferricyanide in the presence of a reducing agent such as potassium iodide or sucrose,
the compound formed is truly iron(II) hexacyanoferrate(III). 95 However, it reverts to
Prussian blue rapidly upon warming with dilute hydrochloric acid or standing in humid
air.
A particularly interesting example of linkage isomerism was reported by Shriver
and coworkers. 96 Mixing solutions of iron(ll) salts and potassium hexacyanochro-
mate(III) results in a brick-red precipitate which turns dark green on heating:
Fe 2+ + K + + [Cr(CN) 6 ] 3- - KFe“[Cr“ , (CN) 6 ] brick red (12.21)
KFe[Cr(CN) 6 ] KCr lll [Fe II (CN) 6 ] dark green (12.22)
93 Cosgrove, J. G.; Collins, R. L.: Murty, D. S.; J. Am. Client. Soc. 1973. 95, 1083-1086. Robinette,
R.; Collins. R. L. J. Coord. Chem. 1974, i, 333.
96 Shriver, D. F.; Shriver, S. A.; Anderson, S. E. Inorg. Chem. 1965, 4, 725-730.
Other Types of Isomerism
521
This has been interpreted in terms of linkage isomers of the type:
... Fe"—NC—Cr" 1 —CN—Fe n —NC—Cr 111 ... ... Fe"—CN— Cr ,n —NC— Fe n -CN
brick red dark green ' '
in which the linear arrays shown in Eq. 12.23 represent portions of the cubic arrays
shown in Fig. 12.47. The initial product is C-coordinated to the chromium(III) since
that was the arrangement in the original hexacyanochromate(III). The iron(Il) coordi¬
nates to the available nitrogen atoms to form the Prussian-blue-type structure. As in
the case of Prussian blue discussed above, however, there will be preferential LFSE
favoring the coordination of the strong field C-linkage to the potential low spin / 2g
configuration of iron(ll), approximately twice as great as that of the configuration
of chromium(III) (see Table 11.3).
Other Types 1° general the other types of isomerism for coordination compounds are less interest-
of Isomerism
ing than those discussed previously, but will be listed briefly to show the variety of
possibilities.
Ligand Isomerism Since many ligands are organic compounds which have possibilities for isomerism, the
resulting complexes can show isomerism from this source. Examples of isomeric
ligands are 1,2-diaminopropane (“propylenediamine," pn) and 1,3-diaminopropane
(“trimethylenediamine,” tn) or ortho-, meta -, and para-toluidine (CH 3 C 5 H 4 NH 2 ).
Ionization Isomerism The ionization isomers [Co(NH 3 ) 5 Br]S0 4 and [Co(NH 3 ) 5 S0 4 ]Br dissolve in water to
yield different ions and thus react differently to various reagents:
[Co(NH 3 ) 5 Br]S0 4 + Ba 2+
-» BaS0 4 (s)
(12.24)
[Co(NH 3 ) 5 SOJBr + Ba 2 +
- No reaction
(12.25)
[Co(NH 3 ) 5 Br]S0 4 + Ag +
- No reaction
(12.26)
[Co(NH 3 ) 5 S0 4 ]Br + Ag +
- AgBr(s)
(12.27)
Solvate Isomerism This is a somewhat special case of the above interchange of ligands involving neutral
solvate molecules. The best known example involves isomers of "chromic chloride
hydrates," of which three are known: [Cr(H 2 0) 5 ]CI 3 , [Cr(H 2 0) 5 CI]CI 2 H 2 0, and
[Cr(H 2 0) 4 CI 2 ]Cl-2H 2 0. These differ in their reactions:
dchY<Jr. over H»S04
[CriH^gClj —-[CrfH-jOJCIj (no change) (12.28)
[Cr(H 2 0) 5 CI]Cl 2 H 2 0 ' u>ct [Cr(H 2 0) 5 CI]CI 2 (12.29)
[Cr(H 2 0) 4 CI 2 ]CI-2H 2 0 — [Cr(H 2 0) 5 CI]CI 2 (12.30)
[CriH^yClj [Cr(H 2 0) 6 ] 3+ + 3AgCl(s) (12.31)
[Cr(H 2 0) 5 CI]CI 2 — [Cr(H 2 0) 5 CI] 2+ + 2AgCl(s) (12.32)
[Cr(H 2 0) 4 Cl 2 ]CI — [Cr(H 2 0) 4 CI 2 ] + + AgCI(s) (12.33)
Coordination
Isomerism
Salts that contain complex cations and anions may exhibit isomerism through the
interchange of ligands between cation and anion. For example, both hexaam-
514
12 Coordination Chemistry: Structure
This was disputed by Shibata, 77 who claimed that they had quite different spectra!
Lecompte and Duval compared the X-ray powder patterns of the two forms and found
that they were "rigorously identical.” 78 They suggested that the red color in the
supposed nitrito complex was a result of some unreacted starting material, namely
[Co(NH 3 ) 5 CI]CI 2 , in the product.
Adell 79 measured the rate of conversion of the red form to the yellow form
photometrically and found it to be a first-order reaction. This is to be expected if the
conversion is an intramolecular rearrangement involving no other species (with the
possible exception of the solvent). On the other hand, if the red isomer is actually
unreacted starting material in the form of [Co(NH 3 ) 5 CI]CI 2 , the reaction might be
expected to be second order:
[Co(NH 3 ) 5 CI] 2+ + NO, - -» [(NH 3 ) 5 CoNO,] 2+ + Cl - (12.13)
-<f[C 0 (NH 3 ) 3 Cl 2 ] = fc[Co(NHj)5C1 2 + -j [NO --| ( i 2 . 14) 8o
Murmann and Taube 81 showed that the formation of the nitrito complex occurs
without the rupture of the Co—O bond. They used 18 0-labeled [(NH 3 ) 5 CoOH] 2+ as a
starting material and found that all of the l8 0 remained in the complex. This argues in
favor of reaction 12.15 in preference to 12.16:
(NH 3 ) 5 Co I8 0—H -| 2+
0—N—6noJ ^ (,2 -' 5)S2
[(NH 3 ) 5 Co ,8 ONO] 2+ + HONO
[(NH 3 ) 5 Co ,8 OH] 2+ + NO; -► [(NH 3 ),CoN0 2 ] 2+ + ,8 OH- (12.16) 82
The labeled nitrite complex may be caused to rearrange by heating. In this process
no loss of l8 0 is found even in the presence of excess nitrite, confirming Adell’s
hypothesis that the reaction is an intramolecular rearrangement:
,8 o 1 2+
[(NH,) J Co l8 ONO] 2+ - (NH,),Co' | - [(NH,),CoNO' 8 01 2+ (12.17)
S N—oJ
[(NH,) 5 Co I8 OH] 2+ + n 2 o 3
77 Shibata. Y. J. Coll. Sci. Imp. Univ. Tokyo 1915. 37. 15.
78 Lccomptc, J.; Duval, C. Bull. Soc. Cliim. 1945. 12. 678. Powder patterns are determined by the type
of crystal lattice and by the spacings in the lattice. They are useful as ''fingerprinting'' devices for
the identification of crystals.
79 Adell, B. Z. Anorg. Chem. 1944. 252. 272.
811 First-order kinetics is to be expected for an intramolecular reaction, but its presence is not proof
that the reaction takes place by such a mechanism. The brackets in this equation represent the
concentrations of the various species (in mol L~') rather than indications of structural moieties.
81 Murmann. R. K.; Taube. H. J. Am. Chem. Soc. 1956. 78. 4886-4890.
82 The differences in these reactions of N,Oj versus NOf. the presence or absence of OH - as a
product, etc., arc more apparent than real since these species will interact with each other to form
an equilibrium mixture. The general argument docs not depend upon the exact nature of the
reactants and products.
Linkage Isomerism
515
Electronic Effects
Finally, the l8 0 can be quantitatively removed by the basic hydrolysis of the nitro
isomer:
[(NH 3 ) 5 CoNO i8 0] 2+ + OH" - [(NH 3 ) 5 CoOH| 2+ + 0N I8 0- (12.18)
All these experiments are consistent with the original hypothesis of J0rgensen and
Werner of linkage isomerism. It is difficult to rationalize the “rigorous” contrary
evidence of some of the early workers except by the general phenomenon that it is
deceptively easy to obtain the experimental results that one expects and desires.
Werner knew of two other examples of linkage isomerism, both nitro-nitrito
isomers, and they underwent the same period of skepticism and confirmation as the
compounds discussed above although considerably less work was done with them. A
period of more than 50 years passed before Basolo and coworkers attacked the
problem with rather amazing results. 83 Linkage isomerism, once relegated to a few
lines as an "exceptional” situation in discussions of isomerism, now boasts an
extensive chemistry which continues to develop. The first new linkage isomers pre¬
pared were nitro-nitrito isomers of Cr(lII), Rh(lII), Ir(III), and PitIV). In all cases
except Cr(lII). the nitrito isomer converts readily to the more stable nitro isomer.
The first thiocyanate linkage isomers were isolated after it was noted that the struc¬
tures of cis complexes containing thiocyanate and either ammonia or phosphine were
S- or N-linked, respectively (Fig. 12.43). The hypothesis provided was that these
isomers were more stable than the alternatives (i.e., S-bonded in the phosphine
complex, N-bonded in the ammine complex) because of the competition lor it bonding
orbitals on the metal. The phosphine forms the best it bonds and hence tends to
monopolize the tt bonding <1 orbitals of the platinum, reducing the stability of the
weaker sulfur a- bond, hence the thiocyanate ion bonds through the nitrogen atom. In
the absence of competition for a-orbitals (ammonia cannot form a Trbond). the sulfur
atom is preferentially bonded. Using this hypothesis as a basis. Basolo and cowork-
ers 84 attempted to find complexes in which the ir bonding tendencies were balanced,
Fig. 12.43 Structures of [Pt(SCN),(NHj),J and [Pt(NCS),(PRj),| illustrating the competition
for 7r bonding </ orbitals of the metal. One set of 7r bonds has been omitted for clarity. The
"</ orbital" left-right symmetry has been lost due to polarization. Compare with Fig. 11.30.
In addition to empty J orbitals, sulfur has filled p orbitals. See Table I l.l I and accompanying
discussion.
83 Basolo. F.; Hammakcr, G. S. J. Am. Chem. Soc. 1961, 82. 1001-1002: Inorg. Chem. 1962, I. 1-5,
84 Basolo, F.; Burmcister, J. L.; Poii, A. I.J. Am. Chem. Soc. 1963, S5, 1700-1701. Uurnieister. J. L.t
Basolo, F. Inorg. Chem. 1964, 3. 1587-1593.
518
12 • Coordination Chemistry: Structure
Linkage Isomerism
519
Symbiosis
nitrogen atom. Rearrangement of one of the thiocyanate ligands occurs. Significantly,
it is the group that is trans to the nitrogen atom that isomerizes. Furthermore, the
ubiquitous presence of the Irons influence 88 in these complexes indicates the per¬
vasive consequences of electronic effects. We may therefore conclude that if either
the electronic or the steric factor in a series of complexes is held constant, it is
possible for the other factor to determine the nature of the resulting linkage isomer.9
Jprgensen 90 proposed the principle of symbiosis with respect to hard and soft
acid-base behavior. This rule of thumb states that hard species will tend to increase
the hardness of the atom to which they are bound and thus increase its tendency to
attract more hard species. Conversely, the presence of some soft ligands enhances the
ability of the central atom to accept other soft ligands. In terms of the electrostatic
versus covalent picture of Pearson’s hard and soft or Drago's E A E B and C A C B
parameters (see Chapter 9), the best ‘‘strategy” of a complex is to “put all its eggs in
one basket,” i.e., form all hard (“electrostatic") or all soft (“covalent”) bonds to
ligands. There are many examples that could be given to illustrate this tendency in
metal complexes:
All ligands hard
[Co(NH 3 ) s NCS] 2+
[Rh(NH 3 )jNCS] 2+
(Fe(NCSe)J 2 -
All ligands soft
[Co(CN),SCN] 3 ~
[RhtSCNVJ 5 -
(CpFe(COMSeCN)l
In the first example the hard ammonia ligands tend to harden the cobalt, and so the
thiocyanate bonds preferentially through the nitrogen atom. Conversely, the soft
cyanide ligands soften the cobalt, making it bond to the soft end of the thiocyanate (the
sulfur atom). Similarly, in the case of Rh(IlI) five ammonia ligands result in preference
for nitrogen at the sixth position; if all six soft sulfur atoms can ligate, they will.
Iron) 11) appears to prefer the hard nitrogen atom unless softened by the presence of
carbonyl groups.
The symbiotic theory adequately covers most of the linkage preferences observed
for octahedral complexes. Unfortunately, it contradicts exactly the 7r bonding theory
applied above to square planar complexes. Pi bonding can be equated with softness. <r
bonding with hardness. In the case of octahedral complexes we say that the presence
of soft, 7 r bonding ligands favors the addition of more soft, n bonding ligands (symbi¬
otic theory), but in the case of square planar complexes we say that soft, n bonding
ligands discourage the presence of other n bonders and favor the addition of hard,
rr-only ligands (n competition theory). Obviously the situation is somewhat less than
perfect if the two theories are applicable only in limited areas and appear to contradict
each other in their basic raisons d'etre. Nevertheless they have heuristic value and
W1 See Chapter 13 for a discussion of the electronic factors operative in the trans influence.
See Meek. D. W.; Nicpon. P. E.; Meek. V. I. J. Am. Client. Sue. 1970. 92. 5351-5359. Huhecy.
J. E.; Grim. S. 0. Inorg. Nttcl. Chem. Lett. 1974. 10. 973-975. These papers analyze the electronic
and steric factors in compounds of this type.
90 Jorgensen. C. K. Inorg. Chem. 1964. 3. 1201-1202.
Prussian Blue and
Related Structures
serve to emphasize that there are many factors involved, both electronic and steric, in
determining which of the possible isomers will be preferred.91
Pearson, in elaborating upon these ideas, distilled the essence of the n competi¬
tion theory to two soft ligands in mutual trans positions will have a destabilizing
influence on each other when attached to class (b) (soft) metal atoms. He also
provided additional examples of the rule that symbiosis prevails in octahedral com¬
plexes. antisymbiosis in square planar complexes. Tetrahedral complexes are ex¬
pected to show antisymbiosis but on a much reduced scale compared with the square
planar complexes. 92
Linkage isomerism is but a special case of ambidentate behavior in ligands. The
cyanide ion provides good examples of such behavior. In discrete complexes it almost
always bonds through the carbon atom because of the stronger ir bonding in
that mode. It has also been reported to form a few linkage isomers such as
c£r-[Co(trien)(CN) 2 ] + and c7.v-[Co(trien)(NC),] + .
A large number of polymeric complexes is known containing ambidentate cyanide
bridging groups. These are related to “Prussian blue," which is formed by the addition
of ferric salts to ferrocyanides:
Fe ,+ + (Fe"(CN) h ] J - -♦ Fe 4 [Fe(CN)J 3 (12.19)
Addition of ferrous salts to ferricyanides produces "Turnbull's blue”:
Fe :+ + [Fe m (CN)„] 3 - - Fe 4 [Fe(CN)J 3 (12.20)
It has been shown that the iron-evanide framework is the same in Prussian blue,
Turnbull's blue, and other related polymeric cyanide complexes (Fig. 12.47), differing
only in the number of ions necessary to maintain electrical neutrality. Various quan¬
tities of water molecules may also be present in the large cubic holes. Prussian blue
has a structure with hexacoordinate, low spin Fe(Il) bonded through the carbon atoms
and hexacoordinate. high spin FedII) bonded through the nitrogen atoms of the
cyanide. To achieve this stoichiometry, one-fourth of the Fe(II) sites are occupied by
water molecules. This reduces the number of bridging cyanide groups
(Fe 11 —C=N—Fe m ) somewhat, and water molecules occupy the otherwise empty
ligand positions thus created. There is also one water molecule in each cubic site. 9 -'
Although prepared from different starting materials, Turnbull's blue is identical.
Although X-ray and magnetic data support this identity, the best evidence comes from
the fact that the Mdssbaucr spectra of Prussian blue and Turnbull's blue are the
same. 94 Since Mossbauer spectra are extremely sensitive to the electron density and
91 For further discussion of this problem, see DcStcfano. N. J.: Burmeister, J. L. Inorg. Chem. 1971,
10. 998-1003. It should also be noted that although the phenomenon of symbiosis is very real, the
choice of the word symbiosis to describe it is unfortunate. As DcStcfano and Burmeister point out.
symbiosis in biology refers to the "flocking together" of different species, rather than the same
species, in intimate association. Nevertheless inorganic chemists will undoubtedly continue to use
the term in its current sense.
« Pearson. R. G. Inorg. Chem. 1973, 12. 712-713.
Uuscr. H. J.: Schwarzenbach. 13.: Pcllcr. W.; Ludi, A. Inorg. Chem. 1977. 16. 2704-2710.
" J Duncan. J. F.; Wiglcy. P. W. R. J. Chem. Soc. 1963. 1120-1125. Fluck. E.; Kcrlcr, W.; Neuwirth,
W. Angcw. Chem. Ini. Ed. Engl. 1963. 2. 277-287. Bonnettc, A. K., Jr.; Allen. J. F. Inorg. Chem.
1971. 10. 1613-1616. The method of preparation and precipitation may affect the nature of the
precipitate (colloidal, etc.) and so varying amounts of water and cations may be incorporated in the
structure.
516
12 Coordination Chemistry: Structure
allowing the isolation of both isomers. Examples of the complexes thus isolated are
[(Ph 3 As) 2 Pd(SCN) 2 J, [(Ph 3 As) 2 Pd(NCS) 2 ], and [(bpy)Pd(SCN) 2 ], [(bpy)Pd(NCS) 2 J. In
both cases, on warming, the S-bonded isomer is converted to the N-bonded isomer,
which is presumably slightly more stable.
The competition for rr bonding is indicated in the behavior of the selenocyanate
group, SeCN~. This group readily bonds to the heavier group VIIIB (8) metals via the
selenium atom to form complexes such as [Pd(SeCN) 4 ] 2 ~ and rra/w-[Rh(PPh 3 ) 2 -
(SeCN) 2 ]~. However, in a closely related complex, /ra/tj-[Rh(PPh 3 ) 2 (CO)(NCSe)], the
presence of a trans carbonyl group apparently favors coordination via the non-vr-
bonding nitrogen atom. 85
Another example of apparent electronic (i.e., n bonding) control of linkage
isomerism comes from bidentate chelates having one strong and one weak donor atom
(Fig. 12.44). The presence of an S-bonded thiocyanato group trans to the non-?r-
bonding nitrogen atom, but an N-bonded isothiocyanato group trans to the n bonding
phosphine donor is indicative of n competition in this complex. 86
Sferic Effects Steric factors may play an important role in stabilizing one or the other of a pair of
linkage isomers. Thus nitro-nitrito, thiocyanato-isothiocyanato, and seienocyanato-
isoselenocyanato pairs differ in steric requirements (Fig. 12.45).
One or more factors may be operating simultaneously to provide a delicate
balance of counterpoising effects. An interesting series of compounds illustrates the
competing effects in linkage isomers of square planar palladium(II) complexes
(Fig. 12.46a-d). 87 The six-membered chelate ring in Figure 12.46c allows an essen-
Fig. 12.44 The molecular
structure of isothiocyanato-
thiocyanatod-diphenyl-
phosphino-3-dimethyl-
aminopropane)palladium(ll).
Note: (I) trans arrangement
of P—Pd—N and
N—Pd—S bonds and (2)
linear vs. bent arrangement
of the NCS group. [From
Meek. D. W.; Nicpon,
P. E.; Meek, V. I. J. Am.
Chem. Soc. 1970, 92.
5351-5359. Reproduced
with permission.]
83 Burmcister, J. L.; DcStcfano. N. J. Chem. Commun. 1970, 1968.
86 Meek. D. W.; Nicpon. P. E.; Meek. V. I. J. Am. Chem. Soc. 1970. 92. 5351-5359. Clark. G. R.;
Palenik, G. J. Inorg. Chem. 1970. 9. 2754-2760.
87 Palenik. G. J.; Steffen, W. L.; Mathew. M.; Li, M.; Meek. D. W. Inorg. Nucl. Chem. Led. 1974. 10.
125-128.
586
14- Some Descriptive Chemistry of the Metal:
Fig. 14.1 Molecular
structure of diiodobis[o-
phenylenebisfdimethyl-
arsine)]nickel(II).
I
the octahedral complexes. If there is sufficient disparity between the positions in the
spectrochemical series of L and L' [e.g., diiodobis(diars)nicke!(II), Fig. 14.1), the
resulting adduct is diamagnetic. The complex may be viewed as a square planar
complex that has not been sufficiently perturbed by the tetragonal field produced by
the weak iodo ligands to cause unpairing of the electrons.
Only a few simple copper(lll) salts, e.g., KCu0 2 and Cs 3 CuF 6 , are known, but
numerous complexes containing organic ligands exist. These behave as might be
expected trom their analogy to Ni(II), forming square planar structures. However,
unlike Ni(II), all purely inorganic Cu(lII) species are strong oxidizing agents. Rela¬
tively stable Cu(III) complexes have been found in biological systems.
A few Co(I) complexes are known, and although they must be considered excep¬
tional. vitamin B l2 depends on this oxidation state for its action (Chapter 19).
The d 9 This configuration is found in copper(II) compounds but is otherwise unimportant. It
Configuration has neither the closed subshell stability of cl 10 nor the LFSE possible for d. Cop-
per(II) may be fairly easily reduced to copper(I) (see Eqs. 14.10 and 14.11).
Six-coordinate complexes are expected to be distorted from pure octahedral
symmetry by the Jahn-Teiler effect and this distortion is generally observed (Chap¬
ter 11). A number of five-coordinate complexes are known, both square pyramidal and
trigonal bipyramidal. Four-coordination is exemplified by square planar and
tetrahedral species as well as intermediate configurations.
The d'° For the first transition series this configuration is limited to Cu(l) and Znlll). but it is
Configuration also exhibited by the posttransition metals in their highest oxidation states (Ga(IIl).
GotIV)). The copper(I) complexes are good reducing agents, being oxidized to Cu(ll).
They may be stabilized by precipitation with appropriate counterions to the extent
that Cu(I) may form to the exclusion of Cu(ll):
Cu 2+ (aq) + 2I-(aq) - Cul(s) + ±I 2 (s) (14.10)
Cu 2+ (aq) + 2CN"(aq) -♦ CuCN(s) + i(CN) 2 (g) (14.11)
Zinc(II), gallium(III), and germanium(IV) are the most stable oxidation stales for
these elements, but the later nonmetals (arsenic, selenium, and bromine) show a
reluctance to assume their highest possible oxidation state.
The spherically symmetric d 10 configuration affords no LFSE, so the preferred
coordination is determined by other factors. For Cu(I) the preferred coordination
appears to be linear (sp), two-coordination, although three-coordinate complexes are
known as well as several tetrahedral complexes. Zinc(II) is typically either tetrahedral
(e.g., [ZnCIJ 2 ) or octahedral (e.g., [Zn(H 2 0) 6 ] 2 ). but both trigonal bipyramidal and
The Chemistry of the Heavier Transition Metal:
587
square pyramidal five-coordinate complexes are known (see Chapter 12). The post-
transition metals form tetrahedral (e.g., (GaCIJ - ) and octahedral (e.g. [Ge(acac),J +
[GeC! 6 ] 2 , and [AsF 6 ] ) complexes.
The Chemistry of
the Heavier
Transition Metals
A detailed account of the descriptive chemistry of the heavier transition metals is
beyond the scope of this book. Many aspects of the chemistry of these elements such
as metal-metal multiple bonds, metal clusters, organometallic chemistry, and coordi¬
nation chemistry are discussed in other chapters. The present discussion will be
limited to a comparison of the similarities and differences of the heavier metals and
their lighter congeners.
In general, the coordination numbers of the elements of the second and third
transition series lend to be greater than for the first series because the ionic radii are
larger by about 15-20 pm (0.15-0.20 A) for corresponding species. 211 Thus tetrahedral
coordination is considerably less frequent although observed in species such as
[WOJ-", [ReOJ ”, and Os0 4 . Square planar coordination is found in d H species such
as Rh(I), Pd(Il), Pt(II). and Au(lII), which are especially stabilized by LFSE. Oc¬
tahedral species are quite common, and the occurrence of coordination numbers 7. 8,
9, and 10 is fairly common.
The heavier congeners show a pronounced tendency toward higher oxidation
states. Whereas the + 2 state is known for all elements of the first transition series, it is
relatively unimportant for the heavier metals. Cadmium is nearly restricted to the +2
oxidation state (Cd; + is known 2 ') and Hg(II), Pd(II), and Pt(II) are the only other
important dipositivc species. Although cobalt is known as both Co(II) and Co(III), its
congeners rhodium and iridium are essentially limited to the +3 oxidation state or
higher. Chromium(III) is the most stable oxidation stale of chromium, but both
molybdenum and tungsten are strongly reducing in that oxidation state with the +6
oxidation state being much more important. In general, the stability of the highest
possible oxidation state (i.e.. the group number oxidation state) is considerably
greater in the heavier metals. 1 hus |ReOj| . unlike |Mn0 4 ) ~, is not a strong oxidizing
agent. The trend is extended further along the series as well, culminating in ruthenium
letroxide and osmium tetroxide (both powerful oxidants), two of the few known cases
of valid +8 oxidation states. 22 Further examples are the stabilities of Pd(IV). Pt(IV),
and Au(III) relative to their lighter congeners. Gold is even able to achieve the
unexpected oxidation state of +5 (see Chapter 18).
As discussed in Chapter II, there is a much greater tendency toward spin pairing
in the heavier transition metals and consequently the existence of high spin complexes
is much less common than among the earlier metals. Thus, in contrast to Ni(II), which
forms tetrahedral, square planar, square pyramidal, trigonal bipyramidal. and oc¬
tahedral complexes, Pd(II) and Pt(II) form complexes that are almost universally low
spin and square planar. A few weakly bonded live-coordinate adducts are known and
Differences in coordination number and in spin slate complicate a direct comparison. The above
range was taken from comparison of six-coordinate Sc ,+ and Y ,+ [Ar = 15 pm (0.15 A)), and Zn 2 +
and Cd~ |Ar = 20 pm (0.20 A)|. Because of the lanthanide contraction, the radii of the third scries
arc very similar to those of the second series. See page 579.
21 Faggiani. R.; Gillespie. R. J.; Vekris. J. E. J. Chan. Site. Client. Commun. 1986. 517-518.
-- The +8 oxidation state also is found in the amine complexes of Os0 4 (e.g.. Os0 4 py). See Kobs,
S. F.; Bchrman. E. J. Inorg. Cltim. Ada 1987. 128. 21-26.
588
14- Some Descriptive Chemistry of the Metals
often appear as intermediates in substitution reactions of square planar complexes
(see Chapter 13). More recently it has been shown that this geometry can be stabilized
with a ligand combination of chelating amines and n accepting ligands.- 3 The effect of
ligand field strength on the instability of the d 9 configuration in silver(II) and gold(II) is
pronounced. The splitting of the 5 d orbitals in a d 9 Au(II) complex, for example, is
about 80% greater than that found in analogous Cu(II) complexes (see Chapter II).
The ninth electron of the gold complex would have to reside in the highly unfavorable
d x i_ 2 orbital and this would lead to an extreme tetragonal distortion. Thus the odd
electron of Au(Il) is easily ionized and disproportionation to Au(I) and Au(III) results.
There is also an important difference displayed by heavier transition metals in
their magnetic properties. Because of extensive spin-orbit coupling, the spin-only
approximation (Chapter 11) is no longer valid. The simple interpretation of magnetic
moment in terms of the number of unpaired electrons cannot be extended from the
elements of the first transition series to their heavier congeners.
Finally, the heavier posttransition metals have group number oxidation states
corresponding to d [0 configurations: indium(III), thallium(III), tin(IV), lead(IV), anti-
mony(V), bismuth(V), etc. However, there is an increasing tendency, termed the
“inert pair effect,” for the metals to employ p electrons only and thus to exhibit
oxidation states two less than those given above (see Chapter 18).
Oxidation States
and EMFs of
Groups 1-12
Having compared in general terms the properties of transition metals both on the basis
of the {/-electron configuration and the properties of the light versus heavier metals,
we shall now look more specifically at the stabilities of the various oxidation states of
each element in aqueous solution. Every oxidation state will not be examined in detail,
but the emf data to make such an evaluation will be presented in the form of a Latimer
diagram.
If you are not thoroughly familiar with the principles of electrochemistry, you
should review Chapter 10 and the Latimer diagram derived there (below) before
considering the following discussion for determining the stability of oxidation states:
MnO; HMnO'Mn0 2 Mn 3 Mn 2 Mn
I _ ♦ 1,70 _ I |_ ♦ 23 _I
Stabilities of
Oxidation States
A table of emf values appears in Appendix F.
There are three sources of thermodynamic instability fora particular oxidation state of
an element in aqueous solution: (I) The element may reduce the hydrogen in water or
hydronium ions; (2) it may oxidize the oxygen in water or hydroxide ions; or (3) it may
disproportionate.
The emf values for reduction of hydrogen in water are given in Eqs. 10.116 to
10.118. These determine the minimum oxidation emfs necessary for a species to effect
n Albano, V. G.; Braga. D.; De Felice, V.; Panunzi. A.; Vitagliano, A. Organometallics 1987. 6.
517-525.
Oxidation States and EMFs of Groups 1-12 589
reduction of hydrogen: 1 M acid. £° > 0.000 V; neutral solution, E > +0.414 V; 1 M
base, E° > +0.828 V. For manganese the only oxidation state that is unstable in this
way is Mn(0), which readily reacts with acid:
Mn(s) - 4
Mrr + (aq) + 2e~
E° = +1.18 V
(14.12)
2H + (aq) + 2e _ --
H 2 (g)
E° = 0.00 V
(14.13)
Mn(s) + 2H + (aq) -
Mn :+ (aq) + H 2 (g)
E° = + 1.18 V
(14.14)
The emf values for oxidation of the oxygen in water are given in Eqs. 10.119 to
10.121. These determine the minimum reduction emf necessary fora species to effect
oxidation of the oxygen: 1 M acid, E° > + 1.229 V; neutral solution, E > +0.815 V;
1 M base, E° > +0.401 V. There are several oxidation states of manganese that are
reduced by water, but the protonated manganate ion is typical:
HMn0 4 (aq) + 3H(aq) + 2e" - MnO,(s) + 2H,0 £° = +2.09 V (14.15)
H,0 -♦ iO,(g) + 2H + (aq) + 2e~ E° = -1.23 V (14.16)
HMn0 4 (aq) + H + (aq) - MnO,(s) + H,0 + J0 2 (g) E° = +0.86 V (14.17)
Species that reduce or oxidize water can be spotted rapidly in emf diagrams such
as the one given above for manganese. For example, in acid solution all negative emfs
result in reduction of H + ion by the species to the right of that emf value. 24 All values
more positive than + 1.23 V result in oxidation of water by the species to the left of
that value. Examination of the manganese diagram for acid solution reveals that the
following species are unstable: Mn° (oxidized to Mn 2 ), Mn ,+ (reduced to Mn 2 '),
and Mn0 4 (reduced to Mn0 2 ). One should also examine the skip-step emf values for
possible reactions leading to instability. Thus, although water will not reduce Mn0 4 "
to HMnO^, the skip-step emf for Mn0 4 _ to Mn0 2 (+1.70 V) is sufficiently large to
make the reaction proceed:
2Mn0 4 (aq) + 2H + (aq) + 2e -♦ 2HMn0 4 (aq)
H 2 0- 4 $0 2 (g) + 2H + (aq) + 2e"
E° = +0.90 V
£° = -1.23 V
(14.18)
(14.19)
2Mn07(aq) + H z O-♦ 2HMnO^(aq) + i0 2 (g)
C77,
ll
1
c
<
(14.20)
2Mn0 4 "(aq) + 8H + (aq) + 6e _ - 4 2MnO,(s) + 4H 2 0
3H,0- 4 iO,(g) + 6H + (aq) + 6e~
£° = 1.70 V
£° = -1.23 V
(14.21)
(14.22)
2Mn0 4 _ (aq) + 2H + (aq) - 4 2MnO,(s) + |0 2 (g) + H 2 0
E° = +0.47 V
(14.23)
24 Predictions based on emf values are not always borne out in the laboratory. For example, pure
Mn(s) would be predicted to react with neutral water but no reaction is observed. In some cases
reactions are extremely slow and are not observed for kinetic reasons. In others, products of the
reaction, such as oxide coatings, protect the reactant surfaces. Furthermore, reactions are usually
not run at standard conditions and then E° values do not reflect the true spontaneity of the reaction.
590
1 4 • Some Descriptive Chemistry of the Metals
Disproportionation occurs when a species is both a good reducing agent and a
good oxidizing agent. In basic solution, for example, CI-, disproportionates to CP and
CIO" ions:
CI 2 (g) + e~ -► CP(aq) £° = +1.40 V (14.24)
Cl 2 (g) + 20H~(aq) - C10~(aq) + H 2 0 + e - £° = -0.89 V (14.25)
Cl 2 (g) + 20H _ (aq) -> CP(aq) + CICT(aq) + H,0 E° = +0.51 V (14.26)
Species susceptible to disproportionation are readily picked out from an emf diagram
such as that given for manganese. The “normal" behavior of an element (i.e, when
uncomplicated by disproportionation) is for the emf values to decrease steadily from
left to right. Good reducing agents are on the right, good oxidizing agents are on the
left, and stable species are toward the middle. Whenever this gradual change from
more positive to more negative is broken, disproportionation can occur. For man¬
ganese in acid solution such breaks occur at two species: Mn 3+ and HMnOJ. As it
turns out, both ions are also unstable because they are reduced by water, but even if
they were stable in this regard they would be unstable as a result of disproportionation
reactions:
Mn VH (aq) + e - -♦ Mn 2+ (aq) £° = |.56 V- (14.27)
Mn 3+ (aq) + 2H,0-♦ MnO,(s) + 4H + (aq) + e~ £° = -0.90 V (14.28)
2Mn 3+ (aq) + 2H,0-♦ Mn z+ (aq) + Mn0 2 (s) + 4H + (aq) £° = +0.66 V (14.29)
Other applications of emfs include the prediction of thermodynamically possible
redox reactions [e.g., will Sn 4+ oxidize Fe 2+ to Fe 3+ ?] and the stabilization of oxida¬
tion states through the formation of complexes. The former is a straightforward
application of thermodynamics and will not be discussed further here. The second is of
great importance. It was introduced in Chapter 11 and will be discussed further below.
The Effect of The Nernst equation was given before (Eq. 10.115), and in this chapter the effect of pH
Concentration on the reduction potential of the hydrogen ion has been mentioned, but the effect in
on Stability general should be emphasized. There are several types of reactions in which con¬
centrations of the reactants and products affect the stability of various oxidation
states. This can be understood through application of the Nernst equation. The
reduction potential of hydrogen will vary with the concentration of the hydrogen ion;
hence the commonly known fact that many reasonably active metals dissolve in acid
but not in base.
Perhaps even more important is the effect of hydrogen ion concentration on the
emf of a half-reaction of a particular species. Consider the permanganate ion as an
oxidizing agent in acid solution (as it often is). From the Latimer diagram above we
can readily see that the reduction emf is 1.51 V when all species have unit activity.
What is not shown is the complete equation:
MnO^(aq) + 8H + (aq) + 5e“ -► Mn 2+ (aq) + 4H 2 0 (14.30)
which makes it clear that the concentration of the hydrogen ion enters the Nernst
equation to the eighth power—the oxidizing power of the permanganate ion is strongly
Oxidation States and EMFs of Groups 1-12
591
PH dependent. If the hydrogen ion concentration is reduced to I0 _l4 M (I M OH"), a
different set of values is obtained:
0.59 -0044
Thus there are oxidation states of manganese that are unstable in acid but stable in I M
base. In the above discussion we have seen that the tendency for a species to accept or
provide electrons, as quantified by emf values, may be strongly dependent on pH. Our
examination has been restricted to aqueous solutions of Mn0 4 ~ in which the pH is
cither 0 or 14. A fuller picture of the equilibrium chemistry of manganese, showing a
broad range of pH and £ values, is given by a Pourbaix diagram (sometimes called a
predominance area diagram) (Fig. 14. 2 ).25 Diagrams of this type are temperature and
concentration specific; in this case concentrations are 1.0 M for all species but H and
OH and the temperature is 25 °C.26 Dotted lines representing the oxidation and
reduction of water have been added. Any species above or below these dotted lines
will, in principle, oxidize or reduce water, respectively. In practice the range of
stability in water is larger than that depicted because of overvoltage.
It is instructive to examine Fig. 14.2 in some detail. In this diagram we see
horizontal, slanted, and vertical lines. The solid lines arise from values of £ and pH at
which two different oxidation states can exist in equilibrium. The equation for each
solid line is given by:
P _ p0 ( 0 . 0592 )( m )( pH )
t ~ L (I4.3D
where n is the number of electrons required to reduce the higher oxidation slate and m
is the number of hydrogen ions consumed. For example, at all points on solid line A.
an equilibrium exists between Mn(s) and Mn 2+ (aq). The line is horizontal because
there is no pH dependence for the reduction:
Mn 2+ (aq) + 2e - -► Mn(s) (14.32)
At some higher pH. Mn(OH),(s) becomes the predominant species and as is shown by
solid line B. the voltage varies with pH. consistent with the half-reaction:
Mn(OH) 2 (s) + 2e‘ - Mn(s) + 20H-(aq) (14 . 33)
The vertical dashed line C shows the pH at which Mn 2+ (aq) and MnlOHUs) exist in
equilibrium at unit activity:
Mn(OH),(s) Mn 2+ (aq) + 20H"(aq) (1434)
This, of course, is an equilibrium between two species of the same oxidation state and
therefore does not involve oxidation or reduction.
-'Campbell. J. A.; Whitekcr. R. A. J. Cltem. EJttc. 1969. 46. 90-92. Pourbaix. M. Allas of Elec¬
trochemical Equilibria in Aqueous Solution (English translation by Franklin. J. A.) Pcrgamon:
Oxford. 1966. Liang. C. C. In Encyclopedia of Electrochemistry of the Elements', Bard. A. J Ed -
Marcel Dekkcr: New York. 1973.
Geologists often construct their Pourbaix diagrams based on very dilute solutions lo correspond
more closely to that found in nature.
592
14 - Some Descriptive Chemistry of the Metals
Fig. 14.2 A potential-pH
(Pourbaix) diagram for
manganese. [Modified from
C. C. Liang In Encyclo¬
pedia of Electrochemistry
of the Elements: Bard.
A. J.. Ed.; Marcel
Dekker; New York, 1973;
p 360. Reproduced with
permission.)
Species existing a. high, voltages (e.g., MnO.) are good oxtd.xmg «««
those at low voltages (metallic manganese) are good reducing agents. It is clear from
Fig 142 that manganese in the + 2 oxidation state is the predominant spec.es over a
wide area of pH and potential combinations. The half-filled d shell for Mn is oug
to be largely responsible for this stability. .. D) jf
Manganese(Ill) oxide. Mn,0 3 ( s >- exists in equilibrium with Mn(OH),(s) (line D).
conditions are sufficiently basic or with Mn- + (aq) (line E) at somewhat lower pH
values:
Mn 2 0 3 (s) + 3H 2 0 + 2e" -♦ 2Mn(OH) 2 (s) + 20H-(aq) (14.35)
Mn 2 0 3 (s) + 6H + (aq) + 2e“ -♦ 2Mn 2+ (aq) + 3H 2 0 (14-36)
Lines G and F show equilibria between Mn0 2 (s) and Mn 2+ (aq) (acidic conditions) and
between Mn0 2 (s) and Mn 2 0 3 (s) (more basic conditions), respectively.
Mn0 2 (s) + 4H + (aq) + 2e“ -♦ Mn 2+ (aq) + 2H 2 0 (14.37)
2MnO z (s) + H 2 0 + 2e“ -♦ Mn 2 0 3 (s) + 20H"(aq) (14.38)
Above line H the predominant species is Mn0 4 -(aq), while below line H Mn0 2 (s) is
dominant. At the far right side of the diagram, line H intersects with lines I and J
creating a small triangle of stability for MnOT<«I> and making it the predominant
species in that area.
Oxidation States and EMFs of Groups 1-12
593
Group IA (1)
It should be kept in mind that only species that can exist in concentrations
corresponding to an activity of one are shown on this particular diagram. For example,
the solubilities of Mn(OH) 2 (s) and Mn 2 0 3 (s) increase in strong base presumably due to
the formation of some Mn(OH)j“(aq) [or Mn(OH) 3 ] and Mn(OH) 4 . respectively, but
these anions are not shown in the diagram. The hypomanganate ion, Mn0 4 (aq), has
been detected in basic solution, although not at concentrations approaching 1.0 M.
Known species, such as Mn(OH) 3 (s) and Mn 2 0 7 are absent because of their ther¬
modynamic instability. 27 Some species are just outside the range depicted. For exam¬
ple, Mn 3+ becomes the predominant species in 15 N H 2 S0 4 at a potential of 1.5 V.
Finally, some species such as Mn 3 0 4 (s), have been omitted from the diagram in order
to minimize its complexity.
The subtlety of concentration effects may well be illustrated by the puzzlement
once occasioned by the inclusion of dry ice in a list of ingredients for the preparation
of potassium permanganate—all the more so because it was obvious that the dry ice
was a true reagent, not a coolant. The preparation takes advantage ot the tact that the
oxidation emfs of manganese are more favorable in basic than in acid solution: One
can oxidize the readily available manganese dioxide to the green manganate ion,
[MnOJ 2- , with an emf of only -0.60 V to overcome.
40H~(aq) + Mn0 2 (s) -♦ [Mn0 4 ] 2 -(aq) + 2H 2 0 + 2e“ E° = -0.60 V (14.39)
Since this half reaction is so highly hydrogen-ion sensitive, we can force the reaction
even more by increasing the hydroxide ion concentration above I M, say, by using
fused potassium hydroxide (for an example of a nonaqueous, fused-sall reaction, see
Chapter 10). Now, how can we get manganate oxidized the remainder of the way to
permanganate? By increasing the hydrogen ion concentration and gradually shifting
over from the basic toward the acidic Latimer diagram. As shown by this diagram,
[Mn0 4 ] 2_ (aq) disproportionates in acidic solution, forming two moles of permanga¬
nate for every one of manganese dioxide.
4H+(aq) + 3[Mn0 4 ] 2 "(aq) - Mn0 2 (s) + 2[Mn0 4 ]-(aq) + 2H 2 0 (14.40)
But how can this acidification be effected without adding large amounts of strong acid
(recall that permanganate is unstable in concentrated acid)? Simple: Dissolve the
potassium manganale/potassium hydroxide mixture in water, throw in a few chunks
of dry ice. and in the ••witches' cauldron" effect, watch the solution turn from green to
deep purple!
These active metals lose one electron readily, but the loss of a second is energetically
very unfavorable (see Chapter 4). Thus the chemistry of the group is nearly defined in
terms of the + I oxidation state. As these metals are powerful reducing agents, it is
understandable that the reduction of M + is very unfavorable:
K + (aq) + e“ - K(s) £° = -2.94 V (»4.41)
The emf values for Li + , Na + , Rb + , and Cs + are -3.04, -2.71,-2.94. and -3.03 V,
respectively. Although lithium metal is the most easily oxidized in the thermodynamic
sense, it is less reactive in water than the other alkali metals because of its relatively
high melting point (180 °C). The others become molten from the heat of reaction,
melting at less than 100 °C, and as a result expose a much greater surface area.
27 Adding concentrated sulfuric acid to cold permanganate solutions give Mn 2 0 7 , a brownish oily
substance, that explodes violently.
594
14 • Some Descriptive Chemistry of the Metal:
!
Group 11A (2)
| Group IIIB (3)
|
Group IVB (4)
Group VB (5)
I
II
Alkalide anions, M~, discussed in Chapter 12, may be stabilized by various
macrocylic ligands. 28 The dissolution of sodium and heavier alkali metals in ethers
gives not only solvated M + and e~, but also solvated M~, which results from
disproportionation of the metal atom. 29
2M(s) - M + (solv) + M~(solv) (14.42)
The relative ease with which both j electrons are lost from atoms of these elements
leads to compounds in which only the +2 oxidation state is found. In general, as
discussed in Chapter 4, M is unstable with respect to disproportionation. The metals
of this group are less strongly reducing than the alkali metals, but still must be
considered strongly reducing.
Ca 2+ (aq) + 2e" -♦ Ca(s) E° = -2.87 V (14.43)
The emf values range from - 1.97 V for Be to -2.91 V for Ba.
Scandium, yttrium, and lanthanum are all quite active, resembling the alkaline earth
metals (IIA. 2) to a certain degree. For example, they reduce water, react with
oxygen, and dissolve in strong acids to give soluble salts. The +3 oxidation state is the
only important one for this group, and aqueous M 3 cations have been extensively
studied. Scandium(III), with its high charge and small radius, resembles Al 3+ in its
chemistry. The lanthanides will be discussed later in this chapter.
Sc 3 (aq) + 3e- -► Sc(s) E° = -2.09 V (14.44)
Titanium has a more extensive redox chemistry than either zirconium or hafnium. In
addition to the +4 oxidation state, the most stable for all three elements of this group,
titanium! Ill) and titanium(II) compounds are known. Titanium(III) is a good reducing
agent and exists in aqueous solution as [Ti(H 2 0) 6 ] 3+ under acidic conditions. Ti¬
tanium! II) reduces water, but in some instances the reaction appears to be sufficiently
slow to allow this oxidation state to be detected. Significant hydrolysis of the +4
cations occurs, more so for the small titanium(IV) than for the other two members of
the group. Hydrolysis of Ti 4+ leads to a mixture of species, including TiO 2 .
lTi(GH),f\ and various oligomers. 30
-1.00
Potentials in parentheses are estimated values.
A wide range of oxidation states are known for all of the elements of this group, but
only vanadium has an extensive redox chemistry.
28 Dawes. S. B.; Ellaboudy, A. S.; Dye. J. L. J. Am. Chem. Soc. 1987. 109. 3508-3513.
39 Dvc. J. L. Pro K . Inorg. Chem. 1984. 32. 327-441.
30 Gratzel, M.; Rotzingcr, F. P. Inorg. Chem. 1985. 24. 2320-2321. Comba. P.; Mcrbach, A. I rung.
Chem. 1987. 26. 1315-1323.
Oxidation States and EMFs of Groups 1 — 12
595
Group VIB (6)
-0.23
Oxidation states of vanadium are found as low as -3 and as high as +5 (see
Table 14.2). Vanadium(II) is strongly reducing and although aqueous solutions of the
violet [V(H 2 0) 6 ] 2+ ion can be prepared, they are inherently unstable with respect to
water reduction. Vanadium(IIl). though stable in water, is a fairly strong reducing
agent as well. Vanadium (V) is a good oxidizing agent, but only in concentrated acid.
The dependency on the hydrogen ion concentration is such that in neutral solutions
the reduction potential is lowered to such an extent that reduction of V(V) is difficult.
The +4 oxidation slate, which is generally the most stable one for vanadium, is best
represented by the vanadyl ion. VO 2 , which is stable in aqueous solution and
undergoes complexation with a wide range of ligands. Vanadium has a rich isopoly¬
anion chemistry, which is discussed in Chapter 16.
The lower oxidation states of niobium and tantalum are unimportant compared to
the +5 state. Because of the general insolubility of the oxides and the lack of stable
lower oxidation states, there is little solution redox chemistry. Niobium!Ill) does
appear to form upon the reduction of niobium(V) with zinc, and is stable in the cold in
the absence of air. but if the solution is heated, decomposition occurs with precipita¬
tion of mixed oxides.
Chromium continues the pattern we have seen for vanadium: The highest oxidation
stale is strongly oxidizing, the lower ones are strongly reducing, and some intermedi¬
ate states disproportionate. Blue aqueous solutions of chromium!II) can be prepared
by dissolution of pure metal in oxygen-free acid solutions or by reduction of chro-
mium(III). These solutions are oxygen sensitive and unstable with respect to water
reduction. The +2 oxidation state can be stabilized through complexation or forma¬
tion ol insoluble salts. A vast number of compounds is known for chromium(III), the
most stable lorm lor this element. Chromium! VI) is a powerful oxidizing agent in acid
solution (as dichromate), but because of the hydrogen ion dependence, it is much less
so in basic solution (as chromate).
♦ II .11
597
596 14 • Some Descriptive Chemistry of the Metals
The heavier congeners, molybdenum and tungsten, have a less interesting redox
chemistry. The emfs are small and the differences relatively unimportant. The chem¬
istry of these elements in iso- and heteropoly acids, multiple bonds, etc. is generally ot
more interest (see Chapter 16).
Group VIIB (7)
Group VIIIB
(8, 9, 10)
The first member of this family, manganese, exhibits one of the most interesting redox
chemistries known; thus it has already been discussed in detail above. Technetium
exhibits the expected oxidation states, and associated with these are modest emi
values. All of the isotopes of technetium are radioactive but Tc has a relatively long
half-life (2.14 x 10 5 years) and is found in nature in small amounts because of the
radioactive decay of uranium. Oxidation states of rhenium range from +7 to - 3, with
some species (e.g., Re0 3 and Re 3+ ) unstable with respect to disproportionation.
TcO;
TcO,
■ Tc 3
Tc 2
“1
-.Tc
Re0 4
ReO,
ReO,
Re 3
1
Re
J
Historically the triads of Fe. Ru, Os; Co. Rh, Ir; and Ni. Pd, Pt have been called
collectively Group V1I1B. This heterogeneous assortment of elements was combined
into a single family more from a desire not to have any group number exceed eight, a
"magic number” in chemistry even before Lewis formalized it in his octet theory,
than from any compelling logic. This, of course, ignored the fact that the set of hve d
orbitals has a capacity of ten electrons, and thus there should be ten families ot
transition elements. Although not fully agreed upon by all chemists. The Commission
on Nomenclature of Inorganic Chemistry (1990) has recommended that numbers 1-18
be used instead of Roman numerals followed by A and B designations. Thus the three
triads now appear in separate groups (8, 9. and 10) and this perhaps is as it should be
since the chemistry of iron is not more similar to that of nickel than it is to that ot
chromium.
HFeOj —ff^->Fe 3 ——^—►Fe 2 —► Fe
_--1
| -
(♦1.4)
.
(♦1.5)
1 „ (+0.5)
, (+0.60) j'
RuO., -
-► RuO., —
—KuU 3
|
(+1.5)
♦0.83
1
(+U>
(-0.1)
Os 3 < ° ,, -->
i
Os
0s0 4
^ OSO2
1 _
1.02
J L
♦0.65
J
Group IB (11)
Oxidation States and EMFs of Groups 1-12
CoO, W» —»Co 3 -^-♦ Co 2 ———♦ Co
RhO,
• RhO,
Rh 3
• Rlri-
Rh -
•Rh
IrO,
• IrO,
->Ir ’
NiO, —^- Ni 3 ,+2 ’ 3) Ni 2 - fty —» Ni
Pd0 3 1+1,1 ->Pd(OH) 2 l+ '- 54 --» Pd 2 092 ~> Pd
PtO, <‘- s > -> Pt(OH) 2 l - u —» Pt 2 " 8 > Pt
The pattern we have seen in the immediately preceding elements continues with
iron and its congeners—the metal and +2 oxidation stale are reducing, the higher
oxidation states are oxidizing species. Members of the cobalt and nickel families,
however, tend to be stable only in the +2 oxidation state unless stabilized by
complexation. The reader may readily apply the methods illustrated previously to
examine the relative stability of the individual oxidation states.
The Group VI11B (8. 9, and 10) metals illustrate well the point made previously
that heavier congeners more readily assume higher oxidation states. Thus iron, cobalt,
and nickel are effectively limited to +2 and +3 oxidation states, but all ot their
congeners have reasonably stable higher oxidation states.
The elements copper, silver, and gold show such anomalies that there sometimes
appears to be little congruence as a family, with the member that is least reactive as a
metal (Au) being the only one that has an appreciable chemistry in the +3 oxidation
state and also the only one to reach the - 1 and + 5 oxidation states (CsAu and AuF ? ),
although both copper and silver may be oxidized to +4. The members ol the family
more or less routinely (silver less frequently) violate the very useful rule of thumb you
have seen earlier: The maximum oxidation state of an element is equal to or less than
its group number (IB, IVB, V11B, etc.). Thus we have CuS0 4 , AgF 2 , and [AuCIJ .
Each member of the family has a different preferential oxidation state (Cu, +2; Ag,
+ I; Au, +3). The one property they do have in common is that none has a positive
emf for M —» M" + ; therefore, the free metals are not affected by simple acids, nor are
they readily oxidized otherwise, leading to their use in materials intended to last. 11
31 Copper and silver will dissolve in nilric acid or hot sulfuric acid. A mixture of hydrochloric and
nitric acids (aqua regia) will dissolve gold. It has recently been reported (hal mixtures of halogens
and quaternary ammonium halides in organic solvents dissolve gold faster than does aqua regia: sec
Nakao. Y. J. Chem. Soc. Chem. Commttn. 1992. 426-427.
598
1 4 • Some Descriptive Chemistry of the Metals
This together with their market value, has led to the term "coinage metals" for the
members of this family. 32
•0.34
Although copper forms compounds in any of four different oxidation states, only
the +2 state enjoys much stability. The +3 state is generally too strong an oxidizing
agent, though Cu(III) has been found in biological systems. Complexation by peptides
can lower the reduction emf to the range 0.45-1.05 V. 33 The free +1 ion will
spontaneously disproportionate (+0.52 V > +0.16 V). Copper(l) compounds are
known.^ however, in the form of complexes such as [Cu(CN),] - , [CuCU] - .
[CuClj] -- . 34 or as the sparingly soluble halides.
Silver forms stable compounds only in the + I oxidation state, all higher states
being strong oxidizing agents. Even silver(I) is not overly stable, as shown by the large
reduction potential (0.80 V), and has become a common oxidizing agent in inorganic
and organometallic synthesis. 35 The photosensitized reduction of silver halides is. of
course. Ihe basis of photography.
None of the oxidation states occurring in gold compounds can really be said to be
thermodynamically stable. Gold (II) and gold(I) are subject to disproportionation. The
reduction potential ofgold(III) togold(I) is marginally above that necessary to oxidize
water, but the presence of complexing agents can stabilize + I and +3. with the latter
usually being the more stable.
Group 11B (12) The c/'V configuration of this family is not conducive to an extensive redox chem¬
istry. The overwhelming tendency is to lose the s electrons to form stable + 2 cations;
indeed, this essentially describes the entire redox chemistry of zinc.
These metals occur in elementary form in nature and were probably the first metals known. Gold
and silver arc now considered too valuable to use in coins and have been replaced by less expensive
metals, Alloys of copper (e.g., the “nickel" coin is composed of 75% copper and 25% nickel) arc
still widely used.
" Margerum, D. W.; Owens. G. D. In Metal Ians in Biological Systems'. Sigel. H.. Ed.; Marcel
Dekker: New York. 1981; Vol. 12. p 75.
34 Stevenson. K. L.; Grush. M. M.: Kurtz. K. S. Inorg. Chem. 1990, 29. 3150-3153.
" Sec. for example, Cotton, F. A.; Feng. X.: Matusz. M. Inorg. Chem. 1989. 2B. 594-601.
The Lanthanide and Actinide Elements
599
The ability of mercury to form an Hg— Hg bond (cadmium lo a much less extent)
plus a greater tendency to form coordination compounds compared to the other
members of the group increases the complexity of ils Latimer diagram somewhat, but
not much. Its electrochemistry is straightforward, with both mercury!I) and mer-
cury(Il) being stable in aqueous solution.
This section includes the chemistry of the elements La to Lu. and Ac to Lr. 3fi In
addition, some speculations are made concerning heavier elements that may be
synthesized in the future.
The lanthanides are characterized by gradual filling of the 4/ subshell and ihe
actinides by filling of the 5/subshell. The relative energies of the ml and (// - 1/
orbitals are very similar and sensitive lo the occupancy of these orbitals iFig. 14 .3). 32
The electron configurations of the neutral atoms isee Table 2.1) thus show some
irregularities. Notable is the stable/' configuration found in Eu. Gd. Am. and Cm. For
the + 3 cations ol both the lanthanides and actinides, however, there is strict reg¬
ularity; all have 4/”5«/‘W or Sf"M a 7s° configurations.
In many ways the chemical properties of the lanthanides are repeated by the
actinides. Much use of this similarity was made during the early work on the chemistry
ol the synthetic actinides. Given that these elements were often handled in very small
quantities and are radioactive, prediction of their properties by analogy to the lan¬
thanide series proved very helpful. On the other hand, it should not be thought that the
actinide series is merely a replay of the lanthanides. There are several significant
differences between the two series related principally to the differences between the 4/
and 5/ orbitals.
Stable Oxidation The characteristic oxidation state of the lanthanide elements is +3. The universal
States preference for this oxidation state together with the notable similarity in size led to
great difficulties in the separation of these elements prior lo the development of
lh Most chemists would consider the elements La and Ac to he of Group 3 and not lanthanides or
actinides. However, it has been argued, based on electronic configurations ihat La and Ac are more
properly placed in the periodic table as the first members of the lanthanide and actinide series and
that Lu and Lr arc best placed in Group 3. See Chapter 2 and Jensen. W. I). J. Chem. Ethic. 1982,
59. 634-636.
” Compare the discussion of a similar problem ul versus s orbitals) in transition metals. Chapter 2.
The Lanthanide and
Actinide Elements
6(J0
14 Some Descriptive Chemistry of the Metals
The Lanthanide and Actinide Elements
601
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Mv No
Fig. 14.3 Approximate relative energies of the/"~W and/V electron configurations.
[From Fred. M. In Lanthanide/Actinide Chemistry. Fields. P. R.; Moeller, T.. Eds.;
Advances in Chemistry 71; American Chemical Society: Washington. DC. I%7. Reproduced
with permission.]
chromatographic methods. 38 Despite their propensity to form stable +3 cations, the
lanthanides do not closely resemble transition metals such as chromium or cobalt. The
free lanthanide metals are more reactive and in this respect are more similar to the
alkali or alkaline earth metals than to most of the transition metals. They all react with
water with evolution of hydrogen. One difference lies in the sum of the first three
ionization energies—from 3500 to 4200 kJ mol - ' (36 to 44 eV) for the lanthanides,
compared with 5230 kJ mol -1 (54.2 eV) for Cr 3+ and 5640 kJ mol 1 (58.4 eV) for
Co 3+ . A second factor is the heat of atomization necessary to break up the metal
lattice: Transition metals with d electrons available for bonding are much harder and
have higher heats of atomization than the alkali, alkaline earth, and lanthanide metals.
Two lanthanides, europium and ytterbium, are particularly similar to the alkaline earth
elements. They have the lowest enthalpies of vaporization and the largest atomic radii
of the lanthanides (Fig. 14.4), making them more similar to barium in their properties
than to typical lanthanides. Presumably these elements donate only two electrons to
the bonding orbitals (“bands") in the metal and may be said to be in the "divalent"
state in the metal unlike their congeners. These same two elements resemble the
3 For a discussion of Ihe difficulties and confusion surrounding the early chemistry of these elements,
see Systematics and the Properties of the Lanthanides; Sinha. S. P., Ed.; Reidel: Boston. 1983. and
Kremers, H. E. J. Chem. Educ. 1985. 62. 665-667.
Atomic number
Fig. 14.4 Atomic radii of
barium, lanthanum, the
lanthanides, and hafnium.
[From Spedding, F. H.;
Daane. A. H. The Rare
Earths; Wiley: New York,
1963. Reproduced with
permission.]
alkaline earth metals in another respect—they dissolve in liquid ammonia to yield
conducting blue solutions (see Chapter 10).
Although +3 is the most characteristic oxidation state of the lanthanides (the only
one found in nature), the +2 oxidation state is of some importance. 3 ’ As might be
anticipated from the above discussion, Eu 2+ and Yb 2+ are the most stable dipositive
species. These ions are somewhat stabilized by 4/ 7 and 4/ 14 configurations (from
exchange energy) enjoying the special stability of half-filled and filled subshells. 40
Aqueous solutions of Eu 2+ , Yb 2+ . and Sm 2+ can be prepared, but all reduce water
over time (Yb 2+ and Sm 2+ rapidly) and all are readily oxidized by oxygen. Other
lanthanides (Nd, Dy, Tm, Ho) form M(II) compounds which are stable as solids
(Table 14.3). Not all "divalent" lanthanide compounds are truly such; i.e., some
do not contain M(II) ions (see above). For example. LaU, Cei-,, etc. have been
formulated as M 3+ (I _ ) 2 e“. Although this formulation appears strange because
of the free electron, it is not more so than Na + e _ encountered in Chapter 10 or
[K(crown-6)][e(solvent)] in Chapter 12. However, in contrast to the electrolytic be¬
havior of these electrides in ammonia or crown ether solution, the lanthanide diiodides
have delocalized electrons and are actually considered to be metallic phases. 41
Oxidation states higher than + 3 are exhibited by Ce, Pr, and Tb, but only Ce J ' is
stable (kinetically) in water. It is a very strong oxidizing agent in aqueous solution
(£° = 1.74 V) and is used as a volumetric standard in redox titrations. Some of its
salts [e.g., cerium(IV) ammonium nitrate, cerium(IV) sulfate] find application in
3V Johnson. D. A. Adv. Inorg. Chem. Radiochern. 1977, 20. 1-132. Meyer. G. Chem. Rev. 1988, 88.
93-107. Mikheev, N. B.; Kamenskaya, A. N. Coord. Chem. Rev. 1991. 109, 1-59.
40 Exchange energy associated with nearly half-filled and filled configurations ol samarium(II) and
thulium(U) is not thought to contribute much to the stability of halide salts. See Johnson, D. A.
J. Chem. Educ. 1980. 57, 475-477 for thermodynamic considerations,
at Corbett. J. D. In Solid State Chemistry: A Contemporary Overview; Holt. S. L.; Milstein, J. B.;
Robbins, M.. Eds.; Advances in Chemistry 186; American Chemical Society: Washington DC,
1980; Chapter 18.
602 14 • Some Descriptive Chemistry of the Metals
Table 14.3
Oxidation states of
Symbol
2+
3+
4 +
Symbol
1 +
2+
34-
4+
5+
6+
7+
La
Ac
Ce
Th
( + )
Pr
( + )
Pa
( + )
Nd
( + )
( + )
U
Pm
Np
Sm
Pu
( + )
Eu
Am
( + )
(?)
Gd
Cm
( + )
(?)
(?)
Tb
( + )
Bk
Dy
( + )
( + )
Cf
( + )
( + )
(?)
Ho
( + )
Es
(?)
(?)
Er
Fm
Tm
( + )
Md
(?)
Yb
No
Lu
Lr
•' Abbreviations: + , exists in solution; ( + ), found in solid state only; (?), claimed but not
substantiated. Bold face represents the most stable oxidation stale. For lanthanides, sec
Meyer. G. Chan. Rev. 1988 . 88, 93-107. For actinides, sec Katz. J. J.; Morss. L. R.; Scaborg,
G. T. In The Chemistry of the Actinide Elements', Katz. J. J.. Seaborg, G. T.: Morss, L. R.,
Eds.; Chapman and Hall: London. 1986; Vol. 2, Chapter 14.
organic chemistry as oxidizing agents. 42 Although all of the actinides exhibit a +3
oxidation state, it is not the most stable one for several of them. Thorium(lll) and
protactinium(III) exist in the solid state only, and although uranium(III), nep-
luniumllll). and plutonium(III) have an aqueous chemistry, greater stability is found
in higher oxidation states. In contrast to the lanthanides, the actinides utilize their ./'
electrons more readily and thus exhibit positive oxidation states equal to the sum of
the 7s, 6d, and 5/electrons: Ac(III). Th(IV). Pa(V), U(VI). and Np(VII). As in the lirst
transition series, this trend reaches a maximum at +7, and thereafter there is a
tendency toward lower maximum oxidation states (see Table 14.3). A reduced ten¬
dency to use 5/electrons as one progresses along the actinide series is apparent: U(III)
may be oxidized with water, Np(III) requires air. and Pu(III) requires a strong
oxidizing agent such as chlorine. The +4 state is the highest known for curium,
berkclium. and californium, and beyond these elements only +2 and +3 oxidation
states have been substantiated. Nobelium is actually more stable in solution as No 2
than No 3 (cf. Yb 2+ ).
The aqueous chemistry of the +3 and +4 actinide ions is complicated by their
tendency to hydrolyze and polymerize. Higher oxidation states are represented by
stable actinyl ions (e.g., MOT, MO;, and MO, - ).
j:: Kagan. H. B.; Namy. J. L. Tetrahedron 1986. 42. 6573-6614.
The Lanthanide
and Actinide
Contractions
The Lanthanide and Actinide Elements 603
0 I 2 3 4 5 6 7 8 9 10 II 12 13 14
Number of/electrons
Fig. 14.5 Ionic radii (C.N. = 6) of Sc’, Y 3 , La’, Ln’, and An’ ions.
As a consequence of the poor shielding of the 4/and 5/electrons, there is a steady
increase in effective nuclear charge and concomitant reduction in size with increasing
atomic number in each series. 43 Although this trend is apparent from the atomic radii
(Fig. 14.4), it is best shown by the radii of the +3 cations (Fig. 14.5). There are two
noticeable differences between the two series of ions: (I) although the actinide con¬
traction initially parallels that ol the lanthanides, the elements from curium on are
smaller than might be expected, probably resulting from poorer shielding by 5/
electrons in these elements; (2) the lanthanide curve consists of two very shallow arcs
with a discontinuity at the spherically symmetrical Gd 3 (4/ 7 ) ion. A similar discon¬
tinuity is not clearly seen at Cm 3
A consequence of the lanthanide contraction is that when holmium is reached, the
increase in size from n = 5 - n = 6 has been lost and Ho 3 is the same size as the
much lighter Y 3 (104 pm) with correspondingly similar properties. 44 The contraction
does not proceed sufficiently far to include Sc 3 (88 pm), but its properties may be
extrapolated from the lanthanide series, and in some ways it provides a bridge
between the strictly lanthanide metals and the transition metals.
43 See Chapter 18 for relativistic considerations.
44 For other consequences, see Chapter 18.
604
14 Some Descriptive Chemistry of the Metals
The f orbitals The / orbitals have not been considered previously except to note that they are
ungerade (Chapter 2) and that they are split by an octahedral field into three levels,
t lfl , t 2g , and a lf , (Chapter II). A complete set of seven 4/ orbitals is shown in Fig. 14.6.
As with the d orbitals, there is no unique way of representing them, nor is there even a
way which is optimum for all problems. Thus Fig. 14.6 presents two sets, a “general
set” and a “cubic set.” The latter is advantageous in considering the properties of the
orbitals in cubic (i.e., octahedral and tetrahedral) fields.
Differences between
the 4 f and 5 f
Orbitals
As with other orbitals of the same type (same /), the 4/and 5/orbitals do not differ in
the angular part of the wave function but only in the radial part. The 5/orbitals have a
radial node which the 4/ orbitals lack, but this is not likely to be of chemical
significance. The chief difference between the two seems to depend on the relative
energies and spatial distributions of the orbitals. The 4/ orbitals populated in the
lanthanides are sufficiently low in energy that the electrons are seldom ionized or
shared (hence the rarity of lanthanide +4 species).
Furthermore, the 4/ electrons seem to be buried so deeply within the atom that
they are unaffected by the environment to any great degree. This point will be
discussed further below. In contrast, the 5/electrons, at least in the earlier elements of
the series, Th to Bk, are available for bonding, allowing oxidation states up to +7. In
this respect these electrons resemble d electrons of the transition metals. Because of
the higher oxidation states in the early actinides, it was once popular to assign these
elements to transition metal families: thorium to IVB (4), protactinium to VB (5), and
uranium to VIB (6). In 1944 Seaborg suggested that this arrangement was incorrect
and that the elements following actinium form a new “inner transition" series analo¬
gous to the lanthanides. 45 This suggestion, known as the "actinide hypothesis,” was
useful in elucidating the properties of the heavier actinides and was fully substantiated
by their behavior (notably their lower oxidation states) and electron configurations.
Nevertheless, we should not lose sight of the fact that in the earlier actinides the 5/
electrons are available for use and that these elements do show certain resemblances
to the transition metals.
Absorption Spectra
of the Lanthanides
and Actinides
The absorption spectra of the lanthanide +3 cations are shown in Fig. 14.7. These
spectra result from /-/ transitions analogous to the d-d transitions of the transition
metals. In contrast to the latter, however, the broadening effect of ligand vibrations is
minimized because the 4/orbitals in the lanthanides are buried deep within the atom.
Absorption spectra of the lanthanide cations are thus typically sharp and line-like as
opposed to the broad absorptions of the transition metals.
The absorption spectra of a number of trivalent actinide ions are shown in
Fig. 14.8. They may be conveniently divided into two groups: (1) Am 3+ and heavier
actinides which have spectra that resemble those of the lanthanides; and (2) Pu J+ and
lighter actinides which have spectra that are similar to those of the lanthanides in some
ways but exhibit broadening resembling that seen in the spectra of the transition metal
ions. Apparently the greater "exposure" of the 5/ orbitals in the lighter actinide
elements results in a greater ligand-metal orbital interaction and some broadening
45 Seaborg was warned not to publish his new periodic table because it would ruin his scientific
reputation. He is quoted as saying sometime later, "I didn’t have any scientific reputation so I
published it anyway." For a discussion of this and other interesting historical developments in
actinide chemistry, see George Kauffman's review. "Beyond Uranium" in Client. Eng. News 1990,
63(47), 18-29.
Fig. 14.6 The/orbitals: (a) plots of the angular part of the wave functions of the/orbitals; (b) contours of a 4/orbital. Dots indicate maxima in
electron density. The lines are drawn for densities which are 10%, of maximum. |(a) From Friedman, H. G.; Choppin, G. R.; Feurerbacher, D. G.
J. Chem. Educ. 1964, 41, 354-358. (b) From Ogryzlo, E. A. J. Cliem. Educ. 1965, 42, 150-151. Reproduced with permission.]
1 4 • Some Descriptive Chemistry of the Metal:
Note change in scale
Note change in scale
Frequency (kK)
Fig. 14.7 Absorption spectra of Pr'\ Nd 5 . Pm 5 . Sm 5 , Eu 5 , Tb 5 . Dy 5 , Ho 5 ,
Tm . Yb' in dilute acid solution. Compare the sharpness of these with that of Ti 5
(Fig. 11.8), a first-row transition element. [Modified from Carnall. W. T.; Fields. P. R
Lanthanide/Actinide Chemistry, Fields. P. R.; Moeller. T.. Eds.; Advances in Chemi
American Chemical Society: Washington. DC. 1967. Reproduced with permission.)
The Lanthanide and Actinide Elements
607
Magnetic Properties
of the Lanthanides
and Actinides
Fig. 14.8 Absorption spectra of Irivalent actinide ions in dilute acid solution. [From Carnall,
W. T.; Crosswhite. H. M. In The Chemistry of the Actinide Elements', Katz. J. J.; Seaborg,
G. T.; Morss. L. R., Eds.: Chapman and Hall: New York. 1986: Vol. 2, Reproduced with
permission.)
from vibrational effects. As the nuclear charge increases, the S/orbitals behave more
like the 4/orbitals in the lanthanides and the spectra of the heavier actinides become
more lanthanide-like.
We observed in Chapter II that the paramagnetic moment of the lanthanide ions
(Table 11.25) could be calculated from the expression. g[JU + |)JI (Eq. 11.41). This
approach is successful because spin-orbit coupling is large and only the ground state is
populated. Ligand field effects are small because (he 4/orbitals do not effectively
interact with the ligands of the complex. For Sm ,+ and Eu 5 . spin-orbit coupling is
not large enough to prevent occupation of the first excited state at room temperature,
but if one includes this occupation in the calculation, good results are obtained.
The magnetic properties of the actinides are quite complex. Whereas the spin-
only formula (ligand field effects large compared to spin-orbit coupling) gives reason¬
able results for the first-row transition compounds, and Eq. 11.41 (spin-orbit coupling
large compared to ligand field effects) gives good results for the lanthanides, neither
formula is adequate for (he actinides. The 5/electrons of the actinides interact much
more with ligands than do the 4/electrons of (he lanthanides. As a result the spin-orbit
coupling and ligand field effects are of comparable magnitude. Experimental values of
the paramagnetic moment vary wilh temperature and in general are lower than those
of the corresponding lanthanides.
698
14 • Some Descriptive Chemistry of the Metals
Coordination
Chemistry 46
Comparison of Inner
Transition and
Transition Metals
The lanlhanides behave as typical hard acids, bonding preferentially to fluoride and
oxygen donor ligands. In the presence of water, complexes with nitrogen, sulfur, and
halogen (except F _ ) donors are not stable. The absence of extensive interaction with
the 4/orbitals minimizes ligand field stabilization energies. The lack of LFSE reduces
overall stability but on the other hand provides a greater flexibility in geometry and
coordination number because LFSE is not lost, for example, when an octahedral
complex is transformed into trigonal prismatic or square antiprismatic geometry. Fur¬
thermore, the complexes tend to be labile in solution. Table 14.4 presents a summary of
these differences based on properties of typical transition metal complexes.
One noticeable difference is the tendency toward increased coordination numbers
in the lanthanide and actinide complexes. This is shown most readily by the early (and
hence largest) members of the series when coordinated to small ligands. The struc¬
tures of the crystalline lanthanide halides, MX 3 , exhibit this effect. For lanthanum,
coordination number 9 is obtained for all of the halides except Lal 3 whereas for
lutetium, only the fluoride exhibits a coordination number greater than 6. The coordi¬
nation number of the lanthanide ions in hydrated salts in which the anion is a poor
ligand tends to be 9 as shown by many X-ray studies. The nine water molecules in
[M(H 2 0) 9 ] 3+ are typically found in a tricapped trigonal prismatic arrangement. The
degree of hydration in solution, however, has long been debated and many early
Table 14.4
Comparison of transition
metal ions and lanthanide
ions 0
Lanthanido ions
First series
transition metal ions
Metal orbitals
4/
3 d
Ionic radii
106—85 pm (1.06-0.85 A)
75-60 pm (0.75-0.6 A)
Common
coordination
numbers
6, 7, 8, 9
4, 6
Typical
Trigonal prism, square
Square plane, tetrahedron.
coordination
antiprism, dodecahedron
octahedron
polyhedra
Bonding
Little metal-ligand
orbital interaction
Strong metal-ligand
orbital interaction
Bond direction
Little preference in
bond direction
Strong preference in
bond direction
Bond strengths
Bond strengths correlate
with electronegativity,
decreasing in the
order: F~, OFT, H 2 0,
NOj, cr
Bond strengths determined
by orbital interaction,
normally decreasing in
following order: CN“,
NHj, H 2 0, OH~, F"
Solution
Ionic; rapid ligand
Often covalent; covalent
complexes
exchange
complexes may
exchange slowly
■' Karrakcr, D. G. J. Chem. Educ. 1970, 47, 424-430.
46 For a discussion of the coordination chemistry of the lanthanides, see Hart, F. A. In Comprehensive
Coordination Chemistry; Wilkinson. G.; Gillard. R. D.: McCleverty. J. A.. Eds.; Pergamon: Oxford,
1987; Chapter 39. The Chemistry of the Actinide Elements; Katz, J. J.; Seaborg, G. T.; Morss,
L. R., Eds.; Chapman and Hall: New York, 1986; Vol. 2.
Coordination Chemistry 609
experiments led to the conclusion that the degree of hydration decreases in progres¬
sion along the series. Evidence came from several kinds of data such as the partial
molal volumes of the hydrated lanthanide +3 ions. 47 As the central ion decreases in
size, the partial molal volume decreases as expected until crowding of the ligands
becomes too intense. At this point (Sm), a water molecule is expelled from the
coordination sphere and the molal volume increases temporarily before resuming (Tb)
a steady decrease (Fig. 14.9). Luminescence lifetime studies have been interpreted to
give formulations for hydrated Eu 3 + and Tb’^ ions in solution of [Eu(H 2 0) 9 6±0 5 ] 3+
and [Tb(H 2 0) 90±05 ] 3+ . 48 In these studies the experimental reciprocal lifetimes of
ions in their excited states (r _l ) can be correlated to the number of water molecules in
the first coordination sphere (Fig. 14.10). It was suggested that the larger early
lanthanides have coordination numbers of 10, while the smaller, later lanthanides have
coordination numbers of 9. As shown in Fig. 14.10, the number of coordinated water
molecules diminishes as they are replaced by oxygen chelating ligands such as nitri-
lotriacetate (nta) and ethylenediaminetetraacetate (edta). Recent neutron diffraction
work, however, is in agreement with coordination numbers of 9 and 8 for the early
lanthanides and 8 for the later ones. 49
Separation of the The early separation of the lanthanides was beset by difficulties as a result of the
Lanthanides and similarity in size and charge of the lanthanide ions. The separations were generally
Actinides based on slight differences in solubility, which were exploited through schemes of
fractional crystallization. The differences in behavior resulting from a decrease in ion
Fig. 14.9 Partial molal
volumes of hydrated Ln 3 .
Lines represent suggested
nine- and eight-coor¬
dination. The hydrated
Sm 3+ and Gd 3+ ions
represent equilibria between
the two species. [From
Spedding, F. H.; Pikal,
M. Ayers, B. O.
J. Phys. Chem. 1966 , 70,
2440-2449. Reproduced
with permission.]
— r tpm)
47 Spedding, F. H.; Pikal. M. J.; Ayers, B. O. J. Phys. Chem. 1966 . 70, 2440-2449. For a current
discussion, sec Rizkalla. E. M.; Choppin. G. R. In Handbook on the Physics and Chemistry of Rare
Earths; Gschncidncr. K. A.; Eyring. L„ Eds.; North-Hoiland: Amsterdam, 1991; Vol. 15,
pp 393-442.
4 Horrocks. W. DcW.; Sudnick. D. R. J. Am. Chem. Soc. 1979, 101, 334-340. Horrocks, W. DcW,;
Sudnick. D. R. Acc. Chem. Res. 1981. 14, 384-392.
49 Helm. L.; Merbach. A. E. Eur. J. Solid State Inorg. Chem. 1991, 28. 245-250.
610
14- Some Descriptive Chemistry of the Metals
Fig. 14.10 A plot of reciprocal luminescence lifetime (t _ i ) vs. mole fraction of H 2 0 for
D 2 0/H,0 Tb(III) solutions. Here we see a coordination number of 9 for Tb(lII) in contrast to
a coordination number of 8 shown in Fig. 14.9. Experiments like these suggest that perhaps
the early lanthanides have coordination numbers of 10 while the later ones have coordination
numbers of 9. In the presence of a quadridentate ligand, nitrilotriacetatc (nta). four water
molecules are lost and the number of coordinated water molecules drops to five. When
ethylenediaminetctraacetate (edta), a hexadcntate ligand, is added, the number of water
molecules drops to three. [From Horrocks, W. DeW.; Sudnick, D. R. Acc. Cliern. Res. 1981.
/■/, 384-392. Reproduced with permission.]
radius along the series are commonly attributed to a decrease in basicity, reflected by
a decrease in solubility of the hydroxides, oxides, carbonates, and oxalates. The
fractional crystallization and fractional precipitation methods are extremely tedious
and have been replaced by more efficient techniques.
The decrease in basicity (or more realistically, the increase in acidity) of the
lanthanides provides an opportunity for employing coordinating ligands to effect
separation. Other things being equal,' 1 ' the more acidic a cationic species the more
readily il will form a complex. In practice, the lanthanides are placed on an ion-
exchange resin and eluted with a complexing agent, such as citrate ion or a-hydroxy-
isobutyrate ion. 51 Ideally, the complexes should come ofl'the column with minimum
overlapping of the various bands (Fig. 14.11). Such processes have increased the
amounts of lanthanides available and opened up many possibilities for commercial use
fe.g.. rare-earth phosphors for color television). The initial separations of many of
the actinide elements as they were synthesized were effected by similar methods.
Lanthanide Chelates The stability of lanthanide complexes can be increased by means of the chelate effect,
and much early work was directed toward the elucidation of the stability of the lan¬
thanide chelates. 52 The results arc only partially interprctable in terms of simple
As we shall see below, these "other things" may be extremely complicated.
51 Ka,z - J - J ' : Morss, L. R.; Scaborg, G. T. In The Chemistry of the Actinide Elements-. Katz. J. ].:
Morss. L. R.; Seaborg. G. T.. Eds.: Chapman and Hall: New York. 1986; Vol. 2, pp 1131-1133.
Moeller. T.; Martin. D. F.: Thompson. L. C.: Ferrus. R.; Fcistel. G. R.: Randall, W. J. Cliern. Rev.
1965. 65. 1-50.
Coordination Chemistry 611
trivalent lanthanide and
actinide ions on a Dowex
50 cation-exchange resin
with an ammonium
a-hydroxy isobutyrate
eluant. The band for Lr 3
is predicted. (From Katz.
J. J.; Morss, L. R,;
Seaborg, G. T. In The
Chemistry of the Actinide
Elements-. Katz, J. J.;
Morss, L. R.; Seaborg,
G. T., Eds.; Chapman and
Hall: New York. 1986;
Vol. 2. pp 1131—1133.
Reproduced with
permission.]
models. Figure 14.12 portrays the relative stabilities of various lanthanide chelates.
Two types of behavior may be noted: 5 ' (I) "ideal'’ behavior exemplified by chelates of
ethylenediamineteiraacetate (edta) and the closely related traits-i .2-cyclohexanedi-
aminetetraacetate (edta); 54 and (2) "nonideal" behavior as exemplified by diethylene-
triaminepentaaceiate (dtpa) complexes. The former conforms to our expectations
based on simple electrostatic or acid-base concepts of size and charge (a more or less
uniform increase in stability accompanying the decrease in ionic radius). The discon¬
tinuity at gadolinium (the "gadolinium break") could be attributable to the discon¬
tinuity in crystal radii at this ion or. more plausibly, both may reflect small LFSEs
associated with splitting of the partially filled / orbitals. The position of yttrium on
these stability curves is that expected on the basis of its size—il falls very close to
dysprosium.
Unfortunately, about half of the ligands that have been studied in complexes wilh
all of the lanthanides show discrepancies from the simple picture presented above and
must be considered type 2 ligands. In general, these may be characterized as having
stability/atomic number curves similar to type 1 for the lighter lanthanides, usually
with a break at gadolinium. The behavior of the heavier lanthanides is variable,
however, often showing essentially no change in stability, sometimes even showing
decreased stability wilh increasing atomic number. Furthermore, the placement of
53 This classification is adequate for the present discussion, but some authors further subdivide the
second category into two classes: sec Footnote 52.
54 This behavior is ideal in the sense that it follows our preconceived notions of what should occur.
612
14• Some Descriptive Chemistry of the Metals
Fig. 14.12 Formation
constants at 25 °C for 1:1
chelates of Ln 3+ ions
with various aminepoly-
carboxylate ions (ida,
iminodiacetate; nta. nitrilo-
triacetate; , AMiydroxy-
ethylethylenediaminetri-
acetate; edta, ethylene-
diaminetetraacetate; cdta.
lrans-l ,2-cyclohexane-
diaminetetraacetate; dtpa.
diethylenetriaminepenta-
acetate). [From Moeller. T.
J. Chem. Educ. 1970. 47,
417-430. Reproduced with
permission.]
Atomic number
yttrium on these curves is variable, often falling with the pre-gadolinium elements
rather than immediately after gadolinium as expected on the basis of size and charge
alone.
Several factors have been advanced to account for the unusual behavior of the
type 2 complexes. First, ligand field effects might be expected to influence the position
of yttrium, since it has a noble gas configuration with no d or/electrons to provide
LFSE, in contrast to all of the lanthanide ions except Gd 3 ' 1 ' and Lu 3 \ Obviously,
however, this is insufficient to account for the variable results for the Tb 3f -Lu 3 '
complexes. A second factor is the possibility of coordination numbers greater than 6,
which may also vary along the series. Thus it is entirely possible that an effect similar
to that seen previously for the degree of hydration is taking place. At some point along
the series the decrease in metal ion size might cause the expulsion of one of the donor
groups from a multidentate ligand and decreased stability. This point could be reached
at different places along the series depending upon the geometry and steric require¬
ments of the multidentate ligand. It should be remembered that the thermodynamic
stability of complexes in aqueous solution reflects the ability of the ligand to compete
with water as a ligand and that the observed trends will be a summation of the effects
of the lanthanide contraction, etc., on the stability of both the complex in question and
the aqua complex. For this reason it is not surprising that the situation is rather
complicated.
Lanthanide and actinide complexes, Ln(L-L) 3 and An(L-L),,, of sterically hin¬
dered /3-diketonates [e.g., [Me 3 CC(0)CHC(0)CMe 3 r (dpm) and [F 3 CCF 2 CF 2 -
C(0)CHC(0)CMe 3 ]~ (fod)], are of considerable interest because of their volatility.
Despite their high molecular weights, they have measurable vapor pressures at tem¬
peratures below the boiling point of water. This volatility has been exploited in
The Transactinide Elements
613
Fig. 14.13 Proton NMR
spectra (60 MHz) of
l-heptanol. (a) Spectrum
of 0.3 M l-heptanol in
CDCIj. (b) Spectrum
of 0.78 molar ratio of tris-
(dipivaioylmethanato)-
europium(lll)/l-heptanol in
CDC1 3 . [From Rabenstein,
D. L. Anal. Chem. 1971.
43, 1599-1605, as modified
by Mayo, B. C. Chem. Soc.
Rev. 1973. 2, 49-74.
Reproduced with
permission.]
separating the lanthanides by means of gas chromatography. 55 In addition they have
applications as antiknock additives; in trace analysis, solvent extraction, and vapor
plating of metals; and as homogeneous catalysts. 56
Perhaps /3-diketonates such as dpm and fod have attracted greatest attention as
NMR shift reagents. 57 In 1969 Hinckley discovered that the complicated proton NMR
spectrum of cholesterol is greatly simplified in the presence of Eu(dpm) 3 (py) 2 . 5
Simplification occurs because chemical shifts are induced by the paramagnetic lan¬
thanide ion. An example of the effect is shown in Fig. 14.13. The prolon NMR
spectrum of heptanol, without a shift reagent, is very complicated because of acciden¬
tal overlap of signals (Fig. 14.13a). In contrast, a spectrum of Eu(dpm) 3 solution in
heptanol shows the resonance of each set of equivalent nuclei as independent signals
(Fig. 14.13b). Although the availability of high-field NMR instruments has reduced the
need for NMR shift reagents in organic chemistry, applications to biological systems
(lipid bilayers, proteins, ion mobility) continue to grow. 59
The Transactinide At one time il was considered extremely unlikely that there would be any significant
-- chemistry for elements with atomic numbers greater than about 100. The nuclear
Elements ___ stability of the transuranium elements decreases with atomic number, so that the half-
lives for the heaviest elements (Table 14.5) become too short for fruitful chemical
studies (i.e., ^ = seconds). 60 However, advanced chemical techniques have helped
55 Sievers, R. E.; Brooks. J. J.: Cunningham. J. A.; Rhine. W. E. In Inorganic Compounds with
Unusual Properties-. King. R. B.. Ed.; Advances in Chemistry 150; American Chemical Society:
Washington. DC, 1976; pp 222-231.
5 Wenzel. T. J.; Williams. E. J.; Haltiwanger. R. C.; Sievers, R. E. Polyhedron, 1985, 4, 369-378.
57 Han, F. A. In Comprehensive Coordination Chemistry, Wilkinson. G.; Gillard, R. D.; McCIcverty,
J. A., Eds.; Pcrgamon: Oxford, 1987; pp 1100-1105.
58 Hinckley, C. C. J. Am. Chem. Soc. 1969. 91, 5160-5162.
it Williams, R. J. P. Struct. Bonding (Berlin) 1982. 50, 79-119. Lanthanide Probes in Life, Chemical
and Earth Sciences: Theory and Practice', Bunzli, J.-C. G.; Choppin. G. R., Eds.; Elsevier:
Amsterdam, 1989.
60 Not only are the half-lives short, but in some instances it is not possible to produce more than about
one atom per week!
614
14- Some Descriptive Chemistry of the Metals
Table 14.5
Half-lives of selected Element
Atomic number
Mass number
Half-life
actinide nuclides 0 , . .
Actinium
89
227
21.8 yr
Thorium
90
232
1.41 x lO 10 yr
Protactinium
91
231
3.28 x I0 4 yr
Uranium
92
238
4.47 x 10 s yr
Neptunium
93
237
2.14 x I0 6 yr
Plutonium
94
239
24,150 yr
Americium
95
241
433 yr
Curium
96
248
3.4 x I0 5 yr
Berkelium
97
249
320 days
Californium
98
249
350 yr
Einsteinium
99
253
20.5 days
Fermium
100
257
100 days
Mendelevium
101
256
1.27 h
Nobelium
102
255
3.1 min
Lawrencium
103
256
31 s
Rutherfordium
104
257
4.3 s
Hahnium
105
260
1.5 s
Unnilhexium
106
263
0.9 s
Unnilseptium
107
262
4.7 ms
“ The Chemistry of the Actinide Elements ; Katz. J. J.; Seaborg, G. T.; Morss, L. R.. Eds.;
Chapman and Hall: New York. 1986; Vol. 2.
elucidate information about these elements. More promising is the outlook for further
synthesis of transactinide elements. With the synthesis of l0I Md, l02 No, and , ()J Lr in
the years 1955 to 1961. the actinide series was completed. Since then six transactinide
elements. l04 Rf (1969), lo5 Ha(1970). l06 Unh (1974). 107 Uns (1981). l08 Uno (1984). and
U)y Une (1982), which are congeners of hafnium, tantalum, tungsten, rhenium, os¬
mium, and iridium, have been synthesized. 61 Claims for element 110 have not been
confirmed. 63 There has been much speculation over the possibility of stable species of
even higher atomic number. Theoretical calculations on the stability of nuclei predict
unusual stability for atomic numbers 50, 82, 114. and 164. 63 The prediction is borne
out for , 0 Sn. which has more stable isotopes than any other element, and for H2 Pb and
ktU i. which tire the heaviest elements with nonradioactive isotopes. The stability of
nuclei in the regions of the “magic numbers" has been described allegorically by
Seaborg as mountains in a sea of instability, as shown in Fig. 14.14. The expected
stability is proportional to the elevation of the islands above "sea level." The penin¬
sula running "northeast" from lead represents the decreasing stability of the actinide
elements. The predicted island of stability at atomic number 114 and 184 neutrons may
American names are nobelium (102). lawrcncium (103). rutherfordium (104). and hahnium (105).
Soviet workers have suggested jolioiium (102). kurchatovium (104). and neilsbohrimn (1051. Their
German counterparts have proposed neilsbohrium (107), hassium (108). and meitnerium (109). No
name has yet been suggested for element 106. Dispute over who first discovered elements 102. 103.
104. and 105 continues. Dagani. R. Cliem. Eng. News 1992. 7(7(37). 4-5.
6: Kauffman. G. Client. Eng. News 1990. 6/1 (47). 18-29.
63 See Seaborg. G. T.: Keller. O. L.. Jr. fit The Chemistry of the Actinide Elements ; Katz. J. J.;
Seaborg. G. T.; Morss, L. R.. Eds.; Chapman and Hall: New York. 1986; Vol. 2. Chapter 24.
“Future Elements."
The Transactinide Elements
Mountain
Submanr
-Submarine-
. Mountain
Fig. 14.14 Known and predicted regions of nuclear stability surrounded by a sea of
instability. (Front Seaborg. G. T. J. Chem. Educ. 1969, 46. 626-634. Reproduced with
permission.]
be accessible with new methods that make it possible to "jump" the unstable region
and form these nuclei directly. Thus one might expect to find a group of relatively
stable nuclei in the region of elements 113-115. The possibility of jumping to the next
island (not shown) at atomic number 164 provides even more exciting (and improba¬
ble) possibilities to extend our knowledge of the chemistry of heavy elements.
Lawrencium completes the actinide series and fills the 5/set of orbitals. Rutherford-
ium and hahnium would be expected to be congeners of hafnium and tantalum and to
be receiving electrons into the 6 d orbitals. This process should be complete at element
112 (eka-mcrcury). and then the Ip orbitals would fill from element 113 to element 118,
which should be another noble gas element. Elements 119. 120. and 121 should belong
to Groups IA (IK IIA (2). and IIIB (3), respectively. The former two will undoubtedly
have 8,v' and 8,v configurations. If the following elements parallel their lighter con¬
geners. we might expect 121 (eka-actinium), to accept one Id electron and the
following elements (eka-actinides) to proceed with the filling of 6/orbitals. Unfortu¬
nately, we know very little about the relative energy levels for ihese hypothetical
atoms except that they will be extremely close. Thus, although Fig. 2.10 would predict
as the order ol filling: 8.t, 5g, 6/. Id. 8p. etc., it is not known whether this would be
followed or not. Calculations indicate that the levels are so close together that
"mixed" configurations (analogous to the 5 d [ 4f n configurations found in the lan¬
thanides) such as 8 j'6/ '5 g" or $s~7d l 6f i 5g" may occur. 64 For this reason it does not
seem profitable to speculate on the separate existence of the 5g lx and 6/ 14 series.
Seaborg has suggested that the two series be combined into a larger series of 32
elements called the superactinides . 6S His revised form of the periodic chart is shown
in Fig. 14.15. The 6/and 5g elements form an extra long "inner transition" series
followed by (presumably) a series of ten transition elements (154-162) with filling of Id
orbitals, etc. Magic number nucleus 164 would thus be a congener of lead (dvi-lead). 66
Periodicity of the
Translawrencium
Elements
M Mann. J. B.; Waber. J. T. J. Chem. Phys. 1975, S3. 2397-2406.
65 Seaborg. G. T. J. Chem. Educ. 1969. 46. 626-634.
66 Since element 164 would have an atomic number exactly twice that of lead (82), it has been
suggested that it be dubbed “zwei Blei"!
The Transactinide Elements
617
Fig. 14.16 Extrapolated
heat of sublimation of eka-
lead (114), from which, by
application ofTrouton's
rule, a boiling point of
147 °C is obtained. [From
Seaborg, G. T.; Keller,
O. L., Jr. In The Chemistry
of the Actinide Elements:
Katz. J. J.; Seaborg, G. T.;
Morss, L. R., Eds.;
Chapman and Hall: New
York, 1986; Vol. 2,
Chapter 24. Reproduced
with permission.]
3 4 5 6 7
Row in periodic table
While much of the preceding is speculative, it is no more speculative chemically
than Mendeleev’s predictions of gallium (eka-aluminum) and germanium (eka-silicon).
The speculation centers on the possible or probable stability of nuclei with up to twice
as many protons as the heaviest stable nucleus. The latter falls outside the realm of
inorganic chemistry, but the synthesis and characterization of some of these elements
would be most welcome.
Several workers have predicted the properties of certain translawrencium ele¬
ments. 67 For example, the "inert pair effect" should be accentuated making the most
stable oxidation states of eka-thallium, eka-lead, and eka-bismuth +1, 4-2. and 4-3,
respectively. Relativistic effects become so important for these elements that eka-
lead, with its Ip 1 configuration, may be thought of as having a closed shell. Its boiling
point has been predicted to be 147 °C, based on heat of sublimation extrapolations
(Fig. 14.16) and application ofTrouton's rule. The possibility of forming elements with
atomic numbers in the region of 164 is of considerable interest. As discussed in
Chapter 18, elements following the completion of each new type of subshell (e.g., 2p fi ,
3 d'°, 4/ 14 ) show “anomalous" properties, and thus the chemical properties of dvi-
lead should be equally interesting.
67 Seaborg, G. T.; Keller, O. L., Jr. In The Chemistry of the Actinide Elements ; Katz, J. J.; Seaborg,
G. T.; Morss, L. R., Eds.; Chapman and Hall: New York, 1986; Vol. 2, Chapter 24 and references
therein.
618
1 4 • Some Descriptive Chemistry of the Metals
Problems
14.1 Francium has a smaller atomic radius than cesium and radium is smaller than barium.
Explain. (See Pyykko, P. Client. Rev. 1988. 88. 563-594.)
14.2 Account for the fact that a 23% contraction of radii occurs for 3d M 3 ' - ions (Sc 3+ to Cu 3 '’),
but only a 1% contraction occurs in the 5 d series (Lu 5 to Au 3 ). (See Mason, J. J. Client.
Educ. 1988. 65, 17-20.)
14.3 Arrange the following complex ions in order of increasing stability: [AuF 2 ]~, [AuCI 2 ]~,
[AuBrJ - , [Aul 2 ] _ , [Au(CN) 2 l - . Provide a rationale for your series.
14.4 Pyridinium chlorochromate and pyridinium dichromate are widely used in organic syn¬
thesis as oxidizing agents. What are the formulas and structures for these reagents? What
advantages might they have over more conventional oxidizing agents such as potassium
dichromate or potassium permanganate?
14.5 Ruthenium and osmium, unlike iron, form compounds with the metal in the +8 oxidation
state. Can you think of a nonmetal that achieves a +8 oxidation stale in some of its
compounds?
14.6 Compounds of iron exist in which there are 0. 1,2, 3, 4. and 5 unpaired electrons. Find an
example for each spin state. Classify these complexes as low spin or high spin.
14.7 The tendency of a metal ion to form compounds of high coordination numbers decreases
across the first row of the transition elements. Explain.
14.8 Chromium(VI) oxide is strongly acidic, vanadium(V) oxide is amphoteric, titanium(IV)
oxide is inert, and scandium(III) oxide is basic with some amphoteric properties.
a. Explain the relative acidities of these oxides.
h. Write chemical equations which show the amphoteric nature of vanadium(V) and
scandium(III) oxides.
14.9 The powerful oxidizing strength of [Fe0 4 |'~ is shown by its ability to liberate oxygen
from water and to produce dinitrogen from ammonia. Write balanced equations for these
two reactions.
14.10 The following high spin complexes of 1,2-bis(diisopropylphosphino)ethane (dippel have
been prepared: [CrCI 2 (dippe)J 2 , MnClj(dippe). FeCI 2 (dippe), and CoCU(dippe). Suggest a
structure for each and predict its magnetic moment. (See Hermes. A. R.; Girolami. G. S.
Inorg. Chem. 1988. 27, 1775-1781.)
14.11 a. The intense red of an [Fe(phen)jl : '’ solution is replaced by pale blue when cerium(IV)
sulfate is added to it. Explain.
b. Whereas [Fe(phen)(H 2 0) 4 |' and |Fc(phen)4H-0) : |‘‘ are paramagnetic,
|Fclphen)j] k is diamagnetic. Explain.
14,12 The complex, Ni(Ph 2 PCH 2 CH 2 PPh 2 )Br 2 , is diamagnetic. Suggest a structure. The reac¬
tion of diphenylvinylphosphinc with nickel bromide represents one of the more unusual
methods for its preparation:
Suggest a mechanism for the formation of this complex. (Sec Rahn. J. A.: Delian, A.:
Nelson. J. H. Inorg. Chem. 1989. 28. 215-217.)
Problems
619
14.13 The reaction of coball(II) chloride thf in pentane with l-norbornyllithium affords
tctrakis( I-norbornyIIcobalt. What is the oxidizing agent in this reaction? Express this
reaction with a chemical equation. Draw the structure of the complex. Rationalize the
observed magnetic moment (2.0 BM).
14.14 The anions [Ni(Se 4 ) 2 |" and [Zn(Se 4 ) 2 ]-~ have been characterized. hS The nickel complex
has a square planar geometry, but the zinc complex is tetrahedral. Offer explanations as to
why the nickel complex is not tetrahedral and why the zinc complex is not square planar.
14.15 The reaction of Ni(enl 2 I : with I 2 yields a diamagnetic complex of formulation Ni(en) : I ft .
What is the oxidation state of nickel in this complex? Is your answer consistent with "the
absence of a magnetic moment? The complex. lNi{«-C r ,H J (PMc.,i.} ; |I lll . has also been
isolated.' What is the oxidation stale of nickel in this complex? Explain,
14.16 It is possible to isolate /«i#u-Mn(L—L) 2 CI 2 . f/wt.v-|Mn(L— L) : CT : |'. and irtmx-
|Mn(L —L) : CI ; | : ". which have magnetic moments of 6.04. 3.10. and 3.99 BM. respec¬
tively. Classify these complexes as low spin or high spin. (L—L is a bidentute ligand.)
How might you expect the Mn—Cl bond lengths to vary in this series? Why? (.See
Warren. L. F.: Bennett. M. A. Inorg. Chem. 1976. 15. 3126-3140.1
14.17 The crystal structure of pyrazinium chlorochromate has been reported.'" Draw structures
of the cation and anion. Hydrogen bonding exists between the two ions. Predict the atoms
involved and compare your prediction with the observed pattern.
14.18 Titanium reacts with chlorine to form TiCl.i and TiCI,. How might you separate these two
products? (See Groshcns. T J.: Klabundc. K. J. Inorg. Chem. 1990. 29. 2979-2982.)
14.19 How would you prepare the following compounds?
a. |R 4 N||VCIjlMcCNljl b. VCHlpyt,
c. VOCl.tpv). ( 1 . |Ph 4 P|j|VO : CI.|
c. (K ,N);| VOCIj|
(See Zhang. Y . Holm. R. H. Inorg. Chem. 1990. 39. 911-917, lor leading references.)
14.20 The purple dimer. ITiCId 1.2-bi<(diisopropylphosphino)ethnne)|.. exhibits octahedral co¬
ordination about titanium. Suggest a structure for this complex. Is it paramagnetic or
diamagnetic? (See Hermes. A. R.: Girolami. G. S Inorg. Chem. 199(1, 29. 313-317.)
14.21 For many years Pel, was thought to be nonexistent. Suggest reasons for its instability.
The compound has now been prepared in nonaqueous solvents.' 1 How could you prove
that it is not a mixture of I'eI• and l 2 ?
14.22 Iron!Illi chloride reacts with Iriphcnvlphosphine to form trigonal bipyramidal trims-
PeCl.iPI'hib but with Iricyclohe.xylphosphine to form pseiidnieirahedral FeCl.lPCy.t).
OtVcr an explanation for these structural differences. (See Walker. .1. I).; Puli. R. Inorg.
Chem. 1990. 39. 756-761.)
14.23 Potassium hexalluoronickelate(IV) reacts with sodium pentasullide to form nickel
disulfide.' 1
K.Nil-',, + 2Na 2 S< -♦ NiS ; f 4NaP - 2KF + XS
' Ansari. M A : Mahler. C II : Chorghadc. Ci. S.; I .it. Y.-J: liters. .1. A. Inorg. Clinn. 1990, ?9.
3832-3839.
'••’Gray. 1,. K : Higgins. S. J.; l.evason. W.: Webster. M. J, Chem. Soe. Dalton Irons. 1984,
1433-1439.
7,1 Prcssprich. M. R . Willett. R. I).: Paudler. W. W.: Card. G. I.. Inorg. Chem. 1990. 3V. 2872-2873.
71 Yoon. K II.. Kochi. J. K. Inorg. Chem. 1990. 29. 869-874.
7; Bonne.ui. P. K : Shihao. R, K.; Kaner. R. U. Inorg. Chem. 1990. 29. 2511-2514.
620
14 - Some Descriptive Chemistry of the Metals
Do you think that Ni(IV) is reduced to Ni(ll) or do you think that S; disproportionates to
S 2 ~ and S? What is the oxidation state of nickel in NiS 2 ? How do you know .’
14.24 Explain the emfs of the silver halides in terms of their solubilities. What about silver
acetate, which is soluble?
14.25 How can one verify, just by looking at the Latimer diagram of silver, that sodium
thiosulfate (hypo) is useful in photographic processes that require the removal ot excess,
unreacted silver halide? Is this process (fixing) actually a redox reaction .’ Explain.
14.26 When citing the Sandmeyer reaction, organic chemistry textbooks frequently write the
needed copper(I) halide as Cu 2 CI 2 , Cu 2 Br 2 , etc. Comment.
14.27 Consider the complex ions dibromoaurate(I) and tctrabromoaurate(III). Which is more
stable in aqueous solution ? Explain.
14.28 Explain in terms of redox chemistry how the formation of chloro complexes stabilizes
rhenium(III).
14.29 Predict whether each of the following reactions will proceed to the left or the right:
a. 2Fe 3 + Sn 2 - 2Fe 2 + Sn 4
b. 2Cu 2 + 41" - 2CuI(s) + I 2 (s)
c. 2Cu 2 + 4Br" - 2CuBr(s) + Br 2 (l)
d. V0 2 + 2H + Cu - VO 2 + Cu 2 + H 2 0
14 30 Studies of radioisotopes, both natural and from fallout, in Mono Lake. California, showed
that “Sr 226 Ra and 2l,, Pb occur at lower levels than might have been expected.™ Some
actinides'such as 234 Th. 23l Pa, 238 U. and 240 Pu occurred at higher levels than expected, but
others such as 227 Ac and 241 Am did not. The most notable characteristic of Mono Lake is
its high alkalinity (pH = 10) caused by large amounts of carbonate ion (~0.3 M). Suggest
factors that may be responsible for these relative abundances.
14.31 If you did not answer Problem 10.8 when you read Chapter 10. do so now.
14.32 In this chapter we have referred to the synthesis of heavy actinides. Take a trip to the
library and answer the following questions:
a. Americium-243 can be obtained in kilogram quantities. Write an equation for a nuclear
reaction for its preparation.
b. Nuclear reactors allow 239 Pu to be produced by the ton. Write a nuclear equation for its
synthesis.
c. Elements 107. 108. and 109 have been produced by a method known as "cold fusion.
Describe this method.
14.33 Plutonium-239 is an extremely radioactive alpha emitter. Shielding from alpha radiation,
however, is easily accomplished with thin paper. Why then is - 3 Pu considered to be a
dangerous isotope?
™ Simpson. H. J.; Trier, R. M.: Toggweiler. J. R.: Malhicu. G.: Deck, B. L.; Olsen. C. R.; Hammond.
D. E.; Fuller. C.: Ku. T. L. Science 1982. 216. 512-514. Anderson. R. F.; Bacon. M. P.: Brewer.
P. G. Science. 1982. 216. 514-516.
Problems
621
14.34 Actinide ions often form acidic solutions as a result of hydrolysis:
M(H,0>r - M<H 2 0),_,(OH)<"-'> + + H
Arrange the following sets of cations in order of their tendency to undergo hydrolysis:
a. Pu 4 . Pu 3 , Pu 2
b. Ac 3 , U 3 . Pu 3 . Cm 3
14.35 There is no physiological process for plutonium removal from the body. Various chelating
agents have been used as therapeutic reagents for its removal. 74 One of these, shown
below, is a tetracatechol ligand.
Speculate on how this ligand binds to Pu 4 .
14.36 Paramagnetic ions may alter NMR chemical shifts by what is known as a contact shift or
by what is known as a pseudocontact shift. Consult an NMR book and explain the
difference.
14.37 Anhydrous LnX, can be prepared from the reaction of Ln metal and HgX 2 . Write the
balanced equation for this reaction. Can you name an attractive feature of this reaction?
When aqueous solutions of Cel, and CeClj are concentrated, CeI,(H 2 0) 9 and
CeCI 3 (H 2 0) 6 , respectively, crystallize. Suggest structures for both salts. Do you think
heating these salts mighl be a good route for preparing anhydrous CeCI 3 and Cel 5 ?
14.38 Goggles made of didymium (a mixture of Pr and Nd) are preferred for glassworking to
absorb the glare from sodium. Explain why these goggles are more suitable than a pair
made with Ti 3 , for example.
14.39 Biochemists sometimes think of Ln 3 ions as analogues of Ca 2 . Discuss reasons why
there mighl be a resemblance between Ln 3 and Ca 2 . Any notable differences? In what
ways might biochemists exploit a resemblance?
74 Raymond, K. N.; Smith. W. L. Struct. Bonding IBerlinI 1981, 43. 159-186.
622
14 • Some Descriptive-Chemistry of the Metals
La Cc Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Fig. 14.17 Variations in
standard enthalpy changes.
A//", at 298.15 K for
reactions of lanthanides: (a)
M 3 (aq) + edta J- (aq) —
M(edta)“(aq): (b) MCMs)
5CI 2 (g) -» MCI,(s): (c)
M(s) + lCI 2 (g) — MCIi(s).
Open and closed circles
represent estimated and
experimental values,
respectively. [From
Johnson. D. A. J. Chew.
Educ. 1980. 57. 475-477.
Reproduced with
permission.!
14.40 There has been a tendency to view the lanthanide elements as having nearly identical
chemistry. In recent times this view has been criticized. 75 Standard enthalpy changes for
three reactions are plotted in Figure 14.17. How do you account for the dramatic
differences shown in plots (a) and (b)? Can you provide an explanation for the "bumps" at
Eu and Yb in plot (c)'.’
75 Johnson. D, A. J. Client. Ethic. 1980. 57. 475-477.
Chapter
Organometallic
Chemistry
An organometallic compound is generally defined as one that possesses a metal-
carbon bond.' The bonding interaction, as delineated by the journal Organometallics.
must be ionic or covalent, localized or delocalized between one or more carbon
atoms of an organic group or molecule and a transition, lanthanide, actinide, or main
group metal atom." Despite this rather rigorous definition, the borderlines that dis¬
tinguish organometallic chemistry from other branches are sometimes unclear. For
example, all chemists would undoubtedly characterize nickel tetracarbonyl. Ni(CO).,,
as an organometallic compound even though carbon monoxide is hardly a typical
organic compound. Likewise organoboron, organosilicon, organoarsenic, and organo-
tellurium compounds are included in organometallic chemistry even though boron,
silicon, arsenic, and tellurium tire borderline metals. Traditional inorganic chemicals
such as sodium cyanide, although possessing a metal-carbon bond, are not normally
categorized as organometallic compounds.
Organometallic chemistry can be viewed as a bridge between organic and in¬
organic chemistry. On the practical side, nearly 25 billion dollars was realized from
industrial processes utilizing homogeneous catalysis based on organometallic chem¬
istry in 1985. and it is predicted that the role of organomctallics in the production of
pharmaceuticals, agrichemicals, flavors, fragrances, semiconductors, and ceramic
Precursors will continue to expand during the next decade.- Organometallic catalysts
1 There are a number of valuable sources which can provide varied perspectives on a definition for
organometallic chemistry as well as expanded coverage of virtually every topic included in this
chapter. See: Elschenbroich. C.; Salzer. A. Organometallics. 2nd ed.: VCH: Weinheim. 1992.
Crabtree. R. L. The Organometallic Chemistry of the Transition Metals ; Wiley: New York. 1988.
Powell. P. Principles of Organometallic Chemistry. Chapman and Hall: London. 1988. Thayer J S
Urnanoinetallic Chemistry. VCH: New York. 1988. Cullman, J. P.; Hegcdus. L. S.; Norton. J. R.;
Finke, R. G. Principles anil Applications of Organotransition Metal Chemistry. 2nd ed.; University
Science Books: Mill Valley. CA. 1987. Parkins. A. W.: Poller, R. C. At, Intro,lac,ion to
Organometallic Chemistry. Macmillan: London. 1986. Yamamoto. A. Orpanotransition Metal
Chemistry- Wiley: New York. 1986. Lukcharl. C. M, Fundamental Transition Metal Organometallic
Chemistry: Brooks/Cole: Monterey. CA. 1985. Comprehensive Organometallic Chemistry. Wilkin¬
son. G.; Stone. F. G. A.; Abel. E. A.. Eds.: Pergumon: Oxford. 1982.
Parshall. G. W. Organometallics 1987, 6. 687-692 and private communication.
622
624
15 • Organometallic Chemistry
will become increasingly important in an age when temperature (and hence fuel) needs
to be minimized in chemical processes. As petroleum reserves are depleted, it is likely
that such catalysis will play a major role in converting synthesis gas. derived from
coal, into useful organic intermediates.
The 18-E!ectron The first attempt to account for the bonding in transition metal complexes was made
-by Sidgwick, 3 who extended the octet theory of G. N. Lewis to coordination com-
Rule _pounds. Ligands were considered to be Lewis bases which donated electrons (usually
one pair per ligand) to the metal ion which in turn acted as a Lewis acid. Stability was
assumed to be attendant to a noble gas configuration for the metal. The sum of the
electrons on the metal plus the electrons donated from the ligands was called the
effective atomic number (EAN), and when it was equal to 36 (Kr), 54 (Xe), or 86 (Rn).
the EAN rule was said to be obeyed. An alternate and more general statement is that
when the metal achieves an outershell configuration of ns 2 (n - 1 )d'°np t> , there will be
18 electrons in the valence orbitals and a closed, stable configuration. This rule of
thumb, which is referred to as the 18-electron rule, has the advantage of being the
same for all rows of the periodic chart, eliminating the need to remember a different
EAN for each noble gas. Furthermore, the number is an easy one to recall since it is
merely the total capacity of nine orbitals, one set each of .v, p, and d orbitals. Because
the rule is obeyed with rather high frequency by organometallic compounds, es¬
pecially those having carbonyl and nitrosyl ligands, it has considerable usefulness as a
tool for predicting formulas of stable compounds.
Molecular Orbital
Theory and the
18-Electron Rule
As with most rules of thumb, the 18-electron rule is not always strictly obeyed: Stable
complexes with both more than and fewer than 18 outershell electrons are fairly
common. 4 Insight into the connection between stability of organometallic compounds
and the 18-elcctron rule—and a basis for rationalizing the exceptions—can be gained
by reviewing the molecular orbital description of bonding in complexes (Chapter II).
For an octahedral complex (Fig. 11.20), the most stable arrangement will be that in
which all of the bonding orbitals (a, A „ e K . and t 2/ .) are fully occupied and all of the
antibonding orbitals are empty. Since there are nine bonding molecular orbitals, this
will require 18 electrons, as predicted by the 18-electron rule. Complexes will there¬
fore tend to adhere to the rule if they have large A„ values, making occupation of the
antibonding e orbital unfavorable. Included in this category are complexes ofsecond-
and third-row transition metals, which are never found to have more than 18 electrons
beyond the core MOs. There may well be fewer than 18 electrons, however, if the
ligands do not provide stabilization of the t 2x level by tt bonding. This is observed for
complexes such as [WC1J 2- (14 electrons), [TcF h ] 2 " (15 electrons), [OsCIJ" (16
electrons), and [PtFJ - (17 electrons). Ligands such as CO and NO. which are high in
the spectrochemical series because they are good it acceptors, are very effective at
stabilizing the t 2x orbitals. This leads to a larger A„ value and an increase in the total
bonding energy~(Figs. 11.27 and 11.28). As a result, octahedral carbonyl and nitrosyl
complexes are found to seldom depart from the 18-electron rule.
If A„ is small, as is the case for first-row transition metal complexes, occupation
of the weakly antibonding e orbitals is easily possible. As a result, stable com-
3 Sidgwick, N. V. The Electronic Theory of Valency, Cornell University: Ithaca. 1927.
4 Mitchell, P. R.; Parish. R. V. J. Chem. Educ. 1969. 46, 811-814.
The 18-Electron Rule
625
plexes with^ 19 electrons ([Co(H-,0) f ,] ::+ >- 20 electrons ([Ni(en) 3 ] 2+ ), 21 electrons
([Cu(NH 3 )J 2+ ), and 22 electrons ([Zn(NH 3 ) 6 ] 2+ ) are well known. Transition metals
on the left side of the periodic table have few outershell electrons to begin with and to
reach a total of 18 may require more ligands than is sterically possible (the nonexistent
[TiF 9 ] 5- would obey the 18-electron rule). For these metals, stable complexes having
fewer than 18 electrons are thus fairly common: [TiF f) ] 2- (12 electrons), [VCI 6 ] 2-
(13 electrons), [Cr(NCS) 6 ] 3 ~ (15 electrons), etc.
The picture is somewhat more complicated for complexes of other geometries. In
the case of tetrahedral tetracarbonylnickel(O), the four a bonds from the carbonyl
groups result in four strongly bonding molecular orbitals (a, and /,), accommodating
eight electrons (Fig. 11.21). The remaining ten electrons must occupy the e and
orbitals, which are formally nonbonding and antibonding, respectively. Since A, is
relatively small, occupation of the antibonding level is not energetically costly and the
complex is stable. With only four ligands (capable of contributing two electrons each),
any tetrahedral complex in which the metal has fewer than ten electrons available
obviously will have fewer than 18 electrons in total in the molecular orbitals. Thus
tetrahedral exceptions to the 18-electron rule, such as the stable 13-electron species
[FeClJ~, are quite common.
Square planar d s transition metal complexes are consistent exceptions to the 18-
electron rule. The combination of eight metal d electrons and two electrons from each
of the four ligands gives a total of 16. Yet these complexes possess such high stability
that it is often said they obey a 16-electron rule. With 16 electrons, all of the bonding
molecular orbitals in a square planar complex are occupied (Fig. 11.22); any additional
electrons would have a destabilizing effect because they would occupy antibonding
orbitals. The addition of one ligand (donating two electrons) could convert a square
planar species into a five-coordinate. 18-eleclron complex, and in fact, five-coordinate
complexes such as [Ni(CN) 5 J 3_ are well known (see Chapter 12). Yet in many
instances the added ligand leads to a less stable complex.
In general, the conditions favoring adherence to the 18-electron rule are an
electron-rich central metal (e.g.. one that is in a low oxidation state) and ligands that
are good 7r acceptors. 5
Counting Electrons The 18-electron rule has remarkable utility for predicting stabilities and structures of
in Complexes organometallic compounds. By counting the number of outershell electrons surround¬
ing each metal atom in a complex, it is possible not only to predict whether the
complex should be stable, but in some cases, whether there will be metal-metal
bonds, whether the ligands will be bridging or terminal, etc. There are two popular
procedures for electron counting, the so-called neutral atom and oxidation state
methods, each with their ardent supporters. Either method may be used quite success¬
fully. but care must be taken not to mix the two. In other words, a strict loyalty to one
procedure or the other is required when counting electrons in a particular molecule or
ion. The neutral atom method is perhaps more foolproof because it does not require
correct assignment of oxidation states, which can sometimes be difficult for organo¬
metallic compounds.
To use either electron-counting procedure, it is necessary to know how many
electrons each ligand in a complex donates to the metal. Table 15.1 gives electron
contributions for a variety of ligands for both the neutral atom and oxidation state
5 Chu and Hoffmann have discussed some interesting exceptions to the 18-electron rule in (he context
of molecular orbital descriptions. Chu. S-Y.; Hoffmann R. J. Phys. Client. 1982. 86, 1289-1297.
-.£. y. .ii. ■iJLi. :
. , : ; ■■
■•• •' ■ : • - - .
v -1 -- .
f' cs n ■ m i
Mt
111- I
g -s-i S s
iljfi-
eS if I c
, . •', .
. _
\ _ / _ y
u T c \~ a T u “ s
=S,
. £ •.$
< ... 22
^ z
I
i H s
: 8 I £ o'
: S S e I
: _• c g 2
! fills
Z X X <
626
_ Sometimes counted as a four-electron donor if it functions as a w donor.
The 18-Electron Rule
627
methods. The electron count for neutral ligands is the same by either method; thus
phosphines and CO are listed as two-electron donors in both columns of the table. The
electron count for three complexes involving these ligands, Cr(CO) ft , Ni(PF,) 4 . and
Fe(CO) 4 PPh,, would be
Fe
8e~
Cr
6e
Ni
10e"
4CO
8e“
6CO
12e"
4PF 3
8e~
Ph 3 P
2e"
Cr(CO) 6
18e“
Ni(PF 3 ) 4
18e~
Fe(CO) 4 PPh 3
18e“
The electrons counted for the metal atom in each of these complexes are those in its
valence .v and d orbitals. Metals having odd numbers of electrons obviously cannot
satisfy the 18-electron rule by simple addition of CO (or other two-electron) ligands
because the resulting moiety will necessarily also have an odd number of electrons.
For example. Mn(CO), and Co(CO) 4 are both 17-electron species and. consistent with
prediction, do not exist as stable molecules. However, their corresponding anions,
[Mn(CO) 5 ]“ and [ColCOlJ - , are stable species and conform to the 18-elec-
tron ruie:
Mn
7e"
Co
9e“
5CO
lOe"
4CO
8e“
charge
le _
charge
le“
[Mn(CO) s ]-
18c-
[Co(CO) 4 ] “
18e"
(The count shown is for the neutral atom method. In the oxidation state procedure,
each metal would be considered as an M~ species and given an additional electron,
and there would be no entry for the overall charge.)
The dimeric species. Mn : (CO) m and Co 2 (CO) 8 , also are stable and are diamag¬
netic. If it is assumed that each compound has a metal-metal single (two-electron)
bond, the electron count yields a total of 18 for each metal:
2Mn
14c -
2Co
18c-
IOCO
20c-
8CO
16e
Mn—Mn
2e _
Co—Co
2e"
Mn,(CO) ia
, 36e"
Co 2 (CO) 8
36e'
or 18c",
(Mn
or 18e"/Co
The molecular structure of the manganese dimer clearly reveals that there is an
Mn—Mn bond (Fig. 15.la, b). In the cobalt structure, two of the CO ligands are
bridging , i.e.. they are simultaneously bound to both Co atoms (Fig. 15. Ic, d). This
does not affect (he electron count, however, because CO and other neutral ligands
donate two electrons to a complex whether they are terminal or bridging (Table 15.1):
630
1 5 • Organomotallic Chemistry
more extended tt systems exhibit a greater number of binding modes. For example,
the cyclopentadienyl ligand may be pentahapto, tri-, or mono-:
The power of the 18-electron rule for predicting structures of complexes involving
unsaturated ligands can be illustrated with W(CO),(C s H 5 ) 2 . If both C 5 H, ligands were
pentahapto, the compound would have 20 electrons, two "more than the optimum for
stability. However, if one of the ligands is presumed to be pentahapto and the other
trihapto, we have an 18-electron complex:
I his predicted structure is indeed the one observed. 6
Finally, in applying any electron-counting procedure to organometallic com¬
plexes. we must remember that it is merely a formalism—albeit a very useful one. We
must resist any tendency to presume that either method presented here, or the 18-
electron rule for that matter, reveals anything about electron distribution in a
metal-ligand bond or mechanistic details about how a complex forms. What an
electron-counting exercise does provide is a good first approximation of the stability
and structure of an organometallic complex, thus serving much the same function as a
Lewis diagram does for main group compounds.
Metal Carbonyl
Complexes
Almost all of the transition metals form compounds in which carbon monoxide acts as
a ligand. There arc three points of interest with respect to these compounds: (I) Car-
-bon monoxide is not ordinarily considered a very strong Lewis base and yet it forms
strong bonds to the metals in these complexes; (2) the metals are always in a low
oxidation state, most often formally in an oxidation state of zero, but sometimes also
in a low positive or negative oxidation state; and (3) as already discussed, the 18-
electron rule is obeyed by these complexes with remarkable frequency, perhaps 99%
of the time.
6 Huttner. G.; Brintzinger. H. H.: Bell. L. G.; Friedrich. P.; Benjenkc. V.; Neugcbaucr. D
1. Organomet. Chan. 1978. 145. 329.
Metal Carbonyl Complexes
631
Formulas for stable carbonyl complexes formed by metals in the first transition
series are given in Table 15.2. Several arc polynuclear species, which will be discussed
in a later section. Among the mononuclear compounds, the only exception to the 18-
electron rule is hexacarbonylvanadium. V(CO) 6 , which is a paramagnetic, 17-electron
molecule. Interestingly, it does not dimerize to form the 18-electron analogue to
Mn 3 (CO), 0 and Co : (CO) s . If the V,(CO) l2 dimer did form, it would give each metal a
coordination number of 7. which may present too much steric hindrance to allow
stability. Ligand repulsions may overcome a weak metal-metal bond or may provide a
kinetic barrier to dimerization. In any event. V(CO) 6 is less stable than carbonyl
complexes obeying the 18-electron rule. It decomposes at 70 °C and, if given the
opportunity, readily accepts an electron to form the 18-electron anion:
Na + V(CO) 6 - Na + + [V(CO) 6 ]- ( is.2)
The molecular structures adopted by simple carbonyl complexes are generally
compatible with predictions based on valence shell electron pair repulsion theory.
Three representative examples from the first transition series are shown in Fig. 15.2.
The second- and third-row transition metals form a number of compounds analo¬
gous to those in Table 15.2. e.g.. Mo(CO)„. Tc,(CO) l0 . and Re,(CO) 10 . However,
there are also some distinct contrasts between the first row and the heavier elements.
For example, the binary carbonyl complexes of niobium and tantalum, unlike van¬
adium. are unknown and. whereas tetracarbonylnickel is stable, Pd(CO) 4 has only
been observed at low temperatures. Likewise Os(CO) 5 and Ru(CO) 5 are less stable
than Fe(CO) 5 . The dinuclear nonacarbonyl complexes of ruthenium and osmium.
Ru,(CO),, and Os-dCO),,. are also less stable than their iron congener. Fe,(CO) y .
Increasing metal-metal bond strength as one descends a column of the periodic table
leads to much greater stability for Os,(CO), ; and Ru,(CO) i: relative to their mono¬
nuclear M(CO), units than is the case for Fe. 7 In fact, for both Ru and Os. the most
stable binary carbonyls are the trinuclear complexes. Despite the differences enumer¬
ated here, the resemblance of the first-row to the heavier elements is probably greater
for carbonyl complexes than for other classes of coordination compounds.
Table 15.2
Stable carbonyl complexes
of the first-row transition
metols
Mononuclear VtCO) Cr(CO),, Fe(CO), Ni(CO)„
Dinuclear « Mn,(CO)„, Fe,(CO)„ Co^CO)
Trinuclear Fe,(CO) l2
Telranuclear ' Co 4 (CO) l2
Hexanuclear _____ Co ( ,(C O)
" Evidence for the existence of V 2 (CO) l2 at 10 K has been reported. Ford. T. A.; Huber, H.;
Klotzbiicher. W.; Moskovits. M.; Ozin. G. A. Inorg. Chan. 1976. 15. 1666-1669.
7 Unfortunately, the strength of metal-metal bonds in transition metal complexes is hard to determine.
For instance, the Mn — Mn bond strength in Mn : (CO»i„ has been reported to be as low as 67 kJ
mol 1 and as high as 172 kJ mol '. Pulsed lime-resolved photoaeoustie calorimetry places the value
at 159 ± 21 kJ mol -1 and is consistent with recent kinetic results. Goodman. J. L.; Peters. K. S.;
Vaida. V. Organomeltillics 1986. 5. 815-816.
784
16- Inorganic Chains, Rings, Cages, and Clusters
P 5 )J, [(TjJ-CjMc^FejInS-Pj)!.'®' and [( 7 j 5 -C 5 Me,) 2 Cr 2 (^-P 5 )]'°2 A triple decker, with
an As, ring, can be made from pentamethylcyclopentaarsine :"' 3
[CpMo(CO ) 3 ] 2 CpMo(As),MoCp (16.71)
There is an interesting series of oxocarbon anions of general formula [(CO),,] 2-
(Fig. 16.35). The croconate ion, C 5 O 3 - . was the first member of the series to be
synthesized. From a historical point of view it is especially interesting: (I) It was
isolated in 1825 by Gmelin and thus shares with benzene (isolated from coal tar by
Faraday the same year) the honor of being the first aromatic compound discovered. (2)
It was the first “inorganic” substance discovered that is aromatic, although its
importance was unrealized until later. (3) It is a bacterial metabolic product and was
possibly the first “organic" compound synthesized, predating Wohler's synthesis of
urea by three years, although here too the significance was unappreciated at the lime.
All of these oxocarbon anions are aromatic according to simple molecular orbital
calculations. The aromatic stabilization of the anion is apparently responsible for the
fact that squaric acid (H 2 C 4 0 4 ) is about as strong as sulfuric acid ." 34 There is a
considerable and growing body of knowledge of the chemistry of these systems, but
most of it is probably more appropriate to a discussion of organic chemistry . 105
In a formal sense, silicon might be expected to parallel the extensive alicyclic and
aromatic chemistry of carbon, and to some extent it does. Substitution of hydrogen
atoms by methyl groups seems to stabilize these systems. A large series of per-
methylcyclosilanes can be synthesized by treatment of chlorosilancs with an active
metal over a prolonged period of lime:
Me,SiCI, + Na/K (Me,Si) r + (Me,Si)„ (16.72)
L|„H|
n» Scherer, O. J.; Briick, T. Angew. Client. Ini. Ed. Engl. 1987, 26. 59.
i«- Scherer. O. J.: Schwalb, J.; Wolmershauser. G.; Kaim. W.; Gross. R. Angew. Chem. Ini. Ed.
Engl. 1986, 25. 363-364. Chamizo, J. A.; Ruiz-Maz6n. M.: Salcedo, R.; Toscano. R. A. Inorg.
Chem. 1990. 29. 879-880.
no Rhcingold, A. L.; Foley. M. J.; Sullivan. P. J. J. Am. Chem. Soc. 1982. 104. 4727-4729. See also
DiMaio. A.-J.; Rhcingold. A. L. Inorg. Chem. 1990. 29. 798-804.
" M Note that oxalic acid containing carbon in a comparable oxidation state but not aromatic has a K\
approximately equal to Kj for squaric and sulfuric acids, and AC, for oxalic acid is three orders of
magnitude smaller. For a difference of opinion concerning the aromaticity of the oxocarbon anions,
see Aihara. J. J. Am. Chem. Soc. 1981. 103. 1633-1635.
'05 Oxocarbons-, West. R.. Ed.; Academic: New York, 1980. Seitz. G.; Imming. P. Chem. Rev. 1992.
92, 1227-1260.
Cages
785
The product consists of various amounts of high polymer U is very large) and discrete
cyclostlanes with n - 5-35. This is (he largest homologous series of cyclic compounds
now known except tor the cycloalkanes. Although these compounds are formally
saturated, they behave in some ways as aromatic hydrocarbons. They can be reduced
to anion radicals, and EPR spectra indicate that the unpaired electron is delocalized
over the entire ring . 106
Cage structures range from clathrate compounds on the one hand to metal-metal
clusters and boranes on the other. These classes are discussed elsewhere,'°7 and this
section will be restricted to certain nonmetal compounds having cage structures
I he simplest cage-type molecule is white phosphorus. P 4 . Although P, molecules
tsoelectromc with N 2 . are found in phosphorus vapor at higher temperatures P is
more stable at room temperature. »>« This molecule is a tetrahedron of phosphorus
atoms: r r
222.3 pm
.-•P)
P-t P
Such a structure requires bond angles of 60°. Inasmuch as the smallest interorbital
ang e available using only and p orbitals is 90° (pure p orbitals), the smaller bond
angle in P 4 must be accomplished either through the introduction of d character or
through the use ot bent bonds. Ab initio calculations show the importance of d orbital
participation.' 0 '' In spite of ring strain, the P 4 molecule is stable relative to P, or the
nonexistent P„. Nevertheless, the molecule is quite reactive. It can be stored under
water, but it reacts readily with oxygen to form P 4 0 1( „ often called phosphorus
pentoxide, based on the empirical formula. P,0,:
p -t + 50, ► P 4 O lo (( 6 . 7 3 )
Other Phosphorus oxides <P 4 0„. P 4 0 7 . P 4 0. P 4 0„) are known but no. easily pre¬
pared. For example, P 4 0„. a liquid at room temperature (mp 23 °C), can be obtained
by controlled oxidation of P 4 . followed by distillation:'"
P 4 + 30, ► Pj0 6 (16 . 74)
This molecule'" (T d symmetry) has four lone pairs of electrons (one for each phos¬
phorus) which can be donated to one. two, three, or four oxygen atoms to form other
y o TlW-3m PPl ' C " em - ,9S2 ' m '~ Hm Br0Ugh ' L - F: Wcsl ' R - y Chem. Soc. 1981.
'o' Claihnuc compounds are discussed bo.h in Chapter 8 and earlier in .his chapier. while mc.al
clusters and boranes arc found later in this chapter.
" w Bock. H.; M Oiler, H. Inorg. Chem. 1984. 2J. 4365-4368.
" N Schmidt. M. W.; Gordon. M. S. Inorg. Chem. 1985, 24. 4503-4506.
"° Al currcnI P rices y° u cao buy P 4 O 10 for $ 0 .10/g. while P 4 O h cosls aboul $250/g.
Heinze, D. Pure Appl. Chem. 1975 .44. 141-172.
1,2 Jansen - M.: Moebs. M. Inorg. Chem. 1984. 23. 4486-4488.
632
15 Organometallic Chemistry
Fig. 15.2 Structures of the simple carbonyl complexes of chromium, iron, and nickel.
Preparation and Some carbonyl complexes can be made by direct interaction of the finely divided
Properties of metal with carbon monoxide:
Carbonyl Complexes Ni + 4C0 Ni(CO) 4 (bp 43 °C) (15.3)
200 aim
Fe + 5CO Fe(CO) 5 (bp 103 °C) (15.4)
Nickel tetracarbonyl is a highly toxic volatile colorless liquid that is shipped in
cylinders pressurized with carbon monoxide. 8 Its vapor is about six times as dense as
air. Purification of nickel by the Mond process is based on the decomposition of
Ni(CO) 4 , the reverse of Eq. 15.3. The yellow-red iron pentacarbonyl slowly decom¬
poses in air and is sensitive to light and heat. In fact. Fe 2 (CO) 9 , an orange solid, is
prepared by photolysis of Fe(CO) 5 .
Fe(CO), Fe 2 (CO) 9 (mp 100 °C) (15.5)
For most carbonyl complexes, however, the metal must be reduced in the pres¬
ence of carbon monoxide:
CrCI 3 + Al + 6CO AlClj + Cr(CO) 6 (mp 154 ”C) (15.6)
Re 2 0 7 + I7CO ” ™ » 7C0 2 + Re 2 (CO) 10 (mp 177 °C) (15.7)
In Eq. 15.7 the carbon monoxide itself is acting as a reducing agent.
Infrared spectroscopy is a particularly informative technique for characterizing
carbonyl complexes because of the direct connection between the number of C—O
absorptions and molecular structure. Another advantage to the method is that there
are few absorptions in the 1800-2200 cm -1 window except for those arising from
C—O stretching vibrations. We saw in Chapter II that the frequencies of these
absorptions can provide information regarding the carbon-oxygen bond strength in
coordinated CO. It is a straightforward matter to predict the number of C—O stretch¬
ing bands expected for a carbonyl complex based on its symmetry. For a complex
such as Fe(CO) 5 (0 3/l ), we would treat each C—O stretching vibration as a vector
(Fig. 15.3) and then determine the symmetries of the set of vibrations by application of
methods outlined in Chapter 3. The result is that the vibrational modes include two of
A\ and one each of A 2 and E' symmetries, of which only the A'i and £' are infrared
active, according to the D 3/ , character table (Appendix D). Since one of these is
For a historical account of Ludwig Mond's discovery of Ni(CO) 4 . the first metal carbonyl complex,
see Roberts, H. L. J. Orguitoinet. Chem. 1989, 372. 1-14; Abel, E. Ibid. 1990. 383. 11-20.
Metal Carbonyl Complexes
633
O
It
| £ 2C J 3C > 2S 3 H
o — C— Fe r= 5 2 13 0 3
c n % T + + r
o
Fig. 15.3 Identification of the symmetries of the C—O stretching vibrations (represented as
vectors) for Fe(CO) s . The reducible representation, T r , is derived by counting the number of
vectors remaining unmoved during each operation of the D v , point group (see Appendix D for
character table). Its irreducible components are obtained by application of Eq. 3.1.
Polynuclear
Carbonyl
Complexes
nondegenerate and the other doubly degenerate, we expect the IR spectrum for
Fe(CO), to show two C—O stretching absorptions of unequal intensity (Fig. 15.4).
Infrared C—O stretching frequencies for mononuclear carbonyl complexes of the
first-row transition metals are given in Table 15.3.
The dinuclear cobalt complex. Co 2 (CO) 8 (Fig. 15.1) represents a large number of
polynuclear carbonyl species containing bridging carbonyl ligands in addition to the
terminal carbonyl ligands found in all binary metal carbonyl compounds. The —C=0
representation implies that a CO ligand bound lo a single metal is relatively unchanged
from free carbon monoxide, i.e., that the carbon-oxygen bond remains a triple bond.
That the C—O bond order is approximately three in a terminal CO ligand is reflected
in carbon-oxygen stretching frequencies: 2143 cm -1 for free carbon monoxide and
2125-1850 cm -1 for terminal carbonyl groups.
Bridging CO ligands, on the other hand, are electronically much closer to the
carbonyl group of organic chemistry with a carbon-oxygen bond order of approx¬
imately two as found, for example, in ketones. Again, we may infer the bond order
from carbon-oxygen stretching frequencies, which are typically between 1850 and
Fig. 15.4 Infrared
spectrum of the carbonyl
region for Fe(CO) s showing
the absorptions associated
with the A'j and E’
stretching vibrations.
Wavenumbers (cnT )
1889
782
16- Inorganic Chains, Rings, Cages, and Clusters
Tej + , SJ + , Se x + , Seft. Sfc. Sj,;. and Te-.Se],;. The siruclures ofS;\ Se; + . and Te;-
ions have been shown to be square planar.
r -a+
s—s
Se — Se
f Te—Te T
b
s—s
|Q|
Se—Se
1
i-
Note that these are isoelectronic with the previously mentioned S,N 2 . All three are
thought to be stabilized to a certain extent by a Hiickel sextet of t r electrons.
Many cyclopolyphosphines are known. 96 The simpler ones, (RP)„ - 3-6) I Fig.
16.34), are prepared by pyrolysis or elimination reactions.
RPC1 2 + RHP—PHR
^ rpci 2 + - rph 2
(RP) 3 + 2HC1
(RP)„ + nHCl
In addition to the +3 oxidation state seen in the homocyclic rings discussed
above, phosphorus rings exist in which the +5 oxidation state is exhibited:
H N>V°
0- P X OH
H °x p p ^°
0^ ^ X 0H
HO X 0
The anion of this acid results from the oxidation of red phosphorus with hypohahtes in
alkaline solution:
-P„ + 9X0“ + 60 H
(P0 2 )ft“ + 9X“ + 3H 2 0
R n R R
^ m t? ,
R R R R R
(RP), (RP) 4 (RP) 5
. R Fig. 16.34 Structures of
(RP)„ (n = 3-6) molecules.
Woolins. J. D. Non-Metal Rians. Cafes, anil Clusters'. John Wiley: New York. 1988.
Rings
783
There is a series of analogous cyclic thiophosphoric acids with the formula
(HS 2 P)„ that may be prepared by the oxidation of red or while phosphorus with
polysulfides under a variety of conditions. For example, the reaction of white phos¬
phorus with a mixture of sulfur and hydrogen sulfide dissolved in triethylamine (which
acts as a base) and chloroform opens the phosphorus cage to form the tetrameric
cyclic anion:
P 4 + 4Et 3 N + 2H,S + 6S - [Et 3 NH] 4 [P 4 S x ] (16.67)
The square structure of the anion has been confirmed by X-ray crystallography. 97
A series of cyclopolyarsines is known. They may be prepared by a generally
useful reaction that is reminiscent of the Wurtz reaction of organomelallic chemistry:
RAsCl 2 —(RAs) 0 (16.68)
Although four- and five-membered rings can be made in this way. the three-membered
ring requires a special, though related, reaction:
r-Bu r-Bu
\ /
(/-Bu) As — As (r-Bu) + (r-Bu)AsCI,- As —As + 2KCI (16.69)
K K As
r-Bu
This compound is stable only at -31) °C in the dark and in Ihe absence of air. It
spontaneously ignites on exposure to air. 9K
Alkali metal peniaphosphacyclopentudienides (Li and Na) have been obtained in
solution from reactions of red phosphorus and dihydrogenphosphide in dimethyl-
formamide: 99
p —p
P„+ [PH,]"- pXZXp (16-70)
\ p/
Similarity to the C,H; anion is apparent if you allow each phosphorus a lone pair of
electrons, which gives live p orbitals (with six electrons) available for 7r bonding.
Transition metal complexes containing P, rings, however, were known prior to the
synthesis of the free P; ligand (see Chapter 15). 100 Examples are [(Tj s -C J Me,)Fe(7j-‘ i -
97 Falius. H.; Krause. W.: Sheldrick. W. S. Anne"-. Client, lot. Ed. Ennl. I9HI. 21), 103-104.
w Baudlcr. M.; Etzbach. T. Client. Her. 1991. 124, 1159-1160.
99 Baudlcr. M.; Akpapoglou, S.; Ouzounis. D.; Wasgestian. F.; Meinigkc. B.; Diidzikicwicz. H,:
Miinstcr. H. Angew. Client. Ini. Ed. Engl. 1988. 27. 280-281. Hamilton. T. P.; Schaefer. H. F.. III.
Angew. Client, hit. Ed. Engl. 1989, 28. 485.
a 1 Baudlcr. M.: Etzbach. T. Angew. Cheat. Ini. Ed. Ennl. 1991. JO. 580-582.
634
1 5 • Orgonometallic Chemistry
Tobl e 15.3 _
Infrared C — O stretching
frequencies of mononuclear
carbonyl complexes"
1700 cm -1 for bridging CO ligands, compared with 1715 ± 10 cm 1 for saturated
ketones. It would be false, however, to think of a bridging carbonyl group as a ketone.
The M—C—M bond angle is 90° or less, compared to a C—C—C bond angle ot
about 120° for an organic ketone. Furthermore, a bridging CO ligand is almost always
accompanied by a metal-metal bond, which leads to extensive delocalization in the
M—C — M moiety involving overlap of metal orbitals with both tr and rr orbitals of CO
(Fig. 15.5). 9 The infrared spectrum of a metal carbonyl compound provides important
information on the nature of the carbonyl groups present and may allow one to
distinguish between a structure with only terminal carbonyls, such as Mn : (CO) m . and
one containing one or more bridging carbonyl groups, such as Co : (CO) s . Note that
alternative 18-electron structures may be drawn for these two compounds—diman¬
ganese decacarbonyl with bridging groups and dicobalt octacarbonvl with none, as
shown in Fig. 15.6. Although the bridged dimanganese compound is unknown, there is
infrared evidence that in solution the dicobalt compound exists as an equilibrium
mixture of three isomeric forms, one of which is the structure shown in Fig. 15. Id.
Structures of the other two isomers are uncertain but neither contains bridging
Fig. 15.5 Overlap of tr and
ir orbitals of bridging
carbonyl ligands with the d
orbitals of metal atoms. The
tr orbital of CO can donate
electron density to the
metal orbitals and the
empty tt orbital of CO can
accept electron density
from the d orbitals. [From
Kostic. N. M.: Fenske.
R. F.; Inorg. Client. 1983.
666-671. Reproduced
with permission.)
Compound
Geometry
Point
group
C — O stretching
mode symmetries
IR active
modes
Observed
frequencies
(cm - ')
Ni(CO) 4
Tetrahedral
T d
a„ r.
■4|
2125 (CCI 4 )
72,
2045
Fe(CO),
Trigonal
Dy,
2a; , A". E'
A"
2002 (liquid)
bipyramidal
E'
1979
Ru(CO),
Trigonal
Dy,
2 a; . A", E'
A“
1999 (C 7 H i6 )
bipyramidal
E'
2035
Os(CO),
Trigonal
Dy,
2 A . A", £'
A'!,
2006 (vapor)
bipyramidal
E'
2047
Cr(CO) 6
Octahedral
o h
A tv
Tu,
2000 (vapor)
Mo(CO) 6
Octahedral
o,
E g , T,„
T u ,
2003 (vapor)
W(CO)„
Octahedral
o h
A ht' E g' T ln
Ti„
1998 (vapor)
“ Bratcrman. P. S. Metal Carbonyl Spectra', Academic Press: London, 1975.
9 Kostid. N. M.; Fenske. R. F. Inorg. Client. 1983, 22. 666-671.
Metal Carbonyl Complexes
635
(a) (W
Fig. 15.6 Alternative structures for dimanganese decacarbonyl and dicobalt octacarbonyl.
Structure (a) is unknown, but there is infrared evidence for the existence of (b) in solution.
carbonyl ligands. One possibility is shown in Fig. 15.6b. 10 The bridged form is the only
one observed in the solid state, however.
The dinuclear manganese and cobalt carbonyl complexes, as well as a number of
similar compounds, may be rationalized on the grounds that the 17-electron mono¬
nuclear units must form a metal-metal bond in order to provide each metal atom with
18 electrons. There is another group of polynuclear carbonyl complexes that may be
regarded as "carbon monoxide deficient" in the sense that they can be constructed
from the simple binary complexes by replacing one or more carbonyl groups with
metal-metal bonds. For example, in addition to Fe(CO) 5 , Table 15.2 shows two other
complexes of iron: diiron nonacarbonyl. Fe : (CO)„ and triiron dodecacarbonyl.
Fe,(CO) P . These compounds, as well as the tctranuclear Co 4 (CO) l2 , obey the 18-
electron rule if metal-metal bonds are included in the formulations:
2Fe
16e “
3Fe
24e“
4Co
36c-
9CO
18e
12CO
24e~
I2CO
24e"
M—M
2e"
3M—M
6e~
6M—M
12e -
Fe,(CO) 9
36c-
Fe 3 (CO) 12
54e“
Co 4 (CO) 12
72e
or 18e"
/Fe
or 18e"
/Fe
or I8e“
/Co
Without metal-metal bonds, these complexes would have 17. 16. and 15 electrons per
metal atom, respectively. The deficiencies are compensated by one, two. and three
M — M bonds per metal. Structures for the two iron complexes are shown in Fig. 15.7.
In the telracoball complex, the four metal atoms are arranged in the form of a
tetrahedron with the Co atoms occupying the corners and the six metal-metal bonds
forming the edges. Larger clusters such as Co h (CO)| h are also known, but the 18-
electron rule breaks down for complexes with more than four metal atoms. In these
species, electron delocalization is extensive, and other models (described in Chapter
16) are of greater value. 11
111 Bor. G.: Dietlor. U. K.: Noack. K. J. ( hem. Soc. Client. Common. 1976. 914-916. Lichtcnbergcr.
D. L.: Brown. T. L. Inorg. Client. 1978. 17. 1381-1382. lilliot, D. J.: Mirzu. H. A.; Puddephntl,
K. J.; Holnh. O. G.: Hughes, A. N.: Hill, K. H.; Xia. W. Inorg. Client. 1989. 2H. 3282-3283.
Johnson. B. F. G.: Parisini. H. Inorg. Chim. Acta 1992. 198-200, 345-349.
M Wade. K. Adv. Inorg. Client. Rudiochem. 1976. lb. I Mingos, D. M. P. Ate. Client. Res. 1984. 17.
311-319, Mingos. D. M. P.: Johnston. R. L. Struct. Bonding IBerlin/ 1987. 68. 29-87 Too. B. K.
Inorg. Client. 1985. 2J. 1627-1638. Johnston. R. L.: Mingos. D. M. P. Inorg. Client. 1986. 25.
1661-1671. Wales. D. J.: Mingos. D. M. P.: Slee, T.: Zhcnyang. L. Aft . Client. Res. 1990. 23.
780
16- Inorganic Chains, Rings, Cages, and Clusters
When we consider that complexes with chelating ligands (see Chapters 11 and 12)
are compounds with rings, we realize that rings with metal atoms are quite common.
In addition to these, however, new classes of compounds are appearing regularly in
which metal atoms have replaced nonmetal atoms of traditional nonmetallic heterocy¬
cles. Replacing a PPh 2 group of a trimeric cyclophosphazene with C1 3 W yields planar
[CI 3 WN 3 (PPh,),] and replacing two PPh 2 groups of a tetrameric cyclophosphazene
with VCU gives planar [CI 2 VN 2 PPh 2 ] 2 . 92
Homocyclic
Inorganic Systems
Isolobal relationships between metal and nonmetal fragments will undoubtedly con¬
tinue to be exploited and intense activity can be expected in this area for some time to
come.
Several elements form homocyclic rings. Rhombic sulfur, the thermodynamically
stable form at room temperature, consists of S x rings in the crown conformation.
Unstable modifications, S„, are known which include n = 6 through n = 36. In fact,
sulfur has more allotropes than any other element. 93 Selenium also forms five-, six-,
seven-, and eight-membered rings, but they are unstable with respect to the chain
form.
Organometallic chemistry has become an important player in the rings of sulfur. 94
The existence of an isolobal relationship between S and Cp 2 Ti leads to the prediction
that it should be possible to substitute the latter for the former in sulfur rings. The
formal replacement of one sulfur atom in S ft by Cp 2 Ti gives Cp 2 TiS 5 [or replacing two
sulfur atoms in S„ by two Cp 2 Ti units gives l ,5-(Cp 2 Ti) 2 S 6 J. In practice one takes these
93 Witt. M.; Rocsky, H. W.; Noltemcyer. M; Shcldrick. G. M. Angew. Cliem. hit. Ed. Engl. 1988. 27.
850-851, and references therein.
93 Stcudcl, R. Top. Carr. Client. 1982, 102, 149. Steudcl. R.; Steidel. J.; Sandow, T. Z. Notitrforscll..
B: Anorg. C/iem.. Org. Client. 1986. 41. 958-970. For rings containing both sulfur and selenium, sec
Steudcl. R.; Papavassiliou, M.; Strauss. E.-M.: Laitinen. R. Angew. Client. Int. Ed. Engl. 1986.25.
99-101.
94 Draganjac. M.; Rauchfuss. T. B. Angew. Client. Int. Ed. Engl. 1985. 24. 742-757.
Ring:
781
reactions in Ihe reverse direction. For example, S 4 , which is unstable with respect to
S K . may be prepared from the readily available Cp 2 TiS, complex.
2NH 3 + H,S + tS 8 -► (NH 4 ) 2 S 5 Cp 2 TiS 5 (16.62)
SC1 = — + c <’= Ticl = (i«.o)
The Sj anion that forms in the first step is one of several polysulfides (S; - , Sj“, S^~,
Sfi )• oil of which have open chain structures. The versatility of Cp,TiSj as a precur¬
sor to other rings and sulfur-carbon compounds is shown in Fig. 16.33.
Oxidation of several nonmetals in strongly acidic systems produces polyatomic
cationic species of the general type Y” + .« Among those characterized are Sj + , Se^,
Fig. 16.33 Reactions of Cp : TiS s . [From Draganjac. M.; Rauchfuss, T. B. Angew. Client. Int.
Ed. Engl. 1985. 24. 742-757. Reproduced with permission.]
M Burtord. N.; Passmore. J.; Sanders. J. C. P. In From Atoms to Polymers: Isoelectronic Analogies',
Liebman. J. F.; Greenberg. A.. Eds.: VCH: New York. 1989; Chapter 2. Collins. M. J.; Gillespie,
3L5 J 352 aWyCr ’ 1 F Chem ' 1987 ‘ 26 ’ l476_l481 - Gillcs P ic - R - J- Chem. Soc. Rev. 1979, 8 .
636
15 • Organometaliic Chemistry
(c) (d)
Fig. 15.7 Stereoviews (left) and conventional computer plots (right) of Fe 3 (CO)| 2 (a. b) and
Fc;(CO)i) (c, d). [From Johnson, B. F. G.; Benfield, R. E. In Topics in Inorgunic and
Organometaliic Stereochemistry, GeolTroy. G. L., Ed.: Wiley: New York. 1981. Reproduced
with permission.]
Whereas the 18-electron rule is of great aid in predicting metal-metal bonds, it
does not assist us in distinguishing between bridging and terminal CO ligands, in¬
asmuch as the electron count is the same for either mode of bonding. Polynuclear
carbonyls both with and without bridging CO units are common. Among those lacking
CO bridges are M 2 (CO) l0 (M = Mn, Tc. Re). M 3 (CO) 12 (M = Ru. Os), and lr 4 (CO) l2 .
New carbonyls of osmium [e.g., Os.,(CO) l4 , Os 4 (CO)u, Os 4 (CO) l6 , Os 5 {CO),l,
Os 5 (CO)| 9 , Os ft (CO) ls , Os 6 (CO) 2l . Os 7 (CO) 2l , Os„(CO) 23 ] are being discovered with
regularity and all found thus far are essentially nonbridging.
A beautiful series of tetranuclear osmium complexes, Os 4 (CO) lh Os 4 (CO),,. and
Os 4 (CO), 4 has been prepared (Fig. 15.8). 13 Consistent with the 18-electron criterion,
these complexes have 4, 5, and 6 metal-metal bonds, respectively. In Os 4 (CO) l4 , four
of the carbonyl ligands are weakly semibridging and ten are nonbridging. The term
semibridging is used to describe CO ligands that are unequally shared between two
metal centers, making them intermediate between terminal and bridging in their
ligation. 14
12 Pomeroy. R. K. J. Organomel. Chem. 1990, 383. 387-411.
13 Johnston. V. J.; Einstein. F. W. B.; Pomeroy, R. K. Organometallics 1988. 7, 1867-1869.
1,1 Cotton, F. A.; Wilkinson, G. Advanced Inorganic Chemistry, 5th ed.; Wiley: New York, 1988;
pp 1028-1032.
Metal Carbonyl Complexes
637
o
° V C0 >,6
(a)
Fig. 15.8 Structures of three tetranuclear osmium carbonyl complexes: (a) Os 4 (CO) l6 , (b) Os 4 (CO) l5 , and (c) Os 4 (CO), 4 .
Distances are in picometers. Broken lines indicate semibridging CO ligands. [From Johnston, V. J.; Einstein. F. W. B.;
Pomeroy. R. K. J. Am. Chem. Soc. 1987. 109, 8111-8112 (a); J. Am. Chem. Soc. 1987, 109, 7220-7222 (b); and
Organometallics 1988, 7, 1867-1869 (c). Reproduced with permission.]
c K M a = b > C X 3
a ,| P «-p
a b
a (3
symmetric p,-CO semibridging p-.-CO
Weakly semibridging carbonyls, such as those in Os 4 (CO) l4 , are nearly linear but
show some distortion toward the second metal. Of course it is not always clear
whether departures from linearity arise from interaction with the metal or because of
packing forces in the crystalline solid. All CO ligands are fully nonbridging in the
puckered Os 4 (CO)| h , described as a metal carbonyl analogue of cyclobutane. 15 The
dark red crystals of Os 4 (CO) l5 possess nearly planar Os 4 units with only terminal CO
ligands.
The existence of both bridged and nonbridged forms of Co 2 (CO) g in solution
reveals how little difference in energy exists between the two isomers. The factors
causing carbonyl ligands to bridge or not bridge are not at all well understood. Steric
factors are probably important, for just as vanadium hexacarbonyl refuses to dimerize
(it would become seven-coordinate), a bridged dimanganese decacarbonyl would have
a coordination number of 7 or more (depending upon the number of bridges) rather
than 6. Not even a trace of a bridged dimanganese isomer has been detected in
equilibrium with the nonbridged species, in contrast to the observations for cobalt and
iron. However, steric considerations do not account for the common observation that.
15 Johnston. V. J.; Einstein. F. W. B.; Pomeroy. R. K. J. Am. Chem. Soc. 1987, 109, 8111-8112.
778
16- Inorganic Chains, Rings, Cages, and Clusters
(a)
Fig. 16.30 Eighi-membered sulfur-nitrogen rings: (a) molecular structure ofS 4 N 4 ; (b)
diagrammatic structure of N 4 S 4 Fj illustrating alternating bond lengths. [From Sharma. B. D.:
Donohue, J. Acta Crystallogr. 1963. 16, 891. Reproduced with permission.]
Fig. 16.31 Molecular
structure of S 4 N : . [From
Olivers. T.; Codding.
P. W.; Oakley. R. T. J.
Client. Soc. Client.
Cumntnn. 1981. 584-585.
Reproduced with
permission.!
An even larger number of binary sulfur-nitrogen cations and anions are known.
Reduction of S 4 N 4 (with metallic potassium or sodium azide) yields the planar stx-
membered ring. S 3 Nj~, (Fig. 16.32a). At first glance one might think that this is another
benzene analogue. An electron count dispels that notion as there are ten r electrons
instead of six. Still, the Hiickcl 4n + 2 rule is obeyed and the system satisfies the
requirement for aromaticity. However, four of the rr electrons occupy antibondmg
orbitals, which has the effect of weakening the S—N bond (Fig. 16.32b).
Sulfur-nitrogen compounds often have unpredictable structures. One example is
the sulfur diimide. PhSN=S=NSPh. for which the following three configurations
could be envisioned:
Configuration (a) is preferred rather than the more open structures (b) and (c). 1 his
runs counter to our intuition that ihe most hindered structure would be least stable.
The separation between the end sulfur atoms is only 329 pm. compared with the van
Rings
779
I amibonding
electrons
} bonding
elecirons
(b)
Fig. 16.32 (a) Structure of
S,N,' and (b) qualitative
molecular orbital diagram.
[From Chivers, T.; Oakley.
R. T. Top. Ctirr. Client.
1982. 102, 117. Reproduced
with permission.]
der Waals sum of 360 pm, suggesting significant sulfur-sulfur interaction. In fact one
can write resonance structures such as:
R + + R
S—S''
/ \
~N N"
R v
^S —S'
// \
N N
R + ,R
^S —S'
/ \
N^-^N
which might lead us to believe that a ring structure is a belter description. However,
ab initio self-consistent field calculations do not support the ring description but rather
reveal that Ihe unusual conformation is the result of electronic interactions between
nitrogen and sulfur lone pairs. 1,1
Reduction of telrasulfur telranitride with tin(ll) chloride produces letrasulfur
tetraimide. S 4 (NH) 4 . isoelectronic with sulfur. S„. Like sulfur, the letraimide also
exists in a crown configuration. As S and NH are isoelectronic. it has been possible to
produce a series of ring compounds. S v |NH) x _ t . that includes all possible isomers
except those with N—N bonds.
Not only can S 4 N 4 be reduced as illustrated in the preceding example, but it can
also be oxidized. When it is subjected to chlorine, trilhiazyl trichloride is produced:
3S 4 N 4 + 6C1, -> 4N 3 S 3 CI 3 (16.59)
This compound may be converted into the corresponding lino title or oxidized to
sulfanuryl chloride:
Bannister. R. M.: Rzcpa. H. S. J. Client. Soc. Dalton Trans. 1989. 1609-1611.
638
15 • Organometallic Chemistry
Meta! Carbonyl Complexes
639
within a given family, the heavier congeners tend to have fewer bridges than the
lighter ones. Thus Fe 3 (CO), 2 [or Fe 3 (/4-CO) 2 (CO)| 0 ] (Fig. 15.7a. b) and Co 4 (CO), 2 [or
Co 4 (/x-CO) 3 (CO) 9 ] (Fig. 15.9) are bridged and Os 3 (CO), 2 and Ir 4 (CO), 2 (Fig. 15.9) are
unbridged. Although steric hindrance cannot be the explanation here, it may be that a
size effect of a different sort is operating. As the M—M bond lengthens (Co < Ir). the
M—C bond must lengthen even more or the M—C—M angle must open up. or both.
Either of these structural changes could destabilize the structure, though it is not clear
that this is the explanation.
The remarkable carbon monoxide ligand can also bridge three metal atoms. 16
M— M
Four of the eight triangular faces in octahedral Rh 6 (CO) l6 [Rh 6 (/z 3 -CO) 4 (CO) l2 ]
(Fig. 15.10) contain carbonyl groups bridging three metal atoms. The corresponding
cobalt compound is thought to have a similar structure. 17
The elucidation of the structures of polynuclear carbonyls is a challenging area.
Bridging carbonyl ligands can often be detected via their infrared absorptions, but in
the case of complicated structures, spectral interpretation can be difficult because
sometimes other absorptions, such as overtones or combination bands, appear in the
carbonyl region. Infrared spectroscopy does have one advantage over methods such
as X-ray crystallography in the ability to study structures in solution, which may differ
from those found in the solid state. X-ray crystallography provides unambiguous
information on structure but occasionally is beset with difficult problems arising from
disorder (see Chapter 3). A classic example is triiron dodecacarbonyl. for which
disorder in the solid prevented a complete solution to the crystal structure for many
years. 18 The available X-ray data were compatible with a triangular array of iron
atoms and so a structure similar to that of the isoelectronic Os 3 (CO) l2 , which has only
Fig. 15.9 Structures of Co 4 (CO) )2 , Osj(CO)i 2 , and Ir 4 (CO), 2 .
/ I \
OC CO t o
1,1 CO has even been shown to bind simultaneously to four metal atoms. In the IHFc 4 (CO)|s|~ ion. for
example, three Fc atoms arc bound to a single CO through the carbon atom while a fourth Fe
interacts with the 7r system. Carbon monoxide in this instance functions as a four-electron donor.
Horwitz. C. P.: Shriver. D. F. Adv. Organo/net. Chan. 19X4, 23, 209-305. Adams. R. D.: Babin.
J. E.: Tasi. M. Inorg. Chan. 198R. 27. 2618-2625.
17 Chini. P. Inorg. Chan. 1969 .8. 1206-1207.
18 A fascinating historical account of the process is provided by Dcsiderato. R.. Jr.: Dobson. G R.
J. Chan. Educ. 1982, 59. 752-756.
Carbonylate Ions
Fig. 15.10 The Rh,,(CO), 6
molecule. Note that four of
the faces of the octahedron
formed by the Rh atoms
have carbonyl ligands
bridging three rhodium
atoms. [From Johnson. B,
F. G.; Benfieid, R. E.
In Topics in Inorganic
and Organometallic
Stereochemistry, Geoffroy,
G. L., Ed.; Wiley: New
York. 1981. Reproduced
with permission.]
terminal CO ligands, was suggested. A weak band at 1875 citT 1 in the solid state
infrared spectrum of Fe 3 (CO), ; was dismissed as Fermi resonance or as arising from
crystalline interactions rather than fundamental C—O vibrations. The simple model
was finally overthrown by Mossbauer evidence which proved that the three atoms are
not in identical environments. |y Subsequent refinement of the X-ray structure showed
that the iron atoms do indeed form a nearly equilateral triangle, with one Fe—Fe pair
being bridged by two CO ligands (Fig. 15.7a. b). 211 The solution structure, however, is
still under discussion. A recent EXAFS study of frozen solutions suggests that
whether Fe,(CO) l2 exists as a fully unbridged species or as a mixture of bridged and
unbridged forms depends on the solvent. 21 An 1R study also supports the existence of
a mixture of bridged and nonhridged structures in solution. 2 - That the molecule is
lluxional in both solution and solid stales seems clear from a number of spectroscopic
investigations, including IR and variable-temperature NMR. Among the attempts that
have been made to elucidate the nature of this lluxional process are studies of various
derivatives of the dodecarbonyl compound. 21
Numerous anionic carbonyl complexes, also called carbonylate ions, are known.
These anions generally conform lo the 18-electron rule and arc of interest because of
the information they provide regarding bonding and structure as well as tor their
usefulness in the synthesis of other carbonyl derivatives. They are often electronically
and structurally related to neutral carbonyl complexes (Table 15.4). For example,
i‘< See Grandjcan. F.; Long. G. J.: Benson. C. G.: Russo, U. Inorg. Chan. 1988. 27. 1524-1529. fora
recent Mossbauer analysis of l ; e 3 (CO)|j.
-" For recent observations on the structure of Fet(COI| 2 and comparisons with that of FcjOstCOIn,
sec Churchill. M. R.: Fetiinger. J. C. Organometallic: 1990. 9, 446-452.
21 Biasted, N.: Evans. J.; Greaves. G. N.: Price. R. J. J. Chan. Soc. Chan, Cornmnn. 1987,
1330-1333. For a discussion of the EXAFS method, see Chapter 19.
Dobos. S.: Nunziante-Cesaro. S.; Maltese. M. Inorg. Chiin. Acta 1986, 113. 167,
’’ For example, see Lentz, D.; Marschall. R. Organomeialllcs 1991. It). 1487-1496.
776
16 - Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.28 Diagrammatic
structures of the
metasilicate anions: (a)
trisilicate. [SijO^f - ; (b)
hexasiiicate, (Si ft O| 8 ] l:_
Polymeric chain, band, and sheet silicate structures have been discussed previ¬
ously (page 742), and it should not be surprising to learn that cyclic silicate anions,
such as [SijOc,] 6 ' and [Si 6 O lx l l2_ (Fig. 16.28) are known. These anions are sometimes
referred to as metasilicates in line with the older system of nomenclature, which
assigned ortho to the most fully hydrated species [as in “orthosilicic acid," Si(OH) 4 ]
and meta to the acid (and anion) from which one mole of water has been removed
[either in fact or formally; for example, "metasilicic acid,” OSi(OH) 2 ].
Isoelectronic with cyclic silicates are cyclic metaphosphates. The simplest mem¬
ber of the series is the trimetaphosphate anion, [P 3 0 9 ] 3- . The tetrametaphosphate
anion. [P 4 O l2 J 4- , is also well known. By careful chromatographic separations of the
glassy mixture of polymeric phosphates and metaphosphates known as Graham’s salt
it is possible to show the existence not only of tri- and tetrametaphosphates, but also
penta-, hexa- hepta-, and octametaphosphates. The separation is effected and some
qualitative knowledge of structure is gained from the fact that two factors play a role in
the mobility of phosphate anions: (1) Higher molecular weight anions move more
slowly than do lower members of the series; and (2) the ring or metaphosphate anions
move more rapidly in basic solution than do the straight-chain anions of comparable
complexity (Fig. 16.29).
In progressing from silicon to phosphorus, the increase of one in atomic number
results in a corresponding decrease of one per central atom in the anionic charge of the
rings. Further progression from trimetaphosphate to sulfur trioxide results in a neutral
molecule, trimeric sulfur trioxide. This form is known as y-S0 3 and is isoelectronic
and isostructural with the analogous trimetasilicate and trimetaphosphate anions. It
exists in a chair form and is thermodynamically unstable with respect to two other
forms: ft- SO-,, which consists of infinite chains, and a-S0 3 , which probably consists of
infinite sheets caused by cross-linking. Traces of moisture convert the y-S0 3 form into
the u-SOj and 0-SO 3 forms.
Compounds which contain sulfur-nitrogen rings were known in the last century,
but many new ones have been prepared in the last decade. It is currently an area of
considerable interest. The ammonolysis of sulfur monochloride. S,C1 2 , either in
solution in an inert solvent or heated over solid ammonium chloride, yields tetrasulfur
tetranitride:
S 2 C1 2 S 4 N 4 + S 8 + NH 4 C1 (16.58)
1 The name hexametaphosphate has caused confusion over the years. On the basis of erroneous
reasoning concerning the nature of double salts, the term hexametaphosphate was assigned to
Graham's salt of empirical composition NaPOj. It has also been applied to the related commercial
product (Calgon) in which the Na/P ratio is 1:1. The true metaphosphate contains a twclve-
membered phosphorus-oxygen ring and is but a very minor component of the mixture known as
Graham's salt.
Ortlio
Octamcta
Heptameta
Hexamcta
Tri poly
Tetrapoly
Pentapoly
Hexapoly
Tnmeta
Telrameta
Penlameta
Fig. 16.29 Separation of polyphosphate anions by paper chromatography. The ions arc first
allowed to migrate in a basic solvent and subsequently in an acidic solvent. The straight-
chain polyphosphates lie on the ascending branch of the "Y," the metaphosphates on the
lower branch. [From Van Wazer. J. R. Phosphorus and Its Compounds', Wiley: New York.
1958; Vol. I. p 702. Reproduced with permission.]
Banister. A. J. Inorg. Synth. 1977. 17, 197-199.
Chivers, T.; Codding, P. W.; Oakley. R. T. J, Client. Soc. Client. Commun. 1981. 584-585.
640
15 Organometallic Chemistry
Tabic 15.4
Isoelectronic and isosiructural carbonyl and carbonylate complexes 0
Group
IVB (4)
Group
VB (5)
Group
VIB (6)
Group
VI IB (7)
Group
VIIIB (8)
Group
VIIIB (9)
Group
VIIIB (10)
|Ti(CO) 6 ] 2-
wcoy-
[V(CO) 5 ] 3-
Cr(CO) 6
[Cr(CO) 5 ] 2 -
[Cr(CO) 4 ] J -
[Mn(CO) 5 r
[Mn(CO) 4 ] 3_
Fe(CO) s
[Fe(CO) 4 ] 2 -
[Co(CO) 4 J-
[Co(CO) 3 ] 3 -
Ni(CO) 4
“ Complexes in the same column are isoelectronic; those in the same row are both isoelectronic and isosiructural.
[Cr(CO) 4 ] 4 , [Mn(CO).,] 3 , [Fe(CO) 4 ] 2- , and [Co(CO)J are isoelectronic and iso-
structural with Ni(CO) 4 (Fig. 15.2).
A common method of preparing carbonylate ions is reduction of a neutral car¬
bonyl complex.
Mn 2 (CO) 10 + 2Na - 2Na + + 2[Mn(CO) 5 ) _ (15.8)
Fe 3 (CO)| 2 + 6Na -♦ 6Na + + 3[Fe(CO) 4 ] 2 ~ (15.9)
Both of these reactions involve use of a strong reducing agent. In recent years,
reduction of carbonyl complexes has been pushed to its limit with the synthesis of
highly reduced anions such as [Mn(CO) 4 ] 3_ , [Cr(CO) 4 ] 4 ", [V(CO) 5 ] 3- , and
[Ti(CO) 6 ] 2 . 24 Since [Mn(CO) 6 ] + , Cr(CO) 6 , and [V(CO) 6 J~ were known to be rela¬
tively stable, there was some expectation, based on the 18-electron rule, that it would
be possible to synthesize [Ti(CO) 6 ] 2 ~, even though Ti(CO) 7 is unknown. Often the
expectation that a product should exist is a long way from synthetic success. The
involved synthesis of [Ti(CO) ft J 2 ^ illustrates the point. 25 The overall reaction is
(l)KC|||H,+cryp(anil222
Ti(CO) 3 (dmpe) 2 - [K(C222)],[Ti(CO) 6 ] (15.10)
dmpe = Me 2 PCH 2 CH 2 PMe 2
Potassium metal, which is a powerful reducing agent, reacts with naphthalene to give
potassium naphthalenide, another powerful reducing agent, which has the advantage
24 Ellis. J. E. Adv. Organomet. Client. 1990, 31. 1-51. Beck, W. Angew. Client. Int. Ed. Engl. 1991.
30. 168-169. Ellis, J. E.; Barger, P. T.; Winzcnburg. M. L.; Warnock. G. F. J. Organomet. Client.
19-8). 383. 521-530.
23 Chi. K.-M.: Frerichs, S. R.; Philson. S. B.; Ellis J. E. J. Am. Chem. Joe. 1988. 110, 303-304. Also
see Ellis. J. E.; Chi. K.-M. J. Am. Chem. Soc. 1990. 112. 6022-6025. for preparation of [HftCOU 2-
by a similar method.
Metal Carbonyl Complexes
641
of being dispersed throughout the reaction medium. The cryptand is added to coordi¬
nate the potassium ion, creating a large cation which allows isolation of the large
anion. There is evidence that uncoordinated alkali metal ions prevent formation, or
stimulate decomposition, of [Ti(CO) 6 ] 2- . The reaction is run under Ar instead of N-,
since the former is less reactive. Only carefully purified carbon monoxide can be used
because traces of impurities such as H 2 . N 2 , orCO ; compete in the reaction. Finally,
the reduction must be carried out at low temperatures (-70 °C) to minimize side
reactions. The product of this reaction is air sensitive (it reacts with 0 2 ), as are most
carbonylate anions. Such reactions require special equipment such as Schlenk glass¬
ware or a glove box, which are standard items in any modern organometallic labora¬
tory. 26 It should also be noted that highly reduced anions may be shock sensitive, as
are Na 4 M(CO) 4 (M = Cr, Mo. W), K,[V(CO),J. and Cs 3 [Ta(CO) 5 ]. 27
Not all reactions leading to carbonylate anions require strong reducing agents.
Some involve reduction of the metal by carbon monoxide already present in the metal
carbonyl or disproportionation of the complex. In fact, the first synthesis of a metal
carbonylate involved the former procedure;
Fe(CO) s + 40H~ - [Fe(CO) 4 ] 2 “ + CO 2- + 2H z O (15.11)
Often a Lewis base will effect disproportionation of a complex:
3Mn,(CO) 10 + 12py -♦ 2[Mn(py) 6 ] 2+ + 4[Mn(CO) s ]" + I0CO (15.12)
Mn 2 (CO) 10 + 2dppe - [Mn(CO) 2 (dppe) 2 ] + + [Mn(CO) s ]- + 3COOS.13)
dppe = Ph 2 PCH 2 CH 2 PPh 2
The reaction with the relatively hard base, pyridine, has long been recognized as
disproportionation but reaction 15.13 with the softer phosphine base has only recently
been fully elucidated. 2S
Numerous bimetallic carbonylate anions are also well known. Among them are
[Cr 2 (CO) l0 ] 2- and (Fe 2 (CO) 8 ) 2- , which are isoelectronic and isosiructural with
Mn 2 (CO),„ and Co 2 (CO) 8 , respectively.
Carbonyl Hydride Acidification of carbonylate ions often results in the formation of carbonyl hydrido
Complexes complexes, which may be regarded as the conjugate acids of the carbonylales:
[Co(CO) 4 ]“ + H 3 0 + - HCo(CO) 4 + H,0 (15.14)
[Re(CO) s ]“ + H 2 0 -♦ HRe(CO) 5 + OH~ (15.15)
[Fe(CO)J 2- [HFe(CO)J" — H 2 Fe(CO) 4 (15.16)
The stronger bases can be protonated with acids that are even weaker than water. 29
[Cr(CO) 5 ] 2- + MeOH -♦ [HCr(CO) 5 ]- + MeO" (15.17)
26 Shrivcr. D. F.; Orczdzon, M. A. The Manipulation of Air-Sensitive Compounds. 2ml ed.; John
Wiley: New York. 1986.
27 Warnock, G. F. P.; Sprague, J.; Fjarc, K. L.; Ellis. J. E. J. Am. Client. Soc. 1983, 105, 672.
28 Kuchynka. D. J.; Kochi. J. K. Inorg. Chem. 1988. 27. 2574-2581.
29 Darensbourg, M. Y.; Deaton. J. C. Inorg. Chem. 1981. 20. 1644-1646. Lin, J. T.; Hagen, G. P.;
Ellis. J. E. J. Am. Chem. Soc. 1983, 105. 2296-2303.
774
16 - Inorganic Chains, Rings, Cages, and Clusters
moisture. However, it has been found that trimeric chlorophosphazene can be poly¬
merized thermally:
Cl Cl
\ /
Cl J |>
/ P X^ P \
Cl N Cl
If this is done carefully, extensive cross-linking does not take place and the polymer In
= 15,000) remains soluble in organic solvents. The reactive chlorine atoms are still
susceptible to nucleophilic attack and displacement:
[PNC1,]„ + 2/iNaOR
[PNC1,]„ + 2/iR-jNH
[PN(OR),]„ + 2/iNaCI
[PN(NR-,),]„ + 2//HCI
By varying the nature of the side chain, R, various elastomers, plastics, films, and
fibers have been obtained. These materials tend to be flexible at low temperatures, and
water and fire resistant. Some fluoroalkoxy-substituted polymers (R = CH : CF,) are
so water repellent that they do not interact with living tissues and promise to be useful
in fabrication of artificial blood vessels and prosthetic devices.
Although the hydrolytic stability of some phosphazene polymers makes them
attractive as structural materials, it is possible to create hydrolytically sensitive
phosphazenes that may be useful medically as slow-release drugs. Steroids, antibiot¬
ics, and catecholamines (e.g., dopamine and epinephrine) have been linked to a
polyphosphazene skeleton (Fig. 16.27) with the intention that slow hydrolysis would
provide these drugs in a therapeutic steady state. XA
Materials containing an inorganic polymeric backbone often have useful elec¬
trical, optical, and thermal properties. In addition they are being explored for use as
precursors to ceramics. One way to alter the properties of a polymer is to make
changes in the backbone. Recently, Manner and Allcock S7 have shown that a C—Cl
group may be substituted for one of the PCU groups in (PCI ; N), to give a ring that
forms a polymer with carbon in its backbone:
C1 0 II / C1
p p
n N n
C=N — P=N — P=N
I I I
Cl Cl Cl
This polymer is the first example of a poly(carbophosphazene).
Allcock. H. R. In Rings. Clusters, and Polymers of the Main Croup Elements : Cowley. A. H.. Ed.;
ACS Symposium Scries 232; American Chemical Society: Washington. DC. 1983.
K7 Manner. Allcock. H. R. J. Am. Chem. Soc. 1989. III. 5478-5480.
Rings 775
Other Heterocyclic
Inorganic Ring
Systems
(c)
Fig. 16.27 Polymeric phosphazenes: (a) steroid-bound; (b) sulfadiazine-bound:
(c) catecholamine-bound. [From Allcock. H. R. In Rings. Clusters, anil Polymers of the Main
Group Elements-. Cowley. A. H.. Ed.; ACS Symposium Series 232; American Chemical
Society: Washington. DC. 1983. Used with permission.]
The reaction of MeXiCU with water at elevated temperatures gives polymeric
(Me-.SiO)„ (Eq. 16.23). but if hydrolysis is carried out at room temperature, the
mixture which results also includes cyclic siloxanes, (Me 2 SiO)„ {it = 3. 4, etc.).
The trimer has a chair structure analogous to cyclohexane and the tetramer has a
crown structure analogous to S„. In the previous section we saw that cyclophos-
phazenes could be turned into linear polymers thermally. The same is true for
siloxanes. In order to achieve high molecular weights (the production of silicone
7<Me-SiO)
4
Kllll
pressure
(Me,SiO),
(16.57)
1 5 • Organometallic Chemistry
[Cr(CO) 4 ] 4
(OC) 4 C< )Cr(CO).
The reaction in Eq. 15.18 leads to a product with bridging hydride ligands: the nature
of the bonding in complexes of this type will be discussed later in this section.
In addition to the protonation route, hydrido complexes may be prepared by
reaction of carbonyl complexes with hydride donors:
Fe(CO).C
Fe(CO) 4 C
[HFe(CO) 4 ]’ + CO
In a mechanistically similar reaction hydroxide ion can serve as the source of
hydrogen:
Fe(CO), + OH"
Fe(CO) 4 C
FelCO) 4 C
|HFe(CO) 4 r+ CO,
Addition of excess base in these reactions generates the dianionic starting material in
Eq. 15.16. The intermediate shown in Eq. 15.19. which has a coordinated formyl
group IC(0)H], has been isolated and characterized. Its counterpart in the OH"
reaction is a hydroxycarbonyl complex (see Problem 15.45). Both are inherently
unstable, decomposing to the hydrido complex by elimination of CO or CO,. When
these reactions are attempted with Cr(CO) 6 , the final product is not the expected
[HCr(CO),]" anion, but rather a bridging hydride complex: 1
I'll. i
Cr(CO)„-» Cr(CO) 5 C
10 Darensbourg. M. Y.; Deaton. J. C. Inorg. Client. 1981. 20. 1644-1646.
Metal Carbonyl Complexes
643
Complexes with a coordinated hydrogen atom are called hydrides whether or not
they exhibit the chemical properties of a hydride. Certainly compounds such as NaH
or LiAIH 4 . which serve as a source of H . are legitimately called hydrides, But many
of the transition metal carbonyl hydrides are. in fact, quite acidic. Both HCo(COl 4 and
H\ lCO) h have acid strengths similar to that of HCI in water, i.e.. they are essentially
completely dissociated in aqueous solution. At the other extreme. Re(CO) 5 H is such a
weak acid that its conjugate base is readily hydrolyzed by water (Eq. 15.15). The
dihydride. H-,FeiCO) 4 . was the first carbonyl hydride complex to be synthesized,
having been prepared by Walter Hieber in 1931. It is a yellow liquid, unstable above
- 10 J C. and behaves as a dibasic acid in water (pAf, = 4.4: pA\ = I4t. The large
difference in the two ionization constants provided the first evidence that the hydrogen
atoms in the complex were both bound to the same atom land hence to the iron atom).
The pA', values of a variety of metal carbonyl hydride complexes are given in
Table 15.5.
Although in many respects carbonyl hydride complexes may be regarded as acids,
they also show some similarities to the basic hydrides of the main group metals le.g..
Table 15.5
Acidities of common
Compound 1
P„(H 2 0)s
Compound 1,
pKJ H 2 0)s
transition metal hydrides"
HV(CO>„
strong
I I . Rut CO) ,
11.2
HV(CO),PPh,
6.8
HRuCptCO),
12.7
HCrCp(CO),
5.4
H,Os(CO) 4 "
13.3
HMoCp(CO),
6.2
HColCO)j
strong (0,9)
HWCp(CO),
8.1)
HCo(CO),(PPh,)
7.0
HMiKCO),
7.1
HCo(CO)j|P(OPh)J
4.9
HMn(CO) 4 (PPh,»
12.9
HCo(CO),(PPh 3 ),
very weak
HRet CO),
13.6
HNi|P(OMe),j; ’
5.9
H,Fc(CO) 4
4.4
HPd|P(OMe),] 4 +
3.1
HEeCplCO),
11.9
HPUPtOMe),);
11
HFeCp’lCO),
18.8
Kristjansdottir. S. S.; Norton. J. R. In Transition Metal llytlrldex: Recent
Advances in Theory and Experiment'. Dcilicu. A., lid.; VCH: New York. 1992;
Chapter 9.
Cp = C,H<. Cp = C«Mc,.
1 Because water is a poor solvent for most transition metal hydrides, acidities
were determined from equilibrium measurements in acetonitrile and pA‘., values
in water were estimated from the equation: pA' ; ,(H 2 0) = pAj.tCHrCN) - 7.5.
772
Rings
773
Fig. 16.25 Theory of Dewar for n bonding in phosphazenes: (a) the relation of the
orthogonal phosphorus d" and d h orbitals to the nitrogen p : orbitals as seen perpendicular and
parallel to the c axis, respectively; (b) the three-center bond model for PjNjCIj. [From
Schmulbach. C. D. Prog. Inorg. Cliem. 1965, 4. 275. Reproduced with permission.)
Dewar and coworkers offered an alternative view. 79 In their model the d x , and d vz
orbitals are hybridized to give two orbitals which are directed toward the adjacent
nitrogen atoms (Fig. 16.25). This allows for formation of three-center bonds about
each nitrogen. 80 This scheme, sometimes called the “island" model, results in de-
localization over selected three-atom segments of the ring, but nodes are present at
each phosphorus atom since the two hybrid orbitals of phosphorus are orthogonal to
each other. Evidence has been offered in support of both models, but neither theory
has been confirmed to the exclusion of the other. A third viewpoint holds that d orbital
participation is relatively unimportant in the bonding in these molecules. 81
The structures of tetrameric phosphazenes are more flexible than those of the
trimers. The structure of (NPF 2 ) 4 is planar, but others are found in a variety of
conformations (tub, boat, chair, crown, saddle, and structures in between). The
particular structure adopted is not very predictable and suggests that intermolecular
forces play a major role. The tetrameric chlorophosphazene has been isolated in two
forms (Fig. 16.26). 82 the most stable of which assumes a chair arrangement (some¬
times called the T form). The other form (metastable K) has a tub conformation. An
interesting feature of these compounds is that the nonplanar structures do not militate
against extensive delocalization in the rings. The corresponding organic compound,
cyclooctatetraene, C 8 H 8 , is nonaromatic for two reasons: (I) Its nonplanar, chair
structure precludes efficient p„-p„ overlap; and (2) it does not obey the Hiickel rule of
(4/i + 2)n electrons. The Hiickel rule was formulated on the basis of p„-p„ bonding
and holds for cyclic organic compounds from n = I (benzene) to n = 4 ([ISJannulene).
The use of d orbitals removes the restrictions of the Hiickel ruie and also allows
greater flexibility of the ring since the diffuse d orbitals are more amenable to bonding
in nonplanar systems. Both the Craig/Paddock and Dewar models predict that the
79 Dewar, M. J. S.; Lucken, E. A. C.: Whitehead. M. A. J. Cliem. Sue. I960. 2423-2429.
811 See page 791 for a discussion of thrcc-ccntcr bonding.
81 Trinquier. G. J. Am. Chem. Soc. 1986. 108. 568-577.
82 Krisnamurthy. S. S.; Sau. A. C.; Woods. M. Adv. Inorg. Cliem. Radiochem. 1978. 21. 41-112.
Phosphazene
Polymers
< NPN = 117°
(a) (b)
Fig. 16.26 Structures of tetrameric phosphazene. P.,N 4 CI K : (a) tub conformation; (b) chair
conformation. [Structure (a) from Hazekamp. R.; Migchelsen. T.; Vos, A. Acta Crystallogr.
1962. 15. 539; structure (b) from Wagner, A. J.: Vos. A. Acta Crysudlogr. 1968. 24. 707.
Reproduced with permission.)
tetramer is stabilized by delocalization (unlike cyclooctatetraene) and the stabilization
is either equal to (Dewar) or more than (Craig/Paddock) that of the trimer.
Our discussion has dealt with trimeric and tetrameric phosphazenes, but many
other ring sizes have been synthesized. For example, all of the compounds. (NPMe,)„
(" = 3-12), have been studied crystallographieully. 8 -'' Furthermore, the first cycio-
diphosphazene has been prepared; 8 - 1
(/-Pr),N
^ P = N
[(/-Pr) 2 N],PN 3 “TF (i-Pr) 2 N
N(t-Pr),
N(t-Pr),
(16.51)
Diphosphazcnes were long thought to be too unstable for isolation because of ring
strain.
Phosphazenes can be polymerized and in many instances their polymers have advan¬
tages over the carbon-based polyolefins and polyesters.However, commercial
application is not as well developed as for the silicones (R 3 SiO)„, (see page 749). Early
studies were hampered by the sensitivity of the phosphorus-chlorine bond to
Oakley. R. T.: Reltig. S. J.: Paddock. N. L.; Trotter. J. J. Am. Chem. Soc. 1985. 107. 6923-6936.
84 Bacciredo. A.; Bertrand. G.: Majorat, J.-P.; Sicard. G.; Jaud. J.; Guly, J, J. Am. Chem. Soc. 1984
106. 6088-6089.
85 Inorganic and Organometallic Polymers-. Zeldin, M.; Wynne. K. J.; Allcock, H. R.. Eds.; ACS
Symposium Scries 360; American Chemical Society: Washington. DC. 1988. Mark. J. E.; Allcock.
H. R.; West. R. Inorganic Polymers'. Prentice-Hull: Englewood ClilTs, NJ, 1992.
644
15-Organometallic Chemistry
Metal Carbonyl Complexes
645
NaH, LiAIH 4 ). For instance, they can act as reducing agents toward many organic
compounds and are capable of hydrogenating alkynes and alkenes. Thus their H li¬
gands are intermediate between the strictly hydridic hydrogens in the saline hydrides
and the protonic hydrogens in compounds with nonmetals (e.g.. HCI, NH 3 ).
Hydride ligands bound to transition metals generally give proton NMR signals
that are considerably upheld from TMS (0 to -60 ppm), a feature that has been quite
useful in characterizing hydride complexes. 31 For example, 'H chemical shifts for
H 2 Fe(CO) 4 and [HFe(CO) 4 ]~ are - 11 and -9 ppm, respectively, while typical values
for organic compounds are downheld from TMS (0 to + 15 ppm). One should not
conclude, however, that the upheld chemical shifts in hydride complexes arise be¬
cause of high electron density about hydrogen. It is rather thought to be the result of
paramagnetic contributions from the nearby transition metal.
The reaction of hydrogen gas with a transition metal complex can lead to a
monohydride complex, as shown in Eq. 15.26, or to a complex with two coordinated
H atoms. One of the more famous examples of the latter is the reversible reaction of
Vaska's complex, Ir(CO)L 2 Cl (L = PPh 3 ), with hydrogen:
L
\
/
OC
H,
L
Cl
H
C
O
(15.28)
This reaction has the characteristics of an oxidative addition (page 689): The formal
oxidation state of Ir increases from +1 to +3 and the coordination number increases
from 4 to 6. The process is believed to proceed via a concerted mechanism:
M + H 2
H
/
M
\
H
(15.29)
The proposed intermediate in this scheme represents an alternate mode of coordina¬
tion for two H ligands—as a coordinated H 2 molecule. Although the intermediate in
Eq. 15.28 has not been isolated, a number of other dihydrogen complexes have been.
The first was prepared in I983: 32
Mo(CO) 3 (cht) + 2PPH, -► MofCO^PPr'j), -I- cht (15.30)
Mo(CO) 3 (PPr') 2 + H 2 - Mo(CO) 3 (PPr'),(H 2 ) (15.31)
(cht = cycloheptariene)
The dihydrogen (rf-H-,) ligand in this complex exists with its H—H bond intact
(Fig. 15.11). Since the discovery of the first H, complex, many others have been
51 Crabtree, R. H. The Organometallic Chemistry of the Transition Metals; John Wiley: New York,
1988.
32 Kubas, G. J. Acc. Client. Res. 1988, 21. 120-128.
Fig. 15.11 Structure of
Mo(CO) 3 (PPr;,) 2 H,, the first
nonclassical dihydrogen
complex to be discovered.
[From Kubas, G. J. Acc.
Chem. Res. 1988. 21.
120-128. Reproduced with
permission.]
characterized, including some which for years had been presumed to be cis dihydrido
complexes.
It was not a simple matter to prove that the complex in Eq. 15.31 should be
viewed as M(H 2 ) rather than c/i-M(H), (see below). Even though it was possible to
grow a crystal and subject it to X-ray and neutron diffraction analysis, some questions
remained unresolved. The X-ray and neutron data gave H —H bond lengths of 75(16)
pm and 84 pm, respectively, compared to 74 pm in uncoordinated H 2 . These results
strongly suggested that the H—H bond had not been broken, but the inherent
uncertainty associated with locating hydrogen atoms left some room for skepticism.
Synthesis of an tj 2 -HD complex eliminated all doubt. The proton NMR spectrum of the
complex revealed an H —D coupling constant of 33.5 Hz comparable to 43.2 Hz for
uncoordinated H—D. The significance of these values is appreciated when they are
compared to hydride-deuteride coupling (less than 2 Hz) in complexes in which bond
distances are too great for an H—D bond to exist. In this instance, it was spec¬
troscopy instead of crystallography that provided the final confirmation of structure!
Complexes which contain the tj 2 -H 2 ligand are now referred to as nonclassical,
while those in which the H—H bond has been severed are called classical.
H H
M—|
H "'H
Nonclassical Classical
The difficulties encountered in firmly establishing the structure of the first di¬
hydrogen complex still have not been entirely overcome. The H—D coupling constant
experiment described above offers a method of distinguishing between the classical
and nonclassical forms, but it is not applicable if the system is rapidly lluxional.
770
16-Inorganic Chains, Rings, Cages, and Clusters
separate is the trimer, n = 3. Smaller amounts of the tetramer and other oligomers up
to /; = 8 have been characterized and higher polymers exist as well (see below).
Analogous bromo compounds may be prepared in the same manner, except that
bromine must be added to suppress the decomposition of the phosphorus penta-
bromide:
PBr 5 PBr, + Br 2
NH.Br
PBr, -— [NPBr,],
The fluoride must be prepared indirectly by fluorination of the chloride:
[NPCI 2 ] 3 + 6NaF -♦ [NPF 2 ] 3 + 6NaCl (16.50)
The corresponding iodide is unknown, but a phosphazene with a single phos¬
phorus-iodine bond, NjtPCUKPfR)!. has been reported. 7h
The halide trimers consist of planar six-membered rings (Fig. 16.23). 77 The bond
angles are consistent with sp 2 hybridization of the nitrogen and approximately sp 2
hybridization of the phosphorus. Two of the sp 2 orbitals of nitrogen, containing one
electron each, are used for a bonding and the third contains a lone pair of electrons.
This leaves one electron left for the unhybridized p. orbital.
The four sp 2 hybrid orbitals (housing four electrons) of phosphorus are used for a
bonding leaving a fifth electron to occupy a d orbital. As shown in Fig. 16.23,
resonance structures can be drawn analogous to those for benzene indicating aromat¬
icity in the ring. However, the situation is more complex than these simple resonance
structures indicate. The planarity of the ring, the equal P—N bond distances, the
shortness of the P—N bonds, and the stability of the compounds suggest delocaliza¬
tion. However, not all phosphazenes are planar, and the absence of planarity does not
appear to make them any less stable. Furthermore, the phosphazenes yield UV
spectra unlike those of aromatic organic compounds and they are much more difllcult
to reduce. Thus the extent of delocalization and the nature of the aromaticity have
been debated for years. Unlike in benzene, tr bonding in cyclophosphazenes involves
(/and p orbitals. There have been several descriptions offered for such (/„-/?„ bonding.
Craig and Paddock suggested the following model: 71 The d x; orbital of the phosphorus
atom overlaps with thep. orbitals of the nitrogen atoms adjacent to it (Fig. 16.24a). As
a result of the gerade symmetry of the e/ orbitals, an inevitable mismatch in the signs of
the wave functions occur in the trimer (see Fig. 5.10) resulting in a node which reduces
the stability of the delocalized molecular orbital. The d y . orbital, which is perpen-
Allcock, H. R.; Harris, P. J. hwr K . Chan. 1981. 20. 2844-2848.
77 Small deviations from planarity found for the chloride, bromide, and lluoridc may be the result of
packing effects. The P—N bonds appear to be flexible and angle changes lead to little stability loss.
7H Paddock. N. L. Q. Rev. Clian. Sac. 1964. IS. 168.
(16.48)
(16.49)
Rings
771
Fig. 16.23 Structure of trimeric phosphazene, PjN 3 CI,: (a) contributing resonance
structures: (b) molecular structure as determined by X-ray diffraction. (From Bullen. G. J. J.
Chem. Soc. (A) 1971. 1450. Reproduced with permission.)
dicular to the d r .. can also overlap with the p. orbitals of nitrogen, but in this case no
nodal surface results (Fig. 16.24b). There may also be in-plane rr bonding between the
sp 2 nonbonding orbital of nitrogen and the d xy and/or c/ r :„ v : orbitals of phosphorus
(Fig. 16.24c,d).
(O
Fig. 16.24 Theory of Craig and Paddock for -n bonding in phosphazenes: (a) interaction of
p_.(N) and «/,.(P); (b) interaction of p.( N) and (/,..(P); (c) interaction of ,vp 2 (N) and (/„.(P);
(d) interaction of rp 2 (N) and [From Corbridge, D. E. C. Phosphorus-. Elsevier:
Amsterdam. 1978; p 235. Reproduced with permission.]
646
1 5 • Organometallic Chemistry
Infrared analysis is difficult because M—H absorptions tend to be weak. Neutron
diffraction requires large crystals, and X-ray diffraction is not precise enough for
locating H atoms. An NMR method has been developed for differentiating classical
and nonclassical structures that is based on the assumption that hydrogen nuclei
involved in nonclassical coordination will have significantly faster relaxation rates
than those involved in classical coordination. 33 In this approach, the measured spin-
lattice relaxation time. 7j, is correlated with the H—H bond distance and hence with
mode of coordination. Application of the method to the hydrogen adduct of Vaska’s
complex (Eq. 15.28) and to H 2 Fe(CO) 4 discussed earlier led to the conclusion that
both are best regarded as classical complexes. Recent reports have urged caution in
applying the 7, criterion, based on the finding that the ranges of relaxation rates for
classical and nonclassical formulations overlap. 34
The metal-H-, bond may be profitably compared with a metal-carbonyl bond
since both involve it donation to the metal by the ligand and both ligands can accept
t r electron density into antibonding orbitals. The accepting orbitals for CO are empty
tt orbitals, whereas for H, they are a orbitals (Fig. 11.23). Like the C—O bond,
the H — H bond is weakened as a result of this metal-ligand 7r interaction. A strong
il-<r interaction can sever the H — H bond and lead to formation of a classical
complex.
Some special note should be made of the structure and bonding in complexes
containing a bridging hydride ligand. 33 Probably the most famous bridged hydride in
inorganic chemistry is diborane, B-.H,, (discussed more extensively in Chapter 16). in
which two of the H atoms bridge the pair of boron atoms. Useful parallels may be
drawn between the hydride bridges in borohydrides and those in metal complexes.
The complex [(OC) 5 Cr—H—Cr(CO),| - is similar to [BH,—H — BH,] in the sense
that both can be said to involve donation of a bonding pair of electrons (those in the
B — H and Cr—H bonds of [BHJ - and [HCr(CO) s | _ ) to a Lewis acid (BH, or
Cr(CO) 5 ]:
H H
/ / \ (15 32 )
L„M + ML,,- L„M ML,,
In molecular orbital terms, the donation can be viewed as a HOMO-LUMO interac¬
tion (Chapter 9). Double hydrido bridges, as found in B : H,,. also are exhibited by
bimetallic species such as the chromium anion formed in Eq. 15.18. The similarity
between these two dibridged species is underscored by the fact that their Lewis acid
and Lewis base fragments can be interchanged: [HCr(CO) s | reacts with BH, to
form: 3 '’
33 Hamilton. D. G.; Crabtree. R. H. J. Am. Clicn,. Site. I0KK. 110. 4126-4133.
3J Dcsrosiers, P. J.; Cai. L.; Lin. Z.; Richards. R.: Halpcrn. J. J. Am. Clivm. Soc. 1991. IIJ,
4173-4184.
•” Dahl. L. F. Ann. NY Acad Sri. 1983. 4/3. 1-26.
• u ’ Darensbourg. M. Y.: Bau. R.: Marks. M. W.: Burch. R. R., Jr.: Deaton. J. C. Slater. S. J. Am.
Chcm. Soc. 1982. 104. 6961-6969.
Parallels with
Nonmetal Chemistry:
Isolobal Fragments
Metal Carbonyl Complexes 647
The bridging hydride interaction, whether it involves two boron atoms, two metal
atoms, or one boron and one metal atom, is best described in terms of a three-center,
two-electron bond (see Chapter 16).
Many of the reactions of metal carbonyl complexes parallel closely those of certain
nonmetal elements and compounds. For example, the Mn(CO) 5 fragment has 17
valence electrons, one short of the total necessary to fulfill the 18-electron rule. It is
analogous to the chlorine atom and the methyl free radical, each with seven valence
electrons, one short of a noble gas configuration. The compounds and reactions of the
pentacarbonyl fragment may thus be related to similar ones for chlorine or the methyl
group. All three are formally free radicals and much of their chemistry derives from
pairing the odd electron: The manganese carbonyl normally exists as a dimer,
Mn : (CO),„ (cf. CL. C : H (1 ), but it may be reduced to the anion, [Mn(CO),] - (cf. Cl - ,
CH^ in CH,MgX). which is the conjugate base of an acid. HMn(CO), (cf. HCI. CH,);
furthermore, it will combine with other species having a single unpaired electron, for
example. R' and F. to form neutral molecules. RMntCO), (cf. RCI. RCH,| and
Mn(CO) 5 I (cf. IC1. CHjI). The three fragments may be considered as electronically
equivalent groups 37 or as isolobal fragments . 3X The concept is an outgrowth of
equating the Lewis octet rule of organic and main-group chemistry with the 18-
electron rule of transition metal organometallic chemistry. 39 Of course one should not
push these ideas too far. For example, pentacarbonylhydridomanganese is a much
weaker acid than is HCI. and methane is normally not considered to be an acid at all.
The isolobal formalism is more concerned with structural predictions based on elec¬
tronic similarities than with topics like polarity.
Isolobal fragments have relationships that go beyond simple electron counting.
The calculated electron density of the MnH, fragment (isolobal with MntCOL. but
simpler for calculations) may be compared with that of the methyl radical. CH, t Fig.
15.12). When one examines the overlap integrals of these two isolobal fragments with
respect to an incoming probe such as a hydrogen atom, the results are remarkably
similar (Fig. 15.13). The manganese fragment always has a somewhat greater overlap,
but the dependence on distance is essentially identical.
Table 15.6 lists a number of examples of transition metal fragments that are
isolobal with main-group fragments. Metal fragments with 16 electrons will behave as
Group VIA (16) elements. Thus Fe(CO)., may form H : Fe(CO) 4 (cf, H ; S) and
|Fe(CO).,| : (cf. S : 'l. Fifteen-electron fragments such as IrlCO), are isolobal with
CH and with Group VA (15) elements, such as phosphorus. Each of these is three
electrons short of a closed shell and each has three directed orbitals which form a
triangular face. Hence the complex lr,(CO), : is isostruclural with the P 4 molecule
(Fig. 15.14) us well as with tetrahedrane. (CH) 4 .
Earlier in this chapter, (he polynuclear carbonyl complex OsjlCO),,. was referred
to us an analogue of cydobulane. it should now be clear that the connection between
37 Foust. A. S.: Foster. M. S.; Dahl. I.. F ./. Am. Clicw. Soc. 1969. 9/. 563M633. F.llis. J E, ./.
Clwm. Eil iic. 1976. 53. 2-6.
3K Hoffmann. R. Angew. Clicm. lot. Ed. Engl. 1982. 21. 711-724.
39 The isolobal concept, like most of Sherlock Holmes's explanations, seems obvious in hindsight, but
it was first proposed in its present form by Halpcrn. It has since been elaborated by Dahl. Mingos.
Wade, and Ellis, and extensively developed by Hoffmann. (See p 42 of Mingos. D. M. P.: Johnston.
R. L. Struct. Bonding I Berlin ) 1987. 29-87.) It was Hoffmann's theoretical work in this area,
much of which is beyond the scope of this book, that formed the basis Tor his Nobel laureate address
(see Footnote 38).
768
16-Inorganic Chains, Rings, Cages, and Clusters
R R
\ /
N-B
/ \
R-B x O N —R
N —:—'&
/ I \
R Cr R
Borazine analogues of naphthalene and related hydrocarbons have been made by
pyrolyzing borazine or by passing it through a silent discharge. Related four-mem-
bered rings. R 2 B 2 M,R 2 , and eight-membered rings. R 4 B 4 M 4 R 4 , are also known, but
considerably less work has been done on them than on borazine.
Renewed interest in borazine derivatives has resulted from their possible applica¬
tion as precursors to boron nitride ceramics. For example, the inorganic analogue of
styrene, (H-,C=CH)B 3 N 3 H 5 , has been polymerized and decomposed to produce
BN. 77
H N -C Borminc
^B^ B H HvT
H B H
H \ / B \ H / H
lO'
„- b V- b -h
BN (16.43)
Borazines have the correct B-to-N ratio for the production of this ceramic and its
polymeric precursor may be used to deposit a uniform surface coating.
Benzene may be hydrogenated to produce the saturated compound cyclohexane.
Hydrogenation of borazine results in polymeric materials of indefinite compositions.
Substituted derivatives of the saturated cycloborazane, B 3 N 3 H I2 , form readily by
addition to borazine (Eqs. 16.39 and 16.41), but special techniques are necessary to
prepare the parent compound. It was first synthesized by the reduction of the chloro
derivative:
2B 3 N 3 H 6 + 6HCI
2C1 3 B 3 N 3 H 9
2B 3 N 3 H, 2 + 3B,H 6 + 6NaCI (16.44)
72 Lynch. A. T.; Sneddon, L. G. J. Am. Client. Soc. 1989, III. 6201-6209. Boron nitride exists in two
forms, one analogous to graphite and the other to diamond. The graphite form has layers in which
the boron atoms lie above nitrogen atoms in the layer below. Although this material shares with
graphite the property of being a lubricant, unlike graphite, it is an electrical insulator. The cubic
form, second only to diamond in hardness, is an excellent synthetic abrasive.
Rings
769
Isoelectronic with borazine is boroxine. H 3 B 3 0 3 . It can be produced by the
explosive oxidation of or B S H„. Boroxine is planar but has even less it de-
localization than borazine. It is also less stable, decomposing at room temperature to
diborane and boron oxide.
A boron-phosphorus analogue of borazine has been synthesized rather recently. 7 - 1
Ph o Ph
PI / i N'' , Sh
The electronegatitives of B and P are similar, unlike those of B and N. As a result,
polarization should be less extensive in this compound than in borazine. The B 3 P 3 ring
is planar, with equal BP bond lengths and shortened BP bonds, suggesting significant
aromaticity. Even more recently the boron of borazine derivatives has been replaced
with aluminum to give "aiumazenes." 744
Phosphazenes 75 Early workers noted the extreme reactivity of phosphorus pentachloride toward basic
reagents such as water or ammonia. With the former the reaction is reasonably
straightforward, at least for certain stoichiometries:
PC1 5 + H 2 0
PC1 5 + 4H 2 0
OPCl 3 + 2HC1
H 3 P0 4 + 5HCI
For reactions with ammonia analogous products such as HN=PC1, and HN=P(NH,) 3
were proposed, but characterization was hampered by incomplete reactions, separa¬
tion-resistant mixtures, and sensitivity to moisture. Furthermore, gradual polymeriza¬
tion occurred with loss of ammonia to yield “phospham", a poorly characterized solid
of approximate formula (PN,H) r as the ultimate product. If instead of free ammonia its
less reactive conjugate acid is used, reaction with PCI, proceeds at a moderate rate
and the results are more definitive:
NH < C1 + rc, 3 ~Rc nuing chcuchci , 1 “PNCI," (16.47)
If the product were a monomer, its structure could be drawn as CUP=N, which is
analogous to organic nitriles. R—C=N. For this reason the original names used for
these compounds were phosphonitriles, phosphonitrilic chloride, etc. However, the
products are actually either cyclic or linear polymers of general formula [NPC1,],,.
Ihus, by analogy with benzene, borazine, etc., these compounds have become known
as phosphazenes. The major product of the reaction in Eq. 16.47 and the easiest to
73 Dias. H. V. R.; Power. P. P. Angew. Client. Ini. Ed. Engl. 1987. 26. 1270-1271. Dias, H. V. R.;
Power. P. P. J. Am. Client. Soc. 1989. III. 144-148.
7J Waggoner. K. M.; Power. P. P. J. Am. Chem. Soc. 1991. 113. 3385-3393.
75 Allcock. H. R. Phosphorus-Nitrogen Compounds’, Academic: New York. 1972.
■-—Ij
Fig. 15.12 Calculated
contour diagrams for the
isolobal a | orbitals of
[MnH,|'~ (left) and (CH 3 )'
(right). The contours are
plotted in a plane passing
through manganese and
three hydrogen atoms and
through carbon and one
hydrogen. [From
Hoffmann, R. Angew.
Chem. hit. Ed. Engl. 1982,
21. 711-724. Reproduced
with permission.)
Fig. 15.13 Overlap
integrals for the interaction
between the a, frontier
orbital of [MnH s ]' _ or
(CHj)' and a l.v orbital on
H at a distance R from the
Mn or C. [From Hoffmann.
R. Angew. Chem. Ini. Ed.
Engl. 1982, 21. 711-724.
Reproduced with
permission.)
Table 15.6
Some isolobal transition
metal and main group
fragments
Co(CO) 3
Mn(CO) 4
Cr(CO) 2 Cp
NiCp
Fe(CO).
Cr(CO)j
Ni(CO) :
CoCp
CH +
Fe(CO)„
Cr(CO) 5
Ni(CO) 3
Co(CO)Cp
Mn(CO) 5
Co(CO) 4
Fe(CO),Cp
Ni(CO),
Fe(CO),
Cr(CO),
CH7
Metal Carbonyl Complexes
649
(a) (b)
Fig. 15.14 Comparison of (a) Ir 4 (CO), 2 and (b) P 4 . Both are tetramers. composed of the
isolobal fragments Ir(CO) 3 and P, respectively, each of which is trivalent. [Structure (a) from
Wilkes, G. R.; Dahl, L. F. Perspectives in Structural Chemistry 1968, 2, 71. Reproduced with
permission.)
the two compounds goes beyond the fact that both possess a four-membered ring. The
building blocks of the two structures. CH, and Os(CO) 4 , arc isolobal fragments. In
fact the isostructural series has been extended to include Os 2 (CO) 8 (CH 2 ) 2 , which has
two Os(CO) 4 and two CH 2 fragments in the ring, and Os(CO) 4 (CH,) 3 , with a ring
consisting of one Os(CO) 4 and three CH, units. 40 Our earlier comparison of the
bridging hydrides of boron and transition metals also could have been couched in
terms of isolobality. The Lewis acids. BH-, and Cr(CO) 5 , are isolobal fragments, as are
BH 4 " and [HCr(CO) 5 ] _ .
Less obvious than the above examples are isolobal relationships that exist be¬
tween fragments which appear to have different numbers of frontier orbitals. The
addition or subtraction of H + to an organic fragment does not change the number of
electrons in its frontier orbitals. As a consequence, CH 2 , CH 3 , and CH~ are isolobal
fragments. This result is less surprising if you consider that there are a variety of ways
in which the carbon atom may be hybridized. It is worth noting that if two fragments
are both isolobal with a third, they are isolobal with each other as well.
We shall encounter further examples of isolobal fragments later in this and
subsequent chapters, but for now we can sum up their essential features as follows:
Two fragments are isolobal if the number, symmetry properties, approximate ener¬
gies, and shapes of their frontier orbitals and the number of electrons in them are
similar.
40 Anderson. O. P.; Bender. B. R.; Norton. J. R.; Larson. A. C.; Vergamini, P. J. OrganometulHcs
1991, 10. 3145-3150. Lindner. E.; Jansen. R.-M.: Hiller. W.; Fawzi, R. Chem. Ber. 1989. 122.
1403-1409.
766
1 6- Inorganic Chains, Rings, Cages, and Clusters
H H H
Fig. 16.21 Electronic structures of (a) benzene; (b) borazine.
density is localized on the nitrogen atoms (Fig. 16.22). 7 " This partial localization
weakens the ir-bonding in the ring. Each nitrogen receives more a-electron density
from neighboring boron than it gives away as a tr-donor. The net effect is that the
charge density on nitrogen increases. In addition, nitrogen retains its basicity and
boron its acidity. Polar species such as HCI can therefore attack the double bond
between nitrogen and boron. Thus, in contrast to benzene, borazine readily undergoes
addition reactions:
[^^j] + HCI - No reaction
(16.40)
7 " Boyd. R. J.; Choi. S. C.: Hale. C. C. Client. Pltys. Leu. 1984. 112. 136-141. Fink. W. H.; Richards.
J. C. J. Am. Client. Sue. 1991. 113. 3393-3398.
Rings 767
Fig. 16.22 Contour map of the charge density in the molecular plane of la) benzene
lb) borazine. [From Boyd. R. J.; Choi. S. C.; Hale. C. C. Client. Phys. Leu. 1984. 112.
136-141. Reproduced with permission,)
The contrasting tendencies of the two compounds toward addition vs. aromatic
substitution is illustrated by their reactions with bromine:
H Br Br
Br \/ Br H i H
h~-n n N'-H
h 7 b ^ / B \ H
Br Br Br N Br
H Br H
The electronic difference between benzene and borazine is further supported by
the properties of compounds of the type (R (l B,N,)Cr(CO),. Although these are for¬
mally analogous to (r; < '-C h R h )Cr(CO),, the bonding is not nearly so strong in the
borazine complex—its ring-metal dissociation energy appears to be about one-half
that of the arene complex. In addition, there is considerable evidence that the borazine
molecule is puckered in these complexes. 71 The actual structure appears to be inter¬
mediate between a true 7r complex and the extreme a-only model:
71 Scotti. M.: Valdcrrama. M.; Ganz. R.: Werner. H. J. Orauiwmel. Client. 1985. 2H6. 399-406.
650
1 5 • Organometallic Chemistry
Nitrosyl Complexes Few complexes containing only nitrosyl ligands are well characterized, but many
-mixed carbonyl-nitrosyl complexes are known. 41 They may be formed readily by
replacement of carbon monoxide with nitric oxide:
Fe(CO) 5 + 2NO -♦ Fe(CO) 2 (NO) 2 + 3CO (15.33)
Co 2 (CO) g + 2NO- 2Co(CO) 3 (NO) + 2CO (15.34)
Unlike carbon monoxide, which can be used in excess at high temperatures and
pressures, nitric oxide in excess can cause unfavorable oxidation, and at high pres¬
sures and temperatures it decomposes. Many of the current syntheses avoid the use of
nitric oxide by substituting nitrosyl chloride, nitrites, or nitrosonium salts: 4 -
[Mn(CO) s ]" + NOC1 -» Mn(CO) 4 NO + CP + CO (15.35)
(T, 5 -C 5 H,)Re(CO) 3 + NO + - [(r, 3 -C 5 H 5 )Re(CO) 2 NO| + + CO (15.36)
[Co(CO) 4 ]- + NO, + 2C0 2 + H 2 0 -♦ Co(NO)(CO) 3 + 2HC0 3 - + CO
(15.37)
Although the nitrosyl group generally occurs as a terminal ligand, bridging nitro-
syls are also known:
As in the case of the corresponding carbonyl complexes, infrared stretching frequen¬
cies are diagnostic of the mode of coordination. 43 For the product in Eq. 15.38. v
(terminal NO) = 1672 cm -1 and v (bridging NO) = 1505 cm' 1 .
Since the nitrosyl cation. NO\ is isoelectronic with CO. it is not surprising that
there is a great similarity in the behavior of the two ligands. They each have three
bonding pairs between the atoms and lone pairs on both atoms. Although either atom
in NO is a potential donor, the nitrogen atom coordinates preferentially icf. carbon
monoxide), avoiding a large formal positive charge on the more electronegative
oxygen atom. However, in one important respect the nitrosyl group behaves in a
manner not observed for carbon monoxide. Although most nitrosyl ligands appear to
be linear, consistent with sp hybridization of the nitrogen, a few cases of distinctly
bent species are known. A bent nitrosyl ligand is an analogue of an organic nitroso
group or the NO group in Cl—N=0. where the nitrogen can be considered to be sp~
hybridized and bears a lone pair. It is this lone pair that causes the nitrosyl group to be
41 Except for Cr(NO)j. no binary metal nitrosyl complexes have been obtained in pure form. Guest.
M. F,: Hillier. t. H.; Vincent. M.; Rosi. M. J. Clion. Soc. Chon. Common. 1986. 438-439.
42 Caution. K. G. Coord. Chon. Rev. 1975. 14. 317-355.
44 The stretching frequency of the nitrosyl group is sensitive to other bonding parameters and is found
over a range of about 1000-2000 cm' 1 . For carbonyl-nitrosyl complexes, the N—O stretch is
typically found between 1500 and 1900 cm '.
Nitrosyl Complexes
651
I
bent. In the neutral atom method of electron counting, a linear nitrosyl ligand is
regarded as a three-electron donor and a bent nitrosyl as a one-electron donor. The
count for the linear case includes the nonbonding electron pair on nitrogen as well as
the unpaired antibonding electron in NO:
M + :N = 0: - M — N =6: 4 - M = N = 0 (15.39)
In the oxidation state method, the ligand is viewed as a coordinated nitrosyl ion, NO + ,
when linear and a coordinated NO" when bent: it is a two-electron donor in both
forms.
The first well-characterized example of a bent nitrosyl ligand was that found in a
derivative of Vaska’s complex:
Ir(Ph 3 P) 2 (CO)Cl + NO + BF; - [Ir(Ph 3 P) 2 (CO)(NO)Cl] + BF; (15.40)
The product is square pyramidal with a bent nitrosyl ligand (L Ir—N—O = 124°) at
the apical position (Fig. 15.15a). 44 Other complexes with a bent M—NO have been
found, including the remarkable example [Ru(Ph,P)-,(NO),Cl| h [PF,71. which contains
both linear and bent nitrosyl groups 45 (Fig. 15.15b).
The question of whether a nitrosyl ligand will be linear or bent resolves itself into
whether the pair of electrons in question will be forced to reside in an orbital on the
nitrogen atom (bent group) or whether there is a low-lying metal-based molecular
orbital available to it. If there are available nonbonding MOs on the metal (an electron-
poor system), 46 the pair can reside there and allow the nitrogen to function as an sp </
donor with concomitant 7r back bonding. On the other hand, if all the low-lying
orbitals on the metal are already filled (an electron-rich system), the pair of electrons
tbt
Fig. 15.15 Complexes containing bent nitrosyl groups: (a) (Ir<PPh.,) 2 (COK NO)CI 1 and
(b) [Ru(PPhj) 2 (NO) 2 CI]. Phenyl groups have been omitted for clarity: distances are in
picometers. (Structure (a) from Hodgson. D. J.; Payne. N. C.; McGirtnety, J. A.; Pearson. R.
G.; Ibers. J. A. J. Am. Chem. Soc. 1968. 90. 4-186-4488. Structure (b) from Pierpont. C. G.:
Eisenberg, R. Inorg. Chem. 1972, II, 1088-1094. Reproduced with permission.!
44 Hodgson. D. J.; Ibers. J. A. Inorg. Chon. 1968. 7. 2345-2352,
44 Pierpont. C. G.; Eisenberg. R. Inorg. Chon. 1972. II. 1088-1094. See also Bell, L. K.; Mingus.
D. M. P.; Tew, D. G.; Larkworthy. L. F.; Sandell. B.; Povey, D. C.; Mason. J. J. Chon. Soc.
Chon. Common. 1983. 125-126.
This is a relative term. All metals in carbonyl-nitrosyl systems tire relatively electron rich compared
with, for example, those in fluoride complexes.
764
16 Inorganic Chains, Rings, Cages, and Clusters
(b)
Fig. 16.20 l? 0 NMR spectra of (a) [W 6 0|»] 5- and (b) [(Cp)Ti(W,0, B )] 3_ . Labels on spectral
lines indicate assignments: line a to site A, line b to site B. etc.; = impurity. [From Day.
V. W.; Klemperer. W. G. Science 1985, 228, 533-541. Reproduced with permission.]
one signal (bl for the twelve bridging oxygen atoms, one signal (a) for the encapsulated
oxygen, and one for the terminal oxygens (c). The organometailic derivative
[(Cp)Ti(W 5 0|„)] 3- gives a spectrum (Fig. 16.20b) considerably more complicated.
Now there are six different kinds of oxygen atoms, each giving rise to a separate
signal, but by comparison to the spectrum of [W 6 O l9 ] 2- , reasonable assignments can
be made for terminal oxygens E and F, encapsulated oxygen A, and bridging oxygens
C and B. The remaining signal is assigned to the D oxygens.
It can be only mentioned here that isopoly and heteropoly anions also give rise to
highly colored mixed oxidation state species: the tungsten bronzes 64 and the hetero¬
poly blues. 65
64 Whiltingham, M. S.; Dines. M. B. Surv. Prog. Client. 1980. 9. 66-69.
65 Jcannin, Y.: Launay. J. P.; Seid Sedjadi. M. A. Inorg. Client. 1980, 19. 2933-2935. Casan-Pastor.
N.: Gomez-Romcro, P.; Jameson, G. B.; Baker. L. C. W. J. Am. Client. Soc. 1991. 113, 5658-5663.
Rings
765
Rings 66 The most important ring system of organic chemistry is the benzene ring, either as a
separate entity or in polynuclear hydrocarbons such as naphthalene, anthracene, and
Borazines phenanthrene. Inorganic chemistry has two (at least) analogues of benzene: borazines.
BjNjRf,. and trimeric cyclophosphazene compounds. P 3 N 3 X 6 .
Borazine has been known since the pioneering work of Alfred Stock early in this
century. Stock's work was important in two regards: He was the first to study
compounds such as the boranes, silanes, and other similar nonmetal compounds, and
he perfected vacuum line techniques for the handling of air- and moisture-sensitive
compounds, invaluable to the modern inorganic chemist. 67 Slock synthesized
borazine by heating the adduct of diborane and ammonia: 68
3B 2 H 6 + 6NH 3 - 3[BH 2 (NH 3 ) 2 ][BH 4 ] 2B 3 N 3 H 6 + 12H 2 (16.34)
More efficient synthesis are: 69
NH 4 C1 + BC1 3 -♦ C1 3 B 3 N 3 H 3 B 3 N 3 H 6 (16.35)
NH 4 C1 + NaBH 4 -> B 3 N 3 H 6 + H, + NaCl (16.36)
N- or B-substituted borazines may be made by appropriate substitution on the starting
materials prior to the synthesis of the ring:
[RNH 3 ]C1 + BC1 3 - C1 3 B 3 N 3 R 3 H 3 B 3 N 3 R 3 (16.37)
or substitution after the ring has formed:
C1 3 B 3 N 3 R 3 + 3LiR' -♦ R' 3 B 3 N 3 R 3 + 3LiCl (16.38)
Borazine is isoelectronic with benzene, as B=N is with C=C, (Fig. 16.21). In
physical properties, borazine is indeed a close analogue of benzene. The similarity of
the physical properties of the alkyl-substituted derivatives of benzene and borazine is
even more remarkable. For example, the ratio of the absolute boiling points of the
substituted borazines to those of similarly substituted benzene is constant. This
similarity in physical properties led to a labeling of borazine as "inorganic benzene."
This is a misnomer because the chemical properties of borazine and benzene are quite
different. Both compounds have aromatic it clouds of electron density with potential
for delocalization overall of the ring atoms. Due to the difference in electronegativity
between boron and nitrogen, the cloud in borazine is "lumpy" because more electron
66 The Chemistry of Inorganic Ring Systems'. Steudel, R., Ed.: Elsevier: New York. 1992. Wollins,
J. D. Non-Metal Rings. Cages and Clusters'. Wiley: New York. 1988, Haiduc. I.; Sowerby, D. B.
The Chemistry of Inorganic Homo- and Heterocycles'. Academic: New York. 1987; Vols. I and 2.
Heal. H. G. The Inorganic Heterocyclic Chemistry of Sulfur. Nitrogen and Phosphorus'. Academic:
New York, 1980. Massey. A. G. Main Group Chemistry. Ellis Horwood: New York, 1990.
67 See Shrivcr, D. F. The Manipulation of Air-Sensitive Compounds', McGraw-Hill: New York, 1983,
and Experimental Organometailic Chemistry. Wayda. A. L.; Darcnsbourg, M. Y., Eds.; ACS
Symposium Scries 357; American Chemical Society: Washington. DC. 1985.
68 Stock. A.; Pohland. E. Chem. Ber. 1926, 59. 2215, Stock. A. Hydrides of Boron and Silicon'. Cornell
University: New York. 1933.
69 Borazine syntheses are reviewed in Gmelin Handhuch Der Anorganishen Cliemie: Springer-Vcrlag:
New York. 1978; Vol. 17.
652
15 • Organometallic Chemistry
must occupy an essentially nonbonding orbital on the nitrogen, requiring trigonal
hybridization and a bent system. A comparison of the bonding possibilities for NO and
CO is shown in Fig. 15.16.
The metal-nitrogen bond lengths in the ruthenium complex containing both types
of nitrosyl ligands (Fig. 15.15b) are in accord with the viewjust presented. In the linear
system there is a short metal-nitrogen bond (173.8 pm) indicating substantial -n
bonding (as in metal carbonyls). The bent system, in contrast, shows a relatively long,
essentially er-only metal-nitrogen bond (185.9 pm). It would be expected that the
N—O bond of a bent nitrosyl would be longer than that of a linear nitrosyl. How¬
ever, the insensitivity of the NO bond length to small changes in bond order coupled
with systematic errors in the crystallographic analysis make evaluation of such
data difficult. Within experimental error, the bent and linear N—O bonds of
[Ru(PPh 3 ) 2 (NO),CI] + are the same length (117.0 and 116.2 pm). The N—O bond
lengths in NO^, NO, and NO - are 106, 115, and 120 pm, respectively. We can only
conclude that the NO bond order, for both the apical and basal arrangements, lies
between two and three.
We can actually see the process of electron pair shift with a resultant change in
structure in the complex ion [Co(diars) 2 NO] 2 + (where diars is a bidentate diarsine
ligand) (Fig. 15.17). The 18-electron rule predicts that the nitrosyl group will be linear
(a three-electron donor), as indeed it is. Reaction of this complex with the thiocyanate
ion (a two-electron donor) would violate the 18-electron rule unless a pair is shifted
from a molecular orbital of largely metal character to an orbital on nitrogen. This is in
fact what happens and "stereochemical control of valence" results. 47 As NO goes
from being a three-electron to a one-electron donor, a coordination site capable of
accepting a pair of electrons becomes available.
Fig. 15.16 Geometry (linear vs. bent) of nitrosyl ligands correlated with the hybridization of
the nitrogen atom and parallel correlations for analogous compounds containing carbonyl
ligands.
47 Encmark, J. H,; Feltham. R. D. J. Am. Chem. Soc. 1974, 96. 5002-5004. 5004-5005. Encmark.
J. H.; Feltham, R. D. Coord. Chem. Rev. 1974, 13. 339-406. Feltham. R. D.; Enemark. J. H. In
Topics in Inorganic and Organometallic Stereochemistry, Geoffroy, G. L.. Ed.; Wiley: New York
1981.
Dinitrogen Complexes
653
Fig. 15.17 Stereochemical control of valence. Note localization of the lone pair on the
nitrogen atom and bending of the nitrosyl group upon addition of thiocyanate ion to the
coordination sphere. Bond lengths are in picometers. (From Enemark. .1. H.; Feltham R D
Proc. Natl. Acad. Sci. USA 1972, 69. 3534. Used with permission.)
Molecular nitrogen. N 2 , is isoelectronic with both carbon monoxide and the nitrosyl
ion but. despite the numerous complexes of CO and NO, for many years it proved to
be impossible to form complexes of dinitrogen. This difference in behavior was
usually ascribed to the lack of polarity of N 2 and a resultant inability to behave as a
tt acceptor. 49
The first dinitrogen complex, characterized in 1965, resulted from the reduction of
commercial ruthenium trichloride [containing some Ru(lV)] by hydrazine hydrate.
The pentaammine(dinitrogen)ruthenium(II) cation that formed could be isolated in a
variety of salts. 5 " Soon other methods were found to synthesize the complex, such as
the decomposition of the pentaammineazido complex, [Ru(NH 3 ) 5 N 3 ] 2+ , and even
direct reaction with nitrogen gas:
[Ru(NH 3 ) 5 CI] 2+ [Ru(NH 3 ) 5 H 2 0] 2+ (15.41)
[Ru(NH 3 ),H 2 0] 2+ + N, -► [Ru(NH 3 ) 5 N,] 2+ ( 15 . 42 )
4 Kaul - B 8 : Hayes. R. K.; George. T. A. J. Am. Chem. Soc. 1990. 112, 2002-2003, Birk. R.; Berkc
H.; Huttncr, G.; Zsolnai. L. Chem. Ber. 1988. 121, 1557-1564. Henderson, R. A.; Leigh. G. J.;
Pickett. C. J, Adv. Inorg. Chem. Radiochem. 1983. 27, 197-292. Pelikdn. P.; Boia, R. Coord.
Chem. Rev. 1984 . 55. 55-112. Leigh. G-J. Acc. Chem. Res. 1992. 25. 177-181. Collman. J. P.;
Hutchison. J. E.; Lopez, M. A.; Guilard, R. J. Am Chem. Soc. 1992, 114. 8066-8073.
Note, however, that the dipole moment of carbon monoxide is extremely small: 0.375 x I0 -30 C ni
(0.112 D).
50 Allen - A - D - ; Senoff. C. V. J. Chem. Soc. Chem. Commun. 1965. 621-622.
Dinitrogen
Complexes 48
762
16- Inorganic Chains, Rings, Cages, and Clusters
763
(a) (b)
Fig. 16.16 (a) Planar ring of Mo atoms surrounding heteroatom in 6-heteropoly acids,
(b) Puckered ring of Mo atoms surrounding seventh Mo atom in [M 07 O 24 ] 6- .
Fig. 16.17 Relation between the structures of 12-hcleropoly and dimeric 9-heteropoly acids:
la) the lPW l2 0 4 ol ,_ anion (see Fig. 16.13); (b) the [PWvOjiJ 3- half-unit formed by removal of
shaded octahedra from (a); (c) the dimeric (PjWHiOm] 6- ion formed from two half-units, (b).
|From Wells. A. F. Structural Inorganic Chemistry, 5th ed.; Oxford University: London.
1986; p 522. Reproduced with permission.]
in which many polyanions have been shown to function as ligands.'’- In some in¬
stances the cation simply binds to bridging or terminal oxygen atoms found on the
surface of the polyoxometalatc anion. For example, Mn 2+ binds weakly to a terminal
oxygen atom of [H 2 W l2 0 4( ,] 6_ ; thus the anion functions as a monodentate ligand. If.
however, the polyoxometalate anion has vacancies created by missing metal units
(Fig. 16.18). coordination may occur by incorporating the cation into the vacancy (Fig.
16.19). Polyoxometalate structures with vacancies are referred to as lacunary (a space
where something has been omitted) species. These species may function as pentaden-
tate ligands (e.g., six-coordinate Co in [(SiW, l O v ,)Co(H 2 0)) h -) or tetradentate
ligands (e.g.. four-coordinate Cu in [(PW, l 0 39 )Cu] 5- ).
All facets of study have been greatly aided by the ease with which crystal
structures may be obtained and by the availability of sensitive Fourier transform NMR
spectrometers which allow nuclei such as l7 0, 5I V, v3 Nb, 95 Mo. and "°W to be used
for structural studies. Oxygen-17 NMR spectroscopy has proved to be particularly
useful because l7 0 chemical shifts are very sensitive to environment. As a result it is
possible to distinguish between terminal and various kinds of bridging oxygen sites.
The l7 0 spectrum of [W 6 O iy ] 2- and its structure are shown in Fig. 16.20a. M We see
Chains
Fig. 16.18 Examples of the polyoxometalate (r-IXM 12 O.n 1 ]'' - ion losing both one metal
fragment (XMhOvj) and three metal fragments (isomers A-XM v O w and B-XMyOjj), The
structures formed (called lacunary structures) have vacant sites which can be filled by other
metal fragments. (From Pope. M. T.; Miiller. A. Angtrw. diem. lilt. Ed. Engl. 1991. JO.
34-48. Reproduced with permission.]
h - Pope. M. T.; Muller. A. Aitgew. Client, hit. Ed. Engl. 1991. JO. 34-48.
« Day. V. W.; Klemperer. W. G. Science 1985. 22S. 533-541.
Fig. 16.19 The lacunary polyoxometalate ligand, XM||Oj y . reacting with variety of Lewis
acids. |From Pope. M. T.; Muller. A. Angew. Client, hit. Ed. Engl. 1991. JO. 34-38.
Reproduced with permission.!
654
15- Organometallic Chemistry
The unexpectedly strong nucleophilicity of dinitrogen shown by its displacement of
water in Eq. 15.42 is also exhibited in the formation of a bridged complex:
[Ru(NH 3 ) 5 N 2 ] 2t + [Ru(NH 3 ) 5 H 2 0] 2+ -»
[Ru(NH 3 ) 5 N 2 Ru(NH 3 ) 5 ] 4+ + H 2 0 (,5 ‘ 43)
There are two structural possibilities for terminal dinitrogen ligands and two for
bridged cases: 51
M—N—N M—N—N—M
end-on terminal end-on bridging
.N
M | M x I
N N N
side-on terminal side-on bridging
An X-ray study of the original ruthenium-dinitrogen complex 52 indicated that the
nature of the Ru—N—N linkage was end on. but disorder in the crystal prevented
accurate determination of bond lengths. Since then, many structures of other dim-
trogen complexes have been determined, including the bis(dinitrogen) crown thioether
complex shown in Fig. 15.18. 53 The results for this complex are typical and show that
N, greatly resembles CO in its bonding to metals. Back donation of electron density
from the metal into tt antibonding orbitals is apparent from the short Mo—N bond,
which is more similar in length to an Mo-CO bond than to the Mo-NH 3 bond m
ammine complexes. The mean N-N bond length (110.7 pm) is slightly greater than
[hat found in molecular nitrogen (109.8 pm), suggested a weakening of the nitro¬
gen-nitrogen triple bond from the donation of electron density into the tt antibonding
orbitals of nitrogen (cf. the value of 111.6 pm for a bond order of 2.5 in N,). No
examples of side-on binding by N 2 have been confirmed in nonbridging complexes. 54
The Raman stretching frequency of free N 2 is 2331 cm . Upon coordination, this
vibration becomes infrared active and shifts to a lower frequency. For example, strong
N—N stretching bands appear at 2105 cm 1 for [Ru(NHjh(N ; )]CI, and at 1955 and
1890 cm' 1 for //wt.v-Mo(N-,) 2 Me K [I6]aneS 4 .
It appears that the metal-ligand bonds in carbonyl and dinitrogen complexes arc
similar but somewhat weaker in the dinitrogen complexes. Carbon monoxide is not
only a better cr donor but also a better tt acceptor. This is to be expected on the basis
of whatever polarity exists in the CO molecule (see Fig. 5.18) and the fact that the i r
antibonding orbital is concentrated on the carbon atom (see Fig. 5.20). which favors
overlap with the metal orbital. The superior ir accepting ability of CO also accounts
for the instability of carbonyl dinitrogen complexes. Both Cr(CO) 5 N 2 and cis-
51 Chalt. J.; Dilworth, J. R.: Richards, R. L. Ghent. Rev. 1978. 78. 589-625.
57 Bouomlcy, F.; Nyburg. S. C. Acta Crystullogr. Sect. B 1968. 24. 1289-1293.
53 Yoshida. T.; Adachi. T.; Kaminaka. M.; Ucda. T.; Higuchi. T. J. Am. Client. Soc. 1988. 110.
4872-4873.
m Although there have been no sidc-on structures verified by X-ray analysis, fairly good spectroscopic
evidence supports that mode of coordination in (tj-C 5 H 5 ) 2 Zr (CHtSiMejlJtN;). Jeffery. J.: Lap-
pert. M. F.; Riley. P. I. J. Organomet. Chem. 1979. 181. 25-36.
Fig. 15.18 Structure of
Mol Me K l 16 ]uneS 4 )( N 2 );.
Bond lengths are in
picometers. [From Yoshida.
T.; Adachi. T.; Kaminaka.
M.; Ueda, T.; Higuchi. T.
J. Ant. Client. Soc. 1988.
lit). 4872-4873. Used with
permission. I
Me,| l6|ancS 4
Metal Alkyls,
r nes, a,
Carbides
Alkyl Complexes
Metal Alkyls, Carbenes, Carbynes, and Carbides 655
Fig. 15.19 Structure of
Sm(i|'-C 5 Me 5 ) 2 (N 2 ). [From
Evans. W. J.; Ulibarri.
T. A.; Ziller, J. W. J. Am.
Chem. Soc. 1988, HO.
6877-6879. Used with
permission.]
Cr(CO) 4 (N 2 ) 2 have been investigated at low temperatures, but decompose when
warmed. 55 Replacing some of the carbonyls with phosphines can provide sufficient
electron density lo significantly enhance stability; hence. Mo(CO) 3 (PCy 3 ) 2 N 2 can be
isolated at room temperature. 56
Mo(CO) 3 (PCy 3 ) 2 + N, -► Mo(CO) 3 (PCy 3 ) 2 N 2 (15.44)
When dinitrogen functions as a bridging ligand, it usually exhibits end-on coordi¬
nation: this is the case in the diruthenium complex of Eq. 15.43. for example. Bridging
side-on complexes are also known, however, and a recently reported example is also
the first dinitrogen complex of an/element (Fig. 15.19). 57 In this samarium complex,
obtained from the reaction of Sm(C<Mc,H and N : , ihe two samarium atoms and the
two nitrogen atoms are in a planar arrangement. The Sm—N bond distances suggest
the presence of Smtllll. implying a reduced N — N bond, but strangely enough the
N —N bond distance (108.8 pm) is even shorter than that found in free dinitrogen.
The ability to synthesize complexes containing dinitrogen. especially those with
considerable alteration of the electronic state of nitrogen, opens up possibilities of
direct fixation of nitrogen from the atmosphere, a long-standing challenge to the
chemist. 5 ll also provides insight into the closely related process of biological fixation
of nitrogen and the enzyme systems involved (see Chapter 19).
Single, double, and triple bonds between carbon and nonmetals such as carbon,
nitrogen, and oxygen have long occupied a central position in organic chemistry. The
chemistry of metal-carbon single bonds in main group compounds (e.g., Grignard
reagents and organomercury compounds) dates back to the 19th century. Transition
metal compounds containing metal-carbon single, double, and triple bonds have come
to be understood much more recently:
M—CR 3 M=CR 2 M=CR
Although there are some early examples of complexes in which M—C single
bonds are present (e.g.. (Me 3 PtI] 4 synthesized in 1907), (he prevailing view for many
55 Upmacis. R. K.; Poliakoff. M.; Turner. J. J. J. Am. Client. Soc. 1986. 108. 3645-3651.
5 Wasserman. H. J.; Kubas. G. J.: Ryan. R. R. J. Am. Client. Soc. 1986. 108. 2294-2301.
’ 7 Evans. W. J.; Ulibarri. T. A.: Ziller. J. W. J. Am. Client. Soc. 1988. I/O. 6877-6879.
58 Colquhoun. H. M. Acc. Client. Res. 1984. 17. 23-28. George. T. A.; Tisdale. R. C. Inorg. Cltem.
1988. 27. 2909-2912. See Chapter 13 for a photolytic conversion of N 2 lo NH,.
Fig. 16.13 (a) The structures of two heteropoly anions, 12-molybdophosphate or
12-tungstophosphate. (b) and (c) Details of coordination of three M0 6 octahedra with one
corner of the heteroatom tetrahedron.
It has been noted that there is a cavity in the center of the metatungstate ion. This
cavity is surrounded by a tetrahedron of four oxygen atoms (Fig. 16.13) that is
sufficiently large to accommodate a relatively small atom, such as P(V), As(V), Si(IV),
Ge(IV), Ti(IV), or Zr(IV). 56 The 12-tungstoheteropoly anions 57 are of general formula
[X 1 W i; O 40 ]i ff - n )- - 58 Analogous molybdoheteropoly anions are also known. For
example, when a solution containing phosphate and molybdate is acidified, the ion
[PMo I 2 O 40 ] 3- is formed. Obviously phosphorus-oxygen bonds are not broken in the
process so we can view the product anion as the incorporation of P0 4 ~ into an
M°, ; 0 V) cage. Molybdoheteropoly anions of this type are of some importance in the
qualitative and quantitative analytical chemistry of phosphorus and arsenic.
Between 35 and 40 heteroatoms are known to form heteropoly anions and their
corresponding acids. Large heteroatoms such as Ce(lV) and Th(IV) arc found ico-
sahedrally coordinated in salts such as (NH 4 ) 2 H fi CeMo, : 0 42 (Fig. 16.14). It is unique
inasmuch as pairs of Mo0 6 octahedra share faces to form Mo 2 0 9 groups which are
corner connected to each other. 59
Of the many other heteropoly acids, the 6-molybdo species are of some interest.
These form with heteroatoms Te(VI) and I(VII) and tripositive metal ions such as
Rh(III). All of these heteroatoms prefer an octahedral coordination sphere, which can
be provided by a ring of six Mo0 6 octahedral (Fig. 16.15). Note that formally the 6-
Heteropoly Anions
36 The resulting structure, which has Tj symmetry, has come to be known as the Keggin slriiclure,
named after its discoverer.
57 The prefix 12- may be used to replace the more cumbersome “dodeca’'- to indicate the number of
metal atom octahedra coordinated to the helcratom.
3K One or more protons may be affixed to the anion with corresponding reduction of anionic eh,.
59 Dexter. D. D.; Silverton. J. V. J. Am. Client. Soc. 1968, 90. 3589-3590.
Fig. 16.14 Idealized sketch
of the [CeMoijOjj] 11 " ion
showing the linkage of the
MoO ( , octahedra.
heteropoly formulation can be applied to the heptamolybdate species discussed earlier
if the seventh molybdenum atom is considered to be a pseudo-heteroatom. At one
time it was felt that the 6-heleropoly acids should be isomorphous with the hep¬
tamolybdate. This is not the case as may be seen by comparison of Figs. 16.10c and
16.15. The structures are more similar than might be supposed, however, the principle
difference being whether the heteroatom is surrounded by a planar ring of molyb¬
denum atoms (6-heteropoly species. Fig. 16.16a) or a puckered ring (heptamolybdate.
Fig. 16.16b). There are also more complicated heteropoly acids, including di¬
heteropoly acids such as [P 2 W ]K 0 62 ] 6 ", which has been found to have a structure
(sometimes called the Dawson structure) related to the 12-hetcropoly acids (Fig.
16.17).
As a class, the isopoly and heterpoly anions offer several interesting facets for
study. 60 They may be considered small chunks of metal oxide lattices. As such they
provide insight into catalysis by heterogeneous oxides, an approach that is currently
enjoying strong interest for selective oxidation of organic molecules. As anions they
show very low surface charge densities and low basicities. For example, we generally
think of the perchlorate ion, CI0 4 , as having a very low basicity. One study has shown
that the hexamolybdatet -2) ion and l2-tungstophosphate(-3) ion have lower
basicities than perchlorate, and the l2-molybdophosphate(-3) ion is only slightly
more basic than perchlorate. 61 Nevertheless, a rich coordination chemistry is evolving
Fig. 16.15 The siructurc of
the 6-molybdotellurate
anion, |TeMo h 0 24 )'’'
[Adapted, in part, from
Kepert, D. L. Prog, htorg.
Client. 1965, 4. 199.
Reproduced with
permission.)
“ Day. V. W.; Klemperer. W. G. Science 1985. 228. 533-541.
6 ‘ Barcza. L.: Pope. M. T. J. Phys. Client . 1975, 79. 92-93.
656
15-Organometallic Chemistry
T ab le 15.7
Bond dissociation
enthalpies, D, for
metal-carbon bonds"
years was that transition metal alkyls, unlike the main group alkyls, are ther¬
modynamically unstable. This conclusion was reached because synthetic attempts to
obtain compounds such as diethyliron or diethylcobalt (e.g., by reactions between
FeBr 2 and EtMgBr) were unsuccessful. In fact transition metal-carbon bonds are in
general no less strong than main group metal-carbon bonds (Table 15.7). However, it
should be noted that, although metal-carbon bonds decrease in strength as the atomic
number increases for the main group metals, they increase in strength as the atomic
number increases for transition metals. Thus the early focus on the first-row transition
series was least favorable from a thermodynamic point of view. However, the prin¬
cipal difficulty in obtaining transition metal-carbon bonds was not thermodynamic but
kinetic. There are a number of favorable pathways available to metal alkyls for
decomposition. One of the most important is f) elimination (page 699):
Metal Alkyls, Carbenes, Carbynes, and Carbide:
657
The syntheses of transition metal alkyls can be accomplished in several ways. A
common approach is to take advantage of the nucleophilicily of a carbonylale ion. 59
For example:
[Mn(CO) 5 ]~ + Mel -+ MeMn(CO) 5 + I - (15.46)
In this reaction [Mn(CO) 5 ] _ , which is quite nucleophilic, increases its metal coordina¬
tion number by one. The reaction may be viewed as an electrophilic attack by R h on
the metal. Similarly, it is possible to prepare bridging alkyl complexes by this
method: 60
H,
[Fe 2 (CO) 8 ] 2 “ + CH 2 I 2 -♦ (OC) 4 Fe-Fe(CO) 4 + 21" (15.47)
A second approach involves a nucleophilic attack on the metal. This is seen in the
reaction of methyl lithium with tungsten! VI) chloride:
WCI 6 + 6MeLi - WMe 6 + 6LiCl (15.48)
The red crystals which form in this reaction melt at 30 °C and are reasonably stable.
Two other important routes to transition metal alkyls are oxidative addition and
insertion, topics that are discussed more in the section on organometallic reactions.
Carbene, Carbyne,
and Carbide
Complexes
I he first of these are called carbene complexes and the latter are referred to as
carbyne complexes. 6 ' The first carbene complex was reported in 1964 by Fischer and
Maasbol 6 - and was prepared by reaction of hexacarbonyltungsten with methyl or
phenyl lithium to generate an acyl anion which was then alkylated with diazomethane.
W(CO) 6 + RLi- [W(CO) 5 (COR)]-Li + (OC) 5 W=C /
N -OMe
(15.49)
59 Nucleophilicily is a kinelic term used to describe the rate at which a nucleophile reacts with a
substrate. Relative nuclcophilicities of [Re(CO),]“. (MnlCO),]". and |Co(CO)d' as measured by
their reaction with Mel are 22,900, 169. and I. respectively. Pearson, R. G.; Figdore P E J Am
Chem. Soc. 1980. 102, 1541-1547.
“ Summer. C. E.. Jr.; Riley. P. E.; Davis. R. E.; Pettit. R. J. Am. Chem. Soc. 1980, 102. 1752-1754.
Holton, J.; Lappcrt, M. F.; Pearce, R.; Yarrow, P. I. W. Chem. Rev. 1983, S3. 135-201.
61 Carbene complexes arc also referred to in Ihe literature as alkylidencs and carbyne complexes are
sometimes called alkylidynes.
« Fischer. E. O.; Maasbol. A. Angew. Chem. Ini. Ed. Engl. 1964. 3, 580-581.
It has been relatively recent in chemical chronology that compounds containing formal
metal-carbon double and triple bonds were discovered:
M=C M=C—R
R
758
16 Inorganic Chains, Rings, Cages, and Clusters
competing for the same vacant t 2u metal orbital. Instead they are found opposite a
bridging or internal oxygen. The effect is that the metal ion is displaced in the direction
of the terminal oxygen, away from the oxygen opposite it (trans effect), just as you
would predict based on a metal ion-metal ion repulsion model. Metal ions such as
AI(III) or Ga(III) are poor -tr acceptors. Thus their terminal oxygen atoms are not
stabilized and can repeatedly attack other units to give continuing polymerization. The
terminal oxygen atoms of the transition metal polyanions, however, are stabilized by
it bonding and have less affinity for adjacent metal units. 47
Although elucidation of various molybdate species continues, four appear to be
most important: (I) the simple molybdate, MoOj - , stable at high pH; (2) the- hep-
tamolybdate (also known as paramolybdate), [Mo 7 0 24 ] 6 ~ (Fig. 16.10c), formed in
equilibrium with molybdate down to pH 4-5; (3) octamolybdate, [0-Mo g O 26 ] 4- (Fig.
16.10b), 48 formed in more acidic solutions; (4) [Mo 36 0,| 2 (H 2 0) I6 ] 8- , the largest
isopolyanion known, present in solutions at about pH I.8. 49 From strongly acidic
solutions can be precipitated polymeric Mo0 3 -2H 2 0 consisting of sheets of corner-
shared MO fi octahedra.
The formation of isopolytungstates is similar to that described for the molybdates
although the chemistry is even more difficult. The simple tungstate, WOj - exists in
strongly basic solution. Acidification results in the formation of polymers built up from
WO fi octahedra. The nature of the tungsten species present depends not only on the
present conditions (e.g., pH) but also on the history of the sample since some of the
conversions are slow. Upon acidification of W0 4 _ , "paratungstate A”, 50 [W 7 0 2 J 6- ,
forms rapidly. Its protonated form, [HW 7 0 24 ] 5_ , has also been detected. 51 From these
solutions are precipitated salts of the dodecameric anion, [H 2 W 12 0 42 J 10- (paratung-
state B), whose framework is shown in Fig. 16.1 la. From more acidic solutions it is
possible to crystallize a second dodecatungstate ion, (H 2 W 12 O 40 l 6 ~ ("metatung¬
state"). The structure of this ion, although built of the same W0 6 octahedra, is more
symmetrical, resulting in a cavity in the center of the ion (Fig. 16.11b). 52
In recent times many advances in isopoly anion chemistry have been made by
shifting reaction chemistry from aqueous to aprotic solution. This can often be done
by employing a solubilizing cation such as tetrabutylammonium ion. For example,
when [(/j-Bu) 4 N]OH and [(/i-Bu) 4 N|[H 3 V, (l 0 2g ] are mixed in acetonitrile a new iso¬
polyvanadate forms: 53
[H, v „,0 2R ] 3 - + 30H" -► 2[V s O I 4 J j - + 3H,0 (16.33)
47 Pope. M. T. Heteropoly ami Isopoly Oxomettilates: Springer-Vcrlag: New York. 1983.
J,< Both a and 0 isomers of [MogOjfti - are known and isomcrizc intramolccularly in solution.
Klemperer, W. G.; Shum. W. J. Am. Chem. Soc. 1976. 98. 8291-8293. Masters. A. F.; Gheller. S.
F.; Brownlee, R. T. C.; O'Connor, M. J.; Wedd. A. G. Inorg. Chem. 1980. 19. 3866-3868.
Klemperer, W. G.; Schwartz, C.; Wright, D. A. J. Am. Chem. Soc. 1985, 107. 6941-6950.
4V Krebs, B.; Paulat-Boschen. I. Ada Crystallogr.. Sect. B 1982, 38. 1710-1718.
50 Prior to any definite knowledge of the structure or even of the empirical formula of each of the
various paratungstate ions, they were arbitrarily assigned letter labels such as A. B. X. Y. and Z.
Much early confusion in this field occurred because workers referred to "paratungstate” without
specifying which of the many possible species was being studied.
31 Cruywagen, J. J.; van der Merwe. I. F. J. J. Chem. Soc. Dalton Trans. 1987, 1701-1705.
52 Although it might appear that there is a similar, but smaller, cavity in the paratungstate B ion. the
van der Waals radii of the oxygen atoms on the inner apices of the octahedra forming the structure
effectively till the cavity.
53 Day. V. W.; Klemperer. W. G.; Yaghi. O. M. J. Am. Chem. Soc. 1989. III. 4518-4519.
Chains
759
Fig. 16.11 The structures of two apex-shared dodecatungstate isopoly anions; (a) the
paratungstate B ion. [H ; W| ; O 4 ;] l0 “; (b) the metatungstate ion. [H : W| 2 0. lu r'. [From
Lipscomb, W. N. Inorg. Chem. 1965, 4. 132. Reproduced with permission.)
This anion (Fig. 16.12a) is of special significance because it is the first example of a
transition metal polyoxoanion cage that is built from corner-shared tetrahedra. In a
similar vein, refluxing [/?-Bn 4 N][HV l0 O 2s J in acetonitrile leads to [MeCNCV l2 0, 2 | J_
(Fig. 16.12b) in which, remarkably. MeCN is found suspended into a [V| 2 0 : , 2 J 4-
basket. 54
Undoubtedly we can expect many interesting and unusual isopolyanions to be
isolated and characterized in the years ahead, The number of practical applications for
these materials and their derivatives is impressive and extends to medicine, catalysis,
and solid state devices. 55
1
(a) lb)
(/) = vanadium
0 = oxygen
Fig. 16.12 Structures of two oxoanions of vanadium prepared from nonaqueous solvents.
(a) [VjO| 4 |' _ . (b) [CHjCNCV^OjjJ 4- . (The acetonitrile molecule is suspended in the basket
denoted by the symbol C.) [From Day, V. W.; Klemperer. W. G.: Yaghi. O. M. J. Am.
Chem. Soc. 1989, III. 4518 and 5959. Reproduced with permission.!
54 Day. V. W.: Klemperer. W. G.; Yahgi. O. M. J. Am. Chem. Soc. 1989. III. 5959-5961.
55 Pope, M. T.; Muller, A. Angew. Chem. hit. Ed. Engl. 1991. 30. 34-48.
658
15 • Organometallic Chemistry
Substantial improvement in the convenience and scope of carbene synthesis followed
by replacing diazomethane with other alkylating agents such as R,O r BF, or
MeOS0 2 F.« (See Eqs. 15.148-150 for synthesis from [Cr(CO),] : .) Hundreds of
carbene complexes of the type shown in Eq. 15.49 are known. They are characterized
by having a metal in a low oxidation state, tr-accepting auxiliary ligands, and substi¬
tuents on the carbene carbon capable ot donating 7r electron density. When they
participate in reactions, the carbene carbon behaves as an electrophile. Complexes
having these properties are known as "Fischer" carbenes. 64 The usefulness ot these
complexes in organic synthesis is presently under intense investigation (see
page 705). 65
Free carbenes exist in both triplet and singlet slates but those containing a
heteroatom (e.g., O or N), as found in Fischer carbenes, tend to be of the latter
variety. Thus the free ligand may be represented as follows:
p.
The pair of electrons in the s/r orbital may be donated to a metal to form a ir bond, and
an empty />. orbital is present to accept tt electron density. Filled d orbitals of the
metal may donate electrons to the p. orbital, to give a metal-carbon double bond, and
electrons from filled p orbitals of the oxygen atom may also be donated to form a
carbon-oxygen double bond (Fig. 15.20). Resonance torm I5.2()b appears to be
dominant and. although the M— C bond is shorter than expected for a single bond, it is
too long for an M—C double bond, leading to the conclusion that the bond order is
between one and two.
Just ten years after the discovery of Fischer's electrophilic carbene. Schrock
discovered a class of carbenes which are nucleophilic. 6
Ta(CH 2 CMc 3 ) 3 Cl 2 + 2LiCH 2 CMe 3 -♦
CMe 4 + (Mc 3 CCH 2 ) 3 Ta=C x + 2LiCI (15.50)
N CMe 3
These nucleophilic carbenes are composed of early transition metals in high oxidation
states. non-Tr accepting auxilliary ligands, and non-7r donating substituents on carbon.
Mcihyllluorosiilfonatc is a powerful and very toxic alkylating agent. See Client, Enp. News 1976.54
136). 5.
m They arc called carbenes even though they are not made from carbenes and carbenes arc not
synthesized from them. See Dotz. K. H.; Fischer. F.. O.: Hofmann. P.: Kreissel. F. R.: Schubert.
U.; Weiss. K. Transition Metal Carbene Complexes: Verlag Chemie: Deerfield Beach. FL. 1983.
M Dotz. K. H. Anpew. Chem. Ini. Etl. Enpl. 1984 . 23. 587-608. Casey. C. P. In Transition Metal
Orpanometallics in Orpanic Synthesis: Alpcr. H.. Ed.: Academic: New York. 1976: Chapter 3.
WulIf. W. D. In Advances in Metal-Orpanic Chemistry: Licbcskind. L. S.. Ed.: JAI: Greenwich.
1987; Vol. I.
«• Schrock. R. R. J. Am. Chem. Soe. 1974. 96. 6796-6797.
. ■!, ... ' -V:- T a TTtf -.'I- - • : • ' ■ - ■" ”
Metal Alkyls, Carbenes, Carbynes, and Carbides
659
Fig. 15.20 Resonance forms for a transition metal carbene complex. Form (a) shows
metal-carbon double bond character which results from donation of metal d electron density
to an empty p orbital of carbon. Form (b) shows oxygen-carbon double bond character which
results from donation of oxygen p electron density to an empty p orbital of carbon. Form (b)
provides the dominant contribution.
They are called "Schrock” carbenes to distinguish them from the Fischer carbenes.
One way to view these complexes is in terms of two orbitals on the carbene. each
housing an unpaired electron (triplet state) overlapping with two metal orbitals, each
of which provides an electron.'’ 7
mc
($“
The electrophilic nature of a Fischer carbene is illustrated in the following
(OC),CrC
(OC)«Ct— C — NH
(OC)jCrC + iVlcOI-l (15.52)
X Mc
In this reaction the nucleophile. NH,, attacks the carbene carbon to form an inter¬
mediate which eliminates methanol. The reaction is favorable because nitrogen is not
as electronegative as oxygen and its u donating ability exceeds that of oxygen
[stabilizing resonance form (b) in Fig. I5.20|.
The nucleophilic nature of a Schrock carbene is seen in its reaction with Me,AI:
(r, 5 -C 5 Hj) 2 MeTa=CH, + AIMe, -» (Tj 5 -C 5 H,) 2 MeTn—CH,AIMe 3 (15.53)
67 Taylor. T. E.; Hall. M. B. J. Am. Chem Sac. 1984. /06. 1576-1584.
756
16-Inorganic Chains, Rings, Cages, and Clusters
Chains
757
pH
Fig. 16.9 Dominant
oxoanions of vanadium
present in aqueous solution
as a function of concen¬
tration and pH. [From
Pope. M. T. Heteropoly
and Isopoly Oxometalates:
Springer-Verlag: New
York. 1983. Reproduced
with permission.]
Other metals such as vanadium have more complicated chemistry. The vanadate
ion. VO 4 , exists in extremely basic solution (Fig. 16.9). Under very dilute conditions
as the pH is lowered, protonation occurs to give monomers:
VOJ- VO 3 OH 2 - V0 2 (0H) 2 " VO(OH) 3 (?) - V0 2 4
(16.32)
When solutions are more concentrated, however, protonation and dehydration occur
to form [V,0 7 ] 4- and higher vanadates . 44 Further polymerization occurs until hydrous
V 2 0 5 precipitates at low pH. The precipitation of vanadium(V) oxide from aqueous
solution as well as the similar behavior of other metal oxides, such as MoOj and W0 3 ,
stands in sharp contrast to the extremely hygroscopic behavior of the analogous
nonmetal compounds P,0 5 and S0 3 .
The polymerization of vanadate, molybdate, and tungstate ions forming isopoly
anions has received a great deal of attention. Early in the condensation process the
coordination number of the metals changes from 4 to 6 . and the basic building unit in
the polymerization process becomes an octahedron of six oxygen atoms surrounding
each metal atom. Unlike tetrahedra, which can only link by sharing an apex, the
resulting octahedra may link by sharing either an apex or edge (rarely a face) due to
the relaxation of electrostatic repulsions in the larger octahedra. As a result, the
structures tend to be small clusters of octahedra in the discrete polyanions, culminat¬
ing in infinite structures in the oxides. When the edge sharing takes place, the
structure may be stabilized (relative to electrostatic repulsions) if some distortion
occurs such that the metal ions move away from each other. As the polymerization
increases, it becomes more and more difficult to have all metal ions capable of moving
to assist in this reduction in electrostatic repulsion. Ultimately the sharing of edges
ceases since Ihe requisite distortion becomes impossible. It might be expected that the
smaller the metal ion, the less the repulsion and the larger the number of edge-sharing
octahedra per unit. This expectation is borne out in a general way. For example, the
metal radii (Table 4.4) are V 5 (68 pm), Mo A+ (73 pm). W 64 (74 pm). Nb 54 (78 pm) =
Ta 54 (78 pm) and the most common corresponding edge-shared polyanions are
(V|oO :[i ]''~, [Mo 7 0 ; J 6 -, LMo s 0 2 J 4 -. [W.OJ 2 -. [W 7 0 2 /“, [Nb h O,,/-, and
[Ta 6 0 | 9 ] 8 . 4S To form larger polyanions such as 1W I 2 0 42 ] 12- or [H 2 W l 2 O 40 j'’~, edge
sharing must give way to apex sharing.
The isopoly anions may be considered to be portions of a closest packed array of
oxide ions with the metal ions occupying the octahedral holes. The edge-sharing array
found in [V| 0 0 28 ] 6- consists of ten octahedra stacked as shown in Fig. 16.10a. This
seems to be the largest stacked-octahedral isopoly anion cluster compatible with
metal-metal repulsions, and the remaining edge-shared structures represent portions
of this unit . 46
However, explanations for growth limitation based on repulsion of metal ions may
be somewhat oversimplified. Elements other than vanadium, niobium, tantalum,
molybdenum, and tungsten do not form isopoly anions. Other ions which have
appropriate radii (e.g., Al 34 , 67 pm; Ga 34 , 76 pm; I 74 . 67 pm) for discrete isopoly
anion formation instead form chains, sheets, or three-dimensional frameworks. Why
does polymerization stop for isopoly anions? An oxygen atom in a terminal position in
an isopoly anion is strongly it bonded to a transition metal such as Mo(V|) or W(VI).
These terminal oxygen atoms are never found trans to one another because they avoid
(c)
Fig. 16.10 The structures of edge-shared isopoly anions showing their relation to the M m O:«
structure: (a) M = V; (b) M = Mo; (c) M = Mo; (d) M = Nb. Ta. [From Kepert. D. L.
Inorg. Client. 1969. 8. 1556. and Day. V. W.; Klemperer, W. G.; Maltbie, D. J. J. Am.
Client. Soc. 1987. 109 , 2991-3002. Reproduced with permission.!
45 The isolation of [Nb, n O M |' , “ (Graebcr. E. J.; Morrison. B. Acta Cryslallogr.. Sect. II 1977. J3.
2137-2143), which has a structure analogous to [VioO-kI'’”. weakened the validity of this
relationship.
46 Kepert. D. L. Inorg. Cliem. 1969. 8. 1556-1558.
Pope. M. T. Heteropoly and Isopoly Oxometalates: Springer-Verlag: New York. 1983.
660
15’Organometallic Chemislry
Metal Alkyls, Carbenes, Carbynes, and Carbides
661
In many of their reactions, these carbenes behave like the familiar Wittig reagent,
Ph,PCH 2 . Schrock carbenes are important intermediates in olefin metathesis. 68
Classification of carbene complexes as Fischer or Schrock perhaps focuses too
much on their differences and too little on their similarities. Both contain a metal-
carbon bond of order greater than one. Whether the carbene carbon tends to seek or
provide electrons will depend on the extent of it bonding involving the metal and the
carbon substituents. Some carbene complexes lie between the Fischer/Schrock ex¬
tremes, behaving in some reactions as nucleophiles and in others as electrophiles. 69
A decade after the announcement of the metal-carbon double bond, Fischer's
group reported the first complex containing a metal-carbon triple bond. 70
OMe
(OC) 5 W=C [(OC) 5 WEE=CMe] + BCl 4 “ + BCl 2 (OMe)
X Me
Cl(OC) 4 W=CMe (15.54)
The carbyne ligand may be viewed as a three-electron donor, similar to the nitrosyl
ligand, with a pair of electrons in an sp orbital and a single electron in a p orbital.
Donation of the sp electrons and pairing the p electron with one from the metal atom
gives a a bond and a n bond, respectively. The second tr bond results from donation of
an electron pair from the metal atom to the empty p orbital of the ligand.
(15.55)
Shortly after the preparation of the first carbyne, Schrock’s group provided a high
oxidation state complement. 71
Me,CCH,MBCI
W(OMe) 3 CI 3 --- (Bu—CH 2 ) 3 W=CC(Me) 3 (15.56)
It also proved possible to put alkyl, carbene, and carbyne ligands into the same
molecule. 72
Me 2 c
^P^J/CH 2 CMe 3
^ ^CHCMe 3
Me 2
M Schrock. R. R. Science 1983, 219, 13-18,
69 Gallop, M. A.; Roper, W. R. Adv. Organomet. Client. 1986, 25, 121-198.
7U Fischer, E. O.; Schubert, U. J. Organomet. Chem. 1975, 100. 59-81.
71 Clark. D. N.; Schrock. R. R. J. Am. Chem. Soc. 1978, 100. 6774-6776. Schrock, R. R. Acc. Chem.
Res. 1986, 19. 342-348.
72 Churchill, M. R.; Youngs, W. J. Inorg. Chem. 1979, 18, 2454-2458.
The tungsten-carbon single, double, and triple bond lengths in this compound are
225.8, 194.2, and 178.5 pm, respectively, and the accompanying W—C—C bond
angles are 124°, 150°. and 175°, all of which is quite consistent with tungsten-carbon
bond orders of 1, 2. and 3.
Systematically removing hydrogen atoms from a methane molecule would leave
us with a carbon atom in the final step.
CH 4 -CH 3 -CH, -CH -6 (15.57)
We have thus far seen complexes containing alkyl, carbene, and carbyne ligands, and
if you have speculated about the possibility of atomic carbon functioning as a ligand,
your thoughts have been well placed. Complexes in which carbon is bound only to
metal atoms are known as carbido complexes (Fig. 15.21). The first example was
reported in 1962, before carbene and carbyne complexes were discovered, but until
recently, carbido complexes remained chemical oddities synthesized by a variety of
serendipitous routes. 73 A carbon atom surrounded by metal atoms is not very reac¬
tive, but if it can be exposed by removal of one or more metal atoms, it becomes a
reactive species. Oxidation of [Fe 6 C(CO), h ] 2 ^ (Eq. 15.185) removes two iron atoms as
Fe- + and uncovers a positively charged carbon atom which can react with nu¬
cleophiles such as carbon monoxide (see Eq. 15.186). In effect, this sequence creates a
carbon-carbon bond, which is always of interest to the organic chemist, and further¬
more the added carbon can be easily functionalized. Thus carbido complexes show
potential in organic synthesis.
Fig. 15.21 Structural examples of carbido complexes: (a) Fe 5 (CO)| 5 C, (b) [Fe„(CO)|„C] 2 ",
and (c) (AuJPPhdftCJ 2 . (From Braye. E. H.; Dahl. L. F.; Hubei, W.; Wampler, D. L.
J. Am. Chem. Soc. 1962, 84, 4633-4639 (a): Bradley. J. S. AJv. Organomet. Chem. 1983. 22,
1-58 (b); Scherbaum. F.: Grohmann. A.; Huber. B.; Kruger, C.: Schmidbaur, H. Angetv.
Chem. hit. Ed. Engl. 1988, 27, 1544-1546 (c). Used with permission.]
73 Braye. E. H.; Dahl, L. F.; Hubei, W.; Wampler. D. L. J. Am. Chem. Soc. 1962. 84. 4633-4639.
Bradley, J. S. Adv. Organomet. Chem. 1983. 22. 1-58. Hriljac. J. A.; Holt. E. M.; Shriver, D. F.
Inorg. Chem. 1987.26. 2943-2949. Hayward. C-M. T.; Shapley, J. R.; Churchill, M. R.; Bueno. C.;
Rheingold, A. R. J. Am. Chem. Soc. 1982. 104, 7347-7349.
754
16 Inorganic Chains, Rings, Cages, and Clusters
N - S N - S N - S Fig. 16.7 Polymerization of
| | i | I j I S : N : to form (SN). r chains
l_I with minimal movement of
S - N S - N S -N atoms.
N - S N - S N - S
N - S N - S N - S
N -S N- S N - S
JS N --- S N —^ S N.
illustrates many features: a conjugated single-bond-double-bond resonance system
with nine electrons on each sulfur atom rather than a Lewis octet; every S—N unit
will thus have one antibonding u electron. The half-filled, overlapping tr orbitals will
combine to form a half-filled conduction band in much the same way as we have seen
half-filled 2 j orbitals on a mole of lithium atoms form a conduction band (see Chapter
7). Note, however, that this conduction band will lie only along the direction of the
(SN) t fibers; the polymer is thus a "one-dimensional metal."
Similar to (SN) r in their one-dimensional conductivity properties are the stacked
columnar complexes typified by [Pt(CN) 4 | : “. These square planar ions adopt a closely
spaced parallel arrangement, allowing for considerable interaction among the cl .2
orbitals of the platinum atoms. These orbitals arc normally filled with electrons, so in
order to get a conduction band some oxidation (removal of electrons) must take place.
This may be readily accomplished by adding a little elemental chlorine or bromine to
the pure tetracyanoplatinate salt to get stoichiometries such as K ; [Pt(CN) 4 ]Br 0 3 in
which the platinum has an average oxidation state of +2.3. The oxidation may also be
accomplished electrolytically. as in the preparation of Rb 2 [Pt(CN) 4 )(FHF) n4 (Fig.
16.8), which has a short Pt—Pt separation. The Pt—Pt distance is only 280 pm.
almost as short as that found in platinum metal itself (277 pm) and in oxidized platinum
"pop” complexes (270 to 278 pm; see Chapter 15). 4 - Gold-bronze materials of this
type were discovered as early as 1842, though they have been little understood until
recent times. The complexes behave not only as one-dimensional conductors, but
•« Roundhill. D. M.;Gray. H. B.:Chc. C-M. Aec. Client. Res. 1989. 22. 55-61. Schultz. A. J.: Coffey.
C. C.: Lee. G. C.; Williams. J. M. Inorg. Client. 1977. 16. 2129-2131.
Chains
755
Fig. 16.8 Perspective view of the tinii cell of Rb : [Pt(CN) 4 |(FHF)o. 4l |. One-dimensional chains
of staggered lPt(CN) 4 ] :_ ions occupy the corners and center of the unit cell. The triad of
small circles represents the partially occupied positions of the FHF' ions. Note the very
short Pt—Pt distance (279.8 pm). [From Schultz. A. J.: Coffey. C. C.: Lee. G. C.; Williams.
J. M. Inorg. Client. 1977. ft. 2129. Reproduced with permission.!
their electrical properties are considerably more complex. Detailed discussion of ihe
many interesting aspects of these materials is beyond the scope of this hook, but
fortunately there are thorough reviews of the subject. 4 ''
Transition metals in their higher oxidation states are formally similar to nonmetals
with corresponding group numbers: V (VB)and P(VA)in V0 4 '~ and P0 4 ", Cr (VIB)
and S (VIA) in CrO^' and SOj - . Mn (VIIB) and Cl (VIIA) in Mn0 4 " and CIOJ. The
analogy may be extended to polyanions, such as dichromate, Cr-0:~; however, the
differences in behavior between the metal and nonmetal anions are often more impor¬
tant than their similarities. Whereas polyphosphoric acids and polysulfuric acids form
only under rather stringent dehydrating conditions, polymerization of some metal
anions occurs spontaneously upon acidification. For example, the chromate ion is
stable only at high pHs. As the pH is lowered, protonation and dimerization occur:
CrOj- + W - IllOiCrO," (16.28)
(HO)CrO, + H" - (HOjX'rO, (16.29)
2(HO) ; CrO, -► Cr, 05 “ + H,0 + 2H‘ (16.30)
Treatment with concentrated sulfuric acid completes the dehydration process and red
chromium(Vl) oxide ("chromic acid") precipitates:
^Cr : 0=~ + «H + - (CrO,)„ + ^1-1,0 (16.31)
The structure of Cr0 3 consists of infinite linear chains of Cr0 4 lelrahedra.
4 ' Extended Linear Chains. Miller. J. S.: lid.: Plenum: New York. 19X2. Vols. I-II; 1983, Vol. III.
Ibers. J. A.: Pace. L. J.: Martinscn. J.. Hoffman. B. M. Slrnel. Bantling I Berlin ) 1982. 50, 3-55.
Gliemann. G.: Yersin. H. Slrnel. Bantling (Berlin I 1985. 62. 87-153.
662
1 5 • Organometallic Chemistry
Nonaromatic
Alkene and Alkyne
Complexes
Alkene Complexes
Complexes between metal salts and alkenes have been known since 1827 but they
were not understood until the latter half of this century. For example, Zeise isolated
stable yellow crystals after refluxing an alcoholic solution of platinum tetrachloride.
Zeise's salt is now made from K 2 PtCl 4 and C 2 H 4 :
[PtCl 4 ] 2 “ + CH, = CH 2 — [Pt(C 2 H 4 )CI 3 ] + Cl + H 2 0 (15.58)
Silver ions form similar alkene complexes which are soluble in aqueous solution and
may be used to effect the separation of unsaturated hydrocarbons from alkanes.
Catalysts for the polymerization of alkenes also form metal-alkene complexes which
lead to polymerized product.
A structural investigation of the anion in Zeise's salt has shown that the ethylene
occupies the fourth coordination site of the square planar complex wth the C C axis
perpendicular to the platinum-ligand plane (Fig. 15.22). 75 Relative to free ethylene,
the C—C bond is lengthened slightly (from 133.7 pm to 137.5 pm), and the hydrogens
are slightly tilted back from a planar arrangement.
The bond between the ethylene molecule and the metal ion may be considered as
a dative
C,H 4 ) significantly influences the metal-carbon bonds in these compounds. As with
carbon monoxide, usually only metal atoms in low oxidation states are sufficiently
good it donors to stabilize alkene complexes.
In extreme cases, such as Pt(Ph 3 P) 2 (CH,=CH 2 ) in which the metal is very
electron rich, tt back donation is significant anil as a result, the carbon-carbon bond
lengthens considerably (to 143 pm) and the hydrogen atoms are bent considerably out
of the plane:
Fig. 15.22 The structure of
the anion of Zeise's salt.
trichloro(ethylene)platinatc-
(II) ion.
In this molecule ethylene lies in the same plane as the other ligands, unlike the case in
Zeise's salt where planarity is sterically prevented. By analogy to a three-membered
ring of carbon atoms, this compound may be viewed as a mctallacyclopropane.
74 For a translation of Zeise's original paper, sec Classics in Coordination Chemistry: Purl 2;
Kauffman. G.. Ed.; Dover: New York. 1976; pp 21-37.
Love. R. A.: Koctzle. T. F.; Williams. G. J. B.; Andrews. L. C.: Bau. R. hwrg. Chem. 1975. 14.
2653-2657.
76 Guggenbcrger. L. J.; Cramer. R. J. Am. Chem. Sue. 1972. 94. 3779-3786. Evans. J. A.; Russell.
D. R. Chem. Comrnun. 1971,197-198.
Nonaromatic Alkene and Alkyne Complexes
663
Fig. 15.23 Representation
of 77 coordination of an
ethylene ligand and a
transition metal.
suggesting that we can consider the two bonding extremes, 77 donation and metallacy-
cle. as resonance structures:
(a)
(b)
A similar and even more extreme case of bond lengthening is found in complexes of
C : (CN) 4 . in which the 77 accepting ability of the alkene is enhanced by iis electro¬
negative substituents. Electron-rich metals and strongly electron-withdrawing alkene
substituents favor structure (b) while opposite conditions favor structure (a). Struc¬
ture (b) suggests that the metal has been formally oxidized with the loss of two
electrons. For further insight into the mode by which the C 2 (CN) 4 ligand binds to
metals, it is interesting to compare the structure of letracyanoethylene oxide with that
of a letracyanoethylene nickel complex.77
0
/\
(NC),C — C(CN) ;
K >
Ni
/\
(NC),C — C(CN),
(i)
( 2 )
Both the C—C bond lengths [(I) = 149.7 pm: (2) = 147.6 pmj and the bending of the
substituents out of the plane [(I) = 32.2°: (2) = 38.4°] are nearly the same. Although
we can draw an alternative resonance form for the nickel complex, the bonding model
shown is the only one applicable to the oxide. In view of the strong structural
similarities, we can feel justified in using the cyclic structure as an approximation for
certain complexes as well.
When a ligand is bound to a metal, its chemistry typically changes. For alkenes
the change is particularly dramatic. Free alkenes are susceptible to electrophilic attack
77 Matthews. D. A.: Swanson. J.: Mueller. M. H.; Stucky. G. D. J. Am. Chem. Sue. 1971, 93.
5945-5953.
O u
752
16- Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.5 Filling of
available hexagonal sites in
each layer of graphite: For
the limit of C«K. K = • +
O; for C|;„K. K = O only.
(From Whittingham, M. S.:
Dines. M. B. Surv. Prog.
Client. 1980. 9. 55.
Reproduced with
permission.]
The mention of Ihe K ' ion presupposes knowledge of the nature of the potassium
species present. Because of the similarity in energies of the valence and conduction
bands, graphite can be either an electron donor or acceptor. Intercalation of potassium
atoms into graphite results in the formation of K + ions and free electrons in the
conduction band. Graphite will react with an electron acceptor such as bromine to
form C„Br in which electrons have transferred from the valence band of the graphite to
the bromine. Apparently, simple bromide ions are not formed, but polybromide chains
form instead (see Chapter 17 for similar polyhalide chains). The expected Br—Br
distance in such a polyhalide 17 (254 pm) compares well with the distance between
centers of adjacent hexagons in graphite (256 pm). In contrast, the expected Cl—Cl
(224 pm) and I —I (292 pm) distances do not fit well with the graphite structure, and in
fact these two halogens do not intercalate into graphite! In contrast, iodine mono¬
chloride (expected average 1—Cl distance = 255 pm) does.
Both of the potassium and polybromide intercalation compounds are good con¬
ductors of electricity. In the potassium intcrcalant. the electrons in the conduction
band can carry the current directly, as in a metal. In the compounds of graphite with
polybromide, holes in the valence band conduct by the mechanism discussed previ¬
ously for semiconductors (Chapter 7).
Recently, it has proved possible to intercalate a variety of organic molecules into
transition metal dichalcogcnides. 1,, Unlike the above examples these do not usually
involve electron transfer. When single molecular layers of MoS,. suspended in water,
arc shaken with water-immiscible organic molecules such as ferrocene, the latter is
adsorbed onto the former. A highly oriented, conducting ferrocene-MoS, film results
when exposed to a glass substrate. The interlayer spacing of MoS; increases by 560
pm upon ferrocene inclusion. It has been suggested that the best chance for producing
a useful high-temperature superconductor may lie with incorporating an organic
superconductor into a layered inorganic compound.
ne-Dimensional There is an unusual hetero chain, (SN) t , discovered in 1910, which did not receive
inductors detailed attention until the 1970s. Interest centers on the fact that although it is
composed of atoms of two nonmetals, polymeric sulfur nitride (also called polythiazyl)
has some physical properties of a metal. The preparation is from tetrasulfur
tetranitride (see page 776):
S 4 N 4 S 2 Nj -► (SN) X (16.27)
17 See Wells. Footnote 33.
1,1 Divigalpitiya. W. M. R.; Frindt. R. F.; Morrison. S. R. Science 1989. 246. 369-371.
Chains
753
«
The S 4 N 4 is pumped in a vacuum line over silver wool at 220 °C. where it polymerizes
slowly to a lustrous golden material."’The resulting product is analytically pure, as is
necessary for it to show metallic properties to a significant degree; it has a conduc¬
tivity near that of mercury at room temperature, and it becomes a superconductor at
low temperatures (below 0.26 K).
X-ray diffraction studies show that the SN chains have the structure shown in Fig.
16.6. This chain can be generated from adjacent square planar S-.N-, molecules. The
S—N bonds in this starting material have a bond order of 1.5 and a bond length or
165.4 pm. intermediate between single (174 pm) and double (154 pm) sulfur-nitrogen
bonds. A free radical mechanism has been suggested- 1 " leading to the linear chains of
the polymer (Fig. 16.7). Since polymerization can take place with almost no move¬
ment ol the atoms, the starting material and product are pseudomorphs and the
crystallinity of the former is maintained. 4 '
If one attempts to draw a unique Lewis structure for the (SN) V chain, one is
immediately frustrated by the odd number of electrons available. Many resonance
structures can be drawn and they contribute to the hybrid structure, but the single
structure:
:S=N: :S = N:
I I I I I
:S=N: :S=N:
Fig. 16.6 (SN), chains in one layer of polymeric sulfur nitride. (From MucDiarmid. A. G.:
Mikulski. C. M.; Saran. M. S.; Russo. P. J.; Cohen. M. J.: Bright. A. A.; Garito. A. F.;
Heoger. A. J. In Compounds with Unusual Properties'. King. R. B., Ed.: Advances in
Chemistry Series 150. American Chemical Society: Washington. DC, 1976. Reproduced with
permission.]
• w Labes. M. M.; Lowe. P.; Nichols. L. F. Chem. Rev. 1979. 79. 1-15.
40 Mikulski. C. M.: Russo. P. J.; Saran. M. S.: MacDiarmid. A. G.; Garito. A. F.: Hceger. A. J, J.
Am. Chem. Soc. 1975. 97. 6358-6363.
41 This is in contrast to the more commonly observed result of solid state reactions: beautiful crystals
turning into an amorphous powder.
664
1 5 • Organomelallic Chemistry
Alkyne Complexes
but not to nucleophilic attack. When coordinated to a metal, the carbon atoms become
somewhat more positive and a reversal of reactivity occurs, i.e., the alkene becomes
susceptible to nucleophilic attack and loses its susceptibility to electrophilic attack. 78
The chemistry of alkyne complexes is somewhat more complicated than that of alkene
complexes because of the greater possibilities for ir bonding by alkynes and the
tendency of some of the complexes to act as intermediates in the formation of other
organomelallic compounds.
The simplest alkyne complexes, the metal acetylenes, resemble those of ethylene.
For example, there are analogues of Zeise's salt in which an acetylene molecule is
bound to platinum(II) and occupies a position like that of ethylene in Zeise’s salt. In
addition, there are L 2 Pt(RC=CR) complexes that have structures paralleling that of
L 2 Pt(H 2 C=CH 2 ) (Fig. 15.24). For both of these Pt(0) complexes, an approximate
square planar arrangement around the metal is found. Alkynes are more electro¬
negative than alkenes and are therefore better -n acceptors. Thus it is appropriate to
view them as metallacyclopropenes: 79
R R
\ /
C = C
\ /
M
Alkynes have two t r and two it orbitals that can potentially interact with metal
orbitals, and in some instances, it is thought that all of these are involved at the same
time in a mononuclear complex. An extended Huckel calculation on Mo(/neso-
tetra-p-tolylporphyrin)(HC=CH) supports this view (Fig. 15.25). 811 Thus both bond¬
ing orbitals of the alkyne (6, and a,) can donate electron density to molybdenum to
form the lb, and la, MOs, and both antibonding orbitals ( b 2 and a 2 ) can accept
electron density to form the I b 2 and la 2 MOs. Notice that both tt bonding orbitals (a,
and b ,) of acetylene interact significantly with metal d orbitals of the same symmetry.
Fig. 15.24 Molecular
structure of bis-
(triphenylphosphine)-
diphenylacetylcneplati-
num(O); bond lengths are in
picometers. [From
Glanville, J. 0.; Stewart,
J. M.; Grim. S. O. J.
Organomet. Chem. 1967, 7,
P9-PI0. Reproduced with
permission.]
7K For a further discussion of this reversal and others (referred to as "umpolung"), sec Crabtree, R. H.
The Organomelallic Chemistry of the Transition Metals-. Wiley: New York, 1988: p 91.
79 Cotton, F. A.; Shang. M. Inorg. Chem. 1990. 29. 508-514.
80 Tatsumi. K.; Hoffmann. R.; Templeton, J. L. Inorg. Chem. 1982. 21. 466-468.
W" '
Nonaromalic Alkene and Alkyne Complexes
665
Fig. 15.25 Diagram
showing interaction of the
frontier orbitals of
Mo(meso-letra-p-toiylpor-
phyrin) with those of
acetylene. Strong inter¬
actions are represented
with solid lines and weak
ones with dashed lines.
[From Tatsumi, K.;
Hoffmann, R.; Templeton,
J. L. Inorg. Chem. 1982.
21, 466-468. Reproduced
with permission.]
Both pairs of tt electrons in an alkyne ligand are more likely to be involved in the
bonding if it is coordinated to two metal atoms. If acetylene is allowed to react with
dicobaltoctacarbonyl, two moles of carbon monoxide are eliminated:
Co 2 (CO) 8 + HC=CH -► (H 2 C 2 )Co 2 (CO) 6 + 2CO ( 15 . 59 )
The production of two moles of carbon monoxide and the 18-electron rule lead us to
predict that the acetylene molecule is acting as a four-electron donor. In fact this is
just one of many complexes in which alkynes bind in this fashion. 8 ' For example, the
structure of the diphenylacetylene complex in Fig. 15.26 shows that the positions of
the two rhodium atoms are such as to allow overlap with both n orbitals in the
carbon-carbon triple bond. 82 The extent of back donation into the antibonding orbitals
determines the lengthening of the C—C bond and the extent to which the C—H bonds
are bent away from the complex. Bond length values vary greatly from system to
81 Raithby. P. R.; Rosales. M. J. A civ. Inorg. Radiochem. 1985, 29. 169-247. Reger. D. L.; Huff,
M. F.; Wolfe, T. A.; Adams, R. D. Organometallics 1989, 8, 848-850.
82 Berry. D. H.; Eisenberg, R. Organometallics 1987. 6. 1796-1805.
750
16- Inorganic Chains, Rings, Cages, and Clusters
Intercalation
Chemistry 36
I
The analogous silicon compounds are also unstable, but the "dimethylsilicone" that
forms is a mixture of linear polymers (and cyclic products to be discussed in the next
section):
Me Me Me
I I I
— O— Si— O—Si—O— Si—O—
I I I
Me Me Me
(16.23)
Hundreds of thousands of tons of pure Si are produced every year by the reduc¬
tion of Si0 2 in an electric furnace:
Si0 2 + C -► Si + C0 2 (16.24)
Although some of this is used for the production of ultra-pure silicon for semiconduc¬
tors and for alloys with iron, aluminum, and magnesium. 98% goes for the production
of methyl silicon chlorides:
3CH,C1 + Si - CHjSiClj + C 2 H fi (16.25)
2CH 3 CI + Si -► (CH 3 ) 2 SiCI 2 (16.26)
These products are separated by distillation and used to make over 500 million kg per
year of silicone rubbers, oils, and resins. All of these materials repel water and are
electrical insulators. The rubbers are flexible and the oils are liquids over a wide range
of temperatures.
Intercalation compounds consist of layers (“sandwiches") of different chemical spe¬
cies. The name comes from that describing the insertion of extra days (such as
February 29th) into the calendar to make it match the solar year. Most work on
intercalation compounds has been on synthetic systems in which atoms, ions, or
molecules have been inserted between layers of the host material. However, some
aluminosilicates that we have encountered above provide useful examples. Thus talc
and micas form layered structures with ions between (he silicate sheets (Fig. 16.3).
Some minerals, including all clays, have water molecules intercalcated between the
framework sheets. In some, such as vermiculite, the water may rapidly and dramat¬
ically be evacuated by heating. The water molecules leave faster than they can diffuse
along the layers—exfoliation occurs. The result is the familiar expanded vermiculite
used as a packing material and as a potting soil conditioner.
Another example of this type of intercalation compound is sodium beta alumina
where the sodium ions are free to move between the spinel layers. The sodium ions
can be replaced by almost any + I cation such as: Li. K. Rb. Cs + . NH^ . H,0.
Tl + , Ga + , NO, etc. The conductivity of these materials varies with the size of the
ions moving between the fixed-distance (A!—0—Al) layers.
Graphite is perhaps the simplest layered structure. The intralayer C—C distance
(142 pm) is twice the covalent radius of aromatic carbon (cf. 139 pm in benzene) and
the interlayer C—C distance is 335 pm, twice the van der Waals radius of carbon. The
sheets are held together by weak van dcr Waals forces. Many substances can be
• ,h Whittingham, M. S.; Dines. M. B. Surv. Proa. Cliem. 1980. 9. 55-87. Bcnsenhard. J, O.: Fritz.
H. P. Angeiv. Client. Int. Ed. Engl. 1983. 22. 950-975.
Chains
751
Fig. 16.3 Layered silicate structures: (a) talc. Mg 3 (0H) 2 (Si.,0,»); (b) muscovite (a mica)
KAI 2 (OH) : (Si,AIOm). [Note: (I) Electroneutrality is maintained by balance of Ktl), Mg(ll).
AI(III). and Si(IV). (2) The repealing layers in muscovite are bound together by the K
cations.| [From Adams. D. M. Inorganic Solids: Wiley: New York. 1974. Reproduced with
permission.!
intercalated between the layers of graphite, but one of the longest known and best
studied is potassium, which can be intercalated until a limiting formula of C 8 K is
reached. This is known as ihe first-stage compound. The earlier, lower stages have the
general formula ol C, 2 „K. The stages form stepwise as new layers of potassium are
added, giving well-characterized compounds with n = 4. 3. 2. (Fig. 16.4). The final
step (yielding stage I) includes filling in all of the remaining available sites (Fig. 16.5) in
addition to forming the maximum number of layers. Presumably, further intercalation
cannot take place because of electrostatic repulsion.
Upon intercalation, the graphite layers move apart somewhat (205 pm), though
less than expected as estimated from the diameter of the potassium ion (304 pm or
greater). I his indicates that the K ‘ ion “nests" within the hexagonal carbon net. and
one can even speculate about weak complexing to the carbon rr-electron cloud.
Fig. 16.4 Staging in
graphite intercalation
compounds, C| 2 „K.
Addition of potassium
proceeds through n = 4. 3.
2, ... to the limit in stage
I: C i: „K — C„K. [From
Whittingham. M. S.; Dines.
M. 8. Surv. Prog. Chem.
1980. 9. 55. Reproduced
with permission.)
666
1 5 - Organomelallic Chemistry
Fig. 15.26 Molecular
structure of
Rh 2 (CO) 2 (PPh 2 CH 2 PPh 2 ) 2
(PhCssCPh). All but one
carbon atom of each phenyl
group on the phosphine
ligands have been omitted
for clarity. [From Berry.
D. H.; Eisenberg, R.
Organometallics 1987, 6,
1796-1805. Reproduced
with permission.]
Allyl and
Pentadienyl
Complexes
system, 83 but for the rhodium complex of Fig. 15.26, the C—C bond length of 1.329 A
indicates it is best described as a double bond.
Few ligands are as common and important to organometallic chemistry as the allyl
(CjH,) group. A recent organometallics text devotes an entire chapter to the use of its
complexes in organic synthesis. 84 It can function as a one-electron donor (monohapto)
or as a three-electron donor (trihapto).
H
J Z \
H,C CH,—M
1, --► H,C I CH, s CH,
The trihapto arrangement, which is by far the most common, usually has equal C—C
bond lengths in accord with delocalization over the three -ir orbitals. The three MOs of
the allyl radical (Fig. 15.27) can overlap with wand v orbitals of the metal, but the rr
interaction of the ligand with the metal d xt orbital is not very significant.
A variety of preparations for these complexes have been developed, one of which
is the reaction of allyl bromide with an organometallic anion:
Na[Mn(CO) s ] + CH 2 =CHCH 2 Br-♦ (CO) s Mn(»; 1 -C 3 H 5 ) + NaBr (15.60)
(OC) 5 Mn(t 7 ‘-CH 2 CH=CH 2 ) -♦ (OC) 4 Mn(r/ 3 -C 3 H 5 ) + CO (15.61)
M Gcrvasio, G.; Rossetti, R.; Stanghcilini. P. L. Organometallics 1985. 4. 1612-1619.
KJ Coltman. J. P.; Hegcdus, L. S.l Norton. J. R.; Finke, R. G. Principles and Applications of
Organotransition Metal Chemistry, 2nd ed.; University Science Books: Mill Valley. CA. 1987;
Chapter 19.
Nonaromatic Alkene and Alkyne Complexes
667
Fig. 15.27 Diagram showing interactions between metal d orbitals and the n orbitals of an
allyl ligand.
syn
H
H
Fig. 15.28 Molecular
structure of (tj’-allyl^Ni as
established by neutron
diffraction. [From Goddard,
R.; Kruger, C.; Mark, F.:
Stansfield. R.; Zhang, X.
Organometallics 1985. 4.
285-290. Reproduced with
permission.]
The first reaction lakes place at ambient temperature and the second occurs when the
system is heated. Initial formation of the monohapto complex is believed to be typical,
but often this intermediate is not observed.
A second synthetic approach is to use a Grignard reagent and a metal halide:
NiCl, + 2CH,=CHCH,MgCl -♦ Ni(f/ 3 -C 3 H 5 ) 2 + 2MgCl, (15.02)
The bis(7/'-allyl)nickel complex is a yellow, pyrophoric compound that melts at I °C.
Its structure, which has been determined by neutron diffraction, illustrates some of the
important features of these complexes (Fig. I5.2S). 8 ' The two terminal carbon atoms
are further from the metal atom (202.9 pm) than is the central carbon atom (198.0 pm).
Even so. the terminal carbon atoms are tilled toward the metal to provide better rr
overlap. The C—C—C bond angle is 120.5° and the mean C—C bond length is 141.6
pm. all of which is in agreement with a conjugated tr system. The anti (trans to meso)
hydrogen atoms are bent away from the metal and the syn (cis to meso) and meso
hydrogen atoms are bent toward the metal.
In addition to the stereochemistry possible for allyl complexes as a result of
different syn and anti substituents, geometrical isomers are known which arise be¬
cause of the position of the central allyl carbon atom relative to other ligands. Both
e.to- and c/M/«-Rulrr , -C J H J )(i 7 3 -C 4 H 7 )(CO) have been isolated (Fig. 15.29). 8 '
ss Goddard. R.; Kruger. C.: Mark. F.: Stansfield. R.; Zhang. X. Organometallics 1985, 4, 285-290.
Hsu. L-Y.; Nurd man. C. E.: Gibson. D. H.; Hsu. W-L. Organometallics 1989. 8, 241-244.
748
16 • Inorganic Chains, Rings, Cages, and Clusters
Quite remarkably the reaction occurs without the formation ol the more ther¬
modynamically favored trimethylamine. Although some dimethylamine is produced in
the reaction, the channel size in which the reaction takes place favors the formation of
methylamine. The process, as of this writing, is about to be commercialized by Du
Pont. . .
We have seen previously shape-selective catalysis by ZSM-5 in the conversion ot
methanol to gasoline (Chapter 15). 27 Other commercial processes include the forma¬
tion of ethylbenzene from benzene and ethylene and the synthesis ofp-xylene. The
efficient performance of ZSM-5 catalyst has been attributed to its high acidity and to
the peculiar shape, arrangement, and dimensions of the channels. Most of the active
sites are within the channel so a branched chain molecule may not be able to diffuse in.
and therefore does not react, while a linear one may do so. Ot course, once a reactant
is in the channel a cavity large enough to house the activated complex must exist or
product cannot form. Finally, the product must be able to diffuse out. and in some
instances product size and shape exclude this possibility. For example, in the methyla-
tion of toluene to form xylene:
The "linear" p-xylene can escape from the catalyst much more easily than the bent
in- or o-xylene (see Figs. 1.4 and l.5).=» The o- and m-xylenes are trapped but not
wasted. Under the acidic conditions of the catalyst they continue to rearrange, and
whenever a p-xylene molecule is formed, it can pop out and leave the system.
Conversion is thus essentially complete. Catalytic zeolites have been compared to
enzymes because shape and size are crucial for the catalytic action ot both.- 7
Zeolites also provide convenient framework sites for activating transition metal
ions for redox catalysis. Iwamoto 30 has described a Cu(Il)/Cu(I) exchanged zeolite
that holds promise for the high-temperature conversion of NO, (in diesel and auto
exhaust) to N ; and 0 2 :
Reducing capacity is enhanced by hydrocarbons (unburnt tuel) which provide a source
of hydrogen.
Transition metal ions in zeolites behave much as expected for ions in a weak oxide
field, but often the metal ions are found in trigonal sites, so their spectra and magnetic
properties are somewhat different from those of the more common octahedral and
tetrahedral fields. 31
:7 Hdldcrich. W.; Hesse, M.; Naumann. F. Angew. Chem. Ini. Ed. Engl. I98R. 27. 226-246.
:k Csicscry, S. M. Chem. Hr. 1985. 21. 473-477. Nagy. J. B.: Dcrouanc. E. G.: Rcsing. H. A.; Miller.
G. R. J. I’hys. Chem. 1983. 87. 833-837.
2 v Thomas. J. M. Angew. Chem. Ini. Ed. Engl. 1988. 27. 1673-1691.
'■> Iwamoto. M.; Yahiro. H.: Tanda. K.: Mizuno. N.: Mine. Y.; Kagawa. S. J. Pliy.i. Chem. 1991. 95.
3727-3730. Sato. S.: Yu-u. Y.: Yahiro. H.: Mizuno. N.; Iwamoto. M. Appl. Culal. 1991. 70. LI-L5.
71 Klier. K. Langmuir 1988.4. 13-25.
I •' -
Another class of framework aluminosilicates is the ultramarines. They are charac¬
terized by an open framework and intense colors. They differ from the previous
examples by having "free" anions and no water in the cavities. Ultramarine blue,
which is the synthetic equivalent of the mineral lapis lazuli, contains radical anions.
S7 and ST. The dominant Sf gives rise to its blue color. Ultramarine green also
contains these two anions but in comparable amounts. Although these two anions are
also found in ultramarine violet and pink, the characteristic color is due to a third
species, perhaps S 4 or S 7. 32 Structurally related, but colorless, minerals such as
sodalite (containing chloride anions) and noselite (containing sulfate anions) are some¬
times included in the broad category of ultramarines.
The study of silicaceous minerals is important, not only with respect to better
understanding of the conditions of formation and their relation lo the geochemistry of
these minerals, but also with respect to structural principles and the synthesis of new
structures not found in nature (synthetic zeolites and ultramarines ). 33
In all of the silicates discussed above, the sharing of oxygen atoms between
tetrahedra is by an apex only:
°\ r
Si—O—Si
°o' So
No cases are known in which edges or faces are shared:
°\ /°\ /°
Si Si
0^
O— Si Si—O
V>rV
0 o
Pauling has listed a set of rules for predicting stability in complex crystals based on an
ionic model. 34 Although no one now accepts a purely electrostatic model for silicates
and similar compounds, Pauling's rules are still reasonably accurate as long as the
partial charges on the atoms are sufficiently large to make electrostatic repulsions
significant. Such repulsions militate against the sharing of edges or faces by tetrahedra
since this places positive centers too near each other.
No section on helerocatenation would be complete without a discussion of sil¬
icones. (R-jSi'O),,. 33 The term silicone was coined by analogy to ketone under the
mistaken belief that monomeric R : Si=0 compounds could be isolated. Silicon com¬
pounds that are formally analogous to carbon compounds are found to have quite
different structures. Thus carbon dioxide is a gaseous monomer, but silicon dioxide is
an infinite singly bonded polymer. In a similar manner, gem-diols are unstable relative
to ketones:
Me,C(OH), -► MeC(0)Me + H,0 (16.22)
32 Clark. R. J. H.; Dines. T. J.; Kurmoo, M. Inorg. Chem. 1983, 22. 2766-2772.
33 For a more thorough discussion of silicate minerals, sec Wells. A, F. Si natural Inorganic Chem¬
istry. 5th ed.; Oxford: London. 1984: Chapter 23.
34 For a complete discussion see Pauling, L. The Nature of the Chemical Hand. 3rd cd.: Cornell:
Ithaca. I960; pp 544-562.
35 Rochow. E. G. Silicon and Silicones: Springer-Verlag: New York. 1987.
668
Metallocenes
669
15 Organometallic Chemistry
(a) (b)
Fig. 15.29 Structures of (a) endo- and (b) exo-Rufr^-CjHvMT^-CjHsMCO). [From Hsu. L-Y.;
Nordman, C. E.; Gibson. D. H.; Hsu, W-L. Orgtinometallics 1989, 8, 241-244. Reproduced
with permission.!
The allyl group can be extended to create a pentadienyl group, a tlve-carbon
system with two double bonds, CH 2 =CHCH=CH—CH-,. Its many modes of bond¬
ing (tj 5 , tj 3 , V) have led to a rich chemistry, including the synthesis of metallabenzene
complexes which are thought to be aromatic (Fig. I5.30). 87
V. „ f El .t
c —r~C
A ' N \ I
C ' , Ir
N c—c/
PE«,
'PHt,
/
Fig. 15.30 Square pyramidal structure of tris(triethylphosphine)-(2,5-dimethylpentadienyI)-
iridium. The six-membered ring is nearly planar and the carbon-carbon distances
(137-140 pm) are consistent with extensive delocalization. The complex is an example of a
metallabenzene. [From Bleeke, J. R.; Xie. Y-F.: Peng. W-J.; Chiang, M. J. Am. Client. Soc.
1989, III. 4118-4120. Reproduced with permission.!
Organometallic chemistry leaped forward in the early 1950s when the structure of
ferrocene, Fe(Tj 5 -C 5 H 5 ) 2 , was elucidated. 88 Prior to that, ideas regarding metal-
ligand interactions included only the coordinate covalent bond (e.g., M—CO) and the
covalent bond (e.g.. M—CH 3 ). It was revolutionary in bonding theory to propose a
metal-ligand bond between a metal and the v orbitals of C 5 H V Ferrocene was the first
of many complexes which came to be known as metallocenes, a name which arose
because they participated in reactions similar to those of aromatic molecules. For
obvious reasons complexes in which a metal atom was found between two parallel
carbocyclic rings became known as “sandwich" compounds. Some of these are
shown in Fig. 15.31. All of the complexes in Fig. 15.31, except the last two, obey
the 18-electron rule. Depending on the electron counting method adopted, the cyclo-
pentadienyl ligand may be viewed as either a five-electron donor (neutral atom) or a
six-electron donor (oxidation state) (Table 15.1).
The 18-electron rule is not obeyed as consistently by these types of organome¬
tallic compounds as by the carbonyl and nitrosyl complexes and their derivatives.
For example, in addition to ferrocene. M(ij 5 -CjH 5 ) 2 compounds are known for most of
the other elements of the first transition series (M = V, Cr. Mn. Co, Ni) and these can¬
not obey the 18-electron rule. However, only ferrocene shows exceptional thermal
stability (stable to 500 °C) and is not oxidized by air. Furthermore, cobaltocene,
a 19-electron species, is readily oxidized to the 18-electron cobaltocenium ion,
[Co(tj 5 -C 5 H 5 ) 2 ] + , which reflects much of the thermal stability of ferrocene. Mixed
cyclopentadienyl carbonyl complexes are common: [(tj 5 -C s H 5 )V(CO),,|, [(tj 5 -C 5 H 5 )-
Cr(CO) 3 ] 2 , [(7, 5 -C,H,)Mn(CO) 3 ]. [(T ? 5 -C 5 H 5 )Fe(CO) 2 ) 2 . [(t, 5 -C 5 H 5 )Co(CO) 2 ],
and [Tj 5 -C 5 H 5 )Ni(CO)] 2 . Of interest is the fact that among these compounds, the odd-
atomic-number elements (V, Mn, and Co) form monomers and the even-atomic-
number elements (Cr. Fe, and Ni) form dimers, which is in direct contrast to the
behavior shown by the simple carbonyl complexes. Cyclopentadienyl derivatives are
now known for every main group and transition metal of the periodic table and for
most of the /-block metals. 89
Fig. 15.31 Some examples of known metallocenes containing four-, five-, six-, seven-, and
eight-membered rings.
88 Wilkinson, G.; Roscnblum. M.: Whiting, M. C.; Woodward, R. B. J. Am. Client. Soc. 1952, 74,
2125-2126. E. O. Fischer and G. Wilkinson received the Nobel prize in 1973 for work done
independently on metallocenes. For a personal account of Wilkinson's early work, sec
J. Organomet. Client. 1975, 100, 273-278.
89 Marks, T. J. Prog. Inorg. Chem. 1979, 25. 223-333. Poli. R. Clicin. Rev. 1991, 9/, 509-551. Bursten,
B. E.; Strittmatter, R. J. Angew. Clwm. Ini. Ed. Engl. 1991, 30, 1069-1085.
87 Ernst, R. D. Struct. Bonding IBerlin) 1984. 57. 1-53. Bleeke, J. R.; Xie, Y-F.: Peng. W-J.; Chiang,
M. J. Am. Chem. Svc. 1989. Ill, 4118-4120. Bleeke. J. R. Acc. Chem. Res. 1991. 24, 271-277.
16- Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.2 Stereoviews of: (a) socialite. NaAUSi«0 2 4-2H 2 0 [ring sizes are 4 and 6 (220 pm)];
tb) ZSM-5, Na,AljSi., } 0|. )2 'l6H 2 0 [ring sizes are 4. 5, 6. 7. 8. 10 (580 pm)|; (c) mordemte.
Na„ A l„SijoO.-24H 2 0 [ring sizes arc 4. 5. 6. 8. 12 (760 pm)|. See also Fig. 1.3. Lines
represent oxygen bridges; intersections of lines show positions of the aluminum and silicon
atoms. Note increasing size of pore aperture (largest diameter given in parentheses). [From
Meier. W. M.; Olson. D. H. Allas of Zeolite Structure Types, 2nd ed.; Buttcrworths:
London. 1987. Reproduced with permission.]
cage formed about them. Molecules small enough to enter, yet large enough to fit with
reasonably large dipolar and London forces will be selectively adsorbed.
Zeolites may also behave as acidic catalysts. The acidity may be of the Bronsted
type if hydrogen ions are exchanged for mobile cations (such as Na + ) by washing with
acid. If the zeolite is heated, water may then be eliminated from the Bronsted sites
leaving aluminum atoms coordinated to only three oxygen atoms:
•xm9
Chains
747
Table 16.1
Some natural and synthetic
Name
Formula (idealized composition)
Ring sizes
zeolites 0
faujasite
N a js A1 , g S i, 34 0 3K4 • 240H 2 0
6, 12
natrolite
N a i6 / ^if.Si 2 ,,C) (i0 -I6H,0
8
stilbite
Na 4 Ca 8 AL 0 Si,,O| 44 -56HiO
5, 6. 8. 10
Linde A r
Na l2 AI |2 Si 12 0 4x -27H,0
4, 6, 8
ZSM-5<-
Na 3 AI 3 S'i 9 36 l 92 l 6 H ,6
10
boggsite
Na 2 9 Ca 7 h AI| X ,Si 77 ,O l9 ,-70H,O
4, 5, 6. 10, 12
sodalite
Na 6 AI 6 Si 6 0 24 -2H,0
6
mordenite
Na 8 Al lj Si 40 O 96 -24H,O
4, 5. 6. 8. 12
rho'’
Na l2 Al 12 Si 3fi 0 % -44H 2 0
8
" Smith, J. V. Chew. Rev. 1988, 88. 149-182. Meier, W. M.; Olson. D. H. Allas of Zeolite
Structure Types. 2nd ed.; Builcrworths: London, 1987.
Natural; substitution of ions often occurs in natural zeolites.
r Synthetic.
Na Na r
°\ /C
Si A1 Si Si Al Si
/ \ / \ / \ / \ / \ / \
O 00 00 00 00 00 0
Si Al Si Si Al Si
/ \ / \ / \ / \ /\ / \
O 00 OO 00 oo oo o
-H,0 °\
/ Al x / S \ f\ / AI X f\ (16 - 18 >
0 0 0 00 00 00 00 o
These will act as Lewis acids. The catalytic siles occur at high density and are uniform
in their activity (as opposed to amorphous solids) because of the microcrystalline
nature of the zeolites.
Heterogeneous catalysis by acidic zeolites is one of the most intensely investi¬
gated topics of chemistry. 25 The reaction of ammonia with methanol to give methyl-
amine can be catalyzed by acidic zeolite rho (Table 16.1 ):-<>
McOH + NH, MeNH : + H,0 (16.19)
u Thomas. J. M. J. Client. Soc. Dalton Trans. 1991. 555-563. Rabo. J. A.; Gajda. G. J. Catal. Rev.
1989-1990. 31. 385.
s Corbin. D. R. J. Mol. Catal. 1989, 29. 271.
670
1 5 • Organometallic Chemistry
Molecular Orbitals The importance of metallocenes and the complexity of their bonding make it worth-
of Metallocenes while to describe them with molecular orbital theory. The properties of the molecular
orbitals of the tt system of cyclic polyenes may be briefly summarized as follows.'
There is a single orbital at lowest energy that consists of an unbroken, i.e.. nodeless 91
‘■doughnut” of electron density above and below the plane of the ring. At slightly
higher energy there is a doubly degenerate set of orbitals each of which has one nodal
plane containing the principal axis. This is followed by another doubly degenerate set
with two nodal planes and yet higher energy. This pattern continues with doubly
degenerate orbitals of increasing energy and increasing number of nodal planes until
the number of molecular orbitals is equal to the number of atomic p orbitals, i.e., the
number of carbons in the ring. If this number is odd. the highest antibonding orbital is
doubly degenerate; if the number is even, the highest antibonding orbital is nondegen¬
erate:
The increasing number of nodes will result in molecular orbitals with symmetries (as
viewed down the ring-metal-ring axis in the metallocenes) of <r (complete cylindrical
symmetry), n (one nodal plane), <5 (two nodal planes), etc. These orbitals on the two
ligands can be added and subtracted to form ligand group orbitals (LGOs). which in
turn can be combined with atomic orbitals of matching symmetry on the metal to form
MOs. For example, consider the lowest energy ligand bonding orbital. If the wave
functions for this orbital on the two rings in the metallocene arc added, a gerade ligand
group orbital of the same symmetry (a, ft ) as an atomic $ orbital is produced. On the
other hand, if the two wave functions are subtracted, an tingereide orbital LGO of the
same symmetry as an atomic p orbital (</ 2 „) is obtained. In the same manner, other
LGOs can be constructed by cither adding or subtracting the higher molecular orbitals
of the two rings. The resulting combinations are shown in Fig. 15.32.
Although symmetry considerations allow us to decide what molecular orbitals tire
possible, knowledge of relative energies and overlap integrals is necessary in order to
estimate the nature of the resulting energy levels. 92 The ordering of the energy levels
has been the subject of much discussion. 93 Photoelectron spectroscopy studies in
conjunction with ligand field theory support the energy level diagram for ferrocene
This is noi ihe place to delve into the nature of organic ring systems. For a discussion of these, see
Lowry. T. H.; Richardson. K. S. Mechanism and Theory in Organic Chemistry ; Harper and Row:
New York. 1987.
1,1 This refers to nodal planes perpendicular to the plane of Ihe ring. The ring itself must be a nodal
plane since the tt system is constructed from atomic p orbitals,
l.auher. J. W.; Hoffmann, R. J. Am. Cheat. Stic. 1976. 98. 1729-1747.
Grebenik. P.; Grinter, R.; Perutz, R. N. Cheat. Sot. Rev. 1988. 17. 453-490.
Metallocenes
671
Fe Ferrocene
AOs molecular
orbitals
e>o
'oo ^
3d,.(3I --
e>o
oo‘
GO 4 '" 1 -
Fig. 15.32 Ligand group orbitals and matching atomic orbitals on iron for ferrocene.
shown in Fig. 15.33. 94 The «, v (cr) orbitals of cyclopentadiene arc so stable relative to
the metal orbitals that they interact but little, i.e.. the ligand is a poor cr donor. On (he
other hand, the e^ir) orbitals are so high in energy compared to Ihe metal orbitals of
the same symmetry that they also interact very liltle. which is another way of saying
that these empty orbitals arc not good tt acceptors. The e Ul and tt 2ll (4 p) orbitals on Ihe
iron atom are at a high energy, and so these orbitals likewise do not contribute much to
the bonding. The only orbitals that arc well matched are e iK ring and metal (3d)
94 Green. J. C. Struct. Bonding IBerlinI 1981. 43. 37-112.
744
16- Inorganic Chains, Rings, Cages, and Clusters
example, crocidolite, Na 2 Fe 5 (OH) 2 [Si 4 O n ] 2 (also known as blue asbestos), and
amosite (Mg, Fe) 7 (0H) 2 [Si 4 0, ( ] 2 , a gray-brown asbestos.- 1
Further linkage by the complete sharing of three oxygen atoms per silicon (analo¬
gous to the edge bonding between many amphibole bands) results in layer or sheet
structures (Fig. 16. le). This yields an empirical formula of [Si 2 0 5 ] 2- . By itself, it is a
rather unimportant structure. However, if we interweave layers of gibbsite, -y-
AI(OH) 3 , or brucite. Mg(OH) 2 , we obtain important mineral structures: (I) A structure
of repeated silicon layers bonded to aluminum layers with bridging —O— and
— 0(H)— is present in the kaolin (china clay) minerals. AI 2 (0H) 4 Si 2 0 5 . (2) A struc¬
ture of repeated pairs of silicon layers with aluminum layers between (and bridged
with) —O— and —0(H)— is present in pyrophyllite. AI 2 (OH) 2 Si 4 O, 0 . (3) If the Al in
the kaolin structure is replaced by Mg, the serpentine structure, Mg 3 (OH) 4 Si,0 5 . is
formed. The dimensions of brucite, Mg(OH) 2 , are slightly larger than those of the
Si 2 0 5 sheet, so the composite layers tend to curl. Fibers from the curled layers form
chrysotile or white asbestos. (4) Similarly, talc. Mg 3 (OH) 2 Si 4 O l0 . is the magnesium
analogue of pyrophyllite. These minerals tend to be relatively soft and slippery.
Further substitution can occur with one out of four silicon atoms in each Si 4 0, 0
unit replaced by aluminum. Because of the difference in charge between Al 3+ and
Si 4 \ a +1 cation must also be added. This muscovite (white mica), KAI 2 (OH) 2 -
Si 3 AIO l0 , is related to pyrophyllite. Phlogopite (Mg-mica), KMg 3 (OH) 2 Si 3 Al6 10 , and
biotite (black mica), K(Mg,Fe) 3 (OH) 2 Si 3 AIO| 0 , are similarly related to talc. The micas
are hard since the layers are composed of strong Al—0—Si bonds. However, they
are relatively easily cleaved between the layers which are held together by the
electrostatic interactions with the potassium cations.
There are more complicated structures intermediate between pyrophyllite and talc
with variable substitution of Al 3+ and Mg 21- . Electroneutrality is maintained by
hydrated cations between layers. Thus the montmorillonites are unusual clays forming
thixotropic aqueous suspensions that are used as well-drilling muds and in nondrip
paints. They are derived from the formulation Al 2 (0H) 2 Si 4 O, 0 -.vH,O with variable
amounts of water, Mg 2+ (in place of some Al 3+ ), and compensating cations. M' ,+
(M = Ca in fuller’s earth, which is converted to bentonite, M = Na). Vermiculite
likewise has variable amounts of water and cations. It dehydrates to a talc-like
structure with much expansion when heated (see page 750).
The ultimate in cross-linking and sharing of oxygen atoms by silicon is the
complete sharing of all four oxygen atoms per Si0 4 tetrahedron in a framework
structure. Silicon dioxide can exist in several forms such as quartz (ther¬
modynamically stable at room temperature), tridymite. and cristobalite. as well as
more dense varieties such as coesite and stishovite that form under high pressure.
With the exception of the latter, all of these contain silicate tetrahedra with complete
sharing but with different linking arrangements of the tetrahedra. 2 - Finally, silicon
21 Asbestos is a commercial term applied to a variety of minerals which can be woven into mate¬
rials which are heat and lire resistant. The most common asbestos (95% of that in use) is chrysotile.
It does not persist in lung tissue and is much less dangerous than crocidolite. which is known to
cause asbestosis and malignant mesothelioma (Mossman. B. T.; Bignon. J.; Corn. M,; Seaton. A.;
Gee. J. B. L. Science 1990. 247. 294-301.)
23 Stishovite. a high-pressure SiO; polymorph, has six-coordinate silicon. Although four-coordinate
silicon is the common building block of inorganic silicates, there are now a number of examples of
silicates with six-coordinate silicon. See Weeding. T. I..; dc Jong. B. H. W. S.: Vccman. W. S.;
Aitken. B. G. Nature 1985. 318 . 352-353.
Chains
745
dioxide also occurs as a glass with (he tetrahedra disordered so that no long-range
order exists.
We have seen that Al 3 may replace Si 4+ so long as electroneulrality is main¬
tained by compensating cations. Three classes of aluminosilicate framework minerals
are of importance: the feldspars, zeolites, and ultramarines.
The feldspars, of general formula MAI 2 _ v Si 2Kt O s , are the most important rock¬
forming minerals, comprising some two-thirds of igneous rocks, such as granite,
which is a mixture of quartz, feldspars, and micas. Feldspars weather to form clays:
4K(AISi 3 0 ) «) + 4C0 2 + 6H : 0-♦ 4KHCO, + 2AI,(0H) 4 Si,0 5 + 8SiO, (16.14)
, The zeolites are aluminosilicate framework minerals of general formula
M "/«[AI,Siv0 2r+2v r~zH 2 0. 23 They are characterized by open structures that permit
exchange of cations and water molecules (Fig. 16.2). In the synthetic zeolites the
aperture and channel sizes may sometimes be controlled by a sort of template
synthesis the zeolite is synthesized around a particular organoammonium cation.
This yields channels of the desired size. The zeolite framework thus behaves in some
ways like a clathrate cage about a guest molecule (Chapter 8). The synthesis of
zeolites also involves several other factors such as the Al/Si ratio, the pH. the
temperature and pressure, and the presence or absence of seed crystals: 24
Na 2 Si0 3 + AI(OH) 3 sodalite (16.15)
high SVAl.di lV NiOH
Na,Si0 3 + AI(OH), —-- ZSM-5 (16.16)
low SI/AI.
Na 2 SiO, + AI(OH), -[^77- mordenite (16.17)
In some instances attempts are made to synthesize naturally occurring zeolites.
Boggsitc. shown on the cover of this book and discussed in Chapter 1, is one such
example. Its low abundance in nature restricts studies which could demonstrate ils
usefulness.
Both natural and synthetic zeolites (Table 16.1) lind wide application as cation
exchangers since (he ions can migrate rather freely through the open structure. Some
cations will lit more snugly in the cavities than others. In addition, certain zeolites may
behave as molecular sieves if the water adsorbed in the cavities is completely re¬
moved. Various uncharged molecules such as C0 2 , NH„ and organic compounds can
be selectively adsorbed in the cavities depending upon their size. Again, the zeolite
framework behaves similarly to a clathrate cage except that the adsorbed molecules
must be capable of squeezing through the preformed opening rather than having the
2:1 Smith. J. V. Chan. Rev. 1988. 88. 149-182. Thomas, J. M.: Callow. C. R. A. Proa. /"‘"T. Cltew.
1987. 35. 1-49. The word zeolite, meaning boiling stone, was first used by the Swedish scientist.
A. F. Cronstcdt, in 1756 after he observed that water was evolved from stilbile when it was healed
in a flame.
24 Barrer. R. M. In Zeolites: Drtaj, B.: Hocevar. S.: I’cjovnik. S.. Eds.: Elsevier: Amsterdam. 1985;
pp 1-26. Vansant. E. F. Pore Size Ennineerinn in Zeolites: Wiley: New York, 1990.
672
15-Orgonometallic Chemistry
Ligand Metallocene Metal
orbitals orbitals orbitals
e iu
u
Fig. 15.33 Qualitative molecular orbital diagram for a metallocene. Occupation of the
orbitals enclosed in the box depends on the identity of the metal; for ferrocene it is e K a J.
[From Lauher, J. W.; HolTmann. R. J. Am. Client. Soc. 1976. 98, 1729-1742. Reproduced
with permission.)
orbitals which form two strong n bonds. These Cp—M 7r bonds are believed to supply
most of the stabilization that holds the ferrocene molecule together.
If we supply enough electrons to fill all the bonding and nonbonding molecular
orbitals but none of the antibonding orbitals in ferrocene, nine pairs will be required.
Once again we see that the 18-electron rule is a reflection of filling strongly stabilized
MOs. But how can we rationalize the existence of metallocenes which do not conform
to the rule? Table 15.8 shows that manganocene, chromocene, and vanadocene are
one, two, and three electrons short, respectively, of 18, while cobaltocene and
Metallocenes
673
Table 15.8
Properties ol
Electron
Unpaired
Melting
bis(pentahaptocyclo-
Compound
configuration
electrons
Color
point (°C)
pentadienyl) complexes of
the first-row transition
Cp 2 V
44
3
Purple
167-168
metals
Cp 2 Cr
44 ^
2
Scarlet
172-173
Cp 2 Mn
444
5
Amber
172-173
Cp 2 Fe
4°Tr
0
Orange
173
Cp 2 Co
444
1
Purple
173-174
Cp,Ni
444
2
Green
173-174
Structures of
Cyclopentadienyl
Compounds
nickelocene have one and two electrons in excess of 18. We also see that except for
ferrocene all are paramagnetic. The molecular orbital diagram of Fig. 15.33 allows us
to understand both the rule violations and the magnetic properties. The highest
occupied molecular orbitals (e 2l! and a l/; ) are only slightly bonding and therefore
removing electrons from them does not greatly destabilize the complex. The e lf!
LUMO is not significantly antibonding so when electrons are added to create 19-
eiectron and 20-electron species the stability loss is minimal. Allhough it is probable
that there is some change in the relative energies of the molecular orbitals in going
from complex to complex,' self-consistent results can be obtained with the ferrocene
model.
As we saw in ligand field theory, so long as the difference in energy between
and e 2fl is less than the pairing energy, both levels can be occupied. Thus the three
unpaired electrons in Cp 2 V are accounted for with its J A , configuration. To be
discussed later is Cp 2 Mn which has an u-’u- configuration consistent wilh its five
unpaired electrons. In this instance two of these electrons are found in the e if ,
antibonding level. Substituents on the Cp ring can be very influential in altering the
energy differences between MOs. Wilh five methyl groups on the ring (pentamelhyl-
cyclopentadienyl, Cp), spin pairing occurs and the electronic configuration becomes
44’ suggesting that a crossing of the e 2K and a [f! levels has occurred.
The cyclopentadienyl metallocenes of the elements of the first transition series are
isomorphous and have melting points which are remarkcdly constant at or near 173 °C
(Table 15.8). The structure of ferrocene in the solid slate, originally described as
staggered, is now viewed as nearly eclipsed (a rotational angle of 9° between rings), 95
as are its heavier analogues ruthenocene and osmocene. The eclipsed configuration of
ferrocene is also favorable in the gas phase, where the rotational barrier is only 4 ± I
kJ mol -1 , allowing the rings essentially free rotation. Neutron diffraction studies
reveal that, in the solid state, the hydrogen atoms of the ring are tilted toward the iron
atom. 96 The staggered arrangement is found for the Cp and decabenzyl derivatives
(Fig. 15.34) and results because of van der Waals interactions between the methyl or
benzyl groups of the two rings. The same interactions also cause the methyl and
phenyl groups to tilt away from the iron atom.
In typical metallocene compounds. Cp 2 M, all of the C—C bonds are of the same
length and the rings are parallel. However, there are several cyclopentadienyl com-
95 Seiler. P.; Duniiz. J. D. -4cm Crystallogr., Seel. B 1979, 35, 1068-1074.
96 Takusagawa. F.; Koetzle. T. F. Ac la Crystallogr., Seel. B 1979. 35. 1074-1081.
742
1 6- Inorganic Chains, Rings, Cages, and Clusters
0 O
II II
H 2 S0 4 + SO, -► HO —S —0 —S—OH
0 0
(16.11)
(16.12)
Condensed polyphosphates such as sodium triphosphate are of great industrial impor¬
tance since they are used in large tonnages as "builders in the manufacture ot
detereents. As such they function to adjust the pH and to complex water-hardening
ions such as Ca 2+ and Mg 24- .'’ Industrially, sodium triphosphate is not made from the
reaction of Eq. 16.12 but from dehydration of sodium hydrogen phosphate and sodium
dihydrogen phosphate mixtures:
2Na 2 HP0 4 + NaH 2 P0 4 -^ Na 5 P 3 O 10 + 2H,0 (,6 - ,3)
Silicate Minerals'® Silicon forms a very large number of compounds containing heterocatenated anions.
These are of great importance in the makeup of various minerals since about three-
fourths of the earth's crust is silicon and oxygen. Simple silicate anions. Si0 4
Fig. 16.1 Various silicate structures: (a) Si0 4 tetrahedron. When carrying a -4 charge, this
is the orthosilicatc ion. (b) The disilicale anion, (c) Portion of an infinite single chain.
(SiO,) 2 "". (d) Portion of an infinite double chain or band. |Si 4 0u]„ . (e) Portion of a sheet
or layer structure, [SijOjJn •
,7 Nicuwcnhuizcn. M. S.: Peters. J. A.; Sinncma. A.; Kicboom. A. P. G.: van Bckkum. H. J. Am.
Chem. Soc. 1985. 107. 12-16.
Liebau. F. Structural Chemistry of Silicates-. Springcr-Veriag: New York. 1985.
Chains
743
(orthosilicates; Fig. 16.1a), are not common in minerals, although they are present in
olivine, (Mg, Fe)->Si0 4 , an important constituent of basalt which, in turn, is the most
voluminous of the extrusive rocks formed from outpouring of magma. Although not
occurring in nature, both Na 4 Si0 4 and K d Si0 4 have also been shown to be orthosili¬
cates. 19 Other minerals containing discrete orthosilicate ions are phenacite (Be 2 Si0 4 ),
willemite (Zn 2 Si0 4 ), and zircon (ZrSi0 4 ). The large class of garnets is composed of
minerals of the general formula MjMi'^SiOjJj, where M 11 can be Ca 2+ . Mg 2 . or
Fe 2+ . and M 1 " is Al 3 . Cr 3 . or Fe 3 '
Minerals containing the pyrosilicate or disilicate anion. Si 2 0 7 - (Fig. 16.lb), are
not common: thortveitite, Sc 2 Si 2 0 7 . hemimorphite, Zn 4 (0H) 2 Si 2 0 7 Ti 2 0 (does not
contain discrete Si 2 0$“ ions), and barysilite. MnPb„(Si 2 0 7 ),, as well as vesuvianitc
and epidote which contain both SiO 4- and Si 2 0 7 - ions. Linear tri- and tetrasilicates
are almost unknown.
Although the term "discrete" is universally applied to the orthosilicate anion and
sometimes to the disilicate anion, they cannot be considered analogues of perchlorate,
CIOJ. The metal-oxygen bond in all silicates contains considerable covalent character
(just as the silicon-oxygen bond contains considerable ionic character). The orthosili¬
cates contain no Si—O—Si linkages such as are present in the disilicale. chain
silicates, and the cyclic compounds. Although one can formulate them as
Na 4 (Si0 4 ) 4- . Zr 4r (Si0 4 ) 4- , or M;(Si0 4 ) 4- , as the more electropositive metal
(M" = Na. Mg 2 , Fe 2t \ Zn 2 , Al 3 ) becomes more like silicon in character, it
becomes more difficult to discern discrete silicate anions. When M = Si. it becomes
impossible and the result is SiO, (page 744). Thus the aluminosilicates usually arc
treated as large covalent structures, and the hemimorphite mentioned above is better
viewed as a 3-D structure related to the aluminosilicates than as 4Zir\ 20H~,
(SLO/ - , H,0.
Alternatively, one can treat silicates as closest packed arrays of oxide ions with
Si 4 " ions fitting into tetrahedral holes and other metal ions fitting into either
tetrahedral holes as in phenacite or octahedral holes as in olivine. (See Fig. 7.3 for the
alternative ways of describing olivine.) Transition metal ions in these structures
behave as they do in complexes: Olivine gets its name from the greenish color caused
by partial substitution of Fe : ' for Mg 2 ions in the octahedral holes. The hexa-
aquairon(II) ion has a similar green color. The hue of garnets also comes from
transition metal ions.
The next higher order of complexity consists of the so-called metasilicate anions,
which arc cyclic structures-" of general formula (SiO,) 2 "' occurring in benitoilc.
BaTiSi,0,,. catapleilc, Na,ZrSi 3 0„H 2 0, dioptase. Cu (> Si,,0 IK -6H,0. and beryl,
Be,AI 2 Si h O| X . This is the most important mineral source of beryllium, and also may
form gem-quality stones (see Problem 16.9).
Infinite chains of formula (SiO,) 2 " - are found in minerals called pyroxenes. In
these chains the silicon atoms share two of the four tetrahedrally coordinated oxygen
atoms with adjacent atoms (Fig. I6,lc). Examples of pyroxenes include enstatite.
MgSiO,, diopside, CaMg(SiO,) 2 . and the lithium ore. spodumene, LiAI(SiO,) 2 . If
further sharing of oxygen atoms occurs by half of the silicon atoms, a double chain or
band structure is formed. This is the structure found in amphiboles (Fig. 16.Id).
Amphiboles contain the Si 4 0,Yrepeating unit as well as metal and hydroxide ions, for
'v Barker. M. G.; Good. P. G. J. Chem. Res.. 1981. 274.
-" The cyclic silicates are considered later in the section on inorganic ring systems.
1 5 • Organometallic Chemistry
Fig. 15.34 Molecular
structure of staggered
decabenzylferrocene. (From
Schumann. H.; Janiak. C.;
Kohn. R. D.; Loebel. J.:
Dietrich. A. J. Organomet.
Chem. 1989. 365. 137-150.
Reproduced with
permission.]
pounds in which the rings are tilted with respect to one another. Examples are
Cp : ReH, Cp-,TiCI 2 , and Cp ; TaH, (Fig. 15.35), in which the steric requirements of
additional ligands prevent parallel rings. Lone pair requirements in Sn(II) and Pb(II)
result in similar tilting of the rings in Cp 2 Sn and Cp 2 Pb.97 Less clear are explanations
for bent structures in Cp?Sr and CpBa (Fig. 15.36). although packing forces may be
responsible.Finally, there are compounds with more than two cyclopentadienyl
Fig. 15.35 Some cyclopentadienyl complexes containing lilted rings.
1,7 Connolly, J. W.; Hoff. C. Adv. Organomet. Chem. 1981. 19. 123-153. Jutzi, P.: Hiclschcr. 13.
Organomelallics 1986,5, 1201-1204.
w Williams. R. A.; Hanusa, T. P.; Huffman. J. C. J. Clian. Site. Chem. Coni mini. 1988. 1045-1046:
Orguiwmetallics 1990 . 9. 1128-1134. Blom. R.; Faegri. K.. Jr.: Volden. H. V. Orgtmomeitillics
1990. 9. ill-379. Mosges. G.; Hampel. F.; von Ragud Schleycr, P. Organomelallics 1992. II.
1769-1770.
Metallocenes
675
Fig. 15.36 Solid-state structure of (Me<C,):Ba. Metallocene units are arranged in
quasipolymeric chains and a ring-Ba-ring tilt of 131° is observed. Lines connecting Ba and
methyl groups indicate the shortest intermolecular contacts. [From Williams, R. A.: Hanusa,
T. P.: Huffman. J. C. J. Chem. Soc. Chem. Common. 1988. 1045-1046. Reproduced with
permission.]
rings. Examples in which several rings are attached to the same metal atom are
(tetrakiscyclopentadienyl)titanium (two rings are tj 5 and two are r/ 1 ) and tetrakis-
(cyclopentadienyl)uranium (all rings arc ry and are arranged tetrahedrally. Fig. 15.35).
A different type of structure is the layered arrangement of nickel atoms and cyclopen¬
tadienyl rings in lCp 3 Ni ; p, as shown in Fig. 15.37a. Complexes having this arrange¬
ment are often called "triple deckers" and have been described with molecular orbital
theory.99 A variety of center slices have been used in place of Cp. as shown in Fig.
15.37 .km Progress also continues in the synthesis and bonding theory of tetradecker
sandwich complexes. 101
There are a few compounds known having only one cyclopentadienyl ring per
metal atom. Sodium cyclopentadienide. cyclopenladienylthallium(l) (vapor), and
cyclopenladienylindium(l) (vapor) have structures that may be described as "open-
laced sandwiches." 1 "- In the solid, cyclopentadienylindium(l) polymerizes to form an
infinite sandwich structure of alternating cyclopentadienyl rings and indium atoms.
In addition to the monocyelopcniadienyl compounds just described, there are
many compounds which contain it single ring per metal atom with a variety ol
additional ligands, such as carbon monoxide, to complete the coordination sphere.
These include monomers and dimers, with and without metal-metal bonds as neces¬
sary to obey the 18-electron rule. Some examples are shown in Fig. 15.38.
Lauher. J. W.. Elian. M.; Summerville. R. II.: Hoffmann. R. J. Am. Chem. Soc. 1976. 9H.
3219-3224. Chcsky. P. T.; Hall. M. 13. J. Am. Chem. Soc. 1984. 106. 5186-5188. Jemmis. E. D.;
Reddy. A. C. Organontetiillics 1988, 7, 1561-1564.
Davis. J. H.. Jr.: Sinn E.: Grimes. R. N, J. Am. Chem. Soc. 1989, III. 4776-4784. Edwin. J.:
Geiger. W. E.; Bushwellcr. C. H. Organomelallics 1988. 7. 1486-1490.
"" Jemmis. E. D.: Reddy. A. C. J. Am. Chem. Soc. 1990. 111. 722-727.
IU - Beachlcy. O. T.. Jr.: Blom. R.; Churchill. M. R.: Faegri. K.. Jr.; Fettinger, J. C.; Pazik. J. C.:
Vicloriano. L. Organonietallics 1989. 8. 346-356.
740
16- Inorganic Chains, Rings, Cages, and Clusters
Many different organic R groups have been incorporated into these polymers and n
may be as large as 750.000.' Low molecular weight polymers in which one of the R
groups is H have also been produced. '<> Their wide range of solubilities and electronic
properties (electron delocalization occurs along the Si chain) have stimulated much
commercial interest in recent times with possible applications in the areas of thermal
precursors to silicon carbide, photoinitiators, photoconductors, and microlithogra¬
phy. 11 Fluorinated and chlorinated long-chain compounds are known up to and
including Si lft F 34 and Si 6 Cl l4 . It is worth mentioning that bulky substituents may en¬
hance the stability of silanes relative to alkanes. For example, Si 2 Br h can be distilled
without decomposition at 265 °C, but C 2 Br h decomposes to C 2 Br 4 and Br 2 at 200 °C.
The chemistry of germanes is similar to that of silanes. Heavier congeners of
carbon, however, show severely restricted catenation properties. Distannane, Sn 2 H 6 .
is known, although it is unstable. Plumbane, PbH 4 , is of marginal stability itself, and
hence a large number of heavier analogues is not expected, although the interesting
compound Pb(PbPh 3 ) 4 has been synthesized.
Some other nonmetals such as nitrogen, phosphorus, and sulfur form chains, but
their chemistry is less important than that of the polymers of Group IV (14). Although
chain lengths for nitrogen up to eight atoms are known (most of which are extremely
explosive), only hydrazine, H-,NNH 2 , and hydrazoic acid, HN 3 , are stable at room
temperature, and chains longer than 2-tetrazene, H 2 N—N=N NH 2 , require
organic substituents. 13 The series of sulfanes, HS„H. is fairly extensive, and chains up
to n = 8 have been obtained in pure form. Diphosphine, P 2 H 6 (very air-sensitive), is
well known, and triphosphine, H : PPHPH,, has been fully characterized. Tetraphos-
phine. H 2 PPHPHPH 2 . and higher analogues have been identified spectroscopically in
mixtures. M The open-chain structures become increasingly unstable relative to cyclic
structures (of which there arc many) as the number of phosphorus atoms in the chain
increases:
2P 4 H„ - P 5 Hj + P : H 4 + PHj (16.3)
Oxygen forms no chains longer than three atoms, and besides the familiar ozone. 0 3 ,
and its anion, O^, few compounds are known; all of them are bis(perfluoroalkyl)
trioxides such as F,COOOCF,.
Allotropes of both sulfur and selenium are known in which helical chains of great
length are present. While the sulfur chains are unstable with respect to cyclic S x . the
chain form for selenium is most stable. Red phosphorus is polymeric and is thought to
involve chains of pyramidal phosphorus atoms.
“'West. R.: Maxka. J. In Inorganic and Organometallic Polymers ; Zeldin. M.; Wynne. K. J.;
Allcock. H. R.. Eds.; ACS Symposium Series 360: American Chemical Society: Washington. DC.
1988.
1,1 West. R. J. Organomel. Client. 1986. 300. 327-346.
11 The band gap in polysilanes approaches 4 eV compared to nearly 8 eV in saturated carbon
skeletons. The polysilanes arc insulators in pure form but often can be doped to give semi¬
conductors.
'= As with Si. high molecular weight polymers—(R 2 Ge)„ —have been recently characterized. Sec
Footnote 8.
13 lieck. J.: Striihle. J. Angew. Cltcm. Ini. Ed. Engl. 1988. 27. 896-901.
IJ Baudler. M. Angew. Cliem. Ini. Ed. Engl. 1982. 21. 492-512.
Chains
741
The halogens are known to form reasonably stable chains in polyhalide anions,
the best-known example being the triiodide ion. 1,. They are considered separately in
Chapter 17. A great number of nonmetals form simple X—X bonded molecules:
B 2 CI 4 , N 2 H 4 . P 2 I 4 . As,R 4 . H 2 0 2 . S 2 C1 2 . X 2 (halogens), etc. In general these are
relatively stable, although all are susceptible to attack by reagents that can cleave the
X—X bond:
R,AsAsR, + X 2 -♦ 2R,AsX
(16.4)
C1 2 BBC1 2 + CH 2 =CH 2 -♦ CI 2 BCH,CH 2 BC1 2
(16.5)
Cl 2 + OH" - HCIO + Cl
(16.6)
|H| (reducing agent)
HOOH - H 2 0
(16.7)
NHilll
H 2 PPH 2 -» PH 3 + (PH),
(16.8)
H 2 NNH, + 0 2 N 2 + 2H : 0
(16.9)15
Heterocatenation
The extensive chemistry of metal-metal bonds is sufficiently interesting to war¬
rant separate treatment (page 807).
Although there is a paucity of inorganic compounds exhibiting true catenation, the
phenomenon of heterocatenation, or chains built up of alternating atoms of different
elements, is quite widespread. While homocatenated polysilanes are nearing commer¬
cialization (page 739). there are two classes of inorganic heterocatenated polymers
that already enjoy wide application. 16 These are the silicones, (R 2 SiO)„, and the
polyphosphazenes. [PN(OR) 2 ]„. discussed on pages 749 and 773. respectively.
The simplest heterocatenated compounds are those formed by the dehydration of
acids or their salts:
O
2 0 = P — OH
o-
Such dehydrations were first effected by the action of heat on the simple acid
phosphate salts and hence the resulting product was termed pyrophosphate (Gr. mtp,
fire). With an increased number of polyphosphates |P„0 3fl + ,] ( " ' known, however,
the preferred nomenclature has become diphosphate {n = 2), triphosphate (n = 3),
etc.
For many heterocatenated compounds there are other, sometimes simpler, syn¬
thetic routes than the thermal elimination of water.
13 The formation of N 2 rather than cleavage as in the other examples can be attributed to the extremely
strong triple bond in molecular nitrogen.
■ h Inorganic and Organonielullic Polymers'. Zeldin. M,; Wynne. K. J.; Allcock. H. R.. Eds.; ACS
Symposium Scries 360. American Chemical Society: Washington. DC. 1988.
heat
O'
n — p_ n
O'
_p — n
VJ — r — u
O'
O'
676
15 • Organometallic Chemistry
<») lb) (c) (d)
Fig. 15.37 Some known “triple deckers" containing cyclopentadienyl ligands.
There has been much discussion over the years regarding the C—C bond dis¬
tances in the Cp ring. Five equidistant bonds is consistent with the electron delocaliza¬
tion found in the D Sh cyclopentadienyl anion. If however, there are two intermediate,
one short, and two long C—C bonds, a structure closer to cyclopentadiene (C.) is
Red-Orange Solid
mp I38C
Air Sensitive
Yellow Solid
mp 77"C
Air Sensitive
Dark Brown
Solid
mp (dec) I I9'C
Dark Red Liquid
bp 38C
Air and Heat Sensitive
Dark Red Liquid
bp 7|5 145-C
Purple-Red Solid
mp 2I5’C
Air Sensitive
Red-Purple Solid
mp (dec) I94‘C
Air Stable
Red Solid
mp (dec) I36‘C
Air Sensitive
Fig. 15.38 Some commercially available cyclopentadienyl complexes containing carbonyl
and nitrosyl ligands.
Metallocenes 677
present and suggests localized bonding. It would seem that an X-ray analysis would
answer questions of this sort unequivocally but positional uncertainties caused by
thermal ring motion often present a problem. Low temperature measurements mini¬
mize these effects and in some instances it is dear that not only are there different
C—C bond lengths in the Cp ring but there are also small deviations from planarity. In
Rh(Cp)(CO),. C—C bond lengths of 138.4(8), 144.5(8), 144.7(7), 141.2(8), and
141.0(7) pm have been measured and a small distortion from planarity has been
noted. 103 It follows that not all carbon atoms of the Cp are equidistant from Rh.
Distortions of Cp have mechanistic implications as well. When ReCpMe-
(NO)(PMe 3 ) reacts with two moles of PMe 3 , Tj 5 -Cp rearranges to 7j'-Cp:" M
(15.63)
Me
MCjP | _PMe 3
‘‘Re’'’ c.h;
^| N, 55
Mc 3 P ^ PMc 3
0
As the five-electron donor converts to a one-electron donor. Re accepts an additional
four electrons from the two trimethylphosphine ligands. One additional mole of PMe 3
causes loss of Cp” from the metal. It is highly likely that the rj 3 -Cp complex is an
intermediate in the reaction although it was not detected. When Cp ligands are found
coordinated in arrangements other than the symmetric tj 5 , we sometimes refer to them
as “slipped" ring systems. The rhenium reaction is an extreme example in which the
ring has been induced to slip completely off the metal atom (tj 5 — ij J ^ tj 1 =: tj 0 ). Thus
one, two, and three vacant coordination sites become successively available on the
metal in this reaction. Indenyl complexes have attracted much attention because they
undergo substitution reactions much faster than Cp complexes. 105 It is believed that
an tj 3 intermediate is stabilized by formation of the aromatic benzo ring. 106
10:1 Lichtenberger, D. L.; Blevins. C. H. II; Ortega. R. B. Orgunometallks 1984. .), 1614-1622,
Fitzpatrick. P. J.; Lc Page, Y.; Scdman, J.; Butler, 1. S. Inorg. Client. 1981, 20, 2852-2861.
104 O'Connor. J. M.; Casey. C. P. Client. Rev. 1987, 87. 307.
105 A ring slippage mechanism was first proposed in l%6. For a discussion of this work, see Basolo, F.
Polyhedron 1990. 9. 1503-1535.
Habib. A.; Tanke. R. S.; Holt. E. M.; Crabtree, R. H. Organometallics, 1989. 8. 1225-1231.
Inorganic Chains, Rings,
Cages, and Clusters
From lopics discussed previously in Chapters 7 and 15 it should be obvious that there
is no sharp distinction between inorganic and organic chemistry. Nowhere is the
i borderline less distinct than in the compounds of the nonmetals. Some, such as the
halides and oxides, are typical inorganic compounds, but others, such as compounds
of nonmetals with organic substituents, are usually called organic compounds. 1 The
situation is further complicated by the tendency of some nonmetals to resemble
I carbon in certain properties. This chapter discusses the chemistry of nonmetals in
terms of one such property: their propensity to form chains, rings, and cages. Most
metals show less tendency to form compounds of this type, and the length of the
chains and size of the rings thus formed are restricted. However, the ease with which
both metals and nonmetals and combinations of the two form clusters has only been
recognized in the last decade and an explosive growth in this branch of chemistry is
underway. 2
Chains If there is a single feature of carbon which makes possible a unique branch of
chemistry devoted to a single element,' it is its propensity to form extensive stable
Catenation chains. This phenomenon is not common in the remainder of the periodic table.
1 In this area the nomenclature of the chemist is imprecise, to say the least. Thus it is common to
distinguish between "organophosphorus compounds," such as PPh 3 and Me 2 P(S)P(.S)Mc 2 . and
"inorganic phosphorus compounds," such as PCI 5 and Na 3 P 3 0.>. While the latter undoubtedly
belong to the province of inorganic chemistry, a chemist interested in the former is as apt to be an
organic as an inorganic chemist.
There is not agreement on what compounds should be called clusters. Some believe that the term
should be restricted to compounds which have metal-metal bonds (Cotton. F. A.; Wilkinson. G.
Advanced Inorganic Chemistry, 5th ed.; Wiley: New York, 1988; p 1053). Others take a broader
view that includes, for example, borancs (no metal atoms present) or polyoxometal systems (only
long-range metal-metal bonding possible).
•' This is a poor definition of organic chemistry, of course. Even the simplest organic compounds
involve carbon and hydrogen, and inorganic chemists cannot resist pointing out that the most
interesting parts of organic chemistry usually involve "inorganic" functional groups containing
oxygen, nitrogen, sulfur, etc.
738
Chains
739
although the congeners of carbon (especially silicon) and related nonmetals exhibit it
to a reduced extent. Despite the fact that there appears to be no thermodynamic
barrier to the formation of long-chain silanes, Si„H-,„ + ,, their synthesis and char¬
acterization are formidable tasks. Although silicon-silicon bonds are weaker than
carbon-carbon bonds, the energy differences do not account for the observed stability
differences. The explanation lies with the low-energy decomposition pathway avail¬
able to silanes which is not available to alkanes. 4 In addition to their inherent kinetic
instability, silanes are difficult to handle because they are very reactive. Their reac¬
tions with oxygen, as shown in Eq. 16.2, appear similar to the alkane reactions of
Eq. 16.1:
2 n i j
C„H, n + 2 + —-— 0 2 -• «CO, + (n + l)H,0 ( 16 . 1 )
Si„H 2n+2 + 3n ^~ 1 0 2 - nSi0 2 + (n + 1)H 2 0 (16.2)
In fact both reactions are thermodynamically favored to proceed to the right. The
important difference, not apparent in the stoichiometric equations, is the energy of
activation which causes the paraffins to be kinetically inert in contrast to the reactive
silanes. 5
Further complications with silanes arise from lack of convenient syntheses and
difficulties in separation. Nevertheless, compounds from n = I to n = 8 have been
isolated, including both straight-chain and branched-chain compounds,We should
not judge silicon's tendency to catenate by looking at these hydrides, however, since a
much different result is obtained when substituents other than hydrogen are present. 7
Factors other than inherent Si—Si bond strength must be involved because it is
possible to isolate a large number of polysilane polymers:
R,
4 At 400 "C. SI.H„ decomposes It) 1 ’ limes raster than C 2 H h . I( appears that hcciiu.sc of die low-lying
unoccupied 4.t or 3 d orbiials of Si. migration of a hydrogen atom wiili simultaneous cleavage of an
•Si—Si bond is possible, iking. M. A. In Homontomic Kings. Chains and Mncronwlccttles of Main-
Group Elements; Rheinyold. A. L.. Ed.: Elsevier: New York. 1977.)
5 The difference in activation energy can be qualitatively described as follows: The reaction of Eq. lb. I
can be initialed by striking a match while the reaction of F.q. lb.2 merely requires allowing oxygen to
contact the silane. Il has been stated that one of the prerequisites for the synthesis of higher silanes is
a large amount of courage.
Henegge. F„ In Silicon Chemistry; Corey. J. Y.: Corey. E. R.; Caspar. P, I’,. Eds.: Wiley: New
York. 1988.
7 To quote Robert West. "Although (he myth that silicon is not capable of extensive catenation still
persists, very large cyclic polysilanes have been synthesized as well as polysilane polymers with
molecular weights in the hundreds of thousands." I'nre Appl. Client. 1982. 54. 1041-1050.
Miller. R. D.: Michl. J. Cliem. Rev. 1989. ,W. 1359-1410. Silicon-Rased Polymer .Science; Ziegler.
J. M.: Fcaron. F. W. G.. Eds.: Advances in Chemistry 224; American Chemical Society; Wash¬
ington. DC. 1990.
678
15-Organometallic Chemistry
Covalent versus
Ionic Bonding
We have already seen that metallocene complexes often violate the 18-electron
rule and that stability of these violators can be attributed to the nature of the molecular
orbitals to which electrons are added or from which they are removed. In view of this
electronic flexibility, it is perhaps not surprising that Cp ligands are capable of
stabilizing a wide range of oxidation states. Normally we think of organometallic
chemistry as being the domain of low oxidation state complexes, 107 but increasing
interest in high oxidation state complexes is apparent and often the cyclopentadienyl
group is present as a stabilizing ligand. 108 Oxidation of cyclopentadienyi complexes
may lead to isolable oxo complexes as shown by the following hydrogen peroxide
reaction.
CpRe(CO) 3 —^ CpRe(0) 3 (15.65)
Some other oxo complexes are shown in Fig. 15.39. Although the oxide ligand is
generally thought of as Cr~ rather than neutral O, its similarity to the carbene ligand.
CR 2 (which could also be viewed as CR 2 ~), tends to be obscured as a result. Interest
in oxo complexes is tied to their relationship to metal oxide catalysts which are widely
used in organic synthesis.
Metallocenes such as Cp 2 Cr, Cp,Fe, and Cp 2 Co are considered to have strong
covalent bonding between the metal and the rings. Although not all of these metal¬
locenes are stable with respect to oxidation, etc., all have strong bonding with respect
to dissociation of the rings from the metal atom (Table 15.9). The bonds between the
metal and rings certainly have some polarity, but these compounds do not react like
polar organometallic compounds (as exemplified by the well-known Grignard reagent):
RMgX + H 2 0 -► RH + MgXOH (15.66)
Cp 2 Fe + H 2 0 -► No reaction (15.67)
Fig. 15.39 Cyclopen¬
tadienyl complexes in
which the metal is in a
high oxidation state.
11,7 R ven CO. which normally has little affinity for metals in high oxidation slates, has been found in
WCI:(PMePh 2 ) 2 (C0)(0). Su. F.-M.: Cooper. C.: Gcib. S. J.t Rhcingold. A. L.: Mayer. J, M. J.
Am. Client. Sue. 1986, 108. 3545-3547. Alkyl groups may also stabilize high oxidation states, e.g..
WMe„.
I(W Boltomley. F.; Sutin. N. Ailv. Orgonomel. Chan. 1988, 28. 339. Lcgzdins. P.; Phillips. F.. C.;
Sanchez. L. OrganometaUics 1989. 8. 940-949. Marshman. R. W.; Shusta, J. M.; Wilson. S. R.:
Shapley. P. A. OrganometaUics 1991. 10, 1671-1676.
736
737
15-Organometallic Chemistry
15.49 Upon encountering Ni 3 Cp,(CO) : for the first time, you might suppose that a typographical
error had occurred. Why?
15.50 Metals in low oxidation states are usually strong reducing agents. Give an example of a
metal in a zero oxidation state acting as an oxidizing agent.
15.51 Consider the carbonylation of di-CHjMn( l5 CO)(COU with unlabeled CO. Assuming that
methyl migration occurs, predict the products and their ratios.
15.52 In the hydrogenation of an alkene using Wilkinson's catalyst. Rh(PPh 3 ) 2 <RCH=CH 2 )-
CI(H)(H) reacts to give Rh(PPh 3 ) 2 CI(solvent) and an alkane. In this reaction rhodium is
reduced from + 3 to +1 and an alkene is reduced to an alkane. What is oxidized?
15.53 Using data given in this book, calculate the standard enthalpies of reaction at 25 °C for:
a. C(s) + HjO(g) - CO(g) + H 2 (g)
b. H 2 0(g) + CO(g) -► C0 2 (g) + Hj(g)
15.54 In this chapter we have examined examples of polynuclear metal carbonyl complexes as
well as simple metal carbonyl hydrides. Consider now the polynuclear carbonyl hydride
complex, H 3 Osj(CO)| 2 . Rationalize the formulation of this species. From your application
of the 18-electron rule, what can you say about the structure of this molecule? How is it
similar to or different from the complex Os 3 (CO), 2 shown in Figure 15.9? (See Churchill
M. R.: Wasserman, H. J. Inorg. Chem. 1980. 19. 2391-2395.)
15.55 On page 643 the following statement was made, somewhat casually, in reference to
H 2 Fe(CO) 4 : "The large difference in the two ionization constants provided the first
evidence that the hydrogen atoms in the complex were both bound to the same atom (and
hence to the iron atom)." Present support for this line of thinking.
15.56 Why do alkenes and aromatics compose such a large fraction of the products from the
reaction of methanol with ZSM-5 (Eq. 15.181)?
15.57 Suggest a reaction mechanism for the following base-induced migration:
o o
15.58 When (r)'-Me,C.,H,) : Nb(H)(CO) reacts with AIMe 3 . a single nonionic product (I) is
obtained which shows a carbonyl stretching frequency of 1721 cm"' compared to
1875 cm" 1 for the starting material. The 'H NMR hydride chemical shift changes from
-5.59 ppm to -4.72 ppm in going from reactant to product. However, when (Cp) (Cp)-
Nb(H)(CO) reacts with AIEl, to give a single neutral product (II). the CO stretching
frequency changes from 1863 cm -1 to 1900 cm -1 and the 'H NMR hydride chemical shift
changes from -5.48 ppm to -10.2 ppm. Suggest structures for products I and II and
rationalize their infrared and NMR data. (Sec McDade, C.; Gibson. V. C.; Santarsiero.
B. D.; Bcrcaw, J. E. Orgntwmetallics 1988. 7. 1-7.)
15.59 The solid state magic angle spinning "C NMR spectrum of Fe(CO), shows a single
absorption at -26 °C but two signals (relative intensity = 2:3) at - 118 °C. Provide an
explanation for these spectral observations. (See Hanson. B. E.t Whitmire. K. H. J. Am.
Chem. Soc. 1990. 112 . 974-977.)
15.60 The first bis(hydroxymethyl) transition metal complex has been prepared. The synthesis
was achieved by an interesting sequence of reactions. Diazomethane converted (1.5-
Problems
cyclooctadiene)diiodoplatinum(ll) into (l.5-cyclooctadiene)bis(iodomethyl)platinum(ll).
Treatment of this complex with silver trifiuoroacelate led to replacement of both iodine
atoms with trifiuoroaeetate. Methanolysis gave (1.5-cycIooctadiene)bis(hydroxy-
methyl)platinum(ll). Reaction of (his complex with diethylazodicarboxylate resulted in
loss of water and intramolecular cyclization to form an oxametallacyclobutane complex of
platinum from which the diene was replaced by two triphenylphosphine ligands. Draw
structures of each of the complexes and reagents in this sequence. (See Hoover. J. F.:
Stryker. J. M. J. Am. Chem. Soc. 1989, III. 6466-6468.)
15.61 The first example of a four-electron donor, side-on bridging thiocarbonyl, [HB(pz) 3 ]-
(CO) 3 W(7| : -CS)Mo(CO) : (indcnyl). has been reported. Draw the structure of this complex.
Why is the bridging CS group in this complex called side-on instead of semibridging? (See
Doyle. R. A.; Daniels. L. M.; Angelici. R. J.: Slone. F. G. A../. Am. Chem. Soc. 1989,
III. 4995-4997.)
15.62 Amino alkenes such as H : NCH 2 CH 3 CH : CH=CH ; are catalytically converted to five-
membered heterocycles by [(Me.sCjHjhLaHfe. Give the steps of the reaction sequence
and incorporate them into a Tolman catalytic cycle. Present arguments against formation
of an La-alkene bond. (See Gagne, M. R.; Marks. T. J. J. Am. Chem. Soc. 1989. III.
4108-4109.)
80
15- Organometallic Chemistry
One form of manganocene, Cp 2 Mn, consists of infinite chains of CpMn fragments
bridged by cyclopentadienyl rings in the solid. 111 Upon being warmed to 159 °C the
color changes from brown to orange and the product is isomorphous with ferrocene. It
has been formulated as ionic, high spin d 5 , Mn 2 2Cp~. Evidence for ionic bonding is
of three types: (1) Manganocene reacts instantaneously with ironll chloride in tetra-
hydrofuran to form ferrocene and is hydrolyzed immediately by water; (2) the dis¬
sociation energy (Table 15.9) is closer to that of magnesium cyclopentadienide than to
those of the other transition metal metallocenes; and (3) the magnetic moment of
manganocene is 5.86 BM, corresponding to five unpaired electrons. All these data are
consistent with a d 5 Mn 2+ ion. The evidence is not unequivocal, however. Other
metallocenes such as chromocene react with iron(II) chloride to yield ferrocene as
well, although admittedly not so rapidly. Assignment of “ionic bonding” to man¬
ganocene on the basis of its being high spin Mn(ll) is reminiscent of similar assign¬
ments to other high spin complexes. The presence of high spin manganese and a lower
dissociation energy for manganocene indicate the absence of strong covalent bonding
(i.e., a small ligand field stabilization energy from the eig^lg'Tg configuration) but do
not prevent the possibility of some covalent bonding. In any event, the stability of the
half-filled subshell, d 5 configuration is responsible for the anomaly of manganocene.
This anomalous behavior disappears when methyl groups replace the hydrogen atoms
of the ring (page 673). The Cp Mn complex is low spin (one unpaired electron) and
has chemistry similar to that of other metallocenes.
Synthesis of The first metallocene was discovered by accident independently by two groups. In one
Cyclopentadienyl group, an attempt was made to synthesize fulvalcne by oxidation of the cyclopenta-
Compounds llicn y' Grignard reagent:"2
O-^-x-^O-O
Table 15.9
Dissociation energies, D, of
Metallocene
D(M—Cp) (kJ mol' 1 ) 6
metallocenes 0
Cp 2 V
369
Cp 2 Cr
279
Cp,Mn
212
Cp 2 Fe
302
Cp,Co
283
Cp 2 Ni
258
“ Chipperfield, J. R.; Sneyd, J. C. R.;
Websler. D. E. J. Organomet. Chem.
1979, 178, 177-189.
These are average dissociation
energies based on one-half of the
energy required for the reaction.
Cp 2 M(g) - M(g) + 2Cp(g).
111 Blunder, W.; Weiss, E. Z. Naturforsch. B: Anorg. Chem., Org. Chem. 1978. 33, 1235-1237.
"2 Kealy, T. J.; Pauson. P. L. Nature 1951. 168. 1039-1040.
Metallocenes
681
The synthesis of fulvalene was unsuccessful, 113 but a stable orange compound was
isolated which was subsequently characterized and named ferrocene. The iron(lll) is
first reduced by the Grignard reagent to iron(Il) which then reacts to form ferrocene:
pe 3 + pe2+ Fe( ^ 5Hs)2 (, 5 . 72 )
Sodium and thallium cyclopentadienide provide more versatile syntheses of ferrocene
and other metallocenes.
MX, + 2NaC 5 Hj -» M(tj 5 -C s H 5 ), + 2NaX (15.73)
Thallium cyclopentadienide is often used when the reducing power of the sodium salt
is too great. The limited solubility of the thallium complex in organic solvents and its
toxicity are its chief disadvantages.
Ferrocene, the most stable of the metallocenes, may be synthesized by methods
not available for most of the others. Iron will react directly with cyclopenladiene 114 at
high temperatures (ferrocene was discovered independently by this method 115 ). Use
of amines facilitates the removal of the acidic hydrogen of cyclopentadiene. allowing
the synthesis to be accomplished at lower temperatures:
Fe + 2(R 3 NH)C1 -► FeCl 2 + 2R 3 N + H 2 (15.74)
FeCl 2 + 2C 5 H 6 + 2R 3 N -f Fe(r/ 5 -C 5 H 5 ) 2 + 2(R 3 NH)CI (15.75)
Net reaction: Fe + 2C 5 H 6 -> Fefa 5 -C 5 H 5 ) 2 + H 2 (15.76)
Arene Complexes Although the cyclopentadienyl group is the best-known aromatic ligand, there are
several others of considerable importance. None leads to complexes as stable as the
most stable metallocenes, however, and the chemistry of the complexes that do form
is more severely limited.
Benzene and substituted benzenes normally act as six-electron donors, although
dihapto and tetrahapto complexes are also known. Dibenzenechromium was prepared
early in this century but was not characterized until 1954. It was first synthesized via a
Grignard synthesis. When PhMgBr reacts with CrCl 3 in dielhylether solvent, a mono-
hapto complex [CrPh 3 (Et,0) 3 ] forms which rearranges, 116 presumably by a free radi¬
cal reaction to give, among other products, [Cr(Tj 6 -C 6 H 6 ) 2 ] + . This ion can be reduced
to the neutral metallocene.
CrCI, , . Nji.S.O.
PhMgBr — [CrPh,(Et,0) 3 ] -> [Cr(V’-C 6 H 6 ),] + Cr(7, h -C h H ( ,),
(15.77)
Arene complexes are synthesized more cleanly by the Fischer-Hafner adaptation of
the Friedel-Crafts reaction. In this reaction aluminum is used to reduce the metal salt
115 Fulvalene is highly unstable and exists only as a transient species; however, it is stabilized when
coordinated to a variety of transition metal atoms. For a leading reference, sec Moulton, R. D.;
Bard, A. J. Organometallics 1988, 7, 351-357.
"•'Typically instead of starting with cyclopentadiene, one begins with its Diels-AIdcr product,
dicyclopentadiene. This compound cracks when heated to give cyclopentadiene.
115 Miller, S. A.; Tebboth. J. A.; Tremain. J. F. J. Chem. Sue. 1952. 632-635.
116 Please note that whereas monohapto and pentahapto Cp are isomers, phenyl and benzene ligands
differ by a hydrogen atom.
734
15 • Organometallic Chemistry
15.31 Isocyanides. RNC. and nitriles. RCN. are both well-known ligands. Complexes such as
Cr(CNR),, exist, but not complexes such as Cr(NCR)„. Compare the bonding charac¬
teristics of these two ligands and account for the relative stabilities of these two com¬
plexes.
15.32 Predict the product of the following reaction:
[(r, 5 -C 5 H 5 )Mo(CO) : (CN) : )' + 2MeI -
15.33 Nitriles arc almost invariably bound to metals through nitrogen and exist in a linear
arrangement. An exception to this is (Ty'-CsHsHPPh-dlrf^-NCQ.HjCI). Draw the struc¬
ture for this complex. Does it obey the 18-electron rule? (See Chetcuti, P. A.; Knobler.
C. B.: Hawthorne. M. F. Organomeiallics 1988, 7. 650-660.)
15.34 Thermodynamically, c/.s-Mo(CO) 2 (PPh 2 CH 2 CH 2 PPh 2 ) 2 is more stable than irans-Mo
(CO) 2 (PPh 2 CH 2 CH 2 PPh 2 ) 2 , but [rra/i.v-Mo(CO) 2 (PPh 2 CH 2 CH 2 PPh 2 ) 2 ] is more stable than
[m-MofCOJjCPPhiCHjCHjPPhi)})'. Provide an explanation for the relative stabilities of
these isomers. (See Kuchynka, D. J.; Kochi. J. K. Organomeiallics 1989. 8. 677-686.)
15.35 The conversion of/«c-Mo(bpy)(CO)jlP(OMe),] to /<7c-Mo(bpy)(CO);i[P(OMe) 2 F] by ac¬
tion of BF 3 -OEt 2 is thought to take place by passing through an intermediate containing a
phosphenium ligand. Draw structures of the reactant, the intermediate, and the product.
(See Nakazawa, H.; Ohta, M.; Miyoshi. K.; Voneda. H. Organomeiallics 1989. 8,
638-644.)
15.36 Metal-phosphine complexes are ubiquitous in organometallic chemistry. There are also
meiallated phosphoranes (L„M—PR.,), phosphides (L„M — PR 2 ), and phosphinidenes
(L„M — PR). Give specific examples of each.
15.37 When CO becomes coordinated to BH, its stretching frequency increases, but when CO
becomes coordinated to Ni(CO), its stretching frequency decreases. Explain.
15.38 Does the CO stretching frequency increase or decrease when
a. L of L„M—CO becomes more electron withdrawing?
b. CO of L„M—CO becomes coordinated to a Lewis acid. A. to become L„M—CO—A?
15.39 Suppose you were directing a research student who came to you and stated that he or she
had isolated a compound that was either (a) or (b):
Ph : Ph,
Me = S k I MC = S k LcH,
• N —— Ir PPh.CH^ :.N —lr^
Mo .Si j, MCjSi^ I, PPh ;
^ Ph, Ph,
(a) tbi
What experiments (other than X-ray analysis) might you suggest to clarify the situation?
(See Fryzuk. M. D.; Joshi. K. Organomeiallics 1989. 8, 722-726.)
15.40 Using Fig. 15.33. predict the number of unpaired electrons in
a. (Cp 2 Tir b. [Cp 2 Crr c. Cp 2 Cr
15.41 Organometallic chemistry seems especially prone to the development of descriptive
words and phrases. Although much of this language is sometimes considered jargon and is
not allowed to enter the formal literature, it is common in oral usage. See if you can define
the following terms:
a. open-faced sandwich h. supersandwich
c. club sandwich d. ringwhizzer
e. molecular broad jump f. piano stool molecules
Problems
735
g. A-frame complex h. triple-decker sandwich
i. butterfly cluster
15.42 The synthesis of a neutral homoleptic uranium complex, URj, has finally been
achieved. 219 Why do you think R = [CH(SiMe 2 ) 3 ]~ was chosen for this synthesis?
15.43 The complexes of Fig. 15.38 are commercially available. Starting with complexes which
contain only carbonyl ligands, suggest syntheses for the following:
a. CpV(CO) 4 b. CpFe(CO) 2 I
c. CpNiNO d. Cp 2 Mo 2 (CO) ft
15.44 The average bond dissociation energy for ferrocene is large (302 kj mor', Table 15.9) as
you might expect, but it is even larger for vanadocene (369 kJ mol" 1 ). Can you suggest
any reasons that vanadocene is more reactive even though it is more stable?
15.45 The reaction of Fe(CC »5 with (Et 4 N)(OH) in methanol at -78 °C gives a mixture of
[Fe(C0) 4 (C0 2 H))~ and [Fe(C0) 4 (C0 2 Me)]~as shown by infrared spectroscopy. When
LiOH is used, however, the sole product is [Fe(C0) 4 (C0 2 Me)|“. Explain. (See Lee,
S. W.: Tucker. W. D.; Richmond, M. G. Inorg. Chem. 1990, 29. 3053-3056.)
15.46 Predict the product of the following nucleophilic addition:
15.47 Discuss the diflercnce between formulating Cr(C„H„) : as dibenzenechromium (a) and as
bisthexahaptocyclohexairienelchromium (b):
tal (b)
Do these represent different chemical species or merely semantic variations? Suggest
experimental methods that might distinguish between (a) and (b).
15.48 When one mole of (Ph,P) 2 Pl(C 2 H 4 ) is treated with two moles of BF,. ethylene is quan-
titively released and the BF 3 is completely consumed. The product consists of a single
compound, which is monomeric in dichloromelhane solution. Formulate the product and
describe the bonding in it. (See Fishwick, M.; Noth. H.: Petz. W.: Wallbridge, M. G. H.
Inorg. Chem. 1976. 15. 490-492.)
219 Van Der Sluys. W. G.: Bums, C. J.; Sattelberger, A. P. Organomeiallics 1989. S. 855-857.
682
15 Organomelollic Chemistry
(o a lower oxidation state and with the assistance of AlClj, benzene is coordinated to
chromium:
3CrCl 3 + 2A1 + A1C1 3 + 6C 6 H 6 - 3 [Cr(r; 6 -C 6 H 6 ) 2 ][AICl 4 ] (15.78)
The cation can be reduced to dibenzenechromium with sodium dithionite. Na 2 S,0 4 .
The dibenzene complexes of Cr, Mo. and W are all air sensitive and those ot Mo and
W are especially so. Dibenzenechromium is a black solid that melts at 280
The success of the Fischer-Hafner method depends upon the particular arene and
on the skill of the experimentalist. Several decades ago Timms introduced a chem¬
ically innovative approach for making a variety of organometallic complexes." t is
based on the premise that if you wish to synthesize a zerovalent complex, it is logical
to begin with metal atoms rather than with salts that must be reduced. Highly active
metal atoms can be created by vaporizing metals under vacuum with resistive heating
and then condensing the atoms on low-temperature walls of a specially designed
vessel (Fig. I5.4I)." S A ligand, benzene for example, can be introduced into the
vessel where it will react with the metal atoms to form a complex.
M(s) -♦ M(g) — M(t, 6 -C„H 6 ) 2 (15 ' 79)
Fig. 15.41 An inexpensive apparatus constructed from a modified rotating evaporator and
used to vaporize metals for condensation with ligands. [Adapted from Markle. R. J-t
Pcttijohn. T. M.: Lagowski. J. J. Organome,allies 1985, 4. 1529-1531. Used wuh perm.ss.on.]
Modified
sample
Solium ll.i>k
Timms. P. L.: Turney. T. W. Adv. Organomet. Chan. 1977. 15. 53.
"» Klabunde. K. J. Acc. Chen,. Res. 1975.«. 393-399. Markle. R. J.; Pettijohn. T. M.; Lagowski. J. J.
Organometallics 1985,4. 1529-1531.
Metallocenes
683
The rings of the dibenzene chromium, molybdenum, and tungsten complexes are
eclipsed and have a small rotational barrier. Unlike ferrocene, these complexes have
labile rings which can be displaced:
Cycloheptatriene
and Tropylium
Complexes
The benzene rings can be removed completely by reaction with a more active ligand:
Cr(r? 6 -C 6 H 6 ) 2 + 6PF 3 -> Cr(PF 3 ) 6 + 2C fi H 6 (15.81)
There are a number of heteroatom six-membered aromatic rings which are analo¬
gous to benzene in that they can donate six electrons to a metal. These include
phosphabenzene. borabenzene anion, borazine. and arsabenzene shown in Fig.
15.42."'
Cycloheptatriene is a six-electron donor that can form complexes similar to those of
benzene but differing in the localization of the n electrons in C 7 H„. The alternation in
bond length in the free cycloheptatriene is retained in the complexes (Fig. 15.43).
Furthermore, in C 7 H x Mo(CO) 3 the double bonds are located trans to the carbonyl
groups, providing an essentially octahedral environment for the metal atom.
Fig. 15.42 Complexes of phosphabenzene. borabenzene. borazine. and arsabenzene. These
six-membered rings are analogous to benzene, i.c.. they are six-electron donors and are
aromatic.
mv Elschenbroich. C.: Nowolny. M.: Metz. 13.; Massa. W.: Graulich. J.; Biehler. K.: Sauer, W.
Angew. Clwm. Ini. Ed. Engl. 1991. 30. 547-550.
732
15 • Organometallic Chemistry
(CH 3 C=CCH 3 )J, for example, contains p,. if, and X. Draw the structure of this complex
and explain the meaning of each of prefix.
15.13 Substitution reactions of polynuclear metal carbonyls with tertiary phosphines often
induce the formation of bridging carbonyls. Provide an explanation.
15.14 Ligands are described with a variety of adjectives, some not used in this chapter. Using
other sources, try to determine what is meant by
a. an ancillary ligand
b. an amphoteric ligand
c. a sterically noninnocent ligand
15.15 Suggest reasonable syntheses for
a. Mo(V-C„TW(CO)j b. (OC) 4 CoMn(CO)j
c. (i 7 5 -C,H.,)Re(CO) : C(OMe)Ph d. Pt(Ph 3 P) 2 (H 2 C=CH 2 )
e. <7, 5 -C 5 Hj),ReH f. (r, 5 -C 5 H 5 ) : Fe 2 (CO) 4
15.16 Which one of the following complexes do you think you would most likely be able to
isolate: W 2 (CO),„. [W,(CO) lo f', [W 2 (CO) l0 F, [W 2 (CO)„J\ or [W 2 (CO),„r?
15.17 The mechanism of CO replacement by L in Mn 2 (CO), 0 was controversial for a number of
years. Some believed that substitution first occurred by homolysis of the Mn—Mn bond
while others interpreted their data in favor of CO dissociation. Crossover experiments
with Mn 3 ( l3 CO)| 0 and Mn 2 ( ,3 CO)| 0 2 ' 6 and also with IH5 Re 2 (CO)| 0 and ls7 Re 2 (CO) m 217
support the second interpretation. Discuss this mechanistic problem and explain how the
isotopic experiments rule out homolysis of the Mn—Mn bond.
15.18 Determine the number and symmetry designations of the infrared-active C—0 stretching
modes in the following derivatives of Mo(CO) h .
a. Mo(CO)jPRj b. c«-Mo(CO) 4 (PR,) 2
c. r«mj-Mo(CO) 4 (PRj )2 d. foe- Mo(CO).,(PR,),
e. mcr- Mo(CO)j(PRj ) 2
15.19 Sulfur dioxide may function as a ligand by forming a metal-sulfur bond ora metal-oxygen
bond. Draw Lewis structures for these interactions. It is possible to distinguish between
the two modes of ligation with S—O stretching frequencies. Explain. (See Green. L. M.:
Meek. D. W. Organometallics 1989, 8, 659-666: Wojcicki. A. Adv. Organomel. Client.
1974. 12. 32.)
15.2(1 Identify a reagent(s) that will effect each of the following transformations:
Fe(s)-♦ Fe(CO),-» [Fc(CO) 4 1 3 ~ - [MeFe(CO) 4 |' -♦
lMeC(0)Fe(C0)j(PPh.,)r - MeC(0)H
15.21 Transition metal hydrides participate in a variety of reactions involving loss of hydrogen.
In some instances it may be lost as H~ (hydride transfer), in others H loss occurs (proton
transfer), and in yet others hydrogen is lost as H'lhydrogen atom transfer). Complete the
following reactions and categorize each as one of the three types.
Mn
P ' PPh,.
Ph, H
[Ph,C|[AsF 6 ]
(Sec Kuchynka. D. J.; Kochi. J. K. Organometallics 1989, 8. 677-686.)
216 Coville. N. J.; Stolzenberg, A. M.; Muettcrties. E. L. J. Am. Chem. Soc. 1983. 105. 2499-2500.
2,7 Stolzenberg. A. M.; Muetterties, E. L. J. Am. Chem. Soc. 1983. 105. 822-827.
Problems
733
b. OsH 4 (PMc 2 Ph)j + KH -
(See Huffman. J. C.; Green M. A.: Kaiser, S. L.; Caullon. K. G. J. Am. Chem. Soc.
5111-5115.)
c. HW(CO)j(tj s -C s HU 4- Ph,CCI -
(See Hoffman, N. W.; Brown. T. L. Inorg. Chem. 1978. 17. 613-617.)
15.22 Explain why W(CO) 5 [C(OMe)Me] is far more stable than W(COK(CMe 2 ).
15.23 When CpFe(CO) 2 (CF 3 ) reacts with BF 3 , the product is [CpFc(CO) 2 (CF 2 )||BF 4 |. but when
it reacts with BCI 3 , CpFe(CO) 2 (CCI 3 ) forms. Propose a mechanism for these reactions.
What is the driving force for each reaction? (Sec Crespi. A. M.; Shriver. D. F. Orgaiw-
metallics 1985. 4. 1830-1835.)
15.24 Isolobal considerations suggest that it should be possible to replace the CH fragments of
letrahedrane. C.,H.,, with ML,, fragments. What ML„ fragments would you suggest? Draw
structures of the complexes.
15.25 What metal fragment, ML,,, might you suggest for creating L„MsML„. analogous to
HC=CH?
15.26 The carbidoheptarhenium carbonyl cluster. [Re 7 C(CO) 2 |]'’ - . is analogous to CiHT and
forms a set of bicapped octahedral complexes of formulation [RejCICOhtML,,] 2- . Sug¬
gest some [ML,,) 7 fragments which should be capable of capping [Re 7 C(CO) 2 ,]'~. (See
Henly, T. J.: Shapley. J. R.: Rheingold. A. L,; Geib, S. J. Organometallics 1988. 7,
441-448.)
15.27 A typical C—P—C bond angle in a PMe, complex is 102°, but the cone angle for this
ligand is given as 118° in Table 15.10. Explain.
15.28 Isomerization of c7.v-Mo(CO) 4 (PPh 3 ) 2 to /r«/t.r-Mo(CO) 4 (PPh 3 ) 2 proceeds by a dissociative
process while isomerization of i7.v-Mo(CO) 4 (PBuj) : proceeds by an intramolecular non-
dissociative process. How do you explain these results? (See Darensbourg. D. J. Inarg.
Chem. 1979. 18. 14-17.)
15.29 Bond lengths and angles have been determined by gas phase electron diffraction for two
similar complexes. Fe(CO) 4 (C 2 H 4 ) and Fe(CO) 4 (C 2 F 4 ). 2,, ‘ Identify which sot of data (I or
II) belongs to each complex.
15.30 Nitric oxide loses an electron rather easily to form a nitrosonium cation yet its tendency to
undergo reduction often causes difficulties when it is used as a reagent for synthesis of
nitrosyl carbonyl complexes. How can you rationalize the ease of oxidation of NO with its
ease of reduction?
;l Deeming. A. J. In Comprehensive Organometallic Chemistry, Wilkinson. G.: Stone. F. G. A.:
Abel, E. A.. Eds.; Pergamon: Oxford. 1982; Vol. 4, p 388.
684
15 Organometallic Chemistry
II
O
Fig. 15.43 Molecular
structure of tricarbonyl-
cycloheptatrienemolyb-
denum(O) illustrating
alternation in bond lengths
and location of double
bonds trans to the carbonyl
ligands. Bond lengths are in
picometers.
Cycloheptatriene complexes can be oxidized (hydride ion abstraction) to form
cycloheptatrienyl (sometimes called tropylium) complexes: 120
The tropylium ring is planar with equal C—C distances. Like benzene and the
cyclopentadienide anion, the tropylium cation is an aromatic, six-clectron species.
Cyclooctatetraene
and Cyclobutadiene
Complexes
In accord with the Huckel rule of 4/t + 2 electrons, both cyclobutadiene and cyclooc-
latetraene (cot) are nonaromatic. Cyclooctatetraene contains alternating bond lengths
and has a tub-shaped conformation:
This nonplanar molecule becomes planar on reaction with an active metal to produce
the cyclooctatetraenide anion (cot) 2- . 121
2K + ChH, - 2K + + (C 8 H 8 ) 2 - (15.83)
42 « Tropylium salts, such as (C 7 H 7 )Br. are not used directly for organometallic synthesis because of
their oxidizing power.
121 Dry KiC«Hk explodes violently upon contact with air. Gilbert. T. M.; Ryan. R. R.: Sattelberger,
A. P. Organometallics 1988. 7. 2514-2518.
Metallocenes
685
Fig. 15.44 Structure of
uranocene showing the two
eclipsed cyclooctatetraenyl
rings.
The cot dianion. likeCp - , is aromatic but has ten electrons in eight 7r orbitals. When it
is allowed to react with tetrapositive actinides, such as U 4+ , Np J + , Th J+ , Pa J+ , and
Pu J+ , a neutral metallocene results. 122
UC1 4 + 2cot 2 “ -<• U(cot) 2 (15.84)
The uranium compound was the first metallocene of cot 2- to be synthesized. 123 A
sandwich structure was proposed for it and later verified (Fig. 15.44). 124 By analogy to
ferrocene it was called uranocene. The extent of covalency and the 5/orbital contribu¬
tion to the bonding in these complexes has long been debated. Recent photoelectron
spectroscopy results’ 23 and ab initio quantum mechanical calculations' 2 - support
significant covalency and/orbital involvement. A molecular orbital scheme suitable
for uranocene is shown in Fig. 15.45. Missing from the diagram are the low-lying
nondegenerate a >K and a 2ll orbitals, which house four of the 22 valence electrons.
These are primarily ligand orbitals and do not contribute significantly to metal-ligand
bonding. The remaining 18 electrons fill the e, ff . e lir e 2x . and e 2ll levels and leave the
<? 3 „ level half filled (two unpaired electrons). The e 2ll LGO donates electron density to
the e 2ll (5f xyz , 5/. (r i_ v :,) orbitals and the e 2f! LGO donates electron density to the e-,,
(6cJ x 2_ y 2, 6d xy ) orbitals of the uranium atom. Neptunocene (5/ 3 ) and plutonocene (5/ J )
have one and no unpaired electrons, respectively, as predicted from Fig. 15.45.
The lanthanides might be expected to form similar complexes, using 4/orbitals in
place of the 5/ orbitals of the actinides. They do so, generally forming [Ln(cot),P
complexes (D„ (/ ), although Ln(cot)CI species are also known.' 27 These complexes are
viewed as essentially ionic with minimal 4/covalent participation.
Cyclobutadiene, C 4 H 4 , eluded synthesis for many years because of its reactivity.
Although simple Huckel theory predicts a square molecule with Iwo unpaired elec¬
trons, infrared studies carried out at low temperatures (it dimerizes at 35 K) have
shown that it is rectangular with alternating single and double bonds. In addition, all of
its electrons are paired. More sophisticated MO treatments arc in accord with these
results. In 1965. a great deal of excitement was generated when a complex containing
coordinated C 4 H 4 was synthesized. 128
Cl
”, + Fe : (CO)„
H
Thus cyclobutadiene, which was nonexistent at the time, was shown to be stabilized
by complexation. Oxidation of the complex liberated free cyclobutadiene which was
trapped by ethyl propynoate to give a cycloadduct. The experiments established that
122 Marks. T. J. Prog. Inorg. Chem. 1979, 25. 224-333.
123 Strcilwicscr. A.. Jr.; Mailer-Wcstcrhofl', U. J. Am. Chem. Soc. 1968. 90. 7364.
124 Zalkin, A.; Raymond. K. N. J. Am. Cliem. Soc. 1969, 91. 5667-5668.
125 Brennan. J. G.; Green J. C.; Redfcrn C. M. J. Am. Chem. Soc. 1989, III, 2373-2377.
12,1 Chang. A. H. H.; Pilzcr. R. M. J. Am. Chem. Soc. 1989. III. 2500-2507. Burstcn. B. E.; Burns,
C. J. Comments Inorg. Chem. 1989. 9. 61-100.
127 Marks. T. J. Prog. Inorg. Chem. 1978, 24. 51-107.
128 Emerson. G. F.; Watts. L.; Pettit. R. J. Am. Chem. Soc. 1965. 87. 131-133.
O
730 15-
Conclusion
Organometollic Chemistry
Fig. 15.58 Energy barriers (in kJ mol -1 ) for three separate types of rotational motion
involving a bridging vinyl carbyne ligand. The values were obtained by line shape analysis of
variable temperature proton NMR spectra. [From Casey. C. P.; Konings. M. S.; Marder.
S. R.; Takezawa, Y. J. Organomet. Chem. 1988. 358, 347-361. Reproduced with permission.1
and eight lines in the cyclopentadienyl region for the two ABCD sets of ring protons
(all ring protons are inequivalent due to the overall chirality of the complex). Coales¬
cence to give a single methyl peak and two peaks for the ring protons occurs as the
temperature is raised to 70 °C, a result that is ascribed to Ru—Ru bond mobility. A
rearrangement pathway involving a D Zll intermediate has been proposed. 211
A final example of stereochemical nonrigidity, illustrating the application of
sophisticated NMR line shape analysis for obtaining rotational barriers, is provided by
a cyclopentadienyl diiron complex containing a bridging vinylcarbyne (Fig. 15.58).-'-
Barriers for three distinct types of rotational motion involving the bridging ligand have
been obtained: (I) rotation of the entire vinylcarbyne ligand (44.3 kJ mor 1 ), (2) rota¬
tion of the dimethylamino group (45.6 kJ mor 1 ). and (3) rotation of the aryl group
(54.3 kJ mor 1 ). 2 ' 2
Problems
Organometallic chemistry, as its name implies, has links to both organic chemistry and
inorganic chemistry. The ties to organic chemistry, at first in compounds that were
more or less laboratory curiosities, have burgeoned into indispensable components of
the petroleum chemicals industry. Transition metal organometallic chemistry is
largely an extension of the coordination chemistry of strongly bonded ligands, but
cluster chemistry anticipates the polyhedral chemistry of boranes with extreme de-
localization in molecular orbitals (Chapter 16). The concept of isolobal fragments also
ties the two disciplines together, with fragments as seemingly disparate as CH ? C and
(OC),Co being isolobal. The entire field continues to expand as seen by the American
Chemical Society’s decision to establish a new journal [Organometallics) in 1982 and
by the appearance of at least eight textbooks on organometallic chemistry since then.
It will be especially interesting to observe the impact of organometallic chemistry on
production of pharmaceuticals, semiconductors, and ceramic materials, as well as its
continued connection to organic feedstocks and energy production.
311 Houser, E. J.; Amarasckera, J.; Raiichluss. T. 13.: Wilson. S. R. J. Am. Chan. Sue. 1991. 113,
7440-7442.
313 Casey. C. P.; Konings. M. S.; Marder. S. R.: Takezawa. Y. J. Organomet. Cltem. 1988. 358.
347-361.
Problems 731
15.1 A cyclopentadienyl phosphorus compound. (Me^COIBu'lPCI. has been identified spec¬
troscopically. 313 Is this an organometallic compound? Explain your answer. By electron
counting, draw conclusions about the hapticity of the cyclopentadienyl group. How many
electrons does phosphorus need in order to obey the effective atomic number rule?
15.2 Although 18 valence electrons are found in IFelH^OJh] 2 . the effective atomic number
rule is violated. Explain.
15.3 Formulate neutral. 18-electron complexes of chromium which contain only
a. cyclopentadienyl and nitrosyl ligands
b. cyclopentadienyl. carbonyl, and nitrosyl ligands
15.4 Postulate geometries for bis-, tris-. and tetrakis(triphenylphosphine)palladium. Which of
these obey the 18-electron rule? What is the geometry of |PdCU :_ ? Why is it different
from that of the tetrakis(triphenylphosphine) complex?
15.5 Predict the metal-metal bond order for neutral complexes having the formula |(OC) 4 M(/j.-
PR : );M(CO) a ] when M = V. Cr. and Mn.
15.6 Complexes of CS are known, but homoleptic (all ligands the same) examples such as
Fe(CS) 5 have not been synthesized. Why do you think efforts to prepare these complexes
have failed?
15.7 a. Complexes containing Ph-PCH^CH^PPh;. sometimes known as diphos. are abundant.
Using the 18-electron rule as guide, draw structures of
i. (OC),W(PPh-CH : CH 2 PPh : )
ii. (OC)aW(PPh;CH : CH-PPh ; )
iii. (OC) 4 W(PPh;CH;CH : PPh : );
iv. (OC), u W : IPPh : CH;CH : PPh ; )
v. (OC) K W ; (PPh : CH : CH : PPh : )-
b. It has not been possible to synthesize (OC)|nW ; (PPh : CH : PPh : ). but (OC)|»W r
(PMe-CH’PMe;) is known. Provide an explanation.
15.8 Postulate a monometallic manganese complex that obeys the 18-electron rule and contains
only the ligands
a. hydrogen, acyl, cyclobutadiene, and cyclopentadienyl
h. thiolate. alkene. benzene, and phosphine
c. alkyl, ourhene. carbyne. and nitrosyl
(I. cyclohcpiatrienyl. dinilrogen. and isocyanide
15.9 Trimctallie complexes containing phosphide bridges are well known. 314 Assume that the
18-electron rule is obeyed and postulate a structure for Mn(/x-PH-).
15.11) Dixneuf and coworkcrs have prepared Fe(rj : -Ph : PCH=C(R)S(PPhi)(CO)
which obeys the 18-electron rule. 31 ' Confirm that it does and draw its structure. What is
the oxidation state of iron in this compound?
15.11 Both Me,P and P(OMe), have been widely used as ligands. Compare their reactivities
toward O- and H : 0. Which of these two ligands is generally considered to be the better
v-donor? jr-acceptor? Why?
15.12 Prefixes abound in organometallic nomenclature. The complex Fe,(CO).,[Mr'r--l--
3 ' 3 Cowley, A. H.: Mchrotra. S. K. J. Am. Chem. Soc. 1983. 105. 2074-2075.
314 Powell. J.: Sawyer. J. F.t Shiraiian, M. Organometallics 1989. 8. 577-583.
315 Samb. A.; Dcmerscman. B.; Dixneuf. P. H.; Mealli. C. Organometallics 1988. 7. 26-33.
686
1 5 • Organometallic Chemistry
I
1
I
Fig. 15.45 (a) Molecular
orbital diagram for
actinocene, An(i)' 1 -C(iH s ) : .
(An = actinide metal).
(b) Interaction of/orbital
of the metal with a ligand
e lu orbital, (c) Interaction
of d orbital of the metal
with a ligand e 2j , orbital.
[From Brennan. J. G.;
Green. J. C.; Redfern.
C. M. J. Am. Chem. Soc.
1989, III . 2373-2377. Used
with permission.]
o o
cyclobutadiene could exist, however briefly, and led eventually to its low temperature
isolation.
Reactions of
Organometallic
Complexes
Substitution
Reactions in
Carbonyl
Complexes 129
The earliest methods of replacing one or more carbonyl ligands from a complex relied
on brute force (heat or light) to break the M—CO bond. 130 The idea was that once the
gaseous CO had dissociated, it would escape easily from solution and thus have
minimal chances of recombining with the metal. The departure of CO from a complex
leaves a vacant coordination site and in general an unstable metal fragment which is
electron deficient. The fragment can then react with a nucleophile such as a phos¬
phine, R,P. to produce a substituted metal carbonyl. The entire dissociative process
can be described as follows: 131
Albers. M. O.: Covillc. N. J. Coord. Chan. Rev. 1984. 5J. 227-259.
13,1 The use of ultrasound as an energy source for breaking the M—CO bond also has been investi¬
gated. Suslick, K. S. Adv. Organomet. Cliem. 1986. 25. 73-119.
131 !_• ;• f^tes 0 f substitution reactions for carbonyl complexes are nearly independent of the
' ■■?. of incoming ligand, which supports a dissociative mechanism.
" •" -■ i
Reactions of Organometallic Complexes
687
L„M—CO L„M + CO (15.86)
L„M + R 3 P -» L„M—PR 3 (15.87)
Of course some complexes lose CO more readily than others. For example, it is rather
easy to displace all four CO groups of Ni(CO) 4 with L (L = R,P) in stepwise fashion:
Ni(C0) 4 —^ Ni(CO) 3 L —Ni(CO) 2 L 2 —±-» Ni(CO)L 3 —NiL 4 (15.88)
The task is much more difficult for Fe(CO) s , which has a large energy of activation for
substitution and requires high temperatures. At these temperatures side reactions are
significant and yields of substituted products are low:
Fe(CO)j —^ Fe(CO) 4 L —U Fe(CO) 3 L 2 (15.89)
Notice that Eq. 15.89 shows only two CO ligands being displaced. Each time CO is
replaced by R,P, the complex becomes more electron rich and the remaining CO
groups receive more 7r electron density. This means that in general the M—CO bond
strength increases and CO becomes more resistant to dissociation. Of course the steric
requirements of the phosphine may limit the degree of substitution as well (see cone
angles, page 688).
The thermal and photolytic reactions described above usually give a mixture of
products and therefore are not as popular as they once were. Reactions have now been
developed which give a good yield of the particular product of interest. For example, if
one wishes to prepare W(CO),PR,. one would not heal W(CO) ( , with PR, at high
temperatures or irradiate the reaction mixture with ultraviolet light because both
of these methods would give mixtures of W(CO),PR,. c7.v-W(CO) 4 (PR,) 2 .
/r«/t.v-W(CO) 4 (PRj) 2 . and perhaps facial or meridional trisubstituled products as well.
A preferable approach would be to first prepare W(CO),thf by photolysis of W(CO) 6
in thf and then, without isolation of this complex, displace the thf with l he phosphine
in a subsequent room-temperature reaction:
W(CO) h + thf —^ W(CO) 5 thf + CO (15.90)
W(CO),thf + PR, -- W(CO),PR, + thf (15.91)
Tetrahydrofuran is a sufficiently poor ligand that it seldom displaces more than one
CO group in the photolysis step and thus the reaction yields the monosubslitutcd
product exclusively.
Another twist is to add Me,NO which attacks the carbon of a coordinated CO,
leading to eventual loss of CO : and formation of an unstable trimethylamine complex.
The phosphine easily displaces the amine to form the final product:
(OC) 5 WCO + ONMe,-» (CO),WNMe, + CO, <OC),WPR, + Me,N (15.92)
The preparation of pure Fe(CO) 4 PR, and /raHS-Fe(CO)j(PR,) 2 have long been
frustrating because thermal and photolytic methods give mixtures of products which
are not easy to separate. The monosubstituled complex may now be prepared by
several routes, one of which involves cobalt(II) chloride as a catalyst:
Fe(CO) s + PR, — Fe(CO) 4 PR, + CO (15.93)
The exact role of the catalyst in this reaction is unknown. 132 Alternatively, the
reaction may be catalyzed by polynuclear iron anions, such as [Fe 2 (CO) 8 ]“ _ or
Albers. M. O.; Covillc. J. H.; Ashworth. T. V.; Singleton. E. J. Organomet. Chem. 1981. 2/7,
385-390.
728
15-Organometallic Chemistry
Observed Calculated
la) <>>
Fig. 15.55 Observed (a) and simulated (b) variable temperature proton NMR spectra of
Os,(CO)|,|lMr , )’-CH 1 CH 3 C=CCH : CH,). The spectrum at -65 °C. characterisiic of two
equivalent ABX, spin systems, is produced when the movement of the alkync on the metal
surface is slowed sufficiently that it is fixed in one position on the NMR time scale. At
+ 82 °C the alkyne molecule is moving rapidly on the metal surface giving an averaged
spectrum in which all methylene protons arc equivalent. [From Rosenberg. E.; Bracker-
Novak. J.; Gellert, R. W.; Aimc, S.; Gobetto. R.; Osclla. D. J. Organomet. Client. 1989,
163-185. Reproduced with permission.)
methylene protons, which are diastereotopic and therefore nonequivalent). As the
temperature is increased, the multiplet signals broaden and collapse into a single
quartet as a result of the free movement of the alkyne. By simulating the observed
spectra (Fig. 15.55b). it was possible to determine rate constants and an energy of
activation (60.4 ± 2 kJ mol -1 ) for the process.
A different type of dynamic process involving a polynuclear metal system has
been identified in [(CH,C J H 4 ) 4 Ru 4 S 4 ] :+ . which has a distorted cubane-like structure
with three Ru—Ru bonds in the crystalline state (Fig. 15.56). By following its methyl
and ring proton resonances over a temperature range from +70 to -43 °C
(Fig, 15.57), the complex is shown to undergo a dynamic process involving the
metal-metal bonds. At the low-temperature limit, the spectrum contains features
predicted for the static structure: two lines of equal intensity for the methyl protons
688
15-Organometallic Chemistry
[Fe 3 (CO),,] 2 J 33 A successful method for producing the trans disubstituted complex
is to create [HFe(CO) 4 ]~ and allow it to react with PR-, in refluxing I-butanol: 134
T
Fe(CO) s Fe(C0) 4 C - [HFe(CO)J - + CO (15.94)
L X H_
[HFe(CO) 4 ] _ + 2PR 3 + BuOH-► //wij-Fe(CO) 3 (PR 3 ), + CO + H-, + BuO~
(15.95)
The counterion is quite important to the outcome of this reaction. Ion pairs, which
form between alkali metal ions and the complex, induce CO lability which aids in the
substitution process. 135 Large charge-delocalized cations such as PPN + are much less
effective in forming ion pairs. The substitution process shown in Eq. 15.95 occurs
readily when the counterion is Na + but fails when it is PPN + . Another good example
of this effect can be seen by comparing Na[Co(CO)J and PPN[Co(CO)J. 136 The
former is readily substituted by l3 CO, phosphines, and phosphites, while the latter is
inert with respect to substitution.
Ligand Cone Angles It is intuitively obvious that the space occupied by a ligand can influence not only how
many will fit around a metal atom but also the effectiveness of overlap between metal
and ligand orbitals. If ligands are too crowded, repulsion between them forces
metal-ligand distances to increase, weakening the metal-ligand bond, and enhancing
the overall lability of the complex. Though steric effects have been long discussed in
the literature, only after Tolman introduced the concept of the cone angle in 1970 did
inorganic chemists have a quantitative means of expressing these ideas. 137 It should be
clear that for a ligand such as Ph 3 P, the C—P—C bond angle does not give a
satisfactory measure of its space requirements. It is the volume of space taken up by
the three phenyl groups on phosphorus that is crucial and a simple bond angle does not
reflect this information. From a study of zerovalent nickel complexes, NiL, (L =
phosphine or phosphite), Tolman observed that the binding ability of a ligand de¬
pended strongly on its steric needs. He concluded that he needed a way of measuring
the cone created by the phosphorus ligand. Not all good ideas require an expensive
piece of equipment. With a block of wood, a nail, a space-filling model of each ligand,
and an attachment for measuring angles, he was able to do what was required. The
measuring device is shown in Fig. 15.46a and the cone angle, defined as the apex angle
of a right cyclindrical cone, is outlined in Fig. 15.46b. The distance from the center of
the phosphorus atom to the tip of the cone, 228 pm. was chosen on the basis of the
nickel-phosphorus bond distance. Cone angle values for a number of ligands are
shown in Table 15.10.
Ligands such as CO are sterically undemanding and as many as required by the
18-electron rule can easily arrange themselves around a metal atom. When bulky
115 Butts, S. B.: Shriver. D. F. J. Organonwl. Chan. 1979. 169. 191-197.
134 Keitcr, R. L.: Keitcr. E. A.: Hcckcr. K. H.; Boecker. C. A. Organon,elallics 1988. 7. 2466-2-169.
135 Ash. C. E.; Dclord. T.: Simmons, D.; Darcnsbourg, M. Y. Orgunon,elallics 1986. 5. 17-25.
I3A Ungviiry, F.: Wojcicki. A. J. Am. Chan. Soc. 1987. 1119. 6848-6849.
137 Tolman. C. A. Chan. Rev. 1977, 77.313-348. For some new developments in this area, see Brown,
T. L. Inorg. Chem. 1992 .31. 1286-1294.
Reactions of Organometallic Complexes
689
Fig. 15.46 Device lor
determining cone angles.
[From Tolman, C. A.
Chem. Rev. 1977, 77.
313-348. Used with
permission.]
phosphines are part of the coordination sphere, however, it becomes increasingly
difficult to satisfy the 18-electron rule simply because there is not enough space for the
required number of ligands. It is not difficult to replace three CO groups of Mo(CO)„
with three phosphines having modest cone angles. When PPri,, which has a cone angle
of 160°. is chosen, however, it is only possible to replace three CO groups with two
phosphines. The result is that a 16-electron species is sterically stabilized. We saw
earlier (Eq. 15.31). that H : may be added to Mo(CO) 3 (PPi j 3 ), to give an 18-eleclron
complex, and this illustrates an important point, namely that complexes containing
bulky ligands may have vacant coordination sites, but they are available to only (he
smallest donors. Often this idea can be used to advantage in designing syntheses for
organometallic complexes.
A further example of large ligands blocking out smaller ones to give low coordina¬
tion numbers is provided by the work of Otsuka. '3 Phosphines with small cone angles
such as PEt 3 (132°) allow isolation of Pt(PEt,) 4 which obeys the 18-electron rule, but
ligands with cone angles greater than 160° make it possible to isolate complexes with
just two ligands, e.g.. Pt(PBu',L.'- w
It is also observed that bulky ligands avoid being cis to one another. 1411 In a series
ol octahedral W(CO) 4 L, (L = phosphine) complexes, the percentage of trans isomer
isolated from a substitution reaction increases as the cone angle of the phosphine
increases (Table 15.11). Electronic factors favor cis arrangements, but steric factors
become dominant as the cone angle increases. Attempts to synthesize Fe(CO) 3 (PBu$) 2
have failed because the large cone angle of the phosphine ligand prevents coordination
of more than one of them.
Oxidative Addition One of the most important classes of reactions in organometallic chemistry is med
and Reductive oxidative addition. In these reactions a coordinatively unsaturated complex in a
Elimination relatively low oxidation state undergoes a formal oxidation by two units (loss of two
lw Otsuka. S. J. Organonwl. Chem. 1980. 200. 191.
m Thc reccnl synthesis of(Ph,P) 2 Pd (Urata. H.; Suzuki. H.: Moro-oka. Y.; Ikawa T. J. Organomet.
Chan. 1989. 364. 235-244) shows that a coordination number of 2 may be induced wilh somewhat
smaller cone angles (Ph,P: 145").
140 Shaw. B. L. J. Organomet. Chem. 1980. 200. 307-318.
726
1 5 • Organometallic Chemistry
Fig. 15.52 Proton NMR
spectra of (V-CsHsWrj 5 -
C 5 H 5 ) ; Ti from -27 to
+62 °C. The separate
resonances observed at low
temperature for the two
types of Cp rings gradually
broaden and collapse,
giving a single line at 62 °C.
[From Cotton, F. A. In
Dynamic Nuclear Magnetic
Resonance Spectroscopy ;
Jackman. L. M., Cotton.
F. A., Eds.; Academic:
New York. 1975; Chapter
10. Used with permission.]
ppm
single resonance for the average environment of the twelve hydrogens as the iron
presumably migrates over the allene ir system.
In organometallic clusters, ligands frequently appear to move over the surface
of the metal framework. The triosmium complex, Os 3 (CO), 0 (/z 3 -t} 2 -CH 3 CH ; C=C-
CH,CH 3 ), in Fig. 15.54 provides an example in which the alkyne moves over the
Fig. 15.53 Migration of
iron in tctracarbonyltetra-
methylallenciron(O).
Stereochemically Nonrigid Molecules
727
It
Fig. 15.54 Fluxional behavior of OsjtCOluitMr^'CHjCHsCsCCHiCHi). In Stage I
terminal carbonyl groups are exchanging. In Stage 2 the alkyne ligand moves about the
(riosmium surface and as it does so. the bridging carbonyl shills to remain trans to it. [From
Rosenberg. E.: Bracker-Novak. J.: Gellert. R. W.; Aime, S.; Gobelto. R.: Osella, D. J.
Organomet. Chem. 1989. 365, 163-185. Reproduced with permission.]
face of the metal triangle at elevated temperatures but remains fixed at low tem¬
peratures. 210 Variable temperature proton and carbon NMR studies show that the
alkyne movement is accompanied by exchange of the bridge and terminal carbonyl
ligands. In addition, axial and radial carbonyl groups are undergoing exchange. The
variable temperature proton NMR spectra are shown in Fig. 15.55a. At -65°C we see
a spectrum characteristic of two equivalent ABX 3 spin systems (the AB spins are the
3111 Rosenberg. E.: Bracker-Novak. J.: Gellert, R, W.; Aime. S.: Gobetto, R.: Osella. D. J. Organo¬
met. Client. 1989. 365, 163-185.
Table 15.10
Reactions of Organometallic Complexes
691
Cone angles, II, for various
ligands, L°
L
8
L
8
P(OCH,) 3 CEt
101
P(p-CF 3 C„H 4 ) 3 145
PPhH,
106
PBzlPh 2
152
P(OMe) 3
107
PCyPh,
153
P(OEt) 3
109
P(/-Pr) 3
160
P(0CH 2 CH,C1) 3
110
PCy,Ph
162
PMe(OEt),
112
PBzl,
165
P(CH,CH,CN).H
117
P(//i-tol) 3
165
PMe 3
118
PCy,
170
P(OMe),Ph
120
P(/-Bu),Ph
170
P(OEt),Ph
121
P(/-Bu) 3
182
PMe„Ph
122
NH 3
94
PPh,H
126
NH 2 Me
106
P(OPh),
128
NH,Et
106
P(O-i-Pr),
130
NH,Ph
ill
P(OMe)Ph,
132
NH,Cy
115
PEt 3
132
NHMe,
119
PBu,
132
Piperidine
121
P(CH,CH,CN),
132
NHEL
125
P(OEt)Ph,
133
NHCy 2
133
PMe(/-Buj 2
135
NHPh,
136
PEuPh
136
NEt 3
150
PMePh,
136
NPh,
166
PPh(CHXHXN),
136
NBz1 3
210
PEtPh,
140
H
75
P(0-o-tol) 3
141
Me
90
PPh,(CH,CH 2 CN)
141
F
92
P(OCy) 3
141
CO
95
P(/-Bu)j
143
Cl
102
PCy,H
143
Et
102
PPh 3
145
Br
105
P(p-tol) 3
145
Ph
105
P(p-MeOC 6 H 4 ) 3
145
I
107
P(p-CIC„H 4 ) 3
145
/-Pr
114
P(p-FC A H 4 ) 3
145
/-Bu
126
P(p-Mc,NC fl H 4 ) 3
145
" Rahman. M. M,; Liu. H-Y: Eriks. K.: Prock, A.;
Gicring, W. P. OrganometaUies 1989, 8. 1-7. Tolman,
C. A. Chart. Rev. 1977. 77. 313-348. Sdigson. A. L.
Troglcr, W. C. J. Am. Chan. Soc. 1991. 113, 2520-2527.
Table 15.11
Cone angles and ratios of
% cis
% trans
cis/trons isomers in
L
Cone angle
isomer
isomer
W(CO),,L 2 complexes
isolated from the reaction: 0
PPh,H
PPhXt
126
140
100
38
0
62
W(CO) 4 (tmpa) + 2L -
W(CO) 4 L 2
PPh 2 (/-Bu)
P(p-tol) 3
P(o-tol) 3
157
145
194
21
21
0
79
79
100
" Lukchart, C. M. Fundamental Transition Metal
Organometallic Chemistry. Brooks/Colc: Monterey. CA.
1985. for all but PPhjH. which is from the authors'
laboratories, (tmpa = Me 2 N(CH 2 ) 3 NMe 2 )
electrons) and at the same time increases its coordination number by two. An example
is the reaction of Vaska's complex with molecular hydrogen (Eq. 15.28). In this
instance, iridium is oxidized from +1 to +3 and at the same time the coordination
number of the complex increases from 4 to 6. The reverse reaction, in which H-, is lost
from the complex, involves reduction of iridium from +3 to +1 and a decrease in
coordination number from 6 to 4. This process is called reductive elimination. This
specific example of oxidative addition/reductive elimination may be generalized as
follows:
ML, + X—Y
oxidative
addition
reductive
elimination
L,M
/ A
\y
(15.96)
In order for oxidative addition to occur, vacant coordination sites must be available. A
six-coordinate complex is not a good candidate unless it loses ligands during the
course of the reaction making available a site for interaction. A further requirement is
that suitable orbitals be available for bond formation. An 18-electron complex such as
[FelCOlJ 2- has only four ligands but addition of X—Y would require the use of
antibonding orbitals, which of course is not energetically favorable.
Mechanisms for oxidative additions vary according to the nature of X—Y. If
X—Y is nonpolar, as in the case of H 2 , a concerted reaction leading to a three-
centered transition state is most likely.
(15.97)
Nonclassical complexes of dihydrogen (page 645) may be thought of as complexes in
an arrested transition state and their existence provides strong support fora concerted
reaction mechanism. Dioxygen, another nonpolar molecule, also adds reversibly to
Vaska's complex, but in this case the X—Y bond is not completely broken. The bond
order of O, is reduced from two to essentially one.
OC
Cf
L
O
o
L
(15.98)
If X—Y is an electrophilic polar molecule such as CH,I. oxidative addition
reactions tend to proceed by S N 2 mechanisms involving two-electron transfer (Eq.
15.99) or via radical, one-electron transfer mechanisms (Eq. 15.100).
H s H
L„VI:+CH 3 I -► (L„M---'C"-I| -► IL„M —CH,|I -► L„M(CH,)I
i (15.99)
L„M + CH 3 I -► L n IM' + ‘CHj -► L„M(CH 3 )I
(15.100)
Other factors besides a vacant coordination site are important in determining the
tendency for a complex to undergo oxidative addition. The ease of oxidation (usually
724
15-Orgonometallic Chemistry
Stereochemically Nonrigid Molecules
725
0
(b)
Energy
Position
Fig. 15.51 Energy and position of the hydrogen atom in (a) the symmetric HF : system and
(b) an unsymmetric O-H - O system. In (a) the average position of the hydrogen atom is
midway between the fluoride ions. In (b) the positional sketch on the left represents the
average position of the hydrogen in the left potential well and the dotted sketch represents
the average position for the other potential well. The height of the energy barrier is
qualitative and not meant to represent any particular system.
If the barrier between the two configurations is thermally accessible, the system is
fluxional.
In choosing a method for studying a particular molecular structure or dynamic
process, it is essential that the time scale of the method be appropriate for the lifetimes
of the species involved (Table 15.12). Diffraction methods have lime scales of 10 18 to
IO -20 s, which is rapid compared to the frequency of molecular motions. Obviously
this time period relates to a single interaction between the diffracted wave and the
molecule, and the actual experiment must take a considerably longer period of time for
collection of the data. The resulting structure is thus a weighted average of all the
molecular configurations present, and this is commonly encountered as thermal ellip¬
soids of motion (Chapter 3). Consider again the molecular fragment in Fig. 15.5lb. If
the higher vibrational levels were occupied, the hydrogen atom would appear to be
smeared over the entire vibrational amplitude. If only the lowest vibrational level is
occupied, the hydrogen will show up as a half atom at one minimum and a half atom at
the other.
Like diffraction, spectroscopic methods using ultraviolet, visible, or infrared light
are generally much faster than molecular vibrations or interconversions, and the
Toblo 15.12
Time scales for structural
techniques
Technique
Approximate time scale (s)
Electron diffraction
|0 -2O
Neutron diffraction
IO -18
X-ray diffraction
IO -18
Ultraviolet
10 13
Visible
IO -14
Infrared-Raman
10 13
Electron spin resonance"
10 4 to 10 8
Nuclear magnetic resonance"
10 1 to 10 9
Quadrupole resonance"
10 ' to IO -8
Mossbauer (iron)
10 7
Molecular beam
10 6
Stop-flow kinetics
10 3 to 10 2
Experimental separation of isomers
10 2
" Time scale depends on chemical system under investigation.
spectra reflect weighted averages of the species present (cf. the broad absorption
bands in the visible spectra of transition metal complexes. Chapter 11). The remaining
spectroscopic methods are slower, and the time period of the interaction may be
compared to that of the lifetimes of individual molecular configurations. The nature of
interconversions between configurations can be studied by such techniques.
Nuclear magnetic resonance techniques have proved to be particularly valuable in
the study of fluxional molecules.- 07 The most common experimental procedure in¬
volves analyzing the changes in NMR line shapes that occur with variations in
temperature. The simplest dynamic process is one involving two molecular configura¬
tions that have equal probability. If the interconversion process between them is slow
on the NMR time scale, as might be the case at a low temperature, two separate sets of
equal-intensity resonances, one for each configuration, will be observed in the spec¬
trum. If we can raise the temperature of the sample sufficiently so that the process
becomes rapid on the NMR time scale, the result will be a single set of spectral lines,
and they will appear at the midpoint of the two sets observed at lower temperature. At
this high-temperature limit, the molecule is undergoing changes so rapidly that NMR
cannot distinguish the two separate molecular configurations, only an average. As an
example of a system of this type, consider (V-C 5 H 5 ) 2 (77 5 -C 5 H 5 ),Ti.- (W The crystal
structure of the molecule shows two monohapto and two pentahapto cyclopentadienyl
rings. At 62 °C, the ‘H NMR spectrum consists of a single line (Fig. 15.52). consistent
with a dynamic process that renders the four ligands equivalent. As the temperature is
lowered, the signal broadens and gradually splits into two lines which sharpen into
equal-intensity singlets at -27 °C. At this point, the process which inlerconverts
mono- and pentahapto ligands is occurring slowly enough that both configurations are
observable in the spectrum. However, even at this temperature, the monohapto rings
are involved in a dynamic process that averages the signals for the three types of ring
protons. Instead of separate resonances in a 1:2:2 ratio, only a single fairly sharp line
is observed due to rapid migration of the metal from one site to another within each
ring. When the temperature is decreased further, this process also is slowed so that the
peak for the monohapto ligands broadens and then collapses, eventually reemerging in
the expected pattern for three nonequivalent ring protons at -80 °C. Similar NMR
behavior has been observed for (C 5 H,),Sc. showing that at 30 °C a tr-tr exchange
occurs between two rj 5 and one rj' ring. Separate peaks in a 2:1 ratio for the two types
of rings are observed at -30 °C. with individual resonances for the monohapto ring
protons becoming apparent as the temperature is decreased further.- 00
Fluxional processes involving acyclic unsaturated hydrocarbon ligands, such as
allyls and allenes. are also common. For the tt complex formed between tetramethyl-
allene and tetracarbonyliron (Fig. 15.53), the proton magnetic resonance spectrum
below - 60 °C shows three peaks in the ratio 1:1:2. representing the three cis hydrogen
atoms, three trans hydrogen atoms, and six hydrogen atoms in a plane perpendicular
to the carbon-iron bond. With an increase in temperature, the spectrum collapses to a
207 Sundslrom. J. Dynamic NMR Spectroscopy. Academic: New York. 1982. Mann. B. E. In Compre¬
hensive Organomelallic Chemistry: Wilkinson, G.: Stone, F. G. A.; Abel. E. W.. Eds.: Pcrgamon:
Oxford. 1982; Vol. 3. Chapter 20. Orrcll. K. G.: Sik. V. Ann. Rep. NMR Spectroscopy 1987. IV.
79-173. Dynamic Nuclear Magnetic Resonance Spectroscopy ; Jackman. L. M.; Colton. F. A..
Eds.; Academic: New York. 1975.
208 Cotton. F. A. In Dynamic Nuclear Magnetic Resonance Spectroscopy: Jackman. L. M.: Cotton.
F. A.. Eds.: Academic: New York. 1975: Chapter 10.
209 Bougcard. P.: Mancini. M.; Sayer. B, G.; McGlinchey. M. J. Inorg. Chem. 1985, 24. 93-95.
692
1 5 • Organometallic Chemistry
j to </' With formal loss of two electrons), the relative stability of coord.nat.on
number 4 compared to 5 or 6. and the strength of new bonds created <M-X and
M —Y) relative to the bond broken (X-Y) all must be considered. Oxidation of the
metal is easier for electron-rich systems than for electron-poor ones; hence oxidative
addition is more likely for low-valent metals. The ease of oxidation increases from top
to bottom within a triad [Co(I) < Rh(l) < W)] and the tendency toward five-
coordination decreases from left to right across a transition series [Os(O) > Ir(l) >
1 Cleavage of the H —H bond by transition metal complexes suggests that similar
reactions may be possible with C-H and C-C bonds. In fact it has been known for a
number of years that coordinated triphenylphosph.ne can undergo intramolecular
This reaction is also called orthometallation because it is the ortho carbon of the
phenyl group that participates. Although most common with phenyl groups of phos¬
phines or phosphites, examples involving alkyl groups are also known.
The recognition that H 2 is capable of nonclassical coordination to a metal, which
may be regarded as an interaction involving incomplete rupture of the H H bond,
was preceded by the discovery that C-H bonds can interact with metal atoms
without bond cleavage. It was first observed that in the structures of some complexes
a hydrogen atom attached to a phenyl ring was abnormally close to the metal atom.
This was viewed at the time as a form of hydrogen bonding. A subsequent neutron
diffraction study of [Fe(n 3 -C l ,H l3 ){P(OMe)j} 3 ] + revealed a strong C-H • • ■ Fe inter¬
action. 142
Studies of a number of polymetallic complexes also provided evidence for C-H
coordination. Particularly notable was a beautiful NMR study by Calvert and Shapley
of a triosmium complex, which was shown to exist as an equilibrium mixture of two
structures, one of which included a C—H Os linkage. 141
mi Bailey. N. A.: Jenkins, i. M.; Mason. R.; Shaw. B. L. J. Chen,. Soc. Chen. Common. 1965.
237-238. La Placa, S. J.; Ibers. J. A. l/wrg. Chrm. 1965. 4. 778-783.
.« Brown. R. K.; Williams. J. M.: Schultz. A. J.: Stucky. C. D.: l«el. S. D.; Harlow. R. L. J. Am.
Chen. Soc. 1980. 102. 981-987.
M) Calvert. R. B.; Shapley. J. R.: Schultz. A. J.: Williams. J. M.; Suib. S. L.; Stucky. G. D. J. An,.
Chen,. Soc. 1978. 100. 6240-62-11.
Reactions of Organometallic Complexes
693
Of added significance in this work was the fact that a methyl C—H bond is broken,
which has positive implications for activation of alkanes by transition metals
(page 694).
As more and more examples of a carbon-hydrogen bond acting as a ligand were
confirmed, the term agostic was coined to describe the interaction. Agostic. as
originally proposed, referred to “covalent interactions between carbon-hydrogen
groups and transition metal centers in organometallic compounds, in which a hydro¬
gen atom is covalently bonded simultaneously to both a carbon atom and to a
transition metal atom.” 144 These interactions, like those of B —H — B and
W—H—W (page 646). are currently described as three-center, two-electron bonds
(see Chapter 14).
As with all other forms of ligand-metal coordination, an agostic interaction
requires that a vacant coordination site be available and donation of C—H electron
density to a metal contributes toward satisfaction of the 18-electron rule. The cation
|Fe(Tj 5 -C,jH, 3 ){P(OMe).,}-,] ' shown above, is an 18-electron system if one counts the
C—H bonding pair of electrons. Structural tip-offs indicating the existence of an
agostic interaction include C—H bond lengthening by 5-10% relative to a nonbridging
C—H bond and M —H distances longer than a terminal M — H distance by 10-20%.
The term agostic has now been extended to generally include M — H — Y systems,
in which Y may be B. N. Si. Cl. or F, as well as C. The structure of Ti(BH 4 ),(PMe 3 ) 2
reveals one bidentate BHj - and two monodentate BHj ions, each of which possesses
one B—H bond attached to titanium in a side-on. agostic manner. 145
The preceding examples of C—H coordination are all intramolecular in nature. Of
greater interest, especially with commercial goals in mind, are intermolecular reac¬
tions involving hydrocarbons. Functionalizing the alkane constituents of petroleum
under mild conditions is a major challenge in organometallic chemistry and one that
144 Brookhart. M.; Green. M. L. H.; Wong, L.-L. Prog. Inorg. Chen. 1988. 36. 1-124. The term
agoslic is drived from ihc Greek word iiyorrr6 (and translates "to clasp, to draw towards, to hold lo
oneself."
145 Jensen. J. A.: Wilson. S. R.: Girolami. G. S. J. An,. Chen. Soc. 1988. 110. 4977-4982.
722
1 5 • Organometallic Chemistry
where RX is a halogen, alkyl halide, or hydrogen. In other words oxidative addition
leads to the formation of a metal-metal bond. Unlike the usual oxidative addition, two
metal centers are involved simultaneously. The reverse reaction, reductive elimina¬
tion, may also occur.
We can follow our suppositions regarding the formation of an M—M bond upon
oxidative addition by watching the change in bond length:
[Pt 2 (pop) 4 ] [Br Pt(pop) 4 Pt Br] 4
£/ Pl _ Pl = 292 pm r/p t _p, = 272 pm
(15.195)
The greater the overlap of the metal d.i orbitals, the greater the repulsion in the
reduced. </ 8 state, and the stronger the metal-metal bond in the oxidized, cl h state. We
should expect oxidative addition to be least favored at very long metal-metal dis¬
tances, and much more strongly favored at those distances corresponding to the
normal metal-metal single bond distance, and this is what is found. Conversely, the
longer distances favor the reverse, reductive elimination reaction. Indeed, with some
bimetallic complexes reversible addition/elimination of H 2 can be accomplished, not
with molecular dihydrogen, but with a hydrogen donor molecule such as a secondary
alcohol:
Me 2 CHOH + “M 2 ” -♦ Me 2 C=0 + H—M—M—H (15.196)
Although the reaction in Eq. 15.196 is potentially catalytic, true catalysis has rarely
been achieved. However, when "M 2 " is [Pt 2 (pop) 4 ] 4- , solar-driven catalytic oxida¬
tion has been observed. 201 In monometallic Pt(ll) complexes the HOMO-LUMO
transition is in the UV region with perhaps a tail into the violet. The complexes are
thus white or perhaps yellowish. The interaction present in bimetallic complexes, as
shown in Fig. 15.50, narrows the HOMO-LUMO gap such that this transition may be
shifted into the visible region. The electron may thus be excited by solar radiation:
[PUpoplJ 4 - — [Pt 2 (pop) 4 ] 4 - (J5 19?)
The excited diplatinum complex is now predisposed to react further. The incipient
Pt—Pt bond is partially formed and addition of hydrogen is enhanced: 202
[Pt 2 (pop) 4 ] 4_ + Me 2 CHOH -» [H—Pt(pop) 4 Pt—H] 4 ~ + Me 2 C=0
(15.198)
2111 Roundhill. D. M. J. Am. Chem. Soc. 1985. 107. 4354-1356.
2,12 Sweeney. R. J.; Harvey. E. L.; Gray H. B. Coord. Chem. Rev. 1990. 105 . 23-34.
Stereochemically Nonrigid Molecules
723
Reductive elimination of dihydrogen may then take place to complete the catalytic
cycle:
[H Pt(pop) 4 Pt H] 4 “ -> [Pt 2 (pop) 4 ] 4 " + H 2 (15.199)
In addition to converting secondary alcohols into ketones, the platinum catalyst also
converts toluene and other benzyl hydrocarbons into dimers.
Similarities between [Ru(bpy)-J 2+ (discussed in Chapter 13) and [Pt 2 (pop) 4 ] 4- are
apparent. Reactive excited states are produced in each when it is subjected to visible
light. The excited state ruthenium cation, [Ru(bpy) 3 ] 2+ , can catalytically convert
water to hydrogen and oxygen. The excited state platinum anion, (Pt,(pop)J 4_ , can
catalytically convert secondary alcohols to hydrogen and ketones. An important
difference, however, is that the ruthenium excited state species results from the
transfer of an electron from the metal to a bpy ligand, while in the platinum excited
state species the two unpaired electrons are metal centered. As a consequence,
platinum reactions can occur by inner sphere mechanisms (an axial coordination site is
available), a mode of reaction not readily available to the 18-electron ruthenium
complex. 203
Chemists are prone to think in terms of static molecular structures. This viewpoint is
initiated by stick-and-ball models and reinforced by regular inspection of molecular
structures determined by “instantaneous" methods such as X-ray diffraction. 204 In
fact many molecules are stereochemically nonrigid. If the rearrangement leads to
configurations which are chemically equivalent, we say that the molecule is fluxion-
a/- 205 On the other hand, if the rearrangement gives rise to chemically distinguishable
molecules, we simply say that isomerism has occurred. Fluxional molecules differ
from other stereochemically nonrigid molecules in possessing more than a single
configuration representing an energy minimum. Several such minima may be present
and may be accessible with ordinary thermal energies. As a very simple example,
consider symmetric and unsymmetrie hydrogen bonds (Chapter 8). If in the symmetric
HFJ ion the fluoride ions are considered relatively immobile with respect to the much
lighter hydrogen, the motion of the latter can be considered, to a first approximation,
as a vibration in a potential well with an average position midway between the
fluorides (Fig. 15.5la). In contrast, unsymmetrie hydrogen bonds possess two poten¬
tial wells in which the hydrogen can vibrate (Fig. 15.5lb), occasionally being suf¬
ficiently excited thermally to jump to the other well. 20 ' In such a system the hydrogen
would be found in one potential well or another by rapid methods such as diffraction.
2U -' Roundhill. D. M.; Gray, H. B.; Che. C-M. Acc. Chem. Res. 1989. 22. 55-61.
2,1J Obviously u struciurc is nol determined instantly with X-ray diffraction techniques. Rather the
structure is instantaneous in the sense that the time period over which the diffracted wave interacts
with the electrons of the molecule is infinitesimally short with respect to the frequency of atomic
motions.
2,15 This definition includes all stereochemically nonrigid molecules having identical energy minima
and configurations at those minima including molecules such as the pscudorotational PFj and
Fe(CO)j (Chapter 6).
2,K ’ For simplicity's sake, both potential wells arc shown to be the same depth. In general for a
hydrogen bond such as —O—H N this is not true. All of the configurations of a truly fluxional
system are energetically equivalent, however, and have equivalent potential wells.
Stereochemically
Nonrigid Molecules
694
1 5 • Organometallic Chemistry
has received a great deal of attention during the past decade. 146 As you know from
courses in organic chemistry, alkanes are quite unreactive and their inertness is
attributed to high C—H bond energies (typically about 400 kJ mol). A plausible
sequence for functionalizing hydrocarbons begins with coordination ot an alkane:
L„M + R—C—H
R
L,M
reagents -'
LM + hydrocarbon derivatives
(15.103)
(15.104)
(15.105)
Eq. 15.103 shows formation of a three-center, two-electron bond, presumed to be a
first step in C—H activation. The next step (Eq. 15.104) completes the oxidative
addition of the alkane to the metal. The process may be thought of as insertion of the
metal into the C—H bond. Further reaction with a functionalizing reagent gives the
desired organic product (Eq. 15.105). An ultimate goal is to accomplish this process
catalytically.
In 1982. Janowicz and Bergman at Berkeley 147 and Hoyano and Graham at
Alberta 1411 reported the first stable alkane intermolecular oxidative addition products.
The Berkeley group photolyzed (Tj 5 -Me J C J )(Me 3 P)lrH 2 with the loss of H,. while the
Alberta group photolyzed (yy-MejCjJIriCOK with the loss ol CO to give highly
reactive iridium intermediates which cleave C—H bonds in alkanes:
O
Bergman. R. G. Science 1984. 223. 902-908. Crabtree. R. H. Client. Rev. 1985.85. 245-269. Shilov.
A. E. The Activation of Saturated Hydrocarbons by Transition Metal Complexes'. Rcidel: Dor¬
drecht. 1984. Stoutland. P. O.; Bergman. R. G.; Nolan. S. P.: HolT. C. D. Polyhedron 1988. 7.
1429-1440. Jones. W. D.: Feher. F. J. J. Am. Cltem. Soc. 1984. 106. 1650-1663.
147 Janowicz. A. H.: Bergman. R. G. J. Ant. Chem. Soc. 1982. 104, 352-354.
14,1 Hoyano, J. K.; Graham. W. A. G. J. Am. Client. Soc. 1982. 104. 3723-3725.
Insertion and
Elimination
Reactions of Organometallic Complexes 695
Even methane, under the right experimental conditions, can be activated by an
organometallic complex. This is particularly important not only because methane is
one of the most abundant hydrocarbons but also because its C—H bond is the
strongest among the alkanes (434 kJ mol -1 ). Success in activating this molecule
suggests that all alkane activation barriers are surmountable. The first homogeneous
(solution) reaction between methane and an organometallic complex was reported in
1983 and described the exchange of i3 CHj for CH V 149
(7j 5 -C s Me 5 ) ; Lu(CH,) + l3 CH 4 -►(r, 5 -C 5 Me 5 ) ; Lu(' : 'CH 1 ) + CH 4 (15.108)
Somewhat similar reactions with (7j'-C 5 Me 5 ) 2 ThR : have also been investigated. 1 ' 0 It
is likely that neither of these two reactions proceeds by oxidative addition because
lutetium and thorium are in high oxidation stales in the reactants.
The activation of methane in solution by an organometallic complex presents
some experimental difficulties because any solvent that is likely to be chosen will be
more reactive than methane. In addition, insolubility of the complex in liquid methane
may preclude reaction with the pure hydrocarbon. These problems were overcome in
the case of the reaction of CH 4 with the iridium complex of Eq. 15.106 by taking
advantage of the fact that the desired hydrido methyl complex is thermodynamically
more stable than other hydrido alkyl complexes. The methyl complex was produced
by first creating a hydrido cyclohexyl complex and then allowing it to react with
methane. 151
■'(T) 5 -CjMe 5 )( PMejJlr” C " -->
, / H IK. , / H
(T, , -C 5 Me,)(PMe n )lr (ij 5 -C,Me,)(PMe,)lr (15.109)
X C h H l( • 1 N CH,
The reactant "(ry-C 5 Me 3 )(PMe 3 )Ir" in Eq. 15.109 represents the presumed reactive
intermediate that forms when (T) 5 -Me 5 C 5 )(PMej)IrH 2 is photolyzed.
Another complex to which methane, as well as other hydrocarbons, will ox¬
idatively add is [bis(dicyclohexylphosphino)ethane)platinum(0).
The very electron-rich, two-coordinate. 14-electron platinum reactant was not spec¬
troscopically observed, but its existence was inferred from the reaction products. 152
Oxidative addition reactions lead to products that appear to have had a metal atom
inserted into a bond, but the term insertion has generally been reserved for reactions
which do not involve changes in metal oxidation stale. These reactions are enor¬
mously important in catalytic cycles (see page 708). Special emphasis in this section
Watson. P. L. J. Am. Chem. Soc. 1983, 105. 6491.
,5H Kendrick. C. M.: Marks. T. J. J. Am. Chem. Soc. 1986. 108. 425-437.
151 Bergman. R. G. Science 1984. 223. 902-908.
,5: Hackctl. M.; Whitesides. G. M. J. Am. Chem. Soc. 1988. 110. 1449-1462. Photolysis of
FeH 2 (dppe) 2 in a xenon/methane solution also leads to methane activation. Field. L. D.; George.
A. V.; Messerle. B. A. J. Chem. Soc. Client. Comm. 1991. 1339-1341.
720
15 • Organometallic Chemistry
(where -» represents the lone pair on the ligating carbon atom). 198 Note that the
system must be linear throughout the C—N=C—M system. We thus have lour
r 2 C—N=C groups on each metal spaced by (CH 2 )„ bridges which may vary in length
depending upon n.
A second type of bidentate ligand holding the two metal atoms close together is
pyrophosphite. It forms from diphosphorous ("pyrophosphorous ) acid and. because
of the P—O—P linkage, is often abbreviated “pop”:
O
4 '0—P —O
H
Coordination of the pyrophosphite occurs through the phosphorus, which is preterred
over the harder oxygen by the soft Pt(II) ion. Thus there is a tautomeric relationship
analogous to that in phosphorous acid:
OH
P —OH (15.192)
I
OH
The formulation of the tetrakis(/x-diphosphito)diplatinate(II) ion is thus [Pt ; (pop)J J
(Fig. 15.49).i" The metal in these complexes is ideally a cl 8 species such as Ir^ or
Pt 2 \ The van der Waals radii of these metals (Table 8.1) are about 350 pm. so we
should expect metal-metal interaction to begin at about this distance and become
more significant as the distance decreases. What will this interaction consist ot)
Figure 15.50 illustrates the energy levels of a d 8 square planar complex together with
Fig. 15.49 Crystal
structure of K 4 (Pt:(fj.-
P 2 0 ? H,)J. [From
Roundhill, D. M.; Gray.
H. B.; Che. C-M. Acc.
Chem. lies. 1989. 22.
55-61. Reproduced with
permission.)
iw Complexes of these ligands have been extensively studied by Gray and coworkers. Sec. for
example. Maverick. A. W.; Smith. T. P.; Maverick. E. F.; Gray. H. B. Inorg. Chan. 1987. 26.
4336-43-11.
i->9 This ion was first obtained from the reaction of potassium letrachloroplalinatc(ll) with molten
phosphorous acid. The ion gives an intense green luminescence when subjected to light. Zipp.
A P. Coord. Chan. Rev. 1988. 84, 47-83. The ion undergoes oxidation in the presence of air and
chloride ion (Che. C.-M.t Butler. L. G.; Grunthancr. P. J.; Gray, H. B. Inorg. Chan. 1985. 24.
4662-4665).
the interaction of two filled d.i and two emply p z orbitals on adjacent metal atoms. 20,1
As the distance between the two metal atoms decreases, the orbitals along the ; axis
will interact more strongly with each other. For both the d.: orbitals and the p.
orbitals, this will give rise to a bonding and an anlibonding interaction. Since the </.:
orbital gives rise to the HOMO and the p. orbital to the LUMO. this will decrease the
HOMO-LUMO gap with interesting consequences (see below). More important for
the present discussion, the increase in energy for the HOMO will destabilize the
molecule and make it more reactive. Another way of looking at this enhanced reac¬
tivity is to say that if the two electrons were removed from the antibonding «, 1(
HOMO, there would be an M —M single bond, stabilizing the molecule. One way to
remove the two HOMO electrons without costing energy—indeed, by providing
further energy that actually drives the reaction—is to form two new bonds with them:
RX
(15.193)
:o " Roundhill. D. M.; Gray. H. B.; Che. C.-M. Acc. Chem. Res. 1989, 22. 55-61. Isci. H.; Mason.
W. R. Inorg. Chem. 1985 , 24, 1761-1765. Fordycc, W. A.: Brummcr, J. G.; Crosby, G. A.
J. Am. Chan. Soc. 1981, 103. 7061-7064.
696
15 • Organometallic Chemistry
At first glance, these two processes may seem to be indistinguishable. However,
careful consideration of the results of the infrared study will reveal otherwise. The
reaction of l3 CO with CH,Mn(CO) 5 yields c«-(CH 3 CO)Mn( ,, CO)(CO) 4 as the ex¬
clusive product. None of the tagged CO is found in the acetyl group, which establishes
that the reaction is not an intermolecular insertion, i.e., no reaction occurs between
gaseous CO and the M—C bond. Furthermore, none of the l3 CO ends up trans to the
acetyl group. This is an extremely important observation because it establishes that
the CO ligands in the product do not scramble to give a statistical distribution. In other
words, the outcome of the reaction is kinetically. not thermodynamically, controlled.
Although this is an important result, it does not allow a firm distinction to be made
between CO insertion and methyl migration since the product would be the same for
either mechanism. However, additional mechanistic information can be gained by
studying the reverse reaction, i.e.. decarbonylation of ciA--(CHjCO)Mn( l3 CO)(CO) 4 ,
because the mechanisms of the two reactions must be the reverse of each other
according to the principle of microscopic reversibility.
Consider the possible products that can form if decarbonylation takes place by the
reverse of CO insertion.
Noack. K.; Calderazzo. F. J. Organomet. Client. 1967. 10. 101-104. Caldcrazzo. F. Ange. Client,
hit. Ed. Engl. 1977. 16. 299-311.
Reactions of Organometallic Complexes
697
/,C = 0 CH, CH,
OC/( I Ys, l3 CO OC^ I ‘,.CO OC,, I ,. l3 CO
i Mn i -= £2 -» 'Mn' + ~Mn' (15.114)
oc"" I ^co oc^ I ^co oc^ I ^co
c c c
O 0 0
25% 75%
(predicted)
The CO of the acetyl ligand has a choice of four cis positions into which it may
shift, displacing the CO that is already there. One of these sites is occupied by l3 CO.
Thus we would predict that 25% of the product would have no l3 CO and the other 75%
would have a l3 CO ligand cis to the methyl group. Experimentally it is found that 25%
of the product is devoid of tagged CO. 25% of the product has l3 CO trans to CH,. and
50% of the product has l3 CO cis to CH,. Therefore CO insertion must be eliminated as
a mechanistic possibility. A methyl migration mechanism, however, is consistent with
these experimental results.
(15.115)
O 0
C C
oc. j ,ch, oc. s j j;co h 3 c, v
Mn + Mn +
OC^ I % CO H,C^ I ^CO OC^
C 3 C
0 0
O
C
J 3 C0
o
25%
25%
50%
The methyl group as it migrates may displace the l3 CO ligand to give product
containing no l3 CO (25%); it may displace either of the two CO ligands adjacent to the
tagged CO to give the product with CH , cis to l3 CO (50%); or it may displace the CO
ligand trans to the tagged CO to give the trans product (25%). This result has been
further supported by carbon-13 NMR.'>
The validity of this mechanism has been demonstrated for a number of “CO
insertion" reactions. Thus when chemists use the term CO insertion, they usually
mean alkyl migration. Several things to keep in mind when considering a reaction of
this type are (I) it involves ligands which are cis to one another. (2) in the course of the
reaction a vacant coordination becomes available, and (3) the reverse reaction cannot
occur unless a ligand is first eliminated.
A chiral metal center, as is found in a pseudoletrahedral iron complex with
cyclopentadienyl. carbonyl, triphenylphosphine, and ethyl ligands, has also been used
to address the question of alkyl migration versus carbonyl insertion. Inversion of
154 Flood. T. C.; Jensen. J. E.; Statler. A. J. J. Am. Client. Soe. 1981. 103. 4410-4414.
718 1 5- Organometollic Chemistry
Ziegler-Natta
Catalysis ' 93
Immobilized
Homogeneous
Catalysts
One of [he great discoveries of organometallic chemistry was the catalyzed poly¬
merization of alkenes at atmospheric pressure and ambient temperature. Vast quan¬
tities of polyethylene and polypropylene (over 15 million tons annually) are made by
Ziegler-Natta catalysis. Ziegler and Natta received the Nobel prize in chemistry in
1963. and the importance of their work in stimulating interest in organometallic
chemistry should not be underestimated.
The Ziegler-Natta catalyst, which is heterogeneous, is made by treating titanium
tetrachloride with triethylaluminum to form a fibrous material that is partially alky¬
lated (Et,AlCI is used as a cocatalyst). Third generation catalysts (introduced about
1980) use"a MgCl, support for the TiCI 4 . The titanium does not have a filled coordina¬
tion sphere and acts like a Lewis acid, accepting ethylene or propylene as another
ligand. The reaction is thought to proceed somewhat after the manner of Wilkinson's
catalysts discussed above except that the alkyl group (instead of a hydrogen atom)
migrates to the alkene:
H,C = CH,
„ ' 1 /
— Ti — CH,—CH, —— Ti—CH,—CH, -
CH,— CH,
I l I '
— Ti' CH, —CH,
^ i
CH,
i ch,
CH,
The heterogeneous nature of the reaction makes it a difficult one to study, but it has
been modeled homogeneously with a lutetium complex which undergoes oligomer¬
ization: |,w
. 1I,C—CHlCHi)
(17 -MejCjLLuCH,
M.C-CMICII,)
(t) , -Mc 5 C,),LuCH,CH(CH,), —-
(i 7 5 -Me,C 5 ),LuCH,CH(CH J )CH,CH(CH,), < 15.189)
Since homogeneous catalysts tend to offer fast reaction with high selectivity and
heterogeneous catalysts offer ease of separation, it is not surprising that efforts have
been made to combine the advantageous properties of both. One way to effect this
combination is to attach the "homogeneous’' catalysts to the surface of a polymer
such as polystyrene. Wilkinson's catalyst, for example, can be treated as follows : 195
'« Goodall. B. L. J. Clicm. EJuc. 1986. 63. 191-195.
|,,J Watson. P. L. J. Am. Chtm. 6'<« 1982. It «. 337-339.
W Grubbs. R. H. Chemteeh 1977. 7. 512-518.
Catalysis by Organometallic Compounds
719
CH, CH,
A Photodehydro¬
genation Catalyst
("Platinum Pop")
We have seen in Chapter 11 that the </ x square planar configuration is particularly
stable because four of the mostly metal nd molecular orbitals are stabilized relative to
the d,.:_ v :-derived orbital . 196 Three of the four orbitals (d xy . d x: . and d v .) may partici¬
pate in metal-ligand n bonding, but to a first approximation, the d.: orbital does not
interact with any other orbitals in an isolated square planar complex . 197 Suppose,
however, we configure the complex so that the d.:. is forced into interaction. By
bridging two square planar complexes, the two systems are forced to lie parallel (face-
to-face) at a fixed distance:
Many possible ligands could span the distance, but if we carefully design the bridge,
we can control this metal-metal distance to some extent. Two types of ligands that
have been extensively studied are isocyanides and pyrophosphite. Consider, for
example, isocyanide ligands of the type:
R,C—N=C-
/
(CH,)„
r 2 c—n=c-
l ' , ' , Iii the simple crystal field theory, the d x : ,.s orbital is very strongly repelled and empty: in MOT, it
interacts with the . 1 . -.r. y, —y ligands to form a tilled, strongly bonding MO and an empty,
strongly antibonding MO.
197 Since the </.: orbital docs have some electron density in the xy plane, there is some interaction with
the x and y ligands.
698
1 5 • Organomefallic Chemistry
configuration is expected for ethyl migration, but retention is expected for carbonyl
insertion. 155
alkyl migration
(inversion of configuration)
CO inserlion
(retenlion of configuration)
(15.116)
"' C c°
v -- / El
When this reaction is carried out in nitromethane, inversion of configuration is
observed, consistent with ethyl migration. In some solvents both stereochemical
products are obtained, which may mean that both pathways are operative or that
chiral integrity is lost in the intermediate step.
At this point it is fair to ask, what is the driving force for carbonyl insertion?
These reactions involve breaking a metal-carbon bond and formation of a car¬
bon-carbon bond. In addition, a bond is formed between the metal and the incoming
Lewis base (CO in the foregoing examples, but frequently a phosphine or an amine).
The enthalpy change, AH, for the reaction
Mn(CO) 5 CH 3 + CO -♦ Mn(CO) 5 (COCH 3 ) (15.118)
has been calculated as -54 ± 8 kJ mol -1 . 156 The less energy required to break the
M — R bond and the more energy released when the C—C and M— CO bonds are
formed, the more favorable will be the reaction. Since gaseous CO is captured, it
would be expected that the entropy change would inhibit spontaneity, but even so. the
larger negative enthalpy term is dominant. Insertions are not always thermodynami-
lw Flood. T. C.; Campbell, K. D. J. Am. Client. Soc. 1984. /CM. 2853-2860.
I?h Connor. J. A.; Zafarani-Moattar. M. T.; Bickcrton. J.: El Saied, N. I.; Suradi. S.; Carson. R.;
Al Takhin, G.; Skinner, H. A. Organometallics 1982. /. 1166-1174.
Reactions of Organomefallic Complexes
cally favorable, however, as is illustrated by the absence of reaction when Mn(CO) 5 H
is subjected to CO.
Mn(CO) 5 H + CO Mn(CO) 5 (COH) (15.119)
The calculated enthalpy change for this reaction is approximately 20 kJ mol -1 . This
result is of considerable consequence because it suggests that reduction of CO with a
transition metal hydride, is not a useful route to organic products (see Fisher-Tropsch
catalysis, page 715).
The organometallic chemistry of actinides, ignored in the early development of
the field, is currently receiving a great deal of attention. 157 In many instances the
chemistry of this group of elements is unlike that of the transition metals. For
example, it has been shown that a thorium hydride, in conlrast to the manganese
hydride shown above, does undergo CO insertion. 15
(^ 5 -C 5 H 5 ),Th + CO
OR
O
/ \
(T, 5 -C 5 H 5 ),Th-C-H
OR
A driving force for this reaction is the strong interaction of the oxygen of the inserted
CO with the thorium atom.
Of equal importance to carbonyl insertion into a metal-carbon bond is olefin
insertion into a metal-hydrogen bond.
M—H + R 2 C=CR 2 -► M-CR 2 CR 3 H (15.121)
Catalytic hydrogenation and hydroformylalion are just two of the many important
processes in which these reactions are fundamental (see page 711). The first step in
the reaction is coordination of the alkenc to the metal, followed by rapid insertion into
the M—H bond. The transition state involves a four-center planar structure.
— C —C —
The reverse of alkenc insertion, fi elimination, represents the chief pathway for
decomposition of a transition metal alkyl complex.(Eq. 15.45). The process begins with
deinsertion of the alkyl ligand to yield a metal hydrido alkenc complex, which then
eliminates the alkene. 15 ' 7 The decomposition process can be thwarted by designing
alkyl complexes in which fi elimination is not possible either because the ligands have
no hydrogens on carbon atoms to the metal [e.g.. PhCH 2 , CH 3 . or (CH 3 ) 3 CCH 3 ] or
the fi hydrogen is too far from the metal to allow deinsertion to occur (e.g.. C=CH).
Note also that fi elimination cannot lake place unless there is a vacant site for
Marks. T. J.: Slrcilwicser. A.. Jr. In The Chemistry of the Actinide Element .«; Kao., J. J.; Seahorg,
G. T.: Morss. L. R.. Eds.; Chapman and Hall: New York. 1986, Marks. T, J. Ace. Client. Ilex.
1992. 25. 57-65.
oh Fagan. P. J.; Moloy. K. G.; Marks. T. J. J. Am. Client. Soc. 1981. I0J. 6959-6962.
I5V Cross. R. J. In The Chemistry of the Menil-Cnrhon Bond ; Hartley. F. R.; Patai. S-. Eds.; Wiley:
New York. 1985.
716
1 5 ■ Organometallic Chemistry
Alcohols and alkanes are also primary products and are no. stow" in thet simpMed
Eq is 182 The overall reaction is complicated and, as a result, its mechanism has
been the subject of considerable debate.- The reaction may be vtewed as the
reductive polymerization of carbon monoxide, with molecular hydrogen as the reduc-
mg agent. A variety of heterogeneous catalysts, such as me.alhc iron and coMt on
alumina, have been used. It is believed that carbon monoxide dissociates on he
catalytic surface to give carbides and that these are in turn hydrogenated g.
surface carbenes: 189
co
(15.183)
Carbene insertion into a metal-hydrogen bond gives a methyl group that can undergo
carbene insertion in a propagating manner:
CH 3
etc. (15.184)
Although it is always somewhat risky to draw conclusions about surface reactions
from solution experiments, a number of such studies support the carbide/carbene
mechanistic proposal. A model compound for the carbide proposal is a butterfly
cluster formed from an unusual six-coordinate caibide.
The exposed carbide carbon reacts successively with carbon monoxide and
methyl alcohol:
. W Rofcr-DePoortcr. C. K^CItem. Rev. 1981. 81. 447-474. Herrmann. W. A. Angew. Chen,. In,. Ed.
Engl. 1982. 21. 117-130.
aw Brady. R. C.. Ill; Pettit. R- J- Am. Client. Soc. , 980. 102. 6181 -618 .
,9n Bradley. J. S, Ansell. G. B, Leonowicz. M. E, Hill. E. W. 7. Am. Chen. Soc. .98,. ,03.
4968-4970.
Catalysis by Organometallic Compounds
717
Fc Fe
Fe F C
(15.186)
followed by hydrogenation to yield methyl acetate. Note that a carbon-carbon bond is
formed in this reaction. The net synthesis produces an organic oxygenate from
synthesis gas, which may also occur in the Fischer-Tropsch reaction.
Support for the propagation step of the Fischer-Tropsch reaction is provided by
the homogeneous reaction: 191
CH, + CH, = CHCH, (15.187)
The bridging carbenes of the bimetallic complex, which parallel the surface carbenes
of the Fischer-Tropsch catalyst, are involved in C—C bond formation.
Both the Fischer-Tropsch reaction and the Mobil process enable one to convert
synthesis gas into hydrocarbons. Since synthesis gas may be obtained from coal, we
have in effect a means of converting coal to gasoline. Germany moved its Panzer
Korps in World War II with synthetic fuels made from the Fischer-Tropsch reaction,
and improved technological developments have enhanced the atlractiveness of the
process. South African Synthetic Oil Limited (SASOL) currently operates several
modern Fischer-Tropsch plants. Many organometallic chemists refer to both the
Fischer-Tropsch and Mobil processes as "political processes" 192 because they are
heavily subsidized by countries that find it important to be independent of foreign oil.
191 Isobe. K.; Andrews. D. G.; Mann. B. E.; Maillis. P. M. J. Client. Soc. Chen. Common. 1981.
809-810.
192 Collman. J. P.; Hcgcdus, L. S.; Norton. J. R.; Finke. R. G. Principles ami Applications of
Organotrtmsition Metal Chemistry. 2nd cd.; University Science Books: Mill Valley. CA. 1987;
p 620.
700
15 • Organometallic Chemistry
Nucleophilic and
Electrophilic Attack
of Coordinated
Ligands
hydrogen to occupy in the deinsertion step. Thus 18-electron complexes with ligands
that remain attached to the central metal, such as the dicarbonyicyclopentadienyl-
ethyliron(ll) complex shown below, are kinetically inert with respect to elimination.
Organometallic complexes frequently are susceptible to nucleophilic attack by an
external reagent. In some instances the attack takes place on the metal center (see
substitution reactions, page 686). while in others it occurs on a bound ligand. Already
in this chapter we have seen many instances in which coordinated carbon monoxide
undergoes nucleophilic attack. Examples include reactions with H - to produce a
formyl complex (Eq. 15.19), with R~ to form an acyl complex (Eq. 15.49), and with
OH~ to give a hydroxycarbonyl complex (Eq. 15.21).
M—C=0 + Nu‘
[M—C(Nu)0]'
We have also observed that the carbon atom of a Fischer carbene is subject to reaction
with nucleophiles (Eq. 15.52).
Coordinated unsaturated hydrocarbons are particularly susceptible to nucleo¬
philic attack even though as free organic molecules they tend to resist such reactions
because they are relatively electron rich. Upon coordination, they yield some electron
density to the metal and thereby lose some resistance to reaction with nucleophiles. A
metal fragment with good 7r-accepting ligands and/or a positive charge (i.e.. one that is
more electronegative) will therefore be an especially good candidate for activating an
unsaturated hydrocarbon toward nucleophilic attack. Of course not all coordinated
unsaturated hydrocarbons are equally reactive. The following order of nucleophilic
susceptibility in 18-electron cationic complexes has been established. 1 ' 10
The usefulness of such a series is twofold: (I) If two different unsaturated ligands are
found in the same complex, one can predict which ligand will react, and (2) it is
possible to estimate how activating a metal fragment must be in order to cause a
reaction to occur. Notice that hydrocarbons of even hapticity are more reactive than
those with odd hapticity. In addition, acyclic ligands are more reactive than cyclic
ones.
Reactions illustrating nucleophilic attack on coordinated olefins and allyls are
shown in Eqs. 15.124 and 15.125.
INI Davies. S. G.; Green. M. L, H.: Mingos. D. M. P. Tetrahedron 1978. Jd. 3047-3077. (Tollman,
J. P.: Hegcdus. L. S.: Norton. J. R.: Finke. R. G. Principles and Applications of Orpanotransition
Metal Chemistry. 2nd ed.; University Science Books: Mill Valley. CA. 1987: p 409. Crabtree.
R. H. The Organometallic Chemistry of the Transition Metals'. Wiley: New York. 1988: p 148.
Reactions of Organometallic Complexes
701
710
1 5 • Organometallic Chemistry
can be altered with the water-gas shift reaction, which can be catalyzed by a variety of
heterogeneous and homogeneous catalysts:
CO + H 2 0 - C0 2 + H 2 (15.161)
There are several reasons for wishing to alter the hydrogen concentration. First,
hydrogen is a more versatile industrial chemical than water gas. Second, small organic
molecules tend to have roughly three to four times as many hydrogen atoms as carbon
atoms, so if the H 2 /CO mole ratio can be changed to about two, a good feedstock is
obtained.
Commercially, the water-gas shift reaction is usually carried out over Fe 3 0 4 . 173
However, current interest centers on homogeneous catalysts. Metal carbonyl com¬
plexes such as [HFe(C0 4 ]~. [Rh(CO) 2 I 2 J - . and [Ru(bpy) 2 (CO)Cl] are effective and
although all the mechanisms have not been worked out completely, the reactions may
be viewed in general terms as beginning with a nucleophilic attack on coordinated
carbon monoxide:
co -co,
M M—C=0 - M— C=0 - [M—H] _ (15.162)
OH
The hydridic hydrogen can then attack water:
[M—H]“ + H 2 0 ->M + OH- + H 2 (15.163)
Alternatively (and equivalently) a water molecule can attack in Eq. 15.162 (freeing an
H " ion) followed by attack of a proton on the hydridic ion in Eq. 15.163. A scheme for
the reaction catalyzed by [Ru(bpy) 2 (CO)Cll + can be presented as:
[RuL,(CO)CI| +
H,0
[RuL 2 (C0)(H 2 0)] 2
n'A K h,o
|RuL : (CO)H] + [RuL,(CO)(OH)RuL,(CO),| : - (15.164)
co.
[RuL 2 (CO)(COOH)l
[RuL,(CO)(COO")|
This particular cycle is significant because all of the key intermediates have been
isolated. 174 Substitution of H 2 0 for Cl - in [Ru(bpy) 2 (CO)CI]" gives |Ru(bpy),(CO)-
(H 2 0)] :+ which exists in equilibrium with (Ru(bpy) 2 (CO)(OH)|. Carbon monoxide
l7J Ford. P.C. Acc. Client. Res. 1981. 14, 31-37.
174 Ishida. H.; Tanaka, K.; Morimoto, M.; Tanaka, T. OrganometuUks 1986. 5, 724-730.
Catalysis by Organometallic Compounds
711
displaces water to give [Ru(bpy) 2 (CO) 2 ] :+ which undergoes nucleophilic attack to
form a hydroxycarbonyl complex. Decarbonylation occurs with hydride formation
followed by liberation of hydrogen gas and reformation of the catalyst.
The preferred route to synthesis gas currently is by reforming methane (principal
component of natural gas):
CH 4 + H 2 0 - CO + 3H 2 (15.165)
Whatever the source of synthesis gas, it is the starting point for many industrial
chemicals. Some examples to be discussed are the hydroformylation process for
converting alkenes to aldehydes and alcohols, the "Monsanto process” for the pro¬
duction of acetic acid from methanol, the synthesis of methanol from methane, and the
preparation of gasoline by the Mobil and Fischer-Tropsch methods.
Hydroformylation 175 The reaction of an alkene with carbon monoxide and hydrogen, catalyzed by cobalt or
rhodium salts, to form an aldehyde is called hydroformylation (or sometimes the o.xo
process ):
CojlCOI,
2RCH=CH, + 2CO + H, —- RCH 2 CH 2 CHO + RCH,(CHO)CH, (15.166)
It was discovered by Roelen in 1938 and is the oldest and largest volume catalytic-
reaction of alkenes, with the conversion of propylene to butyraldehyde being the most
important. About 5 million tons of aldehydes and aldehyde derivatives (mostly alco¬
hols) are produced annually making the process the most important industrial syn¬
thesis using a metal carbonyl complex as a catalyst. 17 '’ The name hydroformylation
arises from the fact that in a formal sense a hydrogen atom and a formyl group are
added across a double bond. The net result of the process is extension of the carbon
chain by one and introduction of oxygen into the molecule.
The most widely accepted mechanism for the catalytic cycle is the following one
proposed by Heck and Breslow:' 77
CO
l7< Pruclt. R. L. J. Client. Ethic. 1986. 63. 196-198.
I7h Orchin. M. Acc. Client. Res. 1981. 14, 259-266.
177 Hcck, R. F.; Brcslow, D. S. J. Ant. Client. Soc. 1961. 83. 4023-4027.
702
15 • Organometallic Chemistry
The nei result is that benzene has undergone a substitution reaction that is not possible
for the free molecule.
Although the coordinated cyclopentadienyl group resists nucleophilic attack, it
does react with electrophiles. Ferrocene resembles free benzene in that it reacts with
many electrophilic reagents, but it does so at an even faster rate than benzene. The
aromatic character of ferrocene was recognized soon after the complex was identified
and has led to a rich literature. Among the numerous reactions that have been studied
is acylation in the presence of a Friedel-Crafts catalyst.
The AICI, catalyst reacts with CH 3 COCI to generate the electrophile. CH 3 C + =0.
CHjCOCl + AlClj - CHjC + =0 + AICI (15.129)
The reaction of acetic anhydride with phosphoric acid will generate the same elec¬
trophile and offers the advantage that only the monoacyl product results. Acylation of
the first ring deactivates the second and the concentration of CH 3 C + =0 from the
phosphoric acid reaction is too small to produce the diacyl product.
A second example of the reactivity of the ferrocene rings is their condensation
with formaldehyde and amines (the Mannich condensation):
Fe(T 7 5 -C 5 H 5 ) 2 + CH 2 0 + HNMe 2 • (T, 5 -C J H 5 )Fe(r 7 5 -C J H 4 CH 2 NMe 2 )
(15.130)
Ferrocene thus resembles the more reactive thiophene and phenol rather than benzene
which does not undergo Mannich condensation.
Other reactions typical of aromatic systems, such as nitration and bromination.
are not feasible with metallocenes because of their sensitivity to oxidation. IM How¬
ever. many of the derivatives that would be produced in these types of reactions can
be made indirectly by means of another reaction typical of aromatic systems: metalla-
tion. Just as phenyllithium can be obtained from benzene, analogous ferrocene com¬
pounds can be prepared:
IW Ferrocene loses an electron rather reluctantly since it involves disrupting an 18-elcctron configura¬
tion. but it does so when subjected to strong oxidizing agents like nitric acid or bromine. Co-
baltocenc is readily oxidized to the very stable cobaltocenium ion. losing the 19th and only
antibonding electron. Nickelocenc loses one of its two antibonding electrons to form the relatively
unstable nickclocenium ion.
These lithio derivatives are useful intermediates in the synthesis of various ferrocenyl
derivatives. Some typical reactions are:
(15.132)
(15.133)
Carbonylate Anions Carbonylale complexes have many useful synthetic applications. Typical reactions
as Nucleophiles involve nucleophilic attack of the metal anion on a positive center (alternatively
viewed as an electrophilic attack on the metal). The synthesis of metal alkyl com¬
plexes has been referred to earlier (Eqs. 15.46 and 15.47). Other examples include:
O O
II II
RCX + [Co(CO) 4 ]“ - RCCo(CO) + X" (15.134)
Mn(CO) s Br + [Mn(CO) s ]- -♦ Mn 2 (CO), 0 + Br‘ (15.135)
Although the reaction in Eq. 15.135 is of little importance in the manufacture of
Mn 2 (CO), 0 (the reactants typically are synthesized from the dimangancse complex), it
illustrates a general and useful method of forming metal-metal bonds that can be
applied to cases in which the metals are different:
Ethylene is commonly chosen to illustrate homogeneous hydrogenation with
Wilkinson’s catalyst, but the process is actually very slow with this alkene. The
explanation lies with the formation of a stable rhodium ethylene complex, which does
not readily undergo reaction with H : . Ethylene competes effectively with the solvent
for the vacant coordination site created when triphenylphosphine dissociates from
Wilkinson's catalyst and thus serves as an inhibitor to hydrogenation.
(Ph 3 P) 3 RhCl + H 2 C=CH 2 -♦ (Ph 3 P) 2 RhCl(^ 2 -C 2 H 4 ) + Ph 3 P (15.158)
Toiman Catalytic
Loops 169
A reaction involing a true catalyst can always be represented by a closed loop. Thus
we may combine Eqs. 15.152-15.157 into a continuous cycle with the various catalytic
species forming the main body of the loop and reactants and products entering and
leaving the loop at appropriate places:
■ w Toiman. C. A. Chan. Sue. Rev. 1972. /. 337-353.
Catalysis by Organometailic Compound;
709
Hydrogen migration
Synthesis Gas 170 The history of organic industrial chemistry can be organized around the relative
reactivities of the organic feedstocks used during a particular era.' 7 ' The years
1910-1950 can be characterized as the "acetylene period" since readily available,
highly reactive, but rather expensive acetylene was employed. From 1950 to the
present, alkenes have predominated. It now appears that alkenes in turn will be
replaced by synthesis gas (H : /CO) as the raw material of choice, and we shall enter
the era of one-carbon feedstocks. The evolution of homogeneous catalysis for indus¬
trial use has been extended to ever cheaper and less reactive feedstocks. Synthesis
gas 17 - can be produced from coal and was one of ihe reasons interest in coal rose
markedly during the oil embargos of the 1970s. Enthusiasm toward coal as a resource
waxes and wanes as the price of petroleum rises and falls.
I he first H,/CO mixture to be of commercial importance was obtained from the
action of steam on red-hot coke and. because of its origin, became known as water
gas:
H,0 + C >CO + H 2 (15.160)
In the 19th-century days of gas lamps, water gas was frequently used for domestic
purposes, a practice fraught with danger because of the extreme toxicity of carbon
monoxide (see Chapter 19). The ratio of hydrogen to carbon monoxide in water gas
17,1 Sheldon. R. A. Chemicals from Synthesis Gas: D. Rcidcl: Dordrcchl. 1983.
171 Parshall. G. W. Homogeneous Catalysis-, Wiley: New York. 1980: p 223.
172 The term syngas is a broad term used to cover various mixtures of carbon monoxide and hydrogen.
704
1 5 • Organometallic Chemistry
Mn(CO) 5 Br + [Re(CO) s ]" -♦ (OC) 5 MnRe(CO) 5 + Br" (15.136)
HgS0 4 + 2[Mn(CO) 5 ]" -► (OC) 5 MnHgMn(CO) 5 + SO 2 " (15.137)
PhSnCl 3 + 3 [Co(CO) 4 ]- -► PhSn[Co(CO) 4 ] 3 + 3C1" (15.138)
(Ph 3 P) 2 NiCl 2 + 2[Co(CO) 4 ]“ -► (OC) 4 CoNi(PPh 3 ) 2 Co(CO) 4 + 2C1"
(15.139)
Applications of carbonylate reactions in organic synthesis are numerous. Particularly
noteworthy are schemes involving tetracarbonylferrate( -11) (referred to as (Tollman s
reagent), which can be isolated as a sodium salt, Na-,Fe(CO) 4 -1.5 dioxane, and is
commercially available. The highly nucleophilic [Fe(CO) 4 ]‘ reacts readily with alkyl
halides to yield alkyl iron carbonylates:
RX + [Fe(CO) 4 ] 2 ' -► [RFe(CO) 4 ]- + X" (15-140)
These alkyl complexes do not undergo /3 elimination (the stable 18-electron complex
does not provide the necessary vacant coordination site) and optically active R groups
do not undergo racemization. Migratory insertion reactions (page 695) do occur in the
presence of Ph 3 P or CO to give the corresponding acyl complexes.
O
[RFe(CO) 4 r + L-► [RCFe(CO) 3 L]- (L = Ph 3 P or CO) (15.141)
Although the alkyl and acyl products shown in Eqs. 15.140 and 15.141 have been
isolated and characterized, they are frequently allowed to simply form as intermedi¬
ates. which arc then treated directly to produce aldehydes, carboxylic acids, ketones,
esters, or amides.
(15.142)
(15.143)
(15.144)
(15.145)
(15.146)
(15.147)
Thus Collman's reagent functions much like a Grignard reagent in its ability to convert
alkyl halides into a wide variety of organic compounds.
Catalysis by Organometallic Compounds
705
A rich chemistry has also developed for the chromium dianion, [Cr(CO) 5 ) : . IM The
expected displacement of Cl” occurs when this reagent reacts with an acid chloride:
[Cr(CO) 5 J-~ + RCCI - [RCCr(CO),r + Cl" (15.148)
The acylate complex may be alklyated directly to give an alkoxycarbene or the same
end may be achieved by acetylation followed by alcoholysis:
C = Cr(CO).
|RCCr(CO),
ch,cocN (OC),Cr = C
(OC),Cr = C
The resulting transition metal carbenes have been used to synthesize a wide variety of
organic compounds such as furanocoumarins IM , pyrroles. 165 and /3-lactams. Ifih
Catalysis by A thermodynamically favorable reaction may be slow at modest temperatures and
—--- therefore not of value for synthesis. Increasing the temperature of the reaction may
Urganometallic significantly accelerate its rate, but providing the energy to do so is expensive and
Compounds 167 higher temperatures may induce competing side reactions that will greatly reduce
product yields. A more attractive approach to increasing the rate of a reaction is to use
a catalyst. Catalysts are classified as homogeneous if they are soluble in the reaction
medium and heterogeneous if they are insoluble. Each type has its advantages and
disadvantages. Heterogeneous catalysts are easily separated from the reaction prod¬
ucts (a very positive feature) but tend to require rather high temperatures and pres¬
sures and frequently lead to mixtures of products, i.e.. they have low selectivity.
Scmmclhack, M. F.; Lee. G. R. OrgaiMinetallics 1987. 6. 1839-1844,
,M WullT. W. 0.: McCallum. J. S.: Kunng. F.-A. J. Am. Chvm. Sue. 1988. HO. 7419-7934. Diitz,
K. H. Angew. Client. Ini. Ed. Engl. 1984. 23. 587-608.
,w Dragisich. V.; Murray. C. K.; Warner. B. P.: WulfT. W. D.; Yang. D. C. J. Am. Chem. Sot:. 1990,
112. 1251-1253.
imi Hegedus. L. S.: dc Week. G.: D'Andrca. S. J. Am. Chem. Soe. 1988, HO. 2122-2126.
Ih7 For more thorough discussions of this topic, see (Tollman. J. P.; Hegedus, L. S.: Norton. J. R.;
Finke. R. G. Principles and Applications of Organotrunsition Metal Chemistry. 2nd ed.; Univer¬
sity Science Books: Mill Valley. CA. 1987; Crabtree, R. L. Principles of Organometallic Chem¬
istry: Wiley: New York 1988: Lukehart. C. M. Fundamental Transition Metal Organometallic
Chemistry: Brooks/Cole: Monterey. CA 1985; Parshall, G. W. Homogeneous Catalysis: Wiley:
New York. 1980: Bond. G. C. Heterogeneous Catalysis. 2nd ed.; Clarendon Press: Oxford. 1987.
706
1 5 - Organometallic Chemistry
Homogeneous catalysts must be separated from the product (a negative feature) but
operate at low temperatures and pressures (a very positive aspect), and usually give
good selectivity (another very positive aspect).
Many important chemicals are produced commercially by reactions which are
catalyzed by organometallic compounds and this fact provides one of the motivating
forces for studying organometallic chemistry. Much of the focus in this section will be
on homogeneous catalysis because solution reactions are better understood than are
the surface reactions of heterogeneous systems. It is also easier to modify an
organometallic compound and evaluate the effects of the modification than it is to alter
and study a surface.
Alkene Although the reaction of hydrogen gas with ethylene is thermodynamically favorable,
Hydrogenation it does no1 lake P lace at room temperature and pressure.
H 2 C=CH 2 + H 2 -■ H 3 C—CH 3 (15.151)
AH' 1 = -136 kJ mol -1 AG° = - 101 kJ mol -1
However, in the presence of metallic nickel, copper, palladium, or platinum, the
reaction is fast and complete. The metal may be deposited on an inert solid support
such as alumina or calcium carbonate, but the reaction is with the metal surface and
therefore is heterogeneously catalyzed.
The first effective homogeneous catalyst to be discovered for hydrogenation was
the square planar 16-electron </ H complex chlorotris(triphenylphosphine)rhodium(l).
(Ph,P),RhCI (Fig. 15.47). which is known as Wilkinson's catalyst. In Chapter II we
saw that this geometry and electron configuration are an especially favorable combina¬
tion. These species also have wide possibilities for oxidative addition (page 689). They
can become five-coordinate through simple addition of a ligand or six-coordinate
through addition combined with oxidation. In either case they become isoelectronic
with the next noble gas. i.e., they achieve an 18-electron valence shell configuration. It
Catalysis by Organometallic Compounds 707
is not surprising then that 16-electron square planar complexes have been regarded as
very attractive catalyst candidates on the premise that they may oxidatively add two
reactant molecules and thereby enhance their reactivity.
Wilkinson’s catalyst is thought to behave as follows: In solution one of the
phosphine ligands dissociates, leaving (Ph,P)-,RhCl. This tricoordinate complex is
very reactive and has not as yet been isolated, but the closely related [(Ph 3 P) 3 Rh]'\
which could form from the dissociation of a chloride ion from Wilkinson’s catalyst has
been studied and found to have an unusual structure (Fig. 15.48). Unlike most three-
coordinate complexes (Chapter 12). it is more T-shaped than triangular. The evidence
for dissociation of a Ph,P ligand from (Ph 3 P) 3 RhCI is indirect but persuasive: (I) For
complexes with less sterically hindered phosphines (e.g., Et 3 P), the catalytic effect
disappears—apparently steric repulsion forcing dissociation is necessary; and (2) with
the corresponding iridium complex in which the metal-phosphorus bond is stronger,
no dissociation takes place and no catalysis is observed.
To return to the catalysis, the (Ph,P) : ,RhCl molecule, possibly solvated, can
undergo oxidative addition of a molecule of hydrogen. An alkene can then coordinate
and react with a coordinated hydrogen ligand to form an alkyl group. This reaction will
result from a migration of a hydrogen from the metal to a carbon in the coordinated
alkene. Although the hydrogen atom does essentially all of the moving, this reaction is
often called an alkene insertion reaction (page 699).
The reactions involved in hydrogenation with Wilkinson's catalyst thus can be
represented as follows (L = Ph,P. S = solvent molecule). 168
L / S
L—Rh — L+S -► L— Rh — L+L (15.152)
Cl'" Cl
Fig. 15.48 Stereoview of the structure of the ((PfhPhRh] cation, showing the planar,
approximately T-shaped coordination about the rhodium atom. Note the unusual manner in
which the phenyl ring at the lower right is drawn toward the rhodium atom, (From Yared,
Y. W.; Miles. S. L.: Bau. R.: Reed. C. A. J. Am. Chan. Sue. 1977, 99, 7076-7078.
Reproduced with permission.!
iM For a more detailed mechanistic view, see Halpcrn. J. I nor a. Cliim. Arm 1981, 50. 11-19. It has
been suggested that the phosphine ligands may be cis in the octahedral intermediates. Brown.
J. M.: Lucy. A. R. J. Client. Soc.. Client. Comntun. 1984. 914-915.
786
1 6 • Inorganic Chains, Rings, Cages, and Clusters
\ / \ /
0 0
o
II145
I60,P
o
\ / \ /
O 0
0
II141
J58^P^
O
7 | I63l 0 ^ V
P p P
o o L
Fig. 16.36 Phosphorus
cage molecules: (a) P 4 0 (l ;
(b) P 4 0 7 : (c) PjO»; (d)
P 4 0 9 ; (e) P 4 0,„. [Daia
(distances in pm) taken
from Jansen. M.; Moebs,
M. Inorg. Cliem. 1984. 23.
4486-4488. Reproduced
with permission.]
p-^p p.
V^o' 0 o'H'o
oxide cages (Fig. 16.36). All are anyhydrides that react readily with water to form the
corresponding acids:
P 4 O h + 6H,0 -♦ 4H,PHO, (,6 ' 75)
P.,O in + 6H 2 0-► 4H 3 P0 4 < 16 - 76 >
The P 4 0 7 molecule reacts to form both phosphoric acid and phosphorous acid. In
addition to the discrete cage molecule pictured in Fig. 16.36e. phosphorus pentoxide
also exists in several polymeric forms. 113
White phosphorus can be converted readily to its more stable allotropes:
•ip . - --> p (red phosphorus) (16.77)
4 or heal •
ip -pressure^ p (black phosphorus) <6- 78 >
4 4 or Hr catalyst
Crystalline black phosphorus has a corrugated layer structure." 4
"Red phosphorus” does not appear to be a well-defined substance but differs
according to the method of preparation. It probably consists of random chains. The
rate of formation is increased by certain substances such ns iodine which appear to be
incorporated into the product.
The chemistry of phosphorus and sulfur is considerably more complicated than
phosphorus-oxygen chemistry." 5 Only two phosphorus sulfides, P 4 S I0 and P 4 S 9 , are
isoelectronic and isostructural with phosphorus oxides. The former may be prepared
by allowing stoichiometric amounts of phosphorus and sulfur to react:
i" Sharma. B. D. Inorg. Client. I9H7 , 26 . 454-455.
i" For this and several other interesting elemental structures, see Donahue. J. The Structures of the
Elements-. Wiley: New York. 1974.
i" Hoffmann, H.: Beckc-Goehring. M. Top. Phosphorus Client. 1976. S. 193-271.
Cages
787
4P 4 + 5Sj, - 4P 4 S m (16.79)
By mixing phosphorus and sulfur in appropriate stoichiometric quantities, P 4 S 3 and
P 4 S 7 may be obtained. Slow oxidation of P 4 S 3 with sulfur yields P 4 S 5 :
4P 4 S 3 + S 8 - 4P 4 Sj (16-80)
Two cage phosphorus sulfides may be synthesized by the formation of sulfide bridges
through the action of bis(trimethyltin) sulfide." 6
(Me 3 Sn) 2 S +
P^X^S
A
P^ \^s
-► I x c s I +2Me 3 Snl
(16.81)
(o-P 4 S 3 I 2 )
(a-P 4 S 4 )
sAs
1 / S 1
p \ Ok
' x p^
sAs
-► | oS | + 2Me 3 SnI
(16.82)
(NWj)
(P-p 4 s 4 )
(Me,Sn),S +
The structures of all of these sulfides are known (Fig. 16.37). They are all derived from
a tetrahedron of phosphorus atoms with sulfur atoms bridging along various edges. All
except P 4 S 1(I and P 4 S., retain one or more P—P bonds.
The heavier congeners of phosphorus resemble it in a tendency lo form cages.
Both arsenic and antimony form unstable tetrameric molecules which readily revert to
polymeric structures. Cage molecules as well as polymeric forms are also known for
As 4 0 (> and Sb 4 0,,. In addition there are a number of sulfides, some of which arc known
to exist as cages (Fig. 16.38).
By extension of the reactions involved in the formation of cyclopolysilanes. West
and Carberry" 7 synthesized bicyclic and cage permethylpolysilanes such as:
Me,Si > SiMe,
2 1 Me,St I
I Me,Si I
Me,Si / SiMe,
• '"'Sr '
" 6 Griffin. A. M.: Minshall. P. C.: Sheliirick. G. M. J. Cliem. Soe. Client. Common. 1976. 809-810.
117 West. R.: Carberry, E. Science 197S, 189. 179-186.
A-88
Index
Symmetry ( Continued )
translational, 74
Symmetry allowed transitions,
64-65
Symmetry elements, 46-53
Symmetry forbidden transitions.
64-65
Symmetry groups, 53, A13-A22
Symmetry operations, 46-53
Synthesis gas, 709-711
Synthetic gasoline, 715-717
Systematic absences, 79
Talc, 750
Tanabe. Y.. 443-447, A38-A39
Tanabe-Sugano diagrams,
443-447, A38-A39
Taube, H., 387
Teller, E.. 449
Term symbols, 26-27, A7-AI2
Tetrachloroiodate anion. 209
Tetragonal crystal system, 75, 78
Tetragonal symmetry, 403-404
Tetrahedral complexes, 401-403,
418-420, 441. 448-455,
474-477
Tetranuclear clusters, 815-816
Thermal ellipsoids, 234. 724
Thermochemical calculations,
predictive power of. 127-129
Thermochemical radii, 117
Thermodynamics, and chelate
efTect. 523
Thermodynamic convention, 379
Thermodynamic stability,
547-548
Time scales for structural tech¬
niques, 238-239, 724
Tolman catalytic loops, 708
Toxicity, of biological elements,
943-948
Trace elements, in biological
systems, 941-953
Transactinide elements, 599-607,
613-617
Trans effect, 543-545
Transferrins, 937
Transition metal hydrides, acid¬
ities of, 643
Transition metals, 27-28
heavy, 587-588
and inner transition, 608-609
oxidation states of, 580-582
Transitions, and Laporte rule,
438-440
Translational symmetry, 74
Translawrencium elements, peri¬
odicity of, 615-617
Transport, dioxygen, 895-910
Triatomic molecules and ions,
molecular orbitals in,
175-182
Triclinic crystal system, 75, 78
Trigonal bipyramidal (TBP)
structure, 480-483
Trigonal crystal system, 75, 78
Trigonal hybrid, 150-153
Trigonal prism, 489-491
Trimethylborane, 205
Trinegative ions, 332-333
Trinuclear clusters, 813-815
Tropylium complexes, 683-684
Tutankhamen, 458
Twist angles, 490
Ultraviolet (UV) radiation, 244
Ungerade symmetry, 48, 160
Ungerade orbitals, 17-18, 160,
438
Uninegative ions, 332-333
Unit cell, 74
Units and conversion factors,
A3-A6
Unsold's theorem, 19
Unstable complexes, 547-548
Unsymmetric hydrogen bonding,
302
Usanovich, M., 325
Usanovich acid-base definition,
325-326
Vacancy mechanism, 266
Valence bond (VB) theory.
139-153, 391-394. 474
Valence shell electron pair repul¬
sion (VSEPR) model,
203-206, 217-218
Valence state, promotion to,
148-149
Valence state values, 191
van der Waals forces, 252, 299.
See also London disperson
forces
van der Waals radii, 290-291,
301, 720. See also London
disperson forces
Van Vleck, J. H., 395
Vaska's complex, 691
Vertical plane of symmetry, 51
“Vibrational Spectra and
Structure of Xenon Tetra-
fluoride," 70
Vibronic transitions, 438
Violeo complex, 388. 491. 493
Vitamin B,,, and coenzymes,
929-931
Volumes of activation, 542, 553
Wacker process, 714-715
Wade's rules, 798, 805
Walden inversion, 243
Walsh diagrams, 218
Water, physical properties of,
360
Water-gas shift reaction, 710
Werner, Alfred. 387. 388, 389.
390, 491,495
Wilkinson, G., 387, 669
Wilkinson's catalyst, 707
Wurtzite, 96-97
Xenon. 5, 70. 832-836
X-ray diffraction. 233-235, 238,
497, 645
p-Xylene, 5-6
Yatsimirskii, K. B., 117
Zeolites, 3-5, 7, 378, 715,
745-748
Ziegler, K., 718
Ziegler-Natta catalysis, 718
Zinc. 582-587
Zinc blende, 96-97
Zintl salts. 527, 817
ZSM-5, 747-748
788
16 Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.37 Molecular
structures of some
phosphorus suifides.
Distances in picometers.
Fig. 16.38 Molecular
structures of two arsenic
sulfides. Distances in
picometers.
In order to get branching to form cages (bridgehead silicon atoms) some meth-
yltrichlorosilane is added to the dimethyldichlorosilane in the reaction.
In the limited space allowed here, it has been possible to mention only a few of the
many nonmetallic inorganic cages. If we consider those which also include carbon
atoms, we have an even larger group from which to choose. One of the more
remarkable cages to be synthesized recently is (/-BuCP) 4 which has a cubane struc¬
ture. 118 Two molecules of tert-butylphosphaacetylene (1) undergo a head-to-tail di¬
merization to give an intermediate (2) which is thought to dimerize once again or react
"« Welding, T.: Schneider. J.; Wagner, O.: Kricicr. C. G.; Rcgitz, M. Angew. Chem. Ini. Ed. Engl.
1989. 28. 1013-1014.
Boron Cage Compounds
789
with two additional molecules of the starting material to give (3) which "zips up" to
give the final product (4).
" V a discussion of the synthetic chemistry of B —H compounds, sec Shore. S. G. In Rings.
Cto/erv. and Polymers of tin- Main Group Elements', Cowley. A. H.. Ed.; ACS Symposium Series
232. American Chemical Society; Washington. DC. 1983.
Trimcthylboron. unlike BH } . shows no tendency to dimerize.
A-86 Index
Periodicity ( Continued)
of elements, 27-29
first- and second-row anoma¬
lies, 858-861
fundamental trends, 857-858
size effects in nonmetals,
859-860
of translawrencium elements,
615-617
Perovskite, 253, 285
Perutz mechanism, 903
pH, and solvents, 322
Phorphorus pentafluoride, 205
Phosgene, 205
Phosphazene polymers, 773-775
Phosphazenes, 769-773
Phosphine complexes, and
oxides, 871
Phosphorus, and nitrogen,
864-865
Phosphorus oxyfluoride, 205
Phosphorus trihalides, 211
Photodehydrogenated catalyst,
718-723
Photoelectron spectroscopy, 165,
431-433
Photoexcited semiconductors,
272-274
Photosynthesis, 916-917
and chlorophyll, 917-919
electron transfer, and respira¬
tion. 911-919
Physiology, of myoglobin and
hemoglobin. 900-902
Pi bonding. 420-433, 872
in coordination compounds,
421-433
d orbitals in, 868-875
p orbitals in, 861-866
Platinum anticancer drugs, 559,
957
Platinum pop. 718-723
Point groups and molecular sym¬
metry, 53-59
Point symmetry, 48
Poisoning, and enzyme inhibi¬
tion, 925-929
Polar bonds, solid-state materials
with, 276-288
Polarizability tensor, 67
Polarization of ions, 129-134
Politzer, P., 849
Polyatomic zintl anions and
cations, 817
Polyelectronic atom, 20-43
Polyhalide ions, 839-843
Polyhalogen cations, 848
Polymers, and phosphazene
polymers, 773-775
Polynuclear carbonyl complexes,
633-639
Porterfield, W. W.. 295
Positive oxidation states,
halogens in, 837-848
Posttransition metals, 28, 876
Potassium, 309, 582-587
Potentials, electrode, 378-383
Pourbaix diagram, 591-592
Praseo complex, 388, 491, 493
Predominance area diagram, 591
Prewitt, C. T.. 116-117
Principal axis, 51
Prism, trigonal prism, 489-491
Probability function, 13
Prosthetic group, 919
Proteins, and blue copper pro¬
teins, 912-916
Proton
affinities, 332-333
and hydron, 318, A49
Protonic solvents, summary of,
367-369
Proton sponges, 342-343
Prussian blue, and linkage iso¬
merism, 519-521
Pseudohalogens, 852-853
Pseudorotation, 240-243
Racah parameters, 443
Racemization, and isomerization,
555-557
Radial wave function (R), 11-14
Radii
atomic, 290-296
covalent, 291-296
ionic, 114-117, 291
of polyatomic ions, 117
thermochemicals, 117
van dcr Waals, 290-291, 301,
720
Radius ratios, and coordination
number, 123, 473
Radon, 836
Raimondi, D. L., 31, 32
Raman spectroscopy, 65-71, 235,
239
Rate law, for nucleophilic sub¬
stitution, 540-543
Ray, P. C., 556
Ray-Dutt twist, 556
Rayleigh, Lord, 825
Reaction rates
influenced by acid and base,
553-555
and ligand field effects,
550-551
Reactions
of coordination compounds,
537-538
excited state outer sphere elec¬
tron transfer, 561-565
of organometallic complexes,
686-705
redox. See Redox reactions
water-gas shift, 710
Reactivity, 5-7
and d orbitals. 875-876
of molecules, 203
Redox reactions
in biological systems, 569-572,
891-895
mechanisms of, 557-572
Reducible representations, 62-63
Reduction, 2
Reduction potentials of elements,
A35-A37
Reductive elimination, and
oxidative addition, 689-695
Relativistic effects, 579, 879-880
Repulsive forces, 102, 229-230,
291, 299-300
Resonance, 141, 142-146, 340
Respiration
electron transfer, and photo¬
synthesis, 911-919
and metalloporphyrins,
891-895
Ribonucleic acid (RNA), 958-960
Rings, 738, 765-785
Room-temperature molten salts,
375-376
Rotational axis, 49-51
Rube Goldberg effect, 903
Rubredoxins, 911-912
Rutile. 97
Salts
molten, 374-378
zintl, 527
Sanderson, R. T„ 190, 195, 198
s block elements, 27
Scanning tunneling microscopy
(STM), 4
Schoenflies system, 77, 78
Schottky defect, 263, 265
Schottky-Wagner defect, 263,
265
Schrock carbenes, 659
Schrodinger equation, 10, 20
Screening constant, 31
Selection rules for electronic
transitions, 437-438
Self-consistent field (SCF), 20
Semantics, and the periodic
table, 29-30
Semibridging ligands, 636
Semiconductors
impurity and defect, 274-276
intrinsic and photoexcited,
272-274
Seppelt, K.. 866
Sepulchrate ligands, 530
Shannon, R.. 116-117
Shapley. J. R.. 692
Shibata, Y„ 514
Shielding. 30-33
Sickle cell anemia (SCA). 907
Silicate minerals. 742-750
Silicon, and carbon, 861-863
Silver halides, 309
Sixteen-electron rule, 625
Slater. J. C„ 31. 32. 140
Slipped ring complexes. 677
Sodium chloride. 94-95
Soft acids and bases, 344-355
Solid catalysts, acid and base, 378
Solids
complex, 253-263
held together by covalent
bonding, 269-276
Solid-state materials, with polar
bonds. 276-288
Solid superacids, 378
Solubility, 310-314
Solutions
of metals, 377
of metals in ammonia, 362-364
in molten salts, 374-378
Solvate isomerism, 521
Solvents
aprotic, 367-369
aqueous and nonaqueous,
359-386
molten salt, 374-378
and pH. 322
protonic, 367-369
Solvent system acid-base defini¬
tion. 320-324
South African Synthetic Oil Lim¬
ited (SASOL), 717
Space group, 78
Spectrochemical series, 405
Spectroscopy, 7, 8, 65-71, 165,
235, 238. 239, 331. 428,
431-433
sp hybrid, 150-153
Spinels, 411-412
Spin quantum number, 21-23
Square planar complexes.
403-404, 418-420. 477-479,
538-547
Square pyramidal (SP) com¬
plexes. 480-482. 484-485
Stability
and concentration. 590-593
of oxidation states. 588-590
thermodynamic and kinetic.
547-548
Stable complexes. 547-548
Stable oxidation stales. 599-607
Standard reduction potentials of
elements, A35-A37
States, density of, 280
Stereochemically nonrigid mole¬
cules. 723-730
Stereochemical models, A40-A45
Stereoselectivity, and conforma¬
tion of chelate rings, 498-502
Stereoviews, A44-A45
Steric effects, 229-231, 341-343,
513, 516-518
Stereopsis, A40-A45
Stern, K. H., 133
Structural techniques, time scales
for, 238-239, 724
Index A-87
Structure
of complex solids, 253-263
of coordination compounds.
472-536
of cyclopentadienyl com¬
pounds, 673-678
and function in enzymes,
919-924
and function in hemoglobin,
902-908
and hybridization, 220-231
layered. 260-262
of molecules, 203-220
of molecules with lone electron
pairs, 206-217
and nonbonded repulsions,
229-230
prediction for heteroboranes
and organometallic clusters,
802-807
Subhalogen, 850
Subnodal maxima, 14
Substitution reactions
in carbonyl complexes,
686-688
in octahedral complexes,
548-557
in square planar complexes.
538-547
Sugano, S.. 443-447, A38-A39
Sulfur hexafluoride, 205
Sulfuric acid. 364-367
Sulfur telrafluoride. 208-209
Superacids, 378
Superconductors, high-
temperature, 285-288
Superhaiogen, 850
Supersonic Iranport (SST), 244
Swaddle, T. W„ 553
Symbiosis, 348-350, 518-519
Symmetric hydrogen bonding,
302
Symmetry
and group theory, 46-91
of molecular orbitals, 160
octahedral, 397-399, 488-489
and optical activity, 63-74
of orbitals, 17-18
and overlap, 157-160
tetragonal. 403-404
tetrahedral, 401-403, 418-420
790
16 Inorganic Chains, Rings, Cages, and Clusters
agents and show no tendency to accept electrons when offered by reducing agents. A
number of approaches have been used to rationalize the bonding in these compounds.
The most successful and extensive work in this area, as we shall see. has been that of
William N. Lipscomb. 121 Before proceeding with an examination of the bonding in
diborane, it will be helpful to examine its structure (Fig. 16.39). Each boron atom is
surrounded by an approximate tetrahedron of hydrogen atoms. The bridging hydrogen
atoms are somewhat further from the boron atom and form a smaller H—B — H bond
angle than that for the terminal hydrogen atoms.
The earliest attempt at rationalizing the dimerization of borane invoked resonance
in a valence bond (VB) context:
H H H
y b«L
H H H
H H H
+:B ^B^ (16.88)
H H H
Although adequate from a formal point of view, it suffers from the usual unwieldiness
of VB terminology when extensive delocalization exists. A second attempt considers
the B,Hj" anion as isoelectronic and isostructural with ethylene. C : H 4 . Such an ion
would have a cloud of electron density above and below the B—H plane. 122 The
neutral B,H ( , molecule could then be formally produced by embedding a proton in the
electronic cloud above and below the plane of the B 2 Hj ion. Although this may
appear to be somewhat farfetched, it is but a simplistic way of describing the bonding
model which is currently accepted as best—the three-center, two-electron bond.
Fig. 16.39 Molecular
structure of diborane.
B : H„.
121 Professor Lipscomb's work has been of such value that he received the 1976 Nobel Prize in
chemistry (sec Science 1977, 196. 1047-1055 for his Nobel Laureate address). It has been aptly
said, "Boron hydrides are the children of this century, yet the discovery of polyhedral borancs.
carborunes, and mctalloborancs and the subsequent elaboration of chemistry, structure, and
theory, and the incredibly rapid one considering the small number of investigators, arc among the
major developments in inorganic chemistry" (Boron Hydride Chemistry. Muettcrtics. E. L.. Ed.;
Academic: New York. 1975).
122 Note that this is true whether the it-tt model or a bent-bond model is employed for the double
bond.
Boron Cage Compounds
791
Consider each boron atom to be sp hybridized. 123 The two terminal B — H bonds
on each boron atom presumably are simple cr bonds involving a pair of electrons each.
This accounts for eight of the total of twelve electrons available for bonding. Each of
the bridging B—H—B linkages then involves a delocalized or three-center bond as
follows. The appropriate combinations of the three orbital wave functions. d> Bl .
(approximately sp hybrids), and d> H (an j orbital) result in three molecular orbitals:
B> + ^b 2 + v!^h (16.89)
!>„ = ^d> B , “ (16.90)
= i^B, + -d’B; + v^H (16.91)
where <// h is a bonding MO. t !>„ is an antibonding MO, and i/>„ is. to a first approxima¬
tion, a nonbonding MO. 124 The diagrammatic possibilities of overlap together with
sketches of the resulting MOs and their relative energies are given in Fig. 16.40.
Each bridging bond thus consists of a bonding MO containing two electrons.
Although the nonbonding orbital could conceivably accept an additional pair of
electrons, this would not serve to stabilize the molecule beyond that achieved by the
configuration t/ig. The second B—H — B bridge likewise may be considered to have a
configuration t lij,. This accounts for the total of twelve bonding electrons and provides
the rationale for the existence of the dimer (Fig. 16.41).
Diborane provides examples of two types of bonds found in higher boranes: the
two-center, two-electron B—H terminal bond and the three-center, two-electron,
Fig. 16.40 Qualitative description of atomic orbitals (left), resulting three-center molecular
orbitals (right), and the approximate energy level diagram teenier) for one B—H—B bridge
in diborane.
i: -' This is only approximately correct (sec Problem 16.33). The argument here does not rest upon the
exact nature of the hybridization.
| 2- 1 The exact energy of this orbital varies depending on the nature of the bonding properties of the
atoms involved. For our purposes it does no harm land simplifies the model) if it is assumed to be
nonbonding.
A-84
Index
London dispersion forces
(continued)
van der Waals forces and
van der Waals, radii
Lowest unoccupied molecular or¬
bital (LUMO), 351, 721, 722
Low oxidation state transition
metals, 580-581
Lowry, T. M., 318
Lux, H., 319
Lux-Flood acid-base definition,
319-320
Macrocycles, 525-531
Mudelung constants, 100-101,
262-263, 295, 308, 311,
340
Madelung energy, 199, 291, 311,
340
Magnetic properties
of complexes, 45. 459-468, 485
of lanthanides and actinides,
607
Magnetic quantum number, 19
Magnetic susceptibility
mass. 460-463
molar, 461
volume, 460
Malm, J. G., 70
Map of twist angles, 490
Marcus theory, 571
Mass spectrometry. 239
Maximum multiplicity, 26-27
Mechanisms
inner sphere, 565-567
outer sphere, 558-561
of redox reactions, 557-572
of substitution reactions.
545-547. 551-553
Medicinal chemistry, 954-960
Meissner effect, 285
Melting points, and chemical
forces, 307-310
Metal alkyl complexes, 655-657
Metal carbonyl complexes,
630-649
Metal chains, infinite, 818-819,
820
Metal clusters, 807-819
Metal complexes, as probes of
nucleic acids, 958-960
Metal ions, and nephelauxetic
effect, 413
Melallacarboranes, 801-802
Metallocenes, 669-686
Metalloporphyrins, and respira¬
tion, 891-895
Metallothioneins, 932
Metals, 27-28
descriptive chemistry of,
577- 622
general periodic trends,
578- 579
heavier transition, 587-588
inner transition, 608-609
oxidation states of, 580-581
solutions of, 377
solutions of, ammonia, 362-364
Metal-to-ligand charge transfer
(MLCT), 457
Methyl migration, 696-698
Migration, ion, 266-269
Minerals, silicate, 742-750
Mirror plane, 48
Mixed valence complexes.
568-569
Mobil gasoline synthesis, 711,
717
Models, paper, A40-A43
Molar susceptibility, 461
Molecular orbital (MO) theory,
139, 153-182, 413-433
and ir bonding, 420-425
and eighteen-electron rule,
624-625
and molecular structure,
218-220
Molecular orbitals
in heteronuclear diatomic mol¬
ecules, 167-175
in homonuclear diatomic mole¬
cules, 160-166
of metallocenes, 670-673
in octahedral complexes,
414-418
in square planar complexes,
418-420
symmetry of, 160
in tetrahedral complexes,
418-420
in triatomic molecules and
ions, 175-182
Molecular structure
experimental determination of.
233-237
and hybridization, 220-231
and molecular orbitals,
218-220
Molecular symmetry, and point
groups, 53-59
Molecules
reactivity of, 203
stereochemically nonrigid,
723-730
structure of, 203-220
Molten salts, 374-378
Monoclinic crystal system, 75, 78
Monsanto acetic acid process,
712-714
Morris, D. F. C., 312
Mossbauer spectrascopy, 908,
909
Muetterties, E. L., 223, 241
Muller, K. A., 285
Mulliken-Jaff6 electro¬
negativities, 183-186, 258
Mulliken labels, 61
Multiplicity, 26-27
Myoglobin
and dioxygen, 897-899. 900
physiology of, 900-902
Naples yellow, 458
Natta. G., 718
Negative oxidation state, 580-581
Neon. 163-164
Nephelauxetic effect, 413
Neutral atom electronegativity.
186
Neutron diffraction, 235, 238,
497, 645
Nido structures, 798-800. 807
Nitrite ion, 175-182, 201-202,
210, 249
Nitrogen, 164-166, 864-865
Nitrogen dioxide, 210
Nitrogen fixation, 933-935
Nitrosyl complexes, 650-653
Nitryl ion, 210
Noack, K., 696
Nobel prize winners
Bednorz, J. G., 285
Brown, H. C., 387
Index
A-85
Nobel prize winners (Continued)
Elion. G. B., 929
Fischer, E. O., 669
Hoffmann, R., 387, 647
Lipscomb, W., 387
Muller, K. A.. 285
Natta, G„ 718
Taube, H., 387
Werner, A., 387
Wilkinson, G„ 387, 669
Ziegler, K., 718
Noble gases, 357, 824, 825-836
discovery and early chemistry,
825-827
fluorides, 828-829
Nomenclature of inorganic chem¬
istry, A46-A77
Nonaqueous solvents, 359,
360-373, 381-383
Nonaromatic alkene complexes,
662-668
Nonbonded repulsions, and
structure, 229-230
Nonbonding orbitals, 157
Nonmetals, 28
biochemistry of, 953-954
and d orbitals, 866-875
hybrid orbitals for, 195-196
periodic anomalies of, 876
Nonphotosynthetic processes,
889-891
Nonrigid molecules, 723-730
Normalizing constant, 156
Nuclear magnetic resonance
(NMR) spectroscopy, 7. 8,
239. 335. 469, 471
of dihydrogen complexes,
645-646
of fluctional molecules,
725-730
of hydride complexes, 644
of polyoxymetallates, 762-764
shift reagents, 613
and water exchange reactions,
549-550
Nucleic acids
metal complexes as probes of,
958-960
reaction of cisplatin with, 957
Nucleophiles, carbonylate anions
as. 703-705
Nucleophilic attack of coordi¬
nated ligands. 700-703
Nucleophilic displacement,
243-244
Nucleophilic substitution
in octahedral complexes,
548-556
in square planar complexes,
540-547
Oak Ridge Thermal Ellipsoid
Program (ORTEP) diagram,
234-235, 478
Octahedral complexes, 397-399,
414-418, 441. 448-455,
488-489
geometric isomerism in,
491-492
optical isomerism in, 492-494
Octahedral substitution, kinetics
of, 548-557
Olivine, 253, 256
One-dimensional conductors,
752-755
Optical activity. 63-74, 492-494.
697-698
Optical rotatory dispersion
(ORD), 496, 497, 498
Orbitals
and atomic structure, 11
d. 395-396
d, in it bonding. 868-875
energies of. 18-20
/. 604
ligand group (LGOs), 175-176,
179-182, 414-420
of metallocenes, 670-673
molecular. See Molecular
orbitals
p in ir bonding, 861-866
symmetry of, 17-18, 160
Organometallic chemistry, 7,
623-737
Organometallic clusters, struc¬
ture prediction for, 802-807
Organometallic complexes
catalysis by, 705-723
reactions of, 686-705
Organometallics, 623
Orgel diagrams, 440, 448
Orthorhombic crystal system, 75,
78
Outer sphere mechanisms,
558-561
Overlap
and hybridization, 153
and symmetry, 157-160
Oxidation, 2
Oxidation states
comparison of properties by,
581-582
and emfs, 588-599
of halogens, 837-848
of inner transition metals,
599-602
range of, 581
stabilities of, 588-590
of transition metals, 580-582
Oxidative addition, and reductive
elimination, 538, 689-695
Oxide acceptor, 320
Oxides
acidity of nonmetal. 327
acidity parameters for, 321
basicity of metal, 326-327
halogen, 846-847
phosphine. 868-871
Oxo process, 71 1
Oxyacids. of heavier halogens.
845-846
Oxyfluorides, 846-847
Oxygen. 163, 843-845
Packing, and crystal lattices,
118-122
Paper models, A40-A43
Paramagnetic atoms, 22
Paramagnetism, 460-468
Pauli exclusion principle, 22
Pauling, Linus, 139, 182, 183,
185, 193-194, 391, 811
Pauli principle, 21-23
Pauli repulsive forces, 22, 206
p block elements, 28
Pearson, R. G., 352
Peattie, Donald Culross, 960
Penetration, 14
Pentadienyl complexes, 666-668
Pentafluorotellurate anion, 209
Periodicity, 857-888
diagonal relationship, 860-861
792
16- Inorganic Chains, Rings, Cages, and Clusters
Boron Cage Compounds
793
(a) (b)
Fig. 16.41 (a) Qualitative picture of bonding in diborane. (b) A common method of
depicting B—H — B bridges.
bridging B—H —B bond. Two other bonds are of importance in the higher analogues:
(I) the two-center, two-electron B—B bond, best exemplified by the boron sub¬
halides. X-.B—BX 2 ; and (2) the three-center, two-electron B—B — B bond, which
may be formed by overlap of three orbitals from three comers of an equilateral triangle
of boron atoms (Fig. 16 . 42 ).'25 Like the three-center B—H—B bond, three molecular
orbitals will result, of which only the lowest energy or bonding one will be occupied by
a pair of electrons.
With this repertoire of bonding possibilities at our disposal, we can construct the
molecular structures of various boron-hydrogen compounds, both neutral species and
anions. The simplest is the tetrahydroborate '26 or borohydride ion, BH 4 . Although
borane is unstable with respect to dimerization, the addition of a Lewis base, H ,
satisfies the fourth valency of boron and provides a stable entity. Other Lewis bases
can coordinate as well.
B 2 H„ + 2NaH 2NaBH 4
(16.92)
B 2 H a + 2CO -► 2H 3 BCO
(16.93)
B^H,, + 2R,N -» 2 H 3 BNR 3
(16.94)
Fig. 16.42 The closed three-center, two-electron B—B—B bond: (a) formation from three
boron atomic orbitals: (b) simplified representation.
135 For some heteroatom systems an open or linear B—B — B bond which more resembles the open
B—H — B bond discussed above is required.
136 Tetrahydroborate is the preferred name.
Although the ammonia adduct of BH, is stable, it must be prepared by a method that
does not involve B,H h such as:
Me-jOBHj + NHj - H,BNH, + Me 2 0 ( 1 6 . 95)
Direct reaction of ammonia and diborane results in the “ammoniate" of diborane,
which has been shown to be ionic. 127
B 2 H 6 + 2NH 3 -► [BH 2 (NH 3 ) 2 ] + + BH; (,6.96)
This unsymmetric cleavage:
B
H
H
is typical of reactions with small, hard Lewis bases and further examples will be
discussed below. Larger bases, such as phosphines, promote symmetric cleavage:
H
H
Diborane is very air sensitive, reacting explosively when exposed to air. Allhough
it is said that in extremely pure form, the compound is stable in air at room tem¬
perature, these conditions are rarely met. In general, the higher molecular weight
boranes are much less reactive. For example, decaborane (B I0 H I4 ) is quite stable in
air.
All of the compounds discussed thus far (except diborane and decaborane) con¬
tain only two-center, two-electron bonds. A simple boron hydride containing three
types of bonds is tetraborane, B 4 H in (Fig. 16.43). It is formed by the slow decomposi¬
tion of diborane:
2B,H 6 -► B 4 H 10 + H, (16.97)
In addition to terminal and bridging B—H bonds, this compound contains a direct
B—B bond. Tetraborane undergoes both symmetric and unsymmetric cleavage (Fig,
16.44). Larger Lewis bases tend to split oil' BH, moieties, which are cither complexed
or allowed to dimerize to form diborane:
2B 4 H, 0 + 2Et 2 0
2B 4 H 10 + 2Me 2 S
2Et 2 OB 3 H 7 + B 2 H 6
2Me 2 SB 3 H 7 + B 2 H 6
(16.99)
(16.100)
Small, hard Lewis bases such as ammonia and the hydroxide ion result in unsym¬
metric cleavage, ie., the splitting off of the BH} moiety:
B 4 H io + 2NH, -> [H 2 B(NH 3 ),r + [B 3 H S ]“ (16.101)
2B 4 H io + 40H“ -► [B(OH) 4 r + [BHJ" + 2[B,H 8 r ( 16 . 102 )
137 Further heating gives borazine (see Eq. 16.34).
Gasoline, synthetic, 715-717
Gas-phase proton affinities,
332-333
Geometric isomerism, in octa¬
hedral complexes, 491-492
Gerade orbitals, 17-18
Gerade symmetry, 48, 160
Gillespie, R. J.. 832
Glide plane, 77
Gold, 2, 881-885
Gouy method for magnetic sus¬
ceptibility, 461
Graham, W. A. G.. 694
Grignard reagent. 680, 681, 704
Ground state. 23
Ground terms, 441, A12
Group electronegativity, 196-197,
202
Group IA (I), 593-594
Group IB (I I). 597-598
Group 11A (2), 594
Group I IB (12). 598-599
Group 111B (3), 594
Group 1VB (4). 594
Group VB (5). 594-595
Group VIB (6). 595-596
Group V11B (7). 596
Group(s) VIIIB (8. 9. 10).
596-597
Groups (symmetries), and char¬
acter tables. AI3-A20
Group theory, and symmetry.
46-91
Guest molecules. 305
Guggenberger. L. J.. 223. 241
Gutmann, V., 323
Half-lives, of selected actinide
nuclides, 614
Halides. 308. 309. 848-852
Halogen cations, 847-848
Halogen oxides. 846-847
Halogens, 824
electrochemistry of. 853
heavier, oxyacids of, 845-846
physical inorganic chemistry
of. 848
in positive oxidation stales,
837-848
and pseudohalogens, 852-853
Hapticity, 629, A77
Hard acids and bases. 344-355
Hard-soft acid-base (HSAB)
interaction, 351
Hartree-Fock method, 20
Heavier transition metals,
587-588
Heck, R. F., 711
Heme, and dioxygen, 895-897
Hemerythrin, 908
Hemoglobin
physiology of, 900-902
structure of, 902-908
Hermann-Mauguin system,
77, 78
Hess’s law, 108
Heteroboranes, structure predic¬
tion for, 802-807
Heterocatenation, 741-742
Heterocyclic inorganic ring
systems, 775-780
Heterogeneous catalysts, 705
Heteronuclear diatomic mole¬
cules, molecular orbitals in,
167-175
Heteropoly anions, 760-764
Hexagonal crystal system, 75, 78
Hexanuclear clusters, 816
Hexgonal closest packed (hep)
system. 120
Hieber, Walter. 643
Highest occupied molecular
orbital (HOMO), 351, 428,
721, 722
High-temperature supercon¬
ductors, 285-288
Hinckley, C. C., 613
Hoffman. R., 387, 647
Homocyclic inorganic systems,
780-785
Homogeneous catalysts, 705
Homonuclear diatomic mole¬
cules, molecular orbitals in,
160-166
Horizontal plane of symmetry,
51
Host molecules, 305
Hoyano, J. K., 694
Huckel rule, 772
Hund’s rule, 26-27
Hybridization, 148-153
energetics of, 225-229
and overlap, 153
and structure, 220-231
Hydrates, 304-306
Hydration enthalpy, 312
Hydride complexes, 641-644
Hydroformylation, 711-712
Hydrogenation of alkenes, 503,
706-708
Hydrogen atom, 10-20
Hydrogen bonding, 300-306
Hydrogen-bridged systems, 305,
316, 646. 791
Hydrogen migration. 709
Hydrolysis constants, and
charge-size functions, 328
Hydrometallurgy, 382-383, 384
Hydron, and proton, 318, A49
Hypho structures, 798-800, A77
Identity operation, 51
Immobilized homogeneous cata¬
lysts, 718
Imperfections, in crystals,
263-265
Improper rotation. 52-53
Impurity semiconductors,
274-276
Induced dipole interactions,
298-299
Inert complexes, 547-548
Inert pair effect, 877-879
Infinite metal chains, 818-819,
820
Infrared (IR) spectroscopy,
65-71, 235, 238
of carbonyl complexes,
428-431, 633-634
Inherent atom electronegativity,
186
Inhibition, and poisoning,
925-929
Inner sphere mechanisms,
565-567
Inner transition elements, 29,
599-613
Inorganic chains, 738-764
Inorganic Chemistry, 74
Index
A-83
Inorganic chemistry
of biological systems, 889-964
introduced, 1-9
IUPAC nomenclature of,
A46-A77
literature of, AI-A2
Insertion and elimination reac¬
tions, 695-700
Instantaneous dipole-induced
dipole interactions, 299
Intensity of magnetization, 460
Intercalation chemistry, 750-752
Interchange associative substitu¬
tion reactions, 540
Interchange dissociative substitu¬
tion reactions, 540
Interhalogen compounds,
837-839
International system of sym¬
metry. 77, 78
International System of Units
(SI). A3-A6
International Union of Pure and
Applied Chemistry (IUPAC),
A46-A77
Intemuclear distances, and
atomic radii, 290-296
Interstitialcy mechanism, 266
Interstitial mechanism, 266
Intrinsic semiconductors,
272-274
Inversion, atomic, 237-240
Inversion center. 48-49
In vitro nitrogen fixation.
933-934
In vivo nitrogen fixation, 934-935
Ion-cyclotron resonance spec¬
troscopy, 331
Ion-dipole forces, 297
Ionic bond, 92-99, 129-134,
258-260. 296-297, 678-680
Ionic charge, 146
Ionic compounds, 92-137
thermodynamic calculations
for, 127-129
Ionic radii, 105-108, 291
Ionic resonance energy, 194, 201,
340
Ionic solids, conductivity in,
266-269
Ionization. 38-40
Ionization energy, 35-38
Ionization isomerism, 521
Ion migration. 266-269
Ions, 92
bond lengths and ionization
energies, 166-167
polyatomic, 117
polyhalide, 839-843
size effects, 115-127
Iron, biochemistry of, 935-941
Irreducible representations, and
character tables, 59-63
Irving-Williams series, 348-349,
454
Island model of phosphazenes,
772
Isoelectronic complexes, 640
Isolobal fragments, 647-649
Isomerism, 475-477, 486-488
in five-coordinate complexes,
486-488
geometric, 461-462
linkage isomerism. 513-521
optical, 492-494
types of. 521-522
Isomerization, and racemization.
555-557
Isopoly anions. 755-760
Isostructural carbonyl com¬
plexes, 640
I-strain. 342
Jaffe. H. H.. 183, 185. 258
Jahn. H. A.. 449
Jahn-Tellcr theorem. 403.
449-455. 524
Janowicz, A. H.. 694
Jprgensen. Sophus Mads. 387
Kinetics
of octahedral substitution.
548-557
of redox reactions, 557-572
of square planar substitution,
538-547
Kinetic stability, 547-548
Koopmans' theorem, 39. 166,
351
Labile complexes, 547-548
Lacunary structures, 762
Langford. C. H., 553
Lanthanide chelates, 610-613
Lanthanide elements, 28,
599-607, 609-610
Laporte rule, and transitions. 438
Latimer diagram, 588, 853
Lattice energy, 99-115
Lattice points, 74
Lattices, and efficiency of pack¬
ing, 118-122
Layered structures. 260-262
Lecompte, J., 514
Lewis. G. N„ 324-325
Lewis acid-base interactions,
systematics of. 336-340
Lewis acid-base definition,
324-325
Lewis structure, 138-139
Life, energy sources for, 889-891
Ligand cone angles, 688-689, 690
Ligand field effects, and reaction
rates, 550-551
Ligand field (LF) theory. 391.
434
Ligand field stabilization energy
(LFSE). 474. 578
Ligand group orbitals (LGOs),
175-176. 179-182. 414-418,
421. 670-673
Ligand isomerism, 521
Ligands
and electron counting, 626
and ncphelauxctic effect. 413
nucleophilic and electrophilic
attack or, 700-703
sepulchrate, 530
Ligand-to-metal charge transfer
(LMCT). 457
Linear combination of atomic or¬
bitals (LCAO) method, 153
Linkage isomerism. 513-521
Lipscomb. W., 387
Literature of inorganic chem¬
istry. AI-A2
Lithium. 163
Localized pair, 143
London dispersion forces, 299.
307, 310, 314. See also
794
16- Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.43 Bonding and structure of tetraborane, B 4 H, 0 . [From Muetterties, E. L. The
Chemistry of Boron and Its Compounds', Wiley: New York. 1967. Reproduced with
permission.)
Fig. 16.44 Symmetric
(a) and unsymmetric
l_l (b) cleavage of tetraborane.
X H
The reaction in Eq. 16.102 can be considered as an abstraction of BHt if it is assumed
that [H ; B(OH) 2 r forms and disproportionates:
2[H,B(OH) 2 ]- -► [B(OH)J- + [BHJ- (16.103)
Rather than continue to progress from less complex to more complex
boron-hydrogen compounds, it will be more convenient to jump to a complex bill
highly symmetric borohydride ion. [B, 2 H| 3 ) 2- . It may be synthesized by the pyrolysis
of the [BjHg] - ion:
5[B 3 H 8 ]- - [B 12 H 12 ] 2 - + 3[BHJ- + 8H 2 (16.104)
It is not necessary to use [B 3 H g r directly in this reaction; it may instead be formed in
situ from diborane and borohydride:
2[BH 4 ]“ + 5B 2 H 6 - [B 12 H l2 ] 2 - + 13H 2 (16.105)
The [B I2 H, 2 ) 2- ion is a regular icosahedron of atoms, each of the twenty faces
being an equilateral triangle (Fig. 16.45a). All of the hydrogen atoms are external to
the boron icosahedron and are attached by terminal B—H bonds. The icosahedron
itself involves a resonance hybrid of several canonical forms of the type shown in Fig.
16.45b and c. Both two-electron, two-center B—B and two-electron, three-center
B—B — B bonding are involved.
An icosahedral framework of boron atoms is of considerable importance in boron
chemistry. Three forms of elemental boron as well as several nonmetal borides
Boron Cage Compounds
795
let
Fig. 16.45 (a) Molecular structure of the |B| 2 H| 2 ) S- anion: (b). (c) front and back of the
[B l2 H l2 J'~ framework showing one of many canonical forms contributing to the resonance
hybrid, [(a) From Muetterties. E. L. The Chemistry of Boron and Its Compounds: Wiley:
New York. 1967; (b) from Jolly. W. L. The Chemistry of the Non-Mends: Prentice-Hall:
Englewood Cliffs. NJ. 1966. Reproduced with permission.!
contain discrete B, ; icosahedrn. For example, rr-rhombohedrul boron consists of
layers of icosahedra linked within each layer by three-center B — B—B bonds and
between layers by B—B bonds (Fig. 16.46). /3-Rhombohedral boron consists of twelve
B )2 icosahedra arranged icosahcdrally about a central B i; unit. i.e.. B, 2 (8, 2 ) (2 .
Tetragonal boron consists of icosahedra linked, not only by B—B bonds between the
icosahedra themselves, but also by tetrahedral coordination to single boron atoms.
Several boranes may be considered as fragments of a B| 2 icosahedron (or of the
[BpH,,] 2- ion) in which extra hydrogen atoms are used to "sew up" the unused
valences around the edge of the fragment. For example, decaboranel I4) 1 - 8 (Fig.
16.47a) may be considered a B i; H i; framework from which B, and B (l (Fig. 16.45)
have been removed leaving "dangling" three-center bonds that are completed with
hydrogen atoms to form B—H—B bridges.
Other examples of boranes that are icosahedral fragments tire hexaborane(IO),
which is a pentagonal prism (Fig. 16.47b), pentaboranef 11). similar lo the former with
I2K The prolix gives the number of boron atoms and the number in parentheses gives the number ol
hydrogen atoms. Thus decaboranel 14) is B u>H u-
A-80
Index
Index
A-81
Chloride acceptors and donors,
372
Chlorophyll, and pholosynthetic
reaction center. 917-919
Chromium carbonyl complexes,
bond lengths in, 427
Circular dichroism (CD), 496-499
Claasen, H. H., 70
Clathrate compounds, 304-306
Clathro-chelates, 530
Clays, 750
Clementi. E., 31, 32
Closo structures, 798-800. 807
Clostridium pasteurianum, 934
Clusters, 738, 807-819
Coenzyme, 919
Coenzymes, vitamin B l2 ,
929-931
Collman’s reagent, 704
Complex formation, of molten
salts, 377-378
Complex solids, 253-263
Concentration, and stability,
590-593
Conductivity
of coordination compounds,
389
in ionic solids, 266-269
in metals, 271-272
in semiconductors. 272-276
Conductors, one-dimensional,
752-755
Cone angles. 688-689. 690
Conformation of chelate rings,
498-502
Conjugates, 319
Conjuncto structures, 798-800
Conversion factors and units,
A3-A6
Coordinated ligands, nucleophilic
and electrophilic attack of,
700-703
Coordination complexes
asymmetric syntheses, cata¬
lyzed by. 502-503
bonding, 391-433
conductivity of, 389
electronic spectra of, 433-459
of inner transition metals,
608-613
magnetic properties of,
459-468
pi bonding in, 421-433
reactions, kinetics, mecha¬
nisms, 537-576
structure. 472-536
Coordination isomerism, 521-522
Coordination number 1, 472-473
Coordination number 2, 473
Coordination number 3, 474
Coordination number 4, 474-479
Coordination number 5, 479-488
Coordination number 6, 488-503
Coordination number 7, 503-507
Coordination number 8, 507-509
Coordination numbers
generalizations about, 511-513
higher, 509-511
and radius ratio, 123, 473
Corrin ring, 929
Cotton effect, 496, 497, 498
Coulomb’s law, 100
Coulson, C. A., 140
Covalent bonds, 138-202, 296
and ionic bonds, 129-134,
258-260, 678-680
and reactions of molecules,
237-246
solids held together by,
269-276
Covalent energy, 340
Covalent radii, 291-296
Craig/Paddock model, 773
Cristobalite, 97-98
Crown ethers, 525
Cryptates. 530
Crystal field (CF) theory, 391,
394-413, 474
Crystal field stabilization energy
(CFSE), 399-401, 408-413
Crystal lattices, and efficiency of
packing, 118-122
Crystallography, 74-85
Crystals, imperfections in,
263-265
Crystal systems, 75
Cubic closest packed (ccp)
system, 120
Cubic crystal system. 75, 78
Curie's law, 466
Curie-Weiss law, 467
Cyclobutadiene complexes,
684-686
Cycloheptatriene complexes,
683- 684
Cyclooctatetraene complexes,
684- 686
Cyclopentadienyl complexes
reactions of, 700-703
structures of, 673-678
synthesis of, 680-681
Cytochromes, 569-572, 891-895
d block elements, 27-28, 583-588
Defect semiconductors, 274-276
Degenerate, representations, 62
Delocalization, 143
Density of states (DOS), 280
Deoxyhemerythrin, 909
Deoxyribonucleic acid (DNA),
958-960
Descriptive chemistry, of metals,
577-622
Dewar, M. J. S., 772
Diamagnetic atoms, 22
Diamagnetic susceptibilities,
462-463
Diamagnetism, 467-468
Dichloroiodate anion, 209
Diffraction
neutron, 235, 238, 497, 645
X-ray, 233-235, 238, 497. 645
Dihydrogen complexes. 644-646
Dimension of representations, 61
Dinitrogen complexes, 566,
653-655, 933-935
Dinuclear complexes, 808-813
Dioxygen
binding, transport, use,
895-910
and heme, 895-897
and myoglobin, 897-899, 900
Dipole-dipole interactions, 298,
307
Dipole moments, 64-65 , 201
Disorder, crystallographic, 80-85
Displacement, nucleophilic,
243-244
Dissociative substitution reac¬
tions, 540
Distortions, from perfect octa¬
hedral symmetry, 488-489
Donor number (DN), 370
Drago, R. S., 218, 233 n. 323.
335, 336. 337, 341. 352
Dutt, N. K., 556
Duval, C., 514
Eigenfunction. 10
Eighteen-electron rule, 624-630
Electrochemistry
of halogens and
pseudohalogens, 853
in nonaqueous solutions.
381-382
Electrode potentials, 378-383,
A35-A37
Electron
counting complexes, 625-630
Electron affinities, 40-43
anomalous, 880-884
Electron configuration, 23
Electronegativity, 142, 171.
182-199, 258
Allred-Rochow, 195
alternation of, 884-885
of elements, 187-190
equalization, 198-199. 258
group. 196-197
and hardness and softness of
acids and bases, 350-355
Mulliken-Jaffd, 183-258
Pauling's electronegativity.
193-194
recent advances in theory for,
186-191
variation of. 191-193
Electroneutrality principle,
393-394
Electronically equivalent groups,
647
Electronic spectra
of complexes, 433-459
of lanthamide and actinide
ions. 604-607
Electron-nucleus attraction. 27
Electron paramagnetic resonance
(EPR), 414, 914
Electron spin, and Pauli prin¬
ciple, 21-23
Electron transfer, and respiration
and photosynthesis, 911-919
Electrophilic attack of coordi¬
nated ligands, 700-703
Electrophilic substitution in
square planar complexes,
538
Elements
and atomic radii and multiple
bonding parameters, 292
electron affinities of. 42-43
electron configurations of,
24-25
electronegativities of, 187-190
essential and trace, in biolog¬
ical systems, 941-953
ionic radii of. 105-108
ionization energies of, 35-38
nuclear charges of, 34
periodicity of, 27-29
potassium-zinc comparison by
electron configuration,
582-587
standard reduction potentials
of, A35-A37
Elimination and insertion reac¬
tions. 695-700
Elion. G. B., 929
Enantiomers. 476-477. 495
Energetics of hybridization.
225-229
Energy
bond, 340-341, A2I-A34
bond lengths and ionization
energies. 166-167
crystal field stabilization
(CFSE), 399-401
exchange, 27, 141
ionization, 35-38, 579
lattice, 99-115
of orbitals, 18-20
sources of. for life, 889-891
Enthalpy, 312
Entropy, 311
Enzymes, 919-932
Escher, M. C., 90
Essential elements, in biological
systems, 941-953
Ethylenediamine complexes, and
chelate effect, 523
Eutrophication, 3
Exchange energy, 27, 141
Excited state outer sphere elec¬
tron transfer reactions,
561-565
Excited terms, 441, AI2
Experimental determination, of
molecular structure, 233-237
Experimental evidence for -n
bonding, 425-433
Extended X-ray absorption fine
structure (EXAFS), 238, 639
Eyring, Henry, 391
Facial isomer, 491-492
Fajans, K., 129
Fajans’ rules, 129-130. 169, 258.
309
Faraday. Michael, 460
Faraday method for magnetic
susceptibility, 461-462
/block elements. 28, 599-613
Fermi level, 271
Ferredoxins. 911-912
Ferromagnetism. 467-468
First-stage interclatcs. 751
Fischer carbenes, 658
Fischer. E. O.. 669
Fischer-Hafner adaptation, of
Friedel-Crafts reaction. 681
Fischcr-Tropsch reaction. 711,
717
Fixation, nitrogen. 933-935
Flood. H., 319
Flowchart, and point symmetry.
57-58
Fluorides, of noble gases,
828-831
Fluorine, 163-164. 849-851
Fluorine-oxygen chemistry,
843-845
Fluorite. 97, 288
Fool's gold, 2
/orbitals, 604
Formal charge, 146-148
Free radical mechanisms,
244-246, 963
Frenkel defect, 265. 266
Friedel-Crafts reaction, 681
F-strain, 341-342
796
797
16 Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.46 Structure of a-
rhombohedral boron. The
icosahedra are linked within
the layer via three-center
bonds. This layer is linked
to the layer above by B—B
bonds arising from the
boron atoms marked • and
to the layer below by three
additional boron atoms not
seen on the opposite face.
Fig. 16.47 Molecular structures of boranes related to [B| 2 HiJ 2- ; (a) decaborane(l4) formed
by removal of atoms B, and B h ; (b) hexaborane( 10). Note that the pentagonal pyramid is an
apical fragment of an icosahedron, (c) Octaborane(l2) related to [B^HiJ 2- by removal of B,.
B : , Bj. B h . [From Muettcrties, E. L. The Chemistry of Boron and Its Compounds', Wiley:
New York. 1967. Reproduced with permission.]
Boron Cage Compounds
a basal boron atom missing. octaborane(l2) (Fig. 16.47c), and nonaborarie (15).
Although relating borane structures to icosahedra was the first successful means of
systematizing the structural chemistry of these cages, further experimental work
revealed that the icosahedron of [B, 2 H I2 ] 2- was merely the upper limit of a series of
regular deltahedra. 1 - 9 [B„H„]- - . complete from n = 6 to n = 12. An n = 4 structure
also exists in the form of B 4 CI 4 (Fig. 16.48). If all of the vertices of the deltahedron are
occupied, as in the [B„HJ 3- series, the structure is called a closo (Gr., "closed' - )
structure. It is possible to correlate the structure of a borane or its derivatives with the
number of electrons involved in the bonding in the framework of the deltahedron. 130
The number of vertices in the deltahedron wiil be one less than the number of bonding
pairs in the framework. This approach is sometimes called the polyhedral skeletal
Fig. 16.48 (a) The structure of the (BkH„J 3- anion compared to an idealized dodecahedron.
[From Guggenberger, L. J. Inorg. Client. 1%9. S, 2771. Reproduced with permission.] (b)
Molecular structure of B 4 CI 4 compared to an idealized tetrahedron. [From Muctterties, E. L.
The Chemistry of Boron and Its Compounds', Wiley: New York. 1967. Reproduced with
permission.]
1:9 A deltahedron is a polyhedron with all faces that are equilateral triangles. The deltahedra from n =
4 to n = 12 are tetrahedron (4). trigonal bipyramid (5). octahedron (6), pentagonal bipyramid (7),
bisdisphcnoid (dodecahedron) (8). tricappcd trigonal prism (9). bieappcd square aniiprism (10).
octadccahcdron (II). and icosahedron (12). Most of these are illustrated in Chapters 6 and 12. Sec
also Fig. 16.50.
O'Neill. M. E.: Wade. K. In Metal Interactions with Boron Clusters', Grimes, R. N.. Ed.; Plenum:
New York. 1983. Wade, K. Adv. Inorg. Client. Rudioelietn. 1976. IB. 1-66. Rudolph, R. W, Act'.
Client. Res. 1976. 9, 446-452. Grimes. R. N. Coord. Client. Rev. 1979. 28, 47-96. O'Neill, M. E.;
Wade. K. In Comprehensive Organometallic Chemistry, Wilkinson. G.; Stone, F. G. A.; Abel, E.,
Eds.; Pergamon: New York. 1982; Vol. I. Chapter I.
Index
Absolute configuration of com¬
plexes, 495-496
Absolute electronegativity, 351
Absolute hardness, 351
Absorption spectra of lanthanide
and actinide ions, 604-607
Acceptor number (AN). 370
Achiral molecules, and point
groups, 64
Acids, 2
hard and soft. 344-355
parameters of. 337
reaction rates influenced by,
553-555
solid catalysts, 378
sulfuric acid. 364-367
transition metal hydrides as,
643
See also Acid-base chemistry
and Base
Acid-base chemistry. 318-358
acid and base parameters. 337
bond energies. 340-341
concepts and definitions,
318-332
and electronegativity, 350-355
generalized acid-base concept,
326-332
hard and soft acids and bases,
344-355
measures of acid-base
strength. 330-344
proton sponges, 342-343
solvation effects, and "anoma¬
lies." 343-344
steric effects, 341-342
symbiosis, 348-350
See also Acid and Base
Actinide elements, 28, 599-607,
609-610
Affinities
electron, 40-43, 333-334, 339,
880-884
methyl cation, 353-354
proton, 330-333, 339, 353
Agostic interaction. 693
Alkali halides, 308
Alkali metals, 27
Alkaline earth metals, 27
Alkane activation, 693-695
Alkene complexes, 662-664
Alkene hydrogenation, 503,
706-708
Alkyl complexes, 655-657
Alkyne complexes, 662-666
Allen. L. C„ 191
Allred-Rochow electro¬
negativities, 190, 195
Allyl complexes, 666-668
Alternation of electronegativities,
884-885
Aluminum bromide, 205-206, 247
Amines, basicity of, 329
Ammonia, 360-362
basicity of, 329
and chelate effect, 523
physical properties of, 361
solutions of metals in, 362-364
Ammonium tetrafluoroborate.
205
Anemia, 907
Angular wave functions, 14-17
Anions
carbonylate, as nucleophiles,
703-705
heteropoly, 760-764
isopoly, 755-760
nucleophiles, 703-705
polyatomic zintl, 817
Anomalous ionization energies,
and electron affinities,
880-884
Antibiotics, 954-955
Antibonding molecular orbital,
154
Antiferromagnetism, 467-468
Aprotic solvents, 369-373
Aqueous solvents, and non-
aqueous solvents, 359-386
Arachno structures, 798-800, 807
Aragonite, 98, 953
Arene complexes, 681-683
Arrhenius acid-base definition,
319
Associative reactions substitu¬
tion, 540
A-78-
Index
A-79
Astatine, 851-852
Asymmetric syntheses, 502-503
Atom
electronic structure of, 10-45
hydrogen, 10-20
polyelectronic, 20-43
sizes of, 33-35
Atomic inversion. 237-240
Atomic radii, 290-296
Atomic states, 26-27, A7-A12
Atomization, 307
Aufbau principle, 23-26. 162
Aurophilicity, 884
Axis of rotatory inversion, 77
Azolohacler vinelandii. 934
Back bonding, 393-394
Bailar twist, 556
Band theory, 270-272
Base, 2
hard and soft, 344-355
parameters of, 337
reaction rates influenced by,
553-555
solid catalysts, 378
See also Acid and Acid-base
chemistry
Basicity, of ammonia and
amines, 329
Bcdnorz, J. G.. 285
Bent bonds, 230-231, 251
Bent’s rule. 225-229, 247
Bergman. R. G.. 694
Bernal. I.. 530
Berry. R. S.. 241-243
Berry pseudorotation. 240-243
Beryllium, 163
Bethe. Hans, 394
B-group elements, 27-28
Bijvoet analysis. 495
Biochemistry
of iron. 935-941
of nonmetals. 953-954
Biological systems
essential and trace elements in,
941-953
inorganic chemistry of,
889-964
Blue copper proteins, 912-916
Body-centered cubic system, 121
Boggsite, 2, 5, 7, 747
Bohr magnetons (BM), 462
Boiling points, and chemical
forces. 307-310
Boltzmann distribution. 266. 272
Bond angles
and hybridization, 220-229
and nonbonded repulsions,
229-231
and tt bonding, 203-217
Bond energies, A2I-A34
Bonding
in coordination compounds,
387-433
and group theory, 71-74
in noble gas fluorides, 829-831
Bonding molecular orbital. 154
Bonding theory, 139
Bond lengths, 232-233, A21-A34
in chromium carbonyl com¬
plexes, 427
Bond multiplicity, 232-233
Boranes, 789-800
Borazines, 765-769
Born exponent, 102-103
Born-Haber cycle. 108-114, 201,
288, 311. 340, 355, 357
Born-Land6 equation, 103. 296.
312, 340. 408
Boron. 164-166
Boron cage compounds, 789-807
Bratsch. S. G.. 196. 853
Bravais lattices, 75. 76
Brcslow. D. S.. 711
Bridging ligands. 305, 512. 627,
636
Bromine trifluoride. 209
Bronsted. J. N.. 318
Brpnsted-Lowry acid-base defi¬
nition. 318-319
Brown. H. C., 387
B-strain, 342
Buckyball. 82, 86
Cages. 738, 785-789
Calcite, 98. 953
Calderazzo, F., 696
Calvert. R. B., 692
Carbene complexes, 655, 657-662
Carbide complexes, 655, 657-662
Carbon, 164-166, 861-863
Carbonylate complexes, 639-641
as nucleophiles, 703-705
Carbonyl complexes, 630-649
bridging, 635-638
polynuclear, 633-639
preparation and properties of,
632-633
substitution reactions in,
686-688
Carbonyl fluoride, 211-213
Carbonyl hydride complexes,
641-647
Carboranes, 800-801
Carbyne complexes, 655,
657-662
Catalysis, by organometallic
compounds, 705-723
Catalysts
and asymmetric syntheses,
502-503
solid acid and base, 378
Catenation. 738-741
Cations
halogen. 847-848
polyatomic zintl, 817
polyhalogen, 848
Center of symmetry, 48-49
Cesium chloride. 95-96
Chains, 738-764
Characters, 61
Character tables. 59-63,
AI3-A20
Charge coefficient in electro¬
negativity, 186
Charge transfer spectra, 455-459
Chelate effect, 522-531
Chelate rings
conformation of, 498-502
and Jahn-Teller effect, 454
Chelates, lanthanide, 610-613
Chelate therapy, 956-958
Chemical forces, 290-317
effects of. 307-314
types of, 296-300
Chemical reactivity, 5-7
Chcrnick, C. L., 70
Chevrel phases, 818
Chiral molecules, and point
groups, 64 , 240, 502
798
16 - Inorganic Chains, Rings, Cages, and Clusters
electron pair theory or more often Wade’s rules. For the closo series, the number of
framework electrons equals 2 n + 2. To count framework electrons in. for example.
[B,-,H i2 ] : ~, one notes that each boron atom has one of its three valence electrons tied
up with the exo B—H bond (an exo B—H bond is one extending radially outward
from the center of the cluster; see Figs. 16.45 and 16.48) and it thus has two to
contribute to the framework, giving a total of In (in this case 24) electrons from the B
atoms. No neutral B„H„ species are known, but we have seen an array of dianions
corresponding to the In + 2 rule. The 26 electrons in [B, 2 H, 2 ] 2 ~ are just the number
required to fill all of the bonding molecular orbitals in [B, 2 H I2 ] 2- and correspond to 13
(« + I, n = 12) electron pairs as expected for an icosahedron.
If we (in a thought experiment) remove a boron atom from a vertex of a closo
structure, a cup-like or nest-like structure remains (Fig. 16.49). Such structures are
termed nido (Latin, "nest”). We have seen that structures such as this contain extra
hydrogen atoms to "sew up” the loose valencies around the opening. The nido
structures obey the framework electron formula In + 4. Consider B 5 H 9 . for example.
Five exo B—H groups will contribute two electrons each and the four "extra"
hydrogen atoms will contribute four electrons for a total of 14 [In + 4, n = 5). This
corresponds to 7 (n + 2) electron pairs and the geometry will be derived from an
octahedron (n - I vertices). The structure is thus a square pyramid nido form derived
from the closo octahedron. The extra four hydrogen atoms form bridges across the
open edges of the nest (Fig. 16.49).
If we remove two vertex boron atoms, the resulting framework is an uruchno
(Gr., “spider’s web") structure. With two vertices missing, the structure is even more
open than is the nido case and the resemblance to the parent closo structure is less
apparent. Arachno structures obey the electronic formula 2n + 6 (or n +3 electron
pairs). Pentaboraned I), B 5 B M , must therefore have an uruchno structure. In the
uruchno series the extra hydrogen atoms form endo B—H bonds (lying close to the
framework) as well as bridges.
The hypho (Gr., "net") series of boranes, with electronic formula 2 n + 8. has
been suggested to augment the closo, nido, and uruchno series. Although no neutral
boranes fit this scheme, some borane derivatives do. It is also possible to construct
units consisting of more than one of the above types. These are called conjuncto
(Latin, "joined subunits”) structures.
The complete structural relationships among the closo. nido. and uruchno species
are shown in Fig. 16.50. The diagonal lines connecting the species represent the
Fig. 16.49 (a) Structure of /ii«/o-penlaborane(9); (b) structure of ni«/«-decaborane< 14). Cf.
closo structures in Fig. 16.50.
Boron Cage Compounds
Arachno
Fig. 16.50 Structural relationships of closo. nido, and arachno
boranes. Structural homologues are connected by the diagonal lines:
theoretical redox reactions arc in horizontal rows. |From Rudolph.
R. W. 4cc. Chan. lies. 1976. 9, -146. Reproduced with permission.)
A-76
I IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
appearing before the ligand name and separated by a hyphen. The whole term, e.g., p,-
chloro. is separated from the rest of the name by hyphens, as in ammine-p.-chloro-chloro,
etc., or by parentheses if more complex ligands are involved. The bridging index, the
number of coordination centers connected by a bridging ligand, is indicated by a right
subscript, p. n . where n > 2. The bridging index 2 is not normally indicated. Bridging ligands
are listed in alphabetic order along with the other ligands, but a bridging ligand is cited
before a corresponding nonbridging ligand, as with di-/x-chloro-tetrachloro.
Examples:
[CoCu 2 Sn(CH 3 )][{/i-(C 2 H 3 0 2 )} 2 (C 5 H 5 )] bis(/x-acetato)(cyclopentadienyl)(methyl)-
cobaltdicoppertin
[|Cr(NH 3 ) J },(//,-OH))CI 5 /x-hydroxo-bis(pentaamminechromium)(5 + )-
pentachloride
[[PtCI{P(C (> H,) 3 }) 2 (/x-CI) 2 ] di-/x-chlorobis[chloro(triphenylphosphine)-
platinum]
[{Fe(NO) 2 } 2 (/x-P(C 6 H 3 ) 2 } 2 l bis(/x-diphenylphosphido)bis(dinitrosyliron)
Metal-metal bonding may be indicated in names by italicized atomic symbols of the
appropriate metal atoms, separated by a long dash and enclosed in parentheses, placed
after the list of central atoms and before the ionic charge.
Examples:
[Br 4 ReReBr 4 ] 2 ~ bisftetrabromorhenateX/fe— Re)( 2-)
[Mn,(CO) l0 ] bis(pentacarbonylmanganese)(/V//i— Mn) or
bis(tetrabromorhenate)(/?e— Re)(2 -)
bis(pentacarbonylmanganese)(/Wn— Mn) or
decacarbonyIdimanganese(A7n— Mn)
Organometallic Species
General. Organometallic entities are usually considered to include any chemical
species containing a carbon-metal bond. The simplest entities are those with alkyl radical
ligands, such as diethylzinc. In general, ligands bound by a single carbon atom to metals are
named by the customary substituent group names, though these ligands must be treated as
anions in order to calculate oxidation numbers. In any case, the designation is arbitrary.
Ligands conventionally treated as having metal-donor double bonds (alkylidcnes) and
triple bonds (alkylidynes) are also given substituent group (radical) names.
Examples:
[Hg(CHj) 2 ] dimethylmercury
MgBr|CH(CH,),] bromo(isopropyl)magnesium
| ri(CN)(C,.H,) 2 ] cyanodiphenylthallium
[Fe(CH,CO)l(CO) 2 (P(CH 3 ) 3 } 2 ] acetyldicarbonyliodobis(trimethylphosphine)iron
Complexes with unsaturated groups. Since the first reported synthesis of ferrocene,
the numbers and variety of organometallic compounds with unsaturated organic ligands
have increased enormously. Further complications arise because alkenes, alkynes. imides,
diazenes, and other unsaturated ligand systems such as cyclopentadienyl, C 5 Hj, 1,3-
butadiene, C 4 H fl , and cycloheptatrienylium, , may be formally anionic, neutral, or
cationic. The structures and bonding in some instances may be complex or ill-defined. For
these cases, names indicating stoichiometric composition, constructed in the usual manner,
are convenient. The ligand names are arranged in alphabetical order, and followed by
central atom names, also in alphabetical order. Bonding notation is not given.
Examples:
[PtCI 2 (C 2 H 4 )(NH 3 )] amminedichloro(ethene)platinum
[Hg(C 5 H,) 2 ] bis(cyclopentadienyl)mercury
3- [FejCu^CjHjJjfKCHjJ-NJCjHJJ tetrakis(cyclopentadienyl)tetrakis[(dimethyl-
amino)cyclopentadienyl]tetracoppertetrairon
I - IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-77
met-X and h c S 0f h y drocarbon a "d other rr-electron systems to
metals and the complex structures of these entities have rendered conventional nomen¬
clature impotent. To accommodate the problems posed by the bonding and structures the
hapto nomenclature symbol was devised - The hapto symbol. „ (Greek eta)! Snumer!
hr lT!i SC T'r V,deS 3 [opologlcal description by indicating the connectivity between
ort oX ofThe itnd ^ ^ ^ 77 is 10 e Ifrand name, or to that
p rtion of the ligand name most appropriate, to indicate the connectivity, as in (r, 2 -
kaMn W C T" ' e K e) and < ethe "yH -cyclopentttdienyl). The right superscript numer-
(ExampTi 4to To) S " ° f ‘ i8aIin8 a '° mS which bind 10 tbe -"etui
Examples:
t' !r e S < X )3 ^ 4H ' ,SO)] « ricarbon y | (r-2,5-dihydrothi 0 phene-l-oxide- K 0)iron
5- |Cr(C 3 H s ) 3 J tns(7? 3 -allyl)chromium
1 !p. r n?r Si „ tetra ca r bonyl(r} 4 - 2 -methyIene-l ,3-propanediyDchromium
/. 11 tU-,(C ; ,H 4 )(NH 3 )] ammmedichloro(7j--ethene)platinum
i nl C 2 ,3 1 < ? 7H,,)] (V- b icyclo[2.2.l]hept a -2.5-diene)tricarbonyliron
. tut^M^j bis(7j -l.3.5,7-cyclooctatetraene)uranium [uranocene]
/As dicarbonyl(rj -cyclopentadienyl)-[(| ,2.3-7j)-2,4,6-
V-i:/ /: \ cycloheptatrienyljmolybdenum
BORON HYDRIDES AND RELATED COMPOUNDS
[The basic principles of the structure and naming of boron hydride cage compounds have
in th rhr T K Pler '?■ SeC P anicular| y f ' 8- 16-50. A very similar figure is presented
rizid in Table°i hT Slmllar cx P |anation ' The name; < of «he structure types are summa-
Summary of polyhedral
polyboron-hydride structure-
typos, according to
stoichiometry and electron-
counting relationships
Closed polyhedral structure with triangular faces only; known only as
with molecular formula (B„H„) J -; („ + |) skeletal electron-pairs for
polyhedron.
Nest-like, nonclosed polyhedral structure; molecular formula B H
(n + 2) skeletal electron-pairs; n vertices of the parent (n + H-atom
c/oso-polyhedron occupied.
Web-like, nonclosed polyhedral structure; molecular formula B H
(n + 3) skeletal electron-pairs; n vertices of the parent (n + 2)-atom
c/oso-polyhedron occupied.
Net-like, nonclosed polyhedral structure; molecular formula B H
(n + 4) skeletal electron-pairs; n vertices of the parent (n + 3)-atom ’
c/oso-polyhedron occupied.
Open bmnch-like polyhedral structure; molecular formula B H
. ( " f 5) skeletal electron-pairs; n vertices of the parent (n +"4) , -atom
c/ojo-polyhedron occupied.
37 (Cotlon. F. A. J. Am. Chem. Soc. 1968. 90. 6230.)
800
16'Inorganic Chains, Rings, Cages, and Clusters
hypothetical transformations discussed above, removal of boron vertices moving from
lower left to upper right. 131 The horizontal series represent structures having the same
number of boron atoms but differing in the total number of framework electrons so as
to conform to the closo (2 n + 2), nido (In + 4), or arachno (2 n + 6) electronic
specifications. In a few cases (see below) the change from one structure to an adjacent
one on the same line can be effected by a simple redox reaction, but in most instances
this is not so. However, we can probably anticipate more examples of this type of
transformation now that the general principles of framework electron count and
structure are belter understood.
What is the source of the 2n 4- 2, 2 n 4- 4, and In 4- 6 rules and their corres¬
pondence with the closo, nido, and arachno structures? Space does not allow deriva¬
tion of the molecular orbitals for deltahedra, but the results may be stated simply: For
a regular deluiltedron having n vertices, there will he n + I bonding molecular
orbitals. The electron capacity of these bonding MOs is therefore 2/t 4- 2. This gives
the highly symmetric closo structure. If there are two more electrons (In + 4), one
bonding MO and one vertex must be used for these extra electrons rather than for a
framework atom, and a nido structure with a missing vertex results. Although the
electrons in boranes are delocalized and cannot be assigned to a specific region in a
particular structure, the parallel between the extra electrons in nido structures and
lone pairs in molecules like NH-, is real. The 2 n + 6 formula of arachno structures
simply extends these ideas by one more electron pair and one more vertex.
Carboranes 132 Carbon has one more electron than boron, so the C—H moiety is isoelectronic with
the B—H“ or BH-, moieties. Note that an isoelectronic relationship also exists
between C and BH or B~. In a formal sense it should be possible to replace a boron
atom in a borane with a carbon atom (with an increase of one in positive charge) and
retain an isoelectronic system. The best-studied system. C,B I0 H 12 , is isoelectronic
with [B| 2 Hand may be synthesized readily from decaborane and alkynes and
diethyl sulfide as solvent.
B I0 H I4 + 2Et,S -♦ B 10 H l2 -2SEt, + H, (16.106)
B l() H l2 -2SEt 2 + RC=CR -♦ R 2 C 2 B 10 H I0 + 2Et 2 S + H, (16.107)
The acetylene may be unsubstituted (R = H) or substituted, in which case the
reaction proceeds even more readily. The resulting compound is known as 1,2-
dicarba-c/«.vo-dodecaborane(I2), or the “ortho" carborane, and is isoelectronic and
isostructural with [B l2 H l2 ] 2- . It is stable to both heat and air, but it isomerizes at high
temperatures to the 1,7 ("meta" or "neo” isomer) and the 1,12 ("para" isomer) (Fig.
16.51). The mechanism of isomerization, thought to be intramolecular, has been
discussed for many years. 133
Other dicaboranes are derived from the corresponding pentaborane(5), hexa-
borane(6), hexaborane(8), heptaborane(7), octaborane(8), nonaborane(9), and deca-
1,1 Only Ihc cxo hydrogen atoms are shown in Fig. 16.50. The number of bridging and endo hydrogen
atoms will vary depending upon whether the species is neutral or ionic, has hctcroatoms (sec
below), has a Lewis base coordinated to it. etc.
132 Grimes, R. N. Adv. Inure. Chem. Radiochem. 1983. 26. 55-117. Onak. T. In Comprehensive
Organomeltillic Chemistry, Wilkinson. G.; Slone, F. G. A.; Abel. E. W., Eds.: Pergamon: Oxford.
1982; Vol. I. Chapter 5.
133 Johnson. B. F. G., J. Chem. Soc. Chem. Common. 1986. 27-30.
Boron Cage Compounds
Fig. 16.51 Structures and isomerizations of the three isomers of dicarba-c/two-
dodecaborane. O = B: • = C.
borane(10) dianions. [B n HJ 2 . The monocarboranes. CB,H 7 and CB S H 9 , are also
known.
The carboranes conform to the electronic rules given above for boranes and are
known in closo. nido, and arachno structures. When applying the formulas to the
carboranes, each C—H group should be regarded as donating three electrons to the
framework count. Some carboranes provide interesting examples of the possible
horizontal transformations of Fig. 16.50 mentioned above. For example: IW
c7«.v<!-C 2 B,,H|, + 2e~ -♦ (/jWo-C 2 B v H,,j : " (16.108)
[/i/W«-C 2 B 9 H, 1 1 2_ -♦ [c7of0-C 2 B 9 H M ] + 2e" (16.109)
Metallacarboranes 135 Strong bases attack 1,2-dicarba-c/o.vo-dodecaborane( 12) with the splitting out of a
boron atom:
c . .... c , L C 2 B I0 H i2 + MeO” + 2MeOH -► [C-,B 1> H r ] - + H, T B(OMc), (16.110)
Fig. 16.52 Structure of the
[C 2 B„H|,]’~ anion. The The resulting anion is the conjugate base of a strong acid which may be obtained by
live orbitals directed toward acidification:
the missing apical boron arc
included. (From Adler.
R. G.: Hawthorne. M. F. J.
Am. Chem. Soc. 1970 . 92,
Reproduced with
permission.]
1C,b.,h i2 |- + hci — c 2 b,,h,, + cr (i6.nu
Conversely, treatment of the anion with the very strong base sodium hydride abstracts
a second proton:
[C,B.,H I2 r + NaH — [C 2 B 9 H n ] 2 - + H, + Na K (16.112)
The structure of the [C 2 B 9 H M ] 2- anion is shown in Fig. 16.52. Each of the three boron
atoms and the two carbon atoms on the open face of the cage directs an orbital (taken
as sp for convenience) toward the apical position occupied formerly by the twelfth
boron atom. Furthermore, these orbitals contain a total of six electrons. They thus
,,J Chowdhry. V.: Prctzer. W. R.; Rai. D. N.: Rudolph. R. W. J. Am. Chem. Soc. 1973 . 95,
4560-4565.
155 Metal Interactions with Boron Clusters-. Grimes. R. N.. F.d.: Plenum: New York. 1982. Grimes,
R. N. In Comprehensive Orpanometallic Chemistry. Wilkinson. G.; Stone, F. G. A.; Abel. E..
Eds.; Pergamon: Oxford. 1982. Sec also Kennedy. J. D. Prog. Inorg. Chem. 1986. 34. 211-434.
A-74
I - IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
a completely systematic treatment. Thus, for a particular diastereoisomer of
[Co(NH,) 2 (NH 2 CH 2 CH 2 NH,)Br 2 ]'''. the polyhedral symbol is OC-6 and the ligating atom
priority numbers are as shown below.
Example:
HjN ? H 3,,NH r
(3)/ /
Co-” I
S\ \ /CH,
Br ' NHX 2
For this case, there are no additional complications, and the configuration index is assigned
in the usual way as OC-6-32.
The classic case of diastereoisomerism that arises among chelate ligand derivatives is
the tris(didentate) complexes in which the two donor atoms of the identical ligands are
different. Glycinate, NH 2 CH,C0 2 . and 2-aminoethanethiolatc, NH 2 CH 2 CH 2 S _ . illustrate
this. For complexes of either ligand, the facial and meridional labels described previously
could be applied, but the more systematic configuration indexes are OC-6-22 and OC-6-21.
Examples:
facial.
or OC-6-22
meridional,
or OC-6-21
Priming convention. The configuration index is especially useful for bis(tridenlate)
complexes and for more complicated cases. Bis(tridentate) complexes exist in three di-
astereoisomeric forms which serve to illustrate the utility of a priming convention. These
isomers are represented below, along with their site symmetry symbols and configuration
indexes. For Examples I, 2, and 3, the two ligands are identical and the ligating-atom
priority numbers are indicated.
Examples:
( 2 ) ( 2 ) (!•)
O’) (I)
(2') (I)
Chirality Symbols
Symbols R and S. There are two established and well-used systems for chirality
symbols and these differ in fundamental ways. The first, the convention for chiral carbon
atoms is equally appropriate to metal complexes and is most often used in conjunction with
ligand chirality. However, it can be applied to metal centers and has been useful for
pseudotctrahedral organometallic complexes when, for example, cyclopentadienyl ligands
are treated as if they were monodentate ligands of high priority.
!• IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-75
Skew-line convention for octahedral complexes. The second is the skew line conven¬
tion and applies to octahedral complexes. Tris(didentate) complexes constitute a general
family of structures for which a useful unambiguous convention has been developed based
on the orientation of skew lines which define a helix. Examples I and 2 represent the delta
and lambda forms of a complex, such as [Co(NH 2 CH 2 CH 2 NH,)] 3+ .
Examples:
I. delta 2. lambda
[There follows an explicit discussion of the chirality of octahedral complexes and chelate
rings as summarized in Chapter 12.]
Chirality symbols based on the priority sequence. The procedure is applied as for
tetrahedra, but modified because there is a unique principal axis. The structure is oriented
so that the viewer looks down the principal axis, with the ligand having the higher priority
closer to the viewer. Using this orientation, the priority sequence of ligating atoms in the
[horizontal] plane is examined. If the sequence proceeds in a clockwise fashion, the
chirality symbol C is assigned. Conversely, if the sequence is anticlockwise, the symbol A
is assigned.
Examples:
I. Chiraliiy symbol = C 2. Chirality symbol = A
Chirality symbol = C 4. Chirality symbol = A
[An example of a real compound to which this system may be applied is found on page 488.
c/.v-dicarbonylchloro(cyclopentadienyl)(methykiiphenylphosphine)molybdenumtll). The
enantiomer shown there has the chirality symbol C.]
Polynuclear Complexes
Polynuclear inorganic compounds exist in a bewildering array of structural types, such
as ionic solids, molecular polymers, extended assemblies of oxoanions both of metals and
nonmetals, nonmetal chains and rings, bridged metal complexes, and homo- and hetero-
nuclear clusters. This section treats primarily the nomenclature of bridged metal complexes
and homo- and hetero-nuclear clusters.
Bridging ligands, as far as they can be specified, are indicated by the Greek letter p.
802
16- Inorganic Chains, Rings, Cages, and Clusters
bear a striking resemblance to the p orbitals in the it system of cyclopentadienide
anion. Noting this resemblance, Hawthorne suggested that the [C 2 B 9 H n ] anion
could be considered isoelectronic with C 5 HJ and should therefore be capable of acting
as a u ligand in metallocene compounds. He and his coworkers then succeeded in
synthesizing metallacarboranes. launching a new area of chemistry which is still being
actively investigated: 136
2[C 2 B 9 H m ] 2 - + FeCU_ - [(C 2 B 9 H n ),Fe] 2 - + 2CP (16.113)
[C,B 9 H n ] 2 - + [CjHj] - + Fed,- [C 2 B 9 H n FeC 5 H 5 ]- + 2CI~ (16.114)
[C 2 B 9 H m ] 2 " + BrMn(CO) s -♦ [C 2 B 9 H,,Mn(CO) 3 ]' + Br~ + 2CO (16.115)
The ferrocene analogues, like ferrocene, are oxidizable with the loss of one electron.
In cases for which structures have been determined, they have been found to corres¬
pond to that expected on the basis of metallocene chemistry (Fig. 16.53).
Other heteroboranes such as
O BH # NH
may also be synthesized. ' 37 This compound may be regarded as one in which NH has
formally replaced a BH^~ moiety of [B I2 H I2 ] 2_ (also see Problem 16.41).
Structure Prediction
for Heteroboranes
and Organometallic
Clusters
oft Baker. R. T.; Delaney. M. S.; King. R. E.. Ill; Knobler. C. B.; Long. J. A.; Murder. T. B.:
Paxson, T. E.; Teller. R. G.; Hawthorne. M. F. J. Am. Client. Sac. 1984, 106. 2965-2978. Long,
J. A.; Mardcr, T. B.; Bchnken. P. E.; Hawthorne. M. F. Ibid. 2979-2989. Knobler. C. B.: Mardcr.
T. B.; Mizusawa. E. A.; Teller. R. G.: Long. J. A.; Behnkcn. P. E.. Hawthorne. M. F. Ibid.
2990-3004. Long. J. A.; Mardcr, T. B.. Hawthorne. M. F. Ibid. 3004-3010.
o 7 Muller. J.: Runsink. J.; Paetsold. P. Angew. Client, hit. Ed. Engl. 1991. 30. 175.
I.' O'Neill. M. E.: Wade, K. In Metal Interactions with Boron Clusters'. Grimes. R. N.. Ed.: Plenum:
New York. 1982.
"v Mingos, D. M. P. Acc. Client. Res. 1984, 17. 311-319. Wales. D. J.: Mingos. D. M. P.. Slee. T.:
Zhenyang. L. Acc. Client. Res. 1990. 23. 17-22.
In Chapter 15 we observed that the 18-electron rule was adequate for predicting
stabilities of small organometallic clusters. In this chapter we have seen that Wade's
rules allow us to make predictions about borane structures based on the number of
framework electrons. These rules also are adequate for most carboranes. metallacar¬
boranes. and other heteroboranes.' 3H Furthermore, organometallic clusters that are
not derived from boranes can be dealt with in a similar fashion. More sophisticated
extensions are required for complex larger clusters. 13 ’
Boron Cage Compounds
803
u)
Fig. 16.53 Structures of some curbollyl metallocene compounds, lal Dicarbollyl species:
M = Fe. n = 2. and M = Co. n = I are isoelectronic with ferrocene and cobaliocenium
ion; M = Fe. it = I is isoelectronic with ferroecnium ion: (b) mixed
carbollyl-cyclopentadienyl analogue of ferrocene, n = I: (c) mixed carbollyl-carbonyl
compound. M = Mn. Re. ((a) and (b) from Hawthorne. M. F.: Andrews. T. D. J. Am.
Client. Soc. 1965. 87. 24%: (c) from Adler, R. G.; Hawthorne. M. F. J. Am. Client, Soc.
Reproduced with permission.]
In the previous section we viewed the [C 2 B,,H n ) 2 ~ anion as a ligand analogous to
[C,H 5 ] . Perhaps a more useful approach is to view [Fe(Tj-C s H s )n as a replacement
for a BH fragment, i.e.. a species which, like BH. provides three orbitals and two
electrons. In other words, we might predict that we can replace the BH unit with any
species that is isolobal with it (Chapter 15). Possibilities include (in addition to
|Fe(T) 3 -C,H,)]") Fe(CO),. Co(tj 5-C,H<). Ni(CO),. AIR. or Sn llhe transition metal
fragments are 14-electron species, four electrons short of 18. and the nonlransition
metal units are four electrons short of an octet). Similarly, one could imagine a CH
unit of a earborane being replaced by a species which can provide three orbitals and
three electrons. Fitting this description are Co(CO),. Ni(i 7 5 -C y Hj), and P. In Table
16.2 are listed organometallic fragments and the number of electrons each can provide
to a framework structure. You can construct your own table by remembering that
each transition metal has twelve electrons associated with it that are reserved for
nonframework bonding. 140 Electrons in excess of twelve can be contributed to the
framework (thus the 14-electron species above contribute two electrons each, and the
15-electron species contribute three electrons, etc.). If there are fewer than twelve
electrons in the fragment, the framework must make up the difference [e.g., Mn(CO) 2 ,
an 11-electron species, is assigned a framework contribution of - l|.
'■“» The transition metal has nine orbitals (one ,v. three p. and live il) available for bonding, but only
three are available for framework bonding. The other six. which house twelve electrons, arc used
for bonding to external ligands.
A-72
I - IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
Examples:
l.
©
Cl N
©X®
-N Cl
w
© % c
a n
©X©
-N Cl
W
I5P-4- l|-(aceioni[nlc)dichloro(pyridine)platinum(II)
l]-(aceioniirile)dichloro(pyridine)platinum( II)
Octahedral coordination systems (OC-6). The configuration index consists of two
digits. The first digit is the priority number of the ligating atom trans to the ligating atom of
priority number I. These two ligating atoms define the reference axis of the octahedron.
The second digit of the configuration index is the priority number of the ligating atom trans
to the ligating atom with the lowest priority number (most preferred) in the plane that is
perpendicular to the reference axis.
Examples:
(OC-6 - 22) - trinmminetrinitrocoball(formerly fac - isomer) 34
(OC-6-2I >-lriumminctrinitrocobal(( formerly /
Formulae and Names for Chelafe Complexes
Designation of the ligating atoms in a polydentate ligand
Donor atom symbol as the index. A polydentate ligand possesses more than one
donor site, some or all or which may be involved in coordination. Thus, dithiooxalate anion
conceivably may be attached through S or O, and these were distinguished by dithioox-
alato-5,5' and dithiooxalato- 0 ,O', respectively.
Examples:
0 s'
s 0
/\ /
c
/\ /
c
<-
Pt
2
1
\ A
s 0
bis(dithiooxa-
lato- 0 , 0 ')nickel(ll)
bis(dithiooxa-
Iato-5,5')platinum(II)]
34 The isomer designators fac and mer may be useful for general discussions but are not recommended
for nomenclature purposes_____
I "IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-73
If the same element is involved in the different positions, the place in the chain or ring to
which the central atom is attached is indicated by numerical superscripts.
Examples:
(CH 3 COCHCOCH 3 ) 2.4-pentanedionato-C 3
O = C — 0. 3. 0=C —O' 4. 0 = C — O
tartratop-FO'.O 2
tartrato(4-)-0 2 ,0 3
tartrato( 2 -)- 0 l , 0 4
The kappa convention. As the complexity of the ligand name increases, a more
general system is needed to indicate the points of ligation. In the nomenclature of polyden¬
tate chelate complexes, single ligating atom attachments of a polyatomic ligand to a
coordination center are indicated by the italic element symbol preceded by a Greek kappa,
k. Monodentate ambident ligands provide simple examples, although for these cases the
kappa convention is not significantly more useful than the 'donor atom symbol' convention
[above ]. 35 Nitrogen-bonded NCS [formerly isothiocyanato] is thiocyanato-KA/ and sulfur-
bonded NCS is thiocyanato-K.S. Nitrogen-bonded nitrite [formerly nitro] is named ni-
trito-«A( and oxygen-bonded nitrite is nitrito-KO, i.e., [0=N—0—Co(NH 3 ),] 2+ is pen-
taamminenitrito-«0-cobalt(III) ion.
For polydentate ligands, a right superscript numeral is added to the symbol « in order
to indicate the number of identically bound ligating atoms in the flexidentute 1 ' ligand. Any
doubling prefixes, such as bis-, are presumed to operate on the k locant index as well.
Examples [I and 2] use tridentate chelation by the linear tetraammine ligand, iV,/V’-bis(2-
aminoethyl)- 1 . 2 -ethanediamine to illustrate these rules.
Examples:
CH-, — CH.
/ \
H.N NHCH.CH.
^ Pt ^ NH
X \ I
Cl NH.CH.CH->
CH-> — CH.
/ ' \
H 2 N NH v,
/ X
Cl NH x
\
(A/,/V'-bis(2-amino-KN-ethyl)-1,2-
ethanediamine-KA/]chloroplatinum(Il)
[A'-(2-amino-KA'-ethyl)-/V'-(2-aminoethyl)-1,2-
ethanediamine-K’A'.N'JchloroplatinumOl)
Stereochemical descriptors for chelated complexes. Stereochemical descriptors can be
provided for compounds containing chelated ligands but they involve considerations
beyond those described above. The polyhedral symbol is determined as in the case
of monodentate ligand derivatives. Also, the priority numbers are assigned to ligaling
atoms as for monodentate ligands. However, a general treatment for the assignmenl
of the configuration index requires the use of priming conventions in order to provide
33 [The chief advantage of the kappa convention seems to be that it unambiguously denotes ligating
atoms in a complex ligand. This distinguishes the ligating atom in an organic ligand that may have
organic descriptors, as in jV.N-dimethylaniline.]
36 Any chelating ligand capable of binding with more than one set of donor atoms is described as
fiexidentate. cf., Stratton. W. J.; Busch, D. H. J. Am. Chem. Soc. 1958, 80. 3191.
£04
16" Inorganic Chains, Rings, Cages, and Clusters
Table 16.2
Electrons available for
Framework electrons 0
framework bonding for
Fragment
Cr,Mo,W
Mn,Tc,Re
Fe,Ru,Os
Co,Rh,lr
Ni,Pd,Pt
various organometallic
fragments
M(7,5-C 5 H 5 )
-1
0
1
2
3
M(CO),
-2
-1
0
1
2
M(CO) 3
0
I
2
3
4
M(CO) 4
2
3
4
5
6
" Framework electrons (F) equal the number of metal valence electrons (M) plus the number of
electrons donated by ligands (L) minus twelve (F = M + L - 12).
One of our goals here is to be able to predict the structure of a cage or cluster from
its molecular formula. We do this by first finding the number of framework electrons.
The structure will then be predicted to be closo, nido , or arachno if the number of
framework electrons is In + 2, In + 4, or In + 6, respectively. As an example let us
consider B,H 7 [Fe(CO) 3 ] 2 . for which n equals five. The three BH units and the two
Fe(CO) 3 units contribute two electrons each and the four extra hydrogen atoms
contribute one electron each to give a total of 14 framework electrons:
2Fe(CO) 3 : 2x2= 4e“
3BH: 3x2= 6e“
4H: 4x1= 4e~
Total = I4e“
Since n = 5. we see that there are In + 4 framework electrons and we predict a nido
structure which is found experimentally. The square pyramidal structure (Fig. 16.54)
can be thought of as resulting from substitution of two BH units with two Fe(CO) 3
units in B 5 H 9 (Fig. 16.49a).
Let us apply these procedures to the nonborane molecule. Rh 6 (C0) l6 , for which
n equals six. Each of the six Rh(CO) 2 units contribute one electron to the framework,
while the four extra CO molecules provide eight electrons:
6Rh(CO) 2 : 6x|= 6e~
4CO: 4x2 = Je^
Total = I4e~
Thus we have 14 framework electrons with the complex fitting the 2n + 2 category
and predicted to have a closo structure (Fig. 15.10). There are two terminal CO groups
per rhodium and four bridging carbonyl groups which span alternate triangular faces.
Another method for obtaining the number of framework electrons starts by counting
the valence electrons of all of the metal atoms and then adds all of the electrons
donated by the ligands:
6Rh: 6x9= 54e~
16CO: 16 x 2 = 32c'
Total = 86e'
Twelve of these electrons per rhodium (a total of 72) will be used for nonframework
• H bonding leaving 14 for framework bonding. Thus there are seven bonding pairs in the
Fig. 16.54 Structure of
BjH 7 [Fe(CO).i)’- (From
Grimes, R. N. In
Comprehensive Organo-
meiallic Chemistry-,
Wilkinson. G.. Slone.
F. G, A.. Abel. E. W..
Eds.: Pcrgumon: Oxford,
1982; Vol. I. p 470.
Reproduced with
permission.]
Boron Cage Compounds 805
framework corresponding to 2 n + 2 electrons and. as above, a closo structure is
predicted. It is worth noting that the 18-electron rule fails for RhJCO),,,. while Wade's
rules are entirely successful.
There are exceptions to Wade's rules, even among modest-sized clusters (see
Footnote 135). In some cases large transition metals cause geometrical distortion. In
others, a kineticaily favored structure may not be able to rearrange to a more
thermodynamically favored one. In still other instances the assumption that transition
metal atoms will use twelve electrons for external ligands is not valid. As with most
rules, one should not expect predictions to be foolproof.
The bonding capabilities of transition metal clusters (no nonmetals in the frame¬
work). based on molecular orbital calculations, has been nicely summarized by
Lauher' 41 (Table 16.3). Within this table we see three structures (tetrahedron, but¬
terfly, and square plane) for tetranuclear metal clusters, The tetrahedron is a 60-
electron cluster, while the butterfly and square plane clusters have 62 and 64 elec¬
trons. respectively. When we go from a tetrahedron to a butterfly, one of the edges of
the tetrahedron is lengthened corresponding to bond breaking.
h C ir
Table 16.3
Relationship between
geometry, molecular
Geometry
No. of metal
atoms
Bonding molecular
orbitals
Cluster
electrons
Examples
orbitals, and cluster valence
electrons 0
Monomer
I
9
18
Ni(CO) 4
Dimer
2
17
34
Fe 2 (CO) y
Trimer
3
24
48
Os 3 (CO), 2
Tetrahedron
4
30
60
Rh 4 (CO) P
Butterfly
4
31
62
Rc 4 (CO);~
Square plane
4
32
64
Pt 4 (0,CMe) 8
Trigonal bipyramid
5
36
72
Os 5 (CO) l6
Square pyramid
5
37
74
Fe 5 (CO) l5 C
Bicapped tetrahedron
6
42
84
Os 6 (CO)| 8
Octahedron
6
43
86
Ru 6 (CO) 17 C
Capped square pyramid
6
43
86
Os 6 (CO) 18 H 2
Trigonal prism
6
45
90
Rh 6 (CO),jC 3 '
Capped octahedron
7
49
98
Rh 7 (CO)-] 6 '
“ Lauher. J. W. J. Am, Chem. Soc. 1978, 100, 5305-5315. All framework atoms are transition
metals.
141 Lauher. J. W. J. Am. Chem. Soc. 1978. 100, 5305-5315.
A-70
I • IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
Table 1-6
Structural prefixes used in
inorganic nomenclature
These affixes are italicized and separated from the rest of the name by hyphens.
antiprismo eight atoms bound into a rectangular antiprism
arachno a boron structure intermediate between nido- and hypho- in degree
of openness. [See Table 1-8.]
asym asymmetrical
catena a chain structure; often used to designate linear polymeric
substances
cis two groups occupying adjacent positions, not now generally
recommended for precise nomenclature purposes
closo a cage or closed structure, especially a boron skeleton that is a
.' • polyhedron having all faces triangular
cyclo a ring structure. Cyclo here is used as a modifier indicating structure
and hence is italicized. In organic nomenclature, ‘cyclo’ is
considered to be part of the parent name since it changes the
molecular formula, and therefore is not italicized
dodecahedro eight atoms bound into a dodecahedron with triangular faces
fac three groups occupying the comers of the same face of an
octahedron, not now generally recommended for precise
nomenclature purposes
hexahedro eight atoms bound into a hexahedron (e.g., cube)
Itexaprismo twelve atoms bound into a hexagonal prism
liypho an open structure, especially a boron skeleton, more closed than a
kWo-structurc, but more open than an arac/mo-structure
icosahedro twelve atoms bound into a triangular icosahedron
klado a very open polyboron structure
mer meridional; three groups occupying vertices of an octahedron in such
a relationship that one is cis to the two others which are
themselves trans, not now recommended for precise nomenclature
purposes - _
nido a ncst-like stmcture, especially a boron skeleton that is almost
closed
octahcdro six atoms bound into an octahedron
pentaprismo ten atoms bound into a pentagonal prism
quadro four atoms bound into a quadrangle (e.g., square)
sym symmetrical
tetrahedro four atoms bound into a tetrahedron
trans two groups directly across a central atom from each other, i.e., in
the polar positions on a sphere, not now generally recommended
for precise nomenclature purposes
triangulo three atoms bound into a triangle
triprismo six atoms bound into a triangular prism
M (mu) signifies that a group so designated bridges two or more centers of
coordination
A (lambda) signifies, with its superscript, the bonding number, i.e., the sum of
the number of skeletal bonds and the number of hydrogen atoms -
associated with an atom in a parent compound
"
I-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-71
parentheses, and separated from the name by a hyphen. The polyhedral symbols for the
most common coordination geometries for coordination numbers 2 to 9 inclusive are given
in (Table 1-7].
Configuration index
Definition of index and assignment of priority numbers to ligating atoms. Having
developed descriptors for the general geometry of coordination compounds, it becomes
necessary to identify the individual coordination positions. The configuration index is a
series of digits identifying the positions of the ligating atoms on the vertices of the
coordination polyhedron. The individual configuration index has the property that it
distinguishes between diastereoisomers. The digits of the configuration index are estab¬
lished by assigning an order of priority to the ligating atoms as described [below].
The procedure for assigning priority numbers to the ligating atoms of a mononuclear
coordination system is based on the standard sequence rule developed for enantiomeric
carbon compounds by Cahn, Ingold, and Prelog. The ligating atom with highest priority is
assigned the priority number I. the ligating atom with the next highest priority. 2. and so
U on.
Table 1-7 _
List of polyhedral symbols
Configuration indexes for particular geometries
Square planar coordination systems (SP-4). The configuration index is a single digit
which is the priority number for the ligating atom trans to the ligating atom of priority
number I. 33
Coordination
polyhedron
Coordination
number
Polyhodral
symbol
linear
2
L-2
angular
2
A-2
trigonal plane
3
TP-3
trigonal pyramid
3
TPY-3
tetrahedron
4
T-4
square plane
4
SP-4
trigonal bipyramid
5
TBPY-5
square pyramid
5
SPY-5
octahedron
6
OC-6
trigonal prism
6
TPR-6
pentagonal bipyramid
7
PBPY-7
octahedron, face monocapped
7
OCF-7
trigonal prism, square face monocapped
7
TPRS-7
cube
8
cus
square antiprism
8
SAPR-S
trigonal prism, triangular face bicapped
8
TPRT-8
trigonal prism, square face bicapped
8
TPRS-8
trigonal prism, square face tricapped
9
TPRS-9
heptagonal bipyramid
9
HBPY-9
dodecahedron
8
DD-8
hexagonal bipyramid
8
HBPY-8
octahedron, trans-bicapped
8
OCT-8
33 [Note that] cis-lrans terminology alone is not adequate to distinguish between the three isomers of a
square planar coordination entity [MabcdJ.
806
16- Inorganic Chains, Rings, Cages, and Clusters
To do this, two additional electrons must be added to the tetrahedron to keep all
electrons paired. In fact this is a general principle: Adding electrons to a closo
complex opens the structure, converting it to one of lower symmetry. The butterfly
structure results when an edge is removed from the tetrahedron.
If we add two electrons to the butterfly structure, another edge is lengthened
(another bond broken) and we end up with a square plane.
These principles apply equally well to heteronuclear clusters which can be illus¬
trated with the trigonal bipyramidal cluster of ruthenium and sulfur. [(/7-cymene)j-
Ru 3 S 2 ]" + . 142 This 48-electron closo cation (24 electrons from three Ru atoms, 18
electrons from three p-cymene molecules, and eight electrons from two S atoms) may
be reduced reversibly to the 50-electron square pyramidal nido cluster by adding two
electrons as shown in Fig. 16.55. Both the closo and nido clusters have been isolated
and characterized crystallographically. The average Ru—Ru bond distance in the
closo structure is 277.8 pm. corresponding to three Ru—Ru single bonds. The nido
structure has two Ru—Ru single bonds (272.3 pm) intact, and one bond severed as
shown by the long Ru—Ru distance (361.2 pm).
As you become more familiar with transition metal clusters (no nonmetals in the
framework) you will come to associate closo structures with numbers of electrons. A
trimer will have 48 electrons, a tetrahedron will have 60 electrons, a trigonal bipyramid
will have 72 electrons, and an octahedron will have 86. Some care is required,
however, as can be illustrated with Os,H ; (CO) 10 . An electron count gives us 46 elec¬
trons rather than 48. If. however, we allow for one Os—Os double bond, the electron
count is as expected. In accord with this expectation, one osmium-osmium bond is
found to be shorter than the other two and the complex shows the reactivity expected
for an unsaturated complex.
S 5
Fig. 16.55 The Ru,S : core in (p-cymene)|Ru,S : (right) and its dicalion [(p-
cymene),RujS : ] 2+ (left). The arenes attached to ruthenium as six-electron donors arc not
pictured. Two-electron reduction converts the complex from a closo to a nido geometry.
[From Lockcmeyer, J. R.: Rauchfuss. T. B.; Rhcingold. A. L. J. Am. Client. Soc. 1989. III.
5733-5738. Reproduced with permission.]
143 Lockemeyer, J. R.; Rauchfuss. T. B.; Rheingold. A. L. J. Am. Client. Soc. 1989. Ill, 5733-5738.
Metal Clusters
807
Fig. 16.56 Structures of
osmium complexes which
have seven pairs of skeletal
electrons. Each capped
triangular face adds twelve
electrons to the total
electron count, but the
number of skeletal pairs
remains seven. Likewise
removing Os(CO)., deletes
twelve electrons without
changing the number of
skeletal pairs. The diagonal
lines show alternate
geometries with the same
total number of electrons.
[From McPartlin. M.
Polyhedron 1984 . 3. 1279.
Reproduced with
permission.)
Metal Clusters 145
Some of the beautiful relationships that exist between closo. nido. and nniclino
osmium complexes are show n in Fig. 16.56. 14 -' For lack of space we have not touched
on many subtleties associated with geometry/bonding/eleetron counting procedures
and the reader is encouraged to consult more advanced sources. 144
Compounds containing metal-metal bonds are as old as chemistry itself (calomel was
known to the chemists of India as early as the twelfth century). The dimeric nature of
the mercurous ion was nol confirmed until the turn of this century and in the next
141 McPartlin. M. Polyhedron 1984.3. 1279-1288.
144 Mingos, D. M. P.: Wales. D. J. Introduction to Cluster Chemistry: Prentice-Hall: Englewood
Cliffs, NJ. 1990. Teo, B. K.; Zhang. I-L: Shi. X. Inory. Chan. 1990 . 29. 2083-2091. Also see
Footnote 139.
I4 ’ Cotton. F. A. J. Client. Edttc. 1983, 60. 713-720.
A-68
l-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
Names for Mononuclear Coordination Compounds
with Monodentate Ligands
Sequence of central atom and ligand names. The ligands are listed in alphabetical
order, without regard to charge, before the names of the central atom. Numerical prefixes
indicating the number of ligands are not considered in determining that order.
Example[ sj:
dichloro( Me,l>ln 1 |
Rc,CI 4 iPMe,Ph) 4
in)
ib)
10
Id)
te)
Fig. 16.60 Qualitative molecular orbital diagram for dinuclear rhenium and molybdenum
complexes. All of the bonding molecular orbitals are filled for [Re 2 CI„J' _ (c) and a bond order
of 4 results from one a. two rr, and one 6 bond. When electrons are added to the 5 level,
the bond order is reduced as shown for (d) and (e). Removing electrons from the S bond also
leads to a lower bond order as shown by (a) and (b). [Taken in part from Cotton, F. A.
Chem. Soc. Rev. 1983. 12. 35. Reproduced with permission.)
,J " Hall, M. B. Polyhedron. 1987. 6. 679-684. Ziegler. T.; Tschinke, V.; Bcckc. A. Ibid. 685-693.
Burstcn. B. E.; Clark. D. L. Ibid. 695-704.
149 Hino. K.; Saito, Y.: Bcnard. M. Acta Crystallogr., Sect. B. 1981, 37. 2164-2170.
150 Pauling. L. Proc. Natl. Acad. Sci. U.S.A. 1975, 72. 3799-3801.
151 Ziegler. T.; Tschinke. V.; Bcckc. A. Polyhedron 1987, 6. 685-693.
A-A4
|. IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
Coordination number. As defined for typical coordination compounds, the coordina¬
tion number equals the number of sigma-bonds between ligands and the central atom. Even
though simple ligands such as CN". CO. N,. and P(CH 3 ) 3 may involve both sigma- and pi-
bonding between the iigating atom and the central atom, the pi-bonds are not considered in
determining the coordination number. 26
The sigma-bonding electron pairs in the following examples are indicated by : before
the ligand formulae.
Examples:
Complex
I. [Co(:NH 3 ) 6 ] J+
[Ru(:NH 3 ) 5 (:N,)] 2
[Cr(:CO)]j -
Coordination
number
6
6
5
Complex
[Fe(:CN) 6 | 3
[Ni(:CO)J
[Co(:CI)J 2 “
Coordination
number
6
4
4
Chelation. Chelation involves coordination of more than one sigma-electron pair
donor group from the same ligand to the same central atom. The number of such ligating
groups in a single chelating ligand is indicated by the adjectives didentate. 27 tridentate.
tetradentate, pentadentate, etc., (see [Table 1-3] for a list of numerical prefixes). The
number of donor groups from a given ligand attached to the same central atom is called the
denticity.
Examples:
CH, — CH,
Cl NH, / - -
\ / A / \
\ / CH, H,N NH
A L ‘ Ap.r ^M 2 ™ 2
didcnlatc chelation
2 . didentate chelation
CH, —CH,
/ ' \h
II, N N
CH, —CH,
h/ \h
N N
tridentate chelation
•I. tciradentate chelation
Oxidation number. The oxidation number of a central atom in a coordination entity is
defined as the charge it would bear if all the ligands were removed along with the electron
pairs that were shared with the central atom. It is represented by a roman numeral.
Coordination nomenclature, an additive nomenclature. According to a useful, histor¬
ically-based formalism, coordination compounds are considered to be produced by addi-
26 This definition is appropriate to coordination compounds, but not necessarily to other areas, such as
crystallography, (Note also the strong dependence of this system on the assumed bonding model.]
27 Previous versions of the Nomenclature of Inorganic Chemistry used bidentate rather than diden-
late. for linguistic consistency. In this edition, a single set of prefixes is used throughout and these
arc listed in [Table 1-3]. [Bidentate has been a perfectly good word in the English language for 250
years or more, independent of chemistry. Unless one proposes to place the entire content of the
language under the purview of the IUPAC, one must view this as a violation of good language that
has as little logic as it does good etymology.]
I • IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-65
tion reactions and so they named on the basis of an additive principle. The name is build up
around the central atom name, just as the coordination entity is built up around the central
atom.
Example:
I. Addition of ligands to a central atom: Ni 2+ + 6H,0 - [Ni(H,0) h ] 2+
Addition of ligand names to a central atom name: hexaaquanickel(II) ion
This nomenclature extends to still more complicated structures where central atoms
are added together to form dinuclear. trinuclear. and even polynuclear species from some
mononuclear building blocks. The persistent centrality of the central atom is emphasized
by the root -nuclear.
Bridging ligands. In polynuclear species it is necessary to distinguish yet another
ligand behavior, the action of the ligand as a bridging group.
A bridging ligand bonds to two or more central atoms simultaneously. Thus, bridging
ligands link central atoms together to produce coordination entities having more than one
central atom. The number of central atoms joined into a single coordination entity by
bridging ligands or metal-metal bonds is indicated by dinuclear. trinuclear. tetranuclear,
etc.
Examples:
I. |(NHj) 5 Co — Cl — Ag] 3 '
°\ / Cl \ y°
Al At
/ \ / \
Cl Cl Cl
Formulae for Mononuclear Coordination Compounds
with Monodentate Ligands
Sequence of symbols within the coordination formula. The central atom is listed first.
The formally anionic ligands appear next and they are listed in alphabetic order according
to the first symbols of their formulae. The neutral ligands follow, also in alphabetical order,
according to the same principle. Polydentate ligands are included in the alphabetical list.
Complicated organic ligands may be designated in formulae with abbreviations [Table 1-5],
Uses of enclosing marks. The formula for the entire coordination entity, whether
charged or not. is enclosed in square brackets. When ligands are polyatomic, their formulae
are enclosed in parentheses. Ligand abbreviations are also enclosed in parentheses. In the
special case of coordination entities, the nesting order of enclosures is as given [on page
A-51 ]. There should be no space between representations of ionic species within a coordi¬
nation formula.
Examples:
I. [Co(NH 3 )JC1 3 2. [CoCI(NH,),]CI, 3. [CoCI(NO,)(NH,) 4 ]CI
[PtCI(NH,CHj)(NH,),]CI 5. K,|PdCI 4 ] ‘ 6. [Co(en),]CI 3
Ionic charges and oxidation numbers. If the formula of a charged coordination entity
is to be written without that of the counterion, the charge is indicated outside the square
bracket as a right superscript, with the number before the sign. The oxidation number of a
central atom may be represented by a roman numeral used as a right superscript on the
element symbol.
Examples:
I. [PtCIJ 2 - 2. [Cr(H,0)J 3+
[Cr ni (NCS) 4 (NH,),] -
312
16 Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.60a) which have less than full occupation of the 6 bonding orbitals have been
prepared. The characterization of W 2 CI 4 (OR) 4 (R), (R = Me.Et) with a W=W double
bond completed a series of ditungsten compounds with bond orders of 4. 3.2. and 1. 152
There are two metals, Cu(II) and Cr(II). in the first transition series which form
acetate complexes similar in structure to the rhenium and molybdenum carboxylate
complexes (Fig. 16.58d). Like their Re and Mo analogues, the Cu and Cr complexes
are diamagnetic, indicating that spins are paired. They differ significantly from the
complexes of the heavier metals, however. The Cu—Cu distance in the Cu(II) (d 9 )
complex is 264 pm, which is actually somewhat longer than the Cu—Cu distance in
metallic copper (256 pm). It appears that the Cu—Cu bond in copper(II) acetate is
only a weak single bond resulting from pairing the odd electron on each copper atom.
The chromium(II) acetate molecule was long thought to have the same metal-
metal bond length as the copper compound and thus have a similar weak bond.
However, its structure was redetermined and the Cr—Cr distance was found to be
236.2 pm, which is considerably shorter than that found in metallic chromium
(249.8 pm). 1 ” In fact, the Cr—Cr bond has been estimated to be about 45 kJ mol -1 ,
which makes it stronger than the Cu—Cu bond. 134 All of this evidence and orbital
symmetry suitability might suggest that the Cr—Cr bond in chromium acetate is a
quadruple bond. Still, not everyone is willing to go that far. The problem is that this
"quadruple" bond is estimated to be only about as strong as a typical Cr—Cr single
bond. 133 It appears that most participants in the debate have chosen to view the bond
as a very weak quadruple bond. Aside from this controversy, the chief interest in
dichromium compounds has been in the wide range of bond lengths they display
(185-254 pm). Some of these are the shortest metal-metal bonds known and have been
dubbed "super-short" bonds. The variation in bond length, which is dependent upon
the nature of the substituent ligands, is in sharp contrast to the relative uniformity in
the length of quadruple bonds in the heavier congeners (Mo—Mo = 204-218 pm;
W—W = 216-230 pm).
Among the more interesting metal-metal multiple bonded complexes are the
hexaalkoxo dinuclear tungsten and molybdenum complexes, [M 2 (OR) h l (M =
Mo. W): 136
RO v ,-OR
RO — M = M'— OR
RO^ 'OR
These complexes are chemically like polynuclear metal carbonyl complexes (class I)
but are included here instead of in Chapter 15 because they do not possess
metal-carbon bonds. A rich chemistry has been developed in which alkoxides lunc-
i-« Anderson, L. B.; Cotton, F. A.; DeMarco. D.: Fang, A.; Ilsley. W. H.; Kolthammer. B. W. S.;
Walton, R. A. J. Am. Clwm. Soc. 1981. 10). 5078-5086.
133 Cotton. F. A.; DeBoer, B. G.; LaPradc. M. D.; Pipal.J. R.; Ucko. D. A. J. Am. Client. Soc. 1970.
92. 2926-2927.
134 Cannon, R. D. Inorg. Client. 1981. 20. 2341-2342.
•33 Hall, M. B. Polyhedron 1987. 6. 679-684.
'«• Chisholm. M. H.; Clark. D. L.: Hampden-Smilh. M. J. J. Am. Clwm. Soc. 1989. III. 574-586.
Metal Clusters
813
lion as stabilizing ligands for 12-electron clusters. 137 The alkoxide group, RO - , has
two filled p orbitals capable of donating -it electron density to the metal centers. Even
so. because these p orbitals are ligand centered, the complexes are looked upon as
coordinatively unsaturated and containing formal metal-metal triple bonds (o^ir 1 ).
The M=M bonds are somewhat analogous to carbon-carbon triple bonds. For
example. the metal-metal bond can undergo addition reactions:
(RO),W=W(OR), + 2X-, -► (X),(RO),W— W(OR) 3 (X),
(X = CI,Br.I) ' - (16.121)
(/- PrO) j M o= M o( O-/'- Pr), + /-PrOO-/-Pr-♦(/-PrO) 4 Mo=Mo(0-/-Pr) 4 (16.122)
It is also possible to prepare (/-BuO)-,W=CR (R = Me. Et, Ph) compounds in which
the isolobality of CR and W(OR), is apparent: 138
(t-Bu0) 3 W=W(0-/-Bu) 3 + RC=CR -► 2(/-BuO),W=CR (16.123)
Recently, it has been shown that W 2 (0-/-Pr) ft dimerizes, existing in equilibrium
with Wj(0-/-Pr)| 2 . a molecule which may be thought of as an analogue of cyclor
butadiene: 139
O. °
V-°
.v/
o'V \
o o
o °
\ />o
w
///
,w
O'V \
o o
0 o
Os\/
~ \ w—o
O-. v // \ / ,o
y —w'
cf/\ /\ % 0
o—w—o
/ \
o o
The tetramer has been shown to be fluxional such that the tungsten-tungsten double
and single bonds migrate about the W., ring. At the same time, the two isopropoxide
groups attached to each wingtip tungsten undergo proximal/distal exchange (Fig.
16.61). All of this motion taken together has come to be known as "The Bloomington
Shuffle" after the city in which it was discovered. 16,1
Trinuclear Clusters The best-known examples of noncarbonyl clusters containing three metal atoms are
the rhenium trihalidcs ((ReCI,),] and their derivatives. The basic structural unit is
shown in Fig. 16.62a. Each rhenium atom is bonded to the other two rhenium atoms
directly by metal-metal bonds and indirectly by a bridging halogen ligand. In addition.
137 Chisholm. M. H. Angew. Clwm. Ini. Ed. Engl. 1986, 25. 21-30. Chisholm. M. H.; Clark. D. L.;
Hampdcn-Smith. M. J.; Hoffman, D. H. Angew. Clwm. Ini. Ed. Engl. 1989. 28. 432-444.
Chisholm. M. H. Acc. Clwm. Res. 1990, 2J. 419-425.
,M McCullough. L. G.; Schrock. R. R.; DcWan. J. C.: Murdzek. J. C. J. Am. Clwm. Soc. 1985, 107,
5987-5998.
1,9 Chisholm. M. H.; Clark, D. L.; Hampdcn-Smith. M. J. J. Am. Clwm. Soc. 1989, III, 574-586.
16,1 ls 11 possible that this name was proposed the same year that the Chicago Bears choreographed the
"Supcrbowl Shuffle"?
A-62
I IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
syllables ‘ur' and ‘orus‘, respectively, from the acid name when it is converted to the anion
name.
Polynuclear Acids
Isopolyacids (hoinopolyacids). These materials are generally referred to in the liter¬
ature as isopolyacids. The name homopolyacids is preferable because the Greek root of
homo- implies ‘the same', in direct contrast to that of hetero- signifying ‘different’, whereas
the root of iso- implies equality. Detailed nomenclature of those compounds has been
presented elsewhere. 24
Acceptable abbreviated names may be given to polyoxoacids formally derived by
condensation (with evolution of water) of units of the same mononuclear oxoacid. provided
that the central atom of the mononuclear oxoacid has the highest oxidation state of the
Periodic Group to which it belongs, that is, VI for sulfur, etc. The names are formed by
indicating with numerical prefixes the number of atoms of central element present. It is not
necessary to state the number of oxygen atoms.
Examples:
H 2 S 2 0 7 disulfuric acid, or dihydrogen disulfate [pyrosulfuric acidj
H 2 Mo h O l9 dihydrogen hexamolybdate
H 6 Mo 7 0 24 hexahydrogen heptamolybdate
H,Pj 0 9 trihydrogen cyc/o-triphosphate [trimetaphosphoric acid]
Hctcropolyacids. A detailed nomenclature of these compounds has been given else¬
where. 24 Names are developed using coordination nomenclature (Section I—10).
Examples:
HjSiW.,,0.,,, tetrahydrogen hexatriacontaoxo(tetraoxosilicato)dodecatungstate(4 -)
H (l P 2 W IK 0 ()2 hexahydrogen tetrapentaconta
oxobis(tetraoxophosphato)octadecatungstate(6 -)
Some abbreviated semitrivial names are retained for present use due to long-standing
usage. This applies if all the central atoms are the same, if the polyanion contains only
oxygen atoms as ligands and only one kind of heteroatom, and if the oxidation state of the
central atoms corresponds to the highest oxidation state of the Periodic Group in which
they occur, in this usage, the Main Group atoms receive specific abbreviated names for
incorporation into the hetero-polyacid name. These are:
B boro Si silico Ge germano P phospho As arseno.
Examples:
H 4 SiW l2 O 40 tetrahydrogen silicododecatungstatc
H () P,W ik O, ) 2 hexahydrogen diphosphooctadecatungstate
Ions Derived from Oxoacids
Anions. The hydrogen nomenclature name described above consists of two parts, the
second of which is an anion name. This can stand alone to represent the anion itself.
Traditional names are still accepted for the exceptions listed in [Table 1-4]. The ending
-ic of the acid name becomes -ate in the anion name, and -ous becomes -ite.
Cations. The cations considered here are obtained by adding formally one or more
hydrogen cations to a neutral molecule of the acid.
Example:
I. (H,SOJ + trihydroxooxosulfur(VI) cation [sulfuric acidium cation]
24 See Pure Appl. Chem. 1987, 59, 1529.
1' IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-63
Note that an extension of the organic style of nomenclature as in (CH 1 CO : H 2 ) + = ethanoic
acidium, is discouraged because it is based on the word ‘acid' and is" often not easily
adaptable to languages other than English. 25
1-10 COORDINATION COMPOUNDS
Coordination entity. A coordination entity is composed of a central atom, usually that
of a metal, to which is attached a surrounding array of other atoms or groups of atoms, each
of which is called a ligand, in formulae, the coordination entity is enclosed in square
brackets whether it is charged or uncharged.
Examples:
I. [CofNHJJ 3 2. [PtCIJ 2 " 3. [Fe,(CO) l2 ]
Central atom. The central atom is the atom in a coordination entity which binds other
atoms or groups of atoms (ligands) to itself, thereby occupying a central position in the
coordination entity. The central atoms in [NiCI 2 (H,0)J, [Co(NHJJ 3+ , and [PtCIJ 3- are
nickel, cobalt, and platinum, respectively.
Ligands. The ligands are the atoms or groups or atoms bound to the central atom.
The root of the word is often converted into other forms, such as to ligate, meaning to
coordinate as a ligand, and the derived participles, ligating and ligated.
Coordination polyhedron. It is standard practice to think of the ligand atoms that are
directly attached to the central atom as defining a coordination polyhedron (or polygon)
about the central atom. Thus [Co(NH,) ft ] 3+ is an octahedral ion and [PtCIJ 2 " is a square
planar ion. In this way the coordination number may equal the number of vertices in the
coordination polyhedron. Exceptions are common among organometallic compounds.
Examples:
I. ucinlieilnil 2. square planar 3. ici railed nil
coordination coordination coordination
polyhedron polygon polyhedron
Historically, the concepts and nomenclature of coordination compounds were unam¬
biguous for a long time, but complications have arisen more recently. According to
tradition, every ligating atom or group was recognized as bringing one lone-pair of elec¬
trons to the central atom in the coordination entity. This sharing of ligand electron pairs
became synonymous with the verb ‘to coordinate.' Further, in the inevitable electron
bookkeeping that ensues upon consideration of a chemical compound, the coordination
entity was dissected (in thought) by removal of each ligand in such a way that each ligating
atom or group took two electrons with it. Coordination number is simply definable when
such a thought process is applied.
25 [However, as long as "sulfuric acid" is the only name used in the literature for H 2 S0 4 . "sulfuric
acidium cation" (based on the previous IUPAC Red Book) will be necessary unless we use
something trivial like "protonated sulfuric acid."]
16- Inorganic Chains, Rings, Cages, and Clusters
"The Bloomington Shuffle"
10 Ot
0<. /
\ w —o
°- \// \ I ..o
w-w'
o'l\ //\^o
°- / Vo
'o b|
o o
\ /
°xi>Ai -°
oVA
o —w —o
A
0 o
|0 ot
\ /, o
O — W''/
o.. |/V^o
W -'. W N.
o^/W /l^o
I'N — O
o<- \
to 0|
Fig. 16.61 Dynamic intramolecular rearrangement of W.,(0-/-Pr)i;. [From Chisholm.
M. H.; Clark, D. L.; Hampden-Smith. M. J. J. Am. Client. Soc. 1989. Ill, 574-586.
Reproduced with permission.!
(c)
Fig. 16.62 Rhenium(III) clusters: (a) The structural unit present in a Re(lll) trinuclear
cluster. The positions marked O are empty in the trihalides in the gas phase but have
coordinating groups in other situations. [From Penfold. B. R. In Perspectives in Structural
Chemistry, Dunitz. J. D.; Ibers. J. A., Eds.; Wiley: New York, 1968; Vol. 2. p 71.
Reproduced with permission.] (b) The structure of solid (ReClj).,. [From Cotton. F. A.;
Mague. J. T. Inorp. Client. 1964. j. 1402. Reproduced with permission.! (c) The [Re,Cl i: l'‘
anion. [From Bertrand. J. A.; Cotton. F. A.; Dollase, W. A. Inorp. Client. 1963. 2. 1166.
Reproduced with permission.!
Metal Clusters
815
Tetranuclear
Clusters
Fig. 16.63 Structure of
Wj(OR)| h .
RO I I OR
w cr w — or
“° IVJ^U
RO — W — — W
RO I § I OR
each rhenium atom in the triangular array is coordinated by two more halide ligands
above and below the plane defined by the three rhenium atoms. Each Ret III) has a d A
configuration which would lead to a paramagnetic complex if only metal-metal single
bonds were present. The complexes are diamagnetic, however, which implies that
each Re atom is doubly bonded to its rhenium neighbors.
In the solid state the halides retain this basic unit, but further bridging between
rhenium atoms by chloro ligands results in a polymeric structure (Fig. 16.62b).
Likewise, dissolving the halides in solutions of the hydrohalic acids leads lo formation
of dodecahalotrirhenate(III) ions. [Re,X,,] J ~ (Fig. 16.62c). in which additional halide
ligands have coordinated lo the empty positions present in the Re,X y units. Other
ligands (such as R,P. Me : SO, or MeCN) can also coordinate to these positions. The
Re, cluster is persistent in many chemical transformations. The bond length is 240-
250 pm. which is indicative of strong bonding although weaker than in IRe^X,,] 2- .
Although common among carbonyl clusters, far fewer examples of tetranuclear clus¬
ters are found among the halides and oxides. One example noted previously is
W 4 (OR) i: which forms by dimerization of W,(OR) h . The tetrameric W 4 (OR) lh has also
been synthesized. Whereas W,(ORV, and W 4 (OR),, may be viewed as unsaturated.
W 4 (OR)| 6 is saturated, containing W—W single bonds (Fig. 16.63). Ihl
Quadruply bonded dinuclear compounds also can dimerize to give telrameric
molecules:
The resulting Ibur-membered ring is not square, and it appears from bond length
measurements that there are alternating single and triple Mo—Mo bonds.
Tetranuclear cluster units (rhomboidal Mo 4 ). connected by oxygen atoms and forming
infinite chains, are found in Ba, u Mo K 0| 6 . l '’ J
,M Chisholm. M H.: Huffman. J. C.: Kirkpatrick. C. C.: Leonclli. J.; Foiling. K. J. Am. Client. Soc.
1981, I0J, 6093-6099.
,M MeCarlcy. R. E.: Ryan. T. R.: Tonirdi. C. C. In Reactivity of Metal-Metal Bonds: Chisholm,
M. H.. Ed.: ACS Symposium Series 155; American Chemical Society: Washington. DC. 1981;
p 41.
"■’ MeCarlcy. R. E. In Inorganic Chemistry Toward the 21st Century: Chisholm. M. H., Ed.: ACS
Symposium Series 211: American Chemical Society: Washington. DC. 1983; p 273.
A-60
l-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
centrated sulfuric acid], are trivial. Later these names were superseded because they were
found to be inconvenient and names reflecting chemical information, in this case the acid
property (as, for example, with sulfuric acid), were coined. Names for the various deriva¬
tives of the parents were developed from these names. This semisystematic approach has
limitations, and has also led to ambiguities and inconsistencies.
Formulae
In a formula, the hydrogen atoms which give rise to the acid property are cited first, then
comes the central atom, and finally the atoms or groups of atoms surrounding the central
atom. These last are cited in the following order: oxygen atoms which are bound to the
central atom only, followed by other atoms and groups of atoms ordered according to
coordination nomenclature rules, that is, ionic ligands precede neutral ligands. Within each
class, the order of citation is the alphabetical order of the symbols of the ligating atom.
Examples:
I. H,S0 4 2. H,S0 3 3. H,SO s 4. H 4 P,O fi or (HO),OPPO(OH) 2
H>,0 7 or (HO),OPOPO(OH) 2 6. HS0 3 CI
Traditional Names
History. Some traditional names (a selection is in [Table 1-4]) were introduced by
Lavoisier. Under his system, oxoacids were given a two-word name, the second word
being 'acid'. In the first word, the endings -ous or -ic were added to the stem of the name,
intended to indicate the content of oxygen, which is known today to be related to the
oxidation states of the central atom. Unfortunately, these endings do not describe (he same
oxidation states in dilferent families of acids. Thus sulfurous acid and sulfuric acid refer
to oxidation states IV and VI, whereas chlorous acid and chloric acid refer to oxidation
states III and V.
An extension of this system became necessary as more related acids were recognized.
The prefixes hypo- (for very low oxidation states) and per- (for very high oxidation states)
were introduced. The prefix per- should not be confused with the syllable in the ligand
name peroxo-. Finally, it became necessary to use other prefixes, ortho-, pyro-, and meta-,
to distinguish acids differing in the 'content of water'. These traditional names do not
provide specific information on the number of oxygen atoms, or the number of hydrogen
atoms, whether acidic or not. The use of prefixes is not always consistent; for instance,
hypo- has been associated with the -ous ending (hyponitrous acid) and with the -id ending
(hypophosphoric acid). In the case of sulfur acids, two classes of acid occur, one with the
stem 'sulfur' and the other with the stem 'thio'. Moreover, in substitutive nomenclature
other names such as ‘sulfonic acid' for —S0 3 H, and "-sulfinic acid' for —SO,H were
developed, thereby forsaking the restriction of -ic to the higher oxidation state.
As discussed above, the important chemical property of acidity is highly solvent-
dependent. but a traditional nomenclature emphasizes this property by using the word
'acid' in the name. The aim of the systematic coordination nomenclature presented here is
to describe a composition and a structure, not a chemical property. Consequently, a
specialized word such as 'acid' has no place in it. This is the hydrogen nomenclature
[shown in Table 1-4]. However, in recognition of current practice, the acid nomenclature is
retained as an alternative. This is only partly systematic.
Allowed traditional names for acids and their derived anions. It is recommended that
retained traditional names be limited to very common compounds having names estab¬
lished by a long practice. Systematic names should be used for all other cases. A list of
these traditional names which are retained for present use is given in [Table 1-4].
The use of -ous, -ic, per-, hypo-, ortho-, and meta- should be restricted to those
compounds and to their derivatives; their anions are named by changing -ous into -ite and
-ic into -ate. In addition, and exceptionally, sulfur and phosphorus compounds lose the
i .a M o . 2 -
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.[ The PrefiX '° nb0 '' Si8nifiCS ,he acid Wilh ,he maximum hydration possible, as H 3 PO< (as opposed to HP0 3 ). H 5 I0 6 (as opposed to HI0 4 , etc.]
[• The prefix 'mela-' signifies a dehydration product, as: nH 3 P0 4 ('ortho') , (HP0 3 )„ ('meta'), etc.)
lsoc y anic acid is HNCO; this acid is not an oxoacid since hydrogen is not bound to an oxygen atom.
tha ‘ Wh3t " n0rma " y referTCd IO 35 '■ phOSph0r ° US add " is ' b lhis «•». a tautomeric mixture of "phosphorous acid" and "phosphonic
816
16 Inorganic Chains, Rings, Cages, and Clusters
Hexanuclear
Clusters
Fig. 16.64 The structure of
the Mo ft Clx H ion: # = Mo,
O = Cl. (b) The structure
of the M A Xl 2 f ions: • =
Nb. Ta;0 = Cl. Br. [From
Cotton. F. A. Act. Chem.
Res. 1969. 2. 240.
Reproduced with
permission.]
(a) (b)
Clusters of six molybdenum, niobium, or tantalum atoms have been known for many
years, predating the work with rhenium. There two types: In the first, an octahedron
of six metal atoms is coordinated by eight chloride ligands, one on each face of the
octahedron (Fig. 16.64a). This is found in "molybdenum dichloride," Mo 6 CI| 2 , better
formulated as [Mo 6 CI 8 ]CI 4 . Each Mo(II) atom can use its four electrons to form four
bonds with adjacent molybdenum atoms and can receive dative bonds from the four
chloride ligands. 1 ' 4
Cotton has pointed out that a metal in a low oxidation state can adopt one of two
strategies in forming clusters. It can form multiple bonds to another metal, as in
[Re 2 X 8 ] 2 -, or it can form single bonds to several other metal atoms, as in the
octahedral clusters. It is interesting that Mo(II) adopts both methods (Fig. 16.65) and
that both structures have a cubic arrangement of chloride ions.
The second class of hexanuclear clusters also contains an octahedron of metal
atoms, but they are coordinated by twelve halide ligands along the edges (Fig. 16.64b).
Niobium and tantalum form clusters of this type. Here the bonding situation is
somewhat more complicated: The metal atoms are surrounded by a very distorted
square prism of four metal and four halogen atoms. Furthermore, these compounds
are electron deficient in the same sense as the boranes—there are fewer pairs of
electrons than orbitals to receive them and so fractional bond orders of 3 are obtained.
O 3 Mo
® ® = a
(a) (b)
Fig. 16.65 A comparison of the two chloro complexes of Mo(IH: (a) quadruply bonded
[Mo ; CI 8 j J_ ; (b) singly bonded [Mo 6 CI 8 r ,_ . [From Cotton. F. A. Arc. Chem. Res. 1969, 2.
240. Reproduced with permission.!
Iw Cotton. F. A. Acc. Chem. Res. 1969. 2, 240.
Metal Clusters
817
Polyatomic Zintl It has been known for nearly 100 years that posttransition metals dissolve in liquid
Anions and Cations ammonia in the presence of alkali metals to give highly colored anions. I6S In the 1930s,
polyatomic anions (Fig. 16.66a.b) such as Sn 9 _ , Pb 2 ~, Pb?,", Sb}~, and Bi^ - were
identified but not structurally characterized. Attempts at isolating crystals were unsuc¬
cessful because they decomposed in solution. This problem was overcome in 1975 by
stabilizing the cation of the salt as a cryptate (see Chapter 12), e.g., [Na(crypt)J,Pb s
and [Na(crypt)] 4 Sn 9 , which reduces the tendency of the salt to convert to a metal
alloy. 166
Salts of polyatomic cations, such as Bi;' + and Te^, are obtained from melts and
stabilized by large weakly basic anions such as AICI 4 :
Bi + BiClj + AICI,(excess)- Bi 5 [AlCI 4 J 3 (16.126)
Since these homopolyatomic ( Zintl) anions and cations are devoid of ligands, they are
sometimes referred to as "naked" clusters. In general there is a good correlation
between electronic structure and geometry as predicted by Wade's rules for these
clusters, though some exceptions are known. Thus whereas Sn 9 " and Bi 9 + are
isoelectronic. they have different structures, the latter violating the rules. Only a small
distortion of the bismuth cation, however, would convert it to the geometry observed
for the tin cluster.
(c) (d)
Fig. 16.66 Some representative Zintl ions: (a) Pb?". (b) SnJ”, (c) BiJ\ (d) TeJ.
165 Corbett. ]. D.; Critchlow, S. C.: Burns. R. C. In Rinus. Clusters, and Polymers of the Main Group
Elements; Cowley, A. H.. Ed.; ACS Symposium Series 232: American Chemical Society: Wash¬
ington, DC. 1983; p 95.
166 Corbett, i. D. Chem. Rev. 1985. 85, 383-397.
A-58
I • IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
the central atom are treated as ligands in coordination nomenclature. The name of the
central atom, where not a metal, may be contracted.
Examples:
I rpF ]“ hexafluorophosphate(V), or hexafluorophosphatefl )
[Zn(OH) 4 ] 2- tetrahydroxozincate(2-) [. or tetrahydroxozincate(II)]
[SO ] 2- tetraoxosulfate(VI), or tetraoxosulfate(2 -)
[HF^] — difluorohydrogenated -) (often named hydrogendifluoride)
Even when the exact composition is not known, this method can be of use. The
number of ligands can then be omitted, as in hydroxozincate, or zincate ion, etc.
Oxoacid anions. Although it is quite practical to treat oxygen in the same manner as
ordinary ligands and use it in the naming of anions by coordination nomenclature, some
names having the suffix -ite (indicating a lower-than-maximum oxidation state) are useful
and therefore are still permitted.
Examples:
I. NOT nitrite 2. CIO - hypochlorite
A full list of permitted alternative names for oxoacids and derived anions can be found
in [Table 1-4).
Substituent Groups or Radicals
Definitions. The term radical is used here in the sense of an atom or a group of atoms
having one or more unpaired electrons.
Systematic names of substituent groups or radicals. The names of groups which can be
regarded as substituents in organic compounds or as ligands on metals are often the same as
the names of the corresponding radicals. To emphasize the kind of species being described,
one may add the word 'group' to the name of the species. Except for certain trivial names
names of uncharged groups or radicals usually end with -yl. Carbonyl is an allowed trivial
name for the ligand CO.
Examples:
I. (CHj)" methanyl or methyl 2. (NO) nitrosyl
Certain neutral and cationic radicals containing oxygen (or chalcogens) have, re¬
gardless of charge, special names ending in -yl. These names (or derivatives ol these
names) are used only to designate compounds consisting of discrete molecules or groups.
Prefixes thio-, scleno-, and tclluro- are allowed to indicate the replacement of oxygen by
sulfur, selenium, and tellurium, respectively.
Examples:
I. HO hydroxyl 2. CO carbonyl 3. NO, nitry! 22
PO phosphoryl 5. SO sulfinyl, or thionyl 23
SO, sulfonyl, or sulfuryl 33
HOO hydrogenperoxyl, or perhydroxyl. or hydroperoxyl
CrO, chromyl 9. UO, uranyl 10. CIO chlorosyl
II. C10 2 ‘ chloryl 12. CIO, perchloryl
2= The name nitroxyl should not be used for Ihis group because of the use of the trivial name mtroxyhe
acid for H,NO,.
23 The former names are preferred, but the latter are allowed. The variant to be used in any particular
case depends on the circumstances. Thus, sulfuryl is used in inorganic rad.cofunctional nomen¬
clature and sulfonyl is used in organic substitutive nomenclature. [That is. organic chemists tend to
speak of methanesulfonyUhloride. MeSO : CI. while inorganic chemists speak of sulfuryl chlonde.
I IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-59
Such names can also be used in the names of more complex molecules or in ionic species.
Examples:
COCI, carbonyl dichloride [phosgene]
PSC1, thiophosphoryl trichloride 3. NOCI nitrosyl chloride
Salts
Definition of a salt. A salt is a chemical compound consisting of a combination of
cations and anions. However, if the cation H,0 + is present the compound is normally
described as an acid. Compounds may have both salt and acid character. When only one
kind of cation and one kind of anion are present, the compound is named as a binary
compound. When the compound contains more than one kind of cation and/or anion, it is
still considered to be a salt, and can be named following the guidelines below.
When polyatomic cations and/or anions are involved, enclosing marks should be used
to avoid possible ambiguity. Thus Tl'l, is thallium (triiodide) and Tl m l, is thallium tri¬
iodide. Alternatively, thallium(I) triiodide and thallium!III) triiodide would suffice.
Salts containing acid hydrogen. Salts containing both a hydron which is replaceable
and one or more metal cations are called acid salts. Names are formed by adding the word
'hydrogen', with numerical prefix where necessary, after the name of cation(s). to denote
the replaceable hydrogen in the salt. 'Hydrogen' is followed without space by the name of
the anion. IH In certain cases, inorganic anions may contain hydrogen which is not easily
replaceable. When it is bound to oxygen and it has the oxidation state of + I, it will still be
denoted by 'hydrogen’, though salts containing such anions cannot be designated acid
salts.
Examples:
NaHCO, sodium hydrogcncarbonutc
LiH,P0 4 lithium dihydrogenphosphate
K-,HP0 4 dipotassium hydrogenphosphate
CsHS0 4 cesium hydrogensulfate. or cesium hydrogentetraoxosull'ate(VI), or
cesium hydrogentetraoxosulfale( I -)
1-9 OXOACIDS AND DERIVED ANIONS
Introduction
Inorganic chemistry is developing in such a way that names based on function are disap¬
pearing. and nomenclature is based preferably on composition and structure, rather than on
chemical properties. Chemical properties such as acidity depend on the reaction medium
and a compound named as an acid might well function as a base in some circumstances.
The nomenclature of acids has a long tradition and it would be unrealistic to system¬
atize acid names fully and alter drastically the commonly accepted names of important and
well-known substances. However, there is no reason to provide trivial names which could
have a very limited use for newly prepared inorganic compounds.
Definition of the Term Oxoacid
An oxoacid is a compound which contains oxygen, at least one other element, at least one
hydrogen bound to oxygen, and which produces a conjugate base by loss of positive
hydrogen ion(s) (hydrons). The limits of this class of compound are dictated by usage rather
than rules.
Oxoacids have been extensively used and studied and many of them therefore have
names established by a long practice. The oldest names, such as 'oil of vitriol' [= con-
818 16 Inorganic Chains, Rings, Cages, and Clusters
Chevrel Phases Ternary molybdenum chalcogenides, M x Mo h X s , are polynuclear clusters ot special
interest. These compounds, often called Chevrel phases, have both unusual structures
and interesting electrical and magnetic properties. An example is PbMo 6 S g , which is a
superconductor at temperatures below 13.3 K. The idealized structure may be thought
of as an octahedral cluster of molybdenum atoms (as in Fig. 16.65b) surrounded by a
cubic cluster of sulfur atoms, which in turn is surrounded by a cubic lattice of lead
atoms. However, in the actual structure, the inner Mo 6 S 8 cube is rotated with respect
to the Pb lattice (Fig. 16.67). 167 It appears that this rotation is the result of very strong
repulsions between the negatively charged sulfur atoms (or sulfide ions) in one S g cube
with those in an adjacent cube. Thus, if lead is replaced by a more electropositive
metal (e.g., Eu 2+ ), the calculated charges on sulfur increase and the turn angle
increases. Since the superconductivity is thought to be dependent upon the overlap ol
the cl orbitals on molybdenum, this property may be "tunable" by appropriate choice
of metals. 168
Infinite Metal Chains Many highly reduced halides of scandium, yttrium, and zirconium have been found to
have infinite metal-metal bonded chains. 169 Zirconium chloride, for example, contains
double metal layers alternating with double chlorine layers (Fig. 16.68). It was dis-
Fig. 16.67 Structure of one of the Chevrel compounds. PbMo h S H . (•) Mo: (O) S: (O) Pb.
(From Cotton. F. A. In Reactivity of Metal-Metal Bonds', Chisholm. M. H., Ed.: ACS
Symposium Series 155; American Chemical Society: Washington. DC: 1981. Reproduced with
permission.]
uo Dclk, F. S., II; Sienko. M. J. Inorg. Client. 1980. IV. 1352-1356. Potcl. M. Chevrel. R.tScrgent. M.
Acta Crystallogr., Sect. B 1980. 36. 1319-1322.
iMt Burden, J. K.; Lin. J. H. Inorg. Client. 1982. 21, 5-10. Corbett. J. D. J. Solid Stale Client. 1981.
39. 56-74. Saito. T.; Yamamoto. N.; Nagasc. T.; Tsuboi. T.: Kobayashi. K.: Yamagata. T.;
Imolo. H.; Unoura. K. Inorg. Chem. 1990, 29. 764-770.
Zicbarth. R. P.. Corbett. J. D. Acc. Client. Res. 1989. 22. 256-262. Rogel. F.; Zhang. J.: Payne.
M. W.; Corbett. J. D. In Electron Transfer in Biology and tlic Solid Slate: Inorganic Compounds
with Unusual Properties'. Johnson. M. K., Ed.; Advances in Chemistry 226: American Chemical
Society: Washington. DC. 1990; pp 369-389.
Synthesis of
Metal Clusters
Conclusion
Conclusion 819
Fig. 16.68 The structure of
ZrCI, a reduced metal
halide system containing
infinite metal-metal bonds,
showing double metal atom
layers alternating with
double chlorine atom
layers. (From Corbett, J. D.
Acc. Client. Res. 1981. 14.
239. Reproduced with
permission.]
covered rather recently that many of the Groups III (3) and IV (4) halides, once
thought to be binary, are in fact stabilized by the presence of interstitial atoms
(introduced unknowingly) such as hydrogen, carbon, or nitrogen. An example is
Sc 5 CI 8 N, once thought to be Sc 5 Cl x . Its structure reveals an interstitial nitrogen atom
and consists of infinite pairs of chains in which Sc 6 CI I2 N clusters are connected by
shared chlorine atoms (Fig. 16.69a) and by shared metal edges (Fig. 16.69b).’ 7 ” By
exploiting the stabilizing role of interstitial atoms, systematic syntheses have been
developed for many new and interesting substances including over 60 zirconium
chloride phases.
In the preceding description of metal clusters, synthetic reactions were given for
some, but not for others. The paucity of reactions reflects in part the fact that the
synthetic chemistry is not fully systematized—often an attempt to make a quadruply
bonded compound, for example, may lead to a doubly bonded one instead. Many
years ago Cotton observed that “. . the student of cluster chemistry is in somewhat
the position of the collector of lepidoptera or meteorites, skipping observantly over
the countryside and exclaiming with delight when fortunate enough to encounter a
new specimen."’ 7 ’ And more recently, he has stated ". . , the most common method
of synthesis is some sort of thermally driven process, quite often a pyrolysis, and thus
design and selectivity tend to be absent much of the lime."’ 77 Even so, progress
toward rational syntheses continues to be made and. in particular, application of the
principles of isolobality are proving to be especially useful.’ 7 -'
An extremely wide variety of compounds, ranging from metal-only to nonmetal-only
to those molecules that are mixed metal/nonmetal, has been encountered in this
,7H Hwu, S-J.: Dudis. D. S. Corbett. J. D. Inorg. Chem. 1987. 26. 469-473.
• 7 ’ Cotton. F. A. Q. Rev. Chem. Sac. 1966. 20. 397.
I7: Cotton. F. A.: Wilkinson. G. Advanced Inorganic Chemistry, 5th ed.: Wiley: New York, 1988: p
1055.
171 Adams. R. D.: Babin, J. E. Inorg. Cliem. 1987. 26. 980-984. GeolTroy. G. L. In Mend Clusters in
Catalysis'. Gates. B. C.. Guczi, L., Knozinger. H., Eds.: Elsevier: Amsterdam 1986: Chapter I.
Vargas. M. D.; Nicholls. H. N. Adv. Inorg. Chem. Rudiochcm 1986. 30, 123-222, Saito. T.;
Yoshikawa. A.; Yamagata. T.; Imoto. H.: Unoura. K. Inorg. Chem. 1989, 2H. 3588-3592. Glad-
feltcr. W. L. Adv. Organomet. Chem. I98S. 24. 41-86. Stone. F. G. A. In Inorganic Chemistry:
Toward the 21st Century: Chisholm. M. H.. Ed.; ACS Symposium Series 211: American Chemical
Society: Washington. DC. 1983: p 383.
A-56
l-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
Examples:
( 0 2 ) + dioxygen(l+) ion [dioxygenyl]
(S.,") 2+ tetrasulfur(2 + ) ion [This would be "cydo-tetrasulfur(2 + )'' in the known
species.]
(Hg 2 ) 2+ dimercury(2 + ) ion. or dimercury(I) cation [mercurous]
Cations obtained formally by the addition of hydrons to binary hydrides - 5 The name
of an ion derived by adding a hydron to a binary hydride can be obtained by adding the
suffix -ium to the name of the parent hydride, with elision of any final ‘e’. For polycations,
the suffixes -diium, -triium, etc., are used without elision of any final "e'.
Examples:
N 2 H
N,H 2
diazanium. or hydrazinium
diazanedium, or hydrazinium (2 + )
Alternative names for cations obtained formally by the addition of hydrous to mono¬
nuclear binary hydrides. Names for these simple cations can be derived as described
above. Alternatively, they may be named by adding the ending -onium to a stem of the
element name. +
The name oxonium is recommended for H-,0 as it occurs in. for example. H 3 0 CI0 4
(hydronium is not approved )' 5 and is reserved for this particular species. If the degree of
hydration of the H ion is not known, or if it is of no particular importance, the simpler
terms hydron or hydrogen ion may be used.
Examples:
I. NH 4 ammonium, or azanium 2. H 3 0 + oxonium
3. PH 4 h phosphonium 4. H 2 S + sulfonium
Coordination cations. The names of complex cations are derived most simply by
using the coordination cation names (see Section 1-10). This is preferred whenever ambigu¬
ity might result.
Special cases. There are a few cases where trivial, nonsystematic or semisystematic
names are still allowed. Some particular examples are shown.
Examples:
I. NO nitrosyl cation
3. NO 2 nitryl cation
2 . OH hydroxylium
[HOC(NH,) 2 ] + uranium
Monoatomic [and homopolyatomic] anions are named by replacing the termination of the
element name by -idc [and adding a numerical prefix as needed]. In many cases, contrac¬
tions or variations are employed, as exemplified below.
Examples:
Systematic
1 . H"
hydride
2 . 0 2 ~
oxide
S 2_
sulfide
I"
iodide
0 2 -
dioxide(l -)
hyperoxide, or superoxide ' 6
13 [Hydronium appears to be used exclusively in American textbooks.]
16 Although OT is called superoxide in biochemical nomenclature, the Commission recommends the
use of the systematic name dioxide! 1 -), because the prefix super- does not have the same meaning
in all languages. Other common names are not recommended.
l-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry A-57
6 . 0 ;-
dioxide( 2 -)
peroxide
oj
trioxided - )
ozonide
8 . C l~
dicarbide (2 -)
acetylide ' 7
Nj"
trinitride(l —)
azide
Pb~
nonaplumbide(4-)
There are several anions for which trivial names used in the past are no longer recom¬
mended. Other anions have trivial names which are still acceptable. A selection follows.
Examples:
1 . OH - hydroxide (not hydroxyl)
HS hydrogensulfided -)' (hydrosulfide not recommended in inorganic
nomenclature ) 19
NH~ imide, or azanediide 5. NCS“ thiocyanate 20
NH7 amide, or azanide 6 . NCO~ cyanate 20
Anions derived from neutral molecules by loss of one or more hydrons. The names of
anions formed by loss of hydrons from structural groups such as acid hydroxyl are formed
by replacing the -ic acid, -uric acid, or -oric acid ending by -ate. If only some of the acid
hydrons are lost from an acid, the names are formed by adding 'hydrogen', 'dihydrogen',
etc., before the name to indicate the number of hydrons which are still present and which
can, in principle, be ionized.
Examples:
I. CO 5 carbonate 2 . HCO^ hydrogencarbonatel I -)'"
S0 4 sulfate 4. HSO^, hydrogensulfated-) or
hydrogcntetraoxosulfate(Vl)
Anions derived formally by the addition of a hydride ion to a mononuclear hydride are
named using coordination nomenclature (see Section 1-10), even when the central atom is
not a metal.
Examples:
BH 4 ~ tetrahydroboratelI-) (not tetrahydroboronate) 2 '
2 . PH hexahydridophosphaled -)
Coordination nomenclature for hcteropolyatomic anions. The names of polyatomic
anions which do not tall into classes mentioned above are derived from the name of the
central atom using the termination -ate. Groups, including monoatomic groups, attached to
17 |This is the "carbide" ("calcium carbide") of "carbide (acetylene) lamps."]
' In lhc Nomenclature of Organic Chemistry. 1979 edition, hydrogen is always used as a separate
word. However, the names used here are of coordination type, and different rules apply. In
inorganic nomenclature hydrogen is regarded as a cation in the names of acids unless the name is
intended to show that it is combined in an anion [as in the examples] above. [Inasmuch as the
hydrogen is bound the same in H 2 COj. dihydrogen carbonate (carbonic acid), as it is in HCOJ, the
hydrogen carbonate anion, American chemists generally put spaces between the words.]
19 Use of the prefix mono-, to give monohydrogcnsullide, avoids confusion with hydrogen sulfide,
H 2 S. Note that hydrosulfide and hydroperoxide is [sic] still allowed in organic nomenclature.
When coordinated in mononuclear complexes, these ions may bind through cither end. This has led
to the use of the names isocyanate [-NCO vs. cyanate, -OCN], etc., to distinguish the donor. This
usage is discouraged, and the italicized donor symbol, namely, cyanato-0. or cyanato-/V, should be
employed. [Sec page A-72.]
Hydro to represent 'hydrido' or 'hydrogen' is sanctioned by usage in boron nomenclature (see
Section I-II). but is not to be used in other contexts.
Problems
821
820 16
Problems
• Inorganic Chains, Rings, Cages, and Clusters
Fig. 16.69 (a) The structure of “Sc 5 CIk" as it was first reported with stacked Sc ft octahedra
(black bonds) and bridging stacked ScCl fc octahedra (black atoms and white bonds). The
poorly scattering nitrogen atoms are not located, (b) The correct structure of Sc$Cl x N
showing Sc 6 C1| 2 N units bridged by ScCI„. The interstitial nitrogen atoms lie at the center of
the Sc„ octahedra. Part (a) shows several layers of a pair of these chains. If the nitrogen
atoms were shown in (a) they would be stacked in the center of the (wo columns of stacked
black parallelograms, a. [From Hwu. S-J.; Dudis, D. S.; Corbett, J. D. Inorg. Client. 1987.
26, 469-473. Reproduced with permission.]
chapter. All have had their descriptive chemistry systematized on the basis of struc¬
tural principles. Perhaps surprisingly (and perhaps noil) metal clusters obey the same
rules as nonmetal borane clusters; metal-metal multiple bonds follow the same sym¬
metry as those in organic chemistry, then go one better by allowing quadruple bond
formation; catenation, long thought to be an almost exclusive province of organic
chemistry has proved to be an extremely important aspect of inorganic chemistry. A
unified view of the chemistry of all of the elements is emerging as we approach the
21 st century.
16.1 Draw all of the structural isomers of P 4 H h . Assume that inversion at phosphorus is slow
and draw all possible stereoisomers.
16.2 Suggest a structure lor P 7 H 3 and its anion, P 7 H 3 ”.
16.3 As indicated in the text, silanes are less stable than alkanes largely because a facile
decomposition pathway is available to them. Suggest a mechanism for the decomposition
of Si 2 H h .
16.4 Compare the relative reactivity of silanes and alkanes toward nucleophilic attack, hydro-
. lysis, and halogenation___ __
16.5 Methods of successfully synthesizing characterizable organic polysilanes were developed
only in the last decade. With the aid of references given in this chapter, present a
synthesis for the homopolymer (McPhSi),, and the copolymer (MePhSi)„(Me,Si),„.
16.6 Draw structures of [Si0 4 ] 4- . [Si 2 0 7 ] h -. [SiOp]„. [Si 4 0fn„ . [Si 4 07,7)„ . and [Si0 2 |„.
Enclose the repeating units in brackets and show that these empirical formulas are cor¬
rect. How do the ratios of oxygen to silicon correlate with the degree of polymerization in
silicates (i.e.. discrete ions compared to chains compared to double chains compared to
infinite sheets compared to three-dimensional frameworks)?
16.7 Both noselile and ultramarine contain AI,,Si ( ,0 24 formula units. What is the charge on this
unit? In addition, noselite contains a sulfate ion and ultramarine has a persulfide ion. How
many sodium ions are present in each overall empirical formula?
16.8 a. Why are such seemingly disparate substances as talc. clay, and graphite slippery and
useful as lubricants?
b. Although the structures of talc and muscovite arc rather similar, the latter is much
harder and unsuitable as a lubricant. Why? Should these minerals have any properties
in common?
16.9 Gem quality beryls are aquamarine (blue), emerald (green), and golden beryl. Likewise,
amethyst is a violet-colored silica, and sapphire (blue) and ruby (red) are alumina. Yet
pure beryl (Be.,AI 2 Si h O| X ). silica (Si0 2 ). and alumina (Al,0,) arc colorless. Explain.
16.10 Although olivine is a common rock-forming mineral and quartz is the commonest of
minerals, they are never found together. Explain.
16.11 Muscovite and biotite have very similar compositions. Why is one "white mica" and the
other "black mica"? In the same vein, talc is white, chrysolite is white asbestos,
crocidolitc is blue asbestos and amosite is a gray-brown asbestos.
16.12 Compared to molybdenum!VI) and tungsten!VI). chromium(VI) does not have an exten¬
sive polyanion chemistry. Suggest an explanation.
16.13 In addition to chromate and dichromate, trichromatc. |Cr,0,„] : ', exists. Postulate a
structure for the trichromate ion. Compare its structure to that of |P,0,„] J ". Trichromate
hydrolyzes in water. Predict the hydrolysis products.
16.14 From an inspection of the figures, or even better, from molecular models, determine the
geometry of the coordination sphere (cavity) in each of the heteropoly anions discussed in
the chapter.
16.15 Write balanced equations showing the following conversions:
a. |V0 4 ] J - to [V,0 M ]'- b. [H;V lo Oj] , ‘~ to [VOjP
16.16 Consider the structure of the anion. [CeMo l 2 0 42 )' , ‘. This structure may be thought of as
consisting of twelve MoO ft octahedra or six Mo ; 0„ groups. The former suggests 72 oxygen
atoms and the latter 54. yet there are only 42 oxygen atoms in the structure. Explain. How
many terminal oxygen atoms per molybdenum arc present? How many bridging oxygen
atoms are present? What is the coordination number of Ce(IV) and of each kind of oxygen
atom? What is the point group of the entire anion?
16.17 Determine the point groups of the cyclic phosphines in Fig. 16.34 and the phosphorus
oxides of Fig. 16.36.
16.18 The iminoborane. /-PrB=N-f-Bu, trimerizes to (/-PrBN-f-Bu), which has a Dewar-
borazine type structure. I7J Compare this structure to that of borazine.
16.19 Boroxincs result from the condensation of boronic acids, RB(OH) 2 . The cyclic Irimeric
anhydride of methylboronic acid is (MeBO)j. Give a balanced equation for the reaction of
(MeBO)j with water. (See Brown. H. C.; Cole, T. E. Orgunometullics 1985. 4. 816.)
174 Paetzold. P.: von Plotho. C.; Schmid, G.; Boesc. R. Z. Naturforsch. B.Anorg. Chem., Org. Client.
1984.39, 1069.
A-54
I IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
directly without space or hyphen. The final vowels of numerical prefixes should not be
elided, except for compelling linguistic reasons. Note that monoxide is an exception.
Where the compounds contain elements such that it is not necessary to stress the propor¬
tions, for instance, where the oxidation states are usually invariant, then indication propor¬
tions need not be provided.
Examples:
Na,S0 4 sodium sulfate, preferred to disodium sulfate
CaCl, calcium chloride, preferred to calcium dichloride
The prefix mono- is always omitted unless its presence is necessary to avoid con¬
fusion.
Examples:
I. N,0 dinitrogen oxide 2. NO, nitrogen dioxide
N,0 4 dinitrogen tetraoxide 4. Fe 3 0 4 triiron tetraoxide
SXU disulfur dichloride [sulfur monochloride]
6 . MnOj manganese dioxide 7. CO carbon monoxide
The use of these numerical prefixes does not affect the order of citation, which
depends upon the initial letters of the names of the constituents.
However, when the name of the constituent itself starts with a multiplicative prefix (as
in disulfate, dichrcJmate, triphosphate, and tetraborate), two successive multiplicative
prefixes may be riecessary. When this happens, and in other cases simply to avoid
confusion, the alternative multiplicative prefixes, bis-, tris-, tetrakis-, pentakis-, etc. are
used [see Table 1-3] and the name of the group acted upon by the alternative prefix is placed
in parentheses.
Example:
Ba[BrF 4 ] 2 barium bis(tetrafluorobromate)
TI(Ij)j thallium tris(triiodide)
U(S,0 7 ) 2 uranium bis(disulfate ) 11
1-6 SOLIDS
[The Red Book has a twelve-page section of recommendations with respect to solid stale
nomenclature, defects, phases, polymorphisms, etc. Very little of this material is applica¬
ble to this book, so the section has been omitted because of space.]
1-7 NEUTRAL MOLECULAR COMPOUNDS
Substitutive Nomenclature
Introduction. This is a method of naming, commonly used for organic compounds, in
which names are based on that of an individual parent hydride, usually ending in -ane, -ene
or -yne. The hydride name is understood to signify a definite fixed population of hydrogen
atoms attached to a skeletal structure.
[Note the difference between d/sulfatc, SjO 2 , and Wsulfatc, an older, unsystematic name for
HS0 4 ".]
I-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-55
Hydride names
Names of mononuclear hydrides. Substitutive nomenclature is usually confined to
the following central elements: B, C, Si, Ge, Sn, Pb, N, P, As, Sb, Bi, 0. S, Se, Te, Po, but
it may be extended to certain halogen derivatives, especially those of iodine.
In the absence of any designator, the ending -ane signifies that the skeletal element
exhibits its standard bonding number, namely 3 for boron, 4 for Group 14 [1VA] elements, 3
for the Group 15 [VA] elements, and 2 for the Group 16 [VIA] elements. In cases where
bonding numbers other than these are exhibited, they must be indicated in the hydride
name by means of an appropriate superscript appended to the Greek letter A, 12 these
symbols being separated from the name by a hyphen.
Examples:
I. PH 5 A 5 -phosphane 2. SH 6 A 5 -sulfane
This use of lambda applies to the -ane names but not to the synonyms for these names.
[That is, PH 3 can be phosphane (systematic) or phosphine (nonsystematic), but PH S can be
only A s -phosphane; A 5 -phosphine is not permitted.]
Names of oligonuclear hydrides derived from elements of standard bonding number.
Names are constructed by prefixing the ane names of the corresponding mononuclear
hydride with the appropriate multiplicative prefix (di-, tri-, tetra-, etc.) corresponding to the
number of atoms of the chain bonded in series.
Examples:
H-.PPH-, diphosphane [diphosphine] 2. HSeSeSel-l triselane
SiHjSiHjSiHjSiHj tetrasilane 4. H 3 SnSnH 3 distannane
1-8 NAMES FOR IONS, SUBSTITUENT GROUPS AND
RADICALS, AND SALTS
Cations
Names of monoutmnic cations. Monoatomic cations are named by adding in paren¬
theses after the name of the element [either the appropriate charge number followed by the
plus sign or the oxidation number (Roman numeralll followed by the words 'cation' or
'ion'.
Examples:
Na sodiuind +) ion, sodium(l) cation 13
Cr 3 chromium(3 + ) ion, chromium(lll) cation [chromic] 1 ' 1
Cu copperd+ ) ion, copper(I) cation [cuprous]
Cu 2 copper(2 + ) ion. copper(II) cation [cupric]
U h uranium(6+) ion. uranium(Vl) cation
Names of polyatomic cations
Homopolyatomic cations. The name for a homopolyatomic cation is [also the name
of the neutral species plus the charge number or oxidation number in parentheses].
13 For a fuller discussion of the lambda convention, see Pure Appl. Client. 1984. J6, 769.
13 When there is no ambiguity about the charge on a cation, it may be omitted, c.g., aluminum ion for
aluminum(3 + ) or sodium ion for sodium(l+).
14 Older names such as chromous for chromium(U), cupric for copper(II). and mercuric for mer-
cury(II) [and the other older names in brackets above] are no longer recommended.
822
16- Inorganic Chains, Rings, Cages, and Clusters
16.20 Phospham, (PN 2 H)„, can be obtained by the reaction of red phosphorus with ammonia.
Write a balanced equation for its production from these two reagents and draw a possible
structure for this cross-linked polymer.
16.21 Monophosphazenes, R,P=NR'. are well known and may be prepared from the reaction of
RjPCI 2 and R'NH 2 . Write a balanced equation for this reaction.
16.22 As discussed in the chapter, trimeric phosphazenes are usually planar but can be forced
out of this geometry. In contrast, benzene derivatives are strictly planar. Discuss the
reasons for the greater flexibility of the phosphazenes.
16.23 Draw all of the possible isomers, excluding those that are N—N bonded, of S,(NH) X -,.
16.24 The classical argument concerning the equivalence of the positions on the benzene ring is
based on the existence of three (ortho, meta. para) isomers of xylene (dimethylbenzene).
How many isomers are there of dimethylborazine?
16.25 Complexes (0C) 4 Fe(P 4 0,,) and (OC),Fe(PjO ft ) 2 form from the reaction of Fe(CO ) 5 and
P 4 O h . Suggest structures for these complexes. Would you expect similar reactions with
P 4 0|„? (See Walker. M. L.; Mills, J. L. Inorg. Client. 1977, 16. 3033.)
16.26 Phosphorus pentoxide is an excellent dehydrating agent. For example, it can be used to
remove water from nitric acid. Write a chemical equation for this reaction.
16.27 Suggest a structure for P 4 O h S 4 . synthesized from P 4 0|„ and P 4 S,„.
16.28 Ethane reacts with oxygen to give carbon dioxide and water. Diborane reacts with oxygen
to give boron(lll) oxide and water. Write balanced equations for these two reactions.
Look up heats of formation for the reactants and products of these reactions and calculate
the heats of reactions. Considerable work was expended in evaluating boranes as high-
energy fuels in the 1950s. Compare ethane and diborane as fuels.
16.29 It has been suggested 17 ’ that Sej’' exists in the endo form rather than an exo "crown"
form with Se(l) flipped down because of reduced lone-pair repulsions between Se(2),
Sc(3). Se( 6 ) and Se(7). Sketch these two forms of Sej” Add lone pairs to your drawing and
indicate how stabilization occurs in the endo form.
16.30 What structures do you predict for the anions. B 2 HJ and B jH ?
16.31 Diborane is widely used in organic chemistry (hydroboration) to convert alkenes into
alcohols. Consult an organic book and show reactions for this conversion. What is the
advantage of this method?
16.32 Complete the following equations:
a. IPNCLJ, + excess Me : NH -♦
b. B ; H„ + 2R,P -»
c. B 2 H ft + 2NH, -►
d. P 2 CI 4 + excess Cl 2 -
16.33 Assuming that the external H —B — H angle in B 2 H„ accurately reflects the interorbital
angle:
a. Calculate the j and p character in these bonds.
b. Calculate the s and p character remaining for the bridging orbitals.
c. Compare the value from (b) with the experimental internal angles.
16.34 Use Wade's rules to predict the structures of the following:
a. B,H 3 (CO) 2 Fc(CO)j b. C 2 B 4 H a Pi(PEi ,) 2
c. C 2 B 7 H 7 Me : Fe(CO ) 3 d. CB.,H ii 1 AsCo(ti’-C,H,)
,7 ’ Corbett. J. D. Prog. In or g. Clie nt. 1976. 21. 129.
Problems
823
16.35 Use the polyhedral skeletal electron-counting rules and show that they are consistent with
the niilo I l-vertcx structure shown below. 17 '’
OBH •C-Gt
16.36 What is the maximum bond order you would predict for neutral W 2 (no ligands)?
16.37 What bond angle would you expect for M—O—R in an alkoxide complex? How might
this bond angle change as it donation from the p orbitals of oxygen increases?
16.38 The chluro groups in [Re 2 Cl s ] : ~ are eclipsed, but the chloro groups in [Os 2 C]„]'~ are
staggered. Offer an explanation. (See Agaskar. P. A.: Cotton, F. A.: Dunbar. K. R.;
Falvello. L. R.. Tetrick. S. M.: Walton. R. A. J. Ant. Client. Soc. 1986. 11)8. 4850.)
16.39 Note that the product of Eq. 16.125 contains Mo-Mo pairs that are doubly bridged by
chlorine and Mo-Mo pairs that are not. If this molecule contains alternating single and
triple bonds, which bonds are which?
16.40 The structure of NijCI 2 S 2 (PPh 4 ) 4 is analogous to that of (p-cymene),Ru,S 2 (Fig.
16.55). 177 Determine if Wade's rules arc satisfactory for this molecule.
16.41 Which of the following do you think would be most likely in view of isolobal considera¬
tions and Wade's rules? (See Little. J. L.: Whilesell, M. A.; Kestcr. J. G.; Foiling. K.;
Todd. L. J. Inorg. Client. 1990. 29. 804-808.)
X Y
a. Bi S
b. S S
c. P CM
d. P Bi
Obh
16.42 In addition to the phosphazenes discussed in this chapter, a large group of heterocyclic
compounds known as phosphazancs has been characterized. These contain P—N single
bonds and may contain phosphorus either in the +3 or +5 oxidation states. Draw
structures of |CLPNMe | 2 and [CI(0)PNMe|.. Phosphorus!III) tri- and letraphosphazanes
have been stabilized by placing o-phenylene groups between adjacent nitrogen atoms.
Give one example of each and draw its structure. (See Barcndt. J. M.: Halliwanger. R. C.;
Squicr. C. A.: Norman. A. D. Inorg. Client. 1991, 30. 2342-2349.)
16.43 Draw structures for the four possible isomers of c/uv«-Et ; C 2 B<Hv (Sec Beck. J, S.:
Sneddon. L. G. Inorg. Cliem. 1990. 29. 295-302).
,7h Swisher. R. G.: Sinn. E.. Butcher. R. J.; Grimes. R. N. Organonteuillics 1985. 4. 882-890.
77 Fenske. D.: Ohmer. J.: Hachgenei. H.: Mcrzweiler. K. Angew. Client. Int. Ed. Engl. 1988. 27.
1277-12%.
A-52
I-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
Addition compounds. In the formulae of addition compounds, the component mole¬
cules are cited in order of increasing number, if they occur in equal numbers, they are cited
in alphabetical order of the first symbols. Addition compounds containing boron com¬
pounds or water are exceptional, in that the water or boron compound is cited last. If both
are present, the boron precedes water.
Examples:
3CdS0 4 -8H,0 2. Na,CO 3 -l0H,O 3. AI,(S0 4 ) 3 K,S0 4 -24H,0
C 6 H h NH 3 -Ni(CN) 2 5. 2CH 3 OH BF 3 6. BF 3 -2H 2 0
1-5 NAMES BASED ON STOICHIOMETRY
Names of Constituents
Electropositive constituents. The name of a monoatomic electropositive constituent is
simply the unmodified element name. A polyatomic constituent assumes the usual cation
name, but certain well established radical names (particularly for oxygen-containing spe¬
cies such as nitrosyl and phosphoryl) are still allowed for specific cases.
Examples:
I. NH 4 C1 ammonium chloride 2. OF 2 oxygen difluoride
U0 2 CI 2 uranyl dichloride 4. 0,F, dioxygen difluoride
FOCI, phosphoryl trichloride 6. NOCI nitrosyl chloride
7, 0,[PtF fi ] dioxygen hexafluoroplatinate [dioxygenyl hexafluoroplatinate]
Monoatomic electronegative constituents. The name of a monoatomic electronegative
constituent is the element name with its ending (-en, -ese, -ic, -ine, -ium, -ogen, -on, -orus,
-urn, -ur. -y, or -ygen) replaced by the anion designator -ide.
Examples:
chloride derived from chlorine
2 . carbide derived from carbon
tungstide derived from tungsten
arsenide derived from arsenic
silicide derived from silicon
6 . hydride derived from hydrogen
oxide derived from oxygen
8 . phosphide derived from phosphorus
Finally, there are monoatomic anions whose names in English, though derived as described
above, are based on the Latin root of the element names. [See table inside back cover.] In
these the ending -um or -ium is replaced by -ide.
Examples:
I. auride—aurum—gold 2. plumbide—plumbum—lead
stannide—stannum—tin 4. natride—natrium—sodium 9
Homoatomic electronegative constituents. These have the name of the monoatomic
parent, but qualified by a multiplicative prefix, if appropriate. It may be necessary to use
parentheses to emphasize subtle points of structure.
9 [Although the Latin natrium has been around fora long time, the IUPAC has not recommended its
use previously, and Na - has been universally called "sodide." In view of the phasing out of terms
like “cuprous" and “ferric," the introduction of hitherto unused latinized names is unexpected.]
I- IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-53
Examples:
I. Na 4 Sn 9 tetrasodium (nonastannide) 2. TI(I 3 ) thallium (triiodide)' 0
3. TICI 3 thallium trichloride 10 4. Na 2 S 2 sodium disulfide
Heteropolyatomic electronegative constituents. The names of these anions take the
termination -ate, though a few exceptions are allowed (see Examples 5-16 below). The
ending -ate is also a characteristic ending for the names of anions of oxoacids and their
derivatives. The names sulfate, phosphate, nitrate, etc., are general names for oxoanions
containing sulfur, phosphorus, and nitrogen surrounded by ligands, including oxygen,
irrespective of their nature and number. The names sulfate, phosphate, and nitrate were
originally restricted to the anions of specific oxoacids, namely SO;; - , P0 4 - , NO]", but this
is no longer the case.
Examples:
I. SO 2- trioxosulfate, or sulfite 2. SO 2- tetraoxosulfate, or sulfate
N0 2 dioxonitrate, or nitrite 4. NOJ trioxonitrate, or nitrate
Many names with -ate endings are still allowed, though they are not completely in
accord with the derivations outlined above. Some of these are cyanate, dichromate,
diphosphate, disulfate, dithionate, fulminate, hypophosphate, metaborate, metaphosphate,
metasilicate, orthosilicate, perchlorate, periodate, permanganate, phosphinate, and phos-
phonate. The exceptional cases where the names end in -ide or -ite rather than -ate are
exemplified below.
Examples:
CN cyanide 6. NHNH, hydrazide 7. NHOH - hydroxyamide
8 . NH 2 amide 9, NH 2- ' imide 10. OH - hydroxide
II. As0 3 arsenite 12. CI0 2 chlorite 13. CIO - hypochlorite
NO, - nitrite 15. SO 2- sulfite 16. S,0^ - dithionite
Indication of Proportions of Constituents
Use or multiplicative prefixes. The proportions of the constituents, be they mono-
atomic or polyatomic, may be indicated by numerical prefixes (mono-, di-, tri-, tetra-,
penta-, etc.) as detailed in [Table 1-3]. These precede the names they modify, joined
Table 1-3_
Numerical prefixes
2 di (bis)
3 tri (tris)
4 tetra (tetrakis)
5 penta (pentakis)
6 hexa (hexakis)
7 hepta (heptakis)
8 octa (octakis)
9 nona (nonakis)
10 dcca (decakis), etc.
11 undcca
12 dodeca
13 tn'dcca
14 tetradeca
15 pentadeca
16 hexadeca
17 heptadeca
18 octadeca
19 nonadeca
20 icosa
21 henicosa
22 docosa
23 tricosa
30 triaconta
31 hentriaconta
35 pentalriaconta
40 tetraconta
48 octatetraconta
50 pentaconta
52 dopentaconta
60 hexaconta
70 heptaconta .
80 octaconta
90 nonaconta
100 hecta
10 The use of oxidation state designators would be appropriate. [The parentheses around “triiodide''
distinguishes TI(I 3 ) - from T1 3 3I - .]
Chap ter
17
The Chemistry of the Halogens
and the Noble Gases
At first thought it might appear incongruous to discuss the chemistry of the halogens
and the noble gases together. The former includes the violently reactive fluorine which
will oxidize all save a half dozen elements, even reacting explosively with a compound
as stable as water, while the latter family contains the inert gases' neon (differing from
fluorine by one proton and one electron per atom) and argon (Gr. argos. lazy, useless).
In one important aspect, however, they are very similar, namely their ionization
energies: F = 1681 kJ mol” 1 (17.42eV). Ne = 2081 kJ mol -1 (2l.56eV), Ar = 1520kJ
mol -1 (15.75 eV).- The difference between these two families lies in the inordinate
disparity in electron affinities. The common tendency to emphasize the differences
between the families and dismiss the similarities results from a lack of recognition of
the two types of behavior (gain vs. loss of electrons) and the fact that the noble gases
are unique only in the discontinuity of their ionization energy-electron affinity (or
electronegativity) function. Thus it is often said or implied that the electronegativities
of the noble gases are low or nonexistent, equating electronegativity with electron
affinity. It is true that the noble gases have no negative oxidation states. But. on the
other hand, electronegativity also means a reluctance to release electrons and in this
regard the noble gases, as a group, are unsurpassed. In fact, the limiting factor with
regard to which noble gas compounds form, or do not form, seems to be the extent to
which a given noble gas atom is willing to share electrons with an atom of another,
more electronegative element.
The halogens (except fluorine) exhibit both electron-accepting and electron-re¬
leasing behaviors. Since they resemble the noble gas elements only in regard to
electron-releasing, this aspect of the chemistry of the two families will be discussed
first. The more familiar acceptance of electrons by halogens is discussed later in the
chapter.
1 Several names have been applied to the Group VIIIA (18) elements. The term ’•inert" is inapplicable
to the group as a whole (it is used above as a specific adjective of neon and argon, not as a group
appellation) because at least three members of the family are nol inert. The name "noble gas"
implies a reluctance to react rather than complete abstention, thus paralleling the use of this term in
describing the chemistry of certain metals such as gold and platinum.
2 One could also draw a comparison between the noble gases and the alkali metals based on their low
electron affinities. However, the electron affinities of noble gases appear always to be endothermic
whereas the alkali metals have small but finite exothermic electron affinities leading to some
chemistry based on acceptance of electrons (Chapter 12).
Noble Gas Chemistry
825
Noble Gas
Although the observation of a line in the sun's spectrum as early as 1868 led workers
to postulate the existence of an unknown element in the sun's atmosphere, (he
isolation in 1889 of helium from the mineral cleveite by heating was not recognized as
a related phenomenon. The first definitive work was by Lord Rayleigh, who noticed a
discrepancy between the density of "chemical nitrogen" and that of "atmospheric
nitrogen." The former was obtained by chemically removing nitrogen from various
nitrogen oxides, ammonia, or other compounds. The latter was obtained by removal
of oxygen, carbon dioxide, and water vapor from air. The difference in density was not
great: 1.2572 x 10 -3 g cm - ' for "atmospheric nitrogen" compared with 1.2506 x
I0 -3 g cm -3 for "chemical nitrogen" under the same conditions. The careful work
necessary to establish this difference has often been pointed to. quite rightly, as an
example of the importance of precise measurements. Unfortunately, loo often the
emphasis has been upon the number of significant figures rather than the realization by
Rayleigh and Ramsay that the difference was chemically significant. Their arguments
concerning the significance of the ratio of specific heats of the noble gases (C /C r =
1 .66) 3 and their rebuttal of various arguments advanced by their critics show as much
chemical insight as the gas-density argument, if nol more. 4
Ramsay and Rayleigh succeeded in isolating all of the noble gases except radon
and in showing that they were inert to all common reagents. They also discovered the
identity of alpha particles and ionized helium.
Chemistry
The Discovery of the
Noble Gases
The Early Chemistry It is often assumed that the noble gases had no chemistry prior to 1962. This is true
of the Noble Gases on ly if one restricts the definition of a chemical compound to (I) something containing
“ordinary" covalent or ionic bonds and (2) something which may be isolated and
placed in a bottle on the reagent shell'. If either of these criteria is dismissed, much
important chemistry of the noble gases can be recognized prior to the 1960s.
If an aqueous solution of hydroquinone is cooled while under a pressure of several
hundred kilopascals (equals several atmospheres) of a noble gas [X = Ar, Kr. Xe], a
crystalline solid of approximate composition [C ( ,H 4 (OH),J,X is obtained. These solids
are /3-hydroqtiinone clathrates with noble gas atoms filling most of the cavities.-'
Similar noble gas hydrates are known (Fig. 17.1). These clathrates are of some
importance since they provide a stable, solid source of the noble gases. They have also
been used to effect separations of the noble gases since there is a certain selectivity
exhibited by the clathrates.
Of particular interest is the effect of noble gases in biological systems. For
example, xenon has an anesthetic effect. This is somewhat surprising in that the
conditions present in biological systems are obviously not sufficiently severe to effect
chemical combination of the noble gas (in the ordinary sense of that word). It has been
proposed that the structure of water might be altered via a clalhrale-type interaction.
Although clathrate formation and dipole interactions are perfectly acceptable
subjects for chemical discussions, chemists feel more at ease when they can find stable
compounds formed from the species being studied. A logical approach would be the
3 (indicating that the gas must be monatomic since energy is absorbed only by translational modes, not
by vibration or rotation (cf. C,,/C,. = 1.40, 1.36. and 1.32 for N : . CL. and I3r : )|.
For a discussion of the earliest work on the noble gases, see Wolfendcn, J. H. J. Client. Ethic. 196V.
46 . 569; Hicbcrt. E. N. In Noble-Gas Compounds: Hyman, H. H.. Ed.; University of Chicago:
Chicago, 1963; p 3.
5 The percentages of available cavities that are filled by noble gas atoms are 67% (Ar), 67-74% (Kr).
and 88% (Xe).
A-50
I - IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
1-4 FORMULAE
[This section contains considerable material on the proper writing of formulae, much of it
routine, some of it highly specialized. Most of the indicators of the number of atoms,
oxidation states, ionic charge, optical activity, and structures are very similar to those for
names and will be discussed in later sections, especially Section I—10.]
Free radicals. IUPAC recommends that the use of the word radical be restricted to
species conventionally termed free radicals. A radical is indicated by a dot as right
superscript 6 to the symbol of the element or group. The formulae of polyatomic radicals are
placed in parentheses and the dot is placed as a right superscript to the parentheses. In
radical ions, the dot precedes the charge.
Examples:
H’ 2. (CN)' 3. (HgCN)' 4. (0 2 )~
Br' 6 . [Mn(CO) s ]' 7. (SnCl 3 ) - 8 . (0 2 ) ’
Structural modifiers. Modifiers such as cis-. Irons-, etc., are listed in [Table 1-6; see
also page A-69J. Usually such modifiers are used as italicized prefixes and are connected to
the formula by a hyphen [cis-, irans-, etc.].
Examples:
m-[PtCI,(NH 3 ),] 2. //wij-[PtCI 4 (NH 3 ),]
Sequence of Citation of Symbols
Priorities
General. The sequence of symbols in a formula is always arbitrary and in any
particular case should be a matter of convenience. Where there are no overriding require¬
ments. the guidelines summarized [in Table 1-2] should be used.
Electronegativities and citation order. In a formula the order of citation of symbols is
based upon relative electronegativities, the more electropositive constituent(s) being cited
first. In the formulae of Brpnsted acids, acid hydrogen is considered to be an elec¬
tropositive constituent and immediately precedes the anionic constituents.
Examples:
I. KCI 2. HBr 3. H,S0 4 4. NaHS0 4
CaSO., 6 . [Cr(H,0)JCI 3 7. H[AuClJ 8 . IBrCI,
Table 1-2 _
Assignment of formulae
of compounds
•; n t •? —t- ~^ T -• y
(i) Assign symbols to the constituents.
(ii) Indicate proportions of constituents.
(iii) Divide constituents into electropositive and electronegative. This requires
decisions concerning compound type. There are special rules for acids,
polyatomic groups, binary compounds, chain compounds, intermetallic
compounds, coordination compounds, and addition compounds [q.v.].
(iv) Assemble the formula.
(v) Insert appropriate modifiers (geometrical, etc.).
(vi) Insert oxidation states, charges, etc., if required.
6 [It may be noted here that very often in the past the free radical dot was set "on line" (H •) rather
than as a superscript.]
I 1 IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
A-51
Chain compounds. For chain compounds containing three or more different ele¬
ments, the sequence should generally be in accordance with the order in which the atoms
are actually bound in the molecule or ion.
Examples:
I- —SCN (not CNS) 2. HOCN (cyanic acid) 3. HONC (fulminic acid)
Polyatomic ions. Polyatomic ions, whether complex or not, are treated in a similar
fashion. The central atom(s) (e.g., 1 in [ICI 4 ] - , U in UO; + , Si and W in [SiW p 0 40 ] 4- ) or
characteristic atom (e.g.. Cl in C1CT, 0 in OH - ) is cited first and then the'subsidiary
groups follow in alphabetical order of the symbols in each class.
Examples:
I. SO]' 2. NO,” 3. OH -7 4. [P 2 W IK 0 62 ] h - 5. [BH 4 ] _
6 . U0; + 7. CIO' 8 . P 2 0 4 “ 9. [Mo ft O l8 ] 2 - 10. [IC1J"
Polyatomic compounds or groups. It is necessary to define the central atom of the
compound or group, and this is always cited first. If two or more different atoms or groups
are attached to a single atom, the symbol of the central atom is followed by the symbols of
the remaining atoms or groups in alphabetical order. The sole exceptions are acids, in the
formulae of which hydrogen is placed first. When part of the molecule is a group, such as
P=0, which occurs repeatedly in a number of different compounds, these groups may be
treated as forming the positive part of the compound.
Examples:
PBrCI 2 2. POCIj or PCI 3 0 3. H 3 P0 4 4. SbCI,F
Coordination compounds. [In the formula of a coordination entity,] the symbol of the
central atom(s) is placed first, followed by the ionic and then the neutral ligands. Square
brackets are used to enclose the whole coordination entity whether charged or not. This
practice need not be used for simple species such as the common oxoanions (NO^, NO,",
S0 4 - , OH”, etc.). Enclosing marks arc nested within the square brackets as follows: [()],
[{()}], [{[()]}]. [{{[())}}]. etc.
A structural formula of a ligand occupies the same place in a sequence as would its
molecular formula.
Examples:
I. K,[Fe(CN)J 2. [AI(0H)(H,0)jJ I+ 3. [Ru(NH,),(N,)]CI,
c-«-[PtCI 2 {P(C 2 H,) 3 },| 5. Nn[PtBrCI(NO)NH,] 6 . [PlCI,(C,H,N)NH>
Abbreviations. These may be used to represent ligands in formulae, and they are
cited in the same place as the formulae they stand for. The abbreviations should be lower
case, and enclosed in parentheses. Some commonly used abbreviations are in [Table 1-5.]
Examples:
I. [Pt(py) 4 ][PtCl 4 J 2. [Fe(en) 3 ][Fe(CO) 4 ] 3. [Co(en),(bpy)]- 1+
The commonly used abbreviations for organic groups, such as Me, Ph, Bu, etc., are
acceptable in inorganic formulae. Note that the difference between an anion and its parent
acid must be observed. Thus acac is an acceptable abbreviation for acetylacetonate.
Acetylacetone (pentane-2,4-dione) then becomes Hacac.
7 The hydroxide ion is represented by the symbol OH", although the recommendations for the
formulae of acids would suggest HO. Example 3 accords with majority practice.
8 [Note that in this example the cis-trans isomer is not specified.]
826
17-The Chemistry of the Halogens and the Noble Gases
Fig. 17.1 The structure of
the xenon hydrate clathrale.
The xenon atoms occupy
the centers of regular
pentagonal dodecahedra of
water molecules
(cf. Fig. 8 . 8 ).
investigation of the noble gases for possible Lewis basicity. Since the noble gases are
isoelectronic with halide ions which can be strong Lewis bases, it seemed reasonable
that noble gas adducts of strong Lewis acids might likewise exist:
F“ + BF 3 -► BF 4 -
(17.1)
Ne + BF 3 -♦ NeBF 3
(17.2)
Xe + BF 3 - XeBFj
(17.3)
Thorough studies of solutions of xenon in boron trichloride and boron tribromide were
undertaken. A phase study of the melting point of these systems as a function of
composition showed no evidence of compound formation. The Raman spectra of these
mixtures are identical to those of pure BX, indicating no noble gas-boron trihalide
interactions.
An alternative approach to the formation of true chemical compounds of the noble
gases is suggested by two lines of thought. (I) From an acid-base point of view, the
strongest Lewis acid is the bare proton. H", so if any of the noble gases is capable of
exhibiting basic behavior it might be expected to do so with W (not H,0‘):
He + H + -► HeH + (17.4)
Ar + H -► ArH + _ ' (17.5)
(2) From a simple molecular orbital diagram such as given in Fig. 5.7. it might be
supposed that four electrons would result in a nonbonding condition but that any
number of electrons less than four would result in some bonding though not neces¬
sarily in tin integral bond order. Spectroscopic evidence for species such as HeH and
ArH + has been obtained from a mixture of hydrogen and noble gases passed through
gas discharge tubes. Similar reactions can take place between two noble gas atoms if
energy is supplied to remove the necessary electron:
He + He
He-
(17.6)
Noble Gas Chemistry
827
The Discovery of
Stable, Isolable
Noble Gas
Compounds
Kr + Kr Kr-T (17.7)
Ne + Xe NeXe + (17.8)
The noble gas hydride ions should have a bond order of one and the diatomic noble gas
ions should have a bond order of one-half. Neither type can be isolated in the form of
salts of the type HeH + X~ or He 2 X~ since the electron affinity of positive helium, etc.
is greater than that of any appropriate species X, and so such salts would spon¬
taneously decompose:
2HeJF" -► 4He + F 2 (17.9)
ArH^Cl - -+ Ar + HCI (17.10)
Although the above reactions may seem to be of little interest to the chemist, it has
been found that in similar gas-phase reactions, xenon behaves as a nucleophile,
forming the methylxenonium ion. CH,Xe. The C—Xe bond in this ion has a strength
of 180 ± 30 kJ mol -1 .
Although there had been suggestions that some of the noble gases might form com¬
pounds. unsuccessful attempts to oxidize krypton and xenon with fluorine in the 1930s
essentially put a halt to such speculation, especially in view of the success of valency
theory in relating stability to filled octets. This worship of the octet is all the more
surprising in view of the fact that compounds with expanded valence shells were
already known for over two-thirds of the remaining nonnielals ! 6
In the early 1960s Neil Bartlett was studying the properties of platinum hex¬
afluoride. an extremely powerful oxidizing agent. In fact, by merely mixing dioxygen
with platinum hexafluoride, it is possible to remove one electron from the oxygen
molecule and isolate the product:
0 2 + PtF 6 -► 0 2 [PtF 0 ] - (17.11)
Bartlett realized that the first ionization energy of dioxygen. I ISO kJ mol -1 (12.2 eV),
is almost identical to that of xenon. 1170 kJ mol -1 (12.1 eV). Furthermore, the
dioxvgen cation should be roughly the same size as a Xe" ion. and hence the lattice
energies of the corresponding compounds should be similar. He mixed xenon and
platinum hexafluoride with the result that an immediate reaction look place with
formation of a yellow solid. Based on the volumes of gases that reacted. Bartlett
suggested the formulation Xe + [PlFJ ", 7 However, the reaction is not so simple as was
once thought, and the product has been formulated variously as Xe(P(FJ +
Xe|PtF h ],. Xc|PtFJ r (I < .r < 2). or FXe + |PtF„J- + FXe + |Pt,F,,r. Despite the
experimental difficulties in characterizing the products of the reaction there was no
doubt whatever that a reaction had taken place and the myth of inertness was
shattered forever.
<• This point will be discussed further later in this chapter and in Chapter 18. but for now it may be
noted that about a dozen nonmctals are known with valence shells (by conventional electron-pairing
formalisms) containing from It) to 14 electrons. B. C. N. O. and F more strictly obey Ihc octet rule.
7 Bartlett. N. Proc. Client. Soc. 1962. 218. For a review of Ihe early chemistry of xenon compounds,
sec Bartlett. N. EnJctivoiir 1963. 88. 3. Note that pure Xe[l>tF,,r forms only if a large excess of
inert SF,. is added as a diluent (Graham. L. Ph.D. Thesis. University of California at Berkeley. 1978.
as cited in Ihc reference in Footnote 8).
A-48
I-IUPAC Recommendations on the Nomenclature of Inorganic Chemistry
T able 1-1 _
A selection of affixes used in
inorganic and organic
nomenclature
Names are constructed by joining other units to these base components. Affixes are
syllables or numbers added to words or roots and they can be suffixes, prefixes, or infixes,
according to whether they are placed after, before, or within a word or root. Representative
examples are listed in [Table 1-1], together with their meanings.
Termination vowel for skeletal replacement nomenclature: -oxa [O], -aza
[N], -carba [C], -thia [S]
Termination for names of neutral saturated hydrides of boron and elements
of Groups 14 [IVA], 15 [VA], and 16 [VIA]: diphosphane [= P 2 H 4 ,
diphosphine]
General suffix for many polyatomic anions in inorganic nomenclature
(including coordination nomenclature) and in organic nomenclature:
nitrate, acetate, hexacyanoferrate.
Termination for name of many acids both inorganic and organic: sulfuric
acid, benzoic acid
Termination for name of certain monoatomic anions: chloride, sulfide
Termination for names of the more electronegative constituent [atom or
group] in binary type names: disulfur dichloride; triiodide; cyanide
Termination for trivial name of certain hydrides such as N 2 H 4 and PH 3 :
hydrazine, phosphine
General termination for radicals and substituent groups of all kinds
containing a metal center from which the linkage is made: cuprio-,
methylmercurio-, tetracarbonylcobaltio-
Termi nation for esters and salts of certain oxoacids having the -ous ending
in the acid name: sulfite [from sulfurous acid]
Termination of names for many elements, and preferred termination for the
name of any new element
Termination for many electropositive constituents in binary type names,
either inorganic or organic, either systematic or trivial
Termination indicating the addition of one hydrogen ion (or a positive alkyl
group) to a molecular hydride or its substitution product: ammonium.
Termination of cations formed from metallocenes: ferrocenium
Termination indicating a negatively charged ligand: bromo-. Usually it
appears as -ido, -ito, -ato
Termination for the names of many inorganic and organic radicals: chloro-,
piperidino-
Tcrmination for infixes to indicate replacement of oxygen atoms and/or
hydroxyl groups: thio-, nitrido-
Suffix for the trivial names of bis(cyclopentadienyI)metals and their
derivatives: ferrocene
Termination indicating a substituted derivative of the type XH 5 :
dichlorotriphenylphosphorane (X = P)
Termination for prefix indicating a group of the type X(O):
phosphoryl (X = P)
Termination of a name of an oxoacid of a central element in an oxidation
state lower than the highest. This nomenclature is not generally
recommended: phosphorous [acid].' . ;
Termination for names of certain radicals containing oxygen: hydroxy,
carboxy ■ ' ' 1 '
Common termination for names of radicals: methyl, phosphanyl, uranyl.
Termination of trivial names of many oxygenated cations: uranyl j
I IUPAC Recommendations on the Nomenclature o! Inorganic Chemistry
A-49
1-3 ELEMENTS, ATOMS, AND GROUP OF ATOMS
Names and Symbols of Atoms
The lUPAC-approved names of the atoms of atomic numbers 1-109 for use in the English
language are listed in alphabetical order in [the table inside the back cover].
Indication of Mass, Charge, and Atomic Number Using Indexes (Subscripts
and Superscripts)
Mass, ionic charge, atomic number, [and molecular formula] are indicated by means of:
left upper index [superscript] mass number
left lower index [subscript] atomic number
right upper index [superscript] ionic charge
[right lower index (subscript) molecular formula] 4
Example:
SSi + represents a doubly [positively] ionized [tetrasulfur cation composed of] sulfur
atom[s] of atomic number 16 and mass number 32.
Isotypes
Isotypes of hydrogen. The three isotopes, 'H, 2 H, and J H, have the names 'protium',
‘deuterium', and 'tritium', respectively. It is to be noted that these names give rise to the
names proton, deuteron, and triton for the cations 'H \ 2 H, 3 H respectively. Because
the name 'proton' is often used in contradictory senses, of isotopically pure 1 H ions on
the one hand, and of the naturally occurring undifferentiated isotope mixture on the other,
the Commission recommends that the latter mixture be designated generally by the name
hydron, derived from hydrogen. 5
Name of an element or elementary substance of definite molecular formula or struc¬
ture. These are named by adding the appropriate numerical prefix [see above] to the name
of the atom to designate the number of atoms in the molecule. The prefix mono is not used
except when the element does not normally exist in a monoatomic stale.
Examples:
Symbol
Trivial name
Systematic name
H
atomic hydrogen
monohydrogen
2 . 0,
oxygen
dioxygen
Oj
ozone
trioxygen
4- Pa
white phosphorus
tetraphosphorus
(yellow phosphorus)
S 6
—
hexasulfur
6 . S 8
a-sulfur, /3-sulfur
octasulfur
i. s„
/i-sulfur (plastic sulfur)
polysulfur
I
i _
4 |Not all of these sub- and superscripts will normally be used at one time. Thus we may have 3 S 2+ ,
the IUPAC example, or we might have S 4 for the telrasulfur(2 + ) cation.]
5 [This is a new recommendation. Note that it implies that all uses with respect to acids and bases
involving the normal isotopic mixture of 'H, 2 H, and 3 H would require the use of hydron, i.e., the
hydron affinity of bases. Brflnsted acids are hydron donors, etc. We have retained the current usage
of proton affinity, etc., because the recommendation came out as this book was going to press. The
reader should note however that hydron is already receiving some European usage, as in: "GS~ =
GSH dehydronated at the thiol group."]
17 "The Chemistry of the Halogens and the Noble Gases
Mixing xenon and fluorine and activating the mixture by thermal, photochemical, or
similar means result in the production of fluorides:
The Fluorides of the
Noble Gases 8
The chief difficulties in these reactions (once the proper equipment for handling
elemental fluorine at high pressure has been assembled) are not the syntheses but the
separations. All three products tend to form (Fig. 17.2). Xenon difluoride can be
obtained either by separating it rapidly before it has a chance to react further (by
freezing it out on a cold-finger, for example) or by keeping a high Xe-to-F, ratio. The
hexafluoride is favored by large excesses of fluorine and low temperatures, but some
XeF 4 is present which must be separated. The best production of XeF 6 is obtained
using gases at low pressure and a “hot wire," a nickel filament at 700-800 “C. The
reactor is cooled in liquid nitrogen. The most difficult compound to prepare pure
is xenon tetrafluoride since even optimum conditions for its formation thermally
(Fig. 17.2) will result in concomitant formation of XeF, and XeF 6 . The use of dioxygen
difluoride at low temperatures and pressures provide XeF 4 in high yield and purity: 9
Xe + 20 2 F 2 -» XeF 4 + 20, (17-15)
The chemistry of krypton is much more limited than that of xenon. Apparently
only the difluoridc forms directly from the elements. Attempts to make helium, neon,
and argon fluorides have been unsuccessful. Radon should react even more readily
than xenon, but its chemistry is complicated by the difficulty of working with a
Fig. 17.2 Equilibrium
pressures of xenon fluorides
as a function of tempera¬
ture. Initial conditions:
125 mmol Xc. 1225 mmol F 2
per liter. At higher Xe to F 2
ratios the XcF„ diminishes
considerably and
the remaining two curves
shift to the left. (From
Selig, H. Halogen Client.
1967, /. 403. Reproduced
with permission.]
Temperature (K)
Scppclt. K.; Lentz. D. Prog. Inorg. Client. 1982, 29. 167-202.
9 Nielsen, J. B.; Kinkead, S. A.; Purson, J. D.; Eller, P. G. Inorg. Client. 1990, 29, 1779-1780.
Noble Gas Chemistry 829
compound of such exceedingly high radioactivity. Nevertheless, the formation of
compounds with fluorine was shown conclusively shortly after the discovery of xenon
compounds, but their exact natures were not elucidated. More recently radon chem¬
istry in solution has been studied (see below).
Bonding in Noble There are currently two approaches to the problem of bonding in noble gas com-
Gas Fluorides pounds. Neither is completely satisfactory, but between the two they account
adequately for the properties of these compounds. The first might be termed a valence
bond approach. It would treat the xenon fluorides by means of expanded valence
shells through promotion of electrons to the 5 d orbitals:
Ground state: Xe = [Kr] 5s 2 4d l0 5p 6
Valence state: Xe = [Kr] 5s 2 4(/‘°5p 6 '"5d"
For XeF,, n = 1 and two bonds form; for XeF 4 , n = 2 and four bonds form; and for
XeF fi , n = 3 and six bonds form. Using the arguments of VSEPR theory (see Chapter
6 and further discussion below) the resulting electronic arrangements and structures
are as follows:
Compound
Electron
pairs
Hybridization
Predicted
structure 10
Experimental
structure
XeF,
5
sp 3 d
Linear (TBP)
Linear
XeF 4
6
spd 2
Square (octahedral)
Square planar
XcF„
7
sp 3 d 3
Nonoctahedral
(capped octahedron?)
Unknown exactly,
but not octahedral
The use of Gillespie's VSEPR theory has allowed the rationalization of these as
well as several other structures of noble gas compounds (Fig. 17.3). One of the signal
successes of this approach was the early prediction that XeF h was nonoctahedral (see
Chapter 6). The most serious objection to it is the required promotion of electrons.
This has been estimated to be about 1000 kJ mol -1 (10 eV) or more for xenon, a large
amount of energy. Furthermore, d orbitals tend lo be diffuse and their importance in
nonmetal chemistry is a matter of some controversy (see Chapter 18).
An alternative approach to bonding in noble gas compounds is the molecular
orbital approach involving three-center, four-electron bonds. Consider the linear
F—Xe — F molecule. A 5 p orbital on the xenon can overlap with fluorine bonding
orbitals (either pure p orbitals or hybrids) to form the usual trio of three-centered
orbitals: bonding, nonbonding, and antibonding (Fig. 17.4). Filling in the four valence
electrons (Xe Jp z + F 2/) i + F 2 p i) results in a filled bonding orbital and a filled nonbond¬
ing (to a first approximation) orbital. A single bond (or bonding MO) is thus spread
over the F — Xe — F system. A second p orbital at right angles to the first can form a
second three-center F—Xe — F bond (XeF 4 ), and the third orthogonal p orbital can
form a third three-center bond (XeF,,). The nature of the p orbitals involved in the
bonding allows one to predict that XcF 2 will be linear, XeF 4 square planar, and XeF 6
10 The first structure is that of the atoms; the geometry in parentheses refers to the approximate
arrangement of all of the valence shell electrons.
A p p e n d i x
in
ii
I!
II
II
I
K
V
I'
II
II
I
II
I
Ii
r
Ii
Ii
w
I
»
I
A-46
i
IUPAC Recommendations on the
Nomenclature of Inorganic Chemistry
The standards of nomenclature in chemistry are proposed by the International Union of
Pure and Applied Chemistry (IUPAC). The current edition (the third)' for inorganic
nomenclature is Nomenclature of Inorganic Chemistry: Recommendations 1990. issued by
the Commission on Nomenclature of Inorganic Chemistry. Part I consists of 289 pages and
it is to be followed by several other parts of specialized nomenclature. It is possible
therefore to include here only a very small fraction as a general guide to good usage. Thus
the following material is intended to guide the reader through the process of using good
nomenclatural practice, but it is not meant to be a substitute for the Recommendations
themselves.
The guidelines presented here (or even the unabridged set in the Red Book) should not
be viewed as a rigid code but as an evolving attempt to clarify the process of naming. Usage
must be by consensus and it is interesting to note that whereas preceding editions of the
Red Book were entitled "Rules," the current edition consists of Recommendations.
The usage in our book has been as close to IUPAC nomenclature as is consistent with
good pedagogy. We have followed the IUPAC Recommendations except in those cases in
which they conflict directly with current American usage. 1 2 3 Students are not served well by
finding one type of nomenclature in the text while being encouraged to read the original
literature in which they find a radically different nomenclature. The prime purpose of this
book is to illustrate inorganic chemistry rather than the details of nomenclatural technique.
The nomenclature has therefore been that which would serve best in teaching inorganic-
chemistry and help the students in reading the original literature.
This appendix consists of short excerpts from Recommendations 1990. Our com¬
ments and other additions have been placed in square brackets [ ]. Deletions have not been
marked, but every attempt has been made to keep the intent of the original. Otherwise the
following are verbatim extracts of the Recommendations with the exception of minor
editing such as changing (1) numbers in series (Footnotes, Tables, and Examples) to make
1 All three editions have been published with red covers and arc nicknamed the “Red Book."
2 A summary of inorganic nomenclature from the American point of view may be found in Block.
B. P.; Powell, W. H.; Fcrnclius, W. C. Inorganic Chemical Nomenclature: Principles and Practice.
American Chemical Society: Washington. DC, 1990. Unfortunately, it went to press too early to
incorporate the changes in the 1990 IUPAC Recommendations.
3 Nomenclature of Inorganic Chemistry: Recommendations 1990: Leigh. G. J.. Ed.: Blackwell Scien¬
tific: Oxford, England, © 1990, International Union of Pure and Applied Chemistry. Reproduced
with permission.
I • IUPAC Recommendations on the Nomenclature of Inorganic Chemistry A-47
them continuous and (2) the spelling to conform to American usage, e.g., aluminum (Engl,
aluminium), center (Engl, centre), cesium (Engl, caesium), etc. Footnotes noi in square
brackets are footnotes from the Red Book; footnotes in square brackets have been added
by us. The Red Book consists of eleven chapters (with numbered subsections), which are
here abstracted as “sections"; cross references have generally been omitted except to
these major sections. Section I-1, Section 1-2, etc.
We have occasionally placed older, often obsolete names in square brackets after the
IUPAC name. In no case are these intended to be recommended alternatives but merely to
be useful guides to the older (and sometimes current) literature.
All chapters in (he Red Book begin with much interesting historical and philosophical
material concerning nomenclature. These make interesting and educational reading, but
they have not been reproduced for space reasons.
1-1 GENERAL AIMS, FUNCTIONS, AND METHODS OF
CHEMICAL NOMENCLATURE
Methods of Inorganic Nomenclature
Systems of nomenclature
Binary-type nomenclature. In this system, the composition of a substance is spe¬
cified by the juxtaposition of element group names, modified or unmodified, together with
appropriate numerical prefixes, if considered necessary.
Examples:
I. sodium chloride [NaCl] 2. silicon disulfide [SiS-,] 3. uranyl difluoride [UO : F,|
Coordination nomenclature. This is an additive system for inorganic coordination
compounds which treats a compound as a combination of a central alom with associated
ligands (see Section 1-10).
Examples:
triamminetrinitrocobalt [Co(NO,),(NH,)j]
sodium pentacyanonitrosylferrate Na ; |Fe(CN) ? NO|
Substitutive nomenclature. This system is used extensively for organic compounds,
but it has also been used to name many inorganic compounds. Ii is often based on the
concept of a pareni hydride modified by substitution of hydrogen atoms by groups (radi¬
cals). [See Section 1-6.J
Examples:
I. bromobutane [C 4 H v Br] 2. difiuorosilane [SiH 2 F,] 3. trichlorophosphane [PCI 3 ]
1-2 GRAMMAR
Introduction
Chemical nomenclature may be considered to be a language. As such, it is made up of
words and it should obey the rules of syntax.
Generally, nomenclature systems use a base on which the name is constructed. This
base can be derived from a parent compound name such as sil (from silane) in substitutive
nomenclature (mainly used for organic compounds) or from a central atom name such as
cobalt in additive nomenclature (mainly used in coordination chemistry).
830
17-The Chemistry of the Halogens and the Noble Gases
Linear molecule with ihree nonbonding
electron pairs at the points of an equilateral
triangle
Square planar molecule with two nonbonding
electron pairs, one above and one below the
plane of the molecule
Distorted octahedron with a nonbonding
electron pair either at the center of a face or the
midpoint of an edge
Square pyramidal molecule with a
nonbonding electron pair protruding from the
base of the pyramid
(D
Trigonal pyramidal molecule with a
nonbonding electron pair protruding from the
apex of the pyramid
Perfect tetrahedron
Pentagonal planar ion with two nonbonding
electron pairs above and below the plane of the
pentagon.
Fig. 17.3 Molecular shapes predicted by simple VSEPR theory. Bond angle values represent
experimental results where known.
warn ■
octahedral. The first two predictions are correct, but the last is nol. On the other hand,
difficulties with promotion energies are avoided. However, in a pure 3-c-4-e model,
the entire molecule is held together by only three bonds.
Structural Data for
14-Electron Species
The number of species isoelectronic with XeF 6 is quite limited. The anions SbBr^ - ,
TeCI - . and TeBrg - are octahedral. Both 1F^ and XeF h are nonoctahedral. 11 Iodine
heptafluoride and rhenium heptafluoride may be considered isoelectronic with these
species if they are all considered to have 14 valence shell electrons of approxi¬
mately equal steric requirements. Both have a pentagonal bipyramidal structure (see
Fig. 6.12). Most interestingly, the XeFJ anion, formed by the Lewis acid XeF 4 :
M + F- 4- XeF 4 -♦ M^XeF; (17.16)
is the first example of a pentagonal planar inorganic ion (Fig, 17.3g). 12 It can be
rationalized in terms of five bonding pairs to fluorine atoms in a plane with a lone pair
above and below the plane. The lone pairs appear to be “locked in" to the axial
positions as the molecule is not fluxional, as is the isoelectronic XeF 6 .
The molecular structure of XeF h has been a vexing problem. In the solid, XeF,
and F~ ions exist. The former has five bonding and one nonbonding pair and is
therefore expected to be a square pyramid with the lone pair occupying the sixth
position of the idealized octahedron. Experimentally. Ihis is found to be the case. In
the gas phase, however, the structure is much more perplexing, from both an experi-
mental and a theoretical view. Electron diffraction studies indicate that the molecule is
a slightly distorted octahedron that is probably “soft" with respect to deformation.
There are no measurable dipole moments, ruling out large, static distortions. The
model that is currently accepted is that of a stereochemically nonrigid molecule that
rapidly passed from one nonoctahedral configuration to another.
The perfectly octahedral species conform to the expectations based on the simple
MO derivation given above. The nonoctahedral fluoride species do not. but this
difficulty is a result of the oversimplifications in the method. There is no inherent
necessity for delocalized MOs to be restricted to octahedral symmetry. Furthermore,
it is possible to transform delocalized molecular orbitals into localized molecular
orbitals. Although the VSEPR theory is often couched in valence bond terms, it
depends basically on the repulsion of electrons of like spins, and if these are in
localized orbitals the results should be comparable.
11 Both the CIF and BrFi" ions are octahedral. Christc, K. O.: Wilson. W. W.; Chirakal, R. V.;
Sanders, J. C. I’.: Schrobilgen. G. J. Irwrg. Chew. 1990, 29, 3506-3511.
•- Christc. K. O.t Curtis. E. C.: Dixon. D. A.; Mcrcier, H. P.; Sanders. J. C. P.; Schrobilgen. G. J. J.
Am. Chem. Soc. 1991. IIJ. 3351-3361.
A-44
H Models, Stereochemistry, and the Use of Stereopsis
Stereoviews and Stereopsis
Closely related to the use of models is the corresponding one of stereoviews to
illustrate molecular perspective. The increased availability of molecular data and the
use of computers to generate stereoviews has made their use routine in journal
accounts of structure determinations. Unlike models, no matter how useful, stereo¬
views have the ability to capture the depth of a three-dimensional structure on a two-
dimensional sheet of paper. An increased number of stereoviews has been included in
this edition.
It is not necessary to view stereoviews three dimensionally: Either half conveys
much information by itself. But every effort should be made to learn the “tricks” of
stereopsis—the amount of insight to be gained is more than rewarding. And although
initial attempts may be frustrating, it is well worth the effort and the process becomes
increasingly simple and routine. Furthermore, in addition to being a purely informa¬
tional routine, it is an aesthetic experience. Don’t be afraid to experiment—each
person has different ways of facilitating the process. However, the following gener¬
alizations should be helpful.
The basic principle (“secret"?) of stereopsis is that one is having the eyes do
something that they would never otherwise do; that is, each eye must look at a
different image (as opposed to the same image from a slightly different angle). The
common frustration in attempting to look at stereoviews is that first both eyes look at
the left image, then both eyes look at the right image, then both eyes. ... It was to
break this coordinated behavior of the human visual system that the old-fashioned and
bulky stereoscope was invented. Today, folding viewers can readily do the job (see
below). By all means, use such a viewer in your initial attempts. It isolates the eyes
from each other. Start with a simple stereoview with lots of visual clues to aid the
perspective; the cubic crystal systems of Chapter 4 are good. Have the image to be
viewed lying perfectly flat. Put the stereoviewer over the images and try to have
everything “squared away." Relax and view the image. You should see a single image
with depth of field.
Once the above has been achieved, it is usually possible for the (human) viewer to
become sufficiently adept at the process that the (optical) viewer can be dispensed
with. The "trick" is to have the eyes looking parallel so that the left eye is looking at
the left image, and the right eye at the right image. The way to have the eyes looking
with parallel lines of sight is to view something "at infinity." It does not have to be the
Andromeda Nebula; across the room will do. Looking over the top edge of the book or
journal containing the stereoview, gaze across the room. Relax. Without making any
particular effort to focus on anything, let your gaze drop to the stereoview. If
successful, you will see three images. The center one is the important one and the one
with stereopsis. The other two are unimportant, except as possible distractions—they
are the left and right images seen "out of the comer" of the opposite eye. If both eyes
"lock" on to one or the other of the images, look up, relax, and try again. Try at a
different time of day, with a different figure, when you’re fresher, or even when you’re
tired, but relaxed.
A third method of stereopsis with which your authors have not been particularly
comfortable, but which many people use quite routinely is the "cross-eyed method." 4
4 Graham. D. M. J. Chent. Educ. 1986, 63. 872.
H Models, Stereochemistry, and the Use of Stereopsis
A-45
Quite simply, it is just that: The lines of sight of the eyes are crossed and the right eye
looks at the left image and vice versa. The method works because most people have a
greater ability to command their eyes to cross than to command them to remain
perfectly parallel. (Hence the repeated suggestions to relax above.) If you find that
you can achieve stereopsis more readily this way, fine: If it works, use it! But please
remember that the image you see will be inverted. For most purposes this makes no
difference, but if the image is a chiral molecule, the cross-eyed method will give the
perception of the other enantiomer from the one portrayed. 5
Problems
H.l Construct a chiral D } model as shown. Through which faces docs the C 3 axis pass? In
viewing the model as an example of D y symmetry (A = B), where are the secondary C 2 axes
that place it in a dihedral group? How is it that the C 3 group lacks these? Which enantiomer
do you have? Do you have any problems constructing the other enantiomer? Does the
"trick" for constructing the second enantiomer give you any insight into symmetry opera¬
tions and chirality?
II.2 Construct a model of buckminsterfullerene, "buckyball", according to the directions in the
reference in Footnote 2.
a. Turn to Chapter 3 and do Problems 3.32 through 3.36 on the basis of the model in your
hands. Use paper "glue-on's" or light pencil drawings to indicate the osmyl groups of
Fig. 3.34.
I). Bromination of buckyball yields a derivative with 12 Br 2 molecules adding across double
bonds:
C w + 12 Br : -» C„,Br : 4 (H.l)
The resultant structure has a very high and unusual symmetry, with the remaining 18
double bonds shielded from addition by the bulky bromine atoms. Suggest a structure.
H.3 Turn to Chapter 12 and do Problem 12.28.
5 For further discussion of stereopsis sec Spcakman, J. C. New Scientist 1978, 78. 827; Chcm. Britain.
1978, 14. 107; Johnstone. A. H.; Lclton, K. M.; Spcakman, J. C. Educ. Client. 1980. 17. 172-173,
177. Jensen. W. B. J. Client. Educ. 1982. 59. 385; Falk. D. S.; Brill. D. R.: Stork. D. G. Seeing the
Light-. Harper & Row: New York. 1986; pp 209-219; Smith. J. V. Chem. Rev. 1988. 88. 149-182.
Tebbe. F. N.; Harlow. R. L.; Chase. D. B.; Thorn. D. L.: Campbell. G. C.. Jr.; Calabrese, J. C.;
Herron, N.; Young. R. J.. Jr.; Wasserman, E. Science 1991. 256. 822-825.
832
17Tlie Chemistry of the Halogens and the Noble Gases
Other Compounds
of Xenon
As with many inorganic systems, the differences between alternative interpreta¬
tions is more apparent than real. Favoring the purely octahedral molecule will be
reduced promotion energies 13 and reduced steric requirements. If these two con¬
straints are relaxed, a stereochemically active (“hybridized”) lone pair is favored,
probably as a result of better overlap and stronger bonds.
Gillespie 14 first discussed the problem presented to the VSEPR theory by the
perfectly octahedral species such as SbBr^~, TeClj; - , and TeBrg". He pointed out that
steric interactions between the large halide ligands will be of considerable importance.
(The Br—Br distance is approximately equal to the sum of the van der Waals radii and
a "seven-coordinate” structure with a large lone pair occupying one position would be
unfavorable.) He therefore suggested that as a result, the seventh pair of electrons
resides in an unhybridized s orbital inside the valency shell. As such it would be
sterically inactive except for shielding the valence electrons and loosening them from
the nucleus. The somewhat lengthened bond in TeBrj; - , 275 pm, compared with that
expected from addition of covalent radii, 250 pm, is consonant with this interpretation
(as it also is with a bond order of less than one from a three-center bond). In most
fluorides the reduction of steric factors allows the lone pair to emerge to the surface of
the molecule, although perhaps less than it would in a four- or five-coordinate
molecule; hence these molecules appear less distorted than might have been expected
(see also Chapter 6).
There are two essentially isostructural cation/anion pairs that are not iso-
electronic; They differ by a pair of electrons that could potentially be stereochemically
active lone pairs. 15 These are BrF^/BrFj" and IF#/IFJ. Simple VSEPR theory would
predict an octahedron and a square antiprism (or closely related eight-coordinate
structure) for the cations. The anions might be expected to be a distorted octahedron
GBrFj"; cf. the isoelectronic :XeF 6 ) and a distorted square antiprism (MFjj - ). How¬
ever. bromine is smaller than xenon, and even the larger iodine atom apparently
reaches (as does xenon in :XeFx) its coordination limit with eight fluorine atoms.
Thus the anions are also a perfect octahedron and a perfect square antiprism; They
differ from the corresponding cations only in having longer X—F bonds, as might be
expected if steric crowding of the fluorine atoms forces the nonbonding pair of
electrons into a shielding, centrosymmetric j orbital.
Attempts to isolate a stable xenon chloride have not been very successful. Two
chlorides have been identified, and both arc apparently unstable species observable as
a result of trapping in a matrix. The radioactive decay of l -' , I in K.ICI.,:
m ICl 4 - -► m XeCl 4 + r (17.17)
has been used to produce xenon tetrachloride, characterized by means of the
Mossbauer effect (see Chapter 18) of the gamma emission from the resulting excited
IJ Use of the .s orbital in stereochemically active ("hybridized") orbitals requires raising the energy of
these electrons to that of the bonding orbitals. The use of empty, high energy if orbitals also requires
an increase in average electron energy ("lowering of holes").
14 Gillespie. R. 3. J. Chem. Etluc. 1970, 47, 18. More recent treatments can be found in Gillespie, R. J.
diem. Soc. Rev. 1992, 21. 59-69 and Gillespie. R. J,; Hargittai, I. The VSEPR Model of Molecular
Geometry, Allyn and Bacon: Boston, 1991.
13 For the determination of the structure of IFh . a discussion of the BrFft/BrF^ and IF»/IFx
problem, and appropriate earlier references, see Mahjoub, A. R.; Scppelt, K. Angew. Chem. hit.
Ed. Engl. 1991, 30, 876-878.
Noble Gas Chemistry
833
state. Mixtures of xenon and chlorine that were passed through a microwave dis¬
charge and immediately frozen on a Csl window at 20 K gave infrared evidence for the
existence of XeCl 2 .
Careful hydrolysis of xenon hexafluoride produces xenon tetrafluoride oxide:
XeF 6 + H 2 0 -> XeOF 4 + 2HF (17.18)
Complete hydrolysis of xenon hexafluoride or hydrolysis or disproportionation of
xenon tetrafluoride produces the trioxide:
XeF 6 + 3H,0 -► Xe0 3 + 6HF ( 17 . 19 )
6XeF 4 + 12H,0 -► 2Xe0 3 + 4Xe + 30 2 + 24HF (17.20)
Xenon trioxide is highly explosive and thus renders any intentional (or unintentional)
hydrolysis of these xenon fluorides potentially hazardous. Alternative sources of
oxygen to form the oxyfluorides have therefore been proposed:
XeF 6 + NaN0 3 -► XeOF 4 + NaF + FN0 2 (17.21)
XeF 6 + OPF 3 -► XeOF 4 + PF S (17.22)
Xenon difluoride dioxide cannot be isolated as an intermediate between the partial
hydrolysis to form XcOF 4 and complete hydrolysis to Xe0 3 . but may be prepared by
combining these two species:
XeOF 4 + Xe0 3 -► 2Xe0 2 F 2 (17.23)
It may also be prepared by treatment of an excess of XeOF 4 with cesium nitrate: '7
XeOF 4 + CsN 0 3 -► Xe0 2 F 2 + CsF + FN0 2 (17.24)
By comparison, hydrolysis of xenon difluoride results only in decomposition:
2XeF 2 + 2H 2 0 - 2Xe + 4HF + 0 2 (17.25)
Huston has systematized much of this type of chemistry in terms of the
Lux-Flood definition of acids and bases (see Chapter 9). The latter is one of those
highly specialized definitions that may be very useful in a restricted area but not in the
more general case. In the present instance ihe xenon fluorides are (he oxide acceptors,
while they in turn fluoridate Ihe oxide donors. It is possible to construct a scale of
relative acidity: XeF 6 > Xe0 2 F 4 > Xe0 3 F, > Xe0 4 > XeOF 4 > XeF 4 > Xe0 2 F 2 >
XeOj > XeF,. wherein any acid can react with any base below it to produce an
intermediate acid. Thus, in general, XeF 6 is the most useful and XeF 2 the least useful
fluoridator. When XeF 4 is used in place of XeF h , the reactions are slower. IX
Xenon fluorides are also excellent fluorinators, though not so reactive as KrF 2
(see below). They are often "clean,” the only by-product being xenon gas:
2(S0 3 ) 3 + 3XeF 2 -► 3S 2 0 6 F 2 + 3Xe (17.26)
(C 6 H 5 ) 2 S -t- XeFj -♦ (C 6 H 5 ) 2 SF 2 + Xe (17.27)
16 Christc. K. 0.; Wilson. W. W. Inorg. Chem. 1988. 27, 1296-1297. Nielsen, J. B.; Kinkcad, S. A.;
Eller. P. G. Inorg. Chem. 1990, 29. 3621-3622.
17 Christe. K. O.: Wilson. W. W. Inorg. Chem. 1988. 27, 3763-3768.
,lt Huston. J. L. Inorg. Chem. 1982. 21, 685.
H-Models, Stereochemistry, and the Use of Stereopsis A-43
Chiral (D 3 ) "octahedral" models. By cutting and pasting in such a way as to
leave "chelate rings" on the model 1 both A and A enantiomers can readily be
constructed. Note that having chelate rings on both triangle 2 and triangle 7 is
redundant. One may be removed or used as a construction flap. If A = B, the
model has D 3 symmetry; if A ^ B. the symmetry is C 3 .
Icosahedral models. Cut out the entire figure drawn with solid lines. Omit the
faces marked "O h only." Construction is facilitated by bending and gluing the
two end sections into the “capping" and "foundation" pentagonal pyramids
first. Then the remaining "equatorial" band of ten faces can be wound around
and fastened to these “end groups." The complete icosahedron represents the
B,,HjT ion or dicarbaclosododecaboranes. Other polyhedral boranes can be
formed by removal of the appropriate faces from the complete icosahedron.
Pentagonal dodecahedral models. This model can be constructed in a similar
way by copying, cutting, and gluing the pentagonal drawing.
A similar outline drawing for the construction of buckminsierfullerene has been
published. 2 Several other polyhedral models constructed from simple materials have
been described in a series of articles. 3
1 Note that this model has been altered somewhat from the general model of Fig. II.I. In previous
editions an attempt was made to avoid such changes but with the availability of photocopying
machines with size-altering adjustments and the student's ubiquitous use of them, we have been less
restrictive in this edition olfcring the student greater flexibility.
2 Vittal. J. D. J. Cheat. Educ. 1989. 66. 282.
5 Yamana. S. J. Cltem. Educ. 1988. 65. 1072 (tetragonal pyramid); 1989, 66. 1019 (triangular
dodecahedron); 1990. 67. 1029-1030 (interpenetrating trigonal pyramids). These are but three exam¬
ples. Additional ones may be found from the references in these or by consulting the author indexes
of the Journal of Chemical Education between the years 1979 and present.
834
1 7 • The Chemistry of the Halogens and the Noble Gases
Sometimes the (luorination occurs with displacement of a by-product:
2R 3 SiCl + XeF 2 -> 2R 3 SiF + Cl, + Xe (17-28)
Even more interesting is the production of the Xe? cation in antimony pentafluoride as
solvent. It was first prepared by reduction of Xe(Il). Many reducing agents are
suitable, including metals such as lead and mercury, or phosphorus trifluoride, lead
monoxide, arsenic trioxide, sulfur dioxide, carbon monoxide, silicon dioxide, and
water. Surprisingly, even gaseous xenon may be used as the reducing agent: 19
Xe + XeF + — Xe? (17-29)
Alternatively, one can view this as an acid-base (instead of a redox) reaction of a basic
xenon atom undergoing a nucleophilic attack on an acidic xenon cation to form the
diatomic cation (cf. the reaction of Xe + CH?, page 827).
As mentioned above, xenon trioxide is an endothermic compound which explodes
violently at the slightest provocation. Aqueous solutions are stable but powerfully
oxidizing. These solutions are weakly acidic ("xenic acid") and contain molecular
XeO,. When these solutions are made basic. HXe0 4 ions are formed and alkali
hydrogen xenates, MHXe0 4 , may be isolated from them. Hydrogen xenate ions
disproportionate in alkaline solution to yield perxenates:
2HXeO? + 20H~ - XeOjJ" + Xe + O, + 2H,0 (17.30)
Xenate solutions may also be oxidized directly to perxenate with ozone. Solid perxe¬
nates are rather insoluble and are unusually stable for xenon-oxygen compounds:
Most do not decompose until heated above 200 °C. X-ray crystallographic structures
have been determined for several perxenates, and they have been found to contain the
octahedral XeO£" ion. which persists in aqueous solution (possibly with protonation
to H XeOJ“).
Treatment of a perxenate salt with concentrated sulfuric acid results in the most
unusual xenon tetroxide:
Ba,Xe0 6 + 2H 2 S0 4 2BaS0 4 + 2H,0 + Xe0 4 (17.31)
The tetroxide is the most volatile xenon compound known with a vapor pressure of
33(H) Pa (25 mm Hg) at 0 °C. The structure of the molecule is tetrahedral as is the
isoelectronic 10? ion.
The xenon-oxygen compounds are extremely powerful oxidizing agents in acid
solution as shown by the following £> values:
2.17
H XeO. XeO Xe
4 0 J
The nature of the species and the values of the potentials are not as well characterized
in basic solution, but the oxidizing power seems to be somewhat less:
1.17
HXeOft” Xe0 3 0H- — Xe
•'» Stein. L.; Henderson. W. W. J. Am. Client. Sue. 1980. 102. 2856.
Noble Gas Chemistry
835
Like the fluorides, they are relatively "clean" reagents. Unfortunately, the explosive
properties of Xe0 3 have resulted in less work being done with them.
Xenon forms stable compounds only with the most electronegative elements:
fluorine (x = 4.0) and oxygen (x = 3.5), or with groups such as OSeF 5 and OTeF 5 that
contain these elements. Reasonably stable, though uncommon, bonds are known
between xenon and both chlorine (x = 3.0) and nitrogen (x = 3.0). Bis(tri-
fluoromethyDxenon, Xe(CF 3 ), (x CFj = 3.3), is known but decomposes in a matter of
minutes.
Xenon hexafluoride can act as a Lewis acid. It reacts with the heavier alkali
fluorides to form seven-coordinate anions, which in turn can rearrange to form eight-
coordinate species:
MF + XeFft -♦ M + [XeF 7 ]”(M = Na, K, Rb, Cs) (17.32)
2M + [XeF 7 ]” —^ M 2 [XeF s ] + XeF fi (M = Rb, Cs) (17.33)
The octafiuoroxenates are the most stable xenon compounds known; they can be
heated to 400 °C without decomposition. The anions have square antiprismatic geome¬
try. They. too. present a problem to VSEPR theory analogous to that of XeF b since
they should also have a stereochemically active lone pair of electrons that should
lower the symmetry of the anion. If the steric crowding theory is correct, however, the
presence of eight ligand atoms could force the lone pair into a stereochemically inert s
orbital.
Xenon fluorides can also act as fluoride ion donors. Strong Lewis acids react with
xenon fluorides to yield the expected compounds, but since both the cationic and
anionic species can form fluoride bridges, the stoichiometries may appear strange at
times:
XeF 6 + PtF 5 -♦ XeFj'PtFft-
(17.34)
2XeF, + AsF 5 -♦ Xe,F?AsF 6 -
(17.35)
XeF 4 + 2SbF 5 -♦ XeFjSbjF?,
(17.36)
Even compounds with deceptively simple stoichiometries may be more complex, as in
the case of XeF : AsF 3 . which has both bridged cations and anions:
r f i
f F f F
/ \
1 / 1 /
Xe
Xe
F — As — F— As— F
/
\
/I /I
F
F
F F F F
Bartlett's compound (page 827). though still incompletely understood, is thought to be
of this type. Krypton difluoride forms analogous compounds such as KrF AsF?,
KrF + As 2 FJ" l . KrFSbF?. KrF + Sb : F? l , and Kr,F?SbF?, and these together with
the parent KrF, were the only known compounds of krypton known until recently. 20
Al-Mukhtar. M.; Holloway. J. H.: Hope. [3. G.: Schrobilgcn. U. J, J. Client. Sue.. Dalton Trans.
1991. 2831-2834. Schrobilgcn. G. J. J. Chan. Soc.. Chan. Ctmrnun. 1988. 863-865. 1506-1508.
Append i x
Models, Stereochemistry,
and the Use of Stereopsis
Paper Models
It is convenient for many purposes to have models available for inspection in order to
realize fully the three-dimensional aspect of molecular and lattice structures. "Ball-
and-stick” models of various stages of sophistication are useful when it is necessary to
be able to see through the structure under consideration. Space-filling models of atoms
with both covalent and van der Waals radii are particularly helpful when steric effects
are important. The space-filling models and the more sophisticated stick models tend
to be rather expensive, but there are several inexpensive modifications of the “ball-
and-stick” type available. It is extremely useful to have such a set at hand when
considering molecular structures.
Simple tetrahedral or octahedral models are useful in connection with basic
structural questions (as, for example, the first time you try to convince yourself that
the two enantiomers of CHFCIBr or of [Co(en) 3 ] +3 are really nonsuperimposable). If
stick models are not available, such simple models can be constructed in a few
minutes from paper. In addition, models having bond angles not normally found in
ball-and-stick kits—for example, the icosahedral boranes and carboranes—can also
be readily constructed from paper. Paper models are especially useful when large
numbers of models are necessary as, for example, in constructing models of the iso-
and heteropolyanions.
On the following pages generalized outlines are given for the construction of
tetrahedra, octahedra, icosahedra (Fig. H.l) and pentagonal dodecahedra (Fig. H.2).
These outlines may be reproduced as many times as desired by means of photocopy¬
ing machines. Instructions for cutting are as follows:
Tetrahedral models. Cut out the four triangles enclosed by the T d brackets
(Fig. H.3) and marked with the vertical lines in the drawing. Glue or tape tabs
onto adjacent faces to form the tetrahedron.
Octahedral models. Cut out the two sets of eight triangles enclosed by the O h
brackets and marked with the horizontal lines in the drawing. Glue or tape tabs
onto adjacent faces to form the octahedron.
A-40
rat
836
17-The Chemistry of the Halogens and the Noble Gases
Reaction of HC=NH + salts of [AsFJ or R F C=N adducts of AsF s with KrF 2 in
nonaqueous solvents lead to salts characterized as [HC^N — Kr —F] + [AsF,~] and
[R F C=sN — Kr— F] + [AsF^] (R F = CF 3 , C 2 F 5 , rt-C 3 F 7 ). Krypton fluoride has proven
extremely useful as a fluorinating agent: It is 50 kJ mol -1 more exothermic than
fluorine, F 2 ! It may be used to raise metals to unusual oxidation states : 21
8 KrF 2 + 2Au - 2KrF + AuFg + 6 Kr + F 2 (17.37)
KrF + AuF^ -^° c > AuF s + Kr + F, (17.38)
Bond Strengths in As might be expected, xenon does not form any strong bonds, but it does form
Noble Gas exothermic compounds with fluorine. Some typical bond strengths are listed in Table
Compounds 17.1. Bartlett 22 has shown that such values might have been expected by extrapolation
of known bond energy in related nonmetal compounds.
The Chemistry Since radon is the heaviest member of the noble gas family it has the lowest ionization
of Radon energy, 1037 kJ mol -1 (10.7 eV), and might be expected to be the most reactive. The
radioactivity of this element presents problems not only with respect to the chemist
(who can be shielded) but also with respect to the possible compounds (which cannot
be shielded). On the other hand, this radioactivity provides a built-in tracer since the
position of the radon in a vacuum line can be ascertained by the y radiation of 2 l 4 Bi,
one of the decay products . 22 It was found that when a mixture of radon and fluorine
was heated, a nonvolatile product was formed, possibly an ionic radon fluoride.
Similar experiments with chlorine mixtures left the radon in volatile form, presumably
unreacted. More recently it has been found that radon reacts with various halogen
fluoride solvents (BrF 3 , BrF 5 , and C1F 3 ) to form a species in solution which remains
behind when the solvent is volatilized. It is quite possible that a Rn 2+ species is
present. Although the charge is not known with certainty, the radon may be present as
a cation in solution since it migrates to the negative electrode under certain circum¬
stances, but also towards the positive anode under others, indicating the possible
formation of Rn 24- , RnF + , and RnFJ.
Table 17.1
Bond strengths in noble gas
Compounds
Bond
Bond energy
compounds (kJ mol -1 )
XeF 2 , XeF 4 , XeF 6
Xe—F
130 ±4
Xe0 3
Xe—O
84
KrF 2
Kr—F
. 50
:l Holloway, J. H.; Schrobilgen, G. J. J. Chem. Soc., Client. Comm. 1975, 623.
22 Bartlett. N. Endeavour 1963, 88. 3. It is interesting to note that Bartlett did not use isocleclronic
scries in his extrapolation. Furthermore, although his extrapolations provide quite reasonable
values for XeF 6 and Xe0 3 , they led to much too high a value for KrFj.
21 The a and p radiation of Rn and the first and second daughters will not penetrate the vacuum line.
Halogens in Positive Oxidation States
837
Halogens in Positive
Oxidation States
Interhalogen
Compounds
T able 17.2 _
Interhalogen compounds
In addition to Ihe dihalogen species. X 2 . known for all of the halogens in the elemental
state, all possible combinations XY are also known containing two different halogen
atoms . 24 In addition there are many compounds in which a less electronegative
halogen atom is bound to three, five, or seven more electronegative halogen atoms to
form stable molecules. The known interhalogens are listed in Table 17.2.
Several trends are noticeable from the data in the table. The bond strengths of the
interhalogens are clearly related to the difference in electronegativity between the
component halogen atoms, as expected on the basis of Pauling's ideas on ionic
character (Chapter 5). Furthermore, the tendency to form the higher fluorides and
chlorides depends upon the initial electronegativity of the central atom . 22 Only iodine
forms a heptafluoride or a trichloride. Not shown in Table 17.2 (except indirectly by
computation from the values) is the instability of certain lower oxidation stales to
disproportionation:
5IF - :I 2 + if j (17.39)
Bond energies: 1390 298 + 1340
Gain in bond energy = 250 kJ mol -1
This tendency towards disproportionation is common among the lower fluorides of
iodine and bromine. The behavior of the four iodine fluorides presents a good picture
of Ihe factors important in the relative stabilities. Both IF and IF, tend to dispropor¬
tionate (the former to the extent that it cannot be isolated), not because of weakness in
XY
xy 3
xy 5
xy 7
X
1.38
IF (277.8)
IF 3 (-272)
IF, (207.8)
IF 7 (231.0)
a>
1.28
BrF (249.4)
BrF, (201.2)
BrF, (187.0)
0.95
CIF (248.9)
CIF, (172.4)
CIF, (-142)
s
0.43
IC1 (207.9)
(1CI 3 ),
0 )
0.33
BrCI (215.9)
0.10
IBr (175.3)
0
F, (154.8)
0
CU (239.7)
0
Br 2 (190.2)
0
I 2 (148.9)
" Based on p orbital electroncgalivities. Table 5,6.
b Values in parentheses are bond energies from Appendix E (kJ moF 1 ).
24 Since Ihe most stable isotope of astatine has a half-life of only 8.3 hours, ihe chemistry of this
halogen has not been studied extensively. In the following discussion generalities made about the
halogens may or may not include astatine. In the present instance AtBr and AtCI have been
prepared. Sec the discussion of astatine chemistry later in this chapter.
25 S,cr,c faclors may also be important since stability also parallels the size of the central atom
compared to that of the surrounding atoms.
A p p e n d i x
Tanabe-Sugano Diagrams
[Originally from Tanabe, Y.; Sugano, S. J. Phys. Soc. [Japan) 1954, 753, 766, these
figures are from Figgis, B. N. Introduction to Ligand Fields ; Wiley: New York. 1966.
Reproduced by permission of John Wiley and Sons, Inc. An extensive set of diagrams
of this sort may be found in Konig, E.; Kremer, S. Ligand Field Diagrams', Plenum:
New York, 1977.]
DqtB
cl\C/B = 4,42
A-38
838
1 7 • The Chemistry of the Halogens and the Noble Gases
bonding (IF has the strongest bond of any of the interhalogens!), but because of the
greater number of bonds in the pentafluoride to which they disproportionate. At the
other extreme, the heptafluoride, while stable, is a reactive species (it is a stronger
fluorinating agent than 1F 5 ) because of the weaker bond energy (resulting from both
steric factors and resistance to the extremely high oxidation state on the part of the
iodine). Bromine fluoride likewise disproportionates, but BrF, and BrF 5 are stable.
Chlorine forms a monofluoride, trifiuoride, and pentafluoride.
The competitive forces tending to stabilize high or low oxidation states can be
readily rationalized. The simplistic statement: "The tendency to stabilize high oxida¬
tion states in compounds XY„ is favored by high electronegativities of Y (usually
lluorine) and low electronegativities of X (the heavier halogens)." is definitely not
wrong; but does it explain for you the relative instability of IF (with the strongest
interhalogen bond, see above )? 26 Consider the following reaction to oxidize a halogen
monohalide to a trihalide:
Y
X— Y + Y— Y - Y — X — Y (17.40)
I bond + I bond 2 X (3-c - 4-e bonds) or
3 VBT bonds minus promoiion energy
It is obvious that no great change in bond order occurs in this process and so any
enthalpic driving force must result from the quality of the bonds. A simple analysis of
ionic resonance energy (£ IRE ), in terms of partial charges, rationalizes the relative
stability of the monohalide and the trihalide. Assume that the ionic resonance energy
that increases the quality of the bonds may be equated to simple Madelung or
coulombic energy:
^IRE.XY = SxSy Eire.yy = 0 £|RE.XY, = 3(5 x <5 y (17.41)
Case I. 5 y is approximately the same in XY and XY, and 8 X = 35 Y . This is a good
approximation for Y = F, an element with high values of both a and h. i.e.. a high
inherent electronegativity but low charge capacity; it becomes saturated with negative
charge easily. It is also a good approximation fora large, soft iodine. X = 1. which can
increase its positive charge to accommodate three lluorine atoms. In this case, clearly.
3(5 X «5 V »S x 8 y , and the trihalide is favored.
Case II. 8 X is similar in XY and XY 3 , and since 8 X = 35 Ylnhii| . dc . S Ynuin . . -
3S Ylnh i|M|e . This becomes an increasingly good approximation as Y becomes larger with
lower values of both a and b, and the large, soft, central atom becomes smaller and
harder. In this case the monohalide is favored.
Exactly the same result is obtained if the initial electronegativity of the central
halogen is assumed to be higher in a higher oxidation state, and Ax and ionic
resonance energy are lower. The same arguments apply equally well to all of the
oxidation states:
X—X + 7Y 2 ?£ 2X—Y + 6 Y-, ^ 2XY 3 + 4Y 2 - 2XY 5 + 2Y : = 2XY 7 (17.42)
26 Sec the discussion of the instability of CaX in Chapter 4.
Halogens in Positive Oxidation States
839
Note that the more ionic the bond, the less important the apparent distinction
between 3-c-4-e bonding and VBT becomes, because the weaknesses of both methods
of modeling the bonding decrease when one goes towards the limit of a "purely ionic
bond.”
The interhalogen compounds obey the expectations based on the VSEPR theory,
and typical structures are given in Chapter 6 . One compound not included there is the
dimeric iodine trichloride, in which the iodine atom of the monomeric species appears
to act as a Lewis acid and accept an additional pair of electrons from a chlorine atom
(Fig. 17.5).
The molecules Br 2 , I 2 , and ICI show an interesting effect in the solid. Although
discrete diatomic molecules are still distinguishable, there appears to be some inter-
molecular bonding, For example, the molecules pack in layers with the intermolecular
distance within a layer 20-80 pm smaller than the distance between layers. Within
layers “molecules" approach each other much more closely than would be indicated
by addition of van der Waals radii but less than that of normal covalent radii (Fig.
17.6). At the same time there is a slight lengthening of the bond between the two atoms
forming the nominally diatomic molecule. It would appear that in the solid some
delocalization of electrons takes place, making a simple, single-bonded molecular
structure no longer completely appropriate.
Polyhalide Ions It has long been known that although iodine has a rather low solubility in water (0.3 g
kg -1 at 20 °C), it is readily soluble in aqueous solutions of potassium iodide. The
Fig. 17.5 Molecular
structure of LCI,,. Distances
in pm. |From Boswijk,
K. H.; Wicbenga. E. H.
Acta Crystallai’r. 1954. 7.
417. Reproduced with
permission.]
Fig. 17.6 Crystal structure
of iodine. The molecules lie
in parallel layers, one of
which is pictured here.
Distances in pm.
A-36
F- An Overview of Standard Reduction Potentials of the Elements
Table F.l ( Continued)
Half-reaction
Reduction of water in
basic solution.
2H 2 0 + 2e~ -+H 2 + 20H~
Zn~ + + 2e —» Zn
Se + 2e~ -» Se 2 ~
S 8 + I6e~ - 8S~
0 2 + e~ — OJ
Fe 2> + 2e' -> Fe
2H,0 + 2e~->H 2 + 20H" } _ _
2H 3 0 + + 2c“ -> H 2 + 2H 2 0 J P
Co 2+ + 2e“ -• Co
Ni 2+ + 2e~ -> Ni
Mn,0 3 + 3H 2 0 + 2e~ — 2Mn (0H) 2 + 20H~
Sn+ + 2e~ — Sn
0 2 + HjO + 2e~ HOj'+ 0H~
0 2 + e" + H 3 0 + -» H0 2 + H 2 0
Mn0 2 + 2H 2 0 + 2e~ -» Mn(OH) 2 + 20H
2H 3 0 + + 2c - —► H 2 + 2H 2 0
Sn' l+ + 2e~ —» Sn 2+
Sn(OH 3 ) + + 3H 3 0 + + 2c - — Sn 2+ + 3H 2 0
S 8 + I6H 3 0 + + 16e" -» 8H 2 S
2Mn0 2 + H } 0 + 2e~ -» Mn 2 0 } + 20H
Mn0 4 ~ + e~— Mn0 4 ~
Cu 2+ + 2e _ - Cu
Hg 2+ + 2e _ — 2Hg
Firsl transition series except Cu ( + 2 -> 0).
See pp 580-587, 594-599.
Reduction of neutral water
Reduction of water (H 3 0 + )
in acidic solution.
+027
Reduction of (oxidation
by) 0 2 in basic solution;
the reverse is the oxida¬
tion (destruction) of
water as solvent.
40 ir
-r u.ow
+ 026
+ 0.60
. +0.588
+ 0.620
+ 0.695
+ 0.771
+ 0.773
+ 0.799
Heavier transition anil post-transition metals
(ts-0). Sec pp587-588. 594-599.
Reduction of (oxidation
by) O z in neutral water;
the reverse is the
oxidation (destruction)
of water as solvent.
0 2 + 4e~ + 4H 3 0 + —» 4H 2 0
O, + 2H.O + 4e~ -+ 40H -
F- An Overview of Standard Reduction Potential of the Elements
Table F.l ( Continued )
Half-reaction
HMnO]
Halogens (0 - -1). see pp 837-853,
Reduction of (oxidation
by) 0 2 in acid solution;
the reverse is the oxi¬
dation (destruction) of
water as solvent.
Mn0 2 + 4 H,0 + + 2e _ —» Mil 21 ' + 2H,0
Oj + H 2 0 + 2e~ ->0 2 + 20H~
Cr 2 0 2 ~ + I4H 3 0 + 6e~ —♦ 2Cr 3+ + 7H 2
CU + 2e _ — 2CI -
Au 3+ + 2e _ —> Au +
MnOJ + 8H 3 0 + + 5e~ -» Mn 2+ + I2H,C
Mn 3+ + e _ — Mn 2+
Au + + e~ -» Au
Mn0 4 ~ + 4H 3 0 + + 3e _ -» MnO, + 2H 2 0
H,0 2 + 2H 3 0 + + 2e“ -+ 4H„0 '
Ag 2 + e~ — Ag +
Rn 2+ + 2e” -» Rn
0 3 + 2H 3 0 + + 2e - —» 0 2 + 3H 2 0
Xe0 3 + 6HjO + + 6e“ Xc + 6H 2 0
HMnO; + 3H + + 2e~ -♦ MnO, + 2H,0
H 4 Xe0 6 + 2H + + 2e" - XeO,‘ + 3H,0
Kr0 3 + 6H 3 0 + + 6e~ -» Kr + 6H 2 0
F 2 + 2e~ — 2F~
F 2 + H 3 0 + 2e - —» 2HF + 2H 2 0
20F 2 + 2e~ -0 } + 2F~
20F 2 + H 3 0 + +2e~ - 0 2 + 2HF
Strong, oxygen-conlaining oxidants.
Noble gases, sec pp 825-836.
Strongest oxidizing agent, OF ;
840
17-The Chemistry of the Halogens and the Noble Gases
Halogens in Positive Oxidation States
841
molecular iodine behaves as a Lewis acid towards the iodide ion (as it does to other
Lewis bases; see Chapter 9):
12 + 1 "
(17.43)
Similar reactions occur with other halogens, and every possible combination of
bromine, chlorine, and iodine exists under appropriate conditions in aqueous solution:
Br 2 + I - IBrJ (17.44)
The participation of fluorine is less common, but several fluoride-containing trihalide
ions have been isolated as crystalline salts (Table 17.3). The triiodide ion presents
exactly the same problem to classical bonding theory as does xenon difluoride, and
although the triiodide ion was discovered in 1819, only eight years after the discovery
of iodine itself, chemists managed to live with this problem for almost a century and a
half without coming to grips with it. The explanation offered most often was that the
interaction was electrostatic—an ion-induced dipole interaction. The existence ot
symmetrical triiodide ions as well as unsymmetrical triiodide ions makes this inter¬
pretation suspect, and the existence of ions such as BrF 4 and IF fi makes it untenable.
Two points of view are applicable to these species, as they also are to the
isoelectronic noble gas fluorides: (1) a valence bond approach with promotion of
electrons to d orbitals; and (2) three-center, four-electron bonds. The same arguments,
pro and con. apply as given previously, so they will not be repeated here. Independent
of the alternative approaches via VB or MO theory, all are agreed that Madelung
energy (“ionic character") is very important in stabilizing both the polyhalide ions and
the polyhalogens. 27
The polyhalide ions may conveniently be classified into two groups (X 3 -type ions
belong to both groups): (I) those that are isoelectronic with noble gas compounds and
Tabic 17.3
Polyhalogen anions in
crystalline salts
X 3
xj-
X?
Xf
'
If
It"
I,Br~
B G"
i„cr
l«Br”
i 2 cr
l 4 Br"
if;
IBrj
I,Br 3 -
Br 6 CI -
ICI 2
I,Br,CI
BrF 6 -
IF 2
I 2 BrCI 2
CIF 6 -
IBrCP
w
IBrF"
lBrCl 3 -
BrJ
ici 4 -
BrjCT
IC1 3 F"
BrClj"
if;
BrF 2 “
BrF 4 "
cij
cif;
CIFJ
27 For calculations, see Wiebcnga, E. H.; Kracht, D. Inorg. Chem. 1969. 8, 738.
have general formulas XY„ . where r — an even number; (II) polyhalide ions, mostly
polyiodide ions, IJ. where .r can have various values, usually odd.
Group I polyhalide ions generally consist of a central atom surrounded by two,
four, six, or eight more electronegative atoms, with linear, square planar, distorted
octahedral, and square antiprismatic structures, respectively. These structures obey
the VSEPR rules and are closely related to polyhalogen structures, differing by a lone
pair in place of a bonding pair. They obey the rules discussed above for polyhalogen
and noble gas compounds. The Group II polyhalide ions present some unusual
bonding situations, and so it is simplest to start with I 2 and the two limiting structures
of the I, ion. The bond length in the iodine molecule is 267 pm. We can imagine that
upon the approach of an I" anion, the charge cloud of the I-, molecule will distort with
a resulting induced dipole. The greater the distortion of the iodine charge cloud, the
weaker the original I —I covalent bond is apt to become, 2 and we might expect that
bond to lengthen somewhat. This is the situation in ammonium triiodide: The I —I
bond lengths are 282 pm and 310 pm. The "intramolecular” bond (original diiodine
bond) has lengthened by 15 pm, while the long "intermolecular" bond is 40 pm longer
than a “normal" covalent bond and only about 80-100 pm shorter than a simple van
der Waals contact. Further approach by the iodide ion eventually will result in a
symmetrical triiodide system. Such is found in tetraphenylarsonium triiodide.
Ph 4 As + I^, in which the two bond lengths are equal at 290 pm (Fig. 17.7). 29 The
question as to why both symmetrical and unsymmetrical triiodide ions are found in
crystal structures has not been completely resolved. In many ways it parallels the
problem of symmetrical and unsymmetrical hydrogen bonding systems. |F—H —F|~
O'
(a) (b) (cl id)
Fig. 17.7 Structural changes as an iodine molecule, L (a), is approached by an iodide ion
(b) and changes to an unsymmetrical (c) or symmetrical (d) triiodide ion. Distances in pm.
2,1 Since the unperturbed I—1 molecule represents the minimum in the energy curve for those two
atoms, any change in the I — I bond length in the molecule must result in a decrease in the original
I —I portion of the bond energy. In other words, if a change in bond length in I —I could strengthen
that bond, it would have already occurred without the influence of the iodide ion.
29 See Popov, A. I. Halogen Chem. 1967, /, 225; Wiebenga. E. H.; Havinga, E. E.; Boswijk. K. H.
Adv. Inorg. Chem. Radiochem. 1961. 3, 133-169, for the original references and further discussion.
Slrongcst reducing ngcnl, Nf
A-34
E'Bond Energies and Bond Lengths
boldface, and Brewer and coworkers (italics). As in the case of the energies the
purpose of the table is for quick reference and the bond length is given as a typical
value that should be accurate for most purposes ±2 pm (0.02 A).
For special purposes and precise computations, the original sources should be
consulted for the value, nature, and source of the bond energies, and for the accuracy,
experimental method, and variability of the bond lengths. Space requirements prohibit
extensive tabulation of information of this type here.
An Overview of Standard Reduction
Potentials of the Elements
Th e following table gives a quick overview and perspective of electrochemical poten¬
tials 1 of the elements. For each group of elements, brackets indicate the limiting
potentials for half-reactions [M" + + ne - — M (metals); X 2 + nt~ - 2X'' ,ri) ~
(halogens)) of the elements [e.g., Li + /Li and Na + /Na for Group IA (1)J for that group.
Other selected half-reactions are also listed. Bold-face type in the table indicates
important limiting potentials for aqueous solutions (see Chapter 10) and italic type
indicates half-reactions in I M OH - . Many elements and oxidation stales are omitted
here for simplicity's sake. More extensive data may be found in the discussions of the
descriptive chemistry of the elements in the text on the pages cited. Potentials in
parentheses are estimated values.
Half-reaction (V)
3N 2 + 2e~ -> 2Nj -3.608
3N + 2HjO+ + 2e - —»2HNj + 2H,0 -3.334
Li + + c - —► Li -3.040
K + + e - —> K -2.936
Ba 2+ + 2e - —» Ba -2.906
Sr 2+ + 2c - — Sr -2.899
Ca 2+ + 2c - — Ca -2.868
Na + + e - -»Na -2.714
La 3+ + 3e - — La -2.379 j
Mg 2+ .+ 2e - —■ Mg -2.360 I
Lu 3+ + 3e - —> Lu -2.28
Al 3+ + 3e“ — Al -1.677
U 3 : +3e--U -1.642 |
Table F.l
Alkali mclali, Group IA (I) (+ I -» 0).
pp 593-594.
Alkaline earth metals. Group IIA (2) (+ 2 — 0),
p 594.
Most lanthanides and actinides ( + 3 - ,+0).
Eu and Yb are exceptional. See pp 599-613.
A-35
All potentials are standard redaction potentials with all species at unit activity and fugacily except
for those values labeled "pH = 7" which have all species at unit activity except (HjO) = [OH - ] =
1.005 x I0 -7 .
842
17-The Chemistry of the Halogens and the Noble Gases
versus [F—H—FT (Chapter 8). We might equate the symmetrical arrangement with
optimal, very strong bonding. Indeed, in hydrogen bonding systems only the very
strongest, F —H —F and O—H —O (and not all of them), exhibit the symmetrical
structure. The unsymmetrical structure could be attributed to the polarization efiects
from cations that may produce an imbalance in the triiodide system and make the
unsymmetrical form more stable. ..
The higher polyiodides provide a more complicated picture than the triiodide. The
pentaiodide ion, I 5 ", is an L-shaped molecule which may be considered to be two
iodine molecules coordinated to a single iodide ion. Alternatively, it can be considered
to be two unsymmetrical triiodide ions sharing a common iodine atom. There are two
bond lengths. 1.2 and 4.5 (282 pm) and 2.3 and 3,4 (317 pm), which correspond very
well to the bond lengths in the unsymmetrical triiodide (Fig. 17.8a). 2 _
The “tetraiodide ion” as found in Csl 4 has been found to be dimeric. 15 • It
consists of another leg added to the I, ion to form a Z arrangement. It can be
considered to be a central I 2 molecule with an asymmetrical triiod.de coordinated to
each end. The "short" bonds of the triiodide groups (bonds 1.2 and 7.8) are 285 pm
and the "long" bonds (2,3 and 6.7) are 300 pm. in fair agreement with the preceding
values. The bonds joining these triiodide moieties to the central duodine - molecule
(3.4 and 5.6) are longer than any encountered thus far: 342 pm (Fig. 17.8b).
The so-called heptaiodide ion is found in (Et 4 N)I 7 . No discrete I 7 ions are
present. The structure is an infinite three-dimensional framework of "diiodine mole¬
cules” (274 pm) and symmetrical "triiodide ions" (290 pm) coordinated through very
long" bonds (344 pm). A discrete If ion. also consisting of one I, and two 1, units, has
recently been characterized in (Ph 4 P)l 7 (Fig. 17.8c).- 10
Finally, the enneaiodide ion. I,;, is a still more complex structure, which has been
characterized as If + I, or If + 21,, but is probably best considered as a three-
/
Fig. 17.8 Structures of polyiodide ions: (a) The pentaiodide ion. 17; (b) the octaiodide ion.
Ij-; (c) the heptaiodide ion as found in Ph 4 PI 7 . Solid lines represent essentially normal
covalenl bonds, dashed lines represent weakened or partial bonds, and dolled lines represent
very long and weak interactions. Distances in pm.
30 Pol,, R. ; Gordon. J. C.; Khanna, R. K.; Fanwick. P. E. Inorg. Chem. 1992. 31. 3165-3167.
Halogens in Positive Oxidation States
843
dimensional structure similar to the so-called heptaiodide but of more irregular struc¬
ture with bond lengths of 267, 290. 318, 324, 343, and 349 pm. Unless the latter is
arbitrarily considered too long to be a true bond, the system must be considered to be
an infinite polymer. A portion of the structure is shown in Fig. 17.9.
Fluorine-Oxygen There is no evidence that fluorine ever exists in a positive oxidation state. This is
Chemistry reasonable in view of the fact that there is no element that is more electronegative and
capable of taking electron density away from it. 11 Certainly the oxygen compounds of
fluorine come the closesl to achieving a positive charge on fluorine, and since their
chemistry is in some ways comparable to that of the other oxyhalogen compounds it is
convenient to include these compounds here.
Few oxygen fluorides are known. The most stable of these is oxygen difluoride. It
is usually prepared by the passage of fluorine through aqueous alkali:
2F, + 20H- OF, + H,0 + 2F" (17.45)
Thermodynamically OF, is a slightly stronger oxidizing agent than mixtures of oxygen
and fluorine. Thus, it occupies the extreme position in the standard electrode potential
table (Appendix F). though this potential may not always be realized as alternative
reductive pathways are available. It is relatively unreactive unless activated by an
electric discharge or similar high-energy source. In contrast to most molecules con¬
taining fluorine, it has a very small dipole moment and thus has one of the lowest
boiling points- 1 - of any inorganic compound. 49 K. Other oxygen fluorides have been
suggested of which only 0,F, has been well characterized. It is an orange-yellow solid
Fig. 17.9 A portion of
the structure of the
■enneaiodide" ion found in
Me 4 NI». The shortest
distance between (his
moiety and the next is only
349 pm.
31 Of course since electronegativity is a function of t'u/emevnm'("hybridization" and "charge"), it is
nol impossible that fluorine might lind itself in a compound with an element in a particular valence
stale lhat was more electronegative than fluorine. And. of course, since the distinction between loss
of electrons and gain in electrons (which may be two different valence stales in the admittedly highly
electronegative noble gasesl is not always made, there is constant skirmishing in the literature over
this.
The dipole moment of OF, is only 0.991 x It) C m (0.297 D). This results from the convergence
of the electronegativity of the hybridized oxygen and unhybridized fluorine and (he elTcct of lone
pair moments. Similar molecules arc NFj. ix = 0.7X3 x It)' 30 C m (0.235 D). bp = 153 K;
<7.v-i-N = NF. tx = 0.53 x It) C m (0,16 D). bp = 167.5 K. Dipole moment values are from
Nelson. R. D.. Jr.; Lide, D. R.. Jr.: Maryott. A. A. Selected Values of Electric Dipole Moments for
Molecules in the Gas Phase: NSRDS-NBS 10. Washington. DC. 1967.
A-32 E
Table E.l
[Continued)
• Bond Energies and Bond Lengths
D 0
r
Bond
kJ mol 1
Iccal mol -1
pm
A
Bi=Bi
Bi—F(BiF 5 )
Bi—F(BiFj)
Bi—Cl(BiCl 3 )
192 ± 4
-297
-393
274.5
46 ± 1
-7/
—94
65.6
248
2.48
Bi—Br(BiBr 3 )
232.2
55.5
263
2.63
Bi—l(BiI,)
Group VIA (16)
0—0(H,0 2 )
HO—OH
168.2
-142
207.1 ± 2.1
40.2
-34
49.5 ± 0.5
148
1.48
0=0
493.59 ± 0.4
117.97 ± 0.1
120.7
1.207
0—F
189.5
45.3
142
1.42
S—S(S 8 )
226
54
205
2.05
S-S(H 2 S 2 )
268 ± 21
64 ± 5
205
2.05
s=s
424.7 ± 6.3
101.5 ± 1.5
188.7
1.887
S—F
284
68
156
1.56
S—CI(S,CI 2 )
255
61
207
2.07
S— Br(S 2 Br,)
217?
52?
227
2.27
S=S(S0)
517.1 ± 8
123.6 ± 2
149.3
1.493
StM)(S0,)
532.2 ± 8
127.2 ± 2
143.2
1.432
Sttt0(S0 3 )
468.8 ± 8
112.1 ± 2
143
1.43
Se—Se(Se 6 )
Se=Se
172
272
41
65
215.2
2.152
Se—F(SeF fi )
Se—F(ScF 4 )
284.9
-310
68.1
-74
Se—F(SeF 2 )
~35l
-84
Se—CI(SeCl,)
-243
-58
Se — Br(SeBr 4 )
-151
-36
Se — Br(ScBr,)
201
-48
Se—I(Sel,)
Te—Te
Te=Tc
Te—FCTeFft)
Te — F(TeF 4 )
Te—F(TeF 2 )
Te—CI(T cCI 4 )
-151
(126)«
218 ± 8
329.7
- 335
- 393
310.9
-36
(30)-
52 ± 2
78.8
-80
-94
74.3
233
2.33
Te—CI(TeCI 2 )
Te—Br(TeBr 4 )
284
~. 176
-68
-42
268
2.68
Te—Br(TeBr 2 )
-243
-58
251
2.51
Te—l(TeI 4 )
~I2I
-29
Te—I(Tel 2 )
~I92
-46
Group V1IA (17)
1—I
148.95 ± 0.04
35.60 ± 0.01
266.6
2.666
I — F(IF 7 )
23I.Q'-
55.2
-183
-1.83
I—F(IF 3 )
267.8'-
64.0'-
175,186°
1.75,1.86°
I — F(IF 3 )
-272'-
-65
-jiUj
. ••
I — F(IF)
277.8 ± 4
66.4 ± 1
191°
1.91°
I—CI(ICI)
207.9 ± 0.4
49.7 ± 0.1
232.1
2.321
1—Br(IBr)
175.3 ± 0.4
41.9 ± 0.1
.
I—0(1—OH)
201
48 : i: >•- -
MS!
E- Bond Energies and Bond Lengths A-33
Bond
kJ mol 1
kcal mol 1
pm
A
Br—Br
190.16 ± 0.04
45.45 ± 0.01
228.4
2.284
Br—F(BrF 5 )
187.0
44.7
Br— F(BrFj)
201.2
48.1
172,184-'
1.72,1.84
Br—F(BrF)
249.4
59.6
175.6
1.756
Br—CI(BrCl)
215.9 ± 0.4
51.6 ± 0.1
213.8
2.138
Br—0(Br—OH)
201
48
Cl—Cl
239.7 ± 0.4
57.3 ± 0.1
198.8
1.988
Cl—F(ClFj)
-142
-34
Cl—F(ClFj)
172.4
41.2
169.8
1.698
Cl—F(CIF)
248.9 ± 2.1
59.5 ± 0.5
162.8
1.628
Cl—0(CI—OH)
218
52
F—F
154.8 ± 4
37.0 ± 1.0
141.8
1.418
At—At
115.9
27.7
Group VlUA (18)
Xe—F(XeF 6 )
126.2°
30.2°
190'
1.90°
Xe—F(XeF 4 )
130.4°
31.2°
195°
1.95°
Xe—F(XeF 2 )
130.8°
31.3°
200°
2.00°
Xe—O(XeOj)
84°
20°
175°
1.75°
Kr—F(KrF 2 )
50°
12°
190°
1.9°
" Huhccy, J. E.; Evans, R. S. J. lnorg. Nncl. Client. 1970, 32, 383.
Slein, L. Halogen Chemistry 1967, /, 133.
° Harshbcrgcr W.; ct al. J. Am. Cliem. Soc. 1967, 89, 6466.
Burbank, R. D.; Beancy, F. N., Jr. J. Client. Phys. 1957, 27, 982.
° Holloway, J. H. Noble-Gas Chemistry: Methuen: London, 1968.
Values in boldface type are from Darwent and represent his estimates of the "best
value" and uncertainties for the energies required to break the bonds at 0 K. Where
values are not available from Darwent. they arc taken from Brewer and coworkers for
metal halides and dihalides (boldface italics) or from Feber for transition metal,
lanthanide, and actinide halides (italics). These values represent enthalpies of atomiza¬
tion at 298 K. The remaining values are from Cottrell (Arabic numerals) and other
sources (Arabic numerals with superscripts keyed to references at end of table).
The table is intended for quick reference for reasonably accurate values for rough
calculations. No effort has been made to convert the values from 298 K to 0 K, and in
many cases the errors in the estimates are greater than the correction term anyway.
The accuracy of the values can be graded in a descending scale: (1) those giving ± un¬
certainties; (2) those giving "exact” values to the nearest 0.1 or 1; (3) those expressed
as “about" a certain value (~); and (4) those which are almost pure guesses, followed
by a question mark. All values are experimental except for a few for which A—A
bond energies are not known but would be helpful (as for electronegativity calcula¬
tions). Estimates for these hypothetical bonds (such as Be—Be) arc listed in paren¬
theses.
The bond lengths are mainly from two sources: Tables of Interatomic Distances
and Configuration in Molecules and Ions (The Chemical Society, Special Publication
No. II, 1958) and Supplement (Special Publication No. 18. 1965) with values given in
844
17‘The Chemistry of the Halogens and the Noble Gases
that decomposes slowly even at low temperatures and is a powerful oxidizing agent.
Little is known of 0.,F 2 which supposedly forms as a red-brown solid at 77 K but
decomposes on warming.
Several workers have investigated the possibility of synthesizing fluorine oxygen
acids such as HOF. HOFO. HOFO,, or H0F0 3 . Despite the various claims to have
formed species such as these, it is the present opinion that only one, HOF, has been
synthesized, and it is also the only hypohalous acid that has been prepared pure. 33 It
may be prepared by passing dilluorine over ice (Fig. 17.10):
F, + H 2 0 HOF + HF (17.46)
It is difficult to prepare and isolate because of its reactivity towards both water and
dilluorine (Fig. 17.11):
HOF + H 2 0 -> HF + H 2 0 2 (17.47)
HOF + F 2 -► HF + OF z (17.48)
Other compounds containing the —OF group are known: CF 3 OF, 0 2 N0F.
FjSOF, OjCIOF, CH 3 C(0)0F, and CH 3 OF. These are all thermodynamically unsta¬
ble compounds with strong oxidizing properties, though some such as CF 3 OF and
FjSOF are sufficiently inert kinetically that they can be stored in cylinders.
Fig. 17.10 Apparatus for carrying out the reaction of fluorine with cold ice. All parts arc
made from Kel-F, Teflon, Viton plastics, or Monel metal. Reaction tube B is packed with
Raschig rings cut from Teflon spaghetti tubing, and wet with 1-2 mL water that is frozen
with dry ice. Traps C and E are filled with dry Tetlon Raschig rings. Tube A is held at
- 196 °C with liquid nitrogen, traps B and C at -40 and —50 °C. respectively, with chilled
ethanol, trap D at -78 °C with dry ice. and traps E and F at - 183 °C with liquid oxygen.
Gases are re-circulated with pump G. The pressure is measured with gauge I, and vacuum
pumps and a fluorine cylinder are attached through valve J. The HOF collects in Tube E; the
OF. in Tube A. [From Appelman. E. H.; Jachc. A. W. J. Am. Chem. Soc. 1987. 109.
1754-1757. Reproduced With permission.!
A B C D E F G
33 Appelman. E. H.; Jache. A. W. J. Am. Chcm. Soc. 1987. 109. 1754-1757. Poll. W.; Pawelke. G.;
Mootz, D.; Appelman, E. H. Angew. diem. Ini. Ed. Engl. 1988, 27. 392-393.
Halogens in Positive Oxidation States
845
Fig. 17.11 Effect of
reaction temperature on
yields of various products
from the reaction of F 2 with
ice. [From Appelman,
E. H.; Jache, A. W. J. Am.
Chem. Soc. 1987. 109 ,
1754-1757. Reproduced
with permission.]
Oxyacids of the
Heavier Halogens
The series of acids HOCI. HOCIO. HOCIO,. and HOCIO, (or HCIO. HCIO,, HCI0 3 ,
HCI0 4 ) is well known, arising from the disproportionation of chlorine and related
reactions:
hoci + h 3 o + cr
- hoci + co 2 + ci-
-—♦ cio; + 5cr + 3H,o
• 3KCIO, + KCI
4KCIO
careful healing.
400-501) "C
Chlorous acid and chlorite salts cannot be formed in this way but must be formed
indirectly from chlorine dioxide which in turn is formed from chlorates:
moist
KCI ° 3 K 2 C 2 0 4 + C0 2 + CIO, (17.53)
In basic solution chlorine dioxide disproportionates with the formation of chlorate and
chlorite, and the latter is used to form the free acid:
cio 2 + cio; + h 2 o
—» 2HOCIO + BaSO,
2CI0 2 + 20H“ —
Ba(C10 2 ) 2 + H 2 SO.
The heavier halogens form similar series of compounds although less complete. In all
probability neither HOBrO or HOIO exists. The periodate ion exhibits a higher
A-30
E- Bond Energies and Bond Length:
Table E.l
(Continued)
D 0
r
Bond
IcJ mol 1
kcal mol 1
pm
A
In—Br(lnBr)
In—I(Inlj)
In— I(Inl)
Tl—F(T1F)
406 ± 21
225.1
439 ± 21
97 ± 5
53.8
105 ± 5
254.3
275.4
2.543
2.754
TI—CI(TICl)
Tl—Br(TIBr)
364 ± 8
326 ± 21
87 ± 2
78 ± 5
248.5
2.485
TI—I(TU)
280 ± 21
67 ± 5
281.4
2.814
Group IVA (14)
C—C
345.6
82.6
154
1.54
C=C(C 2 )
602 ± 21
144 ± 5
134
1.34
c=c '
835.1
199.6
120
1.20
C—F
485
116
135
1.35
C—Cl
327.2
78.2
177
1.77
C—Br
285
68
194
1.94
C—I
213
51
214
2.14
C—0
357.7
85.5
143
1.43
c=o
798.9 ± 0.4
190.9 ± 0.1
120
1.20
c=o
1071.9 ± 0.4
256.2 ± 0.1
112.8
1.128
c—s
272
65
182
1.82
c=s
573 ± 21
137 ± 5
160
1.60
C—N
C=N
304.6
615
72.8
147
147
1.47
CsN
887
212
116
1.16
C—P
264
63
C—Si(SiC)
318
76
185
1.85
C—Ge(GcEt„)
213?
51? -
194
1.94
C—Sn(SnEt 4 )
226
54
216
2.16
C—Pb(PbEt 4 )
130
31
230
2.30
Si—Si
222
53
235.2
2.352
Si—F
565
135
157
1.57
Si—Cl
381
91
202
2.02
Si—Br
310
74
216 K
2.16
Si—I
234
56
244
2.44
Si—0
452
108
166
1.66
Si—S
293? .
70?
-200
-2.0
Ge — Ge
Ge=Ge
188
272 ± 21
45
65 ± 5
241
2.41
Ge—F(GeF 4 )
-452
-108
168
1.68
Ge— F(GeF 2 )
481
115
•; -
Ge—Cl(GeCI 4 )
348.9
83.4
210
2.10
Ge—CI(G cC1 2 )
-385
-92
Ge—Br(GeBr 4 )
Ge—Br(GeBr 2 )
276.1
325.5
66.0
77.8
230
2.30
Ge—I(GeI 4 )
211.7
50.6
Ge—I(GeI 2 )
264.0 ■. '
63.1
.. }.
Sn—Sn
146.4
35.0
Sn—F(SnF 4 )
-414
-99
Sn—F(SnF 2 )
-48!
-115
I’t- ■ 1
Sn—CI(SnCI 4 )
323.0
77.2
233
2.33
Sn—Cl(SnCl 2 )
385.8
:: k
msi . ....
242 .
2.42
E-Bond Energies and Bond Length:
A-31
Table E.l
Do
r
Bond
kJ mol 1
kcal mol -1
pm
A
Sn—Br(SnBr 4 )
272.8
65.2
246
2.46
Sn—Br(SnBr,)
329.3
78.7
255
2.55
Sn— KSng
-205
-49
269
2.69
Sn—I(SnI 2 )
261.5
62.5
273
2.73
Pb—F(PbF 4 )
-331
-79
Pb—F(PbF,)
394.1
94.2
Pb—CI(PbCl 4 )
-243
-58
Pb—Cl(PbCU)
303.8
72.6
242
2.42
Pb—Br(PbBr 4 )
-201
-48
Pb—Br(PbBr 2 )
260.2
62.2
Pb— I(PbI 4 )
-142
-34
Pb—I(PbI 2 )
205.0
49.0
279
2.79
Group VA (15)
N—N(N,H 4 )
-167
-40
HjN—NH 2
247 ± 13
59 ± 3
145
1.45
N=N
' 418
100
125
1.25
N=N
941.69 ± 0.04
225.07 ± .01
109.8
1.098
N—F
283 ± 24
68 ± 6
136
1.36
N—Cl
313
72
175
1.75
N—O
201
48
140
1.40
N=0
607
145
121
1.21
N = N
678
162
115
1.15
P—P(P 4 >
201
48
221
2.21
CUP—PCI,
239
57
PsP
D_FT/PE \
481 ± 8
115 ± 2
189.3
1.893
4'AJ
117
154
1.54
P—Cl(PCIj)
326
78
203
2.03
P—Br(PBr,)
264
63
P— I(PIj)
184
44
P—O
335?
80?
163
1.63
P=0
-544
-130
-150
-1.5
P=S
-335
-80
186
1.86
As—As(Asj)
146
35
243
2.43
As=As
380 ± 21
91 ± 5
As—F(A sF 3 )
-406
-97
As—F(A sF 3 )
484.1
115.7
171.2
1.712
As—Cl(AsClj)
321.7
76.9
216.1
2.161
As—Br(AsBr,)
458.2
61.7
233
2.33
As—1 (AsI 3 )
200.0
47.8
254
2.54
As—O
301
72
178
1.78
As=0
-389
-93
Sb—Sb(Sb 4 )
121?
29?
Sb=Sb
295.4 ± 6.3
70.6 ± 1.5
Sb—F(SbF 5 )
-402
-96
Sb—F(SbFj)
-440
-105
Sb—CI(SbCl 5 )
248.5
59.4
Sb—CI(SbCi 3 )
314.6
75.2
232
2.32
Sb—Br(SbBr 5 )
184
44
Sb—BrfSbBr 3 )
259.8
62.1
251
2.51
195.0
46.6 .
^5 ' - -• -3
846
17-The Chemistry of the Halogens and the Noble Gases
Halogen Oxides and
Oxyfluorides
Table 17.4 _
Halogen oxides and
oxyfluorides
coordination number-' (resulting from the increase in radius of iodine over chlorine) of
six. as well as four, 10;. For many years it proved to be impossible to
synthesize the perbromate ion or perbromic acid. The apparent nonexistence of
perbromate coincided with decreased stability of other elements of the first long
period in their maximum oxidation states. This reluctance to exhibit maximum valence
has been correlated with promotion energies and with stabilization through n bonding
(see Chapter 18). The first synthesis of perbromate resulted from the in situ production
of bromine by /3~ decay (compare to the synthesis of XeCI 4 above). It was soon found
that it could be readily synthesized by chemical means:
NaBrO, + XeF 2 + H 2 0 -» NaBr0 4 + 2HF + Xe] (17.56)
providing a good example of the use of noble gas compounds as oxidizing agents. They
are extraordinarily "clean,'’ providing a convenient source of fluorine with only the
“inert” xenon given off as a gas.
A more practical synthesis of perbromate is to use fluorine directly as the oxidiz¬
ing agent:
NaBrO, + F 2 + 2NaOH - NaBrO + 2NaF + H 2 0 (17.57)
Once formed, perbromate is reasonably stable. Although perbromate is a stronger
oxidizing agent than either perchlorate or periodate, the difference is not great:
cio; cio;
(17.58)
BrO; BrO,"
(17.59)
l0 ; 10 ,-
(17.60)
The crystal structure of potassium perbromate has been determined, and it was
found that the perbromate ion is tetrahedral as expected from the isoelectronic CI0 4 ,
10 and Xe0 4 species.
The heavier halogens form a large number of oxides and oxyfluorides (Table 17.4).
Most are rather strong oxidizing agents and some are extremely unstable. These will
Cl,0
Br,0
CIO,
Br0 2
CIA
Br 2 0 4
1A
Cl,0 4
BrA?
W
IA
CIO,F
B rO,F
IO,F
CIOF,
IOF,
CIA
ciA
cio,f
BrOjF
IO,F
CIO,F,
IO,F
I OF,
• u The chemistry of the orthoperiodate ion is more complicated than implied by the formula 10 k
Periodic acid often behaves as a dibasic acid forming salts of H,IO;‘ Furthermore, pyro-type salts
are known with the IO ft octahcdra sharing edges and faces. Sec Wells. A. F. Structural Inorganic
Chemistry, 5th ed.; Oxford University: London, 1984: pp 405-406.
Halogens in Positive Oxidation States
847
not be discussed here except to call attention to the use of chlorine dioxide above
(Eqs. 17.54 and 17.55) and to its use commercially as a bleaching agent. One exception
to the general reactivity of this class of compounds is perchloryl fluoride, CIO,F.
Although it is inherently a strong oxidizing agent, it behaves as such only at elevated
temperatures. It has a dipole moment of 0.077 x HT 30 C m (0.023 D), lower than any
other polar substance. Perbromyl and periodyl fluorides are also known and share the
lessened reactivity and low dipole moment but are somewhat less stable.
In addition to the polyhalide ions discussed previously, which were all anionic, there
are comparable cationic species known, 33 although they have been studied consider¬
ably less. Many pure interhalogen compounds are thought to undergo autoionization
(see Chapter 10) with the formation of appropriate cationic species:
2ICI 3 ^ ICI 2 + ICi; 07.61)
2IF 5 ^ if; + IF- (17.62)
In many cases these cationic species have been postulated on the basis of chemical
intuition coupled with the knowledge that the pure interhalogen compounds are
slightly conducting (see Problem 17.4). Homoatomic halogen cations can be prepared
in highly acid media. For example when iodine is dissolved in 60% oleum (S0 3 /
H 2 S0 4 ), it is oxidized (by the SO, present) to yield a deep blue species. Conduc¬
tometric, spectrophotometric. cryoscopic. and magnetic susceptibility measurements
are all compatible with l?\ Using stoichiometric amounts of oxidizing agents [such as
bis(fiuorosu!furyl)peroxide or arsenic pentafluoride] allow various polyiodine cations
to be prepared:
21, + FSO,OOSO,F 21, 4- 2FSO;
31, + FSO,OOSO,F 21 ; + 2FSO;
51, + FSO,OOSO,F 21, + 2FSO;
21, + 3AsF 5 I 3 + 2AsF; + AsF,
31, + 3AsF s 2i; + 2AsF; + AsF,
51, + 3AsF, -SgU 2i; + 2A.sF,V + AsF,
The use of liquid sulfur dioxide as solvent allows the crystallization of the hexfiuoroar-
senate salts of these cations including the very interesting tetraiodine dication. I 3+ .
The latter forms through dimerization of U as its bright blue solutions are chilled.
Dimerization is accompanied by a color change to red.
Although the presence of the monatomic cation, I + , has not been demonstrated in
the systems discussed above, there are conditions under which it is stabilized through
coordination and is well characterized. The dichloroiodale(I) anion. ICIJ. is a simple
example. Other complexes of I may be prepared, as through the disproportionation
of iodine in the presence of pyridine:
I 2 + 2py + Ag - I(py) 2 + Aglj (17.69)
33 See Gillespie. R. J.; Passmore. J. Adv. Inorg. Chcm. Radiodiem. 197S. 17. 49-87: Shamir, J. Struct.
Bonding IBerlin ) 1979. 37, 141-210.
(17.63)
(17.64)
(17.65)
(17.66)
(17.67)
(17.68)
A-28
E-Bond Energies ond Bond Length
Table E.1
( Continued )
Lanthanides and actinides (kcal mol ’)
Metal MF 3
MCI 3 MBr 3
mi 3 mf 4 mf 5 mf 4
Yb ~ 124
-92 -77
-60
Lu -145
-114 -98
-82
Th
153.2
Pa
U -148
118.4 101.4
-82 143.0
135.0 124.8
Np -140
-110 -95
-78 -134
-124 -114
Pit 134.5
105.7 90.5
-74 -128
-116 -102
Am -137
-108 -93
-76 -124
Do
r
Bond
Id mol - '
kcal mol -1 pm
A
Transition metals
Ti—F(TiF 4 )
584.5
139.7
Ti—CI(TiCI 4 )
429.3
102.6 218
2.18
Ti—CI(TiCl,)
460.2
110.0
Ti—Cl(TiCI 2 )
504.6
120.6
Ti—Br(TiBr 4 )
366.9
87.7 231
2.31
Ti—I(TiI 4 )
296.2
70.8
Zr—F
646.8
154.6
Zr— Cl
489.5
117.0 232
2.32
Zr—Br
423.8
101.3
Zr—I
345.6
82.6
Hf—F
-649.4
-155.2
Hf—Cl
494.9
I18J
Hf—Br
-431
-103
Hf—I
-360
' -86
Mn—F(MnFj)
457.7
109.4
Mn—CI(MnCl 2 )
392.5
98.8
Mn— Br(MnBr,)
332.2
79.4
Mn— I(Mnlj)
267.8
64.0
Fe—Fe
156 ± 25
37.5 ± 6
Fe—F(FeF,)
-456
-109
Fe—F(FeF 2 )
481
115
Fc—CKFeCI,)
341.4
81.6
Fe—Cl(FeCI 2 )
400.0
95.6
Fe— Br(FcBrj)
291.2
69.6
Fe—Br(FeBr 2 )
339.7
81.2
Fe— I(FeI 3 )
233.5
55.8
Fe—I(FeI 2 )
279.1
66.7
Ni—Ni
228.0 ± 2.1
54.5 ± 05
Ni-F(NiF 2 )
462.3
110.5
Ni—CI(NiCI 2 )
370.7
88.6
Ni—Br(NiBr 2 )
Y 312.5
74.7
Ni—I(Nilj)
254.8
60.9
Cu—Cu
190.4 ± 13
45.5 ± 3
Cu—F(CuF 2 )
365 ± 38
88 ± 9
Cu—F(CuF)
-418
-100
Cu—CI(CuCI 2 )
293.7 ,
70.2
• ■•o
; cu— d(c'uci)
bsbmmhnhhhi
360.7
86.2
Table E.l
( Continued )
848
17 The Chemistry of the Halogens and the Noble Gases
Halides
849
Table 17.5 _
Polyhalogen cations
Halides
Physical Inorganic
Chemistry of the
Halogens
Table 17.6
Radii, ionization energy,
electron affinity, and
electronegativity of the
halogens
X 2 +
Xa
v+ v 2+ V+
A4 A4 A5
X7 XYJ-
X 2 Y +
XYZ +
XY 4
xy:
ci 3 +
ci;
CIFj
BrF 2 +
ci 2 f +
cif 4 +
cif;
Br 2 +
Br 3 +
Br;
BrF 4
BrF ;
Iz +
I 3 +
ir i 5 +
If? IFj
if;
IF 6 +
ICIj
I,C1 +
IBr 2
I 2 Br +
BrICl +
This brief overview has not included all of the poly halogen cations known, but merely
discussed a few typical examples. See Table 17.5 for a listing.
Although many of the compounds of the halogens discussed thus far have exhibited a
halogen in a positive oxidation state, most of the chemistry of this family involves
either halide ions or covalent molecules in which the halogen is the most elec¬
tronegative atom.
The pertinent trends in the Group VI1A (17) elements are size and tendency to attract
electrons (Table 17.6). It is only when both these factors are considered that the
chemistry of these elements can be rationalized. The most obvious trend in the family
is the attraction for electrons. The ionization energy decreases from fluorine to iodine
as expected. There is an apparent anomaly in the case of the electron affinity of
fluorine, which is lower than that of chlorine. The small size of the fluorine atom
causes it to be saturated quickly with electron density, and the addition of a unit
charge causes some destabilization (see Chapter 2). The great electronegativity of
fluorine combined with its small size (which enhances Madelung energy from
5 + + 5“ = r effects) results in a much greater exothermicity of reactions of fluorine than
those of the remaining halogens. In covalent molecules it is exhibited by compounds
of fluorine without other halogen analogues, for example, AsF s , XeF 4 A , and IF 7 . In
aqueous solution it is exhibited by the high emf of the fluorine electrode resulting from
large hydration energy of the small fluoride ion. 3h This much greater reactivity of
fluorine has led to its characterization as a "superhalogen."
Element
r cov
fx-
'Vdw
IE°
EA°
Electronegativity 1
F
71
136
150
17.4
3.4
3.90
Cl
99
181
190
13.0
3.6
2.95
Br
114
195
200
11.8
3.4
2.62
I
133
216
210
10.4
3.1
2.52
" Electron volts.
Mullikcn-Jaffd values in Pauling units.
36 Alternatively the electrode potential can be ascribed to case with which the F— F bond is broken.
As shown by the following discussion the interrelation between bond energy, size, electronegativity
energy, etc., is complex and attributing everything to one factor is unwise.
The Anomaly
of Fluorine
Quite often the first member of a periodic group differs from the remaining members of
the group (see Chapter 18). In the case of fluorine the anomaly is quite pronounced.
Politzer 37 has illuminated this odd behavior by documenting the weakening of bonding
by fluorine to other elements compared with that expected on the basis of extrapola¬
tions from the heavier halogens. For example, we have seen that the electron affinity
of fluorine is less than might have been expected from the trend of the other halogens.
If this trend is extrapolated to fluorine, a value of 440 kJ mol -1 is obtained, 110 kJ
mor 1 greater than the experimental values. As a result of the lower electron affinity,
ionic compounds of fluorine have bond energies which are slightly more than 100 kJ
mor 1 weaker than values extrapolated from the other halides: LiF (104 kJ mol -1
lower), NaF (108 kJ mol - ' lower), KF (117 kJ mol' 1 lower), RbF (104 kJ mo!" 1
lower), and CsF (130 kJ mol" 1 lower). This destabilization can be attributed to forcing
a full (or nearly full) electronic charge onto the small fluorine atom. The surprising fact
pointed out by Politzer is that covalent compounds of fluorine seem to show the same
destabilization. In Fig. 17.12 the dissociation energies of the hydrogen halides and of
the C — X bonds in the methyl halides are plotted against the reciprocals of their bond
lengths. The compounds of the three heavier halogens fall on a straight line which,
when extrapolated, predicts values for the fluorine compound that are 113 kJ mol' 1
(HF) and 96 kJ mol 1 (CH 3 F) too high. This indicates that even when sharing an
electron from another atom fluorine is destabilized by its small size. Finally, the
fluorine molecule itself has a notoriously weak bond (155 kJ mol" 1 ) compared with
chlorine (243 kJ mol" 1 ), and it is some 226 kJ mol" 1 (= 2 x 113) weaker than the
extrapolated value. The weak bond in F 2 has traditionally been interpreted in terms of
lone-pair repulsions between the adjacent fluorine atoms. There may be a more
general phenomenon in terms of small size, charge capacity, and electron-electron
repulsion.
Fig. 17.12 Bond
dissociation energies and
bond lengths of the
hydrogen halides, methyl
halides, and halogen
molecules. Note that this
figure, which is taken
directly from Pulitzer's
work, portrays in a
different way relationships
that are closely related to
Fig. 9.7. [From Politzer. P.
J. Am. Client. Soc. 1969,
91. 6235. Reproduced with
permission.)
37 Politzer. P. Inorg. Client. 1977. 16. 3350. Politzer. P. In Homoutomic Rings. Chains an,l Mac-
romolecules of Main Group Elements; Rhcingold. A. L„ Ed.; Elsevier: New York. 1977; pp 95-115.
See also Politzer. P.; Huhccy, J. E.; Murray. J. S.; Grodzicki. M. J. Mol. Structure (THEOCHEM).
1992, 259. 99-120.
A-26 E- Bond Energies and Bond Lengths
Table E.l
Kj Nj Nj
Table E.l
( Continued)
Lanthanides and actinides (kJ mol -1 )
Metal
mf 3
mci 3
MBr 3
mi 3
MF„
mf 5
mf 6
Ce
~644
499.6
-431
-356
Pr
-623
482.4
-418
338.5
Nd
~611
470.7
-406
323.8
Pm
-590
-452
-385
-305
Sm
-565
-427
-360
-285
Eu
-552
-414
-347
-272
Gd
-619
-485
-418
-343
Tb
-615
-477
-410
-339
Dy
-577
-444
-377
-305
Ho
-577
-444
-381
-305
Er
-582
-448
-381
-310
Tm
-548
-418
.. -351
-251
Yb
-519
-385
-322
-251
Lu
-607
-477
-410
-343
Th
641.0
Pa
U
-619
495.4
424.3
-343
598.0
564.8
522.2
Np
-586
-460
-397
-326
-561
-519
-477
Pu
562.7
442.2
378.7
-310
-536
-485
-427
Am
-573
-452
-389
-318
-519
Lanthanides and actinides (kcal
mol -1 )
Metal
mf 3
mci 3
MBr 3
mi 3
mf 4
MF S
mf 6
Ce
-154
119.4
-103
-85
Pr
-149
115.3
-100
80.9
Nd
-146
112 J
-97
77.4
Pm
-141
-108
-92
-73
Sm
-135
-102
, —86
-68
Eu
-132
—99
-83
-65
Gd
-148
-116
-100
-82
Tb
-147
-114
—98
-81
Dy
-138
-106
-90
-73
% V X7
. / 4 j
Ho
-138
-106
-91
-73
Er
-139
-107
-91
-74
Tm
-131
-100
-84
-60
Bond
Do
r
kJ mol 1
kcal mol -1
pm
A
Sc—I
-322
-77
Y—Y
156.1 ± 21
37.3 ± 5
Y—F
-628
-150
Y—Cl
-494
-118
Y—Br
-431
-103
Y—I
362.8
86.7
La—La
241.0 ± 21
57.6 ± 5
850
17- The Chemistry of the Hologens and the Noble Gases
So how do we resolve this apparent paradox: Is fluorine a "superhalogen or a
“subhalogen”? Does it bond better than the other halogens or worse than expected ’
There really is no conflict here; it depends upon what one is using for a reference. In
comparison with the heavier halogens, fluorine is by far the most active, the most
electronegative, and provides the most strongly exothermic reactions. In this regard
the weakening present in any X-F bond is offset by the weak F—F bond, so the
overall enthalpy of the reaction is not affected, and the effect ot a high elec¬
tronegativity on Madelung energy and electronegativity energy terms is dominant.
Fluorine has three factors favoring it over the larger halogens, all resulting trom its
small size: (1) the largest electronegativity, at least for small partial charges: (2) large
Madelung energies in polar molecules; and (3) good covalent bonding resulting trom
the ability to "get in close” to the adjacent atom. 38 Except for F 2 in which factors I
and 2 cannot come into play, fluorine typically forms stronger bonds than chlorine:
H — F = 565, H —Cl = 428: Li—F = 573, Li —Cl = 464; C —F = 485, C —Cl =
327 (in kJ mol -1 ). Thus fluorine displays all the properties expected of the smallest
halogen. The deficit of about 100 kJ mol -1 is simply an example of the law of
diminishing returns; At some point decreasing size no longer provides increasing
bonding benefits in proportion to further decrease in size. Were it not for the satura¬
tion effect coming into play, fluorine would probably be a "super-superhalogen !
A very similar question that depends entirely upon what one takes as normal or
standard is the question of the covalent radius of fluorine. One half of the single
] covalent bond length in difluorine is 143 pm + 2 = 71 pm (and so listed in Table 8.1).
One can then employ the Schomaker-Stevenson relationship and attribute the short-
i ening (and strengthening) of an M — F bond to ionic-covalent resonance (Madelung
energy) and account for the fact that the M —F bond is always shorter than r Mcov +
| /■„ . Alternatively, once can assume that the bonding in most fluorine compounds is
"normal" and that the long (and weak) difluorine bond is the "anomaly" because ol
the lone pair-lone pair repulsions. The two viewpoints are essentially equivalent. For
a careful analysis of the covalent bond radius of fluorine, see Gillespie and
Robinson. 39
Because of its supreme reactivity, the preparation of elemental difluorine was not
accomplished until long after the preparation of the other halogens. 9 " So most chem¬
ists accepted the impossibility of producing difluorine in any practical way other than
the electrolysis of anhydrous HF:
?HF KFi lJL -> H, + F, (17.70)
“ electrolyte
However, Christe 91 has shown that by taking advantage of the difference in stability of
a given oxidation state of a transition metal depending upon whether it is fully
'H Note that with the exception of hydrogen, fluorine is the smallest bonding atom known, and so
almost every bond il makes will be with a larger atom,
w Gillespie, R. J.; Robinson, E. A. Inorg. Client. 1992. 31. 1960-1963.
9 " For an interesting review of the early history of fluorine chemistry, sec Fluorine: Tlw First Hundred
Years (1886-1986); Banks, R. E.; Sharp. D. W. A.: Tatlow. J. C„ Eds.: Elsevier Sequoia: New
York, 1986.
4 . Christe. K. O. Inorg. Client. 1986. 25 . 3721-3722. See also Christe. K. O.: Wilson. R. D. Inorg.
Client. 1987, 26. 2554-2556.
Halides
851
coordinated or not, difluorine gas can be generated chemically. Potassium hex-
afluoromanganate(IV) can be prepared by reduction of potassium permanganate:
2KMn0 4 + 2KF + 10HF + 3H-.0, ~—“ q HF >
2K 2 MnF 6 ! + 8H 2 0 + 30 ; (17.71)
Although K 2 MnF 6 is stable (perhaps because of its insolubility), the free Lewis acid
MnF 4 is not. Preparation of the latter is accomplished by using very strong Lewis
acids such as SbF s , TiF 4 . and BiF 5 (see Chapter 9) which are also redox stable in the
presence of difluorine gas.
K 2 MnF 6 + 2SbF 5 -► 2KSbF 6 + MnF 2 + ^F 2 (17.72)
Similar reactions can be run using nickel or copper as the transition metal.
Whatever one's interpretation of fluorine chemistry, sub-, super-, or super-super¬
halogen, it is obvious that the thermochemistry is extremely important to the under¬
standing the chemistry of fluorine. It is fortunate that good, thorough thermodynamic
data on fluorine compounds are available. 9 -
It was noted above that discussion of astatine together with the other halogens is
inconvenient. Although it is, as expected, the most “metallic" of the halogens, there
are few values or experimental data to cite in support of this. (Note for example that
such fundamental quantities as experimental ionization energies are unavailable.)
Various isotopes of astatine are produced only in trace amounts, with half-lives of a
few hours or less, and therefore the chemistry of astatine is essentially the descriptive
chemistry obtained by tracer methods; macroscopic amounts are not available. The
best known oxidation state of astatine is -I. Astatine may be readily reduced to
astatide:
2Al + SO, + 2H,0 -» 2At“ + 3H f + HSOJ (17.73)
which forms an insoluble silver astatide precipitating quantitatively with silver iodide
as carrier.
Studies of elemental astatine are complicated by the fact that the small amounts of
astatine present are readily attacked by impurities that normally would nol be consid¬
ered important. Most studies of At(0) involve an excess of iodine which lies the
astatine up in All molecules. 93 It behaves much as might be expected from the known
behavior of I 2 : It is readily extractable into CCI 4 or CHCI, and may be oxidized to
positive oxidation states by reasonably mild oxidizing agents.
The best characterized positive oxidation state is At(V). Astatate ions may be
formed by oxidation of At by peroxodisulfate, the ceric ion, or periodate:
Af + 6Ce 4+ + 3H 2 0 -► At0 3 " + 6Ce 3+ + 6H (17.74)
As such it may be quantitatively precipitated with insoluble iodates such as Pb(IO,) 2
and Ba(IOj) 2 .
It appears that perastatate, At0 4 , has not been prepared. When astatate was
treated with very strong oxidizing agents, a negligible amount of the activity precipi-
9: Woolf. A. A. Adv. Inorg. Che in. Radiothent. 1981. 24. 1-55.
93 Strictly speaking one might argue that the astatine in Atl should be considered Aid) instead of Al(0)
since astatine should be less electronegative than iodine. The difference in eleclronegalivilies is
certain to be very small, however, and Atl can probably best be considered as an almost nonpolar
molecule with essentially zero charge on both aloms. behaving like l 2 .
A-24
E- Bond Energies and Bond Lengths
E-Bond Energies and Bond Length:
A-25
None of these represents the breaking of a hypothetical, isolated N—N bond:
• N—N • -> 2 • N •
By using N—H and N—F bond energies from NH 3 and NF 3 it is possible to estimate
the inherent strength of the N—N bond:
Total energy of atomization, N 2 H 4
4N — H bonds (assumed from NH 3 )
Difference (equated with N—N)
Total energy of atomization, N 2 F 4
4N—F bonds (assumed from NF 3 )
Difference (equated with N—N)
1703 kJ mol"
1544 kJ mol"
(407 kcal mol" 1 )
(369 kcal mol -1 )
159 kJ mol (38 kcal-mol ')
1305 kJ mol"' (312 kcal mol" 1 )
1134 kJ mol" 1 (271 kcal mol" 1 )
172 kJ mol -1 (41 kcal mol -1 )
The results of this calculation are gratifyingly congruent, and we feel reasonably
confident in a value of about 167 kJ mol -1 (40 kcal mol" 1 ) for the N—N bond. Similar
results can be obtained for hydrogen peroxide and dioxygen difluoride to obtain an
estimate of about 142 kJ mol” 1 (34 kcal mol” 1 ) for the O—O bond.
Although the calculations for N—N and O—O bonds are self-consistent, there is
always the possibility that a wider series of compounds would show greater vari¬
ability. This is especially probable in molecules in which the electronegativity of the
substituents is believed to affect the bonding by particular orbitals, e.g., overlap by
diffuse d orbitals. Thus on the basis of observed stabilities the presence of elec¬
tronegative substituents such as —F and —CF 3 seems to stabilize the P—P bond
relative to H,P—PH 2 , although there are not enough data to investigate this pos¬
sibility quantitatively.
There is at present no convenient, self-consistent source of all bond energies. The
standard work is Cottrell, T. L. The Strengths of Chemical Bonds, 2nd ed.; Butter-
worths: London, 1958, but it suffers from a lack of recent data. Darwent (National
Bureau of Standards publication NSRDS-NBS 31, 1970) has summarized recent data
on dissociation energies but did not include some earlier work or values known only
for total energies of atomization rather than for stepwise dissociation. Three useful
references of the latter type are: Brewer, L.; Brackett, E. Chem. Rev. 1961, 61, 425;
Brewer, L.; et al. Chem. Rev. 1963, 63, 111; Feber, R. C. Los Alamos Report
LA-3164, 1965. The book by Darwent mentioned above also lists bond energy values
for some common bonds.
Table E. I has been compiled from the above sources. The ordering is as follows:
hydrogen. Group IA (1), Group IIA (2), Group IIIB (3), transition elements, Group
IIIA (13). Group IVA (14), Group VA (15), Group VIA (16), Group VIIA (18), Group
V111A (19). For a given group (such as IA 1) the compounds are listed in the following
order: halides, chalcogenides, etc. Unless specified otherwise, the bond energies are
for compounds representing group number oxidation states, such as BC1 3 , SiF 4 , and
SF 6 . Other compounds are listed in parentheses, such as (T1C1) and (PC1 3 ). Values are
for the dissociation energies of molecules A—B and mean dissociation values for
AB„. For molecules such as N 2 H 4 two values are given: H 2 N—NH 2 represents the
dissociation energy to two amino radicals, and N—N (N 2 H 4 ) represents the estimated
N—N bond energy in hydrazine obtained by means of assumed N—H bond energies
as shown above.
Table E.1
Bond energies and bond
lengths
Bond
Do
r
kJ mol ’
kcal mol” 1
pm
A
Hydrogen Compounds
H—H
432.00 ± 0.04
103.25 ± 0.01
74.2
0.742
H—F
565 ± 4
135 ± 1
91.8
0.918
H—Cl
428.02 ± 0.42
102.3 ± 0.1
127.4
1.274
H—Br
362.3 ± 0.4
86.6 ± 0.1
140.8
1.408
H—I
294.6 ± 0.4
70.4 ± 0.1
160.8
1.608
H—O
458.8 ± 1.4
109.6 ± 0.4
96
0.96
H—S
363 ± 5
87 ± 1
134
1.34
H—Se
276?
66?
146
1.46
H—Te
238?
57?
170
1.7
H—N
386 ± 8
92 ± 2
101
1.01
H—P
-322
-77
144
1.44
H—As
-247
-59
152
1.52
H—C
411 ± 7
98.3 ± 0.8
109
1.09
H—CN
531 ± 21
127 ± 5
106.6
1.066
H—Si
318
76
148
1.48
H—Ge
153
1.53
H—Sn
170
1.70
H—Pb
H—B
389?
93?
119
1.19
H—Cu
276 ± 8
66 ± 2
H-Ag
230 ± 13
55 ± 3
H — Au
285 ± 13
68 ± 3
H—Be
H — Mg
H—Ca
U C-
n — or
H—Ba
H—Li
-243
-58
159.5
1.595
H—Na
197
47
188.7
1.887
H—K
180
43
224.4
2.244
H—Rb
163
39
236.7
2.367
H—Cs
176
42
249.4
2.494
Group IA (1)
Li — Li
105
25
267.2
2.672
Li—F
573 ± 21
137 ± 5
154.7
1.547
Li—Cl
464 ± 13
111 ± 3
202
2.02
Li—Br
418 ± 21
100 ± 5
217.0
2.170
Li—I
347 ± 13
83 ± 3
239.2
2.392
Na—Na
72.4
17.3
307.8
3.078
NaF
•A sfr
477.0
114.0
184
1.84 j
NaCl
407.9
97.5
236.1
2361 3
NaBr
362.8
86.7
250.2
2.502
Nal
304.2
72.7
271.2
2.712
k 2
49.4
11.8
392.3
3.923
KF
490 ± 21
117 ± 5
213
2.13
K.CI
423 ± 8
101 ± 2
266.7
2.667
KBr
'ijfi'ivji f'.
378.7 ± 8
90.5 ± 2
282.1
.2.821
K1
326 ± 13
78 ± 3
304.8
3.048
Rb-Rb
45.2
.. i.
10.8 >
U-'L k %
Electrochemistry of the Halogens and Pseudohalogens
tated with KI0 4 , most precipitating instead with Ba(I0 3 ) 2 . The apparent absence of
At(VlI) is surprising in view of the lower electronegativity and larger size of astatine.
If perastatic acid does exist, it is probably with coordination number 6: H 5 At0 6 .
At least one more oxidation state, presumably At(I) or At(III), is known in
aqueous solution, but it has not been well characterized. It can be produced by
reduction of astatate by chloride ion or oxidation of At(0) by Fe 3+ . Little is known
about it except that it differs from the other oxidation states of astatine. It does not
precipitate with silver (At") or barium (AtO J), nor extract into CC1 4 (Atl), but it does
follow the dipyridineiodine(I) cation.
2HCN
2SCN
In summary, the concept of pseudohalogens proves useful in systematizing the
chemistry of some of these nonmetallic groups, but it should never be followed
blindly.
Electrochemistry Simple Latimer diagrams for the halogens are given below. The data are from
Bratsch. 45
There are certain inorganic radicals which have the properties of existing either as
monomeric anions or as neutral dimers. In many ways these groups display properties
analogous to single halogen atoms, and hence the terms pseudohalogen or halogenoid
have been applied to them. Examples of pseudohalogen behavior may be found in the
chemistry of cyanide, thiocyanate, and azide anions. Some typical reactions are:
Acid solution.
Pseudohalogens
Oxidation of X ions to form dipseudohalogens:
2SCN- + 4H + + Mn0 2 -» (SCN) 2 + 2H 2 0 + Mn :
Disproportionation of the free pseudohalogen by a base:
(CN) 2 + 20H' -► CN“ + OCN“ + H z O
Precipitation by certain metal ions:
Ag + + Nj- - AgNj!
HCIO.
HOC!
HOBr
Formation of complex ions:
Zn 2+ + 4SCN“ - [Zn(SCN) 4 ] 2- (17.78)
Formation of acids with hydrogen, HX. These acids generally are considerably
weaker than the hydrohalic acids, e.g., p/C., for HCN = 9.
Basic solution.
The extent to which the various pseudohalogens resemble halogens is generally
quite high, with some remarkable parallels:
-OJM
il 2 + Cul J
J(CN), + CuCNi
although, however, there are several exceptions. Thus thiocyanogen, (SCN) 2 , is
stable only at low temperatures, and at room temperature polymerizes to (SCN) r .
With respect to division into hard and soft species, most pseudohalogens are com¬
posed of several nonmetal atoms, often with multiple bonding, and so are quite
polarizable. As such they tend to resemble iodine considerably more than fluorine.
Some are ambidentate, however, and can behave as reasonably hard bases by coordi¬
nation via a nitrogen or oxygen atom (see Chapter 12). Finally, pseudohalogens can be
compared with the halogens on the basis of their relative oxidizing power. They tend
to resemble iodine and bromine (see below for the values of these elements): 44
44 Data from Bard. A. J.; Parsons, R.; Jordan. J. Standard Potentials in Aqueous Solution'. Marcel
Dekker: New York. 1985.
45 Bratsch, S. G. J. Phys. Chem. Ref. Data 1989, 18. 1-21
The quantity which is generally more accessible experimentally is the enthalpy.
The enthalpy of dissociation at a given temperature differs from the energy of dissocia¬
tion by PAV work: —2.5 kJ mol -1 ; -600 cal mol -1 at room temperature. Some
examples of the various quantities for H, are:
H,-► 2H 458.1(109.5) 432.00 (103.25) 433.21(103.54) 435.93 (104.19)
Since the last three values—those most often quoted for "bond energies"—differ by
very little, the difference may be ignored except in precise work.
The electronic energy, AU cl , is of interest mainly in connection with bonding
theory since it is not an experimentally accessible quantity. A second quantity of this
type is the “intrinsic bond energy,” the difference in energy between the atoms in the
molecule and the separated atoms in the valence state, i.e., with all of the atoms in the
same condition (with respect to spin and hybridization) as in the molecule. It is a
measure of the strength of the bond after all other factors except the bringing together
of valence state atoms have been eliminated (cf. the discussion of methane. Chapter 5
and McWeeny, R. Coulson's Valence, 3rd ed.; Oxford University: Oxford. 1979;
Chapter 7).
The situation becomes further complicated in polyatomic molecules. The energy
of interest to chemists, generally, is that associated with breaking the bond without
any change in the remaining parts of the mulecule. For example, if we are interested in
the bond energies in CH 4 or CF 4 , we wish to know the energy of the reaction:
x / J, x x x'^x
In general, the quasitetrahedral species CX 3 is not observed. In CF 3 a pyramidal
molecule approaching this configuration is found, but in -CH 3 the resulting species is
A-23
E-Bond Energies and Bond Lengths
planar with sp 2 hybridization instead of sp 3 . The energies associated with various
dissociative steps for methane are:
n
K
1
-» ch 3
H
AU = 421.1 kJ mol -1 (101.6 kcal mol -
ch 3 -
- CH,
H
AU = 469.9 kJ mol -1 (112.3 kcal mol -
ch 2 —
-+ CH
H
AU = 415 kJ mol -1 (99.3 kcal mol -1 )
CH —
— C +
H
AU = 334.7 kJ mol -1 (80.0 kcal mol -1 )
We can associate the greater energy of the second dissociation step with a presumed
greater bonding strength of trigonal: sp 2 hybrids over sp 2 hybrids. Whether we
understand (or at least believe we do) the reasons for each of the quantities listed
above or not. it is obvious that none represents the bond energy in methane. However,
since the summation of these four experimentally observable processes must be
identical to the energy for the nonobservable (but desirable) reaction:
CH 4 - C + 4H
the average of these four quantities (411 kJ mol -1 ; 98.3 kcal mol -1 ) can be taken as the
mean bond energy for the C—H bond in methane.
The mean bond energy is a useful quantity, but it should be remembered that it is
derived from a particular molecule and may not be exactly correct in application to
another molecule. Thus if the total bond energy in dichloromethane does not equal
two times the average bond energy in methane plus two times the average bond energy
in carbon tetrachloride, we should not be surprised. The presence of bonds of one type
may have an effect in strengthening or weakening bonds of another type. As a matter
of fact, there is no unequivocal way of assigning bond energies for molecules such as
dichloromethane by means of thermodynamics. The summation of all of the bond
energies may be determined as above for methane, but the assignment to individual
bonds must be made by secondary assumptions, e.g., the bond C—H energies are
comparable to those in methane. Alternatively, the bond energies in molecules con¬
taining more than one type of bond may be assigned on the basis of some other type of
information such as infrared stretching frequencies.
One of the most serious problems hindering the assignment of bond energies
arises for bonds such as N—N and O—O. The nitrogen triple bond and oxygen
double bond may be evaluated directly from the dissociation of the gaseous N 2 and 0 2
molecules. Single bonds for these elements present special problems because addi¬
tional elements are always present. For example, consider the following dissociation
energies accompanying splitting of the N—N bond:
N 2 H 4 -♦ 2NH 2 D 0 = 247 kJ mol -1 (59 kcal mol -1 )
N 2 F 4 -♦ 2NF 2 D 0 = 88 kJ mol -1 (21 kcal mol -1 )
N 2 0 4 -» 2NO : D„ = 57.3 kJ mol -1 (13.7 kcal mol -1 )
j
854
1 7 • The Chemistry of the Halogens and the Noble Gases
;
I )
Problems 17.1. Consider the formation of 0?[PtF 6 ]". Why is the ionization energy of O, less than that of
0? Is it likely that a compound N 2 [PtF 6 r wil > f° riTI?
17 2 The absence of interaction between noble gases and the Lewis acids BX, was demon¬
strated by Raman spectroscopy. Discuss the nature of the evidence and what would have
been observed if there had been a significant interaction.
17.3. Show how the FHF" ion (Chapter 8) can be treated as a three-center, four-electron bond.
17.4. Suggest autoionization possibilities for BrFj, ICI, and BrF 5 . and probable structures for
the ions formed.
17.5. Pure iodine is purple in color as are its solutions in CCI 4 , and CHCI 3 . and benzene.
Aqueous solutions of Kl, are brown. Solutions of diiodine in acetone, dimethyl sulfoxide,
and diethyl ether are brown. Suggest an explanation.
17.6. Suggest syntheses for
a. Xe0 4
b. HClOj
c. KBrOj
17.7. Suggest probable structures for l 4 CP and ICI 4 ". and give reasons why the two are
probably not isostructural.
17 8 The production of pseudohalogens requires mild oxidizing conditions (F.q. 17.,5). Why
do you suppose that it has never been possible to oxidize the azide ion to hexamtrogen
(diazyl)?
2N - N =N—N=N—N=N < 17 - 83 '
17.9. Why arc the halogen cations Cl,. Br,. and I, best isolated as salts of AsF,T. SbF„\ and
similar anions?
17.10. The melting points of the fluorides MF. MF,. and MF, arc generally somewhat lower
than those of the corresponding oxides. M,0. MO. M 2 0 3 . because of the greater lattice
energy resulting from the dinegative oxide ion. O' . Yet all of the following reactions are
exothermic. Explain. 46
I.i ; 0 + F,
2LiF + JO,
AGsixi “
-602 kJ mol " 1
(17.84)
MgO + F,
MgF, + JO,
AC^h, =
-740 kJ mol " 1
(17.85)
Fe,0 3 + 3F 2
2FeFj + JO,
Afjs’lHI =
1162 kJ mol " 1
(17.86)
ZrO, + 2F,
ZrF 4 + O,
AO 500 =
-740 kJ mol " 1
(17.87)
17.11. If you ask an organic chemist which element can form the largest number ot compounds
(not that more than a fraction have been synthesized yet), you will usually get one of two
answers: carbon from some, hydrogen from others. If you ask an inorganic chemist, you
may get a third answer. What is your answer? Discuss.
17.12. On page 835 it is stated that xenon forms bonds with only the most electronegative
elements such as the very active fluorine and oxygen. How can you reconcile this with
the formation of the Xe—Xe bond in Xe 2 ? (Him: Rethink Problem 5.15.)
17.13. Xe? obviously will have a fairly high electron affinity (see the ionization energy of atomic
xenon), and if it gains an electron, it will dissociate (see Chapter 5). Combine these facts
with the choice of SbF, as solvent and acid-base theory to provide a self-consistent
interpretation.
46 Porticr. J. Angew. Chem. Int. Ed. Engl. 1976. 15. 475.
J H 21 ( uunts per channel
Problems
855
17.14. The OSeF 5 and OTeF, groups are very electronegative as shown by the stability of their
xenon compounds. If you did not do Problem 5.30 when you read that chapter, do so
now.
17.15. The photoelectron spectra of Xe. F 2 . XeF,, XeF 4 . OXeF 4 , and XeF ft are shown in. Fig.
17.13. What information can you obtain from these spectra? (The appearance of two
peaks for Xe is attributable to the ejection of 3 d electrons of different j values and is
irrelevant to the question being asked.)
17.16. Consider the series of xenon oxyfluorides and their relative acidity. Discuss the reasons
for the ordering of these compounds. Can you semi-quanlify your answer?
Kinetic energy (eV)
810 805 800 795 790 785 780
0 675 680 (.85 690 695 700 705
Bimling energy (eV)
g. 17.13 Photoeleetron spectra of Xe. F-. XcF ; . XeF 4 , OXeF 4 . and XeF,,. |From Carroll.
X.; Shaw. R. W.. Jr.; Thomas. T. D.; Kindle. C.: Bartlett. N. J. Am. Chem. Soc. 1974.
. 1989. Reproduced with permission.)
Bond Energies and Bond Lengths
Although the concept of bond energy seems intuitively simple, it is actually rather
complicated when inspected closely. Consider a diatomic molecule A—B dissociat¬
ing. It might be thought that it would be a relatively simple matter to measure the
energy necessary to rupture the A—B bond and get the bond energy. Unfortunately,
even if the experiment is feasible the result is generally not directly interpretable in
terms of "bond energies" without further work. Among the factors to be considered
are the vibrational, rotational, and translational energies of the reactants and products,
the zero-point energy, and pressure-volume work if enthalpies are involved. The
interested reader is referred to books on thermodynamics for a complete discussion
(especially Dasent, W. E. Inorganic Energetics; Penguin: Harmondsworth, England,
1970). The following is meant as a brief outline of the problem.
Consider the energy of a diatomic molecule as shown in the figure. The concept of
bond energy may be equated with the difference between the bottom of the energy
curve and the energy of the completely separated atoms (A(7 C |). However, as a result
of the zero-point vibrational energy of the AB molecule, even at 0 K, the energy
necessary to separate the atoms A U is somewhat less (by a quantity of A/jv). The zero-
point energy is greatest in molecules containing light atoms such as hydrogen (25.9 kJ
mol -1 ; 6.2 kcal mol -1 in H 2 ) and somewhat less in molecules containing heavier
atoms.
There is a corresponding difference between two estimates of the bond distance in
a molecule A—B. One. r c , corresponds to the minimum in the energy dislance curve
(see figure). The second. r 0 , corresponds to the average distance in a molecule
vibrating with zero-point energy. Since the curve is not perfectly parabolic the two
values are not identical.
If the dissociation is to take place at some temperature, T, other than 0 K, the
energy necessary to accomplish the dissociation must include an amount sufficient to
provide the separated atoms with the translational energy at that temperature (iRT).
Compensating in part for this will be the translational, rotational, and vibrational
energy of the molecule AB at temperature T (-6.3 kJ mol -1 ; -1.5 kcal mol -1 for H 2 at
298 K). The difference between the dissociation energy at 298 K (A(/ 298 ) and that at 0
K (A U 0 ) is very small (~I kJ mol -1 ; kcal mol -1 for H 2 ).
A-21
856
17 -The Chemistry of the Halogens and the Noble Gases
17.17. a. Predict the most likely by-products of the reduction of XeF 2 to XeJ" as discussed on
page 834.
b. Write balanced equations for all of these reactions.
17.18. The xenon fluorides are described in this chapter as both fluoridators and fluorinators. Is
this a typographical error? Can these terms be differentiated? Discuss.
17.19. On page 827 the statement is made that "the electron affinity of positive helium, etc. is
greater than that of any appropriate species X . . . ." What is the electron affinity
(numerical value) of He + ?
17.20. A general rule of molecular structure derived from VSEPR theory and Bent's rule is that
lone pairs and substituents of low electronegativity prefer equatorial positions in a
trigonal bipyramid, whereas substituents of high electronegativity prefer axial positions.
The pentafluoroxenate(IV) ion. XeFJ, seems to reverse this rule by having the LPs
preferentially occupying the axial positions. Discuss.
17.21. On page 832 the statement is made that in “most fluorides the reduction of steric factors
allows the lone pair to emerge to the surface of the molecule,. . ." and yet there are two
hexafluoro species mentioned in this chapter that are perfectly octahedral. What are
they? How can they be octahedral in light of the above statement?
17.22. Although the Xe—O bond (-80 kj mol" 1 ) is not as strong as that in the xenon fluorides
(-180 kJ mol" 1 ), it is far from being the weakest bond known. How then is it possible for
XeOj to be so violently exothermic (explosive) when it decomposes?
17.23. If you did not do Problem 14.15 when you read Chapter 14, do so now.
17.24. On page 834, the statement is made: "Hydrogen xcnate ions disproportionate in alkaline
solution to yield perxcnates . . . ." Can you confirm this assertion based on the Latimer
diagrams for xenon?
Chapter
Periodicity
The most fascinating aspect of inorganic chemistry as well as its most difficult
problem is the diversity of reactions and structures encountered in the chemistry of
somewhat over one hundred elements. The challenge is to be able to treat adequately
the chemistry of boranes and noble gas fluorides, transition metals and inner transition
metals, cuprate superconductors and zeolites, all without developing a separate set of
rules and theories for each element or system. Much of this book has been devoted to
establishing relationships that connect various aspects of inorganic chemistry. The
tool that the inorganic chemist uses to systematize elemental relationships is the
periodic table, now somewhat over one hundred years old. 1 It is considered so
essential that no general chemistry textbook would be complete without a discussion
of the trends summed up in Chapter 2. Unfortunately, the impression left by these
textbooks is often simply that all periodic properties vary smoothly.
Fundamental Trends The basic trends of the periodic chart have been discussed in Chapter 2. They may be
summarized as follows. Within a given family there are increases in size and decreases
in ionization energy, electron affinity, electronegativity, etc. Increasing the atomic
number across a given period results in concomitant increases in ionization, electron
affinity, and electronegativity, but a decrease in size. The change in effective nuclear
charge within a period is reasonably smooth, but the various periods differ in length (8.
18, and 32 elements). The properties of an element will depend upon whether it follows
an 8, 18, or 32 sequence. One of the best known examples is the very close similarity
in properties of hafnium, tantalum, tungsten, and rhenium to those of zirconium,
niobium, molybdenum, and technetium, respectively, as a result of the lanthanide
contraction and associated effects. These anomalies continue through the elements
gold, mercury, thallium, and lead. Similar but smaller effects follow the filling of the 3 d
orbitals (rarely referred to as the “scandide" contraction).
Another area of the periodic chart revealing pronounced differences between
similar elements is between the first ten, H-Ne, and those immediately following.
1 Periodic classifications of the elements by Dmitri Mendeleev and by Lotliar Meyer appeared in 1869.
For a centennial-celebrating discussion of the periodic table, see van Spronsen. J. W. The Periodic
System of Chemical Elements', Elsevier: Amsterdam, 1969.
857
Nu-Ar. II is not completely obvious why this is true. Certainly the lighter elements
utilize only the Is, 2s, and Ip atomic orbitals in their ground states, and therefore
simple bonding theory, whether VB or MO. suggests four covalent bonds. In contrast.
VB theory suggests that the presence of d orbitals in elements with n > 3 allows
hybrids with more than four bonding orbitals. However, the use of d orbitals by
nonmetals presents energetic problems, is unnecessary in the simplest molecular
orbital approaches, and has been one of the most controversial topics in bonding
theory. Before entering this theoretical discussion, a brief examination of chemical
differences is appropriate.
In many ways the first ten elements differ considerably from the remaining 99.
Hydrogen is a classic example—it belongs neither with the alkali metals nor with the
halogens although it has some properties in common with both. Thus it has a +1
oxidation state in common with the alkali metals but the bare H + has no chemical
existence^ and hydrogen tends to form covalent bonds that have properties more
closely resembling those of carbon than those of the alkali metals. With the halogens it
shares the tendency to form a - 1 oxidation state, even to the extent of forming the
hydride ion. H - ; however, the latter is a curious chemical species. In contrast to the
proton which was anomalous because of its vanishingly small size, the hydride ion is
unusually large. It is larger than any of the halide ions except iodide ! 5 The source of
this apparent paradox lies in the lack of control of a single nuclear proton over two
mutually repelling electrons. Since the hydride ion is large and very polarizable it
certainly does not extend the trend of T through F" of decreasing size and increasing
basicity and hardness.
The elements of the second row also differ from their heavier congeners. Lithium
is anomalous among the alkali metals and resembles magnesium more than its con¬
geners. In turn, in Group IIA (2) beryllium is more closely akin to aluminum than to
the other alkaline earths. The source of this effect is discussed below. We have already
seen that fluorine has been termed a superhalogen on the basis of its differences from
the remainder of Group V1IA (17).
One simple difference that the elements Li to F have with respect to their heavier
congeners is in electron-attracting power. Thus fluorine is much more reactive than
chlorine, bromine, or iodine; lithium is less reactive than its congeners . 4 The most
electronegative and smallest element of each family will be the one in the second row.
The great polarizing power of the Li + cation was commented upon in Chapter 4.
As a result of its small size and higher electronegativity this ion destabilizes salts that
are stable for the remaining alkali metals:
: Those who disapprove of writing H } 0 + often point out that the hydration number or the H is
uncertain and "all cations arc hydrated in solution." To treat H ’ (rather than HiO ) as a cation
similar to Na. for example, is to equate nuclear particles with atoms, a discrepancy by a factor of
about 10-'.
’ Pauling ( Tin• Nature of the Chemical Hand. 3rd ed.; Cornell: Ithaca. NY. I960: p 514) has provided
an estimate of 208 pm for the hydride ion compared to 216 pm for I”. To be sure, the existence of an
unpolarizcd hydride ion is even less likely than a large unpolnrizcd anion of some other kind, but
insofar as ionic radii have meaning this would be the best estimate of the size of a free hydride ion.
J The inherent unreactivity of lithium is offset in aqueous solution by the exothermic hydration of the
very small Li ion. Nevertheless, in general, lithium is a less reactive metal than Na. K. Rb. orCs.
First- and Second-Row Anomalies
859
2LIOH —— Li 2 0 + H,0
red heal 1 i
2NaOH • No reaction
red heal
2LiSH -► Li 2 S + H 2 S
Li 2 C0 3 - Li,0 + C0 2
(18.D
(18.2)
(18.3)
(18.4)
In contrast, for the large polarizable hydride ion which can bond more strongly by a
covalent bond the lithium compound is the most stable:
Size Effects in
Nonmetals
i
Tabl e 18 .1_
Maximum coordination
numbers of the nonmetols as
shown by the fluorides
LiH • — -> No reaction (18,5)
2NaH Na, + H, (18.6)
One of the most obvious differences between the first ten elements and their heavier
congeners is in their maximum coordination number, usually four or less in simple
covalent molecules (Table 18.1). Arguments of the radius ratio type suggest lhat these
atoms would have lower coordination numbers and that these steric effects would be
relaxed for larger atoms. We have seen that this is true for coordination compounds of
transition metals (Chapter 12). Table 18.1 can be readily interpreted in these terms:
The smallest atoms have a maximum coordination number of four, larger atoms have
coordination number six. and only the largest have coordination numbers as high as
eight, in addition, the highest coordination numbers are found with the small fluorine
atoms as ligand . 5 The hydroxy group. OH. is very similar in size, electronegativity,
and other bonding properties to the fluorine atom, yet there are distinct differences
between the oxygen and fluorine compounds of the nonmetals. The occurrence of
molecules with the maximum number of hydroxy groups (the so-called ortho acids) is
rare. Even the relatively small oxygen atom tends to result in lower coordination
numbers. Thus, in contrast to the fluorides shown in Table 18.1. orlhocarbonic acid.
C(OH) 4 , and orthonitric acid. ON(OH)-,. are unknown, the simple acids of these
elements being three-coordinate, 0=C(0H), and 0,N0H.
The next two series of nonmetals, silicon through chlorine and germanium
through krypton, show a maximum coordination number of six in hexafluoro anions.
SF,,. and TeF h . Even here the oxyaeids and oxyanions typically show a coordination
CF 4 nr; OF" FF(FJ)
SiF l~ PF 6 “ SF„ CIF?
_ 1F 7 (IF H ~)
" N. O, and other elements can achieve
higher coordination in onium salts, c.g.,
NH 4 .
5 This is advantageous not only from the steric viewpoint but also from the fact lhat the redox behavior
of fluorine is well suited to stabilize high oxidation states. To put it another, but equivalent, way:
High oxidation state species are hard acids and the fluoride ion is the hardest possible base.
Groups ( Continued)
(-O’, xz, yz)
860 18 • Periodicity
The Diagonal
Relationship
number ot four as in H 3 PO.,, HC10 4 , and HBr0 4 , and the silicates (see Chapters 16
and 17).
The largest nonmetals show coordination numbers as high as eight in the oc-
tatluoroanions. 1F 8 and XeFg~ (see Chapter 17). The corresponding oxyacids and
oxyanions show a maximum coordination number of six: [Sb(OH)J~, Te(OH) 6 ,
OI(OH) 5 , and (Xe0 6 ] 4 ~. Of these, apparently only iodine shows a maximum oxidation
state with a coordination number as low as four: Periodic acid can exist as either
OI(OH), or HI0 4 .
It was mentioned previously that a strong resemblance obtains between Li and Mg, Be
and Al, C and P, and other "diagonal elements,” and it was pointed out that this could
be related to a size-charge phenomenon. Some examples of these resemblances are as
follows:
Lithium-Magnesium
There is a large series of lithium alkyls and lithium aryls which are useful in organic
chemistry in much the same way as the magnesium Grignard reagents. Unlike Na, K.
Rb, or Cs. but like Mg, lithium reacts directly with nitrogen to form a nitride:
3Li 2 + N, - > 2Li 3 N, (18.7)
6 Mg + 2N, -► 2Mg 3 N 2 (18.8)
Finally, the solubilities of several lithium compounds more nearly resemble those of
the corresponding magnesium salts than of other alkali metal salts.
Beryllium-Aluminum
These two elements resemble each other in several ways. The oxidation emfs of the
elements are similar (£^ c = 1.85; £^, = 1.66), and although reaction with acid is
thermodynamically favored, it is rather slow, especially if the surface is protected by
the oxide. The similarity of the ionic potentials for the ions is remarkable (Be : = 48.
Al 3+ = 56) and results in similar polarizing power and acidity of the cations. For
example, the carbonates are unstable, the hydroxides dissolve readily in excess base,
and the Lewis acidities of the halides are comparable.
The Use of
p Orbitals in
Pi Bonding
Boron-Silicon \
Boron differs from aluminum in showing almost no metallic properties and its re¬
semblance to silicon is greater. Both boron and silicon form volatile, very reactive
hydrides; the hydride of aluminum is a polymeric solid. The halides (except BF,)
hydrolyze to form boric acid and silicic acid. The oxygen chemistry of the borates and
silicates also has certain resemblances.
Carbon-Phosphorus, Nitrogen-Sulfur, and Oxygen-Chlorine
All metallic properties have been lost in these elements, and so charge-to-size ratios
have little meaning. However, the same effects appear in the electronegativities of
these elements, which show a strong diagonal effect: 6
6 These values are Pauling thcrmochcmical electronegativities rather than those based on ionization
energy-electron affinity. This choice of empirical values was made to obviate the necessity of
choosing (arbitrarily) the proper valence state (hybridization).
Carbon-Silicon
Similarities and
Contrasts
The Use of p Orbitals in Pi Bonding 861
C = 2.55 x N = 3.04 0 = 3.44 v F = 3.98
Si =1.90 'P = 2.19 S = 2.58 '' Cl = 3.16
The similarities in electronegativities are not so close as that of the ionic potentials for
Be 2+ and AI J + . The heavier element in the diagonal pair always has a lower elec¬
tronegativity. but the effect is still noticeable. Thus when considering elements that
resemble carbon, phosphorus is often as good a choice as silicon, and the resemblance
is sufficient to establish a base from which notable differences can be formulated. 7
In view of the extensive chemistry of alkenes it was only natural for organic and
inorganic chemists to search for analogous Si=Si doubly bonded structures. For a
long time such attempts proved to be fruitless. The first stable C=Si x and Si=Si 9
compounds were synthesized about a decade ago. One synthesis involves the rear¬
rangement of cyclotrisilane:
It is possible to add reagants across the Si=Si double bond in some ways analogous to
the C=C bond in alkenes:
1 The diagonal relationship, like any other rule-of-thumb, can be seen from several viewpoints: in
terms of unifying known facts (when it works), as a predictor of unknown properties (hoping it
works), or in terms of the significance of its exceptions (when it does not work). Sec, for example,
Fcinstcin. H. I. J. Own,. Educ. 1984, 61, 128. Hanusa. T. P. J. Chem. Educ. 1987. 64, 686-687.
Brook, A. G.; Abdesakcn. F.; Gutckunst, B.; Gutekunst, G.; Kallury, R. K. J. Cliem, Soc., Chem.
Commun. 1981, 191-192.
■'West. R.: Fink. M. J.; Michl. J. Science 1981, 214. 1343-1344. Masamune. S.; Hanzawa. Y.;
Murakami, S.; Bally, T.; Blount. J. F. J. Am. Chem. Soc. 1982, 104, 1150-1153. West, R. Angew.
Chem. Int. Ed. Engl. 1987,26, 1201.
862
1 8 • Periodicity
Compounds that are formally analogous to carbon compounds arc found to have
quite different structures. Thus carbon dioxide is a gaseous monomer, but silicon
dioxide is an infinite single-bonded polymer. In a similar manner, .gem-diols are
unstable relative to ketones:
(CH 3 ) 2 C(OH) 2 -♦ CH 3 C(0)CHj + H,0 (18-13)
and the analogous silicon compounds are also unstable, but the “dimethylsilicone” 10
that forms is a linear polymer:
CHj CHj CHj
211,0 -H.O
(CHj),SiCI 2 —=—► [(CH 3 > 2 Si(OH) 2 | —=— —O —Si—O —Si — 0 —Si— (18.14)
CH, CHj CHj
111 The term "silicone" was coined by analogy to ketone under the mistaken belief that monomeric
R ; Si=0 compounds could be isolated. See Chapter 16.
The Use of p Orbitals in Pi Bonding
863
The contrast between the strengths of bonds and their higher-n con¬
geners is responsible for much of the stability of groups important to organic
chemistry: alkenes, aldehydes, ketones, and nitriles. It also permits doubly bonded
molecules such as carbonic and nitric acids, rather than their ortho analogues. A
source of the greater stability of tt bonds between the smaller atoms could be better
overlap of the 2 p orbitals. The overlap integral f<p A <p B (see Chapter 5) is only poorly
depicted by a drawing such as Fig. 18.1. The overlap is strongly affected by the
magnitude of the wave function in the overlap region and, especially for rr bonds, is
increased by small, "dense" orbitals. The first time a given type of orbital (2p. 3d. 4/)
appears, it is nodeless and anomalously small. The small size results from the absence
of inner shells having the same value of / against which this set of orbitals must be
orthogonal." The 2 p orbitals thus are as small as the 2s orbital, in contrast to the 3 p
orbitals which are larger and more diffuse than the 3.v orbitals.'-
For the heavier congeners in Group IVA (14), the differences are even more
striking. Thus, although carbon is generally tetravalent except as transient carbene or
methylene intermediates, it is possible to prepare divalent germanium, tin. and lead
compounds. For example, if bulky substituents [R = CH(SiMe 3 ) : ] are present, the
compounds R,Ge. R 2 Sn. and R : Pb exist as diamagnetic monomers in solution, al¬
though there is a tendency for them to dimerize in the solid. The molecular structure of
the tin dimer has been determined and found to be in the trans conformation:
R
R \
Sn—Sn.
R
In addition to being bent, in contrast to ethylene, the Cie—Ge and Sn—Sn bonds are
not as short as expected for true double bonds. 1 - 1 Calculations indicate that p„-p„
bonding is less important and other interactions may become increasingly important.
Calculated bond orders are Ge—Ge = 1.61 and Sn—Sn = 1.46. 14
Fig. 18.1 Diagrammatic
representation of the
possibly poorer overlap of
the p orbitals in Si—Si as
compared with C—C.
11 Pyykkfl. P. Client. Rev. 1988. XV. 563-594.
'-Walsh. R. Act. Chon. Res. 1981. Id. 246. West, R. Angov. Cliem. Ini. Ed. Engl. 1987. 26.
1201 - 1211 .
11 Davidson. P. J.; Harris. D. H.: Lapped. M. F. J. Chon. Soc. Dollon Trans. 1976. 2268-2274.
Cowley. A. Ha Norman. N. C. Prog. Inorg. Chon. 1986, Jd. 1-63.
14 Lendvay. G. Chon. Pliys. l.oi. 1991, INI . 88-94. See also data on bond energies on page 865 and
Grev. R. S. Adv. Orgunonwl. Chon. 1991, JJ. 125-170.
A-12
C-Atomic States and Term Symbols
Table C.l _
Terms o( various electron
configurations
Equivalent electrons
s 2 , p 6 , and d'°
'5
p and p s
2 P
p 2 and p 4
iP, 'D. '5
P l
4 5, 2 D. 2 P
d and d?
2 D
d 2 and d 8
3 F, 2 P, 'G, 'D, >5
d 3 and d 1
4 F. 4 F, 2 H, 2 G, 2 F, 2 D, 2 D, 2 P
d and d 6
5 D, 3 W, 3 G, 3 F, 3 F, 3 D, 2 P, iP, ■/, 'G, 'G, 'F, 'D, ' D, <S, 'S
d
4 G, 4 F, 4 D, 4 F, 2 I, 2 H, 2 G, 2 G, 2 F, 2 F, 2 D, 2 D. 2 D, 2 P, 2 S
Nonequivalcnt electrons
s s
'5’, iS
S P
'P, iP
s d
'D, iD
P P
iD, 'D, iP, 'P, 2 S, 'S
P d
3 F, 'F, 2 D, ' D, iP, 'P
d d
3 G, 'G, 3 F, 'F, iD, 'D, iP, 'P, iS, '5
s s s
4 5, 25, 2 S
s s p
P, 2 P, 2 P
s p p
4 D, 2 D, 2 D, P, 2 P, 2 P, 2 P, 4 5, 25, 25
s p d
F, 2 F, 2 F, D, 2 D, 2 D, 4 P, 2 P, 2 P
Hund's Rules
The ground state of an atom may be chosen by application of Hund's rules. Hund’s
first rule is that of maximum multiplicity. It states that the ground state will be that
having the largest value of 5, in the case of carbon the 3 F. Such a system having a
maximum number of parallel spins will be stabilized by the exchange energy resulting
from their more favorable spatial distribution compared with that of paired electrons
(see Pauli principle, Chapter 2).
The second rule states that if two states have the same multiplicity, the one with
the higher value of L will lie lower in energy. Thus the 'D lies lower in energy than the
15. 5 The greater stability of states in which electrons are coupled to produce maximum
angular momentum is also related to the spatial distribution and movement of the
electrons.
5 Hund’s rules arc inviolate in predicting the correct ground state of an atom. There are occasional
exceptions when the rules are used to predict the ordering of excited states.
UJ cf cf I -c
Appendix
864
18- Periodicity
Nitrogen-
phosphorus
Analogies and
Contrasts
The stable form of nitrogen at room temperature is N 2 , which has an extraordinarily
strong (946 kJ mol -1 ) triple bond. In contrast, white phosphorus consists of P 4
molecules (see Chapter 16), and the thermodynamically stable form is black phos¬
phorus, a polymer. At temperatures above 800 °C dissociation to P 2 molecules does
take place, but these are considerably less stable than N 2 with a bond energy of 488 kJ
mol -1 . In this case, too, in the heavier element several single bonds are more effective
than the multiple bond.
The phosphorus analogue of hydrogen cyanide can be prepared:
CH 4 + PH 3 HC=P + 3H 2 (18-15)
In contrast to stable hydrogen cyanide, HCP is a highly pyrophoric gas which poly¬
merizes above - 130 °C. In this decade the number of molecules containing C=P
bonds has increased to over a dozen. 15 One method of obtaining them is by
dehydrohalogenation:
CH 3 PCI 2 HC=P 08.16)
CF 3 PH 2 CF 2 =PH FC=P (18.17)
Kinetically stable phosphaalkynes can be synthesized if a sufficiently bulky
substituent (R) is present: 16
P 4 + 12Na/K + 12ClSiMe 3 -► 4P(SiMe 3 ) 3 + 12(Na/K)CI (18.18)
R-C ♦ P(SiMcj)j
Cl
McjSiO / 3 McjSiO Q
R >=\
One of the first challenges facing chemists attempting to prepare phosphorus analo¬
gues of nitrogen compounds was phosphabenzene. First the phosphorus analogue of
pyridine was synthesized, and now all of the group VA (15) analogues of pyridine have
been prepared.
However, these compounds must be considered the exception rather than the rule as
far as the heavier elements are concerned.
15 Regitz, M.; Bingcr, P. Angew. Client. Ini. Ed. Engl. 1988. 27, 1484-1508. Rcgitz. M. Client. Rev.
1990. 90. 191. Maah. M. H.; Nixon. J. F. In The Chemistry of Organophosphorus Compounds'.
Hartley. F. R.. Ed.; John Wiley: New York. 1990: Chapter 9.
Note that since a phosphaalkyne (or any other triply bonded group) will be linear and “exposed."
the protection afforded by [a] bulky group[s] is less than in the corresponding case of doubly bonded
species.
The Use of p Orbitals in Pi Bonding
865
The isolation of compounds containing simple C=P double bonds parallels the
triple-bond work. The first stable acyclic phosphaalkene was synthesized over fifteen
years ago. 17 Again, base-induced dehydrohalogenation and stabilization by bulky
groups is important:
R- R" R- R"
H — C — P _ b H a > C = P (18.20)
/ \ /
R' Cl R'
The steric hindrance is critical: If R = phenyl or 2-methylphenyl, the bulkiness is
insufficient to stabilize the molecules, but the 2,6-dimethylphenyl and 2,4,6-lrimethyl-
phenyl derivatives are stable.
Summary on the For many years the occurrence of double and triple bonds such as discussed above for
Occurrence of silicon and phosphorus was equally rare among other nonmetals, leading to the
p -p_ Bonding in conclusion that only C=C, C=N. C=0. N==N. etc. were stable multiple bonds.
Heavier Nonmetals This, of course, was taken as challenge and much synthetic work was directed at the
problem. None of the multiple bonds between heavier nonmetals is as strong as those
between the 2 p elements. Some typical estimates of the strength of the n bond (cf. to
H 2 C=CH 2 as a “standard” from organic chemistry) are (values in kJ mol -1 ):
C=C 272
C=Si 159 Si=Si 105
C=Ge 130 Si=Ge 105 Ge=Ge 105
C=Sn 79
It now appears that any X=Y double bond can be prepared, given an energetic
enough research attack: Hundreds of these compounds have now been synthesized.
The general method has been to involve bulky substituents. In this way the multiple
bond chemistry of the heavier nonmetals has resembled attempts to make low-
coordination-number complexes (Chapter 12).
The number of triple bonds of the heavier nonmetals that arc known is consider¬
ably smaller—perhaps a dozen. Ii has already been noted above that protecting a
triple bond stcrically is considerably more difficult than for the case of a corresponding
double bond. One very interesting aspect of the C=S bond is that, in contrast to the
C=C in acetylenes, the triple bond does not ensure linearity at the carbon atom (Fig.
18.2). The reasons are not completely clear but may be related to the nonplanarity of
R 2 Ge=GeR 2 and R 2 Sn=SnR 2 (see page 863).
The successful isolation of all of these compounds is more a tribute to the
persistence with which they were pursued than to any inherent stability of the bonds
themselves. To invert George Leigh Mallory's remark about Ml. Everest, the extraor¬
dinary efforts expended on this class of compounds stemmed from the fact that they
were not there. These efforts and their corresponding successes have caused one
observer to comment: “Finding exceptions to the double-bond rule has become a
17 Becker, G. Z. Anorg. Chem. 1976, 423. 247. Cowley, A. H.; Jones, R. A.; Lasch, J. G.; Norman,
N. C.; Stewart, C. A.: Stuart. A. L.; Atwood, J. L.; Hunter. W. E.; Zhang. H.-M. J. Am. Chem.
Soc. 1984. 106. 7015-7020.
A-10
C-Atomic States and Term Symbol:
f
,l/j= 0 0 0 +1 +! +1 -I -I -1 0 0 0 0 0 0
Fig. C.l The fifteen microstates and resultant values of M L and M s for the electron
configuration of carbon.
values are entered directly into the table, 3 but if the electron configurations
have been carefully worked out, there is no need of this. The fifteen micro¬
states of p 2 yield:
Ms
_ +1 0 -1
-2 x
I X XX X
M l Ox xxx x
+ 1 x xx x
2 x
Resolve the chart of microstates into appropriate atomic states. An atomic
state forms an array of microstates consisting of 2S + 1 columns and 2 L + I
rows. For example, a i P state requires a 3 x 3 array of microstates. A 'D state
requires a single column of 5 and a 5 D requires a 5 x 5 array, etc. Looking at
the arrays of microstates, it is easy to spot the unique third microstate at M L =
0 and M s = 0; this must be a '5. A central column of M s - 0 provides a '£>.
Removing these two states from the table, one is left with an obvious 3x3
array of a 3 F state.
3 To ensure Chat all microstatcs have been writlen. Che total number N. of microstatcs associated with
an electronic configuration, l x , having .r electrons in an orbital set with an azimuthal quantum
number, /, is
" j! (N, - .c)l
where N t = 2 (2/ + I), the number of m,. m, combinations for a single electron in the orbital set.
(From Condon. E. U.; Shortley. G. H. The Theory of Atomic Spectra: Cambridge University:
Cambridge, 1963.]
C • Atomic States and Term Symbol:
A-l 1
Fig. C.2 Term splitting in the ground-state i\s'-2s'-2p'-) configuration of carbon. All energies
are in cm . The ’/>, <D. and '5 terms are split as a result of electron-electron repulsion. The
•R term is further split with J = 0. I. 2 as a result of spin-orbit coupling. The scale of the
latter .s exaggerated in this figure. [From DeKock. R. L.; Gray. H. 13, Chemical Structure
and Bonding: Benjamin/Cummings: Menlo Park. CA. 1980. Reproduced with permission.]
The states of carbon are therefore KS. 'D. >P.-> The >P is further split by
differing 7 values to the terms >P () . 3 P„ and 3 />,. The relative magnitudes of
these splittings can be seen in Fig. C.2.
Although the complexity ol determining the appropriate terms increases with the
number of electrons and with higher L values, the method outlined above (known as
Russell-Saunders coupling) may be applied to atoms with more electrons than the
carbon atom in the foregoing example. States for various electron configurations are
shown in Table C.l. Russell-Saunders coupling (also called LS coupling because it
assumes that the individual values of / and s couple to form L and S, respectively) is
normally adequate, especially for lighter atoms. For heavier atoms with higher nuclear
charges, coupling occurs between the spin and orbit for each electron. (J = / + s) The
resultant coupling is known as Jj coupling. In general, LS coupling is usually assumed
and deviations are discussed in terms of the effects of spin-orbital interactions (see
Chapters I! and 18).
4 For alternative approaches and discussions, see Kiremirc. E. M. R. J. Client Ednc 1987 M
951-953; the inorganic textbooks listed on pp A-l and A-2. and the references therein.
866 18- Periodicity
Fig. 18.2 Molecular
structures of (a)
CFjC=SF, and (b)
SF 6 C=SF 3 . Crystal
structures are above and
gas structures below. Note
strong bending in the gas
phase. (From Seppelt. K.
Angew. Chem. hit. Eel.
Engl. 1991. 30. 361-374.
Reproduced with
permission.]
sport in main-group chemistry," 18 but Seppelt. one of the successful synthesizers,
also notes: "In spite of all of the remarkable success with the synthesis of such
compounds, the fact remains that these double [and triple] bonds still form, in the final
analysis, more unfavorable bonding systems than those of elements of the second
period." 19
The Use (or Not)
of d Orbitals
by Nonmetais
Theoretical
Arguments against d
Orbital Participation
in Nonmetals
Several workers have objected to the inclusion of d orbitals in bonding in nonmetals.
The principal objection is to the large promotion energy required to effect
s'~p"d" -► s'p"~”'d"' +l (18.21)
where m = 0 (P), I (S). or 2 (Cl), to achieve a maximum multiplicity and availability of
electrons for bonding. A second factor which does not favor the utilization of d
orbitals is the poor overlap that they make with the orbitals of neighboring atoms. The
3c/ orbitals of the free sulfur atoms, for example, are shielded completely by the lower-
lying electrons and hence do not feel the nuclear charge as much as the 3,v and 3 p
electrons. As a result they are extremely diffuse, having radial distribution maxima at
a distance which is approximately twice a typical bond distance (Fig. 18.3). This
results in extremely poor overlap and weak bonding. 20
Two alternatives have been suggested to account for the higher oxidation states of
the nonmetals; both reduce the importance of high-energy d orbitals. Pauling has
suggested that resonance of the following type could take place:
-n
n
Cl Cl
I /
+/
Cl — P <—
— Cl—P
1 \
1 \
1 Cl
I Cl
Cl
Cl
(I) (II)
Four more forms
like (II)
(18.22)
Editor, Angew. Chem. Ini. Ed. Engl. 1991 .30. A-69. The double bond rule can be staled as follows:
Elements having a principal quantum number greater than two are not likely to form p n -p„ bonds.
19 Seppelt, K. Angew. Chem. hit. Ed. Engl. 1991, 30. 361-364. For recent reviews of some of these
multiply bonded systems, see Nicckc, E.t Gudat. D. Angew. Cliem. Int. Ed. Engl. 1991. 30.
217-237: Tsumuraya. T.; Batchellcr. S. A.: Masamunc, S. I hid. 1991. 30. 902-930: Barrau. J.:
Eseudie. J,; SatgrS, J. Chem. Rev. 1990, 90. 283; and references to earlier work therein. See also
Footnote 15.
•° These same general arguments apply to all of the heavier nonmetais. The d and/orbitals are heavily
shielded by the more penetrating s and p electrons.
The Use (or No!) of d Orbitals by Nonmetais 867
Fig. 18.3 The 3d orbital distribution functions in d 1 configurations (A) in the l 'D lerm of P
(s'/rd '); (B) in the ’D term of S (,v 2 p'd'). Line represents a typical S—F bond length.
(Modified from Mitchell. K. A. R. Cliem. Rev. 1969, 69, 157. Reproduced with permission.)
Only structure I involves d orbitals, and so the d character of the total hybrid is small.
Each P—Cl bond has 20% ionic character and 80% covalent character from resonance
structures such as II. Pauling has termed ihe "extra" bonds formed (over and above
the four in a noble gas octet or "argononic" structure) as "transargononic" bonds and
pointed out that they tend to be weaker than "normal" or "argononic" bonds and
form only with the most electronegative ligands. Thus the average bond energy in PCI,
is 326 kJ mol 1 . but in PCI, it is only 270 kJ mol~'. The same effect is found in PI-, and
I’F,. but in this case ihe difference in bond energy is only 40 kJ mol -1 , corresponding
to the stabilization of the structure by increased importance of the ionic structures in
the fluorides. The stabilization of these structures by differences in electronegativity is
exemplified by the tendency lo form Ihe higher halogen fluorides. The enthalpies of
fluorination of ihe halogen monofluorides are:
CIF(g) + 2F,(g) -♦ Cll-Vg)
Ml =
152.7 kJ mol -1
(18.23)
BrFlg) r 2F : (g)- BrFdg)
Ml =
-376.1 kJ mol -1
(18.24)
IF(g) 4- 2F,(g) - IF,(g>
Ml =
-751.4 kJ mol"'
(18.25)
The second alternative is the three-center, four-electron bond developed by
simple molecular orbital theory for the noble gas fluorides (see Chapter 17). Since this
predicts that each bonding pair of electrons leach "bond") is spread over three nuclei,
the bond between two of the nuclei is less than that of a normal two-center, two-
electron bond. Furthermore, since ihe nonbonding pair of electrons is localized on the
fluorine atoms, there is a separation of charge ("ionic character"). In both respects,
then, this interpretation agrees with Pauling’s approach and with the experimental
facts.
In the case of nitrogen, experimental work indicates that pentacoordinate 21 and
'■< Christo. K. O.; Wilson. W. W.; Sehrobilgen. G. J.; Chirakul. R. V.: Oltih, G. A. Inorg. Chem. 1988,
27. 789-790. In the present contest, "pentti- (or licxa-) coordinate nitrogen (carbon)" means a
nitrogen (carbon) atom with five (six) atoms bonded lo it with more or less localized sigma bonds,
such as in ihe hypothetical NF,. On the other hand, in the carborane. and azaborane.
NBiiHij. cage compounds (see Chapter 16). Ihe carbon and nitrogen atoms are formally attached to
six other nearest neighbor atoms, as arc the isoclcctronic boron atoms in the ieosahedral boranes.
Likewise, note "hcxavalcnl carbon" in certain cluster compounds (Chapter 15) and in the hedgehog
gold compounds (see page 885).
A-8
C- Atomic States and Term Symbol
C- Atomic States and Term Symbols
A-9
For a given value of 5, there will be 25 + I spin states characterized by M s :
M s = 5, 5 - 1, 5 - 2__ -5 (C.6)
or
M s = m si + in „ + ■ • • + m Sn (C.7)
Now the total angular momentum of an electron is the resultant of the orbital
angular momentum vector and the electron-spin angular momentum vector. Both of
these are quantized, and we can define a new quantum number, j:
j = l + s (C.8)
Since s = ± J, it is obvious that every value of / will have two values of / equal to
/ + •} and / - i. The only exception is I = 0, for which j = ±i; these values are
identical since it is the absolute magnitude of j that determines the angular momentum.
We can now couple the resultant orbital angular momentum (L) with the spin
angular momentum (5). The new quantum number J is obtained:
J = L + S. L + S - I, L + 5 - 2. \L - 5| (C.9)
The origin of the J values can be seen from a pictorial representation of the vectors
involved."
L = 2
5= I
J = I
The number of J values available when L > 5 will be equal to 25 + I. and is
termed the multiplicity of the state. In both of the examples pictured above, the
multiplicity is three. The multiplicity is appended to the upper left of the symbol of
state and J to the lower right. The above examples are thus i P and D states
(pronounced •‘triplet P" and "triplet D"). The individual terms are -V 5 ,. and i P„
(left), and -’D,, 3 D,, and i D 0 (right).
When L < 5, the series in Eq. C.9 is truncated (note the absolute magnitude
symbol in the last term) and there are only 2 L + I values. An example is the /is'
configuration, where L = 0 and 5 = J can have only the single value of + i.2
Despite the fact there is but a single value of J for the ground state of hydrogen, spectroscopists write
its term symbol as 3 5| /2 . The reason for this is that transitions between states of different spin
multiplicities are spin forbidden: thus transitions from a spin-paired singlet (such as •S) to a spin-
unpaired triplet (-'5) are not allowed. However, with hydrogen (and the alkali metals as well) the
ground state has an unpaired electron, and transitions to doublet states with a single unpaired
electron are allowed. For ease in noting which transitions are allowed and which are spin forbidden,
spectroscopists write 2 S. though admittedly it can lead to confusion.
To turn again momentarily from the abstractions of orbitals and quantum numbers
back to the spectra that generated them, consider the transition from a \s orbital in
hydrogen to a 2 p orbital. The terms and transitions are:
The 3 P in term lies slightly higher in energy than the 2 P V2 , and therefore the spectral
line is split into a ••doublet"; hence the origin of the usage. It may be noted that
in respect to these transitions the following selection rules operate: A n arbitrary
A/ = ± I; Aj = ± |, 0.
Assigning Term Symbols
We have seen that the term symbol for the ground state of the hydrogen atom is =5„,
For a helium atom L = 0. 5 = 0. J = 0, and the term symbol for the ground state is
5 () . For an atom such as boron, we can make use of the fact that all closed shells and
subshells (such as the He example just given) contribute nothing to the term symbol.
Hence both the [j and 2 s 2 electrons give L = 5 = J = 0. The 2pi electron has L = I,
5 - j. and J = I ± J, yielding -P U2 and 2 P V2 . For carbon there are two p electrons.
The spins may be paired or unpaired, so L = 2, I, 0; 5 = 1, 0; and J = 3, 2 I 0 To
work out the appropriate states for this atom requires a systematic approach.’Note
however, that when neon is reached we have again a '5 0 : sodium repeats the 25 ,,,
magnesium '5 0 . etc. 1/2
A Systematic Approach to Term Symbols
In Chapter 2 it was shown how m, and m s values could be summed to give M. and A-/.,
values to yield terms for the spectroscopic states of an atom. If there arc two or more
electrons, it is usually necessary to proceed in a systematic fashion in generating these
terms. The following is one method of doing so. The p 2 configuration of carbon is
Determine the possible values of M,_ and M s . For the p 2 configuration, L can
have a maximum value of 2 and M, can have values of -2, -1,0, +1, +2.
The electrons can be paired [M s = 0) or parallel {M s = + 1 , - l).
Determine the electron configurations that are allowed by the Pauli princi¬
ple. The easiest way to do this is to draw up a number of sets of p orbitals as
in Fig. C.l (each vertical column represents a set of three/; orbitals) and fill in
electrons until all possible arrangements have been found. The M, value for
each arrangement can be found by summing m, and M s from the sum of m
(spin-up electrons have arbitrarily been assigned = +$). Each microstate
consists of one combination of M L and M s .
Set up a chart of microstates. For example, the microstate corresponding to
the first vertical column in Fig. C.l has M L = +2 and M s = 0. It is then
entered into the table below under those values. Sometimes the m, and m s
868
18- Periodicity
hexacoordinate nitrogen do not occur. On the other hand, a theoretical case has been
made for the possibility of pentacoordinate nitrogen in a molecule such as NF 5 . 22
Experimental
Evidence for d^-p^
Bonding; the
Phosphorus—Oxygen
Bond in Phosphoryl
Compounds
In the case of d„-p n bonding we again find the old problem of detecting the existence
of a bond. We can infer the presence of a a bond when we find two atoms at distances
considerably shorter than the sum of their van der Waals radii. The detection of a ir
bond depends on more subtle criteria: shortening or strengthening of a bond, stabiliza¬
tion of a charge distribution, etc., experimental data which may be equivocal.
One example of the apparent existence of 7r bonding is in phosphine oxides. Most
tertiary phosphines are unstable relative to oxidation to the phosphine oxide:
2R 3 P + 0 2 - 2R 3 PO (18-26)
This reaction takes place so readily that aliphatic phosphines must be protected from
atmospheric oxygen. The triarylphosphines are more stable in this regard but still can
be oxidized readily:
In contrast, aliphatic amines do not have to be protected from the atmosphere
although they can be oxidized:
R 3 N + HOOH -♦ [R 3 NOH] + OH- R 3 NO (18.28)
However, the amine oxides decompose upon heating:
Et 3 NO Et 2 NOH + CH 2 =CH 2 (18.29)
a reaction completely unknown for the phosphine oxides, which are thermally stable.
In fact, the tertiary phosphine oxides form the most stable class of organophosphorus
compounds. Those oxides with no p hydrogen atom are particularly stable: Tri-
methylphosphine oxide and triphenylphosphine oxide do not decompose below
700 °C. 23 They are not reduced even by heating with metallic sodium. The tendency of
phosphorus to form F->0 or P=0 linkages 24 is one of the driving forces of phos¬
phorus chemistry and may be used to rationalize and predict reactions and structures.
For example, the lower phosphorus acids exist in the four-coordinate structures even
though they are prepared by the hydrolysis of three-coordinate halides:
P
// \
X x x
0
P —H
HO 6h
R / \ X X-^—» R y / >H
X OH
(18.30)
(18.31)
22 Ewig, C. S.; Van Wazer. J. R. J. Am. Client. Soc. 1989, III. 4172-4178.
25 Corbriilgc. D. E. C. Phosphorus. 4th cd.; Elsevier: Amsterdam, 1990; p 320.
24 Whether the P—O bond is essentially a single, a. dative bond. P->0, or has at least some d„-p„.
double-bond character is. of course, the argument here, and to portray the following structures with
cither P—O or P=0 tends to anticipate the question unintentionally: See the resonance forms in
Eq. 18.36.
The Use (or Not) of d Orbitals by Nonmetals
869
,'hV
x
OH
-P — P —OH
O O
H—P—P—H
HO OH
r r -
0
0
RO —P —R'
X- -♦ RO—P
O
0
L R J
R
The tendency to form P=0 bonds is responsible for the Arbusov reaction. The
typical reaction is the rearrangement of a trialkyl phosphite to a phosphonate:
O
kx n
(RO)jP (RO),PR (18.33)
It the catalytic amounts of RX in Eq. 18.33 are replaced by equimolar amounts of R'X,
the role of the alkyl halide in the formation of an alkoxy phosphonium salt is revealed:
(RO),P + R'X
Oxidation of trialkyl phosphites by halogens illustrates the same principle:
(RO) 3 P + Cl, -♦ [(RO) 3 PCI] + Cl- - (R0) 2 P(0)C1 + RCI (18.35)
A final difference between amine oxides and phosphine oxides lies in the polarity
of the molecules. The dipole moment of trimethylamine oxide is 16.7 x I0 -30 C m
(5.02 D) compared with 14.6 x IQ- 30 C m (4.37 D) for triethylphosphine oxide. A
consequence of this polarity is the tendency of the amine oxides to form hydrates,
RjNO-HiO, and their greater basicity relative to the phosphine oxides.
The difference between the behavior of the amine oxides and phosphine oxides
can be rationalized in terms of the possibility of back bonding in the latter. Whereas
amine oxides are restricted to a single structure containing a dative N—O bond.
RjN—O. the phosphine oxides can have contributions from cl n -p„ bonding between
the phosphorus and oxygen atoms:
R 3 P > 0" <- RjP= O (,8.36)
(1) (II)
The double bond character introduced by the latter strengthens the bond and accounts
for the extraordinary stability of the phosphorus oxygen linkage. Note that this extra
stability cannot be attributed to ionic resonance energy (a priori a reasonable sug¬
gestion since the difference in electronegativity is greater in P—O than N—O)
because the dipole moment of the nitrogen compound is greater than that of the
phosphorus compound, a result completely unexpected on the basis of elec¬
tronegativities, unless consideration is taken of canonical form 18.36(11), which would
be expected to lead to a reduced moment.
A comparison of the bond energies also supports the above interpretation. The
dissociation energies of P=0 bonds in a variety of compounds lie in the range of
500-600 kJ mol -1 compared with values for N->0 of about 200-300 kJ mol -1 . The
value for the latter is typical of what we might expect for a single bond, but 600 kJ
A-6
B-Units and Conversion Factors
SI units in their education—they may wonder what all of the fuss is about. They will
know as soon as they stop reading textbooks and start reading the original literature!
Length, molecular dimensions. As mentioned above, this surely is the easiest
conversion to make. Bond lengths in picometers are exactly 100 times greater than
when expressed in angstroms.
Energies. Ionization energies expressed in kJ mol -1 are approximately (31% error)
100 times greater than when expressed in electron volts. Bond energies in kJ mol -1 are
approximately (4% error) four times their values in kcal mol -1 .
The calculation of lattice energies (and other Coulomb's law energies) is compli¬
cated somewhat by the fact that in SI the permittivity (dielectric constant) of a vacuum
is no longer defined as one but has an experimentally determined value. Furthermore,
for reasons we need not explore at present. Coulomb's law is stated in the form:
The calculation may be simplified if the values for e 0 anc n are included in the
conversion factor, 1.389 x I0 5 kJ mol -1 pm (the reader should confirm this value),
which allows direct calculation of the lattice energy using ionic charges and distances
in picometers.
Dipole moments. The differences between the two systems are such that there
appears to be no simple correlation. Nevertheless, since most SI tables will probably
list values as coefficients of IO -30 the following mnemonic and rule of thumb should
help: To get SI values from Debyes, c/ivide by 0.3.
Pressure. Fortunately, an atmosphere is almost (1% error) 10 s Pa. So a standard
atmosphere is about 100 kPa, and when high pressure experiments are presented in
terms of GPa, each gigapascal represents about 10 4 atmospheres.
.
Atomic States and Term Symbols 1
The energy of a spectral transition for the hydrogen atom is given by the Rydberg
formula:
= 109,737 cm 1 - 7 ) where tu > n.
\//v nj/
which consists of two terms. It is common for spcctroscopists to apply the word term
to the energies associated with the states of an atom involved in a transition. Term
symbols are an abbreviated description of the energy, angular momentum, and spin
multiplicity of an atom in a particular state. Although the inorganic chemist generates
the term symbols used from knowledge of atomic orbitals, the historical process was
the reverse: S. P, D. and F states were observed spectroscopically, and named after
sharp (5 — P ), principal (P -» S ), diffuse ( D —> P). and fundamental (F —> D)
characteristics of the spectra. Later the symbols s, p, d, and/were applied to orbitals.
Atoms in 5, P. D, F, . . . states have the same orbital angular momentum as a
hydrogen atom with its single electron in an ,v. p, d.f, . . . orbital. Thus we can define
a quantum number L. which has the same relationship to the atomic stale as / has to an
atomic orbital (e.g.. L = 2 describes a D state). L is given by:
L = /1 + l->, li + L ~ I, /1 + U — 2,
We can also define the component of the total angular momentum along a given axis:
M l = L. L - I, L - 2.0. -L (C.3)
The number of possible values of M L is given by 2 L + 1. M, is also given by:
M l = m tl + m t , + • • • + m, n (C.4)
Likewise we can define an atomic spin quantum number representing the total spin:
S ='Zs, (C.S)
1 For an extensive discussion of terms, symbols, and states, see Gcrloch. M. Orbitals, Terms, and
Slates; Wiley: New York, 1986.
870
18- Periodicity
moP 1 is stronger than any known single bond (see Chapter 8). A closer examination of
the strengths of various P=0 bonds in terms of infrared stretching frequencies shows
some interesting trends. For a series of similar molecules, such as the phosphine
oxides, the stretching frequency provides an indication of the strength of the bond
(Table 18.2)25 The highest stretching frequency among the phosphoryl compounds is
that of F 3 PO, and the lowest of the halides is that of Br 3 PO (the iodo compound is
unknown). When the stretching frequencies are plotted as a function of the sum of the
electronegativities of the substituents, a straight line is obtained:
v PO = 930 + 402 (18.37)
where is the Pauling electronegativity of a substituent atom or group on phosphorus.
The correlation between the electronegativity of substituent groups and the
strength of the P=0 bond provides support for a tr-bonding model but not for the
alternative dative o--only model. A u bond might be expected to be destabilized as
electron density is removed from the phosphorus, requiring it to withdraw electrons
from the P->0 bond, weakening it. In contrast, if the oxygen can back bond to the
phosphorus through a d-p tr bond, the induced charge on the phosphorus can be
diminished and the P=0 bond strengthened.
The bond lengths in phosphoryl compounds are in accord with the concept of
double-bond character. In the simplest case, that of P 4 O 10 , there are two P—O bond
lengths. There arc twelve relatively long ones (158 pm) within the cage framework
proper and four shorter ones (141 pm) between the phosphorus atoms and the oxygen
atoms external to the cage. It is interesting to note that the ratio or these two bond
lengths (0.89) is about the same as C=C to C—C or C—O to C—O.
Isoelectronic with the phosphine oxides are the phosphorus ylids, R,PCH,. As for
the oxides, two resonance forms
R,P — CH, <—
—► R,P=CH,
(I)
(II)
Table 18.2
Infrared stretching
Compound
•Vo (cm-')
2X
frequencies of some
F,PO
1404
11.70
phosphoryl compounds
FXIPO
1358
10.75
CljPO
1295
8.85
CUBrPO
1285
8.52
CIBr,PO
1275
8.19
Br,PO
1261
7.86
Ph,PO
1190
(7.2)
Me,PO
1176
(6.0)
" X - substituent electronegativity
Note that the dissociation energy, K,PO - R,P + O is not a sensitive measure of the P=0 bond
energy because the remaining three bonds may be strengthened or weakened in the dissociation
process. The IR stretching frequency is a function of the force constant, k, and the reduced mass. fi.
of the vibrating groups. If the molecule is assumed to be alight oxygen atom vibrating on a "fixed"
larger mass of the R,P group, the reduced mass is constant, and so changes in frequency will reflect
corresponding changes in the force constant. For similar molecules the force constant will be related
to the total bond energy.
A Comparison
of Pi Bonding
in Phosphine
Complexes and
Oxides
The Use (or Not) of d Orbitals by Nonmetals 871
contribute to the stability of the phosphorus ylids but not the corresponding am¬
monium ylids. RjN + —C - H 2 . This difference is reflected in the reactivity. The
ammonium ylids are generally quite basic and quite reactive; the phosphorus ylids are
much less so. many not being sufficiently basic to abstract a proton from water and, in
fact, not dissolving in water unless strong acids are present.
The controversy over the nature of the P=0 bond is reminiscent of that over the
nature of phosphorus-metal bonds in coordination compounds. In both, interpreta¬
tions have long ranged from a (r-only to a highly synergistic ct-tt model. As we have
seen in Chapters i I and 15, cr' orbitals have also been invoked in more recent
phosphorus n bonding arguments, inasmuch as d- it hybrids may be involved. 2> So
the question turns out not to be simply cr vs. -n but the relative contributions of d and
cr orbitals to the latter. As with so many questions in inorganic chemistry, the answer
is neither black nor white, but gray. If the symmetries and energies of orbitals are
compatible, bonding will occur. The appropriate question is one of relative
importance.
Evidence from
Bond Angles
The trimethylamine molecule has a pyramidal structure much like that of ammonia
with a CH,— N—CH, bond angle of 107.8° ± 1°. In contrast, the trisilylamine
molecule is planar. Although steric effects of the larger silyl groups might be expected
to open up the bond angles, it seems hardly possible that they could force the lone pair
out of a fourth "tetrahedral" orbital and make the molecule perfectly planar (even
Ph,N has bond angles of 116°). It seems more likely that the lone pair adopts a pure p
orbital on the nitrogen atom because orbitals on the three silicon atoms can overlap
with it and delocalize the lone pair over the entire system (Fig. 18.4).
(hi
Fig. 18.4 Delocalization of the lone pair in trisilyl amine, (a) Resonance .structures, (b)
Overlap of <4 and p N orbitals.
:fl Orpen. A. G.: Connelly. N. G. J. Client. Sue.. Client. Common. 1985, 1310-1311. Pacchioni. G.;
Bagus. P. S. Inorg. Chan. 1992. 31. 4391-1398.
A-4
B- Units and Conversion Factors
Prefixes
Prefix
Symbol
Multiply by
atto
a
10 -18
fempto
f
10 -13
pico
P
10 - ' 2
nano
n
10 -9
micro
M
I0 -5
milli
m
10 -3
centi
c
I0 -2
deci
d
10 - '
deka
da
10
hecto
h
I0 2
kilo
k
I0 3
mega
M
I0 6
giga
G
10 9
tera
T
I0 12
peta
P
I0 13
exa
E
10"
Physical and chemical constants"
Electronic charge
e = 1.60217733 x I0 -19 C
Planck constant
h = 6.6260755 x I0 -34 J s. 6.6260755 x 10 - 7 erg s
Speed of light
c = 2.997925458 x 10 8 m s 1
Rydberg constant
R = 1.0973731534 x I0 5 cm 1
Boltzmann constant
k = 1.380658 x 10" 23 J K" 1
Gas constant
R = 8.314570 J K -1 mol" 1
Avogadro number
N a = 6.0221367 x IO 23 mol 1
Faraday constant
9 = 9.6485309 x 10 4 C mol 1
Electronic rest mass
m c = 9.1093897 x 10 28 g
Proton mass
m = 1.672623 x I0 - - 4 g
Bohr radius u 0 = 52.91771249 pm
Bohr magneton \l b = 9.274096 x 10 ~ 4 A nr; 9.274096 x
IO -21 erg gauss -1
Permittivity of vacuum e„ = 8.854187817 x 10 ' C‘ m 1 J 1
Pi tr = 3.1415926536
Base, natural logarithms e = 2.71828
Conversion factors"
Multiply
by
to obtain
Length
A
cm
10 s
cm
I0 7
nm
cm
I0'°
pm
A
100
pm
Energy
kcal mol -1 4.184 kJ mol -1
eV 96.49 kJ mol -1
" Quantities, Units, and Symbols in Physical Chemistry. 1988; Mills. I. M.. Ed.; Blackwell
Scientific: Oxford, 1988.
" Quantities, Units, and Symbols in Physical Chemistry. 1988; Mills.
I. M., Ed.; Blackwell Scientific; Oxford. 1988.
B • Units and Conversion Factors
A-5
Conversion factors" ( Continued )
Multiply
by
to obtain
erg
I0 -7
J
wave numbers (cm -1
) 1.1962 )
< I0 -2
kJ mol“ 1
kJ mol 1
83.59
cm -1
eV
23.06
kcal mol -1
Dipole moments
Debye
3.336
X
IO -30
C m
C m
0.300
X
lO 30
D
Pressure
atmosphere
1.013
X
10 s
Pa
mm Hg (torr)
133.3
Pa
pascal
9.869
X
10 -6
atm
pascal
7.501
X
IO -3
mm Hg (torr)
Notes
SI units are obviously going to diplace older systems and units. We must all familiarize
ourselves with them. Just as obvious is the fact that this displacement is going to take
time, and the contemporary literature is going to exhibit a variety of units. Therefore,
this book has tried to take a middle course: To report all values in SI or Sl-derived
units but also to include frequently that same value expressed in "traditional" units.
We realize that by so doing, we run the risk of being “neither fish nor fowl," and of
alienating both the progressive, who would like to see 100% SI, and the conservative,
who would like to stick to the cgs system. Nevertheless, the decision must be based on
consideration of the students as future chemists who will encounter kilojoules, kilo¬
calories. electron volts, picometers. nanometers, angstroms, coulomb meters, and
Debyes; they had better have some familiarity with each of these units. The irony of it
all is that the easiest change to make and the one that causes no confusion whatso¬
ever is that of angstroms to picometers: I A = 100 pm. No one is apt to confuse
a bond length of Au—Au = 300 pm (in the unusual gold compounds of Chapter 18)
with Au—Au = 300 A. once having observed that molecular dimensions arc
measured in angstroms and hundreds of picometers. Yet this change seems to be the
last that will be made: Crystallographcrs hang on doggedly to the convenient units
of angstroms. On the other hand, kilojoules seem to be replacing kilocalories at a
reasonable rate. This means that old values, learned in passing, as kcal mol -1 , are
being supplemented or replaced by new values as they are reported in the literature,
sometimes kJ mol"'. but more often still kcal mol -1 . Recently, one of the authors was
asked in class about the Au—Au bond energy in the compound mentioned above. He
was sure that it was 30. but 30 what?(!) The factor of four between kilojoules and
kilocalories is just small enough to make the confusion of units easy and. simul¬
taneously. large enough to be disastrous! [Do you think that the Au—Au bond
mentioned above has a bond energy of 30 kJ mol -1 or 30 kcal mol -1 ? Make a
prediction and check it in Chapter 18. Does knowing that it is in the range of strong
hydrogen bonding energies help? Sometimes mnemonics like that help—they saved
the aforementioned author's skin!]
In view of this entropy that changing from one system to another entails, the
following suggestions are made to those familiar with older units and trying to get a
“feel" for SI units. Many readers of this book will have encountered essentially only
872
1 8 • Periodicity
The Use (or Not) of d Orbitals by Nonmetals
873
Pi Bonding in the
Heavier Congeners
Rather similar results are obtained by comparing the bond angles in the silyl and
methyl ethers (Fig. 18.5) and isothiocyanates (Fig. 18.6). In dimethyl ether the oxygen
is hybridized approximately sp 3 with two lone pairs on the oxygen atom as compared
to an approximate sp 2 hybrid in disiloxane with rr bonding. In the same way the methyl
isothiocyanate molecule. CH 3 N=C=S, has a lone pair localized on the nitrogen
atom, hence is bent (N ~ sp 2 ), but the delocalization of this lone pair into a v orbital
on the silicon atom of H 3 SiN=C=S leads to a linear structure for this molecule.
The hypothesized delocalization of lone pair electrons in the above silicon com¬
pounds is supported by the lowered basicity of the silyl compounds as compared to the
corresponding carbon compounds. This reduced basicity is contrary to that expected
on the basis of electronegativity effects operating through the a system since silicon is
less electronegative than carbon. It is consistent with an "internal Lewis acid-base
interaction between the nitrogen and oxygen lone pairs and empty acceptor d orbitals
on the silicon. Experimentally this reduced basicity is shown by the absence of
disiloxane adducts with BF 3 and BC! 3 :
(CH 3 ) 2 0 + BF 3 -♦ (CH 3 ) 2 O^BF 3 (18-39)
(SiH 3 ) 2 0 + BF 3 - No adduct (18.40)
and by the absence of trisilylammonium salts. Instead of onium salt formation tri-
silylamine is cleaved by hydrogen chloride:
(SiH 3 ) 3 N + 4HC1 -► NH 4 C1 + 3SiH 3 Cl (18.41)
In view of the uncertainty with which n bonding is known in the very well studied
phosphorus and sulfur systems, it is not surprising that little can be said concerning
the possibility of similar effects in arsenic, antimony, selenium, tellurium, etc. In
general it is thought that the problems faced in phosphorus and sulfur chemistry
concerning promotion energies and diffuse character may be even larger in the heavier
congeners. In the latter regard it is interesting to note the apparent effectiveness of it
bonding in metal complexes. To the extent that softness in a ligand can be equated
with the ability to accept electrons from soft metal ions in d w -d n "back bonds,
information can be obtained from the tendency to complex with ( b) metal ions (see
Chapter 9): P > As > Sb. This order would indicate that the smaller phosphorus atom
can more effectively tt bond with the metal atom.
Fig. 18.5 Comparison of
the molecular structures of
dimethyl ether and
disiloxane.
Fig. 18.6 Comparison of
the molecular structures of
methyl isothiocyanate and
silyl isothiocyanate.
Theoretical
Arguments in Favor
of d Orbital
Participation
Experimental
Evidence for d
Orbital Contraction
and Participation
In contrast to the arguments presented against participation by d orbitals in the
bonding of nonmetals, several workers have pointed out that the large promotion
energies and diffuse character described above are properties of an isolated sulfur or
phosphorus atom. What we need to know are the properties of a sulfur atom in a
molecule, such as SF 6 or PF 5 . This is an exceedingly difficult problem and cannot be
dealt with in detail here. However, we have seen how it is possible to calculate such
properties as electronegativity on isolated atoms as charge is added or withdrawn
(see Chapter 5) and how this might approximate such properties in a molecular
environment.
It is apparent from the preceding discussions that participation of d orbitals, if it
occurs at all, is found only in the nonmetals when in high oxidation states with
electronegative substituents. The partial charge induced on the central P or S atom
will be large merely from the electronegativity of the fluorine (as in PF 5 , SF 6 ) or
oxygen (as in OPX 3 , CLSX-,) irrespective of any bonding model (such as Pauling's or
the three-center bond) invoked.
We have seen in Chapter 2 that increasing effective nuclear charge makes the
energy levels of an atom approach more closely the degenerate levels of the hydrogen
atom. We might expect, in general, that increasing the effective nuclear charge on the
central atom as a result of inductive effects would result in the lowering of the d
orbitals more than the corresponding s and p orbitals since the former are initially
shielded more and hence will be more sensitive to changes in electron density. The
promotion energy would thus be lowered. A second effect of large partial charges on
the central atom will be a shrinking of the large, diffuse d orbitals into smaller, more
compact orbitals that will be more effective in overlapping neighboring atomic or¬
bitals. For example, sample calculations indicate that in SF h (he d orbitals have been
contracted to an extent that the radius of maximum probability is only 130 pm
compared with the large values of 300-400 pm in the free sulfur atom (Fig. 18.3).
One of the most remarkable molecules is thiazyl trifluoride. NSF, (Fig. 18.7). This
compound is very stable. It does not react with ammonia at room temperature, with
hydrogen chloride even when heated, or with metallic sodium at temperatures below
400 3 C. The S—N bond. 141.6 pm, is the shortest known between these two elements.
The FSF bond angles of 94° arc compatible with approximate sp 2 bonding and the
presence or an sp hybrid a bond and two p-d n bonds between the sulfur and the
nitrogen. The contraction of the d orbitals by the inductive effect of the fluorine atoms
presumably permits effective overlap and 7r-bond formation. The alternative explana¬
tion would require a double dative bond from the sulfur atom, extremely unlikely in
view of the positive character of the sulfur atom.
The bond length is consistent with a tripie bond. Bond lengths of 174 pm for single
S—N bonds (in NH 2 S0 3 H) and 154 pm for double S = N bonds (in N 4 S 4 F 4 ) are
consistent with a bond order of 2.7 in thiazyl trifluoride. This value is also in
agreement with an estimate based upon the force constant. The relative bond lengths
of S—N, S=N, and S=N bonds are thus 1.00:0.88:0.81 compared with similar
shortenings of 1.00:0.87:0.78 for corresponding C—N, C=N, and C=N bonds.
Fig. 18.7 Molecular
structure of thiazyl
trifluoride, NSF } .
A-2
A-The Literature of Inorganic Chemistry
Shriver, D. F.; Atkins, P. W.; Langford, C. H. Inorganic Chemistry, Freeman: New
York, 1990. ■
Wulfsberg, G. Principles of Descriptive Inorganic Chemistry, Brooks/Cole:
Monterey, CA, 1987.
Classical and Comprehensive Reference Works
Comprehensive Inorganic Chemistry, Bailar, J. C., Jr.; Emeleus. H. J.; Nyholm. R.;
Trotman-Dickenson, A. F., Eds.; Pergamon: Oxford, 1973.
MTP International Review of Science: Inorganic Chemistry, Series /; Emeleus, H. J.,
Ed.; Butterworths: London, 1972.
Gmelin, L. HandLbuch der anorganischen Chemie; Verlag Chemie: Weinheim,
1924-1991.
Dictionary of Inorganic Compounds; Macintyre, J. E., Ed.; Chapman & Hall:
London, 1992.
Comprehensive Coordination Chemistry, Wilkinson. G.; Gillard. R. D.; McCleverty,
J. A., Eds.; Pergamon: London, 1987.
Comprehensive Organometallic Chemistry, Wilkinson, G.; Stone, F. G. A.; Abel,
E. W., Eds.; Pergamon: Oxford, 1982.
Units and Conversion Factors
The International System of Units (SI)
SI base units
Physical quantity
Unit
Symbol
Length
meter
m
Mass
kilogram
kg
Time
second
s
Electric current
ampere
A
Thermodynamic temperature
kelvin
K
Amount of substance
mole
mol
Luminous intensity
candela
cd
Common derived units
Physicol quantity
Unit
Symbol
Definition
Frequency
hertz
Hz
s" 1
Energy
joule
J
kg m 2 s -2
Force
newton
N
J m -1
Pressure
pascal
Pa
N m~ 2
Power
watt
W
J s _l
Electric charge
coulomb
C
A s
Electric potential difference
volt
V
J A - ' s“'
Electric resistance
ohm
n
V A" 1
Electric capacitance
farad
F
A s V -1
Magnetic flux
weber
Wb
V s
Inductance
henry
H
V s A -1
Magnetic flux density
tesla
T
V s m -2
874
875
1 8 ■ Periodicity
Two other molecules indicating the influence of fluorine substitution on d orbital
participation are S 4 N 4 H 4 and N 4 S 4 F 4 (see Chapter 16). Tetrasulfur tetraimide is
isoelectronic with the S s molecule and so the structure
H
S —N—S
H —N N —H
I I
S —N—S
H
and corresponding crown conformation appear quite reasonable. The fluoride, how¬
ever, has an isomeric structure with substitution on the sulfur atoms:
F —S = N—S —F
I II
N N
II I
F — S — N=S — F
Double bonding in this molecule is clearly shown by the alternation in S—N bond
lengths in the ring (see Fig. 16.30b). Now both the above electronic structure for
S 4 N 4 F 4 and that for S 4 N 4 H 4 are reasonable but raise the question: Why doesn't
tetrasulfur tetraimide isomerize from the N-substituted form to the S-subslituted form
isoelectronic with the fluoride:
H —S = N—S —H
I II
N N
II I
H—S—N=S—H
retaining the same number of cr bonds and gaining four n bonds? Apparently the
reason the isomerism does not take place is that although 7r bonding is feasible in the
presence of the electronegative fluorine atoms, it is so weak with electropositive
hydrogen substituents that it cannot compensate for the weakening of the <r bonding as
the hydrogen atom shifts from the more electronegative nitrogen atom to the less
electronegative sulfur atom.
Presumably substitution by halogens in the phosphazene series results in con¬
tracted d orbitals and more efficient n bonding in the ring (see Chapter 16). Unsym-
metrical substitution may allow the normally planar ring to bend. A good example of
this is found in l,l-diphenyl-3.3.5.5-tetrafluorotriphosphatriazene:
Ph Ph
\ /
P
/ \ S \
F N
F
F
Reactivity and d Orbital Participation
The three nitrogen atoms and the fluoro-substituted phosphorus atoms are coplanar
(within 2.5 pm), but the phenyl-substituted phosphorus atom lies 20.5 pm above this
plane. The explanation offered is that the more electropositive phenyl groups cause an
expansion of the phosphorus d orbitals, less efficient overlap with the p orbitals of the
nitrogen atom, and a weakening of the ir system at that point. This allows 27 the ring to
deform and the Ph : P moiety to bend out of the plane.
Further examples of the jeopardy involved in casually dismissing d orbitals
participation are the findings of Haddon and coworkers 28 that ff-orbital participation is
especially important in S 4 F 4 , which is nonplanar, and also that it accounts for about
one-half of the delocalization energy in the one-dimensional conductor (SN) r In the
latter case, the low electronegativity of the d orbitals (see Chapter 5) increases the
ionicity of the S—N bond and stabilizes the structure.
Finally, it will be recalled that the existence of strong P=0 bonds in OPF, (see
page 870) is consistent with enhanced back donation of electron density from the
oxygen atom to the phosphorus atom bearing a positive partial charge from the four cr
bonds to electronegative atoms. In light of the above discussion of the contraction of
phosphorus and sulfur d orbitals when bearing a positive charge, better overlap may
be added to the previous discussion as a second factor stabilizing this molecule.
The question of d orbital participation in nonmetals is still an open controversy. In
the case of er-bonded species such as SF (l the question is not of too much importance
since all of the models predict an octahedral molecule with very polar bonds. Par¬
ticipation in 77 bonding is of considerably more interest, however. Inorganic chemists
of a more theoretical bent tend to be somewhat skeptical, feeling that the arguments
regarding promotion energies and poor overlap have not been adequately solved. On
the other hand, chemists interested in synthesis and characterization tend to favor the
use of d orbitals in describing these compounds, pointing to the great heuristic value
that has been provided by such descriptions in the past and arguing that until rigorous
and complete calculations on these molecules show the absence of significant d orbital
participation it is too soon to abandon a useful model.
Reactivity and
d Orbital
Participation
It has been pointed out that the elements of the second row (Li to F) not only resemble
their heavier congeners to a certain extent (as far as formal oxidation state, at least)
hut also the lower right diagonal element (as far as charge, size, and electronegativity
are concerned). For example, both silicon and phosphorus form hydrides that have
some properties in common with alkanes, although they are much less stable. As a
result of the electronegativity relationship the P—H bond more closely approaches
the polarity of the C—H bond than does the Si—H bond. The resemblance of
phosphorus to carbon has even been extended to the suggestion that a discipline be
built around it in the same manner as organic chemistry is built on carbon.
There is one important aspect of the chemistry of both silicon and phosphorus
which differs markedly from that of carbon. Consider the following reactions:
CCI 4 + H,0 -♦ No reaction (18.42)
SiCI 4 + 4H,0 - Si(OH) 4 + 4HCI rapid (18.43)
27 Note that this explanation does not state that the presence of the weakening of the 7r bonding causes
the ring bending but allows it. perhaps resulting from crystal packing forces.
K Haddon. R. C.; Wasserman, S. R.; Wudl, F.: Williams. G. R. J. J. Am. Client. Sue. I9S0, 102.
6687 - 6693 .
964
19-The Inorganic Chemistry of Biological Systems
d. The asbestos roofs on the authors' houses in College Park, Maryland, and Charleston,
Illinois
c. The integrity of the copper and galvanized steel eaves-troughs and downspouting on
those houses
f. The integrity of the brick and sandstone siding of those houses
g. The growth of azaleas planted along the foundations of those houses
h. The integrity of the aluminum siding on a neighbor's house
i. The slate roof on another neighbor's house
j. The longevity of galvanized steel fencing in the neighborhood
k. The ability of an aquatic snail to form its shell in a lake in the Adirondack Mountains
19.30 For each of the above for which you predicted an adverse effect, speculate as to the
likelihood that there actually will be an effect, i.e., whether there will be acid rain at that
particular geographic site or not.
19.31 Page 936 refers to "the rust-red soils of Oahu." What is the chemical origin of the “rust-
red" color? What is the physical source of the color?
19.32 Niebohr and Richardson have written an extremely interesting article entitled, "The
replacement of the nondescript term ‘heavy metals' by a biologically and chemically
significant classification of metal ions." Their abstract states, in part:
, It is proposed that the term "heavy metals" be abandoned in favor of a classification
which separates metals . . . according to their binding preferences . . . related to atomic
properties. ... A review of the roles of metal ions in biological systems demonstrates the
potential of the proposed classification for interpreting the biochemical basis for metal-ion
toxicity. . . .
Discuss in terms of the suggestions provided by the abstract. Propose a theme for the
Niebohr/Richardson article (as though it were your own) and give some illustrative
examples. 160
19.33 Review your knowledge of coordination chemistry with respect to nomenclature: Why is
the molecule shown in Fig. 19.27 the "A" isomer?
19.34 The parallelism of sunlight-driven photosynthesis/respiration and the chemolithotophic
oxidation of sulfide and sulfur by bacteria (page 890), as well as the possibility of metal
toxicity near hydrothermal vents (page 952), has been noted. Suggest other problems and
possible solutions to be expected from hydrothermal vent organisms.' 61
19.35 If you did not work Problem 12.34 when you read Chapter 12. do so now.
19.36 Bioinorganic compounds tend to change structure (bond lengths and bond angles), more
or less, upon changes in oxidation state or coordination number. The spectrum runs from
blue copper proteins (almost no changes) to hemoglobin (considerable rearrangement).
Discuss the chemical reasons for these differences in behavior and how they affect the
biological function of these molecules.
19.37 If C=0 and C=N~ are isoelectronic, why does hemoglobin (Fe 2 ) have the stronger
interaction with C=0, but methcmoglobin (Fe 3 ) binds ChN' more tightly? ( Hint :
Compare Fe"— CO. Fc 111 —CO, Fe"—CN~, and Fc 1 "—CN~.)
19.38 At pH 7.8, the structure of reduced (Cu 1 ) plastocyanin has a structure very similar to that
of Cu" plastocyanin (Fig. 19.15) except for small differences in bond lengths. At pH 3.8.
the copper is trigonally coordinated with the fourth interaction (Cu-imidazole) broken.
Predict and discuss the redox activity of plastocyanin as a function of pH.
160 Niebohr, E.; Richardson, D.H. Environ. Pollul. Series B 1980, I. 3.
161 Childress. J. J.; Felbeck, H.; Somero, G. N. Sci. Amer. 1987, 256 (5), 115-120.
p p e n d
i x
The Literature of Inorganic Chemistry
The following is not meant to be an exhaustive list of all of the books of interest to an
inorganic chemist, but it is a short list of useful titles.
Texts and General Reference Books
Butler, 1. S.; Harrod J. F. Inorganic Chemistry: Principles and Applications ; Ben¬
jamin/Cummings: Redwood City, CA, 1989.
Chambers, C.; Holliday, A. K. Inorganic Chemistry, Butterworths: London, 1982.
Cotton, F. A.; Wilkinson, G. Advanced Inorganic Chemistry. 5th ed.; Wiley: New
York, 1988.
Cotton, F. A.; Wilkinson. G.; Gaus, P. L. Basic Inorganic Chemistry, 2nd ed.; Wiley:
New York. 1987.
Douglas. B. E.; McDaniel. D. H.; Alexander, J. J. Concepts and Models of Inorganic
Chemistry. 2nd ed.; Wiley: New York. 1983.
Greenwood. N. N.; Earnshaw, A. Chemistry of the Elements-, Pergamon: Oxford,
1984.
Jolly, W. L. Modern Inorganic Chemistry. 2nd ed.; McGraw-Hill: New York, 1991.
Miessler, G. L.: Tarr, D. H. Inorganic Chemistry, Prentice-Hall: Englewood ClilTs,
NJ, 1991.
Moeller. T. Inorganic Chemistry: A Modern Introduction-, Wiley: New York, 1982.
Porterfield, W. W. Inorganic Chemistry: A Unified Approach-, Addison-Wesley: Read¬
ing, MA, 1984.
Purcell, K. F.; Kotz. J. C. An Introduction to Inorganic Chemistry, Saunders:
Philadelphia, 1980.
Sanderson, R. T. Simple Inorganic Substances', Krieger: Malabar, FL, 1989.
Sharpe, A. G. Inorganic Chemistry, 3rd ed.; Longman: London, 1992.
A-1
876
18- Periodicity
PCI 5 2HCI + OPCI 3 ^ OP(OH) 3 (18.44)
In contrast to the inertness of carbon halides, the halides of silicon and phosphorus are
extremely reactive with water, to the extent that they must be protected from atmos¬
pheric moisture. A clue to the reactivity of these halides is provided by the somewhat
similar reactivity of acid halides which readily react with water:
O 0~ O
II I II
RCCI + H 2 0 -► R — C — Cl -> R — C — OH HC1 (18.45)
0 +
/ \
H H
The unsaturation of the carbonyl group provides the possibility of the carbon expand¬
ing its coordination shell from 3 to 4, thereby lowering the activation energy. Carbon
tetrahalide cannot follow a similar path, but the halides of silicon and phosphorus can
employ 3 d orbitals to expand their octets:
9 ci
ci 9 ci
1 /
\l /
P + H,0 -
— p
—» OPCL + 2HC1, etc.
1 \
/i \
CI C1
Cl c, OH,
This enhanced reactivity of compounds of silicon and phosphorus is typical of all of
the heavier nonmetals in contrast to the elements of the second row.
Periodic Anomalies
of the Nonmetals
and Posttransition
Metals
It is generally assumed that the properties of the various families of the periodic chart
change smoothly from less metallic (or more electronegative) at the top of the family
to more metallic (or less electronegative) at the bottom of the family. Certainly for the
extremes of the chart—the alkali metals on the left and the halogens and noble gases
on the right—this is true; the ionization potentials, for example, vary in a rather
monotonous way. This is not true for certain central parts of the chart, however.
Reluctance of
Fourth-Row
Nonmetals to Exhibit
Maximum Valence
There is a definite tendency for the nonmetals of the fourth row—As, Se, and Br—to
be unstable in their maximum oxidation state. For example, the synthesis of arsenic
pentachloridc eluded chemists until comparatively recently,although both PCI, and
SbCI, are stable. The only stable arsenic pentahalide is AsF,: AsCI, decomposes at
-50 °C. and AsBr, and Asl, are still unknown.
In Group VIA (16) the same phenomenon is encountered. Selenium trioxide is
thermodynamically unstable relative to sulfur trioxide and tellurium trioxide. The
enthalpies of formation of SF (> , SeF h , and TeF h are - 1210. - 1117, and - 1320 kJ
mol , respectively. This indicates comparable bond energies for S—F and Te—F
bonds (317 and 330 kJ mol -1 , respectively), which are more stable than Se—F bonds
(285 kJ mol -1 ).
29 Seppclt. K. Z. Anorg. Client. 1977, 434. 5.
Periodic Anomalies of the Nonmetals and Posttransition Metals
877
Anomalies of
Groups IIIA (13)
and IVA (14)
The best known exceptions to the general reluctance of bromine to accept a +7
oxidation state are perbromic acid and the perbromate ion. which were unknown prior
to 1968 (see Chapter 17). Their subsequent synthesis has made their “nonexistence"
somewhat less crucial as a topic of immediate concern to inorganic chemists, but
bromine certainly continues the trend started by arsenic and selenium. Thus the
perbromate ion is a stronger oxidizing agent than either perchlorate or periodate.
Before seeking an explanation of the reluctance of As, Se. and Br to exhibit maximum
oxidation states, a related phenomenon will be explored. This involves a tendency for
germanium to resemble carbon more than silicon. Some examples are:
Reduction of halides (X) with zinc and hydrochloric acid. Germanium resem¬
bles carbon and tin resembles silicon:
^ ho ' '/C H
^Si-X-^No >i-H
-^Ge-X ^^Ge-H
^'Sn —X -jg-» No Jj.Sn-H
Hydrolysis of the tetraltydrides. Silane hydrolyzes in the presence of cata¬
lytic amounts ol hydroxide. In contrast, methane, germane, and stannane do
not hydrolyze even in the presence of large amounts of hydroxide ion.
Reaction of organolithium compounds with (C b H s ) 3 MH. Triphenylmethane
and triphenylgermane differ in their reaction with organolithium compounds
from triphenylsilane and triphenylstannane:
(18.47)
(18.48)
(18.49)
(18.50)
PhjCH + LiR -
PhjSiH + LiR
PhjGeH + LiR
PhjSnH + LiR
LiCPh, + RH (18.51)
PhjSiR + LiH (18.52)
■ LiGePhj + RH " ,0,H » Ph,GeGePh, + LiH (18.53)
PhjSnR + LiH (18.54)
Alternation in enthalpies of formation. There is a tendency for the enthalpies
of formation of compounds of the Group IVA (14) elements to alternate from
G-Si-Ge-Sn-Pb. Although closely related to the previous phenomena, this
variation is also related to the “inert pair effect" and will be discussed further
below.
The elements of Group IIIA (13) show similar properties, although, in general the
difierences are not so striking as for Group IVA (14). It may be noted' that the covalent
radius of gallium appears to be slightly smaller than that of aluminum in contrast to
what might have been expected. The first ionization energies of the two elements are
surprisingly close (578 and 579 kJ mol '), and ir the sum of the first three ionization
energies is taken, there is an alternation in the series: B = 6887 Al = 5139 Ga =
5521, In = 5084, Tl = 5438 kJ mol -1 .
962
19-The Inorganic Chemistry of Biological Systems
19.16 Predict which way the following equilibrium will lie:
Hb + Hb(0 2 ) 4 2Hb(0 2 ) 2 (19.42)
Explain.
19.17 Although sickle cell anemia causes problems in many organ systems, the chief cause of
death of children with SCA is bacterial infection. Discuss.
19.18 Using the reduction emfs given in Appendix F, construct a Latimer diagram, complete
with skip-step emfs for one-, two-, and four-electron reduction of oxygen to superoxide,
peroxide, and hydroxide. Discuss the biological significance of these emfs. Recall that a
living cell is basically a reduced system threatened by oxidizing agents. 153
19.19 Biochemists tend to speak of “dismutation reactions" such as:
2H + + 20 2 -► H 2 0 2 + 0 2 (19.43)
that are catalyzed by superoxide dismutase. What term do inorganic chemists use for this
phenomenon? What type of metal do you suppose is in superoxide dismutase?
19.20 Gray and coworkers 15 - 1 have prepared copper(II) carboxypeptidase A, Cu"CPA. and
compared its spectrum, that of the enzyme with inhibitor present, and those of several
other copper(II) complexes with nitrogen and oxygen ligating atoms. Some of these data
together with the geometry about the copper ion are:
Set of ligating atoms' 1
Structure
"max (O" 1 )
N, N, N. N
Planar
19,200-19.600
N, N, O. O
Planar
14,100-17,500
N, N, O, O
Pscudotctrahcdral
12,900-14.700
O. O, 0. 0
Cu"CPA:
Planar
13.500-15.000
N. N, 0. O?
a^CPA-pPP":
?
12,580
N, N, 0. 0?
?
11,400
“ These are the four atoms in the coordination sphere of the
copper(ll) ion.
6 (3PP is p-phcnylpropionate, an inhibitor of carboxypeptidase
A.
a. Account for the trends in the values of v m „ x in the first four rows of the table listing
literature values and known geometries.
b. Predict the geometry of the ligating atoms about the Cu" ion in Cu"CPA and in
Cu"CPA0PP.
c. Comparing the values of v m „, of Cu n CPA with and without the inhibitor present,
suggest what effect the inhibitor may be having on the geometry of the copper ion.
19.21 For simplicity the iron-oxygen interaction in myoglobin and hemoglobin (but not
hemerythrin) was discussed in terms of neutral oxygen molecules binding to Fe". How¬
ever, much of the current literature discusses these phenomena in terms of superoxo and
peroxo complexes and one sees Fe nl -0 2 . Discuss what these formulations and terms
mean, and describe the related consequences in terms of charges, electron spins, etc. 153
153 Fridovich. I. Am. Sci. 1975, 63, 54.
154 Rosenberg, R. C.; Root, C. A.; Bernstein. P. K.; Gray, H. B. J. Am. Chem. Soc. 1975. 97.
2092-2096.
us Vaska, L. Acc. Chem. Res. 1976. 9, 175-183. Reed, C. A.: Cheung. S. K. Proc. Natl. Acad. Sci.
U.S.A. 1977, 74, 1780.
Problems
963
19.22 Using your knowledge of periodic relationships, predict which element might come
closest to reproducing the behavior of molybdenum in nitrogenuse. Recall that nitrogen
fixation involves both complexalion and redox reactions. 15 ''
19.23 In order to study the function of oxygen binding by myoglobin and its effect on muscle
function. Cole (Footnote 16) perfused an isolated muscle with hydrogen peroxide. Why
did he do this?
19.24 Discuss each of the following situations:
a. One problem encountered in the manufacture and preservation of H 2 0 2 is its spon¬
taneous decomposition:
2HjOj -♦ 2H 2 0 + 0 2 (19.44)
which is exothermic (A/7 = - 196 kJ mol -1 ). To reduce this decomposition, copper
ions are carefully excluded or chelated. 157
b. Nature's protection against the destructive oxidative powers of H 2 0 2 are the enzymes
catalase and peroxidase, both of which contain iron.
c. Superoxide, 0 2 - , is perhaps even more dangerous than hydrogen peroxide as an
oxidant and free radical, though otherwise somewhat similar.
d. Superoxide dismutase contains both zinc and copper. Zinc may be replaced by Co(II)
and Hg(!l) and the enzyme retains its activity; no substitution of copper with retention
of activity has yet been found. Discuss. {Hint: You can discuss this on several levels,
from the most offhand conjectures to a quantitative demonstration with numbers.
Choose an appropriate level (or ask your professor) and attack the problem
accordingly.)
e. Procaryotes have a primitive superoxide dismutase with a metal other than copper.
Suggest possible metals. 15,1
19.25 If you did not work Problem 12.16 when you read Chapter 12. do so now.
19.26 Fe(OH) 2 (K sp = 1.8 x I0 -1S ) and Fe(OH), (AT, P = 2.64 x 10 w ) have serendipitous
solubilities: If one can remember the exponents, one can “instantly" estimate the pH
necessary to make a I M solution of Fe :+ or Fe'.
a. Explain. What arc the approximate pHs of these solutions?
b. Calculate the exact pHs assuming ideal behavior.
19.27 Bczkorovainy cites several studies indicating reduced iron absorption during febrile
illnesses. Frame a hypothesis for the adaptiveness of such an effect. 15 ''
19.28 Acid rain has been defined us any precipitation with a pH lower than 5.6. Why 5.6 instead
of 7.0? Can you perform a calculation to reproduce this value?
19.29 Explain what effect acid rain would have on the condition of each of the following, and
why:
a. The Taj Mahal, at Agra, India
b. A limestone bam near Antietam Battlefield. Maryland, dating from the Civil War
c. The Karyatides, the Acropolis, Athens. Greece
15 Slicfcl, E. I. Proc. Nall. Acad. Sci. U.S.A. 1973. 70, 988.
i’ 7 Crampton, C. A.; Faber, G.; Jones. R.; Leaver, J. P.; Schclle, S. In The Modern Inorganic
Chemicals Industry: Thompson, R.. Ed.; Chemical Society Special Publication No. 31; Chemical
Society: London, 1977: p 244.
15 Fridovich. I. Adv. Inorg. Biochem. 1979. /. 67. Fratisto-da Silva, J. J. R.; Williams. R. J. P. The
Biological Chemistry of the Elements ; Clarendon: Oxford. 1991; pp 337, 383. 395-396.
■ 59 Biochemistry of Nonheme Iron ; Bczkorovainy, A., Ed.; Plenum: New York, 1980; p 90.
878 18- Periodicity
The "Inert Pair
Effect"
Table 18.3
Ionization energies of s
electrons in kJ mol - ' (elf)
Among the heavier posttransition metals there is a definite reluctance to exhibit the
highest possible oxidation state. Thus in Group IVA (14), tin has a stable + 2 oxidation
state in addition to +4, and for lead the +2 oxidation state is far more important.
Other examples are stable Tl + (Group IIIA, 13) and Bi 3+ (Group VA, 15). These
oxidation states correspond to the loss of the np electrons and the retention of the ns
electrons as an “inert pair”. 30
It can readily be shown that there is no exceptional stability (in an absolute sense)
of the ,v electrons in the heavier elements. Table 18.3 lists the ionization energies of the
valence shell s electrons of the elements of Groups IIIA (13) and IVA (14). Although
the 6s electrons are stabilized to the extent of—300 kJ mol -1 (3 eV) relative to the 5s
electrons, this cannot be the only source of the inert pair effect since the 4x electrons
of Ga and Ge have even greater ionization energies and these elements do not show
the effect—the lower valence Ga(I) and Ge(li) compounds are obtained only with
difficulty.
The pragmatic criterion of the presence or absence of an inert pair effect can be
taken as the tendency (or lack thereof) for the following reaction to proceed to the
right:
MX, -♦ MX„_2 4- X 2 (18-55)
We might then inquire as to the systematic variation in thermodynamic stability of the
higher and lower halides of these elements. There seem to be two general effects
operating. The combination of the two effects gives irregular changes in covalent bond
energies (see Table 18.4). The simplest is the tendency for weaker covalent bond
formation by larger atoms (see Chapter 9). The second is the “anomalous" properties
of those elements that follow the first filling of a given type of orbital (s. p, d.f. . .).•"
All of these elements exhibit a lower tendency to form stable compounds than do their
lighter and heavier congeners. Both sodium and magnesium form less stable com¬
pounds than would be expected, when compared to lithium and beryllium, or po¬
tassium and calcium. 33 These elements are those that follow immediately after the first
filling of a set of p orbitals (Ne). and the same effects of incomplete shielding (though
less pronounced to be sure) presumably are operating here as well as in the
postlanthanide and postscandide elements. This principle has also been used to
predict some of the chemical properties of the superheavy transactinide elements.
Element
IE, + IE 3
Element
ie 3 + ie 4
B
6,087 (63.1)
c
10.843 (112.4)
Al
4,561 (47.3)
Si
7,587 (78.6)
Ga
4.942 (51.2)
Ge
7.712 (79.9)
In
4.526 (46.9)
Sn
6,873 (71.2)
Tl
4,849 (50.3)
Pb
6,165 (63.9)
This is also related to the fact that "R 2 Sn=SnR 2 " compounds may exist as R 2 Sn units in solution
(page 863). Closely related, but not identical, is the fact that an unoxidized ,v electron pair may or
may not be stcrcochemically active. Sec discussions in Chapters 6 and 17.
31 Huhcey. J. E.; Huheey, C. L. J. Chart. Educ. 1972. 49. 227.
32 Evans. R. S.; Huheey. J. E. J. Intirg. Nucl. Client. 1970, 32. 111.
Periodic Anomalies of the Nonmetals and Posttransition Metals 879
Table 18.4
Bond energies of some
group IVA (14) halides in
kJ mol -1 ( kca1 mo/ -1 )
Element
mf 2
mf 4
mci 2
MCI„
MBr 2
MBr 4
mi 2
mi 4
Si
—
565 (135)
—
381 (91)
—
310 (74)
—
234 (55.9)
Ge
481 (115)
452 (108)
385 (92.0)
349 (83.4)
326 (77.8)
276 (66.0)
264 (63.1)
212 (50.6)
Sn
481 (115)
414 (98.9)
386 (92.2)
323 (77.2)
329 (78.7)
273 (65.2)
262 (62.5)
205 (49.0)
Pb
394 (94.2)
331 (79.1)
304 (72.6)
243 (58.1)
260 (62.2)
201 (48.0)
205 (49.0)
142 (33.9)
1
Relativistic Effects
For the lighter elements these effects can readily be formulated in terms of
ordinary shielding effects as discussed in Chapter 2. For heavier elements, however,
the theory of relativity must be invoked.
Normally the theoretical basis of chemistry is the nonreiativistic Schrodinger equa¬
tion. To this are added the postulate of electron spin and ideas related to it such as the
Pauli exclusion principle. Although the latter are thus seemingly ad hoc "add ons" to
make the theory work, most of the theoretical chemistry has been done on this basis.
The corresponding relativistic approach yields the Dirac equation. 33 This gives four
quantum numbers directly, although only the principal quantum number n is the same
in both treatments. The relativistic treatment results in a number of novel effects, both
descriptive'• and theoretical 35 most of which can usually be neglected with little loss
of accuracy and a great gain in convenience. There are two exceptions to this
generalization however. One is spin-orbit, or jj, coupling (see Chapter 11 and Appen¬
dix C). The second is that neglect of relativistic effects becomes increasingly serious
as the atomic number increases. The .v (and to a slightly lesser extent, p) electrons will
accelerate greatly as they approach the nucleus, and their speed relative to the fixed
speed of light cannot be ignored. It has been estimated that for mercury (Z = 80) the
speed of a l.r electron is over half that of light. This results in an approximately 20%
increase in electronic mass and an approximately 20% decrease in orbital size. 36 In the
simplest case we can say that s and p orbitals wiil contract, and that d and /orbitals
will expand somewhat. The seeming paradox that the d and/orbitals expand instead
of contract is an indirect effect. Direct relativistic effects on d and/orbitals are small
because these orbitals do not have electron density near the nucleus. However, the
31 Dirac. P. A. M. Proc. R. Sue. London. Ser. A 1928. All 7. 610; 1928. AII8, 351.
34 Some of these, such as the facts that gold metal has its familiar color and that mercury metal is a
liquid, fall outside the scope of this text. See Pyykkb. P.; Desclaux. J.-P. Ate, Client. Res. 1979. 12,
276-281. Others will be discussed below.
35 Some of these are: Although the quantum number / still determines orbital type t.v, p.d.f. . .). it no
longer determines orbital shape. All orbitals of given value of n and / no longer have the same
energies. Orbital shape is determined by the angular momentum quantum number./'and the magnetic
quantum number in. The shapes of orbitals arc not the familiar ones given by the Schrddinger
equation, but seemingly “misshapen" nodclcss analogues. See McKelvcy, D. R. J. Chan. Educ.
1983. 60. 112-116: Powell. R. E. Ibid. 1968, 45. 558.
36 This is most readily seen from the inverse relationship between Rohr radius and mass (a„ =
Airell'ImZe').
960
19-The Inorganic Chemistry of Biological Systems
Problems
961
We may thus end this chapter on bioinorganic chemistry and this book on modern
inorganic chemistry by noting that a complex that Werner could have synthesized a
century ago (and resolved a short time later) is being used to answer questions that
neither he nor his contemporary biologists could have conceived.
Summary It is true that many of the facts in this chapter were gathered by biologists, bio-
chemists, and X-ray crystallographers, not only by inorganic chemists. But the inter¬
pretation of these facts and their further exploration falls within the realm of inorganic
chemistry. Such factors as (I) alteration of emfs by complexation; (2) stabilization of
complexes by ligand field effects; (3) hardness and softness of acids and bases; (4) the
thermodynamics and kinetics of both “natural” and “unnatural” (i.e., pollutant)
species; (5) catalysis by metal ions; (6) preferred geometry of metal complexes; and (7)
energetics of (a) complex formation, (b) redox reactions, and (c) polyanion formation
come within the ken of inorganic chemists, and they should be able to contribute fully
to the future study of these systems. The effect is already being felt. One need only
compare a recent biochemistry text with one of a decade ago to note the emphasis on
high spin vs. low spin metal ions, coordination geometry and configuration, and redox
reactions and thermodynamics.
The present convergence of physical and analytical techniques combined with
inorganic theory makes this one of the most exciting times to be involved in this area
of chemistry. One can combine the hard facts and principles of our discipline with the
ever elusive yet fascinating mystery of life.
Postscript "I say that it touches a man that his blood is sea water and his tears are salt, that the
seed of his loins is scarcely different from the same cells in a seaweed, and that of the
stuff like his bones arc coral made. I say that physical and biologic law lies down with
him, and wakes when a child stirs in the womb, and that the sap in a tree, uprushing in
the spring, and the smell of the loam, where the bacteria bestir themselves in
darkness, and the path of the sun in the heaven, these are facts of first importance to
his mental conclusions, and that a man who goes in no consciousness of them is a
drifter and a dreamer, without a home or any contact with reality."' 42
Donald Culross Peattie
Problems
19.1 Why was the covalent radius of the metal used on page 891 instead of that of the +2 ion?
19.2 Why are transition metals such as Mn, Fe, Co, and Cu needed in photosynthesis and
respiration rather than metals such as Zn, Ga, or Ca?
19.3 Calculate the energy available from one photon of light at wavelength 700 nm. If it
generates a potential difference of 1.0 V, what is the conversion efficiency?
19.4 Discuss how the use of simple model systems can aid our understanding of biochemical
systems. Is there any way they might detract? 148
147 Peattie. D. C. An Almanac for Moderns', Nonpareil: Boston. 1980; p 14.
148 Wang, J. H. Acc. Chem. Res. 1970, 3, 90.
19.5 There are two ways in which photosynthesis increases the energy available: (I) by using
two light capturing mechanisms. PS I and PS II; (2) by stacking the chlorophyll in the
grana which are in turn stacked in the chloroplast (see Lehninger, A. L. Sci. Am.,
September. 1961. for the structures). Which corresponds to hooking batteries in parallel
and which to hooking them in series?
19.6 Common ions in enzyme systems are those that have low site preference energies (from
LFSE) such as Co', Zn 2 ', and Mn 21- rather than Fe 2 ', Ni 2 '. or Cu 2 '. Discuss this
phenomenon in terms of the entatic hypothesis. 14 ' 7
19.7 Discuss the probable difference in the pockets present in carboxypeptidase and carbonic
anhydrase.
19.8 The toxicity of metals has been variously correlated with their (I) electronegativity;
(2) insolubility of the sulfides; (3) stability of chelates. Discuss.
19.9 Show how coordination of an 0 3 molecule to a heme group can result in pairing of the
electron on the oxygen molecule when the bonding is
a. through a p. bond
b. through a lone pair of one oxygen atom
19.10 Directions for the use of the antibiotic tetracycline advise against drinking milk or taking
antacids with the medication. In addition, warnings are given concerning its use (teeth
may be mottled in certain cases). Suggest the chemical property of tetracycline that may
be involved in these effects.
19.11 High mercury levels in terminal food chain predators like tuna fish have caused consider¬
able worry. It has been found that tuna contain larger than average amounts of se¬
lenium. 150 Discuss the possible role of selenium with respect to the presence of
mercury. 151
19.12 Although the hypothesis of Eganii may be an oversimplification, it is certainly true that
Fe' ’/Fe 1 ' is widely used in redox systems. Zn 2 ' in hydrolysis, esterification, and similar
reactions, and molybdenum in nitrogenase. xanthine oxidase, nitrate reductase, etc.
Putting abundance aside, discuss the specific chemical properties of these metals that
make them well suited for their tasks.
19.13 Carboxypeptidase A(Co 2 ') not only retains the activity of carboxypeptidase A(Zn 21 ). it is
actually a more active enzyme. This being the case, why do you suppose that Co 21 is not
used in the natural system?
19.14 If you did not answer Problem 14.39 when you read Chapter 14. do so now.
19.15 When patients are treated with D-penicillamine for sehlerodertna. cystinuria, rheumatoid
arthritis, and idiopathic pulmonary fibrosis. 32% show decreased taste acuity (hypo-
geusia). In contrast, only 4% of the patients being treated with D-penicillamine for
Wilson's disease exhibit hypogetisia. Discuss a possible mechanism. How might the
hypogeusia be treated? 153
149 Fratisto-da Silva. J. J. R.; Williams. R. J. P. The Biological Chemistry of the Elements', Clarendon:
Oxford. 1991; pp 180-184.
150 Gunther. H. E.: Goudic. C.; Sunde. M. L.; Kopccky, M. J.; Wagner. P.; Oh. S.-H.; l-lockstra,
W. G. Science 1972. 175. 1122-1124.
151 Chen. R. W.; Whangcr, P. D.; Wcswig. P, H. Bioinorg. Chem. 1975.4, 125-133. Schrauzcr, G. N.
Bioinorg. Chem. 1976. 54, 275-281. Matsumoto, K. In Biological Trace Element Research',
Subramanian. K. S.; Iyengar, G. V.; Okamoto, K.. Eds; ACS Symposium Scries 445; American
Chemical Society: Washington. DC. 1991; Chapter 22.
152 Henkin, R. I.; Bradley, D. F. Proc. Natl. Acad. Sci. V. S. A. 1969. 62. 30.
880
18 Periodicity
Periodic Anomalies of the Nonmetals and Posttransition Metals
881
"Anomalous”
Ionization Energies
and Electron
Affinities
increased shielding of d and /orbitals by relativistically contracted s and p orbitals
tends to cancel the effect of increased Z. 37 So the s and p electrons are moved closer to
the nucleus, their energy is lowered (made more negative), and they are stabilized.
The d and / orbitals are raised in energy (destabilized) and expand. Since the
outermost orbitals are the ns and np rather than the {n - )d or (n - 2 )/, each atom as
a whole contracts. 38
The relativistic effect goes approximately as ZP-. and this is the reason for its
importance in the heavier elements. In terms of energy and size, it starts to become
important in the vicinity of Z = 60-70, contributing perhaps an additional 10% to the
nonrelativistic lanthanide contraction (see Chapter 14). 39 As we have seen, this results
in an almost exact cancellation of the expected increase in size with increase in n from
zirconium to hafnium.
While the contraction resulting from the poor shielding of 4/electrons ceases at
hafnium, the relativistic effect continues across the sixth row of the periodic table. It is
largely responsible for the stabilization of the orbital and the inert s pair effect
shown by the elements Hg-Bi. It also stabilizes one 40 of the 6 p orbitals of bismuth
allowing the unusual +1 oxidation state in addition to +3 and +5. JI
Many introductory chemistry books give simple rules for remembering the periodic
changes of ionization energies and electron affinities. The rules usually follow some
modification of "Ionization energies and electron affinities increase as one moves to
the right in the periodic chart; they decrease as one moves from the top to bottom.”
These generalizations, as well as the shielding rules that account for the atomic
behavior, were discussed in Chapter 2, along with some of the exceptions. Unfortu¬
nately for simplicity, the exceptions are somewhat more numerous than is generally
realized. Many of the problems discussed in the preceding sections result from these
“exceptions.”
The horizontal behavior of atoms follows the general rule with good regularity as
might be expected from adding a single proton at a time with expected monotonic
changes in properties. We have already seen the exception of the inversion of the
ionization potentials of the VA (15) and VIA (16) groups related to the stability
associated with half-filled subshells. A similar inversion of electron affinities takes
place, for the same reason, between groups IVA (14) and VA (15).
The vertical exceptions to the generalizations arc much more widespread: If we
count every time that a heavier element has a higher ionization potential or higher
electron affinity than its next lighter congener, we find that about one-third of the
elements show "electron affinity anomalies" 4 - and a somewhat higher fraction of the
elements show "ionization energy anomalies.” With such a high fraction of excep-
37 Pitzcr, K. S. Acc. Chem. Res. 1979. 12. 271-276. Sec also Footnote 34.
38 Note from Figure 2.4 how the maximum electron density for a 3d orbital lies well under those of the
3.r and 3 p and, by extension, even more so those of the 4.r and 4 p orbitals.
39 For smaller effects in the lighter elements, see Pyykkd. P. Chem. Rev. 1988. 88. 563-594.
40 Recall that one of the differences of relativistic orbitals from the nonrelativistic ones that we arc
accustomed to handling is that the orbitals of a given value of / are not degenerate: The 6 p\ K orbital
lies below the bpyz-
41 Jorgensen. C. K. Z. Anorg. Allg. Chem. 1986, 5-10/541. 91-105.
42 The experimental values of atomic electron affinities have been reviewed and plotted as periodic
functions. Chen. E. C. M.; Wentworth. W. E. J. Chem. Educ. 1975, 52. 486.
lions, one wonders why the rules were formulated as they were originally. The answer
seems to lie in the lack of data available until recently; most of the good data were for
familiar elements, such as the alkali metals and the halogens. For these main-group
elements, with the exception of the lower electron affinity of fluorine resulting from
electron-electron repulsion (and paralleled by oxygen and nitrogen), the rules work
fairly well; however, the poorer shielding d and/electrons upset the simple picture.
For the transition metals, higher ionization energies with increasing atomic number in
a group are the rule, not the exception. As we have seen in the preceding discussion,
this carries over somewhat into the posttransition elements, causing some of the
problems associated with families IMA (13) and IVA (14).
The increased ionization energies of the heavier transition metals should not be
unexpected by anyone who has had a modicum of laboralory experience with any of
these elements. Although none of the coinage metals is very reactive, gold has a well-
deserved reputation for being less reactive than copper or silver; 43 iron, cobalt, and
nickel rust and corrode, but osmium, iridium, and platinum are noble and unreactive
and therefore are used in jewelry; platinum wires are the material of choice for flame
tests without contamination; and one generates hydrogen with zinc and simple acids,
not with mercury.
Although the increased electron affinity associated with the heavier elements
usually manifests itself only indirectly (via electronegativity, etc.), it is directly re¬
sponsible for the fact that cesium auride, Cs Au~. is an ionic salt rather than an alloy.
Both the increased ionization energy and increased electron affinity in these elements
result from relativistic effects.
We have seen the use of macrocyclic ligands to aid in the isolation of auride salts
(Chapter 12). The characterization of this unusual oxidation state, both in
[Cs(C222)JAu and CsAu, was accomplished by the use of photoelectron spectroscopy
(Chapter 5). When the binding energy of gold d electrons is plotted as a function of the
lormal oxidation state of gold (0. +1. +2. 4-3). a straight line is obtained (Fig. 18.8).
The fact that CsAu lies on this line at a point corresponding to an oxidation slate of - I
is good evidence for the formation of the auride. Au', ion. 44
In the same way. the |,,7 Au Mossbauer spectrum of CsAu is very similar to that of
[Cs(C222)| Au ‘ (Chapter 12). indicating that the Au~ anion is present in both (Fig.
18 .9). 43 Since the Mossbauer effect is a nuclear one. it is very sensitive to electron
density at the nucleus and therefore to both atomic charge and s character. 4
Gold exhibits other interesting anomalies. For example some Au(l) compounds
with an expected coordination number of two (Chapter 12) and a filled core of 5 d w
43 The legend of aqua regia seems to persist even in the absence of student contact with this powerful
elixir. However, admiration for its chemical reactivity is usually misplaced. Thermodynamically, it
can be no stronger an oxidizing agent that nitric acid itself. It is the complexing ability of the soft
chloride ions on the soft Au 3 ion that allows Au — AuCI 4 to be effected at E a = - 1.00 V. Rather
more astonishing is the lacl that similar interaction of very soft I - ions causes the relatively
innocuous tincture of iodine (found in many medicine cabinets) to dissolve metallic gold readily
(Nakao. Y. J. Chem. Soc.. Chem. Commun. 1992. 426-427).
44 Knccht, J.: Fischer, R.; Ovcrhof. H.; Hcnscl, F. J. Chem. Soc.. Chem. Commun. 1978, 905-906.
43 Batchelor. R. J.: Birchall. T.; Burns. R. C. Inorg. Chem. 1986. 25. 2(8)9-2015.
“ Ehsworth. E. A. V.; Rankin. D. W. H.; Cradock. S. Structural Methods in Inorganic Chemistry.
2nd ed.; CRC: Boca Raton. FL, 1991; Chapter 7.
958
19-The Inorganic Chemistry of Biological Systems
Medicinal Chemistry
959
Metal Complexes
as Probes of
Nucleic Acids
Fig. 19.31 (a) Structure of the c/j>(NH]) 2 Pt{d(pGpG)} complex, where d(pGpG) = guanine
deoxyribosc phosphate dinucleotide. (b) Numbering system of guanine to indicate N7. [From
Sherman. S. E.; Gibson. D.; Wang. A. H.-J.; Lippard. S. J. Science 1985. 230. 412-417.
Reproduced with permission.]
chloride ion concentration of only 4 mM: Hydrolysis and subsequent reactions with
the appropriate biological targets can then readily take place. 143
An interesting aspect of the chemotherapeutic use of cfr-diamminedichloroplat-
inum(II) and related drugs consists of some negative side effects including nephrotox¬
icity. They are thought to be the result of the inactivation of enzymes by coordination
of Pt(ll), like Hg(II), to thiol groups. Application of the ideas of HSAB theory would
suggest the protection of these thiols by the use of competitive "rescue agents" that
have soft sulfur atoms. These include the diethyldithiocarbamate, Et 2 NCS 2 . and
thiosulfate, SiO 2- , ions. 144
The coordination of ci'j-diammineplatinum(II) to guanine bases in DNA is only one
example of a large number of possibilities. The Mg 2 - ion has several important
functions with respect to DNA and RNA structure and action. Nature has also
anticipated the chemist through the use of "zinc finger" proteins as DNA transcrip¬
tional factors. They have a protein chain coordinated tetrahedrally to a zinc atom by
143 Martin, R. B. In Platinum. Golil. and Other Metal Chemotherapeutic Agents; Lippard, S. J.. Ed.;
ACS Symposium Series 209; American Chemical Society: Washington. DC. 1983; Chapter II.
IJJ See discussion by Lempers, E. L. M.; Rcedjik, J. Adv. Inorg. Chem. 1991. 37. 175-217.
two cysteines and two histidines and provide specific structural information for site
recognition on DNA.
Transition metal complexes may be used to probe specific sites on DNA and RNA
chains. Such interactions may yield, information concerning the structure at those sites
or may induce specific reactions at them. Only one example will be given here. 14 '’
DNA helices are chiral. They would thus be expected to interact with chiral metal
complexes in an enantioselective manner. This is illustrated in Fig. 19.32. The inter¬
calation of the A enantiomer of tris(o-phenanthroline)ruthenium(Il) into the right-
handed helix of B-form DNA 146 is more favorable than that of A-[Ru(phen),] 2 + . This
is a necessary result of the interaction of the orientation of the "righl-handed" ligands
with the right-handed helical groove of the DNA. Obviously the chirality of the meial
complex is predominant in its interaction with the DNA. We can expect further
progress in the use of such enantioselective probes.
Lombdo Delta
Fig. 19.32 Illustration of [Ru(phen) 3 ] 2 ‘ enantiomers bound by intercalation to B-DNA. The
A-enantiomer (right) fits easily into the right-handed helix, since the ancillary ligands are
oriented along the right-handed groove. For the A-enantiomer (leftI. in contrast, steric
interference is evident between the ancillary phenanthroline ligands and the phosphate
backbone, since for this left-handed enantiomer the ancillary ligands are disposed contrary to
the right-handed groove. [From Barton. J. K.; Danishefsky. A. T.: Goldberg. J. M. J. Am.
Chem. Sac. 1984. 106. 2172-2176. Reproduced with permission,]
IJ ' The reader's attention is drawn to the pioneering work in this area by Jacqueline Barton: Pyle.
A. M.; Barton. J. K. Progr. Inorg. Chem. 1990. 3H. 413-475.
A discussion of the structures of A. B. and Z DNA is beyond the scope of this text. See either the
reference in Footnote 145 or any modern biochemistry text.
'882
Fig. 18.8 Binding energies of the Au (4/ 7/2 ) levels of gold aloms in various oxidaiion slates.
Note that the value for RbAu and CsAu corresponds to that expected for a - 1 oxidation
state. [From Knecht, J.; Fischer. R.; Overhof. H.: Hensel. F. J. Client. Soc., Client.
Comm tin. 1978. 906. Reproduced with permission.]
Velocity (mms—)
Fig. 18.9 A comparison of
the chemical shifts in the
l97 Au Mdssbuuer spectra of
CsAu and (Cs(C222)| Au .
[From Batchelor. R. J.;
Birchall. T.; Burns. R. C.
Inorg. Cltem. 1986, 25,
2009-2015. Reproduced
with permission.]
Periodic Anomalies of the Nonmetals and Posttransition Metals 883
Fig. 18.10 Two examples of Aud)-Au(I) interactions, (a) Intramolecular. Au—Au = 300
pm: tbi Intermolecular. Au—Au = 344 pm. See Table 18.5 for identification of compounds.
[From Schmidbaur. H.: Graf. W.; Muller. G. Angew. Client, hit. Ed. Eitgl. 1988. 27. 417-419;
Schmidbaur. H.; Weidenhiller, G.: Steigelmann. O.: Muller, G. Client. Ber. 1990. 123.
285-287. Reproduced with permission.]
electrons, nevertheless show additional weak Au-Au interactions. 47 For example the
Au—Au distances in a series of compounds like those shown in Fig. 18.10 are given in
Table 18.5. These distances are ai or below ihe expected van der Waals contacts of
about 350 pm (Table 8.1) and they are indicative of some weak bonding in these
compounds.
It has been estimated that the energy of the Au(l)-AuG) interaction is about 30 kJ
mol' 1 , which is about the order of magnitude of a hydrogen bond and the results are
similar. In some cases the two gold atoms are in the same molecule and cause ring
formation tFig. I8.l0u); in other cases the Au-Au interaction causes two molecules to
dimerize (Fig. 18.10b).
In addition to the Audi—Au(l) bonds, there tire similar interactions between
aloms that are a) congeners of [e.g.. Agd)]. 4 h) isoelectronic with |e.g., Hg(li)j, J ‘' or
Table 18.5
Some examples of
Au(l)-Au(l) interactions at
less than the expected
distance of van dor Waals
contacts
Compound
Infra- or
intermolecular bond
Au—Au (pm)
KNC),C,S,Au]-“
Intra-
279.6“
Mc,P=C(PPh.AuCI), (Fig. 18.10a)
Intra-
300.0' 1
[PhAs(CH,PPh\Au,Cl)|,
Inter-
3I4,I<
[2.4.6-(f-Bu),C ft H,PH,AuCI), (Fig. 18.10b)
Inter-
344.0'
" See Footnote 47(a).
Sec Footnote 47(b).
r See Footnote 47(e).
•' Sec Footnote 47(d).
47 (a) Khan. N. I.: Wang. S.: Fttekler. J. P.. Jr, Inorg. Client . 1989. 23. 3579-3588. (b) Schmidbaur. H.t
Graf. W ; Muller. G. Attune. Client. Ini. Ed. Engl. 1988, 27. 417-419. (c) Balch. A. L.; Fung. E. Y.;
Olmstcad. M. M. J. Am. Cltem. Soc. 199(1. 112. 5181-5186. (d) Schmidbaur. H.t Weidenhiller. G.;
Steigelmann. ().; Muller. G. Client. Her. 1990. 123. 285-287.
4K Wang. S.: Fackler. J. P.. Jr : Carlson. T. F. Organometallies 1990, 9 . 1973-1975.
Wang. S.: Fackler. J. P.. Jr. OrgaitometalliCs 1990, 9, 111-115; Acta Crystallogr. 1990. 065.
2253-2255.
956
19 The Inorganic Chemistry of Biological System;
Chelate Therapy We have seen previously that chelating agents can be used therapeutically to treat
problems caused by the presence of toxic elements. We have also seen that an
essential element can be toxic if present in too great a quantity. This is the case in
W.lson s disease (hepatolenticular degeneration), a genetic disease involving the
buildup of excessive quantities of copper in the body. Many chelating agents “have
"f" dSed t0 re move the excess copper, but one of the best is D-penicillamine
HSC(CH 3 ),CH(NH 2 )COOH. This chelating agent forms a complex with copper ions'
that is colored an intense purple and. surprisingly, has a molecular weight of 2600.
Another surprising finding is that the complex will not form unless chloride or bromide
ions are present and the isolated complex always contains a small amount of halide.
These puzzling facts were explained when the X-ray crystal structure was done.'36
I he structure (Fig. 19.30) consists of a central halide ion surrounded by eight cop-
peri I) atoms bridged by sulfur ligands. These are in turn coordinated to six copper(II)
atoms. Finally, the chelating amino groups of the penicillamine complete the coordina¬
tion sphere of the copper(II) atoms.
As we have seen, the body has essentially no means of eliminating iron, so an
excessive intake of iron causes various problems known as siderosis. Chelating agents
are used to treat the excessive buildup of iron. In many cases the chelates resemble or
are identical to the analogous compounds used by bacteria to chelate iron. Thus
desfemoxamine B is the drug of choice for African siderosis.'37 The ideal chelating
Os. 0 Cu'.
& cr
O N. © Cu"
(a) (b)
, F i 9 'J 9 n 30 Moiecular structure of copper complex or rvpenicillamine. The
Cu^lpemc.llammateJ.jCI] ion: (a) the central cluster of Cu and ligating atoms only (b)
the entire ion with the central cluster oriented as in (a). [From Birker. P. J. M W L
Freeman. H. C. Chem. Common. 1976. 312. Reproduced with permission.)
Birker. P. J. M. W. L.; Freeman. H. C.: Chem. Common. 1976. 312.
£ ndcrscn - W F I" Inorganic Chemistry in Biology and Medicine-. Martell. A E Ed ■ ACS
V y R P °Bacon C B C R 40 Bnr n h Can Washin ? ,0n ' ^ m0 - Chapter 15. Gordcuk.
v. k., tiacon, B. R., Bnttenham. G. M. Ann. Rev. Nutr. 1987, 7. 485.
Medicinal Chemistry
957
agent will be specific for the metal to be detoxified since a more general chelating agent
is apt to cause problems by altering the balance of other essential metals The
concepts of hard and soft metal ions and ligands can be used to aid in this process of
designing therapeutic chelators. '38
A ,? g !!! i r.^ fferent mode of thera Py involves the use of c/j-diamminedichloroplat-
tnum(II), Pt(NH 3 ),Ci 2 , and related bis(amine) complexes in the treatment of cancer,
lhe exact action of the drug is not known, but only the cis isomer is active at low
concentrations, not the trans isomer. It is thought that the platinum binds to DNA
with the chloride ligands first being replaced by water molecules and then by a DNA
base such as guanine.'39 Studies in vitro with nucleotide bases as well as theoretical
ca dilations 40 indicate that the N7 position of guanine is the favored site for platinum
coordination. The m-diammine moiety can bind to groups about 280 pm apart, and in
vitro studies with di- and polynucleosides, as well as in vivo studies on DNA support
the hypothesis that the most important interaction is intrastand linking of two adjacent
guanine bases on the DNA chain by the platinum atom (see Fig. 19.31).'-" The trans
isomer can bond to groups about 400 pm apart that approach the platinum atom from
opposite directions, and it is chemotherapeutically inactive. The binding of cisplatin to
DNA would seriously interfere with the ability of the guanine bases to underao
r»M a°"u Cn v Ck faaSC painng ' Thus when a self-complementary oligomer (a portion of a
DNA chain) reacts with the cis isomer, two adjacent guanines are bound and Wat-
son-Lnck base pairing is disrupted:
m-Pt(NH 3 ),CI 2 + H ,0 ;=i cu-[Pt(NH 3 ) 2 CI(H 2 0 )) + + Cl“ (19.40)
A-p-G-p-G-p-C-p_c_p_T
T-p-C-p-C-p-G-p-G-p-A
rlj.|Pl(NH,l ; CI(H.Ol|
A-p-G-p-G-p-C-p-C-p-T
Pi
/ \
nh 3 nh 3
For cu-diamminedichloroplatinum(II) to work according to the proposed mecha¬
nism, it must hydrolyze in the right piece: if it hydrolyzes in the blood before it gets to
the chromosomes within the cell, it will be more likely to react with a nontarget
species. Fonunatdy for the stability of the complex, the blood is approximately 0.1 M
in chloride ion. forcing the hydrolysis equilibrium (Eq. 19.40) back to the chloro
complex. Once the drug crosses the cell membrane into the cytoplasm, it finds a
P A "rc C <: C ' : Martel i- A - E - in W omr Chemistry in Biology and Medicine. Martell. A. E., Ed.-
R,a "TTT SCn ^ M0: Amcncan Chemical Society: Washington. DC. 1980; Chapter 17.
DUlman, R. A. Struct. Bonding (Berlin ) 1987. 67, 91.
" 9 The kinetics of this substitution reaction is discussed in Chapter 13.
IJ " w an ^ V ' S : C . , ’ U - Y ' H - : Dun can. R. E.; Tobias. R. S. J. Am. Chem. Soc. 1978. 100. 607. Basch,
H.. Krauss. M.: Stevens. W. J.; Cohen. D. Inorg. Chem. 1986 , 25 . 684.
' 4I T’mT' S ’ Lippard - S - J - Chem - Rev - »87. 87. 1153. Recdjik, J.; Fichiinger-Schcpman.
m -,; " I Va £°° slcrom - A - T - : van dc Pane. P. Struct. Bonding (Berlin) 1987. 67. 53. Fouls. C. S ;
Marallt. L. G.: Byrd, R. A.: Summers. M. F.; Zon, G.; Shinozuka, K. Inorg. Chem. 1988. 27. 366.,
Lippert. B. Progr. Inorg. Chem. 1989, 37. 1-97.
IJ - Carradonna. J. P.; Lippard. S. J. Inorg. Chem. 1988 , 27 . 1454, Bruhn. S. L.; Toney. J H ■
lappard, S. J. Prog. Inorg. Chem. 1990. 38. 477-516. Lippert, B. Prog. Inorg. Chem. 1989, 37.
884
18 - Periodicity
Alternation of
Electronegativities
in the Heavier
Nonmetals
c) isoelectronic plus an inert pair with [e.g., TKI), Pb(ll)] 50 the gold(I) atom. Some
examples are listed in Table 18.6. The number of these bonds is limited, but the
subject is still a very new one.
The general tendency for atoms (including other gold atoms) to exhibit greater
than expected valences toward gold atoms is often termed aurophilicity. This is a
useful descriptive name for a bonding behavior that is not completely understood. It
appears to result from relativistic effects and the fact that the gold 5 d'° electrons do
not act as “good” core electrons but mix with low lying excited states. 51 To ra¬
tionalize, if not truly explain, one can consider promotion of electrons from the 5d'°
configuration and their involvement in the bonding.
Schmidbaur’s group 5 - has synthesized some gold(I) compounds with unusual
coordination numbers for small nonmetals. For example, the "hedgehog cation,”
[C(AuPRj)J 2+ , has carbon with the unusual covalency of six (Fig. 18. II). While
carbon has no low energy d orbitals, there is nothing to prevent it from forming ri, A .(2j)
and t|„(2p) MOs and forming three-center bonds. So why should it do so in this
compound and never in "organic chemistry"? Ordinarily the better overlap of
hybridized sp" carbon orbitals ensures carbon's tetracovalency. Perhaps the pos¬
sibility of a dozen Au(I)—Au(I) aurophilic bonds could provide another 300 - 400 kJ
mol -1 , commensurate with the energy of a C—C bond, compensating for weaker
C—Au bonding.
We have seen above the unusual properties of the nonmetals following the first row of
transition metals. This is usually described as “a reluctance to exhibit maximum
oxidation state,” but it may also be stated in terms of an increased electronegativity in
these elements.™ Indeed, gallium, germanium, arsenic and perhaps selenium seem to
have higher electronegativities than their lighter congeners.
In the same way, it has been suggested that the heavier member of each family,
thallium, lead, perhaps bismuth, has a greater electronegativity than its lighter con-
Table 18.6
Some further examples of
Compound
M,, M 2
Mi—M 2 (pm)
Group IB (11) metal-motal
interactions
[AgCH,P(S)Ph,] 2
Au[CH,P(S)Ph,],Hg
[Cl Au P(S) Ph jCH 2 1, Hg
Au[CH,P(S)Ph,],Ti
[Au(CH 2 P(S)Ph,) 2 ],Pb
Ag 1 , Ag 1
Au', Hg"
Au', Hg"
Au', Tl'
Au'. Pb"
299.0-
308.5
331.0, 336.I
295.9
289.6, 296.3‘-.<'
Sec Footnote 48.
See Footnote 49.
•' There are two Au-M 2 interactions per molecule.
■' See Footnote 50.
5“ Wang. S.: Garzbn, G.; King. C.; Wang. J-C.; Facklcr. J. P.. Jr. Inorg. Client. 1989. 28. 4623-4629.
•’t Jansen. M. Angew. Cliem. hit. Ed. Engl. 1987. 26. 1098-1110. Rosch, N.; Gorling. A.; Ellis. D. E.;
Schmidbaur. H. Ibid. 1989, 28. 1357-1359. Pyykkd, P.; Zhao. Y. Ibid. 1991. JO. 604-605.
Scherbaum, F.: Grohmann, A.: Huber. B.; Krflgcr, C.; Schmidbaur. H. Angew. Cliem. Int. Ed.
Engl. 1988. 27. 1544-1546. Scherbaum, F.; Grohmann. A.; Muller. G.; Schmidbaur. H. Ibid. 1989.
28. 463-465. Steigclmann. O.: Bissingcr, P.; Schmidbaur. H. Ibid. 1990. 29. 1399-1400.
55 Indeed, in many ways, these statements are equivalent. See Problem 18.17.
Problems
885
Conclusion
Problems
Fig. 18.11 Structure of the "hedgehog" dication. [(Ph,PAu) 6 C] lH \ The inferred Au—Au
bonds are not shown. [From Scherbaum. F.; Grohmann. A.; Huber, B.; Kruger. C.;
Schmidbaur. H. Angew. Client. Int. Ed. Engl. 1988. 27. 1544-1546. Reproduced with
permission.]
gener, indium, tin. and antimony. When first proposed the explanation rested on the
lanthanide contraction acting on these elements. We now know that relativistic effects
foi these elements are at least as important as the lanthanide contraction and perhaps a
better way of staling the premise is that all of the elements from about platinum on
(there is no sharp demarcation, of course) are more electronegative than would
otherwise be expected.
I he periodic chart is the inorganic chemist's single most powerful weapon when faced
with the problem ol relating the physical and chemical properties of over 100 ele¬
ments. In addition to knowing the general trends painted in broad brush strokes by the
simple rules, the adept chemist should know something of the "fine structure” that is
at the heart of making inorganic chemistry diverse and fascinating.
18.1 Compare Figs. 16.36 and 16.37. The difference in P—O bond lengths in P.,0, 0 is 158 -
141 = 17 pm (11%), but the difference in the P—S bonds in P.,S m is only 209 - 196 =
13 pm (6%). Explain.
18.2 With regard to nuclear engineering, the separation of zirconium and hafnium has been of
considerable interest because of the low neutron cross section of zirconium and the high
neutron cross section of hafnium. Unfortunately, the separation of these two elements is
perhaps the most difficult of any pair of elements. Explain why.
j
-_ 3
954
19-The Inorganic Chemistry of Biological Systems
name "scouring rushes.” Some Protozoa (radiolarians), Gastropoda (limpets), and
Porifera (glass sponges) employ silica as a structural component. Silicon is an essential
trace element in chicks and rats 131 and is probably necessary for proper bone growth
in all higher animals.
The third type of compound used extensively as a structural component is apatite,
Ca ? (P0 4 ) 3 X. Hydroxyapatite (X = OH) is the major component of bone tissue in the
vertebrate skeleton. It is also the principal strengthening material in teeth. Partial
formation of fluorapatite (X = F) from application of fluorides strengthens the struc¬
ture and causes it to be less soluble in the acid formed from fermenting organic
material, hence a reduction of caries. Fluorapatite is also used structurally in certain
Brachiopod shells.
Medicinal Chemistry The suggested antibiotic action of transferrin is typical of the possible action of several
antibiotics in tying up essential metal ions. Streptomycin, aspergillic acid, usnic acid.
Antibiotics and the tetracyclines, and other antibiotics are known to have chelating properties. Pre-
Related Compounds sumably some antibiotics are delicately balanced so as to be able to compete success¬
fully with the metal-binding agents of the bacteria while not disturbing the metal
processing by the host. There is evidence that at least some bacteria have developed
resistance to antibiotics through the development of altered enzyme systems that can
compete successfully with the antibiotic. 132 The action of the antibiotic need not be a
simple competitive one. The chelating properties of the antibiotic may be used in
metal transport across membranes or to attach the antibiotic to a specific site from
which it can interfere with the growth of bacteria.
The behavior of valinomycin is typical of a group known as "ionophore antibiot¬
ics." 133 These compounds resemble the crown ethers and cryptates (Chapter 12) by
having several oxygen or nitrogen atoms spaced along a chain or ring that can wrap
around a metal ion (Fig. I9.29a). These antibiotics are useless in humans because they
are toxic to mammalian cells, but some of them find use in treating coccidiosis in
chickens. The toxicity arises from the ion-transporting ability. Cells become "leaky"
with respect to potassium, which is transported across the cell membrane by val¬
inomycin. In the absence of a metal ion, valinomycin has a quite different conforma¬
tion (Fig. 19.29b), one stabilized by hydrogen bonds between amide and carbonyl
groups. It has been postulated 1 34 that the potassium ion can initially coordinate to the
four free carbonyl groups (A) and that this can provide sufficient stabilization to break
two of the weaker hydrogen bonds (B). This provides two additional carbonyl groups
to coordinate and complete the change in conformation to that shown in Fig. 19.29a.
Such a stepwise mechanism would indicate that the whole system is a balanced one
and that the reverse process can be readily triggered by a change in environment such
as at a membrane surface or if there is a change in hydrogen bonding competition.
Carlisle, E. M. Science 1972. 178, 619; Fed. Proc.. Fed. Am. Soc. Exp. Biol. 1974. JJ. 1758.
133 Woodruff. H. 8.; Miller. I. M. In Metabolic Inhibitors-, Hochstcr, R. M.; Quaslcl. J. H.. Eds.;
Academic; New York. 1963; Vol. II. Chapter 17.
133 Ochiai. E-I. General Principles of Biochemistry of the Elements'. Plenum. New York. 1987;
pp 254-265.
. ^.J 3 Smith, G. D.; Dual. W. L: Langs. 0. A.; DeTilla. G. T.; Edmonds. J. W.; Rohrer, D. C.; Weeks.
... . „ CM. J. Am. Chem. Soc. 1976. 97. 7242.
Medicinal Chemistry
955
• Nitrogen
(b)
Fig. 19.29 la) Molecular structure of valinomycin coordinated to the K‘ ion. lb) Molecular
structure of the free valinomycin molecule. The carbonyl groups marked A arc free to
coordinate to K". Hydrogen bonding is shown by dashed lines, with (hose marked 3 thought
to be most susceptible to breaking. |From Neupert-Laves. K.; Doblcr. M. Hdv. Chitn. Acta
1975. SB. 432; Smith G. D.. Duax. W. L; Langs. D. A.; DeTitta. G. T.: Edmonds. J. W.;
Rohrer. D. C.; Weeks. C. M. J. Ant. Cliern. Soc. 1976, 97, 7242. Reproduced with
permission.)
The tetracyclines form an important group of antibiotics. The activity appears to
result from their ability to chelate metal ions since the extent of antibacterial activity
parallels the ability to form stable chelates. The metal in question appears to be
magnesjum or calcium since the addition of large amounts of magnesium can inhibit
the antibiotic effects. In addition, it is known that in blood plasma the tetracyclines
exist as calcium and magnesium complexes. 135
i 33 Lambs. L.; Dccock-Le Reverend. B.: Kozlowski. H.; Berthon, G. Inorg. Chem. 1988. 27, 3001.
886
18 • Periodicity
18.3 Carbon tetrachloride is inert towards water but boron trichloride hydrolyzes in moist air.
Suggest a reason.
18.4 Below are some conclusions that an average general chemistry student (certainly not
you!) might have after reading about the periodic table in a general chemistry textbook
written by average authors (not us!). Please rewrite each statement to clarify possible
misconceptions (if any).
a. Electron affinities increase toward the upper right of the periodic table.
b. Ionization energies decrease toward the bottom of the table.
c. Atomic radii increase toward the bottom of the table.
d. Atomic radii decrease toward the right of the table.
e. Electronegativity decreases toward the left and toward the bottom of the table.
18.5 Gallium dichloride. GaCL, is a diamagnetic compound that conducts electricity when
fused. Suggest a structure.
I8.fi If a major breakthrough in nuclear synthesis were achieved, two elements that are hoped
for are those with atomic numbers 114 and 164. both congeners of lead. Look at the
extended periodic table in Chapter 14 and suggest properties (such as stable oxidation
states) for these two elements. How do you suppose their electronegativities will compare
with those of the other Group 1VA (14) elements? 54
18.7 The small F—S—F bond angles in F,S=N can be rationalized by
a. Bent's rule
b. Gillespie-type VSEPR rules
c. Bent bonds
Discuss each and explain their usefulness (or lack thereof) in the present case.
18.8 Either look up the article by Chen and Wentworth 55 or plot the electron affinities from
Table 2.5 onto a periodic chart. Discuss the reasons for the “exceptions' - that you
observe.
18.9 Write the first ionization energies from Table 2.3 on a periodic chart. Discuss the reasons
for the "exceptions" that you observe.
18.10 On page 863 the statement is rather casually made that “R : Ge. R ; Sn. and R : Pb exist as
diamagnetic monomers in solution." What experiments must an inorganic chemist per¬
form to substantiate these statements? 5 ' 1
18.11 The compound R 4 Sn : shown on page 863 is diamagnetic. Draw out the most reasonable
electronic structure for it. and compare it with the geometric structure. Discuss.
18.12 Lithium carbonate is often administered orally in the treatment of mania or depression or
both. From what you have learned of the diagonal relationships in the periodic chart,
predict one possible unpleasant side effect of lithium therapy.
18.13 Zinc is a much more reactive metal than cadmium, as expected from the discussion on
pages 877-879. Yet both are used to protect iron from rusting. How is this possible?
18.14 Sodium hypophosphite. NaFLPO;, has been suggested as a replacement of sodium nitrite.
NaNCK, as a meat preservative to prevent botulism. Draw the structure of each anion.
54 See Seaborg. G. T. J. Cltem. Educ. 1969, -16. 626; Katz. J. J.; Morss. L. R.; Seaborg. G. T. In Tlw
Chemistry of the Actinide Elements-. Katz. J. J.; Morss. L. R.; Seaborg. G. T.. Eds.; Chapman and
Hall: New York. 1986. See also Footnote 31.
55 Footnote 42.
56 See Footnote 13.
Problems
887
18.15 The simplest relationship between electronegativity and dipole moments is a linear one:
The greater the difference in electronegativity, the greater the dipole. How can you
reconcile this with the N—O and P—O dipoies cited in this chapter (page 889)?
18.16 The P 4 molecule in white phosphorus has extremely strained bonds: The bond angles in
the Tj molecule are only 60°. Therefore the bonds are weak, only 201 kJ mol -1 for each
one. The total bond energy of two moles of P 2 (962 kJ. Appendix E) is 244 kJ less than that
of one mole of P 4 (1206 kJ. Appendix E). In contrast, the total bond energy of two moles
of N : has been calculated 57 to be 777 kJ greater than that of one mole of N 4 (a hypothetical
tetrahedral molecule isostructura! with P 4 ).
a. Explain the disparity of bond energies of these isoelectronic and isostructural
molecules.
b. If you could manufacture N., and keep it as a metastable material, can you think of any
uses for it?
18.17 Footnote 53 suggests that a reluctance to exhibit maximum oxidation slate may be
equivalent to an increased electronegativity in the posttransition elements. Discuss.
18.18 Of the five data points in Fig. 18.8. that of AuCH : Et : PCH : AuCH : El : PCH ; fits most
poorly. Suggest a possible reason.
18.19 Assuming that CsAu is ionic, what is its most probable structure? Estimate an enthalpy of
formation for it.
18.20 Two variables affecting the isomer shift in Mdssbauer spectroscopy are the s character of
the orbitals involved and the partial charge on the atom being studied. Explain this
phenomenon cliemicnlly. i.e.. in terms of how these two quantities afiect the electron
density at the nucleus.
18.21 Despite the half-century history of Scaborg's hypothesis with respect to the placement of
the lanthanides and actinides in the periodic table, many bench chemists proceed success¬
fully with the working hypothesis that uranium is a congener of molybdenum and tung¬
sten. Does this indicate that Seaborg is wrong? That the bench chemists are wrong?
Discuss.
18.22 How many parallels can you find between [Fe„(CO)„,C| ; (Fig. 15.21b) and
|(AuPI‘hi).,C|" (Figs. 15.21c and 18.11) that contribute to the unusual hexacoordinate
carbon atom?
18.23 Explain the electronic and structural changes involved in the following reaction:
(18.56)
57 Lee. T. J.; Rice. J. E. J. Cltem. Phys. 1991. 94. 1215-1221.
952
19 The Inorganic Chemistry of Biological Systems
Biochemistry of the Nonmetals
953
Adaptations
to Natural
Abundances 125
terrestrialism) and perhaps considering the microabundance of the various elements in
different habitats.
When the abundance of an element is unusually high or unusually low. organisms
develop mechanisms to handle the stress. The first documented examples were the
presence of “indicator species” (plants) that grow where soils contain an usually high
concentration of a metal. For example, the sea pink. Armeria maritima, has been used
in North Wales as an indicator of copper deposits. In one extreme case, the drainage
from the copper deposits has concentrated in a bog to an extent of 20,000-30,000 ppm,
and the sea pink flourishes. Closely related is the adaptation of various plants to
exceptionally high concentrations of various heavy metals in mine dumps and tailings.
Not only have some species adapted to extremely high concentrations of normally
toxic metals, but they have also evolved a high level of self-fertilization to prevent
pollination and gene exchange with nearby populations that are not metal tolerant.
Some of the chemolithotropic bacteria discussed earlier in this chapter illustrate
these ideas. In undisturbed situations their habitat is extremely restricted. During the
process of mining, however, large surfaces of the appropriate metal sulfide are created
and oxidized both in the mine and in the tailings with resultant leaching. This creates a
favorable habitat for exploitation by the bacteria. One unfavorable result is the
lowering of the pH and the solubilization of metals, usually toxic, into the drainage
system. On the other hand, the isolation and selection of productive strains from such
sites, and their controlled application, may lead to useful biometallurgical methods of
extraction of metals from low-grade ores (see Chapter 10). 126
The hydrothermal vents discussed previously provide a parallel, natural environ¬
ment with unusually large amounts of various metals—iron, copper, zinc—dissolved
from the crustal rocks by the superheated water. It will be of interest to learn how the
animals in the hydrothermal ecosystem have developed mechanisms to avoid toxicity
from these metals. Another source of possible toxicity, hydrogen sulfide, is somewhat
better understood. Hydrogen sulfide is comparable to the cyanide ion in its toxicity
towards respiration. The vent organisms have evolved a variety of mechanisms to
prevent sulfide toxicity. One of the more interesting is that of the tube worm, Riftia
pachyptila. Its hemoglobin has a molecular weight of about two million, with an
extremely high affinity for dioxygen (recall that the vent waters are anoxic) and a
second, high-affinity site to bind sulfide. This second site serves the dual purpose of
protecting the tube worm's cytochrome c oxidase from sulfide poisoning and protect¬
ing the sulfide from premature oxidation. Instead, both the dioxygen and sulfide are
transported to symbiotic bacteria that metabolize them to drive the synthesis of ATP
and carbohydrates. 127
At the other extreme are adaptations to very low concentrations of a particular
element. We have already seen mechanisms directed towards the sequestration of iron
when it is present in small amounts. The ability to detect extremely small amounts of
an element can be a useful adaptation for an animal if that element is important to it.
For example, hermit crabs recognize shells suitable for occupation not only by tactile
123 Farago. M. E. In Frontiers in Bioinorganic Chemistry; Xavier, A. V., Ed.; VCH: Weinheim. 1986;
pp 106-122. Ochiai. E.-I. General Principles of Biochemistry of the Elements ; Plenum: New York
1987; pp 379-395.
124 Rossi. G. Biohydrometallurgy; McGraw-Hill: New York, 1990.
127 Childress, J. J.; Felbeck, H.; Somero, G. N. Sci. Am. 1987, 25615). 115-120.
stimuli but apparently also by the minute amount of calcium carbonate that is dis¬
solved in the water around a shell. They can readily distinguish natural shells
(CaC0 3 ), calcium-bearing repiicas (CaSOJ, and naturally containing calcium minerals
(calcite, aragonite, and gypsum) from non-calcium minerals (celestite, SrS0 4 ; rho-
dochrosite. MnC0 3 ; siderite, FeC0 3 ; and quartz, SiO,). l2S Inasmuch as the solubility
product of calcium carbonate is only KT 8 . the concentration of calcium detected by
the hermit crab is of the order of 4 ppm or less. Almost nothing is known about the
chemical mechanisms used by organisms in detecting various elements.
Biochemistry
of the Nonmetals
Structural Uses
of Nonmetals 129
Many of the nonmetals such as hydrogen, carbon, nitrogen, oxygen, phosphorus,
sulfur, chlorine, and iodine are essential elements, and most are used in quantities far
beyond the trace levels. Nevertheless, most of the chemistry of these elements in
biological systems is more closely associated with organic chemistry than with in¬
organic chemistry.
There are three important minerals used by organisms to form hard tissues such as
bones and shells. The most widespread of these is calcium carbonate, an important
structural component in animals ranging from Protozoa to Mollusca and Echinoder-
mata. It is also a minor component of vertebrate bones. Its widespread use is probably
related to the generally uniform distribution of dissolved calcium bicarbonate. Ani¬
mals employing calcium carbonate are most abundant in fresh waters containing large
amounts of calcium and magnesium (“hard water") and in warm, shallow seas where
the partial pressure of carbon dioxide is low (e.g., the formation of coral reefs by
coelenterates). The successful precipitation of calcium carbonate depends upon the
equilibrium;
Ca 2 + 2HC0 3 ~ — CaC0 3 + CO, + H,0 ( 19 . 39 )
and is favored by high [Ca - ] and low [CO,]. Nevertheless, organisms exhibit a
remarkable ability to deposit calcium carbonate from hostile environments. A few
freshwater clams and snails are able to build reasonably large and thick shells in lakes
with a pH of 5.7-6.0 and as little as l.l ppm dissolved calcium carbonate. 130
It is of interest that two thermodynamically unstable forms of calcium carbonate,
aragonite and vaterite. are found in living organisms as well as the more stable calcite.
There appears to be no simple explanation for the distribution of the different forms in
the various species.
Tissues containing silica are found in the primitive algal phyla Pyrrhophyta
(dinoflagellates) and Chrysophyta (diatoms and siiicoflagellates). One family of higher
plants, the Equisetaceae. or horsetails, contains gritty deposits of silica—hence their
r - Mescc. K. A. Science 1982. 215. 993.
129 Vincent. J. F. Structural Biomalerials; Wiley; New York. 1982. Williams. R. J. P. In Frontiers in
Bioinorganic Chemistry-. Xavier. A. V., Ed.; VCH: Weinheim, 1985; pp 431-440. Webb, J.; St.
Pierre. T. G.; Dickson, D. P. E.; Mann. S.; Williams. R. J. P.; Perry. C. C.; Grime. C. C.; Watt, F.;
Runham. N. W. Ibid, pp 441-452.
130 For a discussion of this point as well as other examples of organisms living on limited concentra¬
tions of nutrients, see Allee, A. C.; Emerson. E. E.; Park. O.: Park. T.; Schmidt. K. P. Principles
of Animal Ecology. W. B. Saunders: Philadelphia, 1949; pp 164-167; pp 189-206. Pennak, R. W.
Freshwater Invertebrates of the United States'. Ronald: New York, 1953; p 681; p 705f.
888
18 Periodicity
Chapter
18.24 Discuss any similarities and differences between Eq. 18.56 above and the behavior of
“Pt(pop)” catalysts in Chapter 15.
18.25 From the discussions in this chapter and those in Chapter 14, plus any further data you
find in standard reference works, write a Journal of Chemical Education-lype article
entitled: “Alchemy reversed: The remarkable chemistry of turning gold into even more
interesting substances!" 58
The Inorganic
Chemistry of
Biological Systems
Th e chemistry of life can ultimately be referred to two chemical processes: (I) the use
of radiant solar energy to drive chemical reactions that produce oxygen and reduced
organic compounds from carbon dioxide and water, and (2) the oxidation of the
products of (1) with the production of carbon dioxide, water, and energy. Alter¬
natively, living organisms have been defined as systems capable of reducing their own
entropy at the expense of their surroundings (which must gain in entropy). 1 An
important feature of living systems is thus their unique dependence upon kinetic
stability for their existence. All are thermodynamically unstable—they would burn up
immediately to carbon dioxide and water if the system came to thermodynamic
equilibrium. Life processes depend upon the ability to restrict these thermodynamic
tendencies by controlled kinetics to produce energy as needed. Two important aspects
of life will be of interest to us: (1) the ability to capture solar energy; (2) the ability to
employ catalysts for the controlled release of that energy. Examples of such catalysts
are the enzymes which control the synthesis and degradation of biologically important
molecules. Many enzymes depend upon a metal ion for their activity. Metal-contain¬
ing compounds are also important in the process of chemical and energy transfer,
reactions which involve the transport of oxygen to the site of oxidation and various
redox reactions resulting from its use.
58 Lest we get carried away by the rhetoric, this would simply be a review article of gold chemistry
with the provocative thesis that it has a more varied and interesting chemistry than any other
element.
Energy Sources
for Life
It may be somewhat surprising that most of the reactions for obtaining energy for
living systems are basically inorganic. Of course, the reactions are mediated and made
possible by complex biochemical systems.
Nonphotosynthetic Even though almost all living organisms depend either directly (green plants) or
Processes indirectly (saprophytes and animals) upon photosynthesis to capture the energy of the
sun. there are a few reactions, relatively unimportant in terms of scale but extremely
1 These reductionist definitions of life are not meant to imply that life processes or living organisms are
simplistic or any the less interesting. A similar definition of physics and chemistry might be "the
study of the interactions of matter and energy." None of these definitions hints at the fascination of
some of the problems presented by these branches of science.
950
951
1 9 • The Inorganic
Table 19.4 (Continued )
Abundances of the elements
in the earth's crust, rivers,
and sea water 0
4 X JO-W
| 2 x 10-iw.
i; 0.0033
° Riley, J.P.; Chester, R. Introduction to Marine Chemistry ; Academic: New York, 1971, except—
as noted. v ' -
y ; ; • - ]:t
c Combined nitrogen; about 15 mg L 1 dissolved N ; .;> y;..t ..(J ...
d Considerable' variation occurs. ';
? 1979 ^ % 1
Cha p-- . i S i % M
Chemistry of Biological Systems
Earth's crust
Element
__River water .. — Ocean water
atoms/10 4 atoms Si mg L -1 •’ ; mg L -1
Tin
Antimony
Tellurium
Iodine
Xenon
Cesium
Barium
Lanthanum
Cerium
Praseodymium
Neodymium_J
Promethium
Samarium
Europium
Gadolinium
Terbium
Dysprosium
Hoimium
Erbium
Thulium - -
Ytterbium
Hafnium 3
Tantalum
Tungsten
Rhenium
Osmium
0.002
0.0002
4 x 10 _6r
0.0005
0.003
0.425 ‘
0.03 J .
0.06
0.0082
0.028
0.006
0.0012
0.0054
0.0009
0.003
0.003
0.003
0.002
0 . 02 —''
0.002
4 x I0- 5 °
0.004
-0.005
0.02
.. .I' 3.08
0.2
0.4
0.058
0.19
0.04
0.08
0.034
0.006
x 10-5
.01
.0002 4.
1 x 10-5 ~
0.0002
0.06 c
5 x I0- 6
0.0005
0.03
3.4 x 10- 6
1.2 x 10-4
6 x 10-7 Vs
x 10-4
4.3 X 10-7
1.3 X 10-7
X 10-7
1.4 x 10-7
X 10-7
X 10-7
X 10-7
X 10-7 ..
x 10-7 ■
2 x 10-5
0.00012
X 10-4 ^
Gold 4 x 10-« -r- 2 x 10-5 ... .2 x 10-
Mercury .8 x 10-5 4 x 10- .. 7 X 10-
Thallium 0.00045 . 0.0022 »
Lead 0.0125
Bismuth 0.00014 - 7 x 10—
Polonium
Astatine
Radon
r” : 4 x ,0
Actinium ; •• .. -• ' y.tt
Thorium - 0.0096 0.041J . --- 0.0001
Protactinium yu: ...Jr 2 ,. 5 i.
Uranium ;£ br.a 0.0027 0.011.? 4x10
x 10-» 3
X 10-5rf
X 10-4 r:
x io-5rfT
X 10-5
x 10-'
x 10-'4^
x 10-'"?
Essential and Trace Elements in Biological Systems
crustal abundances: Pb (0.08), Cd (0.0018), and Hg (4 x 10 5 ). 121 The conclusion is
inescapable: Life evolved utilizing those elements that were abundant and available to
it and became dependent upon them. Those elements that are rare were not used by
living systems because they were not available; neither did these systems evolve
mechanisms to cope with them.
A closely related corollary of this thesis is that many elements that are essential
when occurring at ambient concentrations are toxic at higher concentrations (and, of
course, cause deficiency symptoms at lower concentrations). Interesting examples are
copper, selenium, and even sodium—all oceanic organisms are adapted to live in 0.6
M NaCl and our blood has been described as a sample of the primeval seas. Yet too
high concentrations of NaCl are toxic through simple hypertonicity, i.e., osmotic
dehydration. Selenium is a problem when it is either too rare or too abundant in the
environment: Livestock grown on selenium-deficient pasture suffer from “white mus¬
cle disease"; when grazing plants (Astragalus, "locoweed”) that concentrate se¬
lenium from the soil, they suffer central nervous system toxinosis. Copper is essential
to many of the redox enzymes necessary to both plants and animals; yet too mpch
copper is severely toxic to most green plants.
There is an interesting group of trace elements, called ultratrace elements because
they are needed, if at all, at not more than 1 ppm in food, probably less than 50 ppb.
These ultratrace elements include arsenic and nickel, certainly essential at these low
concentrations, and cadmium and lead, probably not essential. Many of these ele¬
ments (e.g., Ni, As. Cd, and Pb) are quite toxic at any concentration much above an
ultratrace level. Naturally, determination of the essentiality of an ultratrace element is
even more difficult than for ordinary trace elements. 122 Life used and adapted to those
elements and those concentrations available to it (see next section). When humans
started mining, using, and releasing these elements into the cn ronment, the eco¬
system was faced with hazards it had never before encountered, and to which it had,
therefore, never adapted.
A slightly different view of this idea has been presented by t-gami, 1 -’ who has
pointed out that the three enzyme systems in the most primitive bacterium,
Clostridium, are involved in electron transfer (e.g., ferredoxin). reduction of small
molecules (e.g., nitrogenase). and hydrolysis (e.g., carboxypeptidase and carbonic
anhydrase), and employ, respectively, iron, molybdenum, and zinc, the three most
common transition elements in sea water. It is postulated that these enzyme systems
arose from protoenzymes that utilized these most common metals in primitive seas.
One puzzle is copper, which is fairly abundant in sea water, and although it has been
thought to be essential for all organisms, apparently no requirement for it has been
found in strict anaerobes. Egami postulates that copper, with a positive standard
reduction potential, was incorporated into living systems only when the atmosphere
shifted from reducing (CH 4 . H 2 , NH 3 ) to oxidizing (0 2 ).' 24 This indicates the impor¬
tance of considering changes that have occurred with time (including the advent of
K' All figures in atoms per 10.000 atoms silicon. For a discussion of Pb. Cd, and Hg in the environ¬
ment, in diet, and their toxicity, sec Choudury, B. A.; Chandra, R. K. Prog. Food Nittr. Sci. 1987.
II, 55.
22 Nielsen. F. H. Ann. Rev. Nu/r. 1984. 4. 21.
'25 Egami. F. J. Mol. Evol. 1974, 4, 113: J. Biochem. 1975, 77. 1165.
' 2 See Broda. E. J. Mol. Evol. 1975, 7, 87, for a discussion of this and related questions concerning
the primitive biosphere.
890
19-The Inorganic Chemistry of Biological Systems
interesting in terms of chemistry, utilizing inorganic sources of energy. Even these
may be indirectly dependent upon photosynthesis, since it is believed that all free
oxygen on earth has been formed by photosynthesis.
Chemolithotrophic 2 bacteria obtain energy from various sources. For example.
iron bacteria produce energy by the oxidation of iron(II) compounds:
2Fe 2+ - J2L f Fe-jOj + energy (19.1)
Nitrifying bacteria are of two types, utilizing ammonia and nitrite ion as nutriments:
2NH, 2NOT + 3H 2 0 + energy (19.2)
NOJ NOj - + energy (19.3)
Though they are photolithotrophs (Gr. photos, "light”) and thus more closely related
to the chemistry of normal photosynthesis (see page 916), the green sulfur bacteria
and the purple sulfur bacteria are included here to demonstrate the diverse bacterial
chemistry based on sulfur paralleling the more common biochemistry involving water
and oxygen. Light energy is used to split hydrogen sulfide into sulfur, which is stored
in the cells, and hydrogen which forms carbohydrates, etc., from carbon dioxide.
To return to the chemolithotrophs, there are species of sulfur bacteria that obtain
energy from the oxidation of various states of sulfur:
8 H,S S 8 + 8H 2 0 + energy (19.4)
S„ + 8H 2 0 -iSU 8SOj“ + 16H + + energy (19.5)
These latter reactions are the source of energy for a unique fauna, one completely
isolated from the sun on the floor of the oceans. These ecosystems have been
discovered at certain rifts in the earth's crust on the ocean's floor, where large
amounts of sulfide minerals are spewed forth from hydrothermal vents. 3 The sulfide
concentration, principally in the form of hydrogen sulfide, ranges routinely up to
100 |iM depending upon the dilution of vent water by surrounding sea water. The H 2 S
has been shown to be depleted, along with 0 2 , in the midst of the aggregated
organisms, and it is the energetic basis of these communities. 4 The sulfide is oxidized
by bacteria as shown above. It is of considerable interest that the enzymes, mecha¬
nisms, and products of this chemically driven synthesis are essentially identical to
those of photosynthesis (page 916), except that the source of electrons for the reduc¬
tion of water to carbohydrates is sulfur(-II) rather than photoactivated chlorophyll.
In addition to free-living bacteria, many of the vent animals contain endosymbiolic
bacteria that serve them as primary energy sources as well as the source of reduced
carbon compounds. The parallel between these endosymbionts in rift animals, such as
tube worms, clams, and mussels, and the chloroplasts of plants is striking. 5 Whether
2 That is. feeding (Gr. trophos) on inorganic (Gr. lithos. “stone") chemicals.
3 Spiess, F. N., Macdonald, K. C.; Atwater. T.; Ballard. R.; Carranza, A.; Cordoba. D.: Cox. C.;
Diaz Garcia, V. M.; Franchetcau. J.; Guerrero. J.; Hawkins. J.; Haymon, R.; Hessler. R.; Juteau.
T.; Kastner, M.; Larson. R.; Luycndyk. B.; Macdougall, J. D,; Miller. S.; Normark, W.; Orcult. J.;
Rangin. C. Science 1980, 207, 1421-1433. Hekinian. R.; Fcvrier, J. L.; Picot, P.; Shanks, W. C. Ibid.
1980. 207, 1433-1444. Edmond. J. M; Von Damm. K. L.; McDulT. R. E.; Measures. C. I. Nulure
(LondonI 1982. 297, 187.
4 Johnson. K. S.; Bcehler. C. L.; Sakamoto-Arnold. C. M.; Childress. J. J. Science 1986. 231.
1139-1141.
5 Childress, J. J.; Felbeck, H.; Somero, G. N. Sci. Am. 1987. 256(5), 114-120.
Metalloporphyrins and Respiration
this parallelism results from an adaptation of the cycle from photosynthetic bacteria,
or whether these chemolithotrophic bacteria are possibly ancestral to photosynthetic
organisms presents the age-old phylogenetic problem—which came first, the chicken
or the egg? The entire community, including predator species such as crabs, is entirely
independent of photosynthesis except for the use of by-product dioxygen. There is
even evidence that some of the animals such as gutless clams can metabolize hydrogen
sulfide independently, simultaneously detoxifying it and using it as an energy source. 6
Metalloporphyrins
Some of the simplest bioinorganic compounds are the various cytochromes. In terms
of overall structure and molecular weight they are anything but simple. However, the
inorganic chemistry of several of them is very simple coordination and redox chem¬
istry. The active center of the cytochromes is the heme group. It consists of a
porphyrin ring chelated to an iron atom. The porphyrin ring consists of a macrocyclic
pyrrole system with conjugated double bonds (Fig. 19.1) and various groups attached
to the perimeter. We shall not be concerned with the nature and variety of these
substituents except to note that by their electron-donating or electron-withdrawing
ability they can “tune" the delocalized molecular orbitals of the complex and thus
vary its redox properties. The porphyrin can accept two hydrogen ions to form the +2
diacid or donate two protons and become the -2 dianion. It is in the latter form that
the porphyrins complex with metal ions, usually dipositive, to form metalloporphyrin
complexes.
From the covalent bond radii (Table 8.1) we can estimate that a bond between a
nitrogen atom and an atom of the first transition series should be about 200 pm long.
The size of the "hole" in the center of the porphyrin ring is ideal for accommodating
metals of the first transition series (Fig. 19. lb). The porphyrin system is fairly rigid.
and Respiration
Cytochromes
Fig. 19.1 (a) The porphyrin molecule. Porphyrins have substituents at the eight pyrrole
positions, (b) A "best" set of parameters for an "average" porphyrin skeleton. Distance in
pm. [From Fleischer, E. B. Arc. Chem. Res. 1970. 3, 105. Reproduced with permission.)
6 Powell. M. A.; Somero. G. N. Science 1986. 233, 563.
icaDie iy.4 n-onrintiea/
Atomic
number Element Biological functions
82 Lead None known.
88-103 Radium None known,
and
Actinides
92 Uranium None known.
Very toxic to most plants;
cumulative poison in mam¬
mals. Inhibits &-
aminolevulinate dehydrase
and thus hemoglobin syn¬
thesis in mammals (see Al).
One of the symptoms of
lead poisoning is anemia.
Toxic to central nervous
system.
May be concentrated in organ¬
isms and toxic as a result of
radioactivity.
U VI is reduced to U lv by iron-
reducing bacteria. May be .
important in the bio¬
geochemical deposit of U. 44
Worldwide pollutant of the at¬
mosphere, concentrated in ‘
urban areas from the com- I
bustion of tetraethyl lead in
gasoline; local pollutant .. i
from mines; some poisoning
from lead-based paint pig¬
ments.
Potential pollutants from use
of nuclear fuel as energy :
source.
Iron-reducing bacteria might
be of use in decontaminat- ,
ing (precipitating) uranium-
,• polluted water. 4
" Toxic effects often are exhibited only at concentrations above those occurring naturally in the environment. See p 952.
Crabtree, R. H. The Organometallic Chemistry of the Transition Metals', Wiley: New York, 1988; pp 400-404.
c Fieve. R. R.; Jamison, K. R.; Goodnick. P. J. In Metal Ions in Neurology and Psychiatry; Gabay, S.; Harris, J.; Ho, B. T., Eds.;
Alan R. Liss: New York. 1985; pp 107-120. : I
d Griggs. K. Science 1973, 181. 842.
' Nielsen, F. H. Ann. Rev. Nutr. 1984,4,21.
J Mohnen, V. A. Sci. Ajner. 1988.259(2), 30. - . . ;
Fridovich. . Am. Sci. 1975, 63. 54. Biological and Clinical Aspects of Superoxide and Superoxide Dismutase; Bannister, W. H.; ;
Bannister. J. V., Eds.; Elsevier-North Holland: New York. 1980.
Schwartz. K. Fed. Proc., Fed. Am. Soc. Exp. Biol. 1974, 33, 1748. ^
‘ Schamschula, R. G.; Barmes, D. E. Ann. Rev. Nutr. 1981, 1, 427.
7 Raloff. J. Sci. News 1988, 133. 356.
Bowen, H. J. M. Environmental Chemistry of the Elements; Academic: New York, 1979. _ i
' Harper. H. A. Review of Physiological Chemistry; Lange: Los Altos, CA, 1971. _
" Eichom, G. L.; Butzow, J. J.;Clark. P.; von Hahn, H. P.; Rao,G.; Heim, J. M.;Tarian. E.;CrappeirD- R.; Kariik.S.XfnT/t^
organic Chemistry in Biology and Medicine; Martell. A. E., Ed.; ACS Symposium Series 140; American Chemical Society:
Washington, DC, 1980; Chapter 4. Wurtman. R. J. Sci. Amer. 1985,252(1), 62. MacDonald. T. L.; Humphreys. W. G.; Martin. R B
Science 1987, 236, 183.
" Gadjusek, D. C. N. Engl.J. Med. 1985.5/2,714. _ ■
® Mertz, W. Science 1981 .213. 1332.
" Volcani. B. E. In Silicon and Siliceous Structures in Biological Systems; Simpson. T. L.; Volcani. B. E., Eds.; Springer-Verlag-
Berlin, 1981.
•i Nielsen. F. H.; Mertz, W. In Present Knowledge in Nutrition, 5lh ed.; Olson, R.E.; Vroquist, H. P.; Chichester. C.O.; Darby, ‘
W. J.; Kolbyc, A. C.. Jr.; Staivey.R. M.. Eds.; Nutrition Foundation: Washington, DC, 1984; Chapter42. . , '
r Naylor, G. J. In Metal Ions in Neurology and Psychiatry; Gabay, S.; Harris. J.; Ho, B. T., Eds.; Alan R. Liss: New York, 1985; £ 7
pp 91-105.
' Haylock. S. J.: Buckley. P. D.: Blackwell, L. F.J. Inorg. Biochem. 1983, 19, 105. . .j
' Rollinson. C. L.; Enig. M. G. In Kirk-Othmer Encyclopedia of Chemical Technology, 3rd ed.; Grayson, M., Ed.; Wiley: New York, i
1981; Vol. 15. pp 570-603. ....... ,• . ' - £’ y
Lovlcy, D. R.; Phillips, E. J. P.; Gorby, Y.‘A.; Landa
1987,236.68.
D. W. Inorg. Chem. 1987,26, 1147.
, E. R. Nature 1991,350,413-416. -
Essential and Trace Elements in Biological Systems
Abundances of the elements
in the earth's crust, rivers,
and sea water 0
Earth's crust
atoms/I 0 4 atoms Si
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminum
Silicon
Phosphorus
Sulfur
Chlorine
Argon
River water
mg L-l
I.II9 x 10 3
0.003
< 1 x 10- 4
0.01
1.2
0.25 £ '
8.8 x I0 3
0.1
Ocean water
mg L-’ _
1.078 x I0 3
7.2 x I0-
0.18
6 x 10-7
4.5
28
0.5<-4
8.56 x I0 3
1.4
0.00012
1.105 x I0 4
1.326 x I0 3
0.0057
17
0.077
928
1.987 x I0 4
Potassium
24
610
2.3
Calcium
42
1040
1.5
Scandium
0.022
0.48
4 x I0-
Titanium
5.7
120
0.003
0.001
Vanadium
0.135
2.64
0.001
Chromium
0.1
2
0.001
0.00067
Manganese
0.95
17
-0.005
0.0027
Iron
56
1000
0.67
0.0037
Cobalt
0.025
0.42
0.0002
8 x 10-37
Nickel
0.075
1.3
0.0003
0.002
Copper
0.055
0.86
0.005
0.0037
Zinc
0.070
1.1
0.01
0.005
Gallium
0.015
0.21
1 x I0- 4
Germanium
0.0015
0.021
6 x lO- 3
Arsenic
0.0018
0.024
0.001
0.0023
Selenium
5 x lO- 3
6 X 10- 4
0.0002
0.00045
Bromine
0.0025
0.031
-0.02
68
Krypton
Rubidium
0.09
1
0.001
0.00021
0.12
Stronu'um
0.375
4.26
0.050
8.5
Yttrium
0.033
0.37
0.04
Zirconium
0.165
1.80
0.003
Niobium
0.02
• 0.2
Molybdenum
0.0015
0.016
0.001
0.01
Technetium
' T T
Ruthenium
I x l0-<
1 x 10- 3 '
7 x 10-7
Rhodium
Palladium
2 x 10 - 7 '
8 X l0-7e
2 x IO-<
8 x lO--
.Silver
7 x 10— 3 -
. ••. 6 x 10-1
0.0003
0.0001
Cadmium •
0.0002 :
' 0.0018 ;? •-
‘v- ’ ’• .'!
5 x 10-J
Indium <
iv? -j:
0.0001
9 x lO-'
«V r . • - >•
1 X 10-7
L -
/■
948
892
19 The Inorganic Chemistry of Biological Systems
and the metal-nitrogen bond distance does not vary greatly from 193-196 pm in nickel
porphyrins to 210 pm in high spin iron(II) porphyrins. The rigidity of the ring derives
from the delocalization of the tr electrons in the pyrrole rings. Nevertheless, if the
metal atom is too small, as in nickel porphyrinates. the ring becomes ruffled to allow
closer approach of the nitrogen atoms to the metal. At the other extreme, if the metal
atom is too large, it cannot fit into the hole and sits above the ring which also becomes
domed (see page 903).
The order of stability of complexes of porphyrins with +2 metal ions is that
expected on the basis of the Irving-Williams series (see Chapter 9), except that the
square planar ligand favors the d H configuration of Ni 2+ . The order is Ni 2+ > Cu 2+ >
Co 2+ > Fe 2+ > Zn 2+ . The kinetics of formation of these metalloporphyrins has also
been measured and found to be in the order Cu 2 > Co 2+ > Fe 2+ > Ni 2+ . 7 If this
order holds in biological systems, it poses interesting questions related to the much
greater abundance of iron porphyrins (see below). What might have been the implica¬
tions for the origin and evolution of biological systems if the natural abundance of iron
were not over a thousandfold greater than those of cobalt and copper?
The porphyrin ring or modifications of it are important in several quite different
biological processes. The reason for the importance of porphyrin complexes in a
variety of biological systems is probably twofold: (1) They are biologically accessible
compounds whose functions can be varied by changing the metal, its oxidation state,
or the nature of the organic substituents on the porphyrin structure; (2) it is a general
principle that evolution tends to proceed by modifying structures and functions that
are already present in an organism rather than producing new ones de novo.
The heme group is a porphyrin ring with an iron atom at the center (Fig. 19.2). The
oxidation state 11 of the iron may be either +2 or +3. and the importance of the
/
CH,
\"
CH,
\
CH.
/ -
CH,
i -
Heme A:
R | =CH = CH,
2-c„h w oh
C ,CH v c
JH.-CVS-CH.
’ \ II I / ■'
C-N N-C
H x / W
CH Fe CH
\ / V /
C-N N=C
" i “ c vnA>- ch -
i i
CH, R,
Heme t!
(protoporphyrin IX):
R | »R,.CH=CH,
Heme C.
R! = R,» CH(CH,)S — Pniicin
Cliloniheme:
R =C(H) = 0
R,= CH = CH,
Fig. 19.2 The heme group: Type A hemes are found in cytochrome a; Type B hemes are
found in hemoglobin, myoglobin, peroxidase, and cytochrome b\ Type C hemes are found in
cytochrome c: chloroheme is found in chlorocruorin.
7 Bishop, D. G.; Reed. M. L. Pholochem. Plioiobiol. Rev. 1976. /. I.
K The lerm heme refers to the neutral group containing Fe(II), either isolated or in a protein. When the
isolated heme is oxidized to Fe(III). there will be a net positive charge and an associated, coordi¬
nated anion, often the chloride ion. When oxidized, the term he min is applied, as in hemin chloride.
Hematin. long thought to be "hemin hydroxide." is actually a p-oxo dimer (see page 8%).
Metalloporphyrins and Respiration
893
cytochromes lies in their ability to act as redox intermediates in electron transfer.
They are present not only in the chloroplasts for photosynthesis but also in mitochon¬
dria to take part in the reverse process of respiration.
The heme group in cytochrome c has a polypeptide chain attached and wrapped
around it (Fig. 19.3). This chain contains a variable number of amino acids, ranging
from 103 in some fish and 104 in other fish and terrestrial vertebrates to 112 in some
green plants. A nitrogen atom from a histidine segment and a sulfur atom from a
methionine segment of this chain are coordinated to the fifth and sixth coordination
sites of the iron atom. 9 Thus, unlike the iron in hemoglobin and myoglobin (see
below), there is no position for further coordination. Cytochrome c therefore cannot
react by simple coordination but must react indirectly by an electron transfer mecha¬
nism. It can reduce the dioxygen and transmit its oxidizing power towards the burning
of food and release of energy in respiration (the reverse process to complement
photosynthesis). The importance of cytochrome c in photosynthesis and respiration
indicates that it is probably one of the oldest (in terms of evolutionary history) of the
chemicals involved in biological processes. An interesting “family tree" of the evolu¬
tion of living organisms can be constructed from the differences in amino acid se¬
quences in the peptide chains between the various types of cytochrome c found, for
example, in yeasts, higher plants, insects, and humans. Despite these differences,
however, it should be noted that cytochrome c is evolutionarily conservative. Cyto¬
chrome c from any eucaryotic species will react with the cytochrome oxidase of any
other eucaryotic species, plant or animal, though at reduced rates. 10
There is quite a variety of cytochromes, most of which have not been as well
characterized as cytochrome c. Depending upon the ligands present, the redox poten¬
tial of a given cytochrome can be tailored to meet the specific need in the electron
transfer scheme, whether in photosynthesis or in respiration. The potentials are such
that the electron flow is b — c — a —» 0 2 . At least some of the a type (cytochrome c
oxidase) are capable of binding dioxygen molecules and reducing them. They are thus
the last link in the respiratory chain of electrons flowing from reduced foodstuffs to
oxygen. Therefore, they must be five coordinate (in the absence of 0 2 ) in contrast to
cytochrome c. They are responsible for the unusually severe and rapid toxicity of the
cyanide ion. CN”. The latter binds strongly to the sixth position and stabilizes the
Fe(IlI) to such an extent that it can no longer be readily reduced and take part in
the electron shuttle. The cyanide ion is isoelectronic with the carbon monoxide
molecule and it might be thought that it could bind lightly to hemoglobin as does CO.
However, cyanide binds well only to Fe(III) hemoglobin (methemoglobin 11 ), an aber¬
rant form usually present only in small quantities. Cyanide poisoning is thus not the
result of lack of hemoglobin function (as is CO poisoning). In fact, the standard
treatment for cyanide poisoning is inhalation of amyl nitrite or injection of sodium
nitrite to oxidize some of the hemoglobin to methemoglobin (see page 907). The latter.
9 For the complete structures of fcrrocytoclirome c (Fe :+ ). sceTakano, T.;Trus. B. L.: Mandcl, N.;
Mandcl, G.; Kallai, O. B.; Swanson, R.; Dickerson. R. E. J. Biol. Chem. 1977, 252. 776-785.
10 See Strycr, L. Biochemistry. 3rd cd.: Freeman: New York. 1981: pp 328-329. Dickerson. R. E. Set.
Anter. 1972. 226(4) pp 58-72. Eucaryotic cells have their DNA in true nuclei, as opposed to
procaryotic cells (bacteria and blue-green algae) which do not.
11 The prefix met- is used to signify that the iron atom, normally in the +2 oxidation state, has been
oxidized to +3.
946
1 9 • The Inorganic Chemistry of 8iological Systems
Table 19.3 (Continued)
Atomic
number Element
Biological functions _
Essential to all organisms. See
text.
Essential for many organisms
including mammals; acti¬
vates a number of enzymes;
vitamin B I2 .
Essential trace element.
Chicks and rats raised on
deficient diet show impaired
liver function and mor¬
phology;' stabilizes coiled
ribosomes. Active metal in
several hydrogenases and
plant ureases.'
Essential to all organisms;
constituent of redox en¬
zymes and hemocyanin."
Essential to all organisms;
used in >70 enzymes; sta¬
bilizes coiled ribosomes.
Plays a role in sexual matu¬
ration and reproduction.
U.S. population marginally
deficient.
Normally only slight toxicity,
but excessive intake can
cause siderosis and damage
to organs through excessive
iron storage (hemo¬
chromatosis).'
Very toxic to plants and mod¬
erately so when injected
intravenously in mammals.
Very toxic to most plants,
moderately so to mammals;
carcinogenic.
Very toxic to most plants;
highly toxic to inverte¬
brates. moderately so to
mammals.
Moderately to slightly toxic;
orally causes vomiting and
diarrhea.'
A very abundant element (5%
of earth's crust); may not be
available at high pHs.
Extensive areas are known
where low soii cobalt af¬
fects the health of grazing
animals.'
Local industrial pollutant of
air and water.
Pollution from industrial
smoke and possibly from
agricultural use. Wilson's
disease, genetic recessive,
results in toxic increase in
copper storage.
Pollution from industrial
smoke may cause lung dis¬
ease: use of zinc promotes
cadmium pollution. Certain
areas (e.g., Iran and Egypt)
are zinc deficient."
Essential ultratrace element in Moderately toxic to plants.
red algae, chick, rat, pig,
goat, and probably humans.
Deficiency results in de¬
pressed growth and
increased mortality.
Essential to mammals and
some higher plants. Compo¬
nent of glutathione
peroxidase, protects against
free-radical oxidant
stressors; protects against
heavy (“soft”) metal
May be essential in red algae
and mammals.
highly toxic to mammals.
Moderately toxic to plants,
highly toxic to mammals.
Nontoxic except in oxidizing
forms, e.g., Br 2 .
Serious pollution problems in
some areas: sources include
mining, burning coal, im¬
pure sulfuric acid,
insecticides, and her¬
bicides.
Livestock grown on soils high
in selenium are poisoned by
eating Astragalus (“loco-
weed”), which concen¬
trates it; sheep grown on
land deficient in selenium
develop “white muscle dis¬
ease.” Deficiency of
selenium involved in Key-
shan disease in China."
Function unknown, but found
in the molluscan pigment,
royal purple.
Essential and Trace Elements in Biological Systems
947
Dioxygen Binding, Transport, and Utilization 895
although useless for dioxygen transport, binds cyanide even more tightly than hemo¬
globin or cytochrome oxidase and removes it from the system. 12
The structure of cytochrome c oxidase is not known completely. It contains two
heme groups of the cytochrome type (<; and u-,) and two copper atoms (Cu A and Cu B ).
When the reduced Fe(ll) oxidase is treated with carbon monoxide, the « 3 moiety binds
it and gives a myoglobin-carbon monoxide-like spectrum. The a site does not bind
carbon monoxide, indicating a six-coordinate, cytochrome-c-like structure. The ox¬
idized. Fe(lII) form binds cyanide at a 3 , but not at a. supporting this interpretation.
(Metmyoglobin and methemoglobin will also bind cyanide, but cytochrome c will not.)
The EPR spectra of iron and copper show that tt 3 and Cu B are antiferromagnetically
coupled, and EXAFS (see page 913) measurements indicate that these Fe and Cu
atoms are about 370 pm apart, compatible with a sulfide bridge. The electron flow is
probably: 13
[cyt c] -♦ [«] - [Cu A ] - [Cu B -fl,(00] — 2H 2 0 (19.6)
Dioxygen Binding,
Transport, and
Utilization
The Interaction
between Heme
and Dioxygen
While all of the biochemical uses of the heme group are obviously important, the one
that has perhaps attracted the most attention because of its central biological role and
its intricate chemistry is the binding of the dioxygen molecule. 0 2 . This has been
mentioned briefly above with regard to the binding and reduction of dioxygen by
cytochrome oxidase. Before this step occurs, vertebrates 14 have already utilized two
other heme-containing proteins: Hemoglobin picks up the dioxygen from the lungs or
gills and transports it to the tissues where it is stored by myoglobin. The function of
hemoglobin in the red blood cells is obvious, that of myoglobin is more subtle. Besides
being a simple repository for dioxygen, it also serves as a dioxygen reserve against
which the organism can draw during increased metabolism or oxygen deprivation. 15
Other suggested functions include facilitation of dioxygen flow within the cell and a
•‘buffering'' of the partial pressure within the cell in response to increasing or decreas¬
ing oxygen supply. 1 ' 1
Dioxygen is far from a typical ligand. It probably resembles the carbon monoxide,
nitrosyl. and dinitrogen ligands more than any others. None of these has a significant
dipole moment contributing to the ir bond, but the electronegativity difference be¬
tween the atoms in CO and NO enhances tt interactions (see Chapter 11). Dinitrogen
and dioxygen lack this advantage, but may be considered soft ligands with some tt-
bonding capacity. Iron!ID. il'\ is not a particularly soft metal cation, but the “soften-
12 See Hanzlik. R. P. Inorganic Aspects of Biolopical tntj Or panic Chemistry; Academic: New York.
1976. p 152. Ochiai. E-l. Ilioinorpanic Chemistry; Allyn & Bacon: Boston. 1977; p 483.
13 Sec various articles on cytochrome oxidase in Electron Transport anil Oxygen Utilization: Ho. C.,
Ed.; Elsevier North Holland: New York. 1982; Karlin. K.; Cultneh. Y, 1‘ropr. Inorp. Cheat. 1987.
J5. 310-311.
14 There is an exception in certain "bloodless" Antarctic fishes (Chacnichthyidae) in which the
metabolism is so low and the oxygen solubility so high, both resulting from the extremely low water
temperatures, that oxygen carriers are not necessary.
15 Diving mammals such as whales have a lurge amount of myoglobin in their tissues which presum¬
ably enables them to remain submerged for extended periods of time.
1(1 See Cole. R. P. Science 1982. 216. 523.
896
897
19-The Inorganic Chemistry of Biologicol Systems
ing” (symbiotic) action of the tetrapyrrole ring system probably facilitates dioxygen
binding. Note that the heme group binds the truly soft ligand carbon monoxide even
more tightly, resulting in potentially lethal carbon monoxide poisoning. 17
However, there is another potentially fatal flaw in the binding of dioxygen by
heme: irreversible oxidation. If free heme in aqueous solution is exposed to dioxygen,
it is converted almost immediately into a p.-oxo dimer known as hematin. The
mechanism of this reaction has been worked out in detail. 18 The reactions are as
follows, where the heme group is symbolized by the circle about an iron atom. The
first step is the binding of the dioxygen molecule, as in hemoglobin: 19
(19.7)
The bound dioxygen can now coordinate to a second heme, forming a p.-peroxo
complex:
(19.8)
Cleavage of the peroxo complex results in two molecules of a ferryl complex with the
iron in the +4 formal oxidation state:
(19.9)
Finally, attack of the ferryl complex on another heme results in the formation of
hematin:
(19.10)
17 Carbon monoxide poisoning may be treated by flooding the system with oxygen. Nevertheless, the
binding of CO is about 500 times stronger than the binding of 0 3 . It could be worse. Carbon
monoxide binds even more strongly (by about two orders of magnitude) to free heme. The stcric
hindrance about the heme in hemoglobin and myoglobin may favor the bent 0 2 over the (optimally)
linear carbon monoxide. (Strycr, L. Biochemistry, 2nd cd.: Freeman: New York. 1981; p 54.)
18 Batch, A. L.; Chan. Y.-W.; Cheng, R. J.; La Mar. G. N.: Latos-Grazynski. L.; Renner, M. W. J.
Am. Chem. Soc. 1984. 106. 7779-7785; Pcnner-Hahn. J. E.: Eble. K. S.: McMurry, T. J.: Renner.
M.: Balch. A. L.: Groves, J. T.; Dawson, J. H.; Hodgson. K. O. Ibid. 1986. 108. 7819.
19 The oxidation stales may occasionally be ambiguous—the adduct in Eq. 19.7 may be formulated as
heme(Il)-dioxygen or as heme (lll)-superoxide. See Problem 19.21.
The Binding of
Dioxygen to
Myoglobin
Dioxygen Binding, Transport, and Utilization
Obviously, living systems have found a way to frustrate reactions 19.7-19.10:
otherwise all of the heme would be precipitated as hematin rather than shuttling
electrons in the cytochromes or carrying dioxygen molecules in oxyhemoglobin (and
storing them in oxymyoglobin). There may be more than one mechanism in effect
here, but certainly the primary one is steric hindrance: The giobin pan of the molecule
prevents one oxoheme from attacking another heme. This was first illustrated over
thirty years ago by embedding the heme group in a polymer matrix that allowed only
restricted access to the iron atom: The embedded heme will reversibly bind dioxy¬
gen. 20 More recently this same result has been achieved by "picket-fence" hemes and
related compounds (Fig. 19.4) that reversibly bind dioxygen 71 and not only confirm the
steric hypothesis with regard to the stability of hemoglobin, but allow detailed struc¬
tural measurements to be made of a heme model compound. Thus the angular or bent
coordination of dioxygen to heme (in hemoglobin and myoglobin) was first indicated
by the structure shown in Fig. 19.4. It has since been confirmed in myoglobin and
hemoglobin (see below).
Myoglobin is a protein of molecular weight of about 17.000 with the protein chain
containing 153 amino acid residues folded about the single heme group (Fig. 19.5).
This restricts access to the iron atom (by a second heme) and reduces the likelihood of
formation of a hematin-like Fe(III) dimer. The microenvironment is similar to that in
cytochrome c, but there is no sixth ligand (methionine) to complete the coordination
Fig. 19.4 Perspective view of picket-fence dioxygen adduct. The apparent presence of four
different 02 atoms results from a four-way statistical disorder of the oxygen atoms on
different molecules responding to the X-ray diffraction. [From Collman. J. P.; Gagne, R. R.;
Reed, C. A.; Robinson. W. T.; Rodlcy, G. A. Proc. Natl. Acad. Sci. U. S. A. 1974. 71,
1326-1329. Reproduced with permission.)
20 w «ng. J. H. J. Am. Cliem. Soc. 1958, 80, 3168; Acc. Chem. Res. 1970. 3, 90.
31 Jameson. G. B.: Molinaro, F. S.: Ibcrs, J. A.; Collman. J. P.; Brauman. J. I.; Rose. F..; Suslick.
K. S. J. Am. Chem. Soc. 1980, 102, 3224-3237. For a review of sterically hindered biomimetic
porphyrins, see Morgan. B.; Dolphin. D. Struct. Bonding (Berlin) 1987, 64, 115.
THE NAMES, SYMBOLS,
TOMIC NUMBERS, AND ATOMIC WEIGHTS OF THE ELEMENTS'
h
ame
Symbol
Atomic
number
Atomic
weight^
Name
Symbol
Atomic
number
Atomic
weight
Actinium
Ac
89
(227)
Germanium
Ge
32
72.61
.luminum
Al
13
26.981539
Gold (aurum)
Au
79
196.96654
Americium
Am
95
(243)
Hafnium
Hf
72
178.49
.ntimony
Sb
51
121.757
Helium
He
2
4.002602
Argon
Ar
18
39.948
Holmium
Ho
67
164.93032
vrsenic
As
33
74.92159
Hydrogen
H
1
1.00794
Astatine
At
85
(210)
Indium
In
49
114.82
arium
Ba
56
137.327
Iodine
I
53
126.90447
Barkelium
Bk
97
(247)
Iridium
Ir
77
192.22
'eryllium
Be
4
9,012182
Iron (/errurn)
Fe
26
55.847
Bismuth
Bi
83
208.98037
Krypton
Kr
36
83.80
'oron
B
5
10.811
Lanthanum
La
57
138.9055
Bromine
Br
35
79.904
Lawrencium
Lr
103
(262)
"admium
Cd
48
112.411
Lead (plumbum)
Pb
82
207.2
Calcium
Ca
20
40.078
Lithium
Li
3
6.941
"aliformum
Cf
98
(251)
Lutetium
Lu
71
174.967
_arbon
C
6
12.011
Magnesium
Mg
12
24.3050
Cerium
Ce
58
140.115
Manganese
Mn
25
54.93805
-Jesium
Cs
55
132.90543
Mendelevium
Md
101
(258)
Chlorine
Cl
17
35.4527
Mercury
Hg
80
200.59
chromium
Cr
24
51.9961
Molybdenum
Mo
42
95.94
Cobalt
Co
27
58.93320
Neodymium
Nd
60
144.24
Jopper (cuprum)
Cu
29
63.546
Neon
Ne
10
20.1797
Curium
Cm
96
(247)
Neptunium
Np
93
(237)
dysprosium
Dy
66
162.50
Nickel
Ni
28
58.6934
Einsteinium
Es
99
(252)
Niobium
Nb
41
92.90638
rbium
Er
68
167.26
Nitrogen
N
7
14.00674
Europium
Eu
63
151.965
Nobilium rf
No
102
(259)
ermium
Fm
100
(257)
Osmium
Os
76
190.2
Fluorine
F
9
18.9984032
Oxygen
O
8
15.9994
'rancium
Fr
87
(223)
Palladium
Pd
46
106.42
Gadolinium
Gd
64
157.25
Phosphorus
P
15
30.973762
"allium
Ga
31
69.723
Platinum
Pt
78
195.08
a From [he Table of Weights 1989 by the Commission on Atomic Weights, IUPAC (Pure. Appl. Chem. 1991, 63, 975-1002).
arc scaled to the relative atomic mass A^'C) = 12.
ft The names in parentheses are the Latin forms used in complex formation (except for wolfram): gold ( aurum ); [AuCUl" is
tetrachloroaurate(III).
c Atomic weights in parentheses arc those of the most stable radioisotope.
d Although the name nobelium has official IUPAC sanction, some Russian workers have used the name joliollum.
' Wolfram is the German word for tungsten.
/ German workers have proposed the names neilsbohrium (107), hassium (108), and meitnerium (109).
Russian workers have suggested the names khurchatovium (104) and neilsbohrium (105).
(therfordium.
American workers have suggested the name rui
Nome b
Symbol
Atomic
number
Atomic
weight
Nome
Symbol
Atomic
number
Atomic
weight
Plutonium
Pu
94
(244)
Tellurium
Te
52
127.60
Polonium
Po
84
(209)
Terbium
Tb
65
158.92534
Potassium
K
19
39.0983
Thallium
T1
81
204.3833
Praseodymium
Pr
59
140.90765
Thorium
Th
90
232.0381
Promethium
Pm
61
(145)
Thulium
Tm
69
168.93421
Protactinium
Pa
91
231.03588
Tin (stannum)
Sn
50
118.710
Radium
Ra
88
(226)
Titanium
Ti
22
47.88
Radon
Rn
86
(222)
Tungsten (wolfram)'
W
74
183.85
Rhenium
Re
75
186.207
Unnilennium^
Une
109
(267)
Rhodium
Rh
45
102.90550
Unnilhexium
Unh
106
(263)
Rubidium
Rb
37
85.4678
Unniloctiun/
Uno
108
(265)
Ruthenium
Ru
44
101.07
Unnilpentium
Unp
105
(262)
Samarium
Sm
62
150.36
Unnilquadrum'' 1
Unq
104
(261)
Scandium
Sc
21
44.955910
Unnilseptiun/
Uns
107
(262)
Selenium
Se
34
78.%
Uranium
U
92
238.0289
Si
14
28.0855
Vanadium
V
23
50.9415
Silver (argentum)
Ag
47
107.8682
Xenon
Xe
54
131.29
Sodium
Na
11
22.989768
Ytterbium
Yb
70
173.04
Sr
38
87.62
Yttrium
Y
39
88.90585
Sulfur
S
16
32.066
Zinc
Zn
30
65.39
Tantalum
Ta
73
180.9479
Zirconium
Zr
40
91.224
Technetium
Tc
43
(98)
—
898
19-The Inorganic Chemistry of Biological Systems
Dioxygen Binding, Transport, and Utilization
899
(a) (b)
Fig. 19.5 The myoglobin molecule: (a) the folding of the polypeptide chain about the heme
group (represented by the disk); (b) close-up view of the heme environment. [Modified from
Kendrew, J. C.; Dickerson, R. E.: Strandberg. B. E.; Hart. R. G.: Davies. D. R.: Phillips.
D. C.; Shore, V. C. Nature 1960. 185. 422-427. Reproduced with permission.]
sphere of the iron atom. Thus there is a site to which a dioxygen molecule may
reversibly bind.
Note how the differences in structure between the dioxygen-binding molecules
(myoglobin, hemoglobin, and cytochrome oxidase) and the electron carriers (various
cytochromes, including cytochrome oxidase which performs both functions) correlate
with their specific functions. In myoglobin and hemoglobin the redox behavior is
retarded, and there is room for the dioxygen molecule to coordinate without electron
transfer taking place. 22
Myoglobin contains iron(ll) in the high spin state. Iron(II) is d 1 ' and. when high
spin, has a radius of approximately 92 pm in a pseudo-octahedral environment (the
square pyramidal arrangement of heme in myoglobin and hemoglobin may be consid¬
ered an octahedron with the sixth ligand removed), and the iron atom will not fit into
the hole of the porphyrin ring. The iron(II) atom thus lies some 42 pm above the plane
of the nitrogen atoms in the porphyrin ring (see Fig. 19.6). When a dioxygen molecule
binds to the iron(II) atom, the latter becomes low spin d h (cf. the extremely stable
Co 3 '" complexes with 2.4A LFSE). The ionic radius of low spin iron(II) with coordina¬
tion number six is only 75 pm, in contrast with the 92 pm of high spin iron. Why the
difference? Recall that in octahedral complexes the t< v orbitals arc those aimed at the
ligands. If they contain electrons, which they do in the high spin case ui K . c~).
they will repel the ligands as opposed to the low spin case (/5 V e"), which allows
unhindered access of the ligands along the coordinate axes. Thus the effective radius
of the iron atom is greater (along the x. y, and z axes) in the high spin state than in the
low spin state. The result is that the iron atom shrinks upon spin pairing and drops into
the hole in the porphyrin ring. All of the ligands (including the proximal histidine) are
able to approach the iron atom more closely. The net effect in myoglobin is minimal,
but the process is an important one for the transmission of dioxygen from the lungs to
Ihe tissues by hemoglobin. The spin pairing of the normally paramagnetic dioxygen
molecule is also of interest, though often overlooked (see Problem 19.9).
22 Indeed, hemoglobin has been dubbed a "frustrated oxidase" [Wintcrbourn. C. C.: French. J. K.
Biochem. Soc. Trans. 1977. 5. 1480; French. J. K.; Wintcrbourn. C. C.; Carrell. R. W. Biochem. J.
1978. 173. 19).
Fig. 19.6 Close-up of the heme group in myoglobin and hemoglobin. Note that the iron atom
does not lie in the plane of the heme group.
A knowledge of the exact molecular arrangement of dioxygen in oxymvoglobin
and oxyhemoglobin has been desirable in order to understand the chemistry of
dioxygen transport and storage. Unfortunately, this has been difficult to achieve
because of the high molecular weight of the molecules and the low resolution of the X-
ray-determined structures. The structure that has been determined to the greatest
resolution is that of oxyerythrocruorin which has been refined to a resolution of 140
pm. 23 The dioxygen is bonded to the iron with an angle of ~ 150° and an Fe—0 bond
length of -180 pm. Oxvmyoglobin (sperm whale) 24 and oxyhemoglobin (human) 25
have not been resolved as highly (210 pm), but the Fe—O bond lengths are similar. All
of these are compatible with the more accurate value of 190 pm in the picket-fence
adduct. 2,1 However, the Fe—O—O bond angles vary considerably, from —115° in
myoglobin to 153° in human hemoglobin (for more details, see Table 19.1. page 905)
with the most accurate value being 131° in the model picket-fence compound. The
source of the differences is not clear, but calculations 27 indicate l hat the bond energy
changes but little with bond angle, and so other factors such as sleric effects or
hydrogen bonding with a neighboring group could be important (Fig. 19.7).
Erythrocruorin is a form of myoglobin found in chironomid midges ((lies). In general, the greater the
resolution (the smaller this value), the more accurate the structural determination, but it should be
realized that the refinement of structures of proteins containing tens of thousands of atoms is far
more complicated than the almost routine determination of structures of molecules containing a few
dozen atoms at most. Often assumptions must be made with a resulting shift of values: The
M—O—O bond angle in erythrocruorin was corrected from 170° (extraordinary!) to 150'’ when such
assumptions were changed (Steigemann. W.; Weber. E. J. Mol. Biol. 1979. 137, 309).
24 Phillips. S. E. V. Nature (London) 1978 , 373 . 247: J. Mol. Biol. 1980. 143. 531.
25 Shaanan. B. Nature (London) 1982, 3%. 683: J. Mol. Biol. 1983. 171. 31.
26 Jameson. G. B.: Rodlcy. G. A.; Robinson, W. T.; Gagne, R. R.; Reed, C. A.: Collman. J. P. Inorg.
Chen,. 1978, 17. 850-857.
27 Hoffmann. R.; Chen. M. M.-L.: Thom. D. L. Inorg. Cltcm. 1977, 16. 503-511. Kirchner, R. F.;
Loew, G. H. J. Am. Chem. Soc. 1977, 99. 4639.
944
19- The Inorganic Chemistry of Biological Systems
Table 19.3 (Continued)
Atomic
number Element
Magnesium
Aluminum
Biological functions
None known.
Important in nerve function¬
ing in animals. Major cation
of extracellular fluid in ani¬
mals.
Essential to all organisms.
Present in all chlorophylls.
Has other electrochemical
and enzyme-activating
functions. U.S. population
may be marginally defi¬
cient/
May activate succinic de¬
hydrogenase and 8-
aminolevulinate de-
hydrase. The latter is
involved in porphyrin syn¬
thesis.'
Relatively harmless except in
excessive amounts (lethal
dose ca. 3 g kg -1 ). Associ¬
ated with some forms of
hypertension.
Silicon Essential element for growth
and skeletal development in
chicks and rats; 0 probably
-r-- essential in higher plants.-
Used in the form of silicon
dioxide for structural pur¬
poses in diatoms, some
protozoa, some sponges,
limpets, and one family of
plants/
Phosphorus Important constituent of
DNA, RNA, bones, teeth,
some shells, membrane
phospholipids, ADP and
ATP, and metabolic inter¬
mediates.
Sulfur Essential element in most pro¬
teins; important in tertiary . j
structure (through S—S
links) of proteins; involved T
in vitamins, fat metabolism,
y ,, and detoxification pro- ,
cesses.'H 2 S0 4 in digestive
fluid in ascidians ("sea
squirts"); H 2 S replaces
H 2 0 in photosynthesis of
some bacteria; H 2 S and S g
• - are oxidized by other bac- ;
teria. A; ■ . • • ://
Moderately toxic to most
plants; slightly toxic to
mammals. Suggested as in-
. volved in the etiology of
Alzheimer's disease and
other neurologic diseases." 1
Not chemically toxic, but
large amounts of finely di¬
vided silicates or silica are
“ injurious to the mammalian
lung.
Inorganic phosphates are rela¬
tively harmless; P 4 and PH 3
are very toxic in mammals
and fish. Phosphate esters
are used as insecticides
(nerve poisons).
Elemental sulfur is highly
. toxic to most bacteria and
fungi, relatively harmless to
higher organisms. H 2 S is
highly toxic to mammals;
S0 2 is highly toxic.
Tolerance of and/or depen¬
dence upon sodium chloride
can be an important consid¬
eration in the survival of
plants and aquatic animals.
This depends upon osmotic
regulation rather than so¬
dium specificity.
May cause deficiencies of
other elements (e.g., Fe) by
the effect of the alkalinity of
dolomite.
Relatively inaccessible except
in acidic media as a result of
insolubility of AI(OH) 3 .
Soils and waters high in
Al 3 and low in Mg 2 and
Ca 2 implicated in neu¬
rologic diseases."
Long-term exposure to finely
divided asbestos from con¬
struction work poses a
health problem. Some evi¬
dence for a negative
correlation between silicon
content of drinking water
and heart disease."
Leached from fertilizers ap¬
plied to agricultural land;
present in detergents and
other sewage sources.
Sulfur dioxide is a serious at¬
mospheric pollutant,
especially serious when it
settles in undisturbed
pockets; oxidized to
H 2 S0 4 ; widespread cause
of acid rain/Sulfide miner¬
als cause acid mine
drainage. - - - - - - -j
, '
■ • S V S
Essential and Trace Elements in Biological System:
945
Table 19.3 (Continued)
Atomic
number Element _ Biological functions _ Toxicity” Comments
17 Chlorine Essential for higher plants and Relatively harmless as CP.
mammals. NaCI electrolyte; Highly toxic in oxidizing
HCI in digestive juices; im- forms; Cl 2 . CIO - , C10 3 - .
paired growth in infants has
been linked to chloride defi¬
ciency.
18 Argon None known.
Potassium Essential to all organisms with Extremely toxic to mammals Pollution problem possible
the possible exception of when injected intra- from leaching of fertilizers
blue-green algae; major cat- venously; emesis prevents from agricultural land,
ion in intracellular fluid in oral toxicity,
animals; essential for trans¬
mission of nerve impulse
and cardiac function.
20
21
22
23
24
Calcium Essential for ail organisms; Relatively harmless,
used in cell walls, bones,
and some shells as struc¬
tural component; important
electrochemically and in¬
volved in blood clotting.
Scandium None known. Scarcely toxic.
Titanium None known, but it tends to Relatively harmless,
be accumulated in siliceous
tissues.
Vanadium Essential to ascidians (“sea Highly toxic to mammals if in-
squirts"), which concen- jected intravenously,
trate in a miilionfold from
sea water. Essential to
chicks and rats. Deficien¬
cies cause reduced growth,
impaired reproduction and
survival of young, impaired
tooth and bone metabolism
and feather development.
May be a factor in
manic-depressive illness/
Chromium Essential; involved in glucose Highly toxic as Cr(VI); car-
metabolism and diabetes; cinogenic; moderately toxic
potentiates effect of in- as Crtlll).
sulin." Presence in glucose
tolerance factor from
brewer's yeast questioned/ -
Manganese Essential to all organisms; ac-
• tivates numerous enzymes;
-■ deficiencies in soils lead to
infertility in mammals, bone
- _ . malformation in growing
• ft chicks. - - ..
Moderately toxic.
May cause deficiencies of
other elements (e.g.. Fe) by
effect of alkalinity of lime¬
stone.
Relatively unavailable be¬
cause of insolubility of
TiO,.
Possible pollutant from indus¬
trial smokes—may cause
lung disease.
Potential pollutant since
amount used industrially is
large compared with normal
biological levels; normally
relatively unavailable be¬
cause of low solubility.
Cr(VI) used in comfort
cooling towers, environ¬
mental hazard.
19•The Inorganic Chemistry of Biological Systems
VALE I!
VALE!I
HISE7
HISE7
PHECDt
PHECD1
Fig. 19.7 Stereoview of superimposed heme environments in oxyhemoglobin and
oxymyoglobin. Solid lines denote HbO, and dashed lines MbO,. Note the difference in the
Fe—01—02 bond angles and the presumed hydrogen bond (dotted line) to the histidine (His
E7). [From Shaanan, B. Nature (London) 1982, 296 , 683. Reproduced with permission.]
The Physiology of
Myoglobin and
Hemoglobin
In vertebrates dioxygen enters the blood in the lungs or gills 28 where the partial
pressure of dioxygen is relatively high [21% oxygen = 0.21 x 1.01 x 10 3 Pa
(760 mm Hg) = 2.1 x 10 4 Pa (160 mm Hg)] under ideal conditions; in the lungs with
mixing of inhaled and nonexhaled gases, the value is closer to 1.3 x I0 4 Pa
(100 mm Hg). It is then carried by red blood cells (Fig. 19.8a) to the tissues where the
partial pressure is considerably lower [of the order of 2.5 x I0 3 to 6.5 x I0 3 Pa (20-
50 mm Hg)]. The reactions are as follows:
Lungs (gills) Hb + 40, -► Hb(0 2 ) 4 (19.11)
Tissues Hb(0 2 ) 4 + 4Mb -► 4Mb(0,) + Hb (19.12)
Note that hemoglobin has an ambivalent function: It should bind dioxygen tightly and
carry as much as possible to the tissues, but once there it should, chameleon-like,
relinquish it readily to myoglobin which can store it for oxidation of foodstuffs.
Hemoglobin serves this function admirably as shown by Fig. 19.9: (I) Myoglobin must
have a greater affinity for dioxygen than hemoglobin in order to effect the transfer of
dioxygen at the cell. (2) The equilibrium constant for the myoglobin-dioxygen com-
plexation is given by the simple equilibrium expression:
[Mb(Q,)|
[Mb] [0 2 ]
If the total amount of myoglobin ([Mb] + [Mb0 2 ]) is held constant (as it must be in the
cell) while the concentration of oxygen is varied (in terms of partial pressure), the
2 " Small organisms require no oxygen transport system beyond simple diffusion. There is a family of
lungless salamanders, the Plcthodontidae, which have neither gills nor lungs (as adults) and rely
upon oxygen exchange through the skin and through buccopharyngeal ("mouth and throat")
exchange. Some worms and mollusks have proteins related to hemoglobin for oxygen transport and
storage. Some polychaete worms employ chlorocruorin which turns green upon oxygenation.
Sipunculid worms and some other species utilize nonheme iron proteins, the hemcrythrins. for these
functions (sec page 908). Lobsters, crabs, spiders, cephalopods, and some snails use a copper-
containing protein (hemocyanin, see page 909) for oxygen transport.
Dioxygen Binding, Transport, and Utilization
Fig. 19.8 Relative scale of (a) red blood cells, the biological unit of dioxygen transport; (b)
the hemoglobin molecule, the biochemical unit of dioxygen transport; and (c) the
dioxygen-heme group, the inorganic unit of dioxygen transport. The relative sizes are given
by the factors over the arrows. [Scanning electron micrograph courtesy of M. Barnhart,
Wayne State University of Medicine. Hemoglobin molecule modified from Perutz. M.;
Rossman. M, G.; Cullis. A. F.; Muirhead. H.: Will, G.: North, A. C. T. Nature (London)
1960. 185, 416. Reproduced with permission.]
Partial pressure of oxygen (I’a)
00 6000 800Q IQOOO |’000 14000
Partial pressure of oxygen (mm Hg)
Fig. 19.9 Dioxygen binding curves for (I) myoglobin and for hemoglobin al various partial
pressures of carbon dioxide: (2) 20 mm Hg; (3) 40 mm Hg; (4) 80 mm Hg. Note that
myoglobin has a stronger affinity for dioxygen than hemoglobin and that this effect is more
pronounced in the presence of large amounts of carbon dioxide. [Modified from Bock. A. V
Field, H.. Jr.; Adair. G. S. J. Biol. Chcm. 1924. 59. 353-378. Reproduced with permission.]
942
1 9 • The Inorganic Chemistry of Biological Systems
Periodic Survey of The biochemistry of iron has just been discussed in some detail including the bio-
F -„ntinl and chemical species involved, bioaccumulation, transport, storage, and toxicity. Space
SS p. . does not permit an extensive discussion of other elements of importance. However, a
Trace tlemen s brjef discussion wi n be presented here with a table summarizing what is currently
known.
The number of elements that are known to be biologically important comprises a
relatively small fraction of the 109 known elements. Natural abundance limits the
availability of the elements for such use. Molybdenum (Z = 42) is the heaviest metal,
and iodine (Z = 53) is the heaviest nonmetal of known biological importance. The
metals of importance in enzymes are principally those of the first transition series, and
the other elements of importance are relatively light: sodium, potassium, magnesium,
calcium, carbon, nitrogen, phosphorus, oxygen, chlorine, and, of course, hydrogen.
Table 19.3 lists elements that have been found to be essential or poisonous,
together with notes on biological functions and leading references that may be fol¬
lowed by the interested reader." It is certain that the information in this list will be
expanded as the present techniques and theory are improved.
Biological
Importance,
Biological Fitness,
and Relative
Abundance
There are at least two ways, maybe more, of looking at the fitness of particular
elements to serve particular biological functions. The more "chemical” approach is to
suggest that iron functions well in cytochromes and terredoxins because the
Fe' + /Fe 2+ couple has a reduction potential in the appropriate range for life processes
and. conversely, that mercury is poisonous because it binds irreversibly with en¬
zymes, destroying their activity. Basically, a given element cannot function m a
biological role unless it has specific properties. Yet chemical properties are fixed,
biological systems are not. and there is the -biological” perspective of deciding how
those" biological systems adapted to the working materials available to them: the
"fitness of the organism" to exploit fixed chemical starting materials. From this point
of view, one is immediately attracted to the question: "What are the starting mate¬
rials''" It is then useful to attempt to correlate biological activity with the crustal
abundance of a given element."'' If we look at some typical essential transition
elements, we find in addition to Fe, Co, Zn, Cu, and Mo mentioned previously, V, Cr
Mn, and Ni. Representative essential metals are Na, K, Mg, and Ca. and essential
nonmetals are C, N, O, P. S, and Cl (see page 953). All of these elements except Mo
are relatively abundant in the earth's crust (Table I9.4).'=« When we look tor abundant
elements that are not essential elements, we find only three—Al. Ti. and Zr—all of
which form extremely insoluble oxides at biologically reasonable pH values. No
common element is toxic at levels normally encountered, though almost anything can
be harmful at too high levels (cf. toxicity of the sodium chloride in sea water to
freshwater plants and animals). When we consider the elements that are currently
causing problems in the environment, we find that they are all extremely rare in their
l IK Two books have been written devoted to this general subject: Ochiai. E.-I. General Principles of
Biochemistry of the Elements', Plenum: New York. 1987. Bowen. H. J. M. Environmental Chem¬
istry of the Elements-, Academic: New York. 1979.
i iv Huhcey J E In REACTS 1973. Proceedings of the Regional Annual Chemistry Teaching Sym¬
posium: Egolf, K.; Rodez. M. A.: Won. A. J. K.: Zidick. C.. Eds.; University of Maryland:
College Park. 1973; pp 52-78.
t:o Because almost all of the earth's crust is silicon dioxide or silicates, silicon and oxygen make up
over 95% of the crust and mantle. Only about two dozen elements occur with a frequency of one
atom per 10.C00 atoms of silicon: these are considered "abundant” or "relatively abundant.
Essential and Trace Elements in Biological Systems
943
Table 19.3
Function and toxicity of the elements in biological systems
Atomic
number
Element
Biological functions
Toxicity 0
Comments
1
Hydrogen
Molecular hydrogen metabo¬
lized by some bacteria.
Constituent of water and all
organic molecules. D 2 0 is
toxic to mammals.
Bacterial hydrogenases are
nickel-containing en¬
zymes.
2
Helium
None known.
Used to replace nitrogen as an
0 2 diluent in breathing mix¬
tures to prevent the
"bends” in high-pressure
work.
3
Lithium
None known.
Slightly toxic.
Used pharmacologically to
treat manic-depressive pa¬
tients.'
4
Beryllium
None known.
Very toxic.
Pollution occurs from indus¬
trial smokes. There are
some fears concerning poi¬
soning from camping
lantern mantles.' 7
5
Boron
Unknown, but essential for
green algae and higher
plants; probably essential
ultratrace element in ani¬
mals.'
Moderately toxic to plants:
slightly toxic to mammals.
6
Carbon
Synthesis of all organic mole¬
cules and of biogenetic
carbonates.
Carbon monoxide is slightly
toxic to plants and very
toxic to mammals; CN~ is
very toxic to all organisms.
Carbon dioxide and CO are
global pollutants from burn¬
ing fossil fuels; CN~ is a
local pollutant of rivers near
mines.
7
Nitrogen
Synthesis of proteins, nucleic
acids, etc. Steps in the ni¬
trogen cycle (organic N
NH,-~ NOj — NOJ —
N 2 - organic N) are impor¬
tant activities of certain
microorganisms.
Ammonia is toxic at high con¬
centrations.
Leaching of nitrogenous fertil¬
izers from agricultural land
and nitrogenous materials in
sewage cause serious water
pollution. Nitrogen oxides
arc widespread source of
acid rain/
8
Oxygen
Structural atom of water and
most organic molecules in
biological systems; required
for respiration by most or-
Induces convulsions at high
P 0j ; very toxic as ozone,
superoxide, peroxide, and
hydroxyl radicals.
gamsms.
9 Fluorine Probably essential element; Moderately toxic, may cause Pollution by fluoride present
used as CaF 2 by some mol- mottled teeth. in superphosphate fertil-
lusks. izers. Ca. I ppm in water
provides cariostatic action/
beneficial in the treatment
of osteoporosis.
902
1 9 ■ The Inorganic Chemistry of Biological System:
Structure and
Function of
Hemoglobin
curve shown in Fig. 19.9 is obtained. Myoglobin is largely converted to oxymyoglobin
even at low oxygen concentrations such as occur in the colls. (3) The equilibrium
constant for the formation of oxyhemoglobin is somewhat more complicated. The
expression for the curve in the range of physiological importance in the tissues is:
[Hb(Q,)J
[Hb][0 : ] 2 ' 8
(19.14)
The 2.8 exponent for dioxygen results from the fact that a single hemoglobin molecule
can accept four-dioxygen molecules and the binding of the four is not independent. It
is not the presence of four heme groups to bind four dioxygen molecules per se that is
important. If they acted independently, they would give a curve identical to that of
myoglobin. It is the cooperativity of the four heme groups that produces the curves
shown in Fig. 19.9. The presence of several bound dioxygen molecules favors the
addition of more dioxygen molecules: conversely, if only one dioxygen molecule is
present, it dissociates more readily than from a more highly oxygenated species. The
net result is that at low dioxygen concentrations hemoglobin is less oxygenated (tends
to release 0 2 ), and at high dioxygen concentrations hemoglobin is oxygenated almost
to the same extent as if the exponent were 1. This results in a sigmoid curve for
oxygenation of hemoglobin (Fig. 19.9). This effect favors oxygen transport since it
helps the hemoglobin become saturated in the lungs and deoxygenated in the capillar¬
ies. (4) There is a pH dependence shown by hemoglobin. This is known as the Bohr
effect. 2 ' 1 Hemoglobin binds one H for every two dioxygen molecules released. This
favors the conversion of carbon dioxide, a metabolite of the tissues, into the hydrogen
carbonate ion (HCOf) promoting its transport back' to the lungs. Likewise, the
production of carbon dioxide from cell respiration and of lactic acid from anaerobic
metabolism favors the release of dioxygen to the tissues.
Hemoglobin may be considered an approximate tetramer of myoglobin. It has a
molecular weight of 64,500 and contains four heme groups bound to four protein
chains (Fig. 19.8b). Two of the chains, labeled beta, have 146 amino acids and are
somewhat similar to the chain in myoglobin: the other two. labeled alpha, have 141
amino acids and are somewhat less like the myoglobin chain. The differences between
hemoglobin and myoglobin in their behavior towards dioxygen (particularly 3 and 4
above) are related to the structure and movements of the four chains. If the tetrameric
hemoglobin is broken down into dimers or monomers, these effects arc lost, and the
smaller units do not exhibit cooperativity. Myoglobin does not exhibit the sigmoid
curve nor a Bohr effect.
Upon oxygenation of hemoglobin, two of the heme groups move about 100 pm
towards each other while two others separate by about 700 pm. Perhaps a better way
of describing the movement is to say that one ap half of the molecule rotates 15°
relative to the other half. 30 These movements are the result of a change in the
quaternary structure of the hemoglobin and are responsible for the cooperative effects
observed. The quaternary structure exhibited by the deoxy form is called the T state.
2V Discovered by Christian Bohr, father of Niels Bohr, the pioneer of quantum mechanics.
30 Dickerson. R. E.; Geis, I. The Structure and Action of Proteins: Harper & Row: New York. 1969:
p 59: Hemoglobin: Structure. Function. Evolution, and Pathology: Benjamin/Cummings: Menlo
Park. CA. 1983. Baldwin. J.; Chothia. C. J. Mol. Biol. 1979, 129. 175.
Dioxygen Binding, Transport, and Utilization
and that of the oxy form the R state. 31 The dioxygen affinity of the R form is about the
same as that of isolated et and p chains, but the dioxygen affinity of the T state is some
12-14 kJ mol -1 lower. This fundamental difference in the energetics of dioxygen
binding is responsible for the cooperativity of hemoglobin. The lower affinity of the T
form is responsible for the slow start of the sigmoidal curve (lower left. Fig. 19.9), and
the higher affinity of the R form causes the rapid rise in the curve (upper right.
Fig. 19.9) until it almost matches that of myoglobin.
Perutz 33 has suggested a mechanism to account for the cooperativity of the four
heme groups in hemoglobin. Basically it is founded on the idea that the interaction
between a dioxygen molecule and a heme group can affect the position of the protein
chain attached to it, which in turn affects the other protein chains through hydrogen
bonds, etc., and eventually the tertiary and quaternary structure of the protein. It has
been dubbed the Rube Goldberg effect after the marvelous mechanisms of ropes,
pulleys, and levers in Goldberg’s cartoons. 33 A simplified illustration of the Perutz
mechanism is shown in Fig. 19.10.
The key or trigger in the Perutz mechanism is the high spin Fe(Il) atom in a
dioxygen-free heme. As we have seen, the radius of high spin Fe 2-1 is too large to fit
within the plane of the four porphyrin nitrogen atoms. The iron atom is thus forced to
sit above the center of the heme group (Fig. 19.6; Fig. 19.1 la) with an Fe—N porp | 1>rin
distance of about 206 pm. Furthermore, the heme group is domed upward towards the
proximal histidine.
Fig. 19.10 A schematic diagram of the Perutz mechanism in hemoglobin. T he three most
important factors tire highlighted: (I) the environment of the heme: (2) the movement
(tension) of the protein chains, and (3) the breaking of hydrogen bonds ("suit bridges").
(Courtesy of Professor M. Perutz. Reproduced with permission.!
11 The terms come from the adjectives tense and relaxed that have been applied to the two structures.
However, since the nature, the extent, and even the existence of “tension" in one form or the other
is a matter of considerable controversy, we shall use T and R as labels for the quaternary structures
without structural or mechanistic implications.
j: Perutz. M. F. Nature ll.ondon) 1970. 22S. 726; Hr. Med. Hull. 1976. .12. 195. Perutz. M. F.; Fermi,
G.; Luisi, H.: Shaanan. B.: Liddinglon. R. C. Air. ('hem. Hes. 1987. 20. 309, Perutz. M. Mecha¬
nisms of Cooperativity and Allosteric Hegulation in Proteins: Cambridge University: Cambridge,
1990.
11 Ituheev. J. E. In HEACTS 197.1. Proceedings of llte Regional Annual Chemistry Teaching Sym¬
posium: Fgolf. K.: Rodcz. M. A.: Won. A. J. K.: Zidiek. C.. Eds.: University of Maryland: College
940
941
19The Inorganic Chemistry of Biological Systems
The excess dietary iron is derived from a traditional fermented maize beverage that is
home-brewed in steel drums." 4 It should be noted in this connection that the body has
no mechanism for the excretion of iron, and except for women in the child-bearing
years, the dietary requirement for iron is extremely low.
The absorption of iron in the gut, preferentially in the +2 oxidation state, was
once thought to be a result of special physiological mechanisms, but now is generally
agreed to be merely another aspect of the differential solubility of Fe(OH) 2 and
Fe(OH)j. However, there is a significant differential in the absorption of heme versus
nonheme iron: Heme iron is absorbed 5-10 times more readily than nonheme iron." 5
Since meat contains large quantities of hemoglobin, myoglobin, and cytochromes, this
difference could be nutritionally significant.
It is conceivable that iron could be stored in the form of a complex such as
transferrin or even hemoglobin, and in lower organisms ferrichrome apparently serves
this purpose. Such storage is wasteful, however, and higher animals have evolved a
simpler method of storing iron as ferritin. If iron(III) nitrate is allowed to hydrolyze in
a solution made slightly basic by the hydrogen carbonate ion (HCOJ), it spon¬
taneously forms spheres of "FeOOH” of about 7000 pm in diameter. The core of a
ferritin particle is similar and contains up to 4500 iron atoms and apparently some
Fig. 19.28 Structural features of apoferritin. The gross quaternary structure of the assembled
molecule is shown in the center and more details on the fourfold channels (left), the threefold
channels (upper right) and the subunits (lower right) are also illustrated. [From Harrison,
P. M.; Trcffry, A.; Lilley, T. H. J. Inorg. Biochem. 1986. 27. 287-293. Reproduced with
permission.]
■ M Rollinson, C. L.; Enig. M. G. In Kirk-Othmer Encyclopedia of Chemical Technology. 3rd cd.;
Grayson. M., Ed.; Wiley: New York. 1981; Vol. 15. pp 570-603. Gordcuk. V. R.; Bacon. B. R.;
Brittenham. G. M. Ann. Rev. Nntr. 1987. 7, 485-508.
i 15 Narins, D. In Biochemistry of Nonheme Iron; Bezkorovainy. A.. Ed.; Plenum: New York, 1980:
Chapter 3. Hallberg, L. Ann. Rev. Nutr. 1981, /, 123-147.
Essential and Trace Elements in Biological Systems
phosphate as well as oxo and hydroxo ligands. This core is surrounded by a protein
covering (called apoferritin ) that allows controlled access to the core through eight
hydrophilic channels (along threefold axes) and six hydrophobic channels (along
fourfold axes) (see Fig. 19.28). It is thought that the iron(III) enters via the hydrophilic
channels and leaves via the hydrophobic channels, but the mechanism of iron transfer
is obscure. In any event, ferritin provides high-density storage of inorganic iron
combined with ready availability. "6
Essential and Trace The discussion of metalloporphyrins and metalloenzymes systems has indicated the
Elements in importance of certain metals in chemical reactions within living organisms. Certain
rr~j--——- elements are essential in that they are absolutely necessary (perhaps in large, perhaps
Biological Systems m small quantities) for life processes. Other elements are nonessential since they play
no positive role in biological systems. Obviously, determining the essentiality of an
element is difficult. The term “trace element” although widely used is not precisely
defined. For example, molybdenum averages about 1-2 ppm in rocks, soils, plants,
and marine animals and even lower in land animals. Yet it is an essential trace metal.
At the other extreme, iron, which averages about 5% in rocks and soils and 0.02-0.04%
in plants and animals, might or might not be considered a "trace” metal.
Although the role of iron in various heme derivatives and zinc in carboxypep-
tidase and carbonic anhydrase is clear, there are many instances in which little is
known of the function of the trace metal. For example, it has been known for some
time that ascidians (“sea squirts") concentrate vanadium from sea water by a factor of
a millionfold, but a satisfactory explanation for its role in these animals remains
elusive."7 There are many elements that are known to be useful but for which no
specific function has yet been proved. The list of known functions is expanding
rapidly, however.
The problem of toxicity is difficult to quantify. There are so many synergistic
effects between various components of biological systems that it is almost impossible
to define the limits of beneficial and detrimental concentrations. There is also endless
variation among organisms. Truly, "one man’s meat is another man's poison." The
phenomenon of an essential element becoming toxic at higher than normal concentra¬
tions is not rare. Selenium is an essential element in mammals yet one of the most
vexing problems is the poisoning of livestock from eating plants that concentrate this
element (page 951).
The importance of trace elements is manifold and, unfortunately, previously
hampered by relatively insensitive analytical methods. Good methods for determining
concentrations of I ppm or less have been available for relatively few elements; yet
these may be the optimum concentrations for a particular trace element. When the
responses of living organisms are more sensitive than the laboratory “black boxes,"
the chemist naturally develops an inferiority complex. Fortunately, the recent devel¬
opment of analytical techniques capable of determining parts per billion has opened
new vistas for the study of these problems. Some of these techniques are atomic
absorption, atomic fluorescence, activation analysis, and X-ray fluorescence.
" h The T E - C. Ann. Rev. Biochem. 1987. 56. 289. Lippard, S. J. Angew. Chem. In,. Ed. Engl. 1988.
27, 344.
Kustin, K.; McLeod. G. C.: Gilbert. T. R.; Briggs. L. R.. IV Struct. Bonding (Berlin) 1983. 53.
139-160, Boyd. D. W.; Kustin, K. Adv. Inorg. Biochem. 1984. 6. 312-365. Wever. R • Kustin K
Adv. Inorg. Chem. 1990. 35. 81-115.
19-The Inorganic Chemistry of Biological Systems
Dioxygen Binding, Transport, and Utilization
905
The coordination ot the dioxygen molecule as a sixth ligand causes spin pairing to
take place on the iron atom. Since the radius of low spin Fe(II) is about 17 pm smaller
than high spin Fe(II). it should fit in the porphyrin hole; we expect the smaller iron
atom to drop into the hole. As a matter of fact, it does move about 20 pm towards the
porphyrin ring (Fig. 19.1 lb) and the Fe-N porphy „ n distance shortens to 198 pm.
However, it stops short of moving all of the way into the plane of the ring. Data for the
heme in myoglobin, hemoglobin, and related species are given in Table 19.1.
Distances (pm) and angles (“)
in various heme adducts'
Compound 4
Fe N pro>
Fo-Pn
D _klO
'porph •'•pro
Whale Mb
203
222
42
267
Mb(0 2 )
195
207
18
228
Human Hb
206
215
38
268
Hb(0 2 ) 2
204
220
18
266
Hb(0 2 )<
198
200
0 ± 5
210
Picket fence
(a) l-Melm
0, adduct
197.9
206.8
(bj 2-MeIm
207.2
209.5
40
252
0 2 adduct
199.6
210.7
8.6
217
Movement (N c
182
176 ± 10
" Dala from Perntz, M. F.; Fermi. G.; Luisi, B.; Shaanan, B.; Liddington, R. C. Acc. Chen,. Res. 1987, 20. 309-321.
6 Average value of several methods of determination. See reference in Footnote a for details.
c Bond lengths.
d Distance between the iron atom and the plane of the nitrogen atoms in porphyrin ring.
^omhyrin e rmg Ween ‘ he ^ ^ Pr ° Ximal <0r im!dazolcs in ,hi: Picket-fence compounds) and the plane of the
/ Movement of the proximal histidine (or imidazole) towards the porphyrin ring upon oxygenation.
1 2 ?^ 19 - 4): "yHmidozole
!r 88era f Ii0n ° f IhC PerUlZ mechanism in hemoglobin, (a) Deoxy-T accepts a dioxygen molecule O, to
form oxy-T (b) with partial movement of the iron atom into the ring, which is strained and unstable Addition of more
into T re l 11 llS 3 '' ea,Tan6eme^, o{o ^ T l ° <0 with the iron atom moving completely
Am, • f',h d The . Confi8UraUon about lh e heme group with respect to Leu FG3 in the T and R forms Note that the
flattening of the porphyrin on going Irom deoxy to oxy exerts a leverage on leucine FG3 and valine FG5 which lie at the
switching contact between the two structures. The vertical bars indicate the distance of N hi , of theproximal" ££ R
from the mean plane of the porphynn nitrogens and carbons. The horizontal bar gives the Fe-N , distance and the
value to the right of the iron atoms gives the displacement of the iron from the plane of the porphyrin nitrogens Note that
porphynn is flat only in oxy-R and that the proximal histidine tilts relative to the heme normaMn the T Structures Note
"set str --
938
19-The Inorganic Chemistry of Biological Systems
CONH ^CONH^
\cH 2 )j (CH 2 ) 3 (CHj)j (CH,)j (CHj), ch,
w w v/
(b)
Fig. 19.26 Three types of bacterial sidcrophores: (a) desfcrrichrome; (b) desferrioxamine B.
(c) enierobactin.
o
piq 19 27 The A-cis isomers of metal entcrobactins. The metal lies at the center of a
distorted octahedron of the six oxygen atoms of the three catechol ligands with approximate
C'j symmetry, (a) The structure or iron(IU) enterobactin as determined by CD spec a. (b)
ORTEP plot of the structure of V(IV) enterobactin as determined crystallographies ly. Note
that although both structures are viewed down the approximate three old axis and the atoms
(except Fe/V) are the same in (b) as in (a), the views are 180 apart. .{fromIsicd. S_ S Kuo.
G • Raymond. K. N. J. Am. Chem. Soc. 1976. 98. 1763; Karp.shin, T. B.; Raymond. K. N.
Angew. Chem. Int. Ed. Engl. 1992. 31. 466-468. Reproduced with permission.
The Biochemistry of Iron
939
Competition for Iron In addition to the transport of iron, the transferrins of higher animals and the sidero-
phores of bacteria show another interesting parallel. It can most readily be shown by
ovatransferrin (called conalbumin in the older literature) of egg white, though we shall
see other examples. There is a large amount, up to 16%, in the protein of egg white,
although it has been impossible to find an iron-transporting function for it there. In
fact, in some 200 species for which ovatransferrin has been studied, 99% were
completely devoid of iron binding to the protein! Ovatransferrin and other transfer¬
rins, in general, have larger stability constants towards iron(III) than do the various
siderophores. It is thus quite likely that they act as antibacterial agents. In the
presence of excess ovatransferrin. bacteria would be iron deficient since the sidero-
phore cannot compete successfully for the iron. 1115108
Lactoferrin, found in mother's milk, appears to be the most potent antibacterial
transferrin and seems to play a role in the protection of breast-fed infants from certain
infectious diseases. It has been claimed that milk proteins remain intact in the infant’s
stomach for up to 90 minutes and then pass into the small intestine unchanged, thus
retaining their iron-binding capacity. In guinea pigs, addition of hematin to the diet
abolishes the protective effects of the mother’s milk. 109
The question of iron chelation as an antibacterial defense is receiving increasing
attention. It appears to be far more general than had previously been supposed. 110 An
interesting sidelight is that the fever that often accompanies infection enhances the
bacteriostatic action of the body's transferrins.
An interesting side effect of the presence of ovatransferrin in egg whites is the
custom, long established before any rational explanation, of beating egg whites in
copper bowls to stabilize the foam (as in meringues, etc.) The copper complex of
ovatransferrin stabilizes the protein of egg white against denaturation and thus sta¬
bilizes the foam. 111
Another interesting example of this sort is the competition between bacteria and
the roots of higher plants. Both use chelators to win iron from the soil. However,
higher plants have one more mechanism with which to compete: The Fe(lII) is
reduced and absorbed by the roots in the uncomplexed Fe(Il) form. When edta and
other chelating agents are used to correct chlorosis in plants due to iron deficiency, the
action is merely one of solubilizing the Fe(III) and making it physically accessible to
the roots—the chelates are not absorbed intact. Indeed, chelates that strongly bind
FedI) may actually inhibit iron uptake from the root medium. 112
Exactly the opposite problem may occur for plants whose roots are growing in
anaerobic media. In flooded soils the roots may be exposed to high levels of iron(ll),
posing potential problems of iron toxicity. Rice plants and water lilies with roots in
anaerobic soils transport dioxygen (from the air or photosynthesis, or both) to the
periphery of the roots where it oxidizes the iron(ll) to iron(UI). In this case the
insolubility of iron!Ill) hydroxide is utilized to protect the plant from iron poison¬
ing. 113 A similar problem from too much iron occurs in parts of sub-Saharan Africa.
mu Webb. J.: van Bockxmeer. F. M. J. Chem. EJuc. 1980, 57. 639.
" r ‘ Biochemistry of Nonlieme Iron; Bczkorovainy. A.. Ed.: Plenum: New York. 1980; pp 336-337.
11,1 Weinberg. E. D. Microbiol. Rev. 1978. 42, 45. Hidcr. R. C. Struct. Bonding (Berlin) 1984, 58. 25.
Iron and Infection'. Bullcn, J. J.; Griffiths, E.. Eds.; Wiley: New York, 1987.
111 McGee. H. J.; Long. S. R.; Briggs. W. R. Nature (London) 1984. 308. 667.
112 Olsen. R. A.: Clark. R. B.: Bennett. J. H. Am. Sci. 1981, 69. 378-384.
1,3 Dacey, J. W. H. Science 1980. 210. 1017.
906
19-The Inorganic Chemistry of Biological Systems
The inhibition of free movement of the iron atom into the porphyrin ring has been
attributed to steric interactions between the histidine ligand (which must follow the
iron), the associated globin chain, and the heme group. 34 This apparently results in
considerable strain on the oxyheme and associated tertiary structure of the globin
within the T form. This discourages the addition of the first molecule of dioxygen, or
more important, it "pushes" the last dioxygen molecule off in the tissues, where it is
needed. Addition of a second dioxygen molecule takes place with similar results and,
in effect, the hemoglobin molecule becomes spring-loaded. The structure of the
bis(dioxygen)-T state has been determined and shows little movement of the iron atom
and negligible movement of the histidine. The addition of a third dioxygen molecule
results in interconversion to the R state. This removes the tension of the intermediate
species and allows the iron atom to move freely into the center of the porphyrin ring
(Fig. 19.11c). The porphyrin ring also flattens, and the histidine is free to follow the
iron atom, some 50-60 pm. This change allows the fourth heme to accept a dioxygen
molecule without paying the price of the protein constraint and accounts for the
avidity of Hb(0,) 3 for the last dioxygen molecule. The relaxation of the globin-heme
interaction in the R state versus the crowding in the oxy-T state is shown in
Fig. 19.1 Id.
Support for the view that the globin portion of the molecule produces a constraint
upon the iron atom (which would otherwise move into the heme pocket) comes from
the behavior of myoglobin and model compounds (such as the picket-fence com¬
pounds with I- or 2 -methylimidazole mimicking the porphyrin and histidine), which
are easier to study than the more complex hemoglobin:
I Mclhylimida/nlc 2-Mcthylmiiilti/nlc
adduct tuklucl
In myoglobin and the sterically hindered 2-methylimidazole complex (as in hemo¬
globin). the iron atom does not move into the plane of the porphyrin nitrogen atoms
(remaining 9 pm displaced in the complex), although it does so in sterically unhindered
imidazole models, indicating that the iron atom does indeed shrink enough to fit were
it not constrained. The data on the Fe—O bond length fit this picture: It is longer (and
presumably weaker) in myoglobin and the 2 -methylimidazole/pickct-fence adduct (as
it is in T hemoglobin) and shorter (and presumably stronger) in the unhindered
l-methylimidazole/picket-fence adduct (and R hemoglobin).
It was mentioned above that the deoxy-T «=s oxy-R equilibrium was affected by
pH (Bohr effect) as well as the partial pressure of dioxygen. Other species such as a
’•< Gelin. B. R.; Karpins. M. Proc. Nall. Acad. Sci. U.S.A. 1977. 77. SOI.
Dioxygen Binding, Transport, and Utilizatioi
907
single chloride ion and 2.3-diphosphoglycerale also influence the equilibrium.” Of
perhaps the greatest interest is the fact that the T-> R transition involves the addition
of about 60 molecules of water to the hemoglobin. This hydration of newly exposed
protein surfaces stabilizes the R form which might not even be capable of existence
without this hydration energy. 36
The above discussion has been somewhat simplified inasmuch as the number of
possible interactions in a molecule as large as hemoglobin is very great. On the other
hand, even as presented in abbreviated form, it is quite complicated. Various workers
have placed varying degrees of weight upon different factors. 37 Nevertheless, one
should not lose sight of the fact that the iron atom does undergo a change in spin state
that causes it to move, and the net result is a change in the quaternary structure from T
to R. And lest we get loo involved in the biomechanical "trees" and forget to look at
the biological "forest." recall that it is the reduced affinity of the T form that is
nature's device that makes it possible for hemoglobin to push the dioxygen molecule
off in the tissues and transfer it to myoglobin. We can thus look at dioxygen transport
at several levels (go back to Fig. 19.8 and review).
Before leaving the subject of the binding of dioxygen to hemoglobin, two mo¬
lecular (genetic) diseases should be mentioned. One is sickle cell anemia (SCA): Upon
stressful deoxygenation of the blood, the hemoglobin (Hb s ) polymerizes and precipi¬
tates. resulting in severe deformation of the red blood cells. 31 The genetic defect
responsible is the replacement of hydrophilic glutamic acid at (3-6 with the hydro-
phobic valine. The exposure of the latter upon R — T conversion reduces the
solubility of hemoglobin S compared to normal adult hemoglobin, hemoglobin A.
It was mentioned above that hemefFe’ ) will not bind dioxygen. Heme is always
susceptible to oxidation when in the presence of dioxygen. This reaction results from
the nucleophilic displacement of superoxide by water, and it is acid catalyzed: 39
IFe(II)—O—O] + H f -► [Fe(II)—O—O—H| (19.15)
[FedI)—O—O—HJ + H,0-► [Fe(lll)H ; 0| " + HO, (19.16)
The globin chain gives some protection by providing a hydrophobic environment, but
still about 3% of the hemoglobin is oxidized to methcmoglobin daily. The enzyme
mcthemoglobin reductase returns the oxidized heme to the +2 state ordinarily.
However, some individuals have an inborn metabolic defect that prevents the rcduc-
" For an excellent discussion of tile action of these hetcrotrophic ligands. see Perm/.. M. Mechanisms
of Cooperativity and Allosteric Herniation in Proteins'. Cambridge University: Cambridge. 1990.
36 Colombo. M. F.: Rau. D. C.: Parsegian. V. A. Science 1992. 256. 655-659.
37 For reviews of the subject from different points of view, see Footnote 52 and Collman. J. P.;
Halpcrt. T. R.: Susliek. K. S. In Metal Ion Activation of Dioxygetc. Spiro. T. G., Ed.: Wiley: New
York. 1980: Chapter 1: Bertini, I.: Barton. J. K.: Ellis, W. R.. Jr.; Forsen, S.: George. G.: Gray,
H. R.: I hers. J. A.: Jameson. G. B.: Kordcl. J,: Lippard. S. J.: Ltlehinat. C.: Raymond, K. N.:
Stiefel. E. I.; Theil. E. C.; Valentine. .1. S. ISioinargtuiic Chemistry. University Science Books: Mill
Valley. CA; in press.
,K Notice that the red blood cell in the lower right-hand corner of Fig. 19.8 is badly sieklcd.
Erythrocytes that tire sickled cannot llow as readily through the capillaries as normal rod blood
cells, and they arc more susceptible to mechanical damage. These factors are responsible for the
symptoms of sickle cell anemia.
39 Shikama. K. Coord. Cheat. Rev. 1988. $J. 73.
936
19-The Inorganic Chemistry of Biological System:
Fig. 19.25 Schematic diagram of nitrogenase activity in a bacterial cell. Carbohydrate
provides reducing capacity (fcrredoxin), energy (MgATP), and organic precursors for the
manufacture of amino acids. [From Skinner, K. J. Chem. Eng. News 1976. 54(41), 22-35.
Reproduced with permission.)
abundant transition element and serves more biological roles than any other metal. It
can therefore serve to illustrate the possibilities available for the absorption, storage,
handling, and use of an essential metal. Iron has received much study, and similar
results can be expected for other metals as studies of the chemistry of trace elements
in biological systems advance.
Availability of Iron Although iron is the fourth most abundant element in the earth's crust, it is not always
readily available for use. Both Fe(OH), and Fe(OH) 3 have very low solubilities, the
latter especially so (/fjp = 2 x 10“ l3 ; = 2 x I0 -39 ). An extreme example is iron
deficiency in pineapples grown on rust-red soil on Oahu Island containing over 20%
Fe, but none of it available because it is kept in the + 3 oxidation state by the presence
of manganese dioxide and the absence of organic reducing agents. 101 Similarly under
alkaline conditions in the soil (e.g., in geographic regions where the principal rocks are
limestone and dolomite) even iron(II) is not readily available to plants. The stress is
especially severe on those species such as rhododendron and azalea that naturally live
in soils of low pH. Under these circumstances gardeners and farmers often resort to
the use of "iron chelate,” an edta complex. The latter is soluble and makes iron
available to the plant for the manufacture of cytochromes, ferredoxins, etc. The clever
application of coordination chemistry by the chemical agronomist was predated by
some hundreds of millions of years by certain higher plants. Some, such as wheat and
oats, adapted to grow on alkaline soils, have evolved the ability to exude various
polyamino-acid chelating agents through the root tips to solubilize the iron so that it
may be absorbed.' 02
"" Brasted, R. C. J. Chem. Educ. 1970. 47. 634.
102 Sugiura. Y.; Nomoto, K. Struct. Bonding (Berlin) 1984, 53, 107.
The Biochemistry of Iron
937
The presence of organic chelates of iron in surface waters has been related lo ihe
red tide, an explosive "bloom ot algae (Gymnodium breve) that results in mass
mortality of fish. It is possible to correlate the occurrence of these outbreaks with the
volume of stream flow and the concentrations of iron and humic acid."» At least one
of the dinoflagellates in the red tide possesses an iron-binding siderophore (see
below).' 04
Within the organism a variety of compiexing agents are used to transport the iron.
In higher animals it is carried in the bloodstream by the transferrins. These iron-
binding proteins are responsible for the transport of iron to the site of synthesis of
other iron-containing compounds (such as hemoglobin and Ihe cytochromes) and its
insertion via enzymes into the porphyrin ring.' 03 The iron is present in the +3
oxidation state (Fe"‘ does not bind) and is coordinated to two or three tyrosyl
residues, a couple of histidyl residues, and perhaps a tryptophanyl residue in a protein
chain of molecular weight about 80,000. 106 There are two iron-binding sites per
molecule.
Most aerobic microorganisms have analogous compounds, called siderophores,
which solubilize and transport iron(III). They have relatively low molecular weights
(500-1000) and, depending upon their molecular structure and means of chelating iron,
are classified into several groups such as the ferrichromes, ferrioxamines, and entero-
bactins. Some examples are shown in Fig. 19.26. It is obvious that these molecules are
polydentate ligands with many potential ligating atoms lo form chelates. They readily
form extremely stable octahedral complexes with high spin FedII). Although the
complexes are very stable, which is extremely important to their biological function
(see below), they are labile, which allows the iron to be transported and transferred
within the bacteria.' 07 The ferrichromes and ferrioxamines are trihydroxamic acids
which form neutral trischelates from three bidentate hydroxamate monoanions. En-
terobactin contains a different chelating functional group, o-dihydroxybenzene ("cate¬
chol"). Each catechol group in enterobactin behaves as a dianion for a total charge of
~ 6 for the ligand. A characteristic of all of these is that, in addition to the natural
tendency of trischelates to form globular complexes, the remainder of the siderophore
molecule consists of a symmetric, hydrophilic portion that presumably aids in trans¬
port across the cell membrane (Fig. 19.27).
It is interesting that biologically functioning iron compounds such as hemoglobin,
myoglobin, cytochromes, and ferredoxins employ iron!II) compounds, but the side¬
rophores and transferrins coordinate iron(III). The reduced iron compounds within
biological systems may reflect an evolutionary history from a primitive reducing
atmosphere on earth (see page 951), whereas the siderophores are the response to the
need to deai currently with iron(III) in an oxidized external environment.
103 Martin. D. F.; Martin. B. B, J. Chem Educ. 1976. 55. 614.
104 Trick. C. G.; Andersen, R. J.: Gillam, A.; Harrison. P. J. Science 1983, 219. 306-308.
105 Biochemistry of Nonheme Iron ; Bczkorovainy, A., Ed.; Plenum; New York. 1980; Chapter 4.
Kochan, I. In Bioinorganic Chemistry-\Y, Raymond, K. N., Ed.; American Chemical Society:
Washington. DC, 1977.
" Limits. M. Struct. Bonding (Berlin) 1973, 17, 135-220.
107 Raymond. K. N.; Milller. G.; Malzanke. B. F. Top. Curr. Chem. 1984, 123. 49. Enterobactin is Ihe
most powerful iron(lll) chelator known with an overall stability constant of Kf 10 49 (Loomis. L-.
D.; Raymond, K. N. Inorg. Chem. 1991, 30, 906).
J]
908
19-The Inorganic Chemistry of Biological Systems
lion of methemoglobin. 40 In addition, any individual, but especially an infant, 41 may
be stressed by nitrite/nitrate intoxication, in which case methemoglobin is produced
faster than it can be reduced. In either case the iron(lll) impairs dioxygen transport
and causes cyanosis disproportionate to its abundance. This is because iron(III) is
small enough to fit into the porphyrin ring without binding a sixth ligand, making
methemoglobin very similar in structure to oxyhemoglobin. The presence of two or
three iron(III) atoms can lock a hemoglobin molecule into the R state so that even the
heme group(s) carrying dioxygen cannot release it readily. Recall that R hemoglobin
has about the same dioxygen affinity as myoglobin, so the cooperativity mechanism
has been defeated. 42
Other Biological Hemerythrin is a nonheme, dioxygen-binding pigment utilized by four phyla of marine
Dioxygen Carriers invertebrates. Its chief interest to the chemist lies in certain similarities to and
differences from hemoglobin and myoglobin. Like both of the latter, hemerythrin
contains iron(ll) which binds oxygen reversibly, but when oxidized to methemerythrin
(Fe 3 + ) it does not bind dioxygen. There is an octameric form with a molecular weight
of about 108.000 that transports dioxygen in the blood. In the tissues are lower
molecular weight monomers, dimers, trimers, or tetramers. 43 And just as hemoglobin
consists of four chains each of which is very similar to the single chain of myoglobin,
octameric hemerythrin consists of eight subunits very similar in quaternary structure
to myohemerythrin. A major difference between the hemoglobins and hemerythrins is
in the binding of dioxygen: Each dioxygen-binding site (whether monomer or octamer)
contains two iron(Il) atoms, and the reaction takes place via a redox reaction to form
iron(IIl) and peroxide (Oj - ). Oxyhemerythrin is diamagnetic, indicating spin coupling
of the odd electrons on the two iron(III) atoms. Mossbauer data indicate that the two
iron(III) atoms are in different environments in oxyhemerythrin. This could result
from the peroxide ion coordinating one iron atom and not the other, or from each of
the iron atoms having different ligands in its coordination sphere. The first evidence
concerning the nature of the ligands came from an X-ray study of methemerythrin. 44 It
indicated that the two iron atoms have approximately octahedral coordination and are
bridged by an oxygen atom (from water, hydroxo, or oxo), aspartate, and glutamate.
The remaining ligands are three histidine residues on one iron atom and two histidines
on the other. 45 This is a rather small difference, but it can be reconciled with the other
411 The classic case is of the "blue Fugates” of Troublesome Creek. KY, descendants of a single
couple, each of whom carried the recessive gene. Mansouri, A. Am. J. Med. Sci. 1985. 289. 200.
41 These "blue babies" result because infants have fetal hemoglobin, hemoglobin F. Hb|.-differs from
Hb A (adult hemoglobin) and has a higher oxygen affinity that facilitates dioxygen transfer in iilero
from the mother's Hb A . Hemoglobin F. which is gradually replaced during the first year of life, is
more susceptible to oxidation than adult hemoglobin, methemoglobin is produced more readily, and
oxygen transport is reduced. A different, curious aspect of Hb F is that it protects the infant
(temporarily) from the SCA problems of hemoglobin SI
42 For a review of methemoglobinemia, see Sonozan, N. M. J. Chem. Educ. 1985. 62. 181.
41 It should be remembered that what the chemist glibly calls "hemerythrin" is not necessarily the
same from one species to the next: The four phyla in which hemerythrin is found comprise
thousands of species. Thus generalizations tend to be difficult and somewhat oversimplified. For
reviews of hemerythrin chemistry, sec Klotz, I. M.: Kurtz. D. M.. Jr. Arc. Chem. Res. 198-1, 17, 16.
Wilkins, P. C.; Wilkins, R, G. Coord. Chem. Rev. 1987. 79. 195.
44 Stcnkamp. R. E.; Sicker, L. C.: Jensen. L. H. Proc. Sail. Acad. Sci. U. S. A. 1976. 73. 349-351.
45 The first studies indicated that the second iron atom also had a tyrosine residue attached to it. Later
refinement of the structure showed that it is actually pentacoordinate, but the argument remains the
same.
Dioxygen Binding, Transport, and Utilization
909
data, and so until recently the consensus has tended to favor a simple peroxo bridge
between the two iron atoms:
where the continuous line connecting (he two iron atoms is a simplified representation
ol the coordination spheres and the protein chain holding the iron atoms in place
Militating against this simple structure is the fact that the Mossbauer speclrum does
not distinguish Ihe iron atoms in deoxyhemerythrin. If the difference in amino acid
environment is sufficient to distinguish Ihe iron atoms in the Mossbauer spectrum of
oxyhemerythrin, why not in deoxyhemerythrin?
Further data on this matter came from the Raman spectrum of oxy( I6 O ls O)
hemerythrin, which shows the two oxygen atoms to be in noneqtdvalent positions. 46
Of the various alternative structures that have been proposed, the Raman data are
compatible with only two:
O
O O data, as well as other spectroscopic evidence. 47 are compatible with structure
III. but the question was still open until the X-ray structure of oxyhemerythrin was
further refined. 4 " The proposed structures of deoxyhemerythrin and oxyhemerythrin
are:
His mi “] Als 77
/A
Glu J8 —C C O —H
Asp 1060^ j
His 25 His 54
/ 7 ° \
Glu 58 C’ C s P-H
Asp 106 Q,0y /
7 l '\ O
His 25 His 54
Note that the hydrogen atoms cannot be located at this level of resolution and so the
hydrogen bond shown is merely one suggestion for the possible stabilization of the
peroxide ion.
Another oxygen-containing pigment is confusingly named hemocyanin. which
contains neither the heme group nor the cyanide ion; the name simply means "blue
Kurtz. D. M., Jr.; Shriver. D. F.: Klotz. I. M. J. Am. Chem. Soc. 1976, 98. 5033-5035' Coord
Chem. Rev. 1977. 24. 145-178.
47 Gay. R. R.; Solomon. E. I. J. Am. Chem. Soc. 1978, 100, 1972.
48 Stenkamp, R. E.: Sieker. L. C.; Jensen. L. H.; McCallum. J. D.; Sanders-Loehr, J. Proc. Nall
Acad. Sci. U. S. A. 1985, 82. 713-716.
934
1 9 • The Inorganic Chemistry of Biological System?
The reduction of mono-co-ordinated
molecular nitrogen to ammonia in
a protic environment Nature (London) (Jan. 3, 1975)
Fuel-saving way to make fertiliser The Times (Jan. 3, 1975)
Fuel break-through The Guardian (Jan. 3, 1975)
More progress in nitrogen fixation New Scientist (Jan. 9, 1975)
Cheaper nitrogen by 1990 Farmers Weekly (Jan. 10, 197:)
Basic life process created in UK lab The Province (British Columb.a,
Jan. 15, 1975)
With each retelling the story grew, until by the time it reached British Columbia, it
appeared that the press was almost able in 12 days to duplicate what is recorded as a
6-day event in Genesis! The resultant disappointment when scientists are not able to
meet expectations benefits neither them nor the public (but that, too, is good copy for
the popular press!).
In Vivo Nitroqen There are several bacteria and blue-green algae that can fix molecular nitrogen in vivo.
In Vwo N^ogen ^ free . livjng species and symbiolic spe cies are involved. There are the strictly
F,XQ anaerobic Clostridium pasteurianum,' facultative aerobes like Klebsiella pneu¬
moniae, and strict aerobes like Azotobacter vinelandii. Even in the aerobic forms it
appears that the nitrogen fixation takes place under essentially anaerobic conditions
(see below). The most important nitrogen-fixing species are the mutualistic species of
Rhizobium living in root nodules of various species of legumes (clover, alfalfa, beans,
peas, etc.).
The active enzyme in nitrogen fixation is nitrogenase. It is not a unique enzyme
but appears to differ somewhat from species to species. Nevertheless the various
enzymes are very similar. Two proteins are involved. The smaller has a molecular
weight of 57 000-73,000. It contains an Fe 4 S 4 cluster. The larger protein is an a 2 [i,
tetramer with a molecular weight of 220,000-240.000 containing two molybdenum
atoms, about 30 iron atoms, and about 30 labile sulfide ions .’’The iron-sulfur clusters
probably act as redox centers. It is possible to isolate a soluble jjrolein-free cofactor
containing molybdenum and iron (ca. 1 Mo. 7-8 Fe. and 4-6 S 2 "). Recombination of
the cofactor with inactive nitrogenase restores the activity.
It seems likely that the active site for dinitrogen binding involves the molybdenum
atom. It has been established by EXAFS™ that the coordination sphere consists of
several sulfur atoms at distances of about 235 pm. An Mo=0 double bond, so
common in complexes of Mo(IV) and Mo(VI), is not present. There are other heavy
atoms, perhaps iron, nearby (-270 pm). The ultimate source of reductive capacity is
pyruvate, and the electrons are transferred via ferredoxin (see page 911) to mtro-
Clostridium includes, in addition to the useful nitrogen-fixing C pasteurianunu the dangerous
anaerobic species C. lelani (causes tetanus, "lockjaw"), C. bomlmnm (causes botul.sm), and C.
wekhi (causes "gas gangrene").
97 Stiefcl, E. I. Prog. Inorg. Chem. 1977. 22. 1-223. Nelson. M. J.; Lindahl. P. A.: Ormc-Johnson
W H. Adv. Inorg. Biochcm. 1982. 4. 1-40. Burgmaycr. S. J. N.; Stiefcl. E. I. J. Chem. Edac. 1985.
62. 943. Note added in proof. The structure of nitrogenase has now been determined. CJeorgiadis.
M. M.: Komiya. H.; Chakrabarti. P.; Woo. D.; Komuc. J. J.; Rees. D. C. Saence 199.. 257. 1653.
Kim. J.; Rees. D. C. Science 1992. 257. 1677; Nature 1992. 360. 553.
w Cramer, S. P.; Hodgson. K. O.; Gillum. W. O.; Mortcnson. L. E. J. Am. Chem.Soc. '™. I00.
3398-3407. Conradson. S. D.; Burgess B. K.; Newton. W. E.; Mortensen. L. E.: Hodgson. K. O.
J. Am. Chem. Soc. 1987. 109. 7507-7515.
The Biochemistry of Iron
genase. There is some evidence, not strong, that Mo(lll) is involved. Two Mo(III)
atoms cycling through Mo(Vl) would provide the six electrons necessary for reduction
of dinitrogen. Alternatively, since the enzyme is rich in ferredoxin-type clusters, there
should be a ready flow of electrons, and the molybdenum may stay in the one or two
oxidation states that most readily bind dinitrogen and its intermediate reductants. The
overall catalytic cycle may resemble that shown in Eq. 19.38.99
Pyruvate
electron
pool
A schematic diagram for the production of fixed nitrogen compounds, including
the sources of materials and energy, and the overall reactions, is given in Fig. 19.25.
Note the presence of leghemoglobin. This is a monomeric, oxygen-binding molecule
rather closely resembling myoglobin. It is fell that the leghemoglobin binds any
oxygen that is present very tightly and thus protects the nitrogenase. which cannot
operate in the presence of oxygen. On the other hand, it allows a reservoir of oxygen
for respiration to supply energy to keep the fixation process going.
It is impossible to cover adequately the chemistry of various elements in biological
systems in a single chapter. Before discussing the salient points of other essential and
trace elements, the biochemistry of iron will be discussed briefly. Iron is the most
The Biochemistry
of Iron 100
w Note added in proof: The structure of Ihc Fc-Mo cofacior cilcd in Footnote 97 has led lliosc
authors to suggest that the molybdenum docs not directly participate in binding Ihc dinitrogen
molecule. The Mo is already six-coordinate with three S atoms, two O aloms from a homocitratc
anion, and one N atom from a histidine in the protein chain. Therefore, in Eq. 19.38 the N 2 is
probably bound to an Fe—S cluster in place of Mo.
100 Crichton. R. R. Inorganic Biochemistry of Iron Metabolism -. Ellis Horwood: New York. 1991.
910
19-The Inorganic Chemistry of Biological Systems
I
blood”: Whereas hemoglobin turns bright red upon oxygenation, the chromophore
(Cu‘/Cu n ) in colorless deoxyhemocyanin turns bright blue. Hemocyanin is found in
many species in the Mollusca and Arthropoda.^ The gross molecular structures of the
hemocyanins in the two phyla are quite different, though both bind dioxygen cooper¬
atively. and spectroscopic evidence indicates that the dioxygen-binding centers are
similar. The dioxygen binding site appears to be a pair of copper atoms, each bound by
three histidine ligands (Fig. 19.12). The copper is in the +1 oxidation state in the
deoxy form and +2 in the oxy form.
The structure of oxyhemocyanin has recently been determined. 50 It presents yet a
third mode of binding between oxygen-carrying metal atoms and the dioxygen mole¬
cule. The latter oxidizes each copper(I) to copper(ll) and is in turn reduced to the
peroxide ion (0^~). The two copper(II) atoms are bridged by the peroxide ion with
unusual p-rf.-rf bonds, i.e., each oxygen atom is bonded to both copper atoms.
The parallels and differences among hemoglobin, hemerythrin. and hemocyanin
illustrate the ways in which evolution has often solved what is basically the same
problem in different ways in different groups of animals. 51
Fig. 19.12 The copper-dioxygen binding site in hemocyanin from Panitlirtts inlemipms. The
copper atoms are indicated by open circles, the histidines by pentagons, and the protein
chains by ribbons. [From Volbcda, A.: Hoi. W. G. J. J. Mol. Biol. 1989. 209, 249-279.
Reproduced with permission.]
49 Karlin, K. D.; Gultneh. Y. Prog. Inorg. Client. 1987. 35. 219-327. The structure of hemocyanin
from an arthropod, the spiny lobster, has been determined by Volbeda. A.: Hoi. W. G. J. J. Mol.
Biol. 1989, 209, 249-279.
50 Note added in proof: The complete structure of oxyhemocyanin has not yet been published, but it
contains the /x—coordination mode shown above (Magnus, K. C.; Tong-That. H. J. Inorg.
Biocliem. 1992, 27. 20). This structure was predicted on the basis of model dicoppcr compounds
(Blackburn. N. J.; Strange, R. W.; Farooq, A.; Haka, M. S.; Karlin. K. D. J. Am. Client. Soc. 1988.
110, 4263-4272. Kitajima, N.; Fujisawa, K.; Fujimoto, C.; Moro-oka. Y.: Hashimolo, S.; Kitagawa.
T.:Toriumi. K.;Tasumi. K.; Nakamura. A. J. Am. Cliem. Soc. 1992. 114. 1277-1291). The Cu—Cu
distance in oxyhemocyanin is 360 pm.
M For an interesting discussion of parallels in function, structure, and possibly evolution of hemo¬
globin. hemerythrin, and hemocyanins, see Volbeda. A.: Hoi. W, G. J. J. Mol. Biol. 1989. 206.
531-546.
Electron Transfer, Respiration, and Photosynthesis
911
Electron Transfer,
Respiration, and
Photosynthesis
Ferredoxins and
Rubredoxins 52
There are several nonheme iron-sulfur proteins that are involved in electron transfer.
They have received considerable attention in the last few years. They contain distinct
iron-sulfur clusters composed of iron atoms, suifhydryl groups from cysteine resi¬
dues, and "inorganic” or "labile" sulfur atoms or sulfide ions. The latter are readily
removed by washing with acid:
(RS) 4 Fe 4 S 4 + 8H + -■ (RS) 4 Fe + + 4H 2 S (19.19)
The cysteine moieties are incorporated within the protein chain and are thus noi labile.
The clusters are of several types. The simplest is bacterial rubredoxin, (Cys-S) 4 Fe
(often abbreviated Fe,S 0 , where S stands for inorganic sulfur), and contains only
nonlabile sulfur. It is a bacterial protein of uncertain function with a molecular weigh!
of about 6000. The single iron atom is at the center of a tetrahedron of four cysteine
ligands (Fig. 19.13a). The cluster in the ferredoxin molecule associated with photo¬
synthesis in higher plants is thought to have the bridged structure Fe,S 2 shown in
Fig. 19.13b. The most interesting cluster is found in certain bacteria] ferredoxins
involved in anaerobic metabolism. It consists of a cubane-like cluster of four iron
atoms, four labile sulfur atoms, thus Fe 4 S 4 . and four cysteine ligands (Fig. 19.13c).
Because of the inherent chemical interest in clusters of this sort, as well as their
practical significance to biochemistry, there has been considerable effort expended in
making model compounds for study (Fig. 19.14). These model compounds allow direct
experimentation on the cluster in the absence of the protein chain. 53
c vs-S -Cys Cys-S S-Cys
^ Fe' Fe' N Fe
Cys--Cys Cys-^ S-Cys
(a) (b)
(c)
Fig. 19.13 Iron-sulfur clusters in ferredoxins: (a) Fe^,, in bacterial rubredoxin: (b) Fe : S ; in
photosynthetic ferredoxin; (c) Fc 4 S 4 in cubane-like ferredoxin.
52 Biochemistry of Nonlieme Iron'. Bezkorovainy, A.. Ed.: Plenum: New York. 1980; Chapter 8. Iron-
Sulfur Proteins'. Spiro. T. G., Ed.; Wiley: New York. 1982. The entire volume of Adv. Inorg. Cliem.
1992. 38. 1-487 is devoted to iron-sulfur proteins. Unfortunately, it was received too late to include
much of it in this volume but it should prove to be very useful to the interested reader,
33 See. for example, Liu. H. Y.; Scharbert. B.; Holm. R. H. J. Am. Cliem. Soc. 1991. 113. 9529-9539:
Holm. R. H.; Ciurli. S.; Weigel. J. A. Prog. Inorg. Cliem. 1990. 38. 1-74. and references therein.
932
19-The Inorganic Chemistry of Biological Systems
Metallothioneins 91 We have seen that heavy metals can replace essential metals in enzymes and destroy
the enzymatic activity. In addition, by coordinating to sulfur-bearing amino acids in
the protein chain they might cause an enzyme to be "bent out of shape" and lose its
activity. Protection of enzymes from toxic metals is thus requisite for their proper
function. Serving this purpose is a group of proteins that have the following charac¬
teristics: (I) The molecular weights are about 6000 with 61 to 62 amino acids. (2) One-
third (20) of these amino acids are cysteine [HSCH 2 CH(NH 2 )COOH] residues,
grouped in Cys—Cys and Cys—X—Cys groups (X = a separating amino acid). (3)
None of the cysteines are linked by S—S bridges (cystine). (4) There are few or no
histidines or amino acids with aromatic side chains. (5) With such a high percentage of
amino acids bearing thiol groups and "clumped" along the protein chain, the thioneins
are able to bind several metal ions per molecule, preferentially the softer metals such
as Zn' -1 , Cu -1- , Cd 2 + , Hg 2 ', Ag‘, etc. Metallothioneins containing Zn 2+ and Cu 2+
might possibly be important in the transport of these essential elements, but the
evidence is mostly negative. On the other hand, the binding of heavy metals such as
cadmium and mercury suggests a protective function against these toxic metals.
Indeed, increased amounts of thioneins are found in the liver, kidney, and spleen after
exposure to them. Furthermore, it can be demonstrated that cell lines that fail to
produce thioneins are extremely sensitive to cadmium poisoning while "over-
producers" have enhanced protection. It has been suggested that the binding of
thioneins to cadmium and other heavy metals, with extremely high stability constants,
is one of protection alone; perhaps the reduced binding of copper, an essential metal
but one toxic in high concentrations, serves a “buffering function" of providing
copper for enzymes but not at levels sufficiently high to be toxic. The question of
whether the weaker binding of the less toxic zinc serves a similar function, or "just
happens,” is moot.
For +2 cations such as zinc(II) and cadmium(II) each metallothionein molecule
contains up to seven metal atoms. X-ray studies indicate that the metal atoms are in
approximately tetrahedral sites bound to the cysteine sulfur atoms. The soft mer-
cury(ll) ion has a higher affinity for sulfur and will displace cadmium from metallothio¬
nein. At first the mercury ions occupy tetrahedral sites but as the number increases,
the geometries of the metal sites and protein change until about nine Hg(II) atoms are
bound in a linear (S—Hg—S) fashion. 92 Up to twelve + I cations such as copper(l)
and silver! I) can bind per molecule, indicating a coordination number lower than four,
probably three (see Problem 12.34).
An intriguing problem about which we know very little is the mechanism of metal
identification by the body that triggers its response, as in the case of the build-up of
metallothioneins upon exposure to toxic metals. Perhaps the best understood of the
metalloregulatory proteins is MerR that protects bacteria from mercurial toxicity. It is
extremely sensitive to Hg 2 , and distinguishes it from its congeners Zn 2+ and Cd 2+ .
There is good evidence that the mercury receptor forms three-coordinate mercury(II)
complexes (see Fig. 12. Ic), making possible this specificity. 92
91 Hamer, D. H. Ann. Rev. Biochem. 1986. 55. 913. Dalgarno. D. C.; Armitage. I. M. Adv. tnorg.
Biochem. 1984, 6. 113. Kojima. Y.; Kagi, J. H. R. Trends Biochem. Sci. 1978. 3. 90.
92 Johnson, B. A.; Armitage. I. A. tnorg. Chem. 1987, 26. 3139-3144.
93 Wright, J. G.; Natan. M. J.; MacDonncll, F. M.; Ralston, D. M.; O'Halloran. T. V. Prog tnorg
Chem. 1990, 38. 323-412.
Nitrogen Fixation
933
Nitrogen Fixation An enzyme system of particular importance is that which promotes the fixation of
atmospheric dinitrogen. This is of considerable interest for a variety of reasons. It is a
very important step in the nitrogen cycle, providing available nitrogen for plant
nutrition. It is an intriguing process since it occurs readily in various bacteria, blue-
green algae, yeasts, and in symbiotic bacteria-legume associations under mild condi¬
tions. However, nitrogen stubbornly resists ordinary chemical attack, even under
stringent conditions.
Molecular nitrogen. N 2 , is so unresponsive to ordinary chemical reactions that it
has been characterized as “almost as inert as a noble gas.” 94 The very large triple
bond energy (945 kJ mol -1 ) tends to make the activation energy prohibitively large.
Thus, in spite of the fact that the overall enthalpy of formation of ammonia is
exothermic by about 50 kJ mol - ', the common Haber process requires about 20 M Pa
pressure and 500 °C temperature to proceed, even in the presence of the best Haber
catalyst. In addition to the purely pragmatic task of furnishing the huge supply of
nitrogen compounds necessary for industrial and agricultural uses as cheaply as
possible, the chemist is intrigued by the possibility of discovering processes that will
work under less drastic conditions. We know they exist: We can watch a clover plant
growing at 100 kPa and 25 °C!
In Vitro Nitrogen The discovery that dinitrogen was capable of forming stable complexes with transition
Fixation metals (Chapter 15) led to extensive investigalion of the possibility of fixation via such
complexes. An important development was the discovery that certain phosphine
complexes of molybdenum and tungsten containing dinitrogen readily yield ammonia
in acidic media: 95
[MoCI,(thf),J + 3e - + 2N : + excessdppe -» [Mo(N 2 ),(dpe) 2 ] + 3CI" (19.36)
[Mo(N,),(dpe) 2 ] + 6H + -♦ 2NH, + N 2 + Mo v ‘ products (19.37)
where thf = tetrahydrofuran and dppe = 1,2-bis(diphenylphosphino)ethane, Ph 2 PCH,-
CH 2 PPh 2 . Both reactions take place at room temperature and atmospheric pressure.
1 he reducing agent is a Grignard reagent. This reaction sequence is important because
it models the in vivo nilrogenase systems that appear lo employ molybdenum.
We should not conclude, however, that ambiem temperature and pressure reac¬
tions are likely to replace the Haber-Bosch process. Despite the fad lhal the latter
requires high temperature and pressure, it is efficient and well entrenched, and it can
produce large volumes of product in short time periods. With respect to the former
processes, it is certain that the chemist will not be able to keep pace with the lively
imagination of the journalist. As an interesting aside on the inherent inability of the
scientist to match ever increasing expectations, the reader is directed to the following
selection of titles and headlines. The first is the title of the initial research report by
Chatt's group in England and the remainder are headlines of various reports of it in the
popular press: 95
94 Jolly, W. L. The Chemistry of the Non-Metals-, Prentice-Hall: Englewood ClilTs. NJ. 1966; p 72,
95 Chatt. J.; Pcarman, A. J.; Richards. R. L. Nature ILondon) 1975. 253. 39-40. For a very readable
account of the early work, including the headlines on page 934. sec Chatt. J, Proc. Roy. Instn. Great
Br. 1976. 49. 281. For a current overview, sec Leigh, G. J. Ace. Chem. Res. 1992, 25. 177-181. For
catalytic reduction in aqueous solution see Shilov, A. E. In Perspectives in Coordination Chem¬
istry-. Williams. A. F.; Floriana. C.; Merbach, A. E.. Eds.; VCH: New York. 1992; pp 233-244.
912
19- The Inorganic Chemistry of Biological Systems
C!
Fig. 19.14 Model compound for a cubane-type Fe 4 cluster with a trithiol ligand, (a)
Stereoview of [Fe 4 S 4 (trithiol)L’]; (b) close-up view of the Fe-S cluster. [From Stack, T. D.
P.; Holm. R. H. J. Am. Chem. Soc. 1988, 110, 2484-2494. Reproduced with permission.)
Blue Copper Perhaps the three most important redox systems in bioinorganic chemistry are:
Proteins (I) high spin, tetrahedral Fe(II)/Fe(III) in rubredoxin, ferredoxin, etc.; (2) low spin,
octahedral Fe(II)/Fe(III) in the cytochromes; and (3) pseudotetrahcdral Cu(I)/Cu(II)
in the blue copper proteins, such as stellacyanin, plastocyanin, and azurin. Grays- has
pointed out that these redox centers are ideally adapted for electron exchange in that
no change in spin state occurs. Thus there is little or no movement of the ligands—the
Franck-Condon activation barriers will be small.
The structure of plastocyanin (Fig. 19.15) is especially instructive in this regard.
Copper(I) is d" 1 and thus provides no ligand field stabilization energy in any geometry.
Because it is relatively small (74 pm), it is usually found in a tetrahedral environment.
In contrast. copper(II) is d 9 and is usually octahcdrally coordinated with Jahn-Teller
distortion, often to the point of square planar coordination. In the case of plasto¬
cyanin, the copper is situated in a “flattened tetrahedron" of essentially C 3v symme¬
try, "halfway" between the two idealized geometries.” This facilitates electron
transfer compared to a system that might be at the tetrahedral extreme or at the square
planar extreme: Energetically, either of the latter would require reorganization to¬
wards the other when electron transfer took place. Such structural changes would
inhibit the process.
The mechanism of electron transfer over the long distances (of the order of
1000 pm or more) necessitated by the large size of redox enzymes is one that is not
completely clear despite much current study. These transfers are critical whether one
is considering the photosynthetic center (page 917) or electron carriers such as the
34 Gray, H. B. Chem. Soc. Rev. 1986. IS, 17.
» Colman. P. M.; Freeman. H. C.; Guss, J. M.; Murata, M.; Norris, V. A.; Ramshaw, J. A. M.:
Venkatappa, M. P. Nature (London) 1978, 272, 319-324.
Electron Transfer, Respiration, and Photosynthesis
913
Fig. 19.15. Stereoview of the copper-binding site of plastocyanin. The four ligand residues
are His 37, Cys 84. His 87, and Met 92. Note that the geometry about the copper is neither
tetrahedral nor square planar, but intermediate. [From Colman. P. M.; Freeman. H. C.:
Guss. J. M.; Murata. M.; Norris. V. A.: Ramshaw. J. A. M.; Venkatappa. M. P. Nature
(London) 1978, 272, 319-324. Reproduced with permission.)
cytochromes and copper redox enzymes (above). The rate of electron transfer falls off
exponentially with distance at long range (Chapter 13). The rale is also dependent
upon the thermodynamic driving force and. as mentioned above, facilitated when
structural changes are minimal. One may readily ask why the iron and copper atoms
are not on the surface of the protein so that such long-range transfer would be
unnecessary. Surely one reason is to prevent their irreversible "corruption" to an
unusable form. And almost certainly the surrounding protein shield serves the purpose
of recognition, as yet poorly understood, that allows cytochrome c, plastocyanin, etc.,
to react with the intended target species and not be "short-circuited" by reacting
uselessly with the wrong redox agent.”
The determination of the structures of biologically important copper-containing
redox systems illustrates the multiplicity of techniques that can be brought to bear on
structural bioinorgamc chemistry. Ultimately, one would like to have a highly refined
accurate structure determined by X-ray crystallography. Yet over and over in this
chapter we shall see that this goal has not been met for many of ihe most interesting
compounds. There is a wide variety of techniques that may be used instead to gain the
desired information. These vary in ease of application and in the quality of the results,
but by combining different techniques much can often be said about the active site '
Often the nature, number, and distances of atoms in the coordination sphere of a
metal can be obtained by a relatively new method called extended X-rav absorption
fine structure (EXAFS). It elaborates upon the long-known fact that X-ray absorption
spectra show element-specific "edges” that correspond to quantum jumps of core
N ! aU ^' A ' G : Gray ' H - B - J - Chem - Educ. 985, 62. 932. Mayo. S. L: Ellis. W R j r •
Crutchlcy. R. j ; Gray, H. B. Science 1986, 233. 948. Bowler. B. E. : Raphael. A. L.: Gray H B.’
rT "r- i 8 -™t~ Liang - N - : picink ' j - ; a - g. ; smith. M , Hom™.
B. M; Froc. Nad. Acad Set. U. S. A. 1987, 34. 1249. Wcndoloski. J. J.; Matthew, j. B. . Weber.
. L., Salemme, F. R. Science 1987 .238. 794. McLendon. G. Acc. Client. Res. 1988. 21. 160. See
also articles by these authors in Struct. Bonding f Berlin) 1991. 73, 1-224. The kinetics of electron
transfer of this sort is discussed in Chapter 13.
930
19-The Inorganic Chemistry of Biological Systems
bi „,„ 6 ic a , systems the»»“ ed ucUo„« toj—
(CH-^-THFVThe'coUnjconinoid (methylcobalam.n) cn then partake in biomethyln-
tion reactions:
CHj-THF
■B r jCo')
methionine
THF ^/ V - homocysteine
Certain bacteria can n ^ lhy ^ lte ^i°"pb"'Sn“ Au.'PcT/and i^a^aerob.c sludges. Thus
resulting in environmental heaUh probk ^ P approxima tely Cu 3 (As0 3 ), with
serious problem in the 19th tent . y wag used as a pigm ent in wallpaper,
copper acetate present to enhance the: ’ wallpaper and form volat.le
Under humid conditions certain molds would grow o ^ ^ premises .
trimethylarsine causing arsenic poisoning triD hosphate (ATP), alkylation takes
When vitamin B„ reacts with bond between
place as in Eq. 19.33 with the formatton of a dmect carb concert with
adenosyl and the metal, forming B l2 coenzyme (Fig. 19.W u
t "a. other enzymes to effect 1 .2-shifts of the general type.
_c_c 2 — ^ —c, c 2
to convert it to succinic acid.
Fi „ „ » A nercovisw „f <K — »' »■ »■=• l"»» °'
... _-, :v 5968, A303. 45. Reproduced with permission.)
Enzymes
931
Until the proposed mechanism is examined, some of these rearrangements appear
unusual:
It is believed that the reaction starts with homolytic cleavage of the cobalt-carbon
bond (at a cost of perhaps 100 kJ mol -1 ) 88 to yield a Co(ll) atom and a 5’-deoxy-
adenosyl radical. This radical then abstracts a hydrogen atom (in Eq. 19.35 from the
methyl group). Migration of the —C(0)SR group takes place, followed by return of
the hydrogen atom from 5'-deoxyadenosine to the substrate. This regenerates the 5'-
deoxyadenosyl radical, which can recombine with the Co(U) atom to form the
coenzyme.
A third type of reaction employing B,-, coenzyme is the reduction of
—CH(OH)— groups to —CH,— groups, as in the reduction of ribonucleic acid
(RNA) to deoxyribonucleic acid (DNA).
B,, is unusual in several ways. The ability to form a metal-carbon bond in a
biological system appears to be unique: These are nature's only organometallic
compounds. w It is the only vitamin known to contain a metal. It appears to be
synthesized exclusively by bacteria. It is not found in higher plants, and although it is
essential for all higher animals, it must be obtained from food sources, hence its
designation as a "vitamin.”
The fitness of cobalamin to serve its biochemical functions has been variously
ascribed to different factors by different authors, 90 Certainly, the existence of three
oxidation states. Co 1 . Co 11 , and Co 111 , stable in aqueous (and hence biological) media
is necessary. This, in'itself, may eliminate the earlier transition metals (without
accessible + I oxidation states) and copper (Cu IM is strongly oxidizing). In addition,
we have seen that J^/J" (I fie “ /18e “) systems arc ideal for oxidative addition/reduc¬
tive elimination reactions. It has also been suggested that the flexibility of the corrin
ring allows changes in conformation that may be beneficial. In this regard it may be
noted that the cobalt porphyrin analogues of B,, cannot be reduced to Co(I) in
aqueous solution. Hence the corrin ring was selected in place of porphyrin in the
evolution of the B l2 cobalt complexes.
»" Halpcrn. J.: .4<r. Chan. Kcs. 1982. 15. 238: Kim. S.-H.; Leung. T.W. J. Am. Chan. Sue. 1984. 106 .
8317-8319.
This truism of yesterday is currently being challenged—there is increasing evidence of other
biological mclal-carbon bonds, but. unlike B ( = , none of the suspected compounds has been isolated
and characterized.
9,1 Schrauzcr. C. N. Angeic. Chan. Ini. LSI. Engl. 1976, 15, 417. Ochiai. E.-l. J■ Chan. Edna. 1978. 55.
631: General Principles of Biochemistry of the Elements : Plenum: New York. 1987; p 221.
914 19- The Inorganic Chemistry of Biological Systems
electrons to unoccupied orbitals or to the continuum. By choosing X-ray frequencies
near the X-ray edge of a particular element, atoms of that element can be excited to
emit photoelectrons. The wave of each electron will be backscattered by the nearest
neighbors in proportion to the number and kind of the ligands and inversely propor¬
tional to the interatomic distance. If the backscattered wave is in phase with the
original wave, reinforcement will occur, yielding a maximum in the X-ray absorption
spectrum. Out-of-phase waves will cancel and give minima. The EXAFS spectrum
consists, then, of the X-ray absorption plotted against the energy of the incident X-ray
photon. The amplitudes and frequencies of the oscillations in the absorption are
related to the number, type, and spacing of the ligands. Thus if one bombards heme
with an X-ray frequency characteristics of an iron edge, it should, in principle, be
possible to learn that there are four atoms of atomic number 7 equidistantly surround¬
ing the iron atom. 37
Some EXAFS data for copper proteins are given in Table 19.2. Confirming data
from X-ray crystallography are also listed where known. Copper is particularly well
suited for study by electron paramagnetic resonance. At the very simplest level, this
Table 19.2
The type, accuracy, and
extent ol information given
by EXAFS and X-ray
crystallography on blue
coppor protoins
Compound
EXAFS (pm)
X-ray (pm)
Azurin
Cu-N
Cu-N
Cu-S
Cu-S
= 205°
= 189
= 223
= 270
Cu-N (His-46)
Cu-N (His-117)
Cu-S (Cys-112)
Cu-S (Met-121)
= 206
= 196
= 213
= 260
Cytochrome
oxidase
Plastocyanin
Cu-S
Cu-N
= 227 c
= 197'
d
Cu-N (His-37)
Cu-N (His-87)
= 204 r
- 210
Cu-S
= 211
Cu-S (Cys-84)
Cu-S (Met-92)
= 213
= 290
Stellacyanin
g
g
" Tullius, T. D.; Frank, P.; Hodgson. K. 0. Proc. Natl. Acad.
Sci. V. S. A. 1978, 75, 4069. Grocncweld, C. M.; Feiters, M.
C.; Hasnain, S. S.; Van Rijn, J.; Rccdijk, J.: Canters. G. W.
Biochem. Biophys. Acta 1986, 873, 214.
t> Norris. G. E.; Anderson, B. F.; Baker, E. N. J. Am. Chem.
Soc. 1986, 108, 2784.
r Scolt, R. A.; Cramer, S. P.; Shaw. R. W.; Bcincrt, H.; Gray.
H. B. Proc. Natl. Acad. Sci. U SA. 1981, 78, 664.
d No crystallographic data available.
<■ Guss, J. M.; Freeman, H. C. J. Mol. Biol. 1983, 169, 521.
/ Guss. J. M.; Harrowell, P. R.; Murata. M.; Norris, V. A.;
Freeman. H. C. J. Mol. Biol. 1986, 192, 361.
No EXAFS or crystallographic data available. Spectroscopy
indicates that three ligands arc the same as in plastocyanin.
57 The present discussion has been greatly simplified to give the general technique as well as the
information obtained without going into the details of the analysis. For the latter as well as the
experimental technique, see Cramer. S. P.; Hodgson, K. 0. Prog. Inorg. Chem. 1979 25 I.
Hav ? W Bio-Inorganic Chemistry, &!!' Horwccd: Chicesler, 1984: pp 51-57; Rehr. J. J.: de
ABws. • f j- A. CherK. So c . 1991, 113, 5135-5140.
Electron Transfer, Respiration, and Photosynthesis 915
method can distinguish between Ihe presence of an odd electron (Cu 2+ , P, EPR
signal) and complete electron pairing (Cu\ c/ 10 , no EPR signal). Ligands with nuclei
having nonzero spins (such as nitrogen) will cause hyperfine splittings proportional lo
the number of such atoms bonded to the copper(II) atom [see page 923 with respect to
Cu(ll)-substituted carboxypeptidase A). Finally, a study of the hyperfine splitting,
some of it resulting from the copper atom's nonzero nuclear spin, can provide
geometric information (see below).
Analysis of ligand field and charge transfer absorption bands can provide informa¬
tion concerning the geometry of the copper site and the nature of the ligand, though
mostly of a qualitative sort; values for bond angles and bond lengths cannot be
quantified. It is often useful in this regard to attempt the synthesis and structural
determination of model compounds and to try to match their properties with those of
the active sites in the metalloproteins. These efforts, combined frequently with the¬
oretical calculations of the same properties, often allow predictions to be made
concerning the nature of the active sites. For example, while the structures of azurin
and plastocyanin have been determined by X-ray crystallography, that of the related
blue copper protein stellacyanm has not because suitable crystals have not yet been
grown. Spectral studies have indicated that three of the four ligands (His, His, Cvs)
are the same in plastocyanin and stellacyanin, but that the latter does not contain the
methionine ligand found in the former. 38 A combination of electron paramagnetic
resonance and electronic spectral data with self-consistent-field calculalions has indi¬
cated that the unknown fourth ligand in stellacyanin provides a stronger field than
does methionine. 39 It presumably results in a shorter Cu—X bond as well as a
flattening of the geometry more towards a square planar arrangement.
The synthesis of model compounds has proved to be an interesting challenge. The
Cu 2 f ion is a sufficiently strong oxidizing agent to couple two sullliydryl groups:
2RSH — RS—SR -t- 2H (19.20)
Thus any simple attempt to let thiols coordinate to Cu(il) will result in persulfides and
Cu(l).
R
S
2L,Cu"RSH -► LjCu"^ )cu"L, -► RS-SR + 2L,Cu' (19.21)
S
R
As shown in Eq. 19.21. it is thought that this reaction takes place through a dimeric,
bridged intermediate of coordination number 5. Copper tends to form a maximum of
five reasonably strong bonds, 6 " so complexation with a nonlabile tetracoordinate
macrocyclic ligand (N 4 ) provides only one additional site for a sulfur ligand. The
reaction in Eq. 19.21 is inhibited, and a thiolate complex can be isolated: 61
' Gewirth. A. A.; Cohen. S. L.; Schugar. H. J.: Solomon, E. I. Inorg. Chem. 1987, 26. 1133-1146.
59 Thomann. H.; Bernardo. M.; Baldwin, M. J.: Lowery. M, D.; Solomon. E. I. J. Am. Chem. Soc.
1991, 113, 5911-5913. This discussion of the study of blue copper proteins has necessarily been
brief. Fora comprehensive discussion, see Solomon. E. I.: Baldwin. M. J.; Lowery, M. D, Chem.
Rev. 1992. 92. 521-542.
,1 " See Chapter 12. Copper!Ill undergoes Jahn-Teller distortion when six-coordinate and tends to form
four strong bonds and two weak ones (Chapter 11).
M John- E.; Bharadwaj. P. K.; Potenza. J. A.; Shugar, H. J. Inorg. Chem. 1986, 25, 3065.
928
929
19-The Inorganic Chemistry of Biological Systems
This enzyme catalyzes the oxidation of xanthine to uric acid:
The electron flow may be represented as:
Xanthine -► Mo -► 2Fe 2 S 2 -♦ FAD -► 0 2 (19.30)
Uric acid is the chief end product of purine metabolism in primates, birds, lizards,
and snakes. An inborn metabolic error in humans results in increased levels of uric
acid and its deposition as painful crystals in the joints. This condition (gout) may be
treated by the drug allopurinol which is also oxidized by xanthine oxidase to allo-
xanthine (dashed line in Eq. 19.29). However, alloxanlhine binds so tightly to the
molybdenum that the enzyme is inactivated, the catalytic cycle broken, and uric acid
formation is inhibited. The extra stability of the alloxanthine complex may be a result
of strong N—H---N hydrogen bonding by the nitrogen in the 8-position:
This structure resembles the hydrogen bonded transition state for the nucleophilic
attack of hydroxide ion (Eq. 19.29) where the hydrogen bond promotes the attack on
Vitamin B 12 and the
B 12 Coenzymes 85
Enzymes
the carbon. With a nitrogen atom at the 8-position there is no way for the alloxanthine
to leave. 8
A closely related enzyme is aldehyde oxidase. It also contains two
(Mo/2Fe 2 S 2 /FAD) units with a molecular weight of about 300,000. It converts
acetaldehyde to acetic acid via electron flow:
Acetaldehyde -» Mo(VI) -» 2Fe 2 S-, - FAD - 0-, (19.31)
When ethanol is consumed, the initial metabolic product is the extremely poi¬
sonous acetaldehyde, which is kept in low concentration by the oxidase-catalyzed
conversion to harmless acetic acid. The drug Antabuse, used for treating alcoholism,
is a sulfur-containing ligand, disulfiram:
S S
II II
Et,N—C—S—S—C—NEt 2
In the body, Antabuse inhibits acetaldehyde oxidase, presumably via the soft-soft
molybdenum-sulfur interaction. 84 Any alcohol ingested will be converted to
acetaldehyde which, in the absence of a pathway to destroy it, will build up with
severely unpleasant effects, discouraging further consumption.
In 1948 an "anti-pernicious anemia factor” was isolated, crystallized, and named
vitamin B (2 or cyanocobalamin. The molecule is built around a corrin ring containing a
cobalt(III) atom. The corrin ring is a modified porphyrin ring in which one of the
=CH— bridges between two of the pyrrole-type rings is missing, contracting the ring.
The fifth and sixth coordination sites on the cobalt are filled by a nitrogen atom from
an imidazole ring and a cyanide ion. The latter is an artifact of the isolation procedure
and is not present in the biological system, where the sixth position appears to hold a
loosely bound water molecule.
Vitamin B l2 may be reduced by one electron ("vitamin B l2 ,’’) or two electrons
(“vitamin B,,,") to form the Co(ll) and Co(I) complexes, respectively. 8(1 The latter is
strongly nucleophilic and readily undergoes alkylation via oxidative addition: 87
[B l2 (Co')I + CHjI -► [B, 2 (Co JI1 )—CHj] + 1“ (19.32)
M Slicfcl, E. I. Prog. Inorp. Cltem. 1977, 22, 1-223. The 1988 Nobel Pri/.c in Physiology and Medicine
was awarded for "rational synthesis," i. e., the tailoring of drugs for specific sites and actions.
Elion. G. B. In Vitro Cell. Dev. Dial. 1989. 25, 321-330.
M The above surmise is based on the known chemistry between molybdenum and sulfur-containing
ligands. It has been suggested that disulfiram inhibits the enzyme by oxidizing essential sulfhydryl
groups to form internal S—S bonds (see Vallari. R. C.: Pietruszko, R. Science 1982. 216, 637).
Disulfiram is also used to prevent renal toxicity from platinum when ru-diamminedichloroplatinum
(II) (see page 938) is used to treat neoplasms and trypanosomiasis (see Wysor, M. S.; Zwelling, L.
A.; Sanders, J. E.; Grcnan, M. M. Science 1982. 217, 434-456). The complcxing agent is thought to
be dicthyldithiocarbamate, a metabolite of disulfiram.
1,5 Ochiai, E.-l. General Principles of Biochemistry of the Elements', Plenum: New York, 1987;
pp 217-221. Crabtree, R. H. The Orpanometallic Chemistry of the Transition Metals: Wiley: New
York. 1988; pp 388-393.
1,4 The r stands for "reduced" and the r for "super-reduced." The latter may seem to be something of
an exaggeration until one recalls that the predominant coordination chemistry of cobalt is that of
Co(III), with less for Co(II), and very little for other oxidation states.
87 See Chapter 15 with respect to the basicity of metals in low oxidation states and oxidative addition
reactions.
916
19-The Inorganic Chemistry of Biological Systems
[N 4 Cu] 2+ + HSCH 2 CH 2 COCT -» [N 4 CuSCH 2 CH,COO] + H + (19.22)
Finally, a pseudotetrahedral complex more closely resembling the copper site in
blue copper proteins, including the presence of two cysteine groups, can be achieved
by using a cysteine derivative of ethylenediamine, [HSCH 2 CH(C0 2 CH 3 )NHCH 2 ] 2 .
It is a softer, polydentate (though not macrocyclic) ligand and will displace N 4 :
[N 4 Cu] 2+ + [HSCH 2 CH(C0 2 CH 3 )NHCH 2 ] 2 -
Cu[SCH 2 CH(C0 2 CH 3 )NHCH,] 2 + N 4 + 2H + (19.23)
The Cu—S bonds are about 10 pm longer than those in plastocyanin but about the
same as those in cytochrome c oxidase. 62
Photosynthesis The photosynthetic process in green plants consists of splitting the elements of water,
followed by reduction of carbon dioxide:
2H,0 - [4H] + 0 2 (19.24)
xC0 2 + |[4H] - (CH 2 0) 4 + |0 2 (19.25)
where [4H] does not imply free atoms of hydrogen but a reducing capacity of four
equivalents. The details of the reactions involved in photosynthesis are not known,
although the broad outlines are fairly clear. In all dioxygen-producing organisms
ranging from cyanobacteria to algae to higher plants, there are two coupled photo¬
synthetic systems, PS I and PS II. The two differ in the type of chlorophyll present and
in the accessory chemicals for processing the trapped energy of the photon. The
primary product of PS I is reduced carbon, and the primary product of PS II is energy
in the form of two moles of ATP 6 -’ with molecular oxygen as a chemical by-product.
In addition to the chlorophyll molecules at the reaction centers of PS I and PS II,
there are several other pigments associated with the light-harvesting complex. Among
these are carotenoids, open-chain tetrapyrrole pigments, and others. These serve dual
roles of protecting the cell from light radiation and at the same time harvesting much of
it for photosynthesis. Some of these compounds are arranged in antenna-like rods that
gather the light energy and funnel it into the reaction centers. 64
The energy of an absorbed photon in either PS I or PS II initiates a series of redox
reactions (see Fig. 19.16). 63 System I produces a moderately strong reducing species
(RED,) and a moderately strong oxidizing species (OX,). System II provides a
stronger oxidizing agent (OX„) but a weaker reducing agent (RED,,).
OX„ is responsible for the production of molecular oxygen in photosynthesis. A
manganese complex, probably containing four atoms of manganese, is attached to a
protein molecule. It reduces OX I( which is recycled for use by another excited
chlorophyll molecule in PS II. In the redox reactions the manganese shuttles between
two oxidation states with each manganese atom increasing (and subsequently decreas¬
ing) its oxidation state by one unit, but it is not known with absolute certainty what
62 Bharadwaj, P. K.; Potcnza, J. A.; Shugar, H. J. J. Am. Chem. Soc. 1986. 108 . 1351.
63 Adenosine triphosphate, an important energy-rich species in metabolism.
64 Deisenhofer, J.; Michel, H.; Huber, R. Trends Biochem. Sci. 1985. 10 , 243-248. Zubcr, H.;
Brunisholz. R,; Sidler, W. In New Comprehensive Biochemistry: Photosynthesis'. Amesz, J., Ed.;
Elsevier: Amsterdam, 1987; pp 233-271.
65 Mathis. P.; Rutherford, A. W. In New Comprehensive Biochemistry: Photosynthesis: Amesz, J..
Ed.; Elsevier: Amsterdam, 1987; pp 63-96.
Electron Transfer, Respiration, and Photosynthesis
917
Ph, ’ lo '- VN,cn ’ 11 PholusyMi
O, + 4H'
Fig. 19.16 Electron flow in photosystems I and II (“Z-scheme"). Vertical axis gives mid-point redox potential with
reducing species (top) and oxidizing species (bottom).
these oxidation states are. 6 ( n the reduced form the oxidation states may be as low as
three Mn(II) and one Mn(III), but they are more likely to be three Mn(III) and one
Mn(IV). A suggested scheme for this redox chemistry is shown in Fig. 19.17 in which
the active site cycles between a cubane-like and an adamantane-like configuration.
I here have been several other suggestions concerning these structures, including
•'butterfly clusters" and other modifications of the Mn 4 configuration. 67
Chlorophyll and the The chlorophyll ring system is a porphyrin in which a double bond in one of
Photosynthelic ihe pyrrole rings has been reduced. A fused cyclopentanone ring is also present
Reaction Center (Fig. 19.18). Bacteriochlorophyll is similar but has a double bond in a second pyrrole
ring reduced, and it has a substituent acetyl group in place of a vinyl group. Chlo¬
rophyll absorbs low-energy light in the red region (-600-700 nm). The exact fre¬
quency depends on the nature of the substituents.-
Based on our knowledge of the structure of chlorophyll as well as the results of
studies on the photo behavior of chlorophyll in vitro, it is possible to summarize some
of the features of the chlorophyll system which enhance its usefulness as a pigment in
photosynthesis. 68 First, there is extensive conjugation of the porphyrin ring. This
lowers the energy of the electronic transitions and shifts the absorption maximum into
66 Dismukcs, G. C. Photochem. Photobiol. 1986, 43. 99. Babcock, G. T. In New Comprehensive
Biochemistry: Photosynthesis ; Amesz, J.. Ed.; Elsevier: Amsterdam, 1987; pp 125-158. Brudvig,
G. W. J. Bioenerg. Biomembr. 1987. 19. 91.
67 See Brudvig, G. W.; Crabtree. R. H. Prog. Inorg. Chem. 1989. 37. 99-142; Quc, L.. Jr - True A E
Ibid. 1990, 38. 97-200.
“ Ma S8' ora - G. M.; Ingraham, L. L. Struct. Bonding (Berlin) 1967. 2, 126. Hindman. J. C.: Kugei. R.;
Svirmickas, A.; Katz. J. J. Proc. Nall. Acad. Sci. V. S. A. 1977. 74, 5 - 9 .
926
19-The Inorganic Chemistry of Biological Systems
and NJ\ none of which is known for exceptional softness, are bound with exceptional
strength. They are, however, isoelectronic and isostructural with the reactants and
products of the enzyme reaction, C0 2 , CO 3 - , and HCO^, respectively. The explana¬
tion appears to be a tailoring of the structure of the enzyme molecule to form a pocket
about 450 pm long next to the zinc ion, perhaps containing an additional positive
center, to stabilize ions of appropriate size.
Although some mechanisms illustrate the carbon dioxide coordinated directly to
the zinc atom, this is highly unlikely. The infrared asymmetric stretch for carbon
dioxide is found at 2343.5 cm -1 in the bound enzyme compared with 2321 cm - ' for the
free molecule, hardly compatible with a strong interaction of one oxygen atom and not
the other. The visible spectrum of the Co 2+ -substituted enzyme shows very small
shifts upon binding C0 2 , again incompatible with strong oxygen-metal interactions.
The zinc atom is thought to be considerably more acidic in carbonic anhydrase than in
carboxypeptidase A. The substitution of a third, neutral, and less basic histidine in
place of the glutamate anion contributes to the greater acidity. In addition, the three
histidines are pulled back making the zinc more electronegative and more acidic
towards the fourth position (see Problem 19.38). This polarizes an attached water
molecule, perhaps to the point of loss of a hydrogen ion to form a coordinated hydroxo
group. The mechanism of the reversible hydration of carbon dioxide to carbonic acid
(actually the hydrogen carbonate ion at physiological pHs) is thought to follow the
pathway shown in Eq. 19.27. Like all truly catalytic processes, it is a closed loop : 80
It may operate either clockwise (as drawn) to hydrate carbon dioxide, or coun¬
terclockwise to release carbon dioxide (from the hydrogen carbonate anion) from
80 Bertini, I.; Luchinat, C.; Scozzafava, A. Struct. Bonding f Berlin) 1982, 48. 45-91. Lindskog. S.
Adv. Inorg. Biochem. 1982, 4. 115.
Enzymes
927
solution (as in the blood in the lungs), depending upon the concentrations of the
reactants.
Ligands that can coordinate to an active center in an enzyme and prevent
coordination by the substrate will tend to inhibit the action of that enzyme . 81 We have
seen that azide can occupy the pocket tailored to fit the carbon dioxide molecule. This
prevents the latter from approaching the active site. Furthermore, the infrared evi¬
dence indicates that the azide ion actually does bind the zinc atom: The asymmetric
stretching mode of the azide ion is strongly shifted with respect to the free ion
absorption. Thus the zinc is inhibited from acting as a Lewis acid towards water with
the formation of a coordinated hydroxide ion. Other inhibitors also bind to the metal
atom. As little as 4 X 10 ~ 6 M cyanide or hydrogen sulfide inhibits the enzymatic
activity by 85%.
Inhibition may also be effected by metal ions. Most prosthetic groups involve
metals of the first transition series (molybdenum seems to be the sole exception).
Coordination of the apoenzyme to a heavier metal ion may destroy the enzymatic
activity. Particularly poisonous in this regard are metal ions such as Hg 2+ . The latter
has a special affinity for sulfur (see HSAB, Chapter 9) and thus tends to form
extremely stable complexes with amino acids containing sulfur such as cysteine,
cystine, and methionine. The inhibition of an enzyme by Hg-+ has been taken as an
indication of the presence of thiol groups but is not infallible. (For example, Hg 2+
completely abolishes the activity of carboxypeptidase A in hydrolyzing amide link¬
ages.) Nevertheless, the affinity of sulfur and mercury is responsible for many of the
poisonous effects of mercury in biological systems. Often these effects may be
reversed by addition of sulfur-containing compounds such as cysteine or glutathione.
Another sulfur donor. 2,3-dimercaptopropanol. has a strong affinity for soft metal
ions. Developed during World War I as an antidote for the organoarsenic war gas,
lewisite, it was dubbed British antilewisite (BAL). It has proved to be extremely
useful as an antidote for arsenic, cadmium, and mercury poisoning.
The inhibition of enzyme systems does not necessarily cause unwanted effects.
Consider the enzyme xanthine oxidase. It contains two atoms of molybdenum, four
Fe 2 S 2 , and two FAD (flavin adenine dinucleotide) moieties, and it has a molecular
weight of 275.000-300.000. There is no evidence that the two units (Mo/2Fe 2 S 2 /FAD)
are near each other or interact in any way. It is believed that the immediate environ¬
ment of each molybdenum atom consists of one oxygen and three sulfur atoms
(additional ligands may be present ): 82
168 pmn S — H 2 38 pm 170 pm-s Sy-215 pm (19.28)
0=Mo—S—R - ^) O—Mo—S—R
S—R +c ~ S—247 pm
Reduced Oxidized
The determination of a single oxygen atom at 168-170 pm (and therefore doubly
bonded) and three sulfur atoms at 238 pm (single-bonded HS— and RS—) in the
reduced form, or two at 247 pm (RS—) and one at 215 pm (S=) in the oxidized form,
can be made from the EXAFS spectrum.
1,1 Compare the action of carbon monoxide on hemoglobin, page 896.
82 Cramer. S. P.; Wahl, R.; Rajagopalan. K. V. J. Am. Cltem. Soc. 1981, 103, 7721-7727.
19 The Inorganic Chemistry of Biological Systems
photoevolution of dioxygen. [Modified from
987, Si, 4586. Reproduced with permission.!
19.17 One proposal for the involvement of Mn centers in the
Ivig. G. W.; Crabtree. R. H. Proc. Nall. Acad. Sci. U. S. A. I
Chlorophyll a IR = CH, I
Chlorophyll hi R = CHOI
Fig. 19.18 Structure of chlorophyll. The long alkyl chain at the bottom is the phytyl group.
the region of visible light. Conjugation also helps to make the ring rigid, and thus less
energy is wasted in internal thermal degradation (via molecular vibrations).
The maximum intensity of light reaching the earth's surface is in the visible
region; ultraviolet light is absorbed in the earth’s atmosphere by species such as
dioxygen and ozone (trioxygen), infrared light is absorbed by carbon dioxide, water
etc The absorption spectra of photosynthetic systems fall nicely within that portion of
the sun's spectrum that reaches the earth. Some of the energy of the light not absorbed
by the chlorophyll itself is captured by accessory pigments. In the octave of light trom
i
Mn.
^Mn
Enzymes
919
about 350 to 700 nm. one or more photosynthetic pigments absorbs at every fre¬
quency. This is ihe portion of the total spectrum that is of highest intensity and
corresponds rather closely to the sensitivity of the human eye, another system
adapted to that portion of light that reaches earth.
Thanks to careful spectroscopic and crystallographic work we have considerable
information about the reaction center in photosynthetic bacteria, and it is probable
that the photosynthetic systems in higher plants are modifications, perhaps partial
duplications, of the bacterial system.® The reaction center is a protein with a mo¬
lecular weight of about 150.000. The heart of Ihe reaction center is a pair of chlo¬
rophyll molecules, often referred to as the “special pair" (Fig. 19.19).™ The special
pair are in contact with each other through the overlap of one of the pyrrole rings in
each molecule. In addition, an acetyl group on each molecule coordinates to the
magnesium atom of the other. The sixth coordination position on each magnesium
atom is occupied by a nitrogen atom from a histidine residue in the protein chain.
Associated with the special pair are pheophytin molecules and quinone molecules that
accept the electron from the reaction center (Fig. 19.20). Near the quinone molecule is
a nonheme iron atom complexed by four histidines and one glutamic acid. The
electron appears to be passed through the iron atom to the redox chain. The “hole"
(vacancy resulting in a cationic charge) in the reaction center is filled with an electron
from a cytochrome molecule (there arc four cytochrome-c-type centers lying near the
special pair). The separation of charge between the electron being passed down the Z-
scheme chain and the positive charge residing on the Fe(!U)-cytochrome represents
potential energy that is utilized in the photosynthesis.
Enzymes are the catalysts of biological systems. They not only control the rate of
reactions but. by favoring certain geometries in the transition state, can lower the
activation energy for ihe formation of one product rather than another. The basic
structure ol enzymes is built of proteins. Those of interest to the inorganic chemist are
composed of a protein structure (called an apoenzyme) and a small prosthetic group,
which may be either a simple metal ion or a complexed metal ion. For example, heme
is the prosthetic group in hemoglobin. A reversibly bound group that combines with an
enzyme lor a particular reaction and then is released to combine with another is
termed a coenzyme. Both prosthetic groups and coenzymes are sometimes called
cofactors. 7 1
To illustrate Ihe structure ol tin enzyme and its relation lo function, considercarboxy-
peptidase A. This pancreatic enzyme cleaves the carboxyl terminal amino acid from a
peptide chain by hydrolyzing the amide linkage:
' ' ' P «>-Leu-Glu-Phe A 1 . . . Pro-Leu-Glu + phenylalanine (19.26)
Youvan. D. C.; Marrs. B. L. Cell 1984. 39. I; Sci. Am. 1987. 256 (6). 42-48.
711 Deiscnhofcr. J.; F.pp. O.; Miki. K.: Huber. R.; Michel. H. Nature (London) 1985.3/a. 618. Parson.
VV. W. In New Comprehensive Biochemistry: Photosynthesis', Amesz, J.. Ed.; Elsevier- Amster¬
dam. 1987: pp 43-61. Budil, D. F..: Gasl. P.. Chang. C-H: Schiller. M.; Norris. J. R Ann Rev
Phys. Client. 1987. Jg, 561-583.
71 These terms are not always used in exactly the same way. Sec Dixon. M.: Webb, E. C En-ymes
3rd ed.: Academic: New York. 1979: Hammcs. G. G. Enzyme Catalysis and Regulation. Academic
New York. 1982: Palmer. T. Understanding Enzymes. 2nd ed.: Ellis Horwood: Chicester, 1985,
Enzymes
Structure and
Function
19-The Inorganic Chemistry of Biological Systems
Enzymes
925
A
Icl
Fig. 19.22 Suggested mode of action of carboxypeptidasc A in the hydrolysis of an amide
linkage in a polypeptide, (a) Positioning of the substrate on the enzyme. Interactions are
(I) coordinate covalent bond, carbonyl oxygen to zinc; (2) hydrogen bonds, arginine to
carboxylate and tyrosine to amide; (3) van der Waals attractions, hydrophobic pocket to
aromatic ring; (4) dipole attraction and possible incipient bond formation, carboxylate
(glutamate) oxygen to carbonyl group (amide linkage). Drawing is diagrammatic portrayal in
two dimensions of the three-dimensional structure, (b) Probable intervention of polarized
water molecule in the incipient breaking of the amide (N—C bond) linkage, (c) Completed
reaction, removal of the products (amino acid and shortened peptide chain). Original
configuration of enzyme returns after proton shift from glutamic acid to tyrosine. For a more
detailed discussion of the mechanism of catalysis by carboxypeptidasc A, see Breslow, R.;
Wernick, D, L. Proc. Natl. Acad. Sci. U. S. A. 1977, 74, 1303.
Inhibition and
Poisoning
The study of the factors that enable an apoenzyme to select the appropriate metal ion
is of importance to the proper understanding of enzyme action. 77 The factors that
favor the formation of certain complexes in the laboratory should also be important in
biological systems. For example, the Irving-Williams series and the hard-soft
acid-base principles should be helpful guides. Thus we expect to find the really hard
metal ions [Group IA (1); Group IIA (2)] preferring ligands with oxygen donor atoms.
The somewhat softer metal atoms of the first transition series (Co to Zn) may prefer
coordination to nitrogen atoms (cf. Fig. 9.5). The important thiol group, —SH,
should have a particularly strong affinity for soft metal ions.
The usual structural principles of coordination chemistry such as the chelate
effect, the preference for five- and six-membered rings, and the stability of certain ring
conformations should hold in biological systems. In addition, however, enzymes
present structural effects not observed in simpler complexes. An interesting example
is carbonic anhydrase which catalyzes the interconversion of carbon dioxide and
carbonates. Like carboxypeptidase, carbonic anhydrase has one zinc atom per mole¬
cule (with a molecular weight of -30.000), in this case coordinated to three histidine
residues (His 94, His 96, His 119) 78 and a water molecule or hydroxide ion. The active
site (Fig. 19.23) contains other amino acids that may function through hydrogen
bonding, proton transfer, etc. The relative binding power of the zinc ion towards
halide ions is reversed in the enzyme (I - > BP > Cl~ > F - ) compared with the free
Zn 2+ ions (F~ > CP > BP > P). This reversal could be interpreted as some sort of
“softening” effect on the zinc by the apoenzyme were it not that the soft ligand CN~
is bound equally well by the free ion as by the complex. 79 Furthermore, NO^. CNCP,
Fig. 19.23 Active site of
carbonic anhydrase. In the
resting enzyme a water
molecule (O = •)
coordinates to the zinc
atom. All hydrogen atoms
have been omitted for
clarity.
77 Sigcl, H.; McCormick. D. B. Acc. Cliem. Res. 1970, 3, 201.
78 There arc three human isoenzymes of carbonic anhydrase that differ slightly in amino acid composi¬
tion. and so the sequencing numbers for the histidines differ among them.
79 It should be noted that this discussion is based on a comparison of the equilibrium constants of
enzymic zinc-cyanide complcxation versus aqueous zinc-cyanide complexations. Cyanide has a
high affinity for the soft zinc ion under both conditions (stability constant of [Zn(CN),,] 2- = 7.7 x
I0 16 ); hence it should not be concluded that there is any lack of affinity for cyanide in the enzyme.
920
19-The Inorganic Chemistry of Biological Systems
O = Mg • = N
O = C 8 = 0
lb)
Fig. 19.19 (a) Stereoview of the special pair in the photoreaction center. Rings I of the
chlorophyll molecules are stacked upon each other, and the magnesium atom of each
chlorophyll is coordinated by an acetate group from the other molecule, (b) Close-up of the
nearer chlorophyll molecule from part (a). The unattached acetate group is from the other
chlorophyll molecule. [Modified from Deisenhofcr, J.; Epp, O.; Miki. K.; Huber, R.; Michel.
H. J. Mol. Biot. 1984, 180, 385-398. Reproduced with permission.]
Enzymes
921
Fig. 19.20 Stereoview of the photosynthetic reaction center. The photoexcited electron is
transferred from the special pair to another molecule of bacteriochlorophyll (BCI). then to a
molecule of bacteriopheophytin (BPh), then to a bound quinone (Q). all in a period of 250 ps.
From the quinone it passes through the nonheme iron (Fe) to an unbound quinone (not
shown) in a period of about 100 ns. The electron is restored to the "hole' - in the special
pair via the chain of hemes (He I, etc.) in four cytochrome molecules, also extremely
rapidly (-270 ps). The special pair here is rotated 90° with respect to Fig. 19.19. [From
Deisenhofcr, J.; Michel, H.t Huber. R. Trends Biochem. Set. 1985. 243-248. Reproduced
with permission.]
The enzyme consists of a protein chain of 307 amino acid residues plus one Zn 2+ ion to
yield a molecular weight of about 34,600. The molecule is roughly egg-shaped, with a
maximum dimension of approximately 5000 pm and a minimum dimension of about
3800 pm (Fig. 19.21a). There is a cleft on one side that contains the zinc ion. the active
site. The metal is coordinated approximately tetrahedrally to two nitrogen atoms and
an oxygen atom from three amino acids (His 69, Glu 72, His 169) in the protein chain:
Active
site
922
19-The Inorganic Chemistry of Biological Systems
Fig. 19.21 (a) Stereoview of about one quarter of the carboxypeptidasc A molecule, showing
the cavity, the zinc atom, and the functional groups (shown with black atoms) Arg-145
(right). Tyr-248 (above), and Glu-270 (left), (b) Stcreoview of the same region, after the
addition of glycyl-L-tyrosinc (heavy open circles), showing the new positions of Arg-145.
Tyr-248, and Glu-270. The guanidinium movement of Arg-145 is 200 pm. the hydroxyl of
Tyr-248 moves 1200 pm. and the carboxylate of Glu-270 moves 200 pm when Gly-Tyr bonds
to the enzyme. [From Lipscomb. W. N. Chem. Sac. Rev. 1972. /. 319. Reproduced w.th
permission.]
Enzymes 923
The fourth coordination site is free to accept a pair of electrons from a donor atom in
the substrate to be cleaved. 77 The enzyme is thought to act through coordination of the
zinc atom to the carbonyl group of the amide linkage. In addition, a nearby hydro-
phobic pocket envelops the organic group of the amino acid to be cleaved (Fig. 19.22)
and those amino acids with aromatic side groups react most readily. Accompanying
these events is a change in conformation of the enzyme: The arginine side chain moves
about 200 pm closer to the carboxylate group of the substrate, and the phenolic group
of the tyrosine comes within hydrogen bonding distance of the imido group of the C-
terminal amino acid, a shift of 1200 pm. 73 The hydrogen bonding to the free carboxyl
group (by arginine) and the amide linkage (by tyrosine) not only holds the substrate to
the enzyme but helps break the N—C bond. Nucleophilic displacement of the amide
group by an attacking carboxylate group from a glutamate group could form an
anhydride link to the remainder of the peptide chain. Hydrolysis of this anhydride
could then complete the cycle and regenerate the original enzyme. More likely, the
glutamate acts indirectly by polarizing a water molecule (Fig. 19.22b) that attacks the
amide linkage.
This example illustrates the basic key-and-lock theory first proposed by Emil
Fischer in which the enzyme and substrate fit each other sterically. However, there is
more to enzymatic catalysis than merely bringing reactants together. There is good
evidence that the enzyme also encourages the reaction by placing a strain on the bond
to be broken. Evidence comes from spectroscopic studies of enzymes containing
metal ions that, unlike Zn 2 , show d-d transitions. The spectrum of the enzyme
containing such a metal ion provides information on the microsymmetry of the site of
the metal. For example. Co 2 can replace the Zn 2+ and the enzyme retains its
activity. The spectrum of carboxypeptidase A(Co") is “irregular" and has a high
absorptivity (extinction coefficient), indicating that a regular tetrahedron is not pre¬
sent. 74 The distortion presumably aids the metal to efi'ect the reaction. It has been
suggested that the metal in the enzyme is peculiarly poised for action and that this
lowers the energy of the transition state. The term entatic 75 has been coined to
describe this state of the metal in an enzyme.
The substitution of a different metal into an enzyme provides a very useful method
for studying the immediate environment of the metal site. In addition to the use of
Co - ^ for spectral studies, appropriate substitution allows the use of physical methods
such as electron paramagnetic resonance (Co 2 . Cu 2 "). the Mdssbuuer effect (Fe 2 ).
proton magnetic resonance relaxation techniques (Mn 2 ‘), or X-ray crystallography
(with a heavy metal atom to aid in the structure solution). 7f >
72 There is probably a loosely bound water molecule at this position when the enzyme is not engaged in
active catalysis.
73 Lipscomb. W. N. Chem. Soc. Rev. 1972. /. 319; Tetrahedron 1974 .JO. 1725; Arc. Chem. Res. 1982.
15. 232.
74 Consider the relation between absorptivity and symmetry. Chapter II. Vallec. B. I.,: Williams,
R. J. P. Proc. Natl. Acad. Sci. U. S. A. 1968, 59. 498. Ulmer, D. D.: Vallee. B. L. Bhmorfianic
Chemistry: Gould, R, F., Ed.; Advances in Chemistry 100; American Chemical Society: Wash¬
ington, DC. 1971; Chapter It).
73 Gr. evreivio. to stretch, strain, or bend.
76 Williams. R. J. P. Endeavour 1967. 26. 96. |
7584 | https://arxiv.org/pdf/1609.06172 | Published Time: Sat, 21 Jan 2023 06:33:53 GMT
OPTIMAL STRETCHING FOR LATTICE POINTS AND EIGENVALUES
RICHARD S. LAUGESEN AND SHIYA LIU
Abstract. We aim to maximize the number of first-quadrant lattice points in a convex domain with respect to reciprocal stretching in the coordinate directions. The optimal domain is shown to be asymptotically balanced, meaning that the stretch factor approaches 1 as the “radius” approaches infinity. In particular, the result implies that among all p-ellipses (or Lam´ e curves), the p-circle encloses the most first-quadrant lattice points as the radius approaches infinity, for 1 < p < ∞.The case p = 2 corresponds to minimization of high eigenvalues of the Dirichlet Laplacian on rectangles, and so our work generalizes a result of Antunes and Freitas. Similarly, we generalize a Neumann eigenvalue maximization result of van den Berg, Bucur and Gittins. Further, Ariturk and Laugesen recently handled 0 < p < 1 by building on our results here. The case p = 1 remains open, and is closely related to minimizing energy levels of harmonic oscillators: which right triangles in the first quadrant with two sides along the axes will enclose the most lattice points, as the area tends to infinity?
Introduction
Among ellipses of given area centered at the origin and symmetric about both axes, which one encloses the most integer lattice points in the open first quadrant? One might guess the optimal ellipse would be circular, but a non-circular ellipse can enclose more lattice points, as shown in Figure 1. Nonetheless, optimal ellipses must become more and more circular as the area increases to infinity, by a striking result of Antunes and Freitas . To formulate the problem more precisely, consider the number of positive-integer lattice points lying in the elliptical region
( xs−1
)2
+
(ys
)2
≤ r2,
where the ellipse has “radius” r > 0 and semiaxes proportional to s−1 and s. Notice that the area πr 2 of the ellipse is independent of the “stretch factor” s. Denote by
s = s(r) a value (not necessarily unique) of the stretch factor that maximizes the lattice point count. The theorem of Antunes and Freitas says s(r) → 1 as r → ∞ ,as illustrated in Figure 2. In other words, optimal ellipses become circular in the infinite limit. (Their theorem was stated differently, in terms of minimizing the n-th
Date : September 24, 2018. 2010 Mathematics Subject Classification. Primary 35P15. Secondary 11P21, 52C05.
Key words and phrases. Lattice points, planar convex domain, p-ellipse, Lam´ e curve, spectral optimization, Laplacian, Dirichlet eigenvalues, Neumann eigenvalues.
1
arXiv:1609.06172v2 [math.MG] 6 Sep 2017 2OPTIMAL STRETCHING FOR LATTICE POINTS
Figure 1. Circle s = 1 and ellipse s = 1 .15, for radius r = 4 .96. The ellipse encloses three more points than the circle, as shown in bold.
Figure 2. Optimal s-values for maximizing the number of lattice points in the 2-ellipse. The graph plots the largest value s(r) versus log r. The plotted r-values are multiples of √3/10, an irrational num-ber chosen in the hope of exhibiting generic behavior. The horizontal axis is at height s = 1. eigenvalue of the Dirichlet Laplacian on rectangles, with the square being asymptoti-cally minimal. Section 10 explains the connection.) The analogous result for optimal ellipsoids becoming asymptotically spherical was proved recently in three dimensions by van den Berg and Gittins and in higher dimensions by Gittins and Larson , once again in the eigenvalue formulation. This paper extends the result of Antunes and Freitas from circles to essentially arbitrary concave curves in the first quadrant that decrease between the intercept points (0 , 1) and (1 , 0). The “ellipses” in this situation are the images of the concave curve under rescaling by s−1 and s in the horizontal and vertical directions, respec-tively. Theorem 2 says the maximizing s(r) is bounded. Theorem 3 shows under a mild monotonicity hypothesis on the second derivative of the curve that s(r) → 1 as
r → ∞ . Thus the most “balanced” curve in the family will enclose the most lattice points in the limit. Marshall recently extended this result to higher dimensions by somewhat dif-ferent methods. We have generalized also to translated lattices in 2-dimensions . OPTIMAL STRETCHING FOR LATTICE POINTS 3
Theorem 4 allows the curvature to blow up or degenerate at the intercept points, which permits us to treat the family of p-ellipses for 1 < p < ∞. In each case the p-circle is asymptotically optimal for the lattice counting problem in the first quadrant. The case p = 1 is open problem. Our numerical evidence in Section 9 suggests that the first-quadrant right triangle enclosing the most lattice points does not necessarily approach a 45–45–90 degree triangle as r → ∞ . Instead one seems to get an infinite limit set of optimal triangles. See Conjecture 15, where we describe the recent proof by Marshall and Steinerberger . If one counts lattice points in the closed first quadrant, that is, counting points on the axes as well, then the results reverse direction from maximization to minimization of the lattice count. Theorem 7 shows that the value s = s(r) minimizing the number of enclosed lattice points will tend to 1 as r → ∞ . In the case of circles and ellipses, this result was obtained recently by van den Berg, Bucur and Gittins (and in higher dimensions by Gittins and Larson , generalized by Marshall ). As explained in Section 10, they showed that the maximizing rectangle for the n-th eigenvalue of the Neumann Laplacian must approach a square as n → ∞ .This paper builds on the framework of Antunes and Freitas for ellipses, with new ingredients introduced to handle general concave curves. First we develop a new non-sharp bound on the counting function (Proposition 10) in order to control the stretch factor s(r). Then we prove more precise lattice counting estimates (Proposition 11) of Kr¨ atzel type, relying on a theorem of van der Corput (Appendix A). Convex decreasing curves in the first quadrant, such as p-ellipses with 0 < p < 1, have been treated by Ariturk and Laugesen by building on this paper’s results.
Spectral motivations and results. This paper is inspired by recent efforts to understand the behavior of high eigenvalues of the Laplacian. Write λn for the n-th eigenvalue of the Dirichlet Laplacian on a bounded domain Ω of area 1 in the plane. (We restrict to 2 dimensions for simplicity.) Denote the minimum value of this eigenvalue over all such domains by λ∗
n
, and suppose it is achieved on a domain Ω ∗
n
.What can one say about the shape of this minimizing domain? For the first eigenvalue, the minimizing domain Ω ∗
1
is a disk, by the Faber–Krahn inequality. For the second eigenvalue, Ω ∗
2
is a union of two disjoint equal-area disks, as Krahn and later P. Szego showed. A long-standing conjecture says Ω ∗
3
should be a disk and Ω ∗
4
should be a union of disjoint non-equal-area disks. For higher eigenvalues (n ≥ 5), minimizing domains found numerically do not have recognizable shapes; see [1, 23] and references therein. Antunes and Freitas remark, though, that the “most natural guess” is Ω ∗
n
approaches a disk as n → ∞ , which is known to occur if the area normalization is strengthened to a perimeter normalization [3, 7]. This conjecture would imply the famous P´ olya conjecture λn ≥ 4πn/ |Ω|, as Colbois and El Soufi [8, Corollary 2.2] showed using subadditivity of n 7 → λ∗
n
.A partial result of Freitas succeeds in determining the leading order asymptotic as n → ∞ of the minimum value of the eigenvalue sum λ1 +· · · +λn (rather than of λn
itself). This result provides no information on the shapes of the minimizing domains. Larson shows among convex domains that the disk asymptotically maximizes 4 OPTIMAL STRETCHING FOR LATTICE POINTS
Figure 3. A concave decreasing curve Γ in the first quadrant, with intercepts at L. The point ( α, β ) on the curve is relevant to Theorem 3. the Riesz means of the Laplace eigenvalues, for Riesz exponents ≥ 3/2. If this Riesz exponent could be lowered to 0, giving asymptotic maximality of the disk for the eigenvalue counting function, then one would obtain the desired conjecture about the eigenvalue minimizing domain Ω ∗
n
.A complete resolution for rectangular domains was found by Antunes and Freitas , using lattice counting methods as explained in Section 10. They proved that the minimizing domain for λn among rectangles approaches a square as n → ∞ .Similarly, the cube is asymptotically minimal in 3 and higher dimensions [6, 11].
Open problem for the harmonic oscillator. Asymptotic optimality of the square for minimizing Dirichlet eigenvalues of the Laplacian on rectangles suggests an anal-ogous open problem for harmonic oscillators. Consider the Schr¨ odinger operator in 2 dimensions with parabolic potential ( sx )2 + ( y/s )2, where s > 0 is a parameter. Write s(n) for a parameter value that minimizes the n-th eigenvalue of this operator. What is the limiting behavior of s(n) as n → ∞ ?The results on rectangular domains (which may be regarded as infinite potential wells) might suggest s(n) → 1, but we think that it is not the case. Instead we believe
s(n) might cluster around infinitely many values as n → ∞ . Indeed, after rescaling, the Schr¨ odinger operator has eigenvalues of the form s(j − 1/2) + ( k − 1/2) /s , which leads to a lattice point counting problem inside right triangles, like in Section 9 for
p = 1, except now the lattices are shifted by 1 /2 to the left and downwards. For the unshifted lattice, numerical work in Section 9 suggests that the optimal stretching parameter s does not converge to 1, and instead has many cluster points as r → ∞ .Recent investigations [19, Section 10] suggest that this clustering phenomenon persists for shifted lattices and hence for the harmonic oscillator. 2. Assumptions and definitions
The first quadrant is the open set {(x, y ) : x, y > 0}.Assume throughout the paper that Γ is a concave, strictly decreasing curve in the first quadrant. Our theorems assume the x- and y-intercepts of the curve are equal, occurring at x = L and y = L respectively, as shown in Figure 3. Write Area(Γ) for the area enclosed by the curve Γ and the x- and y-axes. OPTIMAL STRETCHING FOR LATTICE POINTS 5
We represent the curve Γ by y = f (x) for x ∈ [0 , L ], so that f is a concave strictly decreasing function, and of course f is continuous. In particular
L = f (0) > f (x) > f (L) = 0 whenever x ∈ (0 , L ). Denote the inverse function of f (x) by g(y) for y ∈ [0 , L ], so that g also is concave and strictly decreasing. We define a rescaling of the curve by parameter r > 0:
rΓ = image of Γ under the radial scaling ( r 00 r )= graph of rf (x/r ), and define an area-preserving stretch of the curve by: Γ( s) = image of Γ under the diagonal scaling ( s−1 00 s
)
= graph of sf (sx ), where s > 0. In other words, Γ( s) is obtained from Γ after compressing the x-direction by s and stretching the y-direction by s. Define the counting function for rΓ( s) by
N (r, s ) = number of positive-integer lattice points lying inside or on rΓ( s)= # {(j, k ) ∈ N × N : k ≤ rsf (js/r )}.
For each r > 0, consider the set
S(r) = argmax
s> 0
N (r, s )consisting of the s-values that maximize the number of first-quadrant lattice points enclosed by the curve rΓ( s). The set S(r) is well-defined because the maximum is indeed attained, as the following argument shows. The curve rΓ( s) has x-intercept at rs −1L, which is less than 1 if s > rL and so in that case the curve encloses no positive-integer lattice points. Similarly if s < (rL )−1, then rΓ( s) has height less than 1 and contains no lattice points in the first quadrant. Thus for each fixed r > 0, if s is sufficiently small or sufficiently large then the counting function N (r, s ) equals zero, while obviously for intermediate values of s the integer-valued function s 7 → N (r, s )is bounded. Hence N (r, s ) attains its maximum at some s > 0. For later reference, we write down this bound on optimal s-values.
Lemma 1 (r-dependent bound on optimal stretch factors) . If Γ is a concave, strictly decreasing curve in the first quadrant with equal intercepts (as in Figure 3), then
S(r) ⊂ [(rL )−1, rL ] whenever r ≥ 2/L .Proof. The curve rΓ(1) has horizontal and vertical intercepts at rL ≥ 2. Hence by concavity, rΓ(1) encloses the point (1 , 1), and so the counting function s 7 → N (r, s )is greater than zero when s = 1. On the other hand when s < (rL )−1 or s > rL , we know N (r, s ) = 0 by the paragraph before the lemma. Thus the maximum can only be attained when s lies in the interval [(rL )−1, rL ]. 6 OPTIMAL STRETCHING FOR LATTICE POINTS
Main results
The curve Γ has x- and y-intercepts at L, in the theorems in that follow; see Figure 3. We start by improving Lemma 1 to show the maximizing set S(r) is bounded, and the bounds can be evaluated explicitly in the limit as r → ∞ .
Theorem 2 (Uniform bound on optimal stretch factors) . If Γ is a concave, strictly decreasing curve in the first quadrant then
S(r) ⊂ [s1, s 2] for all r ≥ 2/L ,for some constants s1, s 2 > 0. Furthermore, given ε > 0,
S(r) ⊂ [ 14 + ε, 4 + ε] for all large r.
The proof appears in Section 4. If the concave decreasing curve is smooth with monotonic second derivative, then in addition to being bounded above and below the maximizing set S(r) converges to
{1}, as the next theorem shows. Recall that g is the inverse function of f .
Theorem 3 (Optimal concave curve is asymptotically balanced) . Assume (α, β ) ∈ Γ
is a point in the first quadrant such that f ∈ C2[0 , α ] with f ′ < 0 on (0 , α ] and f ′′ < 0
on [0 , α ], and similarly g ∈ C2[0 , β ] with g′ < 0 on (0 , β ] and g′′ < 0 on [0 , β ]. Further suppose f ′′ is monotonic on [0 , α ] and g′′ is monotonic on [0 , β ].Then the optimal stretch factor for maximizing N (r, s ) approaches 1 as r tends to infinity, with
S(r) ⊂ [1 − O(r−1/6), 1 + O(r−1/6)],
and the maximal lattice count has asymptotic formula
max
s> 0
N (r, s ) = r2 Area(Γ) − rL + O(r2/3).
The theorem is proved in Section 7. Slight improvements to the decay rate O(r−1/6)and the error term O(r2/3) are possible, as explained after Proposition 11.
More general curves for lattice counting. We want to weaken the smoothness and monotonicity assumptions in Theorem 3. We start with a definition of piecewise smoothness.
Definition (P C 2).
(i) We say a function f is piecewise C2-smooth on a half-open interval (0 , α ] if f is continuous and a partition 0 = α0 < α 1 < · · · < α l = α exists such that f ∈ C2(0 , α 1]and f ∈ C2[αi−1, α i] for i = 2 , . . . , l . Write P C 2(0 , α ] for the class of such functions. (ii) Write f ′ < 0 to mean that f ′ is negative on the subintervals (0 , α 1] and [ αi−1, α i]for i = 2 , . . . , l , with the derivative being taken in the one-sided senses at the partition points α1, . . . , α l. The meaning of f ′′ < 0 is analogous. (iii) We label partition points using the same letter as for the right endpoint. In particular, the partition for g ∈ P C 2(0 , β ] is 0 = β0 < · · · < β ` = β.OPTIMAL STRETCHING FOR LATTICE POINTS 7
For the next theorem, take a point ( α, β ) ∈ Γ lying in the first quadrant and suppose we have numbers a1, a 2, b 1, b 2 > 0 and positive valued functions δ(r) and (r)such that as r → ∞ :
δ(r) = O(r−2a1 ), f ′′ (δ(r))−1 = O(r1−4a2 ), (1)
(r) = O(r−2b1 ), g′′ ((r))−1 = O(r1−4b2 ). (2) (The second condition in (1) says that f ′′ (x) cannot be too small as x → 0.) Let
e = min {16 , a 1, a 2, b 1, b 2}.
Now we extend Theorem 3 to a larger class of concave decreasing curves.
Theorem 4 (Optimal concave curve is asymptotically balanced) .
Assume f ∈ P C 2(0 , α ] with f ′ < 0 and f ′′ < 0, and f ′′ is monotonic on each subinterval of the partition. Similarly assume g ∈ P C 2(0 , β ] with g′ < 0 and g′′ <
0, and g′′ is monotonic on each subinterval of the partition. Suppose the positive functions δ(r) and (r) satisfy conditions (1) and (2). Then the optimal stretch factor for maximizing N (r, s ) approaches 1 as r tends to infinity, with
S(r) ⊂ [1 − O(r−e), 1 + O(r−e)],
and the maximal lattice count has asymptotic formula
max
s> 0
N (r, s ) = r2 Area(Γ) − rL + O(r1−2e).
The proof is presented in Section 7.
Example 5 (Optimal p-ellipses for lattice point counting) . Fix 1 < p < ∞, and consider the p-circle Γ : |x|p + |y|p = 1 ,
which has intercept L = 1. That is, the p-circle is the unit circle for the `p-norm on the plane. Then the p-ellipse
rΓ( s) : |sx |p + |s−1y|p ≤ rp
has first-quadrant counting function
N (r, s ) = # {(j, k ) ∈ N × N : ( js )p + ( ks −1)p ≤ rp}.
We will show that the p-ellipse containing the maximum number of positive-integer lattice points must approach a p-circle in the limit as r → ∞ , with
S(r) ⊂ [1 − O(r−e), 1 + O(r−e)] where e = min {16 , 12p }.Theorem 3 fails to apply to p-ellipses when 1 < p < 2, because the second derivative of the curve is not monotonic (see f ′′ (x) below), and the theorem fails to apply when 2 < p < ∞ because f ′′ (0) = 0 in that case. Instead we will apply Theorem 4. To verify that the p-circle satisfies the hypotheses of Theorem 4, we let α = β =2−1/p and choose
δ(r) = r−1/p , (r) = r−1/p ,8 OPTIMAL STRETCHING FOR LATTICE POINTS
for all large r. Then δ(r) = r−2a1 with a1 = 1 /2p. Next,
f (x) = (1 − xp)1/p ,f ′(x) = −xp−1(1 − xp)−1+1 /p ,f ′′ (x) = −(p − 1) xp−2(1 − xp)−2+1 /p ,
so that ∣∣f ′′ (δ(r))∣ ∣−1 ≤ (const.) r1−2/p ,
and hence a2 = 1 /2p in (1). Thus f satisfies hypothesis (1). Further, the interval (0 , α ) can be partitioned into subintervals on which f ′′ is monotonic, because the third derivative
f ′′′ (x) = −(p − 1) xp−3(1 − xp)−3+1 /p ((1 + p)xp + p − 2)
vanishes at most once in the unit interval. The calculations are the same for g, and so the desired conclusion for p-ellipses follows from Theorem 4 when 1 < p < ∞.For p = ∞, the ∞-circle is a Euclidean square and the ∞-ellipse is a rectangle. Many different rectangles of given area can contain the same number of lattice points. For example, a 4 × 1 rectangle and 2 × 2 square each contain 4 lattice points in the first quadrant. All such matters can be handled by the explicit formula N (r, s ) =
brs −1cb rs c for the counting function when p = ∞.The case p = 1 is an open problem, as discussed in Section 9. The case 0 < p < 1has been handled by Ariturk and Laugesen using results in this paper. Incidentally, an explicit estimate on the number of lattice points in the full p-ellipse in all four quadrants was obtained by Kr¨ atzel [16, Theorem 2] for p ≥ 2. See the informative survey by Ivi´ c et al. [14, §3.1].
Lattice points in the closed first quadrant, and Neumann eigenvalues. Our results have analogues for lattice point counting in the closed (rather than open) first quadrant, as we now explain. When counting nonnegative-integer lattice points, which means we include lattice points on the axes, the counting function for rΓ( s) is
N (r, s ) = # {(j, k ) ∈ Z+ × Z+ : k ≤ rsf (js/r )},
where Z+ = {0, 1, 2, 3, . . . }. Define
S(r) = argmin
s> 0
N (r, s ).
In other words, the set S(r) consists of the s-values that minimize the number of lattice points inside the curve rΓ( s) in the closed first quadrant. Notice we employ the calligraphic letters N and S when working with nonnegative-integer lattice points.
Theorem 6 (Uniform bound on optimal stretch factors) . If Γ is a concave, strictly decreasing curve in the first quadrant then
S(r) ⊂ [s1, s 2] for all r ≥ 2/L ,for some constants s1, s 2 > 0.OPTIMAL STRETCHING FOR LATTICE POINTS 9
Theorem 7 (Optimal concave curve is asymptotically balanced) . Under the assump-tions of Theorem 3, the optimal stretch factor for minimizing N (r, s ) approaches 1
as r tends to infinity:
S(r) ⊂ [1 − O(r−1/6), 1 + O(r−1/6)] ,
min
s> 0
N (r, s ) = r2 Area(Γ) + rL + O(r2/3),
and under the assumptions of Theorem 4 we have similarly that:
S(r) ⊂ [1 − O(r−e), 1 + O(r−e)] ,
min
s> 0
N (r, s ) = r2 Area(Γ) + rL + O(r1−2e).
The proofs are in Section 8. Consequently, we reprove a recent theorem of van den Berg, Bucur and Gittins saying that the optimal rectangle of area 1 for maximizing the n-th Neumann eigenvalue of the Laplacian approaches a square as n → ∞ . See Section 10 for discussion, and an explanation of why the approach in this paper is simpler. 4. Proof of Theorem 2
To control the stretch factors and hence prove Theorem 2, we will first derive a rough lower bound on the counting function, and then a more sophisticated upper bound. The leading order term in these bounds is simply the area inside the rescaled curve and thus is best possible, while the second term scales like the length of the curve and so at least has the correct order of magnitude. Assume Γ is concave and decreasing in the first quadrant, with x- and y-intercepts at L and M respectively. The intercepts need not be equal, in the lemmas and proposition below. Recall that N (r, s ) counts the positive-integer lattice points under the curve Γ, while N (r, s ) counts nonnegative-integer lattice points.
Lemma 8 (Relation between counting functions) . For each r, s > 0,
N (r, s ) = N (r, s ) + r(s−1L + sM ) + ρ(r, s )
for some number ρ(r, s ) ∈ [−1, 1] .Proof. The difference between the two counting functions is simply the number of lattice points lying on the coordinate axes inside the intercepts of rΓ( s). There are
brs −1Lc + brsM c + 1 such lattice points, and so the lemma follows immediately.
Lemma 9 (Rough lower bound) . The number N (r, s ) of positive-integer lattice points lying inside rΓ( s) in the first quadrant satisfies
N (r, s ) ≥ r2 Area(Γ) − r(s−1L + sM ) − 1, r, s > 0.10 OPTIMAL STRETCHING FOR LATTICE POINTS
Figure 4. Positive integer lattice count satisfies N ≤ Area(Γ) −
Area(triangles), in proof of Proposition 10(a).
Proof. Notice N (r, s ) equals the total area of the squares of sidelength 1 having lower left vertices at nonnegative-integer lattice points inside the curve rΓ( s). The union of these squares contains rΓ( s), since the curve is decreasing. Hence N (r, s ) ≥
r2 Area(Γ), and so
N (r, s ) ≥ N (r, s ) − r(s−1L + sM ) − 1 by Lemma 8
≥ r2 Area(Γ) − r(s−1L + sM ) − 1.
For the upper bound in the next proposition, remember Γ is the graph of y = f (x), where f is concave and decreasing on [0 , L ], with f (0) = M, f (L) = 0. We do not assume f is differentiable in the next result, although in order to guarantee the constant C in the proposition is positive, we assume f is strictly decreasing.
Proposition 10 (Two-term upper bound on counting function) . Let C = M −
f (L/ 2) .(a) The number N of positive-integer lattice points lying inside Γ in the first quadrant satisfies
N ≤ Area(Γ) − 12C (3)
provided L ≥ 1.(b) The number N (r, s ) of positive-integer lattice points lying inside rΓ( s) in the first quadrant satisfies
N (r, s ) ≤ r2 Area(Γ) − 12Crs
whenever r ≥ s/L .Proof. Part (a). Clearly N equals the total area of the squares of sidelength 1 having upper right vertices at positive-integer lattice points inside the curve Γ. Consider also the right triangles of width 1 formed by secant lines on Γ (see Figure 4), that is, the triangles with vertices (i − 1, f (i − 1) ), (i, f (i)), (i − 1, f (i)), where i = 1 , . . . , bLc.OPTIMAL STRETCHING FOR LATTICE POINTS 11
These triangles lie above the squares by construction, and lie below Γ by concavity. Hence
N + Area(triangles) ≤ Area(Γ) . (4) Since f is decreasing, we find Area(triangles) =
bLc
∑
i=1
12
(f (i − 1) − f (i))
= 12
(f (0) − f (bLc)) (5)
≥ 12
(M − f (L/ 2) ) = 12C, (6) because bLc ≥ L/ 2 when L ≥ 1. Combining (4) and (6) proves Part (a). Part (b). Simply replace Γ in Part (a) with the curve rΓ( s), meaning we replace
L, M, f (x) with rs −1L, rsM, rsf (sx/r ) respectively.
Proof of Theorem 2. Recall the intercepts are assumed equal ( L = M ) in this theorem. Let r ≥ 2/L and suppose s ∈ S(r). Then r ≥ s/L by Lemma 1, and so the upper bound in Proposition 10(b) gives
N (r, s ) ≤ r2 Area(Γ) − 12Crs.
The lower bound in Lemma 9 with “ s = 1” says
N (r, 1) ≥ r2 Area(Γ) − 2rL − 1. (7) The value s ∈ S(r) is a maximizing value, and so N (r, 1) ≤ N (r, s ). The preceding inequalities therefore imply 12Crs ≤ 2rL + 1 ≤ 52rL.
Hence s ≤ 5L/C ≡ s2, and so the set S(r) is bounded above. Interchanging the roles of the horizontal and vertical axes, we similarly find s−1 ≤
5L/ ˜C ≡ s−11 , so that the set S(r) is bounded below away from 0, completing the first part of the proof. The fact that S(r) is bounded will help imply an improved bound in the limit as
r → ∞ . Going back to the proof of Proposition 10(a), we see from (4) and (5) that
N + 12
(f (0) − f (bLc)) ≤ Area(Γ) .
Rescaling the curve from Γ to rΓ( s), so that N and f (x) become N (r, s ) and rsf ( sr x),respectively, and the x-intercept L becomes rL/s , we see the last inequality becomes
N (r, s ) + 12rs (f (0) − f ( sr brL s c)) ≤ r2 Area(Γ) .
Hence
N (r, s ) ≤ r2 Area(Γ) − 12rsL + o(r) as r → ∞ ,12 OPTIMAL STRETCHING FOR LATTICE POINTS
where to get the error term o(r) we used that s ∈ S(r) is bounded above and below (s1 ≤ s ≤ s2) and f (L) = 0. Since s is a maxmizing value we have N (r, 1) ≤ N (r, s ), and so (7) and the above inequality imply 12rsL + o(r) ≤ 2rL + 1 ,
which implies lim sup r→∞ s ≤ 4. Similarly lim sup r→∞ s−1 ≤ 4, by interchanging the axes. 5. Two-term counting estimates with explicit remainder
We start with a result for C2-smooth curves. What matters in the following propo-sition is that the right side of estimate (8) below has the form O(rθ) for some θ < 1, and that the s-dependence in the estimate can be seen explicitly. The detailed de-pendence on the functions f and g will not be important for our purposes. The horizontal and vertical intercepts L and M need not be equal, in this section.
Proposition 11 (Two-term counting estimate) . Take a point (α, β ) ∈ Γ lying in the first quadrant, and assume that f ∈ C2[0 , α ] with f ′ < 0 on (0 , α ] and f ′′ < 0 on
[0 , α ], and similarly g ∈ C2[0 , β ] with g′ < 0 on (0 , β ] and g′′ < 0 on [0 , β ]. Further suppose f ′′ is monotonic on [0 , α ] and g′′ is monotonic on [0 , β ].(a) The number N of positive-integer lattice points inside Γ in the first quadrant satisfies:
∣∣N − Area(Γ) + ( L + M )/2∣∣
≤ 6
( ∫ α
0
|f ′′ (x)|1/3 dx +
∫ β
0
|g′′ (y)|1/3 dy
)
175 ( max
[0 ,α ]
1
|f ′′ |1/2 + max
[0 ,β ]
1
|g′′ |1/2
)
14
(|f ′(α)| + |g′(β)|) + 3 .
(b) The number N (r, s ) of positive-integer lattice points lying inside rΓ( s) in the first quadrant satisfies (for r, s > 0):
∣∣N (r, s ) − r2 Area(Γ) + r(s−1L + sM )/2∣∣
≤ 6r2/3
( ∫ α
0
|f ′′ (x)|1/3 dx +
∫ β
0
|g′′ (y)|1/3 dy
)
175 r1/2( max
[0 ,α ]
s−3/2
|f ′′ |1/2 + max
[0 ,β ]
s3/2
|g′′ |1/2
)
14(s2|f ′(α)| + s−2|g′(β)|) + 3 . (8) Proposition 11 and its proof are closely related to work of Kr¨ atzel [16, Theorem 1]. We give a direct proof below for two reasons: we want the estimate (8) that depends explicitly on the stretching parameter s, and we want a proof that can be modified to use a weaker monotonicity hypothesis, in Proposition 12. OPTIMAL STRETCHING FOR LATTICE POINTS 13
A better bound on the right side of (8), giving order O(rθ+) with θ = 131 /208 '
0.63 < 2/3, can be found in work of Huxley , with precursors in [12, Theo-rems 18.3.2 and 18.3.3]. That bound is difficult to prove, though, and the improve-ment is not important for our purposes since it leads to only a slight improvement in the rate of convergence for S(r), namely from O(r−1/6) to O(r(θ+−1) /2) in Theorem 3.
Proof. Part (a). We divide the region under Γ into three parts. Let N1 count the lattice points lying to the left of the line x = α and above y = β, and N2 count the lattice points to the right of x = α and below y = β, and N3 count the lattice points in the remaining rectangle (0 , α ] × (0 , β ]. That is,
N1 = ∑
0<m ≤α
∑
β<n ≤f(m)
1 = ∑
0<m ≤α
(bf (m)c − b βc),N2 = ∑
0<n ≤β
∑
α<m ≤g(n)
1 = ∑
0<n ≤β
(bg(n)c − b αc),N3 = bαcb βc.
In terms of the sawtooth function ψ, defined by
ψ(x) = x − b xc − 1/2,
one can evaluate
N1 = ∑
0<m ≤α
(f (m) − ψ(f (m)) − 1/2 − b βc).
Then we apply the Euler–Maclaurin summation formula
∑
0<m ≤α
f (m) =
∫ α
0
f (x) d x − ψ(α)f (α) + ψ(0) f (0) +
∫ α
0
f ′(x)ψ(x) d t
(which we observe for later reference holds whenever f is piecewise C1-smooth) to deduce that
N1 =
∫ α
0
f (x) d x − ψ(α)f (α) + ψ(0) f (0) +
∫ α
0
f ′(x)ψ(x) d x
− ∑
0<m ≤α
ψ(f (m)) − b αc(1 /2 + bβc)=
∫ α
0
f (x) d x − ψ(α)β − M/ 2 +
∫ α
0
f ′(x)ψ(x) d x
− ∑
0<m ≤α
ψ(f (m)) − b αc(1 /2 + bβc).
Similarly
N2 =
∫ β
0
g(y) d y − ψ(β)α − L/ 2 +
∫ β
0
g′(y)ψ(y) d y
− ∑
0<n ≤β
ψ(g(n)) − b βc(1 /2 + bαc),14 OPTIMAL STRETCHING FOR LATTICE POINTS
and so
N = N1 + N2 + N3
=
∫ α
0
f (x) d x +
∫ β
0
g(y) d y − b αcb βc − (L + M )/2
− ψ(α)β − b αc/2 − ψ(β)α − b βc/2+
∫ α
0
f ′(x)ψ(x) d x +
∫ β
0
g′(y)ψ(y) d y
− ∑
0<m ≤α
ψ(f (m)) − ∑
0<n ≤β
ψ(g(n))
= Area(Γ) − (L + M )/2 +
∫ α
0
f ′(x)ψ(x) d x +
∫ β
0
g′(y)ψ(y) d y
− ∑
0<m ≤α
ψ(f (m)) − ∑
0<n ≤β
ψ(g(n)) + remainder (9) where remainder = −(α − b αc)( β − b βc) + ( α − b αc + β − b βc)/2. (10) This remainder lies between 0 and 1, since 0 ≤ − xy + ( x + y)/2 ≤ 1 when x, y ∈ [0 , 1]. We estimate the sum of sawtooth functions in (9) by using Theorem 18 (which is due to van der Corput): since f ′′ is monotonic and nonzero on [0 , α ], the thoerem implies
∣∣∣ ∑
0<m ≤α
ψ(f (m))∣∣∣ ≤ 6
∫ α
0
|f ′′ (x)|1/3 dx + 175 max
[0 ,α ]
1
|f ′′ |1/2 + 1 (11) and similarly
∣∣∣ ∑
0<n ≤β
ψ(g(n))∣∣∣ ≤ 6
∫ β
0
|g′′ (y)|1/3 dy + 175 max
[0 ,β ]
1
|g′′ |1/2 + 1 . (12) To estimate the integrals of f ′ψ and g′ψ in (9), we introduce the antiderivative of the sawtooth function, Ψ( t) = ∫ t
0
ψ(z) d z, and observe that −1/8 ≤ Ψ( t) ≤ 0 for all
t ∈ R. By integration by parts and the fact that f ′′ < 0, we have
∣∣∫ α
0
f ′(x)ψ(x) d x∣∣ =
∣∣∣[f ′(x)Ψ( x)]x=αx=0 −
∫ α
0
f ′′ (x)Ψ( x) d x
∣∣∣
≤ 18|f ′(α)| + 18
∣∣∫ α
0
f ′′ (x) d x∣∣
= 18|f ′(α)| + 18
(f ′(0) − f ′(α))
≤ 14|f ′(α)| (13) OPTIMAL STRETCHING FOR LATTICE POINTS 15
since f ′(α) ≤ f ′(0) ≤ 0. The same argument gives
∣∣∫ β
0
g′(y)ψ(y) d y∣∣ ≤ 14|g′(β)|. (14) Combining (9)–(14) completes the proof of Part (a). Part (b). Simply apply Part (a) to the curve rΓ( s) by replacing L, M, f (x), g (y), α, β
with rs −1L, rsM, rsf (sx/r ), rs −1g(s−1y/r ), rs −1α, rsβ respectively.
Remark. Proposition 11 continues to hold if the point ( α, β ) = ( L, 0) lies at the right endpoint of the curve. One simply removes all mention of g, β j and from the hypotheses of the proposition, and removes all such terms from the conclusions, as can be justified by inspecting the proof above. The same remark holds for the advanced counting estimate in the following Proposition 12.
Advanced counting estimate. The hypotheses in the last result are somewhat restrictive. In particular, we would like to handle infinite curvature at the intercepts of the curve Γ, meaning f ′′ must be allowed to blow up at x = 0. Further, we would like to relax the monotonicity assumption on f ′′ . The next result achieves these goals. Two numbers δ and appear in the next Proposition. Their role in the proof is that on the intervals 0 < x ≤ δ and 0 < y ≤ we bound the sawtooth function trivially with |ψ| ≤ 1/2. On the remaining intervals we seek cancellations.
Proposition 12 (Two-term counting estimate for more general curve) . Take a point
(α, β ) ∈ Γ lying in the first quadrant, and assume f ∈ P C 2(0 , α ] with f ′ < 0 and
f ′′ < 0, and that f ′′ is monotonic on (αi−1, α i] for i = 1 , . . . , l . Similarly assume
g ∈ P C 2(0 , β ] with g′ < 0 and g′′ < 0, and that g′′ is monotonic on (βj−1, β j ] for
j = 1 , . . . , ` .(a) If δ ∈ (0 , α ) and ∈ (0 , β ) then the number N of positive-integer lattice points inside Γ in the first quadrant satisfies:
∣∣N − Area(Γ) + ( L + M )/2∣∣
≤ 6
( ∫ α
0
|f ′′ (x)|1/3 dx +
∫ β
0
|g′′ (y)|1/3 dy
)
175
( 1
|f ′′ (δ)|1/2 + 1
|g′′ ()|1/2
)
350
( l∑
i=1
1
|f ′′ (αi)|1/2 +
`
∑
j=1
1
|g′′ (βj )|1/2
)
14
( l∑
i=1
|f ′(αi)| +
`
∑
j=1
|g′(βj )|) + 12
(δ + ) + l + ` + 1 .
(b) If functions
δ : (0 , ∞) → (0 , α ), : (0 , ∞) → (0 , β ),16 OPTIMAL STRETCHING FOR LATTICE POINTS
are given, then the number N (r, s ) of positive-integer lattice points inside rΓ( s) in the first quadrant satisfies (for r, s > 0):
∣∣N (r, s ) − r2 Area(Γ) + r(s−1L + sM )/2∣∣
≤ 6r2/3
( ∫ α
0
|f ′′ (x)|1/3 dx +
∫ β
0
|g′′ (y)|1/3 dy
)
175 r1/2
( s−3/2
|f ′′ (δ(r))|1/2 + s3/2
|g′′ ((r))|1/2
)
350 r1/2
( l∑
i=1
s−3/2
|f ′′ (αi)|1/2 +
`
∑
j=1
s3/2
|g′′ (βj )|1/2
)
14
( l∑
i=1
s2|f ′(αi)| +
`
∑
j=1
s−2|g′(βj )|) + r
2
(s−1δ(r) + s (r)) + l + ` + 1 . (15) The integral of |f ′′ |1/3 appearing in the conclusion of Proposition 12 is finite, be-cause by H¨ older’s inequality and the fact that f ′′ < 0 and f is decreasing, we have
∫ α1
0
|f ′′ (x)|1/3 dx ≤ α2/31
∣∣∣∫ α1
0
f ′′ (x) d x
∣∣∣1/3
= α2/31 |f ′(0 +) − f ′(α−
1
)|1/3 < ∞.
The integral of |g′′ |1/3 is finite for similar reasons.
Proof. Part (a). The lattice point counting equation (9) holds just as in the proof of Proposition 11, and so the task is to estimate each of the terms on the right side of that equation. Estimate (11) on the sum of the sawtooth function is no longer valid, because f ′′
is no longer assumed to be monotonic on the whole interval [0 , α ]. To control this sawtooth sum, we first observe
∣∣ ∑
0<m ≤δ
ψ(f (m))∣ ∣ ≤ 12δ
since |ψ| ≤ 1/2 everywhere. Next, we have δ ∈ (αj−1, α j ] for some j ∈ { 1, . . . , l }, and
∣∣ ∑
δ<m ≤αj
ψ(f (m))∣ ∣ ≤ 6
∫ αj
δ
|f ′′ (x)|1/3 dx + 175 max
{ 1
|f ′′ (δ)|1/2 , 1
|f ′′ (αj )|1/2
}
1 by Theorem 18 applied on the interval [ δ, α j ]. Applying that theorem again on each interval [ αi−1, α i] with i = j + 1 , . . . , l gives that
∣∣ ∑
αi−1<m ≤αi
ψ(f (m))∣ ∣ ≤ 6
∫ αi
αi−1
|f ′′ (x)|1/3 dx+175 max
{ 1
|f ′′ (αi−1)|1/2 , 1
|f ′′ (αi)|1/2
}
+1 .OPTIMAL STRETCHING FOR LATTICE POINTS 17
By summing the last three displayed inequalities, we deduce a sawtooth bound
∣∣∣ ∑
0<m ≤α
ψ(f (m))∣∣∣
≤ 12δ + 6
∫ αδ
|f ′′ (x)|1/3 dx + 175
|f ′′ (δ)|1/2 +
l−1
∑
i=j
350
|f ′′ (αi)|1/2 + 175
|f ′′ (α)|1/2 + l − j + 1
≤ 12δ + 6
∫ α
0
|f ′′ (x)|1/3 dx + 175
|f ′′ (δ)|1/2 +
l
∑
i=1
350
|f ′′ (αi)|1/2 + l. (16) Next, we adapt estimate (13) on the integral of f ′ψ by simply applying the same argument on each interval [ αi−1, α i], hence finding
∣∣∣∫ α
0
f ′(x)ψ(x) d x
∣∣∣ ≤
l
∑
i=1
[18|f ′(αi)| + 18
(f ′(αi−1) − f ′(αi))]
≤ 14
l
∑
i=1
|f ′(αi)|. (17) By combining (9), (10) with (16), (17) and the analogous estimates on g, we complete the proof of Part (a). Part (b). Apply Part (a) to the curve rΓ( s) by replacing L, M, f (x), g (y), α, β, δ,
with rs −1L, rsM, rsf (sx/r ), rs −1g(s−1y/r ), rs −1α, rsβ, rs −1δ(r), rs (r) respectively.
A unified approach
The next proposition provides a unified framework for proving our theorems later in the paper. It adapts the scheme of proof employed by Antunes and Freitas .
Proposition 13. Let A ∈ R, L > 0, and 0 < θ < 1. Consider a real valued function
H(r, s ) (for r, s > 0) such that for each closed interval [s1, s 2] ⊂ (0 , ∞) one has
H(r, s ) = Ar 2 − Lr (s + s−1)/2 + O(rθ), (18)
with s ∈ [s1, s 2] allowed to vary as r → ∞ . Assume the function s 7 → H(r, s ) attains its maximum value, for each r > 0, and write S(r) = argmax s> 0 H(r, s ) for the set of maximizing points. Suppose
S(r) ⊂ [s1, s 2] for all large r > 0, (19)
for some constants s1, s 2 > 0.Then the maximizing set S(r) converges to the point {1} as r → ∞ , with
S(r) ⊂ [1 − O(r−(1 −θ)/2), 1 + O(r−(1 −θ)/2)],
and the maximum value of H has asymptotic formula
max
s> 0
H(r, s ) = Ar 2 − Lr + O(rθ).18 OPTIMAL STRETCHING FOR LATTICE POINTS
The error term O(rθ) in (18) has implied constant depending on the interval [ s1, s 2].
Proof. Since S(r) ⊂ [s1, s 2] by hypothesis (19), the asymptotic estimate (18) implies
H(r, s ) = Ar 2 − Lr (s + s−1)/2 + O(rθ),H(r, 1) = Ar 2 − Lr + O(rθ),
for s ∈ S(r) and r → ∞ . Since s is a maximizing value, we have H(r, 1) ≤ H(r, s )and so
s + s−1 ≤ 2 + O(r−(1 −θ)). (20) Hence s = 1 + O(r−(1 −θ)/2) by Lemma 19, which proves the first claim in the theorem. For the second claim, when s ∈ S(r) we have H(r, s ) = Ar 2 − Lr + O(rθ) as r → ∞ ,by (18) and using also that 1 ≤ (s + s−1)/2 ≤ 1 + O(r−(1 −θ)) by (20).
Proof of Theorem 3 and Theorem 4 Proof of Theorem 3. The theorem follows directly from Proposition 13 with H(r, s )being the lattice counting function N (r, s ). The hypotheses of the proposition are verified as follows. Suppose 0 < s 1 < s 2 < ∞. By Proposition 11(b) with L = M one has
N (r, s ) = Area(Γ) r2 − Lr (s + s−1)/2 + O(r2/3), (21) with s ∈ [s1, s 2] as r → ∞ . Thus hypothesis (18) holds for N (r, s ) with the choices
A = Area(Γ) , θ = 2 /3, and L equalling the intercept value of Γ. The boundedness hypothesis (19) holds by Theorem 2.
Proof of Theorem 4. Again let H(r, s ) be the lattice counting function N (r, s ), take
A = Area(Γ), let L be the intercept value of Γ, and note the boundedness hypothesis (19) holds by Theorem 2. To finish verifying the hypotheses of Proposition 13, we suppose 0 < s 1 < s 2 < ∞ and show that (18) holds. Take θ = 1 − 2e, where the number e = min {16 , a 1, a 2, b 1, b 2} was defined in Theo-rem 4. Hypothesis (18) is the assertion that
N (r, s ) = Area(Γ) r2 − Lr (s + s−1)/2 + O(r1−2e), (22) with s ∈ [s1, s 2] as r → ∞ . To verify this asymptotic, we will estimate the remainder terms in Proposition 12(b) as follows. In that proposition take L = M , and note
δ(r) < α and (r) < β for all large r by assumptions (1) and (2). We will show the right side of estimate (15) in Proposition 12(b) is bounded by
O(r2/3) + s−3/2O(r1−2a2 ) + s3/2O(r1−2b2 ) + ( s−3/2 + s3/2)O(r1/2)+ ( s2 + s−2)O(1) + s−1O(r1−2a1 ) + sO (r1−2b1 ) + O(1) for large enough r, where the implied constants in the O(·)-terms depend only on the curve Γ and are independent of s. Since each one of these O(·)-terms is bounded by
O(r1−2e), and s and s−1 are bounded when s ∈ [s1, s 2], hypothesis (18) will hold as desired. OPTIMAL STRETCHING FOR LATTICE POINTS 19
Examining now the right side of (15), we see the first two terms are obviously
O(r2/3). For the next term, observe by assumption in (1) that
r1/2s−3/2
|f ′′ (δ(r)) |1/2 = s−3/2O(r1−2a2 ),
and similarly for the analogous term involving g′′ . Since f ′′ (αi) and g′′ (βj ) are con-stant, the corresponding terms in (15) can be estimated by ( s−3/2 +s3/2)O(r1/2). Sim-ilarly, the terms in (15) involving f ′(αi) and g′(βj ) can be estimated by ( s2 +s−2)O(1). Next, s−1rδ (r) = s−1O(r1−2a1 ) by the assumption in (1), and similarly for (r). And, of course, l + ` + 1 is constant, which completes the verification of hypothesis (18). 8. Proof of Theorem 6 and Theorem 7
First we need a two-term bound on the counting function in the closed first quad-rant, as provided by the next proposition. The result is an analogue of Proposition 10, although the constant C is slightly different than in that result. Assume f is concave and strictly decreasing on [0 , L ], with f (0) = M, f (L) = 0. The intercepts L and M need not be equal.
Proposition 14 (Two-term lower bound on counting function) . Let C = M −f (L/ 4) .(a) The number N of nonnegative-integer lattice points lying inside Γ in the closed first quadrant satisfies:
N ≥ Area(Γ) + 12C. (23)
(b) The number of nonnegative-integer lattice points lying inside rΓ( s) in the closed first quadrant satisfies (for r, s > 0):
N (r, s ) ≥ r2 Area(Γ) + 12Crs.
Proof. Part (a). Clearly N equals the total area of the squares of sidelength 1 having lower left vertices at nonnegative-integer lattice points inside the curve Γ. The union of these squares contains Γ, since the curve is decreasing. We separate the proof into cases according to the value of L.Case (i): Suppose L ≤ 2, so that L/ 4 ≤ 1/2. Consider a rectangle whose lower left vertex sits on the curve at x = L/ 4, and has vertices
(L/ 4, f (L/ 4) ), (1, f (L/ 4) ), (1, M ), (L/ 4, M ).
By construction, this rectangle lies inside the union of squares of sidelength 1, and it lies above Γ because the curve is decreasing. Hence
N ≥ Area(Γ) + Area(rectangle) = Area(Γ) + (1 − L/ 4) (M − f (L/ 4) )
≥ Area(Γ) + 12
(M − f (L/ 4) )
as desired. 20 OPTIMAL STRETCHING FOR LATTICE POINTS
Figure 5. Nonnegative integer lattice count satisfies N ≥ Area(Γ) + Area(triangles), in proof of Proposition 14(a) when L ≥ 2. Case (ii): Suppose L ≥ 2. Consider the right triangles of width 1 formed by tangent lines from the right on Γ, that is, the triangles with vertices (i, f (i)), (i +1, f (i)), (i + 1 , f (i) + f ′(i+)), where i = 0 , 1, . . . , bLc − 1. These triangles all lie above the horizontal axis, since by concavity f (i) + f ′(i+) ≥ f (i + 1) ≥ 0; the last inequality explains why the biggest i-value we consider is bLc − 1. Thus these triangles lie inside the union of squares of sidelength 1, and lie above Γ by concavity. Hence
N ≥ Area(Γ) + Area(triangles) .
To complete the proof of Case (ii), we estimate Area(triangles) ≥ 12
bLc− 1
∑
i=1
|f ′(i+)|≥ 12
bLc− 1
∑
i=1
(f (i − 1) − f (i)) by concavity = 12
(f (0) − f (bLc − 1) )
≥ 12
(M − f (L/ 4) ),
because bLc − 1 ≥ L/ 2 ≥ L/ 4 when L ≥ 2. Part (b). Replace Γ in Part (a) with the curve rΓ( s), meaning we replace L, M, f (x)with rs −1L, rsM, rsf (sx/r ) respectively.
Proof of Theorem 6. Since N (r, s ) ≤ r2 Area(Γ), taking s = 1 and L = M in Lemma 8 gives that
N (r, 1) ≤ r2 Area(Γ) + 2 rL + 1 .
Now suppose s ∈ S (r) is a minimizing value, so that N (r, s ) ≤ N (r, 1). Since
N (r, s ) ≥ r2 Area(Γ) + 12Crs OPTIMAL STRETCHING FOR LATTICE POINTS 21
by Proposition 14(b), we conclude from above that 12Crs ≤ 2rL + 1 ≤ 52rL,
where the last inequality holds for r ≥ 2/L . Hence s ≤ 5L/ C, and so the set S(r) is bounded above. Interchanging the horizontal and vertical axes and recalling L = M
(i.e. , the intercepts are equal in this theorem), one finds similarly that s−1 ≤ 5L/ ˜C.Hence S(r) is bounded below away from 0, which completes the proof.
Proof of Theorem 7. The theorem will follow from Proposition 13 with the choice
H(r, s ) = −N (r, s ), since maximizing s 7 → H(r, s ) corresponds to minimizing s 7 →N (r, s ). The boundedness hypothesis (19) of the proposition holds by Theorem 6. The other hypothesis (18) is verified as follows. Taking L = M in the relation between N (r, s ) and N (r, s ) in Lemma 8, and calling on the asymptotic for N (r, s ) in either (21) (under the assumptions of Theorem 3) or (22) (under the assumptions of Theorem 4), we deduce
N (r, s ) = Area(Γ) r2 + Lr (s−1 + s)/2 + O(rθ)with s ∈ [s1, s 2] allowed to vary as r → ∞ , where
θ =
{
2/3 under the assumptions of Theorem 3, 1 − 2e under the assumptions of Theorem 4. That is, we have verified hypothesis (18) with H = −N , A = − Area(Γ), and L the intercept value of Γ. 9. Open problem for 1-ellipses — lattice points in right triangles
Lattice point maximization for right triangles appears to be an open problem. Consider the p-circle with p = 1, which is a diamond with vertices at ( ±1, 0) and (0 , ±1). It intersects the first quadrant in the line segment Γ joining the points (0 , 1) and (1 , 0). Here L = M = 1. Stretching the 1-circle in the x- and y-directions gives a 1-ellipse
|sx | + |s−1y| = 1 ,
which together with the coordinate axes forms a right triangle of area 1 /2 in the first quadrant, with one vertex at the origin and hypotenuse Γ( s) joining the vertices at (s−1, 0) and (0 , s ). As previously, we write S(r) for the set of s-values that maximize the number of positive-integer (first quadrant) lattice points below or on rΓ( s), when
r > 0. First of all, the 45–45–90 degree triangle ( s = 1) does not always enclose the most lattice points: Figure 6 shows an example. The open problem is to understand the limiting behavior of the maximizing s-values. Does S(r) converge to {1} as r → ∞ ? We proved the answer is “Yes” for
p-ellipses when 1 < p < ∞ (Example 5), but for p = 1 we suggest the answer is “No”. Numerical evidence in Figure 7 suggests that the set S(r) does not converge to {1}
as r → ∞ . Indeed, the plotted heights appear to cluster at a large number of values, 22 OPTIMAL STRETCHING FOR LATTICE POINTS
Figure 6. The 1-ellipse sx + s−1y = r with r = 4 .96, for s = 1 (solid) and s = √2 (dashed). The dashed line encloses three more lattice points (shown in bold) than the solid line.
Figure 7. Optimal s-values for maximizing the number of lattice points in the 1-ellipse (triangle). The graph plots sup S(r) versus log r.The plotted r-values are multiples of √3/10, and the horizontal axis is at height s = 1. possibly dense in some interval around s = 1. These cluster values presumably have some number theoretic significance. In the remainder of the section we remark that maximizing s-values are ≤ 3 in the limit as r → ∞ , and we describe the numerical scheme that generates Figure 7. Lastly, we explain why s = √2 is a good candidate for a cluster value as r → ∞ .
The bound on maximizing s-values for right triangles ( p = 1 ). Given ε > 0, we claim
S(r) ⊂ [ 13 + ε, 3 + ε] for all large r.OPTIMAL STRETCHING FOR LATTICE POINTS 23
This bound is slightly better than the one in Theorem 2 (which had 4 instead of 3), and can be proved in the same way with the help of a special formula for N (r, 1):
N (r, 1) = # first-quadrant lattice points under the line y = r − x
= 12brcb r − 1c≥ 12(r − 1)( r − 2) = 12r2 − 32r + 1 . (24)
How can one efficiently maximize the lattice counting function for the 1-ellipse? A brute force method of counting how many lattice points lie under the line rΓ( s), and then varying s to maximize that number of lattice points, is simply unworkable in practice. The counting function N (r, s ) jumps up and down in value as s varies, sometimes jumping quite rapidly, and a brute force method of sampling at a finite collection of s-values can never be expected to capture all such jump points or their precise locations. Instead, for a given r we should pre-identify the possible jump values of s, and use that information to count the lattice points. We start with the simple observation that a lattice point ( j, k ) lies under the line rΓ( s) if and only if
sj + s−1k ≤ r,
which is equivalent to
js 2 − rs + k ≤ 0. (25) For this quadratic inequality to have a solution, the discriminant must be nonnegative,
r2 − 4jk ≥ 0, and thus we need only consider lattice points beneath the hyperbola
r2 = 4 xy . For each such lattice point, equality holds in (25) for two positive s-values, namely
smin (j, k ; r) = r − √r2 − 4jk
2j , smax (j, k ; r) = r + √r2 − 4jk
2j .
The geometrical meaning of these values can be understood, as follows: as s increases from 0 to ∞, one endpoint of the line segment rΓ( s) slides up on the y-axis while the other endpoint moves left on the x-axis. The line segment passes through the point ( j, k ) twice: first when s = smin (j, k ; r) and again when s = smax (j, k ; r). The point ( j, k ) lies below the line when s belongs to the closed interval between these two values. Thus the counting function is abad
N (r, s ) = # {(j, k ) : smin (j, k ; r) ≤ s ≤ smax (j, k ; r)}
= ∑
j,k> 0
1smin (j,k ;r)≤s − ∑
j,k> 0
1smax (j,k ;r)<s
where we sum only over positive-integer lattice points with 4 jk ≤ r2.The last formula says that the counting function N (r, s ) equals the number of values
smin (j, k ; r) that are less than or equal to s minus the number of values smax (j, k ; r)that are less than s. To facilitate the evaluation in practice, one should sort the list 24 OPTIMAL STRETCHING FOR LATTICE POINTS
of values of smin (j, k ; r) into increasing order, and similarly sort the list of values of smax (j, k ; r). The numbers in these two lists are the only numbers where N (r, s )can change value, as s increases. In particular, when s increases to smin (j, k ; r), the point ( j, k ) is picked up by the line segment for the first time and so N (r, s ) increases by 1. When s increases strictly beyond smax (j, k ; r), the point ( j, k ) is dropped by the line segment and so N (r, s ) decreases by 1. Note the counting function might increase or decrease by more than 1 at some s-values, if the sorted lists of smin and
smax values have repeated entries (arising from lattice points that are picked up by, or else dropped by, the line segment at the same s-value). After sorting the smin and smax lists, we evaluate the maximum of N (r, s ) by scanning through the two lists, increasing a counter by 1 at each number in the sorted smin list, and decreasing the counter just after each number in the sorted smax
list. The largest value achieved by the counter is the maximum of N (r, s ), and S(r)consists of the closed interval or intervals of s-values on which this maximum count is attained. By this method, we can maximize the lattice counting function for the 1-ellipse in a computationally efficient manner, for any given r > 0. The code is available in [20, Appendix B]. When presenting the results of this method graphically, in Figure 7, we plot only the largest s value in S(r), because the family of 1-ellipses is invariant under the map
s 7 → 1/s and so the smallest value in S(r) will be just the reciprocal of the largest value.
Why is the 1-ellipse not covered by our theorems? For the p-ellipse with
p = 1, Theorem 4 does not apply because f is linear and so f ′′ ≡ 0. Specifically, in the proof we see inequalities (11) and (12) are no longer useful, since their right sides are infinite. The situation cannot easily be rescued, because the left side of (8) need not even be o(r). For example, when s = 1 and r is an integer, by evaluating the number N (r, 1) of lattice points under the curve y = r − x we find
N (r, 1) − r2 Area(Γ) + r(L + M )/2 = 12r(r − 1) − 12r2 + r = 12r,
which is of order r and hence has the same order as the “boundary term” r(L + M )/2on the left side. Thus the method breaks down completely for p = 1. We seek instead to illuminate the situation through numerical investigations.
A cluster value at s = √2 ? Inspired by the numerical calculations in Figure 7, we will show that s = √2 gives a substantially higher count of lattice points than
s = 1, for a certain sequence of r-values tending to infinity. This observation suggests (but does not prove) that √2 or some number close to it should belong to S(r) for those r-values. To be clear: we have not found a proof of this claim. Doing so would provide a counterexample to the idea that the set S(r) converges to {1} as r → ∞ .To compare the counting functions for s = 1 and s = √2, we first notice that for
s = 1 the counting function for the 1-circle is given by
N (r, 1) = brcb r − 1c/2, r > 0.OPTIMAL STRETCHING FOR LATTICE POINTS 25
At s = √2 the slope of the 1-ellipse is −2, and for the special choice r = √2( m + 1 /2) with m ≥ 1 the counting function can be evaluated explicitly as
N (r, √2) = m2.
We further choose m such that r ∈ (n − 1/4, n ) for some integer n, noting that an increasing sequence of such m-values can be found due to the density in the unit interval of multiples of √2 modulo 1. Then, writing r = n − where < 1/4, we have
N (r, √2) − N (r, 1) = m2 − (n − 1)( n − 2) /2= 12(r2 − √2r + 1 /2) − 12(r + − 1)( r + − 2)
≥ 12r − (constant) .
Hence lim sup r→∞
(N (r, √2) − N (r, 1) )/r ≥ 1/2, and so s = √2 can give (for certain choices of r) a substantially higher count of lattice points than s = 1, as we wanted to show. The work above implies that 1 /∈ S(r) for a sequence of r-values tending to infinity. More generally, Marshall and Steinerberger showed that if x > 0 is rational then √x / ∈ S(r) for a sequence of r-values tending to infinity (see [22, Theorem 1]), while if x > 0 is irrational then √x / ∈ S(r) for all sufficiently large r (see [22, Lemma 2] and its associated discussion).
Conjecture for p = 1 . To finish the chapter, we state some of our numerical obser-vations as a conjecture. Let
S = {(r, s ) : r > 0, s ∈ S(r)} ⊂ (0 , ∞) × (0 , ∞),S = closure of S in [0 , ∞] × [0 , ∞],S(∞) = {s ∈ [0 , ∞] : ( ∞, s ) ∈ S}.
Earlier in the chapter we proved that S(∞) ⊂ [1 /3, 3]. The clustering behavior of S(r) observed in Figure 7 suggests the following conjec-ture.
Conjecture 15 (p = 1) . The limiting set S(∞) is countably infinite, and is contained in
[1 /3, 3] ∩ { √x : x ∈ Q, x > 0}.
Marshall and Steinerberger [22, Theorem 2] recently proved that S(∞) contains (countably) infinitely many square roots of rational numbers and is contained in [1 /√5, √5]. For example, they showed the set S(∞) contains 1 and √3/2. Yet a precise characterization of the set remains elusive. One would like a characterization in terms of some number theoretic condition. 26 OPTIMAL STRETCHING FOR LATTICE POINTS
Connection between counting function maximization and eigenvalue minimization
Maximizing a counting function is morally equivalent to minimizing the size of the things being counted. Let us apply this general principle to the case of the circle Γ : x2 + y2 = 1 in the first quadrant, and its associated ellipses rΓ( s). In this section, L = M = 1 and Area(Γ) = π/ 4.
Minimizing eigenvalues of the Dirichlet Laplacian on rectangles. Write
{λn(s) : n = 1 , 2, 3, . . . } = {(js )2 + ( ks −1)2 : j, k = 1 , 2, 3, . . . } (26) so that λn(s) is the nth eigenvalue of the Dirichlet Laplacian on a rectangle of side lengths s−1π and sπ . (The eigenfunctions have the form sin( jsx ) sin( ks −1y).) Then the lattice point counting function is the eigenvalue counting function, because
N (r, s ) = # {(j, k ) : ( js )2 + ( ks −1)2 ≤ r2}
= # {n : λn(s) ≤ r2}.
Define
S∗(n) = argmin
s> 0
λn(s),
so that S∗(n) is the set of s-values that minimize the nth eigenvalue. The next result says that the rectangle minimizing the nth eigenvalue will converge to a square as n → ∞ .
Corollary 16 (Optimal Dirichlet rectangle is asymptotically balanced, due to An-tunes and Freitas [2, Theorem 2.1]; Gittins and Larson ) .
The optimal stretch factor for minimizing λn(s) approaches 1 as n → ∞ , with
S∗(n) ⊂ [1 − O(n−1/12 ), 1 + O(n−1/12 )] ,
and the minimal Dirichlet eigenvalue satisfies the asymptotic formula
min
s> 0
λn(s) = 4
π n +
( 4
π
)3/2
n1/2 + O(n1/3).
The proof is a modification of our Theorem 3. Full details are provided in the ArXiv version of this paper [18, Corollary 10]. In the proof one relies on Proposition 10 to bound the stretch factor s of the optimal rectangle. Proposition 10 is simpler in both statement and proof than the corresponding Theorem 3.1 of Antunes and Freitas , which contains an additional lower order term with an unhelpful sign.
Remark. One would like to prove using only the definition of the counting function that
S∗(n) → 1 if and only if S(r) → 1, or in other words that the rectangle minimizing the nth eigenvalue will converge to a square if and only if the ellipse maximizing the number of lattice points converges to a circle. Then Corollary 16 would follow qualitatively from Theorem 3, without needing any additional proof. Our attempts to find such an abstract equivalence have OPTIMAL STRETCHING FOR LATTICE POINTS 27
failed due to possible multiplicities in the eigenvalues. Perhaps an insightful reader will see how to succeed where we have failed.
Maximizing eigenvalues of the Neumann Laplacian on rectangles. If one considers lattice points in the closed first quadrant, that is, allowing also the lattice points on the axes, then one obtains the Neumann eigenvalues of the rectangle having side lengths s−1π and sπ :
{μn(s) : n = 1 , 2, 3, . . . } = {(js )2 + ( ks −1)2 : j, k = 0 , 1, 2, . . . }.
Notice the first eigenvalue is always zero: μ1(s) = 0 for all s. The lattice point counting function is once again an eigenvalue counting function, because
N (r, s ) = # {(j, k ) : ( js )2 + ( ks −1)2 ≤ r2} = # {n : μn(s) ≤ r2}.
The appropriate problem is to maximize the nth eigenvalue (rather than minimizing as in the Dirichlet case), and so we let
S∗(n) = argmax
s> 0
μn(s).
The corollary below says that the rectangle maximizing the nth Neumann eigen-value will converge to a square as n → ∞ .
Corollary 17 (Optimal Neumann rectangle is asymptotically balanced, due to van den Berg, Bucur and Gittins ; Gittins and Larson ) .
The optimal stretch factor for maximizing μn(s) approaches 1 as n → ∞ , with
S∗(n) ⊂ [1 − O(n−1/12 ), 1 + O(n−1/12 )] ,
and the maximal Neumann eigenvalue satisfies the asymptotic formula
max
s> 0
μn(s) = 4
π n −
( 4
π
)3/2
n1/2 + O(n1/3).
One adapts the arguments used for Theorem 7. A complete proof is in [18, Corollary 11]. Note that our lower bound on the counting function in Proposition 14, which one uses to control the stretch factor s of the optimal rectangle, is simpler in both statement and proof than the corresponding Lemma 2.2 by van den Berg et al. . Further, our Proposition 14 holds for all r > 0, whereas [5, Lemma 2.2] holds only for r ≥ 2s. Consequently we need not establish an a priori bound on s as was done in [5, Lemma 2.3]. Those authors did obtain a slightly better rate of convergence than we do, by calling on sophisticated lattice counting estimates of Huxley; see the comments after Proposition 11.
Appendix A. The van der Corput sum
The next thoerem is due to van der Corput [9, Satz 5], and is central to the proofs of Proposition 11 and Proposition 12. We formulate the theorem as in Kr¨ atzel [15, Korollar zu Satz 1.5, p. 24]. The constants in the theorem are interesting only for being modest in size. Recall the sawtooth function ψ(x) = x − b xc − 1/2. 28 OPTIMAL STRETCHING FOR LATTICE POINTS
Theorem 18. Suppose a < b and h ∈ C2[a, b ] with h′′ monotonic and nonzero. Then
∣∣∣ ∑
a<n ≤b
ψ(h(n))∣∣∣ ≤ 6
∫ ba
|h′′ (t)|1/3 dt + 175 max
[a,b ]
1
|h′′ |1/2 + 1 .
Kr¨ atzel’s result has “+2” as the final term. We obtain the theorem with “+1” by correcting a gap in the proof and arguing carefully in one case, as explained in [20, Theorem A.8]. The constant term is irrelevant for our purposes in this paper, in any case.
Appendix B. An elementary lemma Lemma 19. If s > 0 and 0 < t < 1 then
s + s−1 ≤ 2 + t =⇒ |s − 1| ≤ 3√t.
Proof. By taking the square root on both sides of the inequality (s1/2 − s−1/2)2 = s + s−1 − 2 ≤ t,
and then using that the number 1 lies between s1/2 and s−1/2, we find
|s1/2 − 1| ≤ t1/2.
Hence 1 − t1/2 ≤ s1/2 ≤ 1 + t1/2, and now squaring both sides and using that t < t 1/2
(when t < 1) proves the lemma.
Acknowledgments
This research was supported by grants from the Simons Foundation (#204296 and #429422 to Richard Laugesen) and the University of Illinois Scholars’ Travel Fund, and also by travel funding from the conference Fifty Years of Hearing Drums: Spectral Geometry and the Legacy of Mark Kac in Santiago, Chile, May 2016. At that conference, Michiel van den Berg generously discussed his papers [5, 6] with Laugesen. The anonymous referee made useful suggestions that improved the organization of the paper. The material in this paper forms part of Shiya Liu’s Ph.D. dissertation at the University of Illinois, Urbana–Champaign .
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Department of Mathematics, University of Illinois, Urbana, IL 61801, U.S.A.
E-mail address : Laugesen@illinois.edu sliu63@illinois.edu (shiyaliu.zju@gmail.com) |
7585 | https://www.cdc.gov/mmwr/preview/mmwrhtml/00037085.htm | Management of Patients With Suspected Viral Hemorrhagic Fever
Persons using assistive technology might not be able to fully access information in this file. For assistance, please send e-mail to: mmwrq@cdc.gov. Type 508 Accommodation and the title of the report in the subject line of e-mail. Management of Patients With Suspected Viral Hemorrhagic Fever =============================================================
INTRODUCTION
The term viral hemorrhagic fever (VHF) refers to the illness associated with a number of geographically restricted viruses. This illness is characterized by fever and, in the most severe cases, shock and hemorrhage (1). Although a number of other febrile viral infections may produce hemorrhage, only the agents of Lassa, Marburg, Ebola, and Crimean-Congo hemorrhagic fevers are known to have caused significant outbreaks of disease with person-to-person transmission. Therefore, the following recommendations specifically address these four agents.
The increasing volume of international travel, including visits to rural areas of the tropical world, provides opportunity for the importation of these infections into countries with no endemic VHF, such As the United States. Since most physicians have little or no experience with these viruses, uncertainty often arises when VHF is a diagnostic possibility. Lassa, Marburg, and Ebola viruses are restricted to sub-Saharan Africa, and the differential diagnosis of VHF will most often be made for illness in travelers to this region. Since 1976, no imported cases of VHF have been confirmed in the United States, but every year there are approximately five to 10 suspected cases.
These guidelines review the clinical and epidemiologic features of these diseases; provide recommendations on diagnosis, investigation, and care of patients; and suggest measures to prevent secondary transmission. This document updates earlier recommendations, issued in 1983 (2), for the management of suspected and confirmed cases of VHF. Accumulated evidence shows that transmission of these viruses does not occur through casual contact; thus, some earlier recommendations for preventing secondary transmission have been relaxed. Similarly, therapy recommendations have taken into account recent knowledge of the effects of antiviral drugs.
Further information on investigating and managing patients with suspected VHF, collecting and shipping diagnostic specimens, and instituting control measures is available on request from the following persons at CDC in Atlanta, Georgia. For all telephone numbers, dial 404-639 + extension:
Epidemic Intelligence Service (EIS) Officer, Special Pathogens Branch, Division of Viral Diseases, Center for Infectious Diseases (ext. 1344).
Chief, Special Pathogens Branch, Division of Viral Diseases, Center for Infectious Diseases: Joseph B. McCormick, M.D. (ext. 3308).
Senior Medical Officer, Special Pathogens Branch, Division of Viral Diseases, Center for Infectious Diseases: Susan P. Fisher-Hoch, M.D. (ext. 3308).
Director, Division of Viral Diseases, Center for Infectious Diseases (ext. 3574).
After regular office hours and on weekends, the persons named above may be contacted through the CDC duty officer (ext. 2888).
LASSA FEVER
Lassa virus, named after a small town in northeastern Nigeria, is an enveloped, single-stranded, bisegmented ribonucleic acid (RNA) virus classified in the family Arenaviridae. Its natural host is the multimammate rat Mastomys natalensis. This ubiquitous African rodent lives in close association with humans and is commonly found in and around houses in rural areas. The rats are infected throughout life and shed high levels of virus in their urine. Although the rodent reservoir exists across wide areas of Africa, Lassa virus appears to be restricted to the continent's western part. Closely related viruses are found in other areas, but their potential for causing human disease is poorly understood.
Lassa fever was first recognized in 1969 in northern Nigeria (3) when two of three nurses infected in a rural hospital died. Two persons working in a U.S. laboratory with material from the original outbreak subsequently became infected, one fatally. One person had worked with animals infected with live virus, but it is uncertain how the other person acquired the infection (4,5). Naturally occurring infections, often associated with subsequent nosocomial outbreaks, have been recognized in Nigeria, Sierra Leone, and Liberia (6). On the basis of historical information, as well as serologic testing, sporadic Lassa infection may have occurred also in Guinea, Senegal, Mali, and the Central African Republic (6,7). In at least 10 instances, Lassa fever has been imported into countries outside of Africa (3,8-15). In the United States, the last imported case occurred in 1976 (15). No secondary transmission from these imported infections has been documented, despite intensive surveillance of many potentially exposed people (16).
Under natural circumstances, infection with Lassa virus occurs through contact with M. natalensis or its excreta, probably within the household. Subsequent person-to-person transmission occurs, although it is difficult to distinguish epidemiologically between these two modes of infection (17,18). Person-to-person spread requires close personal contact or contact with blood or excreta. Careful follow-up of household and other close contacts of cases imported into western Europe and North America has not shown any evidence of secondary transmission from casual contact. Early reports of Lassa fever stressed the high infectivity of the condition and the risks of nosocomial transmission. Recent evidence shows that avoiding direct contact with infected tissue, blood, secretions, and excretions, even in poorly equipped rural African hospitals, virtually eliminates the risk of infection (19,20).
In areas where it is endemic, Lassa fever occurs more frequently in the dry than in the rainy season. The clinical spectrum of disease is wide, and the ratio of illness to infection is 9%-26% (18). After an incubation period of 1-3 weeks, illness begins insidiously, with early symptoms of fever, sore throat, weakness, and malaise (21). Pains in the joints and lower back, headache, and nonproductive cough commonly follow. Retrosternal or epigastric pain, vomiting, diarrhea, and abdominal discomfort are also common. Frequent physical signs include fever, exudative pharyngitis, and conjunctival injection. Jaundice and skin rash are rare. Diffuse rales may be heard by auscultating the chest, and pleural and pericardial friction rubs may sometimes be detected. Edema of the face and neck, conjunctival hemorrhages, mucosal bleeding, central cyanosis, encephalopathy, and shock characterize the most severe cases. Some patients experience adult respiratory distress syndrome.
After the first week of illness, the patient begins to recover in milder cases, but starts to deteriorate clinically in more serious ones. The mortality rate for patients hospitalized with Lassa fever is 15%-20% (21), despite higher earlier estimates. The prognosis is particularly poor for women in the third trimester of pregnancy, and a high rate of fetal wastage occurs. Overall, the case-fatality rate is about 1%-2% (18). Various degrees of permanent, sensorineural deafness result in nearly one-fourth
Specific diagnosis of Lassa fever can be made in three ways: by isolating the virus from blood, urine, or throat washings; by demonstrating the presence of immunoglobulin M (IgM) antibody to Lassa virus; or by showing a fourfold rise in titer of IgG antibody between acute- and convalescent-phase serum specimens. Antibodies are measured with the indirect fluorescent antibody technique (IFA), which remains the diagnostic method of choice. Nonspecific laboratory abnormalities include protein- uria and elevated liver, enzymes, with aspartate aminotransferase (AST) levels exceeding those of alanine aminotransferase (ALT).
Adverse prognostic factors are AST elevation above 150 international units/liter, and high levels of viremia during hospitalization (22,23). Treatment is supportive and may require all the modern intensive-care facilities, including renal dialysis and mechanical ventilation. It is essential to pay attention to fluid and electrolyte balance, maintenance of blood pressure and circulatory volume, and control of seizures.
A controlled clinical trial has shown an increased survival rate for Lassa fever patients treated with ribavirin (22). All patients with the disease should now receive this drug. Side effects are largely restricted to reversible hemolysis. Severely ill patients should receive ribavirin parenterally. Lassa fever convalescent plasma has not been shown to be beneficial (22) and currently cannot be recommended, particularly when the potential for transmitting other viruses such as human immunodeficiency virus, hepatitis B virus, and the agent(s) of non-A, non-B hepatitis is considered.
Prevention of Lassa virus infection requires an understanding of the disease and its modes of transmission. Persons who intend to work in areas with endemic disease should be briefed about Lassa fever (20). Currently, no vaccine is available for use in humans.
EBOLA HEMORRHAGIC FEVER
Ebola virus is a single-stranded, unsegmented, enveloped RNA virus with a characteristic filamentous structure. Classification of the virus in the new family Filoviridae has been accepted. The virus is named after a small river in northwest Zaire. It is morphologically similar to, but antigen- ically distinct from Marburg virus. The reservoir of the virus in nature remains unknown.
Ebola hemorrhagic fever was first recognized in 1976. Two epidemics occurred within a short time of each other, the first in southern Sudan (24) and the second in northwest Zaire (25). The index case in the Sudan epidemic occurred in a worker in a cotton factory, who subsequently was the source of hospital transmission. The mortality rate among the 284 recognized cases was 53%. In the Zaire outbreak, which from the beginning centered around a hospital, 88% of the 318 affected persons died. Having close contact with a case and receiving injections at the hospital were strong risk factors for acquiring infection.
Two cases were identified elsewhere in northwest Zaire in 1977 and 1978. Retrospectively, another case was diagnosed in a physician in the same area, who cut himself while performing an autopsy in 1972 and contracted an Ebola-like illness 12 days later (26).
In 1979 another small outbreak occurred in the same area as the 1976 outbreak in Sudan. The index case involved a worker in the same cotton factory (27). The case-fatality rate was 65%. Evidence from serologic studies suggested that Ebola virus may be endemic in certain areas of Sudan and Zaire, as well as in other parts of East and Central Africa (28).
The mode of acquiring natural infection with Ebola virus is unknown. Secondary person-to-person transmission results from close personal contact, which, in the epidemics described above, frequently included the nursing of sick patients. Nosocomial transmission depends on contact with blood, secretions, and excretions. Transmission of infection has been documented in the case of a laboratory worker who experienced a needle-stick inj not suggest that spread occurred through casual contact or by aerosol transmission.
The incubation period ranges from 2 to 21 days; the average is approximately 1 week. In the cases resulting from a needle stick (25,29), the incubation period was 6 days; however, this may not characterize the natural illness. The illness-to-infection ratio for Ebola virus is unknown, but seroepidemiologic investigations suggest that mild or asymptomatic infections can occur.
The onset of illness is abrupt, and initial symptoms resemble those of an influenza-like syndrome. Fever, headache, general malaise, myalgia, joint pain, and sore throat are commonly followed by diarrhea and abdominal pain. A transient morbilliform skin rash, which subsequently desquamates, often appears at the end of the first week of illness. Other physical findings include pharyngitis, which is frequently exudative, and occasionally conjunctivitis, jaundice, and edema. After the third day of illness, hemorrhagic manifestations are common and include petechiae as well as frank bleeding, which can arise from any part of the gastrointestinal tract and from multiple other sites.
Specific diagnosis requires isolating the virus from blood or demonstrating IgM or rising IgG antibodies by IFA. Proteinuria occurs early, and elevation of liver enzymes, AST more than ALT, is typical. Experimental infections in primates have shown that neutrophilia, lym- phopenia, and thrombocytopenia occur early in the illness (30).
Treatment is supportive and may require intensive care. Limited information exists on the efficacy of antiviral drugs or immune plasma to prevent or ameliorate Ebola hemorrhagic fever. Ribavirin shows no in vitro activity. Since the Zaire and the Sudan strains of the virus are distinct (31), if immune plasma is considered for therapeutic use, it must be strain specific. The dangers of transmitting other viral infections through plasma should be remembered. No vaccine exists against Ebola virus.
MARBURG HEMORRHAGIC FEVER
Marburg virus is a single-stranded, unsegmented, enveloped RNA virus that is morphologically identical to, but antigenically distinct from Ebola virus. Classification of the virus in the new family Filoviridae has been accepted. Marburg virus is named after the town in Germany where some of the first cases were described (32). Its reservoir in nature remains unknown.
In 1967, 25 people in Europe became ill after handling material from infected African green monkeys, Cercopithecus aethiops, imported from Uganda (32). The case-fatality rate was 23% for the primary cases, but no deaths were reported for the six secondary cases.
An Australian traveler died of Marburg virus disease in South Africa in 1975, after apparently acquiring his infection in Zimbabwe (33). Two persons with secondary cases -- a female companion and a nurse of the index patient -- survived. The third recognized outbreak of Marburg virus disease occurred in Kenya in 1980 (34). A French engineer contracted the infection in western Kenya, and a physician in a Nairobi hospital became infected while trying to resuscitate the engineer from a terminal bout of hema- temesis. The physician survived. Despite extensive contact with other staff before his illness was diagnosed, the infected physician did not spread the disease further. Another case of Marburg virus disease occurred in South Africa in 1982, with no secondary cases identified (35). The most recent case of Marburg virus disease occurred in Kenya in 1987; it involved a boy visiting a park in the western part of the country near where the engineer had acquired the infection in 1980. The boy died, but no secondary cases occurred.
The mode of acquiring natural infection with Marburg virus is unknown. Secondary spread results from close contact with infected persons or contact with blood or body secretions or excretions. In the original epidemic (32), the only persons primarily infected had direct contact with animal blood or tissues, without taking precautions to prevent infection. Sexual transmission apparently occurred in one instance in Germany (32), and virus has been isolated from seminal fluid up to 2 months after illness (34). Marburg virus was also isolated from the anterior chamber of the eye in a patient who developed uveitis 2 months after the acute illness (33). Although the geographic distribution of Marburg virus is ill-defined, Central and East Africa should be considered endemic areas.
The illness-to-infection ratio is unknown but seems high for primary infections, judging from experience with the original 1967 epidemic. The incubation period ranges from 3 to 10 days, but was typically 5-7 days in the original outbreak (32). The physician infected in the Nairobi hospital had a 9-day incubation period.
Clinical and laboratory features of Marburg virus disease are essentially similar to those described for Ebola virus disease. Diagnosis is confirmed by isolating the virus or demonstrating IgM or rising IgG antibodies by IFA. The treatment is the same as for Ebola virus disease, and the same comments about antiviral drugs and the use of immune plasma apply.
CRIMEAN-CONGO HEMORRHAGIC FEVER
Crimean-Congo hemorrhagic fever (CCHF) virus is an enveloped, single-strande Bunyaviridae. A hemorrhagic fever that had long been recognized in Asia came to international attention after a disease outbreak in the Crimean peninsula in 1944 and 1945 (36). The causative agent was later recognized to be identical to the Congo virus (37,38), isolated in Zaire, hence the name CCHF. Many wild and domestic animals act as reservoirs for the virus, including cattle, sheep, goats, and hares. Ixodid (hard) ticks, particularly those of the genus Hyalomma, act both as a reservoir and vector for CCHF virus. Ground-feeding birds may disseminate infected vectors. Twenty-seven species of ticks are known to harbor the CCHF virus (36).
CCHF is endemic in eastern Europe, particularly in the Soviet Union. However, it may occur in other parts of Europe, especially around the Mediterranean. CCHF has been recognized in northwest China (39), Central Asia, and the Indian subcontinent and may occur in the Middle East and throughout much of Africa. Humans become infected by being bitten by ticks or by crushing ticks, often while working with domestic animals or livestock. Contact with blood, secretions, or excretions of infected animals or humans may also transmit infection. In areas with endemic CCHF, the disease may occur most often in the spring or summer.
Nosocomial transmission is well described in recent reports from Pakistan (40), Iraq (41), Dubai (42), and South Africa (43-48). Available evidence, including recently unpublished experiences, suggests that blood and other body fluids are highly infectious, but simple precautions, such as barrier nursing, effectively prevent secondary transmission (41). Concern has been raised about two nosocomial cases in South Africa that occurred without documented evidence of direct exposure to infectious material (43-47). However, all other evidence rules out airborne trans- mission.
The incubation period for CCHF is about 2-9 days. Initial symptoms are nonspecific and sometimes occur suddenly. They include fever, headache, myalgia, arthralgia, abdominal pain, and vomiting. Sore throat, conjunctivitis, jaundice, photophobia, and various sensory and mood alterations may develop. A petechial rash is common and may precede a gross and obvious hemorrhagic diathesis, manifested by large ecchymoses, bleeding from needle-puncture sites, and hemorrhage from multiple other sources. The case-fatality rate has been estimated to range from 15% to 70% (2), but mild or inapparent infections occur. One study suggested an illness-to-infection
Diagnosis requires isolating the virus from blood during the first week of illness or detecting rising antibody titer by IFA, complement fixation, or one of several other methods. No data are available on the evaluation of IgM antibody response. Nonspecific laboratory abnormalities include progressive neutropenia, lymphopenia, thrombocytopenia, and anemia. Hyperbilirubinemia and elevated liver enzymes are common.
Treatment is supportive and may require intensive care. Ribavirin inhibits CCHF virus in vitro, but its efficacy in clinical practice remains unconfirmed. Although immune plasma has been used its effectiveness has not been evaluated.
APPROACH TO A SUSPECTED CASE OF VHF
General Principles
The patient's travel history, symptoms, and physical signs provide the most important clues to the potential diagnosis of VHF. Under natural circumstances, infection is most often acquired in rural areas, and for most visitors and tourists to areas with endemic VHF, exposure to the causative agents is extremely unlikely. If the patient has visited exclusively urban zones, a diagnosis of VHF is improbable. The diagnosis is realistically excluded if the interval between the onset of symptoms and the last possible exposure exceeds 3 weeks. A careful history must be taken about the patient's possible exposure to ill persons or traveling companions in an area with endemic VHF.
Initial symptoms may include fever, headache, sore throat, myalgia, abdominal pain, and diarrhea. Diagnosis at this stage is difficult, since these symptoms are nonspecific. The differential diagnosis is wide and includes other viral infections -- particularly arbovirus infections -- bacterial infections such as typhoid fever, rickettsial diseases, and parasitic infections such as malaria. Symptoms and signs supporting the diagnosis of VHF are pharyngitis and conjunctivitis, a skin rash (particularly for Marburg and Ebola virus diseases), and later, hemorrhage and shock.
Two critical studies should be done for any patient who has recently returned from the tropics and has fever; these are a blood-film examination for malaria and blood cultures. An experienced technician may need to examine several blood smears to identify malarial parasites, particularly for patients who have taken prophylactic antimalarial chemotherapy. When VHF is a diagnostic possibility, blood cultures must be done in a closed system. These initial specimens should be handled with the same precautions used for samples infected with hepatitis B virus or human immunodeficiency virus. An experienced technician should be briefed about the safe handling of such material. All of the patient's body fluids, secretions, and excre-tions
If clinicians feel that VHF is a likely diagnosis, they should take two immediate steps: 1) isolate the patient, and 2) notify local and state health departments and CDC.
The aim of management is to provide optimal care to the patient with the least hazard to staff. A mobile laboratory capable of performing routine laboratory tests is available on request from CDC (see below). Laboratory tests essential for the patient's immediate care must be done by trained staff using the precautions outlined in this document. Meticulous adherence to barrier-nursing procedures and precautions to prevent contact with blood or other body fluids are fundamental to the effective management of patients with possible VHF and to the protection of the staff.
ISOLATION OF PATIENTS WITH SUSPECTED AND CONFIRMED VHF
General Principles
Extensive experience in West Africa has shown that the ordinary pre-cautions patients infected with hepatitis B virus or human immunodeficiency virus, combined with barrier nursing, effectively prevent Lassa virus transmission in hospitals (19). Ideally, patients should be cared for at the hospital where they are first seen, since patients ill with VHF tolerate the stress of transfer poorly, and a move only increases the potential for secondary transmission. If care at the hospital where the patient presented is not possible, transfer to another local facility is preferable to travel to a more distant center. Personnel involved in the transfer of patients with suspected VHF must follow the same precautions recommended for medical and nursing staff.
The patient should be isolated in a single room with an adjoining anteroom serving as its only entrance. The anteroom should contain supplies for routine patient care, as well as gloves, gowns, and masks for the staff. The Appendix lists suggested supplies for the anteroom. Hand-washing facilities should be available in the anteroom, as well as containers of decontaminating solutions. If possible, the patient's room should be at negative air pressure compared with the anteroom and the outside hall, and the air should not be recirculated. However, this is not absolutely required, and does not constitute a reason to transfer the patient. If a room such as described is not available, use adjacent rooms to provide safe and adequate space.
Strict barrier-nursing techniques should be enforced: all persons entering the patient's room should wear disposable gloves, gowns, masks, and shoe covers. Protective eye wear should be worn by persons dealing with disoriented or uncooperative patients or performing procedures that might involve the patient's vomiting or bleeding (for example, inserting a nasogastric tube or an intravenous or arterial line). Protective clothing should be donned and removed in the anteroom. Only essential medical and nursing personnel should enter the patient's room and anteroom. Isolation signs listing necessary precautions should be posted outside the anteroom.
The patient should use a chemical toilet. All secretions, excretions, and other body fluids (other than laboratory specimens) should be treated with disinfectant solution. All material used for patients, such as disposable linen and pajamas, should be double-bagged in airtight bags. The outside bags should be sponged with disinfectant solution and later incinerated or autoclaved. Disposable items worn by staff, such as gowns, gloves, etc., should be similarly treated. Disposable items used in patient care (suction catheters, dressings, etc.) should be placed in a rigid plastic container of disinfectant solution. The outside of the container should be sponged with disinfectant, and the container should be auto- claved, incinerated, or otherwise safely discarded.
If surgery is required, surgical staff should wear protective eye wear and double gloves. Advice should be sought from CDC.
Disinfectant Solutions
Lipid-containing viruses, including the enveloped viruses, are among the most readily inactivated of all viral agents (50). Suitable disinfectant solutions include 0.5% sodium hypochlorite (10% aqueous solution of household bleach), as well as fresh, correctly prepared solutions of glutaraldehyde (2% or as recommended by the manufacturer) and phenolic disinfectants (0.5%-3%) (50,51). Soaps and detergents can also inactivate these viruses and should be used liberally.
CONFIRMATION OF THE DIAGNOSIS
General Principles
The diagnosis of VHF is confirmed by isolating the virus or by demonstrating IgM antibody or a fourfold rise in IgG antibody in serum, as described earlier. Antibody may not appear in blood until the second week of illness. Virus is usually recovered from blood, although Lassa virus may also be isolated from the throat or urine. Liver tissue collected after death may also be a rich source of virus.
Virus isolation must only be attempted in Biosafety Level 4 facilities (52), such as are available at CDC. Serologic tests can be performed either at CDC or in the mobile laboratory (see below). Serologic tests for antibodies are done with gamma-irradiated antigens and serum samples that have been inactivated with heat or gamma irradiation.
Handling Laboratory Specimens
Collecting Specimens
Recommendations for safely collecting and transporting specimens remain unchanged. The essential specimens to be submitted for virus isolation are a sample of venous blood, a midstream ("clean catch") specimen of urine, and a throat swab. If postmortem specimens are available, serum, liver, spleen, and kidney tissue are desirable. The following procedures should be followed:
Glass containers should not be used. Disposable sharp objects, such
as scalpel blades, also should not be handled unnecessarily after use and should be autoclaved or incinerated.
Venous blood samples must be collected with extreme care to avoid self-inoculation. Ten milliliters of clotted blood should be placed in a sealed plastic container. Needles should not be recapped, bent, broken, removed from disposable syringes, or otherwise handled. Blood-taking equipment should be put in a rigid plastic container filled with disinfectant solution and autoclaved or incinerated.
Midstream urine specimens should be collected by clean catch. Five milliliters of urine should be put in a plastic screw-cap container with one of the following: rabbit serum albumin diluted to a final concentration of 25%, human serum albumin diluted to a 1% concentration, or bovine serum albumin at a final concentration of 10%.
Throat swabs should be placed in plastic screw-cap containers in 1 ml of sterile, phosphate-buffered neutral saline containing 25% rabbit serum, 1% human serum albumin, or 10% bovine serum albumin.
The outside of each specimen container should be swabbed with disinfectant, and a label should be attached bearing the patient's name, hospital identification, the date of collection, and the nature of the suspected infection. Then, the specimens should be double-bagged in secure, airtight and watertight bags, which have been similarly labeled. Bags containing specimens should be sponged with disinfectant before they are removed from the patient's room.
Packaging and Transporting Specimens
The Office of Biosafety at CDC (ext. 3883), the persons listed in the Introduction, or the state health department should be contacted for instructions on packaging, labeling, and shipping diagnostic laboratory specimens since shipment of specimens is subject to the applicable provisions of the Federal interstate quarantine regulations (53). In general, the specimens should be packaged as follows:
Place the specimens for transport in a tightly sealed, watertight
container, such as a screw-cap plastic tube or vial, and seal the cap with tape. Make sure plastic containers are resistant to temperatures as low as -80 degrees C. If the specimen is in a glass or other unsuitable container, it should be carefully transferred using the laboratory precautions listed below.
Wrap the primary container in sufficient absorbent material (for example, tissue) to absorb the entire contents in case the container leaks or breaks.
Place the wrapped, sealed primary container in a durable, watertight screw-cap mailing tube or metal can. This secondary container should be sealed with tape. Several primary containers of specimens, each individually wrapped in absorbent material, may be placed in one secondary container, to a maximum of 50 ml of specimen material.
On the outside of the secondary container, attach the specimen labels and other relevant information.
Place the secondary container in a secure box or mailing tube addressed to one of the individuals listed in the Introduction.
Transport the specimen for virus isolation on dry ice. 7. Since individual commercial and noncommercial carriers or shipping services may apply different regulations for transporting biologic specimens, contact a representative of the chosen carrier beforehand to ensure that all necessary formalities are fulfilled. One person listed in the Introduction must be contacted by telephone about the specimen's nature, the method of shipment, and the expected date and time of arrival at CDC.
Exposure of Laboratory Personnel to Specimens
Laboratory tests should be kept to the minimum required for the immediate care of the patient until the mobile laboratory arrives. Critical investigations, such as examination of a blood smear for malaria and the inoculation of blood cultures, must not, however, be postponed. Laboratory staff dealing with specimens from patients who might have a VHF must take the same personal precautions as patient-care staff. Surgical gloves, gowns, shoe covers, and masks should be worn. When possible, laboratory tests should be performed in biological safety cabinets. Blood cultures should be prepared in a closed system. Every effort should be made to avoid creating an aerosol or splashing, and protective eye wear should be worn if possible. A full-face respirator with an HEPA (high efficiency particulate air) filter is an acceptable, but cumbersome alternative to masks and protective eye wear. Nonessential tests should not be performed, nor should routine automated equipment be used unless the specimen has been inact- ivated. Abundant supplies of disinfectant solutions should be readily available. Safe laboratory work has been done with use of these precautions for many years in VHF-endemic areas with poorly equipped hospitals.
Laboratory personnel accidentally exposed to potentially-infected material (for example, through injections or cuts or abrasions on the hands) should immediately wash the infected part, apply a disinfectant solution such as hypochlorite solution, and notify the patient's physician. The person should then be considered as a high-risk contact and placed under surveillance (see below).
Accidental spills of potentially contaminated material should be liberally covered with disinfectant solution, left to soak for 30 minutes, and wiped up with absorbent material soaked in disinfectant.
CLINICAL CARE OF PATIENTS WITH SUSPECTED VHF
General Principles
The challenge of managing patients with VHF is to provide the highest quality of care with the least risk of transmitting infection. Detailed discussion about therapy is beyond the scope of this document. Patients require close supervision, and some will need modern intensive-care facilities. Since pathogenesis is not entirely understood and antiviral therapy is limited, treatment is largely supportive. It is essential to give careful attention to fluid and electrolyte balance. In severe cases, therapy will be required for shock and blood loss. The supportive care of patients critically ill with VHF is the same as the conventional care provided to patients with other causes of multisystem failure. Adult respiratory distress syndrome, renal failure, seizures, and coma may require specific interventions, such as mechanical ventilation, dialysis, and neurologic intensive care. If surgery is required (for example, obstetric intervention), it should be done.
The prognosis for patients with Lassa fever has been shown to correlate with levels of viremia, but not with the development of IgM or IgG anti-bodies ( virus. Experimental infections with Lassa and Ebola viruses in rhesus monkeys suggest that shock results from platelet and endothelial dysfunction, with subsequent leakage of fluid from the intravascular system and hemorrhage. To date, therapeutic use of heparin or corticosteroids has not proven effective and is probably contraindicated.
Patients with Lassa fever should receive ribavirin (see box). For severely ill persons, treatment may begin while confirmation of the diagnosis is pending. Ribavirin is recommended both therapeutically for patients with Lassa fever and prophylactically for high-risk contacts of such patients. Its use for patients with CCHF and their high-risk contacts may be justified but is unstudied.
Treatment Regimen
Ribavirin 30 mg/kg intravenously (IV) loading dose, then 16 mg/kg IV every 6 hours for 4 days, and then 8 mg/kg IV every 8 hours for 6 days (total treatment time 10 days).
Prophylactic Regimen
Ribavirin 500 mg by mouth every 6 hours for 7 days.
Use of convalescent plasma for Lassa fever Is not currently recommended. Analogues of prostacyclin are being evaluated as to their efficacy in restoring the endothelial cell defect (20). Therapy can be discussed with persons listed in the Introduction.
Clinical experience with Ebola and Marburg virus diseases is limited, and individual judgment must determine whether convalescent plasma or antiviral drugs should be used. Interferon and ribavirin show no in vitro effect against these agents.
Ribavirin has been shown effective against some of the Bunyaviridae in vitro, and its use in patients with CCHF seems reasonable, although no clinical experience is available.
MOBILE LABORATORY
CDC has adapted a mobile isolator that can be used as a portable laboratory to investigate cases of suspected or confirmed VHF safely (54). This facility can be transported immediately to any part of the United States, with an accompanying technician and physician experienced in dealing with hemorrhagic fevers. The mobile laboratory has facilities for routine hematologic and biochemical studies, as well as for basic bacter- iologic and coagulation investigations. Serodiagnostic tests for VHF can be performed in this facility, but cultures for virus isolation cannot. Electrolyte measurements on inactivated serum specimens are also possible, but blood gas analysis is not. Early use of this facility is preferable to delays in investigating the suspected case because of concern about the hazards of handling specimens. Further information about the mobile laboratory and its use can be obtained from the persons listed in the Introduction.
AUTOPSY AND HANDLING OF A CORPSE
Before an autopsy is done on a patient suspected to have died from VHF, the possible risks and benefits must be carefully considered. Autopsies have been conducted safely on these patients, sometimes without prior knowledge of the diagnosis (34), but under some circumstances it may be wiser to forego this procedure. Limited autopsy or postmortem collection of blood and percutaneous liver biopsy material may be appropriate.
The same precautions recommended for clinicians and laboratory staff working with infected patients and specimens must be followed. Double gloves, caps and gowns, waterproof aprons, shoe covers, and protective eye wear are required. Aerosol formation must be avoided (for example, electrical cutting instruments must not be used). All solid and liquid waste should be decontaminated with disinfectant solution or by heating for 1 hour at 60 degrees C. Liquid waste can then be washed down the drain; solid waste should be incinerated.
All unnecessary handling of the body, including embalming, should be avoided. Persons who dispose of the corpse must take the same precautions outlined for medical and laboratory staff. The corpse should be placed in an airtight bag and cremated or buried immediately.
DECONTAMINATION PROCEDURES
Disposable items, such as pipette tips, specimen containers, swabs, etc., should be placed in a container filled with disinfectant solution and incinerated. Clothes and blankets that were used by the patient should be washed in a disinfectant, such as hypochlorite solution.
Nondisposable items such as endoscopes used in patient care must be cleaned with decontaminating fluids (for example, gluteraldehyde or hypochlorite). Laboratory equipment must be treated similarly. All non- disposable materials that withstand autoclaving should be autoclaved, after they have been soaked in disinfectant solution. The patient's bed and other exposed surfaces in the hospital room, or in vehicles used to transport the patient, should be decontaminated with disinfectant solution.
IDENTIFICATION, SURVEILLANCE, AND MANAGEMENT OF PATIENT CONTACTS
A contact is defined as a person who has been exposed to an infected person or to an infected person's secretions, excretions, or tissues within 3 weeks of the patient's onset of illness. Contacts may be subdivided into three levels of risk.
Casual contacts are persons who had remote contact with the ill
patient. These include persons on the same airplane, in the same hotel, etc. Since the agents of VHF are not spread by such contact, no special surveillance is indicated.
Close contacts are persons who had more than casual contact with the patient. They include persons living with the patient, nursing or serving the patient when he or she was ill, shaking hands with or hugging the patient, handling the patient's laboratory specimens, etc. These contact persons should be identified by state and local health departments, in collaboration with CDC, as soon as VHF is considered a likely diagnosis for the index case. Once the diagnosis is confirmed, close contacts should be placed under surveillance. This requires these individuals to record their temperatures twice daily and report any temperature of 101 degrees F (38.3 degrees C) or above or any symptom of illness to the public health officer responsible for surveillance. Surveillance should be continued for 3 weeks after the person's last contact with the index patient.
High-risk contacts are persons who have had mucous membrane contact with the patient, such as kissing or sexual intercourse, or have had a needle stick or other penetrating injury involving contact with the patient's secretions, excretions, blood, tissues, or other body fluids. These individuals should be placed under surveillance as soon as VHF is considered a likely diagnosis in the index case.
Any contact who develops a temperature of 101 degrees F (38.3 degrees C) or higher or any other symptoms of illness should be immediately isolated and treated as a VHF patient. Ribavirin should be prescribed as postexposure prophylaxis for high-risk contacts of patients with Lassa fever. Dosage schedules are given in the box on page 11. Although experience is more limited, postexposure prophylaxis with ribavirin is also recommended for high-risk contacts of patients with CCHF.
Convalescent patients and their contacts should be warned that some of the causative agents of VHF may continue to be excreted for many weeks in semen, as demonstrated with Marburg (32,34) and Ebola (29) viruses, and in urine, as occurs sometimes with Lassa virus (13). It is recommended that the persons listed in the Introduction be contacted about arranging shipment to CDC of seminal fluid and urine specimens from patients in the convalescent period for virus isolation. Convalescent patients must be meticulous about personal hygiene. While data are limited concerning infectivity in the convalescent period, abstinence from sexual intercourse is advised until genital fluids have been shown to be free of the virus. If the patient does engage in sexual intercourse before tests are done, the use of condoms is advised.
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Appendix: Suggested Equipment and Supplies for Anteroom Adjoining Patient's Room
Prescribed medications (analgesics, Containers with solution for throat
antipyretics, antibiotics, etc.) swabs and urine specimens (see Resuscitation equipment text) Material for physical examination Labels Portable X-ray machine Marker pens Electrocardiogram machine Plastic airtight bags (various Intravenous equipment and supplies sizes) Tourniquets Plastic trash bags Dry gauze Disinfectant solutions (see text) Alcohol swabs Chemical toilet Needles (various sizes) Urinals Syringes (various sizes) Nursing supplies Plastic container for disposal of Disposable linen, towels, pajamas,
needles and other sharp equipment etc. Tubes for hematologic and Toilet articles
biochemical investigations Gowns, masks, surgical gloves, shoe Blood-culture bottles covers, and protective eye wear
for staff Housekeeping materials
(absorbent towels for spills, etc.)
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Abstract
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1. Introduction
2. Divergent permutations
3. Related questions
References
Journal of Combinatorial Theory, Series A
Volume 173, July 2020, 105231
Divergent permutations
Author links open overlay panel Emanuela Fachini a, János Körner b 1
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Abstract
Two permutations of the natural numbers diverge if the absolute value of the difference of their elements in the same position goes to infinity. We show that there exists an infinite number of pairwise divergent permutations of the naturals. We relate this result to more general questions about the permutation capacity of infinite graphs.
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Keywords
Permutations
Graph capacity
Extremal combinatorics
1. Introduction
We call an arbitrary linear order of the natural numbers an infinite permutation . Let G be an infinite graph with all the natural numbers as its vertex set. We say that two infinite permutations are G-different if for some integer i they have two adjacent vertices of G in the i'th position. Similar concepts have been introduced for permutations of the first n natural numbers in . Once again, two permutations of [n] are called G-different if they put two adjacent vertices of G somewhere in the same position. If the graph G has a finite chromatic number, the largest cardinality of a set of pairwise G-different permutations of [n], denoted by ω(G n) grows only exponentially, and its n-th root has a limit called the permutation capacity of the graph G. This quantity is hard to determine even in case of very simple graphs. Let L be the graph of the infinite path on the set of natural numbers N where the edges connect consecutive integers {i,i+1}. The permutation capacity of this graph,lim n→∞1 n logω(L n) is known to lie between 0.867 and 1, cf. and . It is conjectured in that ω(L n) equals the middle binomial coefficient(n⌊n/2⌋), implying that the upper bound is tight.
It is very natural to consider capacity problems for a class of graphs. The capacity of a graph is an equivalent formulation in combinatorics of Shannon's problem of the capacity of a discrete memoryless channel for probability of error equal to zero . One of the breakthrough results of Lovász' was the determination of this capacity for the cycle of length 5 and all the other self-complementary graphs having a vertex-transitive automorphism group. In information theory it is interesting to extend the capacity problem to a class of channels. This means that we are interested in the capacity of an unknown channel, of which we only know to belong to a given class. In information theory this model is called the compound channel. The interested reader can consult for details. The zero-error version of the compound channel problem was solved by Gargano, Körner and Vaccaro introducing a novel intertwining construction technique. As it is explained in , the intertwining proof technique can be applied to show that the permutation capacity of a finite family of graphs having a finite chromatic number each, is equal to the minimum of the permutation capacities of the graphs. (A channel with minimum permutation capacity is a bottleneck.) The technique cannot be applied if the family of graphs involved is infinite. It was shown in that in some cases of infinite families of directed graphs the bottleneck theorem fails to hold. (In a directed graph the edges are ordered pairs of vertices.) In this paper we continue the exploration of permutation capacity for infinite classes of graphs. We restrict attention to graphs whose edges are not directed.
2. Divergent permutations
We partition the edge set of the complete graph with vertex set N by the graphs L(k) where L(1)=L and the vertices i and j are adjacent in L(k) if the absolute value of the difference of the numbers i and j is k. It is proved by Körner, Simonyi and Sinaimeri as part of the proof of Proposition 3 in that all these graphs have the same permutation capacity. Let us denote by L the family of all our graphs. We say that two permutations of N are L-different if they are L-different for every L∈L. We say that two permutations of N diverge if the absolute value of the difference between integers in the same position goes to infinity. A seemingly weaker condition for pairs of permutations is to be L-different for an infinite number of the graphs L∈L.
Lemma 1
There is an infinite number of permutations of N so that any two of them differ in an infinite number of the graphs L∈L.
Proof
We will explicitly construct an infinite family of permutations of N such that any pair of them will be L-different for infinitely many member graphs L∈L. For every natural number i we construct an infinite permutation x i as follows. In the 2 j'th position of x i we put ij. This defines the value of x i in even positions. We set the missing natural numbers in the odd positions in the increasing order. Consider now an arbitrary pair, x i and x k of our permutations. Suppose that k>i. The difference in the j'th even position is (k−i)j and all these differences are different. We see that (x 2 j i,x 2 j k) form an edge in the graph L((k−i)j) for every j∈N.□
Further scrutiny of our construction shows that it also proves
Theorem 1
There are infinitely many pairwise divergent infinite permutations.
Proof
Let us take a closer look at an arbitrary pair of permutations x i and x k from our construction in the proof of Lemma 1. If k>i, then, obviously, the sequence of the differences (k−i)j, j=1,2,… goes to infinity. We claim that the sequence of the absolute values of the differences in the odd positions also goes to infinity. This implies that alternating the elements of these two sequences we get a new sequence of differences which also tends to infinity. To verify our statement about the subsequences of the odd positions, notice that in the odd positions of x i we find the natural numbers in increasing order, except that in every position which is a multiple of i there is a shift to the left and we see an increase 1 of the value in the position. This means that x j i=j+⌊j i⌋and x j k=j+⌊j k⌋ Since k>i, we have x j i≥x j k and x j i−x j k=⌊j i⌋−⌊j k⌋. The sequence of these differences is non-negative and goes to infinity. In fact, it is easy to see that the sequence is monotonically increasing, and if j=i k t with t tending to infinity, we have⌊j i⌋−⌊j k⌋=(k−i)t going to infinity. Hence the sequence of all the differences alternates between two sequences each of which goes to infinity, implying the same for the whole sequence.□
This leaves open the question about the cardinality of the largest set of pairwise divergent infinite permutations.
3. Related questions
Infinite permutations can differ in many other interesting ways. In the already cited paper of Körner and Malvenuto two permutations of [n] were called colliding if in some coordinate they feature two consecutive integers. Denoting by L the graph of the infinite path over the natural numbers with edges formed by all the pairs of consecutive integers, we now ask whether there is an infinite set of infinite permutations with the property that any pair of them differ in an edge of L in infinitely many coordinates. We will call a pair of infinite permutations with the latter property infinitely colliding.
We prove
Theorem 2
There are infinitely many infinite permutations any pair of which are infinitely colliding.
Proof
For brevity, we will call an odd integer pure if it is the k'th power of an odd prime with k>1. We will call an integer the successor of the pure number q if it equals q+1. By the Catalan-Mihăilescu theorem all these numbers are different. Let B be the set of all the infinite sequences of zeroes and ones with the exception of the all-zero sequence. We will map the elements of B injectively to infinite permutations in such a way that any pair of the permutations so obtained are infinitely colliding. Let z∈B be arbitrary. We define the corresponding infinite permutation ρ by an infinite sequence of transpositions in the identity permutation. Notice first that there is at least one j∈N for which the j'th coordinate of z is equal to 1. Whenever this happens, we swap the integer p j in the corresponding position in the identity permutation with its successor, for every pure number, i.e. for every odd prime p. If z′ and z″ are two different elements of B, then the two permutations differ in an infinite number of swaps so defined.□
The construction shows that the set of permutations in the proof is uncountable; in fact, it has the cardinality of the set of infinite binary sequences. We observe furthermore, that replacing the graph L by any other member graph of L, an analogous statement can be proved, since all these graphs contain a semi-infinite path. It is easy to see that the construction works for any infinite graph containing an infinite matching, i.e., an infinite set of vertex-disjoint edges.
To conclude, for an arbitrary infinite graph G with vertex set N we will say that two infinite permutations ate completely G-different if in every position their corresponding entries form an edge of G. We have
Theorem 3
There is an infinite number of pairwise completely K-different infinite permutations of N, the complete graph with vertex set N .
Proof
We give an explicit construction. For every natural number i let us divide the sequence of natural numbers into successive disjoint intervals of length 2 i. For every such interval let us exchange the second half with the first half. For every i we obtain a different infinite permutation. We claim that any two of these permutations differ in each of their positions. To see this, let us consider the i'th and the j'th of the permutations with j>i. We look into what happens in the first interval of length 2 j of the j'th permutation. The first half of the interval in the identity permutation takes the place of the second half. However, in the i'th permutation any interval of length 2 i in the defining partition keeps its original place as a whole, and changes occur only inside an interval. Hence, in particular, the interval of the first 2 j−1 positions remains, as a whole, in its original place. This then means that none of these positions in the i'th permutation keeps the same value in both of our permutations. The same reasoning applies to the second half of the initial interval of length 2 j in the two permutations. The same relationship holds for the comparison of subsequent intervals.□
Our result leaves open the question about the largest cardinality of a set of pairwise completely K-different permutations.
It is easy to see that the last theorem can be generalized to graphs that are not complete. As an example, consider the infinite graph in which two vertices are adjacent if they have the same parity. For every i apply the previous construction to the odd coordinates and the even coordinates separately. Then, if we consider the i'th and the j'th permutation, the two are adjacent in every odd coordinate for the very same reasoning as before. Hence, the same holds true also for the even coordinates. We immediately realise that our theorem holds, for any natural number q and the infinite graph in which two distinct vertices are adjacent if they are equal moduloq. This raises the question of how sparse a graph can be for the last Theorem to remain true.
In conclusion, we briefly recall an intriguing problem about infinite permutations relative to finite graphs from . Let G be a finite graph whose vertex set is in N. As before, we will say that two infinite permutations are G-different if they have two adjacent vertices of G somewhere in the same position. What is the maximum cardinality of a set of pairwise G-different permutations? Clearly, this number, denoted κ(G) in and , is finite and can be considered an interesting parameter of graph G. Its value is unknown even for complete graphs on at least 4 vertices.
Recommended articles
References
G. Brightwell, G. Cohen, E. Fachini, M. Fairthorne, J. Körner, G. Simonyi, Á. Tóth Permutation capacities of families of oriented infinite paths SIAM J. Discrete Math., 24 (2010), pp. 441-456 CrossrefView in ScopusGoogle Scholar
I. Csiszár, J. Körner Information Theory: Coding Theorems for Discrete Memoryless Systems (2nd edition), Cambridge University Press (2011) Google Scholar
L. Gargano, J. Körner, U. Vaccaro Capacities: from information theory to extremal set theory J. Comb. Theory, Ser. A, 68 (1994), pp. 296-316 View PDFView articleView in ScopusGoogle Scholar
J. Körner, C. Malvenuto Pairwise colliding permutations and the capacity of infinite graphs SIAM J. Discrete Math., 20 (2006), pp. 203-212 CrossrefView in ScopusGoogle Scholar
J. Körner, C. Malvenuto, G. Simonyi Graph-different permutations SIAM J. Discrete Math., 22 (2008), pp. 489-499 CrossrefView in ScopusGoogle Scholar
J. Körner, C. Nair, D. Ng, On the size of pairwise colliding permutations, in: ISIT 2019, manuscript accepted. Google Scholar
J. Körner, G. Simonyi, B. Sinaimeri On types of growth for graph–different permutations J. Comb. Theory, Ser. A, 116 (2009), pp. 713-723 View PDFView articleView in ScopusGoogle Scholar
L. Lovász On the Shannon capacity of a graph IEEE Trans. Inf. Theory, 25 (1979), pp. 1-7 View in ScopusGoogle Scholar
P. Mihăilescu Primary cyclotomic units and a proof of Catalan's conjecture J. Reine Angew. Math., 572 (2004), pp. 167-195 View in ScopusGoogle Scholar
C.E. Shannon The zero-error capacity of a noisy channel IRE Trans. Inf. Theory, 2 (1956), pp. 8-19 View in ScopusGoogle Scholar
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7587 | https://rentman.io/blog/rental-equipment-depreciaition | Rentman
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How to calculate rental equipment depreciation?
How to calculate rental equipment depreciation?
If you find yourself in the business of rental equipment management, you will face this critical challenge at one point or another: equipment depreciation.
As time marches on, rental property will experience a natural decline in value due to wear and tear. If you are not prepared for it, depreciated equipment poses a significant problem for your business. Let’s look at some real-life examples:
Your equipment broke down unexpectedly, which means you have to spend unforeseen money on repairs.
The same equipment was meant to go out to another project tomorrow. As such, you either have to disappoint your customer and send out an incomplete order, or sub-rent the equipment. Both have a negative impact on your bottom line. Also, you risk losing quality of service and trust from your customers.
Fortunately, there are a few methods you can take to deal with equipment depreciation, or even slow it down. In this article, we will provide solutions to ensure more efficient resource management. You will learn about equipment depreciation life and how to calculate equipment depreciation.
So without further ado, let's delve into the details of equipment rental business depreciation!
1. What is equipment depreciation?
2. What is the value of your rental equipment?
3. 3 ways to calculate rental equipment depreciation
What is equipment depreciation?
Equipment depreciation refers to the reduction in value of your rental equipment over time. This is a natural occurrence, as equipment is not made to last forever. The question is not if it will wear out, but rather when.
In accounting terms, equipment depreciation is used to systematically allocate the initial cost of the equipment over its useful life. This allows businesses in the event rental industry to spread out the cost of the equipment over the years it’s being used, thereby matching the expense with the revenue it generates.
Depreciation is crucial for tax purposes, as it enables the company to deduct a portion of the equipment's cost each year, reducing taxable income.
What is the value of your rental equipment?
Before you can calculate depreciation for an asset, you need to figure out three important things:
Depreciable cost of the asset: This includes all the costs required to get and start using the asset. It involves the purchase price of the asset, any shipping costs, and other fees related to acquiring the asset.
Useful life of the asset: This is the number of years the company expects to use the asset before it becomes too old or outdated.
Estimated salvage value: This is the amount of money the company thinks it can get by selling the asset at the end of its useful life.
As for the depreciable cost, Rentman users can save a lot of time to calculate this. Simply export your registered equipment from the software to Excel or Google Sheets and use the right formula. No need for manual labor!
Even though each piece of equipment is unique, you can still make an educated guess to determine the estimated useful life amount of an asset based on factors such as:
Industry Standards: research how long similar types of equipment tend to be used in your industry. This can give you a rough idea of the typical useful life.
Manufacturer Guidelines: check if the manufacturer provides any recommendations or estimates for the expected useful life of the equipment.
Technological Advancements: consider how quickly technology in your industry evolves. If the equipment is likely to become obsolete relatively quickly, you might estimate a shorter useful life.
Similarly, there are a couple of ways to estimate the salvage value:
Comparable Sales: research if there are any similar equipment that has been sold as used. This can provide insights into what kind of value similar items retain.
Expert Opinions: consult experts in the field who might have experience with similar equipment and can provide their perspective on potential salvage value.
Now that you’ve retrieved this information, it is time to decide which method you should choose to calculate equipment depreciation.
What is the depreciation rate for rental equipment?
There are three common ways to calculate rental equipment depreciation. Make sure it fits your type of rental business:
1. Straight-Line Depreciation
This method involves spreading the cost of the equipment evenly over its useful life. To calculate the annual depreciation, simply divide the original cost of the equipment by its expected lifespan.
Example: A rental business decides to purchase a set of event tents for $20,000. They anticipate being able to rent these tents for about 8 years before they need to replace them with newer ones. At the end of those 8 years, they estimate they can sell the used tents for around $5,000.
With this information, the business can calculate the depreciation of the event tents each year. By dividing the depreciable cost ($20,000) minus the estimated salvage value ($5,000) by the useful life (8 years), they find that the depreciation expense for the tents is $2,125 per year.
2. Declining Balance Depreciation
Also known as the reducing balance method. It allows for higher depreciation in the early years of the equipment's life, gradually reducing the depreciation amount over time.
Example: Let's consider an event rental company that invests in a set of high-quality speakers for $15,000. The company chooses to use the declining balance depreciation method to account for the gradual decrease in value over time. They decide to apply a depreciation rate of 30% to the equipment.
Starting Book Value: $15,000
Depreciation Rate: 30%
Year 1:
Depreciation Expense = Starting Book Value Depreciation Rate
Depreciation Expense = $15,000 0.30 = $4,500
Ending Book Value = Starting Book Value - Depreciation Expense
Ending Book Value = $15,000 - $4,500 = $10,500
Year 2:
Starting Book Value: $10,500
Depreciation Expense = Starting Book Value Depreciation Rate
Depreciation Expense = $10,500 0.30 = $3,150
Ending Book Value = Starting Book Value - Depreciation Expense
Ending Book Value = $10,500 - $3,150 = $7,350
And so on for the remaining years until the book value reaches the estimated salvage value or the value is no longer economically useful.
3. Units of Production Depreciation
This method links value of the asset depreciation to the actual usage of the equipment. The more the equipment is used, the greater the depreciation allocated.
Example: Let's consider a delivery company with a fleet of vans. The company purchases a van for $30,000, and it estimates that the van can run for 200,000 miles before it needs replacement. In the first year, the van travels 30,000 miles.
Depreciation per mile = (Cost of Van - Estimated Salvage Value) / Estimated Total Miles
Depreciation Expense = Depreciation per mile Miles Traveled
Depreciation per mile = ($30,000 - $5,000) / 200,000 = $0.125 per mile
Depreciation Expense = $0.125 30,000 = $3,750
The depreciation expense will vary each year based on the number of miles the van travels. So, if in the second year the van travels 40,000 miles, the depreciation expense for that year would be $0.125 40,000 = $5,000.
Understanding equipment depreciation and knowing how to calculate it, helps you with pricing strategies, making informed decisions about when to replace your inventory, and properly managing finances! But if possible, it’s of course best to apply practices that will prolong the lifespan of your equipment.
Make informed decisions with equipment depreciation
Understanding equipment depreciation and the methods used to calculate it, is essential for businesses and rental property owners. Properly accounting for equipment depreciation ensures accurate financial reporting and assists in making informed decisions about equipment maintenance, replacement, and investments.
Rentman offers a comprehensive solution to manage your rental equipment: you can set up and add different ledger accounts in order to categorize your sources of income, streamline your financial workflows, accurately calculate equipment depreciation, and make well-informed decisions.
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7589 | https://www.oed.com/dictionary/catholical_adj | Oxford English Dictionary
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7590 | https://uomus.edu.iq/img/lectures21/MUCLecture_2024_31857636.pdf | 1 AL-Mustaqbal University College of Science Department of Biochemistry Subject: Analytical Chemistry - Practical Lab. : 4 Title: Solutions Preparation and standardization of 0.1 M(HCl) hydrochloric acid solution Theory Hydrochloric acid is produced in solutions up to 38% HCl (concentrated grade). Higher concentrations up to just over 40% are chemically possible, but the evaporation rate is then so high that storage and handling need extra precautions, such as pressure and low temperature. Laboratory grade hydrochloric acid is not sufficiently pure to be used as a primary standard, because it evaporates easily. In this experiment, a standard solution of sodium carbonate is used to determine the exact concentration of a hydrochloric acid solution. The neutralization reaction that occurs is as follows: Na2CO3 + 2HCl 2NaCl + H2O + CO2 Methyl orange indicator solution is used. At the end-point – when neutralization just occurs – the indicator changes color from yellow to peach pink. Procedure:- 1. Preparing (50 ml) 0.1 M HCl Solution:38 % HCL shows density 1.19 g/mL and we can find M by next : - 2 AL-Mustaqbal University College of Science Department of Biochemistry Subject: Analytical Chemistry - Practical Lab. : 4 Title: Solutions Calculate the volume of HCl (conc.):- We must dilute it to preparing 0.1 M HCl in 50 ml from next: Transfer V ml by cylinder to clean and dry beaker containing 30 ml D.W, transfer the solution to volumetric flask capacity 50 ml, and complete the volume to the mark by D.W. 2.Preparing (50 ml) 0.1 M Na2CO3 Solution:-calculate amount from sodium carbonate for prepare 0.1 M in 50 ml – Weigh 0.53 gm. from Na2CO3 in clean and dry beaker and dilute in 30 ml D.W, transfer solution to volumetric flask capacity 50 ml and complete the volume to the mark by D.W. 3 AL-Mustaqbal University College of Science Department of Biochemistry Subject: Analytical Chemistry - Practical Lab. : 4 Title: Solutions 3.Transfer known volume of 5 ml the sodium carbonate solution, with a pipette, to a conical flask then add one or two drops of methyl orange to this solution. 4. Add the acid unknown solution from the burette gradually with continuous swirling of the solution in the conical flask and near the end point, the acid is added drop by drop. Continue the addition of the acid until the color of the solution passes from yellow to faint red. 5.Repeat the experiment three times and tabulate your results then take the mean of the three readings. 4 AL-Mustaqbal University College of Science Department of Biochemistry Subject: Analytical Chemistry - Practical Lab. : 4 Title: Solutions Calculations: Calculate the molarity of HCl Discussion:- 1.What the difference between primary and secondary standard substances? 2.Calculate the volume of conc. HCl required for preparing 250 ml 0.1 M? 3.Calculate the weight of Na2CO3 required for preparing 100 ml 0.1 M? 4.Why is sodium carbonate primary solution? 5.Why standard solution should be colorless? 6.Why is HCl not primary solution? 7.What is the titration? |
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7592 | https://www.reddit.com/r/mathshelp/comments/1br48lk/why_is_the_answer_d_in_this_question/ | Why is the answer D in this question? : r/mathshelp
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7593 | https://141.bluetangent.org/limits/indeterminate-forms | .md
(L5) Indeterminate Form (0/0)
(L5) Indeterminate Form (0/0)#
In this lesson, we will see how to:
calculate the limit of a function at a point using algebraic methods in the case where we have an indeterminate form.
Lecture Videos#
Example 4#
Compute the following limit:
[ \lim_{x\to 2}\dfrac{x^2-4}{x-2} ]
Click through the tabs to see the steps of our solution.
Since we know that this is a rational function, let’s try plugging in (x=2) and see what we get:
[ x=2 \longrightarrow \dfrac{2^2-4}{2-2} = \dfrac{0}{0} ]
The division by (0) tells us though, that (x=2) is not in the domain of our function. So direct evaluation does not work here.
Actually (\tfrac{0}{0}) is a very special type of limit which we call an indeterminate form. We will be spending a considerable amount of time investigating these forms.
Since we have the indeterminate form (\tfrac{0}{0}), one of our main strategies is factor and cancel: try to factor the numerator and denominator and look for any common factors that can cancel. Hopefully, this will get rid of the terms giving us those (0)’s in both the top and bottom of the fraction.
For this function, we can do a little factoring in the numerator:
[ \lim_{x\to 2}\dfrac{x^2-4}{x-2} = \lim_{x\to 2}\dfrac{(x+2)(x-2)}{x-2} ]
Notice that there is an ((x-2))-factor that shows up in both the numerator and denominator. Canceling this gives us an easier limit to calculate:
[ \lim_{x\to 2}\dfrac{(x+2)(x-2)}{x-2} = \lim_{x\to 2}(x+2) ]
As soon as we do the cancellation, we should try evaluating this new simplified limit.
[ \lim_{x\to 2}(x+2) = 2+2=4 ]
(Note that direct evaluation works here, because (x+2) is a polynomial.) That’s it! There is the value for this limit.
Indeterminate Forms#
We have just encountered our first indeterminate form (\dfrac{0}{0}). As we saw in the last example when we encounter these, more work is required. We cannot determine the value of the limit just from the expression (\tfrac{0}{0}).
For intedeterminate forms, the limit could be infinite (\infty) or (-\infty), it might not exist (DNE), or it might be just some number. We don’t know until we do that additional work.
Note that while (\tfrac{0}{0}) is an indeterminate form, (\tfrac{c}{0}) is not. This usually gives us an infinite limit.
Canceling Variables#
Functions that are (almost) equal
If (f(x)=g(x)) except possibly at (x=a), then:
[ \lim_{x\to a} f(x) = \lim_{x\to a} g(x) ]
This theorem allows us to do what we did in the last example and cancel variable terms while calculating a limit. Let’s recall what we did there:
[ \lim_{x\to 2}\dfrac{(x+2)(x-2)}{x-2} = \lim_{x\to 2}(x+2) ]
This is perfectly correct, because of the theorem above. Normally, canceling variable terms fundamentally changes a function - giving us something that is technically different. For limits though, this okay. We are not actually concerned with what happens directly at the number we are approaching, but rather what happens near this number.
As we can see in the two graphs below, the functions (\tfrac{(x+2)(x-2)}{x-2}) and (x+2) are different. The fraction is undefined at (x=2), whereas (x+2) is defined everywhere. However, the limits of these two functions as (x) approaches (2) are equal. That is:
even though directly at (x=2) the functions themselves are not equal
[ \dfrac{(x+2)(x-2)}{x-2}\neq x+2 ]
their limits as (x) approaches (2) are equal:
[ \lim_{x\to 2}\dfrac{(x+2)(x-2)}{x-2} = \lim_{x\to 2}(x+2) ]
Graph of (\dfrac{(x+2)(x-2)}{x-2})
Graph of (x+2)
Example 5#
Compute the following limit:
[ \lim_{h\to 0}\dfrac{(5+h)^2-25}{h} ]
Warning
Here be Algebra tricks - Factor and Cancel
Click through the tabs to see the steps of our solution.
Since this is a rational function, let’s try plugging in (h=0) and see what we get:
[ h=0 \longrightarrow \dfrac{(5+0)^2-25}{0} = \dfrac{5^2-25}{0}=\dfrac{0}{0} ]
The division by (0) tells us though, that (x=5) is not in the domain of our rational function. So direct evaluation does not work here.
Since we have this indeterminate form (\tfrac{0}{0}), we should look for any terms to factor and cancel.
Nothing immediately jumps out as being factorable, so we try simplifying the numerator first by multiplying out that squared term:
[ (5+h)^2 = (5+h)(5+h) = 25+5h + 5h +h^2 = 25 +10h +h^2 ]
Using this in our limit and subtracting the (25)-terms, we get:
[\begin{split} \lim_{h\to 0}\dfrac{(5+h)^2-25}{h} &= \lim_{h\to 0}\dfrac{(25+10h+h^2)-25}{h} \ &= \lim_{h\to 0}\dfrac{10h+h^2}{h} \ \end{split}]
Now we have an expression that can be factored:
[\begin{split} &= \lim_{h\to 0}\dfrac{10h+h^2}{h} \ & \ &= \lim_{h\to 0}\dfrac{h(10+h)}{h} \ \end{split}]
And canceling the common (h)-factor gives us:
[ \lim_{h\to 0}(10+h) ]
As soon as we do the cancellation, it’s a good idea to try evaluating this new simplified limit.
[ \lim_{h\to 0}(10+h) = 10+0 = 10 ]
That’s it! There is our answer. (Note that direct evaluation works here, because (10+h) is a polynomial.)
Example 6#
Compute the following limit:
[ \lim_{x\to 0}\dfrac{\sqrt{x+25}-5}{x} ]
Warning
Here be Algebra tricks - Rationalize
Click through the tabs to see the steps of our solution.
This isn’t a rational function nor a polynomial, but let’s try plugging in (x=0) anyway:
[ x=0 \longrightarrow \dfrac{\sqrt{0+25}-5}{0} = \dfrac{\sqrt{25}-5}{0}=\dfrac{0}{0} ]
Since we have this indeterminate form (\tfrac{0}{0}), direct evaluation does not work here anyway, so we should look for terms to cancel.
When dealing with expressions involving root functions, one trick is to multiply by their conjugate term.
Here, the root expression is (\sqrt{x+25}-5) and its conjugate is (\sqrt{x+25}+5), which we find by switching the subtraction to addition (or vice versa if necessary). In order to eventually simplify our limit, we are going to mulitiply the numerator and denominator by this conjugate term:
[ \lim_{x\to 0}\dfrac{\sqrt{x+25}-5}{x}\cdot \dfrac{\sqrt{x+25}+5}{\sqrt{x+25}+5} ]
We can do this, since we are effectively multplying by a fancy 1. Now, if we calculate the multiplication in the numerator we get:
[\begin{split} \bigg(\sqrt{x+25}-5\bigg)&\cdot\bigg(\sqrt{x+25}+5\bigg) \ = & (x+25)-5\sqrt{x+25}+5\sqrt{x+25}-25 \ = & x \ \end{split}]
Things simplify down quite nicely in the numerator!
We’re now ready to factor and cancel:
[\begin{split} & = \lim_{x\to 0}\dfrac{\sqrt{x+25}-5}{x}\cdot \dfrac{\sqrt{x+25}+5}{\sqrt{x+25}+5}\ & \ & = \lim_{x\to 0}\dfrac{x}{x\bigg(\sqrt{x+25}+5\bigg)}\ \end{split}]
And canceling the common (x)-factor gives us:
[ \lim_{x\to 0}\dfrac{1}{\sqrt{x+25}+5} ]
As soon as we do the cancellation, it’s a good idea to try evaluating this new simplified limit.
[ \lim_{x\to 0}\dfrac{1}{\sqrt{x+25}+5} = \dfrac{1}{\sqrt{0+25}+5} = \dfrac{1}{10} ]
That’s it!
Example 7#
Compute the following limit:
[ \lim_{x\to 2}\dfrac{\tfrac{1}{x}-\tfrac{1}{2}}{x-2} ]
Warning
Here be Algebra tricks - Common Denominator
Click through the tabs to see the steps of our solution.
This is a rational function, so let’s try plugging in (x=2):
[ x=2 \longrightarrow \dfrac{\tfrac{1}{2}-\tfrac{1}{2}}{2-2} =\dfrac{0}{0} ]
Direct evaluation does not work because of that division by (0), so we should look for some terms to cancel.
When dealing with two or more fractions, one helpful technique is to combine the fractions by finding the common denominator.
Let’s start by working with those two fractions in the numerator:
[\begin{split} \lim_{x\to 2}\dfrac{\tfrac{1}{x}-\tfrac{1}{2}}{x-2} & = \lim_{x\to 2}\dfrac{\tfrac{2}{2x}-\tfrac{x}{2x}}{x-2} \ & \ & = \lim_{x\to 2}\dfrac{\tfrac{2-x}{2x}}{x-2} \\end{split}]
When we have fractions stacked on top of fractions like this, it can also be helpful to rewrite them as a product (rather than a quotient).
[ = \lim_{x\to 2}\dfrac{2-x}{2x}\cdot\dfrac{1}{x-2} ]
Looking for terms to cancel, we see that ((2-x)) and ((x-2)) are very similar.
Notice that ((2-x)=(-1)\cdot (x-2)) which gives us:
[ \lim_{x\to 2}\dfrac{2-x}{2x}\cdot\dfrac{1}{x-2} = \lim_{x\to 2}\dfrac{(-1)\cdot (x-2)}{2x}\cdot \dfrac{1}{x-2} ]
And now we can cancel the common ((x-2))-factor, giving us:
[ \lim_{x\to 2}\dfrac{-1}{2x} ]
As usual, as soon as we do the cancellation, the next step should be to try direct evaluation. If we get a number, well, that’ll be our answer.
[ \lim_{x\to 2}\dfrac{-1}{2x} = -\dfrac{1}{4} ]
Done! (Note that direct evaluation works here, because (-\tfrac{1}{2x}) is a rational function.)
Example 8#
Compute (\displaystyle \lim_{x\to 20} f(x)) where (f) is a piecewise-function given by:
[\begin{split} f(x)=\begin{cases} 100+12x & \quad \text{if} \quad x \leq 20 \ \sqrt{180-4x} & \quad \text{if} \quad x > 20\ \end{cases} \end{split}]
Click through the tabs to see the steps of our solution.
In order to calculate this overall limit, we need to first calculate both of the one-sided limits. We usually need to do this for piecewise-functions if the limit we want to calculate happens to be near where the different pieces come together (in this case (x=20)).
Notice that left of (x=20) corresponds with the inequality (x<20) (smaller values). This will help us identify which piece to use for (f(x)):
[ \lim_{x\to20^-} f(x)= \lim_{x\to20^-} (100+12x) = 100+12(20)=340 ]
Direct evaluation works in this last step since ((100+12x)) is a polynomial.
Right of (x=20) corresponds with the inequality (x>20) (larger values), so we will need to use the square root piece of our function.
[\begin{split} \lim_{x\to20^+} f(x)& = \lim_{x\to20^+} \sqrt{180-4x} \ & = \sqrt{\lim_{x\to20^+}(180-4x)} \ & = \sqrt{180-4(20)} \ & = \sqrt{100} \ & =10\ \end{split}]
Notice that we had to use the Power Limit Law first. This allowed us to bring the limit inside the square root, which then allowed us to use direct evaluation on the polynomial (180-4x).
We calculated the left-sided limit:
[ \lim_{x\to20^-} f(x)= 340 ]
and the right-sided limit:
[ \lim_{x\to20^+} f(x)= 10 ]
Since these limits are different, we can conclude that the overall limit does not exist. That is:
[ \lim_{x\to20} f(x) \quad \text{DNE} ]
(Remember that both the left and right sided limits need to be equal to each other in order for the overall limit to exist.)
Contents |
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c program to generate binary number starting from 1 till n
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Asked 3 years, 11 months ago
Modified3 years, 7 months ago
Viewed 3k times
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i need a output like if number is 5 the output should be the 5 binary numbers till 5 but starting from 1 have given example below. it is to be done without using arrays
input:5 output:1,10,11,100,101
but i am getting output like this input=5 output=1,11,11,111,111
is this error beacause of wrong instruction any hint or solution
```c
include
include
void binary_number(int n)
{
int bin_n = 0;
int i = 0;
while (n > 0)
{
bin_n = n % 2;
n = n / 2;
i++;
}
for (int j = i - 1; j >= 0; j--)
printf("%d", bin_n, j);
printf("\n");
}
void binary_number_generation(int n)
{
for (int i = 0; i <= n; i++)
{
binary_number(i);
}
}
int main()
{
int n;
printf("Enter the end value:");
scanf("%d", &n);
printf("\nBinary Number from 1 to %d:", n);
printf("\n");
binary_number_generation(n);
}
```
here i am with a new code idk what i have done but it does give the results
```c
include
int main()
{
int n,i=0,bin=0,plc,dig=0,j;
printf("Enter the n value : ");
scanf("%d",&n);
printf("\nBinary numbers from 1 to %d : ",n);
for(i=0;i<=n;i++)
{
plc=0;bin=0;dig=0;
for(plc=1,j=i;j>0;j=j/2)
{
dig=j%2;
bin=bin+(digplc);
plc=plc10;
}
printf("%d\n",bin);
}
}
```
c
visual-studio
binary
logic
decimal
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edited Oct 29, 2021 at 21:37
NoobprotagonistprogrammerNoobprotagonistprogrammer
asked Oct 29, 2021 at 21:02
NoobprotagonistprogrammerNoobprotagonistprogrammer
1 1 1 silver badge 2 2 bronze badges
12
1 It's not your fault, but: this is a nearly impossible problem. We've seen several variations on it (perhaps your classmates, under the same sadistic instructor?) over the past couple of weeks.Steve Summit –Steve Summit 2021-10-29 21:09:51 +00:00 Commented Oct 29, 2021 at 21:09
The "nearly impossible" part referenced by @SteveSummit is getting the bits in the right order (e.g. 100 rather than 001 for 4). Recursion with a post-traversal print would do the trick.ikegami –ikegami 2021-10-29 21:11:48 +00:00 Commented Oct 29, 2021 at 21:11
You are probably not going to be able to solve it all at once. I suggest first solving one of these two easier problems: (1) Print the first N binary numbers, but you are allowed to use arrays. (2) Print the first N binary numbers, where each number has its digits backwards. That is, 4 is 001, 5 is 101, 6 is 011, 8 is 0001, 10 is 0101, etc. Once you've got one of those working, then you can tackle your actual problem. And I was about to give you the same hint ikegami did: recursion.Steve Summit –Steve Summit 2021-10-29 21:12:59 +00:00 Commented Oct 29, 2021 at 21:12
The reason the code you posted doesn't work is basically that bin_n is not an array. If it were an appropriately-indexed array, your code would work fine.Steve Summit –Steve Summit 2021-10-29 21:17:24 +00:00 Commented Oct 29, 2021 at 21:17
@SteveSummit ik its not possible without arrays atleast for a beginner like me hope someone does come with a solution :(Noobprotagonistprogrammer –Noobprotagonistprogrammer 2021-10-29 21:18:02 +00:00 Commented Oct 29, 2021 at 21:18
|Show 7 more comments
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Let's take 6 for example.
6 10 = 1 2 2 + 1 2 1 + 0 2 0 = 1102
Normally, when building an binary number, you'd use an array. But we can't. But what if we could convert 110 base 2 (six) to the number 110 base 10 (one hundred and ten).
11010 = 1 10 2 + 1 10 1 + 0 10 0
That looks doable!
```
6 base 10 = 1 2^2 + 1 2^1 + 0 2^0 = 110 base 2
| | |
v v v
1 10^2 + 1 10^1 + 0 10^0 = 110 base 10
```
We'll refer to this format as decimal-coded binary (a reference to binary-coded decimal). What this would allow us to do is use printf %d or the likes to print the number.
Solution follows. You should try to implement it yourself before reading on.
```c
include
include
include
// Converts XYZ (base 2) to XYZ (base 10)
// 8 bit inputs require ceil(log2(11111111)) = 24 bit outputs.
// 16 bit inputs require ceil(log2(1111111111111111)) = 50 bit outputs.
uint64_t to_dcb(uint16_t n) {
uint64_t dcb = 0
uint64_t factor = 1;
while (n > 0) {
dcb += (n & 1) factor;
n >>= 1;
factor = 10;
}
return dcb;
}
int main(void) {
printf("%" PRIu64 "\n", to_dcb(6));
return 0;
}
```
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edited Feb 19, 2022 at 18:25
answered Oct 29, 2021 at 21:45
ikegamiikegami
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Steve Summit
Steve SummitOver a year ago
I like your term "decimal-coded binary". See also this answer for how to do it for inputs up to 32 bits.
2021-10-29T21:57:00.8Z+00:00
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As I understand your code is about to converting each number from 1 to n to binary form
```c
include
include
/ This function is to convert the number of digits and then calculate the rest of division
if it's 1 or 0 and then we will form the new number in binary form /
int binary_number(n)
{
int m=0,q=1,s=0;
while(n!=0)
{
s+=(n%2)q;
q=10;
n/=2;
}
return s;
}
/ This number is using to display each number from 1 to n in binary form,
it's better to use an array in next time /
void binary_number_generation(int n)
{
for (int i = 0; i <= n; i++)
{
printf("\n%d\n",binary_number(i));
}
}
int main()
{
int n;
printf("Enter the end value:");
scanf("%d",&n);
binary_number_generation(n);
printf("\n");
}
```
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answered Oct 29, 2021 at 21:44
MED LDNMED LDN
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Here is an alternative method of generating the binary representation of an integer. It's got several problems, but, it has the advantage that it generates the digits in left-to-right order, so you can print them out right away, without having to store them in an array.
```c
void binary_number(int n)
{
int big = 32768;
while(big > 0) {
if(n >= big) {
putchar('1');
n -= big;
} else {
putchar('0');
}
big /= 2;
}
}
```
Exercise for the student: modify it so it doesn't print leading 0's.
Exercise #2: What happens if you try this function on a number bigger than 65535? Can you fix it?
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edited Oct 30, 2021 at 4:05
answered Oct 29, 2021 at 21:49
Steve SummitSteve Summit
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7595 | https://gustavus.edu/chemistry/documents/documents/Isopropylalcohol.pdf | Aldrich - W292907 Page 1 of 8 SIGMA-ALDRICH sigma-aldrich.com Material Safety Data Sheet Version 4.2 Revision Date 01/19/2012 Print Date 06/19/2012 1. PRODUCT AND COMPANY IDENTIFICATION Product name : Isopropyl alcohol Product Number : W292907 Brand : Aldrich Supplier : Sigma-Aldrich 3050 Spruce Street SAINT LOUIS MO 63103 USA Telephone : +1 800-325-5832 Fax : +1 800-325-5052 Emergency Phone # (For both supplier and manufacturer) : (314) 776-6555 Preparation Information : Sigma-Aldrich Corporation Product Safety - Americas Region 1-800-521-8956 2. HAZARDS IDENTIFICATION Emergency Overview OSHA Hazards Flammable liquid, Target Organ Effect, Irritant Target Organs Nerves., Kidney, Cardiovascular system., Gastrointestinal tract, Liver GHS Classification Flammable liquids (Category 2) Skin irritation (Category 3) Eye irritation (Category 2A) Specific target organ toxicity - single exposure (Category 3) GHS Label elements, including precautionary statements Pictogram Signal word Danger Hazard statement(s) H225 Highly flammable liquid and vapour. H316 Causes mild skin irritation. H319 Causes serious eye irritation. H336 May cause drowsiness or dizziness. Precautionary statement(s) P210 Keep away from heat/sparks/open flames/hot surfaces. - No smoking. P261 Avoid breathing dust/ fume/ gas/ mist/ vapours/ spray. P305 + P351 + P338 IF IN EYES: Rinse cautiously with water for several minutes. Remove contact lenses, if present and easy to do. Continue rinsing. HMIS Classification Health hazard: 2 Chronic Health Hazard: Aldrich - W292907 Page 2 of 8 Flammability: 3 Physical hazards: 0 NFPA Rating Health hazard: 2 Fire: 3 Reactivity Hazard: 0 Potential Health Effects Inhalation May be harmful if inhaled. Causes respiratory tract irritation. Vapours may cause drowsiness and dizziness. Skin May be harmful if absorbed through skin. Causes skin irritation. Eyes Causes eye irritation. Ingestion May be harmful if swallowed. 3. COMPOSITION/INFORMATION ON INGREDIENTS Synonyms : 2-Propanol sec-Propyl alcohol Isopropyl alcohol Isopropanol Formula : C3H8O Molecular Weight : 60.1 g/mol Component Concentration 2-Propanol CAS-No. EC-No. Index-No. 67-63-0 200-661-7 603-117-00-0 - 4. FIRST AID MEASURES General advice Consult a physician. Show this safety data sheet to the doctor in attendance.Move out of dangerous area. If inhaled If breathed in, move person into fresh air. If not breathing, give artificial respiration. Consult a physician. In case of skin contact Wash off with soap and plenty of water. Consult a physician. In case of eye contact Rinse thoroughly with plenty of water for at least 15 minutes and consult a physician. If swallowed Do NOT induce vomiting. Never give anything by mouth to an unconscious person. Rinse mouth with water. Consult a physician. 5. FIREFIGHTING MEASURES Conditions of flammability Flammable in the presence of a source of ignition when the temperature is above the flash point. Keep away from heat/sparks/open flame/hot surface. No smoking. Suitable extinguishing media Use water spray, alcohol-resistant foam, dry chemical or carbon dioxide. Special protective equipment for firefighters Wear self contained breathing apparatus for fire fighting if necessary. Aldrich - W292907 Page 3 of 8 Hazardous combustion products Hazardous decomposition products formed under fire conditions. - Carbon oxides Further information Use water spray to cool unopened containers. 6. ACCIDENTAL RELEASE MEASURES Personal precautions Use personal protective equipment. Avoid breathing vapors, mist or gas. Ensure adequate ventilation. Remove all sources of ignition. Evacuate personnel to safe areas. Beware of vapours accumulating to form explosive concentrations. Vapours can accumulate in low areas. Environmental precautions Prevent further leakage or spillage if safe to do so. Do not let product enter drains. Methods and materials for containment and cleaning up Contain spillage, and then collect with an electrically protected vacuum cleaner or by wet-brushing and place in container for disposal according to local regulations (see section 13). 7. HANDLING AND STORAGE Precautions for safe handling Avoid contact with skin and eyes. Avoid inhalation of vapour or mist. Use explosion-proof equipment. Keep away from sources of ignition - No smoking. Take measures to prevent the build up of electrostatic charge. Conditions for safe storage Keep container tightly closed in a dry and well-ventilated place. Containers which are opened must be carefully resealed and kept upright to prevent leakage. Handle and store under inert gas. hygroscopic 8. EXPOSURE CONTROLS/PERSONAL PROTECTION Components with workplace control parameters Components CAS-No. Value Control parameters Basis 2-Propanol 67-63-0 TWA 200 ppm USA. ACGIH Threshold Limit Values (TLV) Remarks Eye & Upper Respiratory Tract irritation Central Nervous System impairment Not classifiable as a human carcinogen STEL 400 ppm USA. ACGIH Threshold Limit Values (TLV) Eye & Upper Respiratory Tract irritation Central Nervous System impairment Not classifiable as a human carcinogen TWA 400 ppm 980 mg/m3 USA. OSHA - TABLE Z-1 Limits for Air Contaminants - 1910.1000 STEL 500 ppm 1,225 mg/m3 USA. OSHA - TABLE Z-1 Limits for Air Contaminants - 1910.1000 TWA 400 ppm 980 mg/m3 USA. Occupational Exposure Limits (OSHA) - Table Z-1 Limits for Air Contaminants The value in mg/m3 is approximate. TWA 400 ppm 980 mg/m3 USA. NIOSH Recommended Exposure Limits ST 500 ppm 1,225 mg/m3 USA. NIOSH Recommended Exposure Limits Aldrich - W292907 Page 4 of 8 Personal protective equipment Respiratory protection Where risk assessment shows air-purifying respirators are appropriate use a full-face respirator with multi-purpose combination (US) or type ABEK (EN 14387) respirator cartridges as a backup to engineering controls. If the respirator is the sole means of protection, use a full-face supplied air respirator. Use respirators and components tested and approved under appropriate government standards such as NIOSH (US) or CEN (EU). Hand protection Handle with gloves. Gloves must be inspected prior to use. Use proper glove removal technique (without touching glove's outer surface) to avoid skin contact with this product. Dispose of contaminated gloves after use in accordance with applicable laws and good laboratory practices. Wash and dry hands. Eye protection Face shield and safety glasses Use equipment for eye protection tested and approved under appropriate government standards such as NIOSH (US) or EN 166(EU). Skin and body protection impervious clothing, Flame retardant antistatic protective clothing, The type of protective equipment must be selected according to the concentration and amount of the dangerous substance at the specific workplace. Hygiene measures Handle in accordance with good industrial hygiene and safety practice. Wash hands before breaks and at the end of workday. 9. PHYSICAL AND CHEMICAL PROPERTIES Appearance Form liquid Colour colourless Safety data pH no data available Melting point/freezing point Melting point/range: -89.5 °C (-129.1 °F) - lit. Boiling point 82 °C (180 °F) - lit. Flash point 12.0 °C (53.6 °F) - closed cup Ignition temperature 425 °C (797 °F) Autoignition temperature 425.0 °C (797.0 °F) Lower explosion limit 2 %(V) Upper explosion limit 12.7 %(V) Vapour pressure 43.2 hPa (32.4 mmHg) at 20.0 °C (68.0 °F) 58.7 hPa (44.0 mmHg) at 25.0 °C (77.0 °F) Density 0.785 g/cm3 at 25 °C (77 °F) Water solubility completely soluble Partition coefficient: n-octanol/water log Pow: 0.05 Relative vapour density no data available Odour alcohol-like Odour Threshold no data available Evaporation rate 3.0 10. STABILITY AND REACTIVITY Aldrich - W292907 Page 5 of 8 Chemical stability Stable under recommended storage conditions. Possibility of hazardous reactions Vapours may form explosive mixture with air. Conditions to avoid Heat, flames and sparks. Extremes of temperature and direct sunlight. Materials to avoid Oxidizing agents, Acid anhydrides, Aluminium, Halogenated compounds, Acids Hazardous decomposition products Hazardous decomposition products formed under fire conditions. - Carbon oxides Other decomposition products - no data available 11. TOXICOLOGICAL INFORMATION Acute toxicity Oral LD50 LD50 Oral - rat - 5,045 mg/kg Remarks: Behavioral:Altered sleep time (including change in righting reflex). Behavioral:Somnolence (general depressed activity). Inhalation LC50 LC50 Inhalation - rat - 8 h - 16000 ppm Dermal LD50 LD50 Dermal - rabbit - 12,800 mg/kg Other information on acute toxicity no data available Skin corrosion/irritation Skin - rabbit - Mild skin irritation Serious eye damage/eye irritation Eyes - rabbit - Eye irritation - 24 h Respiratory or skin sensitization no data available Germ cell mutagenicity no data available Carcinogenicity This product is or contains a component that is not classifiable as to its carcinogenicity based on its IARC, ACGIH, NTP, or EPA classification. IARC: 3 - Group 3: Not classifiable as to its carcinogenicity to humans (2-Propanol) NTP: No component of this product present at levels greater than or equal to 0.1% is identified as a known or anticipated carcinogen by NTP. OSHA: No component of this product present at levels greater than or equal to 0.1% is identified as a carcinogen or potential carcinogen by OSHA. Reproductive toxicity no data available Aldrich - W292907 Page 6 of 8 Teratogenicity no data available Specific target organ toxicity - single exposure (Globally Harmonized System) May cause drowsiness or dizziness. Specific target organ toxicity - repeated exposure (Globally Harmonized System) no data available Aspiration hazard no data available Potential health effects Inhalation May be harmful if inhaled. Causes respiratory tract irritation. Vapours may cause drowsiness and dizziness. Ingestion May be harmful if swallowed. Skin May be harmful if absorbed through skin. Causes skin irritation. Eyes Causes eye irritation. Signs and Symptoms of Exposure Central nervous system depression, prolonged or repeated exposure can cause:, Nausea, Headache, Vomiting, narcosis, Drowsiness, Overexposure may cause mild, reversible liver effects. Synergistic effects no data available Additional Information RTECS: NT8050000 12. ECOLOGICAL INFORMATION Toxicity Toxicity to fish LC50 - Pimephales promelas (fathead minnow) - 9,640.00 mg/l - 96 h Toxicity to daphnia and other aquatic invertebrates EC50 - Daphnia magna (Water flea) - 5,102.00 mg/l - 24 h Immobilization EC50 - Daphnia magna (Water flea) - 6,851 mg/l - 24 h Toxicity to algae EC50 - Desmodesmus subspicatus (green algae) - > 2,000.00 mg/l - 72 h EC50 - Algae - > 1,000.00 mg/l - 24 h Persistence and degradability no data available Bioaccumulative potential no data available Mobility in soil no data available PBT and vPvB assessment no data available Other adverse effects no data available 13. DISPOSAL CONSIDERATIONS Product Burn in a chemical incinerator equipped with an afterburner and scrubber but exert extra care in igniting as this material is highly flammable. Offer surplus and non-recyclable solutions to a licensed disposal company. Contact a licensed professional waste disposal service to dispose of this material. Aldrich - W292907 Page 7 of 8 Contaminated packaging Dispose of as unused product. 14. TRANSPORT INFORMATION DOT (US) UN number: 1219 Class: 3 Packing group: II Proper shipping name: Isopropanol Marine pollutant: No Poison Inhalation Hazard: No IMDG UN number: 1219 Class: 3 Packing group: II EMS-No: F-E, S-D Proper shipping name: ISOPROPANOL Marine pollutant: No IATA UN number: 1219 Class: 3 Packing group: II Proper shipping name: Isopropanol 15. REGULATORY INFORMATION OSHA Hazards Flammable liquid, Target Organ Effect, Irritant SARA 302 Components SARA 302: No chemicals in this material are subject to the reporting requirements of SARA Title III, Section 302. SARA 313 Components The following components are subject to reporting levels established by SARA Title III, Section 313: 2-Propanol CAS-No. 67-63-0 Revision Date 1987-01-01 SARA 311/312 Hazards Fire Hazard, Acute Health Hazard, Chronic Health Hazard Massachusetts Right To Know Components 2-Propanol CAS-No. 67-63-0 Revision Date 1987-01-01 Pennsylvania Right To Know Components 2-Propanol CAS-No. 67-63-0 Revision Date 1987-01-01 New Jersey Right To Know Components 2-Propanol CAS-No. 67-63-0 Revision Date 1987-01-01 California Prop. 65 Components This product does not contain any chemicals known to State of California to cause cancer, birth defects, or any other reproductive harm. 16. OTHER INFORMATION Further information Copyright 2012 Sigma-Aldrich Co. LLC. License granted to make unlimited paper copies for internal use only. The above information is believed to be correct but does not purport to be all inclusive and shall be used only as a guide. The information in this document is based on the present state of our knowledge and is applicable to the product with regard to appropriate safety precautions. It does not represent any guarantee of the properties of the product. Sigma-Aldrich Corporation and its Affiliates shall not be held liable for any damage resulting from handling or from contact with the above product. See www.sigma-aldrich.com and/or the reverse side of invoice or packing slip for additional terms and conditions of sale. Aldrich - W292907 Page 8 of 8 |
7596 | https://www.hackmath.net/en/calculator/conversion-of-mass-units?unit1=Oz&unit2=g&dir=1 | Conversion Oz to g. Troy ounces to grams. 1 troy ounce is 31.10348 grams.
Conversion troy ounces to grams, Oz to g.
The conversion factor is 31.1034768; therefore, 1 troy ounce = 31.1034768 grams. In other words, to convert a value in Oz to g, multiply by the value by 31.1034768. The calculator answers the questions such as: How many g is 90 Oz? or How to convert Oz to g. This is a mass unit conversion from Oz to g.
Conversion result:
1 Oz = 31.10348 g
1 troy ounce is 31.10348 grams.
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Combination
A combination, sometimes called a binomial coefficient, is a way of choosing objects from a set of where the order in which the objects are chosen is irrelevant. We are generally concerned with finding the number of combinations of size from an original set of size
Contents
1 Video
2 Notation
3 Formula
3.1 Derivation
4 Formulas/Identities
5 Examples
6 See also
Video
This video goes over what Permutations & Combinations are, their various types, and how to calculate each type! It serves as a great introductory video to combinations, permutations, and counting problems in general! Permutations & Combinations Video
This video is a great introduction to permutations, combinations, and constructive counting:
Notation
The common forms of denoting the number of combinations of objects from a set of objects is:
Formula
Derivation
Consider the set of letters A, B, and C. There are different permutations of those letters. Since order doesn't matter with combinations, there is only one combination of those three.( are all equivalent in combination but different in permutation.)
In general, since for every permutation of objects from elements which is .Now since in permutation the order of arrangement matters( is not same as ) but in combinations the order of arrangement does not matter( is equivalent to ).For its derivation see this video.You need combinations when order doesn't matter, and you need permutations when order does matter. Combinations and permutations are useful tools in counting.
Suppose elements are selected out.For permutation elements can be arranged in ways.We have overcounted the number of combinations by times.So We have , or .
Formulas/Identities
One of the many proofs is by first inserting into the binomial theorem. Because the combinations are the coefficients of , and a and b disappear because they are 1, the sum is .
We can prove this by putting the combinations in their algebraic form. . As we can see, . By the commutative property, . Because , by the transitive property, we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't want is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest. This identity is also the reason why Pascal's Triangle is symmetrical.
Examples
2005 AIME II Problem 1
See also
Combinatorics
Combinatorial identity
Permutations
Pascal's Triangle
Generating function
Retrieved from "
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7598 | http://www-sop.inria.fr/members/Yves.Bertot/hulls.pdf | Formalizing Convex Hulls Algorithms David Pichardie ENS Cachan-Bretagne pichardy@eleves.bretagne.ens-cachan.fr Yves Bertot INRIA Sophia Antipolis Yves.Bertot@inria.fr No Institute Given 1 Introduction Abstract. We study the development of formally proved algorithms for computational geometry. The result of this work is a formal descrip-tion of the basic principles that make convex hull algorithms work and two programs that implement convex hull computation and have been automatically obtained from formally verified mathematical proofs. A special attention has been given to handling degenerated cases that are often overlooked by conventional algorithm presentations.
Algorithms to compute the convex hull of a collection of points in two or three dimensions abound. The complexity of this problem is known to be ap-proximately the same as for a sorting algorithm in the worst case, but the average complexity depends very much on the distribution of data: some algorithms ac-tually have a complexity of the form n×h, where n is the total number of points and h is the number of points on the convex hull. These algorithms will be more efficient than nlog(n) when few points are on the convex hull. This makes it useful to have several algorithms around.
We have studied two algorithms. The first is an incremental algorithm, where new points are progressively added to the input data. If the new point is outside the convex hull, then it is necessary to add two new edges to the convex hull and remove some others. The second algorithm is known as the package-wrapping algorithm: it follows the intuition of tying a band around the collection of points.
At the nth step, the band is already in contact with a few points and the al-gorithm proceeds by computing what the next point is going to be: if with the band turning clockwise around the points, the next point in the hull is going to be the leftmost remaining point.
All these algorithms rely on an orientation predicate that expresses whether the points of a triangle are enumerated clockwise or counter-clockwise. This orientation predicate is easily computed using the coordinates of the points, but it is meaningless when the points are aligned. Usual presentation of convex hulls algorithms assume that three points in the input data are never aligned.
The structure of convex hull algorithms rely on the fact that the orienta-tion predicate satisfies some properties: for instance the triangles built with four points have to be oriented in a consistent manner. Knuth describes the min-imal properties of the orientation predicates and calls them axioms.
Knuth’s approach of axioms for convex hulls has the nice property of sepa-rating concerns about the properties of arithmetic expressions containing point coordinates and the control structure of algorithms. It has proved a good orga-nization principle for our description and proof development.
Thus, our work contains distinct parts. The first part is about the axioms, and showing that they hold for an implementation predicate. This part involves some numerical computation. The second part describes the main structure of the algorithms based on the axioms, with numeric computation completely avoided.
In the last part, we revisit the algorithms to make them robust with respect to degenerated data.
All our proofs have been mechanically checked using the Coq system .
1.1 Related work Automatic theorem proving and geometry have been in contact for some time.
The work in this domain that we are better aware of is that of Chou . In this development, theorems about basic geometric figures like straight lines, triangles and circles are proved automatically, but there is no result about computational geometry per se, since algorithms are not the object of the study.
Puitg and Dufourd , used the Coq proof system to describe notions of planarity. Work on finding minimal axioms to describe geometric concepts has also been done in constructive proof theory by Jan von Plato and formalized in Coq by Gilles Kahn .
Last, we should refer to all the work done on computational geometry, but this domain is far too large to be cited entirely, we can only refer to the books we have used as reference, and , and .
2 Knuth’s “axioms” for convex hulls In what follows, we assume that points are taken from a set P and we describe the orientation predicate as a predicate over three points p, q, r in P, which we denote d pqr.
Knuth’s axioms describe the various ways in which triangles can be looked at and arranged on the plane. The first axiom expresses that when enumerating the points of a triangle, one may start with any of them: Axiom 1 ∀p, q, r.
d pqr ⇒d qrp.
As a corollary of this axiom one also has d pqr ⇒d rpq. This axiom is “cyclic” it can be repeated indefinitely on the same data. For this reason, it impedes automatic proof search procedures.
The second axiom expresses that if a triangle is oriented counter-clockwise, then the same triangle with two points transposed is in fact oriented clockwise.
Axiom 2 ∀p, q, r.
d pqr ⇒¬d prq.
An immediate consequence of the first two axioms is that three points are oriented only if they are pairwise distinct: ∀p, q, r.
d pqr ⇒p ̸= q ∧q ̸= r ∧r ̸= p Many of the lemmas appearing in our development therefore rely on the as-sumptions that the points being considered are pairwise distinct. We also have a predicate on lists of points expressing that they contain pairwise distinct points.
The third axiom expresses that a triangle is either oriented clockwise or counter-clockwise.
Axiom 3 ∀p, q, r.
p ̸= q ⇒q ̸= r ⇒p ̸= r ⇒d pqr ∨d prq.
This affirmation voluntarily overlooks the fact that points may be aligned. This is a tradition of computational geometry that algorithms can be studied without taking care of what are called degenerate cases. We will have to study this more carefully later.
The fourth axiom expresses that the four triangles obtained with four arbi-trary points may not be oriented in an arbitrary manner: there has to be some sort of consistency, easily understood by observing figure 1.
p q r t Fig. 1. Axiom 4: Consistent orientation for four triangles.
Axiom 4 ∀p, q, r, t.
c tqr ∧c ptr ∧c pqt ⇒d pqr.
The fifth axiom expresses that the orientation predicate may be used to sort points in some way, with a notion that is similar to transitivity. Still one has to be careful to avoid going in circles, this is the reason why this axiom uses 5 points, see figure 2. Clever algorithms use this property of like-transitivity to avoid looking at all points.
Axiom 5 ∀p, q, r, s, t.
c tsp ∧c tsq ∧c tsr ∧c tpq ∧c tqr ⇒c tpr t s p q r Fig. 2. Axiom 5: sorting points in a half plane.
The first three orientation properties express that the three points p, q, and r are in the same half plan. The other three orientation predicates express transitivity: from the point of view of t, if q is left of p and r is left of q, then r is left of p.
There is a variant to this axiom, that can actually be proved using axioms 1, 2, 3, and 5, and which we will refer to by the name axiom 5′.
Axiom 5′ ∀p, q, r, s, t.
c stp ∧c stq ∧c str ∧c tpq ∧c tqr ⇒c tpr Knuth shows that these axioms are independent. For the case of general positions, these axioms are enough, but if we want our algorithms to be robust for possibly aligned points in the input, there are two possibilities. The first one is to introduce notions that are meaningful for triplets of points that are aligned.
In this case, we also use segments and the property that a point belongs to a segment given by its two extremities. This adds up to 9 more axioms, which we do not describe here for the moment. The second solution is to find a way to extend orientation to aligned points in a way that still satisfies the axioms.
Perturbation methods as described in provide a satisfactory approach to this.
2.1 Proving the axioms When points are given as elements of R2, the orientation predicate is given by observing the sign of a determinant: d pqr ≡ xp yp 1 xq yq 1 xr yr 1 > 0.
In the following we will write |pqr| for this determinant. Knuth’s axioms are not proper axioms anymore, they become simple consequences of the properties of addition and multiplication of real numbers.
The proof of Axioms 1 and 2 is given by two equations: |pqr| = |qrp|, |pqr| = |prq|.
From a mechanical proof point of view, the proofs of these equalities is done painlessly by the usual confluent rewriting system for ring structures, as imple-mented by the Ring tactic in Coq. Axiom 3 simply does not hold in general.
When points are aligned, the determinant is null. A common solution to this problem is to consider only data that are in general position, that is, assume that no three points given in the problem are ever aligned. We will first follow this practice to get a better idea of the algorithms. In a later section, we will see different solutions to the problem of considering degenerated cases.
In man-made proofs, Axiom 4 relies on an equality obtained from a 4-by-4 determinant that is obviously null, since it has 2 identical columns. xp yp 1 1 xq yq 1 1 xr yr 1 1 xt yt 1 1 = |pqr| −|tqr| −|ptr| −|pqt| = 0.
Proving the right-hand side equality is also a simple matter of equality in a ring structure. Now, if |tqr|, |ptr|, |pqt| are all strictly positive, then |pqr| is.
Verifying Axiom 5 also relies on an equation, known as Cramer’s equation, which has the following form: |stq||tpr| = |tqr||stp| + |tpq||str| Here again, automatically proving this equation is a simple matter of ring based computations.
The conclusion of this section on the orientation predicate is twofold. First numeric computation can be done in a ring rather than in a complete field. This is important as one of the reasons for problems in geometric algorithms is the loss of precision in floating point arithmetic. For convex hulls, floating points are not required. The points may be given by integer coordinates in a small grid. If floating points are used, one may notice that, since the determinant basically is a polynomial of degree two in the coordinates of the three points being considered, it is only necessary to compute the orientation predicate with a precision that is double the precision of the input data.
2.2 The specification What is the convex hull of a set of points S? One definition is to describe it as a minimal subset S′ ⊂S, such that all points in S can be described as positive linear combinations of points in S′.
An alternative description of the convex hull uses oriented edges. A possibly open line S′ encompasses a set of points S if for any point p in S and any two consecutive points titi+1 in S′, \ titi+1p or p ∈{titi+1} holds. We consider that a list l is a convex hull of S if l is included in S, l is non-empty if S is not, l contains no duplicates, and if t is the last element of l, then t·l, formed by adding t at the head of l, encompasses S. In other terms, the convex hull is defined as a minimal intersection of half-planes.
When a data set is the empty set, a singleton, or a pair, then the convex hull is simply the list enumerating the set’s elements.
Knuth chooses the second definition in his monograph. We will also, for several reasons. First, lists are a natural data-structure in the functional pro-gramming language we use in our proofs. Second, this definition also relies on the orientation predicate, thus making it possible to forget everything about nu-merical computation in the main structure of algorithm. Third, most algorithms to compute convex hulls naturally rely on the fact that the intermediary data they construct already is a list of points included in S.
The three parts of the specification are given using inductive definitions on lists (actually we also consider that the input data is given as a list of points).
Still, we have made an effort to establish a bridge between the two definitions.
We have mechanically verified a proof that if S′ is a convex hull of S and p is an arbitrary point in S, then there exists three points ti, tj, and tk ∈S′ such that p is inside the triangle formed by ti, tj and tk. This is expressed by the following three orientation predicates: d titjp [ tjtkp [ tktip The proof by induction on the length of a convex polygon S′ encompassing p works as follows. First, S′ cannot be empty, since it encompasses p. If S′ is a singleton or a pair, the degenerate p, p, p will do. If the convex polygon contains at least three points, first check whether p is equal to t1, t2, or t3, then compute [ t1t3p. If any of these conditions holds, then take t1t2t3. If none of these conditions holds, then p is actually inside the convex polygon defined by t1t3 . . . You can use an induction hypothesis on this smaller convex polygon to reach the result.
Once you have the three points encompassing p it is possible to show that the coordinates of p are a positive linear combination of the coordinates of the three points. This is simply because the coordinates of p verify the following equation: p × |titjtk| = ti × |tjtkp| + tj × |tktip| + tk × |titjp| This equation is easily verified using the ring decision procedure.
If d titjp, [ tjtkp, and [ tktip hold, then the three determinants on the right-hand side are positive and by Axiom 4 the determinant on the left-hand side also is.
Dividing by this determinant is legitimate and p really is obtained by a positive linear combination of the t’s. The cases where the convex hull contains less than three points are also easy to prove. Notice that this proof requires that we work in a field.
2.3 Finding initial data Some algorithms require that one should be able to find one or two points that are already present in the convex hull to start with. The package-wrapping algorithm is among them. The usual technique is to look for the minimal point using a lexical ordering on the coordinates.
If t is the minimal point for the lexical ordering on coordinates, and if p, q, and r are points of a triangle encompassing t, where the first coordinates of p, q, and r are such p1 ≥r1 then it is necessary that r1 ≥q1. Using this result again, we have p1 ≥r1 ⇒r1 ≥p1. Thus all three first coordinates must be equal. But if they are, the points are aligned and cannot constitute a proper triangle.
Once we have the point t2 whose coordinates are minimal for the lexico-graphic order, we can find its left neighbour by choosing the only point that has the same second coordinate as t2. If no such point exists, take the point that is maximal for the order ≺defined by: p ≺q ⇔c tpq That ≺is transitive is ensured by Axiom 5′, using the point (tx, ty + 1) as point s.
3 Proving algorithms In this section we review the descriptions and proofs for the abstract setting: positions are general (no three points are aligned) and all numerical computation is hidden behind the orientation predicate and the function producing initial points.
3.1 The incremental algorithm new point red edges blue edges purple point new edge purple point Fig. 3. Red and blue edges in the incremental algorithm The incremental algoritm works by taking the points of the input data one by one and constructing the convex hull of all the points seen so far. At each step, either the new point already lies inside the current convex hull and one can directly proceed to the next, or some edges need to be removed and two edges need to be added. All tests rely on the orientation predicate.
The edges from the previous convex hull are sorted into two categories, called blue and red edges (see figure 3). Blue edges represent edges, for which the new point lies on good side. In our description we also say that an edge is blue if the new point is one of the extremities. Red edges are the other. When all edges are blue, the point is inside. All edges cannot be red at the same time and red edges are contiguous.
When looking for red edges there may be four cases: 1. no red edges (including no edges at all), 2. the edge between the last point of the list and the first point of the list (the closing edge) is red, but the last edge of the list is not.
3. the closing edge is red and the last edge too, 4. the closing edge is blue.
In the first case, nothing needs to be done. In the second case, the list of edges can be viewed as the concatenation of two lists, lr and lb, such that lr contains only red edges lb contains only blue edges, and the edge between the last element in lr and the first element in lb is red. This relies on the property that the list describing the convex hull has no duplicates. In this case, the result is the list p · lb. To prove this, we need to show that p is not already in the list (easy), that the new point p is in the input data (easy), that p is left of all the edges in the convex hull (easy, since the edges that are kept are all blue, and the two new edges have p at the extremity), and that all the data considered so far is left of the edges tp and pt′, where t and t′ are the purple points, the last and first elements of lb, respectively. The point t is different from t′ if there is a red edge.
Let us detail part of the proof for the other cases (see figure 4).
tk red edge blue edge p q r t Fig. 4. Using axiom 5′ to justify adding an edge.
Since t and t′ are distinct, there exists a tk distinct from t that is last-but-one in lb. Let q be the other extremity of the first red edge, and let r be an arbitrary point among those that have already been processed. By construction, we have d tktr, d tktq, and c tqr. Because tq is red we don’t have c tqp. By Axiom 3, this gives c tpq. Because tkt is blue, we have d tktp. All hypotheses are present to use Axiom 5′ and conclude that c tpr holds. The proof for d pt′r has the same structure, but uses Axiom 5.
The two other cases are solved by rotating the convex hull to return to the second case. Rotating a convex hull is simply decomposing this convex hull as the concatenation of two lists. In the following, we will denote concatenation of two lists l1 and l2 as l1 • l2. Since concatenation is a simple structural recursive function (recursive over the first argument), it is a simple matter of structural recursion over l1 to show that rotation preserves inclusion in another list, that it preserves the absence of duplicate elements, and that it preserves the data encompassed. Thus, if l2 • l1 is a convex hull of l, l1 • l2 also is.
For the third case, one traverses down the list representing the convex hull until one finds the first point tj such that tj−1tj is blue and tjtj+1 is red. In that case the rotation is given by l2 = tj · · · t and l1 = t1 · · · tj. If ti is the first point in l1 such that the edge ti−1ti is red and the edge titi+1 is blue, then one has: lr = tj · · · t • (t1 · · · ti−1) and lb = ti+1tj.
The fourth case is symmetric to the third one. One also needs to find the two purple points that are boundaries between blue and red edges but they are in reverse order.
The main algorithm repeats this computation for each point. It is pro-grammed as a simple structural recursion over the list of inputs.
3.2 The package-wrapping algorithm The package-wrapping algorithm relies on the possibility to find an initial edge that is known to be part of the convex hull, then it proceeds by constructing an open encompassing list l of pairwise distinct points, included in the data, finding the right neighbour of the first element in the list at each step. At each step one compares the right neigbour with the last element of the list. When the two points are equal, the algorithm terminates without adding the current list.
When the two points are different, the right neigbour is added to the list and recursion restarts (see figure 5). During the recursion the list always has at least future point t p Fig. 5. Intermediary iteration for the package-wrapping algorithm.
two elements, let t be the head and s be the second element. Finding t’s right neighbour is done by finding the greatest element for the relation ≺defined by p ≺q ⇔c tpq.
This relation is not reflexive, but its reflexive closure is a total order, thanks to Axioms 2, 3, and 5. Finding the least element of a list for an arbitrary order is an easy piece of functional programming, and it is also easily proved correct. If p is this greatest element, c tqp holds for every point in the input data, and by Axiom 1 c ptq also holds. Thus the input data is left of the edge pt and the list p · l also encompasses the input data.
A key ingredient to proving the termination of the algorithm is to ensure that the list being constructed never contains duplicate elements and is included in the input data. The hardest part is to show that checking that p is distinct from the last element of l is enough to ensure that p does not belong to l. In fact, if t is the head of the list, t′ is the last element and q is another element of the list, one needs to show that d tqt′ holds. This is done by recursion over the length of list l but there are many intricacies with base cases, because the proof relies on Axiom 5, which needs 5 distinct points. Let us have a look at this proof.
First assume q is the second element of the list, d tqt′ comes easily from the fact that the list encompasses the input data and t′ is in the input data. Then if q is not the second element of the list, let t1 be this second element. We do the proof by structural recursion over the list q · · · t′. If this list is reduced to 0 or 1 element, this is contradictory with our hypotheses, if this list has two elements, then qt′ is an edge from the list, and d qt′t holds because t is in the input data. The result then comes from Axiom 1. If the list q · · · t′ has more than two elements, let q′ be the second element, one has d tq′t′ by induction hypothesis, d tqq′ because qq′ encompasses the input data and thanks to Axiom 1. Moreover, the properties d tt1q, d tt1q′, and d tt1t′ hold, because q, q′, and t′ are in the input data. This is enough to use axiom 5 and conclude.
Now, the package-wrapping algorithm is not structural recursive. The termi-nation argument is that the size of the list l is bounded by the size of the input data, because all its elements are pairwise distinct, and that this size increases at each step. For this reason, the function is defined using Coq’s solution for defin-ing well-founded recursive functions. This means that we have to exhibit a set containing the principal argument of the recursive function and a well-founded order on this set and show that recursive calls are done only on smaller terms.
The set LS we exhibit is the set of lists that are included in the input data S and contain no duplicates. In Coq’s type theory, this set is described as a type using dependently typed inductive types. We show that the length of a list in in this set is always smaller than the size N of the input data S. We define the order <S to be : l <S l′ ⇔N −length(l) < N −length(l′).
Proving that this order is well-founded is a simple application of theorems about composing well-founded orders with functions provided in Coq’s libraries. In other theorem provers where the termination of recursive functions is ensured by exhibiting measure functions, the measure function to exhibit is the following one: f : l 7→N −length(l).
At each recursive step, the main argument is l and the argument for the next recursive call is p · l such that p · l ∈LS, therefore one has length(p) < length(p · l) ≤N and f(p · l) < f(l).
4 Degenerated cases If we want to allow for aligned points in the input data, we need to review our technique. One solution is to change Axiom 3, since it is the one that assumes that a triple of points can always be viewed as a triangle. If we do that, any decision taken with respect to orientation in the algorithms must be reviewed and replaced by a new decision taking extra cases into account. The structure of the algorithm remains, but all the details of its control-flow are deformed by this constraint.
A second solution is to change the orientation predicate so that it satisfies all five axioms even in the presence of aligned triples. This solution somehow imposes that one changes the definition of convex hulls, since our definition relies on the orientation predicate. We shall see that both solutions actually require that we relax our definition of a convex hull.
4.1 Segments When three distinct points are aligned, one of them lies between the two other.
We introduce a new predicate, written ⌉pqr⌈to express this. Axiom 3 is changed to obtain the following formulation: Axiom 3 ∀p, q, r.
p = q ∨q = r ∨p = r ∨d pqr ∨d prq∨⌉pqr⌈∨⌉qrp⌈∨⌉rpq⌈.
In Knuth’s work, Axioms 1-5 express a form of internal consistency for the orientation predicate. We need to have the same kind of internal consistency for the segment predicate and between segments and oriented triangles. The first two axioms describe internal consistency for three aligned points.
Axiom 6 ∀p, q, r.
⌉pqr⌈⇒⌉rpq⌈.
Axiom 7 ∀p, q, r.
⌉pqr⌈⇒¬⌉prq⌈.
The next four axioms describe consistency between segments and triangles.
Axiom 8 ∀p, q, r, t.
d pqr∧⌉ptq⌈⇒c ptr.
Axiom 9 ∀p, q, r, t.
d pqr∧⌉pqt⌈⇒c ptr.
Axiom 10 ∀p, q, r, t.
d pqr∧⌉ptq⌈⇒c tqr.
Axiom 11 ∀p, q, r, t.
d pqr∧⌉tpq⌈⇒c tqr.
The last two axioms describe internal consistency of segments for four aligned points.
Axiom 12 ∀p, q, r, t.
⌉qrt⌈∧⌉pqt⌈⇒⌉prt⌈.
Axiom 13 ∀p, q, r, t.
⌉qrt⌈∧⌉pqr⌈⇒⌉prt⌈.
With these new notions, we change the specification of a convex hull for a set S to be a list of points t1 . . . tk such that for every p in S and every ti, tj such that tj = ti+1 or tj = t1 if ti = tk one has d titjp∨⌉titjp⌈.
4.2 The working horse predicate In the new setting we use a new predicate that combines orientation and segment membership. For the rest of the paper, we will denote this predicate pqr, with the following meaning: pqr ⇔d pqr ∨r ∈{p, q}∨⌉prq⌈.
Generalizing our algorithms to the degenerated cases works by replacing the orientation predicate by the new predicate wherever it occurs and adding a few equality tests to take care of degeneracies. What matters really is that equivalents of Axiom 5 still hold: ∀p, q, r, s, t.
tsp ∧tsq ∧tsr ∧¬qtp ∧qtr ⇒ptr ∀p, q, r, s, t.
stp ∧stq ∧str ∧¬tqp ∧tqr ⇒tpr.
Thus, if t is known to be on the convex hull, we keep the property that the order ≺′ can be used to find minimal and maximal elements, when this order is defined as follows: p ≺′ q ⇔tpq.
4.3 Perturbation An alternative approach is to change the orientation predicate so that Axiom 3 always holds. This option seems awkward: if we change the orientation predicate how can we be sure that we actually compute the same convex hull? The answer is to make sure that we change the orientation predicate in a manner that re-mains consistent with the general orientation predicate, so that the convex hull computed in non-degenerated cases is unchanged.
The solution we are going to present is based on the idea that if points are aligned, then moving one of the points only slightly will remove the degeneracy.
This make it possible to compute an approximation of the hull. However, points are not actually moved, and all points that should appear on the convex hull will be guaranteed to appear in the approximation we compute. Imprecisions will occur only for points that are on the segment between two legitimate vertices of the convex hull. For these points, this method may or may not include them in the resulting convex hull.
The solution is to consider that all points are moving in a continuous manner and that the configuration that is being studied is the snapshot at date t = 0.
The determinants used to compute the orientation predicate also are continuous functions in t, that may have the value 0 when t = 0. The orientation predicate is continuous for all unaligned triples. However, if the perturbation function can be viewed as a polynomial function in t it is possible to ensure that limt→0+ is either positive or negative. If the points are not aligned, the limit is the same as the value in 0, if there are aligned points, taking a value of t that is small enough will always return the same sign, so that we do not even have to predict the value of t: the sign can be predicted from the signs of the polynomial coefficient.
We use a function taken from to indicate how all points move. The coordi-nates of the point p actually are given by the formula xp+t×yp, yp+t×(x2 p+y2 p).
The orientation predicate for points x, y, z then becomes the following term: xp + t × yp yp + t × (x2 p + y2 p) 1 xq + t × yq yq + t × (x2 q + y2 q) 1 xr + t × yr yr + t × (x2 r + y2 r) 1 = D1 + D2 × t + D3 × t2, where D1, D2, and D3 are the determinants defined as follows: D1 = xp yp 1 xq yq 1 xr yr 1 D2 = yp x2 p + y2 p 1 yq x2 q + y2 q 1 yr x2 r + y2 r 1, D3 = xp x2 p + y2 p 1 xq x2 q + y2 q 1 xr x2 r + y2 r 1. The proof that all three determinants cannot be zero is formalized using the sketch provided in and relies very much on deciding equalities in a ring, plus a lot of painful reasoning on equalities. However, we work on polynomials and all our proofs are done in an ordered archimedian field: the field of rational numbers suffices so that our model is adequate for arithmetic numbers as they are implemented in computers.
If the first derminant is 0, then the three points are aligned. There exists three numbers m1, m2, and m3 such that m1 × y = m2 × x + m3 is the equation of the line containing the three points. As these points are distinct, m1 and m2 cannot both be null. There are two cases, depending on whether m1 = 0.
If m1 = 0, then xp = xq = xr, the second determinant is equal to the following determinant: yp x2 p + y2 p 1 yq x2 q + y2 q 1 yr x2 r + y2 r 1 = yp y2 p 1 yq y2 q 1 yr y2 r 1 = (yp −yq)(yq −yr)(yr −yp).
(1) This value can only be zero if one of the factors is 0, in which case two of the points have both coordinates equal: they are the same.
If m1 = 0, D2 = 0 if and only if m1 × D2 = 0.
m2 1 × yp x2 p + y2 p 1 yq x2 q + y2 q 1 yr x2 r + y2 r 1 = (m2 1 + m2 2) yp y2 p 1 yq y2 q 1 yr y2 r 1 = (m2 1 + m2 2)(yp −yq)(yq −yr)(yr −yp) (2) Here again, at least two of the points must have the same vertical coordinate for the determinant to be 0.
The same reasoning applies to the third determinant, which is symmetric with the second one, replacing x coordinates with y coordinates.
Now for every triple of points, the values D1, D2, and D3 cannot be 0 at the same time. We can then show that for every triple p, q, r in the input data, there exists a number ϵpqr such that the perturbed determinant never becomes 0 in the interval ]0, ϵpqr[. Let abs(x) be the absolute value of x. If D1 is non null, then we can choose ϵpqr to be the following value: ϵpqr = min(1, abs( D1 2 × D2 ), abs( D1 2 × D3 )) If D2 = 0 or D3 = 0, just ignore the corresponding term in the minimal compu-tation.
If D1 = 0 and D3 ̸= 0, then we can choose ϵpqr to be the following value: ϵpqr = min(1, abs(D2 D3 ).
If D1 = 0 and D3 = 0 we can take any positive value for ϵpqr.
This concludes the proof of existence of ϵpqr. This proof was also made only using axioms that are fulfilled by the field of rational numbers.
We could now replace the orientation predicate used in the previous section by a new orientation predicate, noted pqr and defined in the following manner: pqr = xp yp 1 xq yq 1 xr yr 1 > 0 ∨ xp yp 1 xq yq 1 xr yr 1 = 0 ∧ yp x2 p + y2 p 1 yq x2 q + y2 q 1 yr x2 r + y2 r 1 > 0 ∨ xp yp 1 xq yq 1 xr yr 1 = yp x2 p + y2 p 1 yq x2 q + y2 q 1 yr x2 r + y2 r 1 = 0 ∧ xp x2 p + y2 p 1 xq x2 q + y2 q 1 xr x2 r + y2 r 1 = 0 Note that the ϵpqr are never computed, we only show their existence to show that the computation of these two determinants is enough to predict the orientations for a general position that is arbitrarily close to the input data. It is not necessary to return a convex hull composed of perturbed points: the points from the input data are correct, since they can be shown to be arbitrarily closed to a legitimate convex hull (we have not done this proof yet).
It is also interesting that the convex hull computation done in this manner returns exactly the same result as the computation described in previous section if the input data is not degenerated. If the input data is degenerated and the convex hulls are required to contain no aligned points, a trivial traversal of the list of points may be needed to remove the useless points. The cost of this traversal should be proportional to the total cost.
It is remarkable that testing the sign of the D2 can be done by simply com-bining the signs of the terms (yp−yq), (yq −yr), (yr −yp), according to equations (1) and (2). In fact, D2 is strictly positive if and only if the line carrying the three points is not horizontal, the point r is outside the segment pq, and the points p and q are ordered in their second coordinate, similarly for D3. As a re-sult, testing the sign of D2 and D3 is equivalent to testing whether r) is outside the segment pq and whether p and q are ordered lexicographically (using the second coordinate first). Thus, we have a program that computes (pqr) with-out computing the larger determinants above. This algebraic reasoning brings an elegant justification to the use of segment predicates and lexical ordering in combination with traditional orientation.
5 Conclusion The development of certified algorithms is well suited to domains where it is easy to have a mathematical formulation of the data and the specifications. Compu-tational geometry is one of these domains. Euclidean geometry makes it easy to connect the various notions encountered in computational geometry to real analysis and algebra. In our experiment, the presence of a decision procedure for ring equalities has been instrumental, although it has also been quite annoying that this decision procedure was not complete, at least for the version we used.
This decision procedure is the Ring tactic, a tactic based on a reflective approach .
The problem of numeric computation also deserves some thoughts. We have performed our proofs with an “axiomatized” set of numbers for the coordinates.
The axioms we have taken were carefully chosen to make them accurate as a representation of computer numbers used in exact computation. These numbers live in an archimedian field. The model for the field that is needed to express the correction of the algorithm is the field of rational numbers. However, the algorithm does not actually use numbers in the whole field. It only computes a polynomial expression of degree two, so that if the numbers describing input data are bounded, the set of numbers needed for the algorithm to work is much smaller. It could be useful to express the difference between the “concrete” arith-metic used in the algorithm, and the “abstract” arithmetic used in the proof.
This is actually the spirit of the work of Laurent thry and his colleagues in a project to study computer arithmetics.
Perturbation methods are used in other parts of computational geometry, for instance for Delaunay triangulation, where degenerate positions occur when four points are cocyclic. They seem very well adapted as a method to make algorithms robust to degeneracy without changing their structure. Still, it seems a rare and lucky occurrence that perturbations are used in this paper to justify an algorithm that actually does not use any perturbation, since it only boils down to an extension of the orientation predicate with a segment predicate and lexical ordering.
We actually heard about the perturbation method after having done most of the proofs for the algorithm based on the extended axioms combining the tradi-tional orientation and segment predicates. Still both developments are useful in their own right, since the first development can be tuned to include or exclude co-linear points from the convex hull at will. For the second method, we can predict that colinear points on edges of the convex hull that grow lexicographically will be included in the convex hull, while colinear points on edges that decrease lex-icographically will not be included. This information that we have not formally proved at the time of writing this article, may be used when writing a second pass that removes colinear points, but it won’t be easy to recover colinear points that have already been discarded by the convex hull computation.
We would like to acknowledge the help of several researchers from the com-putational geometry group at INRIA Sophia Antipolis. In particular, Mariette Yvinec pointed us to Knuth’s work on axioms and Olivier Devillers described us the perturbation method.
References 1. Pierre Alliez, Olivier Devillers, and Jack Snoeyink.
Removing degeneracies by perturbing the problem or the world. Reliable Computing, 6:61–79, 2000. Special Issue on Computational Geometry.
2. Bruno Barras, Samuel Boutin, Cristina Cornes, Judicael Courant, Yann Coscoy, David Delahaye, Jean-Christophe Filliatre Daniel de Rauglaudre, Eduardo Gimenez, Hugo Herbelin, Gerard Huet, Cesar Munoz, Chetan Murthy, Catherine Parent, Christine Paulin-Mohring, Amokrane Saibi, and Benjamin Werner. The Coq Proof Assistant User’s Guide. Version 6.3.1. INRIA, December 1999.
3. Jean-Daniel Boissonnat and Mariette Yvinec. Algorithmic geometry. Cambridge University Press, 1995.
4. Samuel Boutin. Using reflection to build efficient and certified decision procedures.
In Theoretical Aspects of Computer Science, number 1281 in Lecture Notes in Computer Science. Springer-Verlag, 1997.
5. S.C. Chou, X.S. Gao, and J.Z. Zhang. Machine Proofs in Geometry. World Scien-tific, Singapore, 1994.
6. Thomas H. Cormen, Charles E. Leiserson, and Ronald L. Rivest. Introduction to algorithms. MIT-Press, 1990.
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Elements of constructive geometry group theory and domain theory, 1995.
available as a Coq contribution at
8. Donald Knuth. Axioms and Hulls. Number 606 in Lecture Notes in Computer Science. Springer-Verlag, 1991.
9. Fran¸ cois Puitg and Jean-Fran¸ cois Dufourd.
Formal specification and theorem proving breakthroughs in geometric modeling.
In Theorem Proving in Higher-Order Logics, volume 1479 of Lecture Notes in Computer Science, pages 410–422.
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